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We have the following system in $\mathbb{R}^{2}$

$$\dot{y}_1=2-y_1y_2-y_2^2$$

$$\dot{y}_2=2-y_1^2-y_1y_2$$

i) Calculate the equilibrium points en determine their stability.

ii) Draw the Phase Plot.

I know that I have to use the following to calculate the equilibrium points:

$$\dot{y}_1=0 , \dot{y}_2=0$$

That's all I know. So my question is how do I solve these equations, how do I determine the stability of the equilibrium points and how do I draw the phase plot of this system.

An other question I have is what kind of solution you will get, I'm not able to understand what this system represents. So I would like to know how I should interpret such a system like this. Are we looking for something like $f(y_1,y_2)$ or do you need to find $y_1$ and $y_2$. And if so what would the solution mean?

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Critical Points:

$$2 - y_1 y_2 - y_2^2 = 0, 2 - y_1^2 - y_1 y_2 = 0 \implies (y_1, y_2) = (-1,-1),(1,1)$$

Jacobian:

$$J(y_1,y_2) = \begin{bmatrix} -y_2 & -2 y_1-y_2 \\ -y_1-2 y_2 & -y_1 \end{bmatrix}$$

Jacobian's eigenvalues at each critical point:

$$J(-1,-1) = \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}$$

The eigenvalues are $\lambda_1 = -2, \lambda_2 = 4$, which is a saddle point.

$$J(1,1) = \begin{bmatrix} -1 & -3 \\ -3 & -1 \end{bmatrix}$$

The eigenvalues are $\lambda_1 = -4, \lambda_2 = 2$, which is a saddle point..

Phase portrait (you should learn to roughly draw by hand):

enter image description here

In an ideal world, we would like to find a closed-form solution for $y_1(t)$ and $y_2(t)$. For most nonlinear systems, this is not possible, so we resort to qualitative approaches like the above.

When you cannot find closed-form solutions, you can also resort to numerical methods to solve for $y_1(t)$ and $y_2(t)$. You can plot each of those solutions as functions of $t$ or parametrically plot $y_1(t) vs. y_2(t)$. You should try these both with a system you know the answers to (including drawing the phase portrait).

Now, what conclusions can be drawn from all of this?

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  • $\begingroup$ Beat me to it! Not surprising, since I am trying to answer four questions at once plus respond to email concurrently this fine spring morning here in Oakland, California. Was going to ask you for the phase portrait anyway! Glad you're still around, +1! $\endgroup$ – Robert Lewis Apr 7 '14 at 20:48
  • $\begingroup$ Hey! "Frisco" is what the Hell's Angels call it! Plan to keep churning . . . $\endgroup$ – Robert Lewis Apr 7 '14 at 20:52
  • $\begingroup$ To put it in N. Oaktown cellphone speak, where you at? "Frisco" has traditionally been seen by many of the snobby locals, whose numbers ever grow with the Tech boom I might add, as low-class, uncouth, and sort of derogatory. Which why the Red and White put in on their colors! People like me use it as well . . . ;-)! $\endgroup$ – Robert Lewis Apr 7 '14 at 21:01
  • $\begingroup$ Steak and potatoes, that explains your earlier reference to fat fingers! Me too, steak and potatoes. Plus years of muscling out chords/notes on steel-stringed guitars! By the way, UC Berkeley is not my Alma Mater, see my profile for details. Bezerkeley is still fun, sort of . . . :( ;-) $\endgroup$ – Robert Lewis Apr 7 '14 at 21:08
  • $\begingroup$ You didn't know "Bezerkeley"? What a laugh! $\endgroup$ – Robert Lewis Apr 7 '14 at 21:10
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\left\lbrace\begin{array}{rcl} \dot{y}_{1} & = & 2 - y_{1}y_{2} - y_{2}^{2} \\ \dot{y}_{2} & = & 2-y_{1}^{2} - y_{1}y_{2} \end{array}\right.}$

$$ \dot{y}_{1} + \dot{y}_{2} = 4 - \pars{y_{1} + y_{2}}^{2} \quad\imp\quad \color{#00f}{\large y_{1} + y_{2} = 2\,{\expo{4t} - A \over \expo{4t} + A}} \,,\quad A\ \mbox{is a constant} $$

$$ \dot{y}_{1} - \dot{y}_{2} = \pars{y_{1} - y_{2}}\pars{y_{1} + y_{2}} =\pars{y_{1} - y_{2}}\pars{2\,{\expo{4t} - A \over \expo{4t} + A}} $$

$$ \ln\pars{y_{1} - y_{2}} = 2\bracks{-t + \half\,\ln\pars{A + \expo{4t}}} + \mbox{a constant} $$

$$ \color{#00f}{\large y_{1} - y_{2} = B\expo{-2t}\pars{A + \expo{4t}}}\,,\quad B\ \mbox{is a constant} $$

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  • $\begingroup$ Closed form solution! Nice! +1! $\endgroup$ – Robert Lewis Apr 7 '14 at 20:50

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