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Normal derivative in a sphere.

  1. Sep 8, 2010 #1
    Normal derivative is defined as:

    [tex] \frac{\partial u}{\partial n} = \nabla u \;\cdot\; \hat{n} [/tex]

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

    [tex]\int_{\Gamma} \frac{\partial u}{\partial n} \;dS \;=\; -\int_{\Gamma} \frac{\partial u}{\partial r} \;dS [/tex]

    The book claimed on a sphere:

    [tex] \frac{\partial u}{\partial n} = \nabla u \;\cdot\; \hat{n} \;=\; -\frac{\partial u}{\partial r} [/tex]

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

    Can anyone help?
  2. jcsd
  3. Sep 8, 2010 #2


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    There is no vector [itex]\vec{r}[/itex] in that formula. The [itex]\hat{n}[/itex], in [itex]\nabla u\cdot \hat{n}[/itex] is the unit vector perpendicular to the sphere and so parallel to the "r" direction. The "r" in
    [tex]\frac{\partial u}{\partial r}[/tex]
    is the variable r, not a vector.
  4. Sep 8, 2010 #3
    Thanks for the reply.

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

    You have any pointers regarding my original question?
  5. Sep 8, 2010 #4
    I found the explanation from the PDE book of Strauss.

    [tex]\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} [/tex]

    Where [itex] r=\sqrt{x^2+y^2+z^2}[/itex]

    I don’t get how to go from

    [tex] \frac{x}{r} \frac{du}{dx} \;+\; \frac{y}{r} \frac{du}{dy} \;+\; \frac{z}{r} \frac{du}{dz} \;=\; \frac{\partial u}{\partial r} [/tex]

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

    In Spherical coordiantes:

    [tex]\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}[/tex]

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

    [tex]\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}[/tex]

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

    Please give me your opinion.
    Last edited: Sep 9, 2010
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