Proving Poisson's equation in 2D

1. Oct 3, 2011

1. The problem statement, all variables and given/known data

Proving Poisson's equation in 2D

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

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

2. Relevant equations

(shown above)

3. The attempt at a solution

Given

$$\nabla ^2 \ln \left| \vec{r} - \vec{r}_0 \right| = 2\pi \delta (\vec{r} - \vec{r}_0 )$$

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

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

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

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

Note: I'm using the substitutions

$$\vec{r^{\prime}}=\vec{r}-\vec{r_0}$$

and

$$r^{\prime} = \left| \vec{r^{\prime}} \right|$$

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:

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

right?

For the second part, I only have:

$$\nabla ^2 f(\vec{r}) =t(\vec{r})$$

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

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

Can someone help?