# Toronto Math Forum

## MAT244-2018S => MAT244––Home Assignments => Web Bonus Problems => Topic started by: Victor Ivrii on March 18, 2018, 12:43:58 PM

Title: Phaseportrait
Post by: Victor Ivrii 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}
Title: Re: Phaseportrait
Post by: Darren Zhang 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.
Title: Re: Phaseportrait
Post by: Kexin Sun on March 19, 2018, 12:08:43 PM
Use online technique to plot:
http://www.bluffton.edu/homepages/facstaff/nesterd/java/slopefields.html
Title: Re: Phaseportrait
Post by: Meng Wu on March 19, 2018, 12:17:36 PM
WolframAlpha
Title: Re: Phaseportrait
Post by: Victor Ivrii on March 19, 2018, 01:21:44 PM
Now explain it!!!
Title: Re: Phaseportrait
Post by: Yichen Nie 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.
Title: Re: Phaseportrait
Post by: Victor Ivrii 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).
Title: Re: Phaseportrait
Post by: Kexin Sun 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.
Title: Re: Phaseportrait
Post by: Victor Ivrii on March 21, 2018, 01:55:44 AM
Kexin
See my comment above
Title: Re: Phaseportrait
Post by: Meng Wu 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.
Title: Re: Phaseportrait
Post by: Victor Ivrii 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?
Title: Re: Phaseportrait
Post by: Victor Ivrii 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