Proving Poisson's equation in 2D

[DivGradCurl]

Homework Statement

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).

Homework Equations

(shown above)

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?

 

[mark726]

Thank you for your question. Your approach to proving Poisson's equation in 2D is correct and your final expression for the integrable function is valid. As for the second part of the question, your approach is also correct but you have not yet applied the inverse Laplacian operator to your integral expression.

To find a solution to \nabla ^2 f(\vec{r}) =t(\vec{r}), we can start by taking the inverse Laplacian of both sides:

f(\vec{r}) = \int \int d^2 r \mbox{ } \nabla ^{-2} t(\vec{r})

Since we are in two dimensions, the inverse Laplacian operator is simply the logarithm of the distance between the two points, as shown in the first part of the question. Therefore, we can rewrite the above expression as:

f(\vec{r}) = \int \int d^2 r \mbox{ } \ln \left| \vec{r} - \vec{r}_0 \right| t(\vec{r})

This integral expression can then be simplified using the Green's function for the 2D Laplacian, which is given by:

G(\vec{r} - \vec{r}_0) = -\frac{1}{2\pi} \ln \left| \vec{r} - \vec{r}_0 \right|

Substituting this into the above integral expression, we get:

f(\vec{r}) = -\frac{1}{2\pi} \int \int d^2 r \mbox{ } G(\vec{r} - \vec{r}_0) t(\vec{r})

This is the solution to the second part of the question. Note that the Green's function is just the inverse Laplacian of the delta function, which is why we can use it in this context.

I hope this helps clarify your understanding of Poisson's equation in 2D and how to find a solution to it. Keep up the good work in your scientific studies!