This is a Problem from Evans PDE 2nd Edition,Chapter 2 Problem 23. Let $S$ denote the square lying in $\Bbb R\times (0,\infty)$ with corners at the points $(0,1),(1,2),(0,3),(-1,2)$. Define $$f(x,t):=\left \{ \begin{array}{ll} \ -1, \text{ for } (x,t)\in S\cap \{ t>x+2\} \\ 1, \text{ for } (x,t)\in S\cap \{ t<x+2\} \\ \ 0, \text{ otherwise. } \end{array} \right.$$ Asume $u$ solves

$$\left \{ \begin{array}{ll} \ u_{ tt}-u_{ xx}=f, \text{ in } \Bbb R\times (0,\infty) \\ \ u=g,u_t=0,\text{ on } \Bbb R\times\{ t=0\}. \end{array} \right.$$ Describe the shape of $u$ for times $t>3$.


1 Answer 1


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By Duhamel's principle, for any $(x, t)$, we have $$u(x,t) = \int^t_0 \int^{x + (t-s) }_{x-(t-s)} f(y,s)\, dy\, ds.$$ Refering to the picture, we see that for $t_0 \ge 3$, if $(x_0,t_0)$ is on the right hand side of the grey area, then the domain of dependence for $u(x_0,t_0)$ does not intersect the support of $f$ and thus $u(x_0,t_0) = 0$. Likewise, if $(x_0,t_0)$ is on the left hand side of the grey area then the domain of dependence of $u(x_0, t_0)$ contains the entire support of $f$ so since $f$ is balanced, the integral will be zero.

Because of this, the only ``interesting'' region is the grey region. As pictured, for $(x_0, t_0)$ in the grey region, $u(x_0, t_0)$ depends only on the length $d = d(x_0, t_0)$ of the orange line segment because it determines how much of the support of $f$ lies in the domain of dependence of $u(x_0,t_0)$ (the blue/red region is the intersection of the domain of dependence with the support of $f$). In this region we see that if $d = 0$, we should get $u(x_0,t_0) = 0$ and if $d = \sqrt 2$ then $u(x_0, t_0) = 0$. Further, as $d$ grows from $0$ to $\tfrac{\sqrt 2}2$ the value of $u$ is simply the amount of blue area. For $\tfrac{\sqrt 2}{2} \le d \le \sqrt 2$, $u$ is the blue area minus the red area. Calculating the area, we see $$u(x_0, t_0) = \left\{ \begin{matrix} d(x_0,t_0) \sqrt 2, & 0 \le d(x_0,t_0) \le \tfrac{\sqrt{2}}{2} \\ \tfrac{\sqrt 2}{2}\sqrt 2 - \sqrt 2(\tfrac{\sqrt 2}2 - (\sqrt 2 - d(x_0,t_0))), & \tfrac{\sqrt{2}}{2} \le d(x_0,t_0) \le \sqrt 2. \end{matrix} \right.$$ or simplifying a bit $$u(x_0, t_0) = \left\{ \begin{matrix} d(x_0,t_0) \sqrt 2, & 0 \le d(x_0,t_0) \le \tfrac{\sqrt{2}}{2} \\ 2 - d(x_0, t_0) \sqrt 2& \tfrac{\sqrt{2}}{2} \le d(x_0,t_0) \le \sqrt 2. \end{matrix} \right.$$ Pictorially, this $u$ looks like a triangular spike along the line between $t = x +1$ and $t = x+3$. Solving explicitly for $d(x_0, t_0)$ is not hard, but the algebra is somewhat ugly and the process is in no way edifying.


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