Why are translates of travelling waves again travelling waves? 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?_

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.