MAT244--2018F > Quiz-7

Q7 TUT 0201

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Victor Ivrii:
(a) Determine all critical points of the given system of equations.

(b) Find the corresponding linear system near each critical point.

(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?

(d)  Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.
$$\left\{\begin{aligned}
&\frac{dx}{dt} = 1 - xy, \\
&\frac{dy}{dt} = x - y^3.
\end{aligned}\right.$$

Bonus: Computer generated picture

Yulin WANG:
(a)
\begin{equation} 
\left\{ 
             \begin{array}{**lr**} 
             1-xy=0 &  \\ 
             x-y^{3}=0\\
             \end{array} 
\right. 
\end{equation}
\begin{equation} 
\left\{ 
             \begin{array}{**lr**} 
             xy=1 &  \\ 
             x=y^{3}\\
             \end{array} 
\right. 
\end{equation}
\begin{equation} 
\left\{ 
             \begin{array}{**lr**} 
             x=1 &  \\ 
             y=1\\
             \end{array} 
\right. 
\end{equation} or \begin{equation} 
\left\{ 
             \begin{array}{**lr**} 
             x=-1 &  \\ 
             y=-1\\
             \end{array} 
\right. 
\end{equation}
Therefore, the critical points are (1,1) and (-1,-1)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
-y & -x \\
1 & -3y^{2}
\end{bmatrix}\\
~\\
J(1,1) &= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
J(-1,-1) &= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}
\end{align*}
(c)
\begin{align*}
For (1,1), let A&= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
-1-\lambda & -1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda+1)+1=0\\
~\\
\lambda_{1} &= \lambda_{2} = -2 \\
~\\
Then \ the \ system \ has \ a \ stable \ improper \ node \ at \ (1,1) \\
~\\
For (-1,-1), let A&= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda-1)-1=0\\
~\\
\lambda = -1 \pm \sqrt{5} \\
~\\
Then \ the \ system \ has \ a \ unstable \ saddle \ point \ at \ (1,1) \\
\end{align*}
(d) In the attachment.

Zhuojing Yu:
I think when (1,1), it is node or spiral point, not IN(improper node).

Jingze Wang:
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted

Yulin WANG:

--- Quote from: Jingze Wang on November 30, 2018, 08:31:23 PM ---I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted

--- End quote ---
Thanks for submitting the computer-generated phase portrait!!!
BTW, how do u plot the phase portrait on a computer?

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