Laplace-Beltrami on a sphere

I'm trying to compute the Laplace-Beltrami of the function $u(r,\varphi,\theta) = 12\sin(3\varphi)\sin^3(\theta)$ on a unit sphere. Note that $\varphi$ is the azimuth, i.e. $\varphi \in [0,2\pi]$ and $\theta$ the inclination, i.e. $\theta \in [-\frac{\pi}{2},\frac{\pi}{2}]$. For instructive purposes, I'd like to do this step by step.

The Laplace-Beltrami of $u$ is defined as

$$\Delta u := \mathrm{div} (\mathrm{grad} \; u).$$

Since we're talking about a surface (the sphere), I assume that we should use the surface gradient of $u$, defined as

$$\nabla_S u := \nabla u - \vec{n}(\vec{n} \cdot \nabla u).$$

The gradient operator in spherical coordinates is defined as

$$\nabla := \frac{\partial }{\partial r} \vec{e_r} + \frac{1}{r} \frac{\partial }{\partial \theta} \vec{e_\theta} + \frac{1}{r\;\sin(\theta)} \frac{\partial }{\partial \varphi} \vec{e_\varphi},$$

which results in

$$\nabla u = 0 \vec{e_r} + \frac{1}{r} 36 \sin(3\varphi) \sin^2(\theta) \cos(\theta) \vec{e_\theta} + \frac{1}{r} 36 \cos(3\varphi) \sin^2(\theta) \vec{e_\varphi}.$$

Now, I'm not quite sure about the unit normal $\vec{n}$ on the sphere. I thought it would just be $\vec{e_r}$, but that cannot be right since in that case the inner product $\vec{n} \cdot \nabla u$ is zero (and hence the surface gradient would be equal to the regular gradient). Just to be sure, the inner product for a spherical coordinate setting is defined as $a \cdot b = g_{ij} a^i b^j$ — using Einstein notation — with the metric $g$ defined as

$$\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & r^2 & 0\\ 0 & 0 & r^2 \sin^2(\theta) \end{array}\right),$$

correct? Since the spherical coordinate system is right-handed, taking the cross product of the tangent vectors $\vec{e_\varphi}$ and $\vec{e_\theta}$ again results in $\vec{e_r}$. Could someone point out where I'm going wrong?

Next up is the divergence. I assume there is something like the surface divergence, but I couldn't find much about it (any references are most welcome). This would result in $\Delta_S u = \mathrm{div}_S (\nabla_S u)$.

It would be great if somebody could help to complete this elaboration. The eventual result of $\Delta_S u$ should be $-12 u$.

Using the regular divergence operator for a spherical coordinate setting, defined as

$$\nabla \cdot := \frac{1}{r^2} \frac{\partial}{\partial r} r^2 \vec{e_r} + \frac{1}{r \sin(\theta)} \frac{\partial}{\partial \theta} \sin(\theta) \vec{e_\theta} + \frac{1}{r \sin(\theta)} \frac{\partial}{\partial \varphi} \vec{e_\varphi},$$

we get (without using the metric $g$ as defined above)

$$\frac{1}{r^2 \sin(\theta)} \left( 36 \sin(3\varphi) \left\{ 3 \sin^2(\theta) \cos^2(\theta) - \sin^4(\theta) \right\} \right) + \frac{1}{r^2 \sin(\theta)} \left( -108 \sin(3\varphi) \sin^2(\theta) \right).$$

$$\left( 36 \sin(3\varphi) \left\{ 3 \sin(\theta) \cos^2(\theta) - \sin^3(\theta) \right\} \right) + \left( -108 \sin(3\varphi) \sin^3(\theta) \right).$$

Since the solution should be $-12u = -144 \sin(3\varphi) \sin^3(\theta)$, I'm not sure how I should get rid of the term $108 \sin(3 \varphi) \sin(\theta) \cos^2(\theta)$. Anyone?

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The right hand side on the bottom right should be just sin, not sin cubed. On the left side, you can use a trig identity to change it to a function of phi times 108 sin theta-144sin^3 theta (just using Pythagorean identity) which will cancel the right hand side and leave the desired result. –  Brian Rushton Jun 8 '13 at 3:13
@BrianRushton Thanks, I see it now. So, the metric $g$ is already incorporated in the divergence operator for spherical coordinates. Finally, setting $r$ to $1$ (we're looking at the surface of a unit sphere) then gives the correct result. –  Ailurus Jun 8 '13 at 10:31

Your function is independent of $r$, so the gradient always lies on the surface of the sphere, so in this case the surface gradient is the normal gradient.
Right, but the gradient $\nabla u$ is dependent of $r$, so what about the surface divergence? Or should I just set $r$ to be $1$? –  Ailurus Jun 7 '13 at 0:14
Since the functions are independent of $r$, the divergence only measures growth on the surface. –  Brian Rushton Jun 7 '13 at 0:22
For functions independent of $r$ the operator is just the regular Laplacian. –  Brian Rushton Jun 7 '13 at 0:47