Proving $z''_u+z''_v=(x^2+y^2)(z''_x+z''_y)$

Question

If $x=e^u\cos(v)$, $y=e^u\sin(v)$, $z=f(x,y)$, prove that $$z''_u+z''_v=(x^2+y^2)(z''_x+z''_y).$$

It seems simple at first sight, but it's too much answering, so I think I did not notice the point.

So any idea?

Answer 1

It is a fairly straightforward exercise in multivariable chain rule, and that is likely the intended solution.

A non-intended solution is to declare $Z=x+iy$ and $W=u+iv$, so that $Z=g(W)=e^{W}$ and using Wirthinger derivatives, and complex Laplacian we get for holomorphic $g$:

$$\Delta_W f(g(W)) = \left|\frac{\partial }{\partial W}g\right|^2 \Delta_Z f(Z)$$

which in this case gives $z''_u+z''_v=|e^W|^2(z''_x+z''_y)=(x^2+y^2)(z''_x+z''_y)$ as wanted.