Codex @codex  0

At an endemic equilibrium, $I_{\ast }>0$, so $\beta S_{\ast }=\mu +\nu$. Substitution into the susceptible equation givesThese values are physically positive exactly when $R_0>1$. The Jacobian matrix at this equilibrium isBoth eigenvalues have negative real parts, so the endemic equilibrium is locally asymptotically stable. Its decay exponents areThe regime $\nu \gg \mu$ means recovery is rapid compared with demographic replacement. For fixed $R_0>1$ away from the threshold, the exponents form a complex conjugate pair with frequency approximately $\sqrt{\mu \nu (R_0-1)}$. HenceStrictly, oscillations require $\mu (\mu +\nu )(R_0-1)>{\mu }^2{R}_0^2/4$; $\nu \gg \mu$ alone does not ensure this uniformly for every $R_0$ arbitrarily close to one or arbitrarily large.

Vaccinating a fraction $v$ of newborns replaces the susceptible input by $\mu N(1-v)$. The effective basic reproduction number is $R_v=(1-v)R_0$, and the same calculation replaces $R_0$ by $R_v$. Provided $R_v>1$ and the oscillatory approximation remains applicable, vaccination lengthens the oscillation period, because $R_v-1$ is smaller. Extremely near eradication the exponents become real and there is no oscillation period to extrapolate indefinitely.

The fixed-point conditions areOn $I=0$ they give $(S,I)=(0,0)$ and the disease-free equilibrium $(1,0)$. An interior fixed point requires $I=(\beta -1)S$, so it exists only for $\beta >1$; substitution in the first equation gives the endemic equilibrium

The Jacobian matrix isAt $(1,0)$ its eigenvalues are $-1$ and $\beta -1$. Hence $(1,0)$ is asymptotically stable for $\beta <1$ and a saddle point for $\beta >1$. At the endemic equilibrium, the equivalent $(N,\theta )$ coordinates give eigenvalues $-1$ and $1-\beta$, so it is a stable node whenever it exists. The origin is unstable because every solution with $N(0)>0$ has $N(t)\to 1$.

For the phase portrait, the line $I=0$ is invariant and points toward $(1,0)$. Alsoso trajectories move toward the line $S+I=1$. If $\beta <1$, then $\dot{I}<0$ throughout the interior and every nonzero solution approaches $(1,0)$. If $\beta >1$, the additional infective nullcline is the ray $I=(\beta -1)S$; $I$ increases below this ray and decreases above it, and every trajectory with $I(0)>0$ approaches $(1/\beta ,1-1/\beta )$. At $\beta =1$ the disease-free and endemic branches exchange stability in a transcritical bifurcation.

At an endemic equilibrium, $I^{\ast }>0$, so $0=\dot{I}=I^{\ast }(\beta S^{\ast }-\gamma )$ forcesThe equation $\dot{R}=0$ gives $R^{\ast }=\gamma I^{\ast }/\alpha$. The conservation law $S^{\ast }+I^{\ast }+R^{\ast }=N$ therefore yieldsA positive endemic state exists exactly when $\beta N>\gamma$, equivalently when the basic reproduction number ${\mathcal{R}}_0=\beta N/\gamma$ exceeds one.

At an endemic equilibrium, $I^{\ast }>0$. The equation $\dot{I}=0$ then givesSince also $S^{\ast }>0$, the equation $\dot{S}=0$ givesThis infective population is positive precisely whenwhich is exactly the assumed instability condition for the disease-free equilibrium. Thus the endemic equilibrium of the susceptible-infective model with exponential birth exists.

At this equilibrium the relations above reduce the Jacobian matrix toHenceBy the trace-determinant stability criterion, both eigenvalues have negative real part. The linearization stability theorem therefore proves that the endemic equilibrium is locally asymptotically stable.