A time since onset of injection model for hepatitis C spread amongst injecting drug users

Abstract

Studies of hepatitis C virus (HCV) infection amongst injecting drug users (IDUs) have suggested that this population can be separated into two risk groups (naive and experienced) with different injecting risk behaviours. Understanding the differences between these two groups and how they interact could lead to a better allocation of prevention measures designed to reduce the burden of HCV in this population. In this paper we develop a deterministic, compartmental mathematical model for the spread of HCV in an IDU population that has been separated into two groups (naive and experienced) by time since onset of injection. We will first describe the model. After deriving the system of governing equations, we will examine the basic reproductive number $R_0$, the existence and uniqueness of equilibrium solutions and the global stability of the disease free equilibrium (DFE) solution. The model behaviour is determined by the basic reproductive number, with $R_0=1$ a critical threshold for endemic HCV prevalence. We will show that when $R_0\le 1$, and HCV is initially present in the population, the system will tend towards the globally asymptotically stable DFE where HCV has been eliminated from the population. We also show that when $R_0>1$ there exists a unique non-zero equilibrium solution. Then we estimate the value of $R_0$ from epidemiological data for Glasgow and verify our theoretical results using simulations with realistic parameter values. The numerical results suggest that if $R_0>1$ and the disease is initially present then the system will tend to the unique endemic equilibrium.

Appendix

1.1 A: Derivation of $\varLambda _{jk}$ terms

Let $\lambda _j$ denote the rate at which IDUs in experience group $j$ share needles and syringes, $n_j$ denote the number of IDUs in experience group $j$ and $s_{jk}$ denote the fraction of shared injections that an IDU in experience group $j$ will perform while using a needle and syringe which was last used by an IDU in experience group $k$. Then

$$\begin{aligned} \lambda _j n_j s_{jk}\varDelta t+o(\varDelta t) \end{aligned}$$

denotes the total number of shared injections in a time interval of length $\varDelta t$ made by the entire set of IDUs in experience group $j$ with needle and syringes that were last used by IDUs in experience group $k$. Since there are $m_k$ of these needles and syringes in total, the total rate at which an IDU in experience group $j$ picks up a single one of these needles and syringes is given by

$$\begin{aligned} \varLambda _{jk}=\frac{\lambda _j s_{jk}n_j}{m_k}. \end{aligned}$$

1.2 B: Constraint on $s_{01}$ and $s_{10}$

Define $m_0\,(m_1)$ to be the number of naive (experienced) needles and syringes in circulation (as discussed in Sect. 3). Ignoring the infection status of the needles and syringes we obtain the following equations

$$\begin{aligned} \dfrac{dm_0}{dt}&= \varLambda _{01}m_1-\varLambda _{10}m_0=\lambda _0s_{01}n_0-\lambda _1s_{10}n_1,\end{aligned}$$

(53)

$$\begin{aligned} \dfrac{dm_1}{dt}&= \varLambda _{10}m_0-\varLambda _{01}m_1=\lambda _1s_{10}n_1-\lambda _0s_{01}n_0, \end{aligned}$$

(54)

where $\varLambda _{jk}=\dfrac{\lambda _j s_{jk}n_j}{m_k}$, $j,k=0,1$, denotes the rate at which an IDU in group $j$ picks up a needle and syringe last used by a group $k$ IDU and $n_0$ and $n_1$ are, respectively, the number of naive and experienced IDUs at time t. The above equations imply that at equilibrium

$$\begin{aligned} \varLambda _{01}m_1=\varLambda _{10}m_0, \end{aligned}$$

(55)

which implies that

$$\begin{aligned} \lambda _0s_{01}n_0=\lambda _1s_{10}n_1. \end{aligned}$$

Note that $n_0+n_1=n$, the constant number of IDUs in the population and

$$\begin{aligned} \frac{dn_0}{dt}&= \mu n-(\mu +\eta )n_0,\\ \frac{dn_1}{dt}&= \eta n_0-\mu n_1.\nonumber \end{aligned}$$

(56)

Hence as $t\rightarrow \infty$, $n_0\rightarrow \mu n/(\mu +\eta )$ and $n_1\rightarrow \eta n/(\mu +\eta )$ so this constraint becomes

$$\begin{aligned} \lambda _0s_{01}\mu =\lambda _1s_{10}\eta . \end{aligned}$$

(57)

If this condition is not satisfied then Eqs. (53) and (54) imply that one of $m_0, m_1$ eventually becomes negative. If condition (57) is satisfied then from (56) we have

$$\begin{aligned} \dfrac{d}{dt}\left[n_0 \text{ exp}[(\mu +\eta )t]\right] =\mu n \text{ exp}[(\mu +\eta )t]\nonumber . \end{aligned}$$

Integrating over $[0, t]$ gives

$$\begin{aligned} n_0(t)\text{ exp}[(\mu +\eta )t]-n_0(0)&=\frac{\mu n}{\mu +\eta }(\text{ exp}[(\mu +\eta )t]-1), \end{aligned}$$

which implies that

$$\begin{aligned} n_0(t)=n_{0}^*+(n_0(0)-n_{0}^*)\text{ exp}[-(\mu +\eta )t], \end{aligned}$$

(58)

where $n_{0}^*=\mu n/(\mu +\eta )$. Using Eq. (56) and the fact that $n_1=n-n_0$ we can use a similar argument to that used above to obtain

$$\begin{aligned} n_1(t)=n_{1}^*+(n_1(0)-n_{1}^*)\text{ exp}[-(\mu +\eta )t], \end{aligned}$$

(59)

where $n_{1}^*=\eta n/(\mu +\eta )$. Substituting (58) and (59) into Eq. (53) we find that

$$\begin{aligned} \dfrac{dm_0}{dt} =(\lambda _0s_{01}n_0(0)-\lambda _1s_{10}n_1(0))\text{ exp}[-(\mu +\eta )t]. \end{aligned}$$

Integrating over $[0, t]$ we have

$$\begin{aligned} m_0(t)=m_0(0)-\frac{1}{\mu +\eta }(\lambda _0s_{01}n_0(0)-\lambda _1s_{10}n_1(0))(\text{ exp}[-(\mu +\eta )t]-1). \end{aligned}$$

Similarly

$$\begin{aligned} m_1(t)=m_1(0)-\frac{1}{\mu +\eta }(\lambda _1s_{10}n_1(0)-\lambda _0s_{01}n_0(0))(\text{ exp}[-(\mu +\eta )t]-1). \end{aligned}$$

Hence, as $t\rightarrow \infty$, $m_0(t)$ monotonically tends to

$$\begin{aligned} m_0(0)+\frac{1}{\mu +\eta }(\lambda _0s_{01}n_0(0)-\lambda _1s_{10}n_1(0)). \end{aligned}$$

(60)

Similarly $m_1(t)$ monotonically tends to

$$\begin{aligned} m_1(0)+\frac{1}{\mu +\eta }(\lambda _1s_{10}n_1(0)-\lambda _0s_{01}n_0(0)). \end{aligned}$$

(61)

So provided that both (60) and (61) are positive then $m_0(t)$ and $m_1(t)$ will approach strictly positive values. Furthermore, if the spread of HCV has been going on for a long period of time before the model is started we can expect that

$$\begin{aligned} n_0(0)\approx n_{0}^*(0)=\frac{\mu n}{\mu +\eta }\quad \text{ and} \quad n_1(0)\approx n_{1}^*(0)=\frac{\eta n}{\mu +\eta }. \end{aligned}$$

In this case we see from (60) and (61) that the number of needles and syringes in group will not change much from their initial values.

1.3 C: Derivation of $\psi _0$ and $\psi _1$

At time $t$, the total rate at which needles and syringes in the naive group are used is given by

$$\begin{aligned} (\varLambda _{00}+\varLambda _{10})m_0=\lambda _0s_{00}n_0+\lambda _1s_{10}n_1, \end{aligned}$$

where $n_0$ and $n_1$ are, respectively, the number of naive and experienced IDUs. Also the total rate at which needles and syringes in this group are used or exchanged is given by

$$\begin{aligned} (\varLambda _{00}+\varLambda _{10}+\tau _0)m_0. \end{aligned}$$

Hence the probability of choosing an unexchanged needle and syringe from the needles and syringes in the naive group is given by

$$\begin{aligned} \psi _0=\dfrac{\varLambda _{00}+\varLambda _{10}}{(\varLambda _{00}+\varLambda _{10}+\tau _0)}. \end{aligned}$$

Similarly

$$\begin{aligned} \psi _1=\dfrac{\varLambda _{11}+\varLambda _{01}}{(\varLambda _{11}+\varLambda _{01}+\tau _1)}. \end{aligned}$$

1.4 D: Derivation of K = $\mathbf M$(K)K

Substituting the equilibrium expressions for $\pi _y^{0*}$, $\pi _y^{1*}$, $\pi _h^{0*}$ and $\pi _h^{1*}$ given by (20), (24) and (26) into

$$\begin{aligned} K_0^*=\lambda _0s_{00}(1-\phi _0)\psi _{0}^*(\alpha _h\pi _{h}^{0*}+\alpha _y\pi _{y}^{0*})+\lambda _0s_{01}(1-\phi _0)\frac{\psi _{1}^*\pi ^{0*}}{\pi ^{1*}}(\alpha _h\pi _{h}^{1*}+\alpha _y\pi _{y}^{1*}) \end{aligned}$$

we obtain

$$\begin{aligned} K_0^*&= \left[\frac{\big (\lambda _0s_{00}(1-\phi _0)\psi _{0}^*\big (\alpha _h+\frac{\sigma (1-\delta )}{\mu +\eta }\alpha _y\big )+\lambda _0s_{01}(1-\phi _0)\psi _{1}^*\frac{\pi ^{0*}}{\pi ^{1*}}\frac{\eta \sigma (1-\delta )\alpha _y}{\mu (\mu +\eta )}\big )\frac{1}{\mu +\sigma +\eta }}{1+\frac{K_0^*}{\mu +\sigma +\eta }\big (\frac{\mu +\eta }{\mu }+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )}\right.\\&\!\!\left.+\frac{\lambda _0s_{01}(1-\phi _0)\psi _{1}^*\frac{\pi ^{0*}}{\pi ^{1*}}\big (\alpha _h+\frac{\sigma (1-\delta )}{\mu }\alpha _y\big )\frac{\eta }{\mu +\sigma }\frac{1}{\mu +\sigma +\eta }}{\bigg [1+\frac{K_0^*}{\mu +\sigma +\eta }\big (\frac{\mu +\eta }{\mu }+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )\bigg ]\bigg [1+\frac{K_1^*}{\mu +\sigma }\frac{\mu +\eta }{\eta }\big (1+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )\bigg ]}\right]K_0^*\\&\!\!+\left[\frac{\lambda _0s_{01}(1-\phi _0)\psi _{1}^*\frac{\pi ^{0*}}{\pi ^{1*}}\big (\alpha _h+\frac{\sigma (1-\delta )}{\mu }\alpha _y\big )\frac{1}{\mu +\sigma }\big (1+\frac{K_0^*}{\mu +\sigma +\eta }\frac{\mu +\eta }{\eta }\big )}{\bigg [1+\frac{K_0^*}{\mu +\sigma +\eta }\big (\frac{\mu +\eta }{\mu }+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )\bigg ]\bigg [1+\frac{K_1^*}{\mu +\sigma }\frac{\mu +\eta }{\eta }\big (1+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )\bigg ]}\right]K_1^*\\&= M_{00}({\varvec{K}}^*)K_0^*+M_{01}({\varvec{K}}^*)K_1^*. \end{aligned}$$

Here $M_{00}({\varvec{K}}^*)$ is the first term in the large square brackets multiplying $K_0^*, M_{01}({\varvec{K}}^*)$ is the second term in the large square brackets multiplying $K_1^*$, and ${\varvec{K}}^*=({\varvec{K}}_1^*, {\varvec{K}}_2^*)$. Similarly

$$\begin{aligned} K_1^*&= \left[\frac{\big (\lambda _1s_{10}(1-\phi _1)\psi _{0}^*\frac{\pi ^{1*}}{\pi ^{0*}}\big (\alpha _h+\frac{\sigma (1-\delta )}{\mu +\eta }\alpha _y\big )+\lambda _1s_{11}(1-\phi _1)\psi _{1}^*\frac{\eta \sigma (1-\delta )\alpha _y}{\mu (\mu +\eta )}\big )\frac{1}{\mu +\sigma +\eta }}{1+\frac{K_0^*}{\mu +\sigma +\eta }\big (\frac{\mu +\eta }{\mu }+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )}\right.\\&\left.+\frac{\lambda _1s_{11}(1-\phi _1)\psi _{1}^*\big (\alpha _h+\frac{\sigma (1-\delta )}{\mu }\alpha _y\big )\frac{\eta }{\mu +\sigma }\frac{1}{\mu +\sigma +\eta }}{\bigg [1+\frac{K_0^*}{\mu +\sigma +\eta }\big (\frac{\mu +\eta }{\mu }+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )\bigg ]\bigg [1+\frac{K_1^*}{\mu +\sigma }\frac{\mu +\eta }{\eta }\big (1+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )\bigg ]}\right]K_0^*\\&+\left[\frac{\lambda _1s_{11}(1-\phi _1)\psi _{1}^*\big (\alpha _h+\frac{\sigma (1-\delta )}{\mu }\alpha _y\big )\frac{1}{\mu +\sigma }\big (1+\frac{K_0^*}{\mu +\sigma +\eta }\frac{\mu +\eta }{\eta }\big )}{\bigg [1+\frac{K_0^*}{\mu +\sigma +\eta }\big (\frac{\mu +\eta }{\mu }+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )\bigg ]\bigg [1+\frac{K_1^*}{\mu +\sigma }\frac{\mu +\eta }{\eta }\big (1+\frac{\sigma (1-\delta )}{\mu }+\frac{\sigma \alpha \delta }{\mu }\big )\bigg ]}\right]K_1^*\\&= M_{10}({\varvec{K}}^*)K_0^*+M_{11}({\varvec{K}}^*)K_1^*. \end{aligned}$$

Similarly to the above $M_{10} ({\varvec{K}}^*)$ is the first term in the large square brackets multiplying $K_0^*$ and $M_{11}({\varvec{K}}^*)$ is the second term in the large square brackets multiplying $K_1^*$.

Hence, it is possible to write this system in the form

$$\begin{aligned} {\varvec{K}}^*= \begin{bmatrix} K_0^*\\ K_1^* \end{bmatrix} = \begin{bmatrix} M_{00}({\varvec{K}}^*)&\quad M_{01}({\varvec{K}}^*)\\ M_{10}({\varvec{K}}^*)&\quad M_{11}({\varvec{K}}^*) \end{bmatrix} \begin{bmatrix} K_0^*\\ K_1^* \end{bmatrix} = \mathbf M ({\varvec{K}}^*){\varvec{K}}^*. \end{aligned}$$

If we examine the $M_{00}({\varvec{K}}^*)$, $M_{01}({\varvec{K}}^*), M_{10}({\varvec{K}}^*)$ and $M_{11}({\varvec{K}}^*)$ terms more closely we can see that each term is greater than or equal to zero and strictly decreasing in $K_0^*$ and $K_1^*$. In addition, we see that

$$\begin{aligned} \mathbf M ({\varvec{0}})^{T} = \begin{bmatrix} M_{00}({\varvec{0}})&\quad M_{10}({\varvec{0}})\\ M_{01}({\varvec{0}})&\quad M_{11}({\varvec{0}}) \end{bmatrix}, \end{aligned}$$

where

$$\begin{aligned} M_{00}({\varvec{0}})&= \dfrac{\lambda _0s_{00}(1-\phi _0)\psi _{0}^*}{(\mu +\sigma +\eta )}\bigg [\alpha _h+\dfrac{\sigma (1-\delta )\alpha _y}{(\mu +\eta )}\bigg ] +\lambda _0s_{01}(1-\phi _0)\psi _{1}^*\dfrac{\pi ^{0*}}{\pi ^{1*}}\\&\times \frac{\eta }{(\mu +\sigma +\eta )}\bigg [\frac{\alpha _h}{\mu +\sigma }+\frac{\sigma (1-\delta )\alpha _y}{\mu (\mu +\sigma )} +\frac{\sigma (1-\delta )\alpha _y}{\mu (\mu +\eta )}\bigg ],\\&= \kappa _{00},\\ M_{01}({\varvec{0}})&= \dfrac{\lambda _0s_{01}(1-\phi _0)\psi _{1}^*}{(\mu +\sigma )}\dfrac{\pi ^{0*}}{\pi ^{1*}}\bigg [\alpha _h+\dfrac{\sigma (1-\delta )\alpha _y}{\mu }\bigg ],\\&= \kappa _{10},\\ M_{10}({\varvec{0}})&= \dfrac{\lambda _1s_{10}(1-\phi _1)\psi _{0}^*}{(\mu +\sigma +\eta )}\dfrac{\pi ^{1*}}{\pi ^{0*}}\bigg [\alpha _h+\dfrac{\sigma (1-\delta )\alpha _y}{\mu +\eta }\bigg ]\\&+\,\lambda _1s_{11}(1-\phi _1)\psi _{1}^*\dfrac{\eta }{(\mu +\sigma +\eta )} \bigg [\frac{\alpha _h}{\mu +\sigma }+\frac{\sigma (1-\delta )\alpha _y}{\mu (\mu +\eta )} +\frac{\sigma (1-\delta )\alpha _y}{\mu (\mu +\sigma )}\bigg ],\\&= \kappa _{01}, \end{aligned}$$

and

$$\begin{aligned} M_{11}({\varvec{0}})&= \dfrac{\lambda _1s_{11}(1-\phi _1)\psi _{1}^*}{(\mu +\sigma )}\bigg [\alpha _h+\dfrac{\sigma (1-\delta )\alpha _y}{\mu }\bigg ],\\&= \kappa _{11}. \end{aligned}$$