# change of variable

• Oct 28th 2008, 03:06 AM
marianne
change of variable
I'm learning about the second order PDEs, and I've come across the transformation of variables I don't understand.

We have $u=u(x, y)$ and a, b, c functions of class $C^2$, which are not at the same time all equal to zero.
The equation in question is:
(*) $a \frac{\partial^2 u}{\partial x^2} + 2b \frac{\partial^2 u}{\partial x \partial y} +c \frac{\partial ^2 u}{\partial y^2} + f(x, u, \nabla u)=0$
We introduce the change of variables:
$\xi=x+y$, $\eta=x-y$

How does the equation (*) now looks like?
I've tried googling change of variables, but I'm still lost.

Thank you.
• Oct 28th 2008, 05:41 AM
HallsofIvy
By the chain rule, $\frac{\partial f}{\partial x}= \frac{\partial f}{\partial\xi}\frac{\partial xi}{\partial x}+ \frac{\partial f}{\partial\eta}\frac{\partial\eta}{\partial x}$
Here, it is easy to see that $\frac{\partial\xi}{\partial x}= 1$ and $\frac{\partial\eta}{\partial x}= 1$ so $\frac{\partial f}{\partial x}= \frac{\partial f}{\partial\xi}+ \frac{\partial f}{\partial\eta}$.

Similarly, $\frac{\partial f}{\partial y}= \frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}$

You find the second derivatives by using those in succession:
$\frac{\partial^2 f}{\partial x^2}= \frac{\partial \left(\frac{\partial f}{\partial\xi}+ \frac{\partial f}{\partial\eta}\right)}{\partial\xi}+ \frac{\partial\left(\frac{\partial f}{\partial\xi}+ \frac{\partial f}{\partial\eta}\right)}{\partial\eta}$.

$\frac{\partial^2 f}{\partial y^2}= \frac{\partial\left(\frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}\right)}{\partial\xi}- \frac{\partial \left(\frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}\right)}{\partial\eta}$

and finally
$\frac{\partial^2 f}{\partial x\partial y}= \frac{\partial \left(\frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}\right)}{\partial\xi}+ \frac{\partial\left(\frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}\right)}{\partial\eta}$

Of course, since $\xi= x+ y$ and $\eta= x- y$, adding those equations, $2x= \xi+ \eta$ and so $x= \frac{\xi- \eta}{2}$. Use that on the right hand side of the equation.