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).


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.