Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came across the following boundary value problem that I can't solve. It's the Laplacian on the upper half of an annulus with radius $1 \leq r \leq 2$ in polar coordinates:

$u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta} =0$

$u(1,\theta)=u(2,\theta)=0, ~~~ 0<\theta<\pi$

$u(r,0)=0, ~~ u(r,\pi)=r, ~~~ 1<r<2.$

The problem mentions that one must show in this case the choice of separation constant $\lambda=-\alpha^2<0$ leads to eigenvalues and eigenfunctions. The problem is that in this case, after separating variables and solving

$ \Theta'' + \lambda \Theta = 0$

$ r^2R'' + rR' - \lambda R =0$

we would get $\Theta(\theta)=c_1 \cosh(\alpha \theta) + c_2 \sinh(\alpha \theta)$ which is not periodic in $\theta$. Usually we always have $\alpha^2>0$ which gives us the periodic sine and cosine solution for $\Theta$. Help would be appreciated!

share|improve this question
This prescription of boundary values is not continuous in $(1, \pi)$ and $(2,\pi)$. You sure you cited that correctly? –  user20266 Jul 27 '12 at 19:19
@Thomas According to the author of the book, there is no typo. I copied it correctly. –  Parsa Jul 27 '12 at 20:08
@Thomas I see what you mean though, this is a problem. –  Parsa Jul 27 '12 at 20:20
Are you aware that you can use double dollar signs to turn these into displayed equations? That centres them and makes the fractions come out a lot nicer. –  joriki Jul 27 '12 at 20:54
Also, note that you can't "solve a Laplacian" -- the Laplacian is a differential operator; what you're solving here is the Laplace equation, which results from setting the result of applying the Laplacian to zero. –  joriki Jul 27 '12 at 20:56
show 1 more comment

2 Answers

There's no problem with peridocity here. Periodicity is required when the domain consists of full circles, in which case a lack of periodicity would imply a lack of continuity. If the function is only defined for $0\lt\theta\lt\pi$, there's no reason for imposing periodicity.

share|improve this answer
add comment

Notice that to satisfy the boundary conditions, $R(r)$ must be oscillatory. This forces $\Theta(\theta)$ to have exponential behavior. Assume $\Theta'' - \lambda\Theta = 0$, where $\lambda > 0$, to see this directly. After some work we find we are forced by the boundary conditions on $R(r)$ to take $\lambda < 0$. It is a good exercise to work this out.${}^\dagger$ As joriki notes, we need not (and should not) impose the constraint that $\Theta$ be periodic.

Solve the ODE for $R(r)$ and use the boundary conditions to eliminate one of the solutions and find the eigenvalues $\lambda_n$. Use the boundary condition $\Theta(0) = 0$ to eliminate one of the angular solutions. This gives a solution of the form $u_n(r,\theta) = R_n(r)\Theta_n(\theta)$ that satisfies $u_n(1,\theta) = u_n(2,\theta) = u_n(r,0) = 0$.

The solution satisfying the final boundary condition, $u(r,\pi) = r$, will be a sum $$u(r,\theta) = \sum_{n=1}^\infty a_n u_n(r,\theta).$$ We need $$\sum_{n=1}^\infty a_n u_n(r,\pi) = r.$$ Sturm-Liouville theory tells us the eigenfunctions $R_n$ are orthogonal, $$\int_1^2 \frac{dr}{r} R_m R_n \propto \delta_{mn},$$ where $\delta_{mn}$ is the Kronecker delta. (Note that the weight function for the ODE for $R(r)$ is $1/r$.) To find the $a_n$s, multiply by $u_m(r,\pi)/r$ and integrate from $1$ to $2$. This is totally analogous to the process of finding the coefficients in a Fourier series. There's a nice closed form for the $a_n$.

Below we plot $r$ and the partial sum $\sum_{n=1}^{25}a_n u_n(r,\pi)$.

enter image description here

Figure 1. The sum $\sum_{n=1}^{25}a_n u_n(r,\pi)$.

${}^\dagger$Think of solving Laplace's equation in Cartesian coordinates. After separating variables, we can't have both $X(x)$ and $Y(y)$ oscillatory since we need $\frac{X''}{X} + \frac{Y''}{Y} = 0$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.