From $\delta_tu+a\delta_xu=0 $ , show that $\delta_{tt}u=a^2 \delta_{xx}u$.

Question

Let $\delta_tu+a\delta_xu=0 $ with $a$ being a constant. Show that $\delta_{tt}u=a^2 \delta_{xx}u$.

Very simple question but because I am unfamiliar with partial derivatives I am having a little bit of trouble.

Answer 1

Since $\delta_t u = -a\delta_x u$, then $$\delta_{tt}u = -a\delta_t\delta_x u = -a\delta_x \delta_t u = \cdots$$

Finish the argument using the equation $\delta_t u = -a\delta_x u$ once more.

Answer 2

Notation: $u_s=\delta_su=\partial u/\partial_s$

First take $\partial/\partial_t$, so you have: $$u_{tt}+au_{tx}=0$$ so $$u_{tt}=-au_{tx}$$ Now take $\partial/\partial_x$ to the original equation: $$u_{tx}+au_{xx}=0$$ multiply the last equation by $a$ and you will have that $$-au_{tx}=a^2u_{xx}$$ So we conclude what we wanted $$u_{tt}=a^2u_{xx}$$