7.38, §4 Solve

We will again expand all variables in the problem in a Fourier series. Let's start with the function giving the outflow through the perimeter.

This is the way a Fourier series of a periodic function with period always looks.

Since is supposedly known, we should again be able to find its Fourier coeficients using orthogonality. The formulae are as before.

(the bottom is of course equal to ,)

(the bottoms are equal to .)

Since I hate typing big formulae, allow me to write the Fourier series for much more compactly as

where and .Also, all three formulae for the Fourier coefficients can be summarized as

For n=0, only the value i=1 is relevant, of course; . There is no .

Next, let's write the unknown as a compact Fourier series:

We put this into P.D.E. :

Using the Sturm-Liouville equation ,where was found to be n², this simplifies to

We get the following ODE for uⁱ_n(r):

or multiplying by r²:

r² uⁱ_n(r)'' + r uⁱ_n(r)' - n² uⁱ_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 logaritm of the independent variable, in other words, by rewriting the equation in terms of

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

The ODE becomes in terms of :

This is now a constant coefficient equation, so we can write the characteristic polynomial, k² - n² = 0, or , which has a double root when n=0. So we get for n=0:

while for :

Now both 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 r⁰=1, all coefficients are of the form Aⁱ_n rⁿ, including the one for n=0.

Hence our solution can be more precisely written

Next we expand the boundary condition at r=1 in a Fourier series:

producing

n Aⁱ_n = fⁱ_n

For n=0, we see immediately that A₀ can be anything, but we need f₀=0 for a solution to exist! According to the orthogonality relationship for f₀, this requires:

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

For nonzero n:

and our solution becomes

where A₀ can be anything.