solution of 2d laplace equation

Question

$G(x,\zeta)=\ln\left(x^2 + \zeta^2 \right)$ is a solution of the equation $\left(\partial_\zeta^2 + \partial_x^2 \right) G(x,\zeta) = 0$ if $x,\zeta>0$. Now I thought if I transform $\zeta=Rz$ and $x=\frac{R^2}{2}$ then $\left( \frac{1}{R} \partial_z^2 + \partial_R \frac{1}{R} \partial_R \right) G\left(\frac{R^2}{2},Rz\right)$ would be a solution too, but it is not zero. Why is that? Is it not possible to transform $\zeta=Rz$ in the sense that R is an independent (constant) variable or do I always need to consider the combined transformation, that is the total differential of $\zeta$ ? I would have thought I can transform twice independently of each other.

Answer 1

By the multivariable chainrule

\begin{align} \partial_R &= \partial_x x_R + \partial_\zeta \zeta_R = R\partial_x + z\partial_\zeta \\ \partial_z &= \partial_x x_z + \partial_\zeta \zeta_z = R\partial_\zeta \end{align}

Solving the system of equations gives

\begin{align} \partial_x &= \frac{1}{R^2}(R\partial_R - z\partial_z) \\ \partial_\zeta &= \frac{1}{R}\partial_z \end{align}

The issue is likely with your $\partial_x$ derivation. Your Laplacian can't be right since it would also have to include $\partial_{Rz}$.