change of variable to get a quasi-cartesian laplace equation?

Question

when writing the (vector) laplace equation in cylindrical coordinates in a $(r,\theta)$ plane, we get: $$ \left( r^2\frac{\partial^2}{\partial r^2} + r\frac{\partial}{\partial r} + \frac{\partial^2}{\partial\theta^2} \right)A_z(r,\theta)=0 $$ the term $r^2$ is multiplied by the two sides of the equation to form the cauchy-Euler equation.

now through a change of variable of $v=\ln(r)$: $$ \left( \frac{\partial^2}{\partial v^2} + \frac{\partial^2}{\partial\theta^2} \right)A_z(v,\theta)=0 $$ it is like a cartesian laplace operator, which has its advantages of simplification of computations.

however in $(r,z)$ plane the laplace equation becomes as following: $$ \left( r^2\frac{\partial^2}{\partial r^2} + r\frac{\partial}{\partial r} -1 + \frac{\partial^2}{\partial z^2} \right)A_\theta(r,z)=0 $$ is it even possible through a change of $u=f(r)$ to get a result like the same?: $$ \left( \frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial z^2} \right)A_\theta(u,z)=0 $$

additional notes:it is worth noting that in the above equations, the 3D vector-laplace equation has been reduced to 2D plane due to symmetry of the problem.

Answer 1

it is not possible. a Cauchy-Euler equation in terms of radial component appears in the partial equation which cannot be transformed into a simple second-order derivative.

Dear readers, for proof, please refer to this question:

Reduce the Cauchy-Euler equation to a smaller expression?

Answer 2

$$r^2\nabla^2 = \left( r^2\frac{\partial^2}{\partial r^2} + r\frac{\partial}{\partial r} + \frac{\partial^2}{\partial\theta^2} \right)$$ The Laplacien should be: $$\nabla^2=\frac{\partial^2}{\partial r^2} + \dfrac 1r\frac{\partial}{\partial r} + \dfrac 1 {r^2}\frac{\partial^2}{\partial\theta^2} $$ And: $$\nabla^2=\frac{\partial^2}{\partial r^2} + \dfrac 1r\frac{\partial}{\partial r} + \dfrac 1 {r^2}\frac{\partial^2}{\partial\theta^2} +\dfrac {\partial^2}{\partial z^2} $$