A system of nonlinear differential equations

Question

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?

Answer 1

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):

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?

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\left\lbrace\begin{array}{rcl} \dot{y}_{1} & = & 2 - y_{1}y_{2} - y_{2}^{2} \\ \dot{y}_{2} & = & 2-y_{1}^{2} - y_{1}y_{2} \end{array}\right.}$

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

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

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

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