Homework Help: Poisson Summation in Heat Equation (Polar Coordinates)

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1. Oct 28, 2014

kamion42

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

I'm currently trying to follow a derivation done by Shankar in his "Basic Training in Mathematics" textbook. The derivation is on pages 343-344 and it is based on the solution to the two dimensional heat equation in polar coordinates, and I'm not sure how he gets from one step to another.

2. Relevant equations

The derived equation in question is Equation 10.5.80:

$u(r,\theta) = \frac{1}{2\pi} \int_{0}^{2\pi} [\frac{a^{2}-r^{2}}{a^{2}+r^{2}-2ra\cos(\theta-\theta')}] u(a,\theta') d\theta'$

which he claims to be a straightforward geometric sum over integer $\textit{m}$ from Equation 10.5.79:

$u(r,\theta) = \sum_{m=-\infty}^\infty (\frac{r}{a})^{|m|} \int_{0}^{2\pi} e^{im(\theta-\theta')} u(a,\theta') d\theta'$.

3. The attempt at a solution

Trying to do the infinite sum, I first interchanged the summation and the integral to get:

$u(r,\theta) = \int_{0}^{2\pi} \sum_{m=-\infty}^\infty (\frac{r}{a})^{|m|} e^{im(\theta-\theta')} u(a,\theta') d\theta'$.

And then since the geometric summation is from negative infinity to positive infinity, I broke it up into two sums:

$u(r,\theta) = \int_{0}^{2\pi} u(a,\theta') d\theta' [\sum_{m=0}^\infty (\frac{r}{a} \times e^{i(\theta-\theta')})^{m} + \sum_{m=-\infty}^0 (\frac{r}{a} \times e^{-i(\theta-\theta')})^{-m} ]$

And then changing the negative sign on the second sum:

$u(r,\theta) = \int_{0}^{2\pi} u(a,\theta') d\theta' [\sum_{m=0}^\infty (\frac{r}{a} \times e^{i(\theta-\theta')})^{m} + \sum_{m=0}^\infty (\frac{r}{a} \times e^{-i(\theta-\theta')})^{m} ]$

Next, I use the definition of a geometric sum (assuming that $|\frac{r}{a} \times e^{\pm i(\theta-\theta')}|$ is less than 1,) and get:

$u(r,\theta) = \int_{0}^{2\pi} u(a,\theta') d\theta' [\frac{1}{1-\frac{r}{a}e^{i(\theta-\theta')}} + \frac{1}{1-\frac{r}{a}e^{-i(\theta-\theta')}}]$

And get after algebraic simplification:

$u(r,\theta) = \int_{0}^{2\pi} u(a,\theta') d\theta' [\frac{2(a^{2} -ra\cos(\theta - \theta'))}{a^{2}+r^{2}-2ra\cos(\theta - \theta')}]$

Which is different from the answer I was supposed to get by a factor of $\frac{1}{2\pi}$ and the numerator is off by a significant amount. Does anyone have any advice as to how to get a factor of $2\pi$?

At one point at the end of the explanation, he says that the integral is "due to Poisson" but this has never been previously mentioned in the entire book. Maybe I am missing something?

2. Oct 29, 2014

kamion42

I figured it out! There's an error in the text where they forgot to leave the $\frac{1}{2\pi}$ in the numerator, and also I made an error in the summation earlier (counted zero twice!).