My question:
Consider $u_{tt}-u_{xx}=F(x,t)$ in the first quadrant ($x,t>0$) with boundary conditions $u(x,0)=f(x)$, $u_t(x,0)=g(x)$ and $u(0,t)=0$. Does it necessarily follow that $u\geq 0$?
My attempt:
I am not sure how to deal with this problem restricted to the first quadrant. I know I can use D'Alembert's formula to solve it in the plane though. Any suggestions?
The answer is not, see the following example.
Consider $$u_{tt}-u_{xx}=2e^t\sin x$$ coupled with $u(0,t)=0$, $u(x,0)=-\sin x$ and $u_t(x,0)=0$
In the domain $[0,\pi]\times (0,\infty)$ the problem above has a solution $$u=-e^{-t} \sin (x)\le0$$ If you ask about entire quadrant - the same solution will change signs, so still not positive.