Double differentiation of partial equations

Question

Here's the question I am trying to do below:
Let f = f(u,v) and u = x + y, v = x - y.
1. Assuming that f is twice differentiable, compute $f_x$$_x$ and $f_y$$_y$ in terms of $f_u$, $f_v$, $f_u$$_u$, $f_u$$_v$, $f_v$$_v$
2. Express the wave equation $$ \frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} = 0$$
in terms of the partial derivatives of f with respect to u and v.

With the first question I found $\frac{\partial f}{\partial x}$ to be $\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}$ and $\frac{\partial f}{\partial y}$ to be $\frac{\partial f}{\partial u}-\frac{\partial f}{\partial v}$, after that I'm not quite sure how to progress after this stage to find the double derivatives and go onto the second question (which I'm sure requires answers from q1).

Answer 1

Firstly I assume you mean $$\frac{\partial^{2}f}{\partial x^{2}} - \frac{\partial^{2}f}{\partial y^{2}} = 0$$ since what you have for part 2 is obvious.

Ok So you have $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} = f_{u} + f_{v}$$

Then $$\frac{\partial^{2}f}{\partial x^{2}} = \frac{\partial}{\partial x}\left[\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}\right] = \frac{\partial}{\partial x}\frac{\partial f}{\partial u} + \frac{\partial}{\partial x}\frac{\partial f}{\partial v} = \frac{\partial}{\partial u}\frac{\partial u}{\partial x}\frac{\partial f}{\partial u} + \frac{\partial}{\partial v}\frac{\partial v}{\partial x}\frac{\partial f}{\partial v}$$ $$= \frac{\partial^{2}f}{\partial u^{2}}\frac{\partial u}{\partial x} + \frac{\partial^{2}f}{\partial v^{2}}\frac{\partial v}{\partial x}$$

We know $\frac{\partial u}{\partial x}$ and $\frac{\partial v}{\partial x}$ so substituting this gives:

$$\frac{\partial^{2}f}{\partial x^{2}} = \frac{\partial^{2}f}{\partial u^{2}}+ \frac{\partial^{2}f}{\partial v^{2}} = f_{uu} + f_{vv}$$

I'll let you finish off.