Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 have a wave equation:

$$\frac{\partial^2u}{\partial t^2} = a^2 \frac{\partial^2u}{\partial x^2}.$$

How would I verify that the function $u(x,t)=\sin(x-at)$ satisfies the aforementioned wave equation?

share|cite|improve this question
Any function $u(x,t)=f(x-a t)$ does so which you can check easily by just plugging the expression for $u$ into the wave equation. – Fabian Nov 13 '12 at 10:21
I am afraid I don't quite understand what you mean by "plugging the expression for $u$ into the wave equation". – JTJM Nov 13 '12 at 10:26
You were told that $x=2$ satisfies the equation $x^2-2x=0$. How wuld you verify this? Plug $x=2$ into the equation and the L.H.S. (left hand side) = R.H.S. Violà. – Frenzy Li Dec 1 '12 at 3:42
up vote 3 down vote accepted

Just derive your $u(x,t)$ using the chain rule until you get the result:

$\frac{\partial u}{\partial t} = -a\cos(x-at)$

$\frac{\partial^2 u}{\partial t^2} = -a^2\sin(x-at)$

$\frac{\partial u}{\partial x} = \cos(x-at)$

$\frac{\partial^2 u}{\partial x^2} = -\sin(x-at)$


$\frac{\partial^2 u}{\partial t^2} = a^2\frac{\partial^2 u}{\partial x^2} $

$-a^2\sin(x-at) = a^2\left(-\sin(x-at)\right) $

$-a^2\sin(x-at) = -a^2\sin(x-at) $

share|cite|improve this answer

As mentioned in the comments: Plug u into the wave equation, means calculate the second time and space derivatives and see that they are equal.

Left-hand-side: $\partial_{tt} u=-a^2\sin(x-at).$ (Here, we apply the chain-rule twice).

Right-hand-side: $\partial_{xx} u=-\sin(x-at).$ (Here, the "inner" derivative is 1, so we don't have the factor $a^2$).

This means, multiply the rhs by $a^2$ and you get the lhs, the equation is valid.

As Fabian mentioned in the comments, this works for arbitrary twice differentiable functions $f$.

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.