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In looking over a proof on the mean value property for harmonic functions on spheres I came across the following formula: $$\frac{\partial u}{\partial n}=\textbf{n}\cdot \triangledown u=\frac{\bf{x}}{r}\cdot \triangledown u =\frac{x}{r}u_x+\frac{y}{r}u_y+\frac{z}{r}u_z=\frac{\partial u}{\partial r}$$ Where $r=\sqrt{x^2+y^2+z^2}$ is the spherical distance to the center of the circle and because we are considering a sphere $\textbf{n}=\frac{\bf{x}}{r}$ is the outward pointing unit normal. I can understand everything until the last equality sign. I would be inclined to think that $$\frac{\partial u}{\partial r}=u_r=u_x\frac{\partial x}{\partial r}+u_y\frac{\partial y}{\partial r}+u_z\frac{\partial z}{\partial r}$$ Now $r=\sqrt{x^2+y^2+z^2}$ implies that $x=\sqrt{r^2-y^2-z^2}$ and thus: $$\frac{\partial x}{\partial r}= \frac{2r}{2\sqrt{r^2-y^2-z^2}}=\frac{r}{x} $$ Similarly for $y$ and $z$ leading to the conclusion that $$u_r=\frac{r}{x}u_x+\frac{r}{y}u_y+\frac{r}{z}u_z$$ Which has all the fractions multiplying the partial derivatives inversed. I must be missing something obvious here or be making a stupid mistake, if anybody would point it out to me that would be very nice. I want to understand this proof and I can't continue if my thoughts completely disagree with what it says because that way I can never have confidence in the result!

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If you believe the first equality, then you should believe all else because on the unit sphere $r=n$. – Matt Feb 18 '13 at 23:53
I do believe it and I thought of the same reasoning as you propose. The problem is that my reasoning seems to contradict it. Maybe the solution is simply stating that $x=r$ along the $x$ axis (we are dealing with a sphere centered at 0) and similarly for $y$ and $z$ and thus $\frac{x}{r}=\frac{r}{x}=1$? – Slugger Feb 18 '13 at 23:58
The command for the partial derivative symbol $\partial$ that you replaced by a delta is \partial. – joriki Feb 19 '13 at 0:25
Ah thank you so much I've been looking for that for ages. I thought it was a greek letter that wasn't in Latex or something :P – Slugger Feb 19 '13 at 0:34
@Teun: Note that there's an edit link under the question, so it's never too late to get the formatting right :-) – joriki Feb 19 '13 at 0:41
up vote 3 down vote accepted

This is a very common type of mistake that's caused by our suboptimal notation for partial derivatives. See this answer for a collection of answers I've written in response to similar confusions over partial derivatives.

The symbol $\displaystyle\frac{\partial x}{\partial r}$ in the context in which you're using it is intended to mean $\displaystyle\left.\frac{\partial x}{\partial r}\right|_{\theta,\phi}$, that is, the rate of change of $x$ with $r$ as you hold the angular coordinates $\theta$ and $\phi$ of the spherical coordinates $r$, $\theta$ and $\phi$ constant. What you've calculated is $\displaystyle\left.\frac{\partial x}{\partial r}\right|_{y,z}$, that is, the rate of change of $x$ with $r$ as you hold the Cartesian coordinates $y$ and $z$ constant. Such confusion arises very easily because our standard notation for partial derivatives doesn't show what's being held constant and is therefore ambiguous.

With $x=r\sin\theta\sin\phi$, you immediately find $\displaystyle\left.\frac{\partial x}{\partial r}\right|_{\theta,\phi}=\sin\theta\sin\phi=\frac xr$ as expected.

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