FINAL EXAM MATH303 Theory of Ordinary Differential Equations. Spring dx dt = x + 3y dy dt = x y.

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1 FINAL EXAM MATH0 Theory of Ordinary Differential Equations There are 5 problems on 2 pages. Spring points Consider the linear plane autonomous system x + y x y. Find a fundamental matrix of the system. x0 2 Find the solution satisfying. y0 4 Which one from the following figures shows the solution curves of the system? Give reasons for your choice.

2 2. 20 points Find expression of the solution curves for each of the following plane autonomous systems ex + y 2 yex. 2 y xx2 + y 2 x yx2 + y points Consider the plane autonomous system ax y xx4 + y 4 x + ay yx4 + y 4, where a is a real constant. Determine whether the critical point 0, 0 is stable or unstable points Consider the ODE d2 θ sin θ. 2 Reduce the ODE to an equivalent first order ODE system. 2 Find all critical points of the system when θ [0, 2π. For each critical point found in question 2, determine whether it is stable or unstable points Show that the initial value problem has a unique solution for t 4. 2x2 + sin 2 t + x0 0 Solutions. The coefficient matrix is A. The two eigenvalues are λ 2, λ 2 2. The corresponding eigenvectors are φ and φ 2, respectively. Thus a fundamental matrix is e e 2t, e 2t 2t e 2t e 2t e 2t. 2 The general solution of the system is xt c e 2t yt + c 2 e 2t. 2

3 Since x0 c + c 2 y0 c + c 2 2c c 2 4, we have c 5 4 and c 2 4. Thus the solution of the initial value problem is xt yt 5 4 e2t 4 e 2t 5 4 e2t 4 e 2t 5 4 e2t 4 e 2t. The coefficient matrix has two eigenvalues with opposite sign λ > 0 and λ 2 < 0. By the theorem, the critical point 0, 0 is a saddle point and figure c shows the solution curves. 2. Xx, y e x + y, Y x, y 2 ye x X. x + Y y ex e x 0. Thus we can look for V x, y satisfying V y Xx, y ex + y Integrating Eq. with respect to y, we have V x Y x, y 2 + yex 2 V x, y y2 2 + yex + hx. Putting it into Eq. 2 ye x + h x 2 + ye x. Choose hx 2x, we have Thus the solution curves are the level curves of where C is a constant. h x 2. hx 2x + C. V x, y y2 2 + yex 2x. y yex 2x C, 2 Let { x r cos θ y r sin θ

4 Then dr d x 2 + y 2 2x 2 x 2 +y 2 + x 2y 2 x 2 +y 2 x 2 +y 2 [ y xx2 + y 2 ] + y x 2 +y 2 [x yx2 + y 2 ] dθ x2 x 2 +y x2 + y 2 y2 2 x 2 +y x2 + y 2 2 x 2 + y 2 2 r. d arctan y x + y2 x 2 y x y2 x x 2 y [ y xx x 2 +y 2 + y 2 ] + x [x yx 2 x 2 +y 2 + y 2 ] 2. From dr r, we have dr r. 2r 2 t + c. r 2t + c. From dθ, we have θ t + c 2. Thus the solution is where c and c 2 are constants. x cost + c 2t+c 2 y sint + c 2t+c 2,. The linear system is ax y x + ay. a The coefficient matrix is A. The two eigenvalues are λ a + i, λ 2 a i. a For the nonlinear system, ξx, y xx 4 + y 4, ηx, y yx 4 + y 4. When x 2 + y 2 <, we have x, y <, and ξ + η x x 4 + y 4 + y x 4 + y 4 2x 4 + y 4 2x 2 + y 2. 4

5 When a > 0, both λ and λ 2 have positive real part. 0, 0 is unstable for the linear system. By the theorem, 0, 0 is also unstable for the nonlinear system. 2 When a < 0, both λ and λ 2 have negative real part. 0, 0 is strictly stable for the linear system. By the theorem, 0, 0 is also strictly stable for the nonlinear system. When a 0, the system is y xx4 + y 4 x yx4 + y 4. Consider V x, y x 2 + y 2. V x, y is continuously differentiable, V 0, 0 0. V x, y > 0 when x, y 0, 0. When xt, yt is a solution, dv xt,yt V x x, y + V y x, y 2x[ y xx 4 + y 4 ] + 2y[x yx 4 + y 4 ] 2x 2 + y 2 x 4 + y 4 < 0 when xt, yt 0, 0. Thus V x, y is a strong Liapunov function. By the theorem, 0, 0 is strictly stable. Another solution of Consider V x, y x 2 + y 2. V x, y is continuously differentiable, V 0, 0 0. V x, y > 0 when x, y 0, 0. When xt, yt is a solution, When a 0, dv xt,yt V x x, y + V y x, y 2x[ax y xx 4 + y 4 ] + 2y[x + ay yx 4 + y 4 ] dv xt,yt 2x 2 + y 2 [a x 4 + y 4 ] 2x 2 + y 2 x 4 + y 4 < 0, for xt, yt 0, 0. Thus V x, y is a strong Liapunov function. By the theorem, 0, 0 is strictly stable. When a > 0, in the neighborhood of the origin x 2 + y 2 < min{, a}, we have x, y < and x 4 + y 4 x 2 + y 2 < a. Thus dv xt,yt xt, yt 0, 0. By the theorem, 0, 0 is unstable. 4. Let v dθ. The ODE is equivalent to the system 2x 2 + y 2 [a x 4 + y 4 ] > 0 in this neighborhood for dθ v dv sin θ. 2 A critical point satisfies v 0, sin θ 0. θ 0, π when θ [0, 2π. Thus critical points θ, v 0, 0 and π, 0. 5

6 At θ, v 0, 0: sin θ cos θ, sin θ sin θ. Using Taylor expansion for sin θ near θ 0, we have sin θ sin 0 + cos 0 θ 2 sin θ θ 2 θ 2 sin θ θ 2. The linear system is dθ v dv θ. 0 The coefficient matrix is A. Two eigenvalues λ > 0, λ 2 0. By the theorem, 0 0, 0 is unstable for the linear system. For the nonlinear system, ξθ, v 0, ηθ, v 2 sin θ θ 2. ξ + η 2 sin θ θ 2 2 θ2 2 θ2 + v 2. By the theorem, 0, 0 is also unstable for the nonlinear system. At θ, v π, 0: Let { θ θ π v v. Then θ θ + π, sin θ sinθ + π sin θ, and the system becomes dθ v dv sin θ. θ Consider Eθ, v 2 v2 + sin ξdξ 2 v2 cos θ +. Eθ, v is continuously differentiable, 0 E0, 0 0. Eθ, v 2 v2 + cos θ 0, and when θ π 2, Eθ, v 0 only when v 0 and θ 0. If θ t, v t is a solution deθ t,v t E θ θ, v dθ + E v θ, v dv sin θ v + v sin θ 0. Thus Eθ, v is a weak Liapunov function. By the theorem, θ, v 0, 0 is a stable critical point for the system of θ, v and θ, v π, 0 is a stable critical point for the original system. 5. The function Xx, t 2x 2 + sin 2 t + is continuously differentiable, thus it satisfies a Lipschitz condition in x K, t T π 2. 6

7 Let M sup 2x 2 + sin 2 t x, t π 2 Using the Local Existence Theorem, the given initial value problem has a unique solution in the interval t min{t, K M } min{π 2, 4 } 4. 7