Normal derivative in a sphere.

1. Sep 8, 2010

yungman

Normal derivative is defined as:

$$\frac{\partial u}{\partial n} = \nabla u \;\cdot\; \hat{n}$$

Where $\hat{n}$ is the unit outward normal of the surface of the sphere and for a small sphere with surface $\Gamma$, the book gave:

$$\int_{\Gamma} \frac{\partial u}{\partial n} \;dS \;=\; -\int_{\Gamma} \frac{\partial u}{\partial r} \;dS$$

The book claimed on a sphere:

$$\frac{\partial u}{\partial n} = \nabla u \;\cdot\; \hat{n} \;=\; -\frac{\partial u}{\partial r}$$

Where $r$ is the radius of the sphere. I understand $\hat{n}$ is parallel to $\vec{r}$ but $r$ is not unit length.

Can anyone help?

2. Sep 8, 2010

HallsofIvy

There is no vector $\vec{r}$ in that formula. The $\hat{n}$, in $\nabla u\cdot \hat{n}$ is the unit vector perpendicular to the sphere and so parallel to the "r" direction. The "r" in
$$\frac{\partial u}{\partial r}$$
is the variable r, not a vector.

3. Sep 8, 2010

yungman

I understand r is only a variable, I am trying to say $\hat{n}$ is the same as vector irradia from the center of the sphere.

You have any pointers regarding my original question?

4. Sep 8, 2010

yungman

I found the explanation from the PDE book of Strauss.

$$\frac{\partial u}{\partial n} \;=\;\hat{n} \;\cdot\; \nabla u \;=\; \frac{x}{r} \frac{du}{dx} \;+\; \frac{y}{r} \frac{du}{dy} \;+\; \frac{z}{r} \frac{du}{dz} \;=\; \frac{\partial u}{\partial r}$$

Where $r=\sqrt{x^2+y^2+z^2}$

I don’t get how to go from

$$\frac{x}{r} \frac{du}{dx} \;+\; \frac{y}{r} \frac{du}{dy} \;+\; \frac{z}{r} \frac{du}{dz} \;=\; \frac{\partial u}{\partial r}$$

Let me try this way and please comment whether this make sense.

In Spherical coordiantes:

$$\nabla u \;=\; \frac{\partial u}{\partial r}\hat{r} \;+\; \frac{1}{r}\frac{\partial u}{\partial \theta}\hat{\theta} \;+\; \frac{1}{r^2 sin \theta}\;\frac{\partial u}{\partial \phi} \hat{\phi}$$

We know $\hat{r} \hbox { is parallel the the outward unit normal } \hat{n}$ and therefore $\hat{n} \cdot \hat{\theta} = \hat{n} \cdot \hat{\phi}=0$

$$\Rightarrow \; \nabla u \cdot \hat{n} \;=\; [\frac{\partial u}{\partial r}\hat{r} \;+\; \frac{1}{r}\frac{\partial u}{\partial \theta}\hat{\theta} \;+\; \frac{1}{r^2 sin \theta}\frac{\partial u}{\partial \phi} \hat{\phi}] \;\cdot\; \hat{r} = \frac{\partial u}{\partial r}$$

Where I substute n with r. But I still don't get the "-" sign yet.