Solving Laplace's equation

Given Laplace's equation $u_{xx} + u_{yy} = 0$ and 2 boundary conditions

1. $u(x,y)=x^2y$
2. $u(cos\theta,sin\theta) = 1 + cos\theta$

in the unit disc. I want to solve laplace equation.

After using separation of variables I get

where $f(\theta)$ is the boundary conditions. If use the second condition, I get $a_0=\frac{2\pi}{\pi}=2$ and $b_n=0$ since odd function but the problem is with $a_n$ as I keep getting 0. $\frac{1}{\pi}\int_0^{2\pi}{(1+cos\theta)cos n\theta d\theta}=\frac{1}{\pi}\int_0^{2\pi}{cos\theta cos n\theta d\theta}=\frac{1}{2\pi}\int_0^{2\pi}{cos(n+1)\theta d\theta}$ which equals 0 unless $n=-1$ (integral = $\pi$)but can $n$ be a negative number i.e. $a_{-1}$?

and also for first condition, is the same method i.e. using $cos\theta,sin\theta$?

In polar coordinates, the separated solutions of the Laplace equation are (up to constant coefficients):

• $1$
• $r^n \cos n\theta$, $n\ge 1$
• $r^n\sin n\theta$, $n\ge 1$
• $\log r$
• $r^n \cos n\theta$, $n\le -1$
• $r^n\sin n\theta$, $n\le -1$

The last three have a singularity at the origin, so can't be used in a disk; they are listed here for completeness.

Thus, given $1+\cos \theta$ as the boundary condition in the disk, you can readily write down the solution $1+r\cos\theta$ by picking the functions from the above catalog to match the boundary values.

Your computation with cosines is incorrect because $$\cos\theta \cos n\theta \ne \cos (n+1)\theta$$ See Product-to-sum identities.

Concerning $u(x,y)=x^2y$, I guess this condition is also meant to be enforced on the boundary of unit disk. In polar coordinates it becomes $\cos^2\theta \sin\theta$. Use the above product-to-sum identities to write this as a linear combination of terms like $\cos n\theta$ and $\sin n\theta$. Then match them to separated solutions above.

• hmm that's odd... I didn't get quite those separated solutions... so the ODEs I got were $Θ''+λΘ=0$ and $r^2R''+rR'-λR=0$ and impose periodic BCs for $Θ(θ)$ gives $λ=n^2, n=1,2...$ and $Θ(θ)=Acosnθ + Bsinnθ$ and for $λ=0 Θ(θ)=A$ and for R(r) I get $R(r)=Cr^n + Dr^{-n}$ and for n=0 $R=logr$ is it the same as yours? doesn't look the same... – Sumisu Hiko Oct 18 '14 at 4:58
• I initially ignored some possibilities because of singularities at $0$. Rewritten – user147263 Oct 18 '14 at 5:02
• It's a matter of presentation. I like having a catalog of separated solutions written without coefficients, keeping in mind that I can attach coefficients and form sums as needed. – user147263 Oct 18 '14 at 5:05