Confusion about notation for reflected travelling wave solutions

Question

I have a rather stupid question, I guess.

For some PDE, it is said that there exists a right-moving travelling wave solution $u(x,t)=\phi(\xi)$ with $\xi=\pm x-ct, c>0$ and $\lim_{\xi\to-\infty}\phi(\xi)=2\pi, \lim_{\xi\to\infty}\phi(\xi)=0$. It is also said that, upon reflection, there is a left-moving travelling wave solution.

I am a bit confused about two things.

1.) What about the $\pm$-sign in $\xi=\pm x-ct, c>0$? Of course, if I define $\xi=x-ct$ for $c>0$, then $\phi(\xi)$ is right-moving and if I assume that for this $\xi$, $\lim_{\xi\to-\infty}\phi(\xi)=2\pi$ and $\lim_{\xi\to\infty}\phi(\xi)=0$, then isn't $\phi(-x-ct)$ still right-moving with the only difference that it is now an increasing solution?

2.) What is meant with the reflection? For a left-moving solution $\psi$, I need $\psi(\eta)$ with $\eta=x+ct$ and $c>0$. In which sense is this a reflection of $\phi(\pm x-ct)$?

So if I got it right, for $c>0$, both $\phi(x-ct)$ and $\phi(-x-ct)$ are right-moving, with the difference that $\phi(x-ct)$ is decreasing, while $\phi(-x-ct)$ is increasing (or the other way round, depending on which of both we define to be increasing). — selector, Jul 31 '21 at 10:42

EditPiAf · Answer 1 · 2021-08-03T07:55:43.500

1

You can think about the direction of propagation as follows. Consider that the solution takes the form $u(x,t) = \phi(\pm x - c t)$ at some time $t$. Now, we increase time from $t$ to $t + \Delta t$ with $\Delta t>0$, which gives us the waveform \begin{aligned} u(x, t+\Delta t) &= \phi(\pm x - c (t + \Delta t)) \\ &= \phi(\pm (x \mp c \Delta t) - c t) \\ &= u(x \mp c \Delta t, t) . \end{aligned} In other words, the value of $u$ at $(x, t+\Delta t)$ is the same as that at abscissa $x \mp c \Delta t$ and earlier time $t$. Given that $c\Delta t>0$, this shows that information is traveling towards increasing $x$ with increasing times if the upper sign is used $(\xi = x-ct)$, and that information is traveling towards decreasing $x$ otherwise $(\xi = -x-ct)$ -- see also this article for complements. Note that the left-going wave may not simply be a reflection of the right-going wave. In fact, if $u(x,t) = \phi(x-ct)$ solves the PDE problem $$ F(u, u_x, u_t, u_{xx}, \dots) = 0 $$ with suitable boundary conditions at infinity, then its reflection $u(y,t)$ where $y=-x$ solves $$ F(u, -u_y, u_t, u_{yy}, \dots) = 0 $$ with reflected boundaries.
Setting $\psi(\eta) = \phi(-\eta)$ gives you $\psi(x+ct) = \phi(-x-ct)$.

edited Aug 03 '21 at 07:55

answered Aug 03 '21 at 07:42

EditPiAf

20,898

Suppose $\phi(\xi)$ with $\xi=x-ct, c>0$ is a solution of $F(u,u_t,u_{xx})=0$ with $\lim_{\xi\to-\infty}\phi(\xi)=a$ and $\lim_{\xi\to\infty}\phi(\xi)=b$. This is right-moving. Moreoever, let $\psi(\eta), \eta=x+ct, c>0$ a solution of the same PDE but with $\lim_{\eta\to-\infty}\psi(\eta)=b$ and $\lim_{\eta\to\infty}\psi(\eta)=a$. This is a left-moving solution. - - In which sense is $\psi(\eta)$ a reflection of $\phi(\xi)$? I think it is not a reflection at all. – selector Aug 07 '21 at 15:27
To give any sense to the statement on reflection, the only way is see is the following: Suppose that $\tilde{\phi}(\tilde{\xi}), \tilde{\xi}=-x-ct, c>0$ is a solution of the PDE $F(u,u_t,u_{xx})=0$ with boundary values $\lim_{\tilde{\xi}\to -\infty}\tilde{\phi}(\tilde{\xi})=a$ and $\lim_{\tilde{\xi}\to \infty}\tilde{\phi}(\tilde{\xi})=b$. Then, $\psi(\eta)$ as in the last comment is a reflection of $\tilde{\phi}(\tilde{\xi})$; particularly, both solve the PDE and both are left-moving. – selector Aug 07 '21 at 15:31
@selector When speaking of reflection, it needs to be clear whether the image is taken for the spatial coordinate $x \mapsto {-x}$ or for the retarded coordinate $\xi \mapsto -\xi$ with $\xi = x-ct$. In my answer, I point out the fact that these $\pm$ solutions are not necessarily images through the reflection $x \mapsto {-x}$. – EditPiAf Aug 09 '21 at 06:17

Confusion about notation for reflected travelling wave solutions

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