# Homework Help: Multivariable Calculus Chain Rule Problem: Wave equation

1. Nov 12, 2013

### Agent 47

1. The problem statement, all variables and given/known data

Show that any function of the form

$z = f(x + at) + g(x - at)$

is a solution to the wave equation

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

[Hint: Let u = x + at, v = x - at]

2. The attempt at a solution

My problem with this is not that I haven't been able to solve it. The book's solution is right here:

I began to have trouble when I decided not to use f'(u) and g'(v) instead I used $\frac {\partial z} {\partial u}$ and $\frac {\partial z} {\partial v}$

When I did this I got

$\frac {\partial^2 z} {\partial x^2} = \frac {\partial^2 z} {\partial u^2} + \frac {2 \partial^2 z } {\partial u \partial v} + \frac {\partial^2 z} {\partial v^2}$

and

$\frac {\partial^2 z} {\partial t^2} = a^2 (\frac {\partial^2 z} {\partial u^2} - \frac {2 \partial^2 z } {\partial u \partial v} + \frac {\partial^2 z} {\partial v^2})$

And that does not fulfill the condition stated in the beginning.

So I have a two questions in the end:

1)What is the difference between writing f'(u) and $\frac {\partial z} {\partial u}$?

2)Do $\frac {2 \partial^2 z } {\partial u \partial v}$ and $- \frac {2 \partial^2 z } {\partial u \partial v}$ both evaluate to 0? (That's the only way $\frac {\partial^2 z} {\partial t^2} = a^2 \frac {\partial^2 z} {\partial x^2}$) or did I do something wrong in my calculations?

2. Nov 12, 2013

### brmath

One ordinarily would use the notation f'(u) when f is deemed to be a function of a single variable, and ∂f/∂u when f is a function of several variables including u. However, the notation shouldn't make any difference.

Your problem is that when you went for ∂$^2$z/∂x$^2$ you regarded ∂f/∂u as a function of both u and v, which it is not -- it is a function of u alone. Same with v. So you are not going to get any cross terms.

This may be a good reason why ∂f/∂u isn't a good way to notate f'(u).

3. Nov 12, 2013

### Agent 47

Thank you so much. This has been bothering me for a while. So basically I interpreted this as $f(u,v)$ instead of $f(u)$ and $g(v)$ separately. Right?

4. Nov 12, 2013

### brmath

Exactly right.