Toronto Math Forum

MAT244-2018S => MAT244--Tests => Final Exam => Topic started by: Victor Ivrii on April 11, 2018, 08:48:39 PM

Title: FE-P6
Post by: Victor Ivrii 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.

Title: Re: FE-P6
Post by: Tim Mengzhe Geng on April 12, 2018, 12:41:28 AM

For part(b), we have
\begin{equation}
    (x^2+y^2-1)dx+2xydy=0
\end{equation}
Note that
\begin{equation}
    M_y=N_x=2y
\end{equation}
The equation is exact.
By integration
\begin{equation}
    H=\frac{1}{3}x^3+xy^2-x+h^\prime(y)
\end{equation}
\begin{equation}
    h^\prime(y)=0
\end{equation}
We choose
\begin{equation}
    h(y)=0
\end{equation}
In this way,
\begin{equation}
    H(x,y)=\frac{1}{3}x^3+xy^2-x=C
\end{equation}
I will post solution to other parts later if no one else follows.

Title: Re: FE-P6
Post by: Nikola Elez on April 12, 2018, 01:16:43 AM

I have attached a phase portrait

Title: Re: FE-P6
Post by: Syed Hasnain 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

Title: Re: FE-P6
Post by: Nikola Elez on April 12, 2018, 01:37:12 AM

For part a)
Sorry if poor quality

Title: Re: FE-P6
Post by: Tim Mengzhe Geng on April 12, 2018, 09:22:44 AM

Quote from: Syed_Hasnain 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

Please note that at I state, I just choose $h(y)=0$ for simplification.

Title: Re: FE-P6
Post by: Tim Mengzhe Geng on April 12, 2018, 09:51:12 AM

For Part(a), Note that for stationary points, we should have
\begin{equation}
    x^2+y^2-1=0
\end{equation}
And at the same time
\begin{equation}
    -2xy=0
\end{equation}
Therefore there're totally four stationary points. They are
\begin{equation}
    (x,y)=(1,0), (-1,0), (0,1) or (0,-1).
\end{equation}
\begin{equation}
   J={
\left[\begin{array}{ccc}
2x & 2y \\
-2y & -2x
\end{array}
\right ]},
\end{equation}
At point (1,0),
\begin{equation}
   J[1,0]={
\left[\begin{array}{ccc}
2 & 0 \\
0 & -2
\end{array}
\right ]},
\end{equation}
At point (-1,0),
\begin{equation}
   J[-1,0]={
\left[\begin{array}{ccc}
-2 & 0 \\
0 & 2
\end{array}
\right ]},
\end{equation}
At point (0,1),
\begin{equation}
   J[0,1]={
\left[\begin{array}{ccc}
0 & 2 \\
-2 & 0
\end{array}
\right ]},
\end{equation}
At point (0,-1),
\begin{equation}
   J[0,-1]={
\left[\begin{array}{ccc}
0 & -2 \\
2 & 0
\end{array}
\right ]},
\end{equation}

Title: Re: FE-P6
Post by: Nikola Elez on April 12, 2018, 10:50:25 AM

Ah I seem to have missed two points, thanks for adding the full solution Tim!

Title: FE-P6 Comments
Post by: Victor Ivrii 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