# American Institute of Mathematical Sciences

November  2020, 25(11): 4427-4447. doi: 10.3934/dcdsb.2020106

## Kink solitary solutions to a hepatitis C evolution model

 1 Research Group for Mathematical, and Numerical Analysis of Dynamical Systems, Kaunas University of Technology, Studentu 50-147, Kaunas LT-51368, Lithuania 2 Nonlinear Dynamics, Chaos and Complex Systems Group, Departamento de Física, Universidad Rey Juan Carlos, Tulipán s/n, 28933 Móstoles, Madrid, Spain 3 Department of Applied Informatics, Kaunas University of Technology, Studentu 50-407, Kaunas LT-51368, Lithuania 4 Institute for Physical Science and Technology, University of Maryland, College Park, Maryland 20742, USA 5 Department of Software Engineering, Kaunas University of Technology, Studentu 50-415, Kaunas LT-51368, Lithuania

Received  July 2017 Published  March 2020

The standard nonlinear hepatitis C evolution model described in (Reluga et al. 2009) is considered in this paper. The generalized differential operator technique is used to construct analytical kink solitary solutions to the governing equations coupled with multiplicative and diffusive terms. Conditions for the existence of kink solitary solutions are derived. It appears that kink solitary solutions are either in a linear or in a hyperbolic relationship. Thus, a large perturbation in the population of hepatitis infected cells does not necessarily lead to a large change in uninfected cells. Computational experiments are used to illustrate the evolution of transient solitary solutions in the hepatitis C model.

Citation: Tadas Telksnys, Zenonas Navickas, Miguel A. F. Sanjuán, Romas Marcinkevicius, Minvydas Ragulskis. Kink solitary solutions to a hepatitis C evolution model. Discrete & Continuous Dynamical Systems - B, 2020, 25 (11) : 4427-4447. doi: 10.3934/dcdsb.2020106
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Kink solutions $\widehat{x}, \widehat{y}$ to (112) with $\widehat{c} = 1$. The black line denotes $\widehat{x}\left(t\right)$; the gray line denotes $\widehat{y}\left(t\right)$. In (a), $u = 10, v = -4$; in (b), $u = -2, v = 0$
Kink solutions $x, y$ to (112) with $c = 0$. The black line denotes $x\left(\tau\right)$; the gray line denotes $y\left(\tau\right)$. In (a), $u = 10, v = -4$; in (b), $u = -2, v = 0$
Phase plot of (112). Black lines denote kink solution trajectories. The gray circle denotes the unstable node (110). The gray dashed line denotes the equilibrium line (111). Gray arrows denote the direction field. The dotted line illustrates that perturbations in infected cell population $y$ lead to proportional changes in uninfected cell population $x$. As the solution evolves from point $A$ to $B$, $y$ increases by $0.46$, while $x$ decreases by $1.09$
Plot of error (128) for $c = 0, u = 5, v = 1$. Conditions (126) and (127) hold true. The step size $h$ is $10^{-4}$; error is estimated over $N = 100$ steps. Errors higher than 10 are truncated to 10 for clarity. Note that the error is almost zero on the curve defined by (125)
Plot of error (128) for $c = 0, u = 5, v = 1$. Conditions (125) and (127) hold true. The step size $h$ is $10^{-3}$; error is estimated over $N = 30$ steps. Errors higher than 2 are truncated to 2 for clarity. Note that the error is almost zero on the line defined by (126)
Plot of error (128) for $c = 0, u = 5, v = 1$. Conditions (125), (126) hold true. The step size $h$ is $10^{-3}$; error is estimated over $N = 30$ steps. Errors higher than 100 are truncated to 100 for clarity. Note that the error is almost zero on the hyperbola defined by (127)
Kink solutions to (132) with $\widehat{c} = 1$. The black line denotes $\widehat{x}\left(t\right)$; the gray line denotes $\widehat{y}\left(t\right)$. In (a), $u = 4, v = 1$; in (b), $u = -5, v = 2$
Kink solutions to (131) with $c = 0$. The black line denotes $x\left(\tau\right)$; the gray line denotes $y\left(\tau\right)$. In (a), $u = 4, v = 1$; in (b), $u = -5, v = 2$
Phase plot of (131). Black lines denote kink solution trajectories. The gray diamond denotes the saddle point (129). The gray dashed line denotes the equilibrium line (130). Gray arrows denote direction field. The dotted line illustrates that large perturbations in infected cell population $y$ lead to small changes in uninfected cell population $x$. As the solution evolves from point $A$ to $B$, $y$ decreases by $5.19$, while $x$ increases by $0.39$
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