What Is the Normal Derivative in a Sphere?

[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?

 

[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.

 

[yungman]

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.

Thanks for the reply.

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?

 

[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.

Please give me your opinion.