difficulty proving this theorem

Question

Let $x = s + t$ and $y= s - t$. For any $f(x, y)$, let us define this function in terms of s and t in the usual fashion: $g(s, t): = f(x(s, t), y(s, t))$. Show that $$ \left(\frac {\partial f} {\partial x}\right)^2 - \left(\frac {\partial f}{\partial y} \right)^2 = \left(\frac {\partial g} {\partial s}\right) \left( \frac {\partial g} {\partial t}\right) $$

Having a little trouble figuring this problem out. I've tried a few things, like finding the partial derivatives $f_x$ and $f_y$ in terms of $g_s, g_t, s_x, s_y, t_x,$ and $t_y$, but nothing quite cancels out for me to prove the result on the right-hand side. Any thoughts/guidance on how to approach this?

Answer 1

Is actually not that difficult. Start from the right side, you will notice that $$ \frac{\partial g}{\partial s} = \frac{\partial f}{\partial x} +\frac{\partial f}{\partial y} $$ Do you know how to write the derivative of a composite function right? Then $$ \frac{\partial g}{\partial t} = \frac{\partial f}{\partial x} -\frac{\partial f}{\partial y} $$ Now multipliy the two and you will get the left term, remembering that $(a+b)(a-b)=a^2-b^2$.

EDIT: you write for any function... Of course it should be for any function for which you can calculate the derivatives...

Answer 2

Consider the RHS of the result you are trying to show, in particular let's look at the first term $ \frac{\partial g}{\partial s}$. This partial derivative denotes the change in the function $g$ when $s$ is varied and $t$ is held constant. How does $g$ depend on the variable $s$? It depends on it through the relations $x(s,t)$ and $y(s,t)$ since $g(s,t) = f(x(s,t), y(s,t))$. Thus, the partial derivative in question has two terms, corresponding to the effect changing $s$ has on both $x$ and $y$, and the resulting effect of the changes in $x$ and $y$ on $f$:

$ \left(\frac{\partial g(s,t)}{\partial s}\right)_t = \left(\frac{\partial f(x(s,t), y(s,t))}{\partial s}\right)_t = \left(\frac{\partial f(x,y)}{\partial x}\right)_y \left(\frac{\partial x(s,t)}{\partial t}\right)_s + \left(\frac{\partial f(x,y)}{\partial y}\right)_x \left(\frac{\partial y(s,t)}{\partial s}\right)_t $

$= \left(\frac{\partial f(x,y)}{\partial x}\right)_y + \left(\frac{\partial f(x,y)}{\partial y}\right)_x $

where the subscripts denote the variable being held constant with each partial derivative. This is a multidimensional chain rule, which you should be able to prove from the definitions of partial derivatives, if you so desire. The last line follows from the definitions $x = s + t$ and $y = s - t$ so that the partial derivative of each of these with $s$ at constant $t$ is just 1. If you perform a similar computation for $ \frac{\partial g}{\partial t}$, and take the product indicated on the RHS of your original equation, you should get the desired result.