Toronto Math Forum
MAT2442018S => MAT244Tests => Final Exam => Topic started 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^21
\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.

For part(b), we have
\begin{equation}
(x^2+y^21)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^2x+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^2x=C
\end{equation}
I will post solution to other parts later if no one else follows.

I have attached a phase portrait

there is a small mistake..... in step 5 you have mentioned that h(y) = 0
it is notzero, it is a constant

For part a)
Sorry if poor quality

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.

For Part(a), Note that for stationary points, we should have
\begin{equation}
x^2+y^21=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}

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

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 skewHessian.
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