# Homework Help: Fourier series and differential equations

Tags:
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

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?