# Solving a PDE wave equation with initial conditions and boundaries (from Strauss's PDE, exercise 3.2.6)

The problem is:

Solve $u_{tt}=c^2u_{xx}$ in $0<x<\infty, 0\leq t<\infty,u(x,0)=0,u_t(x,0)=V$, $u_t(0,t)+au_x(0,t)=0$.

So I get how $u(x,t)=tV$ for $0<ct<x$. However, I can't seem to get the correct solution in back of book that for $0<x<ct, u(x,t)=(at-x)V/(a-c)$.

Attempt at solution:

The book gives the general formula for the wave equation as $u(x,t)=\frac{1}{2}[\phi (x+ct) + \phi (x-ct)] + \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s)ds$, where $u(x,0)=\phi, u_t(x,0)=\psi$. So we used some reflection method in which we extend the function $u$ so it covers all of the t-x plane. From the equation above, $\phi = 0$ so we are left with $u(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s)ds$. Now for $0<x<c|t|$, this becomes $u(x,t) = \frac{1}{2c}\int_{0}^{x+ct} V ds + \frac{1}{2c}\int_{x-ct}^{0} -V ds = Vx/c$ Plugging this into the boundary condition, I get that $u(x,t) = (x-at)V/c$. I am very close but somehow messing up. Can someone tell me where I went wrong?

Thanks.

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So we used some reflection method in which we extend the function

But did you use the right extension method? You extended $\psi$ so that $\psi(-t)=-\psi(t)$. This ensures $u_t(0,t)=0$ for all time, but your wanted $u_t+au_x=0$. Setting $\psi=-V$ in the negative half-axis is not the right extension of $\psi$.

$u(x,t) = \frac{1}{2c}\int_{0}^{x+ct} V ds + \frac{1}{2c}\int_{x-ct}^{0} -V ds = Vx/c$. Plugging this into the boundary condition, I get that $u(x,t) = (x-at)V/c$

This does not make sense to me. Once you decided what $\psi$ is at negative values, your function $u$ is determined. You can plug it into the boundary condition just to see that it fails them.

Let's start from the basic principles: differentiating the formula $$u(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s)ds \tag1$$ we get \begin{align} u_x(x,t) &= \frac{1}{2c} (\psi(x+ct)-\psi(x-ct)) \\ u_t(x,t) &= \frac{1}{2c} (c\psi(x+ct)+c\psi(x-ct)) \end{align} \tag2 hence $$au_x(0,t) + u_t(0,t) = \left(\frac12+\frac{a}{2c}\right) \psi(ct)+ \left(\frac12-\frac{a}{2c}\right)\psi(-ct) \tag3$$ Equating (3) to zero and recalling that $\psi(ct)=V$ for all positive $t$, you will find the value of $\psi$ for all negative $t$. It will be some multiple of $V$, but not necessarily $-V$.

Finally, return to (1) to calculate $u$.

It is worthwhile to ponder the physical implication of the failure of this method for $a=c$.

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