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Solve the screened Poisson equation

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve the equation
    [tex]
    \nabla^2\phi-\frac{1}{\lambda^2_D}\phi=-\frac{q_T}{\epsilon_0}\delta(r)
    [/tex]
    substituting the [itex]\delta[/itex] representation
    [tex]
    \delta(r)=\frac{1}{4\pi}\frac{q_T}{r}
    [/tex]
    and writing the laplacian in spherical coordinates. Use as your guess
    [tex]
    \phi=\frac{1}{4\pi\epsilon_0}\frac{g(r)}{r}
    [/tex]
    and show
    [tex]
    g(r)=\exp(-r/\lambda_D)
    [/tex]

    3. The attempt at a solution
    I know how to solve this equation using the Green and Fourier formalism, but here I am asked to solve it "the easy way" (as said in "Introduction to plasma physics - Bellan").

    All I can say is
    [tex]
    \nabla^2\left(\phi-\frac{q_T}{4\pi\epsilon_0}\frac{1}{r}\right)-\frac{\phi}{\lambda^2}=0
    [/tex]
    Using the guess
    [tex]
    \phi=\frac{1}{4\pi\epsilon_0}\frac{g(r)}{r}
    [/tex]
    I have
    [tex]
    \frac{1}{r^2}\partial_r\left[r^2\partial_r\left(\frac{g(r)-q_T}{4\pi\epsilon_0r}\right)\right]-\frac{g}{4\pi\epsilon_0r\lambda^2}=0
    [/tex]
    [tex]
    \frac{1}{r^2}\partial_r\left[r^2\left(\frac{r\partial_r g-(g-q_T)}{r^2}\right)\right]-\frac{g}{r\lambda^2}=0
    [/tex]
    [tex]
    \frac{1}{r^2}(\partial_r g+r\partial_r^2g-\partial_r g)-\frac{g}{r\lambda^2}=0
    [/tex]
    And so the final differential equation for g(r) is
    [tex]
    \frac{\partial_r^2g}{r}-\frac{g}{r\lambda^2}=0
    [/tex]
    which has the physical solution
    [tex]
    g(r)=\exp(-r/\lambda)
    [/tex]

    This result is correct, but my doubt is: the [itex]\delta[/itex] term didn't play any role, I mean, if I solved the equation
    [tex]
    \nabla^2\phi-\frac{1}{\lambda^2_D}\phi=0
    [/tex]
    I would have obtained the same result.
    I suppose that without the [itex]\delta[/itex] term there is some problem in r=0, but I don't fully understand what is the problem
     
  2. jcsd
  3. Dec 29, 2013 #2

    haruspex

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    How do you get that? Isn't ##\nabla^2\left(\frac 1r\right) = 0##?
    If so, don't we get ##\nabla^2\left(\phi-\frac{q_T}{4\pi\epsilon_0}\frac{\lambda^2}{r}\right)-\frac{1}{\lambda^2}\left(\phi-\frac{q_T}{4\pi\epsilon_0}\frac{\lambda^2}{r}\right)=0##
     
  4. Dec 29, 2013 #3
    No, the function 1/r is singular at r=0, so the laplacian is proportional to a delta, i.e., it is zero everywhere except at r=0.
     
  5. Dec 29, 2013 #4

    Dick

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    Right. All you can conclude from ##\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=0## is that ##\phi=C \frac{e^{-\frac{r}{\lambda_D}}}{r}## where C is some constant. If you don't have a ##\delta## term then C=0. The ##\delta## function source let's you fix C. The shortcut way is to consider a sphere around the origin of radius r and let r->0 applying the divergence theorem. Show you can ignore the correction from the exponential factor in that limit.
     
  6. Dec 30, 2013 #5
    But the solution
    ##\phi=C \frac{e^{-\frac{r}{\lambda_D}}}{r}##
    satisfies ##\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=0## for every value of the constant C, not only for C=0
     
  7. Dec 30, 2013 #6
    I tried to do that:
    [tex]
    \int_V\nabla\left(\nabla\phi\right)=\int_S-C\frac{\exp(-r/\lambda)}{\lambda r^2}(r+\lambda)=-C\frac{\exp(-r/\lambda)}{\lambda}(r+\lambda)4\pi\:\:\:\:\:\:\:(1)
    [/tex]
    But from ##\nabla^2\phi-\frac{1}{\lambda^2}\phi ## I have
    [tex]
    \int_V\nabla\left(\nabla\phi\right)=\int_V \frac{1}{\lambda^2}C\frac{\exp(-r/\lambda)}{r}=\int_0^r r^2dr\int_0^\pi \sin(\theta)d\theta\int_0^{2\pi}d\phi\,\left(-\frac{1}{\lambda^2}\right)C\frac{\exp(-r/\lambda)}{r}=
    \\
    -\frac{4\pi}{\lambda}C\exp(-r/\lambda)(\lambda+r)-4\pi C \:\:\:\:\:\:\:(2)
    [/tex]
    For formula (1) and (2) to be the same the constant C must be 0.
    The incorrect term ##-4\pi C ## is indeed the value of (1) for ## r\to 0## and can be eliminated with a delta source.
    Is this reasoning correct?
     
  8. Dec 30, 2013 #7

    Dick

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    That is EXACTLY my point. But you don't want to solve ##\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=0##, that's only true for r>0. You want to solve ##\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=-\frac{q_T}{\epsilon_0}\delta(r)##. Integrate both sides over a sphere centered on the origin of radius r. The right side gives you ##-\frac{q_T}{\epsilon_0}## no matter what r is. The integral of ##\frac{1}{\lambda^2_D}\phi## will go to zero as r->0. That just leaves the integral of ##\nabla^2\phi##. As you showed in part (1) of post 7 that consists of a constant part and another part that vanishes as r->0. Now just equate the parts that don't vanish as r->0. Solve for C.
     
  9. Dec 30, 2013 #8

    vanhees71

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    I'd use spherical coordinates + symmetry to begin with, i.e., setting [itex]\phi(\vec{r})=\phi(r)[/itex] with [itex]r=|\vec{r}|[/itex]. Of course you can do that only for [itex]r>0[/itex], because the spherical coordinates are singular along the entire [itex]z[/itex] axis.

    Then adopt the (two) free parameters of the general solution of the (ordinary!) differential equation for [itex]\phi(r)[/itex]. One is [itex]\phi \rightarrow 0[/itex] for [itex]r \rightarrow \infty[/itex], which leaves you with the solution already found earlier in the tread.

    Then use Dick's method to find the remaining constant [itex]C[/itex].
     
  10. Dec 31, 2013 #9
    Thank you everybody for your help
     
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