I need some verification on my intution of the solution to the wave equation on the $D=(0,\infty)$ subject to Dirchilet boundary conditions.
So the problem is
$\begin{cases} v_{tt}-c^2v_{xx}=0\quad\quad 0<x<\infty ,\;t>0\\ v(x,0)=\phi (x),\quad v_t(x,0)=\psi (x)\quad\quad x>0\\ v(0,t)=0\quad\quad t>0\;. \end{cases}$.
By the method of reflections, the initial datum $\phi$ and $\psi $ can be reflected to the whole line via odd extension which yields the auxiliary IVP
$u_{tt}-c^2u_{xx}\;,\quad\quad -\infty<x<\infty ,\;t>0$
$u(x,0)=\phi_{odd}(x),\quad u_t(x,0)=\psi _{odd}(x)$
whose solution $u$ is given by d'Alembert's formula.
The solution $v$ of the original problem, is then given by the restriction of $u$ for $x\ge 0$, that is
$v(x,t)=\frac{1}{2}[\phi_{odd}(x+ct)+\phi_{odd}(x-ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi_{odd} (s)\;ds$.
Furthermore, after considering the case for $x>c|t|$ and $x<c|t|$, we can then write the solution $v$ as the piecewise function
$v(x,t)= \begin{cases} \frac{1}{2}[\phi (x+ct)+\phi (x-ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi (s)\;ds\;,\quad\quad x>c|t|\\ \frac{1}{2}[\phi (x+ct)-\phi (ct-x)]+\frac{1}{2c}\int_{ct-x}^{ct+x}\psi (s)\;ds\;,\quad\quad \;x<c|t|. \end{cases}$
From what I understand, if $x>c|t|$, then the initial waveform splits in two and each wave travels as it would on the whole line. i.e, $t$ before the the wave $\phi (x+ct)$ reaches $x=0$.
Now, when $x<c|t|$, the wave $\phi (x+ct)$ reaches $x=0$ and begins to experience an "interference" from the imaginary wave created by the extension of the initial datum.
It this interpretation correct?
Thanks.
Your interpretation is correct. The principal at work here is the method of images. The idea is that the "reflecting" constraint $v(0,t)$ can be eliminated by expanding the domain and adding an initial condition which automatically enforces the constraint. The idea is that by using the odd extension for the auxiliary PDE $u$, you introduce the symmetry $u(x,t)=-u(-x,t)$, which is preserved by the evolution of the PDE. This in particular means that $u(0,t)=0$ without having to impose any constants. You can then "forget" the $x<0$ half of the domain and the inteference from the virtual half can be reinterpreted as the reflection off of the barrier at $x=0$.
For a visualization, notice that the centerpoint of the above animation does not move; by covering up the right half, it will instead look like the right pulse is reflecting off this centerpoint.