# Problem of the Week #54 - April 8th, 2013

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#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Let $u:\mathbb{R}^3\rightarrow \mathbb{R}$, and define the Laplacian $\nabla^2u$ in rectangular coordinates $(x,y,z)$ by
$\nabla^2u=\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} + \frac{\partial^2u}{\partial z^2}.$
Show that the Laplacian $\nabla^2u$ in spherical coordinates $(\rho,\phi,\theta)$ is given by
$\nabla^2u=\frac{\partial^2u}{\partial\rho^2} + \frac{2}{\rho}\frac{\partial u}{\partial\rho} + \frac{1}{\rho^2}\frac{\partial^2u}{\partial\phi^2} + \frac{\cot\phi}{\rho^2}\frac{\partial u}{\partial\phi} + \frac{1}{\rho^2\sin^2\phi} \frac{\partial^2u}{\partial \theta^2}$

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Suggestion:
It may be a good idea to convert from rectangular to cylindrical coordinates $(r,\theta,z)$ first, where the Laplacian is
$\nabla^2u=\frac{\partial^2u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2u}{\partial\theta^2} + \frac{\partial^2u}{\partial z^2}$
and then convert from cylindrical to spherical.

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#### Chris L T521

##### Well-known member
Staff member
This week's question was answered by Sudharaka. You can find his answer below.

The gradient of a scalar field in spherical coordinates $$(\rho,\,\phi,\,\theta)$$ is given by,$\nabla u(\rho, \phi, \theta) = \frac{\partial u}{\partial \rho}\mathbf{e}_\rho+\frac{1}{\rho}\frac{\partial u}{\partial \phi}\mathbf{e}_\phi+\frac{1}{\rho \sin\phi}\frac{\partial u}{\partial \theta}\mathbf{e}_\theta$
The Laplacian is the divergence of the gradient. So we get,
\begin{eqnarray}
\nabla^2u &=& \operatorname{div}\left(\frac{\partial u}{\partial \rho}\mathbf{e}_\rho+\frac{1}{\rho}\frac{\partial u}{\partial \phi}\mathbf{e}_\phi+\frac{1}{\rho \sin\phi}\frac{\partial u}{\partial \theta}\mathbf{e}_\theta\right)\\
&=& \frac1{\rho^2} \frac{\partial}{\partial \rho}\left(\rho^2 \frac{\partial u}{\partial \rho}\right) + \frac1{\rho\sin\phi} \frac{\partial}{\partial \phi} \left(\frac{\sin\phi}{\rho}\frac{\partial u}{\partial \phi}\right) + \frac1{\rho\sin\phi} \frac{\partial}{\partial \theta}\left(\frac{1}{\rho \sin\phi}\frac{\partial u}{\partial \theta}\right)
\end{eqnarray}
Simplification yields,
$\nabla^2u=\frac{\partial^2u}{\partial\rho^2} + \frac{2}{\rho}\frac{\partial u}{\partial\rho} + \frac{1}{\rho^2}\frac{\partial^2u}{\partial\phi^2} + \frac{\cot\phi}{\rho^2}\frac{\partial u}{\partial\phi} + \frac{1}{\rho^2\sin^2\phi} \frac{\partial^2u}{\partial \theta^2}$

In the next couple days, I'll edit this post to include a step by step justification of the identity from rectangular coordinates.

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