Differentiating the Spherical Mean w.r.t its radius

Question

this question regards differentiating the spherical mean with respect to its radius.

This is my attempt so far:

Start with the equivalent form of the spherical mean so that we can pass the partial derivative into the integrand:

$$S(v,x,r) = {1 \over d \omega_d} \int_{|\xi|=1)} v(x+r\xi)do(\xi)$$

Then take partial derivative with respect to r:

$${\partial \over \partial r}S(v,x,r) = {1 \over d \omega_d} \int_{|\xi|=1} {\partial \over \partial r} v(x+r\xi)do(\xi)$$

What I understand so far is that this turns into the LHS of the equality below, because of the multivariable chain rule, but I don't know how to go from the LHS to the RHS:

$${1 \over d \omega_d} \int_{|\xi|=1} \sum_{i=1}^{d} {\partial \over \partial x^i} v(x+r\xi)\xi^ido(\xi)=S(v,x,r) = {1 \over d \omega_d r^{d-1}} \int_{\partial B(x, r)} {\partial \over \partial \nu} v(y)do(y)$$

I tried to write the integrand of RHS using the definition of divergence, and get it to look like:

$$\int_{\partial B(x, r)} {\partial \over \partial \nu} v(y)do(y)=\int_{\partial B(x, r)} v(y) \cdot \nu do(y)$$

But I am not understanding what would happen to $\xi ^i$, and how/when to make the substitution back to the original form of the spherical mean, over $B(x,r)$.

Any input/hints are appreciated.

Answer 1

Let's look at the RHS: $$ {1 \over d \omega_d r^{d-1}} \int_{\partial B(x, r)} {\partial \over \partial \nu} v(y)do(y)$$ (I suspect that $do$ should be $d\sigma$, but this is a minor point.)

Change the variable as $y=x+r\xi$. Note that $|\xi|=1$ here. Since the domain of integration is $(d-1)$-dimensional, the Jacobian determinant is $r^{d-1}$. This neatly cancels out with the denominator, and we get $$ {1 \over d \omega_d } \int_{|\xi|=1 } {\partial \over \partial \nu} v(x+r\xi)do(\xi)$$

What does ${\partial \over \partial \nu}$ mean? It's the derivative in the direction of normal vector, that is $\nabla v\cdot \nu$ where $\nu$ is unit normal. Since we are on the unit sphere, the vector $\nu$ is simply $\xi$. Writing the dot product in coordinates, we get $${\partial \over \partial \nu} v(x+r\xi) = \sum_{i=1}^d {\partial \over \partial x_i} v(x+r\xi) \,\xi^i$$ which completes the transformation of RHS to LHS.