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Consider $x'=x^2-1-\cos t$. What can be said about the existence of periodic solutions for this equation?

I'm not sure if periodic solutions exist, but if they do, they must have period equal to $ 2\pi$ and $x(0)=x(2k\pi)$ for $\forall k\in\mathbb Z$.

New Edit:

I guess that may use the following lemma:

Lemma: Consider the differential equation $x' = f (t , x)$ where $f(t, x)$ is continuously differentiable in $t$ and $x$. Suppose that $f (t + T, x) = f (t , x)$ for all t . Suppose there are constants $p$, $q$ such that $f (t , p) > 0, f (t , q) < 0$ for all $t$ then there is a periodic solution $x(t )$ for this equation with $p < x(0) < q$.

Realy, I consider $p=2$ and $q=0$ but $f(t,q)=-1-\cos t\leq 0$ and this inequality is not strictly.

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    $\begingroup$ What kind of technique are you being expected to use here? Using technique from here, I might say that I expect at least one unstable periodic solution in strip $x \in \lbrack 0, \sqrt{2} \rbrack$ and at least one stable periodic solution in strip $x \in \lbrack -\sqrt{2}, 0 \rbrack$. So, yes, there is at least one periodic solution. $\endgroup$ – Evgeny Apr 29 '16 at 9:45
  • $\begingroup$ The period could be a multiple of $2\pi$, no ? $\endgroup$ – Yves Daoust Apr 29 '16 at 10:00
  • $\begingroup$ Do you have a proof of the lemma? And so, is it really necessary that $f(t,q) <0$ in the proof, or is $f(t,q) \le 0$ good enough? $\endgroup$ – Hetebrij Apr 29 '16 at 11:57
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Hint:

As explained in a now deleted answer, the equation can be linearized with

$$x(t)=-\frac{y'(t)}{y(t)}$$

giving

$$y''(t)=(1+\cos(t))y(t).$$

Thanks to linearity and homogeneity, the general solution will be of the form

$$y(t)=C_0f(t)+C_1g(t),$$ and

$$x(t)=-\frac{C_0f'(t)+C_1g'(t)}{C_0f(t)+C_1g(t)}=-\frac{f'(t)+Cg'(t)}{f(t)+Cg(t)}.$$

Then we can achieve $x(2\pi)=x(0)$ and form a periodic function by solving

$$\frac{f'(0)+Cg'(0)}{f(0)+Cg(0)}=\frac{f'(2\pi)+Cg'(2\pi)}{f(2\pi)+Cg(2\pi)},$$ a quadratic equation in $C$.

$$f'(0)f(2\pi)-f'(2\pi)f(0)+((g'(0)f(2\pi)-f'(2\pi)g(0)+f'(0)g(2\pi)-g'(2\pi)f(0))C+(g'(0)g(2\pi)-g'(2\pi)g(0))C^2=0.$$

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  • $\begingroup$ Wow, that's cool! :) I wondered if it is possible to show that we have exactly two periodic solutions. $\endgroup$ – Evgeny Apr 29 '16 at 22:00
  • $\begingroup$ @Evgeny: we are not done, as one needs to show that the equation has real solutions :( $\endgroup$ – Yves Daoust Apr 29 '16 at 22:07
  • $\begingroup$ I'm quite optimistic here: if I'm not confusing anything, the method that I've mentioned and @JuliánAguirre 's method (both are somewhat based on stroboscopic mapping) can show that there are at least two real solutions to your equation. $\endgroup$ – Evgeny Apr 29 '16 at 22:34
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Given $z\in\mathbb R$, let $x(t;z)$ be the unique solution such that $x(0)=z$, defined on a maximal interval $[0,T_z)$.

Let $x_1(t)=-\sqrt2$; then $x_1'(t)=0\le(x_1(t))^2-1-\cos t=1-\cos t$, that is, $x_1$ is a subsolution. Let $x_2(t)=0$; then $x_2'(t)=0\ge(x_2(t))^2-1-\cos t=-1-\cos t$ and $x_2$ is a supersolution. If $-\sqrt2\le z\le 0$, then $x(t,z)$ is defined on $[0,\infty)$ and $-\sqrt2\le x(t,z)\le 0$ for all $t>0$. Define $\phi\colon[-\sqrt2,0]\to[-\sqrt2,0]$ by $$ \phi(z)=x(2\,\pi,z). $$ By the results on continuous dependence of initial data, $\phi$ is continuous. It has a fixed point $z_0$. Then $x(t,z_0)$ is a periodic solution.

Here is the graph of the solutions for $z=-1.4$ and $z=-0.1$. enter image description here

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  • $\begingroup$ @AguirreThanks for your answer. your answer is just what I want. Please see my edit of the question and give me your hint. $\endgroup$ – Bimanifold Apr 29 '16 at 11:49
  • $\begingroup$ The result is also true with non strict inequalities (when there is uniqueness of solution.) You can take $p=-\sqrt2$ and $q=0$. $\endgroup$ – Julián Aguirre Apr 29 '16 at 12:15

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