Final Exam December 20, 2011 Math 420 - Ordinary Differential Equations No credit will be given for answers without mathematical or logical justification. Simplify answers as much as possible. Leave solutions in implicit form when appropriate. Scientific calculators only. No notes or books Part I (35 points) Problem 1. (5 pts) Find the real part of e ln(7) π 4 i. Problem 2. (5 pts) Compute, directly from the definitions, the Laplace transform of u 2 (t). 1
Problem 3. (10 pts) Find the particular solution. dx = 2 t x t 2 + 1 + 1, x(0) = 0. (1) Problem 4. (10 pts) Find the general solution. dx = x2 + 2 2 t x (2) 2
Problem 5. (5 pts) Write the following nonlinear second order equation as a system of first order equations. d 2 y 2 + ( ) 2 dy y = 0 (3) 3
Part II (65 points) Problem 6. (65 points total) Consider the damped harmonic oscillator d 2 y 2 + 2pdy + ω2 0y = cos(ωt) u(t) (4) If ω 2 0 p 2 > 0, meaning the system is underdamped, define ω 1 = ω 2 0 p2. Let z be the fundamental solution of (4). That is, z(t) satisfies d2 z 2 and z(t) = 0 for all t < 0. + 2p dz + ω2 0z = δ(t) 6a) (5 pts) Let Z(s) = L(z(t))(s) be the Laplace transform of z(t). Assume the system is 1 underdamped, and prove that Z(s) =. (s + p) 2 + ω1 2 6b) (10 pts) Find z(t). 4
6c) (10 pts) Let y(t) be the solution of the forced equation (4) with initial conditions y(0) = 0, y (0) = 0, and let Y (s) be its Laplace transform. Prove that Y (s) = Do not attempt to perform partial fraction separation. s (s 2 + ω 2 ) [(s + p) 2 + ω 2 1 ] (5) 5
6d) (10 pts) Using the partial fraction expansion technique, we have [ 1 (ω 2 Y (s) = 0 ω 2 ) s + 2pω 2 (ω0 2 ω2 ) 2 + (2pω) 2 s 2 + ω 2 (ω2 0 ω 2 ) (s + p) + p(ω 2 + ω0) 2 ] (s + p) 2 + ω1 2. Do not attempt to prove this. For this question, simply indicate which part of this expression corresponds to the steady-state solution and which corresponds to the transient solution. State why. 6e) (10 pts) Use the table of Laplace transforms to prove that the steady-state solution is [ ω0 2 ω 2 ] y(t) = (ω0 2 ω2 ) 2 + (2pω) 2 cos(ωt) + 2 p ω (ω0 2 ω2 ) 2 + (2pω) 2 sin(ωt) u(t). (6) 6
6f) (20 pts) Using the result from question (6e), show that the total amplitude of the steady-state solution as a function of ω is A(ω) = 1 (ω 2 0 ω 2 ) 2 + (2pω) 2. (7) Assuming ω 0 = 2 and p = 1 2, sketch the frequency response plot. Indicate clearly any critical points. 7
Part III (100 points) Problem 7. (25 points total) Consider the following non-linear system. dx dy ( = 1 x ) x + 1 6 2 xy ( = 1 y ) y 1 (8) 4 3 xy 7a) (5 pts) If interpreted as a predator-prey system, which is the prey and which is the predator? 7b) (5pts) In the absence of any prey species, what type of equation does the predator species follow? 7c) (5pts) In the absence of any predator species, what type of equation does the prey species follow? 8
7b) (10 pts) Determine all critical points of the system (8). 9
Problem 8. (40 points total) Consider again the system dx dy ( = 1 x ) x + 1 6 2 xy ( = 1 y ) y 1 (9) 4 3 xy 8a) (20 pts) Linearize the system at the critical point (x, y) = (2, 4 3 ). Determine all eigenvalues. Sketch the graph of the linearized system, including any straight-line solutions. 10
8b) (20 pts) Linearize the system at the critical point (x, y) = (6, 0). Determine all eigenvalues. Sketch the graph of the linearized system, including any straight-line solutions. 11
Problem 9. (35 points total) Consider again the system dx dy ( = 1 x ) x + 1 6 2 xy ( = 1 y ) y 1 (10) 4 3 xy Most predator-prey systems are not Hamiltonian. However, set ( x H(x, y) = x y 6 + y ) 4 1. (11) 9a) (5 pts) Verify that the system (10) is a Hamiltonian system. 9b) (5 pts) Prove that H is a Hamiltonian for this system. 12
9c) (10 pts) Sketch the level-sets of H in the first quadrant. To do this, make a precise sketch of the level-set H = 0, then make reasonable approximations for the level-sets H = α for values of α both larger and smaller than 0. 13
9d) (10 pts) Sketch the trajectories of the system (10) in the phase plane. Restrict your diagram to the first quadrant. Clearly label all critical points by type, and label any separatrices you find. 9e) (5 pts) What is the ultimate fate of the system with initial condition (x 0, y 0 ) = (1, 1)? What if initial conditions are (x 0, y 0 ) = (10, 10)? 14
Extra Credit Problem. (20 points total) Consider the following slightly modified version of (8): and again set dx = 7 ( 1 x ) x + 15 8 7 32 xy dy ( = 1 y ) y 1 (12) 4 3 xy ( x H(x, y) = x y 6 + y ) 4 1. (13) a) (10 points) Show that the function H is a Lyapunov function for (12), when the system is restricted to x 0, y 0. It may help to rewrite the first equation as [ dx ( = 1 x ) x + 1 ] 6 2 xy x [ y 8 4 + x ] 3 1 (14) b) (10 points) On a scratch page, sketch the trajectories in the phase plane for this system. What is the ultimate fate of a system with initial conditions (x 0, y 0 ) = (1, 1)? 15
Scratch 16
Scratch 17
Scratch 18
Trigonometric Formulas Basic formulas cos(α ± β) = cos(α) cos(β) sin(α) sin(β) sin(α ± β) = sin(α) cos(β) ± cos(α) sin(β) From these follow the sum-to-product formulas: cos(α β) + cos(α + β) = 2 cos(α) cos(β) cos(α β) cos(α + β) = 2 sin(α) sin(β) sin(α + β) + sin(α β) = 2 sin(α) cos(β) sin(α + β) sin(α β) = 2 cos(α) sin(β) 19
Laplace Transforms Definition F (s) = Table of Common Transforms 0 e st f(t) t-domain function δ(t), δ a (t) s-domain function 1, e as 1 u(t), u a (t) s, e as s t u(t) 1 s 2 t n u(t) n! s n+1 e at u(t) sin(ωt) u(t) cos(ωt) u(t) e at sin(ωt) u(t) e at cos(ωt) u(t) 1 s a ω s 2 + ω 2 s s 2 + ω 2 ω (s a) 2 + ω 2 (s a) (s a) 2 + ω 2 20
Properties of the Laplace Transform Property Name t domain function f(t) s domain function F (s) = 0 e st f(t) Linearity α f(t) + β g(t) α F (s) + β G(s) df sf (s) f(0) f 1 s F (s) t f(t) 1 t f(t) df ds F Time Shift Theorem Phase Shift Theorem Time Scale Theorem Convolution Theorem f(t a) e as F (s) e at f(t) F (s + a) f(at) 1 a F ( s a ) (f g)(t) F (s) G(s) 21
Convolutions Definition (f g)(t) = f(t u) g(u) du Property Symmetry Associativity Expression f g = g f f (g h) = (f g) h Linearity f (αg + βh) = α(f g) + β(f h) Differentiation Antidifferentiation The Identity Integration d df (f g) = g = f dg (f g) = f g = f g f δ = f f u = t f(v) dv 22