1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Green's function representation of electric potential

  1. Jun 25, 2012 #1

    hunt_mat

    User Avatar
    Homework Helper

    Hi,

    I have the following problem, I have an electric field (which no charge) which satisfies the usual Laplace equation:
    [tex]
    \frac{\partial^{2}V}{\partial x^{2}}+\frac{\partial^{2}V}{\partial y^{2}}+\frac{\partial^{2}V}{\partial z^{2}}=0
    [/tex]
    in the region [itex]\mathbb{R}^{2}\times [\eta ,\infty ][/itex]. So basically it is the upper half z-plane where the boundary is some fixed surface [itex]\eta[/itex], I also know that on this surface:
    [tex]
    \frac{\partial V}{\partial x}=\frac{\partial\eta}{\partial x}
    [/tex]

    I can do this in 2D by the use of the Hilbert transform. Any suggestions?
     
  2. jcsd
  3. Jun 25, 2012 #2
    Let me make sure I understand: this is a region for [itex]z > \eta[/itex] for some fixed [itex]\eta[/itex], or is [itex]\eta[/itex] a function?
     
  4. Jun 25, 2012 #3

    hunt_mat

    User Avatar
    Homework Helper

    [itex]\eta[/itex] is a function but with further thought the region could be set to [itex]z\geqslant 0[/itex] and I think that this will make the problem easier.
     
  5. Jun 25, 2012 #4
    And if I read what you said correctly, you only know the value of the x-component of the electric field on this surface?
     
  6. Jun 25, 2012 #5

    hunt_mat

    User Avatar
    Homework Helper

    I think that you can also say that [itex]V=\eta[/itex] due to other considerations too.
     
  7. Jun 28, 2012 #6

    hunt_mat

    User Avatar
    Homework Helper

    So I think I have solved this problem, I took [itex]V[/itex] and [itex]\eta[/itex] to be of the same size but small and reduced the complexity of the problem somewhat and the domain is now: [itex]\mathbb{R}^{2}\times [ 0,\infty )[/itex], using Green's second formula, I can write the solution as an integral over the boundary:
    [tex]
    V(x,y,z)=\int_{\mathbb{R}^{2}}g\partial_{z}V-V\partial_{z}g\Big|_{z'=0}d\Sigma
    [/tex]
    Where [itex]g[/itex] is the Green's function for Laplaces's equation for the half space given by:
    [tex]
    g(x,y,x|x',y',z')=\frac{1}{4\pi\sqrt{(x-x')^{2}+(y-y')^{2}+(z-z')^{2}}}-\frac{1}{4\pi\sqrt{(x-x')^{2}+(y-y')^{2}+(z+z')^{2}}}
    [/tex]
    Then using the boundary condition [itex]V=\eta[/itex], then the solution becomes:
    [tex]
    u=\int_{\mathbb{R}^{2}}u\frac{\partial g}{\partial z}d\Sigma
    [/tex]
     
    Last edited: Jun 28, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Green's function representation of electric potential
  1. Green's function (Replies: 7)

  2. Green function (Replies: 2)

Loading...