Ternary compound ontology matching for cognitive green computing

1.   Introduction

Middle East respiratory syndrome (MERS) is a viral respiratory disease caused by Middle East respiratory syndrome coronavirus (MERS-CoV). The intermediate host of MERS-CoV is probably the dromedary camel, a zoonotic virus ^[1]. Most MERS cases are acquired by human-to-human transmission. There is no vaccine or specific treatment available, and approximately $35\%$ of patients with MERS-CoV infection have died ^[2]. There has been extensive works on infectious disease models and viral infection models associated with MERS that can help in disease control and provide strategies for potential drug treatments ^{[3,4,5,6,7,8]}.

Dipeptidyl peptidase-4 (DPP4) plays an important role in viral infection ^[2]. Based on classic viral infection models developed in ^[9,10,11], a four-dimensional ordinary differential equation model is proposed and studied in ^[8]. The model in ^[8] describes the interaction mechanisms among uninfected cells, infected cells, DPP4 and MERS-CoV.

Recently, taking into account periodic factors such as diurnal temperature differences and periodic drug treatment, the model in ^[8] has been further extended a periodic case in ^[12], and then the existence of positive periodic solutions is studied by using the theorem in ^[13].

It is well-known that CTL immune responses play a very critical role in controlling viral load and the concentration of infected cells. Thus, many scholars have considered CTL immune responses in various viral infection models and have achieved many excellent research results ^{[14,15,16,17,18]}. CTL cells can kill virus-infected cells and are important for the control and clearance of MERS-CoV infections ^[19]. Inspired by the above research works, we consider the following periodic MERS-CoV infection model with CTL immune response:

In model (1.1), $T(t)$, $I(t)$, $v(t)$, $D(t)$ and $Z(t)$ represent the concentrations of uninfected cells, infected cells, free virus, DPP4 on the surface of uninfected cells and CTL cells at time $t$, respectively. CTL cells increase at a rate bilinear rate $q(t)I(t)Z(t)$ by the viral antigen of the infected cells and decay at rate $b(t)Z(t)$; infected cells are killed by the CTL immune response at rate $p(t)I(t)Z(t)$. Except for $p(t)$, $q(t)$ and $b(t)$, all the remaining parameters of model (1.1) have the same biological meanings as in ^[12].

Throughout the paper, it is assumed that the functions $\lambda(t)$, $\beta(t)$, $d(t)$, $d_{1}(t)$, $p(t)$, $M(t)$, $c(t)$, $\lambda_{1}(t)$, $\gamma(t)$, $q(t)$ and $b(t)$ are positive, continuous and $\omega$ periodic ($\omega > 0$); the function $\beta_{1}(t)$ is non-negative, continuous and $\omega$ periodic.

From point of view in both biology and mathematics, it is one of the most significant topics to study the existence of periodic oscillations of a system (see, for example, ^{[12,20,21,22,23,24,25,26]} and the references therein).

In the next section, some sufficient criteria are given for the existence of positive periodic oscillations of model (1.1). It should be mentioned here that, in the proofs of the main results in the following section, a new technique is developed to obtain a lower bound of the state variable $Z(t)$ characterizing CTL immune response in model (1.1).

2.   Main results

For some function $f(t)$ which is continuous and $\omega$-periodic on $\mathbb{R}$, let us define the following notations:

Moreover, for convenience, let us give the following parameters:

The following theorem is the main result of this paper.

Theorem 2.1. If $R^{*} > 1$ and $\omega < \omega^{*}$, then model (1.1) has at least one positive $\omega$-periodic solution.

Proof. Making the change of variables $T(t) = \exp\{u_1(t)\}$, $I(t) = \exp\{u_2(t)\}$, $v(t) = \exp\{u_3(t)\}$, $D(t) = \exp\{u_4(t)\}$, $Z(t) = \exp\{u_5(t)\}$, then model (1.1) can be rewritten as

Thus, we only need to consider model (2.1).

Let us set

with the norm

It can be shown that $X$ and $Y$ are Banach spaces. Define

here ${\rm{Dom}}\; L = \{u\in X, \; \dot{u}\in X\}$. It easily has that ${\rm{Ker}}\; L = \{u\in X\; |\; u\; \in \mathbb{R}^5\}$ and ${\rm{Im}}\; L = \{u\in Y\; |\; \int_{0}^{\omega}u(t)dt = 0\}$. Further, it is clear that ${\rm{Im}}\; L$ is closed in $Y$ and ${\rm{dim\; Ker}}\; L = {\rm{codim\; Im}}\; L = 5$. Hence, $L$ is a Fredholm mapping with index zero.

For $\mu\in(0, 1)$, let us consider the equation $Lu = \mu Nu$, i.e.,

For any solution $u = (u_1(t), u_2(t), u_3(t), u_4(t), u_5(t))^{T}\in X$ of (2.2), it has

From the first two equations in (2.2), it has

and

Hence, by integrating the above two equations on $[0, \omega]$, it has

and

Note that $I(t): = \exp\{u_{2}(t)\}$ satisfies

Then, from (2.4) and (2.5), it has

From the third and the fifth equations of (2.2), it has

Note that $u\in X$, there exist $\xi_{i}, \; \eta_{i}\in[0, \omega]$ $(i = 1, 2, 3, 4, 5)$, such that

From (2.2), $\dot{u}_{1}(\eta_{1}) = 0$ and $\dot{u}_{4}(\eta_{4}) = 0$, it has

which imply that

From the last equation of (2.3), it has

which implies that

Then, from (2.6) and $\omega < \omega^{*}$, it has

Thus, it has

From the third equation of (2.3), it has

which implies that

Then, from (2.7), it has

From the second equation of (2.3), it has

which implies that

Then, from (2.7), it has

From $\dot{u}_1(\xi_1) = 0$, $\dot{u}_4(\xi_4) = 0$, (2.8) and (2.10), it has

Thus, it has

Let us give an estimate of the lower bound of the state variable $u_5(t)$ related to CTL immune response. It should be mentioned here that a completely different method from that in ^[12] has been used.

Claim A If $R^{*} > 1$ and $\omega < \omega^{*}$, then

If Claim A is not true, then it has that, for any $t$, $\exp\{u_5(t)\} \leq \exp\{u_5(\eta_{5})\} < \delta^{*}$. Hence, it has from (2.3), (2.9)–(2.11) that

which implies that

Adding (2.4) and (2.5) together, it has

which implies that

which is a contradiction to (2.12). Thus, the claim holds.

From Claim A and (2.7), it has

Now, for convenience, let us define

Note that if $R^{*} > 1$, then it has

which implies that

Let $(u_{1}, u_{2}, u_{3}, u_{4}, u_{5})^{T}\in \mathbb{R}^{5}$ be the solution of the following equations:

Define $\Gamma : [0, Z_{max}] \rightarrow \mathbb{R}$, via

where

Equation (2.14) can be rewritten as

It is obvious that if there is a solution $(u_{1}, u_{2}, u_{3}, u_{4}, u_{5})^{T}\in \mathbb{R}^{5}$ for (2.14), it must have $0 < \exp\{u_5\} < Z_{max}$. In addition, note that $\Gamma(x)$ is monotonically decreasing with respect to $x$ on $[0, Z_{max}]$. It has from $\Gamma\left(Z_{max}\right) = -\widehat{d_{1}}-\widehat{p}Z_{max} < 0$ and

that there exists a unique positive constant $x = Z^{*}\in(0, Z_{max})$ such that $\Gamma(Z^{*}) = 0$.

The above discussions show that, if $R^{*} > 1$, (2.14) has a unique solution $(u_{1}^{*}, u_{2}^{*}, u_{3}^{*}, u_{4}^{*}, u_{5}^{*})^{T}$, here $u_i^* = \ln(e_{i})$ $(i = 1, 2, 3, 4, 5)$,

Let us define the following set

Moreover, by similar arguments as in ^[12], it has that $N$ is $L$-compact on $\overline{\Omega}$.

Now, let us compute the Leray-Schauder degree ${\rm{deg}}\left\{QN, \partial\Omega\cap Ker\; L, (0, 0, 0, 0, 0)^{T}\right\} : = \Delta$ as follows,

where $\widehat{\lambda} = \widehat{\beta}e_{1}e_{3}e_{4}+\widehat{d}e_{1}$ and $\widehat{\lambda_{1}} = \widehat{(\beta_{1}\beta)}e_{1}e_{3}e_{4}+\widehat{\gamma}e_{4}$ are used.

Finally, it has those all the conditions of the continuation theorem in ^[13] (also see, for example, Lemma 2.1 in ^[12]) are satisfied. This proves that, if $\omega < \omega^{*}$ and $R^{*} > 1$, model (2.1) has at least one $\omega$-periodic solution.

Let us consider the following classical viral infection dynamic model ^[9] with CTL immune response:

where, all the coefficients are the same with that in model (1.1).

Define $R_{1} : [0, \omega^{*}] \rightarrow \mathbb{R}$, via

Obviously, $R_1(x)$ is monotonically decreasing on $[0, \omega^{*}]$ and

Therefore, if $R_{1}(0) > 1$, then there exists a unique constant $\omega^{**}\in(0, \omega^{*})$ such that $R_{1}(\omega^{**}) = 1$, $R_{1}(x) > 1$ for $0\leq x < \omega^{**}$ and $R_{1}(x) < 1$ for $\omega^{**} < x\leq\omega^{*}$.

For model (A), it is not difficult to derive the following result.

Theorem 2.2. If $R_{1}(\omega) > 1$ and $\omega < \omega^{*}$ (i.e. $R_{1}(0) > 1$ and $\omega < \omega^{**} < \omega^{*}$), then model (A) has at least one positive $\omega$-periodic solution.

Remark 2.1. If all the coefficients in model (A) take constants values, i.e., $\lambda(t)\equiv\lambda > 0$, $\beta(t)\equiv\beta > 0$, $d(t)\equiv d > 0$, $d_{1}(t)\equiv d_{1} > 0$, $p(t)\equiv p > 0$, $M(t)\equiv M > 0$, $c(t)\equiv c > 0$, $q(t)\equiv q > 0$ and $b(t)\equiv b > 0$, then model (A) becomes the classical model which is first proposed by Nowak and Bangham in ^[9]. the condition $\omega < \omega^{**}$ in Theorem 2.2 is naturally satisfied. Furthermore, it has $R_{1}(0) = (\lambda\beta M q)/(dcq+\beta d_{1}Mb): = R_{1}$. From \emph{^[9]}, it has that the condition $R_{1} > 1$ implies the existence of a unique positive equilibrium. This shows that the conditions and conclusion in Theorem 2.2 are reasonable.

3.   Conclusions and simulations

In summary, Theorem 2.1 in the paper successfully extends the main result in ^[12]) to a MERS-CoV viral infection model with CTL immune response. In the proof of Theorem 2.1, we use a very different method from that in ^[9] to obtain the lower bound $(\ln(\delta^{*})-2\widehat{b}\omega)$ of the state variable $u_5(t)$. Furthermore, as a special case, Theorem 2.2 gives sufficient conditions for the existence of positive periodic solution of model (A). Model (A) is a natural extension of the classical model in ^[9]. As the end of the paper, let us give a example to summarize the applications of Theorem 2.1. Let us choose the coefficients in model (1.1) as follows (for the values of some parameters, please refer to ^[7,27] for the case of some autonomous models), $\lambda(t) = 45(1+0.1\sin(4\pi t))$, $\beta(t) = 1.4\times 10^{-8}(1+0.1\cos(4\pi t))$, $d(t) = 0.001(1+0.5\cos(4\pi t))$, $d_{1}(t) = 0.056(1+0.5\cos(4\pi t))$, $p(t) = 0.00092(1+0.5\cos(4\pi t))$, $M(t) = 100000$, $c(t) = 2.1(1+0.3\cos(4\pi t))$, $\lambda_{1}(t) = 10(1+0.1\sin(4\pi t))$, $\beta_{1}(t) = 0.001$, $\gamma(t) = 0.01(1+0.1\cos(4\pi t))$, $q(t) = 0.005(1+0.5\sin(4\pi t))$, $b(t) = 0.5(1+0.4\cos(4\pi t))$. Then, with the help of Maple mathematical software, it has $\omega = 0.5 < \omega^{*}\approx1.111111$, $M_{1}\approx11.502875$, $M_2\approx4.976734$, $M_3\approx14.965318$, $M_{4}\approx7.108426$, $M_{5}\approx18.976215$, $L_{1}\approx-0.383569$, $L_{2} = 4.007333$, $L_{3}\approx9.795917$, $L_{4}\approx0.623402$, $L_{5}\approx1.387881$, $R^{*}\approx1.2170332 > 1$. From Theorem 2.1, it follows that model (1.1) has at least one positive $\omega$ ($\omega = 0.5$)-periodic solution. Figure 1 gives the corresponding numerical simulation, and the initial value is chosen as $(T(0), I(0), v(0), D(0), Z(0))^T = (12.5,100, 265000,995.4,423)^{T}$.

Acknowledgments

This paper is supported by National Natural Science Foundation of China (No.11971055) and Beijing Natural Science Foundation (No.1202019).

Conflict of interest

The authors declare there is no conflict of interest.