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Polar BVP in Sturm-Liouville

  1. Apr 13, 2009 #1
    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.
     
  2. jcsd
  3. Apr 14, 2009 #2
    Hi PvtBillPilgri. Welcome to PF.


    Not sure whether I can help here.
    It look like you are using the method of separation of variables. Probably if you can show how you separate the variables then the problem will be much clearer.
     
  4. Apr 22, 2011 #3
    Using seperation of variables, One gets A=b (because the -in the swqrt(lambda) cancels and such) Using the second boundary condition as you said one ends up with 2^sqrt(lambda) = 2^-sqrt(lambda).

    This lets you solve for lambda but you have to know the identity e^i(pi)+1=0 (complex number i) Also using some identities for complex numbers and the cyclic nature of sines and cosines, u can come up with the identity e^(2ni(pi)) =1, where n is an integer.
    Now all you have to do is make your equation look like that.
    Take the natural log of both sides and raise it to the power e:
    e^(sqrt(lambda)ln(2))=e^-(sqrt(lambda)ln(2))
    e^2(sqrt(lambda)ln2)=1.......= e^(2ni(pi)) from before
    The exponents must equal
    2sqrt(lambda)ln2=2niPi cancel the 2 divide by ln2 and square it
    lambda = -n^2Pi^2/ln2^2 remember that i^2= -1
    Once you have the eigenvalues you can solve ending up having to use the S-L coeficients...I think havnt gone that far.

    Its my first time posting here but I sort of remember this stuff.
     
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