Dtjytyk

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
$$u_{rr}+frac{1}{r}u_r+frac{1}{r^2}u_{thetatheta}=0, quad r<a$$
which additionally satisfies the boundary condition
$$u(a,theta)=f(theta)$$
is given by Poisson's formula
$$u = frac{a^2-r^2}{2pi} int_0^{2pi}frac{f(xi)}{a^2-2arcos(theta-xi)+r^2} mathrm{d}xi$$
Find the solution, using Poisson's formula, when $f(theta)=cos theta$, given that one may use without proof:
$$int_{0}^{2pi}frac{cos z}{1-alpha cos z} = 2pifrac{1-sqrt{1-alpha^2}}{alphasqrt{1-alpha^2}}, quad |alpha|<1$$

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?

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

$$ u(r,theta) = A_0 + sum_{n=1}^infty r^{n}big(A_ncos (ntheta) + B_nsin (ntheta)big) $$

The boundary condition reads

$$ u(a,theta) = A_0 + sum_{n=1}^infty a^nbig(A_ncos(ntheta)+B_nsin(ntheta)big) = cos(theta) $$

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}}{alphasqrt{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.

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:$$u(r,theta)=e_0+e_1ln r+sum_{n=1}^{infty} (a_nr^n+b_nr^{-n})cdot(c_ncos ntheta+d_nsin ntheta)$$The boundedness on $0<r<a$ implies that $$e_1=0\b_n=0$$therefore$$u(r,theta)=e_0+sum_{n=1}^{infty}r^n(hat c_ncos ntheta+hat d_nsin ntheta)$$and finally by imposing the boundary conditions we conclude:$$u(r,theta)={rover a}cdotcostheta$$

Edit

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