The Dynamics of the Epidemic Process with the Antibiotic-Resistant Variant of the Pathogen Agent

Abstract

The parasite–host system is considered with two different variants of a pathogen agent with complete cross protection, where one of the variants may transform into another with a certain probability. This system corresponds to the epidemic process initiated by the antibiotic-resistant variant of the pathogen agent. The behavior of solutions is studied. It is proved that the main case has a unique nontrivial steady-state solution that is a global attractor. The speed of the exponential approach of small deviations to the steady-state solution is obtained.

APPENDIX

Proof of Theorem 5. Since at ${{I}_{1}}(0) \ne 0$ we assume that ${{I}_{1}}(t) > 0,{\text{ }}{{I}_{2}}(t) > 0$ for positive times, then, without loss of generality, we assume that ${{I}_{2}}(0)$ > 0. In addition, without loss of generality, we assume that $S(t) > 0,{\text{ }}t \geqslant 0$.

If ${{I}_{1}}(t) \to 0$, then the solutions of system (1) tend to the solutions of a system with zero ${{I}_{1}}(t)$. However, this is the parasite–host system with the single variant of the pathogen agent. If ${{R}_{2}} > 1$, then the solutions of the system tend to the steady-state solution with $S(t) \to {1 \mathord{\left/ {\vphantom {1 {{{R}_{{{\kern 1pt} 2}}}}}} \right. \kern-0em} {{{R}_{{{\kern 1pt} 2}}}}}$, and if ${{R}_{{{\kern 1pt} 2}}} \leqslant 1$, then they tend to the trivial steady-state solution with $S(t) \to 1$. However, if $S(t)$ is fairly close to ${1 \mathord{\left/ {\vphantom {1 {{{R}_{{{\kern 1pt} 2}}}}}} \right. \kern-0em} {{{R}_{{{\kern 1pt} 2}}}}}$ in the first case, then it follows from ${{R}_{1}} > {{R}_{2}}$ that ${{\alpha }_{1}}S - {{\beta }_{1}} - m > 0$, and ${{I}_{1}}(t)$ is an increasing function that can tend to zero. In the second case, this statement follows from ${{R}_{1}} > 1$. Consequently, ${{I}_{1}}(t)$ does not tend to zero, and there is an $\varepsilon > 0$ and an increasing sequence of time moments ${{T}_{k}},k = 1,...,\infty$ such that ${{I}_{1}}({{T}_{k}}) > \varepsilon ,{\text{ }}{{T}_{k}} \to \infty$.

Then it follows from the second equation of system (1) that ${{I}_{2}}({{T}_{k}}) > {{\varepsilon }_{1}},{\text{ }}{{T}_{k}} \to \infty ,\,\,{\text{and}}\,\,{{\varepsilon }_{1}} > 0$. Dividing the first equation of system (1) by ${{I}_{1}}$ and integrating it over time we obtain

$${{\alpha }_{1}}\int\limits_0^{{{T}_{k}}} S (t)dt - ({{\beta }_{1}} + m){{T}_{k}} = \ln {{I}_{1}}({{T}_{k}}) - \ln {{I}_{1}}(1),$$

from which we get

$${{\alpha }_{1}}\int\limits_0^{{{T}_{k}}} S (t)dt - {{\alpha }_{1}}{{T}_{k}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} = \ln {{I}_{1}}({{T}_{k}}) - \ln {{I}_{1}}(1)\,\,{\text{or}}\,\,\left| {\int\limits_0^{{{T}_{k}}} S (t)dt - {{T}_{k}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} } \right| \leqslant \frac{{\left| {\ln {{I}_{1}}(0) - \ln \varepsilon } \right|}}{{{{\alpha }_{1}}}}.$$

(A1)

Adding all three equations of system (1) and integrating we obtain

$$\left( {{{I}_{1}}({{T}_{k}}) + {{I}_{2}}({{T}_{k}}) + S({{T}_{k}})} \right) - \left( {{{I}_{1}}(0) + {{I}_{2}}(0) + S(0)} \right) = \gamma T - {{\beta }_{1}}\int\limits_0^{{{T}_{k}}} {{{I}_{1}}} (t)dt - {{\beta }_{2}}\int\limits_0^{{{T}_{k}}} {{{I}_{2}}} (t)dt - \gamma \int\limits_0^{{{T}_{k}}} S (t)dt.$$

Since the module of the left-hand term does not exceed unity, then

$$\left| {{{\beta }_{1}}\int\limits_0^{{{T}_{k}}} {{{I}_{1}}} (t)dt + {{\beta }_{2}}\int\limits_0^{{{T}_{k}}} {{{I}_{2}}} (t)dt + \gamma \int\limits_0^{{{T}_{k}}} S (t)dt - {{\beta }_{1}}{{T}_{k}}{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}} - {{\beta }_{2}}{{T}_{k}}{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}} - \gamma {{T}_{k}}{{S}_{2}}} \right| \leqslant 1,$$

from which, after taking inequality (A1) into consideration, we obtain that for a constant

(A2)

We denote ${{A}_{k}} = \int_0^{{{T}_{k}}} {{{I}_{1}}} (t)dt - {{T}_{k}}{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}} = \int_0^{{{T}_{k}}} {{{i}_{1}}} (t)dt$. Then it follows from (A2) that the module of ${{A}_{k}} + \frac{{{{\beta }_{2}}}}{{{{\beta }_{1}}}}\int_0^{{{T}_{k}}} {{{i}_{2}}(t)} dt$ is uniformly bounded. Integrating the first equation of system (3) over time, we obtain that the module of the expression

$$\left( {{{\beta }_{1}} + m} \right){{A}_{k}} - {{\alpha }_{1}}\int\limits_0^{{{T}_{k}}} {({{I}_{1}}(t)S(t) - } {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} )dt$$

(A3)

is uniformly bounded.

Since ${{I}_{1}}(t)S(t) = ({{i}_{1}}(t) + {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}})(s(t) + \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} ) = {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} + \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} {{i}_{1}}(t) + {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}{{s}_{1}}(t) + {{i}_{1}}(t){{s}_{1}}(t)$ and $\int_0^{{{T}_{k}}} {{{I}_{1}}(t)} S(t)dt$ = ${{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} {{T}_{k}} + \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} \int_0^{{{T}_{k}}} {{{i}_{1}}(t)} dt + {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}\int_0^{{{T}_{k}}} {s(t)} dt + \int_0^{{{T}_{k}}} {{{i}_{1}}(t)s(t)} dt$, and the second term on the right of equality is $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} {{A}_{k}}$, and it follows from (A1) that the third term is uniformly bounded, then

$$\int\limits_0^{{{T}_{k}}} {{{I}_{1}}(t)} S(t)dt - {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} {{T}_{k}} - \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} {{A}_{k}} - \int\limits_0^{{{T}_{k}}} {{{i}_{1}}(t)s(t)} dt,$$

or

$$\int\limits_0^{{{T}_{k}}} {{{I}_{1}}(t)} S(t)dt - {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} {{T}_{k}} - {{{{A}_{k}}} \mathord{\left/ {\vphantom {{{{A}_{k}}} {{{R}_{{{\kern 1pt} 1}}}}}} \right. \kern-0em} {{{R}_{{{\kern 1pt} 1}}}}} - \int\limits_0^{{{T}_{k}}} {{{i}_{1}}(t)s(t)} dt$$

is uniformly bounded. Dividing expression (A3) by α₁ we obtain that $\frac{{{{\beta }_{1}} + m}}{{{{\alpha }_{1}}}}{{A}_{k}}$ –$\int_0^{{{T}_{k}}} {({{I}_{1}}(t)S(t) - {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} )dt}$ = $\frac{{{{A}_{k}}}}{{{{R}_{1}}}} - \int_0^{{{T}_{k}}} {({{I}_{1}}(t)S(t) - {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} )dt}$ is uniformly bounded. However, these are the first three terms of the obtained expression; consequently, the fourth term is also uniformly bounded and

(A4)

We add the first two equations of system (1), integrate, and obtain that the ${{\alpha }_{1}}\int_0^{{{T}_{k}}} {({{I}_{1}}(t)S(t) - } {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} )dt - {{\alpha }_{2}}\int_0^{{{T}_{k}}} {({{I}_{2}}(t)S(t) - {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} )dt}$ module is uniformly bounded, or, taking (A2) into consideration, that the module of

$$\int\limits_0^{{{T}_{k}}} {({{I}_{2}}(t)S(t) - } {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} )dt - \frac{{{{\beta }_{1}} + m}}{{{{\alpha }_{2}}}}{{A}_{k}}$$

(A5)

is uniformly bounded. Then, from

$$\int\limits_0^{{{T}_{k}}} {{{I}_{2}}(t)} S(t)dt = {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{2}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} {{T}_{k}} + \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} \int\limits_0^{{{T}_{k}}} {{{i}_{2}}(t)} dt + {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{2}}\int\limits_0^{{{T}_{k}}} {s(t)} dt + \int\limits_0^{{{T}_{k}}} {{{i}_{2}}(t)s(t)} dt,$$

taking into account that the third term of the expansion is uniformly bounded, $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} \int_0^{{{T}_{k}}} {{{i}_{2}}(t)} dt = \frac{1}{{{{R}_{{{\kern 1pt} 1}}}}}\int_0^{{{T}_{k}}} {{{i}_{2}}(t)} dt$, and taking into consideration the uniform boundedness of the module of ${{A}_{k}} - \frac{{{{\beta }_{2}}}}{{{{\beta }_{1}}}}\int_0^{{{T}_{k}}} {{{i}_{2}}(t)} dt$, we obtain that

(A6)

Note that, since ${{R}_{{{\kern 1pt} 1}}} > {{R}_{2}},{\text{ }}{{R}_{{{\kern 1pt} 1}}} > 1$, then $\frac{1}{{{{R}_{1}}}}\frac{{{{\beta }_{1}}}}{{{{\beta }_{2}}}} - \frac{{{{\alpha }_{1}}}}{{{{\alpha }_{2}}}} > 0$. Dividing the third equation of system (1) by $S(t)$, we obtain $\frac{d}{{dt}}\ln S = - ({{\alpha }_{1}}{{I}_{1}} + {{\alpha }_{2}}{{I}_{2}}) + \frac{\gamma }{S} - \gamma$. Since $S(t)$ is a variable within the range from 0 to 1, then from the Taylor expansion we have

$$\frac{1}{{S(t)}} = 1 - S(t) + \theta (S){{S}^{2}}(t),$$

where $\theta (S)$ is a positive function bounded from below by a positive constant.

Relation (A1) shows that the integral of $S(t) - \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S}$ over time is uniformly bounded. However, if the integral of $\int_0^{{{T}_{k}}} {(\theta (S){{S}^{2}}(t) - \theta (S){{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} }}^{2}})dt}$ is also uniformly bounded, then it follows from the uniform boundedness of the derivative of $S(t)$ over time that $S(t) \to \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S}$, from which, taking the first equation into consideration, it follows that ${{I}_{1}}(t) \to {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}}$, and then the same follows for ${{I}_{2}}(t)$.

If however $\int_0^{{{T}_{k}}} {(\theta (S){{S}^{2}}(t) - \theta (S){{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{S} }}^{2}})dt}$ tends to plus infinity, then, from (A5), we obtain that the integral of ${{\alpha }_{1}}{{I}_{1}} + {{\alpha }_{2}}{{I}_{2}} - {{\alpha }_{1}}{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{1}} - {{\alpha }_{2}}{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }_{2}}$ tends to plus infinity with an increase in ${{T}_{k}}$. Then, proceeding from relation (A2) and taking into consideration that ${{R}_{{{\kern 1pt} 1}}} > {{R}_{2}}$, we obtain that ${{A}_{k}} \to + \infty$.

Adding the equations of system (1) we obtain $\frac{d}{{dt}}$(I₁ + I₂ + S) = γ − (β₁I₁ + β₂I₂ + γS), or ${{d({{i}_{1}} + {{i}_{2}} + s)} \mathord{\left/ {\vphantom {{d({{i}_{1}} + {{i}_{2}} + s)} {dt}}} \right. \kern-0em} {dt}} = - {{\beta }_{1}}{{i}_{1}} - {{\beta }_{2}}{{i}_{2}} - \gamma s$. Multiplying by ${{i}_{1}} + {{i}_{2}} + s$ we obtain that

$$\begin{gathered} - \frac{1}{2}\frac{d}{{dt}}{{({{i}_{1}} + {{i}_{2}} + s)}^{2}} = \left( {{{\beta }_{1}}{{i}_{1}} + {{\beta }_{2}}{{i}_{2}} + \gamma s} \right)\left( {{{i}_{1}} + {{i}_{2}} + s} \right) \\ = {{\beta }_{1}}i_{1}^{2} + {{\beta }_{2}}i_{2}^{2} + \gamma {{s}^{2}} + ({{\beta }_{1}} + {{\beta }_{2}}){{i}_{1}}{{i}_{2}} + ({{\beta }_{1}} + \gamma ){{i}_{1}}s + ({{\beta }_{2}} + \gamma ){{i}_{2}}s. \\ \end{gathered} $$

Let us integrate over time. Since the integral on the left-hand side is uniformly bounded, then the integral on the right-hand side is also uniformly bounded.

It follows from relation (A4) that the integral of ${{i}_{1}}s$ is finite, and from relation (A6), it follows that the integral of ${{i}_{2}}s$ is either finite or tends to plus infinity. If the integral of ${{\beta }_{1}}i_{2}^{2} + {{\beta }_{2}}i_{2}^{2} + \gamma {{s}^{2}}$ is finite, then, due to the uniform boundedness of the derivatives, it follows that the solution tends to the steady-state form. Hence, all we have to do is to examine the case

$$\int\limits_0^{{{T}_{k}}} {{{i}_{1}}(t){{i}_{2}}(t)} dt \to - \infty .$$

(A7)

Dividing the second equation of system (1) by ${{I}_{2}}$, we obtain $\frac{d}{{dt}}\ln {{I}_{2}} = {{\alpha }_{2}}S - {{\beta }_{2}}$ + ${{m{{I}_{1}}} \mathord{\left/ {\vphantom {{m{{I}_{1}}} {{{I}_{2}}}}} \right. \kern-0em} {{{I}_{2}}}}$. Let us integrate it over time to ${{T}_{k}}$. Since ${{I}_{2}}({{T}_{k}}) > {{\varepsilon }_{1}},$${{T}_{k}} \to \infty ,$${{\varepsilon }_{1}} > 0$, the left-hand side of the equality is uniformly bounded. It follows from (5) that ${{\alpha }_{2}}\int_0^{{{T}_{k}}} {Sdt} - {{\beta }_{2}}{{T}_{k}} + m{{T}_{k}}\frac{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}$ is uniformly bounded; thus, $\int_0^{{{T}_{k}}} {\left( {\frac{{{{I}_{1}}}}{{{{I}_{2}}}} - \frac{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right)dt}$ should also be bounded. However, we have from the Taylor expansion of the function $f(1 + x) = \frac{1}{{1 + x}}$ for $x > - 1$ that $f(1 + x) = \frac{1}{{1 + x}} = 1 - x - h(x){{x}^{2}}$, where h(x) is a positive function. Therefore,

$$\frac{{{{I}_{1}}}}{{{{I}_{2}}}} = \frac{{{{I}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}(1 + {{{{I}_{2}}} \mathord{\left/ {\vphantom {{{{I}_{2}}} {{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right. \kern-0em} {{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}})}} = \frac{{{{I}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}\left( {1 - \frac{{{{I}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}} + h\left( {\frac{{{{I}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right){{{\left( {\frac{{{{I}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right)}}^{2}}} \right) = \frac{{{{I}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}\left( {1 - \frac{{{{I}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right) + \frac{{{{I}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}h\left( {\frac{{{{I}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right){{\left( {\frac{{{{I}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right)}^{2}}.$$

The integral of the last term $\frac{{{{I}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}h\left( {\frac{{{{I}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right){{\left( {\frac{{{{I}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right)}^{2}}$ is nonnegative because it is a nonnegative expression.

$$\frac{{{{I}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}\left( {1 - \frac{{{{I}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right) = \frac{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}} + {{i}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}\left( {1 - \frac{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}} + {{i}_{2}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}} \right) = - \frac{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}} + {{i}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}{{i}_{2}} = - \frac{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}{{i}_{2}} - \frac{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}{{i}_{1}}{{i}_{2}}.$$

However, as (A2) shows, ${{A}_{k}} + \frac{{{{\beta }_{2}}}}{{{{\beta }_{1}}}}\int_0^{{{T}_{k}}} {{{i}_{2}}(t)} dt$ is uniformly bounded; since ${{A}_{k}}$ is either uniformly bounded or ${{A}_{k}} \to + \infty$, then the integral of $- \frac{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}{{i}_{2}}$ is either uniformly bounded or tends to $+ \infty$. Relation (A7) shows that the integral of $- \frac{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{1}}}}{{{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{I} }}_{2}}}}{{i}_{1}}{{i}_{2}}$ tends to $+ \infty$.

Therefore, the right-hand side of the equation ${{d(\ln {{I}_{2}})} \mathord{\left/ {\vphantom {{d(\ln {{I}_{2}})} {dt}}} \right. \kern-0em} {dt}} = {{\alpha }_{2}}S - {{\beta }_{2}} + {{m{{I}_{1}}} \mathord{\left/ {\vphantom {{m{{I}_{1}}} {{{I}_{2}}}}} \right. \kern-0em} {{{I}_{2}}}}$, when integrated to ${{T}_{k}}$, tends to $+ \infty$. We have come to a contradiction; consequently, the solution tends to the steady-state form, which completes the proof.