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Normal derivative of Green's function on a disk.

  1. Aug 23, 2010 #1
    For circular region, why is [tex]\frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) [/tex] ?
    Where [itex]\; \hat{n} \:[/itex] is the outward unit normal of [itex]C_R[/itex].
    Let circular region [itex]D_R[/itex] with radius [itex]R \hbox { and possitive oriented boundary }\; C_R[/itex]. Let [itex]u(r_0,\theta)[/itex] be harmonic function in [itex]D_R[/itex].

    The Green's function for Polar coordinate is found to be:

    [tex] G(r,\theta,r_0,\phi) = \frac{1}{2} ln[R^2 \frac{r^2+r_0^2 -2rr_0 cos(\theta-\phi)}{r^2r_0^2 + R^4 - 2rr_0R^2 cos(\theta-\phi)}] [/tex]

    Where [itex]\; \theta \;[/itex] is the angle of [itex]\; u(r_0,\theta_0) \;[/itex] and [itex]\; \phi \;[/itex] is the angle of the two points used in Steiner Invertion.
    Next I want to solve the Dirichlet problem using Green's function. For any value of a hamonic function [itex]u(r_0,\theta_0) in D_R[/itex]. The standard formular for Dirichlet problem is:

    [tex]u(r_0,\theta_0) = \frac{1}{2}\int_{C_R} u(r,\theta) \frac{\partial}{\partial n}G(r,\theta,r_0,\phi) ds[/tex]

    Where [tex]\frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \nabla G(r,\theta,r_0,\phi) \;\cdot \widehat{n} [/tex]

    But the book just simply use [tex]\frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) [/tex] Which is only a simple derivative of G respect to [itex]\; r_0 \;[/itex] where in this case [itex]\; r_0 = R \;[/itex] !!!

    [tex]u(r_0,\theta_0) = \frac{1}{2}\int_{C_R} u(r,\theta) \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) ds[/tex]

    I don't understant how:

    [tex]\frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) [/tex]

    How can a normal derivative become and simple derivative respect to [itex]\; r_0 \;[/itex] only? I know [itex] \widehat{r}_0 \;\hbox { is parallel to outward normal of }\;\; C_R \;[/itex] but the magnitude is not unity like the unit normal. Can anyone explain to me?


    Last edited: Aug 23, 2010
  2. jcsd
  3. Aug 23, 2010 #2
  4. Aug 31, 2010 #3
    Anyone please? Even if you don't have the answer, point me where to look. I am really out of ideas. I have five PDE book and I can't find any help!!!
    Last edited: Aug 31, 2010
  5. Sep 2, 2010 #4
    I'm not sure I completely understand the problem statement, but in general on a circle, the radius is normal to the curve.

    So if you're interested in "normal" derivatives, you usually only have to consider how things change w/ respect to radius. This is something along the lines of the argument of why (in E&M) electrostatic fields are conservative, why Gauss' law goes like 1/r^2, etc... you can "ignore" the angular components b/c they don't contribute to the overall integral.

    However, you're using a function that reads like:


    rather than simply as:


    and it seems like r_0 is a constant.

    Are you asking why a derivative is being taken with respect to something "constant" like r_0?

    If so, it could perhaps be a typo.

    It would make more sense to take the derivative wrt "r"..
  6. Aug 3, 2011 #5
    Hi Everyone,

    please can you tell me why for a circle we have the normal derivative is equal to the tangential derivative, adn what is the antiderivative of a normal derivative???

  7. Aug 3, 2011 #6


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    I think your using "normal derivative" in the context of a directional derivative in the normal direction.

    In general a directional derivative will be the gradient dotted with a unit normal vector. Consider then the gradient operator in polar coordinates:
    [itex] \nabla u(r,\theta) = \hat{r}\frac{\partial u}{\partial r} + \hat{\theta} \frac{1}{r} \frac{\partial u}{\partial \theta}[/itex]
    Since [itex]\hat{n}=\hat{r}[/itex]
    [itex]\nabla_\hat{n} u = \hat{r}\bullet \nabla u = \partial_r u[/itex].
  8. Aug 3, 2011 #7
    Thanks, and what about the tangential derivative. How can I convence my self that is equal to the normal derivative.

    I need these info to be able to do the following integral by part::
    [itex]\int dq G_0(q,r:k)\frac{\partial G(q,r')}{\partial n}[/itex]
    so this where I want to take the antiderivative of the normal derivative.

  9. Aug 3, 2011 #8


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    I'm confused too. In general the tangential derivative and normal derivative of a function relative to a curve will not be the same. In particular for a circle centered at the origin and the function f(r,theta) = f(r) the tangential derivative will be zero and normal derivative will be f'.
  10. Aug 4, 2011 #9
    Thanks, for the moment let us forget the tangential derivative. I only need to compute this integral by parts

    choosing u=G0(q,r:k) and v=∂G(q,r′)\∂n
    so du=∂G0(q,r:k)\∂q dv=????????

    where ∂G(q,r′)\∂n= n.\grad(G(q,r′))

    so what is the antiderivative of the normal derivative?please help me in doing this??
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