Math 5587 Midterm II Solutions

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1 Math 5587 Midterm II Solutions Prof. Jeff Calder November 3, 2016 Name: Instructions: 1. I recommend looking over the problems first and starting with those you feel most comfortable with. 2. Unless otherwise noted, be sure to include explanations to justify each step in your arguments and computations. For example, make sure to check that the hypotheses of any theorems you use are satisfied. 3. All work should be done in the space provided in this exam booklet. Cross out any work you do not wish to be considered. Additional white paper is available if needed. 4. Books, notes, calculators, cell phones, pagers, or other similar devices are not allowed during the exam. Please turn off cell phones for the duration of the exam. You may use the formula sheet attached to this exam. 5. If you complete the exam within the last 15 minutes, please remain in your seat until the examination period is over. 6. In the event that it is necessary to leave the room during the exam (e.g., fire alarm), this exam and all your work must remain in the room, face down on your desk. Problem Score 1 /10 2 /10 3 /10 4 /10 5 /10 Total: /50

2 1. Answer each question below. No justification is required for (b) and (c). [10 points] (a) [4 points] Explain briefly why separation of variables and Fourier series techniques are unlikely to provide a solution for u t ku xx + u 2 = 0. Solution. The PDE is nonlinear. Even if we found a family of separable solutions of the form u n (x, t) = X n (x)t n (t), the superposition principle does not hold for nonlinear equations, so u(x, t) = A n u n (x, t) will not be a solution of the PDE. (b) [3 points] Give an example of a function f(x) whose Fourier series does not converge uniformly. Solution. A classic example we covered in class is { 1 f(x) = 2, if π < x < 0 1 2, if 0 < x < π. Since f is discontinuous, it cannot be the uniform limit of a sequence of continuous functions. Also, we showed in class that the Fourier series for f exhibits Gibb s phenomenon, which immediately rules out uniform convergence of the Fourier series partials sums. (c) [3 points] Give an example of a sequence of functions f n that converges pointwise to some function f, but does not converge uniformly. Make sure to identify the pointwise limit f of your sequence, and the domain of your functions. Solution. Consider f n (x) = nx n (1 x) on the interval [0, 1]. This sequence converges pointwise to the zero function f(x) = 0, but does not converge uniformly. 2

3 2. [10 points] Solve the heat equation u t u xx = 0 on the rectangle 0 < x < π and t > 0 with initial condition u(x, 0) = 1 x and Dirichlet boundary conditions u(0, t) = u(π, t) = 0 for t > 0. Solution. The eigenvalues are λ n = n 2 for n = 1, 2, 3, 4,..., and the eigenfunctions are X n (x) = sin(nx). Thus the general solution is u(x, t) = Using the initial condition we have B n e n2t sin(nx). 1 x = u(x, 0) = B n sin(nx). Therefore B n = 2 π 0 (1 x) sin(nx) dx = { 2 n, if n is even 4 πn 2. n, if n is odd 3

4 3. [10 points] Solve the wave equation u tt = u xx on the rectangle < x < π, t > 0 with periodic boundary conditions u(, t) = u(π, t) and u x (, t) = u x (π, t), initial position u(x, 0) = 1 and initial velocity u t (x, 0) = 1. Solution. The eigenvalues for periodic boundary conditions are λ = n 2 for n = 0, 1, 2,... and the eigenfunctions are X n (x) = A n cos(nx) + B n sin(nx), for n 0 and X 0 (x) = 1. The ODE for T is T (t) + λt (t) = 0, When λ = 0, T 0 (t) = 1 2 (C 0 + D 0 t). When λ = n 2 for n 1 Therefore the general solution is u(x, t) = 1 2 (C 0 + D 0 t) + T n (t) = C n cos(nt) + D n sin(nt). (C n cos(nt) + D n sin(nt)) (A n cos(nx) + B n sin(nx)). Using the initial conditions we have and 1 = u(x, 0) = 1 2 C = u t (x, 0) = 1 2 D 0 + C n A n cos(nx) + C n B n sin(nx), nd n A n cos(nx) + nd n B n sin(nx). Therefore C n A n = C n B n = D n A n = D n B n = 0 and C 0 = D 0 = 2. Hence the solution is u(x, t) = 1 + t. 4

5 4. [10 points] Use separation of variables to find a family of solutions of u tx u xx = 0 on the entire plane R 2 that satisfy the boundary condition u(0, t) = 0 for all t. Solution. We guess a solution of the separated form u(x, t) = X(x)T (t). Then which is equivalent to u tx u xx = X (x)t (t) X (x)t (t) = 0, X (x) X (x) = T (t) T (t). Therefore, both sides are equal to some constant λ for λ R. This gives and Therefore X (x) = Ce λx and X (x) λx (x) = 0, X(0) = 0, T (t) λt (t) = 0. X(x) = C 1 e λx + C 2, for λ 0 and arbitrary constants C 1 and C 2. If λ = 0, then X(x) = C 1 x + C 2. For λ 0, the boundary condition X(0) = 0 gives C 1 + C 2 = 0 or C 1 = C 2. Therefore X λ (x) = C(e λx 1), for an arbitrary constant C. For λ = 0, X(0) = 0 implies X(x) = Cx for an arbitrary constant C. Dropping the constants, we have a family of solutions {X λ } λ R given by X 0 (x) = x and X λ (x) = e λx 1 for λ 0. The general solution of the ODE for T is T (t) = Ce λt. Therefore, the most general family of solutions of the PDE is {u λ } λ R where u 0 (x, t) = x and u λ (x, t) = (e λx 1)e λt. 5

6 5. [10 points] Let f be a 2π-periodic infinitely differentiable function satisfying f(x) + f (x) + f (x) + + f (m) (x) = 0 for all x R, where m is a positive integer and f (m) denotes the m th derivative of f. Show that { A cos(x) + B sin(x), if m {3, 7, 11, 15, 19,... }, f(x) = 0, otherwise. where A and B are arbitrary constants. The condition on m can also be stated as m = 4k 1 for a positive integer k. [Hint: Equate the Fourier series coefficients on both sides.] Solution. Let c n be the Fourier coefficients of f Since f is 2π-periodic for k 1. Therefore (f (k), 1) = c n = 1 2π f(x)e inx dx = 1 2π (f, einx ), f (k) (x) dx = f (k 1) (π) f (k 1) () = 0 0 = (f + f + f + + f (m), 1) = (f, 1), and we find that c 0 = 0. Recall from homework/practice exam (or compute via integration by parts) that for n 0. Therefore 1 2π (f (k), e inx ) = (in) k c n. 0 = 1 2π (f + f + f + + f (m), e inx ) = 1 2π (f, einx ) + 1 2π (f, e inx ) + 1 2π (f, e inx ) π (f (m), e inx ) = c n + inc n + (in) 2 c n + + (in) m c n m = c n (in) k k=0 = c n ( 1 (in) m+1 1 in ), where the last line follows from the formula for the partial sums of a geometric series. It follows that c n (1 (in) m+1 ) = 0. 6

7 For n ±1, we have 1 (in) m+1 0, hence c n = 0. It follows that For n = ±1 we have f(x) = c 1 e ix + c 1 e ix. c 1 (1 i m+1 ) = 0 and c 1 (1 ( i) m+1 ) = 0. (1) Notice i 0 = 1, i 1 = i, i 2 = 1, i 3 = i, i 4 = 1, i 5 = i,.... In general We have two cases. i 2k = ( 1) k and i 2k+1 = ( 1) k i. (1) If m is even then m + 1 is odd and i m+1 is always purely imaginary. Therefore 1 (±i) m+1 0 and we have c ±1 = 0. Therefore f(x) = 0. (2) If m is odd, then m + 1 is even. Let us write m + 1 = 2p for a positive integer p. Then (±i) m+1 = (±i) 2p = ( 1) p. If p is odd, then 1 (±i) m+1 = 2 and by (1) we have c 1 = c 1 = 0 and so f(x) = 0. If p is even, so m + 1 = 4k for some positive integer k, then 1 (±i) m+1 = 0 and c 1, c 1 are arbitrary, which yields f(x) = c 1 e ix + c 1 e ix. We can choose c 1 = c 1 = 1/2 to get f(x) = cos(x), and c 1 = 1/(2i) and c 1 = 1/(2i) to get f(x) = sin(x). Thus the general solution is f(x) = A cos(x) + B sin(x) when m = 4k 1. 7

8 Scratch paper 8

9 Formula Sheet f(x) = A A n cos(nx) + B n sin(nx) = cos(nx) cos(mx) dx = cos(nx) sin(mx) dx = 0 1 π π n= c n e inx, { π, if n = m sin(nx) sin(mx) dx = 0, otherwise.. { π and e inx e imx 2π, if n = m dx = 0, otherwise.. 1 f(x) 2 dx = 2π f(x) 2 dx = 1 2 A2 0 + n= c n 2. A 2 n + Bn. 2 9

Math 201 Assignment #11

Math 201 Assignment #11 Math 21 Assignment #11 Problem 1 (1.5 2) Find a formal solution to the given initial-boundary value problem. = 2 u x, < x < π, t > 2 u(, t) = u(π, t) =, t > u(x, ) = x 2, < x < π Problem 2 (1.5 5) Find