Verify normal derivative of a sphere.

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September 8th 2010, 09:00 PM
yungman

Verify normal derivative of a sphere.

I’ve search through many books and went online without any luck, I come up with a way to proof this and I need someone here to verify I am correct. Here is the problem:

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 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

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

This is the way I have to proof the identity:

In spherical coordinates:

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

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

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

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

Please give me your opinion. and help.

Thanks

Alan
September 9th 2010, 06:18 AM
Ackbeet

I agree with you. Unless there's something else that we're missing here, the gradient of $u$ dotted with your $\hat{n}$ is the positive radial derivative. Incidentally, the $\phi$ component of your gradient is incorrect; however, that's a moot point since you're taking the dot product of the gradient with the unit radial vector.

[EDIT] The OP has been corrected.

Question: in the book that has the minus sign, how do they define the unit normal vector? Is it some weird thing where it points inward or something? Is this sphere centered at the origin?

It's strange that they would intentionally make two mistakes in a row like that, unless there's some subtle aspect of the problem that we're missing here.

Quote:

I don’t get how to go from

$\displaystyle{\frac{x}{r} \frac{\partial u}{\partial x} \;+\; \frac{y}{r} \frac{\partial u}{\partial y} \;+\; \frac{z}{r} \frac{\partial u}{\partial z} \;=\; \frac{\partial u}{\partial r}}$

I don't either, directly, other than to say that it sort of looks like a chain rule. Your proof is much more direct. I would certainly agree that in cartesian coordinates, you have

$\displaystyle{\nabla u\cdot\hat{n}=\frac{x}{r} \frac{\partial u}{\partial x} +\frac{y}{r} \frac{\partial u}{\partial y} + \frac{z}{r} \frac{\partial u}{\partial z}.}$
September 9th 2010, 09:00 AM
yungman

Quote:

Originally Posted by Ackbeet

I agree with you. Unless there's something else that we're missing here, the gradient of $u$ dotted with your $\hat{n}$ is the positive radial derivative. Incidentally, the $\phi$ component of your gradient is incorrect; however, that's a moot point since you're taking the dot product of the gradient with the unit radial vector.

Question: in the book that has the minus sign, how do they define the unit normal vector? Is it some weird thing where it points inward or something? Is this sphere centered at the origin?

It's strange that they would intentionally make two mistakes in a row like that, unless there's some subtle aspect of the problem that we're missing here.

I don't either, directly, other than to say that it sort of looks like a chain rule. Your proof is much more direct. I would certainly agree that in cartesian coordinates, you have

$\displaystyle{\nabla u\cdot\hat{n}=\frac{x}{r} \frac{\partial u}{\partial x} +\frac{y}{r} \frac{\partial u}{\partial y} + \frac{z}{r} \frac{\partial u}{\partial z}.}$

Thanks for your help. I double check the book by Strauss, $\hat{n}$ is outward normal. As you can see the this given by the same book:

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

It does not have the minus sign!!! But the same book give:

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

BTW, I see nothing wrong with my gradian formula!!!!................................

Just kidding, I went back and corrected it. I copy wrong the first time. Thanks

Again, thanks for all your help.

Alan
September 9th 2010, 09:02 AM
Ackbeet

You're welcome. If it's all clear to you, that's great. Have a good one!