Skip to content
START WITH A QUESTION

What would you like to understand?

Find an answer. See the moment it becomes clear. Follow the idea further.

← Concept directory

Answers for “为什么当 (N-1)β/γ 大于 1 时流行病开始增长?”

3 keyword matches

Understanding your question. You can explore the search results below now.

Understand why

↗

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.

Meet the concept

↗

In the discrete one-infected example, the initial values are S=N−1S=N-1, I=1I=1, and R=0R=0. This corresponds to a small population where exactly one person is infected, everyone else is susceptible, and nobody has yet recovered or been removed.

Conditions: Total population scale is N.; There is one infected person.; Everyone else is susceptible at the beginning.; No one has yet recovered or been removed in this initial example.

Meet the concept

↗

The lecture decides that an epidemic occurs if the time derivative of the infected compartment is positive, i.e., dIdt>0\frac{dI}{dt}>0. This means the number of infected people is growing in the early stages of the outbreak, when the susceptible population is close to the total population and the infected population is small.

Conditions: The model is the displayed SIR system.; The relevant comparison is made in the early stages of the outbreak, where S is close to N and I is small compared with N.