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?


  • $\begingroup$ Just out of curiosity, where did you take the discussion from? $\endgroup$
    – PtF
    Jul 2 '20 at 0:24
  • 1
    $\begingroup$ Hello, thanks for the comment. I'm currently reading An Introduction to Partial Differential Equations by Walter A. Strauss chapter 3.2. This is simply some exposition followed by my interpretation of the behaviour of the solution. $\endgroup$
    – whitenoise
    Jul 2 '20 at 0:28

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 $x=0$.

enter image description here

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.

  • $\begingroup$ Thanks you so much for your comment. Its quite the interesting phenomenon. Btw, what program did you use to create the animation? $\endgroup$
    – whitenoise
    Jul 2 '20 at 3:13
  • 1
    $\begingroup$ You're welcome. The gif was from a google image search; I haven't the slightest idea how to create one. $\endgroup$
    – Kajelad
    Jul 2 '20 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.