MHB What Determines the Local Extrema in Dynamical Systems?

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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.
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Any help would be appreciated
 
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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".
 
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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.
 
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