22.5 Finding the Green's function

The previous section found the solution to the ideal flow in a circle in the form

\begin{displaymath}
u = A_0 + \sum_{n=1}^\infty
\left\{
f^1_n \frac{r^n}{n...
...(n\vartheta) + f^2_n\frac{r^n}{n} \sin(n\vartheta)
\right\}
\end{displaymath}

We can write it directly in terms of the given $f(x)$ if we substitute in the expressions for the Fourier coefficients:

\begin{displaymath}
u = A_0 + \sum_{n=1}^\infty
\int_0^{2\pi} f(\phi) \cos(n...
...) \sin(n\phi) { \rm d}\phi \frac{r^n}{n\pi} \sin(n\vartheta)
\end{displaymath}

We can clean it up by combining terms and interchanging integration and summation:

\begin{displaymath}
u = A_0 + \int_0^{2\pi}
\sum_{n=1}^\infty
\left\{
\f...
...n(n\phi) \sin(n\vartheta)]
\right\}
f(\phi) { \rm d}\phi
\end{displaymath}


\begin{displaymath}
u = A_0 + \int_0^{2\pi}
\left\{\sum_{n=1}^\infty \frac{r^n}{n\pi}\cos(n[\vartheta-\phi])\right\}
f(\phi) { \rm d}\phi
\end{displaymath}

This we can clean up even more by giving a name to the function within the curly brackets:

\begin{displaymath}
u = A_0 + \int_0^{2\pi} G(r,\vartheta-\phi) f(\phi) { \rm d}\phi
\end{displaymath}

Nice, not? We can even simplify $G$ by converting to complex exponentials and differentiating:

\begin{displaymath}
G(r,\vartheta) = \sum_{n=1}^\infty \frac{r^n}{n\pi}\cos(n\...
...in\vartheta} +
\frac{r^n}{2n\pi}e^{-in\vartheta}
\right\}
\end{displaymath}


\begin{displaymath}
2\pi \frac{\partial G}{\partial r} =
\sum_{n=1}^\infty
...
...^{i\vartheta}} +
\frac{e^{-i\vartheta}}{1-re^{-i\vartheta}}
\end{displaymath}

The last equation applies because the sums are geometric series.

Integrating and cleaning up produces

\begin{displaymath}
G(r,\vartheta) = -\frac{1}{2\pi} \ln\left(1-2r\cos(\vartheta)+ r^2\right)
\end{displaymath}

So, we finally have the following Poisson-type integral expression giving $u$ directly in terms of the given $f(\vartheta)$, with no sums:

\begin{displaymath}
u(r,\vartheta) = A_0 -\frac{1}{2\pi}
\int_0^{2\pi} \ln\left(1-2r\cos(\vartheta-\phi)+ r^2\right) f(\phi){ \rm d}\phi
\end{displaymath}

Neat!