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Math Help - Green's function for Neumann problem

  1. #1
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    Green's function for Neumann problem

    Hi, I'll post the question and how far I got with it.

    Using the method of images, construct the Green's function of the Neumann problem for Laplace's equation in the half-space D = {(x,y,z) \in \Re^{3} : z > 0}. Hence solve:

    \nabla^{2}u = 0, \ \ x\in D
    \frac{\partial u}{\partial z} = 1 - x^{2} - y^{2}, \ \ x^{2} + y^{2} < 1, \ \ z = 0
    \frac{\partial u}{\partial z} = 0, \ \ x^{2} + y^{2} > 1, \ \ z = 0

    and evaluate the resulting integral on the z-axis.

    =========

    So the first thing I did was to write down the fundamental solution to Laplace's equation in 3D, and take off the image points in the lower half-space (z<0). i.e.

    G(x,y,z;,x',y',z') = -\frac{1}{4\pi} \left( \frac{1}{\sqrt{(x-x')^{2} + (y-y')^{2} + (z-z')^{2}}}\right) -\frac{1}{4\pi} \left( \frac{1}{\sqrt{(x-x')^{2} + (y-y')^{2} + (z+z')^{2}}}\right)

    Now to solve, we use Green's identity.

    Integrating u\nabla^{2} G - G\nabla^{2} u \ over the upper half-space leaves just u(x',y',z') on one side (since \nabla^{2} u vanishes for x\in D).

    For the other side, we can write \frac {\partial}{\partial n} = -\frac {\partial}{\partial z} \ since the outward normal is going in the opposite direction to increasing z.
    We can also say \frac {\partial G}{\partial z} = 0 at z = 0 \ \ (as it's a Neumann problem..?)

    u(x',y',z') = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} G(x,y,0;,x',y',z')\frac{\partial u}{\partial z}\ dxdy

    We now evaluate this resulting integral on the z-axis to get:

    u(0,0,z') = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} G(x,y,0;,0,0,z')\frac{\partial u}{\partial z}\ dxdy


    I get stuck at this point. In the solutions, the above has been transformed into polars. So on the z-axis, G becomes (leaving the constant -\frac{1}{4\pi} outside the integral),

    \left (\frac{1}{\sqrt{x^{2} + y^{2} + z'^{\ 2}}} + \frac{1}{\sqrt{x^{2} + y^{2} + z'^{\ 2}}} \right) = \frac{2}{\sqrt{r^{2} + z'^{\ 2}}}

    And the dxdy goes to rdrd\theta, as usual

    The next line of the solution is:

    u(0,0,z') = -\frac{1}{4\pi}\int_{\theta = 0}^{2\pi} \int_{r = 0}^{1} \frac{2}{\sqrt{r^{2} + z'^{\ 2}}} \ rdrd\theta


    I don't understand what's happened to the \frac{\partial u}{\partial z} - is it equal to 1? Something to do with the boundary conditions maybe?
    After this step, it looks pretty straightforward, just normal integration.

    Any help is much appreciated!

    Thanks.
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  2. #2
    Super Member Rebesques's Avatar
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    what's happened to the \frac{\partial u}{\partial z} - is it equal to 1? Something to do with the boundary conditions maybe?
    Εxactly. You have \frac{\partial u}{\partial z}(x,y,0)=1-x^2-y^2 for x^2+y^2<1, so \frac{\partial u}{\partial z}(0,0,0)=1.
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