How to prove the asymptotic stability of the trivial equilibrium of this system? I was trying to pr

Audrina Jackson

Audrina Jackson

Answered question

2022-07-12

How to prove the asymptotic stability of the trivial equilibrium of this system?
I was trying to prove the asymptotic stability of the trivial equilibrium ( 0 , 0 ) of the two-dimensional non linear ODE system:
d H d t = μ ( H + F ) 2 K 2 + ( H + F ) 2 d 1 H H ( σ 1 σ 2 F H + F ) d F d t = H ( σ 1 σ 2 F H + F ) ( p + d 2 ) F
where H and F are dependent variable and positive.
All other parameters are non negative with σ 2 > σ 1 .

Answer & Explanation

Jayvion Tyler

Jayvion Tyler

Beginner2022-07-13Added 23 answers

You need K to be strictly positive. Because if K = 0 and μ > 0, then in a neighborhood of the origin, d H / d t μ > 0. So, suppose K > 0.
In the first equation, the terms μ ( H + F ) 2 K 2 + ( H + F ) 2 and H σ 2 F H + F are of second order of smallness near the origin. Thus, the sign of d H / d t is determined by ( d 1 + σ 1 ) H which of course suggests stability.
In the second equation, H σ 2 F H + F is of second order. Neglecting it, we are left with H σ 1 ( p + d 2 ) F. This looks troublesome, but if H goes to zero, F will be forced to follow.
Let's summarize. For every ϵ > 0 there is a neighborhood of the origin in which
d H d t < ( d 1 + σ 1 ) H + ϵ ( H + F ) d F d t < H σ 1 ( p + d 2 ) F + ϵ ( H + F )
(I work in the positive quadrant H , F > 0, which is what you are interested in). Hence,
d ( 2 H + F ) d t < ( 2 d 1 + σ 1 ) H ( p + d 2 ) F + 3 ϵ ( H + F )
which is negative, provided 2 d 1 + σ 1 > 0, p + d 2 > 0, and ϵ is chosen sufficiently small.
I took 2 H + F instead of H + F so that the coefficient of H on the right would have σ 1 , increasing the chance of the coefficient being negative.

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