Symmetry of PDEs

Question

I am trying to solve the following problem:

Show that $$(u,t,x) \rightarrow (u, \frac{t}{c^2t^2 - x^2}, \frac{x}{c^2t^2 - x^2})$$ is a symmetry of the Wave equation $u_{tt} - c^2u_{xx} = 0$.

Some help to start this problem would be great. First off, what is a symmetry of a PDE? Then what is the general way to show that a certain transformation is in fact a symmetry.

Answer 1

You are being asked to show that if $u_{tt} - c^2u_{xx} = 0$, then the function $$v(t,x) = u\left(\frac{t}{c^2t^2 - x^2}, \frac{x}{c^2 t^2 - x^2}\right)$$

has $v_{tt} - c^2 v_{xx} = 0$. You will need to use the chain rule a few times in order show that $v$ solves the wave equation.

Answer 2

Apply the (symmetry) transformation to a solution and show that the transformed solution solves the PDE too. $$ L u = 0 \Rightarrow L T u = 0 $$

Answer 3

Consider $T: \mathbb R^2 \to \mathbb R^2$ in this case, we say that $T$ is a symmetry of the equation $Du=0$ if $v(x,t)=u(T(x,t))$ is also a solution of the equation.

In this case you must prove that the function $$v(x,t)=u\left( \frac{x}{c^2t^2-x^2}, \frac{t}{c^2t^2-x^2}\right),$$ is also a solution of the equation.

Via Chain rule you can obtain the proof.

Regards