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Why does the epidemic start growing when (N-1)β/γβ/γ is greater than 1?

The epidemic starts growing when (N−1)β/γ>1(N-1)\beta/\gamma > 1 because this condition ensures that the initial rate of new infections exceeds the rate of removals. With one initial infective (I=1I=1) and S=N−1S=N-1, the infected equation dIdt=βSI−γI\frac{dI}{dt}=\beta SI-\gamma I becomes positive only if the infection-production term β(N−1)\beta(N-1) is greater than the removal term γ\gamma.

Conditions

  • Applies at the start of the outbreak.
  • Uses the single-infective initialization I(0)=1I(0)=1.
  • Assumes the SIR dynamics shown on the board.
  • γ>0\gamma>0 is assumed implicitly by division by γ\gamma.

Reasoning, step by step

  1. Start from the infected-compartment equation: dIdt=βSI−γI\frac{dI}{dt}=\beta SI-\gamma I.
  2. Substitute the initial condition for a single first infective: S=N−1S=N-1 and I=1I=1.
  3. Evaluate the growth rate at the initial state: dIdt∣t=0=β(N−1)−γ\left.\frac{dI}{dt}\right|_{t=0}=\beta(N-1)-\gamma.
  4. Require the initial infected population to increase: β(N−1)−γ>0\beta(N-1)-\gamma>0.
  5. Rearrange to isolate the dimensionless threshold ratio: (N−1)βγ>1\frac{(N-1)\beta}{\gamma}>1.

Example

The board states 'Epidemic occurs if: dI/dt>0' and writes 'R0=(N−1)β/γ>1R_0=(N-1)\beta/\gamma>1'. The lecturer says that when the factor is greater than one, the epidemic starts growing because more infected people are being produced than removed.

Common misconceptions

  • Thinking the threshold is (N−1)βγ>0\frac{(N-1)\beta}{\gamma}>0 instead of >1>1.
  • Assuming the condition applies to the entire duration of the epidemic rather than just the initial onset.

Watch the explanation

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