Proof functions of the form $f(x + at) + g(x - at)$ satisfies wave equation

Question

I have the following statement to prove:

Prove any $C^2$ function of the form $z=f(x+at)+g(x-at)$ is a solution of wave equation $\frac{\partial ^2z}{\partial t^2} = a^2 \cdot \frac{\partial ^2 z}{\partial x^2}$.

I tried it many times getting the same wrong anwer. My development was:

Let $u(x,t) = x+at, r(x,t) = x-at$

$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial r}\cdot \frac{\partial r}{\partial x}$

$=\frac{\partial z}{\partial u}+\frac{\partial z}{\partial r}$

$\frac{\partial ^2z}{\partial x^2}= \frac{\partial }{\partial x}[\frac{\partial z}{\partial u}] +\frac{\partial z}{\partial x}[\frac{\partial z}{\partial r}]$

$=[\frac{\partial ^2z}{\partial u^2} \cdot \frac{\partial u}{\partial x} +\frac{\partial ^2z}{\partial r \partial u} \cdot \frac{\partial r}{\partial x}] + [\frac{\partial ^2z}{\partial u\partial r} \cdot \frac{\partial u}{\partial x} +\frac{\partial ^2z}{\partial ^2r} \cdot \frac{\partial r}{\partial x}]$

$=[\frac{\partial ^2z}{\partial u^2} +\frac{\partial ^2z}{\partial r \partial u}] + [\frac{\partial ^2z}{\partial u\partial r} +\frac{\partial ^2z}{\partial ^2r} ]$

$=\frac{\partial ^2z}{\partial u^2} +2\frac{\partial ^2z}{\partial r \partial u} +\frac{\partial ^2z}{\partial ^2r} $ (Using Schwarz theorem)

By the other hand, to calculate $\frac{\partial ^2 z}{\partial t^2}$ i got:

$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial t}+\frac{\partial z}{\partial r}\cdot \frac{\partial r}{\partial t}$

$=\frac{\partial z}{\partial u}\cdot a+\frac{\partial z}{\partial r}\cdot (-a)$

$\frac{\partial ^2 z}{\partial t^2} = a\frac{\partial }{\partial t}[\frac{\partial z}{\partial u}] - a\frac{\partial z}{\partial t}[\frac{\partial z}{\partial r}]$

And solving:

$=a\cdot[\frac{\partial ^2z}{\partial u^2} \cdot \frac{\partial u}{\partial t} + \frac{\partial ^2z}{\partial r \partial u} \cdot \frac{\partial r}{\partial t} ]- a[\frac{\partial ^2z}{\partial u \partial r} \cdot \frac{\partial u}{\partial t} + \frac{\partial ^2z}{\partial r^2} \cdot \frac{\partial r}{\partial t}]$

$=a\cdot[\frac{\partial ^2z}{\partial u^2} \cdot a + \frac{\partial ^2z}{\partial r \partial u} \cdot (-a) ] - a[\frac{\partial ^2z}{\partial u \partial r} \cdot a + \frac{\partial ^2z}{\partial r^2} \cdot (-a)]$

$=a^2\frac{\partial ^2z}{\partial u^2} -2a^2\frac{\partial ^2z}{\partial r \partial u} + a^2\frac{\partial ^2z}{\partial ^2r} $

$=a^2[\frac{\partial ^2z}{\partial u^2} -2\frac{\partial ^2z}{\partial r \partial u} + \frac{\partial ^2z}{\partial ^2r} ]$

but the sign of the term $-2a^2\frac{\partial ^2z}{\partial r \partial u}$ does not match, so where is my mistake?

Answer 1

Your computation is correct, but it obscures the important fact that:

$$\frac{\partial^2z}{\partial r\partial u}=0.$$

Your computation works for any: $$z=h(u,r),$$ rather than the specific form of $z$ in the question.

Since $z=f(u)+g(r),$ $\partial z/\partial u=f’(u).$ And the derivative of this with respect to $r$ is $0.$

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You only have to prove it for $z=f(x+at).$

If $z_1,z_2$ satisfy the equation, then so does $z_1+z_2.$

And from the case $f(x+at),$ we get $g(x-at)=g(x+(-a)t)$ satisfies the wave equation for $-a,$ but the wave equation does not depend on the sign of $a.$