An optimal control problem applied to plasmid-mediated antibiotic resistance

Abstract

We formulate a system of ordinary differential equations to model the contribution of antibiotic treatment and immune system to combat bacterial infections. We obtained threshold conditions that determine when the bacteria can be eliminated, which are consistent with biological phenomena. In order to minimize the bacterial population, we formulated an optimal control problem considering the action of both antibiotic and immune system. The optimal control is obtained applying the Pontryagin’s Principle. The results show the relevance of the synergism between the antibiotic treatment and the immune system response.

Appendix 1

Lemma 2

If $R_1<R_{\max }$, then the solution $R_2$ defined in (23), of the equation $\varPhi (R)=0$ satisfies $R_2\in (0,R_{\max })$.

Proof

Substituting $R_{max}$ defined in (19) in the inequality $-R_{max}+R_1 < 0$ we obtain

$$\begin{aligned} -\frac{(R_s-1)K}{R_s\left( 1+\frac{\delta \sigma _pK}{\beta _s\mu _p}\right) }+R_1<0. \end{aligned}$$

(49)

Inequality (49) is equivalent to

$$\begin{aligned} \frac{(R_s-1)K}{R_s}\left( \frac{\delta \sigma _pK}{\mu _p\beta _r}-1\right) -\frac{(R_s-1)K}{R_s\left( 1+\frac{\delta \sigma _pK}{\beta _s\mu _p}\right) }+R_1 -\frac{(R_s-1)K}{R_s}\left( \frac{\delta \sigma _pK}{\mu _p\beta _r}-1\right) <0. \end{aligned}$$

From above inequality we obtain

$$\begin{aligned}&\frac{(R_s-1)K}{R_s}\left( \frac{\delta \sigma _pK}{\mu _p\beta _r}-1\right) -\frac{(R_s-1)K}{R_s\left( 1+\frac{\delta \sigma _pK}{\beta _s\mu _p}\right) }\\&\quad \left[ 1+\left( \dfrac{\delta \sigma _pK}{\mu _p\beta _r}-1\right) \left( 1+\dfrac{\delta \sigma _pK}{\beta _s\mu _p}\right) \right] +R_1<0. \end{aligned}$$

Adding the term

$$\begin{aligned} -\frac{qK}{\beta _r}\left( 1+\frac{\delta \sigma _pK}{\mu _p\beta _s}\right) , \end{aligned}$$

in both sides of the above inequality we obtain

$$\begin{aligned}&-\left[ 1+\left( \frac{\delta \sigma _pK}{\mu _p\beta _r}-1\right) \left( 1+\frac{\delta \sigma _pK}{\beta _s\mu _p}\right) \right] \frac{(R_s-1)K}{R_s\left( 1+\frac{\delta \sigma _pK}{\beta _s\mu _p}\right) } +R_1\\&\qquad \quad +\frac{(R_s-1)K}{R_s}\left( \frac{\delta \sigma _pK}{\mu _p\beta _r}-1\right) -\frac{qK}{\beta _r}\left( 1+\frac{\delta \sigma _pK}{\mu _p\beta _s}\right) <-\frac{qK}{\beta _r}\left( 1+\frac{\delta \sigma _pK}{\mu _p\beta _s}\right) , \end{aligned}$$

or equivalently

$$\begin{aligned} -h_2R_{max}+h_1<-\frac{qK}{\beta _r}\left( 1+\frac{\delta \sigma _pK}{\mu _p\beta _s}\right) . \end{aligned}$$

(50)

Oh the other hand, we observe that

$$\begin{aligned} \frac{qK}{\beta _r}\left( 1+\frac{\delta \sigma _pK}{\mu _p\beta _s}\right) = \dfrac{\frac{qK}{\beta _r}\frac{(R_s-1)K}{R_s}}{\frac{(R_s-1)K}{R_s}\frac{1}{1+\frac{\delta \sigma _pK}{\mu _p\beta _s}}}=\frac{h_0}{R_{max}}. \end{aligned}$$

In consequence, substituting above equation in the inequality (50) we obtain

$$\begin{aligned} -h_2R_{max}+h_1<-\frac{h_0}{R_{max}}. \end{aligned}$$

Therefore $\varPhi (R_{max})=-h_2R^2_{max}+h_1R_{max}+h_0<0$. Since $\varPhi (0)=h_0>0$ and $\varPhi (R_{max})<0$, then from the Intermediate Value Theorem we conclude $R_2\in (0,R_{\max })$. $\square$

Lemma 3

There exists a unique function $g:{\mathbb {R}}^7\rightarrow {\mathbb {R}}$ that satisfies $\sigma _p=g(\beta _r,\beta _s,q,\delta ,\mu _s,\mu _r,\mu _p,K)$.

Proof

Let $\varDelta _2(z,\sigma _p)=a_2(z,\sigma _p)a_1(z,\sigma _p)-a_3(z,\sigma _p)$ where $z=(\beta _r,\beta _s,q,\delta ,\mu _s,\mu _r,\mu _p,K)$, then we verify that $\varDelta _2(z_0,\sigma _p^0)=0$, where $z_0$ is the parameter vector corresponding to the parameter $\sigma _p^0$. In addition, from (29) we verified that $\partial \varDelta _2(z_0,\sigma _p^0)/\partial \sigma _p=\partial \varDelta _2(z_0,\sigma ^0_p)/\partial \sigma _p=-\delta S_2\left[ (q+\delta P_2)\frac{S_2}{R_2}+\mu _p\right] \ne 0$. As a consequence, from the implicit function theorem, there is an open ball $U\in {\mathbb {R}}^9$ containing $z_0$ and an interval $V\subset {\mathbb {R}}$ containing $\mu _0$ such that there is a unique function $\sigma _p=g(z)$ defined for $z\in U$ and $\mu \in V$ which satisfies $F(x(\sigma _p^0),\sigma _p^0)=0$, where F is the right side of system (1) [50]. $\square$