I am trying to solve the following partial differential equation.

\begin{align} u_{tt}&=u_{xx} &t>0, x>0, \\ u_x(0,t) &= b\cdot u_t(0,t) &t \geq 0, \\ u(x,0) &= f(x) &x>0,\\ u_t(x,0) &= 0 &x>0.\end{align} with $b$ a non-negative constant.

This is the wave-equation on a semi-infinite interval. I want to use the fact that the general solution is of the form $u(x,t)= F(x-ct) + G(x+ct)$. With $F(x) = \frac{1}{2}f(x)-\int_0^xg(s)\ \text{d}s$ and $G(x) = \frac{1}{2}f(x)+\int_0^xg(s)\ \text{d}s$ (note: $c=1$, follows from the given PDE.). We note that this would imply that $u(x,t) = \frac{1}{2}[f(x+t)-f(x-t)]$. However, as $x-t$ can take on negative values and our $f$ is only defined for positive input, we need to take a closer look.

I thought of using the boundary condition to find a formula for negative inputs of $f$. \begin{align} u_x(0,t) = \frac{1}{2}[f'(t)-f'(-t)]&=\frac{b}{2}[f'(-t)+f'(t)] = b\cdot u_t(0,t), \\ f'(t)-b\cdot f'(t) &= f'(-t) + b\cdot f'(-t), \\ f'(-t) &= \frac{1}{1+b}[f'(t)-b \cdot f'(t)]. \end{align}

I don't think this is correct, since in the second line of the above equation both extending $f(t)$ odd or even doesn't lead to a correct equality (only if $b=0$). Also, if what I've done is correct up to this point, how can I get an explicit statement for $f(-t)$ and not $f'(-t)$?

Thanks for your time,

K. Kamal


You work is correct. To get $f$, just integrate: $$ -f(-t)=\frac{1}{1+b}\bigl(\,f(t)-b \cdot f(t)\bigr)+C,\quad t>0, $$ that is $$ f(x)=\frac{b-1}{b+1}\,f(-x)+C,\quad t<0. $$ where $C$ is an arbitrary constant.


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