Algebraic simplification of a function arising from Poisson's Formula

Question

In a past-exam for my PDE class, there is a question involving finding a solution to the Laplace equation (in radial terms) at a boundary $f(\theta)=\cos \theta$

Consider the Laplace equation in a circular domain of radius $a>0$. Given that the general solution $u(r,\theta)$ of this equation in polar coordinates $$u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=0, \quad r<a$$ which additionally satisfies the boundary condition $$u(a,\theta)=f(\theta)$$ is given by Poisson's formula $$u = \frac{a^2-r^2}{2\pi} \int_0^{2\pi}\frac{f(\xi)}{a^2-2ar\cos(\theta-\xi)+r^2} \mathrm{d}\xi$$ Find the solution, using Poisson's formula, when $f(\theta)=\cos \theta$, given that one may use without proof: $$\int_{0}^{2\pi}\frac{\cos z}{1-\alpha \cos z} = 2\pi\frac{1-\sqrt{1-\alpha^2}}{\alpha\sqrt{1-\alpha^2}}, \quad |\alpha|<1$$

Substituting the boundary condition and integrating is not difficult, and one reaches to the solution:

$$u=\cos \theta \frac{a^2-r^2}{a^2+r^2} \frac{1-\sqrt{1-\frac{4a^2r^2}{(a^2+r^2)^2}}}{\frac{2ar}{a^2+r^2} \sqrt{1-\frac{4a^2r^2}{(a^2+r^2)^2}}}$$

the annoyance arrives in the following, by simplifying (given that $0<r<a$):

\begin{align} \cos \theta \frac{a^2-r^2}{a^2+r^2} \frac{1-\sqrt{1-\frac{4a^2r^2}{(a^2+r^2)^2}}}{\frac{2ar}{a^2+r^2} \sqrt{1-\frac{4a^2r^2}{(a^2+r^2)^2}}} =&\cos \theta \frac{a^2-r^2}{2ar} \frac{1-\sqrt{1-\frac{4a^2r^2}{(a^2+r^2)^2}}}{ \sqrt{1-\frac{4a^2r^2}{(a^2+r^2)^2}}}\\ =&\cos \theta \frac{a^2-r^2}{2ar} \frac{1-\frac{1}{a^2+r^2}\sqrt{(a^2+r^2)^2-4a^2r^2}}{ \frac{1}{a^2+r^2}\sqrt{(a^2+r^2)^2-4a^2r^2}}\\ =&\cos \theta \frac{a^2-r^2}{2ar} \frac{a^2+r^2-\sqrt{(a^2+r^2)^2-4a^2r^2}}{ \sqrt{(a^2+r^2)^2-4a^2r^2}}\\ =&\cos \theta \frac{a^2-r^2}{2ar} \frac{a^2+r^2-\sqrt{a^4+2a^2r^2+r^4-4a^2r^2}}{ \sqrt{a^4+2a^2r^2+r^4-4a^2r^2}}\\ =&\cos \theta \frac{a^2-r^2}{2ar} \frac{a^2+r^2-\sqrt{a^4-2a^2r^2+r^4}}{ \sqrt{a^4-2a^2r^2+r^4}}\\ =&\cos \theta \frac{a^2-r^2}{2ar} \frac{a^2+r^2-\sqrt{(a^2-r^2)^2}}{ \sqrt{(a^2-r^2)^2}}\\ =&\cos \theta \frac{a^2-r^2}{2ar} \frac{a^2+r^2-\sqrt{(r^2-a^2)^2}}{ \sqrt{(r^2-a^2)^2}}\\ =&\cos \theta \frac{a^2-r^2}{2ar} \frac{a^2+r^2-|r^2-a^2|}{ |r^2-a^2|}\\ \end{align} Since $0<r<a$ then $r^2<a^2$ then $|r^2-a^2|=a^2-r^2$ \begin{align} =&\cos \theta \frac{a^2-r^2}{2ar} \frac{a^2+r^2-(a^2-r^2)}{ (a^2-r^2)} \\ =&\cos \theta \frac{2r^2}{2ar}\\ u=&\cos \theta \frac{r}{a} \end{align} which is the desired solution.

Since is this on an previous exam I should expect to maybe encounter solutions of similar complexity. Is there anyway I can simplify the solution in a much faster manner since there are many terms arising from the binomial $(a^2+r^2)$ such as $2ar$ or $4a^2r^2$ when it is squared.

I don't want to spend too much time on the exam trying to simplify such things at the risk of making a stupid algebraic error causing me to get stuck.

Can anyone give me any quick algebraic realizations one can see to arrive to the solution quicker?

Answer 1

This may not be the answer you're looking for.

There is a way to "see" the solution without doing the integral. If you're familiar with separation of variables, the general solution of Laplace's equation on a disk has the form

$$ u(r,\theta) = A_0 + \sum_{n=1}^\infty r^{n}\big(A_n\cos (n\theta) + B_n\sin (n\theta)\big) $$

The boundary condition reads

$$ u(a,\theta) = A_0 + \sum_{n=1}^\infty a^n\big(A_n\cos(n\theta)+B_n\sin(n\theta)\big) = \cos(\theta) $$

When comparing terms, the only non-zero coefficent is $A_1 = \frac{1}{a}$, so the solution is

$$ u(r,\theta) = \frac{r}{a}\cos(\theta) $$

Edit: Your working is pretty much correct. If you want to be more efficient, consider reducing the square root first:

\begin{align} \sqrt{1-\alpha^2} &= \sqrt{1-\frac{4a^2r^2}{(a^2+r^2)^2}} \\ &= \sqrt{\frac{a^4 - 2a^2r^2 + r^4}{(a^2+r^2)^2}} \\ &= \sqrt{\frac{(a^2-r^2)^2}{(a^2+r^2)^2}} \\ &= \frac{a^2-r^2}{a^2+r^2} \end{align}

Then

\begin{align} \frac{1-\sqrt{1-\alpha^2}}{\alpha\sqrt{1-\alpha^2}} &= \frac{1-\frac{a^2-r^2}{a^2+r^2}}{\frac{2ar(a^2-r^2)}{(a^2+r^2)^2}} \\ &= (a^2+r^2)\frac{(a^2+r^2)-(a^2-r^2)}{2ar(a^2-r^2)} \\ &= (a^2+r^2)\frac{2r^2}{2ar(a^2-r^2)} \\ &= \frac{r}{a}\frac{a^2+r^2}{a^2-r^2} \end{align}

and the rest follows.

Answer 2

Here, the method of Separation of Variables is very efficient and useful. In fact any answer to the above Laplace equation has a general answer as follows:$$u(r,\theta)=e_0+e_1\ln r+\sum_{n=1}^{\infty} (a_nr^n+b_nr^{-n})\cdot(c_n\cos n\theta+d_n\sin n\theta)$$The boundedness on $0<r<a$ implies that $$e_1=0\\b_n=0$$therefore$$u(r,\theta)=e_0+\sum_{n=1}^{\infty}r^n(\hat c_n\cos n\theta+\hat d_n\sin n\theta)$$and finally by imposing the boundary conditions we conclude:$$u(r,\theta)={r\over a}\cdot\cos\theta$$

Edit

If you want to simplify your fraction in a faster manner, note that$$\sqrt{1-{4a^2r^2\over (r^2+a^2)^2}}={r^2-a^2\over r^2+a^2}$$