Write

s = lo g r

, so the two circular boundaries are

s = 0

and

s = 2

. The zero angular mode must interpolate between

- 1

and

1

s - 1 = lo g r - 1

For the

n

th cosine mode, the radial factor has equal value

A_{n}

at both boundaries. The unique harmonic function with those data is

A_{n} \frac{c o s h ( n ( s - 1 ))}{c o s h n} cos (n θ) .

Hence the solution of the annular Dirichlet problem is

ϕ (r, θ) = lo g r - 1 + n = 1 \sum \infty A_{n} \frac{cosh ( n ( lo g r - 1 ))}{cosh n} cos (n θ) .

r = 1

and

r = e^{2}

, the hyperbolic cosine quotient equals one, so the boundary conditions are satisfied term by term.

ii

Solution

The mean of

f

A_{n} = \frac{1}{π} [\int_{0}^{π} (\frac{π}{2} - θ) cos (n θ) d θ + \int_{π}^{2 π} (θ - \frac{3 π}{2}) cos (n θ) d θ] = \frac{2 ( 1 - ( - 1 ) ^{n} )}{π n ^{2}} .

Thus

A_{n} = 0

for even

n

and

A_{n} = 4/ (π n^{2})

for odd

n

. Substitution into part (i) yields

ϕ (r, θ) = lo g r - 1 + \frac{4}{π} n \geq 1 n odd \sum \frac{cosh ( n ( lo g r - 1 ))}{n ^{2} cosh n} cos (n θ) .

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