How to find the equilibria?

Question

(dp/dt) = p(1 − p)[α(1 − p) − βp]

where 0 < α < 1  and 0 < β < 1.

Find all the equilibria and state which are stable and unstable.

Answer 1

dp/dt = p(1 - p)[α(1 - p) - βp] = αp(1 - p)² – βp²(1 - p) … (i)

For general equilibrium,

d/dt(dp/dt) = d²p/dt² = p*(1 - p)*d/dt{[α(1 - p) - βp]} + p[α(1 - p) - βp]*d/dt(1 - p) + (1 - p)[α(1 – p) -   βp]*(dp/dt)

Substitute for dp/dt from equation (i),

d²p/dt² = p(1 - p)[α(-1) - β] + p[α(1 - p) - βp]*(-1) + (1 - p)[α(1 - p) - βp]*p(1 - p)[α(1 - p) - βp]

d²p/dt² = p(1 - p)(-α - β) - p[α(1 - p) - βp] + p(1 - p)²[α(1 - p) - βp]² … (ii)

but, for equilibrium condition, p’(t) = dp/dt = 0, then,

p(1 - p)[α(1 - p) - βp] = 0

p = 0; 1 – p = 0; α(1 - p) - βp = 0

                              α(1 - p) = βp

                              α – αp = βp

                              α = (α + β)*p

p = 0; p = 1; p = α/(α + β). These are all the conditions for equilibria!

Case 1:

Substitute p = 0 into the second rate of change, equation (ii),

d²p/dt² = (0)(1 - 0)(-α – β) – (0)[α(1 - 0) - β(0)] + (0)(1 – 0)²[α(1 - 0) - β(0)]²

d²p/dt² = 0 (Neutral equilibrium/stability)

Case 2:

Substitute p = 1 into second derivative, equation (ii),

d²p/dt² = (1)(1 – 1)(-α – β) – (1) [α(1 - 1) - β(1)] + (1)(1 – 1)²[α(1 - 1) - β(1)]²

               = (1)(0)(-α – β) – [α(0) – β] + (1)(0)²[α(0) – β]²

d²p/dt² = β > 0 (stable equilibrium)



Case 3:

Substitute p = α/(α + β) into second derivative, equation (ii),

d²p/dt² = [α/(α + β)]*[1 – α/(α + β)]*(-α – β) – [α/(α + β)]*{α[1 – α/(α + β)] – β[1 – α/(α + β)]} + [α/(α + β)]*[1 – α/(α + β)]²*{α[1 – α/(α + β)] – β[1 – α/(α + β)]}²

= [α/(α + β)]*[β/(α + β)]*[-(α + β)] - [α/(α + β)]*[αβ/(α + β) - αβ/(α + β)] + [α/(α + β)]* [β/(α + β)]²*[αβ/(α + β) - αβ/(α + β)]

d²p/dt² = -αβ/(α + β) < 0 (unstable equilibrium)

Answer 2

p is at equilibrium if p doesn't grow or decline as time goes on

so if dp/dt=0

Obviously dp/dt=0 if p=0 or if p=1. These are two equilibrium points.

Solve alpha(1-p)-beta p =0 for p. Notice 3 versions:

alpha(1-p)-beta p= -p(alpha+beta)+alpha = (alpha+beta)( -p+ alpha/(alpha+beta) )

So alpha(1-p)-beta p =0 if alpha=p(alpha+beta).

Third equilibrium point: p=alpha/(alpha + beta).

Notice the approximate value of the right side: 0<alpha/(alpha + beta)<1

Now you have all equilibrium points.

Now comes the discussion of stable, unstable: You need to be able to make a sign chart of dp/dt:

Arrange the 3 equilibriums in order: 0< alpha/(alpha+beta) < 1

SInce all three factors p, (1-p), and alpha(1-p)-beta p are linear in p, the sign changes when you cross an equilibrium. Start from the bottom up:

If p<0 then dp/dt<0, (since (1-p)>0 and (alpha(1-p)-beta p)>0. )----> p moves away from equilibrium 0 to - infinity

If 0<p< alpha/(alpha+beta) then dp/dt>0------> p grows towards the equilibrium alpha/(alpha+beta)

If alpha/(alpha+beta)<p<1 then dp/dt<0-------> p declines towards the equilibrium alpha/(alpha+beta)

if p>1 then dp/dt>0 --------> p grows forever to infinity.

Conclusion p=0 and p=1 unstable

p=alpha/(alpha+beta) is stable