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My question is on Exercise 7.3.8 of Chaos and Nonlinear Dynamics (2nd ed) by Strogatz:

7.3.8. Recall the system $\dot{r} = r(1-r^2) + \mu r \cos \theta, \; \dot{\theta} = 1$ of Example 7.3.1. Using the computer, plot the phase portrait for various values of $\mu > 0$. Is there a critical value $\mu_c$ at which the closed orbit ceases to exist? If so, estimate it. If not, prove that a closed orbit exists for all $\mu > 0$.

[Here the ODE is given in polar coordinates, i.e. $x = r \cos \theta$ and $y = r \sin \theta$.]

I just want to make sure my reasoning is correct. Here is my work:

Differentiating the equations $x = r \cos \theta$ and $y = r \sin \theta$ with respect to $t$ gives \begin{align*} \dot{x} &= -r \dot{\theta} \sin \theta + \dot{r} \cos \theta \\[2pt] \dot{y} &= r \dot{\theta} \cos \theta + \dot{r} \sin \theta \end{align*}

Using the above equations I found the following Cartesian equations: \begin{align*} \dot{x} &= -y + x(1 - x^2 - y^2) + \frac{\mu x^2}{\sqrt{x^2 + y^2}} \\[4pt] \dot{y} &= x + y(1 - x^2 - y^2) + \frac{\mu xy}{\sqrt{x^2 + y^2}} \end{align*}

I then plotted the phase portraits for $\mu = 0.5, 1.0, 2.0, 5.0$.

enter image description here

I also experimented with much higher values of $\mu$ (e.g. $20$), and it seems that a closed orbit exists for all $\mu > 0$. So prove this, I tried to apply the Poincare-Bendixson theorem. To find a trapping region, we note that the region where the flow points inward (i.e. $\dot{r} < 0$) is separated from the region where the flow points outward (i.e. $\dot{r} > 0$) by the polar curve
\begin{align*} 1 - r^2 + \mu \cos \theta = 0. \end{align*}

Using the fact that $\dot{\theta} = 1$ we have $\theta(t) = t + \theta_0$, so this curve becomes \begin{align*} 1 - r^2 + \mu \cos(t + \theta_0) = 0 \iff r^2 = 1 + \mu \cos(t + \theta_0). \end{align*}

This equation is $2\pi$-periodic, and hence the equation $1 - r^2 + \mu \cos \theta = 0$ defines a closed (and piecewise-smooth) curve in the plane for all $\mu > 0$. Hence, we can find a region $R$ (homeomorphic to an annulus) which does not contain the origin (the only fixed point) on which the flow on the "inner boundary" points outward and the flow on the "outer boundary" points inwards. Therefore, $R$ contains a closed orbit by Poincare-Bendixson.

Is my reasoning correct? Any feedback would be appreciated.

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  • $\begingroup$ look at what happens to your curve for $\mu\geq1$ $\endgroup$
    – LPZ
    Feb 20 at 17:01
  • $\begingroup$ @LPZ: Which curve are you referring to? $\endgroup$
    – Leonidas
    Feb 20 at 18:58
  • $\begingroup$ your polar curve defining your trapping region. For $\mu\geq1$ it contains the origin and has a cusp on it (for $\mu=1$, it's a cardioid). $\endgroup$
    – LPZ
    Feb 20 at 22:06
  • $\begingroup$ @LPZ: Good point. So since the trapping region contains the origin for $\mu \geq 1$, it seems we cannot apply Poincare-Bendixson. But it still looks to me like there do exist limit cycles for $\mu \geq 1$...any suggestions for how to prove this? $\endgroup$
    – Leonidas
    Feb 21 at 14:56
  • $\begingroup$ Actually, I rather suspect that you do not have a limit cycle for $\mu\geq1$ and that all trajectories converge to the origin. Did you simulate it numerically to build intuition? $\endgroup$
    – LPZ
    Feb 21 at 16:03

3 Answers 3

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For the computation of the return or Poincaré map we can set $\theta=t$. Then the question to solve is if the integration of $r$ from $0$ to $2\pi$ can return to the same value $r_*$.

The radius equation is a Bernoulli equation, giving $$ \frac{d}{dt}r^{-2}=-2r^{-3}\dot r=-2(r^{-2}-1+r^{-2}\cos t) $$ This now is linear in $v=r^{-2}$. The integrating factor is $\phi(t)=e^{2t+2\mu\sin t}$, $$ \frac{d}{dt}(\phi v)=2\phi, \\ \phi(2\pi)v(2\pi)-\phi(0)v(0)=2\int_0^{2\pi}e^{2t+2\mu\sin t}\,dt $$ For a closed solution $v(0)=v(2\pi)=r_*^{-2}$, giving the equation for the fixed point $$ r_*^{-2}=\frac2{e^{4\pi}-1}\int_0^{2\pi}e^{2t+2\mu\sin t}\,dt $$

This will give large positive values $v_0=r_0^{-2}$ for large $\mu$, but has no singularities. Thus a limit cycle always exist, with increasingly small but always positive radius for increasing values of the $\mu$ parameter.

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  • $\begingroup$ Thanks for your answer. However, I am a bit confused. Where did the equation $\frac{d}{dt} r^{-2} = -2r^{-3} \dot{r}$ come from? (I am not familiar with Poincare maps; the book apparently discusses this in chapter 8.) $\endgroup$
    – Leonidas
    Mar 4 at 22:01
  • $\begingroup$ That is simply the chain rule and the derivative of a power function. $\endgroup$ Mar 5 at 5:18
  • $\begingroup$ OK, I think I understand now. A few comments: (1) It wasn't the differentiation step that confused me, but rather the change of variables to $v = r^{-2}$. I think stating the substitution $v = r^{-2}$ prior to the differentiation step would make your work a little easier to follow. (I'm not that familiar with Bernoulli ODEs, so I was initially very confused by how you came up with this substitution.) (2) The integrating factor should be denoted by a different letter since $\mu$ denotes the parameter. (3) In your equation for $r_{*}^{-2}$, I think there should be a factor of 2 on the RHS. $\endgroup$
    – Leonidas
    Mar 5 at 17:29
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    $\begingroup$ Thank you, the corrections should be all done now. $\endgroup$ Mar 5 at 18:12
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Here is the solution in my own words, based on Lutz Lehmann's answer.

We claim that a closed orbit exists for all $\mu > 0$. To show this, we first note that
\begin{align*} \dot{\theta} = 1 \implies \theta(t) = t + \theta(0). \end{align*} Hence, all trajectories (except for the fixed points) rotate through an angle of $2\pi$ every unit of time. In particular, any closed orbit must cross the axis $\theta = 0$ every unit of time. Therefore, a closed orbit exists if and only if there exists an initial condition $(r(0),\theta(0)) = (r_0,0)$ such that $r(2\pi) = r_0$. We show that such an $r_0 > 0$ exists by explicitly solving the IVP $$ (1) \quad \begin{cases} \dot{r} = r(1-r^2) + \mu r \cos \theta, \\[2pt] \dot{\theta} = 1 \\[2pt] (r(0),\theta(0)) = (r_0, 0). \end{cases} $$

From $\dot{\theta} = 1$ and $\theta(0) = 0$ we have $\theta(t) = t$, and so we are reduced to solving a single inhomogeneous ODE: $$ \dot{r} = r(1-r^2) + \mu r \cos t, \quad r(0) = r_0. $$

By rewriting the equation in the form \begin{align*} \dot{r} - r \left(1 + \mu \cos t \right) = -r^3 \end{align*}

we recognize this as a Bernoulli equation. Recall that the general form of a Bernoulli equation is \begin{align*} \quad \dot{r} + p(t)r = q(t) r^a \end{align*}

where $a \in \mathbb{R}$ is a constant. In this case we have $p(t) = - (1 + \mu \cos t), \; q(t) = -1$ and $a = 3$. We solve this by making the substitution $v = r^{1-a} = r^{-2}$. Then \begin{align*} \dot{v} = -2r^{-3} \dot{r} &= -2r^{-3} \left[ r(1 + \mu \cos t) - r^3 \right] \\[4pt] &= -2r^{-2}(1 + \mu \cos t) + 2 \\[3pt] &= -2v(1 + \mu \cos t) + 2. \end{align*}

We now have first order linear ODE in $v$: \begin{align*} \dot{v} + (2 + 2 \mu \cos t)v = 2. \end{align*}

Multiplying through by the integrating factor \begin{align*} F(t) = \exp\left( \int (2 + 2 \mu \cos t) \,dt \right) = \exp \left( 2t + 2\mu \sin t \right), \end{align*}

we obtain \begin{align*} \frac{d}{dt} (Fv) = 2F \end{align*}

which implies that \begin{align*} F(t) v(t) = F(0) v(0) + 2 \int_{0}^{t} F(s) \, ds. \end{align*}

Hence, \begin{align*} v(t) &= \frac{1}{F(t)} \left( F(0) v(0) + 2 \int_{0}^{t} F(s) \, ds \right) = \frac{1}{F(t)} \left( \frac{1}{r_0^2} + 2 \int_{0}^{t} F(s) \, ds \right) \end{align*}

since $v = 1/r^2$ and $F(0) = 1$. Therefore, since $r^2 = 1/v$ we have \begin{align*} (r(t))^2 = F(t) \left( \frac{1}{r_0^2} + 2 \int_{0}^{t} F(s) \, ds \right)^{-1}. \end{align*}

Hence, the solution to (1) is a closed orbit if and only if \begin{align*} r(2\pi) = r_0 &\iff (r(2\pi))^2 = r_0^2 \\[4pt] &\iff e^{4\pi} \left( \frac{1}{r_0^2} + 2 \int_{0}^{2\pi} F(s) \, ds \right)^{-1} = r_0^2 \\[4pt] &\iff \frac{e^{4\pi}}{r_0^2} = \frac{1}{r_0^2} + 2 \int_{0}^{2\pi} F(s) \, ds \\[4pt] &\iff \frac{e^{4\pi} - 1}{r_0^2} = 2 \int_{0}^{2\pi} F(s) \,ds \\[4pt] &\iff r_0 = \sqrt{\frac{e^{4\pi} - 1}{2} \left( \int_{0}^{2\pi} e^{2t + 2\mu \sin t} \,dt \right)^{-1}}. \end{align*}

Since $e^{4\pi} - 1 > 0$ and $\int_{0}^{2\pi} e^{2t + 2\mu \sin t} \,dt > 0$ for all $\mu$, the expression under the square root is clearly positive. Therefore, this initial condition does indeed give rise to a closed orbit trajectory. Hence, a closed orbit exists for all $\mu > 0$. Moreover, this work shows that the closed orbit is unique.

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Fix $0<\mu<1$. If $r>1+\mu$, then $$ \dot{r} = r(1-r^2) + \mu r \cos \theta\le r(1-r^2)+\mu r=r(1+\mu-r)<0. $$ If $r<1-\mu$, then $$ \dot{r} = r(1-r^2) + \mu r \cos \theta\ge r(1-r^2)-\mu r=r(1-\mu-r)>0. $$ By Bendixson's Theorem, there is a closed limit cycle in $1-\mu<r<1+\mu$. The following graph is shown for a closed orbit for $\mu=0.5$. enter image description here

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  • $\begingroup$ Thanks for your post. However, the textbook already treats the case $0 < \mu < 1$ in Example 7.3.1. I am specifically interested in proving the existence/nonexistence of limit cycles for $\mu \geq 1$. $\endgroup$
    – Leonidas
    Feb 26 at 13:30
  • $\begingroup$ If $\mu\ge1$, there is a limit cycle. However, Bendixson's theorem is not applicable. $\endgroup$
    – xpaul
    Feb 27 at 20:06
  • $\begingroup$ OK. Do you have any suggestions for how to prove this without Poincare-Bendixson? $\endgroup$
    – Leonidas
    Feb 28 at 14:52

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