If we have a travelling wave solution $y(\xi)$ and translate it to $y(\xi-\xi_0)$, this is again a travelling wave.
It is claimed the following:
If we linearize along a travelling wave, the associated Jacobian therefore has eigenvalue $0$. This is because of the fact that a translation of a travelling wave is again a travelling wave.
Why is this the case? Do not see why the Jacobian has to have eigenvalue $0$.
Could you please explain that to me?