Equation $$\frac{\partial^2 p}{\partial t^2} = \frac{\partial^2 p}{\partial x^2} + \frac{\partial^2 p}{\partial y^2}$$
Boundary conditions:
$p = 0$ for $x = 0, \pi$
$p = 0$ for $y = 0, \pi$
Initial conditions:
$p =\sin x$ for $t = 0$
$\frac{\partial p}{\partial t} = 0$ for $t = 0$
In my notes for this problem it says that
the initial condition isn't compatible with the boundary condition and hence seeking the solution in the form $p = f(t)\sin x$ won't work because it wouldn't satisfy the boundary condition either.
- Firstly, why is the initial condition not compatible with the boundary condition? It looks compatible to me...
- And then can someone explain the "seeking the solution in the form..." bit..I don't understand why is meant to be implied here?
As Jose remarked, the initial value $p(x,y,0)=\sin x$ does not match the boundary conditions $p(x,0,t)=0$ and $p(x,\pi,t)=0$.
Any function of the form $p(x,y,t)=f(t)\sin x$ would fail to satisfy the boundary conditions for the same reason (whenever $f(t)\ne 0$). This is why we can't expect to find a solution that has the above form.
The phrase "seeking the solution in the form..." means looking for solutions that have some specific structure, for example separated solutions $u(x,y,t)=X(x)Y(y)T(t)$. This is often done because assuming this extra structure leads to simplification of the PDE, BC, and IC. On the other hand, if there are no solutions of this kind, then the attempt to find them will be a fruitless waste of time, best avoided.