d'Alemberts's solution on a semi infinite domain with a strange boundary condition

Question

I want to use d'Alembert's solution of the wave equation to find the solution of $$\frac{\partial ^2 u}{\partial x^2} = \frac{\partial ^2 u}{\partial t^2} \qquad (0\leq t, 0\leq x)$$

$$\frac{\partial u}{\partial t}(0,t) = \alpha\frac{\partial u}{\partial x}(0,t) \qquad (0 \leq t)$$

$$ u(x,0) = \phi_0(x) $$ $$ \frac{\partial u}{\partial t}(x,0) = \phi_1(x) $$

for all $\alpha \neq -1$. I've been substituting $u(x,t) = f(x-t) + g(x+t)$ in an attempt to solve for $f$ and $g$ in terms of the initial conditions, but this doesn't seem to be going anywhere. What would be a good way to approach this problem? For example, would it make any sense to allow $f(-x)= -f(x)$, and $g(-x)= -g(x)$?

Answer 1

d'Alembert Solution reads

$$u(x,t) =\frac{\phi_{0}(x+t)+\phi_{0}(x-t)}{2}+ \frac{1}{2} \int_{x-t}^{x+t} \phi_{1}(\xi) d\xi$$

which satisfies the initial conditions.

Now $$u_{x}(x,t) =\frac{\phi_{0}'(x+t)+\phi_{0}'(x-t)}{2}+ \frac{1}{2} [\phi_{1}(x+t)-\phi_{1}(x-t)]$$

$$u_{t}(x,t) =\frac{\phi_{0}'(x+t)-\phi_{0}'(x-t)}{2}+ \frac{1}{2} [\phi_{1}(x+t)+\phi_{1}(x-t)]$$

Boundary conditions: $$u_{x}(0,t) =\frac{\phi_{0}'(t)+\phi_{0}'(-t)}{2}+ \frac{1}{2} [\phi_{1}(t)-\phi_{1}(-t)]$$

$$u_{t}(0,t) =\frac{\phi_{0}'(t)-\phi_{0}'(-t)}{2}+ \frac{1}{2} [\phi_{1}(t)+\phi_{1}(-t)]$$

The BCs are not necessarily satisfied.

If there's freedom of $\phi_{0}$ and $\phi_{1}$, then

$$\phi_{0}'(t)-\phi_{0}'(-t)+\phi_{1}(t)+\phi_{1}(-t) =\alpha [\phi_{0}'(t)+\phi_{0}'(-t)+\phi_{1}(t)-\phi_{1}(-t)]$$

Since these functions are defined when $x \geq 0$ we can say that $-\phi_{i}(-t) = \phi_{i}(t)$ by an odd extension of our domain and go from there.

Hence, in this case:

$$\phi_{0}'(t)=\alpha \phi_{1}(t)$$