- Thread starter
- #1

- Thread starter mt91
- Start date

- Thread starter
- #1

- Jan 30, 2018

- 426

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".

- Thread starter
- #3

I got 3 steady states, as u=0, u=1 and u=-a?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".

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?

- Jan 30, 2018

- 426

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.