Fourier series and differential equations

  • #1
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Homework Statement


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

Homework 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


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)
 

Answers and Replies

  • #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...)
 
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  • #3
vela
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My problem is this expression: y''(t)+(n2 + (-1n))y(t)=y(t+π)
If a=0, we get n=1 or -1, and y''(t)= y(t+π)
In my book, this solution is a constant solution, but how we say that this solution is constant ?!
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.
 
  • #4
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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.


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....
 
  • #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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