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Proving Poisson's equation in 2D

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Proving Poisson's equation in 2D

    Prove that in two dimensions:
    [tex]\nabla ^2 \ln \left| \vec{r} - \vec{r}_0 \right| = 2\pi \delta (\vec{r} - \vec{r}_0 ) [/tex]
    and use your findings to find a solution to:
    [tex]\nabla ^2 f(\vec{r}) =t(\vec{r})[/tex]

    where [tex]t(\vec{r})[/tex] is a an integrable function of continuous derivatives in (a,b).

    2. Relevant equations

    (shown above)

    3. The attempt at a solution

    Given

    [tex]\nabla ^2 \ln \left| \vec{r} - \vec{r}_0 \right| = 2\pi \delta (\vec{r} - \vec{r}_0 ) [/tex]

    Multiply both sides by [tex]t(\vec{r} - \vec{r}_0 ) [/tex] and integrate both sides in the 2D plane. On the left-hand side:

    [tex]\int \int t(\vec{r} - \vec{r}_0 ) \nabla ^2 \ln \left| \vec{r} - \vec{r}_0 \right| d^2 r = \int \int t(\vec{r ^{\prime}} ) \nabla ^2 \ln \left| \vec{r ^{\prime}} \right| d^2 r ^{\prime} = \int \int t(\vec{r ^{\prime}} ) \nabla ^2 \ln r ^{\prime} d^2 r ^{\prime} = \int \int t(\vec{r ^{\prime}} ) \frac{1}{\left( r^{\prime} \right) ^2} \frac{\partial}{\partial r^{\prime}} \left\{ \left( r^{\prime} \right) ^2 \left[ \frac{\partial}{\partial r^{\prime}} \ln r \right] \right\} d^2 r ^{\prime} = \int \int t(\vec{r ^{\prime}} ) \left[ \frac{1}{ \left( r^{\prime} \right) ^2 } \right] d^2 r ^{\prime} =\int _0 ^{\infty} \int _0 ^{2\pi} t(\vec{r ^{\prime}} ) \left[ \frac{1}{ \left( r^{\prime} \right) ^2 } \right] \left( r ^{\prime} d\phi ^{\prime} \right) d r ^{\prime} = 2\pi \int _0 ^{\infty} t(\vec{r ^{\prime}} ) \frac{d r ^{\prime}}{r ^{\prime}} = 2\pi \left[ \left. t(\vec{r ^{\prime}} ) \ln ( r ^{\prime} ) \right| _0 ^{\infty} - \int _0 ^{\infty} \ln ( r ^{\prime} ) t ^{\prime} (\vec{r ^{\prime}} ) \mbox{ } d r ^{\prime} \right] = 2\pi \left[ t(\vec{r _0}) - (-t(\vec{r _0})) \right] = 4\pi \mbox{ } t(\vec{r _0} )[/tex]

    we get an integrable function. On the right-hand side:

    [tex]\int \int t(\vec{r} - \vec{r}_0 ) \left[ 2\pi \mbox{ } \delta (\vec{r} - \vec{r}_0 ) \right] d ^2 r = \int \int t(\vec{r ^{\prime}} ) \left[ 2\pi \mbox{ } \delta (\vec{r ^{\prime}} ) \right] d ^2 r^{\prime} = \int _0 ^{\infty} \int _0 ^{2\pi} t(\vec{r ^{\prime}} ) \left[ 2\pi \mbox{ } \delta (\vec{r ^{\prime}} ) \right] \left( r ^{\prime} d\phi ^{\prime} \right) d r ^{\prime} = \int _0 ^{\infty} \int _0 ^{2\pi} t(\vec{r ^{\prime}} ) \left[ 2\pi \mbox{ } \frac{ \delta ( r ^{\prime} )}{ \pi \mbox{ } r ^{\prime}} \right] \left( r ^{\prime} d\phi ^{\prime} \right) d r ^{\prime} = 4\pi \int _0 ^{\infty} t(\vec{r ^{\prime}} ) \delta ( r ^{\prime} ) d r ^{\prime} = 4\pi \mbox{ } t(\vec{r _0} )[/tex]

    Note: I'm using the substitutions

    [tex]\vec{r^{\prime}}=\vec{r}-\vec{r_0}[/tex]

    and

    [tex]r^{\prime} = \left| \vec{r^{\prime}} \right|[/tex]

    The integration by parts on the left-hand side expansion assumes non-zero r-primes vanish, and I guess my main question at the moment is if that expression and its remainder are correct. In other words, is this:

    [tex]2\pi \int _0 ^{\infty} t(\vec{r ^{\prime}} ) \frac{d r ^{\prime}}{r ^{\prime}} = 2\pi \left[ \left. t(\vec{r ^{\prime}} ) \ln ( r ^{\prime} ) \right| _0 ^{\infty} - \int _0 ^{\infty} \ln ( r ^{\prime} ) t ^{\prime} (\vec{r ^{\prime}} ) \mbox{ } d r ^{\prime} \right] = 2\pi \left[ t(\vec{r _0}) - (-t(\vec{r _0})) \right] = 4\pi \mbox{ } t(\vec{r _0} )[/tex]

    right?

    For the second part, I only have:

    [tex]\nabla ^2 f(\vec{r}) =t(\vec{r})[/tex]

    [tex] f(\vec{r}) = \int \int d^2 r \mbox{ } t(\vec{r}) = \int _0 ^{\infty} \int _0 ^{2\pi} t(\vec{r}) (r \mbox{ } d\phi) \mbox{ }dr = 2\pi \int _0 ^{\infty} r \mbox{ } t(\vec{r}) \mbox{ } dr [/tex]

    and I think I'm supposed to get a delta function. I'm pretty sure I'm undoing the Laplacian incorrectly.

    Can someone help?
     
  2. jcsd
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