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by: Moshe Swift III

DifferentialEquations MATH262

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Moshe Swift III
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This 187 page Class Notes was uploaded by Moshe Swift III on Wednesday September 23, 2015. The Class Notes belongs to MATH262 at Drexel University taught by AnatoliiGrinshpan in Fall. Since its upload, it has received 6 views. For similar materials see /class/212287/math262-drexel-university in Mathematics (M) at Drexel University.


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Date Created: 09/23/15
Notes on Differential Equations Robert E Terrell Preface Students of science and engineering often encounter a gap between their calculus text and the applied mathematical books they need to read soon after What is that gap A calculus student might have learned some calculus without learning how to read mathematics A science major might be faced with an applied text which assumes the ability to read a somewhat encyclopedic mathematics and its application 1 often nd myself teaching students who are somewhere between these two states These notes are mostly from lectures I have given to bridge that gap while teaching the Engineering Mathematics courses at Cornell University The notes could be used for an introductory uni ed course on ordinary and par tial differential equations However this is not a textbook in the usual sense and certainly not a reference The stress is on clarity not completeness It is an introduction and an invitation Most ideas are taught by example The notes have been available for many years from my web page For fuller coverage see any of the excellent books 0 Agnew Ralph Palmer Di erential Equations McGrawiHill 1960 0 Churchill Ruel V Fourier Series and Boundary Value Problems Mc Graw Hill 1941 Hubbard John H and West Beverly HDi erential Equations a Dynamical Systems Approach Parts 1 and 2 Springer 1995 and 1996 Lax P Burstein S and Lax A Calculus with Applications and Computing Vol 1 Springer 1983 Seeley Robert T An Introduction to Fourier Series and Integrals WABenjamin 1966 Some of the exercises have the format What s rong with thisquot These are either questions asked by students or errors taken from test papers of students in this class so it could be quite bene cial to study them Robert E Terrell January 1997 September 2005 April 2009 Contents 1 Introduction to Ordinary Differential Equations 11 The Banker s Equation 12 Slope Fields 2 A Gallery of Differential Equations 3 Introduction to Partial Differential Equations 31 A Conservation Law 32 Traveling Waves 4 The Logistic Population Model 5 Separable Equations 6 Separable Partial Differential Equations 7 Existence and Uniqueness and Software 71 The Existence Theorem 72 Software 8 Linear First Order Equations 81 Newton s Law of Cooling 82 Investments 9 oiler7s Euler7s Nurnerical Method 10 Second Order Linear Equations 101 Linearity 102 The Characteristic Equation 103 Conservation laws7 part 2 1 1 2 12 14 17 30 35 35 36 38 11 The Heat Equation 12 Complex Numbers and the Characteristic Equation 121 The Fundamental Theorem of Algebra 13 Forced Second Order 14 Systems of ODE7s part 1 141 A Chemical Engineering problem 142 Phase Plane 15 Systems of ODE7s part 2 151 Linear Systems 16 Linear Algebra part 1 161 Matrix 162 Geometric aspects of matrices 17 Linear Algebra part 2 18 Linear Algebra part 3 181 Matrix Inverse 18 2 Eigenvalues 19 Linear Algebra part 4 20 More on the Determinant 201 Products 21 Linear Differential Equation Systems 22 Systems with Complex Eigenvalues iii 41 43 47 49 67 71 71 73 75 80 81 83 87 23 Classi cation Theorem for Linear Plane Systems 24 Nonlinear Systems in the Plane 25 Examples of Nonlinear Systems in 3 Dimensions 26 Boundary Value Problems 261 Signi cance of Eigenvalues 27 The Conduction of Heat 28 Solving the Heat Equation 281 Insulation 282 Product Solutions 29 More Heat Solutions 30 The Wave Equation 31 Beams and Columns 32 Power Series 321 Review 322 The Meaning of Convergence 33 More on Power Series 34 Power Series used A Drum Model 35 A new Function for the Drum model J0 351 But what does the drum Sound like 36 The Euler equation for Fluid Flow and Acoustic Waves 361 The Euler equations 91 96 104 106 112 113 114 150 151 362 Sound 153 37 Exact equations for Air and Steam 156 38 The Laplace Equation 158 39 Laplace leads to Fourier 160 40 Fourier7s Dilemma 163 41 Fourier answered by Orthogonality 165 42 Uniqueness 168 43 selected answers and hints 170 1 Introduction to Ordinary Differential Equations TODAY An example involving your bank account and nice pictures called slope elds How to read a diHerential equation Welcome to the world of differential equations They describe many pro cesses in the world around you but of course we ll have to convince you of that Today we are going to give an example and nd out what it means to read a differential equation 11 The Banker7s Equation A differential equation is an equation which contains an unknown function and one of its unknown derivatives Here is a differential equation dyi dt 028 It doesn t look too exciting does it Really it is though It might for example represent your bank account where the balance is y at a time t years after you open the account and the account is earning 28 interest Regardless of the speci c interpretation let s see what the equation says Since we see the term dydt we can tell that y is a function of t and that the rate of change is a multiple namely 028 of the value of y itself We de nitely should always write yt instead ofjust y and we will sometimes but it is traditional to be sloppy For example if y happens to be 2000 at a particular time t the rate of change of y is then 0282000 56 and the units of this rate in the bank account case are dollarsyear Thus y is increasing whenever y is positive PRACTICE What do you estimate the balance will be roughly a year from now if it is 2000 and is growing at 56 dollarsyear This is not supposed to be a hard question By the way when I ask a question don t cheat yourself by ignoring it Think about it and future things will be easier I promise Later when y is say 587533 its rate of change will be 028587533 164509 which is much faster We ll sometimes refer to y 028y as the banker s equation Do you begin to see how you can get useful information from a differential equation fairly easily by just reading it carefully One ofthe most important skills to learn about differential equations is how to read them For example in the equation 3 028y 7 10 there is a new negative in uence on the rate of change due to the 710 This 710 could represent withdrawals from the account PRACTICE What must be the units of the 710 Whether the resulting value of y is actually negative depends on the current value of y That is an example of reading a differential equation As a result of this reading skill you can perhaps recognize that the banker s equation is very idealized It does not account for deposits or changes in interest rate It didn t account for withdrawals until we appended the 710 but even that is an unrealistic continuous rate of withdrawal You may wish to think about how the equation might be modi ed to include those things more realistically 12 Slope Fields It is signi cant that you can make graphs of the solutions sometimes In the bank account problem we have already noticed that the larger y is the greater the rate of increase This can be displayed by sketching a slope eld77 as in Figure 11 Slope elds are done as follows First the general form of a differential equation is 2 at where yt is the unknown function and f is given To make a slope eld for this equation choose some points y t and evaluate 1 there According to the differential equation these numbers must be equal to the derivative 3 which is the slope for the graph of the solution These resulting values of y are then plotted using small line segments to indicate the slopes For example at the point t 6 y 20 the equation 3 028y says that the slope must be 02820 56 So we go to this point on the graph and place a mark having this slope Solution curves then must be tangent to the slope marks This can be done by hand or computer without solving the differential equation Figure 11 The slope eld for y 028y as made by dfield ln Lecture 7 we will discuss where to get dfield Some solutions are also shown Note that we have included the cases y 0 and y lt 0 in the slope eld even though they might not apply to your bank account Let us hope PRACTICE Try making a slope eld for y y t There is also a way to explicitly solve the banker s equation Assume we are looking for a positive solution Then it is alright to divide the equation by y getting 1 dy m Then integrate7 using the chain rule 028 lny 028t c where c is some constant Then Mt e028tc 616028 Here we have used a property of the exponential function7 that eab eael 7 and set 01 50 The potential answer which we have found must now be checked by substituting it into the differential equation to make sure it really works You ought to do this Now You will notice that the constant 01 can in fact be any constant7 in spite of the fact that our derivation of it seemed to suggest that it be positive This is how it works with differential equations It doesn7t matter how you solve them7 what is important is that you check to see whether you are right Some kinds of answers in math and science can t be checked very well7 but these can7 so do it Even guessing answers is a highly respected method if you check them Your healthy scepticism may be complaining to you at this point about any bank account which grows exponentially 1f so7 good It is clearly impossible for anything to grow exponentially forever Perhaps it is reasonable for a limited time The hope of applied mathematics is that our models will be idealistic enough to solve while being realistic enough to be worthwhile The last point we want to make about this example concerns the constant 01 further What is the amount of money you originally deposited7 y0 Do you see that it is the same as 017 That is because y0 0150 and 50 1 If your original deposit was 300 dollars then 01 300 This value y0 is called an initial condition 7 and serves to pick the solution we are interested in out from among all those which might be drawn in Figure 11 EXAMPLE Be sure you can do the following kind x 73x x0 5 Like before we get a solution Mt 06 Then MO 5 ceo c so 0 5 and the answer is xt 567 Check it to be absolutely sure PROBLEMS 1 Make slope elds for x x x 7x x x t x 7x cos t x t 2 Sketch some solution curves onto the given slope eld in Figure 12 below 3 What general fact do you know from calculus about the graph of a function y if y gt 0 Apply this fact to any solution of y 7 consider cases where the Values of y lie in each of the intervals foo 7171 O O 1 and 1 For each interval7 state whether y is increasing or decreasing 4 continuing 3 If y gigs and y0 what do you think will be the llmt oo Mt Make a slope eld if you re not sure 5 Reconsider the banker s equation 11 0281 If the interest rate is 3 at the beginning of the year and expected to rise linearly to 4 over the next two years7 what would you replace 028 with in the equation You are not asked to solve the equation 6 In y 028y710 suppose the withdrawals are changed from 10 per year continuously to 200 every other week What could you replace the 710 by This is a dif cult rhetorical question onlyi something to think about Do you think it would be alright to use a smooth function of the form acos bt to approximate the withdrawals Figure 12 The slope eld for Problem 2 as drawn by dfield 2 A Gallery of Differential Equations TODAY A gallery is a plane to look and to get ideas and to nd out how other people View things Here is a list of differential equations as a preview of things to come Unlike the banker s equation 3 0028y not all differential equations are about money 1 y 028600 7 y This equation is a model for the heating of a pizza in a 600 degree oven Of course the 0028 is there just for comparison with the banker s interest The physical law involved is called believe it or not Newton s Law of Cooling We ll do it in Lecture 8 2 Newton had other laws as well one of them being the F ma77 law of inertia You might have seen this in a physics class but not realised that it is a differential equation That is because it concerns the unknown position of a mass and the second derivative a of that position In fact the original differential equation the very rst one over 300 years ago was made by Newton for the case in which F was the gravity force between the earth and moon and m was the moon s mass Before that nobody knew what a differential equation was and nobody knew that gravity had anything to do with the motion of the moon They thought gravity was what made their physics books heavy You have probably heard a garbled version of the story in which Newton saw or was hit by a falling apple while thinking about gravity What is frequently omitted is that he looked up to see whence it had fallen and saw the moon up there behind the branches Thus it occured to him that there might be a connection Maybe his thought process was something like gravity gravity gravity apple gravity gravity apple moon gravity apple moon gravity moon gravity moon gravity Ohl Moon Gravity77 There is a joke by the famous physicist Richard Feynmann who said that once people thought there were angels up there pushing against the side of the moon to keep it moving around the earth This is of course ridiculous he said After Newton everyone knows that the angels are in fact on the far side of the moon pushing it toward us Anyway Newton s law often leads to second order differential equations PRACTICE The simplest second order equations look like g 7x You ought to be able to guess some solutions to this equation What are they 3 There are also Partial Differential Equations This does not mean that somebody forgot to write out the whole thing It means that the unknown functions depend on more than one variable so that partial derivatives show up in the equation One example is at um where the subscripts denote partial derivatives This is called the heat equation and um t is the tem perature at position x and time 25 when heat is allowed to conduct only along the z axis as through a wall or along a metal bar It concerns a dif ferent aspect of heat than does Newton s law of cooling and we will discuss it more in Lectures 11 and 27 4 utt um looks sort of like the heat equation but is very different be cause of the second time derivative This is the wave equation which is about electromagnetic waves wireless music and water waves in decreas ing order of accuracy The equation for vibrations of a drum head is the two dimensional wave equation utt um uyy ln Lecture 34 we will derive a special polar coordinate version of that utt uM u We ll use that to describe some of the sounds of a drum 5 There are others which we won t study but some of the ideas we use can be applied to them The reason for mentioning these here is to convince you that the earlier statement about many processes77 in the world around you is correct If you bang your st on the table top and the table top is somewhat rigid not like a drum head then the sound which comes out is caused by vibrations of the wood and these are described by solutions of utt 704 214 Here Mm y t is the vertical de ection of the table top while it bends and vibrates Can you imagine that happening at a small scale We will in fact do a related equation utt fun for vibrations of a beam in Lecture 31 At a much smaller scale the behavior of electrons in an atom is described by Schrodinger s equation wt 7w uyy um Vm yzu In this case um y 2 t is related to the probability that the electron is at z y 2 at time 25 At the other end of the scale there is an equation we won t write down but which was worked out by Einstein Not E mcz but a differential equation That must be why Einstein is so famous He wrote down a differential equation for the whole universe PROBLEMS l Newton s gravity law says that the force between a big mass at the origin of the x axis and a small mass at point t is proportional to 72 How would you write the F ma law for that as a di erential equation 2 We don t have much experience with the fourth derivatives mentioned above Let ux t t2x2 7 x3 x4 ls um positive or negative Does this depend on the value of x 3 In case you didn t do the PRACTICE item What functions do you know about from calculus that are equal to their second derivative the negative of their second derivative 3 Introduction to Partial Differential Equations TODAY A rst order partial di erential equation Here is a partial differential equation sometimes called a transport equation and sometimes called a wave equation 8710 871070 at 8x7 PRACTICE We remind you that partial derivatives are the rates of change holding all but one variable xed For example 6 6 Exiy2x2yt Qy x7y2x2yt 172 What is the y partial Our PDE is abbreviated wt l 310m 0 You can tell by the notation that w is to be interpreted as a function of both if and x You can t tell what the equation is about We will see that it can describe certain types of waves There are water waves electromag netic waves the wavelike motion of musical instrument strings the invisible pressure waves of sound the waveforms of alternating electric current and others This equation is a simple model PRACTICE You know from calculus that increasing functions have positive derivatives In Figure 31 a wave shape is indicated as a function of x at one particular time t Focus on the steepest part of the wave ls wag positive there or negative Next look at the transport equation Is wt positive there or negative Which way will the steep pro le rnove next Remember how important it is to read a differential equation 31 A Conservation Law We ll derive the equation as one model for conservation of mass You might feel that the derivation of the equation is harder than the solving of the equation We imagine that w represents the height of a sand dune which moves by the wind along the z direction The assumption is that the sand blows along the surface crossing position mwzt at a rate proportional to w The proportionality factor is taken to be 3 which has dimensions of velocity like the wind Figure 31 The wind blows sand along the surface Some enters the segment a ah from the left and some leaves at the right The net difference causes changes in the height of the dune there The law of conservation of sand says that over each segment a a h you have d ah a wm if all 3wa t 7 3wa h t That is the time rate of the total sand on the left side and the sand ux on the right side Divide by h and take the limit PRACTICE 1 Why is there a minus sign on the right hand term 2 What do you know from the Fundamental Theorem of Calculus about a l dx The limit we need is the case in which f is wtc t We nd that wta t 73wma 25 Of course a is arbitrary That concludes the derivation 3 2 Traveling Waves When you rst encounter PDE it can appear because of having more than one independent variable that there is no reasonable place to start working Do I try if rst x or what In this section we ll just explore a little If we try something that doesn t help then we try something else PRACTICE Find all solutions to our transport equation of the form wx t ax bt In case that is not clear it does not mean derive ax bt somehow It means substitute the hypothetical wx t ax bt into the PDE and see whether there are any such solutions What is required of a and 7 Those practice solutions don t look much like waves Figure 31 Graphs of cosx cosx 7 15 and cosx 7 3 If you think of these as photographs superimposed at three di erent tithes then you can see that it moves What values of t do you associate with these graphs and how fast does the wave move Lets try something more wavey PRACTICE Find all solutions to our wave equation of the form wxt ccosax t So far we have seen a lot of solutions to our transport equation Here are a few of them wx t m 7 3t wm t 721 6375 wzt 40 cos5x7 1525 2 wz t 77 cos8m 7 2425 7 For comparison that is a lot more variety than we found for the banker s equation Remember that the only solutions to the ODE y 028y are constant multiples of 0025 Now lets go out on a limb Since our wave equation allows straight lines of all different slopes and cosines of all different frequencies and amplitudes maybe it also allows other things too Try wm t 7 3t where we won7t specify the function 1 yet Without specifying 1 any further we can t nd the derivatives we need in any literal sense but can apply the chain rule anyway The intention here is that f ought to be a function of one variable say 5 and that the number x 7 3t is being inserted for that variable 5 z 7 325 The partial derivatives are computed using the chain rule because we are composing f with the function x 7 3t of two variables The chain rule here looks like this 7 8w df 35 7 7 7 wquot 825 dsat 31 3t PRACTICE Figure out why wC f ac 7 31 Setting those into the transport equation we get wt 31 73f3f0 That is interesting It means that any differentiable function 1 gives us a solution Any dune shape is allowed You see it doesn7t matter at all what 1 is as long as it is some differentiable function 10 Don t forget differential equations are a model of the world They are not the world itself Real dunes cannot have just any shape f whatsoever They are more specialized than our model PRACTICE Check the case 22 sins 7 10sin3s That is verify that wx t 22 sinx 7 3t710sin3c 7 9t is a solution to our wave equation PROBLEMS 1 Work all the PRACTICE items in this lecture if you have not done so yet 2 Find a lot of solutions to the wave equation yt 511x 0 and tell which direction the waves move and how fast 3 Check that wx t 11 x 7 302 is one solution to the equation 11 310 O 4 What does the initial Value 101 0 look like in problem 3 if you graph it as a function of 7 5 Sketch the pro le of the dune shapes wx 1 and wx2 in problem 3 What is happening Which way is the wind blowing What is the Velocity of the dune Can you tell the velocity of the wind 6 Solve u ux 0 if we also want to have the initial condition ux O cos2x 7 As in problem 67 but with initial condition ux O cost sin4x Sketch the wave shape for several times 4 The Logistic Population Model TODAY The logistic equation is an improved model for population growth We have seen that the banker s equation 3 028y has exponentially grow ing solutions It also has a completely different interpretation from the bank account idea Suppose that you have a population containing about yt in dividuals The word about is used because if y 3251 then we will have to interpret how many individuals that is Also the units could be say thousands ofindividuals rather than just plain individuals The population could be anything from people on earth to deer in a certain forest to bacte ria in a certain Petrie dish We can read this differential equation to say that the rate of change of the population is proportional to the number present That perhaps captures some element of truth yet we see right away that no population can grow exponentially forever since sooner or later there will be a limit imposed by space or food or energy or something The Logistic Equation Here is a modi cation to the bankers equation that overcomes the previous objection d 7 028y17 y In order to understand why this avoids the exponential growth problem we must read the differential equation carefully Remember that I said this is an important skill Here we go You may rewrite the right hand side as 028y 7342 You know that when y is small y2 is very small Consequently the rate of change is still about 028y when y is small and you will get exponential growth approximately After this goes on for a while it is plausible that the y2 term will become important In fact as y increases toward 1 one thousand or whatever the rate of change approaches 0 For simplicity we now dispense with the 028 and for exibility introduce a parameter a and consider the logistic equation 3 ya 7 Therefore if we make a slope eld for this equation we see something like Figure 41 Figure 41 A slope eld for the Logistic Equation Note that solutions starting near 0 have about the same shape as exponentials until they get near a gure made using dfield The solutions which begin with initial conditions between 0 and a evidently grow toward a as a limit This in fact can be veri ed by nding an explicit formula for the solution Proceeding much as we did for the bank account problem7 rst write 1 ayeiyZ To make this easier to integrate7 we ll use a trick which was discovered by a student in this class7 and multiply rst by y Zy z Then integrate 72 y Aggid a w41y illn zy l 71 t c a dydt The integral can be done without the trick7 using partial fractions7 but that is longer Now solve for yt ayil 7 1 eiatc Cleiat a t4447 y 1 615 These manipulations would be somewhat scary if we had not speci ed that we are interested in y values between 0 and a For example7 the In of a negative number is not de ned However7 we emphasize that the main point is to check any formulas found by such manipulations So let s check it 7 aZCleiat y 7 1 016quot 2 13 We must compare this expression to i a2 a2 i 21cle t 71 7 1 615 7 1 615 7 a 1 615 You can see that this matches 3 Note that the value of 01 is not restricted to be positive even though the derivation above may have required it We have seen this kind of thing before so checking is very important The only restriction here occurs when the denominator of y is 0 which can occur if 01 is negative If you stare long enough at y you will see that this does not happen if the initial condition is between 0 and a and that it restricts the domain of de nition of y if the initial conditions are outside of this interval All this ts very well with the slope eld above In fact there is only one solution to the equation which is not contained in our formula ayiyz PRACTICE Can you see what it is PROBLEMS 1 Suppose that we have a solution yt for the logistic equation y ya 7 y Choose some time delay say 3 time units to be speci c and set zt yt 7 3 Is zt also a solution to the logistic equation 2 The three 7Slshaped solution curves in Figure 41 all appear to be exactly the same shape In View of Problem 1 are they 3 Prof Verhulst made the logistic model in the mid1800s The US census data from the years 1800 1820 and 1840 show populations of about 53 96 and 17 million We ll need to choose some time scale t1 in our solution yt a1 Cleiat71 so that t 0 means 1800 t t1 means 1820 and t 2t1 means 1840 Figure out cl t1 and a to match the historical data WARNING The arithmetic is very long It helps if you use the fact that 2H 6W ANSWER 01 362 t1 0031 and a 197 4 Using the result of Problem 3 what population do you predict for the year 1920 The actual population in 1920 was 106 million The Professor was pretty close wasn t he He was probably surprised to predict a whole century EXAMPLE Census data for 1810 1820 and 1830 show populations of 72 96 and 128 million Trying those it turned out that I couldn t t the numbers due to the numerical coincidence that 72128 962 That is where I switched to the years in Problem 3 This shows that the tting of real data to a model is nontrivial Mathematicians like the word nontrivial 5 Separable Equations TODAY Various useful examples Equations of the form dm 3 fltzgtglttgt include the banker s and logistic equations and some other useful equations You may attack these by separating variables Write gt dt then try to integrate and solve for Remember that this is what we did with the banker s and logistic equations EXAMPLE tdt ln g 0 Mt 816t2 Check 5 Ci t2 tx The skeptical reader will wonder why I said try to integrate and solve for z The next two examples illustrate that there is no guarantee that either of these steps may be completed EXAMPLE dx 1 E 1 312 1 3x2dx dt x x3 t 0 Here you can do the integrals but you can t very easily solve for x EXAMPLE d dt 7 dx 62 dt Here you can t do the integral The Leaking Bucket This example comes from the Hubbard and West book See preface We re about to write out a differential equation to model the depth of water in a leaking bucket based on plausible assumptions Well then try to solve for the water depth at various times In particular we will try to reconstruct the history of an empty bucket This problem is solvable by separation of variables but it has far greater signi cance that the problem has more than one solution Physically if you see an empty bucket with a hole at the bottom you just can t tell when it last held water Here is a crude derivation of our equation which is not intended to be physi cally rigorous We assume that the water has depth yt and that the speed of a molecule pouring out at the bottom is determined by conservation of 15 energy Namely as it fell from the top surface it gained potential energy proportional to y and this became kinetic energy at the bottom propor tional to 32 Actually y is the speed of the top surface of the water not the bottom but these are proportional in a cylindrical tank Also we could assume a fraction of the energy is lost in friction In any event we get 32 proportional to y or 2 fax1 The constant is negative because the water depth is decreasing We ll set a 1 for convenience The slope eld looks like the gure Figure 51 Depth of water in the leaking bucket Separation of variables gives if y gt O 73712 dy dt 23412 it c so here we must have t lt 0 otherwise the square root would be negative c CY y 2 Note that ift c then y 0 so the derivation is no good there Comparing our formula to the slope eld we see what must happen 2 i W 20 1ftltc 0 otherwise 16 The nal point here is that if our initial condition is say y0 0 then any choice of c lt 0 gives an acceptable solution This means that if the bucket is empty now it could have become empty at any time in the past Again we consider this kind of behavior to be unusual in the sense that we don t usually notice or want examples of nonunique behavior Curiously in this problem nonuniqueness ts reality perfectly EXAMPLE For you calculate x73 dx dt 7x 2 1H c x72 7t 7 20 so t lt 720 Then at t O we get 73 2 70 7 20 20 719 x72 7t 19 x i7t 19712 Now recheck the initial conditions to determine the sign 73 i0 19 12 therefore use the minus sign I 77t19 12 for t lt 19 Check it 73 719712 is 0k x 777t rg xa is also 0k PROBLEMS 1 Repeat the previous example x xa for the case MO 3 2 Solve x 7x3 with initial condition x0 73 3 Solve x 7x3 with initial condition x0 3 4 Sketch slope elds for problems 1 and 3 5 Solve y y 7 ya 6 Solve x tzx O 7 Solve x tax O 8 Solve x 513 9 Solve x 0281 cos 0x 10 Solve x atx 11 What s Tong with this x 2 72 dt lnc2 t x2 et c There are at least two errors 6 Separable Partial Differential Equations TODAY Another use of the term separable This time PDE You could wait to read this until after you have read Lecture 28 if you plan to focus on O Traditionally a distinction in terminology is made between ordinary DE with the variables separable and the PDE solvable by separation of variables That sounds confusing doesn t it It is not just the sound The method looks different too Here is an example using a transport equation utcum0 The method of separation of variables goes like this Suppose we look for special solutions of the form ux t XxTt This does not mean that we think all solutions would be like this We hope to nd that at least some of them are like this Substitute this into our PDE and you get XmT t cXmTt 0 Now what The problem is to untangle the z and if somehow Notice that if you divide this equation by Xm then the rst term will at least no longer contain m Lets do that assuming temporarily that we are not dividing by X Tt Xm Now what If you divide by Tt you can clean up the second term without hurting the rst one Lets do that too T t lt gt CX z T05 X 90 Ttc 0 0 The rst term only depends on 25 but the second term does not depend on t at all The more you think about that the stranger it becomes We want to argue that this forces the two terms to be constant Write T t X795 Tt C X95 For example when x 0 the right hand side has a certain value perhaps it is 537 Lets write 18 r where for is 537 or whatever So the left side has to be 537 too Next change z to something else say z 15 This doesn t affect the left side 537 because there is no z in the left side But then the right side has to still be 537 You see the equation means that each side is the same constant Calling that constant for we have that Tt ichOf and Xm TX 18 where r is some number Thus we have split our PDE into two ODEs at least for the purpose of nding a few solutions maybe not all of them Fortunately after all these new ideas these ODEs are equations that we already know how to solve We nd Tt 5 0quot and Xz e or any constant multiples of these We have tentatively found a lot of solutions of the form uxt admid and constant multiples of these for our PDE Among these are em ea m 165250t725z etc It is always important to check any proposed solutions particularly since our derivation of these was so new a method In fact going a little further lets try a linear combination of two of them umt exist 166250t725z We get at 705 25 3165250P25m and um em c j 7 25165250t 25m So at cum 0 right enough Apparently you can add any number of these things and still have a solution That worked really well in a certain sense But on the other hand those are awfully big dunes because of the exponentials If you have read about traveling waves on page 8 you know a lot of other solutions to the transport equation So we won t worry about the fact that the separation argument only gave a few special solutions Our purpose was to introduce the method Two Kinds of Separating Finally lets compare the separable variables7 in ODE and separation of variables7 in PDE This is rather strange but worth thinking about When we separate the ODE y 0y 0 we write di 24 icdt and we still consider the two sides to be dependent on each other because we are looking for y as a function oft This has to be contrasted with our T X 757 T X above for the conservation law case Here we View z and t as independent variables and T and X as unrelated functions so the approach is different But the strangeness of this comparison emphasizes the fact that our deriva tions are a kind of exploration not intended as proof of anything The plan is we explore to nd the form of solutions but then we check them to be sure That way anything bogus won t botherus PROBLEMS 1 Above when we found the ODEs Tt icht and Xx rXx we wrote down some answers Tt 67C and Xx eMC pretty fast lf you aren t sure about those solve these for practice 0 E 7 y o 37 73y 0 T t 5Tt O o f s cfs where c is a constant 2 Try to nd some solutions to the equation ut uw u O by the method of separation of variables Do any of your solutions have the form of traveling waves Do they all travel at the same speed 3 The equation u aux u2 0 can t be solved by separation of variables Try it and explain why that doesn t work 4 In problem 3 however there are some traveling wave solutions Find them if you want a challenge But before you start observe carefully that there is one special wave speed which is not possible Reading the differential equation carefully can you tell what it is 7 Existence and Uniqueness and Software TODAY We learn that some equations have unique solutions some have too many and some have none Also an introduction to some of the available software We have seen an initial value problem the leakey bucket problem for which several solutions existed This phenomenon is not generally desirable be cause if you are running an experiment you would like to think that the same results will follow from the same initial conditions each time you repeat the experiment On the other hand it can also happen that an innocent looking differential equation has no solution at all or a solution which is not de ned for all time As one example consider x2m210 This equation has no real solution See why For another example consider 1 132 By separation of variables we can nd some solutions such as 05 31 but this function is not de ned when t 3 In fact we should point out that this formula i has to be considered to de ne two functions not one the domains being foo 3 and 3 00 respectively The reason for this distinction is simply that the solution of a differential equation has by de nition to be differentiable and therefore continuous We therefore are interested in the following general statement about what sorts of equations have solutions and when they are unique and how long in time these solutions last It is called the Fundamental Existence and Uniqueness Theorem for Ordinary Differential Equations which big title means need I say that it is considered to be important 71 The Existence Theorem The statement of this theorem includes the notion of partial derivative which some may need practice with So we remind you that the partial derivative of a function of several variables is de ned to mean that the derivative is constructed by holding all other variables constant For exam ple if ft z 225 7 cost then 2x25 and 2 sint 21 EXISTENCE AND UNIQUENESS THEOREM Consider an initial value problem of the form 9005 1 75 t0 0 where f to and mo are given Suppose it is true that f and are continuous functions of t and z in at least some small rectangle containing the initial condition 0250 Then the conclusion is that there is a solution to the problem it is de ned at least for a small amount of time both before and after to and there is only one such solution There is no guarantee about how long a time interval the solution is de ned for Now let s see what went wrong with some of our previous examples For the leakey bucket equation we have ft z i and 7 These are just ne and dandy as long as 0 gt 0 but when mo 0 you can t draw a rectangle around the initial conditions in the domain of f and even worse the partial derivative is not de ned because you would have to divide by 0 The conclusion is that the Theorem does not apply Note carefully what that conclusion was It was not that existence or unique ness are impossible but only that the theorem doesn t have anything to say about it When this happens you have to make a more detailed analysis A theorem is really like the guarantee on your new car It is guaranteed for 10000 miles but hopefully it does not just collapse and fall to pieces all over the road as soon as you reach the magic number Similarly some of the conclusions of a theorem can still be true in a particular example even if the hypotheses do not hold To see it you can t apply the theorem but you can make a detailed analysis of your example So here in the case of the leakey bucket we do have existence anyway For the rst example above z 2 2 1 0 the equation is not at all in the form required for the theorem You could perhaps rearrange it as z V71 x but this hurts my eyesithe right side is not even de ned within the realm of real numbers in which we are working So the theorem again says nothing about it For the other example above z 2 the form is alright and the right side is ft z 2 This is continuous and 2x which is also continuous The theorem applies here but look at the conclusion There is a solution de ned 22 for some interval oft around the initial time7 but there is no statement about how long that interval is So our original puzzlement over the short duration of the solution remains If you were a scientist working on something which might blow up7 you would be glad to be able to predict when or if the explosion will occur But this requires a more detailed analysis in each caseithere is no general theorem about it A very important point about uniqueness is the implication it has for our pictures of slope elds and solution curves Two solution curves cannot cross or touch7 when uniqueness applies PRACTICE Look again at Figure 51 Can you see a point at which two solution curves run into each other Take that point to be 0 to Read the Fundamental Theorem again What goes wrong 7 2 Software In spite of examples we have seen so far7 it turns out that it is not possible to write down solution formulas for most differential equations This means that we have to draw slope elds or go to the computer for approximate solutions We soon will study how approximate solutions can be computed Meanwhile we are going to introduce you to some of the available tools There are several software packages available to help your study of differ ential equations These tend to fall into two categories On one side are programs which are visual and easy to use7 with the focus being on using the computer to show you what some of the possibilities are You point and click and try things7 and see a lot without working too hard On the other side are programs in which you have to do some active programing yourself While they often have a visual component7 you have to get your hands dirty to get anything out of these The result is that you get a more concrete un derstanding of various fundamentals Both kinds have advantages It seems that the best situation is not to choose one exclusively7 but to have and use both kinds In the easy visual category there are java applets available at http www math cornell edutherrellde and http mathrice edudfielddfpp html Probably the earliest userfriendly DE software was MacMath7 by John Hub bard MacMath was the inspiration for de The other approach is to do some programing in any of several available lan guages These include matlab its free counterpart octave and the freeware program pr There are also java applets on partial differential equations These are called Heat Equation 1D7 Heat Equation 2D7 Wave Equation 1D7 and Wave Equa tion 2D7 and are available from http www math cornell edutherrell Figure 71 This is the slope eld for the Logistic Equation as made by dfield Question Do these curves really touch Read the Theorem again if you re not sure PROBLEMS 1 Find out how to download and use some of the programs mentioned in this lecture try a few simple things7 and read some of the online help which they contain 2 Answer the question in the caption for Figure 71 8 Linear First Order Equations TODAY Applications and a solution method for rst order linear equations Today we consider equations like N am b or m am I These are called rst order linear equations Linear is here a moreor less archaic use of the word and means that z and m only occur to the rst power We will say later what the word linear means in a more modern sense The coefficients a and b can be constants or functions of if You should check that when a and b are constant the solutions always look like Mt 015 02 where 01 is an arbitrary constant but 02 has to be something speci c When I say check it doesn t mean derive it but just substitute into the equation and verify Make sure you see how similar this is to the case when I O PRACTICE Solve x 2x 3 Among these equations is the very rst one we ever did 3 028y which we solved by separation of variables Some but not all of these equations can be solved the same way The simplest one of them is m 0 which is so easy as not to be useful The next simplest is m b and this one is very important because it holds a key to solving all the others We call this one the easy kind of differential equation because it does not involve x at all and can be solved just by integrating 7 7 b m bdt Wouldn t it be nice if all differential equations could be solved so easily Maybe some can Somebody once noticed that all these equations can be put into the form of the easy kind if you will only multiply the equation by a cleverly chosen integrating factor For example consider again x2x3 Multiply by 52 and you will get EZtm252t362t Then and this is the cool part recognise the left hand side as a total deriva tive like you have in the easy kind e my 352 This uses the product rule for derivatives Now all you have to do is integrate and you are done 52 352t dt 52 C 3 zt 5 05721 What do you think the integrating factor should be for m 7 13m cost In this case which will be a homework problem you still have an integral to do which is a little harder and you probably have to do integration by parts There is a general formula for the integrating factor which is not too important but here it is For the equation above m ax b an integrating factor is efadt This always works at least within the limits of your ability to do the integrals As for where this formula comes from we have devised a homework problem so that you can answer that yourself As to the deeper question of how anybody ever thought up such a scheme as integrating factors in the rst place that is much harder to answer Now let s look at some applications of these equations 81 Newt0n7s Law of Cooling Newton s law of cooling is the statement that the exponential growth equa tion z km applies sometimes to the temperature of an object provided that z is taken to mean the difference in temperature between the object and its surroundings For example suppose we have placed a 100 degree pizza in a 600 degree oven We let Mt be the pizza temperature at time 25 minus 600 This makes z negative while z is certainly positive because the pizza is heating up Therefore the constant k which occurs must be a negative number The solution to the equation does not require an integrating factor it is 05 Ce and C 100 7 600 as we have done previously Therefore the 26 pizza temperature is 600 05 600 7 5005 We don t have any way to get k using the information given It would suffice though to be told that after the pizza has been in the oven for 15 minutes its temperature is 583 degrees This says that 583 600 7 50051 So we can solve for k and then answer any questions about temperature at other times Notice that this law of cooling for the pizza can be written as p 7 kp 600k where pt is the pizza temperature which is a rst order linear equation A different environment arrives if we move the pizza to the 80 degree kitchen A plot of the temperature history under such conditions is in Fig 41 It is not claimed that the solution is differentiable at t 6 when the environment changes Figure 81 The pizza heats and then cools Note the trick used to Change the environment from 600 degrees to 80 degrees at time 6 the equation was written as p pabsignt6 and a and b selected to achieve the 600 and 80 gure by dfield There is another use of Newton s law of cooling which is not so pleasant to discuss but it is real The object whose temperature we are interested in is sometimes of interest because the police want to know the last time it was alive The body temperature of a murder victim begins at 37 C and decreases at a rate dependent on the environment A 30 degree road is one environment and a 10 degree morgue table is another environment Part of the history is known namely that which occurs once the body has been found Measurements of time and temperature can be used to nd k Then 27 that part of the history from the unknown time of death to the known time of discovery has to be reconstructed using Newton s law of cooling This involves functions not unlike those in Figure 81 It should be clear that this involves estimates and inaccuracies but is useful nonetheless 8 2 Investments Moving to a more uplifting subject our bank account equation 3 028y can be made more realistic and interesting Suppose we make withdrawals at a rate of 3500 per year This can be included in the equation as a negative in uence on the rate of change 1 028 7 3500 Again we see a rst order linear equation But the equation is good for more than an idealised bank account Suppose you buy a car at 28 nancing paying 3500 per year Now loosen up your point of view and imagine what the bank sees From the point of view of the bank they just invested a certain amount in you at 28 interest and the balance decreases by withdrawals of about 3500 per year So the same equation describes two apparently different kinds of investments In fact it is probably more realistic as a model of the loan than as a bank account By changing the 73500 to 3500 you can model still other kinds of investments EXAMPLE A car is bought using the loan as described above If the loan is to be paid off in 6 years what price can we afford The price is 110 We need 4y 028 7 3500 with y6 0 Calculate e OZEty7028 T 02 7350067 02 6 02 I 7350067 gt e 02 3500028e 028 y 3500028ce 0m 125000ce 023 116 0 125000 06 0236 This implies a 7105669 y0 125000 7 105669 19331 PROBLEMS 1r Solve 5711313 cos t 2 This problem shows where the formula cladt for the integrating factor comes from Suppose we have the idea to multiply x ax b by a factor f so that the result of the multiplication is fb Show that for this plan to work you will need to require that f af Deduce that cladt will be a suitable choice for 28 3 The temperature of an apple pie is recorded as a function of time It begins in the oven at 450 degrees and is moved to an 80 degree kitchen Later it is moved to a 40 degree refrigerator and nally back to the 80 degree kitchen Make a sketch somewhat like Figure 81 which shows qualitatively the temperature history of the pie 4 A body is found at midnight on a night when the air temperature is 16 degrees C lts temperature is 32 degrees and after another hour its temperature has gone down to 305 degrees Estimate the time of death 5 Sara s employer contributes 3000 per year to a retirement fund which earns 3 interest Set up and solve an initial value problem to model the balance in her fund if it began with 0 when she was hired How much money will she have after 20 years 6 Show that the change of variables x 1y converts the logistic equation y 028y 7 yz of Lecture 4 to the rst order equation x 7028x 7 1 and gure out a philosophy for why this might hold 7 A rectangular tank measures 2 meters eastwest by 3 meters northsouth and contains water of depth xt meters where t is measured in seconds One pump pours water in at the rate of 005 masec and a second variable pump draws water out at the rate of 007 002 coswt masec The variable pump has period 1 hour Set up a differential equation for the water depth including the correct value of w 3 WE Figure 82 Here xt is the length of a line of people waiting to buy tickets Is the rate of change proportional to the amount present Does the ticket seller work twice as fast when the line is twice as long 9 oiler s Euler s Numerical Method TODAY A numerical method for solving di erential equations either by hand or on the computer several ways to run it and how your calculator wor Today we return to one of the rst questions we asked If your bank balance yt is 2000 now and dydt 028y so that its rate of change is 56 per year now about how much will you have in one year77 Hopefully you guess that 2056 is a reasonable rst approximation and then realize that as soon as the balance grows even a little the rate of change goes up too The answer is therefore somewhat more than 2056 The reasoning which lead you to 2056 can be formalised as follows We consider 90 1 75 t0 0 Choose a stepsize h and look at the points 251 to h 252 to 2h etc We plan to calculate values zn which are intended to approximate the true values of the solution 05 at those time points The method relies on knowing the de nition of the derivative t h 7 zt t 1 d z 1m h We make the approximation at i 1 m Then the differential equation is approximated by the difference equation nl 7 mn fltn7 tn EXAMPLE With 11 l the bank account equation becomes 7 yquot 028 yn1 M 0287111 30 This leads to yi yo 1 028hyo 1028yo 2056 as we did in the rst place For a better approximation we may take 11 2 but then 5 steps are required to reach the oneyear mark We calculate successively 11 290 028hyo 10056110 2011200000 112 y1028hy1 10056111 2022462720 ya M 028hy2 10056112 2033788511 114 ya 028hy3 10056113 2045177726 ys 111 028hy4 10056114 20336630721 Look you get more money if you calculate more accurately Here the bank has in effect calculated interest 5 times during the year Continuously Compounded77 interest means taking h so close to 0 that you are in the limiting situation of calculus Now at this point we happen to know the answer to this particular problem It is y1 200050280 2056791369 Continuous compounding gets you the most money Usually we do not have such formulas for solutions7 and then we have to use this or some other numerical method This method is called Euler s Method7 in honour of Leonard Euler7 a math ematician of the 18th century He was great7 really The collected works of most famous scientists typically ll a few books To see all of his in the library you will have to look over several shelves This is more impressive still when you nd out that though he was Swiss7 he worked in Russia7 wrote in Latin7 and was in later life blind DO you know what 5 stands for7 as in 2718 exponential maybe It stands for Euler He also invented some things which go by other people s names S0 show some respect7 and pronounce his name correctly7 oiler EXAMPLE Now we ll do one for which the answer is not as easily known ahead of time Since the calculations can be tedious7 this is a good time to use the computer We will set up the problem by hand and solve it two ways on the machine Assume that pt is the proportion of a population which carries but is not a ected by a certain disease virus7 initially 8 The rate of change is in uenced by two factors First7 each year about 5 of the carriers get sick7 so are no longer counted in p Second7 the number of new carriers each year is about 02 of the population but Varies a lot seasonally The differential equation is p 705p021sin27rt 190 08 31 Euler s approximation is to choose a stepsize h put tn nh p0 08 and approximate Mt by pr Where Pn1 pa M70517 021 sintn To compute in octave or matlab use any text editor to make a function le to calculate the right hand side file myfunan function rate myfunctp rate 05p 021sin2pit 39end file myfuncm Then make another le to drive the computation Xfile oiler m h 131 08 for n 15999 pn1 pn hmyfuncnhpn end plot Oh60hp 39end file oilerm Then at the octave prompt giVe the command oiler You should get a graph like the following Note that We used 6000 steps of size 01 so We have computed for 0 g t g 60 Figure 91 You can see the seasonal Variation plainly and there appears to be a trend to leVel OH This is a dangerous disease apparently gure made by the plot commands shown in the text 32 For additional practice you should Vary h and the equation to see what happens There are also more sophisticated methods than Euler s One of them is built into matlab under the name ode23 and you are encouraged to use it even though we aren t studying the internals of how it works You should type help ode23 in matlab to get information on it To solve the carrier problem using ode237 you may proceed as follows Use the same myfunc m le as before Then in Matlab give the commands tp ode23 myfunc 06008 plottp Your results should be about the same as before7 but generally ode23 will give more accuracy than Euler s method There is also an ode45 We won t study error estimates in this course However we will show one more example to convince you that these computations come close to things you already know Look again at the simple equation z m with z0 1 You know the solution to this by now7 right Euler s method with step h gives n1 mn l hmn This implies that 2 1 hm 1 h2 mn 1 h Thus to get an approximation for z1 e in n steps7 we put h 171 and receive 1 V L Let s see if this looks right With 71 2 we get 322 94 225 With 71 6 and some arithmetic we get 766 i 25216267 and so forth The point is not the accuracy right now7 but just the fact that these calculations can be done without a scienti c calculator You can even go to the grocery store7 nd one of those calculators that only does 7quot7 and use it to compute important things Did you ever wonder how your scienti c calculator works Sometimes people think all the answers are stored in there somewhere But really it uses ideas 33 and methods like the ones here to calculate many things based only on 7 Isn t that nice EXAMPLE We ll estimate some cube roots by starting with a differential equation for Mt tla Then x1 1 and xt t wa These give the differential equation 1 9 y Then Euler s method says In1 xn and we will use x0 171 1 1 1 1 E 1033333 Therefore 11 3 10333 10645 I 1 2 1312 Therefore 12 3 10645 etc For better accuracy7 11 can be decreased PROBLEMS 1 What does Euler s method give for if you approximate it by setting Mt and solving 1 7 4 7 x 7 27 With x1 71 Use 1 2 and 4 steps ie7 h 1 57 25 respectively 2 Solve y cos y2 with y0 O for 0 g t g 3 by modifying the oiler m and myflmcm les given in the text 3 Solve the differential equation in problem 2 by separation of variables 4 Redo problem 2 using ode45 and your modi ed myfunc m as in the text 5 Run dfield on the equation of problem 2 Estimate the value of y3 from the graph 6 Compare your answers to problems 2 and 3 Is it true that you just computed tan 713 using only 7 and cosine Figure out a way to compute tan 713 using only 7 7 Solve the carrier equation 17 705p 021 sin27rt using the integrating factor method of Lecture 8 The integral is pretty hard but you can do it Predict from your solution the proportion of the population at which the number of carriers levels OH after a long time remembering from Figure 81 that there will probably continue to be uctuations about this value Does your number seem to agree with the picture Note that your number is probably more accurate than the computer s 8 Newton s law of cooling looks like u fan when the surroundings are at temperature 0 This is sometimes replaced by the StephanBoltzmann law u ibu l if the heat is radiated away rather than conducted away Suppose the constants a b are adjusted so that the two rates are the same at some temperature say 10 Kelvin Which of these laws predicts faster cooling when u lt 107 u gt 107 34 10 Second Order Linear Equations TODAY 2nd order equations and some applications The characteristic equation Conservation laws The prototype for today s subject is z ix You know the solutions to this already though you may not realize it Think about the functions and derivatives you know from calculus In fact here is a good method for any differential equation not just this one Make a list of the functions you know starting with the very simplest Your list might be 0 1 c 25 Now run down the list trying things in the differential equation In x 7m try 0 Well what do you know It works The next few don t work Then cost works Also sint works Frequently as here you don t need to use a very long list before nding something As it happens cost and sint are not the only solutions to z ix You wouldn t think of it right away but 2 cost 7 5 sint also works and in fact any 01 cost 62 sint is a solution PRACTICE Find similar solutions to x 79x The equations z oz happen to occur frequently enough that you should know all their solutions Try to guess all solutions of z 0 and z 25 for practice 101 Linearity We will consider equations of the forms am bz cm 0 am bi oz dcosft 35 These are called second order linear equations and the rst one is called homogeneous The strict meaning of the word 7linear7 is being stretched somewhat here but the terminology is traditional Really an operation L which may be applied to functions such as Lm ax bx cm is linear if it is true that Lz y Lm Ly and Law This does hold here and the consequence is that if z and y are both solutions to Lm 0 then linear combinations like 2x 5y are also solutions That is true for the homogeneous equation because we have L2z 5y L2m L5y 2Lm 5Ly 0 However the nonhomogeneous second equation does not have this property You can see this yourself Suppose z and y are solutions to the same equation say z 3x 2 and y 3y 2 Then we want to see whether the sum 5 z y is also a solution We have 5 35 z y 3x 3y 2 2 4 This is a di erent equation since we got 5 35 4 not 5 35 2 So it is a bit of a stretch to call the nonhomogeneous equation linear but that is the standard terminology You should also be sure you understand that for an equation like z z2 0 even with 0 on the right side the sum of solutions is not usually a solution 102 The Characteristic Equation Next we look at a method for solving the homogeneous second order linear equations Some of the nonhomogeneous ones will be done in Lecture 13 The motivation for this method is the observation that exponential functions have appeared several times in the equations which we have been able to solve Let us try looking down our list of functions until we come to the ones like 5quot Trying z e in am bz cm 0 we nd arzequot bra csquot ow2 br 35 This will be zero only if arz I c 0 since the exponential is never 0 This is called the characteristic equation For example the characteristic equation of z 4m 7 3x 0 is r2 4r 7 3 0 See the similarity We converted a differential equation to an algebraic equation that looks abstractly similar Since we ended up with quadratic equations we will get two roots usually hence two solutions to our differential equation 36 EXAMPLE x 413 7 3x O has characteristic equation r2 4r 7 3 0 Using the quadratic formula the roots are r 72 i So We have found tWo solutions x e 2 7 t and x elt7z t By linearity We can form linear combinations 1t 016727 F0t 025721 rm EXAMPLE You already know x 7x Very Well right But our method giVes the characteristic equation r21 O This does not have real solutions In Lecture 12 We ll see how the use of complex numbers at this stage will unify the sines and cosines We know With the exponential functions Which our characteristic method produces EXAMPLE The Weirdest thing Which can happen is the case in Which the quadratic equation has a repeated root For example r 52 r2 10r 5 0 corresponds to x 10x 25x O The only root is 75 so We only get one solution xt 6 and multiples of that If these Were the only solutions We could solVe for say the initial conditions x0 3 and x 0 7 but could not solVe for x0 3 and x 0 714 As it turns out We have not made a general enough guess and have to go further down our list of functions so to speak until We get to things like t6quot These Work So We end up With solutions xt 81 6206 5t for this problem This is a special case and not too important but it takes more Work than the others for some reason EXAMPLE x 7 75x x0 1 50 2 General solution is t 01 cosh3t 02 sinEt Use initial conditions x 01 x 302202 o 2 Mt coshgt E s1nEt EXAMPLE yl 210 1 MW 2 Characteristic equation is r2 4 r i2 y ale2t 026722 y0 01 62 1 and yO 281 7 202 2 Then eliminate 82 2y0 y 0 401 401 1 02 17 01 0 yt 6 you check it 37 103 Conservation laws part 2 Sometimes a second order equation can be integrated once to yield a rst order equation Good things happen when this is possible since we have many tools to use on rst order equations and because usually the resulting equation can be interpreted as conservation of energy or some similar thing For example let s pretend that we don t know how to solve the equation x ix You can try to integrate this equation with respect to 25 Look what happens m dt izdt You can do the left side getting z but what happens on the right You can7t do it can you The right side doesn7t integrate to mt or something like that does it I m asking all these questions because this is a common place to make mistakes Really you just can t do the integral on the right because the integrand 05 is not yet known But there is a way to x this problem It is very clever because it seems at rst to complicate the problem rather than simplifying it Multiply the equation x 7m by m and see what happens You get mm imm Now stare at that for a while and see if you can integrate it now mm dt 7mzdt You may notice if you look at this long enough that according to the chain rule you can now integrate both sides That is because mm is the derivative 1 1 of m2 and z is the derivative of z 2 So integrating you get 1 1 59532 C 5902 Now there are several things to observe about this First we don t have a second order equation any more but a rst order instead Second we can make a picture for this equation in the z m plane and it just says that solutions to our problem go around circles in this plane since we just got the equation of a circle whose size depends on the constant 0 Third there is a physical interpretation for the rst order equation which is conservation of energy Not everyone who takes this course has had physics yet so the next bit of this discussion is optional and may not make too much sense if you have not seen an introduction to mechanics Conservation of energy means 38 the following z is the position and m the velocity of a vibrating object Peek ahead to Figure 131 if you like to see what it is The energy of this object is in two forms which are kinetic energy and potential energy The kinetic energy gmvz is the z 2 term The potential energy kmz is the 2 term So what is c It is the total energy of the oscillator and the going around in circles is physically the fact that when something swings back and forth the energy is periodically transferred from to potential to kinetic and back The child on a swing has a lot of kinetic energy at the bottom of the path moving fast and all that becomes potential energy by the top of the swing when he is instantaneously not moving and the cycle repeats This kind of calculation nding the integral of the equation or some multiple of the equation is called nding a conservation law It cannot always be used but is a powerful thing to know about Here is an example of the power It is a little more advanced than the rest of the lecture but I bet you will like it THEOREM There is no other realvalued solution to x 7x than the ones you already know about You probably wondered whether anything besides the sine and cosine had that property Of course there are the linear combinations of those But maybe we just aren t smart enough to gure out others The Theorem says no that s all there are PROOF Suppose the initial Values are x0 a and 50 b and we write down the answer xt that we know how to do Then suppose your friend claims there is a second answer to the problem called yt Set ut xt 7 yt for the difference which we hope to prove is O hen u in We know from the conservation law idea that then What is c The initial Values of u are 0 so 0 0 Do you see why that makes u identically O7 PROBLEMS 1 Suppose x3 1312 O and 2 1322 O and set 5 x1 x2 Calculate s 52 to see why the sum of solutions is not usually a solution 2 Suppose x1 and x2 both solve x 13 0 30 2 and set 5 x1 x2 Find out why 5 is not a solution 39 3 Suppose 11 and x2 both solVe x x cos8t and set 5 x1 x2 Is 5 a solution 4 Solve 4yquot i 32 i y 0 210 1 MW 0 5 Solve the same equation as in problem 47 but With initial conditions 210 0 MW 1 6 Solve the same equation as in problem 47 but With initial conditions 210 2 MW 3 7 Can you see that the answer to problem 6 should be 2 times the ansWer to problem 4 plus 3 times the answer to problem 57 8 Find a conservation laW for the equation x x3 O 9 Do you think there are any conservation laws for x z x O7 10 What s Tong with this y y2 07 r2 1 07 r 71 y e t c There are at least 3 errors 40 11 The Heat Equation TODAY We now have enough information to make a good start on the partial differential equation of heat conduction If you want to save PDE for later you could read this lecture after Lecture 28 The heat equation looks like this 814 8214 E 7 usually abbreviated ut hum and it describes the temperature of an object in which heat energy is allowed to ow by conduction in the z direction Heat can also move with a owing uid but that known as convection is not what this is about Heat can also move as radiation like the warming you feel near a re even though the air itself may be cold It is not about that either It can describe the changing temperature along a metal bar or through a thick wall The k is a material constant For now take k 1 Later we give a derivation of this from physical principles Right now lets see how close we can come to solving it Try the separation idea Look for solutions of the form Mm t XxTt This does not mean we think they are all like this We hope that some are Substituting we need TtXx TtX z Now divide by XT assuming for now that this is not 0 Of course later we might nd that sometimes it is 0 but by then we will have made an independent check without dividing anything that our answer really is a correct answer So right now we are just exploring not really proving anything We get T X T 7 Y Again we see that since the left side only depends on t and the right does not depend on t that neither side depends on t or on m They are the same constant call it 70 We must then solve the ordinary differential equations T icT and X 70X 41 Observe two good things happened simultaneously First the PDE broke into two ODEs Second the ODEs turned out to be some that we already know how to solve EXAMPLE Suppose a 1 We know how to solve T 7T and X 7X We know solutions to those Tt cleft and Xx 02 cosc ca sinc Thus some solutions to the heat equation ought to be ux t Cit cosx uxt 367t sinx ux t e t6 cosc 7 53 sinx Now we must check those For one thing they are 0 in a few places and maybe that matters Or maybe we overlooked something else in our sepa ration argument PRACTICE Plug those into the heat equation and Verify that they really do wor PROBLEMS 1 Suppose c 2 instead of 1 in our separated ODEs T icT and X 1X What solutions T X u do you nd 2 What if c is some other positive number besides 1 or 2 What solutions T X u do you nd 3 What happens if c 0 What solutions T X u do you nd 4 Sketch a graph of the temperature for a few times in the solution ux t 36 t sinx above 5 What happens if c is a negative number say 71 Sketch a graph of the temperature for a few times Could that u be the temperature of an object There is something wrong with that as a object temperature in most cases isn t there So we don t use negative 0 usually 6 Find some solutions to the heat equation u kum where the material constant C has been put back in 42 12 Complex Numbers and the Characteristic Equa tion abi TODAY The complex numbers and a motivation for the de nition of C More second order equations Complex numbers are expressions of the form a bi Where a and b are real numbers You add subtract and multiply them just the way you think you do except that i2 71 S0 for example 25 32 625 2253 912 7325 15 If you plot these points on a plane plotting the point z y for the complex number z yi you will see that the angle from the positive z axis to 25 3i gets doubled when you square and the length gets squared Addition and multiplication are in fact both very geometric as you can see from the gures Figure 121 Addition of complex numbers is the same as for Vectors Mul tiplication adds the angles While multiplying the lengths The left picture illustrates the sum of 3 139 and 1 2139 While the right is for the product gure made by the compass command in matlab PRACTICE Use the geometric interpretation of multiplication to gure out a square root of i 43 Division of complex numbers is best accomplished by using this formula for reciprocals 1 i a 7 bi abiia2b2 PRACTICE Verify that this reciprocal formula is correct ie that using it you get aibia bi 1 You are now equiped to do arithmetic such as solving 27 27 5i2 3i for 2 To do algebra correctly we should point out that two complex numbers are equal by de nition when the real and imaginary parts are equal If 2 a bi and a and b are real then we write rez a and imz b not bi Warning If you know that a bi 0 di you cannot conclude that a c and b d For example 1 1 0i 0 However if you know in addition that a b c and d are real then you can conclude that a c and The quadratic formula for solving azz bx c 0 is still good if everything in sight is complex but taking square roots can be tedious That is why we practiced it above There is another important construction which is motivated by algebra and differential equations and very useful That is raising e to a complex power We set 5 9 595 by analogy with known properties of real exponentials but this still requires a de nition of e We claim that the only reasonable choice is cost isint The reason is as follows The whole process of solving second order equations by the characteristic equation method depends on the formula t Let s require that this hold also when r i Writing 5 ft igt this requires 1quot 2 g 2 f2 g 79 so 1quot g and g 1 These should be solved using the initial conditions 50 1 f0 ig0 These give f0 1 f 0 0 and f if The only solution is ft cost and gt sint Therefore our de nition becomes eltstigt 55cost i sint Now it turns out that we actually get 3 re for all complex r You should verify this for practice using our de nition This is a case of pulling 44 oneself up by one s own bootstraps we solve a simple differential equation to make a de nition of e to a complex power so that we can use that to solve more complicated differential equations Now we can do more examples EXAMPLE x QI 3x O The characteristic equation which works by all that has been done above is r2 2r 3 O The solutions are r 71 i 2 71 i So there are solutions to the diHerential equation of the form Mt 81571i t Czeklii Given some initial conditions we can nd the constants For instance if we want to have x0 5 and x 0 0 then we must s0 ve 900 5 01 C2 50 0 71i c1717i c2 The sum of these gives 5 ix ci 7 02 S0 C 78 5 78 7 5i 1 2 N2 2 The rst equation then gives 5139 520 77 2 Finally C2 5217 and c1 5 7 C2 52717 so 7 E 7 7 lt71mgtt E 7 717mgtt t 7 2 lt 1 e 2 1 6 You may notice that the answer besides being messy looks very complex Now the last time you went into the lab all the instruments were probably reading out real numbers weren t they So it is common to rewrite our solution in a way which shows that it really is real after all To do this the following new idea is required If 05 is a complex solution to a second order linear differential equation with real coe icients then the real and imaginary parts of z are also solutions For example 53 is a solution to z 79 The real and imaginary parts are respectively cos3t and sin3t and these are certainly solutions also To see why this works in general suppose that z uiv solves az bm cx 0 This says that au iv bu iv Cu iv 0114 bu 014 ow bv cu 0 Assuming that a b c u and v are all real you can conclude that au bu 014 and av bv cu are also 0 But note that if some of the coefficients had not been real or if the equation had been something like z 2 0 then the argument would not 45 have worked Equations with explicit complex coe icients sometimes occur in physics in spite of what we all noticed above about lab instruments Schrodinger s Equation is an example In such cases you can t simply use the real and imaginary parts So let s rework the previous example slightly After nding the values of r above we wrote down among other things a complex solution 05 e 1i t e tcos t isin t Using the real and imaginary parts we form solutions x05 ale cos t age sin t If we apply the same initial conditions again we need z0 5 a1 and m 0 0 7a1 ag Solving these we end with 5 7 7 75 It is important to realise that this is equal to the complex looking solution we found previously 05 55 i cos t t sin t EXAMPLE Solve for a 3 7 7 2 ia 7 z 7 7 2739 71476397393392 717 72i 2z Z 73l22i2 Z5 2 Z 5 Z EXAMPLE Solve y 2y 2y0 The Characteristic equation is r2 2r 2 0 so by the quadratic formula 7 L H 7 i 139 One solution is eHM e tcost isint Taking the real and imaginary parts the solution is yt C167 cost 8267 sint Don t forget to Check it EXAMPLE x 2x x O r2 2r 1 r 12 Or 7171 x 81 020672 Check x 026 t 7 01 0206 t 026 t 7 x x 7026 t 7 x 7026 t 7 026 t 7 x 7282 7t x x 2x x 720267t x 28267t 7 x x O 46 EXAMPLE x 2x 7 x O r2 2r 71 07 r 71 i m xt 816717 t Cgelt71 gtt PROBLEMS 1 Solve r2 76r100 r3 76r2 1 107 O7 and r2 76T7 10 O 2 Solve y 3171 4y O 3 Finda7 ifa0 ifai 4 Sketch the graphs of the functions e t cost e t cos3t and 67 sin3t 5 Verify that 23 requot for r a bi using the de nition 5amp1 t e cosbt i sinbt 6 For each t7 elit is a complex number7 which is a point in the plane So as t Varies a curve is traced in the plane Sketch it 7 A polynomial r2 br 1 c O has roots 7 72 i 139 Find 1 and c 8 A characteristic equation r2 brc O has roots 7 72 ii What was the differential equation 9 What Tong with this x 4x 07 r2 4 07 r iQi x 0162it 672 121 The Fundamental Theorem 0f Algebra Talk about lifting yourself by your own bootstraps A new number writ ten i was invented to solve the equation 2 1 0 Of course you have noticed that the people who invented it were not very happy about it con trast complex and imaginary with real and rational But soon the following theorem was discovered By introducing the new number i not only can you solve 2 1 07 but you can also solve at least in principle all these 2 2 07 z3 7i 07 m6 7 3 7 20x4 7 m3 Mm 7 391778i 432 O7 etc7 etc FUNDAMENTAL THEOREM OF ALGEBRA Let 2990 a0 aim 12962 an71n71 as be a polynomial of degree n gt 0 with any complex coefficients ak Then there are complex numbers n 72 76 which are roots of p and p factors as WE 907 W7 72 95 7m 47 As a practical matter we can handle the cases 10 x and a0 11 x2 easily We all know the quadratic formula There is a cubic formula that most people don t know for solving the n 3 case and a quartic formula that hardly anybody knows for the n 4 case which takes about a page to write Then something interesting happens ABEL7S THEOREM Let n 2 5 Suppose you gure out eVery formula that could eVer be written in terms of the coef cients ak and the operations of addition subtraction multiplication division and extraction of roots Then none of those formulas giVe you the roots of the polynomial So the roots exist but if you need them then you will usually have to approximate them numerically It is tough PROBLEMS 1 In the Fundamental Theorem we took the coef cient of xquot to be 1 just for conVenience But if you don t do that the factorization must be written differently For example 76xx2x72x3 is correct Figure out the number 0 in the case 730 5x 5x2 CI 7 2c 3 n 2 How must the factorization be written in general if you don t assume the coef cient of x is 17 3 You know that in the factorization 76xx2 x72x3 you have 76 733 In general what is the product of the roots 7 17 2 Tn in terms of parameters appearing in the Fundamental Theorem 4 You know that in the factorization 76 x x2 x 7 2c 3 the x term comes from x 72x 3x when you multiply the right hand side In general what is the sum of the roots r1 r2 7 in terms of parameters appearing in the Fundamental Theorem 48 13 Forced Second Order TODAY Forced equations and how to solve them using the undetermined coef cients method Natural frequency Consider the equations m 2m 0 y 2y 9sin3t The rst one is called the homogeneous form of the second one or the second is called a forced form of the rst and there are variations on this terminology Mechanically what they mean is as follows Since we know the solutions to the rst one don t we are t cl cos t 02 sin t it seems reasonable that this rst equation is about something Vibrating or oscillating It can be interpreted as a case of Newton s F ma law if you write it as 72m 1x Here x is the position of a unit mass m is its acceleration and there is a force 72m which seems to oppose the displacement m We call this a springimass system It can be drawn as in Figure 131 where z is measured up gt gt ltgt ltgt Figure 131 Unforced and forced springimass systems The 72m is interpreted as a spring force because it is in the direction opposite x if you pull the spring 154 units up then x 154 and the force is 7308 or 308 units downward This is also a system without friction as we see from the fact that there are no other forces except for the spring force and that the oscillation continues undiminished forever Note that the natural frequency77 of this system is 2 cyclessecond in the sense that the period 27r ofz1s ztzt 49 The soicalled forced equation involves an additional force as you can see if you write it as 72y 7 9 sin 3t 1y The picture in this case is like the right side of Figure 131 Now we turn to solution methods for the forced equation We are guided by the physics That is ask yourself what will happen if you start with a system which wants to vibrate at a frequency of 27 and somebody reaches in and shakes it at a frequency of What could happen If you think about it it is plausible that the motion might be rather complicated but that a part of the motion could be at each of these frequencies Let7s try that Assume yt zt Asin3t where z is the solution given above for the unforced equation Then y 2y m 7 32Asin3t 2zt Asin3t 7814311675 We want this to equal 9 sin3t Notice how the terms involving z dropped out We need 78A 9 so A 798 Our solution becomes yt c1 cos t 2 sin t 7 gsin3t In particular our physics tyle thinking did lead to a solution Some people who take this course have had physics and some have not had physics In any case we hope you can see that a little physics or banking or later bio logical knowledge goes a long way in helping to set up and solve differential equations Notice also the coe icients of the terms in this solution There are terms at the natural frequency 272 and a term at the forced frequency The coefficient of the forcedifrequency term is xed at 7 while the naturali frequency terms have arbitrary coefficients If you have initial conditions to apply now is the time EXAMPLE To solve y 2y 9 sin3t 210 4 MW 5 we use the solution found above yt C1COS t 02 sin t 7 g sin3t 50 We get y0 4 cl y 0 5 02f7 3 The answer becomes yt 4cos t sin t 7 g sin3t H This is graphed in the gure 132 Figure 132 Can you see tWo frequencies at Work here They are not the obvious ones Do problems 173 to see what they are There is a terminology that traditionally goes with this solution One calls 7 sin3t a particular solution to y 2y 9 sin3t It is also true that 293 cos t 7 gsin3t is a particular solution7 as are many other things One says that the solution is the sum of a particular solution and a solution of the homogeneous equation Other equations can be solved by the same method For example7 it is possible to solve y y y 3 at by guessing a particular solution of the form A Bet This will work However this type of forcing is really not very important for us Sines and cosines are the important things because they are the most practical if you are sitting in class and happen to notice a fan vibration in the ceiling of the room7 that is an example which can be modeled by one of these equations With sinusoidal forcing The spring and mass correspond to some of the ceiling materials7 and the sine or cosine forcing term represents the fan motor running at a constant speed EXAMPLE WWm Try y A sintB cost7 since sine alone Will not Work considering the 31 term Then y 31 2y 7A 2A 7 3B sint 73 QB 31 cost 51 We need A 7 3B 1 B 3A 0 Substituting the second of these into the rst gives A 7 373A 10A 1 Then and B 7 73A 73 Our particular solution is 1sint 7 3 cost For the homogeneous equation the characteristic equation is r23r2r1r20 so 7 71 or r 72 The answer is then yt 6167t C2 T2t 1sint 7 3cost This is shown in gure 1337 for the case Cl 1 62 71 Figure 133 A solution to y 3y 2y sin t Only the forced solution remains after some time the 6 t and 67 terms are transient There is one special case which requires more work to solve than the others This is the case which occurs when the forcing is at exactly the natural frequency7 and is called resonance Like z 4x cos2t Think about what happens then If your little brother is on the swing going back and forth every 3 seconds7 and you help by pushing once every 3 seconds7 pretty soon he will be going higher than he wants to go Mathe matically7 if you try a particular solution Acos2t it won7t work because you get 0 when you substitute it into the left side of the equation Taking these things together7 it is reasonable to try something like At cos2t Bt sin2t These actually work7 giving solutions which grow with time 52 PRACTICE Verify that the solution to x 4x cos2t is t itsin2t 01 cos2t 02 sin2t PROBLEMS 1 Use one of the addition formulas sinab sina cosb cosa sinb or cosab cosa cosb7 sina sinb to show that any function of the form A cosft B sinft can be written in the form Csinft 4 for some numbers 0 and Note that here we were combining sinusoids having the same frequency f27r 2 Similarly to problem 1 show how a combination of two sinusoids of dz erent frequencies can be written as a sum of products of sinusoids derive for example the identity sin8t 23 sin7t 33sin75tcos5t713cos75tsin5t 3 Combine the ideas of problems 1 and 2 to explain the two obvious frequencies in Figure 132 in particular that they are not 27 and 4 Solve x x x 7sin6t 5 Solve x 4x sin 3t and x 4x sin2t and discuss the difference 6 Show that the addition formulas for the sine and cosine are equivalent to the complex exponential rule ea eaeb 7 A challenge After a lecture involving 7frequency7 one day I noticed a crane at the physics building lifting a heavy oa he main load didn t swing much but an interme diate pulley seemed to oscillate too fast for safety on such large equipment Figure 134 A model for a crane holding a heavy load with an intermediate pulley in the cable Doesn t the main load rise twice as much as the pulley Verify that the pulley rises l 7 lcos from its lowest position If you can work out the physics try to see that the energy ignoring friction is mg QMgl 7 lcos 77109 constant Form a time derivative of that There is an approximation sin 6 for small angles which is explained better in problem 1 on page 144 Using that derive a second order linear equation 9 1 6 O Deduce that the period of oscillation of the pulley l m 1a is 27r What that means you can estimate how much mass M they are lifting by just timing the swing of the pulley and making a reasonable estimate of the pulley mass 8 What s Tong with this 1 4x O x cos2t sin2t C 53 14 Systems of ODE S part 1 TODAY Why would anybody want to study more than one equation at a time Find out Phase plane 141 A Chemical Engineering problem Jane s Candy Factory contains many things of importance to chemical en gineers One of the processing lines contains two tanks of sugar solution Pure water and sugar continually enter the rst one where they are mixed and an equal volume of the solution ows to the second where more sugar is added The solution is taken out at the same rate from the second tank The ow rates are as shown in the table tank 1 tank 2 sugar input 10 lbhr 6 lbhr water input 25 gal hr 0 solution out 25 galhr 25 galhr tank contents 100 gal 125 gal weight of sugar z lb y lb Jane s control system tracks the weight of sugar present in the tanks in the variables x and y We can gure out the rates of change easily The basic principle here is conservation of mass m lb z lbhr the rate inf the rate out 10 lbhr 7 m25 galhr See you just keep an eye on the units and everything works out Similarly i L y7610025 12525 Jane would like to predict the length of time required to start up this system ie beginning with pure water in brand new tanks how long before a steady state77 condition is reached if ever This is an initial value problem z 107 25m 3 6 25m 7 2y 950 07240 0 54 Let s solve it the quickest way we can7 and then we ll do more of an overview of systems in general Look7the z equation is rst order linear Jane s Candy Factory is in luck We know the solution is of the form x 15quot25 b Using the equation and the initial condition yields 05 40 7 405 25t Then the y equation reads 3 6 254074Oequot25t 7 21 which is also rst order linear The integrating factor is 5 22 which brings the y equation to gagY 1652 7 105703 Then E39Zty 8052i 7 2005705 c The initial condition y0 0 gives nally 0 120 and our solution is xt 40 7 405251 yt 80 7 20062 120521 The gure shows a plot of these solutions It was made using the matlab commands t 02130 o this was a guess that 30 hours would be enough x 4 40exp25t y 8 200exp25t 120exp2t plottx hold plotty OO Figure 141 For Jane s Candy Factory See whether you can determine which curve is for the rst tank gure made by plot commands as given in the text Finally we can answer Jane s question about how long the plant is going to take to start up As 25 7 007 Mt approaches 40 and yt approaches 80 55 Using the graph we will be within 10 of these values after about 8 hours and well over half there in only 3 That kind of information could be useful if you run a candy factory or a chemical plant 142 Phase Plane There is an easier way to produce a plot for Jane s system although reading it requires a bit of practice The pplane utility makes the following picture Figure 142 Jane s Candy Factory t hidden The limit point 4080 corre sponds to the limiting Values in Figure 141 gure made by pplane In order to interpret this gure one has to realise that the time is not shown so that the relation between z and y may be displayed This is called the phase plane77 for the system Here we explain the idea of a phase plane The vectors shown are made by using the expressions for z and y as components This can be done for any system of the form 90 f 90 y y 90 24 It means that we are thinking of the functions f and g as components of a vector eld We also think of 05 as parametric equations for a curve in the z y plane Then from this point of view look at what the differential equations say The rightihand sides are the vector eld while the leftihand sides give the velocity Consequently the differential equation system just 56 says that the velocity of a solution curve must agree with the given vector eld This is the entire content of the diffential equation You may observe in Figure 142 that the solution curves are tangent to the vectors You do observe that right PRACTICE Choose a point in the x y plane and work out the right hand sides of Jane s system for these coordinates Then draw a vector with these components starting from your point x y See if it is the same as the one pplane drew Converting a System to a Higher order equation It is occasionally of interest to convert a system of differential equations into a single higher order equation This cannot always be done It can be carried out for Jane7s system as follows The y equation gives 25 77 6 2y or z 437 24 8y Substituting this into the z equation gives z 4y 7 0 83 10 7 2543 7 24 1 8y 16 7 y 7 2y Rearranging we nd that 4y 181 2y 16 The initial conditions become y0 O and y 0 6 25z07 2y0 6 Ordinarily as we ll see it is more useful to convert higher order equations to lower That always can be done PROBLEMS 1 Find a second order equation for m if A x1 7 x2 52 1 22 2 Find a second order equation for 2 for the same system as in problem 1 Is it the same as the equation for 1 3 Make a sketch of the phase plane for the system of problem 1 You should do this rst by hand and then let the computer do the work for you by running pplane on the system 4 Here you will nd another system for the second order equation of problem 1 Set Z1 y zz y 7 y and nd the equations for the derivatives of Z1 and Z2 Did you get the same system as in problem 17 5 Make a sketch of the phase plane for the system and make a sketch of the slope eld for the equation 2yit 57 How do these two sketches compare and why 6 Set up a system for a modi ed Version of Jane s Candy Factory in which there are now third and fourth tanks in series with the rst two with parameters as shown tank 3 tank 4 sugar input 0 lbhr 5 lbhr water input 15 galhr 0 solution out 25 galhr 35 galhr tank contents 220 gal 165 gal weight of sugar z lb 11 lb You might notice that the Volume of solution is no longer constant for all of the tanks 15 Systems of ODE S part 2 TODAY Systems of diHerential equations Newton nonlinear springs predatori prey linear systems Preview of algebra to come There are several reasons for studying systems of differential equations One is typi ed by James Candy Factory of the previous lecture where of course if we were accounting for 7 tanks we would need 7 equations Another case comes from Newton s F ma law Each mass m needs a second order equation and there are many masses around EXAMPLE Two Newton s Laws giVe 4 rst order Suppose we have two masses moVing on the x axis pushing each other according to the system mix1 f1x1x2 7712932 f2I17I2 In fact according to another of Newton s proclamations f2 ifi but we don t need this for our present purpose Set 3901 x3 3902 Then we have four Variables representing the positions and Velocities of the two masses The system becomes x1 3901 5 v2 mml f iw z 7nsz f2I1I2 EXAMPLE A nonlinear spring gives two rst order Consider the equation x QI x O is equation looks somewhat like the secondorder equations we studied for springmass systems but it is not linear You can think of the x3 term as a nonlinear spring force Speci cally in order to stretch this spring twice as far you must pull eight times as hard In many respects this is closer to the behavior of a real spring We now show how to convert this equation to a system of rst order equations Set y x Then I x 72x 7 x3 72y 7 13 This gives us the rst order system x y y igyixa There must be some reason for doing this right First order systems do have some advantages They are much easier to think about for one thing to imagine what happens with z 2x 3 0 involves thinking about the graph of t and how the position x slope z and curvature z all interact This can be pretty complicated On the other hand thinking about an equivalent system just involves the idea that the solution curves must follow the vector eld as described previously For emphasis consider again a differential equation system 27 957972 y 99072472 2 9672472 The left sides give the velocity of a solution curve while the right sides say that this velocity must agree with the vector eld Whose components are 1 g and h In addition to this relatively simple description most of the available software is also set up to deal with rst order systems PREDATOR PREY EXAMPLE Suppose we make a model of two populations where 05 and yt are the population sizes We assume that these two species depend in each other in very different ways As a rst approximation we assume that if there were none of the y species present that the z species would grow exponentially and that the exact opposite would happen to the y species if there were no m s around we assume the y s would die out Our rst approximation is z z rst approximation exponential growth 3 7y rst approximation exponential decay This ts if we are thinking of the y s as predators and m s as prey wolves and mice for example The next step is to add some terms for the interaction 59 of these species We assume m z 7 my wolves are bad for mice 3 7y my mice are good for wolves It should be clear that we have neglected to put in various coef cients this does not affect the ideas involved In a homework problem we ask you to run pplane on this system to see what happens For now we will just show that it is possible to begin with a system like this one and convert it to a single second order equation which is however quite a complicated equation sometimes We calculate from the rst equation y m 7 z m 1 7 z lz This gives 3 z 2m 2 7 z lz Then from the second equation 3 71 zy 71 z17 m lm The next step would be to equate these expressions and simplify As you can see this will give a very nasty second order equation The rst order system de nitely makes more sense here 151 Linear Systems A linear system of differential equations means a system of the form 17 2m3y y 4m5y The constants 2 3 4 and 5 might of course be replaced by any others There could also be more than the two unknown functions z and y You may notice that this seems to be a fairly simple system since it does not include the predator7prey example nor the nonlinear spring We study these systems because 1 They are easy and fun to solve and make pretty pictures in the phase plane 2 We learn some new algebra called linear algebra in the process of solving these which is used for many things other than differential equations and 3 They are the foundation for the harder equations We7ll attack the linear system above with exponential functions since these have been useful so many times already Try z as y begt The system becomes are 2015 l 3best bsest 4aert5best 60 It is not quite clear what to do next7 is it But notice what happens if we take 5 r Then all the exponentials cancel an 2a3b br 4a5b You may wish to compare this to what happened with second order linear differential equations As in that case7 we have gotten rid of the differential equations7 and now have algebraic equations to solve It should be easier this way These are easier to solve if you rearrange them as 27 a3b 0 4a57rb 0 What happens next can be confusing7 if you don t see it coming We are not going to solve this algebraic system right now In fact7 we are going to leave differential equations for a while to study the algebra connected with systems like these It happens to be the same algebra which is used for many purposes not connected at all with differential equations7 too Once we know the algebra7 we ll come back to the differential equation problem PROBLEMS 1 Find a rst order system corresponding to x 7 x x3 0 You may use y x or something else that you choose as the second function 2 Using y x nd a rst order system corresponding to x 7 x x O 3 For the system you obtained in problem 2 substitute x a6quot and y beTt and nd the algebraic system which results You do not have to solve the algebraic system 4 Make a sketch of the phase plane for the system of problem 2 You should do this both by hand and by using pplane 5 Run pplane on rst order systems for the equation x 2x xp O for p 1 and p 3 What di erences do you observe 6 Find a rst order system corresponding to x x3 O and try to sketch the phase plane by hand This is too hard to do accurately7 so don t worry about it Next multiply the equation by x and integrate to nd a conservation law as in Lecture 10 Now it should be easy to sketch the phase plane because the conservation law shows you what curves the solutions run along 7 Consider the predator7prey equations again Our attempt to nd a second order equation for this system was a mess7 but it suggests something which actually works you may notice the term a in part of our derivation This suggests lnx7 doesn t it since dlnxdt Show that the equation x y 7 lnx 7 lny c is a conservation law in the sense that dcdt O for all solutions to the predator7prey system Sketch a graph of the function fx y x y 7 lnc 7 lny It may help to use octave for this sketch type help or help plotxyz to nd out how 61 8 Here is Way to discover the conservation laW of problem 7 Imagine that the para metric equations for Mt yt are solved so that y is expressed as a function of 1 Then by the chain rule you get i This gives 37 yTW Show how to solve this by x7acy separation of variables 9 Run pplane on the predator7prey system and on the predator7prey system With mi gration x x 7 xy 7 2 2 7y 96y Which species is migrating here7 and are they moving in or out Write a verbal description of What the diHerences are for the two phase planes Do you think there is a conservation laW for the migratory case 10 What s Tong with this I x y 3 y 796 n x 7 d2 7x3 7 dt2 f dtC lnx2 t x it2 y x37 W 12et2 C There are at least 3 mistakes 16 Linear Algebra part 1 TODAY Systems of linear algebraic equations Matrices A system of linear algebraic equations is something like 2m 7 3y 52 0 7x 7 y 32 0 7y 7 22 7 You might think based on your past experience in math courses that there is always exactly one answer to every problem The situation here is a bit more subtle This system can be thought of as three planes If you plot these planes in an x y z coordinate system you might nd that all three happen to intersect in a single point That point is the solution in that case m m Wkk si m quot t s x m w 3m mmm 7m m 75 75 Figure 161 Three planes on the left and three on the right When two planes in 3D intersect they do so in a line unless they coincide In one case the three planes have only one point in common In the other something else happens That is subtle in the picture Can you see it It may also happen that three planes meet in a line rather than a point Then there are in nitely many solutions Again it is possible that two of the planes are parallel Then there is no solution PRACTICE 1 There is another possibility Can you think what it might be 2 Which of these possibilities holds for the system x y 0 2x 2y 3 z57 161 Matrix A matrix is an array of numbers such as the coef cients in our rst system 2 73 5 A 7 71 3 0 71 72 A matrix can also be formed to hold the variables C H NQER as well as the numbers on the rightihand side 190 7 Matrices are sometimes written using bold type or with underlines over lines overarrows or rounded brackets or square brackets around the name of the matrix We won t use any of those That means that when we use the symbol A or b you have to keep straight from the context whether we are talking about a number or matrix or what The size of a matrix is given as number of rows by number of columns A is 3 by 3 while u and b are 3 by 1 We introduce a new form of multiplication so that our system can be written 2 73 5 x 0 7 71 3 y 0 0 71 72 Z 7 or Aub You might notice that to do this multiplication it is helpful to move your left hand along the row of A and your right hand down the column of u The same concept is used to multiply matrices of any sizes except that to multiply AB you must have the same number of columns in A as rows in B Otherwise some poor number is left without anybody to multiply EXAMPLE A matrix multiplication 2 3 x 7 2x 3y 4 5 y 7 4x By 64 l5 3 El El El 1 2 a b 71a20 1b2d 3 4 c d 7 3a4c 3b4d 1 0 1 0 0 I is the special matrix or 0 1 0 or one for each size It is 0 1 0 0 1 the identity matrix and you should check that always AI IA A u u It is also useful to abbreviate the system Au I further by not writing the variables at all the 3 by 4 matrix A b is called the augmented matrix of the system In general if you have a system of m equations in n unknowns Au b then A is m by n b is n by 1 u is m by 1 and the augmented matrix A b is m by 71 1 162 Geometric aspects of matrices Before continuing with equationisolving there are several remarks Matrices like C H mesa which consist of one column are sometimes called column vectors and pic tured as an arrow from the origin to the point with coordinates x yz In other words u here has exactly the same meaning as mil yf 213 and you can think of it in exactly the same way Of course we are now allowing our column vectors to contain more than three coordinates so the thre dimensional vector notation doesn t carry over Welcome to the fourth di mension There are some very nice geometric ideas connected to this form of multiplication These seem at rst very different and unrelated to equationi solving We ll give one example now just so you realise that there are several different reasons for studying matrices Consider the simple matrix 1 0 A lo al A can be applied to any point z y in the plane by using multiplication A lZl l l 65 This means that every point moves vertically closer to the z axis In spite of how easy this multiplication is the effect on the plane is dramatic as you can see in the next gure an as y 9 Figure 162 Matrices can change little smiley faces What Would happen if the face Were not centered at the origin PROBLEMS 010 1LetA 11 B 010 0 01 D 5 0 andletx 01 10 001 05 b 71 y 2 Calculate AB A2 AA 03 DB BD A1 3Ax 7 QDZ D13 and Cy 2 Suppose Au b and Av b and set 5 71 2 Is it true that As b How about if Av 0 instead What then 3 An airline company plans to make x ights per month from its main hub airport to Chicago and y ights to New York New York ights require 6500 lbs of fuel and 200 softdrinks While Chicago ights require 5400 lbs of fuel and 180 softdrinks The available monthly supply of materials at the hub consists of 260000 lbs of fuel and 9000 softdrinks and all these are to be used if possible Write a system of equations expressing all this Discuss faults of this model 2 1 2 2 4 l 39 39 7 4 What 5 Tong 102th thzs 4 3 4 7 9 12 17 Linear Algebra part 2 TODAY Row Operations Now that our matrix notation is established we turn to the question of solving the linear system Au b You may have solved such things in algebra courses in the past by eliminating one variable at a time This works ne for systems of 2 or maybe 3 equations For our purposes though it is important to learn a very systematic way of accomplishing this We will describe a procedure which applies equally well to systems with any number of variables In case you are wondering about the need for that let me reassure you that linear algebra of all the things you study might be one thing that gets used in real life Consider an airline company which needs to keep track of 45000 passengers 200 planes 300 crews 8500 spare parts 25 cities 950000 pounds ofjet fuel 85600 soft drinks etc etc Then there are the airframe engineers who compute the air ow at 10000 points around the plane and the engine people who compute the temperature and gas ow at 3000 points inside the engines and you get the picture Before we describe row operations you should see what the system of equa tions looks like in a special case like 1 0 0 5 0 1 0 7 0 0 1 9 This is the augmented matrix for the system z 5 y 7 and z 9 There is nothing to solve in this case Consequently the object of these row operations is to put the matrix into the form 11 if possible We now describe row operations for solving the system Au b There are three row operations which are applied to the augmented matrix Ab and have the effect of manipulating the various equations represented by the augmented matrix 1 A row may be multiplied by a nonizero number 2 Two rows may be interchanged 3 A row may be replaced by its sum with a multiple of another row Row operations do not change the solutions of the system This is clear for the rst two and requires a bit of thought for the third one They only change the appearance of the equations 67 A system may have many solutions 2 3 I 7 1 4 6 y 2 represents the tWo lines 2x 3y 1 and 4x6y 2 You might notice that these are tWo equations for the same line Therefore any point on this line should be a solution to both equationsi Let s do the row operations and see What happens We have the augmented matrix 2 3 1 2 3 1 4 6 2gtrow272row1gt0 0 0 Here We have replaced row 2 by row 22IOW 1 Look at the bottom line It says that Ox 1 0y 0 That is true but not helpful The top line says that 2x3y 1 Since this is the only requirement for a solution7 there are indeed in nitely many solutions EXAMPLE The system 1 3 x777 2 2y y anything A system may have no solutions EXAMPLE The system has augmented matrix 230 002 2 3 0 4 6 2gtrow272rowlgt Look at the bottom line It says that Ox 0y 2 Is that possible No So this system has no solution It describes tWo parallel lines The next example involves a bit more arithmetic7 and is our original system It shows that a system may have a unique solution EXAMPLE 2 73 5 0 1 715 25 0 Ab 7 71 3 O H5IOW1 gt 7 71 3 Hrow277row1gt 7 0 71 72 7 o o 01 t to 01 o gt interchange row and row 3 and IOWQ a O 1 715 25 O 1 2 77 7gt roW1 15 row and row 7 05 row 7 0 05 125 0 1 0 5 5 7105 0 1 2 77 7gt roW3115 7gt 0 0 115 35 1 0 5 5 7105 0 1 77 7 row 7 2roW3 and roW1 7 55 roW3 7gt 0 0 1 73043 O 0 7121737 1 0 776086 0 1 3043 The solution to our original system is therefore 7121737 71 776086 3043 Notice there is only one solution7 which says that these three planes intersect in just one point You may notice that this is the system graphed in Figure 1717 on the left A matrix is row7reduced if it looks like one of 1001001405103558 0 010 1 00 0170 1 217 8etc 0 0 1 0 0 0 0 ie7 the leading non7zero in each row is 17 the column of that 1 contains no other non7zeroes7 and these leading 1 s move to the right as you read down successive rows of the matrix Any row of zeros is at the bottom A row7reduced augmented matrix is easy to solve For example the 3 by 4 matrix represents the system m4y 5 z 7 0 0 which has solutions 2 7 y anything m 574y The downright pattern of 1 s allows you to solve the reduced system from the bottom up We point out again the term linear Multiplication of vectors by a matrix is linear in the sense that Aclu1 Cguz clAul CgAuz where the 0 s are any constants Consequently if you have two solutions u1 and U2 to Au 0 it follows that some other solutions are the soi called linear combinations clul cgug So we say that Au 0 is a linear equation and this is linearity in the strictest sense of the word We also say that Au b is a linear equation but here you have to be careful to see that a linear combination of solutions is usually not a solution PROBLEMS 1 Let A 2f and B What single row operation has been done to A to get B 7 2 What row operation can be done to A of problem 1 to produce the identity matrix I 1 0 3 7 3 The matrix 0 0 2 1 needs two more row operations done to put it into the reduced form What are they and what is the reduced form Note that the reduced form is unique but there are two di erent sets of row operations which are capable of producing it 4 Write out the system of algebraic equations corresponding to the augmented matrix 1 1 71 O 1 71 O 3 O O O 2 Is there something odd about the third equation How many solutions does this system have 5 Write the augmented matrix for the following system row reduce it and nd all the solutions xiy O xy 1 7x73 72 6 The following is a reduced augmented matrix for a system of 4 equations in the ve unknowns x1 us Find the solutions to the system 0 3 O O O 1 4 O 5 O O OOOH O O O OHO H00 70 7 Same instructions as problem 5 for the system x 7 y z 7 2w O x y z 7 w 1 7x 7 y 2 8 Whats Tong with this x 2y 0 3y 0 gives y 07 then x 0 Therefore there are no solutions 18 Linear Algebra part 3 TODAY Matrix inverse solutions to linear sytems in matlab7 and the Eigen value concept 181 Matrix Inverse We learned last time about solving a system Am b by row operations 7 you row reduce the augmented matrix A b and read the solutions7 if there are any There is another approach which is harder to compute but helpful to know about sometimes It goes like this We try to solve Am b by analogy with the arithmetic problem 2x 7 In arithmetic the answer is z 2 17 Here there is sometimes a matrix which deserves to be named A l7 so that the solution might be expressed as x A lb A 1 is pronounced A inverse Wouldn t that be nice Yes7 it would be nice7 but we ll see some reasons why it can t work in every case It can t work in every case because7 among other things7 it would mean that there is a unique answer z and we have already seen that this is not at all true Recheck the examples in the previous sections if you need a reminder of this point Some systems have no solutions7 some have one7 and some have in nitely many It turns out that for some matrices A7 the inverse A 1 does exist All of these matrices are square7 and satisfy some additional restrictions which we will list below Before getting to that list though7 return to the phrase used above7 a matrix which deserves to be named A l What does that mean We ll say that a matrix B is an inverse of matrix A if AB BA I This is the key property In fact7 there can only be one matrix with this property if B and C both have this property7 then B BI BAC IC C Since there is only one7 we can name it A l 71 EXAMPLE For A andB 0 2WehaVeAB 1 11 andBAIi S0 01 10 1710 og OQ OH too l H OH mHo 3 75 71 2 i 371 if i 35l 1li i It turns out that there is a formula for the inverse of a 2 by 2 matrix but that there is no really good formula for the larger ones It is 2 5 EXAMPLE For A 1 3 and B 1 Wehavez lB1J 0 IandBAISo 71 a b 1 d 7b C d ic a 117011756750 You can derive this formula yourself try to solve Z i I for z y z and w in terms of a b c and d and you will get it eventually The number ad 7 be is called the determinant of the matrix because it determines Whether or not the matriX has an inverse PRACTICE You should Check that this formula for the inverse agrees With the earlier examples We also are going to record the method used to nd inverses of larger matri ces They may be found when they exist by row operations You write the large augmented matrix A I which is n by 271 if A is n by n Then do row operations attempting to put it into the form I B If this is possible then B A If this is not possible then A 1 does not exist The rationale for this method is left for a later course in linear algebra but you can consider for yourself what equations are really being solved when you use this large augmented matrix 72 71 0 0 EXAMPLE For A 2 1 O we do operations 0 0 4 71 0 0 1 0 0 AI 2 1 O O 1 O gt 7row1 and 14row3 and row 2row1 a 0 0 4 0 0 1 1 0 0 71 0 0 0 1 0 2 1 0 0 0 1 0 0 25 7 0 0 71 0 0 You can Check that 2 1 O 2 1 0 0 0 4 0 0 25 Have you heard about people who only know enough to be dangerous It means they have some special knowledge of something but without enough depth to use it properly octave and matlab provide a tool to put you into this category Isn t that exciting So we hope that you will not get hurt if you know the following the program can sometimes solve Au b if you simply type u Ab That is a backward division symbol intended to suggest that one is moreioriless dividing by A on the left It mimics the notation u A lb In the problems we suggest trying this to see what it produces in several cases 182 Eigenvalues Usually when you multiply a vector by a matrix the product has a new length and direction as we saw with the little smiley faces If u is a vector from the origin to the left eye on page 66 then Au points in a different direction from u This is typical On the other hand let u be the vector from the origin to the top of Smiley s head In this case we see that Av v This is not typical it is special and turns out to be quite important Try to locate other vectors with this special property that A U U A vector 1 3A 0 which has this property for some number lambda is called an eigenvector of A and is called an eigenvalue Eigen is a German word which means here that these things are characteristic property of the matrix You may notice that we excluded the zero vector That is because it would work for every matrix and therefore not be of any interest It is ok 73 for the eigenvalue to be 0 though That just means that Av 0 for some nonzero 1 Since this is not true for all matrices it is of interest Note that this eigenvector equation Av v is super cially like the system equation Am b but there is a big difference The difference is that b is known whereas the rightihand side of the eigenvector equation contains the unknown 1 and the unknown So the eigenvector equation is probably harder Let s write out the eigenvector equation for the matrix Hi 2 If we write a and b for the components of u then Av v becomes 2 3 a a l4 51 M A M 2a3b a 4a5b b Now go back to reread the discussion of the linear system of differential equations near the end of Lecture 15 Do you recognize the equations EXAMPLE Reconsider the matrix A which squeezes smiley faces as on page 66 Any Vector along the 1 axis is unchanged and is therefore an eigenvector with eigenvalue 1 A 1 What happens to Vertical Vectors PROBLEMS 1 Solve 2 3 6 4 sl u H three ways by row operations matrix inverse and the Matlab command 2 Same as problem 1 for gtJgtOgtJgtlo 00100 HHOO OHOO HOOO 74 3 Find the inverses of the matrices g and OOHO OHOO Is there any restriction on a in either case 12 0 l 4 GivenA 2 40u 22 0 05 0 2 3 0 l 6w0x y1and 0 l 1 z 71 Which of u v w x y and z are eigenvectors of A7 and what is the eigenvalue 0 in each case 5 Find all the eigenvalues and eigenvectors for all the identity matrices I 6 Run matlab on the three examples of Lecture 17 to solve Ax I set x Ab Warning We know ahead of time that this must fail in at least two of the three cases Discuss the results 7 Write down a matrix which stretches smiley faces to 4 times their original height Find all the eigenvectors and eigenvalues of this matrix 1 1 8 Whats Tong with this 2 4 x b I g i b 6 8 g g 19 Linear Algebra part 4 TODAY Eigenvectors and Eigenvalues We have seen the eigenvector concept a non zero vector x is an eigenvector for a matrix A with eigenvalue if Am Am and we have seen some exam ples In this lecture we will outline some methods for nding eigenvectors For our purposes in this course7 we need to know the 2 by 2 case very well7 the 3 by 3 somewhat7 and the larger sizes just a little We begin with a 2 by 2 matrix A f I Consider the following statements 1 A does not have an inverse 2 det A 0 3 One row of A is a multiple of the other 4 0 is an eigenvalue of A7 ie there is a non zero z with Am O 75 5 Solutions to Am b7 if they exist for a particular b7 are not unique These statements are all equivalent7 in that if one of them is true of a partic ular matrix A7 then they all are true of A We discussed the equivalence of 1 and 2 on page 72 We won t prove all the equivalences7 but the implication 4 implies 1 is the most interesting for our purpose7 so lets think about that one assume that there is a non zero vector m such that Am 0 Now the claim is that A cannot have an inverse under this circumstance The proof is very easy Suppose A 1 exists Multiply the equation Am 0 by A 1 to get A lAm A 10 or m O This contradicts the fact that m is non zero Therefore A 1 does not exist Now we pull ourselves up by our own bootstraps7 using these facts about eigenvalue 0 to help us nd all eigenvalues So suppose now that B is any 2 by 2 matrix and we want to solve B m Am Note rst that this may be rewritten as B 7 AIm 0 where I is the identity matrix Now we just apply the previous discussion7 taking A to be B 7 AZ The conclusion is that detB 7 AI 0 This is our equation for A You can see7 or will see7 that it is a quadratic equation Once it is solved7 we solve the previous one for m EXAMPLE 1 2 B lo 731 1 2 1 0 1 7 A 2 detB7AI detO 73 7A 0 1 det 0 7374 17 A73 7 A 7 O O This is already factored so we nd A 1 or A 73 These must be the eigenvalues of B The next step is to nd the eigenvectors Take the case A 1 We must solve B 7 1Im O Ebrplicitly this says 8 3112 21 These require m2 O7 and note that is the only requirement So you can take m 01 where m1 is anything other than 0 Usually we take the simplest 76 possible thing Which is We have found tentatively BHH and you should Check this Then do the case A 73 We must solve B3Iy00r B m These require yz 7211 So you can make a simple Choice y We BHZWH Again you must Check this Does it Work When you multiply it out Ok Note that We took simple cases but that there are also other Choices of have found eigenvectors Which are multiples of these such as 710 and 6344 Check these too if you re not sure Note that you can make a picture of the plane showing all the eigenvectors of B and you will nd that they lie on the two lines y 0 and y 72 excluding 0 These are called the eigenspaces for the eigenvalues 1 and 73 respectively except that the 0 vector is included in the eigenspaces even though it is not an eigenvector Everything on the z axis remains unmoved by B While everything on the line y 72m gets stretched by 3 and ipped to the other side ofthe origin Note that having all this eigenvector information gives us a very complete understanding ofwhat this matrix does to the plane PRACTICE What would happen to a smiley face under the action of B7 EXAMPLE 0 l del We get detC 7 AI A2 9 This is zero only if A i3i For the case A 3139 We solve C 7 3iIx 0 Which reads 73139 1 x1 7 Lgidtlio So I2 31km We take x Then Check 1 l C 3 77 For the case A 73139 the only change is to replace 139 by 7i everywhere and check that 1 1 C 73139 3 73139 Note that for C one has run into complex numbers and so the nice ge ometric interpretation which was available in the plane for B is no longer available Two complex numbers are about the same as four real numbers so you might have to imagine something in four dimensions in order to visualize the action of C in the same sense as we were able to visualize B PRACTICE It does make sense nevertheless that not all 2 by 2 matrices can be thought of as stretching vectors in the plane gure out what what 0 does to smiley faces if you want to understand this point better Larger Matrices The procedure for 3 by 3 and larger matrices is similar to what we just did for 2 by 2 s The main thing which needs to be lled in is the de nition of determinant for these larger matrices We do not want to get involved with big determinants in this class To nd eigenvalues for a 3 by 3 matrix you can look up determinants in your calculus book or remember them from high school algebra Also see Lecture 20 There is also a possibility of using some routines which are built into octave These work as follows Suppose we have a matrix 1 2 3 A 4 5 6 7 8 9 To approximate the eigenvalues of A you can type the following commands A1 2 34 5 67 8 9 eigA To get eigenvector approximations also you may type V D eigA This creates a matrix V Whose columns are approximate eigenvectors of A and a matrix D which is a diagonal matrix having approximate eigenvalues along the main diagonal For example D in this case works out to be 161168 0 0 D 0 711168 0 0 0 0 78 PROBLEMS 1 Sketch a smiley face centered at the origin and its image under the matrix B of the text 2 Repeat problem 1 for the matrix C of the text 3 The last example 1 2 3 A 4 5 6 7 8 9 seems to have 0 as an eigenvalue Find a corresponding eigenvector 4 Find eigenvalues and eigenvectors for the matrix X 3 J 5 Find eigenvalues and eigenvectors for the matrix W 31 1 J 71 1 1 1 1 2 2 72 2 6 Which of the vectors 2 1 72 2 are eigenvectors for the matrix 1 2 1 71 0 0 1 0 0 0 0 1 Z 7 1 0 0 0 0 1 0 0 and What are the eigenvalues 7 Which of the vectors in problem 6 are eigenvectors for I the 4 by 4 identity matrix and What are the eigenvalues 8 Which of the vectors in problem 6 are eigenvectors for 7f and What are the eigenvalues 9 Which of the vectors in problem 6 are eigenvectors for the matrix 0 O 1 0 H000 O 1 O O OOOH and What are the eigenvalues 10 If you don t know how to do the determinant of a 3 by 3 matrix7 look it up somewhere or see page 80 and nd the eigenvalues and eigenvectors for 300 AOOO 050 11 What s mg with this A 7 detA7I det 3 g A 7 E A 34x77 A 21 7 7 7 3 A2 A2 710 21 so use the quadratic formula 79 20 More on the Determinant TODAY The determinant determines whether the matrix is invertible Unlike most of the lectures we aren t going to explain everything here but will indicate most critical ideas 2 x 2 You know that for 2 by 2 matrices the determinant is b det ad7bc c d and it determines whether the matrix is invertible It is used in the formula for the inverse 1 a b 7 7 1 d 7b c d 7 ad 7 be 7C a It also tells geometric effects for example if detA 73 then A expands areas by a factor of 3 and reverses the orientation The determinant was discovered when someone like you you could try this solved a linear system symbolically and realized that ad7bc always shows up in the denominator 3x3 If you solve the system a b c u 1 w E b r s t symbolically then you nd the expression wt bun Gus 7 mos 7 but 7 CU in the denominators of the m That expression is then de ned to be the determinant of the 3 by 3 matrix It is so complicated that people have found various ways to express it such as adetr wl7bdetu wcdetu s t r t r s a b det u v r 5 680 80 3 0 0 EXAMPLE det 0 5 0 3 This matrix stretches or compresses the 0 three axes by various factors and it multiplies volumes by 3 The matrix is invertible The determinant of the inverse is g Any Size You can nd the determinant of any square matrix iteratively as in this example note the alternating signs 2 23 det a det 4 6 7 72 4 6 7 0 5 8 73 0 5 8 0 2 0 0 1 0 0 1 2 7b det 2 6 7 c det 2 4 7 7 d det 72 4 6 73 5 8 73 0 8 73 0 5 Then do the 3 by 3 s etc like 1 2 0 det 4 6 7 1det6 7 72det4 7Odet4 6 0 5 8 5 8 0 8 0 5 148 7 35 7 232 7 0 020 7 0 751 201 Products Another generally useful fact detAB detA detB For example if A is invertible then 1 detI dam1 1 detA detA l s0 detA 1 m To prove the product rule is for another course But we can use it to prove something good CRAMER S RULE M Cramer cleverly noticed for example that a b 0 m1 P a b 0 m1 0 0 P b c If u U w m2 D then u U w m2 1 0 D U w r s 25 m3 Q r s 25 m3 0 1 Q s t 81 and taking determinants you get a formula for the answer P b c detAx1 det D v w Q s t Similary for 2 and 3 and for any size system Cramer s rule is an inter esting theoretical tool these days7 but not computationally efficient because it takes too long to work out determinants PROBLEMS 1 Is this matrix invertible O O 1 0 H000 O 8 1 O O O O 0 2 If a 2 by 2 matrix A has columns 2 and u we will write A v w and think of the determinant as a function of the two columns f0 11 detA Convince yourself that an 2w am w w w whenever u v and w are Vectors and a and b are numbers We say that det is a linear function of the rst column of the matrix 3 Of course det is a linear function of the second column too Figure out why that makes this kind of identity hold fau b2 cw dz acfu w adfu7 z bcf39u7 w bdfv z 82 21 Linear Differential Equation Systems TODAY Just when you thought it was safe back to linear di erential equar tions The solution by eigenvectors At this point you have seen those essential parts of linear algebra which are needed for solving linear systems of differential equations of the form mAx Here7 A may be an n by 71 matrix and m an n by 1 column vector function of 25 As in any differential equation system z m7 the right side represents a vector eld which exists in space7 and the mission of any curve 05 which intends to solve the system is to be sure that it s tangent velocity vector z t agrees at all times with the given vector eld In this lecture we deal only with vector elds of the form x Ax7 and these may be solved rather completely using our new eigenvector knowledge Figure 211 A Vector eld in 3 dimensions We think of it as unchanging unless we have a need for explicit time dependence in the eld But there are also two curves shown which you might think of dynamically They are doing their Very best to follow the arrows not only in direction but also to have the correct speed at every point they go through If they succeed in this7 they are solutions to an ODE The eld shown is more complicated than a linear Ax eld For example7 we know that sometimes you nd an eigenvector v for which Av is a positive multiple of 1 In such a case7 the vector eld points away 83 from the origin and so any solution passing through point 1 must be headed away from the origin Similarly negative eigenvalues give rise to solutions headed for the origin The effects of complex eigenvalues are a little harder to guess So let s proceed with the analysis in the following way Suppose by analogy with many things which have gone before that we are looking for a solution of the form Mt Gequot where r is a number as usual but now 0 must be a vector for the assumption to make any sense Then substitute into the differential equation to get z Tee Ax Ace These will agree if Ac TC Now stop and take a deep breath and think about what you learned before about linear algebra This equation is the same except for notation as A U U isn t it We have reached the following fact lf Av Av then one solution to m Am is we EXAMPLE From the example on page 76 33 1 So one solution to x x 2 y i321 is e1t ie x et y 0 Check these in the di erential equation to be sure Note that we certainly don t stop with one solution For each eigenvector you get a solution EXAMPLE Also for the same example we found that is an eigenvector 2 with eigenvalue 73 Consequently we have another solution 6 or x 6 y 726733 Check these too In general we hope for a system z Am where A is n by n that there will be n such solutions Usually there are but not always More on this 84 in a moment The most important thing now is to point out how you can combine these solutions which we have found Our equation m Am is linear in the strictest sense of the word linear combinations of solutions are solutions ie if z l Am and z z Azg then for any constants cl and 02 it is true that 01ml CZmZ Aclm1 02mg This is because differentiation and multiplication by A are both linear operations You should write out the oneline proof of this yourself if you are not sure about it 1 2 i 0 73 for any choice of constants Cl and 82 In order to meet a given initial condi EXAMPLE xt 01 5 et 02 32 67st is a solution to x tion we have to solVe for the 8739 For example if we want to have x0 then it is necessary to solve Cl c2 The answer is 62 l 01 15 and we get the solution Mt 1 et 15 32 fat Back to the question of nding 71 solutions for the case of n by n matrices A Certainly if we nd 71 eigenvectors 127 we can form solutions like Mt clvleAlt onvneAm The question is whether these are all of the solutions to the differential equa tion Equivalently is it possible to meet all the possible initial conditions using 0 01111 any Note that this is a system of linear algebraic equations in the unknown 0739 like we saw in the preVious sample What is needed here is that the u should be linearly independent This concept is slightly beyond the level of these notes but basically it means that the vectors really point in 71 different directions or that none of them may be written as a linear combination of the other ones You will study this concept in a linear algebra course we re only doing an introduction to it here 3 0 0 EXAMPLE Let A O O O This was problem 10 on page 79 We have 0 5 0 l 0 Am 33901 and Avg 0 where 3901 O v2 O and there are no other 1 eigenVectors except for multiples of these So some solutions to x Ax are 85 1 clvleat 82 02 This is one of the exceptional cases Where you do not get n 3 eigenvectors nor the general solution to the diHerential equation See if you can nd another form of solution to these di erential equations for this system it is actually easier to Write the equations out and deal With them explicitly rather than to think about eigenvectors PROBLEMS 1 Solve 2 Solve the system of problem 1 With the initial conditions x0 0 210 5 3 Sketch the phase plane for the system of problem 1 4 Solve the system of problem 1 with initial conditions MO 1 210 1 Sketch your solution onto the phase plane or run pplane on it Show by eliminating t between your formulas for xt and yt that x y This should help your sketch 5 Solve x 2x 2 i321 Show that solutions lie on curves x cy g or lie along the positive or negative axes or are zero Sketch the phase plane 6 Sketch the phase plane for the system 7 1 2 x 7 O 73 x Which Was solved in the text Does it look more like the phase plane for problem 3 or problem 5 qualitatively 7 Solve x x 2y 2 496 i y and sketch the phase plane 8 Solve x x 7 y I y I 5y and sketch the phase plane 1 0 0 9 You are given that a matrix A has eigenvectors u 0 v 3 w 2 0 and that Au 5n Av 42 and Au 3w Solve x Ax if x0 6 Note that NO u 22 10 What s Tong with this 3 x0 6 1 77 7 and A 2 IS given So x 3e C 714 22 Systems with Complex Eigenvalues TODAY A complex example Today s lecture consists entirely of one example having complex eigenvalues It is not really that hard but we want to be careful and complete the rst time we work through one of these Our system is mAx 1 2 A 7 L2 J We know what has to be done to solve this you nd eigenvalues then eigen vectors of A then the solution 05 is in the form of linear combinations of the 125 things This is what we have done before and here we will nd a bit of complex arithmetic to go along with it Here we go where First plot the phase plane to see what to expect You can plot a little bit of it by hand or you can run pplane on our system 3 y 22 z 72y2 87 Figure 221 The phase plane for our example gure made by pplane Now let the analysis begin First we nd the eigenvalues of A We have 2 detAiI det12 7A 2725 This is zero by the quadratic formula if w Zilzim 1 i 22 So we have our eigenvalues Now let s look for an eigenvector for 1 22 We must solve A7 1 2iIv 0 Writing a and b for the components of 127 this says 72ia2b 0 7201722 0 The rst equation says I ia The simplest choice we can make is to take a 17 b 2 Now something strange happens If you check these numbers in the second equation7 you will nd that they work there too In fact7 you can just forget about the second equation7 because it is a multiple of the rst one I know it might not look like a multiple of the rst one7 but that is because it is an imaginary multiple of the rst one Anyway we have found that 32 fl El lt1gtlil Read that through again to check it7 because there are lots of places you can make mistakes in complex arithmetic 88 Next let us state the solution we have found so far and expand it so we can see what it looks like l1 elt12igtt etcos2t isin2t t cos2t t sin2t e 7 sin2tl 5 cos2tl That7s a mess isn t it It is neater to leave it in the original complex exponential form but we expanded it out so you can see exactly what the real and imaginary parts are Now we need a second solution so we can go through all this again with the other eigenvalue 17 22 The only thing which changes if you read through what we just did is that you must replace i by 7i wherever it occurs This happens to work because A is a real matrix The general solution to the system can then be expressed as 95t c1 7i e12it CZ 1 50721 There is an alternate to this sometimes preferred because it emphasises real solutions This goes back to our discussion in Lecture 12 about real and imaginary parts of solutions to real equations The same statement holds here that if z u m is a solution to 1 Ax where u v and A are real then u and v are also solutions You can prove this yourself pretty easily Taking the real and imaginary parts we found for the rst solution above allows us to express solutions in the alternate form w ale cos2t M25 shim 7 sin2t cos2t The relations between the constants are al 01 02 and a2 i01 7 02 It is important to think about the phase plane and see whether it matches the solutions we found You can see that we got the increasing exponential 5t and this has to do with the fact that solutions are moving away from the origin in Figure 221 Also we got sines and cosines which oscillate positive and negative That has to do with the rotational motion in the picture PROBLEMS 72 1 SolvexAcifA 1 Find the solution if x0 7 0 1 2Solvec779 0 J x What is the second order equation for the rst component 1 in this case 89 3 Write out the system corresponding to x 73 using x y and solVe it7 nd a conservation laW for the equation and plot its leVel curves and plot the phase plane for the system How are these plots related 4 Solve x 7 73 2 z z 79y and plot some solutions 5 What s Tong with this x lt12igtt etcos2t isin2t7 so rex et cos2t7 irn et sin2t7 and x etcl cos2t 02 sin2t 23 Classi cation Theorem for Linear Plane Sys terns TODAY Classi cation of linear planar systems Spirals centers saddles nodes and other things more rare Today we have a nice treat for you Rather than doing a lot of computa tion which we have been doing for several lectures we are going to state a classi cation theorem If this sounds like when the doctor says It won t hurt at all that s because we rst have to explain the signi cance of what we are about to do A classi cation theorem is one step higher in the world than computation because it is something which lists all the possible things which could ever happen in a given circumstance Having such a theo rem is considered by mathematicians to be a good thing There are not very many classi cation theorems and mathematicians wish they could nd more Here is what this one says THEOREM ON CLASSIFICATION OF LINEAR SYSTEMS IN THE PLANE Suppose you have a system m Am where A is a 2 by 2 real matrix Draw the phase plane Then the picture you just drew has to be essentially one of the seven pictures shown below and there are no other possibilities Essentially means that you might have to rotate or stretch your picture a little to make it t or reverse the arrows but there will be no more difference than that There now that was not so bad was it Here are the pictures The most important of these should look familiar from homework The classi cation proceeds by considering what the various possibilities are for the eigenvalues of A Figure 231 This is called a node It is What happens When the eigen values are both real numbers and have the same sign The case shown has both eigenvalues positive For both negative the arrows would be reversed Important solutions do not go through the origin Figure 232 This is called a saddle because it is the pattern of rain running OH a saddle as seen from above It occurs when the eigenvalues are real but of opposite sign Figure 233 This is called a center It occurs when the eigenvalues are ibi for some nonzero real number I Figure 234 This is a spiral for obvious reasons The eigenvalues are aibi with a 7 O The case a gt O is shown7 but if a lt O the arrows are reversed Also this can appear clockwise or counterclockwise Figure 235 A rare case in which one eigenvalue is O Figure 236 A rarer case in which both eigenvalues are 0 Figure 237 The rarest case7 in which the entire matrix is O PROBLEMS 1 If you did problems 1 2 and 3 of the previous lecture7 determine which of our phase planes in this lecture applies to each of t em 2 Make up your own system x ax by y at dy by choosing the coef cients at random Draw the phase plane for your system by hand or pplane and classify which type it is by trying to match it with one of the gures in this lecture The theorem says that it should match 3 Is there any other classi cation theorem that you can think of anywhere in mathemat ics 4 Why are there just dots in Figure 237 instead of curves 5 Suppose you re solving a system x Ax and nd that the eigenvalues are 42 i 3139 Which phase plane applies Is the motion toward or away from the origin How can you determine whether the motion is clockwise or counterclockwise just from the diHerential equation itself 6 What s Tong with this Figure 142 and Figure 241 don t match any of the 7 types listed in our classi cation theorem So there must be a mistake in the theorem 24 Nonlinear Systems in the Plane TODAY Nonlinear systems in the plane including things which cannot happen in a linear system Critical points Limit cycles We now know about all the possible phase planes for systems of two linear equations The next step is to see how this helps us analyze more realistic systems There are two new ideas which enter into this study and the rst one is well illustrated by an example we looked at previously the predator prey system of Lecture 15 We will see that the phase portrait for the predator prey system looks as though it has been made by selecting two of the linear phase planes from Lecture 23 and pasting them together in a certain way In this sense the linear systems prepare us pretty well for some of the nonlinear systems Then we will show a second example containing entirely new behavior which can t happen at all in a linear system So from this perspective it seems that perhaps the linear systems don t prepare us well enough at all for some of the nonlinear ones We begin with the predator prey system z m 7 my 2 7y my which we have discussed in lecture 15 The phase plane for this system is shown in the gure Figure 241 The predatorprey system Of course you have already noticed the part which looks like a center and the part which looks like a saddle 96 You may recognize some of its elements immediately if you have done the homework problems for Lecture 23 already It shows a cycling between the sizes of the mouse and wolf populations The points about which the action seems to be centered are known as critical points and you may notice in Figure 241 that these are located at 00 and at 11 There is an easy way to nd the critical points for any system and that is to notice that they are just the constant solutions to the equations In other words all you have to do is set z 0 and y 0 and solve the resulting equations Next we modify the system to include a little bit of emigration thinking that perhaps the mice have discovered the inadVisability of liVing in this neighborhood The system is now z m 7 my 7 2 2 7y my Observe the interesting effect this emigration has on the phase plane What do you think it means when that solution curve spirals outward for a while and then runs into one of the coordinate axes It is not usually signi cant when a curve crosses an axis but here we are talking about populations When a population reaches zero that is signi cant The result of the emi gration can therefore be extinction or at least absence Figure 242 The predatorprey system with emigration of the prey Note the part which looks like a saddle and the part which looks like a spiral 97 In both of these situations we have shown negative values of z and y This is because the system exists independent of its interpretation as a population model and we wanted to look at the whole phase plane If your interest is in the population aspects though you don t care about the negative values and the conclusion is that as soon as a solution curve touches an axis the model no longer applies Our second main example is a classic known as the van der Pol equation It is important for several reasons partly historical and partly because it shows behavior which is impossible in a linear equation The historical aspect is that this was originally a model of an electrical circuit whose understanding was important in radio communications The equation is xH 71m2xm0 You should notice that it looks just like our second order equations from Lecture 10 except that the coef cient of z is not a constant So our char acteristic equation method will do no good here It converts to the system 1 y y 179639790 which is not of the form 2 A2 so our eigenvector methods will also do no good So we let pplane make a picture for us to study Figure 243 The Van der Pol system contains a limit cycle This is completely new signi cant behavior not possible in a linear system 98 Notice what is happening in this phase plane There are solutions spiraling outward from the origin That certainly can happen in a linear system as well There are solutions coming inward also But what is new is what happens between these There seems to be a solution which is periodic and which the others approach In the radio circuit what would that mean It must mean that there is a voltage or current which cycles back and forth repeatedly Now you recognize that this kind of thing happens when you have a center in a linear system too7 but there is a difference here Here there is only one periodic solution This means that this radio is going to settle down near this one special solution no matter what the initial conditions are That is very important Such behavior deserves a new name it is called a stable limit cycle If you run this machine in your laboratory7 the only thing you will observe is the limit cycle PROBLEMS 1 Run pplane on the system x sinx y 1 sirmy How many saddles and spirals can you nd visually 2 Consider the system H r 7 r2 9 1 The rst one is the logistic equation Look back at Lecture 2 to remind yourself that all positive solutions approach 1 What do the solutions to this system do then if you take r 9 as polar coordinates in the plane Convert the system to cartesian coordinates and run pplane on it Did you nd a limit cycle 3 Study a predatorprey system in which the predators are moving out What happens This can be used as a model of a shing industry in which the predators and prey are two sh species7 and move out77 means get caught Are the results reasonable or unexpected If the plan is to harvest the sharks so that the tuna population will increase what would your advice be 4 Run pplane on the system x cosx y y may How many limit cycles can you nd visually 5 Find all critical points for the system I y1 I2 2 I 2 Sketch the phase plane and try to gure out what type of linear behavior occurs near each of them 6 Find all the critical points of the Van der Pol system 7 Try to sketch a phase plane which contains two saddles You are not asked for any formulas7 but just to think about what such a thing could look like Remember the uniqueness theorem7 that solutions cannot run into each other 8 Try to sketch a phase plane containing two limit cycles 9 What s Tong with this Jane s Candy Factory used the system 95 10 7 2595 y 625x 7 2y The critical points are the constant solutions so we solve O 10 7 2513 getting x 40 and O 6 2513 7 2y getting y 32 25 Examples of Nonlinear Systems in 3 Dimen sions TODAY Nonlinear systems in space Chaos Why we can t predict the weather We have seen lots of linear systems now7 and we have also seen that these provide a background for some7 but only some7 things which can happen in nonlinear equations7 in two dimensions Today we move to three dimensions We will see that nothing you learned about two dimensions can prepare you for three dimensions There are aspects of systems of three equations which can be understood using the prior knowledge7 but there are also things called chaos which pretty well suggests how different and intractable they are This lecture is only the briefest kind of introduction to this subject7 but is here because it is important for you to know about these wonders We begin with an example in which your knowledge of plane linear systems does help you understand a 3D system Consider the system 7 72 72 NJ 8 l These equations may be read as follows The equation for z does not involve y or 2 and vice versa In fact the x equation is a simple equation for exponential decay The y and 2 equations are a system corresponding to our favorite second order equation y 7y whose solutions are Sines and cosines We are therefore dealing with a center in the y 2 plane Consequently we 100 can think of this system as having a circular motion in the yz plane combined with an exponential decay in the z direction So the solutions are curves in 3 space moving around circular cylinders centered along the z axis and which approach the y 2 plane as time goes by Any solution already in that plane remains there circling In contrast with that system let us look at the Lorentz system z 10y 7 m y 7m228m7y 2 my 7 32 This very interesting system was made by Edward Lorentz who was a meteo rologist at MIT as a very simpli ed model of circulation in the atmosphere The system is nonlinear because of the 2 and my terms and is derived from much more complicated equations used in uid mechanics You may be interested to know that the weather predictions we all hear on the radio and TV are to some extent derived from calculations based on observational data and the real uid mechanics equations Here is a phase portrait for the Lorentz system Figure 251 The Lorentz system gure made using the Matlab commands shown in problem 2 This gure does not show a phase plane but rather a projection of a phase portrait in 3 dimensions onto a plane How do you suppose it occured to Lorentz to think of the e ect of a butter y on the weather 7 101 You may notice that there seem to be two important critical points about which the solutions circulate One nds that the solution turns a few times around the left one then moves over to the right one for a while then back and so on The only problem is that it is not easy to predict how many times it will circle each side before moving to the other Lorentz came upon this fact as follows He was computing this on whatever kind of computers they had in 1963 and noticed the pattern of randomilooking jumps from one side to the other These were very interesting solutions so he reran the computation from a point part way along to study it in more detail In the process he typed in as initial conditions the numbers which had been printed out by his program These numbers were slightly rounded by comparison with the precision to which they had been computed The result made history the qualitative features were still there but the solution in detail was very different in that after a few turns the numbers of cycles around the left and right sides came out different just because of rounding off a few decimal places in the initial conditions That means that if you try to make a picture like the one above it will look qualitatively the same but will be different in detail It also means something about weather prediction Lorentz asked Does the Flap of a Butter y s Wings in Brazil Set Off a Tornado in Texas77 This is a restatement of the effect of rounding off ie slightly changing the initial conditions Everybody has known for years that weather prediction is an art that you really can7t predict the weather very well Part of what Lorentz did was to explain why you can t predict the weather you can never know enough about the initial conditions no matter how many weather stations you build when the equations of motion are so sensitive to initial conditions We also have an enormous shift in the way the universe is conceived as a result of studying this and other chaotic systems On the one hand there is the uniqueness theorem which says that the future is determined uniquely by the initial conditions This is the Newtonian approach On the other hand we now know about chaotic systems in which yes solutions are uniquely determined by their initial conditions but they may be so sensitive to these conditions that as a practical matter we cannot use the predictions very far into the future Isn t that interesting It is a really major idea PROBLEMS 1 Check out the java applet de which is described in Lecture 7 and run some chaotic systems on it Several are built in as examples 2 Solve a 3D system in Matlab You can t use pplane of course but you can set up a function le to compute your vector eld like 102 file efm function vecef t w vec110w2 W1 vec 2w1 w328w1 w2 vec 3w1 w2 8w3 3 39 end file ef 111 Then giVe commands like t w ode45 ef 7 O20 5 7 9 plotw 1 w 3 This Will compute the solution to the Lorentz system for 0 g t g 20 With initial conditions x 5 7 9 and make a plot The symbol u was used here for y so What Was plotted Was z x Versus z Naturally you can play With the plot command to get di erent Views The hardest part is understanding Why w 1 means x You have to realize that w is a list of three lists before it makes any sense Change these les to solVe a system of your choice Note that you are not restricted to just three dimensions either 3 Change the plot command in problem 2 to plot x Versus t 4 Change the ef 111 le of problem 2 to solVe the linear system described in the text You Will have to play With the plot command to get a nice perspective View7 for example you can plot x Versus y 2z or something like that Does the picture t the description giVen in the text 5 What s Tong with this According to this lecture if you have a system of 3 or more Variables you can get chaos And according to Lecture 157 if you have a system of 2 springmasses you get 2 Newton s laws or a system of 4 rstorder equations Therefore 2 springmasses are alWays chaotic 103 26 Boundary Value Problems TODAY A change from initial conditions Boundary Values We nd that we don t know everything about y 7y after all You have by now learned a lot about differential equations or to be more speci c about initial value problems for ordinary differential equations You have also seen some partial differential equations In most cases we have had initial conditions At this time we are prepared to make a change and consider a new kind of conditions called boundary conditions These are interesting both for ODE and also in connection with some problems in First you have to know what a boundary is It is nearly the same thing in mathematics as on a map the boundary of a cube consists of its six faces the boundary of Puerto Rico is its shore line the boundary of the interval 11 consists of the two points a and b The concept of a boundary value problem is to require that some conditions hold at the boundary while a differential equation holds inside the set Here is an example EXAMPLE 2 2 210 0 WW 0 We are asked to solve a Very familiar differential equation but under very unfamiliar conditions The function y is supposed to be 0 at O and 7r The differential equation here has solutions y A cost B sint We apply the rst boundary condition giving y0 O A So we must take A 0 That was easy Now apply the second boundary condition giving y27r O Bsin27r Well it just so happens that the sin27r is zero So the second boundary condition is ful lled no matter what B is Answer yt B sint B arbitrary Notice how different this example was from our experience with initial con ditions that there were in nitely many solutions Just the opposite thing can happen too EXAMPLE o This seems close to the previous example since only 27r has been changed to 6 This begins as before with A 0 Then it just so happens that the sin6 is not zero So unlike the previous case we have to set B 0 Answer yt 0 There is no other possibility What is a physical interpretation of these problems We know that the equation y 7y describes an oscillator a rock hanging from a Slinky vibrating up and down at frequency If you require the condition y0 0 that just means you want the rock at the origin at time 0 That alone is not a real problem because you can start the stopwatch when it is there Since the frequency of oscillation is i the rock will be back at the origin after 27139 seconds If you then require the condition y27r 0 then there is no problem because that happens automatically But if you require instead that y6 0 then you are asking for the impossible The rock takes 27139 6283 seconds to get back period The only exception is the special function 0 signifying that the rock never moved at all In that case it will certainly be back in 6 seconds since it is there already That is Why there is no solution in the second case except for the 0 solution A more down to earth example of a boundary value problem You can go out with your friends at 800 but you have to be home by 120077 There is a tendency to use x as the independent variable when one is dis cussing boundary value problems rather than t because usually in practice it is position rather than time in which one is interested We ll do that in the next example There is also the concept of eigenvalue in connection with boundary value problems EXAMPLE y icy Finding c is part of the problem too 210 0 WV 0 The eigenvalue here is c We will assume we are looking for positive values of c to keep things simple and a homework problem will be to analyze the cases in which 0 might be 0 or negative To solve this equation y is we recognize that we are again looking at sines and cosines yc A cosEx B sinEz That is where our previous knowledge of differential equations comes in If you have any doubts about these solutions go no further Go back study second order linear equations a little more You can t build a house on a foundation of sand All set now The boundary condition y0 0 Acos0 A determines A 0 as before Then we have to deal with the other boundary condition y7r 0 B sinE7r Certainly if B 0 this is satis ed and we have found a solution yx 0 But we have 105 another parameter c to play with7 so let s not give up too easily Question Is it possible that sinE7r O for some values of c If so B can be anything and such 0 will be our eigenvalues Well what do you know about where sine is zero Certainly sin0 0 but if c 0 then we are back at yx O which is uninteresting by now Where else is the sine zero The sine is zero at 7r for examp e That gives w 7r or c 1 There is a good solution to our problem yc Bsinc7 B arbitrary c 1 But there are other places where the sine is zero For example sin27r 0 That gives w 27r or c 4 So there is another solution yx Bsin2x B arbitrary c 4 Also sin37r 07 giving a solution yx Bsin3x B arbitrary c 9 In general we have solutions yc Bsinnc B arbitrary c n2 n123 So the eigenvalues for this problem are 1 47 97 167 25 261 Signi cance of Eigenvalues Until now you have known eigenvalues as something connected with matri ces Suddenly the same word is being used in a different way7 although in both cases there is some operationsomething equal to the eigen value the same something There is an abstraction here7 that the equations Am Am and y icy have something in common We need to suggest that there is much more to eigenvalues than this They are very important You have gotten along somehow until now without knowing about them7 but they have been all around you For example7 the color of light from your ourescent lights as you read this is determined by the frequency of the light7 which is mainly a difference of atomic energy levels for the atoms in the light bulb These energy levels are eigenvalues of something7 not of a matrix7 but of Schrodinger s equation which was mentioned earlier The music that you might be listening to as you read this comes in pitches which are eigenvalues of some equation describing the guitar strings7 motion The room you are sitting in may occasionally vibrate a little when somebody drops something heavy7 and the frequency is an eigenvalue of something too We will discuss just some of these things in this course7 including the guitar strings but not the atoms There is a drum vibration model in Lecture 34 PROBLEMS 106 1 Find all the solutions to the c g 0 cases of 2 Solve 2 7 icy 210 0 WV 0 yquot icy 210 215 Finding the possible Values of c is part of the problem7 but to keep things simple you may assume that c gt O 3 Find all the solutions to 4 Find all the solutions to y icy 210 0 213 0 y 7 icy y7r0 O a 0 107 27 The Conduction of Heat TODAY A derivation of the heat equation from physical principles Two in fact Our survey of ordinary differential equations is now complete and we have done some work with partial differential equations too As you know they are also somewhat different from ordinary differential equations For exam ple you could make up an ordinary differential equation almost at random and nd that many of the methods we learned will apply to it and the software will solve it approximately and draw pictures of the approximate solutions You could do that although the motivation for doing so might not be clear In the case of partial differential equations almost nobody would just make one up and the reason is that the equations people are already working on which tend to be about things in the real world are hard enough to solve as it is There are no cut and dried software packages for partial differential equations either since to some extent each problem needs its own methods What happens is that rather than solving a lot of equations as we did with ordinary differential equations we study just a few partial differential equations with many different initial and boundary conditions So partial differential equations are about real things and have names to re ect this fact The one in this lecture is called the heat equation The heat equation looks like ut aumz It is an abbreviation for 2 mt 1x t The equation concerns the temperature umt of a metal bar along which heat is conducted You can imagine one end of the bar in the re and the other in the blacksmith s hand to emphasize that the temperature may change with time and with position along the bar The number a is a physical constant which we will explain while we derive the heat equation Now let s read the heat equation carefully to see whether it makes any sense or not Suppose you graph the temperature as a function of x at a particular time Maybe the graph is concave up as in Figure 271 on the left What do you know from calculus about functions which are concave up 108 Isn t the second derivative positive The heat equation contains the second derivative and says that the more positive it is the bigger the time derivative will be What do you know from calculus about the rst derivative being positive Doesn t it mean that the function is increasing So we have to imagine what that means and conclude that the temperature must be rising at any point of the bar where the temperature graph is concave up Similarly for the right side of Figure 271 If the temperature graph is concave down the temperature must be decreasing with time Does that seem correct There is no hope whatsoever of understanding partial differential equations unless you think about things such as that W I Figure 271 Temperature Versus position along a metal bar The arrows show the time dependence according to the heat equation On the left the temperature graph is concave up and the bar is warming and on the right it is cooling The little slab will be used in the derivation below As you will see from our discussion it is not necessary to think of a metal bar It is also possible to think of the conduction of heat in the z direction where this axis passes through a door or wall or any similar situation in which heat energy ows in one dimension only If you want to talk about heat ow along the surface of a sheet of metal you will have to use the two dimensional heat equation Ut um 142121 If you want to talk about heat ow throughout a solid block of metal or though still air you use the three dimensional heat equation u awn uyy um Our plan is to derive the onedimensional heat equation by showing what physical principles are behind it and by using what you might call elemen tary mathematics ie calculus Later we will show a different derivation for the threedimensional heat equation using the exact same physical prin ciples but more sophisticated mathematics the Divergence Theorem In both cases we need to point out that we are assuming some familiarity with 109 physics for these derivations On the other hand it is not so much a speci c knowledge of physics as a willingness to think about heat that you really need for the next few paragraphs Actually solving the equation later does not make so many demands as the derivation does but thinking about heat will help a lot there too We begin Consider a little slab cut from the bar between coordinates m and m Am and suppose the density of the bar is p mass per length We are going to account for the heat energy contained within this little slab We have Am length of the slab pAm mass of the slab physical principle 1 There is something called speci c heat c which re ects the experimental fact that a pound of wood and a pound of steel at the same temperature do not contain the same amount of heat energy According to this principle we have cupAm the heat energy content of the slab physical principle 2 Heat ows from hot to cold and more speci cally there is something called conductivity k which re ects the fact that copper conducts heat better than feathers do According to this principle we have ikum m t the rate heat enters the left side of the slab kumm Am 25 the rate heat enters the right side These two principles give us two different expressions for the rate at which heat energy enters the slab This is wonderful Whenever you have two different expressions for the same thing you are about to discover something important Equate them cupAmt ku m l Am 25 7 kum m t Then ku m Am t 7 kum m t Am Cput Finally take the limit as Am a 0 to get 6014 kum This gives the heat equation and shows us that 17 0p 110 Figure 272 A region R inside the main block of material with some heat energy owing between R and the rest of the block Energy conducts into R through its boundary surface S There is no heat source or sink chemical reaction etc inside R You can place R anywhere within the material A second derivation may be accomplished in three dimensions by using the divergence theorem We imagine heat conduction in a solid block of some thing Fix any region R inside the solid and account for heat ow into R The heat content is fffR Cpu 1V and its rate of change is fffR Cput dV The rate of heat ow through the boundary surface S is ffs k u dA By the divergence theorem this last integral is equal to fffR dV Thus we have chuthRV kvudV or RcputikumuyyuudV0 for all R Now if a function happens to integrate to zero it means nothing except that here it is true for all R This turns out to be enough to insure that the integrand is zero and we get CPU Mum um um PROBLEMS 1 Suppose the same temperature distribution is present in two metal bars at a particular time where the rst bar has u um and the second has u 471 due to di erent material Which bar cools faster 2 A new kind of physics is discovered on Mars Their heat equation says u an 7 u Do Mars bars cool faster or slower than Earth bars 3 Amazingly enough we have discovered that the heat equation used by the Klingons is di erent from either Earth or Mars They have u mm u u Mum uw u etc Sometimes their pizza doesn t cool 01 at all Why 111 4 There is a way of thinking about continuous functions which is a little beyond the level of these notes but you might be interested to hear it Suppose continuous function f is not zero at some particular point a fa 7 0 Maybe it is 0266 there Then you can take some other points x sufficiently close to a that is near 0266 say gt 0200 This will hold for any point x in at least some small neighborhood of 1 Knowing this we take R in Figure 272 to be located right in that neighborhood Then what can you say about MR NV This shows that if these integrals are zero for all R then f must be indentically O 5 What s Tong with this The heat equation can t be right because we already learned Newton s law of cooling in Lecture 8 and it doesn t say anything about the temperature depending on x 28 Solving the Heat Equation TODAY The meaning of boundary Values Steady state solutions Product solutions There are many many solutions to the heat equation by comparison with any ordinary differential equation For example all constants are solutions This makes sense if you think about what the heat equation is about a metal bar can certainly have a constant temperature Thereis another special class of solutions which are also easy to nd These are the steady state solutions ie the ones which do not depend on the time Everyday experience suggests that there are such cases consider heat conduction through the wall of a refrigerator The kitchen is at a constant temperature the inside of the refrigerator is at a constant temperature and there is a continual ow of heat into the refrigerator through the walls To nd such solutions analytically we assume that u is a function of z only What happens to the heat equation then We have at 0 and um u z so the heat equation becomes u 0 That is an ordinary differential equation and a pretty easy one to solve Integrating once we get u a Integrating again gives u ax b All straight lines are the graphs of steady state temperature distributions Check by substituting into the heat equation and you will see that these indeed work Note that these straight line solutions are a generalization of the constant solutions So far we have not said much about boundary conditions or initial con ditions It is typical of partial differential equations that there are many possible choices of conditions and it takes a lot of work to decide what the reasonable ones are Let s say for now that we are interested in heat con 112 duction along a bar of length l which is located in 0 g x g 1 Then one set of conditions which speci es a solvable problem is as follows You may specify the temperature at the ends of the bar and the initial temperature along the bar For example here is one such problem at um the heat equation u0 t 40 a boundary condition at the left end u5 t 60 boundary condition at right end the length is 5 um 0 40 4x the initial temperature You may consider whether these conditions seem physically reasonable or not The bar is going to have one end kept at 40 degrees the other at 60 degrees and the initial temperature is known Should this kind of informa tion be adequate to determine the future temperatures in the bar This is a hard type of question in general but in this case the answer turns out to be yes In fact we can solve this particular problem with one of our steady state solutions PRACTICE Find the steadystate solution to this problem Next consider the problem Ut U11 u0 t 0 u5 t 0 ux 0 40 Here we have a bar which begins uniformly at 40 degrees and at time zero somebody presses ice cubes against both ends of the bar and holds them there forever What should happen In this case the initial condition does not match the boundary conditions so the solution if it exists cannot be a steady state solution In this case it is not possible to just quickly reason out whether the problem has any solution or whether it might have more than one Physically it does seem as though the bar should gradually cool off to zero probably faster at the ends where the ice cubes are pressed 281 Insulation There is another popular boundary condition known as insulation What does insulation do If you put a hot pizza into one of those red insulated 113 containers that the delivery people use the pizza still eventually cools off it just takes longer the better the insulation is So the key to understanding insulation is that we model it by the assumption that no energy passes through it From our derivation of the heat equation we know that the rate of energy conduction is a multiple of um You might want to reread the derivation on this particular point Bottom line The boundary condition at an insulated end is um 0 Another way to think about insulation is like this heat ows from hot to cold So for heat to ow there must be a temperature difference On the other hand if um 0 somewhere then the graph of temperature has slope 0 and microscopically there is no temperature difference for small position changes there Now what is the following problem about Ut U11 um 0 t ul t 6O ux 0 40 This is a bar of length l insulated at the left end and held at 60 degrees at the right end The initial temperature is 40 degrees all along its length 282 Product Solutions There is another class of solutions to the heat equation which can be found by a method we have used many times before We try a solution of the form it 65 If you have read Lecture 3 then you have seen a less speci c assumption about u This trick worked the rst time on rst order ordinary differential equations and indeed the heat equation is rst order with respect to time The trick worked the second time with systems of linear equations This time we need to allow 0 to be a function of z Substituting into the heat equation 14 aum gives mequot LC 5quot These will agree provided that ac TC Whenever you try something like this you have to take a few steps along the path to see whether it is going to lead anywhere In this case we have run across something that we know so things look hopeful What we know is how to solve ac TC It is a second order linear ordinary differential equation for c and it is one of the easy kind with no 0 term To simplify it 114 a little let s write r iawz then what we need is c 71020 The solutions are c Asinwz Bcoswz Thus we have found many solutions to the heat equation of the form it A sinwz Bcoswze aw2t We call these product solutions77 because they have the form of a function of z times a function of t EXAMPLE Determine the Values of 11 such that there is a product solution to m an having frozen boundary conditions u0 t u7r t 0 We compute u0 t Be awzt in the product solution above This will be zero for all t only if B 0 So far we have u Asinwxe aw2t Then we compute u7rt Asinw7re mu2t This will be zero for all t ifwe choose w7r to be any of the zeros of the sine function These make u 1 2 3 e haVe found a list of solutions un t t AT sinnxeim2t n 1 2 3 PROBLEMS 1 Describe what the following problem is about u um 71x0 t O uxl t O ux O 400 Determine whether there is a steadystate solution to this problem 2 Describe what this problem is about u 7 um u0 t O ul t O u LO 400 Determine whether there is a steadystate solution to this problem 3 Find a product solution to u 7 um u0 t O u7r t O ux0 3sin 4 Find a solution to ut Sum 115 u0 t 7 O u7r t O ux O 7 2 s1nx 5 Find a product solution to ut Sum 71x0 t O ux7r t 0 ux O 3 ut aws ux0t O u 7rt O ux 0 3 7 Suppose that M1 and 712 are solutions to t was u0 t O u7r t 0 Note that no initial condition has been speci ed Set 5 ul 712 Is it true that s is also a solution to this system 8 Find a solution to u um u0 t O u7r t O ux0 3sinx5sin2x Which is a sum of tWo product solutions If you need help look ahead to the next lecture for Which this problem is intended to be a preVieW 9 Plot the graphs of the product functions unxt sin739we mb2t Versus x for seVeral Values of t for the cases 77 1 and n 2 Describe What is going on in your graphs in terms of temperature in a metal bar 10 In problem 7 suppose ux t eimyw ie an exponential function of t times a function of x What boundary Value problem of the type considered in Lecture 26 must 1 solve 29 More Heat Solutions TODAY More heat equation solutions Superposition 116 Today we are going to analyze a heat conduction problem which happens to be solvable using some of the product solutions found last time Our emphasis is not on formulas but on understanding what the solutions mean and thinking about the extent to which this heat equation corresponds with our daily experience of temperature Consider the problem in which a metal bar with frozen ends has temperature initially as in Figure 291 We want to nd out how this temperature changes as time goes by What do you think will happen We know from daily experience that heat ows from hot to cold and in fact a quantitative version of that statement was a major part of our derivation of the heat equation So probably the hot area to the left will supply energy to the cooler parts Figure 291 Our initial conditions Note the spot at which the bar is hottest What do you think will happen to the hot spot as time goes by ie will it cool OH will it move to the left to the right etc7 gure made by the matlab commands x O O314153 1415 plot 2 sinx 13 sin2x Our problem is as follows ut umm u0 t 0 u7r t 0 Mm 0 sinm sin m Before attacking the problem analytically lets just read it carefully to see what it is about That may be an obvious suggestion but it is also a way 117 to prevent ourselves from doing something idiotic We see from the rst line that we are solving the case of the heat equation in which the physical constant a has been set to 1 for convenience We can t see what the domain is until we read the boundary conditions The second two lines say that we are dealing with the ice cube case where the bar ends are at 0 and 7139 on the z axis and are kept frozen Dl Figure 292 Ice cubes at the ends of the bar You might be worried about heat escaping from all around the sides of the bar If so think about heat conduction through a wall rather than along a bar or else think of the bar surrounded by perfect insulation so that the x direction is the only one of any concern Finally the fourth line of the problem says that the initial temperature is the sum of two terms Graphed seperately they look like Figure 293 Figure 293 The initial conditions are a sum of these two terms gure made by x O 03141531415 plot xsinx hold on plot 2 13 sin2x hold off If you did problem 7 of Lecture 27 you may suspect why we have included Figure 293 Solutions to the heat equation may be added to produce new solutions If you didn t do that problem yet go do it now In mathematics 118 this is called linearity as we have emphasized several times in these notes In physics it is sometimes called superposition to crystallize the idea that two or more things are going on at the same time and place Now we are ready to solve this problem There are product solutions to the heat equation of the form u Asinwme w2t for all values of w Perhaps the ones we need are the cases of the sine function which appear in our initial conditions Let s try that Our candidates are u1z t A1 sinme t and u2m t A2 sin2ze 4t Now check to be sure that these solve the heat equation and the boundary conditions Do that now Does it work Ok Notice that it doesn t matter what the coefficients Aj are yet and that the sum um t A1 sinme t 1 A2 sin2ze 4t is also a solution to the heat equation with these frozen boundary conditions The next step is to attack the initial conditions We need to compare um 0 A1 sinme0 A2 sin2me0 from our tentative solution with the desired initial condition Mm 0 sinm sin m Now the question is can this be made to work Well yes it is really not hard to see at this point since we did all the hard work already You just take A1 1 and A2 This makes the initial conditions work and we have checked everything else So we have a solution to our problem It is Mm t sinme t sin2me 4t The next step is to understand what this solution means and how it behaves as time goes by The best way to do this is for you to sketch the graphs of the solution for various times thinking about what happens to each of the two terms in the solution If you do that yourself and it is a good idea to do so you should get something like Figure 294 for the individual terms and Figure 295 for the solution 119 Figure 294 The two terms in our solution plotted for t O and t 3 Notice how much each of these has moved in particular that the term involving sin2x has moved a lot more than the sinc term Why is that Figure 295 The solution temperature at times t O 3 6 9 12 Now you see that the hot spot moves toward the center Why If you study the solution formula and the pictures you will see how signi cant the n2 is in the time exponential part of the product solution PROBLEMS 1 Think about what should happen to a metal bar which has an initial temperature distribution consisting of alternating hot cold hot etc in a lot of narrow bands along 120 the bar Think of a bar Which is redhot in six places and cooler in between Do you think it Will take Very long for the temperature to change Solve 71 um u0 t O u7r7 t O ux7 O sin12x and sketch the solution for t O7 3 6 2 Solve u 7 um u07 t O u7r7 t O ux7 O sinc 5 sin3x 25 sin5x Sketch the initial condition 3 Solve u 7 um um07 t O ux7r t O ux7 O cosc 5 cos3c 25 cos5x Note that the initial temperature satis es the insulated boundary conditions 4 Run the Heat Equation 1D applet mentioned in Lecture 7 Try Various initial and boundary conditions to develop some feeling for What this equation is about 121 30 The Wave Equation TODAY Traveling Waves We have now solved several heat conduction problems and are about to turn to another partial differential equation known as the wave equation You can gure out what it is about Notice that as we were working on the heat equation we solved a fairly large number of problems distinguished from one another by boundary and initial conditions while the equation itself never changed This is typical of partial differential equations that there are relatively few of them which are considered to be important and the conditions can be changed to permit these few important equations to apply to many different settings For the heat equation we only dealt with heat ow on a nite interval 0 3 z 3 1 Our wave equation can apply to many things also such as the vibrations of a guitar string but we are mainly going to talk about waves on an in nitely long string the whole z axis The reason for this choice is that guitar strings can be solved by methods similar to those we used for the heat equation and we prefer to introduce new ideas here The wave equation is 2 utt C umm 2 Utt C Um Uyy 2 Utt c um uyy U22 in one two or three dimensions respectively Note that it is second order with respect to the time like Newton s law In fact it is Newton s law in disguise sometimes We know from Lecture 10 that second order equations can oscillate so that at least is reassuring In one dimension u may represent the vertical position of a vibrating string Figure 301 The one dimensional wave equation describes vibrations of a string or less ancurately water waves 122 In two dimensions or three the wave equation can represent vibrations of a drum head or of sound in the air and other things like electromagnetic waves It is a sound and light show all by itself If you have read Lecture 34 on drums you will have seen the equation utt may That is one version of the two dimensional wave equation written in polar coordinates for the case where there is circular symmetry Let s read the one dimensional equation again If u is position what is utt Think about it Did you think about it yet It is acceleration Also when you draw the graph of u as a function of m as in Figure 301 what does um represent Just as for the heat equation it is the curvature of the graph The wave equation says that when the graph is concave up the acceleration must be positive Does that seem to t with reality Think about swinging a heavy jumprope Better still get a rope and try it See whether you think the acceleration direction matches the curvature in this way The wave equation for a string may be derived as follows We apply Newton s law to a bit of the string accounting for vertical forces Figure 302 The tension in the string pulls at dj erent angles on the two ends of any little piece of it depending on the curvature The tension T in the string has a vertical component On the left it is approximately Tumm 25 since the slope um is the tangent of the angle which is near the sine of the angle if the angle is small On the right it is Tum m Am 25 The difference of these must be the mass times the acceleration so Tum m Am t 7 Tum m t pAmutt where p is the mass per length of the string Dividing by Am and taking the limit as Am approaches 0 gives Umm Utt p 123 which is the wave equation with 02 Now let s try to solve the wave equation7 using our usual method We try it fmequot in utt 0214 It becomes r2fmequot 02f ze Cancelling the time exponentials and setting r ac gives 1 a2 Wow Another easy second order equation has popped out Again we nd that we must remember those second order equations Go back and review if you don t remember We nd solutions x 5 x 5quot and linear combi nations For example if a 1 we have found the solutions it ewe f and u e med To understand what these are7 we graphed the second one in Figure 303 Figure 303 The wave uxt af ict graphed for various times Note that to keep the expression x fat constant as t increases you have to increase x also so this wave is moving to the right It could make a surfer cry So7 we have found product solutions to the wave equation of the form it eaw t and u eawia for all values of a7 even complex What does this mean It certainly means that there are an awful lot of solutions which are functions of z i ct Now lets get crazy and generalize Suppose you have a function uzt gm 7 ct Mm ct 124 not necessarily built from exponentials at all If you have read Lecture 3 you have seen waves traveling in one direction there These go left or right Taking derivatives by the chain rule we nd ut 7cgz 7 ct chx ct utt ngN 62h um gx 7 ct hx ct umm 9 l h Woahl Look what happened there7 we got utt 0214 without assuming anything about 9 and h except that they be differentiable Doesn t that mean you can have waves of nearly any shape moving left and right Yes it does They are called traveling waves PROBLEMS 1 Show that cosx 7 2t and 113 202 are solutions to u 0271 for some 0 Find 0 Graph these waves for t O7 1 and 2 2 The function sinx 3t sinc 7 3t consists of 2 traveling waves and solves Ht 971m u0 t 0 u7r7 t O u7 O 2 sinc utc O 0 Sketch the two traveling waves both separately and together7 and try to see why the sum has the O boundary condition at x 07 even though neither of the traveling waves does so separately 3 Use the addition formula for the sine to show that the function in problem 2 is equal to 2sinxc0s3t It therefore represents a vibrating string having frequency Then so ve t 971m u0 t O u7r7 t O ux O 2 sin2x utx O 0 Sketch the motion for this case7 and convince yourself that this vibration produces a sound one octave up from that of problem 2 ie has twice the frequency 4 Continuing in the tradition of problems 2 and 3 now use initial condition ux0 2 sin3x The note you get is the fth above the octave musically 5 This problem is about using the wave equation to model an echo Consider a function of the form ux t ct 7 at where that is the same f traveling in both 125 directions Show that u satis es the condition ux0 t 0 Think of sound waves moving in 0 g x and a wall at the point x O ketch a graph of u for several times using a function f of your choice and try to convince yourself that what you have sketched is a representation of waves bouncing off a wall 6 Repeat problem 5 for ux t ct 7 7 at This time you should get the boundary condition u0t O which involves a different kind of bouncing at the wall Try to see it in your sketch Which of problems 5 or 6 is more like a string vibration with the string tied down at the origin Which one is more like water waves at the edge of a swimming pool 7 Try the Wave Equation 1D and 2D applets and explore for a while to get some feeling for how the solutions behave Try various boundary and initial conditions 8 What s Tong with this The function uxt cost sinc couldn t be a solution to the wave equation because it isn t a traveling wave Figure 304 A wave traveling initially to the left is shown at six times t increasing from top to bottom in the picture It is re ected downward and reversed leftright when it bounces off a wall at the left side Does this t Problem 5 or Problem 6 better 126 31 Beams and Columns TODAY A description of some di erential equations for beams and columns Figure 311 A beam with some exaggerated de ection A beam is a bar loaded transverse to the axis7 while a column is loaded along the axis Of course the orientation doesn t matter even though we think of a column as being vertical7 Figure 312 Left A segment of the beam with shearing force V and moment M acting R is the radius of curvature Right A column buckled by an applied force For either case suppose z axis left to right gives position along the bar7 and we have a segment of the bar as shown with shear force V Newtons and bending moment M Newton meters as functions of z Also F Nm is 127 an applied loading Let be the downward displacement of the beam measured to the centerline say and R the radius of curvature at each point also measured to the centerline In mechanics class you will use centroid which is more correct than centerline Geometric fact not included in most calculus courses y They do tell you about concave up and down up is shown for the segment but this quanti es it PRACTICE Write a function whose graph is the top part of a circle and Verify the second derivative at the topmost point Does it matter if you add some slope 7711 b to your function The hardest part of the analysis is to see that y is also proportional to M Here is how that works For the bending shown the top portion of the beam must stretch and the bottom compress Try it with an eraser Mechanical fact steel has the prop erty that for small stretching and compressing the stress a forcearea and the strain 6 length changeoriginal length are related by Young s modulus E 210 X 109 Newtonssquare meter as 039E For aluminum and granite E is about 13 as much Next geometric fact let yl measure the distance from the centerline of the beam away from the center of curvature So yl 0 at the centerline yl 2 at the top of the beam if the whole size of the beam is 4 units top to bottom etc Then the geometric fact is that the length at level yl changes by the factor 1 Exercise check that out Thus 6 LEI So the bending moment at a particular cross section must cause the length changes and we have stress 0y1 E pulling or pushing normal to the cross section Then the moment is the sum of force times moment arm 1 1 M 0y1y1dAE gang E1 EIy crossisection crossisection This concludes the hardest part 128 For example a q x q square cross section beam has moment of inertia I 2 2 jfgzcgZ y dyldyz q412 The easier parts are to see that M is proportional to the shear V and that V is proportional to F the downward load Nm applied to the beam PRACTICE For M surn moments about the left end point of the segment giving 7Mx Mx Ax AxVx Ax O For V sum downward forces on the segment giving 7Vx Vx Ax 0 Then divide by Ax and take limits as Ax a 0 Putting all that together we summarize beams sign conventions may differ in various texts Ely M Ely 7V ElyW F These are differential equations and in addition they tell us how to interpret various boundary conditions For example a cantilever beam embedded solidly into concrete at one end probably has y y 0 at the concrete end and y y 0 at the free end Two more applications For a column which is de ected into a bent shape by an axial force P you get a bending moment yxP at each cross section and taking account of signs the DE becomes Ely iPy Finally we mention vibrating beams Suppose y depends on x and t and that there is no applied load F If the mass of the beam is p kgm then you nd a PDE Elyxxxx ipytt for the vibrations PROBLEMS 1 Find a type of boundary value problem in Lecture 26 which could be applied to the column Take 1 if that helps What is the smallest eigenvalue at which the column buckles 2 Try to nd some beam vibrations of the form yx t ac0sbt sincx That does not mean derive this formula from scratch It means plug it into the beam equation Pym Efymm 129 and see what must be true of the numbers a b and c if this is to be a solution But it is alright to try a separation of Variables too7 as one way of deriving it We are allowed to try more than one thing 3 Find a list of beam Vibrations from Problem 2 which have the boundary Values 210 t 0 Wm t 0 4 Use your list in Problem 3 to nd a solution to ytt iyacacacac y0 t O y7r7 t O yx O sin1 7 T10 sin3x 5 The minus sign in the beam equation ytt Elyxwxac looks strange if you have been solVing the heat equation recentl Suppose that at some time the shape of the beam is y x4 Do you think the acceleration ought to be toward the y or the 7y direction 6 We know that the rst oVertone of a string Vibrates twice as fast as the fundmental What about a beam Verify that it goes 4 times as fast when Vibrating in two equal parts This is why guitars and Xylophones don t sound the same 130 32 Power Series TODAY How to make ugly functions beautiful 321 Review Polynomials are nice aren t they f17 990 2 7x27 hm m6 7 32353 4 WE 200 718 They are the simplest functions you can build by repeatedly adding and multiplying numbers and one variable People can use these as soon as they are have some idea of a symbol z being used to represent a number These are beautiful functions7 and we know a lot about them But we also know that there is not enough variety in those to do science We need fractional powers7 exponentials7 logarithms7 and trigonometric func tions at least With all those7 we have a list of functions just long enough to include most of the buttons on a calculator In other words7 those are the functions which are so commonly used by every working scientist that it is worth the time of some manufacturing company to create a pro table product7 the calculator7 which includes them They are not polynomials Other functions are needed too7 many ofthem actually de ned by differential equations7 that don t have their own button yet7 so to speak Maybe by the year 2035 the solution to some new equation will be so important that everybody has a calculator for it7 built into their hat or something All these functions wish they were beautiful too7 like the polynomials It turns out that some of them are nearly polynomials after all7 though they may not know it For example7 it is sometimes possible to give pretty good approximations by polynomials Here is one 7 1 2 1 4 cosz71 2x 24m E where lt m if lt That is a nice result For example7 0016 1 i cos2 1 7 02 T i m 7 098007 to four decimals 131 You might remember doing such estimates in your calculus course under the topic Taylor s Theorem Sometimes it seems disturbing that you don t ever nd out exactly what E is because we are more accustomed to equations than to inequalities But if you know E exactly don t you know the cosine exactly too And that is what we don7t know Functions that can be expressed as power series are known as analytic func tions Like the cosine they can be approximated by polynomials not only of degree 4 but of any degree and to whatever accuracy you need Here is what some power series look like coclmch2cgm3 2 m 717z2 43m3 7 5m6 9m coclz71062x7 10263m7103 cm lmm23im3m4 Lets look at both the form and the meaning of these expressions THE FORM The rst and third examples are generic ways to write a power series The rst is said to be expressed in powers of z while the third is in powers of m 7 10 The coe icients ck represent various real or perhaps complex numbers The 10 can be replaced by any other number and the letter x can be replaced by any other letter but no other change is allowed For example does not occur because we only allow nonnegative whole number powers The second example is a polynomial That is a power series because it is in the form of the rst example having most of the ck O which is alright But the other way around a power series not considered to be a polynomial Polynomials have a nite degree the largest exponent which occurs Also if you have an expression involving say m2 7 23 that is not a power series either but it might be possible to rearrange it as one For instance in the identity x2 7 23 m6 7 6354 1235 7 8 both sides are polynomials but only the right hand side is considered a power series The fourth example is known as the power series for the exponential function and we will say more about it later For now just notice that whatever the meaning is the equation seems to be giving us a beautiful almost polynomial for em 132 PRACTICE Did you ever have a formula for 6 0 before Did your formula allow you to calculate say CO 573 without a calculator So far we see what a power series looks like but we haven7t said a word about what you are wondering what does the dot dot dot mean How can we calculate anything that goes on forever 322 The Meaning of Convergence I m going to explain the idea of convergence of power series by using one principal example You can nd a more traditional approach in your calculus book Everything we know about power series comes from one example known as the geometric series It is the most important You need to know everything about it and how to recognize it even in disguise It looks like this 1mz2x3m4z511 iflmllt1 One disguised version of it occurs just below As a rst observation note the restriction that lt 1 Most formulas you are familiar with have had few restrictions on the variable But that needs to change now and we ll have to think about the applicability of our formulas For example in the geometric series if z 1 you would have complete nonsense on both sides of the equation so that has to be excluded Again if z 2 for example the right side is de ned but the left says add the numbers 1 2 4 8 16 etc and never stop adding Obviously that doesn t work either But lets try a small number 01 Then it says 1 1 01 001 0001 00001 000001 m Could that be true It ought to be interpreted as 11111111111 which seems to be correct PRACTICE If you aren t sure about that divide 10 by 9 and see what you get both by hand and on a calculator 133 We still haven t assigned any exact meaning to the dot dot dot expressions Here we will rst nd out exactly what it means for the geometric series We start with an algebraic identity 1m17x17z2 1mz217m17x3 1mm2z317x17z4 PRACTICE Are those true Is there any restriction on the Values of 7 In general we have 1zx2m3xnliz lizwrl and so if z 7S 1 1 7 n1 1mx2z3mni 1 7 m Now that is still a nite sum Notice that if n is 1000 you have a very long expression on the left hand side much too long to write out But the right hand side is very short to write and pretty short to compute no matter what 71 is What happens when we try to allow 71 a 00 On the left side we don t have a clue because that is really what we are trying to de ne After we make sense out of it then it will be alright to write the left side as 1mm2m3 but we don t know what it means yet So look at the right side What happens to 1 7 n1 1 7 m as n a 00 PRACTICE Try Various numbers x and convince yourself that there is no 1 limit unless lt 1 and then the limit is m 134 So we understand the meaning of the geometric series as follows If you stop after adding 71 1 terms you have 1 n1 1zz2z3z 7 1 7 m 1 7 m which is not exactly 1 1 7 m but the error is n1 1 7 m which approaches 0 as n 7 oo PRACTICE What error does that giVe for the approximation 1 002 00004 7 1 1 7 002 In general we make a de nition7 that the meaning of any series a0a1a2a3 is as follows7 power series or not Suppose there is a number L so that whenever you stop at the n th stage you have some remainder Rn aoa1a2a3anLiRn and Rn approaches 0 as n 7 00 Then we write aoa1a2a3 L and we say that the series converges to L I hope you remember from calculus that a0 a1 a2 a3 1 an is called a partial sum of the series In the next lecture we will see many uses of the geometric series PROBLEMS 1 Ebrpress the repeating decimal a 0306306306m as a multiple of a geometric series Express 1 as a rational fraction 1 mn with integers m and n 2 More Challenging Consider the series 1 1 1 1 51273747quot39 135 There is no number x so that the terms in s are each less than the terms of the geometric series 1 1xx2x3 1 7 x Try it7 and you ll see Why But you can group the terms in an interesting Way 1 1 1 1 1 1 1 7 7 7 7 7 7 5 lt2232gtlt42 72gtlt82 152gt so that eVery grouping is less than the corresponding geometric term With some carefully chosen number x Try to gure out x What can you conclude about the size of the number 57 This series Was a research problem somewhere around the 18th century EVentually Euler found that it is a special case of a Fourier series7 and he Was able to Work out the Value exactly7 On page 167 you will Work that out yourself Eken now you can see that there is something advanced about it7 from the 7r Nobody squares 7r in eVeryday life It is common in calculus books to condense this interesting story into the statement that the ratio test fails for this series 136 33 More on Power Series TODAY We use the geometric series as the spring from which a river of functions ows We have learned that 1zm2m3 1 Ma and we have reviewed what the idea of convergence means We have not been very quantitative yet about other series And we have not justi ed why this is so important So why on earth would anybody want to replace the perfectly simple ex pression 11 7 z with a complicated limit which doesn t even work for every value of z I m so glad you asked Let me give you eight excellent reasons 1 We will quote a theorem below which says more about the convergence of power series but for now lets just integrate our geometric series from 0 to m We get 1 1 1 z7m27m37x47ln1im iflmllt1 2 3 4 Did you ever have a formula for the natural logarithm before What was it Did it allow you to calculate ln097 Lets try that 711109 01 001 0001 01 0005 00003333 S0 ln09 701053 Nice eh Look Ma no calculator We have turned the ugly ln into a beautiful almost polynomial 2 Next make a change of variables in the geometric series to get 1 17u2u47u67m lullt1 and integrate from 0 to z to get 1 1 m7 z3gz5 7 tan 1m lt 1 137 Gee We never had that before did we 3 As a limiting case in item 2 take z a 1 to see 1 1 1 1 7 7r 3 5 7 T 4 We have always been told what a strange number 7139 is irrational nonalge braic transcendental in nite nonrepeating decimal expansion People have written books about the hidden messages of the universe in the mysterious decimals of 7139 and other such nonsense Here we see that somehow 7139 is related to the odd integers That is unexpected 4 Here is a standard textbook example Determine whether the series 00 1 X m n0 converges Lets write it out 1 1 L 1 12 12311234139 The biggest single question is whether this could be nite because of a fundamental fact about the real numbers Write the n th partial sum as 1 H 1111 5 7 2 23 THEOREM An increasing sequence of real numbers bounded above has a limit ie if 81 52 83 Llt00 then the 5k have a limit which is less than or equal to L It is certain that our 51 S 52 What about some bound L Here is a way to check our sequence Since 3 4 5 etc are all larger than 2 you have an inequality for each partial sum lt 1 1 1 1 1 1 5n 22r2232n You see the geometric series lurking there in powers of So we can see that all of our partial sums 1 1 snlt1 13 E 138 That is enough to establish convergence to something by the theorem It does not tell us exactly what the sum is just that it is less than 3 Of course we are supposed to know it is e 2718 but that comes from item 5 next Also you can compare this approximation of e to the one on page 33 I hope you see that a knowledge of the geometric series gives information about many other series In fact if you have had a course in which the various convergence tests77 were discussed it will be important for you to realize that most of those tests such as the ratio test are based on comparison with a geometric series just as in this example 5 The same kind of discussion as in item 4 reveals that the series 1 2 1 3 1 4 fz71m m x k z converges for any value of m Again it does not tell us that it converges to em as was claimed at the beginning of this section We need a way to check that We are going to use the uniqueness theorem for differential equations to do that It is on page 22 if you need to look Note rst that the initial condition is f0 1 so it has that in common with e1 at least What characterizes em Isn t it the derivative If we are allowed to calculate f x from the series we get 1 1 1 Hm01E2m 3z2 4m3m 1zz2x3wf That is exactly what we expect of em We have the uniqueness theorem for solutions of differential equations so it looks strongly as though x cm In fact it is true There is only one fault in the logic We still need to know whether it is alright to differentiate the terms of a power series to calculate its derivative This is similar to the problem we had in item 1 above where we didn t know whether it was ok to integrate a series Here is the theorem we need from another course FUNDAMENTAL THEOREM OF POWER SERIES For each power series 0 Z w n0 139 there is a number R called the radius of convergence If 7R lt z lt R the series converges If x lt 7R or R lt z the series di verges In fact the coefficients on are allowed to be complex num bers and the series converges for all complex numbers z inside a circle of radius R in the complex number plane and diverges for every number outside that circle Within the circle of convergence it is allowed to differentiate and integrate the series term by term and the resulting series still has the same radius of convergence The special cases R 0 and R 00 are used to announce that the series converges only for z 0 or converges for all numbers respectively The radius of convergence can usually be found by comparing the series with a geometric series or by using any of the convergence tests most of which are proved by comparison with a geometric series 6 Now that we have the 51 series and the Theorem we get a lot of others almost free 1 1 1 GOSH ew 67m1 5902 1954 1 d 1 1 sinhz 509 7 5 1 Ecoshz 27 Ems if r 7 And if we are comfortable with complex numbers we also get the trig functions cosltzgt gem 1 7 7 t 7d 7 13 15 s1nz7coszimigm tax 7 8 How does a calculator work Some of the functions it has buttons for can be calculated by series Perhaps the buttons cause a partial sum of a series to be computed Is that what it really does Maybe you know if you are in computer engineering But what about functions that we don t know yet The ones needed in science but not used enough to make them pro table as consumer products as buttons on calculators That is what the next section is about PROBLEMS 140 1 What is the radius of convergence of 1II2I3quot397 Whatever you do don t apply the ratio test to this series The ratio test is proved based on what we already know about this series Does it converge when x 7 To what 2 What does the power series theorem give for the radius of convergence of the derivative series 12x3x24x31xx2x3x47 Does it converge when x 7 To what 3 What is the radius of convergence of 1xi10x7102x71037 Does it converge when x 837 4 Use your knowledge of the cosine function to gure out the value of 2 4 6 1 L L L 2 4 6 It really wouldn t pay to evaluate the rst few terms of that would it 34 Power Series used A Drum Model TODAY We set up a model for the vibrations of a drum head The Drum Our model of a drumhead has radius 1 and it only vibrates in circular symmetry That means you can only hit it in the center so to speak Using polar coordinates there is dependence on r and 25 but not on 0 We write ur t for the upward displacement of the drumhead out of its horizontal resting plane and assume that u is small and that u1 t 0 since the edges of the head are pulled down against the rim of the drum We will assume that there is a tension T uniformly throughout the drumhead Convince yourself that this is reasonable as follows Imagine the tension forces on a small triangle of material located anywhere in the drumhead 141 Suppose a triangle with Vertices 00 10 and 01 has a force applied uniformly along each edge perpendicular to the edge in the plane of the triangle with magnitude T times that edgelength Show that the triangle is in equilibrium ie that the sum of forces on it is 0 assuming T is a constant Now write Newton s F ma law for vertical forces on any annular region of the drumhead We take the annulus to have radius running from r to r dr In working out the forces due to the tension T it is best if you have already worked on the wave equation for string vibrations The reason is that while the forces here are quite similar to those acting on the string the geometry here is a little more involved because of the curved shape of the annulus rdr Figure 341 The annulus of drumhead material Viewed from above The length of the inner curved edge of the annulus is 27m There are forces directed inward all along that edge We are interested only in the vertical resultant of those forces That is T27rr times the sine of the small angle with the horizontal plane The slope is 147025 The slope u is nearly the same as the sine of the angle for small displacements See Problem 1 if you aren t sure about that Similarly for the outer edge 142 A r rdr Figure 342 A cut through the annular region7 Viewed on edge The Ver tical displacement is greatly exaggerated We re accounting for the Vertical components of the tension forces The net force vertically is then T27rr drurr dr t 7 T27rrurr t The acceleration of the annulus is utt Mr Newton says that we also need its mass The mass of the segment is its area times its density p kg77127 or p7rr dr2 7 WV By Newton then T27rr drurr dr t 7 T27rrurr t p7r27 dr dr2utt Divide by dr and take dr 7 0 getting 32 prutt So our equation is rutt ru7r or equivalently We will nd some solutions to this having u1t 0 at the rim of the drum The solutions will be vibrations of our drum model7 and hopefully will resemble vibrations of a real drum PRACTICE Polar coordinates are awkward at the origin What should the slope uTO t be at the center7 so that the shape of the drumhead is smooth there Separation of Variables Maybe our wave equation has some product solutions of the form ur t RrTt Since this is supposed to represent music7 lets try ur t RU coswt 143 or sine rather than cosine Set that into the PDE to nd T 1 iwz coswtR 7 RN 713 coswt p r We need 2 1 R 7R QRO r T For any such function we can nd there will be drum Vibrations RU coswt and RU sinwt In the next Lecture we plan to nd a nice solution B However it will be easier if we don t have to deal with the parameters sz all the time Let a new variable x M and choose a to clean up the R equation Write RU We have ptT ifm and R r ffm The DE for R becomes 2 aZfEaf f039 m T Of course we set a w Our new DE becomes 1 f7ff0 m PROBLEMS 1 Suppose a Very acute triangle has Vertices 00 10 1 771 so that the slope is m 11 00 10 Convince yourself that m is the tangent 0f the acute angle exactly and the sine 0f the acute angle is nearly equal to 771 You could ddle with the geometry or try your calculator or try the power series for the sine and cosine or think of something else to try That is why we feel it is justi ed to replace the sine 0f the angles by the m Values in our derivation Of course if it turns out that the model doesn t act like a real drum then we have to rethink this decision 2 If you are not clear about the limit we found try this First tell what this limit is hm fltz h 7 we hHO h Then how about hm 96 hgx h 9 906 7 hHO h Finally the one we need hm r druTr dr t 7 ruTO t Mao dr 144 35 A new Function for the Drum model J0 TODAY We learn a new function which might not be on your calculator7 and use it to describe Vibrations of the drum head So far we have derived a wave equation T 1 Utt Um W p r for the displacement of our drumhead7 and we know that we need to solve f f f 0 Having such a function f we will get many drum vibrations ur t f w gr coswt A new function J0 Try a power series ff0f22f4964m If you did the PRACTICE item on page 1437 you know that we don t want a term flz We are guessing that we don t need any of the odd degree terms You can check later that we truly don t To keep things as clean as possible lets also assume f0 1 We can rescale things later as needed We have 1 f f f lt2f243f4x265f6x487f8x6gt 2f2 4m 6M4 8M6 r 1 f z m4 f6z6 4f2 14394f4 f2962 6 61 f4964 8 81 f6906 For that to be identically 07 we need all the coe icients to be 07 so 1 f2 1 f4 1 1 f2 7 f4 EW7 f6 67 m7 va etc So 2 4 6 8 13 13 13 13 fltgt1 tm mfm Isn t that a beautiful series This function is sufficiently useful that it has a name7 even though you won t nd it on every calculator It is called the Bessell function J0 145 Figure 351 Graph ofBessellJoz17 g 243 7 7 So x J0z or if we need a multiple of that x f0J0z for some constant f0 We need to know what this function is like to see whether it gives us plausible drumhead vibrations urt Mom rcoswt The series tells us some of the properties that we need and the differential equation 1 J6 7J6J00 13 itself gives other properties First the series The series is reminiscent of that for the cosine function so maybe J0 has many zeros and oscillates and is periodic Well 2 out of 3 isn t bad J0 has in nitely many zeros and oscillates somewhat like the cosine function but is not periodic In a homework problem you will use the series to nd that J04 is negative Since we know J00 1 there must be a number ml between 0 and 4 where J0z1 O In fact the rst root 1 is roughly 25 and the second root 2 is near 55 3 is about 85 and there are in nitely many others From the DE we see this suppose you have an interval like 12 where J0 is 0 at each end and nonzero between At any local min or max where J6 0 the DE then tells us that Jg 7J0 there The graph can t be concave up at a local max nor concave down at a minimum So the graph can only have simple humps like the cosine does no complicated zig zags between the roots 146 After analyzing J0 like this for some time people eventually published tables of its values and many years later someone built it into matlab and octave so that we could look at the graph in Figure 351 Isn t that nice 351 But What does the drum Sound like We don t know an yet We don t know what sounds the drum can make We have two boundary conditions The rst is that 1470 25 0 so that the drum shape is not pointy in the center This is automatic since it is a property of Jo PRACTICE How do you know from the series 2 Jox1i that 150 07 The second boundary condition is that at the rim we have u1t f0J0w1coswt 0 To acheive this we need w B m1 T or 2 or some root of Jo Then the lowest sound we hear will correspond to the case T LLT t f0J01T COS gmlt The ordinary cosine function has period 27139 so this one will go through one cycle as T i1t p 27r goes from 0 to 27139 One cycle in every f seconds The frequency is m1 371 For larger roots 2 etc you get shorter cycles higher pitched sounds Any linear combination of these various solutions for different roots zn and using both cosines and sines is a solution to our wave equation So our model predicts what sounds the drum will make Good 147 PRACTICE When you tighten up the drum tension T7 do the frequencies go up or down Also the drummer might change to a heavier material so that p increases What does that do to the frequencies EXAMPLE Take p T for simplicity We have a Vibration 711073 Jox1rcosx1t where x1 is approximately 25 This is a solution to the boundary Value problem um um m 0 lt r lt1 u1 t O m0 t 0 having initial position u1r O Jox1r Another solution is u2r t Joc2r sinx2t where M is approximately 55 This has a different initial shape7 and a different frequency7 and a different initial Velocity PRACTICE What is the initial Velocity of U1 PROBLEMS 1 In this problem you can con rm part of what is shown in the graph of Jo that Jo4 lt 0 so there must be a root x1 somewhere in the interVal 0 4 Write the partial sum of Jew through sixth powers as 5613 2 4 6 x x x 56x1 27m m First check that 564 is negative It saVes a little work if you notice that the second and third terms cancel 2 ln Problem 1 the tail of the series is so small that Jo4 itself is also negative That means7 Jolt4gt 564 E where E is not Very big and the sum is negative You can estimate E by obserVing that the tail is an alternating series7 if you remember those from calculus 3 Make a sketch of the graphs of Joc1r and Jox2r for 0 g r g 17 so that you can see what some of the drum anes look like You can gure these out from Figure 351 by rescaling 4 For a particular real drum it might be feasible to measure p if you could weigh a sample of the drumhead material but T could present more problems Suppose you compare the 148 sound of the drum to a piano or guitar and decide that the fundamental tone is near the note A2 110 cyclessec7 the next to lowest string on a guitar Figure out for this drum 5 Remind yourself that the rst three tones of a string have frequencies in the proportion 1 2 3 ie there are solutions cost sinx cos2t sin2x7 and cos3t sin3x to the Wave equation y ya But What about a drum Check the graph of Jo and nd out approximately how the lowest three frequencies of a drum are related This is Why pianos and drums don t sound the same 149 36 The Euler equation for Fluid Flow and Acous tic Waves TODAY A model for compressible air ow including sound This example is a little more advanced than the rest of the lectures so there are several PRACTICE items here to guide your reading I bet you can handle it The purpose of this lecture is to derive three PDEs which describe compress ible ow of air in one dimension say in a pipe You might imagine pumping air into a bicycle tire through a hose for example The main assumptions are c There is no conduction of heat through the air There is no friction of the air with the pipe or any exchange of heat between the air and the pipe 0 You know about Newton s law F ma and about the idea of conser vation of mass And of course you know calculus Two of the three equations are known as the Euler equations for homentropic How You do not have to know what homentropic means Then having the Euler equations we can further derive that certain approximate solutions ought to also give us solutions to the wave equation These solutions are very familiarithey are sound waves Thus we also get almost free a derivation of the wave equation If you have already derived the wave equation for vibrations of a string you will see that there are two very different situations giving rise to the same mathematics The method used in this derivation is rather common in science There are many assumptions behind every statement So to understand the deriva tion you have to ask yourself at every step Why should that be true What is being assumed This careful questioning can lead to good insights The biggest differences between this lecture and what you might nd in a more advanced treatment are that 1 we work in one dimension rather than three and 2 we make some of that careful questioning explicit But you can still nd plenty of additional questions to ask 150 361 The Euler equations The pipe is full of air The air velocity is um t msec and density pz t kgm3 For air the density is typically around 12 near the surface of the earth but the velocity could vary over a large range We can think of z and u positive toward the right L R Figure 361 A portion of air moving in the pipe The pipe has crosssectional area A The ends L and R don t have to move at the same speed because this is compressible ow We focus on a portion W of the air The letters L and R in the gure refer to the left and right ends of the mass W W is mathematically just a moving interval The left coordinate L is moving at velocity 7 and to be more precise Lt uLt t and uL is an abbreviation for this and similarly for the right end B at velocity uR Besides velocity umt and density pmt we must keep track of the air pressure px t Nmz It is very easy to mistake p and p typographically but this won t happen to you because you are reading very slowly and think ing hard right The pressure provides forces pLA and pRA on the sides of W The homentropic assumption is that the pressure is related to the den sity by p 107 where 39y 14 and a is constant throughout the ow Aside from this thermodynamic assumption the rest of our derivation will consist of ordinary mechanics Newton The rate of change of momentum of the matter in W is equal to the sum of applied forces d R 7 puAdm ipgA pLA dt L 151 PRACTICE Is this really F ma or is it more like 0770 F7 When does the m in Newton s law have to be included inside the derivative Of the quantities and expressions p u A7 and fdx which constitute v and which 7717 Why do we multiply pressure by area Why is there a minus sign on p314 In the left side of Newton s law we need to remember that d N N a he 2 dz Mm dz we t t e we tgtg t W W PRACTICE Problem 1 Derive that from the Fundamental Theorem of Calculus and the Chain Rule So7 if we cancel all the A s7 Newton s law and the homentropic assumption give R L Pu 190 PURUR pULUL PR PL iapla aPZ Now we may be able to study a limit of this for very short intervals If it is possible to divide by R 7 L and take L a R7 then we get W 9142 19 0 This is a partial differential equation which of course assumes the various functions are differentiable We assume this But it is important to know that the limit has been found not in agreement with all experiments We won t go into shock waves7 at all7 but you should know that there are such possibilities We notice that we have two unknowns u and p Probably one equation isn t going to be enough7 so we look for another equation Turning to conservation of mass7 we gure the mass in W to be fLR pAdm This doesn t change as the uid moves because our set W is moving with the uid We have conservation of mass d R i pAdz 0 dt L Doing the same transformations as we did for Newton gives R ptA dm 7 puRA puLA 0 L 152 and in the limit Pt T Plow 0 PRACTICE Problem 2 Use 17 mm 0 in the Newton PDE to nd u uux implapw These two equations are the results of our work7 the onedimensional home ntropic Euler equations PRACTICE Problem 3 We want to identify a physical meaning for the combination of terms ut uuw which occurs in the equation of motion We can show that it is the acceleration of the particle which is passing through the point x at time t Suppose a particle of uid has position giVen by a function Then we have two expressions for the Velocity of this particle7 xt and uxt7 t These must be equal DiHerentiate to show that the composition ut uuacxt t is equal to the acceleration of the air particle PRACTICE Can there be any acceleration at a point where u 07 362 Sound Here we start from the Euler equations7 and imagine a small disturbance superimposed over an ambient stillness PRACTICE Is it true that the ambient stillness u 07 p constant is a solution to the Euler equations of Problem 27 We then look for approximate solutions of the form umt evmt pxt P1 6wmt Where 6 is supposed to be a small number The constant p1 could be taken to be a typical air density at sea level on earth PRACTICE Problem 4 Substitute our assumed u and p into the Euler equations and ignore everything containing 52 or higher powers Show t at you nd for the 51 terms 7 7 2 vt 7 quot1771 was wt lvx 0 153 Combining the two PDEs in Problem 4 we have wtt 701 KY3177110 or Wt Czwmm where 02 a ypfl So the pressure disturbance satis es the wave equation PRACTICE What did we just now assume about the second derivatives This subtlety is omitted routinely in scienti c discussions Are solutions 0 an approximate equation necessarily approximate solutions to the right equation This point is also usually omitted Air pressure at sea level on the earth is about 105 Nm2 Using this and previous information estimate the value of the coef cient 02 occur ing in our wave equation Then recognizing that traveling waves such as fx 7 bt gx bt satisfy a wave equation what do we learn about the speed of sound If lightning strikes at a distance of 3 football elds from 7 you how long before you hear it Finally we have to ask Have we proved anything about the behavior of real air What is the criterion for correctness in Physics in Mathematics Have these notes strictly conformed to either Loose ends a We decided to look for 2 equations for our 2 variables u and p How reliable is the idea that one needs 71 equations in n unknowns before you can solve anything Have you seen examples in linear algebra A5 1 where two equations in three variables might have no solution where ve equations in three variables might have a unique solution Think of the eigenvalue problem A15 A5 which is nonlinear because we view both A and f as unknowns Suppose A is 8 X 8 how many variables are there equations solutions A solutions 5 There is some truth in this idea but few guarantees b You might be wondering whether we need to have another equation about conservation of energy Ordinarily we would but our homentropic assumption p up with no heating and no friction has the consequence that all energy changes are accounted for already by Newton s law For a simpler example a frictionless mass on a spring has Newton s law 771 71613 After multiplying by 5 you can integrate once to get the conservation of energy mx 2 kx2 constant A similar thing happens here If we were to allow heating of the air then we would need another variable which could be taken to be temperature and another equation There is plenty of room for misunderstanding about heating the word heating means transfer of heat energy as through the walls of the pipe The temperature which is not the same as the heating does in fact change even without heating because air obeys the ideal gas law p 287pT mks d We showed that the density disturbance w satis es the wave equation You can show similarly that the velocity disturbance 2 satis es the wave equation with the same coe 39i cient 02 What is the signi cance of the fact that the velocity and pressure disturbances 154 satisfy the same wave equation Speci cally is it physically plausible that these two disturbances might travel at di erent speeds PROJECTS Project 1 There are at least two ways to understand our earlier statement about the mass integral that the mass in W is If pA dx You probably know that you can think of dx as a bit of length Adx as a bit of volume pA dx as a bit of mass and the integral adds them Another interesting approach is explained by P Lax in his calculus book Suppose we write Sf I for the mass contained in interval I 11 when function f gives the density in I S has two physically reasonable properties 1 If I is broken into nonoverlapping subintervals I I1 U I2 1 c U 0b then 5051 30511 50512 2 If there are numbers 771 and M such that m g g M then mb7 a g 5f 1be S Mba Under these conditions you can actually prove that Sf 1 b f dx The project is to look up whatever de nitions you need and gure out why this works The idea applies to many applications other than mass Project 2 Our derivations tracked the momentum and mass of a moving portion of uid Some people prefer instead to track the momentum and mass inside a xed interval accounting for stuff going in and out Do you think one of these approaches might be more correct physically than the other The project is to redo the Euler equation derivation by considering a nonrnoving interval N 11 instead of the moving W The mass conservation is easiest The rate of change of the mass which is currently contained in N is d b 7 Ad dt A p I and this ought to be equal to the rate in at 1 plus the rate in at a ipbAub aAua Of course this leads to the same mass PDE as before Try to redo the Newton law in this context 155 37 Exact equations for Air and Steam TODAY Some Vector elds are gradients and some are not So some differ ential equations are said to be exact Here is the history It was in the early days of steam engines when people rst found out that there was a new invention on which they could travel at 25 miles per hour No human had ever gone nearly that fast except on a horse or on ice skates Can you imagine the thrill Figure 371 How fast can it go It was an outgrowth of the coal mining industry of all things People used coal to stay warm and unfortunately for the miners the mines tended to ll with water A pump was made to x this problem and it was driven by an engine which ran on well it ran on coal But people being as they are it wasn7t long before somebody attached wheels to the engine and they started competing to see who could go fastest At about this time people noticed that every new train went faster than the last one The natural question was whether there was any limit to the speed So M Carnot studied this and found that he could keep track of the temperature and pressure 0fthe steam but that neither 0fthose was equal to the energy of the moving train Eventually it was worked out that the heat energy added to the steam by the fuel was indeed related to the temperature and pressure They called the new rule the rst law of thermodynamics It looked something like this although the numbers I m using here are for air not steam heat added 717 dT 287 T Joulekg is supposed to hold whenever a process occurs that makes a small change in the temperature T Kelvin and the speci c volume V mgkg of the gas Here the pressure comes in again for air through the ideal gas law P 2870T V 10 A main point discovered that expression is not a differential This was so important that they even made a special symbol for the heat added qu which survives to this day in some books 156 PRACTICE Using simpler numbers and Variables check that 7 dx 2 3 dy df for any function fx y If it were df fa dI fy dy then you would have 7 fan and 2E y y See why that can t be true What do you know about mixed partial derivar tiVes fxy and fw This relates to the gradient because an alternate way to express that is For every function f 1 7H 275 mey Anyway the big disappointment to the steam engine builders was that the energy added in the process was not a differential which meant that you could not make a table of values for how fast you are going to go based only on temperature and pressure But the big discovery was that if you divide the heat added by T you can make a table of that PRACTICE d d 7 l 2 l dsomething x y and so for air you also have dT dV heat added 717 T 287 7 7 dsomething The something is called entropy S and for air 717 287 15 In Problem 1 you can gure out S from that There are tables of the entropy of steam in the back of your thermodynamics book So what did Carnot come up with Is there a limit to how fast the train can go Well he thought of an idealized engine cycle where for part of the time you have dT 0 and for the other part 15 0 Since it is possible to relate the V changes to the mechanical work of the engine that allows a computation to proceed He worked out how fast that ideal train can go You ll have to read about it in your thermo book PROBLEMS 1 Do the PRACTICE items if you haven t yet Work out the entropy of air as a function of T and V 2 Using P 287 pT work out the entropy of air as a function of T and P 157 38 The Laplace Equation TODAY Two roads lead here The heat conduction and drum Vibration mod els both morph into the Laplace equation under certain conditions Many roads lead away The one we take will be the Fourier road The Laplace equation looks like this in two dimensions um uyy 0 or in one u 07 or in three um uyy u 0 In two dimensions using polar coordinates not at the origin it looks like 1 1 U77 14 71499 0 7 7 Solutions are called harmonic functions In every case it says that the divergence of the gradient of u is 07 divVu 0 It is abbreviated Au 0 by mathematicians and Vzu 0 by engineers But what does it mean We approach that from two directions7 as follows I The Laplace equation can apply to a steady state temperature We know from Lecture 27 that the temperature in a region is modeled by the heat equation 14 Au which keeps track of the conduction of heat energy from hot places to cold places If the temperature is not changing with time7 that gives the Laplace equation It doesn t mean there is no energy ow A refrigerator can maintain a constant temperature but only by forcing out the energy that seeps in through the walls We will consider the polar coordinate case in a disk with the temperature u speci ed on the boundary circle Figure 381 A membrane stretched or the graph of a steady temperature II Our drum discussion in Lecture 34 can be modi ed In that case we had a dynamic situation uM 14 utt with circular symmetry If you instead analyze a circular membrane stretched statically over a varying rim height7 158 you nd for small angles that the same Laplace equation urruru99 0 does the job for the displacement u What is added is the t9 dependence including nonzero boundary conditions7 and what is removed is the time dependence So we are solving the boundary value problem 1 1 um7u77u990 0ltrlt10 0 27r 7 7 with a speci ed value at the boundary circle 7110 m PRACTICE What can we try We have traveling waves product solutions power series Anything else Lets try separation ur 0 RrF0 You get 1 7 We need to divide by F at least 1 R F R F 72RF 0 7 1 1 F R R R770 That is not quite separated Multiply by g TZRH TR F R 0 Now what do we have The R part doesn t look at all familiar How about the F Well7 there is no doubt about that We need F to be a constant multiple of F7 and we know that equation very well by now The possibilities for F09 are cosou97 sinat97 5 19 5quot and a0 b7 for various numbers 17 b7 or certain linear combinations of those Since we want u continuous in 9 if possible7 it has to connect smoothly when 9 loops around to 27139 which is the same angle as 0 So we better not use 00 1 unless a O A constant is ok7 say I 1 for de niteness Also any of the sinm9 or cosm9 for n 2 1 would be alright So far our list of acceptable functions F is 17 cost97 sint97 cos2t97 sin2197 cos3t97 sin30 and these will be enough for our purposes We7ll abbreviate the list as H cosm97 sinm97 n 2 0 Then FT 7712 159 PRACTICE 1 Convince yourself of that 7712 2 Would F sin be continuous at 0 if we are thinking of running 6 from O to 27L7 With those choices for F7 the requirement on R is that W 7 n2 0 0r 7 HR TR 7 nZR 0 n 2 0 Now what This is a second order linear differential equation7 but look at the coefficients They are not constants So we can t use the characteristic equation method that is based on trying exponentials and it won7t work with variable coe icients PRACTICE Check that in Lecture 10 if you don t remember We can still try something else We haven7t tried a power series for a while As it happens7 the solution for this is the easiest power series you ve ever tried It is left for the Problems In the next lecture we ll continue with u PROBLEMS 1 Having no other good ideas about the unfamiliar R equation r2R TR 7 n2 07 we could try a power series Rraoa1ra2r2 But wait It works out so nicely I would like to save you some work Try just Rr rq If that doesn t work you can always try something harder You will like this one Trust me 2 Combine your solution for R in Problem 1 with its corresponding F that we found and check it in the Laplace equation to make sure that you really do have a solution Try it for a couple of di erent values of n 39 Laplace leads to Fourier TODAY Putting together our Laplace solutions we are lead to challenging questions We are solving the boundary value problem 1 1 um7ur7u990 0ltrlt10 0 27r r r 160 with a speci ed value at the boundary circle M17 t9 1 09 and so far we have a list of solutions to the Laplace equation r cosm9 r sinm9 n gt 0 and also 1 Now we have to think about the boundary conditions PRACTICE What does the boundary look like for the case ur 9 r sin 7 What is rsin anyway EXAMPLE Suppose the boundary condition is that u1 6cos3 Of our list of solutions to Laplace the one which stands out is r3 cos3 It is almost what we need Would it be alright to multiply it by 67 yes see Problem 1 So one solution to the BVP is ur 9 6r3 cos3 Some people call the graph in Figure 381 a saddle and this r3 cos30 a monkey saddle Can you see why EXAMPLE If the boundary condition is now u1 9 6 cos3 3sin6 what will the solution be We can try a linear combination 7109 clr3 cos3 827 6 sin6 At least that will be a solution to the Laplace equation see Problem 2 Then we have to choose the coef cients ck if possible to get the boundary Values It will work to take 01 6 and 02 3 Answer 7109 6r3 cos3 1 37 s sin6 Look you just throw the rquot factors in After all this buildup does that seem strange 161 Fourier M Fourier observed the examples we ve seen He felt that something is missing Of course you probably feel too that our boundary conditions are somewhat arti cial made to t the product solutions we found Fourier wondered if you could work with some such boundary conditions as u10 f0 02 or something like that which has nothing to do with the cosines and sines It is an interesting question to ask Could it be possible that you could somehow expand f in terms of cosines and sines Could a function 1 09 which is not expressed in terms of the cosm9 and sinm9 actually have those hidden within it The question is more strange too because it is nearly the opposite of what we are used to like 02 04 0 1 7 i 7 n cos 2 24 That goes the wrong wayiwe wanted 02 in terms of trig functions Eventu ally we will achieve this on page 168 But if the answer to the question is yes then you can solve the membrane problem by just putting in the powers of r as we did above If 1 09 co a1 cos0 1 sin0 a2 cos20 2 sin20 then you can solve the EV as follows assuming convergence w 0 co ml cos0 51 sin0 ag cos20 b2 sin20 Fourier wanted to know how to extract the frequencies hidden in all func tions Figure 391 What frequencies could be hidden in this function It is plotted on the interval 0 107r 162 PROBLEMS 1 We need to check carefully that when u is a solution to the Laplace equation7 then so is 6n or an for any constant c How can that be checked Start with the rst order derivative Bu 5 Is it true that 6 0 Then how about the second derivatives say C7099 ls that the same as Cugg That is the essential idea behind the fact that V2cu CV27 Why does that prove what we need 2 We also need to know that a sum of harmonic functions is harmonic Reason that out similarly to Problem 1 3 We also need to understand that a linear combination of harmonic functions is harmonic Part of the calculation goes like this V2Clu1 02712 V2Clu1 V202u2 81V2u1 C2V2u2 Which part of that is by Problem 1 2 Why does that prove what we need 40 Fourier s Dilemma TODAY We look into pictures to help Fourier nd his hidden frequencies We now know that it is easy to solve the boundary value problem 1 1 um7ur7u990 0ltrlt10 0 27r r r with a speci ed value at the boundary circle 7417 0 1 09 provided that we can express 1 09 in terms of a series made from our list cosm97 sinm97 n gt 0 and also 1 Fourier wanted to know how to extract the frequencies hidden in f Figure 401 What frequencies could be hidden in this function on O7 107r 163 Figure 402 How about now The function has been modi ed a little to make some features stand out Compare with the next gure Figure 403 Two plain sinusoids cos and cos4 The graphs are drawn together but the functions are not added One is nearly right and the other is perfect Which is which You see from the gures that you can sometimes Visually estimate the hidden information if it is not too deeply hidden Fourier wanted to dig further We ll dig all the way to the bottom in the next lecture For now we take special identities that you know about EXAMPLE 1 You probably remember cos2 g 7 cosQQ 2 From the complex exponential identity eais cw3 we can deriVe that cos3 icosW icos3 So there are at least two functions not originally giVen as a linear combination of the cosn and sin7 L 7 for which we have a Fourier series There are just two terms in each We can think of the numbers 1 and 3 as the hidden circular frequencies and the coef cients and i are hidden amplitudes in the function cos30 We got those examples sort of by luck7 by trig identities In the next lecture we will nd out a systematic method for getting this information PROBLEMS 1 Try to answer the captions in the gures 2 Sketch a graph of the cos3 and sketch the two terms in its Fourier series separately to see how they t together A linear combination of sinusoids is not usually a sinusoid 164 41 Fourier answered by Orthogonality TODAY Finally How to nd the hidden frequencies We have seen only two cases where a function which was not originally expressed as a linear combination of sinusoids in fact is one One example is cos30 Ecosw icos30 But we also know there are good reasons to nd these hidden frequencies if they exist to solve the Laplace equation for example If there were only two examples we wouldn t bother After Fourier s time a concept was developed to extract the hidden informa tion from other functions It is called orthogonality of functions It works like this People knew that 7r cos0 sin0 d0 0 because an antiderivative is sinzw PRACTICE Check that And there are a lot of other similar integrals which are 0 EXAMPLE You don t really need an antiderivative because c0s sin is an odd function the graph has the kind of symmetry where each positive part to the right of O is balanced by a negative part to the left of O and Vice Versa So it must integrate to 0 By the same argument 7r c0sn sinm d9 0 7w for all n and m Functions 1 and g are called orthogonal if 7r f090d0 0 77quot You can compare this to the dot product of vectors a1 a2f ang and b1 by bgk are orthogonal if aibi azbz labs 0 165 The integral is a sum or at least a limit of sums and the values of the functions sort of play the role of coordinates so there is the comparison This was a very new idea to use geometric language to discuss functions EXAMPLE We saw above that all the cosn are orthogonal to all the sinm It is also true that all the cosines are orthogonal to each otheri That is harder to prove but can be done using trig identities Also all the sines are orthogonal to each other and all those trig functions are orthogonal to the constant function 1 So there is a long list of functions all orthogonal to each other 1 cos0 sin0 cos20 sin20 cos30 sin30 cos40 sin40 sort of like i j I in 3 dimensional space We have a lot more than 3 orthogonal functions though since in effect we are now working in an in nite number of dimensions This is a new concept part of our new language But what else does that get you It gets you everything Why Suppose it is possible to write a function f as 1 09 co a1 cos0 1 sin0 a2 cos20 2 sin20 but that we do not yet know what the hidden coef cients 00 an bn are Here is a way to nd them Think about the vector case where a vector 17 can be written 17 a1 a2f as but somehow we do not know what the coef cients are yet Dot with f for example You nd the number a2 by doing a dot product Uja1fa2fj ag j0a20a2 by the orthogonality and the fact that fis a unit vector So we can try the same thing with functions for example I f cos29d9 7r co cos2 16 7r a1 cos cos2 19 7r 1 sin cos2 19 a2 cos2 cos2 d b2 sin2 cos2 d a3 cos3 cos2 d OOOa2 cos2 cos2 d OO 166 We still need to know that remaining integral fir cos20cos20d0 Its value is 7r7 as can be worked out using trig identities These ideas lead to the Fourier coe icient formulas 1 7r an A m cosm9 d0 9 S i f f0sinm9d0 1 7r 0 7 0 d0 0 2W 77r f Of course we only derived the ag case but the ideas work for all We have shown that if f has a Fourier series then these must be the coefficients It is a uniqueness argument It is for another course to study which functions actually have Fourier series PROBLEMS 1 Try a short Fourier series such as f cos 03 sin5 Sketch the graph of the two terms and of f to see how they t together and how the sum no longer has a graph like that of a trig function 2 Same as Problem 1 for f cos9 cos10 This one is hard to do by hand but is interesting because you get to see an example of amplitude modulation7 AM 3 If you are a musician you might know that cos2t sounds an octave higher than cost and cos3t is a fth above that Try to solve the wave equation for a Vibrating string ya yam 2107 t 0 Wm t 0 WE 0 fI supposing the initial string shape has a Fourier series consisting only of the sinnx terms You may use the product solutions cosnt sinnc that we found in Lecture 30 for the wave equation 4 Work out the Fourier coef cient integrals for the function 62 You ll need to integrate by parts twice The result ought to be the series 2 62 7 7 4cos 2742 cos2 7 cos3 4 cos4 7 5742 cos5 167 It is for another course to discuss the convergence of Fourier series but this one is alright for 77r g g 7r Of course this series cannot be correct for all Values of 6 because the series is periodic and 62 is not This series answers Fourier s question on page 162 5 ln problem 4 eValuate the series at 6 7r to discover the Value of Z 51 that we have needed eVer since page 135 Can you nd other interesting series 42 Uniqueness TODAY How do we eVer know that we have found the only solution to a problem Some problems have only one solution EXAMPLE If 3x 7 2 5 then 3x 7 so x That is a proof of uniqueness Most people in reading that example think they have found the solution But that is not quite true We have to do this EXAMPLE If x then 3x 7 so 3x 7 2 5 The problem is so easy that we automatically know all the steps are re versible and we feel no need to check it EXAMPLE If there is an identity of the form 1x1 axx714xi 16x718 then take x 1 You get a 1 That is a proof of uniqueness Hmm As far as that goes it is ok but something is badly wrong isn7t it There is no such identity of course as you see by trying z O Other examples 0 We have a uniqueness theorem on page 21 about some ordinary differ ential equations That is for another course 168 o On page 39 we proved by an energy argument that our favorite second order equation y 7y has only those solutions that we know from calculus c There is no choice about Fourier coefficients as shown in Lecture 41 If a function has a Fourier series there is only one But we avoid the question of when a function has a series Too advanced a Wave Equation example Here is a case where you can gain a good tool for the toolbox This is an outline of an elegant argument which is usually considered to be outside the scope of this course But some of you asked about it and it is a good challenge This is the wave equation for a string of length L utt um 7107 t f0 71L7 73 973 ux O utc O The initial shape 11 and Velocity k are both given and you can even shake both ends of the string by f and g I claim that uniqueness holds if u1x t and u2x t are solutions then they are identically equal How can this be proved Let u be the difference 11 712 7 ul You ll need to convince yourself that subtracting wipes out all the boundary and initial conditions so that w solves the problem wtt 10m with w0 t wLt wx0 wx0 0 Once we prove that w 0 then we will have ul 712 like we claim New idea Let E UL dx which we call the energy of the string In a nutshell here is what happens i The energy is conserved ie constant in time ii E starts out being 0 at t 0 iii The only way E can be identically O is if u is identically 0 Of those items only i needs a calculation and the others just need some careful thinking Here is the calculation without any explanatory commen s dE L L i wtwn wxth dx wth wxwm dx dt 0 O L wthh dx wthlf 0 O PROBLEM Fill in as many details of this argument as you can 169 43 selected answers and hints page 4 3 A function increases where the derivative is positive so y is increasing on 71 0 and 17 00 4 1f y0 12 then y decreases as long as it is positive so probably y 7gt 0 5 y 03 005ty page 7 1 719x72 771x the minus because the force acts toward the large mass at the origin the k as a generic constant and m the small mass Or if you know the physics better k mMG where M is the large mass and G the gravity constant C is about 10 10 mks page 11 2 On a left face of one of these 21 gt 0 The equation tells you then y gt 0 So these waves must move left 3 The idea here is to nd the derivatives This is not strictly necessary because we already proved that waves traveling right at speed 3 are solutions no matter what shape they are But it helps to check speci c cases when you are learning something new 5 You can tell the velocity of the model of the dune because that is how we interpret our solutions But looking at the derivation of the model we never connected the wind speed speci cally with the dune speed So you can t tell how fast the wind is It is not part of the model Naturally the real wind is going the same way as the real dunes but not at the same speed page 14 1 Yes because z t y t 7 3 by the chain rule and because ya 7 y evaluated at t is equal to za7z evaluated at t73 The point here is that no explicit it occurs in the logistic equation This time translation does not occur with an equation such as x It 7 x2 3 Here is one way to outline the arithmetic Write x cf 1 as suggested Starting with y a1 cle atfl you have in years 1800 1820 and 1840 531 C1 a 961 01x 1 171 1 01x2 a The rst and second expressions are both a so they are equal and this will give you x in terms of cl Similarly the rst and third give you x2 in terms of cl Consequently 01 solves a quadratic equation which you can solve Then it is easy to get a x and t1 in that order page 17 2 xt71vt19 170 3 at Wm 10 xt acme amml page 20 2 Trying ux t in u ux u 0 gives fg f g fg 0 It will separate if you divide by fg i J 1 0 g f Since 558 depends only on t but it is equal to 7 8 7 1 which does not depend on t it must be a constant Call it k Then 9 kg and f 7k Now apply what you know from problem 1 The waves don t all travel at the same speed 3 It is important to see cases where a method does not work in addition to cases where it does Proceeding as in problem 2 you get 97 f7 f9 7 0 g f which is not separated There doesn t seem to be any way to separate this one So try something else This is normal trying something else 4 We already know what happens to u 8717 when u is a wave traveling right at speed 0 So that isn t possible unless u 0 But if you try ux t 7 at you nd that you must have c 7 af f2 O a separable equation One of the traveling wave solutions is 71x 7 c 1t Actually there are waves traveling right at every speed except 0 page 24 1 Once you get octave downloaded or matlab bought and installed start it Once you re in type 7help7 with no quotes press the enter key and follow the information given There is also 7help demo 7help plot and many other things which the rst 7help7 will tell you about It is important to keep trying At rst just type simple things like 7237 or 7cospi7 until you get comfortable with it After that learn how to make an m le that is a le called something like 7my lem7 which contains just the text of some commands For example dfield is a large m le for matlab which doesn t seem to work in octave but there is the dfield applet too Here is a much shorter example 23 tO 135 ycost plot t y Then use the le by typing 7my le7 at the prompt All the commands will be executed If an error occurs just edit your m le This is easier usually than xing any errors at the prompt Once you can do this successfully you are over the hurdle and you can use the software to help you learn It is a nice tool to have once you get past the rst stage The freeware xpp or xppaut is also excellent page 28 171 1 This is tough because you have to integrate 6 1 atx 6 1 3 cos t The answer is x eel at 713cost sint1 132 5 x 0313 3000 x0 0 Multiplying by integrating factor 6 03t you nd e oatx 30006 W So e 0 7100 000e 03 c I 7100000 sew Then 950 7100000 c 0 so 8 100000 and m 10000071 e0 Finally 9020 10000071 e 6 82 212 Not bad 6 Here is one philosophy We have x 702813 028 where xt is the balance on an account earning 28 interest and deposits of 53028 per year and y 028y 7 yz where yt is a population with a natural growth rate of 28 and a limiting size of 1 million or whatever Change the units of x to dollars per person with the idea that the analysis certainly could be applied to one individual at a time Change the units of y to people per dollar to acknowledge the fact that the economy determines how many people can be supported Then the substitution x 1y is suggested by the units This argument doesn t proVe that a solution to one equation goes oVer into a solution to the other Only the substitution can show whether that is true if y 028y 7 12 and x 1y then x 7y 2y 7y 2028y 7 m 7028y 171 7028x 71 page 34 3 seczydy dt7 so y tant 6 Yes To get rid of the cosine you would have to start with a differential equation for tanil which contains only arithmetic operations 7 We know that dtan 1tdt 11 t2 so all you have to do is solVe that numerically 8 a 10 b 104 So a 1000b Then Newton says u 71000bu and Stefan says u ibu l The ratio is 10007173 so if u lt 10 Newton cools faster If u gt 10 Stefan cools faster page 39 1 s 52 I1I2HI1I22 x l x 2 xf2x1x2x 002x112 Uhoh 3 No 5 solVes s s 2 cos 8t instead 2 8 x c x3 0 integrates to ac2 g c 10 Somebody tried to use the characteristic equation for a nonlinear equation which is Very rong Then eVen if r2 1 0 had been correct the roots should have been ii Finally you should neVer just add 0 to something at random page 42 1 Tt 816722 Xx Cgcos x casin x uxt e 2t04cos x cssin x 3 You have T 0 which giVes the steady state solutions page 47 2 Trying y eTt we need r2 3r 4 0 r 3i x97162 732 i So yt fatRm cos t 02 sin t 3 1f 1 0 then a2 1 0 So a ii Checking those both work since 7i 172 If a i i then a2 7 ia 1 O Quadratic formula giVes a iii 9174 6 There is a factor of et so this grows and a factor of cost i sint which goes around the unit circle counterclockwise so this is a spiral 7 r2 br 0 r 2 ir2 7 i so multiply it out page 48 4 If you aren t sure try some examples but the identity px aoa1xan1xquot71xn x7r1x7r2x7rn giVes r1 r2 1 rn 7an71 page 53 1 Csinft Csinft cos gt Ccosft sin gt So to match this with the giVen A cosft B sinft you have to solVe C cos gt B and C sin gt A for C and That is like saying that BA is a point in the plane and you want to nd polar coordinates r 9 C for this point That is always possible 2 This is a little harder because the only clue you have is that in the giVen identity the frequencies 727 and are half the sum and difference of the frequencies an e have sina b sina cosb cosa sinb and sina 7 b sina cosb 7 cosa sinb so A sina b B sina 7 b A B sina cosb A 7 B cosa sinb For example sin8t 23 sin7t 33 sin75t cos5t 7 13 cos75t sin5t 3 From problem 2 we see that when you graph something like sin75t cos5t you see a slow and a fast frequency in spite of the fact that you have started with the two near 4 7 8 4 4 frequencies and F In the case of the Figure the near frequencies are H the corresponding slow and fast frequencies are 1ltQiigt 2 27r 27r a and so which are about 12 and 35 The situation is more complicated than the identities in problem 2 though because you really need the awful identity A smabB sinab AB cos gt sina cosb A7B cos gt cosa sinb B sin gt sinbcosa7sina The only important thing is that the slow and fast frequencies still show up page 57 1 x2 x1 7 3 from the rst equation then the second giVes 3 7 x x1 2x1 7 3 or ill73x3 3I1 O 2 It is the same as for 1 Either calculate it or recognise that the formula x2 x1 7 x3 expresses x2 as a linear combination of two solutions to the equation for 1 4 Assuming y 73y 3y O we have Z2 y 7 y z17z1 z z gr7 11 y 7 y 31 7 3y 3Z2 So this system is Zl Z1 Z2 173 zZ 3Z1 This is different from the system of problem 1 eVen though they have the same second order equation associated With them The point is that there is nothing unique about a system associated to a higher order equation page 61 1 If you use y x then 11 x x3 7 x so the system is 3 x y y x 7 x But ifyou take sayy x x then y x x x3IyIx372xy 6 A conserved quantity is 2x 2 x4 The level sets are noncircles Whose shape changes depending on the Value page 66 2 Au v Au 1 Av b I So a sum of solutions to the equation Ax b is not usually a solution Usually 2b 7 I But if Av 0 you get Au v Au Av b 0 I So adding a solution of Ax 0 to a solution of Ax I gives a solution to Ax I And as a special case adding tWo solutions of Ax 0 giVes a solution to Ax 0 3 All the drinks must go to New York or Chicago7 so 9000 20011 180x All the fuel must be used on the Way so 260000 6500y 5400x These are our system It seems fairly clear that this is too simpli ed to be realistic If We really wanted to model costs of an airline We Would certainly not consider drinks as important as fuel The next largest expenses after fuel Would have to be identi ed They might be the cost of the planes or maintenance or personel salaries and fringes 4 This kind of product has not been de ned page 70 1 roW 1 has been replaced by row 1 7 16 roW 2 1 0 0 55 3 The reduced form is 0 0 1 5 J 4 The third equation says 0 2 Which is not true So some incorrect assumption must have been made When the equations Were rst used Be sure you get this When We started doing row operations We assumed there Was a solution So there are no solutions 8 That is one solution not no solutions page 74 1 1 1 J 3 0 g 0 12 With no restriction on a For the 4 by 47 if a 7 0 then 5 0 1 0 0 1 0 0 0 1a 0 0 1 0 1 0 0 0 O O O 1 7 0 1 0 0 If a 7 0 then the 4 by 4 Is not 1nVert1ble a 0 0 0 0 0 1 0 174 For example it can t be invertible because OOOO OOOH OOHO OHOO OOOH l 0000 5 Ix x for all x but 0 is not an eigenvector So all vectors except 0 are eigenvectors of I with eigenvalue 1 6 The octave command works correctly on the case which has a unique solution and seems to give some answer for the other two cases page 79 x 1 2 3 1 2 3 3 WehavetosolveA y 0 RowoperationsgiveAH O 73 76 gt O 1 2 z 0 76 712 0 0 0 1 0 71 1 a O 1 2 soy72zandczSoA 7 0 0 0 0 1 8 For this problem and for problem 77 note that A and 7A have the same eigenvectors7 but the eigenvalues of 7A are 7 times as large 11 The calculation here is correct but misguided As soon as you know that detA7AI 3 7 7 7 A stop because this is already factored There is no need to multiply it out and use the quadratic formula page 82 8x4 112 1 This matrix gives y Ax 1 and it is easy to invert that x Z But if 3 111 you want to calculate the determinant as a check the determinant is 78 7 0 page 86 2 xt O7 and yt Seat 5 x ae2t and y beiat If some of a or b are 0 you get solutions running along a half axis or stationary at the origin Otherwise y Za conste723 3 00715062t const 9 x0 u 22 so Mt uest 23906 10 Don t add C without a reason This is correct otherwise page 89 3 The two plots are identical 5 The real and imaginery parts were taken incorrectly because somebody forgot that the vector is also complex page 95 175 3 All rational numbers have either terminating or repeating decimal expansions while the irrational ones have the other kind All integers are either positive negative or O 1 math teachers are mean 4 The system of equations is just 5 y 0 so the solutions are constant and in the picture this means they don t go anywhere 5 Don t forget that the differential equation tells you a vector eld so it shows you which way the spiral goes page 99 3 I tried the system x x 7 xy y 7y xy 7 2 1 found that solutions can make increasing oscillations until the prey becomes extinct This is not what one would hope for My advice is not to fool with Mother Nature 6 00 only page 102 5 The correct conclusion is that chaos is possible in a system of two or more masses page 106 1 For c 0 there is the solution y O For c lt O the general solution involves linear c combinations of eiF and these all get wiped out by the boundary conditions except or y 4 If C g 0 then the only answer is y 0 If c gt 0 write it as c d2 where you may as well assume 1 gt O and the general solution to y 7d2y is yx acosdx bsindx The boundary condition y0 O forces a 0 Then the condition y 3 O db cos3d forces 3d g 37quot 57quot Thus yc bcos kg for k 13 5 No restriction on Z page 111 2 ln u an in the in term certainly seems to be a negative in uence on the time rate of change So it seems fairly reasonable that Mars bars will cool faster than Earth bars That argument doesn t work if u is ever negative Suppose that u is bounded below ie u gt 7m for some number 771 Then replace u by um to get positive temperatures only That xes the argument So Mars bars cool faster than earth bars unless they have some with in nitely negative cold temperatures page 115 1 This problem concerns heat conduction along a bar which is initially 400 degrees and which has insulation all around including the ends The steady state temperature is what you think it is 5 3cosce t 10 the heat equation for u eimy says that u facingx Hanan ef gy1 So we must have 2quot all with 110 WV 0 page 120 3 ux t e t sinc 56732t sin3t 256 52t sin5t 176 page 125 3 sinc 3t sinx 7 3t sinc cos3t cosx sin3t sinc cos3t 7 cosx sin3t 2 sinc cos3t has the initial condition 2 sinx of problem 2 For this problem the only difference is that we want 2sin2x instead so the solution is 2sin2xcos23t You have to multiply the 3t by 2 because otherwise you would not be creating traveling waves which are functions of x i 3t as required by the wave equation 74 3 um 8 We have said only that sums of left and right traveling waves are solutions In this case the function is seen to be a solution to u um by differentiating it or by using the idea of problem 3 to write it as a sum of left and right traveling waves cost sinc 12sinc t sinc 7 page 129 3 coskn2t sinnx where k n 12 3 4 the answer is not unique because of insuf cient BC page 135 2 x works so 5 is no more than 2 page 140 3 The radius of convergence is no affected by the 710 so it is 1 just like in Problem 1 But when x 83 kc 7 10 gt 1 so you are outside the radius of convergence You are trying to add powers of 17 which diverges page 148 1 Jolt4gtsslt4gtE17 EigE 3 10 3 2 4 7 4 7 4 I I I 7 4 7 4 4 7 I I I 2 The tall E 22426232 22426232102 1 22426232 1 102 10 12 If you 4 4 4 3 remember about alternatlng ser1es that is less than 2 446 8 So Jo4 lt 7 g lt O as needed page 157 1 Integrating 717 g 287 dS you nd 111T717V287 3 0 page 164 1 In the gure there seem to be at least two frequencies at work The larger one has 10 cycles in the interval 0 107r and seems to start positively at 6 0 So a multiple of cos is a good guess for that one The smaller one is harder to count It has less amplitude and about 4 cycles for each of the larger ones It is probably cos or sin of 49 or 59 with a smaller amplitude than the other one page 167 3 We know about the product solutions cosnt sinnc for the wave equation with these string boundary conditions And we know that you can form linear combinations of these solutions So the idea If we could write f as a series of sines then we could use those 177 same coef cients in forming our linear combinations If b1 sinc 2 Sinlt2gt 3 sin3x try yx t b1 cost sin1 2 cos2t Sinlt2gt 3 0033t sin3x 178 Index air 151 156 answers 170 applets 23 approximation of e 33 139 augmented matrix 67 banker s equation 1 beam 7 127 Bessell 145 boundary conditions 161 boundary value 104 bounded 138 bucket 15 butter y 101 calculator 34 145 Carnot 157 census 14 center 93 chaos 103 characteristic equation 36 check 3 14 25 chemical 54 classi cation theorem for plane linear sys tems 91 column 127 column vector 65 complex eigenvalue 87 complex numbers 43 compressible 151 conductivity 110 conservation laW 8 38 convection 41 convergence 133 Cramer 81 de 23 density 110 determinant 80 dfield 23 differential equation says velocity equals vec tor eld 56 83 divergence 111 dot product 166 drum 6 141 eig 78 eigenvalue 73 105 eigenvector solution to linear differential equar tion 84 emigration 97 energy 154 entropy 157 Euler 31 136 150 exact 157 existence and uniqueness theorem 21 exponential function 3 exponential growth 12 Feynmann 6 rst order linear equation 25 ux 9 forced 49 Fourier 136 162 Fourier coe 39icient 167 frequency 10 51 frozen boundary conditions 118 fundamental theorem of algebra 47 fundamental theorem of poWer series 139 179 geometric series 133 gravity 5 guess 4 heat heat equation 6 41 108 hidden frequencies 162 homogeneous equation 51 hot spot 117 ice cube 118 identity matrix 65 initial condition 4 insulation 113 integrating factor 26 interest 1 intersection of planes 63 inverse 71 investments 28 Jane s Candy Factory 54 java applets 23 Laplace 158 Lax 155 limit cycle 99 linear 25 35 linear combination 35 linear algebra 63 linear differential equation system 83 logistic 12 Lorentz 101 matlab 33 matrix 65 matrix inverse 71 Mars 111 membrane 158 model 11 moon 6 more than one solution 15 music 6 natural frequency 52 Newton s Fma 5 49 83 122 141 151 Newton s laW of cooling 5 26 node 92 nonlinear 96 100 nonuniqueness 17 numerical method 30 octave 33 oiler 156 orthogonal functions 165 partial derivative 6 21 partial differential equation 6 7 partial sum 135 particular solution 51 phase plane 56 pizza 26 plane 63 plot 32 polar coordinates 158 population 12 power series 131 pplane 56 predatorprey 59 96 pressure 154 product solutions 114 radiation 41 radius of convergence 140 180 rate of change 1 transient 52 reading a diHerential equation 1 transport equation 7 18 real part of solution 45 traveling Wave 9 122 125 repeated root 37 trying 35 resonance 52 uniqueness 21 139 168 roW operation 67 Van der Pol 98 rowreduced 69 Variable coe 39icients 160 saddle 92 Vector eld 59 83 Schrodinger 7 Wave equation 6 7 122 second order 6 35 Weather 103 sensitiVe to initial conditions 102 xpp 24 separable 14 18 41 sequence 138 slope 10 144 slope eld 2 smiley face 65 software 21 solution not de ned for all t 21 sound 6 149 153 speci c heat 110 spiral 93 spring and mass 49 steady state 112 158 stearn 156 stepsize 30 string 122 sum of solutions 39 163 systems 54 Taylor s Theorem 133 temperature 5 26 158 tension 123 time of death 28 181


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