Lagrange tracking-based long-term drift trajectory prediction method for Autonomous Underwater Vehicle

1.   Introduction

HIV spreads through either cell-free viral infection or direct transmission from infected to healthy cells (cell-to-cell infection) ^[1]. It is reported that more than 50$\%$ of viral infection is caused by the cell-to-cell infection ^[2]. Cell-to-cell infection can occur when infected cells encounter healthy cells and form viral synapse ^[3]. Recently, applying reaction-diffusion equations to model viral dynamics with cell-to-cell infection have been received attentions (see, e.g., ^[4,5,6]). Ren et al. ^[5] proposed the following reaction-diffusion equation model with cell-to-cell infection:

In the model (1.1), $T(t, \, x)$, $T^*(t, \, x)$ and $V(t, \, x)$ denote densities of healthy cells, infected cells and virus at time $t$ and location $x$, respectively. The detailed biological meanings of parameters for the model (1.1) can be found in ^[5]. The well-posedness of the classical solutions for the model (1.1) have been studied. The model (1.1) admits a basic reproduction number $R_0$, which is defined by the spectral radius of the next generation operator ^[5]. The model (1.1) defines a solution semiflow $\varPsi(t)$, which has a global attractor. The model (1.1) admits a unique virus-free steady state $E_0 = (T_0(x), \, 0, \, 0)$, which is globally attractive if $R_0 < 1$. If $R_0 > 1$, the model (1.1) admits at least one infection steady state and virus is uniformly persistent ^[5].

It is a challenging problem to consider the global stability of $E_0$ in the critical case of $R_0 = 1$. In ^[6], Wang et al. studied global stability analysis in the critical case by establishing Lyapunov functions. Unfortunately, the method can not be applied for the model consisting of two or more equations with diffusion terms, which was left it as an open problem.

Adopting the idea in ^[7,8,9], the present study is devoted to solving this open problem and shows that $E_0$ is globally asymptotically stable when $R_0 = 1$ for the model (1.2). For simplicity, in the following, we assume that the diffusion rates $d_1(x)$, $d_2(x)$ and $d_3(x)$ are positive constants. That is, we consider the following model

with the boundary conditions:

where $\Omega$ is the spatial domain and $\nu$ is the outward normal to $\partial\Omega$. We assume that all the location-dependent parameters are continuous, strictly positive and uniformly bounded functions on $\overline{\Omega}$.

2.   Main result

Let $\mathbb{Y} = C\left(\overline{\Omega}, \, \mathbb{R}^3\right)$ with the supremum norm $\parallel\cdot\parallel_{\mathbb{Y}}$, $\mathbb{Y}^+ = C\left(\overline{\Omega}, \, \mathbb{R}^3_{+}\right)$. Then $(\mathbb{Y}, \, \mathbb{Y}^+)$ is an ordered Banach space. Let $\mathcal{T}$ be the semigroup for the system:

where $T_0(x)$ is the solution of the elliptic problem $d_1\Delta T+\lambda(x)-d(x)T = 0$ under the boundary conditions (1.3). Then $\mathcal{T}$ has the generator

Let us define the exponential growth bound of $\mathcal{T}$ as

and define the spectral bound of $\widetilde{A}$ by

Theorem 2.1. If $R_0 = 1$, $E_0$ of the model (1.2) is globally asymptotically stable.

Proof. We first show the local asymptotic stability of $E_0$ of the model (1.2). Suppose $\zeta > 0$ and let $v_0 = (T^0, \, T^*_0, \, V_0)$ with $\left\Vert v_0-E_0 \right\Vert\leq \zeta$. Define $m_1(t, \, x) = \frac{T(t, \, x)}{T_0(x)}-1\ \text{and}\ p(t) = \max_{x\in \overline{\Omega}}\left\lbrace m_1(t, \, x), \, 0 \right\rbrace.$ According to $d_1\Delta T_0(x)+\lambda(x)-d(x)T_0(x) = 0$, we have

Let $\widetilde{T}_1(t)$ be the positive semigroup generated by

associated with (1.3) (see Theorem 4.4.3 in ^[10]). From Theorem 4.4.3 in ^[10], we can find $q > 0$ such that $\left\Vert \widetilde{T}_1(t) \right\Vert\leq M_1e^{-qt}$ for some $M_1 > 0$. Hence, one gets

where $m_{10} = \frac{T^0}{T_0(x)}-1$. In view of the positivity of $\widetilde{T}_1(t)$, it follows that

where $T_m = \min_{x\in \overline{\Omega}}\lbrace T_0(x) \rbrace.$ Note that $(T^*, \, V)$ satisfies

It then follows that

From Theorem 3.5 in ^[11], we have that $s\left(\widetilde{A}\right) = \sup\left\lbrace Re \lambda, \, \lambda\in \sigma \left(\widetilde{A}\right) \right\rbrace$ has the same sign as $R_0-1$. If $R_0 = 1$, then $s\left(\widetilde{A}\right) = 0$. Then we easily verify all the conditions of Proposition 4.15 in ^[12]. It follows from $R_0 = 1$ and Proposition 4.15 in ^[12] that we can find $M_1 > 0$ such that $\left\Vert \mathcal{T}(t)\right\Vert\leq M_1$ for $t\geq 0$, where $M_1$ can be chosen as large as needed in the sequel. Since $p(s)\leq \frac{\zeta M_1e^{-qs}}{T_m}$, one gets

where

By using Gronwall's inequality, we get

Then $\frac{\partial T}{\partial t}-d_1\Delta T > \lambda(x)-d(x)T-M_1\zeta e^{\frac{\zeta M_2}{q}}\left(\beta_1(x)+\beta_2(x) \right) T.$ Let $\widehat{u}_1$ be the solution of the system:

Then $T(t, \, x)\geq \widehat{u}_1(t, \, x)$ for $x\in \overline{\Omega}$ and $t\geq 0$. Let $T_{\zeta}(x)$ be the positive steady state of the model (2.1) and $\widehat{m}(t, \, x) = \widehat{u}_1(t, \, x)-T_{\zeta}(x)$. Then $\widehat{m}(t, \, x)$ satisfies

For sufficiently large $M_1$, from Theorem 4.4.3 in ^[10], we have $\Vert F_1(t)\Vert\leq M_1e^{\overline{\alpha}_0t}$, where $\overline{\alpha}_0 < 0$ is a constant and $F_1(t):\, C\left(\overline{\Omega}, \, \mathbb{R}\right)\rightarrow C\left(\overline{\Omega}, \, \mathbb{R}\right)$ is the $C_0$ semigroup of $d_1\Delta -d(\cdotp)$ subject to (1.3) ^[5]. Hence, we have

Let $\mathbf{K} = M_1^2\zeta e^{\frac{\zeta M_2}{q}}\left(\Vert\beta_1\Vert+\Vert\beta_2\Vert \right)$. By employing Gronwall's inequality, one gets

Choosing $\zeta > 0$ sufficiently small such that $\mathbf{K} < -\frac{\overline{\alpha}_0}{2}$, then

and

Since $p(t)\leq \frac{\zeta M_1}{T_m}$, one gets $T(\cdot, \, t)-T_0(x)\leq \zeta M_1\frac{\Vert T_0(x) \Vert}{T_m},$ and hence

From

by choosing $\zeta$ small enough, for $t > 0$, there holds $\Vert T(\cdot, \, t)-T_0(x)\Vert, \ \left\Vert T^*(\cdot, \, t)\right\Vert, \ \left\Vert V(\cdot, \, t)\right\Vert\leq \varepsilon,$ which implies the local asymptotic stability of $E_0$.

By Theorem 1 in ^[5], the solution semiflow $\varPsi(t): \mathbb{Y}^+\rightarrow \mathbb{Y}^+$ of the model (1.2) has a global attractor $\Pi$. In the following, we prove the global attractivity of $E_0$. Define

Claim 1. For $v_0 = (T^0, \, T^*_0, \, V_0)\in \Pi$, the omega limit set $\omega(v_0)\subset \partial \mathbb{Y}_1$.

Since $\frac{\partial T}{\partial t}\leq d_1\Delta T+\lambda(x)-d(x)T$, $T$ is a subsolution of the problem

It is well known that model (2.2) has a unique positive steady state $T_0(x)$, which is globally attractive. This together with the comparison theorem implies that

uniformly for $x\in \Omega.$ Since $v_0 = (T^0, \, T^*_0, \, V_0)\in \Pi$, we know $T^0\leq T_0$. If $T^*_0 = V_0 = 0$, the claim easily holds. We assume that either $T^*_0\neq0$ or $V_0\neq0$. Thus one gets $T^*(t, \, x) > 0$ and $V(t, \, x) > 0$ for $x\in \overline{\Omega}$ and $t > 0$. Then $T(t, \, x)$ satisfies

The comparison principle yields $T(t, \, x) < T_0(x)$ for $x\in \overline{\Omega}$ and $t > 0$. Following ^[7], we introduce

Then $h(t, \, v_0) > 0$ for $t > 0$. We show that $h(t, \, v_0)$ is strictly decreasing. To this end, we fix $t_0 > 0$, and let $\overline{T^*}(\cdot, \, t) = h(t_0, \, v_0)\phi_2$ and $\overline{V}(\cdot, \, t) = h(t_0, \, v_0)\phi_3$ for $t\geq t_0$. Due to $T(\cdot, \, t) < T_0(x)$, one gets

Hence $\left(\overline{T^*}(t, \, x), \, \overline{V}(t, \, x)\right)\geq (T^*(t, \, x), \, V(t, \, x))$ for $x\in \overline{\Omega}$ and $t\geq t_0$. From the model (2.3), one gets $h(t_0, \, v_0)\phi_2(x) = \overline{T^*}(t, \, x) > T^*(t, \, x)$ for $x\in \overline{\Omega}$ and $t > t_0$. Similarly, we get $h(t_0, \, v_0)\phi_3(x) = \overline{V}(t, \, x) > V(t, \, x)$ for $x\in \overline{\Omega}$ and $t > t_0$. Since $t_0 > 0$ is arbitrary, $h(t, \, v_0)$ is strictly decreasing. Let $h_{*} = \mathop {\lim }\limits_{t \to +\infty }h(t, \, v_0)$. Then we have $h_{*} = 0$. Let $\mathcal{Q} = (Q_1, \, Q_2, \, Q_3)\in \omega(v_0)$. Then there is $\left\lbrace t_k \right\rbrace$ with $t_k\rightarrow +\infty$ such that $\varPsi(t_k)v_0\rightarrow \mathcal{Q}$. We get $h(t, \, \mathcal{Q}) = h_{*}$ for $t\geq 0$ due to $\mathop {\lim }\limits_{t \to +\infty }\varPsi(t+t_k)v_0 = \varPsi(t)\mathop {\lim }\limits_{t \to +\infty }\varPsi(t_k)v_0 = \varPsi(t)\mathcal{Q}$. If $Q_2\neq 0$ and $Q_3\neq 0$, we repeat the above discussions to illustrate that $h(t, \, \mathcal{Q})$ is strictly decreasing, which contradicts to $h(t, \, \mathcal{Q}) = h_{*}$. Thus, we have $Q_2 = Q_3 = 0$.

Claim 2. $\Pi = \left\lbrace E_0 \right\rbrace$.

Since $\left\lbrace E_0 \right\rbrace$ is globally attractive in $\partial \mathbb{Y}_1$, $\left\lbrace E_0 \right\rbrace$ is the only compact invariant subset of the model (1.2). From the invariance of $\omega(v_0)$ and $\omega(v_0)\subset \partial \mathbb{Y}_1$, one gets $\omega(v_0) = \left\lbrace E_0\right\rbrace$. By Lemma 3.11 in ^[9], we get $\Pi = \left\lbrace E_0 \right\rbrace$.

The local asymptotic stability and global attractivity yield the global asymptotic stability of $E_0$.

Acknowledgments

The research is supported by the NNSF of China (11901360) to W. Wang and supported by the Fundamental Research Funds for the Central Universities and the Research Funds of Renmin University of China (2019030196) to X. Lai. We also want to thank the anonymous referees for their careful reading that helped us to improve the manuscript.

Conflict of interest

The authors decare no conflicts of interest.