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Solve Laplace equation

  1. Aug 6, 2014 #1
    1. The problem statement, all variables and given/known data
    I got to this while solving a physical problem, therefore it is hard for me to write the problem statement if there is none. I can write the whole physical problem here but it wouldn't make any sense.
    So, here is how it goes.

    I got to a Laplace equation ##\triangle \phi =0## and I also have boundary conditions that say:

    ##\phi (x=-r/2,y=0, z=d) = -U/2##

    ##\phi (x=r/2,y=0,z=d)= U/2## and

    ##\frac{d}{dz} \phi (x,y,z=d)=0##
    ##\frac{d}{dz} \phi (x,y,z=0)=0##


    2. Relevant equations



    3. The attempt at a solution

    We have actually never done this before, so I don't really know how to start. Any ideas? :/
     
  2. jcsd
  3. Aug 6, 2014 #2
    [itex]\phi=\frac{Ux}{r}[/itex]?
     
  4. Aug 6, 2014 #3
    Ammmm. ? :D
     
  5. Aug 6, 2014 #4

    SteamKing

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    It would probably be better to write the whole problem out. As it is, what you have provided raises as many questions as it answers.
     
  6. Aug 6, 2014 #5

    Orodruin

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    Your boundary conditions are not sufficient to uniquely define the solution, they are all fulfilled by Chester's solution but I can add an arbitrary linear function of y to this and still have a solution fulfilling all your boundary conditions. What is the region you are trying to solve in? As it is you have just given boundary conditions in two planes (z=0 and z=d) as well as two points (which happen to also lie in the plane z = d ...).
     
  7. Aug 6, 2014 #6
    Problem statement:
    On a thin infinite metal plate with thickness ##d## and specific conductivity ##\sigma ## two electrodes are attached. The distance between the two electrodes is ##r## and the potential is ##U##.
    Describe magnetic field anywhere in space, due to the currents in stationary state on the metal plate. (Ignore the wires which bring the current to the electrodes)
    End of problem

    We already discussed this on this forum, and we found out that:

    ##\nabla \cdot {\bf J}=-\frac{\partial \rho}{\partial t}## since we have stationary conditions, this is than ##\nabla \cdot {\bf J}=0##.

    Now because ##J=\sigma E## than ##\nabla E=0## but ##E=-\nabla \phi## so ##\nabla ^2 \phi =0##.

    Now the boundary conditions:
    ##\phi (x=-r/2,y=0, z=d) = -U/2##
    ##\phi (x=r/2,y=0,z=d)= U/2##

    and also ##\frac{d}{dz} \phi (x,y,z=d)=0## beacuse there is no component of electric field perpendicular to the surface of the plate. It might be inside the metal plate, but I don't care about that, it is Important that there is none on the surface of the plane. And I guess the same goes for the other side of the plate, so for ##z=0## than ##\frac{d}{dz} \phi (x,y,z=0)=0##


    Now I have to solve this.
     
  8. Aug 6, 2014 #7
    Here is how I started:

    ##\nabla ^2 \phi =0##

    Now let's say that ##\phi =X(x)Y(y)Z(z)##, so ##\nabla ^2 \phi = \frac{\partial^2 }{\partial x^2}\phi +\frac{\partial^2 }{\partial y^2}\phi +\frac{\partial^2 }{\partial z^2}\phi=##
    ##=X^{''}(x)Y(y)Z(z)+X(x)Y^{''}(y)Z(z)+X(x)Y(y)Z^{''}(z)=0## dividing whole expression with ##X(x)Y(y)Z(z)## leaves me with

    ##\frac{1}{X(x)}X^{''}(x)+\frac{1}{Y(y)}Y^{''}(y)+\frac{1}{Z(z)}Z^{''}(z)=0## which can only be true if each term is in fact a constant, so

    ##\frac{1}{X(x)}X^{''}(x)=a^2##
    ##\frac{1}{Y(y)}Y^{''}(y)=b^2## and
    ##\frac{1}{Z(z)}Z^{''}(z)=-c^2##

    One of them obviously has to be negative, because ##a^2+b^2-c^2=0##.

    Now solving each DE above, leaves me with
    ##X(x)=C_1sin(ax)+C_2cos(ax)##
    ##Y(y)=C_3sin(bx)+C_4cos(bx)## and
    ##Z(z)=C_5sinh(cx)+C_6cosh(cx)##

    So now I know that ##\phi =X(x)Y(y)Z(z)=[C_1sin(ax)+C_2cos(ax)][C_3sin(bx)+C_4cos(bx)][C_5sinh(cx)+C_6cosh(cx)]##

    I guess the next step (if so far everything looks fine) would be to determine the value of all those constants. Right?
     
  9. Aug 7, 2014 #8
    Nobody? :/
     
  10. Aug 8, 2014 #9
    Ok, the way I started, the problem has no solution, beacuse ##\nabla ^2 \phi =0## everywhere except on the electrodes, unless also ##U=0## which is nonsense of course.

    So I guess I have to finally define the shape of the electrodes.
     
  11. Aug 8, 2014 #10

    HallsofIvy

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    In what sense are these "boundary conditions"? These are at single points.

     
  12. Aug 8, 2014 #11
    Yes, these boundary conditions were written when I thought that it is ok to imagine that the electrodes are infinitesimally small - points.
    But if that would be true, than the problem has no solution; unless also ##U=0##.

    So I guess I have to consider some dimension to the electrodes, which will of course also change my boundary conditions.
     
  13. Aug 9, 2014 #12
    For example, let's say that electrodes are circles with radius ##R##, where ##R<<r##. ##r## is the distance between the electrodes.
    I will also use notation ##a=\frac r 2 - R##

    This gives me a new set of boundary conditions:

    ##\phi (x=\pm a,y=0)=-\frac U 2##
    ##\phi (x=\pm \infty , y=\pm \infty )=0##

    Where I also ignored the ##z## component. I think I can do that since the plate is infinite.

    Than ##\nabla ^2 \phi =0## or ##\frac{1}{X(x)}X^{''}(x)+\frac{1}{Y(y)}Y^{''}(y)=0##

    So ##\frac{1}{X(x)}X^{''}(x)=-\frac{1}{Y(y)}Y^{''}(y)=-\lambda ^2##

    Therefore ##X(x)=C_1sin(\lambda x)+C_2cos(\lambda x)##

    Now how does one continue?
     
  14. Aug 9, 2014 #13

    Orodruin

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    The problem here is that this separation of variables is not going to lead you to the eigenfunctions of the problem as these are dependent on a geometry that is not described as simple coordinate surfaces. You could perhaps try some Green's function methods with mirror charges.
     
  15. Aug 9, 2014 #14
    Can you be more specific? Maybe some links, so I can educate myself about that method?
     
  16. Aug 9, 2014 #15

    Orodruin

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    I would start with the Wikipedia page on Green's functions. It should also be described in any textbook on solving PDE problems in physics.
     
  17. Aug 13, 2014 #16
    Ok, I have never worked with the Green's function before so I might need even more help than I normally would. But already normally I need quite a lot of help :D

    So, If I understood wikipedia right, the goal is to solve
    ##\phi(x)=\int _V G(x,{x}')\rho({x}')d^3{x}'+\int _S\left [ \phi({x}'){\nabla }'G(x,{x}')-G(x,{x}'){\nabla }'\phi({x}') \right ]d{\hat{\sigma }}'##

    Where ##\phi ## is electric potential, ##\rho ## is the charge density, ##G## is Greens function.

    But I have to do that in two dimensions, somehow.

    So, following the example on wiki: http://en.wikipedia.org/wiki/Green's_function

    ##Lu(x,y)=\nabla ^2u(x,y)=f(x)=0##

    ##{g}''(x,y)=\delta (x-s,y-t)##

    etc.

    Are things so far ok?
     
  18. Aug 13, 2014 #17

    Orodruin

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    Well, the Green's function for the two-dimensional Poisson equation all of R^2 is well known and you should be able to find it in any reference on the subject. However, you should aim to find one that is also zero on the boundary of your region as this will allow you to eliminate one of the surface terms. This can be done mirroring techniques.
     
  19. Aug 13, 2014 #18
    Ok, why are we talking about Poisson equation now? You meant that in general or does my Laplace equation somehow become Poisson?

    According to http://www-users.math.umn.edu/~olver/am_/leq.pdf on page 24

    ##G(x,y,\xi ,\eta)=-\frac{1}{4\pi }log[(x-\xi )^2+(y-\eta )^2]+z(x,y)## where ##z(x,y)=\frac{1}{4\pi }log[(x-\xi )^2+(y-\eta )^2]## is harmonic with the same boundary conditions as the logarithmic potential function.

    Noooow, my conditions sat that ##\phi (x=\pm \infty, y=\pm \infty)=0##. Buuuut this Greens function is not 0 on the boundary. :/
     
  20. Aug 13, 2014 #19

    Orodruin

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    Because the Poisson equation is the Laplace equation with an inhomogeneity. The inhomogeneity is the ##\rho## (charge density) that will be put into the volume integral so this also disappears and you will end up with only one of the surface integrals.

    It is impossible to impose zero potential at infinity in two dimensions (simply because the Green's function is logarithmic). However, as in your case if you pick the potentials of the two electrodes to U/2 and -U/2, you will end up having 0 at infinity anyway because of the symmetry. In fact, it will show already in your Green's function once you have performed the correct mirroring arguments.
     
  21. Aug 18, 2014 #20
    Hmm... I read some online literature, but... Can you be more specific about "correct mirroring arguments"?

    Or a good link would hopefully do. I really have to solve this Laplace. -.- Honestly I am embarrassed how people knew this in 19th century and I am having so much trouble with it in 21st century.
     
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