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Homework Help: Doesn't this series diverge?

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Currently working on a heat equation problem for an applied math class. I got a solution

    v(x,t) = Ʃn≥1(2/n∏)sin(n∏x)cos(n∏ct), x ε [0,1], t≥0,​

    which seems wrong since, for any fixed x and t, v(x,t)=∞.

    2. Relevant equations

    Governing equation is utt=c2uxx, where c is the speed of sound and u(x,t) is the longitudinal pressure distribution along a tube that is open at x=0 and has a rigid surface at x=1. The boundary condition at x=0 is u(0,t)=g(t)=cos(ωt) where ω is the frequency of the emitted sound. On the other hand, the rigid surface at x=1 implies a vanishing pressure at that point, and thus the boundary condition u(1,t)=0. Assume the initial conditions of utt=c2uxx are both zero.

    3. The attempt at a solution

    I'm supposed to "Use the transformation u(x,t) = v(x,t) + ψ(x)g(t) to homogenize the boundary conditions by enforcing linearity on ψ." Then I'm supposed to "Solve for the homogeneous solution v1(x,t) of the transformed equation."

    I've checked over my work several times and I keep getting

    v1(x,t) = Ʃn≥1(2/n∏)sin(n∏x)cos(n∏ct),
    what has to be wrong.

    Thoughts? Concerns?
  2. jcsd
  3. May 22, 2012 #2
    I don't think that's a problem at all. You see, the solution you have there is a superposition of all possible solutions, and since you don't have a momentum cutoff in your system, there are infinitely many modes that can live in your system in the ultraviolet (short wavelength) limit. This is quite typical for physical problems, especially in quantum field theory, where most of your time gets spent on trying to talk these infinities down.
  4. May 22, 2012 #3
  5. May 22, 2012 #4
    I have a question on my next homework problem. The setup is:

    utt = uxx

    I used separation of variables u(x,t)=X(x)T(t), so that we have XT'' = X''T = k. In the case of k > 0 and k = 0, when I attempted to solve for u I reached an inconsistency with u(x,0)=xe-x2. In the case where k = -λ2 < 0, I have found it impossible to determine the eigenvalue λ. We have X(x) = Asin(λx) + Bcos(λx) and T(t) = Csin(λt) + Dcos(λt). Since ut(x,0) = 0, we need T'(0) = -λC = 0 if we want a nontrivial solution; hence u(x,t) = [Asin(λx) + Bcos(λx)]cos(λt). Then u(0,t) = Bcos(λx)cos(λt) = 0 implies B = 0; hence u(x,t) = Asin(λx)cos(λt). The last thing we could use would be u(0,t) = 0, but this gives us no information.

    What can I do?
  6. May 23, 2012 #5
    You need to use the full solution - you certainly can't have u = x*exp(-x^2) with just one sine and cosine term!
    [tex] X(x) = \sum_\lambda A_{\lambda} \sin(\lambda x)[/tex]
    [tex] T(t) = \sum_\lambda D_{\lambda} \cos(\lambda t) [/tex]

    A Fourier series of x*exp(-x^2) might be useful next!
  7. May 23, 2012 #6


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    Science Advisor

    It is quite obvious that this does converge for x= 0.

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