Agenda Sections 2.4, 2.5

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1 Agenda Sections 2.4, 2.5 Reminders Read 3.1, 3.2 Do problems for 2.4, 2.5 Homework 1 due Friday Midterm Exam I on 1/23 Lab on Friday (Shapiro 2054) Office hours Tues, Thurs 3-4:30 pm (5852 East Hall)

2 Theorem: Existence and Uniqueness for Linear DE s If the functions p and g are continuous on an open interval I =(, )containingthepointt = t 0, then there exists a unique function y = (t) thatsatisfiesthede y 0 + p(t)y = g(t) for each t in I,andthatalsosatisfiestheinitialcondition y(t 0 )=y 0, where y 0 is an arbitrary prescribed initial value.

3 In other words, suppose you have a DE of the form y 0 + p(t)y = g(t), and suppose p and g are continuous functions on some interval (, ). Now, choose a point (t 0, y 0 )such that <t 0 <,asshowninthe figure. Then, the IVP y 0 + p(t)y = g(t), y(t 0 )=y 0 is guaranteed to have a solution that is well defined in the interval (, ), and that solution is the only one that satisfies the IVP.

4 The proof for this theorem is based on the idea that an integrating factor can obtain every solution to the IVP y 0 + p(t)y = g(t), y(t 0 )=y 0 as long as the continuity conditions on p(t) andg(t) are satisfied. Uniqueness is proven by showing that the initial condition determines the constant of integration uniquely.

5 Example Without solving the the problem, determine the largest interval in which the solution of the IVP is guaranteed to exist. (a) (t 3)y 0 +(lnt)y =2t, y(1) = 2 (b) t(t 4)y 0 + y =0, y(2) = 1 (c) y 0 +(tant)y =sint, y( ) =0 (d) (4 t 2 )y 0 +2ty =3t 2, y( 3) = 1 (e) (4 t 2 )y 0 +2ty =3t 2, y(1) = 3 (f) (ln t)y 0 + y =cott, y(2) = 3

6 Example Without solving the the problem, determine the largest interval in which the solution of the IVP is guaranteed to exist. (a) (t 3)y 0 +(lnt)y =2t, y(1) = 2 (0, 3) (b) t(t 4)y 0 + y =0, y(2) = 1 (0, 4) (c) y 0 +(tant)y =sint, y( ) =0 ( /2, 3 /2) (d) (4 t 2 )y 0 +2ty =3t 2, y( 3) = 1 ( 1, 2) (e) (4 t 2 )y 0 +2ty =3t 2, y(1) = 3 ( 2, 2) (f) (ln t)y 0 + y =cott, y(2) = 3 (1, )

7 Partial Derivatives To use the next theorem, we will need to be able to compute partial derivatives. Partial derivatives are used when a function depends on two or more independent variables. To take a partial derivative, we must choose which variable we re di erentiating with respect to. For example, if we want to di erentiate f (x, y) withrespecttoy, (x, To compute a partial derivative, di erentiate with respect to the chosen variable while treating all other independent variables as constants. (exy + x) =xe xy.

8 Theorem: Existence and Uniqueness for Nonlinear DE s Let the functions f /@y be continuous in some rectangle <t <, <y < containing the point (t 0, y 0 ). Then, in some interval t 0 h < t < t 0 + h contained in <t <,thereexists a unique solution y = (t) of the IVP y 0 = f (t, y), y(t 0 )=y 0.

9 In other words, suppose you have a DE of the form y 0 = f (t, y), and suppose f (t, y) and@f /@y are continuous functions in the rectangle <t <, <t <.Now,choose apoint(t 0, y 0 )inthatrectangle,as shown in the figure. Then, the IVP y 0 = f (t, y), y(t 0 )=y 0 is guaranteed to have a solution that is well defined for t near t 0,and that solution is the only one that satisfies the IVP.

10 The full proof for this theorem is much more di cult than the proof for the linear case and is considered beyond the scope of this course. Section 2.8 of the book goes over some of the main ideas behind the proof. We ll talk about it more in the next lecture.

11 Example Using the existence and uniqueness theorem for nonlinear DE s, state where in the ty-plane the DE is guaranteed to have a unique solution. (Refer to the chalkboard.) (a) y 0 = t y 2t+5y (b) y 0 =(1 t 2 y 2 ) 1/2 (c) y 0 = ln ty 1 t 2 +y 2 (d) y 0 =(t 2 + y 2 ) 3/2 (e) y 0 = 1+t2 3y y 2 (f) y 0 = (cot t)y 1+y

12 Example Using the existence and uniqueness theorem for nonlinear DE s, state where in the ty-plane the DE is guaranteed to have a unique solution. (Refer to the chalkboard.) (a) y 0 = t y y 6= 2t 2t+5y 5 (b) y 0 =(1 t 2 y 2 ) 1/2 y 2 + t 2 < 1 (c) y 0 ln ty = t 2 y 2 6=1, t 6= 0, y 6= 0 1 t 2 +y 2 (d) y 0 =(t 2 + y 2 ) 3/2 everywhere in the ty-plane (e) y 0 = 1+t2 3y y 2 y 6= 0, y 6= 3 (f) y 0 = (cot t)y 1+y y 6= 1, t 6= n for all n 2 Z

13 Example Solve the IVP 2y 0 y = t, y( 1) = 0 to determine the number of solutions. Is the result consistent with the existence and uniqueness theorem? We can solve the IVP using separation of variables. Z Z 2 ydy = tdt y 2 = 1 2 t2 + c Imposing the initial condition gives 0 2 = 1 2 ( 1)2 + c =) c = 1 2

14 Example Solve the IVP 2y 0 y = t, y( 1) = 0 to determine the number of solutions. Is the result consistent with the existence and uniqueness theorem? We can solve the IVP using separation of variables. Z Z 2 ydy = tdt y 2 = 1 2 t2 + c Imposing the initial condition gives 0 2 = 1 2 ( 1)2 + c =) c = 1 2

15 Plugging c into the general solution gives y 2 = 1 2 t y 2 = 1 2 (1 t2 ) r 1 t 2 y = ± 2 We have two solutions.

16 To use the existence and uniqueness theorem, we need to write the DE in the form y 0 = t/(2y). The right-hand side function f (t, y) = t/(2y) is not continuous at y =0. Sinceourinitialconditionis y( 1) = 0, the IVP does not satisfy the hypotheses of the existence and uniqueness theorem. Therefore, our result is consistent.

17 Example Solve the IVP y 0 = y 1/3, y(0) = 0 to determine the number of solutions. Is the result consistent with the existence and uniqueness theorem? We can solve the DE using separation of variables. Z Z dy y = dt 1/3 3 2 y 2/3 = t + c 2 y = ± 3 t + c 3/2

18 Example Solve the IVP y 0 = y 1/3, y(0) = 0 to determine the number of solutions. Is the result consistent with the existence and uniqueness theorem? We can solve the DE using separation of variables. Z Z dy y = dt 1/3 3 2 y 2/3 = t + c 2 y = ± 3 t + c 3/2

19 3/2 2 y = ± 3 t + c Imposing the initial condition, y(0) = 0, gives 3/2 2 0=± 3 0+c =) c =0. Therefore our solution is y = ± 3/2 2 3 t. However, this is not the only solution. It turns out that y =0 is also a solution to the IVP. We lost this solution when we divided by y 1/3 in order to separate variables.

20 In fact, it can be shown that, for any arbitrary positive t 0,the function 0, if 0 apple t < t0 y = ± 2 (t t 3 0) 3/2, if t t0 satisfies the IVP. (Notice that the function is continuous and di erentiable at t 0.) This is an example of an IVP with an infinite number of solutions. If we look at the hypotheses for the existence and uniqueness theorem, f (t, y) =y (t, y), = 3y 2/3 we see /@y is discontinuous at y =0. Sinceourinitial condition was y(0) = 0, the existence and uniqueness theorem does not apply. Therefore, our results are consistent.

21 Recall that if an IVP satisfied the hypotheses of the existence and uniqueness theorem for linear DE s, then we could conclude that a solution existed on the entire interval (, ). Unfortunately, this is not the case for nonlinear DE s. In the nonlinear case, the solution y = (t) iscertaintoexistaslong as the point (t, (y)) remains within a region in which the hypotheses of the theorem are satisfied. However, since the solution is usually not known, it may be impossible to be sure that the solution remains in this region. The next example demonstrates this scenario.

22 Example Solve the IVP y 0 = y 2, y(0) = 1. Is the result consistent with the existence and uniqueness theorem? This IVP can be solved using separation of variables. Z Z dy y = dt 2 1 y = t + c y = 1 t + c

23 Example Solve the IVP y 0 = y 2, y(0) = 1. Is the result consistent with the existence and uniqueness theorem? This IVP can be solved using separation of variables. Z Z dy y = dt 2 1 y = t + c y = 1 t + c

24 y = 1 t + c Imposing the initial condition y(0) = 1 gives 1= 1 0+c =) c = 1 The solution is then y = 1 1 t Notice that the solution is not defined at t =1. However, the hypotheses for the existence and uniqueness theorem are satisfied everywhere.

25 Still, this result does not contradict the existence and uniqueness theorem for nonlinear DE s. It would be a mistake to conclude that the solution to the IVP exists for all t. Thetheoremonlyguaranteesauniquesolutionexists.It does not specify the interval where the solution exists.

26 Definition: Autonomous DE (Review) ADEthatcanbewrittenas is said to be autonomous. dy dt = f (y) Example Determine if the statement is true or false. (a) An autonomous DE isn t necessarily a seperable DE. (b) An autonomous DE can have solutions that oscillate. (c) Equilibrium solutions are constant solutions. (d) There s a di erence between a stable equilibrium point and an asymptotically stable equilibrium point.

27 Definition: Autonomous DE (Review) ADEthatcanbewrittenas is said to be autonomous. dy dt = f (y) Example Determine if the statement is true or false. (a) An autonomous DE isn t necessarily a seperable DE. F (b) An autonomous DE can have solutions that oscillate. F (c) Equilibrium solutions are constant solutions. T (d) There s a di erence between a stable equilibrium point and an asymptotically stable equilibrium point. T

28 Exponential Growth Model Assumes that a population grows at a rate proportional to the current population. That is, dy dt = ry, y(0) = y 0 where r is called the rate of growth or decline. Ifr > 0, the population is growing.

29 Logistic Growth Model Assumes that a population grows exponentially when the population y is small. The model includes the assumption that the growth rate decreases as the population approaches the carrying capacity K. TheDEforthismodelis dy dt = r y 1 y, y(0) = y 0, K where r is called the intrinsic growth rate and K is called the saturation level or the carrying capacity.

30 Logistic Growth with a Threshold Similar to the logistic model, except there s an extra assumption that a population below a certain threshold T will die o. The DE has the form dy dt = r 1 y T 1 where T is the threshold level. y y, y(0) = y 0, K

31 Example Without solving the DE, provide a qualitatively accurate sketch of the solutions to dy dt = y(1 y 2 ). We begin by finding the critical points. That is, we re looking for constant solutions. 0=y(1 y 2 ) =) y = 1, y =0, y =1arecriticalpoints. Drawing a phase line may be helpful in understanding the behavior of solutions.

32 Example Without solving the DE, provide a qualitatively accurate sketch of the solutions to dy dt = y(1 y 2 ). We begin by finding the critical points. That is, we re looking for constant solutions. 0=y(1 y 2 ) =) y = 1, y =0, y =1arecriticalpoints. Drawing a phase line may be helpful in understanding the behavior of solutions.

33 At this point, we know the critical points and when the solutions are increasing or decreasing. However, we don t know anything about the concavity (second derivative) of the solutions. To find this, we can use the chain rule to compute d 2 y/dt 2.SupposewehaveaDEoftheform dy dt = f (y), then d 2 y dt = d 2 dt f (y) =f 0 (y) dy dt =f 0 (y)f (y). To find the inflection points, we need to know when d 2 y/dt 2 =0. That can either happen when f 0 (y) =0or f (y) =0. In the latter case, f (y) =0correspondstothe equilibrium solutions. Therefore, we should focus on finding solutions to f 0 (y) =0.

34 Recall that the DE was Computing f 0 (y) gives Setting f 0 (y) =0gives dy dt = y(1 y 2 ). f 0 (y) =1 y 2 2y 2 =1 3y 2. 0=1 3y 2 =) y = ± p 1/3 Therefore, our inflection points are at y = ± p 1/3. By checking the sign of d 2 y/dt 2 = f 0 (y)f (y), we can determine the concavity of the solutions.

35 f (y) =y(1 y 2 ), f 0 (y) =1 3y 2 The dotted lines are there to remind us of the change in concavity.

36 Bifurcation Abifurcationoccurswhenasmallsmoothchangemadetoa parameter in a DE causes a sudden qualitative change in the behavior of the solutions. The value of the parameter (or point in parameter space) where the bifurcation occurs is called the bifurcation point. Examples Critical points changing stability Critical points appearing or disappearing

37 Example Consider the DE y 0 = a y 2 ( ) (a) Find the critical points when a < 0, a =0,anda > 0. (b) Draw the phase line for each case. (c) Plot the location of the critical points as a function of a in the ay-plane. This is called the bifurcation diagram of ( ).

38 y 0 = a y 2 (a). Suppose a < 0, then 0=a y 2 =) no critical points Suppose a =0,then 0=a y 2 =) y =0 is a critical point Suppose a > 0, then 0=a y 2 =) y = p a and y = p a are critical points

39 (b). For a < 0, there are no critical points. For a =0,the critical point at y =0issemistable. For a > 0, the critical point at y = p a is unstable and the critical point at y = p a is asymptotically stable.

40 (c). Bifurcation Diagram (saddle-node bifurcation) Asymptotically stable critical points are shown with a solid line. Unstable critical points are shown with a dashed line.

41 There are additional types of bifurcations that can occur.

42 Bifurcation Diagram (transcritical bifurcation) Asymptotically stable critical points are shown with a solid line. Unstable critical points are shown with a dashed line.

43 Transcritical Bifurcation Govindarajan & Marathe

44 Bifurcation Diagram (pitchfork bifurcation) Amzoti Asymptotically stable critical points are shown with a blue line. Unstable critical points are shown with a red line.

45 Pitchfork Bifurcation For r apple 0, there s one asymptotically stable critical point For r > 0, the single critical point splits into three critical points

46 For each type of bifurcation, the reverse situation can occur where the critical points have the opposite stability. Either way, the names for the bifurcations are the same.

47 Example Consider the DE dy dt = ay y 3 (a) Find the critical points for a < 0, a =0,anda > 0. (b) Sketch several solutions to the DE in the ty-plane. (c) Draw the bifurcation diagram for the DE. What type of bifurcation is it?

Sample Questions, Exam 1 Math 244 Spring 2007

Sample Questions, Exam 1 Math 244 Spring 2007 Sample Questions, Exam Math 244 Spring 2007 Remember, on the exam you may use a calculator, but NOT one that can perform symbolic manipulation (remembering derivative and integral formulas are a part of