Where does Laplace's equation in spherical polars come from

[latentcorpse]

Where does Laplace's equation in spherical polars come from

\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}=0

?

i can derive from scratch the expression for the laplacian in spherical polars but this bears no resemblance to the above, even if i decide to ignore the \phi dependence?

 

[robphy]

Those are in 2D-polar coordinates.

 

[latentcorpse]

hmm, I am guessing that's meant to be easier but we haven't covered that in class (despite doing the 3D case) - could you possibly run over the derivation or post a link to a website (whenever i google it i only get the 3d case) please?

 

[robphy]

http://www.google.com/search?q=laplacian+polar

 

[Ben Niehoff]

When deriving it from scratch, remember that in spherical coordinates, the unit vectors are not constants anymore!

The gradient operator is

\vec \nabla = \hat r \frac{\partial}{\partial r} + \hat \theta \frac1{r} \frac{\partial}{\partial \theta} + \hat \phi \frac1{r \sin \theta} \frac{\partial}{\partial \phi}

what you want to find is

\vec \nabla \cdot \vec \nabla

but you must take into account the fact that the differential operators also act on the unit vectors.

 

[ber70]

From my weblog
http://buyanik.wordpress.com/2009/05/02/laplacian-in-spherical-coordinates/"