Ordinary and Partial Differential Equations
Ordinary and Partial Differential Equations MATH 251
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Differential Equations LECTURE 28 IVPS with Laplace Transforms Now that we have a good grasp of how to take Laplace transforms and inverse transforms let s return to differential equations First we should recall the following formula with H denoting the nth derivative of f E W s Fltsgt 7 s 1flt0gt7 st m 7 m7 sf 20 7 f 10 We ll be dealing exclusively in this lecture with second order differential equations so in particular we ll need E 2 8Y8 7 240 5 2 8238 7 8M0 7 MO You should be familiar with the general formula however REMARK Notice that we must have our initial conditions at t 0 to use Laplace transforms EXAMPLE 281 Solve the following VP using Laplace transforms y 7 52 7 6y 5t 240 1 20 2 We begin by transforming both sides of the equation 3 24 5 2 i 9 5 t 3213 7 sy0 7 yO 7 5SYS 5340 7 6Ys mm mm m 3275376Yss7275 As we ve already begun doing now we solve for Ys 5 77s YS3232753768275376 7 5 77s 73376Xs1s76XsD 5717732733 7 n We now have an expression for Ys which is the Laplace transform of the solution yt to the initial value problem We ve simpli ed it as much as we can now it s time to take the inverse transform The partial fraction decomposition is A B C D Y 7 7 7 S 332s76s1 Setting numerators equal gives us 6 732 7 33 Ass 7 6s 1 B3 7 6s 1 1 0323 1 1 D32s 7 6 Differential Equations Lecture 28 IVPs with Laplace Transforms We can nd constants by choosing key values of 3 30 676B B71 36 422520 0 571 1477 5 7 77 1 31 12 4014 143 D EXERCISE Solve the initial value problem in the previous example using Undetermined Coef cients Do you get the same thing Which method took less work EXAMPLE 282 Solve the following initial value problem y 23 5y cost 7 101 y0 0 yO 1 We begin by transforming the entire equation and solving for Ys E 34 2E 3 SE cost710 t 10 8 7 8210 7 we we 7 210 ms 8217 2 i s 10 s 2s5Ys717 8217 So we have s 10 1 YltSgt 321822s5 7 32322s5 322s5 Y1s Y2s Y3s Now we ll have to take inverse transforms This will require doing partial fractions on the rst two pieces Let s start with the rst one s A8 B Cs D 321822s5 821 322s5 After putting everything over a common denominator7 we set numerators equal 3 As32 23 5 B82 23 5 0332 l Ds2 1 ACs32ABD325A2BCs 5BD Y18 This gives us the following system of equations7 which we solve AC 2ABD 1 1 1 1 5A2BC1 AE BE 0 13 5BD Differential Equations Lecture 28 IVPs with Laplace Transforms Thus our rst term becomes 1 3 1 1 1 3 1 1 gmimm g sums 5 3223539 We ll hold off on taking the inverse transform for the time being Now7 let s deal with Y23 Y1S YZltSgt W We put everything over a common denominator and set numerators equal 710 A332 23 5 332 23 5 033 D32 A 133 2A B Ds2 5A 2B3 513 This gives the following system of equations AC0 2ABD0 A B 2 C D 5A2B0 7 7 7 7 5B710 Thuswehave 41 2 4 3 2 1 Y 777777 7 28 53 32 532235532235 Let s return to our original function Y8 Y18 368 Yams 1 3 1 1 41 2 1 4 3 1 2 1 gs mu gg ilt 3 3gtmlt731gtm 1 3 1 1 41 2 3 9 1 ESTN 7quotTWWW Now we have to adjust the last two terms to make them suitable for the inverse transform Namely7 we need to have 3 1 in the numerator of the second to last7 and 2 in the numerator of the last 1i 1313 75321 10321 53 32 3124 103124 1i 1 3 L 75321 10321 53 32 3124 103124 71 3 1 1 41 2 31 19 2 75321 10321 53 32 3124 203124 So our solution is yt cos sint g 7 2t 7 e cos2t e tsin2t D We could have done both of the preceding examples using Undetermined Coefficients ln fact7 it would have been a lot less work Let s do some involving step functions7 which is where Laplace transforms really shine Differential Equations Lecture 28 IVPs with Laplace Transforms EXAMPLE 283 Solve the following initial value problem 17 51 6y 2 7 U2t62t 4 MO 0 MO 0 As before we begin by transforming everything Before we do that however we need to write the coefficient function of u2t as a function evaluated at t 7 2 y 7 53 1 6y 2 7 u2t52t72 Now we can transform E y 7 5E 3 SE 2 1 E u2t52t72 3mg 7 sy0 7 yO 7 53143 5y0 7 6Ys 7 g 7 5293 521i 2 1 2 7 7 7 725 s 536Ys S e 872 So we end up with 2 725 1 33 7 3s 7 2 5 s 7 3s 7 22 Y1s eist s Since one of these terms has an exponential we ll need to deal with them separately I ll leave it to you to check all of the partial fractions Ys 11 2 1 1 Y1 32 gm a 1 1 1 Y2Sis73 872s722 Thus we have 7 11 2 1 1 725 1 1 1 YltSgt72ggs73is72e 7373s72s722gt and Mt g 7 53 7 52 7 1205 753072 7 52t72 7 t7 252t72gt 1 2 5 7 353 7 52 7 1205 763t76 7 6274 7552741 once we observe that E 1 teat D EXAMPLE 284 Solve the following initial value problem y 4y8tu4t y0 0 yO 0 We need to rst write the coefficient function of u4t in the form ht 7 4 for some function ht So we write ht 7 4 t t7 4 4 and conclude ht t 4 So our equation is y 4y 8 t 7 4 4U4t Now we want to Laplace transform everything 5 34 M y SE 1 E t 7 4 4U4t 8 32Ys 7 sy0 7 yO 4Ys g 5 4SE t 4 S Differential Equations Lecture 28 IVPs with Laplace Transforms So we have 8 1 4 y 7 49 7 8 332 4 1 6 8282 1 4 1 332 4 7 L i 745 8824 5 32824 YISH 5 YZlt where we ve consolidated the two fractions being multiplied by the exponential to reduce the number of partial fraction decompositions we need to compute After doing partial fractions leaving the details for you to check7 we have 23 Y 77 18 3 327174 and 1 11 s 1 1 Y 7 777 77 28 31432 327174 43271747 so and the solution is yt 7 2 7 2 cos2t u4t lt1 7 g 7 4 7 cos2t 7 4 7 sin2t 7 4 2 7 2cos2t u4t it 7 cos2t 7 8 7 sin t 7 8 Differential Equations LECTURE 37 Applications of Nonlinear Systems In today s lecture we ll look at some particular nonlinear systems generally inspired from physics or biology We ll then use the linearization method we ve been discussing for the last few lectures to try to understand the behavior of these systems 1 Competing Species One of the classical examples of a nonlinear system is the Lotka Volterra model of competition between two species let s take them to be rabbits and kangaroos Both species are competing for the same food supply the grass and there s a limited amount of this resource We ll assume these populations exist in an environment without predators What are the main considerations we ll need to keep in mind to write down this system 1 When we discussed the logistic equation as an example of an autonomous rst order equation we mentioned the existence of an environmental carrying capacity that is given a species which consumes resources at a certain rate there is a certain upper limit for the population of this species that the environment can support Here since we re assuming that we ve got a kangaroo population and a rabbit population and no others we can assume that in the absence of the other population each would grow to its carrying capacity Thus we ll incorporate logistic growth into the equations for each species We ll also assign rabbits a higher intrinsic growth rate due to the well known ability of rabbits to reproduce What happens when the two species encounter each other while grazing Sometimes the rabbit might get to eat but the kangaroo being signi cantly larger they will normally push the rabbit aside and start eating We can assume that these encounters will occur at a rate proportional to the size of each population In light of the previous comment we ll also assume that the con icts reduce the growth rate for each species but that the rabbit population is more severely affected A to V In particular if mt is the rabbit population at time t and yt is the kangaroo population at time t the model m m37z72y 371 2 y27907y 372 incorporates these assumptions Our rst task is to nd the xed points of this system We solve z y 0 and obtain four xed points 00 0 2 30 and 11 Next let s use the linearization method to try to classify them The Jacobian of the system 371 is 7 3 7 2m 7 2y 72m A T lt 7y 2 7 m 7 2y 39 Let s now consider each xed point to determine the nearby behavior 0 00 Here A 0 This has eigenvalues 1 3 and 2 2 so we can conclude that 3 0 2 00 is an unstable node as non degenerate or star nodes are preserved in the original Differential Equations Lecture 37 Applications of Nonlinear Systems nonlinear system Recall that right at a node7 typical trajectories are tangential to the slow eigensolution7 which in this case is the eigensolution corresponding to 2 2 The eigenvector corresponding to 2 is i lt1 7 so trajectores will leave the node along the y axis 07 2 Here A This matrix has eigenvalues 1 71 and 2 72 hence the xed point is a stable node A typical trajectory approaches it along the eigensolution corresponding to A1 which is in the direction of the eigenvector i lt72 30 Here A lt70 The eigenvalues are 1 3 and 2 71 so this is also a stable node The slow eigensolution is in the direction of the eigenvector of A2 namely 3 77 72 1 72 1 71 We can further note that the m axis is a straight line trajectory7 as z 0 when x O Similarly7 the y axis is a straight line trajectory Lines where z 0 or y 0 are called nullclines7 and in this case these nullclines are also solutions Putting this all together and using some common sense to ll in the rest of the trajectories we obtain a phase portrait that looks something like Figure 371 o 11 Here A gt and 12 71i This is a saddle point J 111 i KKlKA Ti ALlA ff KKKK f KKKKK sn KKKK gt 444 I FIGURE 371 Phase portrait for the rabbits vs kangaroo model 371 Notice the xed points and their stability This has an interesting biological interpretation it shows that in general7 one species drives the other to extinction If we start above the stable solution of the saddle point7 the kangaroos drive the rabbits to extinction7 but if we start below it7 the rabbits drive the kangaroos to extinction This phenomenon shows up in more complex models of competing species as well7 and it has led to the formulation of the principle of competitive exclusion which states that in general7 two species competing for the same resources cannot coexist This is why releasing pets into the wild is a bad idea its species might drive native populations competing for the same resources out Differential Equations Lecture 37 Applications of Nonlinear Systems 2 Nonlinear Pendulum In the absence of any damping or external driving the motion of a pendulum is governed by the equation 2 d 0 i w s1n0 0 where 0t is the angle from the downward vertical 9 is acceleration due to gravity and L is the length of the pendulum This is derived from the rotational formulation of Newton s Second Law 739 Ia where 739 7mgLsin0 is the torque I mL2 is the moment of inertia and Oz is the rotational acceleration Let s write we Then our equation is s wg sin0 0 373 For very small angles this equation will be linearized using sin0 z 9 this is generally done in high school for example but using phase plane methods we can study the nonlinear equation even for large angles Writing 373 as a system yields 0 z 37 4 1 flag sin0 where 1 is the angular velocity The xed points are of the form mr 0 where n is any integer The Jacobian of the system is A lt7w030slt0gt 5 Notice that there is no difference between angles that differ by 27139 either physically or formally so we ll focus on the two xed points 00 and 7r0 Near 0 0 the linearized system is 117 011 77w0039 The eigenvalues of the coef cient matrix are 12 iwoi which says that the linearized system has a center As we discussed last class however that doesn t mean we have a nonlinear center at the xed point we could have a spiral So what do we do It turns out that the system 374 is an example of a reversible system that is if we reverse time by changing t to it and 1 to 71 since reversing time will reverse the velocity as well the system stays the same Physically this should make sense if we tape a pendulum s motion and play it backwards we won t see any physical absurdities It turns out that for reversible systems if the origin is a linear center it will be a nonlinear center as well The idea is that we take a trajectory close to the origin which swirls around the origin as the origin will have to be a spiral or a center Reversing time and velocity re ects this to a twin trajectory with the same endpoints but with the arrow reversed which closes the orbit This is illustrated in Figure 372 Thus 00 and hence 2k7r0 are all centers What happens near 7r0 The linearized system is 701 illiwooill The coef cient matrix has eigenvalues 12 iwwo Hence it and any xed point of the form 2k 17r0 is a saddle point Differential Equations Lecture 37 Applications of Nonlinear Systems FIGURE 372 Why an origin that is a linear center is also a nonlinear center in a reversible system the trajectory to the left also has its mirror image as a trajectory Now we can draw the phase portrait near the xed points How do we physically interpret the classi cation we just found The centers correspond to a state of neutrally stable equilibrium the pendulum is at rest and is hanging straight down as 9 21m If we move the pendulum slightly away from there and possibly give it a little initial velocity we ll get oscillation back and forth The saddles on the other hand correspond to the cases where the pendulum is at rest but is balanced perfectly up They re unstable as if we move the pendulum slightly from this balance they ll swing back down What happens away from the xed points This corresponds to giving the pendulum a lot of initial velocity We ll actually end up with the pendulum rotating around and around the axis so that the phase portrait looks like Figure 373 EXERCISE Free Damped Pendulum If we add a damper with damping coef cient 39y gt 0 to the nonlinear pendulum modeled by 373 the new equation governing the motion of the pendulum is 0 39y0 I we sin0 0 After writing this equation as a system of rst order equations nd and classify the xed points for all 39y gt 0 and plot the phase portraits for each qualitatively different case FIGURE 373 Phase portrait for the free undamped nonlinear pendulum 374 Differential Equations Lecture 37 Applications of Nonlinear Systems 3 Next Steps We ve only dipped our toes into the theory of nonlinear systems7 but hopefully we ve gotten the sense that there s a lot of interesting phenomena occuring here Let s brie y list some of the directions there are go in from here most of these would get at least a brief discussion in a second course in differential equations 31 Limit Cycles and Periodic Solutions After equilibria7 the next most important fea ture of nonlinear systems is the possibility of periodic solutions These also very strongly affect the overall phase portrait There are several global methods that can be brought to bear on the problem of detecting and understanding periodic solutions There are many questions that can be asked here If we have a closed trajectory7 must it always circle a xed point What kinds of xed points are permitted inside of one ls there a restriction on the number of closed trajectories we can have or where they can be located ln fact7 we ve seen examples where closed trajectories occur in nested families They can also be isolated such a solution is called a limit cycle Nearby solutions7 which aren t closed7 can limit to or away from the limit cycle Stable limit cycles are very important scienti cally7 as they model self sustained oscillations7 such as in the human heart These are7 as we ve seen7 inherently nonlinear phenomena if a linear system has closed solutions7 so are nearby solutions 32 Bifurcat ions We ve only looked at what happens if we have a single system of nonlinear differential equations What if we have a family of them7 differing by some parameter For example7 what if we have a nonlinear pendulum or an electrical system driven by a constant external force7 which we proceed to increase Can the behavior change as we vary the parameter The answer is yes Many different things can happen here the number or stability of xed points or closed orbits can change7 for example Such a change is called a bifurcation This is very important if you think about an externally driven electrical circuit7 for example understanding where and what bifurcations can occur will tell us how much force we can safely apply to the circuit Bifurcations can also occur for onedimensional equations7 as well7 and some very interesting ones occur for discrete systems7 where we take some function and iterate it repeatedly7 looking for stable and unstable xed points 33 Chaos It can be shown that for a two dimensional system7 things behave generally nicely and limit cycles are typical in some sense However7 in three dimensions7 all bets are off In 19637 while modeling convection rolls in the upper atmosphere7 Edward Lorenz wrote down the following system of differential equations m7amby y 7mzrmiy z zyibz where 011 are some constants and r gt This looks like a relatively simple system7 but it turns out to be surprisingly complicated The solutions don t ever settle down to a periodic orbit or xed point7 but they also don t run off to in nity lnstead7 they wrap around two equilibria in a fairly crazy manner The manner in which they wrap around these equilibria depends very strongly on the initial conditions small changes lead to solutions which behave very differently This is known as sensi tivity to initial conditions7 and it s one of the hallmarks of a chaotic system It s become known as the butter y effect a butter y aps its wings in Beijing7 and in New York it rains a few days Differential Equations Lecture 37 Applications of Nonlinear Systems later instead of being sunnyl It s why we can t predict the weather very far in advance since in struments and computers only have a certain number of decimal places they can measurecompute to but down the line the small bits that are lost in the process become very important 1There s another possibility for why this is called the butter y e ect77 There s a classic science ction story and an episode of The Simpsons parodying it about a time traveler who goes back in time to hunt dinosaurs and steps on a butter y When he returns to his time everything has changed in a substantial way Killing that one butter y changed the initial conditions77 of the universe and upon moving forward a signi cant amount of time everything evolved in a di erent way Differential Equations LECTURE 40 Separation of Variables and Heat Equation IVPS 1 Initial Value Problems Partial differential equations generally have lots of solutions To specify a unique one we ll need some additional conditions These conditions are usually motivated by the physics and come in two varieties initial conditions and boundary conditions An initial value problem for a PDE consists of the equation initial conditions and boundary conditions An initial condition speci es the physical state at a given time to For example an initial condition for the heat equation would be the starting temperature distribution ux 0 This is the only initial condition required because the heat equation is rst order with respect to time The wave equation which we will look at later is second order with respect to time and so needs two initial conditions PDEs are also generally only valid on a certain domain Boundary conditions specify how the solution is to behave on the boundary of this domain These need to be speci ed because the solution isn t on the one side ofthe boundary meaning we might have problems with di erentiabiltiy there Our heat equation was derived for a one dimensional bar of length i so the relevant domain in question can be taken to be the interval 0 lt z lt l and the boundary consists of the two points z 0 and z Z We could have derived a two dimensional heat equation for example in which case the domain would be some region in the zy plane with boundary some closed curve It will usually be clear from a physical description of the problem what the appropriate boundary conditions are We might know that at the endpoints z 0 and z L the temperatures u0t and ult are xed Boundary conditions that give the value of the solution are called Dirichlet conditions Or we might insulate the ends meaning there should be no heat ow out of the boundary this would yield the boundary conditions um0t umlt 0 If the boundary conditions specify the derivative of the solution they re called Neumann conditions We could also specify that we have one insulated end and at the other we control the temperature this is an example of a mixed boundary condition As we ve seen changing boundary conditions can signi cantly change the character of a prob lem lnitially to get a feel for our solution method we ll work with the homogeneous Dirichlet conditions u0 t ul t 0 giving us the following initial value problem DE u 1w 401a BC u0t 14125 0 40110 10 limo my 401c After we ve seen the general method we ll see what happens with homogeneous Neumann condition leaving condition for a later discussion Differential Equations Lecture 40 Separation of Variables and Heat Equation IVPs 2 Separation of Variables In Section 1 we derived the heat equation 401a for our bar of length L Suppose we ve got an initial value problem such as 401 How do we proceed We ll want to try to build up a general solution out of smaller solutions that are easier to nd A rather straightforward approach is to start by assuming these mini solutions have some nice form For example we might suppose we have a separated solution where uzt XmTt That is our solution is the product of a function that depends only on x and a function that depends only on t We can then try to write down an equation depending only on x and another depending only on 25 before using our knowledge of ODEs to try to solve them It should be noted that this is very special and should not be expected in general In fact this method cannot always be used and even when it can be used often it s hard to move past the rst step However it works for all of the equations we will be considering in this class and is a valuable starting point How does the method work We begin by plugging our separated solution into the heat equation 401a 8 82 a XTtl kg lXTtl XzT t lacam Now notice that we can move everything depending on x to one side and everything depending on t to the other T t 7 XW kTt T X This equation should look a little funny to you On the left we have an expression which depends only on 25 while on the right we have an expression that depends only on x Yet these two sides have to be equal for any choice of z and t we make The only way this is possible is if both sides of the equation are the same constant In other words T t 7 X 7 kTt T X 7 We ve written the minus sign explictly for convenience it will turn out that gt 0 but these expressions should be negative The equation above really contains a pair of separate ordinary differential equations X X 0 402a T AkT 0 40210 Notice that with these separated equations our boundary conditions 394b become X0 0 and Xl 0 Now 402b is easy enough to solve we have T ikkT so that Tt A54 What about 402a This just gives us the boundary value problem X X 0 X0 0 Xl 0 This should look familiar it s very similar to Example 386 The only difference is instead of our second condition occuring at z 27139 it s at some z 1 As in that example it will turn out that Differential Equations Lecture 40 Separation of Variables and Heat Equation IVPs our eigenvalues have to be positive Letting 32 for B gt 0 our general solution is Xz Bcos z Csin m The rst boundary condition says B 0 The second condition says that Xl Csin l 0 To avoid only having the trivial solution we must have Bl 717139 In other words 2 An and Xnz sin for n 123 So we end up having found an in nite number of solutions to our boundary value problem 401a and 401b one for each positive integer n and no other values of n since our eigenvalues are all positive They are WK 2 unm t Ane 7 ktsin The heat equation is linear and homogeneous As such the Principle of Superposition still holds a linear combination of solutions is again a solution So any function of the form N u mt A e 2kt sin 403 lt gt n lt gt is also a solution to 401a and 401b Notice that we haven t used our initial condition 401c yet which is why we referred to 396 as a solutions to just the boundary value problem How does our initial data come into play We have N 7171391 m m 0 ZAnsm n0 So if our initial condition has this form 403 works perfectly for us with the coef cients An just being the associated coef cients from EXAMPLE 401 Find the solution to the following heat equation problem on a 7011 of length 2 Ut Umm u0 t u2 t 0 uz0 sin 7 5sin37rz In this problem we have k 1 Now we know that our solution will have the form of something like 403 since our initial condition is just the difference of two sine functions We just need to gure out which terms are represented and what the coef cients An are This isn t too hard to do our initial condition is 35 sin 7 5sin37rx Looking at 403 with l 2 we can see that the rst term corresponds to n 3 and the second n 6 and there are no other terms Thus we have A3 1 A6 75 and all other An 0 Our solution is then Wat l sin 7 559W2t sin 37m Differential Equations Lecture 40 Separation of Variables and Heat Equation IVPs This isn t very satisfying there s no reason to suppose that our initial distribution is a nite sum of sine functions Physically7 such situations are clearly very special What do we do if we have a more general initial temperature distribution Let s consider what happens if we take an in nite sum of our separated solutions Then our solution to 401a and 401b is 00 Mfkt i mm umt Z Ame l 811171 gt n0 Now the initial condition speci es that the coef cients must satisfy 00 mm 35 Z A s1n 404 n This idea is due to the French mathematician Joseph Fourierl7 and 404 is called the Fourier sine series for There are several very important questions raised by this7 however Why should we believe that our initial condition x ought to be able to be written as an in nite sum of sines Why should we believe that such a sum would converge to anything We ll come to these in good time7 but keep them in mind 21 Neumann boundary conditions Now let s consider a heat equation problem with homogeneous Neumann conditions7 DE ut kn 405a BC um07 t nzl7 t 405b lC uz0 405c We ll start by again supposing that our solution to 405a is separable7 so we have umt XzTt and7 as we re using the same differential equation as before7 we obtain the pair of ODEs 402a and 402b The solution to 402b is still Tt Ae wt Now we need to determine the boundary conditions for 402a Our boundary conditions 405b are on iim07 t and umlt thus they are conditions on X 0 and X l7 as the t derivative isn t controlled at all So we have the boundary value problem X X 0 X 0 0 X z 0 Along the lines of the analogous computation last lecture7 this has eigenvalues and eigenfunctions rm 2 A lt7 7747171 am ltTgt for n O7 17 27 So individual separable solutions to 405a and 405b have the form Wm B unmt Ane l coslt l 1Fourier 17681830 was a big promoter of the French Revolution and traveled with Napoleon to Egypt where he was made governor of Lower Egypt Fourier has been credited with being one of the rst to understand the greenhouse e ect and more generally planetary energy balance There s a great though almost certain apocryphal story about what led Fourier to study the heat equation Reportedly what deeply concerned Fourier was being able to nd the ideal depth to build his wine cellar so that the wine would be stored at the perfect temperature yearround and so he proceeded to try to understand the way heat propagated through the ground Differential Equations Lecture 40 Separation of Variables and Heat Equation IVPs Taking nite linear combinations of these work similarly to the Dirichlet case and is the solution to 405 when x is a nite linear combination of constants and cosines in direct analogy to 403 but in general we re interested in knowing when we can take in nite sums ie l 00 M 2k mm umt 5A0 ZAWA l tcos ri1 Notice how we wrote the n 0 case as A0 the reason for this will be made clear in the future The initial condition 405c means that we need 1 00 mm 35 5A0 Ancos 406 An expression of the form 406 is called the Fourier cosine series of 22 Other boundary conditions It s also possible for certain boundary conditions to re quire the full Fourier series of the initial data this is an expression of the form 1 00 mm mm x 5A0 21 An cos Bn s1n 407 7 but for most of our purposes we ll want to solve problems with Dirichlet or Neumann conditions However in the process of learning about Fourier sine and cosine series we ll also learn how to compute the full Fourier series of a function Differential Equations LECTURE 8 Exact Equations The nal category of rst order differential equations we will consider are the so called exact equations These nonlinear equations have the form at 81 Mz7yN7vy0 dm where y is a function of z and 8M 7 8N 8y 7 8m 7 where these two derivatives are partial derivatives discussed more below We ll begin this lecture by brie y discussing how to compute the partial derivatives of a mul tivariable function Then we ll see a motivating example of how to solve an exact equation which will in turn lead to a discussion of why what we did in the example worked Finally we ll see some examples done with the solution method in mind 1 Multivariable Differentiation Suppose we have a function m y depending on two variables What does it mean to differ entiate this function Recall that if we have a function of a single variable 9a we can interpret the derivative as the rate of change of the output of 9a as x increases We d like to have a similar notion of differentiation for our multivariable function m The situation is of course complicated We no longer just have to worry about what happens as x increases we have to worry about how to measure the rate of change of m y as y increases or even as x and y vary in different ways Geometrically we can think of the graph of fzy as specifying some surface in three dimensional space so talking about the slope at a point is vague in this way So what do we do It turns out and this is a topic that is explored more in a multivariable calculus class that if we can account for how m y varies as only z or only y vary we don t have to worry about mixtures of the two Let s consider how we want to express the derivative of fz y at a point mo yo Without loss of any generality we can consider what happens only as x varies since the y situation will be analogous and we ve already punted on considering mixing the two to a class that is more suited to discussing that Fixing y yo reduces our function of two variables fzy to a function of a single variable 9a m yo So we de ne the partial derivative of f with respect to x at 0343 denoted by 3 8900790 Ma avo to be g m0 with 9a fmy0 Geometrically this is the slope of the curve on the graph of 1 corresponding to the line y yo at z me What does this mean computationally Our function g is de ned by treating y as the constant yo So its derivative will be equivalent to the derivative of 1 taken with respect to z while treating y as a constant The following examples should clarify the above discussion Differential Equations Lecture 8 Exact Equations EXAMPLE 81 Let ag y 952g yz Then 81 a 7 2mg 81 m2 2y D EXAMPLE 82 Let ag y ysinz 8f 7 a 7 y cosz 8f 7 i 87y 7 s1nz D We will also need to be able to recognize the multivariable chain rule The relevant version of this says that if we have a function ltIgtmyz depending on some variable x and a function y depending on x then dltIgt 7 81 81 dy M g an 11 Elly 2 Exact Equations Before we talk in more detail about how to solve an exact equation7 let s work one example to get a feel for what an exact equation is and why the solution method works We ll rework this later after a more general discussion to reveal all the details EXAMPLE 83 Consider d 2my79m2 2yz21i 0 dm The rst step in solving an exact equation is to nd a certain function Mm We ll see how to nd this function later and in fact this computation is where most of the work lies7 but for this example it will turn out that the desired function is May y2 x2 1 y 7 3x3 Notice that if we compute the partial derivatives of I we obtain Emwmww Pgmy 2y m2 1 Looking back at the differential equation7 we can see that it can be rewritten as dy Pm PgE 0 Then7 thinking back to the chain rule as expressed in Equation 827 we see that the differential equation is just dltIgt 7 0 1 This tells us that our function I must be equal to some constant7 since its ordinary derivative is zero7 and thus the general solution is 342 z21y73z3c for some constant c which is just the constant ofintegration If we had an initial condition7 at this point we could solve for c and get the particular solution to the initial value problem Differential Equations Lecture 8 Exact Equations Let s think about what we saw in the previous example An exact equation has the form dy M N i 0 79 7ydm with Myz y Nmzy The key to nding the solution to an exact equation is to construct a function Qm y such that the differential equation turns into dQ 7 0 dm by using the multivariable chain rule 82 as we did above Thus we require that Q satisfy ltIgtm7 y MW 24 Pg719 NW7 REMARK One of the standard facts of multivariable calculus is that mixed partial derivatives commute This is why we want My Nm My me and Nm Qym and these should be identical if such a function Q can exist lt s imperative to check that a function is actually exact by computing Mm and Ny and seeing that they coincide before proceeding with the solution process or there s no way it can work Once we have this function Q then we know O and hence 907 y 0 yielding an implicit general solution to the differential equation So we can see that the real work involves computing Qm How can we do it Let s revisit Example 83 this time lling in all the details We ll also add an initial condition into the mix EXAMPLE 84 Solve the initial value problem dy 2 7 9 2 2 2 1 7 0 0 2 my xydm M Let s begin as we always should be checking that this differential equation is actually exact Comparing the equation to Equation 81 we have Mzy 2mg 7 9m2 and Nmy 2yz2 1 Then My 2x and Nm 2m hence the equation is exact Now how do we nd Qm y We have that Qm M and Qy N Thus we could compute Q in one of two ways Qzy Mdm or Qzy Ndy We ll need to be slightly careful here as we ll see but rst let s just note that it doesn t usually matter which of M or N you choose to integrate to get Q There are some examples in which one is clearly easier but it s a judgement call integrate whichever seems easier to you In this case they re equally easy so let s use the rst one Qzy 2xy 7 9m2 dm xzy 7 3m3 Notice the presence of the function Hg in the integral This is the equivalent of the constant of integration that we obtain when we integrate a single variable function since if we differentiate Q with respect to m any function that just depends on y will vanish So instead of tacking on a constant to the end we need to have a function of y Differential Equations Lecture 8 Exact Equations If we had integrated N with respect to y to get Q we would have needed an equivalent function of m something like Mm to play the role of the constant of integration It s very important that this not be forgotten Now all we need to do is to nd Mg and we ll have our Q How do we do this We know that if we differentiate Q with respect to m h will vanish which is utterly unhelpful However if we differentiate Q with respect to y we re in good shape since 71 will hang around and we know that Q N So we ll compute Qy and any terms in N that aren t in Q must be h y Analogously if we had started by integrating N with respect to y to nd Q we would want to differentiate Q with respect to z and compare with M Ok so we can see that Q 2 h y and N 2 2y 1 So since these are equal we must have h y 2y 1 and so hyh ydyy2y REMARK As we ve tended to do we re going to be a little careless with constants In general we ll drop the constant of integration from our computation of h since it will end up combining with the constant c that we get as part of the solution process Thus we have Way 9021 73953 y2 H yz 2 1y 7 3963 which is precisely the Q that we used in Example 83 From here on it s exactly the same as it was there we observe that the differential equation is just dQ 7 dm 7 and thus Qm y y2 m2 1y 7 3x3 c for some constant c To compute c we ll use our initial condition y0 2 0 22 2 c a c 6 and so we have a particular solution of y2 x2 1y 73mg 6 This is a quadratic equation in y so as we ve seen we can complete the square or use the quadratic formula to get an explicit solution which we want to do when we can 342 m21y73m3 6 262 12 4 2 12 63z3x y2 002 1y m212 z412x32x225 y 7 2 4 7x2 1 i m4 12x3 2m2 25 W f Now we use the initial condition to gure which whether we want the or the 7 in that i Since y0 2 we have 2y071 V25277339 Thus we see that we ll want the solution and our particular solution is 7x2 1 354 12z3 2x2 25 f Differential Equations Lecture 8 Exact Equations The following examples should be less long since we won t need to repeat the rationales behind each of the steps EXAMPLE 85 Solve the initial value problem Qxyz 2 23 7 Kay 1471 1 First we need to put the equation into the form of Equation 81 mez 2 7 23 7 mzyy 0 Now we have Mzy 2zy2 2 and Nzy 723 7 y incorporating the minus into N is very important otherwise the derivatives won t work out Thus My 4mg Nm and the equation is exact The next step is to compute May In this example it s equally easy to integrate either M or N so let s use N ltIgtzy Ndy 2z2y 7 6dy xzyz 7 6y 1 To nd Mm we compute Pm 2zy2 h m and notice that for this to be equal to M h m 2 Hence hz 2x and we have an implicit solution of mzyz 7 634 2m c Now we use the initial condition y71 1 17672cc77 So our implicit solution is m2y276y2m70 Again we can solve for the explicit solution by completing the square or using the quadratic formula 6 i M36 74x22x 7 WE 2m2 7 3ix972x377z2 m2 u77 and using the initial condition we see we want the is solution so the explicit particular solution 37972x377z2 2 EXAMPLE 86 Solve the VP 2759 2 t2172t747lnt 1y 0 y20 and nd the solution s interval of validity This is already in the form of Equation 81 so let s start by checking if its exact Mty gigl 7 2t and Nt y lnt2 1 7 4 so My t l Nt Thus the equation is exact Now let s compute I In this case it will be easiest to integrate M 27534 2 2 t2 1 y ltIgty 11122 1 hy lnt2 1 7 4 N Differential Equations Lecture 8 Exact Equations so we conclude h y 7y and thus My 74y So our implicit solution is then ylnt217t2 74y 0 and we use our initial condition to compute c 74 Thus the particular solution is ylnt2 1 7 t2 7 4y 74 and this is very easy to solve explicitly Doing so we obtain 252 7 4 M95 Now let s nd the interval of validity We don t have to worry about the logarithm since 252 1 gt 0 for all 25 Thus we only need to avoid division by zero so we need to avoid the following points lnt21740 1n2 1 4 t2 e4 7 1 t Tm So there are three possible intervals of validity 700 7m 7 M mum 00 The middle one contains 25 2 so our interval of validity is 7 W D EXAMPLE 87 Solve 3y353my 7 1 2y53my 71 3my253my y 0 y1 2 We have Mm y 334353 7 1 and Nz y 23453 Bmyzegmy so My gyzesmy Qxysesmy Nm Thus the equation is exact We ll integrate M since it s a bit easier I Mdz 3y3531y 71 yzegmy 7 z My and by 2y53my 3my253my hy Comparing by to N we see that they are already identical so we must have h y 0 and hence My 0 since we re ignoring constants in h So we have 242631 7 C and using the initial condition gives 0 456 7 1 Thus our implicit particular solution is yzesmy 7 m 456 71 and we re done because we won t be able to solve this explicitly D Partial Fraction Decomposition for Inverse Laplace Trans form Usually partial fractions method starts with polynomial long division in order to represent a fraction as a sum of a polynomial and an another fraction7 where the degree of the polynomial in the numerator of the new fraction is less than the degree of the polynomial in its denominator 5317 7s1 521L178 52139 We7 however7 never have to do this polynomial long division7 when Partial Fraction Decomposition is applied to problems from Chapter 6 Another important fact in Chapter 6 is that we use only the following three types of fractions 5 7 a b 1 1 2 37 s 7a2b27 57 a2 l927 571 7 because we know the corresponding Inverse Laplace transforms 1 1 eatcosbt 2 1 l e s1nbt7 1 57a b 57a2b2 57a2b2 1 1 1 t2 3 71 at 71 t at 71 i at 2 l l 57 lt gt2 57 sear 2 l 1 t3 1 t4 1 t 71 i at 71 i at quot671 i at sia4l 66 7 57a5 246 7 57a 1 n16 We will call fractions 123 as standard fractions The Partial Fraction Decomposition for Inverse Laplace Transform is as follows Step 1 Suitable decomposition The objective of this step is to give the correct format of the partial fraction decomposition for a given fraction Rules of suitable decomposition Numerator does not matter Number of standard fractions equals the degree of the denominator Number of undetermined constants equals the degree of the denominator All standard fractions involved should be different Simplest Scenario When you solve a homogeneous equation ay by cy 0 you always have to solve r5905 Y 7 T as b5 c7 where I put 7 in the numerator because by Rule 1 the numerator does not matter in Step 17 and in the denominator you will have the characteristic polynomial Since the characteristic polynomial is quadratic you will need two different standard fractions Rule 2 and 4 and two undetermined constants Rule 3 There are three cases here a asz b5 c 0 has two distinct real roots Example 7 1 1 A B7 5273574 574 51 where 52 7 35 7 4 s 7 4s 1 As in the example above7 here the rule is that the two different standard fractions should be 87517 87527 where 51 and 2 are the roots of 152 b5 c 0 Further examples 7 1 1 7 1 1 7 A 37717 B7 52s 5 5175271 51L 571 b 152 b5 c 0 has two repeated real roots Example 7 A B 1 5272517 5711L 571239 where 52 7 25 1 s 7 12 As in the example above7 here the rule is that the two different standard fractions should be 1 1 576 5787 where S is the repeated root of 152 b5 c 0 Further examples 7 A1 B 1 77 A 1 B 577 E752659 7 53L 53 Note that for the cases a and b you will need to use the rst formula in 2 for the lnverse Laplace Transform b 152 b5 c 0 has two complex roots Example 7 s 7 1 2 39 4 B 5272s5 57121L41L 57124 where 52 7 2s 5 s 7 12 4 As in the example above7 here the rule is that the two different standard fractions should be 5 7 k b 97277127 97277127 where the denominator s 7 k2 m2 is obtained my completing the squares a52bscas7k2m2 Further examples 7 4 512 1 7 xB 8 717 B7 452455 s1221 s12217525 525 525 More Complicated Scenario When you solve a nonhomogeneous equation ay by By gt you will have to deal with a fraction7 which denominator is of degree 3 or higher The same four rules of suitable decomposition still apply here 1 Numerator does not matter 2 Number of standard fractions equals the degree of the denominator 3 Number of undetermined constants equals the degree of the denominator 4 All standard fractions involved should be different The rst preliminary step we have to take here is to decompose the denominator into a product standard polynomials These standard polynomial are exactly of two types a s 7 a b and 5 7 k2 m2 Examples s7152453 s71s1537 s71524s5 s71s2217 s 71s2 45 7 5 s 71s 71s 5 s 712s 5 Note that in the last example above we have to combine two terms 5 7 1 into one s 7 127 so that all the involved standard polynomials have distinct roots The standard fractions that arise in this more complicated scenario are identical to the ones that we discussed in the simplest case with one exception We may now need to use 1 1 1 8 537 8 5 SiSV WH For example 7 1 1 1 1 544 573B5732 543 573439 For everything else we just use the rule that we need to get the appropriate number of fractions for each standard polynomial in our denominator and just add them up together For example7 suppose you need to nd a suitable form of a partial fractions decomposition for 7 545 1 1s2 1s 712s2 71 You rst observe that there is one further decomposable fraction 5271 s71s1 Therefore 7 77 545 1 1521L1s 712s2 71 545 1 1s2 1s 713s 17 and the denominator is the product of ve standard fractions 547 s 12 17 52 17 s 7 137 s 1 For each one of them individually we know what we should do 77 1 1 1 1 7 A7 B7 of D7 s4 s 52 53 s47 7 51 1 4 19 7 5121 5121 s121 7 s 1 A B 521 521L11L 521 7 Then7 for the product 545 12 152 15 135 17 we should just add the individual suitable partial decompositions 77 7A1B101D1E 51 54512152151351 T 5 52 53 54 5121 1 5 1 1 1 1 1 F G7 H7 I J7 K7 L 51211L 5211L 5211L 511L 512 513 51 Step 2 Evaluation of the undetermined constants Here numerator matters A rst preliminary step here is to reduce the problem to an equality of polynomials The bullet proof way to do it is to multiply by the denominator of your given fraction in its standard form 52 25 3 1 5 1 1 A7 B 5152252 H11 511 Hwy where 5 152 25 2 5 15 12 1 51512152253 i A515121 5152252 T 1 51 51515121 O515121 511 511 7 515121 52253 A5121B5151C51 Then you need to nd the undetermined constants You do that by nding linear equations for these constants The number of equations should be equal to the number of unknowns Apart from small tricks7 there are two major methods evaluation of your expression for some values of s7 and comparison of coef cients for of the Blt polynomials a Evaluation at some points This method is very effective when the denominator has only distinct real roots 52 1 1 1 1 A B7 07 55151 51L 511L 517 52 1 A515 1 B55 1 055 17 5071A 51722Cg51722B A17 B17 01 Hence 52 1 7 1 1 1 55151T5 51 5139 b Comparison of eoe eients This method is bullet proof7 but it is quite long 521 55151 521A5151B551C5517 521A52AB52B5C52C5ABC52CB5A7 1 1 1 A7B7Oi7 5 51 51 On the right hand side the coef cients in front of 52 57 and 1 are 17 07 and 17 respec tively On the left hand side the coef cients in front of 52 57 and 1 are A B C7 0 7 B7 and 7A respectively Therefore 1ABQ 0073 17A Solving the above system we obtain Ai17 B17 01 Hence 52 1 1 1 1 ss1571 5157139 As you see7 in the above example the evaluation method is somewhat better than the comparison method Try to do the evaluation method for the next problem and compare with the following solution 53 7A8B1OSD2 5215247 521 521 524 52439 53 As52 4 Bs2 4 0552 1 Ds2 1 A Os3 B Ds2 4A Cs 4B D 1AQ 0Bu 04AO 04Bu From the second and the fourth equation above we obtain B D 0 The rst and the third equations give A 713 C 43 53 771 s Jr4 s 1x 4 3 1 2m24 Differential Equations LECTURE 36 Linearization Near Critical Points Last lecture we began considering the nonlinear system 90 1 24 361 1 We 24 We saw that near the critical points points 0340 satisfying fm0y0 gz0y0 0 which correspond to equilibrium solutions of the system we can approximate the nonlinear system 361 by the linear system 3 3 z 75900790 if 900790 x 39 362 9 900790 ail900790 y We can then analyze these linear systems to conclude what we can about the behavior of solutions near the equilibria Notice that this is entirely a local method we don t obtain information about what happens far away from the critical points However we can sometimes use common sense to determine other features of the phase portrait A 1quot EXAMPLE 361 Find the critical points of the following systems then determine the lineariza tions near them i xy273m2 ym27y2 The rst task is to nd the critical points We do this by setting z fzy and y gz y and then looking for the values of z and y that make both 1 and 9 zero Looking at the two equations we might decide that it s easy to conclude something from gz y 2 7 y2 0 Indeed we see that we must have m y Thus the equation fm y y2 7 3x 2 0 becomes 352 7 3m 2 0 x 7 2m 71 0 So we must have x 2 or z 1 Now we know 2 yz So ifz 2 y2 4 and so y i2 Similarly if z 1 y2 1 and so we get y i1 Summarizing we have four critical points 1 1 1 71 22 2 72 Now let s nd the linearizations of the system near each of these points The Jacobian matrix of the nonlinear system at a general point z y is e W i 9m 9y 296 72y 39 Differential Equations Lecture 36 Linearization Near Critical Points To get the coef cient matrix of our linearized system near one of our critical points all we have to do is evaluate these terms at the point Thus we get the following linearizations 11 w 3 32w Hr w 3 22 w f f4w w 13 1 ii my y7zm3 Again the rst task is to nd the critical points For z 0 we need to require that 0 Then for y 0 we need z 3 This means z 0 i1 So our critical points are 00 710 and 10 The Jacobian matrix of the system at z y is 0 1 713z2 0 7 so we end up with the following linearizations at each of the critical points 00 113 lt31 1 710 111 1 10 113 1 D So we ve found the critical points and the linearizations of the original nonlinear system near them Now what Normally we would go through our linear systems analysis to determine the type and stability of each critical point remember we can do this from just determining the eigenvalues Ideally the type and stability of the origin in these linear systems would correspond to the type and stability of the associated critical point in the nonlinear system EXERCISE For each of the critical points in both systems given in Example 361 determine the type and stability of the origin in the linearized systems However there s 1 A Caveat Or A Term7s A Term No Matter How Small ln turning our nonlinear system near a critical point into the linear system 362 we disregarded terms of order two or higher on the grounds that they were very small ls it really safe to do that In principle after all those terms could in uence the behavior of the system in ways that make this linearization method unreliable It turns out that as long as linearized system s critical point is not one of several 77borderline77 cases this doesn t affect the qualitative type of the critical point Differential Equations Lecture 36 Linearization Near Critical Points 11 Simple Eigenvalues If the linearized system near 0343 has simple 239e distinct eigenvalues the only problems occur when the linearization predicts we will have a center Spirals nodes and saddles are all preserved Moreover stability is preserved we can t go from having an unstable spiral in the linearized system to having a stable spiral in the nonlinear system Thus if our linearization predicts we have a spiral node or saddle we can conclude that the critical point will be of the same type Here s an example of how this fails when the linearization predicts a center EXAMPLE 362 Show that the two systems z 7ymx2y2 y zym2y2 z 7ymx2y2 y z7ym2y2 have the same linearized systems at the critical point 0 0 but dz erent phase portraits The Jacobian matrix of the rst system is 3m2y2 7l2xy l2my x23y2 39 The Jacobian matrix of the second system is 73m2 7 y2 717 2mg 17 2mg 7m2 7 3342 39 At 00 then both have the same linearization namely 13 7 0 71 1 T 1 0 39 The eigenvalues of this matrix are it and so this linearized system has a center However we can analyze this particular system more directly by using polar coordinates If we do that the two systems become 773 01 7773 01 Thus non zero trajectories for the rst system will expand as r gt 0 for all r gt 0 while non zero trajectories for the second system will decay as r lt 0 We actually have an unstable spiral for the rst system and an asymptotically stable spiral for the second system even though the linearization predicted a center These spirals however rotate in the same direction as the predicted center The previous example demonstrated the sort of thing we would need to do if we actually wanted to determine the phase portrait near a predicted center In general this will require a more direct computation rather than a linearization For our purposes though it will generally suf ce to be aware that we can t trust the prediction of a center even small higher order terms can throw a center to a spiral of either stability 12 Repeated Eigenvalues In the case of a repeated eigenvalue we ended up getting either a star node or a degenerate node depending on whether was complete or defective This is also a delicate case we will however be able to conclude something If lambda is complete and the linearization predicts a star node we can conclude only that the nonlinear system will have a node possibly degenerate or star at the associated critical point On the other hand if is defective we could get a node again it could possibly be degenerate or star or even a spiral point The stability of the linearized system however will be preserved Differential Equations Lecture 36 Linearization Near Critical Points REMARK Throughout this entire discussion of systems7 we ve assumed that our coe icient matrices are nonsingular that is7 they don t have any zero eigenvalues This is an interesting case7 though since it s the only case where you get substantially different behavior between the linearization and the nonlinear system
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