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

Audrina Jackson 2022-07-12 Answered
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 .
You can still ask an expert for help

Want to know more about Inequalities systems and graphs?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Jayvion Tyler
Answered 2022-07-13 Author has 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.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-06-10
Find the smallest possible value of ω such that the system of inequalities
( 1009 1010 2018 ) x + 2018 2018 ω ( 1008 1009 2018 ) x + 2018 2016 ω ( 1007 1008 2018 ) x + 2018 2014 ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 2 2018 ) x + 2018 2 ω
has a real solution in x.
Subtracting the first inequality from the second inequality, we get that x = 1009 2017 . This is also correct when we subtract the second inequality from the third one. When I plugged in x, I got an unruly fractional value for ω. Can someone guide me through the solution? Help is much appreciated.
asked 2022-05-23
What is the domain of 1 / x 2 4 using the set builder notation?
I'm confused when it comes to getting the domain of a function using set builder notation. Please explain also how to do the system of inequalities when getting the domain of the given function
x 1 x 2 4
asked 2022-07-09
Use set notation to describe the set of values of x for which:
x 2 7 x + 10 < 0 and 3 x + 5 < 17
asked 2022-07-15
Consider the following linear system
y = A 1 x 1 + A 2 x 2
subject to the linear constrains
C 1 x 1 + C 2 x 2 d
I am looking for a solution for the above linear system that gives more priority to the coordinates corresponding to x 1 compared to those of x 2 . This is what I precisely mean by priority
If x 1 alone can geenerated the given y while x 2 is kept at its minimum feasible value.
If there are infinite number of solutions for x 1 in step 1 then we take the minimum norm solution.
If x 1 alone cannot generate y then we allow x 2 to participate in generating y with the optimal x 1 obtained in step 1 and 2.
If there are infinite solutions for x 2 , then we take the minimum norm solution.
How to formuate the above described problem as an optimization problem for instance QP?
asked 2022-06-21
Let ( A , m ) be a local Noetherian ring and let x 1 , , x d be a system of parameters, i.e. m = ( x 1 , , x d ). Then
dim A / ( x 1 , , x i ) = d i
i = 1 , , d
I know just a few basic facts about dimension theory. I think I can prove the inequality via Krull's Hauptidealsatz in this way: the maximal ideal of A / ( x 1 , , x i ) is
m i := ( x i + 1 ¯ , , x d ¯ ) .
So it must be h t ( m i ) d i.
But how to prove the other inequality? I think I should do it by induction, but I cannot understand how to begin. So, if what I said so far is right, my question is: how can I prove that
dim A / ( x 1 ) d 1 ?
asked 2022-06-24
Given a system of n linear equations. Prove that the system is inconsistent if and only if you can obtain 0 = 1, by using linear combinations.
I do not want to apply theorems from linear algebra here. Instead of it I want to use Farkas' lemma.
Every equation can be rewrited in form of two inequalities with and , now I got 2 n inequalities instead of n equations.
I think that's the moment for me to apply Farkas' lemma, but why I can always get 0 1 and 0 1?
asked 2022-06-18
Solve the system of linear inequalities with parameters
(*) { 0 2 x + 2 y 3 b + 3 a 2 0 2 x 3 y + 6 b + 3 a 1 0 x 1 0 y 2 0 a 1 0 b 1
Here x,y are unknown variables and a,b are parameters.
My attempt. By adding the inequalities with some coeficients I separated the variables and get the simple system
(**) { 0 y + 6 a 5 , 0 x + 9 a + 3 b 8.
and I am able to solve it. But the solutions of the last system are not solution of the initial system!
Maple and wolframAlpha cant solve the system.
P.S.1 For a = 63 100 and b = 59 100 Maple gives the solutions
{ x = 1 , 9 50 y , y 11 25 } , { x = 3 / 2 y + 127 100 , 9 50 < y , y < 11 25 } , { 9 50 < y , x < 1 , y < 11 25 , 3 / 2 y + 127 100 < x } , { y = 11 25 , 61 100 x , x < 1 } , { x = 3 / 2 y + 127 100 , 11 25 < y , y < 127 150 } , { 11 25 < y , x < 2 y + 47 25 , y < 127 150 , 3 / 2 y + 127 100 < x } , { x = 2 y + 47 25 , 11 25 < y , y < 127 150 } , { x = 0 , 127 150 y , y 47 50 } , { y = 127 150 , x 14 75 , 0 < x } , { 0 < x , 127 150 < y , x < 2 y + 47 25 , y < 47 50 } , { x = 2 y + 47 25 , 127 150 < y , y < 47 50 }