5.5 Finding the Green's function

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

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

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

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

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

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

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

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

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

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

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

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

The last equation applies because the sums are geometric series.

Integrating and cleaning up produces

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

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

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

Neat!