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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?

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3  
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". –  Nathan Wilson 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 DT. Dec 1 '12 at 3:42

2 Answers 2

up vote 2 down vote accepted

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

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

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

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

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

Therefore:

$\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) $

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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 $f$.

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