Laplace Operator in $3D$

Question

I am looking to find the radial part of Laplace's operator in three dimensions. I looked up Laplace's operator in spherical coordinates and from there I guess the radial part is: $\frac{\partial^2}{\partial r^2} + \frac{2\: \partial}{r\:\partial r}$. Is this correct and is there a better, quicker way to find the radial part?

Answer 1

Yes, this is correct and there is a quicker way if you are interested only in the radial part. Let $u(x)=f(|x|)$ be a radially symmetric function. Its gradient is $\nabla u(x)=f'(|x|)\frac{x}{|x|}$. Therefore, the flux of the gradient across the outward-oriented sphere $|x|=r$ is $4\pi r^2 f'(r)$.

Consider the spherical shell $r<|x|<r+h$. Net flux of $\nabla u$ out of this shell is $$4\pi (r+h)^2 f'(r+h)-4\pi r^2 f'(r)$$ By the divergence theorem, this is equal to $\iint_{r<|x|<r+h}\Delta u$. Dividing by $h$ and letting $h\to 0$ we obtain $$ \iint_{|x|=r} \Delta u(x) = 4\pi (r^2 f'(r))' $$ Finally, divide by the area $4\pi r^2$ to find the Laplacian itself: $\Delta u(x) = r^{-2} (r^2 f'(r))' $ where $r=|x|$.

Similarly, in $n$ dimensions the radial part of $\Delta$ is $$ r^{1-n} (r^{n-1} u_r)_r = u_{rr}+(n-1)r^{-1}u_r $$