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hello, it is often said that a conformal mapping preserves the Laplace equetion in 2D. However, if this is true for the cartesian coordinates (x,y), where the laplacian is: $$ \frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}=0 $$

is it true for the cylindrical coordinates (r,z) where:

$$ \frac{\partial^2 \phi}{\partial r^2}+\frac{\partial^2 \phi}{\partial z^2}+\frac{1}{r}\frac{\partial \phi}{\partial r}=0? $$

More precisely, if you have a function $\phi(r,z)$ verifying the previous equation, is it also verified by $\psi(u,v)$ is $u+i v=f(z+ir)$ and f is analytic/holomorphic?

edited Oct 13 2010 at 15:42

Willie Wong
6,987●20●43

asked Oct 13 2010 at 14:43

Mermoz
45●5

How about $$\frac{\partial^2 \phi}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 \phi}{\partial z^2}+\frac{1}{r}\frac{\partial \phi}{\partial r}=0 $$ – Gerald Edgar Oct 13 2010 at 15:15

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Maybe the change of variables you want is $ue^{iv} = f(re^{iz})$ ... – Gerald Edgar Oct 13 2010 at 15:27

To elaborate on Gerald's comment: the coordinate change you wrote down is not a conformal mapping [see en.wikipedia.org/wiki/Conformal_map ] in general. Also, this question has nothing to do with conformal field theory. – Willie Wong Oct 13 2010 at 15:41

but that is suited when you want to apply you conformal mapping to the $(r,\theta)$ plane and not the $(r,z)$ plane, or? – Mermoz Oct 13 2010 at 15:50

wait a second, what do you actually mean by the cylindrical coordinates in 2D? Are you actually interested in functions defined over 3 dimensional space but is constant along the azimuthal direction? – Willie Wong Oct 13 2010 at 18:30

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Denis Serre · Accepted Answer · 2010-10-13 15:17:49Z UTC

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No. Take $\phi=r^2-2z^2$ and $f(z+ir)=(z+ir)^2$, thus $u=r^2-z^2$, $v=2rz$. Then $\phi(u,v)=r^4+z^4-10r^2z^2$ is not cylindrical-harmonic.

answered Oct 13 2010 at 15:17

Denis Serre
18.7k●1●31●93

Invariance of the cylindrical Laplace equation under conformal transform

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