APPM 2360: Midterm 3 July 12, 2013.

Transcription

1 APPM 2360: Midterm 3 July 12, ON THE FRONT OF YOUR BLUEBOOK write: (1) your name, (2) your instructor s name, (3) your recitation section number and (4) a grading table. Text books, class notes, and calculators are NOT permitted. A one page (2 sided, letter sized) crib sheet is allowed. There are 5 questions and each is worth 20 points. Problem 1: (a) (2 points) Write the characteristic equation for x + bx + 4x = 0 and solve its roots r 1,2 for general b using the quadratic formula. The characteristic equation is r 2 + br + 4 = 0. It has roots r 1,2 = b ± b = b 2 ± ( b 2) 2 4. (b) (9 points) For which values of b is the system over-damped, critically-damped, and under-damped? For each case, sketch and describe: (i) the general behavior of x(t) versus t for each situation, and (ii) the corresponding phase plot depicting x (t) versus x(t). The system is overdamped when b > 4, critically damped when b = 4, and underdamped when b < 4. (c) (5 points) Let b = 0 and use your above expression for the roots to derive the general solution in terms of sine and cosine functions. [Hint: use Euler s expansion for e iθ to convert the exponentials to the trigonometric functions. Using that the roots are r 1,2 = ±2i, we have that the general solution is given by x(t) = k 1 e i2t +k 2 e i2t. Because e i2t = cos(2t)+i sin(2t) and e i2t = cos(2t) i sin(2t), we can rewrite the general solution as x(t) = C 1 cos(2t) + C 2 sin(2t), where C 1 = (k 1 + k 2 ) and C 2 = i(k 1 k 2 ). (d) (4 points) Solve the IVP for the initial conditions x(0) = 1 and x (0) = 4. Note that x(t) = C 1 cos(2t) + C 2 sin(2t) has the derivative x (t) = 2C 1 sin(2t) + 2C 2 cos(2t). Using that x(0) = 1, we find that C 1 = 1. Using that x (0) = 4, we find that C 2 = 2.

2 Problem 2: (a) (5 points) Given the second-order differential equation x + 5x + 4x = 0, convert the equation to a system of two first order differential equations and write the system in d matrix form, ( y ) = A y. dt [ y1 Letting y = y 2, where y 1 = x and y 2 = x, we see that y 1 = y 2 and y 2 = x = [ 0 1 4y 1 5y 2. we can write y = A y, where A =. 4 5 (b) (6 points) Find and plot the null clines. Be sure to indicate the direction field for all the regions of the phase portrait as well as on the null clines. The vertical null cline is found by setting y 1 = 0, which occurs when y 2 = 0. The horizontal null cline is found by setting y 2 = 0, which occurs when y 2 = 4 5 y 1. Note that the direction field for the vertical null cline points down when y 1 > 0 and up when y 1 < 0. Note that the horizontal null cline points right when y 2 > 0 and left when y 2 < 0. (c) (5 points) Find the general solution and draw an example trajectory in the phase portrait drawn in (b). The characteristic equation is r 2 +5r +4 = (r +4)(r +1), which has roots at r 1 = 4 and r 2 = 1. the general solution is given by x(t) = C 1 e 4t + C 2 e t. (d) (4 points) Find and classify the stability of all equilibria. The only equilibrium is at y 1 = y 2 = 0. Examination of the characteristic equation shows that all roots of the characteristic equation have negative real part and thus the equilibrium is stable. Problem 3: Let L(y) denote the linear operator L(y) = y 2y + y. (a) (8 points) For each following forcing functions f(t), write down a suitable form for y p to solve L(y) = f(t) using the method of undetermined coefficients. [You do not need to solve the coefficients. (i) f(t) = 2 + t 3 (ii) (iii) (iv) f(t) = e t f(t) = cos(t) + sin(2t) f(t) = e 2t + t cos(t). The homogenous solution is of the form y h = C 1 e t + C 2 te t. Therefore the following are suitable forms for y p :

3 (i) y p = A 3 t 3 + A 2 t 2 + A 1 t + A 0 (ii) y p = αt 2 e t (iii) y p = A cos(t) + B sin(t) + C cos(2t) + D sin(2t) (iv) y p = αe 2t + (At + B) cos(t) + (Ct + D) sin(t). (b) (8 points) Let g(t) = 100 sin(3t) + 5t. Solve y p for L(y) = g(t) using the method of undetermined coefficients (i.e., evaluate all coefficients). A suitable y p is y p = A cos(3t) + B sin(3t) + Ct + D, which has the derivative y p = 3A sin(3t) + 3B cos(3t) + C and second derivative y p = 9A cos(3t) 9B sin(3t). [ [ L(y p ) = 9A cos(3t) 9B sin(3t) 2 3A sin(3t) + 3B cos(3t) + C +A cos(3t) + B sin(3t) + Ct + D = ( 8A 6B) cos(3t) + ( 8B + 6A) sin(3t) + (C)t + ( 2C + D). Noting that L(y p ) = sin(3t) + t, after matching terms find that 0 = 8A 6B 1 = 8B + 6A 5 = C 0 = 2C + D. (A, B, C, D) = (6, 8, 5, 10), giving y p = 6 cos(3t) 8 sin(3t) + 5t (c) (4 points) Using g(t) and your y p from part (b), find the general solution to the initial value problem (IVP) L(y) = g(t) with initial conditions y(0) = 16 and y (0) = 0. The general solution is y G = y h + y p = C 1 e t + C 2 te t + 6 cos(3t) 8 sin(3t) + 5t + 10, which has derivative y G = C 1 e t + C 2 e t + C 2 te t 18 sin(3t) 24 cos(3t) = y G (0) = C , C 1 = 0. 0 = y G(0) = C 1 + C , C 2 = 19.

4 Problem 4: Let L(y) = t 2 y 2ty + 2y, y 1 = t, and y 2 = t 2. (a) (4 points) Show that y 1 and y 2 are solutions to the corresponding homogeneous equation, L(y 1 ) = L(y 2 ) = 0. Since y 1 = t, y 1 = 1, and y 1 = 0, it follows that L(y 1 ) = t 2 (0) 2t(1) + 2(1t) = 0 and thus y 1 is a solution. Since y 2 = t 2, y 2 = 2t, and y 2 = 2, it follows that L(y 2 ) = t 2 (2) 2t(2t) + 2(t 2 ) = 0 and thus y 2 is a solution. (b) (4 points) Prove that {y 1, y 2 } are a basis for the solution space to the homogeneous equation, and thus any solution may be written as y h = C 1 y 1 + C 2 y 2. Because they are solutions to L(y) = 0, y 1 and y 2 are a basis if they are linearly independent, which we can prove by showing that the Wronskian W [y 1, y 2 is nonzero. W [y 1, y 2 = t t2 1 2t = 2t2 t 2 = t 2 0. (c) (8 points) Find y p using variation of parameters. [Hint: you may need to use udv = uv vdu for integration. Using that y h = C 1 t + C 2 t 2 we look for a a particular solution of the form y p = v 1 (t)t + v 2 (t)t 2. v 1 and v 2 can be found using Cramer s method as v 1 = y 2f(t) W [y 1, y 2 = (t2 )(t sin t) t 2 = t sin t v 2 = y 1f(t) (t)(t sin t) = = sin t. W [y 1, y 2 t 2 After integrating, we have that v 1 = t cos t sin t and v 2 = cos t. y p = (t cos t sin t)t + ( cos t)t 2 = t sin t. (d) (4 points) Use your results from (c) to solve the initial value problem L(y) = t 3 sin(t) using that at time t 0 = 1 the initial conditions are y(1) = sin(1) and y (1) = sin(1). Write the full solution y(t) to the IVP. From above, the general solution is which has derivative y G = y h + y p = C 1 t + C 2 t 2 t sin t, y G = C 1 + 2C 2 t sin t t cos t, sin(1) = y G (1) = C 1 + C 2 sin(1) sin(1) = y G(1) = C 1 + 2C 2 sin(1) cos(1).

5 These may be solved to find C 1 = cos(1) and C 2 = cos(1), which leads to y = cos(1)t + cos(1)t 2 t sin t. Problem 5: Let u + bu + 100u = A cos(ω f t). (a) (4 points) Consider the forced system with b = 0 and A = 40. Write the general solution x G for (i) when ω f does not equal the resonant frequency and (ii) when it does. [You do not need to derive the solutions, but simplify as much as possible. Does there exist a forcing frequency ω f for which the general solution x G = x h + x p becomes unbounded? When ω f ω 0, one has A u G = C 1 cos(ω 0 t) + C 2 sin(ω 0 t) + cos(ω ω0 2 ωf 2 f t) 40 = C 1 cos(10t) + C 2 sin(10t) + cos(ω 100 ωf 2 f t). (1) When ω f = ω 0, one has u G = C 1 cos(ω 0 t) + C 2 sin(ω 0 t) + A t sin(ω 0 t) 2ω 0 = C 1 cos(10t) + C 2 sin(10t) + 2t sin(10t) Note that at the resonant frequency, ω f = ω 0, the term A 2ω 0 t sin(ω 0 t) is unbounded. (b) (6 points) Use your solution from part (a) for ω f = 60 to solve the IVP for u(0) = 1 and u (0) = 50. From (b) we have that u G = C 1 cos(10t) + C 2 sin(10t) + cos( 60t). u G = 10C 1 sin(10t) + 10C 2 cos(10t) 60 sin( 60t). From our initial conditions, we find that Therefore we have that 1 = u G (0) = C 1 + 1, C 1 = = u G(0) = 10C 2, C 2 = 5. u = 5 sin(10t) + cos( 60t).

6 (c) (5 points) Now consider the damped system with b = 16, A = 1, and ω f = 1. Find the general solution. You do not need to derive the particular solution y p, but simplify as much as possible. [Hint: try bringing the denominator of the quadratic equation into the square root to find the roots of characteristic polynomial. The characteristic equation is r r = 0, which has roots at r 1,2 = 8 ± 6i. the homogeneous solution is u h = e [C 8t 1 cos(6t) + C 2 sin(6t). The particular solution is u p = C cos(t δ), where C = A = (ω0 2 ωf 2)2 + b 2 ωf 2 1 (99) Therefore tan(δ) = bω f ω 2 0 ω 2 f = u G = e [C 8t 1 cos(6t) + C 2 sin(6t) + C cos(t δ). (d) (5 points) Use your solution from part (c) to solve the IVP with u(0) = u (0) = 0. [You may express the constants in terms of variables defined for your particular solution. From part (c) we have that [ u G = 8e [C 8t 1 cos(6t) + C 2 sin(6t) + e 8t 6C 1 sin(6t) + 6C 2 cos(6t) C sin(t δ). From the initial conditions, [ 0 = u G (0) = (1) C C cos(δ) [ 0 = u G(0) = 8 C (1) [0 + 6C 2 + C sin(δ) = 8C 1 + 6C 2 + C sin(δ). C 1 = C cos(δ) C 2 = C 6 sin(δ) + 4C 3 cos(δ).