Showing there is a limit cycle

• Jun 3rd 2011, 01:29 PM
Borkborkmath
Showing there is a limit cycle
Planar system:

x' = x-y-x^3
y'=x+y-y^3

only has a critical point at origin.

The question asks to show that this system has a periodic orbit.

The hint for this says to convert to polar coordinates and show that for all $\epsilon$>0, r'<0 on the circle r=sqrt(2)+ $\epsilon$ and r'>0 on the circle r=1- $\epsilon$. Then show that this implies that there is a limit cycle in the annulus: {x $\in$R^2 : 1<= |x| <= sqrt(2)}.

I'm not sure how to even convert this to polar coordinates. I know that x=rcos( $\theta$) and y=rsin( $\theta$). But after i plug it in, I just get a equation and I don't know how to get r' or $\theta$' from it.
• Jun 3rd 2011, 10:21 PM
TheEmptySet
Quote:

Originally Posted by Borkborkmath
Planar system:

x' = x-y-x^3
y'=x+y-y^3

only has a critical point at origin.

The question asks to show that this system has a periodic orbit.

The hint for this says to convert to polar coordinates and show that for all $\epsilon$>0, r'<0 on the circle r=sqrt(2)+ $\epsilon$ and r'>0 on the circle r=1- $\epsilon$. Then show that this implies that there is a limit cycle in the annulus: {x $\in$R^2 : 1<= |x| <= sqrt(2)}.

I'm not sure how to even convert this to polar coordinates. I know that x=rcos( $\theta$) and y=rsin( $\theta$). But after i plug it in, I just get a equation and I don't know how to get r' or $\theta$' from it.

Use the formula $r^2=x^2+y^2$ and take the derivative with respect to time

$2r\dot{r}=2x\dot{x}+2y\dot{y} \iff \dot{r}=\frac{1}{r}(x\dot{x}+y\dot{y})$ and

$\theta=\tan^{-1}\left( \frac{y}{x} \right)$

$\dot{\theta}=\frac{1}{1+\left( \frac{y}{x}\right)^2}(\frac{\dot{y}}{x}-\frac{y}{x^2}\dot{x})=\frac{1}{x^2+\left( y\right)^2}(x\dot{y}-y\dot{x})=\frac{1}{r^2}(x\dot{y}-y\dot{x})$

Now we need to compute the "inside" of each of the above

$x\dot{x}+y\dot{y}=x(x-y-x^3)+y(x+y-y^3)=x^2+y^2-(x^4+y^4) = x^2+y^2-(x^2+y^2)^2+2x^2y^2=r^2-r^4+2r^2\sin^2(\theta)\cos^2(\theta)$

$x\dot{y}-y\dot{x}=x(x+y-y^3)-y(x-y-x^3)=x^2+y^2-xy^3+x^3y=x^2+y^2-xy(y^2-x^2)=r^2-r^4\sin(\theta)\cos(\theta)[\cos^2(\theta)-\sin^2(\theta)]$

This gives the system of ODE's

$\frac{dr}{dt}=r-r^3+2r\sin^2(\theta)\cos^2(\theta)$

$\frac{d\theta}{dt}=1-r^2\sin(\theta)\cos(\theta)[\cos^2(\theta)-\sin^2(\theta)]$

either I have made a mistake (very possible) or I can't see how to simplify and solve this system. I will leave this to you at this point.
• Jun 4th 2011, 07:13 AM
Jester
From what TheEmptySet said

$\dfrac{dr}{dt} = r\left( 1 - r^2 + r \dfrac{\sin^2 2 \theta}{2}\right)$

the following identity might help

$1 - r^2 \le 1 - r^2 + r \dfrac{\sin^2 2 \theta}{2} \le 1 - r^2 + \dfrac{r}{2}$
• Jun 6th 2011, 05:26 PM
Borkborkmath

When I expand xx'+yy', I get $x^2+y^2-x^4-y^4$ and just leave it that way. Then i sub in rcos and rsin which leads to: $r-r^3(cos^4(\theta)+sin^4(\theta))$. Is the way I did it wrong?
Since you simplified $x^4-y^4$ to $"-(x^2 + y^2)^2+2x^2y^2"$
• Jun 6th 2011, 05:31 PM
TheEmptySet
Quote:

Originally Posted by Borkborkmath

When I expand xx'+yy', I get $x^2+y^2-x^4-y^4$ and just leave it that way. Then i sub in rcos and rsin which leads to: $r-r^3(cos^4(\theta)+sin^4(\theta))$. Is the way I did it wrong?
Since you simplified $x^4-y^4$ to $"-(x^2 + y^2)^2+2x^2y^2"$

No the two expressions are equivalent by trig identities. When I did this I was trying to find a closed form solution, but you can bound this just like Danny did by finding min and max of the function with respect to theta.
• Jun 6th 2011, 05:37 PM
Borkborkmath
Ah ic.

So now I should just plug in the two given values for r and show the r'<0 and r'>0 respectively.
Then this leads to the annulus having a limit cycle because: "A compact set that is positively or negatively invariant contains either a limit cycle or an equilibrium point", and since the only critical point in at origin we know this to be a limit cycle?

Hm, but I seem to run into a problem that when $r=sqrt(2)-\epsilon$ the equation isn't negative.
• Jun 6th 2011, 08:51 PM
TheEmptySet
Quote:

Originally Posted by Borkborkmath
Ah ic.

So now I should just plug in the two given values for r and show the r'<0 and r'>0 respectively.
Then this leads to the annulus having a limit cycle because: "A compact set that is positively or negatively invariant contains either a limit cycle or an equilibrium point", and since the only critical point in at origin we know this to be a limit cycle?

Hm, but I seem to run into a problem that when $r=sqrt(2)-\epsilon$ the equation isn't negative.

$\dot{r}=r-a \cdot r^3, a \in \left[\frac{1}{2},1 \right]$
$r(t)=\pm \frac{1}{\sqrt{a+ce^{-t}}}$ for since $r > 0$
$r(t)= \frac{1}{\sqrt{a+ce^{-t}}} \lim_{t \to \infty}r(t)=\frac{1}{\sqrt{a}}$