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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!

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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.

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  • $\begingroup$ While finding the solution $x \left( t \right)$, did we use the power series method? $\endgroup$ – Aniruddha Deshmukh May 28 at 8:07
  • $\begingroup$ Yes, I would like some more details. In the meanwhile, even I will try finding the method somewhere. $\endgroup$ – Aniruddha Deshmukh May 28 at 9:05
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    $\begingroup$ @AniruddhaDeshmukh: Yes. The series you wrote is the same I wrote, but in real form. If you expand $c_n=a_n+ib_n$ and $e^{int}=\cos(nt) + i \sin(nt)$ in the series I wrote, you will find yours (up to some multiplicative constants). $\endgroup$ – Giuseppe Negro May 28 at 9:59
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    $\begingroup$ @AniruddhaDeshmukh: Oh, I have to add something. Maybe my previous comment got you confused. Here, I am considering complex solutions, while the formulation of the Fourier series you wrote is for real solutions. This is simply because complex formulas tend to be neater than real ones, but we could easily construct a real example, in the same spirit as the one given in this answer. To pass from complex Fourier series to real Fourier series, via the formulas I wrote in my previous comment, can be confusing. Here you have a nice example. $\endgroup$ – Giuseppe Negro May 28 at 15:34
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    $\begingroup$ I have added a criterion that rules out periodic solutions. Now my answer is complete. If you found it useful, please upvote and accept it. Thanks. $\endgroup$ – Giuseppe Negro May 30 at 8:05

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