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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).

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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.

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  • $\begingroup$ Ah thanks that really helps, I assume you do the same thing for $\frac{\partial^2 f}{\partial y^2}$ $\endgroup$ – ThatITguy Apr 30 '14 at 13:47
  • $\begingroup$ Yes, then substitute. $\endgroup$ – user2850514 Apr 30 '14 at 13:47
  • $\begingroup$ Thanks again and thanks for the correction on my wave equation $\endgroup$ – ThatITguy Apr 30 '14 at 13:49

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