Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working with the chain rule in multivariable calculus and I am having some difficulty with the question:

Let $f,g : \Bbb R \longrightarrow \Bbb R$ where $f$ and $g$ are twice differentiable. Show that $u(x,t)=f(x-at)+g(x+at)$ is a solution to the wave equation $u_{tt} = a^2 u_{xx}$.

Any help would be greatly appreciated! Thanks!

share|cite|improve this question

Let $\phi(x,t) = f(x-at)$. Then $\phi_{tt}(x,t) = a^2 f_{xx}(x-at)$ and $\phi_{xx}(x,t) = f_{xx}(x,t)$. Hence $\phi_{tt} = a^2 \phi_{xx}$, so $\phi$ satisfies the wave equation.

Now let $\eta(x,t) = g(x+at)$, this gives $\eta_{tt}(x,t) = a^2 g_{xx}(x,t)$ and $\eta_{xx}(x,t) = g_{xx}(x,t)$. Hence $\eta_{tt} = a^2 \eta_{xx}$, so $\eta$ satisfies the wave equation.

Since the wave equation is linear, it follows that $u =\phi+\eta$ also satisfies the wave equation.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.