# Double differentiation of partial equations

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$$in terms of the partial derivatives of f with respect to u and v. With the first question I found \frac{\partial f}{\partial x} to be \frac{\partial f}{\partial u}+\frac{\partial f}{\partial v} and \frac{\partial f}{\partial y} to be \frac{\partial f}{\partial u}-\frac{\partial f}{\partial v}, after that I'm not quite sure how to progress after this stage to find the double derivatives and go onto the second question (which I'm sure requires answers from q1). ## 1 Answer Firstly I assume you mean$$\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. • Ah thanks that really helps, I assume you do the same thing for$\frac{\partial^2 f}{\partial y^2}\$ – ThatITguy Apr 30 '14 at 13:47
• Yes, then substitute. – user2850514 Apr 30 '14 at 13:47
• Thanks again and thanks for the correction on my wave equation – ThatITguy Apr 30 '14 at 13:49