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Fourier series and differential equations

  1. Dec 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the values of the constant a for which the problem y''(t)+ay(t)=y(t+π), t∈ℝ, has a solution with period 2π
    which is not identically zero. Also determine all such solutions
    2. Relevant equations
    With help of Fourier series I know that :
    Cn(y''(t))= -n2*Cn(y(t))
    Cn(y(t+π)) = exp(nπ)Cn(y(t)) = (-1n) Cn(y(t))

    We get:
    -n2*Cn(y(t)) + a*Cn(y(t)) = (-1n) Cn(y(t))

    Cn(y) is not zero then we have
    a= n2 + (-1n) , n≠0


    3. The attempt at a solution
    My problem is this expression:
    y''(t)+(n2 + (-1n))y(t)=y(t+π)
    if a =0 n=1 or -1 we get y''(t)= y(t+π)
    In my book this solution is a constant solution but how we say that this solution is constant ?!
    Can't we say that y(t+π)=y(t)?
    Further:
    If n≠1 or -1 or 0 then we get :
    y''(t)+(n2)*y(t)=0 My question for this term is, shall we think y(t)*(-1n)) = y(t+π) ?!
    I know how to solve the rest of this term but I don't get :
    1) y''(t)= y(t+π) and why we say this has a constant solution ? How do we solve this actually ?!
    2) y(t)*(-1n)) = y(t+π)
     
  2. jcsd
  3. Dec 25, 2016 #2

    Delta²

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    Something is fishy here, a is considered to be a constant (as stated by the problem), thus it cannot depend on n , can it?

    EDIT: OK I think I see now, depending on what a is, the solution contains only one fourier term can you see which one? (in other words ##C_k(y)=0## for every k except for ##k=n=f(a)## where f is some function of a...)
     
    Last edited: Dec 25, 2016
  4. Dec 25, 2016 #3

    vela

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    Your book is wrong.

    Could you ask your questions again but use better formatting and proper punctuation so it's clear what you're trying to say? Also, explain your reasoning a bit more, like how you jumped to y''(t)+(n2)*y(t)=0, because some of what you wrote doesn't seem right.
     
  5. Dec 25, 2016 #4

    Well this is the whole solution from my book : If a has the form n2 + (-1n) for some integer n≠0, then the problem has the solution y(t)= A*eint + B*e-int where A and B are constants. If a=1 there are the solutions y(t)=constant. For other values there are no nontrivial solutions.
    I know if a = n2 + (-1n) then we can just enter this term into the equation and we have
    y''(t)+(n2 + (-1n))y(t)=y(t+π) which has the solution y(t)= A*eint + B*e-int when n is not 0 or -1 or 1... but I don't know why....
     
  6. Dec 25, 2016 #5

    vela

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    You need to be more careful. You have ##y(t) = \sum_n c_n e^{int}##. When you plug this into the differential equation, you end up with
    $$\sum_{n=-\infty}^\infty -n^2c_n e^{int} + \sum_{n=-\infty}^\infty ac_n e^{int} = \sum_{n=-\infty}^\infty (-1)^n c_n e^{int}.$$ The orthogonality of the Fourier components then implies that
    $$(-n^2+a) c_n = (-1)^n c_n$$ for all ##n##. There are two ways this relationship can hold: ##a=n^2 + (-1)^n## or ##c_n = 0##.

    Now remember that you're looking at the situation where ##a## is a constant, so it can't vary with ##n##. Suppose ##a=3##. You can see there is no integer value of ##n## such that ##3 = n^2 + (-1)^n##, so the only way the relationship holds is ##c_n = 0## for all ##n##. In other words, ##y=0##, the trivial solution.

    But suppose ##a=5 = 2^2 + (-1)^2##. The solution has to satisfy
    $$\sum_{n=-\infty}^\infty -n^2c_n e^{int} + \sum_{n=-\infty}^\infty [2^2+(-1)^2]c_n e^{int} = \sum_{n=-\infty}^\infty (-1)^n c_n e^{int},$$ which implies
    $$(-n^2+5) c_n = (-1)^n c_n$$ for all ##n##. What happens when ##n=\pm2##? What about when ##n \ne \pm 2##? What does the series look like in this case?
     
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