University:
California State University, Northridge
Course:
MATH 480 | Partial Differential Equations
Academic year:

2024
Views:

272

Pages:

4
Author:

Quintin Palmer

Math 480. Exam 1 Solutions This exam is a closed book/notes exam. Calculators are neither necessary nor allowed. 1. Show that the following sequence is uniformly convergent: f (x) = ∞ X cos nx k=0 Solution: We first observe that the series since | cos x| ≤ 1 for all x ∈ R we see that n2 1 n=1 n2 P∞ converges absolutely. Next ¯ cos x ¯ 1 ¯ ¯ ¯ 2 ¯ ≤ 2, n n therefore by the Weierstrass M-test the given series converges uniformly and absolutely. 2. Consider the initial boundary value problem ut + 2ux = 0; u(x, 0) = x(x ≥ 0); u(0, t) = t(t ≥ 0). Construct a solution u(x, t) on the set [0, ∞) × [0, ∞). Is this solution a classical solution or only a weak solution? Justify your answers. Solution: We know from class that an inital boundary value problem has the solution ( f (x − ct) x ≥ ct u(x, t) = g(t − x/c) x < ct for initial values f and boundary values g. In the given problem c = 2, f (x) = x, and g(t) = t. Hence, ( x − 2t x ≥ 2t u(x, t) = t − x/2 x < 2t This solution is continuous, but not differentiable since ux (x, t) = ( 1 x2t −1/2 x < 2t Therefore, this solution is only a weak solution and not a classical solution. 3. Consider the initial value problem 1 ut + ux = 0; x u(x, 0) = x2 (x > 1). 2 Compute the equation for the characteristics for this problem and sketch characteristics starting at x0 = 1, x0 = 2, x0 = 3. Compute the solution to the initial value problem and sketch snapshots of the solution for t = 1 and t = 2. Solution: The equation for the characteristics is 1 dx = dt x Solving this equation we get x dx = dt and after integration: x2 x20 − =t 2 2 The characteristics with x0 = 1, x0 = 2, and x0 = 3 are shown below. From the equation for the characteristics we get x20 x2 = − t. 2 2 Since the initial values are x2 , 2 this implies u(x, t) = x2 − t. 2 The snapshots are shown below. 4. Consider the following initial value problem: ut + uux = 0;   1     2+x u(x, 0) =  2−x     1 x ≤ −1 −13 1 ≤ x ≤ 3.5 At t = 5 the corner which was at 1(for t = 2) moved to x = 4, the shock at 3.5 moved to x = 8, and the solution is given by:  x≤4   1 u(x.5) = (x − 4)/4 + 1   1 x>8 4≤x≤8 the qualitative graphs of these functions are shown below. 4 3.5 4 3 3.5 2.5 3 2 2.5 2 1.5 1.5 1 1 0.5 0.5 0 −3 −2 −1 0 1 2 3 4 5 Figure 3: The characteristics for problem 4 0 −2 0 2 4 6 Figure 4: The solution at t = 2 and t = 5. 8 10

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