Showing that $v$ also satisfied the wave equation

Question

Suppose $u(x,t)$ satisfies the wave equation $$u_{tt}-c^2u_{xx}=0$$ for some $c \in \mathbb{R}$. Let $\alpha \in \mathbb{R}$, $\alpha \ne 0$ be a constant, and define $$v(x,t) = u(\alpha x, \alpha t)$$ Show that $v$ also satisfies the wave equation.

Answer 1

I believe this type of transformation is called dilation. By the chain rule, $$v_{tt}(x,t) = \alpha^2u_{tt}(\alpha x,\alpha t)$$ and similarly, $$v_{xx}(x,t) = \alpha^2u_{xx}(\alpha x,\alpha t).$$ It follows that $$ v_{tt}(x,t) - c^2v_{xx}(x,t) = \alpha^2(u_{tt}(\alpha x,\alpha t) - c^2u_{xx}(\alpha x, \alpha t)) = \alpha^2\cdot0 = 0. $$ Hence, $v$ satisfies the wave equation. Having seen this, it may be good practice to show that $v(x,t) = u(x - \alpha,t)$ also satisfies the wave equation.