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Homework Help: PDE: Annulus question, Steady State Temperature

  1. Mar 12, 2016 #1


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    1. The problem statement, all variables and given/known data
    Suppose the inner side of the annulus {(r,Φ): r_0 ≤ r ≤ 1} is insulated and the outer side is held at temperature u(1,0) = f(Φ).

    a) Find the steady-state temperature
    b) What is the solution if f(Φ) = 1+2sinΦ ?

    2. Relevant equations

    3. The attempt at a solution

    A = {(rcosΦ,rsinΦ): r_0 ≤ r ≤ 1}
    We have the conditions

    U_rr + r^(-1)*U_r+r^(-2)*U_ΦΦ = 0 in A
    U(1,Φ) = f(Φ)
    U(r_0,Φ) = g(Φ)

    Φ"+v^2Φ = 0
    ϑ^2R"+rR'-v^2R = 0

    Φ => e^(ivΦ) , v= 0,+/-1, +/-2, ...

    r = a)n*r^v+b_n*r^(-v)

    Hence, u(r,Φ) = ∑ c_n * r^(|n|)*e^(inΦ)

    We know that
    C_n = 1/(2*pi) integral from -pi to pi f(Φ)e^(-inΦ)dΦ

    u(r,Φ) = ∑C_n * ( (r^n+r_0^(2n)*r^(-n)) / (1+r_0^(2n)) * e^(inΦ))

    That is the answer for part a. Finding the steady state temperature.

    My question is how did we get to that messy r fraction part. We see what I stated r to equal, and we know cn is right based off fourier coefficients.

    WE went from u(r,Φ) = ∑ c_n * r^(|n|)*e^(inΦ)
    to u(r,Φ) = ∑C_n * ( (r^n+r_0^(2n)*r^(-n)) / (1+r_0^(2n)) * e^(inΦ))

    Must I use the Poisson Kernel?

    b) b is a little tougher for me to understand.

    answer is u(r,Φ) = 1+2*((r^2+r_0^2)/ (r(1+r_0^2))) sinΦ

    I started off by analyzing the c_n by inputting f(Φ) = 1+2sinΦ
    we get that the integral =
    Screen Shot 2016-03-12 at 11.33.05 PM.png
  2. jcsd
  3. Mar 13, 2016 #2


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    Homework Helper

    Does not "the inner boundary is insulated" mean that [itex]\frac{\partial u}{\partial r} = 0[/itex] on [itex]r = r_0[/itex]?

    Umm... no. I'm happy with assuming a solution of the form [tex]
    u(r,\phi) = \sum_{n=-\infty}^\infty c_n R_n(r)e^{in\phi}
    [/tex] where [tex]u(1,\phi) = f(\phi) = \sum_{n=-\infty}^\infty c_ne^{in\phi}[/tex] and the [itex]R_n[/itex] are an as yet unspecified sequence of functions, but [itex]R_n = r^{|n|}[/itex] is not the solution you are looking for as it doesn't satisfy the boundary condition on [itex]r = r_0[/itex].

    If [itex]u(r,\phi) = \sum_{n=-\infty}^\infty c_n R_n(r) e^{in\phi}[/itex] and [itex]f(\phi) = \sum_{n=-\infty}^{\infty} c_ne^{in\phi}[/itex] then [itex]R_n[/itex] is the solution of the boundary value problem [tex]
    \frac{1}{r}\frac{d}{dr}\left(r\frac{dR_n}{dr}\right) - \frac{n^2 R_n}{r^2} = 0[/tex] subject to [itex]R_n(1) = 1[/itex] and [itex]R_n'(r_0) = 0[/itex], obtained by substituting the series definition of [itex]u[/itex] into Laplace's equation and considering the coefficients of [itex]e^{in\phi}[/itex].

    Now there are two points to note:

    (1) The BVP for [itex]n > 0[/itex] is the same as that for [itex]-n < 0[/itex] and therefore they have the same solution: [itex]R_n = R_{|n|}[/itex].

    (2) The BVP for [itex]n = 0[/itex] is a special case.

    Thus we have [tex]
    u(r,\phi) = c_0R_0(r) + \sum_{n=1}^\infty (c_n e^{in\phi} + c_{-n}e^{-in\phi}) R_n(r)[/tex] which is easily rearranged into [tex]
    u(r,\phi) = c_0R_0(r) + \sum_{n=1}^\infty (a_n \cos n\phi + b_n \sin n\phi) R_n(r)[/tex] if that's more convenient, and I'll let you go back and solve the BVPs for [itex]R_n[/itex] for [itex]n \geq 0[/itex].

    The fourier series of [itex]1 + 2 \sin \phi[/itex] is, astoundingly, [itex]1 + 2 \sin \phi[/itex]. Substitute this into the general solution.
    Last edited: Mar 13, 2016
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