Sample Solutions of Assignment 9 for MAT3270B

Transcription

1 Sample Solutions of Assignment 9 for MAT370B. For the following ODEs, find the eigenvalues and eigenvectors, and classify the critical point 0,0 type and determine whether it is stable, asymptotically stable, or unstable. a. b. c. d Answer: a. deta λi = 3 λ λ = λ λ + = 0 λ =, λ = The eigenvectors are X = and X = 0,0 is an unstable critical point of the system.

2 b. deta λi = λ 5 3 λ = λ + λ + = 0 λ = + i, λ = i The eigenvectors are X = 5 i and X = 5 i 0,0 is an asymptotically stable critical point of the system. c. deta λi = The eigenvectors are λ 5 0 λ = λ = 0 X = λ = 0 and 0 X = t ,0 is an unstable critical point of the system. d. deta λi = The eigenvectors are λ λ = λ + = 0 λ = i, λ = i X = i

3 3 and X = i 0,0 is a stable critical point of the system.. Transform the following problems into the form λ µ µ λ a. b Answer: a. deta λi = 3 λ 4 λ = λ λ + = 0 λ = 7 + i, λ = 7 i The problem can be transformed to b. deta λi = 7 7 λ 3 λ = λ λ + = 0

4 4 λ = 7 + i, λ = 7 i The problem can be transformed to Determine the critical points for each of the following systems a. d = 3y, = y y b. d = y, = y 4 y 3y 4 c. d = y, = µ y, µ > 0 d. d = y + y, = + y y Answer: a. The critical points of the system are 0,0, 0, and 3,. 5 5 b. The critical points of the system are 0,0, 0, and,0. c. The critical points of the system is 0,0. d. The critical points of the system is 0,0. 4. Find out an equation for the form H, y = c satisfies for the trajectories: a. d = + y +, = y y b. d = y 3 4y, = y + 6y

5 5 Answer: a. We get d = y y + y + The ODE can be rewritten as y yd + + y + = 0 M, y y So this ODE is eact for M,y y that = N, y = ψ = M, y = y y Integrating above equation, we obtain = N,y. Thus there is a ψ, y such ψ, y = y y + fy Setting ψ y = N gives then we get Hence, f y = y fy = y + c y y y = c b. We get d = y + 6y y 3 4y The ODE can be rewritten as y 6yd + y 3 4y = 0 M, y y N, y = 4y 6 = 4y 6

6 6 So this ODE is eact for M,y y that = N,y. Thus there is a ψ, y such ψ = M, y = y 6y Integrating above equation, we obtain ψ, y = y 3 y + fy Setting ψ y = N gives then we get Hence, f y = 4y fy = y + c y 3 y y = c 5. Show that the trajectories of the nonlinear undamped pendulum equation are given by where = θ, y = dθ. d θ + g l sin θ = 0 g y cos + l = c Answer: Let then = θ, y = dθ d = dθ = y = d θ = g l sin θ = g l sin

7 7 We get Hence, d = g sin l y g l cos + y = c. 6. Prove that for the system dx = fx there is at most one trajectory passing through a given point X 0. Answer: Let φ t, φ t be two trajectories passing through X 0, i.e. φ t = X 0, φ t = X 0, then Set dφ = fφ, φ t = X 0 dφ = fφ, φ t = X 0 φ = φ t + t t, then By uniqueness, d φ = dφ f φ, φ t = X 0 φ t = φ t = φ t + t t Hence, φ and φ are the same trajectories. 7. Prove that if a trajectory starts at a noncritical point of the system dx = fx

8 8 the it can not reach a critical point in a finite length of time. Answer: Assume the contrary. we suppose that 0, y 0 is a critical point of the system, and the solution = φt, y = ψt satisfies φa = 0, ψa = y 0. On the other hand, = 0, y = y 0 is a solution of the given system satisfying the initial condition = 0, y = y 0 at t = a. By the uniqueness, φt = 0, ψt = y 0, t 0, hence, φ0 = 0, ψt = y 0. This contradict the assumption of the problem. 8. Assuming the trajectory corresponding to solution = φt, y = ψt, < t < +, of an autonomous system is closed, show that the solution must be periodic. Answer: Suppose that φt = φt = X 0, t t. By the same proof as in Problem 6, we get φt + t t = φt, t Let T = t t, the φt + T = φt, t Hence φt is periodic with period T. 9. Write the following spring-mass system as a system of equations by introducing = u, u = du m d u + cdu + ku = 0 Find out the critical point and analyze the stability of the critical point.

9 9 Answer: Let then = u, y = du d = du = y = d u = c m y k m 0,0 is a critical point of the system. deta λi = λ c λ m k m = mλ + cλ + k = 0 Case i c 4mk > 0 and k 0 In this case, there are two distinct real eigenvalues, say r, r. Since m, c, k > 0, then r, r < 0 0,0 is asymptotically stable critical point of the system. Case ii c 4mk = 0 and k 0 In this case, there are repeated eigenvalues, say r. Since m, c, k > 0, then r < 0 0,0 is asymptotically stable critical point of the system. Case iii c 4mk < 0 and k 0 In this case, there are two comple eigenvalues, say r = λ + µi, r = λ µi where λ = c. Since m, c > 0, 0,0 is asymptotically stable m critical point of the system if c 0. Similarly, 0,0 is stable critical point of the system if c = 0.

10 0 Case iv k = 0 In this case, r = 0, r = c. 0,0 is stable critical point of the m system. 0. Show that the system is almost linear and 0,0 is a stable critical point of the system d = y, = y y Answer: d = 0 0 Let = r cos θ, y = r sin θ then y y + y lim y r = lim r cos θ sin θ = 0 lim y = 0 r det A =, the system is almost linear. deta λi = λ 0 0 λ λ = = λ + = 0 0,0 is stable critical point of the system.. Show that the system is almost linear and 0,0 is an asymptotically stable critical point of the system d = 3 + y, = y3

11 Answer: Let = r cos θ, y = r sin θ then lim 3 + y r = lim r cos 3 θ + r cos θ sin θ = 0 lim y3 r Hence, the system is almost linear. = limr sin 3 θ = 0 As for the type of the critical point, we will give the solution on Problem 5 of Assignment 0.. Determine all real critical points of the following system of equations a. d = + y, = + y b. d = y, = y Answer: a The critical points of the system are 0,0 and,-. b The critical points of the system are, and,-. 3. Consider the following problem. Determine the eigenvalues and critical points, classify the type of the critical point and determine whether it is stable, asymptotically stable, or unstable. ε = ε Here ε is a real number.

12 Answer: Since det A = ε + > 0, 0,0 is the only critical point of the system. deta λi = Case A: ε = 0 ε λ ε λ = λ ελ + ε + = 0 λ = i, λ = i 0,0 is a stable center. Case B: ε > 0 λ = ε + i, λ = ε i 0,0 is an unstable spiral point. Case C: ε < 0 λ = ε + i, λ = ε i 0,0 is an asymptotical stable spiral point. 4. Consider the following Lienard equation where g0 = 0. d + cd + g = 0 a. Write it as a system of two first order equations by introducing y = d. b. Show that 0,0 is a critical point and that the system is almost linear in the neighborhood of 0,0.

13 c. Show that if c0 > 0, g 0 > 0 then the critical point is asymptotically stable, and that if c0 < 0 or g 0 < 0, then the critical point is unstable. 3 Answer: a.let then y = d = d From the original equation, we get Let = cy g. X = y We get X = 0 0 c X + 0 g Let b. Since g0 = 0, then 0,0 is the critical point of the system. where η 0, 0 where η / 0, 0 The system can written as X 0 = g 0 c0 c = c0 + η g = g 0 + η X + 0 η y η

14 4 Take G T = 0, η y η Then G 0,, y 0, 0. Hence the system is almost linear in the neighborhood of 0,0. c. deta λi = Hence Case A: c0 > 0, g 0 > 0 λ g 0 c0 λ λ = c0 + c 0 4g 0 λ = c0 c 0 4g 0 = λ + c0λ + g 0 = 0 If = c 0 4g > 0, then λ < λ < 0 and 0,0 is an asymptotical stable critical point. If = c 0 4g < 0, then λ = α + βi, λ = α βi where α = c0 < 0 and 0,0 is an asymptotical stable critical point. If = c 0 4g = 0, then λ = λ = c0 < 0 and 0,0 is an asymptotical stable critical point. Case B: c0 < 0 If = c 0 4g critical point. > 0, then λ > λ > 0 and 0,0 is an unstable If = c 0 4g < 0, then λ = α + βi, λ = α βi where α = c0 > 0 and 0,0 is an unstable critical point.

15 If = c 0 4g = 0, then λ = λ = c0 > 0 and 0,0 is unstable critical point. 5 Case C: g 0 < 0 In this case, λ > 0 > λ > 0 and 0,0 is an unstable critical point.