Directional derivative in the boundary of the unit ball in R^3

Question

I have a function $f(r,\theta,\phi)=R(r)\Theta(\theta)\Phi(\phi)$ that is given in spherical coordinates. Consider the set $$\mathbb \Gamma=\left\{(x,y,z)\in \mathbb R^3: x^2+y^2+z^2\leq 1\right\}.$$ I need to calculate $\left.\frac{\partial f}{\partial \boldsymbol\nu}\right|_{\partial \Gamma}$, where $\boldsymbol \nu$ is a unit normal vector to $\partial \Gamma$. I know how to do that in the usual cartesian coordinates but how to compute that derivative without needing to go back to the usual coordinates $(x,y,z)$? I don't know how to represent a unit-vector in $\partial \Gamma$.

Answer 1

Note that I am using the mathematics convention for spherical coordinates, i.e $x=r\cos\theta\sin\phi$.

For a unit vector $\mathrm n$ what we mean by $\frac{\partial f}{\partial \mathrm n}$ is actually $$\nabla f\cdot\mathrm n$$ So, $$(\nabla f\cdot \mathrm n)|_{\partial\Gamma}=(\nabla f\cdot \mathrm n)(1,\theta,\phi)$$ By normal spherical coordinate rules we know $$\nabla f=\partial_r f~\mathrm{e}_r+\frac{1}{r\sin\phi}\partial_\theta f~\mathrm e_\theta+\frac{1}{r}\partial_\phi f~\mathrm e_\phi$$ And, simply $$\mathrm n(r,\theta,\phi)=\mathrm e_r=(1,0,0)$$ Finally, since $\mathrm n$ is a vector and $\nabla f$ is a covector, we can compute the dot product rather simply: $$\mathrm n\cdot\nabla f=n^i(\nabla f)_i$$ Thus $$(\nabla f\cdot\mathrm n)(1,\theta,\phi)=(\partial_r f)(1,\theta,\phi)$$ Very simple.

Answer 2

Your $\Gamma$ is the (closed) unit ball in $\mathbb{R}^3$, so the boundary $\partial\Gamma$ is the surface of said unit ball. A normal vector to some point $P$ on the surface of a ball lies on the line through $P$ and the origin (or the centre of the ball, in general). So the normal vector will point in the radial direction, that is in the $\hat{r}$ direction in spherical coordinates.

So want to calculate the directional derivative of $f$ in the $\hat{r}$ direction; in other words, you want to calculate how the value of $f$ changes if we change $r$. That is just the partial derivative of $f$ with respect to $r$, which in this case is $$ \frac{\partial f}{\partial r} = R'(r)\Theta(\theta)\Phi(\phi) $$ This is your answer.