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.

 

[blue_raver22]

I can provide some clarification on the concept of normal derivative in a sphere. The normal derivative is a mathematical concept that describes the rate of change of a function in the direction of the outward normal vector of a surface. In the case of a sphere, the outward normal vector \hat{n} is always pointing directly away from the center of the sphere, and its magnitude is equal to the radius of the sphere, r.

Now, the equation \frac{\partial u}{\partial n} = \nabla u \cdot \hat{n} is a general expression for the normal derivative on any surface. However, on a sphere, we can simplify this expression to \frac{\partial u}{\partial n} = -\frac{\partial u}{\partial r}. This is because on a sphere, the gradient of a function \nabla u is always parallel to the radius vector \vec{r}, and thus the dot product between them is equal to the magnitude of the gradient in the direction of the outward normal, which is equal to -\frac{\partial u}{\partial r}.

In other words, the normal derivative in a sphere is simply the negative of the radial derivative. This makes sense intuitively, as the normal derivative represents the change of a function as we move along the surface of the sphere in the direction of the outward normal, while the radial derivative represents the change of the function as we move along the radius of the sphere. Since these directions are opposite to each other, their derivatives are also opposite in sign.

I hope this explanation helps clarify the concept of normal derivative in a sphere. It is a useful tool in many areas of science, including fluid dynamics and electromagnetism, and understanding it can help us better understand and model physical phenomena on curved surfaces.