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Given the differential equations: $$ \begin{cases} \frac{dx}{dt} = y-1 \\ \frac{dy}{dt} = -xy \end{cases} $$ I'm trying to find out which solutions are periodic. So I tried to find the solutions $y(x)$ and tried to prove that for some startvalues $x_0$ and $y_0$ we find that $y(x_0) = y(x_0 + p)$, where $p$ is the period.

The solution, $y(x)$, that I found is the following: $$y(x)-\ln{y(x)} = -\frac{1}{2}x^2+\frac{1}{2}x_0^2+y_0-\ln{y_0}$$ Or rewritten as a function of $x(y)$: $$x(y) = \pm(2\ln{y}-2y+2(\frac{1}{2}x_0^2+y_0-\ln{y_0}))^{\frac{1}{2}}$$

I've plotted the solutions and found the following plot

So the periodic solutions are only found when $y_0 > 0$ and $x_0$ is in some range. If $x_0$ is too large or too small the solution goes to $y(x) = -\infty$. But how can I show this correctly?

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    $\begingroup$ You are missing a square, or a square root, in your last equation. $\endgroup$ – N74 Feb 13 '17 at 22:05
  • $\begingroup$ Correct, that is a typo. $\endgroup$ – Whizkid95 Feb 13 '17 at 22:17
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This was a special equation to me as it formed part of my phd. $$ \ddot{x} -x\dot{x} + \mu x = 0 $$ it was found that you get autosave behaviour (periodic) entered about the shift which for you is 1 (for me it was $\mu$)

I used Bendixson Theorem to prove periodic solutions existed.

$$H_x = \dot{x}\\H_y = \dot{y}$$ $\nabla\cdot \mathbf{H}$ does not contain a change in sign for some element in $(x,y)$ then we find a periodic solution.

Which for you implies you need a domain where $x=0$ you can then determine which side of the phase-transition you will have periodic solutions (it is based on $y$)

For the above equation you always have the same form, and there is a relationship between the asymptotic region and bounded solution (it depends on $\mu$ and $\dot{x}$ ). Also, for any parameter in the periodic domain we have closed solutions.

Note that you can transform your equation $$ \ddot{x} + x\dot{x} + x = 0 $$ To my original equation by $t\to -t$ $$ \ddot{x} -x\dot{x} + x = 0 $$ since the solutions are similar in phase space.

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  • $\begingroup$ So is it correct to then say the following: Since $\frac{\partial}{\partial x}H_x = 0$ and $\frac{\partial}{\partial y}H_y = -x$, there are points $(x,y)$ with $x \neq 0$ that satisfy the condition, and therefore periodic solutions van be found? $\endgroup$ – Whizkid95 Feb 13 '17 at 22:33
  • $\begingroup$ basically yes. Quite powerful. $\endgroup$ – Chinny84 Feb 13 '17 at 22:52
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    $\begingroup$ @Chinny84 I think Poincare Bendixson is more useful in the cases when the system is not integrable and one tries to find periodic orbits that are somewhat more scarce (like in the case of Van der Pol). In this case the system can be handled fairly directly by just changing some variables, writing its Hamiltonian as kinetic and potential energy and just looking at the potential. Then the full picture emerges. $\endgroup$ – Futurologist Feb 14 '17 at 6:22
  • $\begingroup$ @futurologist I completely agree - I actually used it to orove a more general equation had limit cycles. But it does no harm including it, and it is nicefor the OP to see one of the great quantitative/qualititative results. $\endgroup$ – Chinny84 Feb 14 '17 at 11:41
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    $\begingroup$ @whizkid95 You are correct in terms the name is wrong (I remembered it wrong from the thesis!) - it is the Bendixson theorem. But you are right that you can not state where the periodic solution will exist, but you can prove that with a change of sign that we have periodic solution. With the assumption we have a simple connected region (which I believe this is satisfied) . $\endgroup$ – Chinny84 Feb 14 '17 at 15:01
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I don't think you need Poincare Bendixson here. This is an integrable system, it has a first integral. Here is what you do.

Observe that all curves $\{(x_0 - t, 0) : t \in \mathbb{R}\}$ starting from a point $(x_0, 0)$ on the $x-$axis trace the $x-$axis, so your phase space, after removal of the orbit $\{(x,y) \, \in \, \mathbb{R}^2 \, : \,\, y=0\}$, is divided into an upper half-plane and a lower half plane.

Next, for $y>0$ (upper half-plane) change the variables as follows: $x = x \, , \,\,\, y = e^z$. Then $\dot{y} = e^z \, \dot{z} = -x \, y = - x \, e^z$ which after canceling out the term $e^z$ on both sides of the latter equality, leads to the system $$ \begin{align} \dot{x} &= e^z - 1\\ \dot{z} &= -x \end{align} $$ Then you cast this system in the form of a second order (Newton's type) equation $$\ddot{z} = - \dot{x} = 1 - e^{z}$$ i.e. $$\ddot{z} = 1-e^z$$ This has a conservation of energy integral of motion $$\frac{\dot{z}^2}{2} + e^z - z = E_0$$ for a constant $E_0$. Here $\frac{\dot{z}^2}{2}$ is the kinetic energy term and $U(z) = e^z-z$ is the potential energy term. Then when you draw the graph of the $U(z) = e^z-z$ you see that it is a convex function with exactly one minimum, equal to $1$ when $z=0$, and growing to $+\,\infty$ when $z \to \pm \, \infty$. The latter means that for each energy level $E_0 > 1$ we have a periodic orbit circling round the equilibrium $x=0, z = 0$ which is the equilibrium $x=0, y=1$ in the original coordinates. Hence, the whole upper half-plane is filled with periodic solutions.

For the lower half-plane $y<0$, change the variables as follows: $x=x\, , \,\, \, y= - e^{z}$. Then proceed analogously. You end up with the second order equation $$\ddot{z} = e^{z} + 1$$ with a conserved total energy $$\frac{\dot{z}^2}{2} - \big(z + e^{z}\big) = E_0$$ If you draw the graph of the potential $U(z) = - \big(z + e^{z}\big)$ you observe that it is a strictly decreasing function so the energy level for each $E_0$ are semi-open trajectories (open on one side, going to infinity). In particular they are never periodic solution.

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    $\begingroup$ Just to add: there is a slightly shorter way to prove this :) Observe that if $(x(t), y(t))$ is a trajectory, then $(-x(-t), y(-t))$ is a trajectory too. From this property, the trajectories that intersect $y$-axis twice are automatically closed (see picture in this answer). The way the nullclines are located explain why and where we have closed trajectories: set $x' = 0$ is located only in upper halfplane and any closed trajectory must pass through it. $\endgroup$ – Evgeny Feb 14 '17 at 14:41
  • $\begingroup$ I like your approach of using the Energy to describe the different waves. I integrated like you did for my thesis, but I have to admit your explanation for the unbounded solution makes more sense, I used qualitative approach to explain this behaviour (though this equation was already studied by my supervisor so I did not really study it in great detail, as I was interested in the extension problem). +1 $\endgroup$ – Chinny84 Feb 14 '17 at 15:12
  • $\begingroup$ @Evgeny well, arguably shorter :)... One needs to justify why the trajectories in the upper half-plane starting from the $y-$axis reach the axis $y=1$ and then from there back to the $y-$axis. So one needs to check the direction of the vector field in various regions. Plus do you really want to compare a method that gives you the full qualitative and quantitative picture and on top of that provides you with all the solutions in implicit form with qualitative methods involving fixed points of Poincre maps? $\endgroup$ – Futurologist Feb 14 '17 at 15:52
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    $\begingroup$ @Futurologist Of course I want, because I mostly do qualitative methods especially for questions of mostly qualitative kind :) Agree that having implicit solution is a bonus when it's needed, but both methods give the same qualitative picture. $\endgroup$ – Evgeny Feb 14 '17 at 17:12
  • $\begingroup$ @Evgeny I understand you, I also use qualitative methods when I work in dynamical systems. After all, most dynamical systems are not explicitly solvable. $\endgroup$ – Futurologist Feb 14 '17 at 18:27

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