20 1. Differential Equations 1, k = 2 and k = 3 on the viewing window −4 x 4 and 0 y 2 for −3 t 3. 6. For an appropriate choice of k, the function u(x, t) = sin(x + kt) is a solution to the nonlinear partial differential equation utux + 2u2 = 2. (a) What value of k makes it a solution? (Show or explain how you determined your answer. Remember that sin2 θ + cos2 θ = 1!) (b) Letting k be as in your previous answer, for what value(s) of the scalar λ is the function U(x, t) = λu(x, t) a solution to the same equation? (c) Answer in one or two complete English sentences: Considering t to represent “time”, what would an animation of this solution u(x, t) look like? (Describe not only what shape its profile has but how it changes in time. Be specific.) 7. Suppose that I have a function u(x, t) that is a solution to the equation ut = (ux)2 2uxx and such that its initial profile looks like u(x, 0) = x2. The point (0, 0) is on the graph at time t = 0. Will this point initially move up or down under the evolution determined by the equation. (Show or explain how you know.) What about the point (1, 1) which is also on the initial profile? 8. (a) Use the SimpleEvolver (with the same evolution equation) to see what would happen to the initial profile u(x, 0) = sin(x) on the interval 0 x π. (Hint: Mathematica calls them Sin[x] and Pi!) (b) Modify SimpleEvolver so that it approximates the dynamics of the evolution equation ut = u3 5uxxx and see what happens to the initial profile u(x, 0) = sin(x) on 0 x π. (c) What happens to the point (0, 0) under the dynamics in (b)? Explain why this makes sense. 9. The equations pxpy + pxyp = 1 (1.12)
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