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## Main Question or Discussion Point

I want to solve the boundary value problem:

Laplace's equation for u(r, theta) = 0, 1 < r < 2, 0 < theta < a

(i) ur(1,theta) = 0

(ii) ur(2,theta) = 0

(iii) u(r,0) = 0

(iv) u(r,a) = f(r)

I let u(r,theta) = R(r)M(theta) and plug this into the DE.

It's too tough to type out, but I get (r/R)d/dr(r*dR/dr) = lambda which implies that d/dr(r*dR/dr) = lambda*R/r, 1 < r < 2

This is in Sturm-Liouville form with p(r) = r, q(r) = 0, and w(r) = 1/r. I have homogeneous S-L conditions making the operator Hermitian. I solve the Euler-Cauchy equation and find that R(r) = Ar^sqrt(lambda) + Br^(-sqrt(lambda)) and with R'(1) = 0, B = -A.

I'm having trouble with R'(2) since this implies 2^sqrt(lambda) = 2^-sqrt(lambda) Can someone help me find the form for R(r) and what this implies about M(theta). I know how to use the S-L series at the end, but I'm having trouble getting there.

Laplace's equation for u(r, theta) = 0, 1 < r < 2, 0 < theta < a

(i) ur(1,theta) = 0

(ii) ur(2,theta) = 0

(iii) u(r,0) = 0

(iv) u(r,a) = f(r)

I let u(r,theta) = R(r)M(theta) and plug this into the DE.

It's too tough to type out, but I get (r/R)d/dr(r*dR/dr) = lambda which implies that d/dr(r*dR/dr) = lambda*R/r, 1 < r < 2

This is in Sturm-Liouville form with p(r) = r, q(r) = 0, and w(r) = 1/r. I have homogeneous S-L conditions making the operator Hermitian. I solve the Euler-Cauchy equation and find that R(r) = Ar^sqrt(lambda) + Br^(-sqrt(lambda)) and with R'(1) = 0, B = -A.

I'm having trouble with R'(2) since this implies 2^sqrt(lambda) = 2^-sqrt(lambda) Can someone help me find the form for R(r) and what this implies about M(theta). I know how to use the S-L series at the end, but I'm having trouble getting there.