May 27th: Understanding a proof

We will examine, in detail (+ some of my interpretations), for the identity for the spherical laplacian, i.e. $\tilde{\Delta} F = r^2 \Delta F - (k^2 + k(n-2)) F$ , where $F$ is a homogeneous polynomial of degree k in $\mathbb{R}^n$ . For definitions of the spherical laplacian, please consult chapter 10 of Arnold’s PDE.

We begin by visualizing what is going on. Let $f$ be the restriction of $F$ to the sphere $S^{n-1}$ . Notice that $\nabla(F)$ can be decomposed into two orthogonal vectors, $\nabla(f)$ and $\frac{\partial{F}}{\partial{r}} \frac{\partial}{\partial n}$ . Key: since $F$ is homogeneous, the tangential component of $F$ to the sphere does not depend on the radius of the sphere, i.e. $\nabla(f)$ never changes when moving along a fixed ray coming from the origin.

The proof of the identity is similar to the proof for showing $\frac{1}{r}$ is harmonic in $\mathbb{R}^3$ in the sense that we want to show a certain ‘property’ of the divergence of $\nabla(F)$ and show that it is true within any bounded region (away from the origin).

Since $\nabla(f)$ does not change on any sphere, and $\frac{\partial{F}}{\partial{r}} = \frac{\partial{r^k f(q)}}{\partial{r}} = kr^{k-1}f(q)$ , it suffices to prove this property on the surface of the unit sphere.

To compute divergence within a region, it suffices to compute the flux across the surface of the region and then apply divergence theorem. Key: how do we pick the test region?

Note (this is where my solution may differ from that of Arnold’s): the region we pick should not be a prism if we want both the top surface and bottom surface to be on the surfaces of the spheres of radius $1 + \epsilon$ and $1$ (which would allow us to more easily compute the flux) AND for the lateral surface to be orthogonal to the top and bottom surfaces. We would need to pick a region, G, that is:

If the area of the bottom surface is $\sigma$ , then by similarity, the area of the top surface would be $\sigma (1 + \epsilon)^{n-1}$ .

The flux of $\nabla(F)$ across the top and bottom surfaces would be the component of $\nabla(F)$ orthogonal to the surface of the sphere (i.e. radial direction); we know this would be $\frac{\partial{F}}{\partial{r}}$ . The flux across the lateral surface would be the component of $\nabla(F)$ tangent to the sphere, i.e. $\nabla(f)$ .

It follows that the total flux is:

$\int_{\partial G} \nabla(F) \cdot \hat{n} dS$

$= \int_{\tau} \nabla(F) \cdot \hat{n} dS + \text{flux across top surface} + \text{flux across bottom surface}$ .

For $\sigma, \epsilon$ small enough, we can assume $f$ within the bottom and top surface to be constant. It follows that the

$\text{flux across top surface} = \text{area} \cdot \text{constant flux} = \sigma (1+\epsilon)^{n-1} k (1 + \epsilon)^{k-1} f$ and

$\text{flux across bottom surface} = \text{area} \cdot \text{constant flux} = \sigma k f$ . (Note: the flux across the bottom is going into the bounded region whereas the flux across the top is coming out of the bounded region).

Hence,

$\int_{\partial G} \nabla(F) \cdot \hat{n} dS = \int_{\tau} \nabla(F) \cdot \hat{n} dS + \sigma (1+\epsilon)^{n-1} k (1 + \epsilon)^{k-1} f - \sigma k f$ (if we define the direction pointing out of the closed surface to be positive).

Now, we can use the binomial approximation to simplify, i.e. $(1+\epsilon)^{n+k-2} = 1 + (n+k-2)\epsilon + o(\epsilon^2)$ . It follows that we have:

$\int_{\partial G}\nabla(F) \cdot \hat{n} dS = \int_{\tau} \nabla(F) \cdot \hat{n} dS + (\sigma k (n+k-2) \epsilon f + \sigma k f + o(\epsilon^2)) - \sigma k f = \int_{\tau} \nabla(F) \cdot \hat{n} dS + \sigma k (n+k-2) \epsilon f + o(\epsilon^2)$ .

By divergence theorem,

$\int_{\partial G} \nabla(F) \cdot \hat{n} dS = \int_{G} div(\nabla(F)) dV$ and

$\int_{\tau} \nabla(F) \cdot \hat{n} dS = \int_{G} div(\nabla(f)) dV$ (because $\nabla(f)$ is tangential to the sphere and so has no flux coming out of the top and bottom surface.

Thus, the above equation becomes:

$\int_{G} div(\nabla(F)) dV \approx \int_{G} div(\nabla(f)) dV + \int_{\tau} \nabla(F) \cdot \hat{n} dS + \sigma k (n+k-2) \epsilon f$ (*).

Notice: the volume of the $G$ is approximately

$\frac{(1+\epsilon)^{n-1}\sigma + \sigma) \epsilon}{2} \approx \frac{(1 + (n-1)\epsilon + o(\epsilon^2))\sigma \epsilon}{2} \approx \sigma \epsilon + o(\epsilon^2) \approx \sigma \epsilon$ .

For $\epsilon, \sigma$ small enough, the divergence at each point within the region is approximately constant. Hence, dividing (*) by the volume of the region gives:

$div(\nabla(F)) = div(\nabla(f)) + k(n+k-2)f$ , which is what we want.

May 27th: Understanding a proof

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