The spherical mean of $u$ is defined as

\begin{align}

\frac{1}{4\pi} \int_{S^2} u(\vec x + r \, \vec y) d\sigma(\vec y).

\end{align}

Now this might be a dumb question, but I don't see how a $r$ derivative, $\partial r$, acts on this. In the literature it says that (if $u$ suff. regular)

\begin{align}

\partial_r \int_{S^2} u(\vec x + r \, \vec y) d\sigma(\vec y) = \int_{S^2} \vec\nabla_x u(\vec x + r \, \vec y) \cdot \vec y \;d\sigma(\vec y).

\end{align}

but why does the multivariable chain rule only produce a gradient for $\vec x$ instead of

\begin{align}

\sum_{i=1}^3 \frac{\partial u}{\partial(x ^i + r \, y^i)} \underbrace{\frac{\partial(x^i + r \, y

^i)}{\partial r}}_{= y ^i}.

\end{align}

Does it have to do with the fact that all this happens under the integral?

## Best Answer

I think the important thing is that the spherical mean is a function of $\vec x$ (and $r$), and $\vec y$ is the dummy variable of integration. The choice of notation $\nabla_{\vec x}$ is just emphasizing that if you differentiate with respect to $r$, the resulting expression is still a function of $\vec x$ (and $r$). I think many authors would not even include the $\vec x$ in the gradient, and instead simply write $$ \partial_r \int_{S^2} u(x+ry)\,d\sigma(y) = \int_{S^2}\nabla u(x+ry)\cdot y\,d\sigma(y). $$ The intermediate step is $$ \int_{S^2}\partial_r\big(u(x+ry)\big)\,d\sigma(y). $$ To evaluate, set $z(r) = x+ry$. By the multivariable chain rule, $$ \partial_r\big(u(z(r))\big) = \nabla u(z(r))\cdot z'(r) = \nabla u(x+ry)\cdot y. $$ Or with the dummy index notation, $$ \partial_r\big(u(z(r))\big) = \sum_i\frac{\partial u}{\partial z_i}(z(r))\frac{dz_i}{dr}(r). $$ In this notation, $z_i$ is just a dummy variable.