Conditions on a coefficient so that the differential equation has no periodic solutions I was reading a paper on existence on periodic solutions of nonlinear differential equations of type
$$x'' + a \left( t \right) x = \dfrac{b \left( t \right)}{x^{\alpha}} + p \left( t \right)$$
where $a, b, p$ are all $T -$ periodic functions
In the proof of existence, the author uses the hypothesis:-
The equation $x'' + a\left( t \right) x = 0$ has no non trivial $T -$ periodic solutions.
Using this hypothesis and a perturbation result, the proof of existence goes quite easily.
However, I would like to know under what conditions on $a$ does this hypothesis hold.
All I have figured out till now is that if $a$ is constant, then it must not be of the form $\left( \dfrac{2n \pi}{T} \right)^2$ otherwise, we get solutions in the form $\cos \left( \dfrac{2n \pi}{T} t \right)$ and $\sin \left( \dfrac{2n \pi}{T} t \right)$ which will be $T$ periodic.
I have the question that if $a$ is non constant and periodic of period $T$, then is it possible to have non trivial $T$ periodic solutions for the equation $x'' + a \left( t \right) x = 0$?
I tried converting it to a system and checking the solutions, but with so general $a$, I am not able to find the matrix solution and hence the periodic matrix which can convert the system to a system with periodic coefficients.
Any help will be appreciated!
 A: The fact that $\ddot{x}+a(t)x=0$ has no nontrivial periodic solutions is generally false, even if $a$ is non-constant. For example, the equation 
$$
\ddot{x}+e^{it}x=0$$ 
has the solution 
$$\tag{1}
x(t)=\sum_{n=0}^\infty \frac{1}{(n!)^2} e^{int},$$
which is $2\pi$ periodic.
The following is a simple criterion that rules out the existence of periodic solutions.

Proposition. If $a(t)<0$, then for any $T>0$ there are no nontrivial $T$-periodic solutions to $\ddot{x}+a(t)x=0$ in the class $C^2$.

Proof. We assume that $x$ is real-valued, but the result is true for complex-valued functions too. Suppose that $x\in C^2$ is $T$-periodic and solves $\ddot{x}+a(t)x=0$. Then in particular $\ddot{x}x+a(t)x^2=0$. Integrating this relation we find 
$$
\int_0^T (\ddot{x}x+a(t)x^2)\, dt=0.$$
We now integrate by parts, noticing that there are no boundary terms because $x$ is $T$ periodic: 
$$
\int_0^T (-|\dot{x}|^2+a(t)x^2)\, dt=0.$$ 
Now, because of the assumption on $a$, the integrand function is the sum of two negative terms. Thus, the vanishing of the integral implies the vanishing of the integrand. We conclude that 
$$
x^2=0\, \quad \forall t\in[0, T], $$
that is, $x$ is trivial. $\Box$
Remark. If $a(t)\le 0$, then the result is still true, with essentially the same proof. 

Appendix. 
Here's a method to compute (1). It is the Fourier series version of the power series method of Frobenius. Write 
$$
x(t)=\sum_{n=-\infty}^\infty c_n e^{int}.$$ 
The task is to find coefficients $c_n$ such that $x(t)$ solves the given differential equation. Now, 
$$
\frac{d^2x}{dt^2}=\sum_{n=-\infty}^\infty -n^2c_n e^{int}, $$
while 
$$
e^{it}x(t)=\sum_{n=-\infty}^\infty c_n e^{i(n+1)t}=\sum_{n=-\infty}^\infty c_{n-1}e^{int}.$$ 
So, the equation $\ddot{x}+e^{it}x=0$ is satisfied if 
$$\tag{2}
n^2c_n=c_{n-1}, \qquad \forall n\in\mathbb Z.$$
The sequence 
$$
c_n=\begin{cases} 0, & n<0, \\ \left(\frac{1}{n!}\right)^2, & n\ge 0, \end{cases}$$ 
is a solution to (2).
Remark. This is not the unique solution. It is just a solution.
