### Author Topic: FE-P6  (Read 5183 times)

#### Victor Ivrii

• Administrator
• Elder Member
• Posts: 2563
• Karma: 0
##### FE-P6
« on: April 11, 2018, 08:48:39 PM »
For the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t = -2xy\, , \\
&y'_t = x^2+y^2-1
\end{aligned}\right.
\end{equation*}

a. Linearize the system at
stationary points and sketch the phase portrait of this linear system.

b.  Find the equation of the form $H(x,y) = C$, satisfied by the trajectories of the nonlinear system.

c. Sketch the full phase portrait.

#### Tim Mengzhe Geng

• Full Member
• Posts: 21
• Karma: 6
##### Re: FE-P6
« Reply #1 on: April 12, 2018, 12:41:28 AM »
For part(b), we have

(x^2+y^2-1)dx+2xydy=0

Note that

M_y=N_x=2y

The equation is exact.
By integration

H=\frac{1}{3}x^3+xy^2-x+h^\prime(y)

h^\prime(y)=0

We choose

h(y)=0

In this way,

H(x,y)=\frac{1}{3}x^3+xy^2-x=C

I will post solution to other parts later if no one else follows.
« Last Edit: April 12, 2018, 09:53:45 AM by Tim Mengzhe GENG »

#### Nikola Elez

• Jr. Member
• Posts: 10
• Karma: 2
##### Re: FE-P6
« Reply #2 on: April 12, 2018, 01:16:43 AM »
I have attached a phase portrait

#### Syed Hasnain

• Full Member
• Posts: 18
• Karma: 3
• mat244h1s-winter2018
##### Re: FE-P6
« Reply #3 on: April 12, 2018, 01:25:37 AM »
there is a small mistake..... in step 5 you have mentioned that h(y) = 0
it is notzero, it is a constant

#### Nikola Elez

• Jr. Member
• Posts: 10
• Karma: 2
##### Re: FE-P6
« Reply #4 on: April 12, 2018, 01:37:12 AM »
For part a)
Sorry if poor quality

#### Tim Mengzhe Geng

• Full Member
• Posts: 21
• Karma: 6
##### Re: FE-P6
« Reply #5 on: April 12, 2018, 09:22:44 AM »
there is a small mistake..... in step 5 you have mentioned that h(y) = 0
it is notzero, it is a constant
Please note that at I state, I just choose $h(y)=0$ for simplification.

#### Tim Mengzhe Geng

• Full Member
• Posts: 21
• Karma: 6
##### Re: FE-P6
« Reply #6 on: April 12, 2018, 09:51:12 AM »
For Part(a), Note that for stationary points, we should have

x^2+y^2-1=0

And at the same time

-2xy=0

Therefore there're totally four stationary points. They are

(x,y)=(1,0), (-1,0), (0,1) or (0,-1).

J={
\left[\begin{array}{ccc}
2x & 2y \\
-2y & -2x
\end{array}
\right ]},

At point (1,0),

J[1,0]={
\left[\begin{array}{ccc}
2 & 0 \\
0 & -2
\end{array}
\right ]},

At point (-1,0),

J[-1,0]={
\left[\begin{array}{ccc}
-2 & 0 \\
0 & 2
\end{array}
\right ]},

At point (0,1),

J[0,1]={
\left[\begin{array}{ccc}
0 & 2 \\
-2 & 0
\end{array}
\right ]},

At point (0,-1),

J[0,-1]={
\left[\begin{array}{ccc}
0 & -2 \\
2 & 0
\end{array}
\right ]},

#### Nikola Elez

• Jr. Member
• Posts: 10
• Karma: 2
##### Re: FE-P6
« Reply #7 on: April 12, 2018, 10:50:25 AM »
Ah I seem to have missed two points, thanks for adding the full solution Tim!

#### Victor Ivrii

• Administrator
• Elder Member
• Posts: 2563
• Karma: 0
##### FE-P6 Comments
« Reply #8 on: April 19, 2018, 11:45:12 AM »
Observe that Hessian of $H(x,y)$ is
$$\begin{pmatrix} 2x &2y\\ 2y &2x \end{pmatrix};$$
compare with the  Jacobi matrix (Jacobian is its determinant). In this particular case (of exact system) sometimes it is called skew-Hessian.

I attach the Contour plot of $H(x,y)$; note that $(-1,0)$ is the local maximum and $(1,0)$ is the local minimum, while $(0,\pm 1)$ are two saddle points
« Last Edit: April 19, 2018, 11:47:06 AM by Victor Ivrii »