Show that the following system of ordinary differential equationsdxdt=0.5x+2.5y−x(x2+y2),dydt=−0.5x+1.5y−y(x2+y2).

Step 1The system has the form v→˙=Av→−r2v→,v→=(xy)r=|v→|A=12(15−13)The stationary points of it are v→=0 and the vectors of eigenpairs (|v→|2, v→), if the matrix had real eigenvalues.Claim: There are no real eigenvalues and the origin is an outward spiral.If you set u=r2=x2+y2 you getu˙=v→TBv→−2u2withB=A+AT=(1223)so thatλminu−2u2≤u˙≤λmaxu−2u2where the λmin,max are the eigenvalues of B. The left side is not helpful, however, the right side is negative at u=λmaxClaim: This makes 0&lt;|v→|&lt;λmax a trapping region with no equilibrium points.

Key to "Show that the following system of ordinary differential..."

komanijuuxw

Answered question

2022-04-25

Show that the following system of ordinary differential equations

\frac{d x}{d t} = 0.5 x + 2.5 y - x (x^{2} + y^{2}),

\frac{d y}{d t} = - 0.5 x + 1.5 y - y (x^{2} + y^{2}) .

Answer & Explanation

Davon Friedman

Beginner2022-04-26Added 13 answers

Step 1
The system has the form

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

\vec{v} = (\begin{matrix} x \\ y \end{matrix})

r = | \vec{v} |

A = \frac{1}{2} (\begin{matrix} 1 & 5 \\ - 1 & 3 \end{matrix})

The stationary points of it are

\vec{v} = 0

and the vectors of eigenpairs

({| \vec{v} |}^{2}, \vec{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} = {\vec{v}}^{T} B \vec{v} - 2 u^{2}

with

B = A + A^{T}

= (\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix})

so that

λ_{min} u - 2 u^{2} \leq \dot{u} \leq λ_{max} u - 2 u^{2}

where the

λ_{min, max}

are the eigenvalues of B. The left side is not helpful, however, the right side is negative at

u = λ_{max}

Claim: This makes

0 < | \vec{v} | < \sqrt{λ_{max}}

a trapping region with no equilibrium points.

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