# Characterisations of second order PDEs by changing the variables.

### Question

If a second order PDE is defined as

a(x, y) uxx + 2b(x, y) uxy + c(x, y)uyy = d(x, y, u, ux, uy)

and the variables are defined as

x, y -> ξ(x,y), η(x,y)

and the transformation is non-singular, how do you show that

ux = uξξx + uηηx

and

uxx = (uξξ ξx + uξη ηxx + uξ ξxx + (uηξ ξx + uηη ηxx + uη ηxx?

I feel like I'm missing something obvious but I just can't seem to wrap my head around how to differentiate ξ and η.

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@doraemonpaul: When doing a massive retagging, please try to edit only a limited amount of questions at a time, say 5 to 10 questions in a day and the rest over subsequent days. Right now the front page is flooded with old PDE questions, and other new questions will not get enough attention. –  Rahul Aug 21 '12 at 1:43

In the new variables, $u$ is a function of $\xi$ and $\eta$, which depend on $x$ and $y$:
$$\begin{matrix} & & u & & \\ & /& & \backslash \\ & \xi & & \eta\\ / & \backslash & & / & \backslash \\ x & y & & x & y \end{matrix}$$
Then the chain rule for functions of several variables gives $$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi}\frac{\partial\xi}{\partial x}+\frac{\partial u}{\partial \eta}\frac{\partial\eta}{\partial x}.$$ Similarly for the derivative with respect to $y$. To find the second derivatives you keep using the chain rule and the product rule.