It is asked to show that the wave equation does not satisfy the wave equation
The wave equation is given by
$$\left\{\begin{matrix} u_{tt}=c^2u_{xx} \\ u(x,0)=\phi(x) \\ u_t(x,0)=\psi(x) \end{matrix}\right.$$
and its solution is given by
$$u(x,t)=\frac{\phi(x+ct)+\phi(x-ct)}{2} + \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s) ds $$
The author suggests to take $\phi(x)=0$ and $\psi(x) > 0$ but I don't see how can I rigorously show that it does not satisfy the maximum principles.
Suppose that $\psi(x)=1, \forall x \in \mathbb{R}$ and $\phi = 0$. Then, the solution will be
$$u(x,t)=t$$
If we only consider as boundary the $x$ axis, then indeed it does not assume maximum. But in diffusion equation, we have as boundary the square delimited by $[0,L]\times[0,T]$. If we need to define something like this for the wave equation, the maximum will be at a square side and I don't know what to do. Should I consider $x \in [0,L]$ as in the diffusion equation or $x \in \mathbb{R}$?
I've seen another examples of functions that shows that the maximum principles doesn't hold in this case, but I would like to follow the author's approach.
Best Answer
For example, let $$ \psi(x) = \begin{cases} 1 ,\quad &|x|<1\\ 0,\quad &|x|\ge 1 \end{cases}$$ The solution is strictly positive when $|x|<1+ct$ and it is zero when $|x|\ge 1+ct$.
Consider a rectangle with base on the $t=0$ line, and such that $u=0$ on the vertical sides. For instance, $0\le t\le 1$, $|x|\le 2+c$.
Not much special about $\psi$: any compactly supported function will do.
One can also use a decaying function, like the Gaussian, by picking a sufficiently wide rectangle so that the values on the vertical sides are small.