The SIR model tracks the susceptible fraction S and the infected fraction I. The ratio R₀ = β·S₀/γ says whether an outbreak grows at first, the infected peak when S falls to γ/β, and every point with I = 0 is an equilibrium, so the equilibria are not isolated. The epidemic ends because the infected run out, not the susceptibles.
Core ideas
Three ideas. The model has two equations, the threshold is one number from them, and the equilibria are a line, not points.
Let S be the fraction who can still be infected and I the fraction who are infected now. Meetings between the two happen at a rate proportional to S·I; each moves someone from S to I at strength β, and the infected recover at rate γ. The recovered need no equation of their own, since nothing depends on them.
Write I′ = I·(β·S − γ). The infected grow exactly when β·S > γ. At the start that is R₀ = β·S₀/γ > 1, and I reaches its peak when S has fallen to γ/β, because S only ever decreases.
Setting I = 0 makes both rates zero for any S. So every point of the S-axis is an equilibrium, and the Jacobian there has zero determinant. The lab reports non-isolated equilibria, and the Jacobian's remaining eigenvalue says whether the point is approached or left.
The material
The model, the threshold worked out exactly, why the equilibria are a line, and the reason the epidemic ends.
With S susceptible and I infected fractions, S′ = −β·S·I and I′ = β·S·I − γ·I, with β the transmission rate and γ the recovery rate.
The assumptions are named: a fixed population, everyone mixing equally, immunity after recovery, no births and deaths. The basic reproduction number at the start is R₀ = β·S₀/γ.
Take β = 1/2, γ = 1/4, and start at S₀ = 9/10, I₀ = 1/10. Then R₀ = (1/2)·(9/10)/(1/4) = 9/5, greater than 1. The factor β·S − γ in I′ = I·(β·S − γ) is positive while S > γ/β = 1/2, so I rises. It is zero at S = 1/2, and negative after, so I peaks exactly when S reaches 1/2. The lab prints this as R₀ = 9/5; I peaks at S = 1/2.
Raise the recovery rate to γ = 1 with the same β and start. Then R₀ = (1/2)·(9/10)/1 = 9/20 < 1, and the peak would be at S = γ/β = 2, which S never reaches. The factor β·S − γ is negative from the start, so I falls at once. The lab prints that I falls from the start.
The same disease with a faster recovery never takes off, which is what the single number R₀ is for.
Every point (S, 0) is an equilibrium, because both rates have a factor I. At (9/10, 0) the Jacobian has rows 0 −9/20 and 0 1/5, so the determinant is zero, the trace is 1/5, and the roots are 0 and 1/5. The positive root is the outbreak: a few infected at S = 9/10 start to grow. The lab's type tile reads non-isolated equilibria, which here means a line, not a point. Switch the grid search on and the status line reports found 65, every one of them on the line I = 0: that is the grid's own count, and the true number is infinite.
Along every solution, S + I − (γ/β)·ln S is constant.
Call the quantity W = S + I − (γ/β)·ln S and differentiate it along a solution. The term ln S changes at the rate S′/S, which is −β·I, by the rate of ln and the chain rule exactly as “Predator and Prey” used them.
W′ = S′ + I′ − (γ/β)·S′/S = −β·S·I + (β·S·I − γ·I) + γ·I.
The terms cancel in pairs, so the quantity is constant. If S tended to 0, then −(γ/β)·ln S would grow without limit while S and I stay between 0 and 1, and the constant could not hold. So S stays above zero.
□
That is the correct reason an epidemic ends. The infected decline once S has fallen below γ/β, because each infected person then passes the disease to fewer than one other before recovering, and I runs down to zero. A positive fraction of the population is never infected. The drawn trajectory in the lab shows S levelling off above zero; it is floating point, and the argument is the conserved quantity.
Interactive
Checked exactly, linearised, and classified
Method
Read the coefficients, compute the threshold, and place the peak.
From S′ = −β·S·I read β as the size of the S·I coefficient; from I′ = β·S·I − γ·I read γ as the size of the coefficient of I on its own. Here β = 1/2 and γ = 1/4.
R₀ = β·S₀/γ. With S₀ = 9/10 it is 9/5. Compare it with 1.
I′ = 0 where S = γ/β. If S₀ is above it, I rises first and peaks there. If S₀ is below it, I falls from the start.
I = 0 for any S. The Jacobian has determinant zero, so the lab prints non-isolated equilibria, and its positive root says an outbreak can start.
I declines after the peak because S is below γ/β, and it never needs S to reach zero. The quantity S + I − (γ/β)·ln S is constant, which keeps S positive.
The first preset. Compute the threshold and the peak by hand, then read them from the tile.
The exact tile reads R₀ = 9/5; I peaks at S = 1/2, as computed. When S is at 1/2, half the population has not been infected, and the infection falls from there.
So the outbreak ends because the infected fraction has run down while many are still susceptible. It did not end because everyone had been through it.
You should be able to read β and γ off the equations, compute R₀ for a start, place the infected peak at S = γ/β, explain why the equilibria form a line, and give the conserved-quantity reason that susceptibles are left at the end.
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