# Fourier series and differential equations

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1. Dec 24, 2016

### Pouyan

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. Dec 25, 2016

### Delta²

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
3. Dec 25, 2016

### vela

Staff Emeritus
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. Dec 25, 2016

### Pouyan

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. Dec 25, 2016

### vela

Staff Emeritus
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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