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A travelling wave solution of a PDE or ODE is a solution that depends on the single variable $\xi=x-ct$.

For example consider the PDE $$ u_t=u_{xx}+f(u)-w,~~~w_t=\epsilon (u-\gamma w).~~~~~(1) $$ Then, a travelling wave $(u(\xi), w(\xi)$ satisfies $$ -cu_{\xi}=u_{\xi\xi}+f(u)-w,~~~~~-cw_{\xi}=\epsilon (u-\gamma w).~~~(2) $$

Now: Why is any translate $(u(\xi-\xi_0), w(\xi-\xi_0))$ with $\xi_0\in\mathbb{R}$ a travelling wave, too?

If we express the original PDE (1) not in coordinates $t$ and $x$ but in coordinates $t$ and $\xi=x-ct$, then we get $$ u_t=u_{\xi\xi}+cu_{\xi}+f(u)-w,~~~~~w_t= cw_{\xi}+\epsilon (u-\gamma w).~~~(3) $$

For a travelling wave, we then have. because of (2), $$ 0=u_{\xi\xi}+cu_{\xi}+f(u)-w,~~~~~0=cw_{\xi}+\epsilon (u-\gamma w), $$

hence, a travelling wave is an equilibria solution for (3).

Does this help to argue why any translate of a travelling wave is also a travelling wave?

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  • $\begingroup$ The translate is a travelling wave since the PDE is invariant under shifts $x\to x + c$, $t\to t + d$, i.e. the PDE using the shifted coordinates is the same PDE as the original PDE. $\endgroup$ – Winther Dec 1 '15 at 16:26
  • $\begingroup$ If $(u(\xi),w(\xi))$ is an equilbrium for (3), why then a translate of it, too? $\endgroup$ – M. Meyer Dec 1 '15 at 16:34
  • $\begingroup$ But whats with the terms $f(u)$ and -w, for example in the first equation? $\endgroup$ – M. Meyer Dec 1 '15 at 16:47
  • $\begingroup$ Doesn't this all follow immediately by the fact that a travelling wave is an equilibrium of (3) and hence time independent? So it does not matter if we consider $\xi=x-ct$ or $\xi-k=x-c(t-(k/c))$, i.e. the time $s:=t-(k/c)$ instead of $t$? $\endgroup$ – M. Meyer Dec 1 '15 at 16:54
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    $\begingroup$ The PDE where the time-independent concept comes from is an abstraction. We know that $\zeta = x - ct$, but lets us just consider the PDE $u_t = u_{\zeta\zeta} + \ldots$ like it was any other PDE and $\zeta$ was just a normal spatial coordinate independent of $t$. From this point of view the solution $u$ is time-independent so the solution can be written $u = f(\zeta)$ independent of $t$. However in the picture we started with we have $\zeta = x-ct$ so this means that $u = f(x-ct)$ so the solution of the problem we started with is not time-independent. $\endgroup$ – Winther Dec 1 '15 at 17:41
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If we perform the translation $t\to t - t_0$ and $x\to x - x_0$, using that derivatives are translation invariant $\frac{d}{d(x-x_0)} = \frac{d}{dx}$, we get that the PDEs

$$\matrix{u_t(x,t) &=& u_{xx}(x,t) + f(u(x,t)) - w(x,t)\\ w_t(x,t) &=& \epsilon[u(x,t) - \gamma w(x,t)]}$$

transforms into

$$\matrix{\hat{u}_t(x,t) &=& \hat{u}_{xx}(x,t) + f(\hat{u}(x,t)) - \hat{w}(x,t)\\ \hat{w}_t(x,t) &=& \epsilon[\hat{u}(x,t) - \gamma \hat{w}(x,t)]}$$

where I have taken $\hat{u}(x,t) = u(x-x_0,t-t_0)$ and $\hat{w}(x,t) = w(x-x_0,t-t_0)$. This is exactly the same PDEs as we started with. If $\{u(x,t),w(x,t)\}$ is a solution then so is $\{u(x-x_0,t-t_0)$, $w(x-x_0,t-t_0)\}$. In terms of the $\zeta$ variable this means that if $\{u(\zeta),w(\zeta)\}$ is a solution then (take $\zeta_0 = x_0 - ct_0$) so is $\{u(\zeta-\zeta_0),w(\zeta-\zeta_0)\}$.

In general any PDE/ODE where we have no explicit coordinate dependence have the property of being translation invariant.

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