komanijuuxw

2022-04-25

Show that the following system of ordinary differential equations

$\frac{dx}{dt}=0.5x+2.5y-x({x}^{2}+{y}^{2}),$

$\frac{dy}{dt}=-0.5x+1.5y-y({x}^{2}+{y}^{2}).$

Davon Friedman

Beginner2022-04-26Added 13 answers

Step 1

The system has the form

$\dot{\overrightarrow{v}}=A\overrightarrow{v}-{r}^{2}\overrightarrow{v},$

$\overrightarrow{v}=\left(\begin{array}{c}x\\ y\end{array}\right)$

$r=\left|\overrightarrow{v}\right|$

$A=\frac{1}{2}\left(\begin{array}{cc}1& 5\\ -1& 3\end{array}\right)$

The stationary points of it are$\overrightarrow{v}=0$ and the vectors of eigenpairs $({\left|\overrightarrow{v}\right|}^{2},\text{}\overrightarrow{v})$ , if the matrix had real eigenvalues.

Claim: There are no real eigenvalues and the origin is an outward spiral.

If you set$u={r}^{2}={x}^{2}+{y}^{2}$ you get

$\dot{u}={\overrightarrow{v}}^{T}B\overrightarrow{v}-2{u}^{2}$

with

$B=A+{A}^{T}$

$=\left(\begin{array}{cc}1& 2\\ 2& 3\end{array}\right)$

so that

$\lambda}_{min}u-2{u}^{2}\le \dot{u}\le {\lambda}_{max}u-2{u}^{2$

where the$\lambda}_{min,max$ are the eigenvalues of B. The left side is not helpful, however, the right side is negative at $u={\lambda}_{max}$

Claim: This makes$0<\left|\overrightarrow{v}\right|<\sqrt{{\lambda}_{max}}$ a trapping region with no equilibrium points.

The system has the form

The stationary points of it are

Claim: There are no real eigenvalues and the origin is an outward spiral.

If you set

with

so that

where the

Claim: This makes

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