# change of variable in a PDE

• Sep 8th 2008, 08:51 AM
chopet
change of variable in a PDE
I'm having some troubles figuring this out:

Given g( x(u,v) , y(u,v) ) = h( u,v),
and that
$\displaystyle u = {\sigma^2 \over 2} (T-x)$
$\displaystyle v = ln {y \over k} - (0.5 \sigma^2 -r)(T-x)$

how do we find $\displaystyle {\partial g \over \partial x}$
and $\displaystyle {\partial g \over \partial y}$?
• Sep 8th 2008, 08:58 AM
chopet
can anyone explain how is it that:
$\displaystyle {\partial g \over \partial x} = - {\sigma^2 \over 2} {\partial h \over \partial u} - (r - 0.5 \sigma^2) { \partial h \over \partial v}$

my guess is:
$\displaystyle {\partial g \over \partial x} = ({\partial \over \partial u}h){\partial u \over \partial x} + ({\partial \over \partial v}h){\partial v \over \partial x}$

normally, we have $\displaystyle {\partial \over \partial u}h$ as the solution, but this time round, we have $\displaystyle {\partial u \over \partial x}$ instead.
Can someone please verify if my reasoning is correct?
• Sep 8th 2008, 10:43 AM
shawsend
If $\displaystyle g=h$ and $\displaystyle h=f_1(u,v)$ with $\displaystyle u=f_2(x,y)$ and $\displaystyle v=f_3(x,y)$, then using the general chain rule:

$\displaystyle \frac{\partial}{\partial x} g=\frac{\partial}{\partial x} h$

and:

$\displaystyle \frac{\partial}{\partial x} h=\frac{\partial h}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial h}{\partial v}\frac{\partial v}{\partial x}$