Class Note for MATH 254 at UA
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Date Created: 02/06/15
MATH 254 Lectures 12 0 Motivation and Examples How to derive a di erential equation Classi cation How to solve rst order separable ODEs What do we mean by a solution to a di erential equation Initial value problems Existence and uniqueness Why do we study differential equations 0 Many natural phenomena for examples the ow of water the motion of the atmosphere the dynamics of the planets the competition between species can be modelled and well described by differential equations We can learn about the behaviors of these systems by studying the solutions of the differential equations which describe theml o The study of the equations themselves leads to beautiful mathematics which re ects many of their obvious and less obvious properties Deriving differential equations It is as important to be able to write down a differential equation for a given situation described to you in English as it is to be able to solve it What follows is a set of problems which we are going to learn to translate into mathematics Example 1 Find a curve y whose tangent at the point zy is everywhere perpendicular to the line joining the point zy to the origin Answer The slope of the line joining 00 to zy is El This is the tangent of z the angle go any The slope of the tangent to the curve at z y is d dig and we are told this is perpendicular to E z z 80 d d Henceiygiloriy7 l 00 d1 1 dzi y The solution to the differential equation diyiz dag 7 7 11 will be a family of curves such that at any point z y the tangent to the curve at z y is perpendicular to the line joining it to the origin Example 2 You nd a dead body at 2ami lts temperature is 88601 An hour later its temperature has dropped to 78601 If you know that the ambient air temperature remained relatively constant at 686013 between midnight and 3am estimate the time of death Tair 22221 Let Tt be temperature of the body in OF at time t measured in hours from the time of death As suming that the person appeared to be in good health before meeting an untimely end T0 98601 How does the body cool We are told by Newton7s law of cooling cooling will occur mainly by radiation loss with some convective loss that the temperature dTt t neous temperature Tt and the background air temperature Let the constant of proportionality be or Then will change at a rate proportional to the difference between the instanta am 7 7 cTt 7 Ta Now since if Tt gt Ta the body temperature will decrease we must have 5 negative We call 5 iki Let t td be the time of discovery which we know to be 2ami Remember t 0 is the time of death Then 597mnwinni am We know T0 986013 Tad 88i6 F Ttd 1 78601 and Tair 686 From this information and the solution of the differential equation we can deter mine k and tag Ansi k ln2hri 1i The time of death was approxi 124 am Example 3 Mixtures of salt solutions I will derive this in class in ow gt 2m valve 1 tank is kept well stirred out ow Given In ow rate Q1 galsmini In ow concentration C1 lbsgall Out ow rate Q2 galsmini Initial concentration in tank C0 lbsgall Initial volume of brine G galsi At time t 0 we open valves one and two We seek to nd 1t7 the amount of salt in pounds in the tank at time t The amount of brine in gallons in the tank at time t is 9 Q1 Q2 The concentration in lbsgal at time t is I G Q1 Q2t Note that if Q1 lt Q2 the problem only makes sense until t h the time in minutes at which the tank is emptiedi We obtain an equation for the unknown variable zt by equating the rate at which salt is added to the tank7 namely to the input rate minus the output rate7 measured in lbs per minute d dix Input gtlt Input 7 Output gtlt Output t ow rate concentration ow rate concentration or dz t Q I t Q101 2 A 13 dt G Q1 7 Q2t To this equation is added the information 10 GOO lbs 3 Equations 1 1li21i3 are equations relating the unknown functions variables yz Tt zt with their rst derivatives Equations which relate unknown functions and their derivatives are known as dif ferential equations Classi cation If the unknown function depends only on one independent variable then the equa tion is known as an ordinary di erential equation or ODE If there is more than one unknown function but each only depends on one independent variable then the set of equations one for each unknown variable or function is called a system of ordinary differential equations or ODE7s If the unknown function depends on more than one independent variable then the equation is called a partial di erential equation or PDEi If there is more than one dependent variable and each depends on more than one independent variable then the set of such differential equations is called a system of PDEs The order of a differential equation for a single unknown function is the order of the highest derivative which appears We will discuss the de nition for the order of a system of ODEs later The differential equation of order n dot d ot F t t 7 0 14 o W W lt gt is said to be linear if F is a linear combination of some given function of t the function yt and its rst n derivatives For example do d2o 7 2d2o do 7 F lttyyyay it Wt 2yis1nt 70 is linear and second order The equation d2o do W 93 is nonlinear because the term y is a product of y and i The ODE 14 is said to be autonomous if F is independent of t For example 2 t y 6 1s hnear and nonautonomous whereas 2715 3 y 0 1s linear and autonomous See Q2 on quiz questions Examples of differential equations 1 The equation dz Q2It i C 7 dt Q1 1 6 Q1 7 Q2t is rst order and linear It is autonomous if Q1 Q2 2 The set of equations dz E am 7 Izzy dy E 7 59 dxy is a system of ODEs for the unknown variables 1t and It is a second order system but we will come to the de nition of the order of a system later 3 The differen 1al equa ions 6 71179 6 uIvy 612 By 0 Laplace 8uzy 82uzt 7 L T 7 HT 7 0 Heat 1 82zt 82uzt L E 62 W 0 We are each linear second order PDEsl 4 The equation d2 d 1 aid1712z 0 is a second order autonomous nonlinear ODE 5 The equation 8uz t 8uz t at 81 is a rst order nonlinear PDEi It is very famous and is known as Burgers7 equationl It is often used as a simple model of the ow of traf c or compressible owsl uzt 0 6 The equations 8p E EJrEpuiO Bu Bu 1 8p m ui 7i7 at 81 p 81 where p pp eigl p p0ltp 07 is given and pzt and uzt are the density and velocity of a uid at point z and time t are a system of partial differential equations which describe gas ow in a tube 5 We will now focus on a class of ODEs F may i i Ml 0 1 5 dtn which we can solve explicitly and uniquely for dnydt as dny 7 do dn ly m 7 f mt d l l l W l 16 What do we mean by a solution of 6 De nition Explicit solution An explicit solution of 16 on the interval a lt t lt B is a function such that t i i i if exist on a lt t lt B and satisfy WW 7 d dni l W dtn 7 f tailgyuww Example The function ce k is a solution to the rst order linear ODE Ly 7k f 7 t all y or 00 lt lt 00 Remark Usually but not always we will be dealing with solutions which are n times continuously differentiable on 15 for some integer n Then we say 45t E C a l It means that in the open interval a lt t lt 645 Hl all exist and are continuous But the existence of solutions is not guaranteed and even when they exist they may not be unique We shall say more about this later Worked Examples 1 Show that yt cosht satis es 22715 7 y 0 on 7007 e equot yt cosht 2 dyi 7 7617671 E7yt7s1nht7T 2 Note we sometlrnes denote 31171 as y and as y h Clearly7 yt satis es y 7 y 0 Also the solution yt exists for all nite t7 is continuous on 700 lt t lt 00 and all derivatives are continuous on 70000 2 Let yt eni Then yy76yenr2r760 When 7 73 or 7 2 Both 6 and 62 are solutions of y y 7 6y 0 for all nite ti 3 Let uzy cosz cosh yi It is easy to check that all derivatives of uz7 y With respect to both I and y exist for all 17y By direct computation check 0 812 By2 7 i 4 Let uzt 67m Bu 7T39 7 2 V7712 7 12 E 72326 m4a2t5 e m Bu 7 7m 7 2 BI 2a2t326 4a t 2 2 QauTT 74722T x 67 a 232 a t 4a2t52 De nition Implicit solution A relation Gzyc 0 5 some constant is said to be an implicit solution to equation 16 on the interval 1 a lt t lt B if it de nes one or more explicit solutions on L In order to make sense of this let us do some illustrated examples by learning how to solve separable equationsi De nition Separable We say the rst order ODE d i NW 17 is separable if we can Write fz y as the product of a function of z and a function of y Suppose 9I The method for solving 17 is simply to get all the 17s on one side the y s on the other and integrate For y not a value for Which My 0 17 becomes dy II ew Then gzdzc solves 1 7 With Note In the book NSS Write fz y as i Example 1 Recall dy z 1 g i 3 Then y dy 71 dz Integrate y22 7122c Gzyc 12y2725 0 The solution of l is a family of curves circles parameterized by the radius Clearly c 2 0 Note the relation Cz y c 0 is a curve on the z y planer It does not uniquely de ne y as a function of 1 We can7 however7 solve for y and obtain two explicit solutions y1z 20712 for 7 Ziclt z lt J2 y2z720712 for 7270ltzlt275 So the implicit solution 12 y2 7 20 0 de nes7 on the open interval 7x20 lt z lt V25 two explicit solutions7 y12z ixQc 7 12 Note y1z V25 7 12 is not a solution on 7x20 S I S V25 as is not de ned it is in nite at z ixZci De nition Initial value problem By an initial value problem for eqn 1 67 we mean Find a solution t on an interval 1 which for to E I satis es the conditions W dnil to 7 0050 7707 E 71739HW 77171 for an arbitrary prescription of 70 71 7n1l Example Find the solution to di 7 1 dz 7 y which passes through the point 37 4 namely for which y3 4 We have found the implicit solution to be 12y220 In order that the curve passes through 34 we must choose 32422025 Then of the two explicit solutions yzi257I2 on 75ltzlt57 25712 it is the solution which satis es y3 4 Existence and uniqueness of solutions to rst order ODEsl This is an important theorem and one of only two or three theorems you will meet during the whole coursel So7 please7 understand what it says Consider the initial value problem d g my 1090 If fz7 y and giany are continuous functions in the rectangle R z a lt z lt 127 c lt y lt d and 107 E Rt Then the initial value problem has a unique solution y z in some interval 10 7 h lt z lt 10 h where h gt 0 about 10 Example Z means positive square root of y Clearly 0 is a solution But if y f 07 we can solve by separation of variables 10 You can also check directly or by using separation of variables that W z c is a solution Suppose we are asked to nd the solution for which yl 0 There are two possible answers 1 WE 0 2 WE I 12 Check 0 and 27 0 exist for all z and satisfy the ODE Likewise I 712 and Z7 2z 71 Z for z 2 l the positive square root of I 712 is z 7 1 if z 2 l and l 7 I if I S l and yz7 27 exist and satisfy the equation for z 2 1 So the solution is not unique But note that 5 i which is not continuous in a rectangle containing the point l7 0 On the other hand7 for the initial condition yl 17 there is a unique solution 12 valid for z gt 0 Note for z lt 07 Jr 7x and twice this is not equal to 27 21 Practice questions for quiz given in Week 2 and based upon rst week lectures The question you will be asked to do will be one of these with numbers slightly changed or one of the set of HW questionsi Qili Consider a tank containing 10 gallons of a brine mixture with a concentration 1 of salt of E lbigalloni At time t 07 a mixture with concentration of l lbgallon begins to ow into the tank at a rate of 2 gallons per minute The tank is drained at the same rate Assuming the tank is kept well stirred7 write down an equation for 1t7 the amount in lbs of salt in tank at time t What is 10 Can you guess lim 1t t7gtoo 11 Qi2i For each ordinary differential equation below7 give its order and state Whether it is linear or nonlinear LNL7 autonomous or nonautonomous ANA ODE Order LNL ANA d2z 17130 12 QiSi For What values of A and 7 is AETE wt 7 71 Am a solution of the differential equation dy i 17 i d1 y y Qi4i Verify that 12 of 17 Where c is an arbitrary nonzero constant7 is a one parameter family of implicit solutions to dy 7 my dz 7 12 7 ll 2 z Hint Differentiate 12 of l Wrt 1 Remove 5 by noting that c 2 9 Draw the member of the family Which passes through the point 07 2 Q5 Use the method of separation of variables to solve the initial value problem dyi 2 7 719 901 Verify your answer Qi6i Find an implicit solution CLy c 0 to the differential equation dy 7 41 7 7 Find the value of the constant c for the curve Which passes through the point 07 2 Use this implicit solution to de ne two explicit solutionsi Over What interval in z are they solutions Write down the branch Which goes through the point l27 namely solve the initial value problem With yl 2J3 13
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