
Originally Posted by
shawsend
I'm a little unsure. Here goes though. You have:
$\displaystyle \frac{\partial^2 z}{\partial v \partial n}=\frac{1}{2n}\frac{\partial z}{\partial v}$
and if the function is well-behaved, then:
$\displaystyle \frac{\partial^2 z}{\partial n \partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}=\frac{\partial}{\partial n}\frac{\partial z}{\partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}$
Let $\displaystyle \frac{\partial z}{\partial v}=g(n,v)$ then we have $\displaystyle \frac{\partial g}{\partial n}=\frac{1}{2n}g$ which is an ordinary DE in the variable n with v kept constant. Then:
$\displaystyle \frac{dg}{dn}-\frac{1}{2n}g=0$ (**)
Solving via integrating factor and noting that the constant of integration is now an arbitrary function of v, I obtain:
$\displaystyle d\left[n^{1/2}g\right]=0$
and so:
$\displaystyle g(n,v)=n^{-1/2} H(v)$ with $\displaystyle H(v)$ arbitrary.
Now:
$\displaystyle \frac{\partial z}{\partial v}=n^{-1/2} H(v)$
so that:
$\displaystyle z(n,v)=n^{-1/2}\int H(v) dv+G(n)$