Impact of Population Recruitment on the HIV Epidemics and the Effectiveness of HIV Prevention Interventions

Abstract

Mechanistic mathematical models are increasingly used to evaluate the effectiveness of different interventions for HIV prevention and to inform public health decisions. By focusing exclusively on the impact of the interventions, the importance of the demographic processes in these studies is often underestimated. In this paper, we use simple deterministic models to assess the effectiveness of pre-exposure prophylaxis in reducing the HIV transmission and to explore the influence of the recruitment mechanisms on the epidemic and effectiveness projections. We employ three commonly used formulas that correspond to constant, proportional and logistic recruitment and compare the dynamical properties of the resulting models. Our analysis exposes substantial differences in the transient and asymptotic behavior of the models which result in 47 % variation in population size and more than 6 percentage points variation in HIV prevalence over 40 years between models using different recruitment mechanisms. We outline the strong influence of recruitment assumptions on the impact of HIV prevention interventions and conclude that detailed demographic data should be used to inform the integration of recruitment processes in the models before HIV prevention is considered.

Appendices

Appendix 1: Proofs of Main Results

1.1 Proof of Proposition 1

Positivity and boundedness of the solutions can be easily proved. Then, periodic solutions can be excluded by Dulac’s criteria. Let $P(S,I):=\frac{dS}{dt}$ and $Q(S,I):=\frac{dI}{dt}$, then $\frac{\partial }{\partial S}\frac{P}{SI}+\frac{\partial }{\partial I}\frac{Q}{SI}=-\frac{\varLambda }{S^2I}<0$.

Model (1) with constant recruitment $f_C(N)$ has two possible steady states $E_{df}=\left( \frac{1}{\mu }\varLambda ,0\right)$ and $E^*=\left( \frac{1}{\beta -d}\varLambda ,\frac{1}{\beta -d}(R_0-1)\varLambda \right)$, where $R_0=\frac{\beta }{\mu +d}$. Notice that $E^*$ (positive steady state) exists if and only if $R_0>1$.

The eigenvalues of the Jacobian evaluated at $E_{df}$ are $\lambda _1=-\mu <0$ and $\lambda _2=(\mu +d)(R_0-1)$ which implies that $E_{df}$ is locally stable when $R_0<1$ and unstable when $R_0>1$.

For the eigenvalues of the Jacobian evaluated at $E^*$, it is true that $\lambda _1+\lambda _2=(\mu +d)\left( 1-R_0-\frac{\mu }{\mu +d}\right) <0$ and $\lambda _1\times \lambda _2=(\mu +d)[\beta (R_0-1)^2+\mu R_0(R_0-1)]>0$ which implies that $E^*$ is locally stable when exists.

Finally by the Poincaré–Bendixson theorem, we have the following results:

• when $R_0<1$, $E_{df}$ is globally stable and $E^*$ does not exist;

• when $R_0>1$, $E^*$ is globally stable and $E_{df}$ is unstable.

1.2 Proof of Proposition 2

The positivity of the solutions of Model (1) with proportional recruitment $f_P(N)$ can be easily proved. The equation for the total population size $\frac{dN}{dt}=\frac{dS}{dt}+\frac{dI}{dt}=rN-\mu S-(\mu +d)I=(r-\mu )N-dI\le (r-\mu )N$ implies that $N(t)\rightarrow 0$ extinction if $r<\mu$. It is clear that the extinction occurs even in the absence of HIV ($I=0)$ when the first equation of Model (1) becomes $\frac{dS}{dt}=(r-\mu )S$. Next, we study the case when $r>\mu$.

We analyze the following fractional form of Model (1) with proportional recruitment $f_P(N)$:

$$\begin{aligned} \frac{ds}{dt}= & {} [r-(\beta -d)s](1-s)\\ \frac{dN}{dt}= & {} [r-\mu -d(1-s)]N\\ i= & {} 1-s. \end{aligned}$$

Notice that the first equation is independent of N which allow us to study s(t) directly in the biologically feasible region $\{0 \le s \le 1\}$. Since $s(0)\in (0,1)$, then $\lim \limits _{t\rightarrow \infty }s(t)=1$ (and $\lim \limits _{t\rightarrow \infty }i(t)=0$) if:

• $\beta <d$ or

• $\beta >d$ and $\frac{r}{\beta -d}\ge 1$.

Combined, these two cases imply that if $\beta <r+d$, then each solution of the fractional model approaches disease-free equilibrium. Under this condition, the population size grows unbounded. Alternatively, if $\beta >r+d$, all solutions of the fractional model approach the endemic equilibrium with $\lim \limits _{t\rightarrow \infty }s(t)=\frac{r}{\beta -d}$ and $\lim \limits _{t\rightarrow \infty }i(t)=1-\frac{r}{\beta -d}$. The equation for the population size N(t) implies that in this case, the population extinction under HIV pressure will be caused if $(\mu +d-r) \beta >(\mu +d)d$. Therefore, the population is endangered only if $r<\mu +d$ and $\beta <\frac{(\mu +d)d}{\mu +d-r}$.

1.3 Proof of Proposition 3

The region $\{S\ge 0,I \ge 0, \, S+I \le K \}$ is positive invariant under Model (1) with logistic recruitment which can be checked by determining the sign of the derivatives on the boundary. Given $S+I > K$ in $\{S\ge 0, I \ge 0 \}$, we have $\frac{dN}{dt}=rN\left( 1-\frac{N}{K}\right) -\mu N-dI \le -\mu N$, and thus, the total population will decrease below carrying capicity. Therefore, we consider only the region $\{S\ge 0,I \ge 0, \, S+I \le K \}$.

Periodic solutions in the region $\{S\ge 0,I \ge 0 ,\, S+I \le K\}$ can be excluded by Dulac’s criteria. Let $P(S,I):=\frac{dS}{dt}$ and $Q(S,I):=\frac{dI}{dt}$, then $\frac{\partial }{\partial S}\left( \frac{P}{SI}\right) +\frac{\partial }{\partial I}\left( \frac{Q}{SI}\right) =-\frac{r}{K}\left( \frac{1}{S}+\frac{1}{I}\right) -\frac{r}{S^2}\left( 1-\frac{S+I}{K}\right) <0.$

The model has three possible steady states:

• 1. Extinction equilibrium $E_{\hbox {ext}}=(0, 0)$ which always exist;

• 2. Disease-free equilibrium $E_{df}=\left( \frac{r-\mu }{r}K, 0\right)$ which exists for $r>\mu$ and

• 3. Endemic equilibrium $E^*=\frac{r-\mu -d\left( 1-\frac{1}{R_0}\right) }{r}K\left( \frac{1}{R_0},\frac{R_0-1}{R_0}\right)$, where $R_0=\frac{\beta }{\mu +d}$. It exist if $R_0>1 \iff \beta > \mu +d$ and $r>\mu +d\left( 1-\frac{1}{R_0}\right)$. The later is equivalent to $\beta (\mu +d-r)<(\mu +d)d$ which is true if:

  • $r>\mu +d$ or

  • $r<\mu +d$ and $\beta <\frac{(\mu +d)d}{\mu +d-r}$

Similar to Proposition 2, we can show that if $r<\mu$, then the population go extinct ($N(t)\rightarrow 0$) and all solutions approach the extinction equilibrium $E_{\hbox {ext}}$. Next, we assume that $r>\mu$.

Eigenvalues of the Jacobian matrix at $E_{df}$ are $-(r-\mu )<0$ and $\beta -(\mu +d)$. It implies that when exists $E_{df}$ is stable when $R_0<1 \iff \beta < \mu +d$ and unstable otherwise.

Eigenvalues of the Jacobian matrix at $E^*$ satisfy $\lambda _1+\lambda _2=-\left[ r-\mu -d\left( 1-\frac{1}{R_0}\right) \right] -\left( 1-\frac{1}{R_0}\right) (\beta -d)<0$ and $\lambda _1\cdot \lambda _2=\left[ r-\mu -d\left( 1-\frac{1}{R_0}\right) \right] [\beta -(\mu +d)]>0$. It implies that $E^*$ is stable when exists.

The proof will be completed when the local stability of the extinction equilibrium $E_{\hbox {ext}}$ is analyzed. We follow an approach similar to one often used in the analysis of ratio-dependent population models (Hews et al. 2010). To avoid singularity at (0, 0), it is studied through the modified model in terms of the fraction of susceptibles $s=\frac{S}{N}$ and total population size N (note that infected fraction satisfies $i=\frac{I}{N}=1-s$):

$$\begin{aligned} \displaystyle \frac{ds}{dt}= & {} \left[ r(1-\frac{N}{K})+(d-\beta )s\right] (1-s)\nonumber \\ \displaystyle \frac{dN}{dt}= & {} \left[ r(1-\frac{N}{K})-\mu -d(1-s)\right] N. \end{aligned}$$

(3)

It possesses two steady states $E_{1}=(1, 0)$, $E_{2}=\left( \frac{r}{\beta -d}, 0\right)$ corresponding to $E_{\hbox {ext}}$.

The Jacobian matrix at $E_{1}$ has an eigenvalue $r-\mu >0$ which implies that $E_{1}$ is unstable. Eigenvalues of the Jacobian matrix at $E_{2}$ are $\lambda _1=d+r-\beta$ and $\lambda _2=-\frac{\mu +d}{\beta -d}(\beta -d-R_0r)$. It implies that $E_{2}$ is stable if $\beta >d+r$ and $\beta >d+R_0r$. As a result, $E_{\hbox {ext}}$ is stable if $\beta >\max \{d+r,d+R_0r\}$ and unstable otherwise. Then,

• when $R_0<1$ ($\beta <\mu +d$), $E_{\hbox {ext}}$ is unstable since $\beta <d+r$. In this case, endemic ($E^*$) equilibrium does not exist, while the disease-free ($E_{df}$) steady state is stable;

• when $R_0>1$ ($\beta >\mu +d$), $E_{00}$ is stable if $\beta >d+R_0r$ ,i.e., when endemic ($E^*$) equilibrium does not exist. In this case, the disease-free ($E_{df}$) steady state is unstable;

Finally the global stability results in the Proposition follow by Poincaré–Bendixson theorem.

1.4 Proof of Proposition 4

Positivity and boundedness of solutions can be easily proved. Then, $\frac{dS^p}{dt}\le k\varLambda -\mu S^p$ implies $\limsup \limits _{t\rightarrow \infty }S^p\le \frac{k\varLambda }{\mu }$ and $\frac{dS}{dt}\le (1-k)\varLambda -\mu S$ implies $\limsup \limits _{t\rightarrow \infty }S\le \frac{(1-k)\varLambda }{\mu }$. Therefore, $\frac{dN}{dt}=\frac{dS^p}{dt}+\frac{dS}{dt}+\frac{dI}{dt}=\varLambda -\mu N-dI\ge \varLambda -(\mu +d)N$ implies $\liminf \limits _{t\rightarrow \infty }N\ge \frac{\varLambda }{\mu +d}$. Then, in the long term we have $\frac{S}{N}\le \frac{(1-k)(\mu +d)}{\mu }$, and $\frac{S^p}{N}\le \frac{k(\mu +d)}{\mu }$, which implies $\frac{dI}{dt}\le I\left[ \beta \frac{(1-k)(\mu +d)}{\mu }+(1-\alpha _s)\beta \frac{k(\mu +d)}{\mu }-(\mu +d)\right] =I\frac{(\mu +d)^2}{\mu }\left[ \beta \frac{1-k}{\mu +d}+(1-\alpha _s)\beta \frac{k}{\mu +d}-\frac{\mu }{\mu +d}\right] =I\frac{(\mu +d)^2}{\mu }\left( R_0-\frac{\mu }{\mu +d}\right)$. Now if $R_0<\frac{\mu }{\mu +d}$, then because of the positivity of the solution, we know $\lim \limits _{t\rightarrow \infty }I=0$. Then, combining this result with the equations in Model (2), implies that $\lim \limits _{t\rightarrow \infty }S^p=\frac{k\varLambda }{\mu }$ and $\lim \limits _{t\rightarrow \infty }S=\frac{(1-k)\varLambda }{\mu }$. Thus, global stability of the infection-free steady state $E_{df}=(\frac{k\varLambda }{\mu },\frac{(1-k)\varLambda }{\mu },0)$ under condition $R_0<\frac{\mu }{\mu +d}$ is proved. For local stability of $E_{0}$, we consider the corresponding eigenvalues $\lambda _1=-\mu <0$, $\lambda _2=-\mu <0$ and $\lambda _3=(\mu +d)(R_0-1)$. Therefore, $E_{df}$ is stable when $R_0<1$ and unstable when $R_0>1$.

Now we consider $E^*$. By setting $F(I)=0$ and dividing by $\varLambda ^2 (\mu +d)$, we obtain that $I^*$ is a root of:

$$\begin{aligned} \begin{array}{lll} \hat{F}(I) &{} \triangleq &{} (R_0-1)+\left( (d-(1-\alpha _s)\beta )-(\mu +d-(1-\alpha _s)\beta )\dfrac{\beta }{\mu +d}-d (R_0-1)\right) \\ &{} &{} \dfrac{I}{\varLambda }- (d-\beta )(d-(1-\alpha _s)\beta )\dfrac{I^2}{\varLambda ^2} \end{array} \end{aligned}$$

Substituting in iN where $i:=\frac{I}{N}$, and $N=\frac{\varLambda }{\mu +d i}$ into $\hat{F}(I)=0$ we notice that $I^*$ is also a root of:

$$\begin{aligned} \bar{F}(i N)\triangleq & {} \mu (\mu +d)(1-R_0)+\beta \left( \mu + d \alpha _s k-(\beta -\mu -d)(1-\alpha _s) \right) i\\&+\, (1-\alpha _s) \beta ^2 i^2 \end{aligned}$$

Now substituting back in for I and $N=\dfrac{\varLambda -d I}{\mu }$, we notice that $I^*$ is a root of:

$$\begin{aligned} \bar{F}(I)= & {} \mu (\mu +d)(1-R_0)+\beta \left( \mu + d \alpha _s k-(\beta -\mu -d)(1-\alpha _s) \right) \dfrac{I}{N}\\&+\, (1-\alpha _s) \beta ^2 \dfrac{I^2}{N^2} \end{aligned}$$

Now we have $\hat{F}(0)=(R_0-1)$ and $\hat{F}(\frac{\varLambda }{d})=-\dfrac{(1-\alpha _s)\beta ^2 \mu }{d^2 (\mu +d)}<0$.

Assume $R_0>1$, then $\hat{F}(0)>0$ and also $\beta >d$. Since $\hat{F}(0)>0$, $\hat{F}(\dfrac{\varLambda }{d})<0$ and $\hat{F}$ is a second-order polynomial, we have existence of a unique endemic equilibrium.

Assume $R_0<1$. Since

$$\begin{aligned} \begin{array}{lll} \mu +d \alpha _s k-(\beta -\mu -d)(1-\alpha _s) &{}\ge &{} \mu +d \alpha _s k - \alpha _s k(\mu +d)\left( \dfrac{1-\alpha _s}{1-\alpha _s k}\right) \\ &{} \ge &{} \alpha _s k(\mu +d)\left( 1-\dfrac{1-\alpha _s}{1-\alpha _s k}\right) >0, \end{array} \end{aligned}$$

we have that $\bar{F}(I)>0$ for all $I \in (0,\dfrac{\varLambda }{d})$, and therefore, we have no solution $I^*$.

Therefore, when $R_0<1$, there is only the disease-free equilibrium, $E_{df}$, and whenever $R_0>1$ there is both the disease-free equilibrium, $E_{df}$, and the endemic equilibrium, $E^*$.

Now, we analyze the stability of the endemic equilibrium, $E^*$. Because of the complexity of the expressions, we will not express the positive steady state explicitly. Now assume that $R_0=\frac{(1-\alpha _sk)\beta }{\mu +d}>1$.

The Jacobian of the system is:

$$\begin{aligned} J=\left( \begin{array}{ccc} -\dfrac{(I+S)(1-\alpha _s)}{N^2}\beta I-\mu &{} \dfrac{(1-\alpha _s)}{N^2}\beta S^p I &{} -\dfrac{(S^p+S)(1-\alpha _s)}{N^2} \beta S^p \\ \dfrac{1}{N^2}\beta S I &{} -\dfrac{S^p+I}{N^2}\beta I-\mu &{} -\dfrac{ S^p+S}{N^2}\beta S \\ \dfrac{I-(I+S)\alpha _s}{N^2}\beta I &{} \dfrac{I+S^p \alpha _s }{N^2}\beta I &{} \dfrac{(S^p+S)(S+S^p(1-\alpha _s))}{N^2}\beta - (\mu +d) \end{array}\right) . \end{aligned}$$

Using

$$\begin{aligned} P=\left( \begin{array}{ccc} 1 &{} -1 &{} -1 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 \end{array}\right) , \end{aligned}$$

we see that the Jacobian is similar to

$$\begin{aligned} H=P^{-1}JP=\left( \begin{array}{ccc} -\mu &{} 0 &{} -d \\ \dfrac{1}{N^2}\beta S I &{} -\dfrac{1}{N}\beta I-\mu &{} -\dfrac{ 1}{N}\beta S \\ \dfrac{I-(I+S)\alpha _s}{N^2}\beta I &{} \dfrac{\alpha _s}{N}\beta I &{} \dfrac{(S^p-I)(1-\alpha _s)+S}{N}\beta - (\mu +d) \end{array}\right) . \end{aligned}$$

Rewriting H evaluated at the endemic equilibrium using $p^*:=\frac{S^{p*}}{N^*}$, $s^*:=\frac{S^*}{N^*}$, $i^*:=\frac{I^*}{N^*}$, $N^*=\frac{\varLambda }{\mu +d i^*}$ and $i^*+p^*+s^*=1$, we obtain:

$$\begin{aligned} H=\left( \begin{array}{ccc} -\mu &{} 0 &{} -d \\ \beta s^* i^* &{} -\beta i^*-\mu &{} -\beta s^* \\ ((s^*+i^*)(1-\alpha _s)-s^*)\beta i^* &{} \alpha _s \beta i^* &{} -(1-\alpha _s)\beta i^* \end{array}\right) \end{aligned}$$

where $s^*=\dfrac{1}{\alpha _s \beta } (\mu +d-(1-i^*)(1-\alpha _s)\beta )$.

We have the characteristic polynomial:

$$\begin{aligned} f(\lambda )=\lambda ^3+A\lambda ^2+B\lambda +C \end{aligned}$$

where

$$\begin{aligned} A= & {} (2-\alpha _s)\beta i+2\mu ,\\ B= & {} (1-\alpha _s)(\beta i^* +d i^*+ 2 \mu )\beta i^*+ \mu (\beta i^*+\mu )+\alpha _s (\beta -d)\beta s^* i^*\\ C= & {} ( (1-\alpha _s)(d \beta i^{*2} + d \mu i^*+ \mu (\beta i^*+ \mu )) +\alpha _s \mu (\beta -d)s^*)\beta i^*. \end{aligned}$$

Clearly if $R_0>1$, then $\beta >d$, which implies that $A>0,B>0$ and $C>0$. Also we have that

$$\begin{aligned} \begin{array}{ll} AB-C=&{}3 \mu ^2 \beta i^*+2\mu ^3+\alpha _s^2 \mu \beta ^2 i^{*2}+\alpha _s \mu (\beta -d)\beta i^* s^*\\ &{} +\,\alpha _s(2-\alpha _s)(\beta -d)\beta ^2 i^{*2} s^*\\ &{}+\,(1-\alpha _s)(\mu d i^*+ \beta ^2 i^{*2}+6 \mu \beta i^*+4 \mu ^2)\beta i^*\\ &{} +\,(1-\alpha _s)^2(\mu +(\beta +d)i^*)\beta ^2 i^{*2}>0.\\ \end{array} \end{aligned}$$

Therefore by the Routh–Hurwitz criteria, the endemic equilibria, $E^*$ is locally stable.

1.5 Proof of Proposition 5

We analyze the fractional model associated with proportional recruitment:

$$\begin{aligned} \frac{dp}{dt}= & {} kr-rp-[(1-\alpha _s)\beta -d]pi\triangleq X(p,i) \end{aligned}$$

(4)

$$\begin{aligned} \frac{di}{dt}= & {} [\beta -(d+r)-\alpha _s\beta p-(\beta -d)i]i\triangleq Y(p,i) \end{aligned}$$

(5)

$$\begin{aligned} s= & {} 1-p-i \end{aligned}$$

(6)

$$\begin{aligned} \frac{dN}{dt}= & {} [r-\mu -di]N. \end{aligned}$$

(7)

Notice that the first two equations of the system (4)–(7) are decoupled from the rest which allow us to study the reduced system (4) and (5).

Positivity of solutions for the fractional system can be easily checked. Further,

$$\begin{aligned} \frac{dp}{dt}+\frac{di}{dt}= & {} kr-rp-\beta pi+\alpha _s pi+dpi+(\beta -d)i-ri-\alpha _s\beta pi-(\beta -d)i^2\\= & {} kr-r(p+i)-\beta pi-(\beta -d)pi+(\beta -d)i-(\beta -d)i^2\\= & {} kr-r(p+i)-\beta pi+(\beta -d)i[1-(p+i)]. \end{aligned}$$

Then at $p+i=1$, $\frac{d(p+i)}{dt}=-r+kr-\beta pi=-r(1-k)-\beta pi<0$ implies that $(p+i)(t)\le 1$ for $t>0$ given that $(p+i)(0)\le 1$.

The periodic solutions for the reduced system, and therefore for the transformed system, can be excluded by Dulac’s criteria: $\frac{\partial }{\partial p}(\frac{X}{pi})+\frac{\partial }{\partial i}(\frac{Y}{pi})=-\frac{kr}{p^2i}-\frac{\beta -d}{p}<0,$ when given $\beta -d>0$.

For the reduced system, there are possibly several steady states: $E_{1}=(k, 0)$ (always exists) and $E_{2}=(p^*_{\text {lin}},i^*_{\text {lin}}):=(p^*, i^*)$ (positive steady state, existence depends on parameter values).

For $E_1$, we have eigenvalues $\lambda _1=-r<0$ and $\lambda _2=(1-\alpha _s k)\beta -(d+r)$. Therefore, if $(1-\alpha _s k)\beta <(d+r)$, then $E_1$ is locally stable; if $(1-\alpha _s k)\beta >(d+r)$, then $E_1$ is unstable.

For $E_2$, we have $A{p^*}^2+Bp^*+C=0$ and $i^*=1-\frac{\alpha _s\beta p^*+r}{\beta -d}$, with $A=\alpha _s\beta [(1-\alpha _s)\beta -d]$, $B=-\{(\beta -d-r)[(1-\alpha _s)\beta -d]+(\beta -d)r\}$ and $C=(\beta -d)kr$. If further $0< p^*<1$ and $0< i^*<1$, then $(p^*,i^*)$ exists as a positive steady state.

Notice that $p^*\in \left( 0, \frac{\beta -(d+r)}{\alpha _s\beta }\right)$ and $i^*\in \left( 0, \frac{\beta -(d+r)}{\beta -d}\right)$. So we assume that $\beta > d+r$. Denote $F(p)=Ap^2+Bp+C$, then we have the following results for F(p) over $\left( 0, \frac{\beta -(d+r)}{\alpha _s\beta }\right)$: $F(0)=(\beta -d)kr>0$, and $F\left( \frac{\beta -(d+r)}{\alpha _s\beta }\right) =-\frac{(\beta -d)r}{\alpha _s\beta }[(1-k\alpha _s)\beta -(d+r)]$. Now if $(1-k\alpha _s)\beta >d+r$, then $F(0)>0$ and $F\left( \frac{\beta -(d+r)}{\alpha _s\beta }\right) <0$ implies a unique solution of $F(p)=0$ over $\left( 0, \frac{\beta -(d+r)}{\alpha _s\beta }\right)$, because F(p) is a parabolic function.

Now consider the case when $(1-k\alpha _s)\beta <d+r$ ($\Rightarrow (1-\alpha _s)\beta -d<r$). If $(1-\alpha _s)\beta \le d$, then $F(0)>0$ and $F\left( \frac{\beta -(d+r)}{\alpha _s\beta }\right) >0$ implies no solution of $F(p)=0$ over $\left( 0, \frac{\beta -(d+r)}{\alpha _s\beta }\right)$, because F(p) is linear or concave down. If $(1-\alpha _s)\beta > d$, then F(p) is concave up and attains its minimum at $\hat{p}=\frac{\beta -(d+r)}{2\alpha _s\beta }+\frac{(\beta -d)r}{2\alpha _s\beta [(1-\alpha _s)\beta -d]}>\frac{\beta -(d+r)}{2\alpha _s\beta }+\frac{(\beta -d)r}{2\alpha _s\beta r}>\frac{\beta -(d+r)}{\alpha _s\beta }$. Therefore, if $(1-\alpha _s)\beta > d$, then $F(0)>0$ and $F\left( \frac{\beta -(d+r)}{\alpha _s\beta }\right) >0$ implies no solution of $F(p)=0$ over $\left( 0, \frac{\beta -(d+r)}{\alpha _s\beta }\right)$, because F(p) is decreasing over $\left( 0, \frac{\beta -(d+r)}{\alpha _s\beta }\right)$.

Therefore, if $\beta \ge d+r$, we have a unique solution of $F(p)=0$ over $\left( 0, \frac{\beta -(d+r)}{\alpha _s\beta }\right)$ when $(1-k\alpha _s)\beta >d+r$ and no solution over $(0, \frac{\beta -(d+r)}{\alpha _s\beta })$ when $(1-k\alpha _s)\beta <d+r$.

Next, we can show that $E_2$ is stable when it exists. Notice that the existence of $E_2$ requires $(1-\alpha _s k)\beta >(d+r)$, when $E_1$ is unstable. Also, we have

$$\begin{aligned}&kr-rp^*-[(1-\alpha _s)\beta -d]p^*i^*=0\Rightarrow -r-[(1-\alpha _s)\beta -d]i^*=-\frac{kr}{p^*};\\&-[(1-\alpha _s)\beta -d]p^*=-\frac{kr-rp^*}{i^*}, \end{aligned}$$

and

$$\begin{aligned} \beta -(d+r)-\alpha _s\beta p^*-(\beta -d)i^*=0. \end{aligned}$$

And we have the following Jacobian matrix for $E_2$:

$$\begin{aligned} J(E_2) = \left| \begin{array}{cc} -r-[(1-\alpha _s)\beta -d]i^* &{} -[(1-\alpha _s)\beta -d]p^* \\ -\alpha _s\beta i^* &{} -(\beta -d)i^*+\beta -(d+r)-\alpha _s\beta p^*-(\beta -d)i^* \end{array} \right| . \end{aligned}$$

Then

$$\begin{aligned} J(E_2) =\left| \begin{array}{cc} -\frac{kr}{p^*} &{} -\frac{kr-rp^*}{i^*} \\ -\alpha _s\beta i^* &{} -(\beta -d)i^* \end{array} \right| . \end{aligned}$$

The corresponding eigenvalues satisfy

$$\begin{aligned} \lambda _1+\lambda _2=-\frac{kr}{p^*}-(\beta -d)i^ *<0, \end{aligned}$$

and

$$\begin{aligned} \lambda _1\cdot \lambda _2=\frac{kr}{p^*}(\beta -d)i^*-\alpha _s\beta (kr-rp^*). \end{aligned}$$

Furthermore,

$$\begin{aligned} \lambda _1\cdot \lambda _2= & {} \frac{kr}{p^*}(\beta -d)\left( 1-\frac{\alpha _s\beta p^*+r}{\beta -d}\right) -\alpha _s\beta (kr-rp^*)\\= & {} \frac{r}{p^*}\left[ k(\beta -d-\alpha _s\beta p^*-r)-\alpha _s\beta (k-p^*)p^*\right] \\= & {} \frac{r}{p^*}\left[ \alpha _s\beta {p^*}^2-2\alpha _s\beta kp^*+k(\beta -d-r)\right] . \end{aligned}$$

Therefore, $\lambda _1\cdot \lambda _2>0$ since $f(p^*)=\alpha _s\beta {p^*}^2-2\alpha _s\beta kp^*+k(\beta -d-r)>0$. We know that $f(p^*)$ attains the minimum value f(k) at $p^*=k$. Now it is sufficient to show that $f(k)>0$. Since $(1-\alpha _s k)\beta>(d+r)\Rightarrow \beta -d-r>\alpha _sk\beta$, then

$$\begin{aligned} f(k)= & {} \alpha _s\beta k^2-2\alpha _s\beta k^2+k(\beta -d-r)\\= & {} k(\beta -d-r)-\alpha _s\beta k^2\\> & {} k\alpha _sk\beta -\alpha _s\beta k^2\\= & {} 0. \end{aligned}$$

Thus, we have proven that $E_2$ is stable when it exists, i.e., require $(1-\alpha _s k)\beta >(d+r)$. Further since $(1-\alpha _s k)\beta >(d+r)$ implies $\beta >d$ (no periodic solutions) and $E_1$ is unstable, then $E_2$ is globally stable when it exists.

If $(1-\alpha _s k)\beta <(d+r)$, then $E_1$ is stable and $E_2$ does not exist. Further $\beta >d$ implies no periodic solutions, and therefore, we conclude that $E_1$ is globally stable provided that $(1-\alpha _s k)\beta <(d+r)$ and $\beta >d$.

Now, for the original system, similar to Proposition 2, we can show that the unique steady state $E=(0, 0, 0)$ is globally stable if $r<\mu$. Next, we will study the cases when $r>\mu$. We know that when $\beta >d$, $E_1=(k,0)$ is globally stable when $(1-k)\beta +k(1-\alpha _s)\beta <d+r$, while $E_2=(p^*,i^*)$ is globally stable when $(1-k)\beta +k(1-\alpha _s)\beta >d+r$. Then by (6), the total population N(t) either approaches 0 when $r-\mu -d\lim \limits _{t\rightarrow \infty }i(t)<0$, or blows up when $r-\mu -d\lim \limits _{t\rightarrow \infty }i(t)>0$. Similarly by (7), the infected population I(t) either approaches 0 when $\beta (1-\lim \limits _{t\rightarrow \infty }p(t)-\lim \limits _{t\rightarrow \infty }i(t))+(1-\alpha _s)\beta \lim \limits _{t\rightarrow \infty }p(t)-(\mu +d)<0$, or blows up when $\beta (1-\lim \limits _{t\rightarrow \infty }p(t)-\lim \limits _{t\rightarrow \infty }i(t))+(1-\alpha _s)\beta \lim \limits _{t\rightarrow \infty }p(t)-(\mu +d)>0$. Thus, periodic solutions for (2) do not exist when $\beta >d$.

Further, if $r > \mu +d$, then $r-\mu -d\lim \limits _{t\rightarrow \infty }i(t)>0$ and the population grows unbounded. Now assume that $\mu<r<\mu +d$. Substituting in $i^*=1-\dfrac{\alpha _s \beta p^*+r}{\beta -d}$ and $p^*$ into the above, we derive conditions for population extinction. We obtain

$$\begin{aligned} \beta (1-p^*-i^*)+(1-\alpha _s)\beta p^*-(\mu +d)=0 \end{aligned}$$

which yields $p^*=\dfrac{(\mu +d)(\beta -d)-r\beta }{d \alpha _s \beta }$. Then, substituting the expression for $p^*$ into the equation $A{p^*}^2+Bp^*+C=0$ and dividing out $\beta -d$ gives us the following conditions for population extinction: if $\beta >\bar{\beta }$, there is population extinction and if $\beta <\bar{\beta }$, the population grows unbounded, where $\bar{\beta }$ is a solution to the equation:

$$\begin{aligned} a \beta ^2+b\beta +c=0 \end{aligned}$$

(8)

with

$$\begin{aligned} a= & {} (r-\mu )(1-\alpha _s)(r-\mu -d)\\ b= & {} d((r-\mu )(\mu +d)(1-\alpha _s)+(r-\mu )\mu +d(r \alpha _s k-\mu ))\\ c= & {} \mu d^2(\mu +d). \end{aligned}$$

Whenever $\mu<r<\mu +d$, we see that $a<0$ and $c>0$. Therefore, by Descartes’ rule of signs, there is exactly one positive solution, namely, $\bar{\beta }=\dfrac{-\bar{b}-\sqrt{\bar{b}^2-4\bar{a}\bar{c}}}{2\bar{a}}$. Moreover, we have $\bar{\beta }>\dfrac{r+d}{1-\alpha _s k}$.

1.6 Proof of Proposition 6

The invariance of the biologically feasible region can be easily proved. Given $S^p+S+I > K$ in $\{S^p\ge 0,S\ge 0, I \ge 0 \}$, we have $\frac{dN}{dt}=rN\left( 1-\frac{N}{K}\right) -\mu N-dI \le -\mu N$, and thus, the total population will decrease below carrying capicity. Therefore, we consider only the region $\{S^p\ge 0,S \ge 0, I \ge 0, \, S^p+S+I \le K \}$. Then, similar to Proposition 2, we can show that the extinction steady state $E_{\hbox {ext}}$ is globally stable if $r<\mu$.

We analyze the fractional model associated with logistic recruitment:

$$\begin{aligned} \begin{array}{lcl} \dfrac{dp}{dt} &{} = &{} k r (1-\frac{N}{K})-r(1-\frac{N}{K})p-((1-\alpha _s)\beta -d) p i \\ \dfrac{di}{dt} &{} = &{} ((\beta -d)(1-i)-r(1-\frac{N}{K})-\alpha _s \beta p)i \\ \dfrac{dN}{dt} &{} = &{} (r(1-\frac{N}{K})-\mu - d i)N. \end{array} \end{aligned}$$

(9)

From now on, we assume $r>\mu$. For the eigenvalues for $E_{df}$, we have $\lambda _1+\lambda _2=-r<0$, $\lambda _1 \cdot \lambda _2=\mu (r-\mu )>0$ and $\lambda _3=(\mu +d)(R_0-1)$. Therefore, $E_{df}$ is stable when $R_0<1$ and unstable when $R_0>1$.

Next, we prove that when $\mu<r<\mu +d$, the extinction steady state is stable if $\beta >\bar{\beta }$ and unstable if $\beta <\bar{\beta }$, where $\bar{\beta }$ is a solution of Eq. (8); and that the positive steady state also exists if $\beta <\bar{\beta }$. Furthermore when $r>\mu +d$, we prove that the positive steady state exists if $R_0>1$.

For Model (9), there are possibly several steady states: $E_{1}=(k, 0, 0)$ (always exists), $E_{2}=(p^*, i^*, 0)$ (existence depends on parameter values), $E_{3}=(k, 0, \frac{r-\mu }{r}K)$ and $E_{4}=(p^*_{\text {log}},i^*_{\text {log}},N^*)$ (existence depends on parameter values). Notice that $E_{3}$ is equivalent with $E_{df}=(k\frac{r-\mu }{\mu }K, (1-k)\frac{r-\mu }{\mu }K, 0)$ for (2) and $E_{4}$ is equivalent with the positive steady state $E^*$ for (2) when it exists, while $E_{1}$ and $E_{2}$ are both corresponding to $E_{\hbox {ext}}=(0, 0, 0)$ for Model (2). Let us assume that $r>\mu$.

For $E_1$, we have eigenvalues $\lambda _1=-r<0$, $\lambda _2=(1-\alpha _sk)\beta -(d+r)$ and $\lambda _3=r-\mu >0$. So $E_1$ is unstable.

For $E_2$, we have $A{p^*}^2+Bp^*+C=0$ and $i^*=1-\frac{\alpha _s\beta p^*+r}{\beta -d}$, with $A=\alpha _s\beta [(1-\alpha _s)\beta -d]$, $B=-\{(\beta -d-r)[(1-\alpha _s)\beta -d]+(\beta -d)r\}$ and $C=(\beta -d)kr$. If further $0\le p^*\le 1$ and $0\le i^*\le 1$, then $(p^*,i^*,0)$ exists as a steady state. This is similar to the case $E_2=(p^*,i^*)$ in Proposition 5 (with proportional recruitment) when taking $N^*=0$. We obtain that $E_2$ is stable whenever $\mu<r<\mu +d$ and $\beta >\bar{\beta }$ and unstable otherwise.

For $E_4=(p^*_{\text {log}},i^*_{\text {log}},N^*):=(p^*,i^*,N^*)$, we have $A{p^*}^2+Bp^*+C=0$, $i^*=\frac{\beta -\mu -d-\alpha _s\beta p^*}{\beta }$, $N^*=\frac{r-\mu -di^*}{r}K$, with $A=\alpha _s(1-\alpha _s)\beta$, $B=-[(1-\alpha _s)(\beta -\mu -d)+\mu +k\alpha _sd]$ and $C=kd\frac{\beta -\mu -d}{\beta }+k\mu$. Therefore, positive steady states may exist but are too complicated to be expressed explicitly.

Notice that $i^*\in (0, \frac{r-\mu }{d})$ and $p^*\in \left( \frac{\beta -(\mu +d)-\frac{r-\mu }{d}}{\alpha _s\beta }, \frac{\beta -(\mu +d)}{\alpha _s\beta }\right)$. So we assume that $\beta > \mu +d$ and $r>\mu$ so that $E_4$ may exist as an endemic steady state. Denote $F(p)=Ap^2+Bp+C$, then we have the following results for F(p) over $(0, \frac{\beta -(\mu +d)}{\alpha _s\beta })$: since $A>0, B<0$ and $C>0$, we have

$$\begin{aligned} 0<\dfrac{-B-\varDelta }{2A}<\dfrac{-B+\varDelta }{2A} \end{aligned}$$

where

$$\begin{aligned} \varDelta =\sqrt{B^2-4AC}. \end{aligned}$$

(10)

Since $F(0)=\dfrac{k(\mu +d)(\beta -d)}{\beta }>0$ and $F\left( \dfrac{\beta -(\mu +d)}{\alpha _s \beta }\right) =- \dfrac{\mu (\mu +d)}{\alpha _s \beta } (R_0 -1) <0$, then there is only one solution on the interval $\left( 0,\dfrac{\beta -(\mu +d)}{\alpha _s \beta } \right)$, and we have that when it exists, $p^*=\dfrac{-B-\varDelta }{2A}$.

Also we have that $p^*$ is a solution on the interval if and only if

$F\left( \dfrac{\beta -(\mu +d)-\beta \frac{r-\mu }{d}}{\alpha _s \beta }\right) >0$. That is:

$$\begin{aligned}&\dfrac{(1-\alpha _s)\beta ^2 (r-\mu )^2+d^3\mu -d \beta (r-\mu )(\beta -\mu )(1-\alpha _s)+d \beta \mu (r-\mu )}{d^2 \alpha _s \beta }\\&\qquad +\,\dfrac{d^2(r(1-(1-k)\alpha _s)\beta +\mu (\mu -\beta (2-\alpha _s)))}{d^2 \alpha _s \beta }>0. \end{aligned}$$

Given $R_0>1$ and $r>\mu$, we have $\beta >\dfrac{\mu +d}{1-\alpha _s k}$. This provides the following conditions for existence:

• $r > \mu +d$

• $r<\mu +d$ and $\beta \le \dfrac{(\mu +d)d}{\mu +d-r}$

• $\dfrac{(\mu +d)\mu }{\mu +d k} \le r<\mu +d$ and $\bar{\beta }>\beta >\dfrac{(\mu +d)d}{\mu +d-r}$

• $r<\dfrac{(\mu +d)\mu }{\mu +d k}$ and $\bar{\beta }>\beta >\dfrac{(\mu +d)d}{\mu +d-r}$

where $\bar{\beta }$ is a solution to the Eq. (8). The above conditions reduce to:

• $r > \mu +d$ and $\beta >\dfrac{\mu +d}{1-\alpha _s k}$

• $\mu<r<\mu +d$ and $\bar{\beta }>\beta >\dfrac{\mu +d}{1-\alpha _s k}$.

Now consider the case when $R_0<1$($\Rightarrow \beta <\frac{\mu +d}{1-k\alpha _s}$). F(p) is concave up and attains its minimum at $\hat{p}=\frac{\beta -(\mu +d)}{2\alpha _s\beta }+\frac{\mu +k\alpha _sd}{2\alpha _s(1-\alpha _s)\beta }>\frac{\beta -(\mu +d)}{\alpha _s\beta }$, since

$$\begin{aligned}&\frac{\beta -(\mu +d)}{2\alpha _s\beta }+\frac{\mu +k\alpha _sd}{2\alpha _s(1-\alpha _s)\beta }-\frac{\beta -(\mu +d)}{\alpha _s\beta }\\&\qquad =\frac{\mu +k\alpha _sd-(1-\alpha _s)[\beta -(\mu +d)]}{2\alpha _s(1-\alpha _s)\beta }\\&\qquad>\frac{\mu +k\alpha _sd-(1-\alpha _s)[\frac{\mu +d}{1-k\alpha _s}-(\mu +d)]}{2\alpha _s(1-\alpha _s)\beta }\\&\qquad =\frac{(2-\alpha _s-\frac{1-\alpha _s}{1-k\alpha _s})\mu +k\alpha _s(1-\frac{1-\alpha _s}{1-k\alpha _s})d}{2\alpha _s(1-\alpha _s)\beta }>0. \end{aligned}$$

Therefore, $F(\frac{\beta -(\mu +d)}{\alpha _s\beta })>0$ implies no solution of $F(p)=0$ over $\left( \frac{\beta -(\mu +d)-\frac{r-\mu }{d}}{\alpha _s\beta }, \frac{\beta -(\mu +d)}{\alpha _s\beta }\right)$, because F(p) is decreasing over $\left( \frac{\beta -(\mu +d)-\frac{r-\mu }{d}}{\alpha _s\beta }, \frac{\beta -(\mu +d)}{\alpha _s\beta }\right)$.

Thus, when $R_0>1$, if either $r>\mu +d$ or both $\beta < \bar{\beta }$ and $\mu<r<\mu +d$, we have a unique solution of $F(p)=0$ over $\left( \frac{\beta -(\mu +d)-\frac{r-\mu }{d}}{\alpha _s\beta }, \frac{\beta -(\mu +d)}{\alpha _s\beta }\right)$. Also there is no solution over $\left( \frac{\beta -(\mu +d)-\frac{r-\mu }{d}}{\alpha _s\beta }, \frac{\beta -(\mu +d)}{\alpha _s\beta }\right)$ whenever either $R_0<1$ or if $R_0>1$ we have both $\mu<r<\mu +d$ and $\beta > \bar{\beta }$.

Now we give partial results for the stability of $E_4$. The Jacobian evaluated at $E_4$ can be expressed as:

$$\begin{aligned} J=\left( \begin{array}{ccc} -\mu -(1-\alpha _s)\beta i^* &{} -((1-\alpha _s)\beta -d)p^* &{} \dfrac{r}{K}(-k+p^*) \\ -\alpha _s \beta i^* &{} -(\beta -d)i^* &{} \dfrac{r}{K}i^* \\ 0 &{} -d N^* &{} -\dfrac{r}{K}N^* \end{array}\right) \end{aligned}$$

which has characteristic polynomial

$$\begin{aligned} f(\lambda )=\lambda ^3+A \lambda ^2+B\lambda +C \end{aligned}$$

where

$$\begin{aligned} A= & {} (\beta -d) i^* + (1-\alpha )\beta i^*+\dfrac{N^* r}{K}+\mu ,\\ B= & {} i^{*} (\beta -d)(i^* \beta (1-\alpha _s)+\mu )+i^* p^* \alpha _s \beta (d-\beta (1-\alpha _s))\\&+\,\dfrac{N^* r}{K} (i^* \beta + i^* \beta (1-\alpha _s)+\mu ), \end{aligned}$$

and

$$\begin{aligned} C=\dfrac{N^* r i^* \beta }{K}(d k \alpha _s + \mu + (i^* -p^* \alpha _s) \beta (1-\alpha _s)). \end{aligned}$$

Given $R_0>1$ and $i^* < \frac{r-\mu }{d}$, we have $A>0$. Substituting in for $i^*$ and $p^*$, we obtain:

$$\begin{aligned} C=\dfrac{N^* r i^* \beta }{K}\varDelta >0, \end{aligned}$$

where $\varDelta$ is as in (10). Therefore, the equilibium is locally stable if and only if $AB-C>0$. We have the following

$$\begin{aligned} \begin{array}{rcl} AB-C &{}= &{} (i^* (1-\alpha _s)\beta +i^*(\beta -d)+\dfrac{N^* r}{K}+\mu )\\ &{}&{}\times (i^* d(\beta (p^* \alpha _s -i^*(1-\alpha _s))-\beta k \alpha _s-\mu )) \\ &{}&{} +\, \dfrac{N^* r (i^* \beta +i^*(1-\alpha _s)\beta +\mu )}{K}(i^*(1-\alpha _s)\beta +i^*(\beta -d)+\dfrac{N^* r}{K}+\mu )\\ &{}&{} +\, i^* \beta \varDelta (i^*(1-\alpha _s)\beta +i^*(\beta -d)+\mu ). \end{array} \end{aligned}$$

Due to the complexity of the above expression, we have only numerically verified for plausible parameter values, that the equilibrium, $E_4$, is locally stable whenever it exists.

Now we show that the HIV prevalence associated with Model (2) under logistic and constant recruitment is the same, i.e., $i^*_{\mathrm {const}} = i^*_{\mathrm {log}}$, whenever both equilibria exist and are stable. Assume $R_0=\dfrac{(1-\alpha _s k)\beta }{\mu +d}>1.$ We have shown already that with constant recruitment, the infected population, $I^*$, is a root of:

$$\begin{aligned} \begin{array}{l} \hat{F}(I) \triangleq A I^2 + B I + C \end{array} \end{aligned}$$

where

$$\begin{aligned} \begin{array}{l} A=- (d-\beta )(d-(1-\alpha _s)\beta )\dfrac{1}{\varLambda ^2},\\ B=\left( (d-(1-\alpha _s)\beta )\left( 1-\dfrac{\beta }{\mu +d}\right) -\mu \dfrac{\beta }{\mu +d}-d (R_0-1)\right) \dfrac{1}{\varLambda },\\ C= R_0-1. \end{array} \end{aligned}$$

Next, we calculate $I^*$. $R_0>1$ implies that $\beta>\dfrac{\mu +d}{1-\alpha _s k}>\mu +d>d$. Thus, we have

$$\begin{aligned} \begin{array}{l} d-(1-\alpha _s)\beta<d-\dfrac{(1-\alpha _s)(\mu +d)}{1-\alpha _s k}< d-(\mu +d) = -\mu \end{array} \end{aligned}$$

and therefore, $A<0$ and $C>0$. By Descartes’ rule of signs, there is exactly one positive root of $\hat{F}(I)$. Since $I^*>0$, we must have that

$$\begin{aligned} \begin{array}{l} I^*=\dfrac{-B-\sqrt{B^2-4 A C}}{2 A}. \end{array} \end{aligned}$$

The total population is given by $N=\dfrac{\varLambda -d I}{\mu }$, so that

$$\begin{aligned} N^*=\dfrac{\varLambda -dI^*}{\mu }. \end{aligned}$$

Therefore, the HIV prevalence under constant recruitment is

$$\begin{aligned} \begin{array}{l} i^*_{\mathrm {const}}=\dfrac{I^*}{N^*}=\dfrac{\mu I^*}{\varLambda - d I^*}. \\ \end{array} \end{aligned}$$

First, notice that $\varLambda$ factors out of $I^*$ and thus will be absent in the final expression. Substituting in for $I^*$, we have that after rationalizing the denominator

$$\begin{aligned} \begin{array}{l} i^*_{\mathrm {const}}=\dfrac{(1-\alpha _s)(\beta -\mu -d)-\mu -\alpha _s k d+ \varDelta _{\mathrm {const}}}{2(1-\alpha _s) \beta }, \end{array} \end{aligned}$$

where

$$\begin{aligned} \varDelta _{\mathrm {const}}=\sqrt{\dfrac{\varLambda ^2 (\mu +d)^2}{\beta ^2}(B^2-4AC)}. \end{aligned}$$

We already have shown that the HIV prevalence given logistic growth is

$$\begin{aligned} i^*_{\mathrm {log}}=1-\alpha _s p^*-\dfrac{\mu +d}{\beta }, \end{aligned}$$

where

$$\begin{aligned} p^*= & {} \dfrac{-\bar{B}-\varDelta _{\mathrm {log}}}{2\bar{A}},\\ \varDelta _{\mathrm {log}}= & {} \sqrt{\bar{B}^2-4\bar{A}\bar{C}}, \end{aligned}$$

and

$$\begin{aligned} \begin{array}{l} \bar{A}=\alpha _s(1-\alpha _s)\beta ,\\ \bar{B}=-\left( (1-\alpha _s)(\beta -\mu -d)+\mu +k\alpha _s d\right) ,\\ \bar{C}= k d \dfrac{\beta -\mu -d}{\beta }+k \mu . \end{array} \end{aligned}$$

Substituting $p^*$ into $i^*_{\mathrm {log}}$ and simplifying, we obtain

$$\begin{aligned} \begin{array}{l} i^*_{\mathrm {log}}=\dfrac{(1-\alpha _s)(\beta -\mu -d)-\mu -\alpha _s k d+ \varDelta _{\mathrm {log}}}{2(1-\alpha _s) \beta }. \end{array} \end{aligned}$$

It can be checked that $\varDelta _{\mathrm {const}}=\varDelta _{\mathrm {log}}$, and therefore, $i^*_{\mathrm {const}}=i^*_{\mathrm {log}}$.

Appendix 2: Demographic Data and Simulations for the Republic of South Africa

Age-structured population data about the Republic of South Africa for year 2001 and from year 2003 to 2011, in thousands, are presented in Table 5. The data have been adapted from Statistics South Africa (Africa 2012). In the table, T(15–49) refers to total population of individuals from age 15 to 49. Similarly, P(15–49) refers to HIV prevalence among people of age 15–49. HIV-T and SUS-T refer to total number of infected and susceptible individuals ages 15–49, respectively.

Parameter values generated from data fitting are listed in Table 6, and the corresponding simulations using fitted parameter values are presented in Figs. 5 and 6.

Table 6 Parameter values generated from data fitting