5.4 An Example with Periodic Boundary Conditions

In this section the method of separation of variables will be applied to a problem in polar coordinates. The selected problem turns out to have two eigenfunctions for each eigenvalue other than the lowest.

5.4.1 The physical problem

The problem is to find the ideal flow in a unit circle if the normal (radial) velocity on the perimeter is known.

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5.4.2 The mathematical problem

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• Finite domain $\bar \Omega$: $0\le r \le 1, 0 \le \vartheta < 2 \pi$
• Unknown velocity potential $u=u(r,\vartheta)$
• Elliptic equation
  
  $$\nabla^2 u = u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\vartheta\vartheta} = 0$$
  
  

• One Neumann boundary condition at $r=1$.

5.4.3 Outline of the procedure

We will try to find a solution of this problem in the form

$$u = \sum_n R_n(r) \Theta_n(\theta)$$

Here the $\Theta_n$ will be the eigenfunctions.

The reason to take the $\Theta_n$ as the eigenfunctions and not the $R_n$ is because separation of variables needs homogeneous boundary conditions. The $r$ direction has an inhomogeneous boundary condition $u_r(1,\theta)=f(\theta)$ at $r=1$.

5.4.4 Step 1: Find the eigenfunctions

This follows the same procedures as in the first example. We substitute a single term $u=R(r) \Theta(\vartheta)$ into the homogeneous partial differential equation

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

That gives:

$$R''\Theta + \frac{1}{r} R'\Theta + \frac{1}{r^2} R \Theta'' = 0$$

which separates into

$$r^2 \frac{R''}{R} + r \frac{R'}{R} = - \frac{\Theta''}{\Theta} = \hbox{ constant } = \lambda$$

Make sure that all $r$ terms are at the same side of the equation! Some students leave an $r^2$ in the $\theta$ side.

Now which ordinary differential equation gives us the Sturm-Liouville problem, and thus the eigenvalues? Not the one for $R(r)$; $u$ has an inhomogeneous boundary condition on the perimeter $r=1$. Eigenvalue problems must be homogeneous; they simply don't work if anything is inhomogeneous.

We are in luck with $\Theta(\vartheta)$ however. The unknown $u(r,\vartheta)$ has ``periodic'' boundary conditions in the $\vartheta$-direction. If $\vartheta$ increases by an amount $2\pi$, $u(r,\vartheta)$ returns to exactly the same values as before: it is a ``periodic function'' of $\vartheta$. Periodic boundary conditions are homogeneous: the zero solution satisfies them. After all, zero remains zero however many times you go around the circle.

The Sturm-Liouville problem for $\Theta$ is:

$$- \Theta'' = \lambda \Theta$$

$$\Theta(0) = \Theta(2\pi) \qquad \Theta'(0)=\Theta'(2\pi)$$

Note that for a second order ordinary differential equation, we need two boundary conditions. So we wrote down that both $\Theta$, as well as its derivative are exactly the same at $\vartheta=0$ and $2\pi$.

Pretend that we do not know the solution of this Sturm-Liouville problem! Write the characteristic equation of the ordinary differential equation:

$$k^2 + \lambda = 0 \quad\quad\Rightarrow\quad\quad k = \pm {\rm i} \sqrt{\lambda}$$

Lets look at all possibilities:

1. Case $\lambda = 0$:

   Since $k_1 = k_2 = 0$:
   

   $$\Theta = A + B \vartheta$$
   
   

   Boundary conditions:
   

   $$\Theta(0) = \Theta(2\pi) \quad\quad\Rightarrow\quad\quad A = A + B 2 \pi$$
   
   
   That can only be true if $B=0$. Then the second boundary condition is
   
   $$\Theta'(0) = \Theta'(2\pi) \quad\quad\Rightarrow\quad\quad 0 = 0$$
   
   
   hence $\Theta=A$. No undetermined constants in eigenfunctions! Simplest is to choose $A=1$:
   
   $$\Theta_0(\vartheta) = 1$$
   
   

2. Case $\lambda \ne 0$:

   We will be lazy and try to do the cases of positive and negative $\lambda$ at the same time. For positive $\lambda$, the cleaned-up solution is
   

   $$\Theta = A \cos\left(\sqrt{\lambda} \vartheta\right) + B \sin\left(\sqrt{\lambda} \vartheta\right)$$
   
   
   This also applies for negative $\lambda$, except that the square roots are then imaginary.

   Lets write down the boundary conditions first:

   $$\begin{eqnarray*} \Theta(0) = \Theta(2\pi) & \quad\Rightarrow\quad & A = A... ...ght) + B \sqrt{\lambda} \cos\left(\sqrt{\lambda} 2\pi\right) \end{eqnarray*}$$
   
   

   These two equations are a bit less simple than the ones we saw so far. Rather than directly trying to solve them and make mistakes, this time let us write out the augmented matrix of the system of equations for $A$ and $B$:
   

   $$\left( \begin{array}{cc\vert c} 1 - \cos\left(\sqrt{\l... ...\left(\sqrt{\lambda} 2\pi\right) & 0 \end{array} \right)$$
   
   
   Any nontrivial solution must be nonunique (since zero is also a solution). So the determinant of the matrix must be zero, which is:
   
   $$1 - 2 \cos\left(\sqrt{\lambda} 2\pi\right) + \cos^2\left... ...} 2\pi\right) + \sin^2\left(\sqrt{\lambda} 2\pi\right) = 0$$
   
   
   or
   
   $$\cos\left(\sqrt{\lambda} 2\pi\right) = 1$$
   
   
   A cosine is only equal to 1 when its argument is an integer multiple of $2\pi$. Hence the only possible eigenvalues are
   
   $$\sqrt{\lambda_1} = 1 \quad \sqrt{\lambda_2} = 2 \quad \sqrt{\lambda_3} = 3 \quad \ldots$$
   
   

   If $\lambda$ is negative, $\cos\left(i\sqrt{-\lambda} 2\pi\right) = {\rm cosh}\left(\sqrt{-\lambda} 2\pi\right)$ which is always greater than one for nonzero $\lambda$.

   For the found eigenvalues, the system of equations for $A$ and $B$ becomes:
   

   $$\left( \begin{array}{cc\vert c} 0 & 0 & 0\\ 0 & 0 & 0 \end{array} \right)$$
   
   
   Hence we can find neither $A$ or $B$; there are two undetermined constants in the solution:
   
   $$\Theta_n = A \cos(n \vartheta) + B \sin(n \vartheta)$$
   
   

   We had this situation before with eigenvector in the case of double eigenvalues, where an eigenvalue gave rise two linearly independent eigenvectors. Basically we have the same situation here: each eigenvalue is double. Similar to the case of eigenvectors of symmetric matrices, here we want two linearly independent, and more specifically, orthogonal eigenfunctions. A suitable pair is
   

   $$\Theta^1_n(\vartheta) = \cos(n\vartheta)$$
   
   
   
   
   $$\Theta^2_n(\vartheta) = \sin(n \vartheta)$$
   
   

We can now tabulate the complete set of eigenvalues and eigenfunctions now as:

$$\begin{array}{ccc} \lambda_0 = 0 & \multicolumn{2}{c}{... ...\sin(4\vartheta) \\ \vdots & \vdots & \vdots \end{array}$$

5.4.5 Step 2: Solve the problem

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We will again expand all variables in the problem in a Fourier series. Let's start with the function $f(\vartheta)$ giving the outflow through the perimeter.

$$f(\vartheta) = f_0 + \sum_{n=1}^\infty f^1_n \cos(n\vartheta) + \sum_{n=1}^\infty f^2_n \sin(n\vartheta)$$

This is the way a Fourier series of a periodic function with period $2\pi$ always looks.

Since $f(\vartheta)$ is supposedly known, we should again be able to find its Fourier coefficients using orthogonality. The formulae are as before.

$$f_0 = \frac{\int_{\vartheta=0}^{2\pi}f(\vartheta) 1 { \rm... ...vartheta} {\int_{\vartheta=0}^{2\pi}1^2 { \rm d}\vartheta}$$

(the bottom is of course equal to $2\pi$,)

$$f^1_n = \frac {\int_{\vartheta=0}^{2\pi}f(\vartheta)\cos... ...cos^2(n\vartheta) { \rm d}\vartheta} \quad (n=1,2, \ldots)$$

$$f^2_n = \frac {\int_{\vartheta=0}^{2\pi}f(\vartheta)\sin... ...sin^2(n\vartheta) { \rm d}\vartheta} \quad (n=1,2, \ldots)$$

(the bottoms are equal to $\pi$.)

Since I hate typing big formulae, allow me to write the Fourier series for $f(\vartheta)$ much more compactly as

$$f(\vartheta) = \sum_{n,i}^\infty f^i_n\Theta^i_n(\vartheta)$$

where $\Theta^1_n=\cos(n\vartheta)$ and $\Theta^2_n=\sin(n\vartheta)$. Also, all three formulae for the Fourier coefficients can be summarized as

$$f^i_n = \frac {\int_{\vartheta=0}^{2\pi}f(\vartheta)\The... ...artheta) { \rm d}\vartheta} \quad (n=0,1,2, \ldots; i=1,2)$$

For $n=0$, only the value $i=1$ is relevant, of course; $\Theta^1_0=\cos(0\vartheta)=1=\Theta_0$. There is no $\Theta^2_0=\sin(0\vartheta)=0$.

Next, let's write the unknown $u(r,\vartheta)$ as a compact Fourier series:

$$u(r,\vartheta) = \sum_{n,i} u^i_n(r) \Theta^i_n(\vartheta)$$

We put this into partial differential equation $u_{rr} + u_r/r + u_{\vartheta\vartheta}/r^2 = 0$:

$$\sum_{n,i} u^i_n(r)'' \Theta^i_n(\vartheta) + \frac 1r \... ...+ \frac1{r^2} \sum_{n,i} u^i_n(r) \Theta^i_n(\vartheta)'' = 0$$

Using the Sturm-Liouville equation $\Theta^i_n(\vartheta)'' = -\lambda \Theta^i_n(\vartheta)$, where $\lambda$ was found to be $n^2$, this simplifies to

$$\sum_{n,i} u^i_n(r)'' \Theta^i_n(\vartheta) + \frac 1r \... ...\frac1{r^2} \sum_{n,i} n^2 u^i_n(r) \Theta^i_n(\vartheta) = 0$$

We get the following ordinary differential equation for $u^i_n(r)$:

$$u^i_n(r)'' + \frac1r u^i_n(r)' - \frac{n^2}{r^2}u^i_n(r) = 0$$

or multiplying by $r^2$:

$$r^2 u^i_n(r)'' + r u^i_n(r)' - n^2 u^i_n(r) = 0$$

This is not a constant coefficient equation. Writing down a characteristic equation is no good.

Fortunately, we have seen this one before: it is the Euler equation. You solved that one by changing to the logarithm of the independent variable, in other words, by rewriting the equation in terms of

$$\rho \equiv \ln r$$

instead of $r$. The $r$-derivatives can be converted as in:

$$\frac{{\rm d}u^i_n}{{\rm d}r} = \frac{{\rm d}u^i_n}{{\rm... ...}\rho}{{\rm d}r} = \frac{{\rm d}u^i_n}{{\rm d}\rho} \frac1r$$

$$\frac{{\rm d}^2 u^i_n}{{\rm d}r^2} = \frac{{\rm d}}{{\rm... ...ght] \frac1r - \frac{{\rm d}u^i_n}{{\rm d}\rho} \frac1{r^2}$$

$$= \frac{{\rm d}}{{\rm d}\rho} \left[\frac{{\rm d}u^i_n}{... ... \frac1{r^2} - \frac{{\rm d}u^i_n}{{\rm d}\rho} \frac1{r^2}$$

The ordinary differential equation becomes in terms of $\rho$:

$$\frac{{\rm d}^2 u^i_n}{{\rm d}\rho^2} - n^2 u^i_n = 0$$

This is now a constant coefficient equation, so we can write the characteristic polynomial, $k^2 - n^2 = 0$, or $k = \pm n$, which has a double root when $n=0$. So we get for $n=0$:

$$u^1_0 = A^1_0 + B^1_0 \rho = A^1_0 + B^1_0 \ln r$$

while for $n\ne 0$:

$$u^i_n = A^i_n e^{n\rho} + B^i_n e^{-n\rho} = A^i_n r^n + B^i_n r^{-n}$$

Now both $\ln r$ as well as $r^{-n}$ are infinite when $r=0$. But that is in the middle of our flow region, and the flow is obviously not infinite there. So from the `boundary condition' at $r=0$ that the flow is not singular, we conclude that all the $B$-coefficients must be zero. Since $r^0=1$, all coefficients are of the form $A^i_n r^n$, including the one for $n=0$.

Hence our solution can be more precisely written

$$u(r,\vartheta) = \sum_{n,i} A^i_n r^n \Theta^i_n(\vartheta)$$

Next we expand the boundary condition $u_r(1,\vartheta) = f(\vartheta)$ at $r=1$ in a Fourier series:

$$\sum_{n,i} n A^i_n \Theta^i_n(\vartheta) = \sum_{n,i} f^i_n \Theta^i_n(\vartheta)$$

producing

$$n A^i_n = f^i_n$$

For $n=0$, we see immediately that $A_0$ can be anything, but we need $f_0=0$ for a solution to exist! According to the orthogonality relationship for $f_0$, this requires:

$$\int_0^{2\pi} f(\vartheta) { \rm d}\vartheta = 0$$

Are you surprised that the net outflow through the perimeter must be zero for this steady flow?

For nonzero $n$:

$$A^i_n = \frac{f^i_n}n$$

and our solution becomes

$$u = A_0 + \sum_{n,i} f^i_n \frac{r^n}{n} \Theta^i_n(\vartheta)$$

where $A_0$ can be anything.

5.4.6 Summary of the solution

Let's summarize our results, and write the eigenfunctions out in terms of the individual sines and cosines.

Required for a solution is that:

$$\int_0^{2\pi} f(\vartheta) { \rm d}\vartheta = 0$$

Then:

$$f^1_n = \frac1{\pi} \int_{\vartheta=0}^{2\pi}f(\vartheta)\cos(n\vartheta) { \rm d}\vartheta \quad (n=1,2, \ldots)$$

$$f^2_n = \frac1{\pi} \int_{\vartheta=0}^{2\pi}f(\vartheta)\sin(n\vartheta) { \rm d}\vartheta \quad (n=1,2, \ldots)$$

$$u = A_0 + \sum_{n=1}^\infty \left\{ f^1_n \frac{r^n}{n... ...(n\vartheta) + f^2_n\frac{r^n}{n} \sin(n\vartheta) \right\}$$

where $A_0$ can be anything.