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Fix some even functions $f$ and $g$, differentiable, such that $f(0)=g(0)=0$ and $f'(0)=g'(0)=0$, and consider the autonomous differential system $$\left\{\ \begin{array}{lcr}x'&=&-y+f(x)\\ y'&=&x+g(y)\end{array}\right.$$

My question is whether every solution $t\mapsto(x(t),y(t))$ of this differential system which passes close enough to its fixed point $(0,0)$, is periodic.

If this helps, one can assume that the functions $f$ and $g$ are smooth, or polynomials, and/or that their sign is constant in a neighbourhood of $0$.

By hypothesis, $f(u)$ and $g(u)$ are negligible with respect to $u$ when $u\to0$. Thus, near the origin $(0,0)$, the differential system above is a perturbation of the linear differential system $$\left\{\ \begin{array}{lcr}x'&=&-y\\ y'&=&x\end{array}\right.$$ Obviously, the solutions of this linear differential system are the circles centered at $(0,0)$, oriented positively.

Simulations based on the cases $f(u)=au^{2n}$ and $g(u)=bu^{2m}$, for various real constants $(a,b)$ and various (small) positive integers $n$ and $m$, seem to support the conjecture (but counterexamples would be welcome, naturally).

A recent question on the site is related to the case $f(u)\propto u^4$ and $g(u)\propto u^6$. Below is a simulation of the phase diagram when $f(u)=u^4$ and $g(u)=3u^6$, which seems to support the conjecture.

enter image description here

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  • $\begingroup$ hmm if it's the case shouldn't there be some potential energy function that's invariant ? i tried to write down the differential equation it should have but i'm bad at it and don't see anything i can do $\endgroup$ – mercio Jun 21 '15 at 20:18
  • $\begingroup$ @mercio That there is an invariant energy function is not saying much. There is one, we cannot compute it (except in very special cases) and its existence does not prove the existence of cycles. $\endgroup$ – Did Jun 21 '15 at 21:06
  • $\begingroup$ If there exists an invariant energy function, does the theorem of "nonlinear center for conservative systems" apply? math.ucdavis.edu/~jlirion/course_notes/… page 2 $\endgroup$ – KittyL Jun 22 '15 at 11:17
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    $\begingroup$ @KittyL ?? There always exists an invariant energy function. Yes if the hypotheses of the theorem hold, the theorem apply. Amazing, eh? $\endgroup$ – Did Jun 22 '15 at 15:20
  • $\begingroup$ The first Lyapunov value is f'''(0)+g'''(0), which is zero, and which also supports your conjecture. It is not difficult for calculate the second Lyapunov value, and maybe even third one. But this is not a proof of course. $\endgroup$ – Artem Jun 25 '15 at 13:48
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The conjecture seems to be false. I write "seems" because still there is nonzero chance that I made a mistake in my calculation. However, I will present both numerical and analytical evidence for my conclusion.

First, analytically, to distinguish center from focus in a general situation one must compute the so-called Poincaré mapping that sends solutions starting at, say, polar angle $\theta=0$ and distance $r_0$ to $\theta=2\pi$. In general it has the form $$ r=f(2\pi,0,r_0)=\alpha_1 r_0+\alpha_2 r_0^2+\alpha_3r_0^3+\ldots $$ It is easy to compute $\alpha_1$ here, which is simply $\alpha_1=1$. Moreover, Lyapunov found that the first nonzero coefficient $\alpha_i$ with $i>1$, if any, must be such that $i$ is odd.

If one considers the function $f(2\pi,0,r_0)-r_0$, then this theorem is available, which I state following this book (Методы и приемы качественного исследованииа динамическикх систем на плоскости (Methods and techniques of the qualitative study of dynamical systems in the plane) – 1990, by N. N Bautin, I am not aware of any English translation):

Theorem: If $\alpha_i\ne0$ for some $i>1$ odd then the equilibrium is a focus. If $\alpha_i=0$ for every $i>1$ odd then the equilibrium is a center.

So, this theorem is practically useless to prove that something is a center, but can be used to prove that the equilibrium is a focus. One calls $\alpha_3$ the first Lyapunov value (this is what is used in the Hopf bifurcation theorem) and $\alpha_5$ the second Lyapunov value.

Simple calculations show, as I mentioned in the comments, that $\alpha_3=0$. Furthermore, I found that $$ \alpha_5=\frac{\pi}{12} \left(3a_2b_2(b_2^2-a_2^2)+11(a_2b_4-a_4 b_2)\right), $$ where $a_j$ and $b_j$ are the coefficients of the Taylor series for $f$ and $g$ respectively, hence $\alpha_5\ne0$ in general. (More details on the computation of the coefficients $\alpha_i$ are in my second answer.)

So what about the StreamPlot function? It seems that, due to the fact that $\alpha_3=0$, the software does not distinguish between a center and a highly nonlinear focus (i.e., the convergence to the equilibrium is very far from being exponential).

So I took this system:

StreamPlot[{-y + x^2 + 2 x^4, x + y^2 + y^4}, {x, -1, 1}, {y, -1, 1}]

And got the expected picture of the center:

enter image description here

However, actually solving by Mathematica,

sol = {x[t], y[t]} /.

NDSolve[{x'[t] == -y[t] + x[t]^2 + 2 x[t]^4, y'[t] == x[t] + y[t]^2 + y[t]^4, x[0] == 1/5, y[0] == 1/5}, {x[t], y[t]}, {t, 0, 50}, AccuracyGoal -> 20, PrecisionGoal -> 20, WorkingPrecision -> 35];

ParametricPlot[Evaluate[sol], {t, 0, 50}]

I got the following figure:

enter image description here

This confirms that the origin is a stable focus, as predicted by our computations that $\alpha_3=0$ and $\alpha_5=-\frac{11}{12}\pi<0$.

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  • $\begingroup$ Very impressive. Could you explain the reason why the first nonzero $\alpha_i$ after $\alpha_1=1$ must correspond to some odd $i$, and how to compute $\alpha_3$ and $\alpha_5$ in general? And, if I understand correctly, $\alpha_3\ne0$ when $f(u)=au^2$, $g(u)=bu^2$, and $a+b\ne0$, right? $\endgroup$ – Did Jul 28 '15 at 20:58
  • $\begingroup$ @Did Honestly, I have no idea why nonzero $\alpha$ must have an odd index, I just saw an attribution to Lyapunov, who proved it. I tried to find a proof in one of the compendium volumes by Andronov et al., but failed. $\alpha_3$ can be calculated (up to a scaling factor) as $f'''(0)+g'''(0)$, which is always zero for your $f$ and $g$. I will add how to compute $\alpha$-s in a separate answer, it requires some tedieous (but easy with a computer algebra system) calculations. $\endgroup$ – Artem Jul 28 '15 at 21:03
  • $\begingroup$ This pair of posts of yours make for a rather excellent answer. Thanks again. $\endgroup$ – Did Jul 28 '15 at 21:47
  • $\begingroup$ By the by... for the system $x'=-y+x^2$, $y'=x+y^2$, or, more generally, when $f=g$, then $\alpha_5=0$--but in this case I guess one expects $\alpha_7\ne0$... $\endgroup$ – Did Jul 28 '15 at 21:52
  • $\begingroup$ @Did. You are welcome. Actually, for $f(u)=g(u)=u^2$ the answer is the center, hence all $\alpha_j=0$. It can be proved by first rotating the coordinate system $\pi/4$ and then notice that the transformation $(t,y)\mapsto(-t,-y)$ (for the new coordinate $y$) leaves the system invariant. It looks like exactly the same argument works for arbitrary case $f=g$ and both are even. $\endgroup$ – Artem Jul 28 '15 at 22:58
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To calculate the Lyapunov values, one must initially transform the system into the polar coordinates, which for this example takes the form $$ \frac{dr}{d\theta}=\frac{f(r\cos \theta)\cos \theta+g(r\sin \theta)\sin \theta}{1+\frac{f(r\cos\theta)\sin\theta-g(r\sin\theta)\cos\theta}{r}}=R(r,\theta)=r R_1(\theta)+r^2R_2(\theta)+\ldots.\tag{1} $$ This equation makes sense since for sufficiently small $r$ the denominator is strictly positive for $0<\theta<2\pi$.

Let $$ r=f(\theta,\theta_0,r_0) $$ be the solution to this equation with the initial conditions $r(\theta_0)=r_0$. Consider the special case $\theta_0=0$ and get (assuming that the right hand sides are analytic) $$ r=u_1(\theta)r_0+u_2(\theta)r_0^2+\ldots\tag{2} $$ This function must solve $(1)$. After plugging $(2)$ into $(1)$ we find $$ \dot u_1=R_1(\theta)u_1,\\ \dot u_2=R_1(\theta)u_2+R_2(\theta)u_1^2,\\ \ldots $$ Clearly we must have $u_1(0)=1,u_i(0)=0$. Hence we can find all $u_j$ if given $R_j(\theta)$. In particular, for the example in the question $R_1(\theta)=0$, and hence $u_1(\theta)=1$.

By taking $\theta=2\pi$ we find $\alpha_j=u_j(2\pi)$. So to find $\alpha_5$ one needs to find and consecutively solve 5 simple (but cumbersome) ODE.

By the way, in the same book there is an expression for $L_2=\alpha_5$ for the system $$ \dot x=-y+\sum_{ij}a_{ij}x^iy^j\\ \dot y=x+\sum_{ij}b_{ij}x^iy^j $$

enter image description here

enter image description here

but it seems to be easier to calculate $L_2$ with the help of diff equations (plus, while having fun working with your problem I found several typos in the book I referenced, so I am not sure whether this expression is 100% is correct).

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