MHB What Determines the Local Extrema in Dynamical Systems?

  • Thread starter Thread starter mt91
  • Start date Start date
Click For Summary
The discussion focuses on determining local extrema in dynamical systems, particularly through the analysis of a separable differential equation. Key points include identifying steady states at u = -a, 0, and 1, with stability assessments indicating that u = -a and u = 1 are local maxima, while u = 0 is a local minimum. The parameter 'a' is crucial, with different cases (a < 1, a = 1, a > 1) affecting the stability and behavior of the system. The integration of the equation using partial fractions is also highlighted as a method for analysis. Overall, the conversation emphasizes the importance of understanding stability and equilibrium points in dynamical systems.
mt91
Messages
13
Reaction score
0
1596153990630.png


Any help would be appreciated
 
Mathematics news on Phys.org
what do you need help with? Do you know what "steady state" means? Do you know what stability is? An interesting, non-mathematical, phrase here is "biological relevance". What do you think that means? There is the parameter, a, that is undefined. You might want to consider the cases a<1, a= 1, and a> 1 separately.

Do you know what a "separable differential equation" is? This equation is "separable"- it can be written $\frac{du}{u(1- u)(a+ u)}= dt$. The left side can be integrated using "partial fractions".
 
Last edited:
Country Boy said:
what do you need help with? Do you know what "steady state" means? Do you know what stability is? An interesting, non-mathematical, phrase here is "biological relevance". What do you think that means? There is the parameter, a, that is undefined. You might want to consider the cases a<1, a= 1, and a> 1 separately.

Do you know what a "separable differential equation" is? This equation is "separable"- it can be written $\frac{du}{u(1- u)(a+ u)}= dt$. The right side can be integrated using "partial fractions".
I got 3 steady states, as u=0, u=1 and u=-a?

and then by plotting u(1-u)(a+u), indicated from the graph that u=-a is stable, u=0 is unstable and u=1 is stable. Not sure what you mean by looking at the cases for a, so are you ok to explain that?
 
That was because I had misread the problem and was looking at u= -1, 0, and -a!

Yes, the equilibrium points are at u= -a, 0, and 1. We can write u'= (u+ a)u(1-u)= (-1)(u+ a)u(u- 1). if u< -a then all four of those are negative so the product, and so u', is positive. u is increasing up to u(-a). If -a< u< 0 then -1, u, and u-1 are negative but u+ a is positive so the product, and so u', is negative. u goes down from u(-a) to u(0). If 0< u< 1, both u+ a and u are positive while -1 and u are negative so the product, and so u', is positive. u goes up to u(1). Finally, for u> 1, all except -1 are negative so u' is negative. u goes down from u(1).

That is, u goes up to u(-a) then down after that so u(-a) is a local maximum. u goes down to u(0) then up so u(0) is a local minimum. u goes up to u(1) then down so u(1) is a local maximum.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K