Problem:
$u_{tt} - c^2u_{xx}=f(x,t) \quad 0<x<\infty, t>0$
$u(x,0) = 0, u_t (x,0) = 0 \quad (for \quad all\quad 0<x<\infty)$
$u(0,t)=0 \quad (for \quad all\quad t>0)$
Ok so on the half-line this is a nonhomo. wave equation, with homo. IC and homo. Dirichelt boundary conditions. c>0 is a constant, f is continuous.
So solving this and sketching the domain of integration, as well as verifying that the BC is satisfied I started using this:
For region $0<ct<x$ I have solution:
$u(x,t) = \frac{1}{2c} \iint_\Delta f$
For region $0<x<ct$ I have solution:
$u(x,t) = \frac{1}{2c} \iint_D f$
I am not sure what exactly the solution to this is. Is this the final answer?
What is the difference between $\Delta$ and D?
I just extrapolated from a worked out problem - so not sure if this is even correct. Please help!
Also what is the domain of integration sketched out?
UPDATE:
I got that D is the domain of dominance?
and $\Delta$ is characteristics?
Please help trying to solve this one!!
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If the problem was posed on the entire line, the solution would be $$ u(x,t) =\frac{1}{2c}\iint_\Delta f \tag{1} $$ where $\Delta$ is the triangle with vertices $(x,t)$, $(x-ct,0)$ and $(x+ct,0)$. It may be called the domain of dependence, or the characteristic triangle, or something else.
In the region $x\ge ct$ this is exactly what you get, since the boundary effect is not felt there.
One way to handle the boundary condition is to use reflection: extend $f$ to the region $x<0$ by $f(-x,t) = -f(x,t)$. The reason for doing this is that the solution with odd source function $f$ will vanish when $x=0$: indeed, the integral in (1) is zero because the half $x<0$ cancels out the by symmetry.
What if $0<x<ct$? Then a part of the triangle $\Delta$ lies in the reflected territory, which cancels out a chunk of the positive part. We can call what's left $D$: so, $$ u(x,t) =\frac{1}{2c}\iint_D f $$
The domain $D$ can be also described in inequalities: $$D=\{(y,s): |x-c(t-s)|\le y\le x+c(t-s)\}$$ but this hardly makes its shape clear than a picture.