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)

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

In case the metric should be used (I'm not sure about this), the result is

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

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?

  • $\begingroup$ 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. $\endgroup$ – Brian Rushton Jun 8 '13 at 3:13
  • $\begingroup$ @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. $\endgroup$ – 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.

For the divergence, try using divergence in spherical coordinates:http://en.wikipedia.org/wiki/Divergence#Spherical_coordinates

The surface divergence for a general vector field on a differentiate manifold is discussed in http://en.wikipedia.org/wiki/Laplace–Beltrami_operator

  • $\begingroup$ Right, but the gradient $\nabla u$ is dependent of $r$, so what about the surface divergence? Or should I just set $r$ to be $1$? $\endgroup$ – Ailurus Jun 7 '13 at 0:14
  • $\begingroup$ Since the functions are independent of $r$, the divergence only measures growth on the surface. $\endgroup$ – Brian Rushton Jun 7 '13 at 0:22
  • $\begingroup$ For functions independent of $r$ the operator is just the regular Laplacian. $\endgroup$ – Brian Rushton Jun 7 '13 at 0:47
  • $\begingroup$ I tried, but I'm not getting the correct result. $\endgroup$ – Ailurus Jun 7 '13 at 22:24
  • $\begingroup$ It should work, because this function is a classic function, one of the spherical harmonics; it should be an eigenvalue of the regular Laplacian. $\endgroup$ – Brian Rushton Jun 8 '13 at 2:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.