# Wave Equation with Non-homogeneous Boundary Condition

Recently I encountered the following problem on a domain $$x>0$$, $$t>0$$:

$$u_{tt} = u_{xx}$$

$$u_x(0, t) = -h(t)$$

$$u(x, 0) = u_t(x, 0) = 0$$

$$h(t)$$ is an arbitrary function. It was given that the solution has the form $$u(x, t) = F(x-t)+G(x+t)$$. Filling out the initial conditions gives you the fact that F and G must be constant for positive arguments.

But for negative arguments, it is more difficult - the problem has to be extended to an infinite domain. I know how to do this for a Dirichlet or Neumann condition, but I struggle with processing such a non-homogeneous boundary condition.

Any help with solving this problem would be highly appreciated!

I think I got as far as is possible with Laplace transforms.

\begin{align} u_{tt} &= u_{xx}\\ u_x(0,t) &= -h(t)\\ u(x,0)&=0\\ u_t(x,0)&=0 \end{align}

Applying a Laplace transform in $$t$$, and treating $$x$$ as a parameter, we get

\begin{align} s^2\hat u(x,s)&=\hat u_{xx}(x,s) \\ \hat u_x(0,s) &= -\hat h(s)\\ \hat u(x,0)&=0 \end{align}

Where the hat indicates Laplace transform. This of course has general solution $$\hat u(x,s)=c_1 e^{sx} + c_2 e^{-sx}$$. Applying the homogenous condition gives us $$\hat u(x,s)=c(e^{sx} - e^{-sx})$$. Then applying the inhomogenous gives us $$x=-\frac{\hat h(s)}{2s}$$, so we then have a simple enough road:

\begin{align} \hat u(x,s) &= \frac{\hat h(s)e^{-sx} - \hat h(s)e^{sx}}{2s} \\ \mathscr{L}^{-1}\{\hat u(x,s)\}&=\mathscr{L}^{-1}\Big\{\frac{\hat h(s)e^{-sx} - \hat h(s)e^{sx}}{2s}\Big\} \\ u(x,t) &= \frac{1}{2} \int_0^t \mathscr{L}^{-1}\{\hat h(s)e^{-sx}\}(v) - \mathscr{L}^{-1}\{\hat h(s)e^{sx}\}(v) dv \\ u(x,t) &= \frac{1}{2} \int_0^t u_x(v)h(v-x) - u_{-x}(v)h(v+x)dv \\ u(x,t) &= \frac{1}{2} \int_x^t h(v-x) dv - \frac{1}{2}\int_0^t h(v+x)dv \\ \end{align}

And without a specific function for $$h$$, I think this is the best we can obtain. If someone else knows how to go further, let me know and I'll edit it in.