### Author Topic: Phaseportrait  (Read 5380 times)

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### Phaseportrait
« on: March 18, 2018, 12:43:58 PM »
Sketch the phaseportrait for the system below. Anyone can post a different solution, but there is a catch: it should be drawn with different s/w than already used and this s/w must be reported
\begin{align}
&x'= \sin(x)\cos(y)\\
&y'=-\cos(x)\sin(y) && -4\le x\le 4, \ -4\le y \le 4
\end{align}

#### Darren Zhang

• Full Member
• Posts: 22
• Karma: 13
##### Re: Phaseportrait
« Reply #1 on: March 19, 2018, 11:48:27 AM »
Attached is the phase portrait.
I use Matlab
The type of this is centre, as is shown in handout 9, 3b http://www.math.toronto.edu/courses/mat244h1/20181/LN/Ch7-LN9.pdf.
It's purely imaginary numbers, and it's stable.
« Last Edit: March 19, 2018, 07:14:19 PM by Junjie Zhang »

#### Kexin Sun

• Newbie
• Posts: 2
• Karma: 0
##### Re: Phaseportrait
« Reply #2 on: March 19, 2018, 12:08:43 PM »

#### Meng Wu

• Elder Member
• Posts: 91
• Karma: 36
• MAT3342018F
##### Re: Phaseportrait
« Reply #3 on: March 19, 2018, 12:17:36 PM »
WolframAlpha
« Last Edit: March 19, 2018, 01:16:36 PM by Meng Wu »

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Phaseportrait
« Reply #4 on: March 19, 2018, 01:21:44 PM »
Now explain it!!!

#### Yichen Nie

• Newbie
• Posts: 1
• Karma: 0
##### Re: Phaseportrait
« Reply #5 on: March 19, 2018, 06:40:31 PM »
The type of critical point is called center, which is neutrally stable.
This occurs when the eigenvalues contain only imaginary numbers.
The trajectories just stay in stable, elliptical orbits as my classmates have shown above.
« Last Edit: March 19, 2018, 06:48:44 PM by Yichen Nie »

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Phaseportrait
« Reply #6 on: March 20, 2018, 06:21:29 AM »
Yichen,
there is the big difference between describe (what you see) and explain (why it is so).

#### Kexin Sun

• Newbie
• Posts: 2
• Karma: 0
##### Re: Phaseportrait
« Reply #7 on: March 20, 2018, 11:59:50 PM »
The origin is a saddle point,and it is stable.
The ponit (π/2,π/2),(-π/2,π/2),(π/2, -π/2),(-π/2, -π/2) are the centers of closed orbits,respectively.Each of them is orbital stablility.
« Last Edit: March 21, 2018, 12:02:20 AM by Kexin Sun »

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Phaseportrait
« Reply #8 on: March 21, 2018, 01:55:44 AM »
Kexin
See my comment above

#### Meng Wu

• Elder Member
• Posts: 91
• Karma: 36
• MAT3342018F
##### Re: Phaseportrait
« Reply #9 on: March 28, 2018, 01:50:57 PM »
$\text{My attempt:}$
\begin{align}x'&=\sin(x)\cos(y)=F(x,y)\\y'&=-\cos(x)\sin(y)=G(x,y)\end{align}
Let \begin{align}\cases{F(x,y)=\sin(x)\cos(y)=0\\G(x,y)=-\cos(x)\sin(y)=0}\end{align}
$$\implies\cases{x=k\pi,y=k\pi\\x={\pi\over2}+k\pi,y={\pi\over2}+k\pi}$$
where $k$ can only be $-1,0,1$, since we have $x\in(-4,4)$ and $y\in(-4,4)$, $\pi\cong 3.14159265359$. $\\$
Thus we have the following critical points:$\\$
Case#1: $(0,0);(\pi,0);(-\pi,0);(0,\pi);(0,-\pi);(\pi,\pi);(-\pi,\pi);(-\pi,-\pi);(\pi,-\pi)$. $\\$
Case#2: $({\pi\over2},{\pi\over2});(-{\pi\over2},-{\pi\over2});({\pi\over2},-{\pi\over2});(-{\pi\over2},{\pi\over2})$.
$$J=\begin{pmatrix}F_x(x,y)&F_y(x,y)\\G_x(x,y)&G_x(x,y)\end{pmatrix}=\begin{pmatrix}\cos(x)\cos(y)&-\sin(x)\sin(y)\\\sin(x)\sin(y)&-\cos(x)\cos(y)\end{pmatrix}$$
When $(x,y)=$ critical points in Case#1: we get diagonal matrices $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ or $\begin{pmatrix}-1&0\\0&0\end{pmatrix}$, wtih eigenvalues $\lambda=\pm1$, eigenvectors $\xi=\begin{pmatrix}1\\0\end{pmatrix}$ or $\begin{pmatrix}0\\1\end{pmatrix}$. $\\$
Thus, the critical points in Case#1 are all $\text{Saddle Points}$. $\\$
When $(x,y)=$ critical points in Case#2: we get matrices $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ or $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ with eigenvalues $\lambda=\pm i$, eigenvectors $\xi=\begin{pmatrix}1\\i\end{pmatrix}$ or $\begin{pmatrix}1\\-i\end{pmatrix}$. $\\$
Thus, the critical points in Case#2 are all $\text{Center}$. $\\$
Therefore based on the phase portraits for saddle point and center, we can conclude the phase portrait for the given system will be like that.

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Phaseportrait
« Reply #10 on: March 28, 2018, 02:02:45 PM »
If you discovered that at the critical point eigenvalues for the linearized system are purely imaginary, for non-linear system it may be either center, or stable focal point, or unstable focal point. However for this system we can exclude focal points. Why?

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Phaseportrait
« Reply #11 on: March 29, 2018, 05:20:42 AM »
One needs to remember, that there are two major cases: General and Integrable. Below subscripts denote partial derivatives (as usual)

In the general case
$$\begin{pmatrix}x\\y\end{pmatrix}'=\begin{pmatrix}F(x,y)\\G(x,y)\end{pmatrix}\tag{1}$$
stationary points are those $(\bar{x},\bar{y})$, where $F(\bar{x},\bar{y})=G(\bar{x},\bar{y})=0$. Then, depending on the eigenvalues of the matrix of the linearized system
$$\begin{pmatrix}F_x & F_y\\G_x& G_x\end{pmatrix},\tag{2}$$
calculated at this point we conclude that it is either a node, or a saddle, or a focal point, or that linearization alone cannot tell us. Right?

In the integrable case we can construct function $H(x,y)$, s.t. it is constant along trajectories of (1), and therefore
$$F=\mu H_y, \qquad G=-\mu H_x \tag{3}$$
where $\mu=\mu(x,y)$ is and integrating factor and the stationary points of (1) are stationary (critical) of function $H(x,y)$. See Calculus II.  Remember that such points are of two types: maxima (or minima) and saddles. Again, some points where matrix of second derivatives (Hess matrix) cannot tell us the answer.

Assume  that this matrix is non-degenerate. Then either only maxima (minima) and level lines form center, or a saddle and level lines form a saddle.  Note matrix (2) at $(\bar{x},\bar{y})$ becomes
$$\mu \begin{pmatrix}H_{xy} & H_{yy}\\-H_{xx}& -H{x}\end{pmatrix},\tag{4}$$
with eigenvalues $\pm \mu \sqrt{H_{xy}^2-H_{xx}H_{yy}}$. Assume that $\mu(\bar{x},\bar{y})\ne 0$. Then

* If $H_{xy}^2-H_{xx}H_{yy}>0$ these eigenvalues are real and it is a saddle.

* If $H_{xy}^2-H_{xx}H_{yy}<0$ these eigenvalues are real and it is a center. In contrast to the general case linearization here gives us an answer (since no focal points are possible).  Only if $H_{xy}^2-H_{xx}H_{yy}=0$ at this point, we need to dig deeper... like in Calculus II.

One should remember this difference (very important). Now, check if the system in question is exact, if it is, find $H(x,y)$ and deal with it

« Last Edit: March 29, 2018, 05:44:14 AM by Victor Ivrii »