The wave equation with a boundary condition that turns on at $t=0$

Consider the wave equation in 1 spatial dimension (in units where $$c=1$$) $$\frac{\partial^2 u(t,x)}{\partial t^2} - \frac{\partial^2 u(t,x)}{\partial x^2} = 0 \ .$$ Now suppose that the solution is $$u(t,x)=0$$ for all $$t<0$$. Then suddenly at $$t=0$$ a boundary condition gets enforced at $$x=x_0$$ $$u(x_0, t) = \Theta(t) b(t,x_0) \ ,$$ for some function $$b$$.

Can progress be made towards a solution for such a boundary condition? For concreteness, I would like to consider the behaviour of the solution $$u(t,x)$$ for $$t>0$$ and $$x>x_0$$ (only to the right of the boundary condition).

For $$t>0$$ and $$x>x_0$$ the solution should have the standard form $$u(t,x) = F(x - t) + G(x + t)$$ for some functions $$F$$ and $$G$$.

My guess is that $$G=0$$ should be true for this set-up (ie. no incoming wave for $$x>x_0$$), since the disturbance caused by the boundary condition should probably cause only an outgoing wave (described just by $$F$$, a function of the retarded time $$t-x$$).

My second guess is that $$F$$ is probably proportional to a $$\Theta(t - x)$$ for $$x>x_0$$, since the outgoing wave needs to take some time to arrive at the point $$x >x_0$$

Is this true? and How to see this?

• Not sure what you have in mind for t<0. if $u(x,t) = 0$ for all x when t<0, then $u(x,t) = 0$ for all x and t – mmesser314 Oct 3 '20 at 17:17
• @mmesser314 Wouldn't that be true only for $b=0$? – QuantumEyedea Oct 3 '20 at 17:18
• see my edited answer. – user45664 Oct 3 '20 at 17:35