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!


1 Answer 1


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.


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