I have the equation:
$\frac{\partial{D}}{\partial{\tau}}-(r-\frac{1}{2} \sigma^2)\frac{\partial{D}}{\partial{Z}}-\frac{1}{2}\sigma^2\frac{\partial^2{D}}{\partial{Z}^2}=0$
I'm than asked to change the coordinates to $y=Z + (r-\frac{1}{2}\sigma^2)\tau$ when this is done it results in the heat equation:
$\frac{\partial{D}}{\partial{\tau}}=\frac{1}{2}\sigma^2\frac{\partial^2{D}}{\partial{y}^2}$
I'm not sure how to go about this. Does the change of coordinates use the same method as a substitution?
Hence using the chain rule:
$\frac{\partial{D}}{\partial{Z}}=\frac{\partial{D}}{\partial{y}}.\frac{\partial{y}}{\partial{Z}}$
$\frac{\partial{D}}{\partial{Z}}=\frac{\partial{D}}{\partial{y}}.(r-\frac{1}{2}\sigma^2)\tau$
and
$\frac{\partial^2{D}}{\partial{Z}^2}=\frac{\partial{}}{\partial{Z}}(\frac{\partial{D}}{\partial{y}}.\frac{\partial{y}}{\partial{Z}})$
$\frac{\partial^2{D}}{\partial{Z}^2}=(\frac{\partial^2{D}}{\partial{y}^2}.(\frac{\partial{y}}{\partial{Z}})^2+\frac{\partial{D}}{\partial{y}}.\frac{\partial^2{y}}{\partial{Z}^2})$
$\frac{\partial^2{D}}{\partial{Z}^2}=(\frac{\partial^2{D}}{\partial{y}^2}.((r-\frac{1}{2}\sigma^2)\tau)^2+\frac{\partial{D}}{\partial{y}}.0)$
subbing in
$\frac{\partial{D}}{\partial{\tau}}-(r-\frac{1}{2} \sigma^2)\frac{\partial{D}}{\partial{y}}.(r-\frac{1}{2}\sigma^2)\tau-\frac{\partial^2{D}}{\partial{y}^2}.((r-\frac{1}{2}\sigma^2)\tau)^2=0$
clearly this doesn't give the desired result.
Can anyone explain how to do this correctly?
I'm not even sure I'm using the correct method.