Author Topic: Q7 TUT 0201  (Read 1583 times)

Victor Ivrii

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Q7 TUT 0201
« on: November 30, 2018, 04:05:34 PM »
(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

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Re: Q7 TUT 0201
« Reply #1 on: November 30, 2018, 04:39:52 PM »
(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.
« Last Edit: November 30, 2018, 05:25:52 PM by Yulin WANG »

Zhuojing Yu

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Re: Q7 TUT 0201
« Reply #2 on: November 30, 2018, 06:29:30 PM »
I think when (1,1), it is node or spiral point, not IN(improper node).

Jingze Wang

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Re: Q7 TUT 0201
« Reply #3 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

Yulin WANG

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Re: Q7 TUT 0201
« Reply #4 on: November 30, 2018, 11:13:57 PM »
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
Thanks for submitting the computer-generated phase portrait!!!
BTW, how do u plot the phase portrait on a computer?

Victor Ivrii

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Re: Q7 TUT 0201
« Reply #5 on: December 01, 2018, 03:57:59 AM »
Sure, it is improper node. Jingze finally got a decent computer generated picture (took a correct range of variables)

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