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PDE Wave EQ in a circle

  1. Feb 15, 2017 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    Solve 2D wave eq. ##u_tt=c^2 \nabla^2u## in a circle of radius ##r=a## subject to $$u(t=0)=0\\
    u_t(t=0)=\beta(r,\theta)\\u_r(r=a)=0\\$$and then symmetry for ##u_\theta(\theta=\pi)=u_\theta(\theta=-\pi)## and ##u(\theta=\pi)u(\theta=-\pi)##.

    2. Relevant equations
    Lot's I'm sure.

    3. The attempt at a solution
    So I find a solution for ##u## after applying the above boundary conditions where I have two double sums and two constants to solve for. To solve for these, I use the initial conditions. However, ##u(t=0)=0## does not allow me to solve for either of the two constants since ##u(t=0)## vanishes. Thus I have one initial condition yet two constants, one for each double sum. Any ideas?

    I was thinking since the boundary was not moving at ##r=a## perhaps there is some energy requirement solvability condition that should be satisfied but I don't know. Any ideas?
     
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  3. Feb 15, 2017 #2

    Orodruin

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    This is similar to saying you cannot solve for A in the equation A=0.
     
  4. Feb 16, 2017 #3

    joshmccraney

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    Can you elaborate? I feel it's asking me to solve ##y=ax## given the line passes through ##(0,0)##.
     
  5. Feb 16, 2017 #4

    Orodruin

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    Why don't you write out what you have and what the condition you get is?
     
  6. Feb 16, 2017 #5

    Orodruin

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    Actually, this is not what it is doing. It is asking you to solve ##x\vec e_1 + y\vec e_2 = 0## where ##\vec e_i## is a set of linearly independent basis vectors.
     
  7. Feb 16, 2017 #6

    joshmccraney

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    Here is what i have!
     

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  8. Feb 16, 2017 #7

    joshmccraney

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    I know, but this is how I see it though, so I need your help:oldbiggrin:
     
  9. Feb 16, 2017 #8

    Orodruin

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    So what would be the solution to that equation?
     
  10. Feb 16, 2017 #9

    joshmccraney

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    Equation (17) I think. But I only have one initial condition for both constants since the other vanishes ##u##!
     
  11. Feb 16, 2017 #10

    Orodruin

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    Just some comments:

    • Most of your solution is fine, but you forgot the inner derivative when doing the time derivative in the end.
    • Note that ##J_m(\sqrt{\lambda_{nm}} r) \cos(m\theta)## and ##J_m(\sqrt{\lambda_{nm}} r) \sin(m\theta)## are all linearly independent functions.
    • You have already used the initial condition ##u(r,\theta,0) = 0## correctly to get rid of the cosine terms in time.
    • You are missing the ##m = 0## term that leads to a constant function.
     
  12. Feb 16, 2017 #11

    Orodruin

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    To be honest, I would write the angular solution on the form ##e^{im\theta}## with ##m\in \mathbb Z## instead - even if these functions are complex and lead to complex coefficients. It is just much easier to keep track.

    I would also save writing some square roots by exchanging ##\lambda_{nm}## for the appropriate expression in terms of the zeros of the Bessel function derivatives.
     
  13. Feb 16, 2017 #12

    joshmccraney

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    Oops, totally spaced the chain rule!
    So in the ##u_t## initial condition are you suggesting I multiply by ##\cos(p \theta):p\in\mathbb{N}## and use orthogonality of cosine and sine to solve for ##A## and then multiply by ##\sin(p \theta):p\in\mathbb{N}## to find ##B##?
    Have I used this initial condition correctly though? Thinking about it now, if I consider the constant function for ##m=0## couldn't the time dependent cosine terms not vanish since those sums could add to the opposite of some arbitrary constant?
     
  14. Feb 16, 2017 #13

    Orodruin

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    Well, ##p\in \mathbb N## for ##\cos(p\theta)## and ##p\in \mathbb N^+## for ##\sin(p\theta)##. You will also need the orthogonality relations for the Bessel functions.

    Consider the eigenvalues carefully. If your eigenvalue is zero, the time solution will not be sines and cosines. Also, there are eigenfunctions of the form ##J_0(\sqrt{\lambda}r)## (for suitable choices of ##\lambda > 0##).
     
  15. Feb 16, 2017 #14

    joshmccraney

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    Thanks, I think you answered everything I had questions about!
     
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