What does "natural" symmetry of the wave equation even mean?
Given a one dimensional wave equation $$\frac{1}{c^2}\frac{\partial ^2 \phi}{\partial t^2}-\frac{\partial ^2 \phi}{\partial x^2}=0$$
I was told to begin with a linear coordinate transformation
Solve for three of the four variables in the matrix above by enforcing that such a transformation is a symmetry of the wave equation. The final variable parameterizes the transformation. What are the allowed values that all parameters (a, b, g, and h) can take?
Also assume $$a=\frac{1}{\sqrt{1-u^2}}$$ for $u \in (-1,1)$ is valid. In what limit for $u$ does the transformation approach this identity? Fix $u$ so that this new transformation reproduces the Galilean transformation in this limit?
I dont understand thee question at all. What does it mean for the wave equation too be symmetric here`?
A symmetry of a PDE is a transformation of the variables (in this case $(x, t) \to (x', t')$) such that when expressed in terms of the new variables, you get an equation equivalent to the original one. For example, $x' = 2 x$ and $t' = 2 t$ would produce $$ \dfrac{1}{c^2} \dfrac{\partial^2 u}{\partial t'^2} - \dfrac{\partial^2 u}{\partial x'^2} = \frac{1}{4c^2} \dfrac{\partial^2 u}{\partial t^2} - \frac{1}{4} \dfrac{\partial^2 u}{\partial x^2} = 0$$