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Problem

Use the general solution to solve the signalling problem with homogeneous wave equation on the half line, homogeneous IC and nonhomogeneous Neumann boundary conditions. Where c>0 is a constant, and h is continuous function. The solution u should be continuous.

$u_{tt} - c^2 u_{xx} = 0, \quad\quad \quad \quad 0 < x < \infty,\quad t>0$

$u(x,0)=0, u_t (x,0)=0 \quad (For \quad All \quad 0 < x < \infty)$

$u_x(0,t)=h(t) \quad \quad \quad \quad \quad(For \quad All \quad t > 0)$

My attempt

General solution: $ u(x,t) = F(x-ct) + G(x+ct)$

Using B.C. : $ u_x(0,t) = h(t) $

$F'(-ct) + G'(ct) = h(t)$

$F'(z) + G'(-z) = h(-z/c)$

$F(z) - G(-z) = \int_{0}^{z} h(-s/c)ds$

$F(x-ct) - G(ct-x) = \int_{0}^{x-ct} h(-s/c)ds$

$F(x-ct) = \int_{0}^{x-ct} h(-s/c)ds + G(ct-x)$

SO:

$u(x,t) = \int_{0}^{x-ct} h(-s/c)ds + G(ct-x) + G(x+ct)$

But we know that $G(ct-x), G(x+ct) =0$ from IC - I think.

So we get:

$u(x,t) = \int_{0}^{x-ct} h(-s/c)ds$

This solution satisfy BC: $ u_x(0,t) = h(t) $

$u_x(x,t) = h(\frac{ct-x}{c})$

$u_x(0,t) = h(t)$ <- Yes?

But it does not satisfy the IC?

Is the answer u(x,t) = 0?

Been at this some time - and the folks over at free math help forum weren't able to assist to get a final answer. Thought I would have better luck here.

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sci-guy
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1 Answers1

4

First of all, you should distinguish the spacetime regions $x\le ct$ (boundary layer) and $x>ct$. When $x> ct$, the influence of the boundary is not felt, and the solution is zero due to the initial condition being zero. From now on assume $x\le ct$.

The formula $$u(x,t) = \int_{0}^{x-ct} h(-s/c)ds + G(ct-x) + G(x+ct)$$ is correct, although the integral is awkwardly written: the upper limit is negative. I would introduce $H(t)=\int_0^t h(s)\,ds$ and write $$ \int_{0}^{x-ct} h(-s/c)ds = -c \int_0^{t-x/c} h(\xi)\,d\xi = -c H(t-x/c) $$

So far, $$u(x,t) = -c H(t-x/c) + G(ct-x) + G(x+ct)\tag{1}$$

But we know that $G(ct-x), G(x+ct) =0$ from IC

Yes. Note that plugging $t=0$ here is problematic because we work in the regime $x\le ct$. It's better to plug $x=ct$, where (by continuity) $u=0$. So, $$ 0 = G(0) + G(2x)$$ hence $G$ is identically zero.

The final answer: $$ u(x,t) = \begin{cases} -cH(t-x/c),\quad &0\le x\le ct; \\ 0,\quad & x>ct\end{cases} $$

Verification: in the boundary layer $x< ct$ we have $u_x(x,t) = h(t-x/c)$, which at $x=0$ gives $h(t)$ as required. The initial conditions are also satisfied, and the function is continuous.