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


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.

  • 8
    $\begingroup$ @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. $\endgroup$ – Rahul Aug 21 '12 at 1:43

It is just the chain rule.

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.


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.