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?

  • $\begingroup$ 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 $\endgroup$
    – mmesser314
    Oct 3, 2020 at 17:17
  • $\begingroup$ @mmesser314 Wouldn't that be true only for $b=0$? $\endgroup$ Oct 3, 2020 at 17:18
  • $\begingroup$ see my edited answer. $\endgroup$
    – user45664
    Oct 3, 2020 at 17:35

1 Answer 1


See D'Alembert's solution for the 1D wave equation. It will provide the solution. But you will need two initial conditions: u(x,0) and u'(x,0) since the wave equation is second order. You will find that F=G and that you have an additional term due to u'(x,0). u is the initial displacement at t=0 and u'(x,0) is the initial speed of displacement at t=0.

See my https://www.researchgate.net/publication/340085346

which shows what is required to cancel the backward wave.


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