Test 2 Practice Problem #2

Any of you night owls capable of explaining how to do the second practice problem? I missed the class where we did Poisson and I know some of you are craving potential real estate for your posts.

For a circle of radius 1, using the code from the PolarLaplacian notebook I got a solution that looked like the following:

For radius 2 it is as follows:

The terms piecewise of Poisson's integral formula for that is not 0 is shown below.
$$u (r, \theta) =
\frac{1}{(2 \pi)} \int_0^\frac{\pi}{2} f(\phi)\frac
{(4 - r^2)}{(
4 + r^2 - 4 r cos(\theta - \phi))} d\phi $$

Where:
$$f(\phi )=
\begin{array}{cc}
\{ &
\begin{array}{cc}
1 & 0<\phi <\frac{\pi }{2} \\
0 & \frac{\pi }{2}<\phi <2 \pi \\
\end{array}
\\
\end{array}$$

The plot for radius 2 is:

PDEDiscourseImage1.JPG847x510 66.8 KB

That's cool! Just a quick nitpick though, even though it doesn't matter (since the integral would be 0) it should be plus the integral from $\pi/2$ to $2\pi$ instead of minus.

@gigaliciousness I think @wbartoli's answer here should help.

Cool thanks, I edited to reflect R=2. I have also included the image of the plot as well.

Isn't it only supposed to be integrated over the second quadrant? So the limits of integration should be from $\frac{\pi}{2}$ to $\pi$, right?

Whoa - that's cool! One comment: Can't you simplify the integral by supressing the $f(\varphi)$? I means you've essentially dealt with the piecewise business by restricting the domain of integration - right?

Just to clarify Dr. McClure. Are the bounds of integration from $\pi/2$ to $\pi$ or are they $0$ to $\pi/2$ ? Since it is the first quadrant we are discussing, I thought that it would be from $\pi/2$ to $\pi$ because the bounds of the circle are $-\pi$ to $\pi$.

Caitlin: I don't have the sheet in front of me but, if it is the first quadrant, then the bounds should be $0\leq\theta\leq\pi/2$.

As it turns out, on the sheet it is the second quadrant so the integration should be from $\frac{\pi}{2}$ to $\pi$. Interestingly the symmetry involved makes it essentially the same plot just with the source portion in the second quadrant.