FINAL EXAM MATH0 Theory of Ordinary Differential Equations There are 5 problems on 2 pages. Spring 2009. 25 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. 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 2.. 25 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. 4. 20 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. 5. 0 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
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 2 2 + yex 2x C, 2 Let { x r cos θ y r sin θ
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 2 + + 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
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
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
Let M sup 2x 2 + sin 2 t + 2 2 + 2 + 4. 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