### Author Topic: Easter challenge  (Read 4835 times)

#### Victor Ivrii

• Elder Member
• Posts: 2566
• Karma: 0
##### Easter challenge
« on: March 28, 2013, 07:36:07 PM »
Draw phase portraits :

\begin{gather}
\left\{\begin{aligned}
&x'=-\sin(y),\\
&y'=   \sin (x);
\end{aligned}\right. \tag{a}\\
\left\{\begin{aligned}
&x'=-\sin(y),\\
&y'= \,2\sin (x);
\end{aligned}\right. \tag{b}
\end{gather}

Explain the difference between portraits and its reason

#### Hareem Naveed

• Newbie
• Posts: 2
• Karma: 2
##### Re: Easter challenge
« Reply #1 on: March 29, 2013, 11:38:46 AM »
Attached are the two phase portraits.

In terms of difference between the two; they are level curves of the following functions:
$$H_{a}(x,y) = \cos(y)+\cos(x) \\ H_{b}(x,y) = \cos(y) + 2\cos(x)$$
Level curves are also attached.

From the level curves, in a, the centres are hemmed in by 2 defined separatrices, not so in b where there is only one.

How could I formalize these statements? Intuitively, I can "see" the answer.
« Last Edit: March 29, 2013, 01:11:45 PM by Victor Ivrii »

#### Victor Ivrii

• Elder Member
• Posts: 2566
• Karma: 0
##### Re: Easter challenge
« Reply #2 on: March 29, 2013, 01:18:24 PM »
Attached are the two phase portraits.

In terms of difference between the two; they are level curves of the following functions:
$$H_{a}(x,y) = \cos(y)+\cos(x) \\ H_{b}(x,y) = \cos(y) + 2\cos(x)$$
Level curves are also attached.

From the level curves, in a, the centres are hemmed in by 2 defined separatrices, not so in b where there is only one.

How could I formalize these statements? Intuitively, I can "see" the answer.

Your solution is definitely correct but I want a bit more observation about dynamics, not only about centers but about all trajectories. Start from the very informal description as I am interested more in an understanding than the formal expression

#### Alexander Jankowski

• Full Member
• Posts: 23
• Karma: 19
##### Re: Easter challenge
« Reply #3 on: March 29, 2013, 06:11:46 PM »
This is a nice problem. I can give a little bit of input: in system (b), the rate of change $y'(x)$ is twice that of the same rate in system (a). Perhaps we can treat (b) as a vertical expansion of (a), which is what the contour maps suggest. We can also note that each system has the same critical points $(2 \pi n,2 \pi n)$, where $n$ is an integer.

By inspection, we see that the separatrices in system (a) are lines with slopes $\pi$ and $-\pi$. In fact, by looking more closely at the contour maps, the separatrices appear to have the equations $$y = Â± \pi x + n \pi,$$ where $n$ is an integer. In system (b), the separatrices are sinusoidal functions that oscillate about an equilibrium line that is parallel to the $y$-axis. Finally, because there are infinitely many critical points in each system, there are also infinitely many separatrices.
« Last Edit: March 29, 2013, 06:20:34 PM by Alexander Jankowski »

#### Victor Ivrii

• Elder Member
• Posts: 2566
• Karma: 0
##### Re: Easter challenge
« Reply #4 on: March 29, 2013, 07:02:27 PM »
This is a nice problem. I can give a little bit of input: in system (b), the rate of change $y'(x)$ is twice that of the same rate in system (a). Perhaps we can treat (b) as a vertical expansion of (a), which is what the contour maps suggest. We can also note that each system has the same critical points $(2 \pi n,2 \pi n)$, where $n$ is an integer.

By inspection, we see that the separatrices in system (a) are lines with slopes $\pi$ and $-\pi$. In fact, by looking more closely at the contour maps, the separatrices appear to have the equations $$y = Â± \pi x + n \pi,$$ where $n$ is an integer. In system (b), the separatrices are sinusoidal functions that oscillate about an equilibrium line that is parallel to the $y$-axis. Finally, because there are infinitely many critical points in each system, there are also infinitely many separatrices.

I know that this is a nice problem, I just invented it You observe interesting things (correctly) albeit separatrices are only sinusoid-looking but not sinusoid lines.

But what about other trajectories which are not separatrices?

#### Alexander Jankowski

• Full Member
• Posts: 23
• Karma: 19
##### Re: Easter challenge
« Reply #5 on: March 29, 2013, 07:43:39 PM »
I made a mistake--the critical points are better described as being $(\pi \ell, \pi n)$, where $\ell$ and $n$ are integers.

For system (a), I can say that each critical point is a stable center. The trajectories are circles when very close to the critical point, but as we zoom out of the neighbourhood of the critical point, the trajectories take on the shape of boundary that is defined by the separatrices. I noticed as well that the orientation of the trajectories depends on $x$ but not on $y$. If $\ell$ is even, then the trajectories have counter-clockwise orientation. However, if $\ell$ is odd, then the trajectories have clockwise orientation. This seems to apply to system (b) as well, and from the orientation of the closed trajectories, we can infer the orientation of the separatrices. Furthermore, there are certain trajectories in (b) that are not closed trajectories about the centers, but are similar in shape and orientation to the sinusoidal separatrices.

I attached stream plots and vector fields for each system to illustrate my point.
« Last Edit: March 29, 2013, 08:12:27 PM by Alexander Jankowski »

#### Victor Ivrii

• Elder Member
• Posts: 2566
• Karma: 0
##### Re: Easter challenge
« Reply #6 on: March 29, 2013, 08:09:47 PM »
I made a mistake--the critical points are better described as being $(\pi \ell, \pi n)$, where $\ell$ and $n$ are integers.

For system (a), I can say that each critical point is a stable center.

All centers are stable; are there any other critical points?

Quote
Furthermore, there are certain trajectories in (b) that are not closed trajectories about the centers, but are similar in shape and orientation to the sinusoidal separatrices.

I attached stream plots and vector fields for each system to illustrate my point.

Thats it! On (a) there are only closed orbits (plus separatrices) but on (b) there are plenty non-closed trajectories! See next post

#### Alexander Jankowski

• Full Member
• Posts: 23
• Karma: 19
##### Re: Easter challenge
« Reply #7 on: March 29, 2013, 08:15:03 PM »
I guess that for certain values of $\ell$ and $n$, we have saddle points, such as $(-\pi,0)$.

#### Victor Ivrii

• Elder Member
• Posts: 2566
• Karma: 0
##### Re: Easter challenge
« Reply #8 on: March 29, 2013, 08:25:35 PM »
OK, let me explain this problem in full. First I consider a bit more general problemâ€”with coefficient $\alpha\ge 1$ instead of $1$ or $2$ (respectively) and correspondingly with

H_\alpha=\alpha \cos (x) + \cos(y).
\label{P}

Then
(i) As $\alpha\ne 0$ there are the same stationary points $(\pi m, \pi n)$ with $\sin(x)=\sin(y)=0$;

(ii) As $\alpha>0$ points with $\cos(x)=\cos(y)=\pm 1$ are extremums (maxima for $+$ and minima for $-$) aka centers and points with $\cos(x)=-\cos(y)=\pm 1$ are saddle points;

(iii) As $\alpha=1$ (problem (a)) all saddle points are on $H_1(x,y)=0$ and these lines are just strait lines with slopes $\pm 1$ (which are separatrices).  All other trajectories are periodic (so picture consists of "whirlwinds" around centers, separated by separatrices);

(iv) As $\alpha > 1$  (or $0<\alpha <1$) saddles with $\cos(x)=-\sin(y)=1$ are on $\{H_\alpha =(\alpha-1)\}$ and saddles with $\cos(x)=-\sin(y)=-1$ are on $\{H_\alpha =-(\alpha-1)\}$ and therefore they are on different curves; in this case separatrices from $(\pi m, \pi n)$ (with $m$ and $n$ having different parities) cannot go to "neighbouring" saddle, but only either to $(\pi m\pm 2\pi, \pi n)$ or to $(\pi m, \pi n\pm 2\pi)$ and from (vi) follows that for $\alpha>1$ it will be the latter case (and for $0<\alpha <1$ one can see that it will be the former one);

(v) Consider what happens as $\alpha>1$ and trajectory starts from point where $| H_\alpha (x,y)|<\alpha (1-\epsilon) -1$ with any $\epsilon >0$. Since $H_\alpha$ is constant along trajectory, $|\cos(x)|$ is confined between $-1+\epsilon$ and $1-\epsilon$ and therefore $\sin(x)$ cannot cross $0$ and remains either in $\{\sin (x) > \epsilon_1=(1-(1-\epsilon)^2)^{1/2}\}$ or in $\{\sin (x) <-\epsilon_1\}$. In the former case $y'>\epsilon_1$ along the whole trajectory (and in the latter $y'<-\epsilon_1$).

(vi) Therefore as $\alpha >1$ we have "whirlwinds" covering zones $\{|H_\alpha |>\alpha-1\}$ and the whirlwinds with the centers at $(\pi m,\pi n)$ ($m$ and $n$ have the same parity) and $(\pi m,\pi n+2\pi)$ are separated just by a saddle at $(\pi m,\pi n+\pi)$. However whirlwinds with the centers at $(\pi m,\pi n)$ and whirlwinds with the centers at $(\pi m+2\pi,\pi n)$ are separated by a vertical "river" floating either up or down (depending on $m$). Sure "shores" of "rivers" are not straight and at saddles the jumper between rivers has width $0$.

(vii) Similarly as $0<\alpha<1$ rivers are horizontal.

Remark. Note the similarity to pendulum (see f.e. http://weyl.math.toronto.edu/MAT244-2011S-forum/index.php?topic=126.msg450#msg450) where there are "normal" oscillations and fast rotations when pendulum goes over top point.
« Last Edit: March 30, 2013, 05:22:58 AM by Victor Ivrii »

#### Alexander Jankowski

• Full Member
• Posts: 23
• Karma: 19
##### Re: Easter challenge
« Reply #9 on: March 29, 2013, 09:20:39 PM »
That is a good explanation. I also noted another of my mistakes... The equations of the separatrices are $$y = Â±x + n\pi,$$ not $y = Â±\pi x + n\pi$. Thank you!