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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!

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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
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@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
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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.

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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 $\frac{X''}{X} + \frac{Y''}{Y} = 0$.

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