# Partial Differential problem

1. Nov 23, 2011

### 54stickers

If $z = f(x,y)$ and $x = e^{u}, y =e^{v}$ Prove:

$x^{2}\frac{\partial^{2}z}{\partial x^{2}} + y^{2}\frac{\partial^{2}z}{\partial y^{2}} + x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{\partial^{2}z}{\partial u^{2}} + \frac{\partial^{2}z}{\partial v^{2}}$

I used $u = ln(x), v = ln(y)$ and the following partial differential set ups:

$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x}$
$\frac{\partial z}{\partial y} = \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}$
$\frac{\partial z}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u}$
$\frac{\partial z}{\partial u} = \frac{\partial f}{\partial y} \frac{\partial y}{\partial v}$

Then:

$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{1}{x}$
$\frac{\partial z}{\partial y} = \frac{\partial f}{\partial v} \frac{1}{y}$
$\frac{\partial f}{\partial u} = \frac{\partial z}{\partial u} e^{u}$
$\frac{\partial f}{\partial v} = \frac{\partial z}{\partial v} e^{v}$

For the second derivatives:

$\frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} f}{\partial u^{2}} \frac{-1}{x^{2}}$
$\frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial^{2} f}{\partial v^{2}} \frac{-1}{y^{2}}$
$\frac{\partial^{2} z}{\partial u^{2}} = \frac{\partial^{2} f}{\partial x^{2}} e^{u}$
$\frac{\partial^{2} z}{\partial v^{2}} = \frac{\partial^{2} f}{\partial y^{2}} e^{v}$

I cannot get these terms to equal the statement that is to be proved. Am I doing something wrong?

Thanks!