Global stability of a diffusive and delayed HBV infection model with HBV DNA-containing capsids and general incidence rate

Ting Guo; Haihong Liu; Chenglin Xu; Fang Yan

doi:10.3934/dcdsb.2018134

Article Contents

2018, Volume 23, Issue 10: 4223-4242. Doi: 10.3934/dcdsb.2018134

This issue Previous Article Carleman estimate for solutions to a degenerate convection-diffusion equation Next Article Upper and lower bounds for the blow-up time in quasilinear reaction diffusion problems

Global stability of a diffusive and delayed HBV infection model with HBV DNA-containing capsids and general incidence rate

1.
Department of Mathematics, Yunnan Normal University, Kunming 650092, China
1.
School of Science, Nanjing University of Science and Technology, Nanjing 210094, China

* Corresponding author: Haihong Liu
* Corresponding author: Haihong Liu

Received: August 31, 2017

Revised: November 30, 2017

Early access: April 2018

Published: September 2018

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Abstract

The aim of this paper is to study the dynamics of a new chronic HBV infection model that includes spatial diffusion, three time delays and a general incidence function. First, we analyze the well-posedness of the initial value problem of the model in the bounded domain. Then, we define a threshold parameter $R_{0}$ called the basic reproduction number and show that our model admits two possible equilibria, namely the infection-free equilibrium $E_{1}$ as well as the chronic infection equilibrium $E_{2}$. Further, by constructing two appropriate Lyapunov functionals, we prove that $E_{1}$ is globally asymptotically stable when $R_{0}<1$, corresponding to the viruses are cleared and the disease dies out; if $R_{0}>1$, then $E_{1}$ becomes unstable and the equilibrium point $E_{2}$ appears and is globally asymptotically stable, which means that the viruses persist in the host and the infection becomes chronic. An application is provided to confirm the main theoretical results. Additionally, it is worth saying that, our results suggest theoretically useful method to control HBV infection and these results can be applied to a variety of possible incidence functions presented in a series of other papers.

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Mathematics Subject Classification: Primary: 92D30, 35K57; Secondary: 34K20.

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Figure 1. Schematic view of the replication process of HBV

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Figure 2. Diagrammatic representation of the mathematical model for HBV infection

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Figure 3. The numerical approximations of system (21)-(23) with parameters $s = 2.6\times10^{7}$, $\mu = 0.01$, $\delta = 0.053$, $a = 150$, $\beta = 0.87$, $c = 3.8$, $d_{v} = 0.01$, $b_{1} = b_{2} = 0.01$, $\alpha_{1} = 0.2$, $\alpha_{2} = 0.28$, $\alpha_{3} = 0.1$, $\tau_{1} = 10$, $\tau_{2} = 0$, $\tau_{3} = 0$ and $k = 3\times10^{-5}$, showing that solution trajectories converge to the infection-free equilibrium $E_{1}: (H_{1}, I_{1}, D_{1}, V_{1}) = (2.6\times10^{9}, 0, 0, 0)$

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Figure 4. The numerical approximations of system (21)-(23) with parameters $s = 2.6\times10^{7}$, $\mu = 0.01$, $\delta = 0.053$, $a = 150$, $\beta = 0.87$, $c = 3.8$, $d_{v} = 0.01$, $b_{1} = b_{2} = 0.01$, $\alpha_{1} = 0.2$, $\alpha_{2} = 0.28$, $\alpha_{3} = 0.1$, $\tau_{1} = 10$, $\tau_{2} = 0$, $\tau_{3} = 0$ and $k = 1.67\times10^{-4}$, showing that solution trajectories converge to the chronic infection equilibrium $E_{2}: (H_{2}, I_{2}, D_{2}, V_{2}) = (1.61\times10^{9}, 2.53\times10^{7}, 4.12\times10^{9}, 9.43\times10^{8})$

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Figure 5. The graphs of the basic reproduction number $R_{0}$ in terms of some parameters: (a) $R_{0}$ in terms of $\alpha_{1}$ and $\alpha_{2}$, (b) $R_{0}$ as a function of $\alpha_{1}$ and $\alpha_{3}$, and (c) $R_{0}$ in terms of $\alpha_{2}$ and $\alpha_{3}$. Here, $s = 2.6\times10^{7}$, $\mu = 0.01$, $\delta = 0.053$, $a = 150$, $\beta = 0.87$, $c = 3.8$, $b_{1} = 0.01$, $\tau_{1} = 5.8$, $\tau_{2} = 6$, $\tau_{3} = 4$ and $k = 2.4\times10^{-3}$

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Figure 6. The plots of the basic reproduction number $R_{0}$ as a function of three delays $\tau_{1}$, $\tau_{2}$ and $\tau_{3}$. Here, $s = 2.6\times10^{7}$, $\mu = 0.01$, $\delta = 0.053$, $a = 150$, $\beta = 0.87$, $c = 3.8$, $b_{1} = 0.01$, $\alpha_{1} = 0.2$, $\alpha_{2} = 0.28$, $\alpha_{3} = 0.1$ and $k = 2.4\times10^{-3}$. (a) $\tau_{3} = 4$, (b) $\tau_{2} = 6$, and (c) $\tau_{1} = 5.8$

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