### Author Topic: Semester End Challenge 2  (Read 3346 times)

#### Victor Ivrii ##### Semester End Challenge 2
« on: April 04, 2013, 06:35:39 AM »
Both parts are separate but related problems

(A) Draw phase
\begin{equation}
\left\{\begin{aligned}
&\frac{dx}{dt}=-6xy,\\
&\frac{dy}{dt}=-3x^2+3y^2.
\end{aligned}\right.
\tag{a}
\end{equation}
s it integrable? Find equilibrium points and try to classify them.

(B) Draw phase
\begin{equation}
\left\{\begin{aligned}
&\frac{dx}{dt}=-\cos (4y)+\cos(2y)\cos (2\alpha x),\\
&\frac{dy}{dt}=\alpha \sin(2y)\sin (2\alpha x).
\end{aligned}\right.
\tag{b}
\end{equation}
Is it integrable? Find equilibrium points and try to classify them. What is connection to (I)? For calculations take $\alpha=1$ but for $\alpha=\sqrt{3}$ picture will be nicer (phase portraits are similar).
« Last Edit: April 04, 2013, 06:40:50 AM by Victor Ivrii »

#### Alexander Jankowski

• Full Member
•   • Posts: 23
• Karma: 19 ##### Re: Semester End Challenge 2
« Reply #1 on: April 04, 2013, 10:15:35 AM »
System $(a)$ has a nice stream plot. The system is, in fact, integrable:
\begin{align*}
\frac{dy}{dx} = \frac{-3x^2+3y^2}{-6xy} & \Longrightarrow -6xydy = (-3x^2+3y^2)dx \\
& \Longrightarrow H(x,y) = x^3 - 6xy^2 = C.
\end{align*}
The contour plot of $H$ is given and it agrees with the stream plot of the system. It looks like the one critical point is $(0,0)$, which can be verified by solving the equations $x'(t) = 0, y'(t) = 0$. The stream plot suggests that it is a saddle point. There are three lines intersecting at $(0,0)$ and they are separatrices. The way that other trajectories interact with them changes as the lines pass through the origin.

The stream plot for system $(b)$ is a little intimidating. Again, the system is integrable:
\begin{align*}
\frac{dy}{dx} = \frac{\sin(2x)\sin(2y)}{-4\cos(4y)+\cos(2x)\cos(2y)} & \Longrightarrow -4\cos(4y)dy + \cos(2x)\cos(2y)dy = \sin(2x)\sin(2y)dx \\
& \Longrightarrow H(x,y) = \cos(2x)\sin(2y) - \sin(4y) = C.
\end{align*}
Again, the contour plot for $H$ agrees with the stream plot. The equilibrium points are difficult to find. However, we will note that there are two main kinds of equilibrium points:

• stable centers, which are surrounded by closed trajectories,
• saddle points, through which the nullclines pass.

The connection between systems $(a)$ and $(b)$ is that by analyzing system $(a)$, we have analyzed the saddle points of system $(b)$. Each saddle point is a point of intersection of three separatrices that separate six surrounding stable centers from one another. This is easy to see at $(0,0)$.

#### Victor Ivrii ##### Re: Semester End Challenge 2
« Reply #2 on: April 04, 2013, 10:35:01 AM »

#### Alexander Jankowski

• Full Member
•   • Posts: 23
• Karma: 19 ##### Re: Semester End Challenge 2
« Reply #3 on: April 04, 2013, 10:49:40 AM »
I am not sure what can be described as unusual. Unlike most other saddle points that I have seen, which have two intersecting lines, these have three. I have also observed no saddle point in system $(b)$ is different from any of the other saddle points. All surrounding trajectories behave in the same way. The saddle points are not identical to the one in $(a)$, but the separatrices are rotated by $\pi/6$ radians.
« Last Edit: April 04, 2013, 10:54:20 AM by Alexander Jankowski »

#### Victor Ivrii ##### Re: Semester End Challenge 2
« Reply #4 on: April 04, 2013, 11:10:35 AM »
I am not sure what can be described as unusual. Unlike most other saddle points that I have seen, which have two intersecting lines, these have three. I have also observed no saddle point in system $(b)$ is different from any of the other saddle points. All surrounding trajectories behave in the same way. The saddle points are not identical to the one in $(a)$, but are rotated by $\pi/6$ radians.

In Calculus II only non degenerate stationary points were considered: Hessian (if $n=2$ it is $\begin{pmatrix} H_{xx} &H_{xy} \\ H_{yx} & H_{yy}\end{pmatrix}$) should have it determinant different from $0$ and if it is positive we get extremumâ€”minimum as $H_{xx}>0$ and maximum as $H_{xx}<0$; $H_{yy}$ automatically has the same signâ€”and if determinant is negative we get a saddle).

Here critical point $(0,0)$ is degenerate (and even all second order derivatives vanish here) and it is not the standard saddle. In fact it is called Monkey saddle (three valleys and three mountains joint here while in the usual saddle two valleys and two mountains join).  While ordinary saddle is represented by $\renewcommand{\Re}{\operatorname{Re}}$ $\Re (x+iy)^2=x^2-y^2$ the monkey saddle by $\Re (x+iy)^3=x^3-3xy^2$. The multiplicity of the types and complexity of degenerate critical points even as $n=2$ is a reason why you have not looked at them in Calculus II.

In fact system (b) is also integrable for any $\alpha\ne 0$. May be centers are difficult to find directly but there are simple geometric observations.

The beauty of $\alpha=\sqrt{3}=\tan (\pi/3)$ is that triangles are regular  (and near critical points it  `almost' coincides with (a)). The sides of triangles are $\pi/\sqrt{3}$ so the whole plane is tiled by regular triangles and their vertices are monkey saddles  and the whole picture is not only periodic but has a rotational (by $\pi/3$) symmetry  and 6 mirror symmetries. Then the rest of stationary points (centers) must be centers of these triangles (and in the regular triangle there is a single center). Here I ignore directions, just purely geometric picture.

Obviously for $\alpha \ne \sqrt{3}$ we need just scale with respect to $x$, so we get equilateral triangles with the base $\pi/\alpha$ and a height $\pi/2$ (the same as before) and the rest of stationary points (centers) must be on the distance $\pi/6$ from the base and they are centroids (points of intersection of medians) of those triangles. Periodic structure remains but only 2 mirror symmetries and rotational by $\pi$ symmetry.

PS Up to a factor
\begin{equation}
H(x,y)=\sin(4y)-\sin( 2y+2\alpha x) -\sin(2y-2\alpha x),
\end{equation}
Then as $\alpha=\sqrt{3}$
\begin{equation}
H(x,y)=\sin(4\mathbf{x}\cdot \mathbf{e}_1 )+\sin(4\mathbf{x}\cdot \mathbf{e}_2 ) +\sin(4\mathbf{x}\cdot \mathbf{e}_3)
\end{equation}
with $\mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix}$, $\mathbf{e}_j=\begin{pmatrix}\sin (2\pi (j-1)/3)\\ \cos (2\pi (j-1)/3)\end{pmatrix}$; those are three unit vectors with angles $2\pi/3$ between. Then rotation by $2\pi/3$ preserves $H(x,y)$ and by $\pi/3$ replaces $H(x,y)$ by $-H(x,y)$ which also preserves geometry.
« Last Edit: April 04, 2013, 11:39:24 PM by Victor Ivrii »