Bang-bang control optimization in infectious disease model with incorporating breakthrough and reinfection

Chen, Ya; Jing, Wenjun; Zhang, Juping; Qin, Peng

doi:10.1038/s41598-026-44921-7

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Published: 31 March 2026

Bang-bang control optimization in infectious disease model with incorporating breakthrough and reinfection

Ya Chen¹,
Wenjun Jing^2,3,
Juping Zhang^3,4,5 &
…
Peng Qin⁶

Scientific Reports , Article number: (2026) Cite this article

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Abstract

Breakthrough infections and reinfection are key factors leading to recurrent epidemic waves. However, sustained control strategies can lead to unnecessary resource wastage when tacking these issues. There is an urgent need to establish dynamic intervention systems capable of rapid response and efficient resource utilization. To address the question of how breakthrough infections and reinfections affect the dynamics of the pandemic, this study develops an infectious disease model that incorporates both breakthrough infections and reinfections, and while introduces bang-bang optimal control as an efficient public health intervention strategy to provide a new perspective and solutions. In theoretical analysis, we derive basic reproduction number via next-generation matrix method, prove the global stability of the disease-free equilibrium, and establish sufficient conditions for the existence of multiple endemic equilibria and the occurrence of backward bifurcation. Numerical simulations further confirm the critical role of breakthrough infections and reinfection in disease persistence and recurrent outbreaks. In control strategy research, we prove the existence of bang-bang optimal solutions based on optimal control theory and demonstrate their distinct advantages in rapidly outbreaks while minimizing operational costs. Simulation results show that a combined strategy implemented under the bang-bang control—reducing transmission rates, expanding vaccine coverage, and enhancing vaccine protection—most effectively contains disease spread. This results provide both theoretical foundation and practical guidance for developing efficient control strategies against recurrent infectious disease outbreaks.

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Data availability

All data generated or analysed during this study are included in this published article and its supplementary information files.

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Acknowledgements

We would like to thank the reviewers for their valuable comments and suggestions which helped us to improve the presentation of this paper significantly.

Funding

This work was supported by the National Natural Science Foundation of China (No. 12471462, No. 12101373, No. 12171291), Fundamental Research Program of Shanxi Province (No. 202403021211154).

Author information

Authors and Affiliations

School of Mathematics, Xi’an University of Finance and Economics, Xi’an, 710100, Shaanxi, China
Ya Chen
School of Statistics, Shanxi University of Finance and Economics, Taiyuan, 030006, Shanxi, China
Wenjun Jing
Complex Systems Research Center, Shanxi University, Taiyuan, 030006, Shanxi, China
Wenjun Jing & Juping Zhang
Shanxi Key Laboratory for Mathematical Technology in Complex Systems, Shanxi University, Taiyuan, 030006, Shanxi, China
Juping Zhang
Key Laboratory of Complex Systems and Data Science of Ministry of Education, Shanxi University, Taiyuan, 030006, Shanxi, China
Juping Zhang
School of Electrical and Control Engineering, North University of China, Taiyuan, 030006, Shanxi, China
Peng Qin

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Ya Chen
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Wenjun Jing
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Juping Zhang
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Contributions

Y. C.: Data curation, Formal analysis, Validation,Writing – original draft, Writing – review & editing. W. J.: Data curation, Formal analysis, Visualization, Writing – original draft, Writing – review & editing. J. Z.: Conceptualization, Investigation, Methodology, Supervision, Writing – review & editing P. Q.: Conceptualization, Data curation, Software, Supervision, Writing – review & editing

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Correspondence to Peng Qin.

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Appendix

The proof of Lemma 1

Proof

Summing the four equations of system (1) up yields $\frac{dN}{dt}=0$. Thus the total population N(t) is a constant. For convenience, let $N(t)=K$. Next, we prove that the solution of system (1) is non-negative.

Solving to the third equation of system (1) yields

$$\begin{aligned} I(t)= I(0)e^{-\int _{0}^{t} \left( -\frac{(S(x)+\sigma V(x)+ \epsilon R(x))\beta }{N}+\eta \right) dx}\ge 0. \end{aligned}$$

Plugging it into the fourth equation of system (1), we can get

$$\begin{aligned} R(t)=\left[ R(0)+\int _{0}^{t}e^{\int _{0}^{x} \left( \frac{\beta \epsilon I(y)}{N}+\gamma \right) dy}\eta I(x)dx\right] e^{-\int _{0}^{t} \left( \frac{\beta \epsilon I(x)}{N}+\gamma \right) dx}\ge 0, \end{aligned}$$

Besides, sloving the first equation of system (1), we obtion

$$\begin{aligned} S(t)=\left[ S(0)+\int _{0}^{t} \left( e^{\int _{0}^{x} \left( \frac{\beta I(y)}{N}+\alpha \right) dy}(\theta V(x)+\gamma R(x))\right) dx\right] e^{-\int _{0}^{t} \left( \frac{\beta I(x)}{N}+\alpha \right) dx}, \end{aligned}$$

Let $t_1>0$ be the first time that $\text {min}\left\{ S(t_1),V(t_1)\right\} =0$. Assuming that $S(t_1)=0$, then S(t) and $V(t)>0$, for$\forall t\in [0,t_1)$. Furthermore we can get

$$\begin{aligned} S(t_1)=\left[ S(0)+\int _{0}^{t_1} \left( e^{\int _{0}^{x} \left( \frac{\beta I(y)}{N}+\alpha \right) dy}(\theta V(x)+\gamma R(x))\right) dx\right] e^{-\int _{0}^{t} \left( \frac{\beta I(x)}{N}+\alpha \right) dx}>0, \end{aligned}$$

this contradicts the assumption $S(t_1)=0$. Similarly, let us assume that $V(t_1)=0$, then according to the second equation of system (1), we can get

$$\begin{aligned} V(t_1) =\left[ V(0)+\int _{0}^{t_1} \left( e^{\int _{0}^{x} \left( \frac{\sigma \beta I}{N}+\theta \right) dx}\alpha S(x)\right) dx\right] e^{-\int _{0}^{t} \left( \frac{\sigma \beta I}{N}+\theta \right) dx}>0, \end{aligned}$$

which contradicts the assumption of $V(t_1)=0$. This mean that all solution of system (1) are non-negative and $\Gamma$ is the positively invariant set of system (1). This completes the proof. $\square$

The proof of Theorem 1

Proof

The Jacobian matrix of system (2) at the disease-free equilibrium $P_0$ is

$$\begin{aligned} K(P_0) = \left( \begin{array}{ccc} -\alpha -\gamma & \theta -\gamma & -\frac{\beta \theta }{\alpha +\theta }-\gamma \\ \alpha & -\theta & - \frac{\sigma \beta \alpha }{\alpha +\theta } \\ 0 & 0 & \frac{\beta \left( \sigma \alpha +\theta \right) }{\alpha +\theta }-\eta \\ \end{array} \right) . \end{aligned}$$

The characteristic polynomial corresponding to matrix $K(P_0)$ is

$$\begin{aligned} \lambda ^3+a_2\lambda ^2+a_1\lambda +a_0=0, \end{aligned}$$

(8)

where

$$\begin{aligned} a_2&=\frac{\alpha ^2+(-\sigma \beta +\eta +\gamma +2\theta )\alpha +\theta (\eta -\beta +\gamma +\theta )}{\alpha +\theta }, \end{aligned}$$

(9)

$$\begin{aligned} a_1&=\frac{\alpha ^2(-\sigma \beta +\eta +\gamma )+\alpha \left[ \theta \left( 2\gamma -(\sigma +1)\beta +2\eta \right) +\gamma (-\sigma \beta +\eta )\right] +\theta \left[ (\eta -\beta +\gamma )\theta +\gamma (\eta -\beta )\right] }{\alpha +\theta }, \end{aligned}$$

(10)

$$\begin{aligned} a_0&=\gamma \left[ (-\sigma \beta +\eta )\alpha +\theta (\eta -\beta )\right] . \end{aligned}$$

(11)

The roots of the characteristic polynomial (8) are $\lambda _1=-\gamma$, $\lambda _2=-\alpha -\theta , \lambda _3=\frac{\beta (\sigma \alpha +\theta )-\eta (\alpha +\theta )}{\alpha +\theta }$. Obviously, when $R_0<1$, $\lambda _3<0$. Therefore, the disease-free equilibrium $P_0$ of system (2) is locally asymptotically stable, if $R_0<1$.

When $R_0=1$, $\lambda _3=0$. At this time, the central manifold theorem is used to prove the stability of the disease-free equilibrium $P_0$ . The following affine transformation is defined

$$\begin{aligned} \left( \begin{array}{c} s \\ v \\ i\end{array}\right) =\left( \begin{array}{c} \frac{\theta }{\alpha +\theta } \\ \frac{\alpha }{\alpha +\theta }\\ 0\end{array}\right) +\left( \begin{array}{ccc} \frac{\theta -\gamma }{\alpha } & -1 & \frac{(-\eta -\gamma )\theta ^2-\sigma (\eta +\gamma )\alpha \theta +\alpha \eta \gamma \sigma }{(\alpha +\theta )(\alpha \sigma +\theta )\gamma }\\ 1& 1 & -\frac{\alpha \left[ (\eta +\gamma )\sigma \alpha +(\eta \sigma +\theta )\gamma +\theta \eta \right] }{(\alpha +\theta )(\alpha \sigma +\theta )\gamma }\\ 0 & 0 & 1 \end{array}\right) \left( \begin{array}{c} \bar{s}\\ \bar{v}\\ \bar{i}\end{array}\right) . \end{aligned}$$

(12)

Furthermore,

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{d\bar{s}}{dt} =-\gamma \bar{s}+G_{11}\bar{s}\bar{i}+G_{12}\bar{v}\bar{i}+O\left( |\bar{s},\bar{v},\bar{i}|^2\right) ,\\ \frac{d\bar{v}}{dt} =-(\alpha +\theta )\bar{v}+G_{21}\bar{s}\bar{i}+G_{22}\bar{v}\bar{i}+O\left( |\bar{s},\bar{v},\bar{i}|^2\right) \\ \frac{d\bar{i}}{dt} =\frac{(1-R_*)\eta (\alpha +\theta )(\alpha \sigma +\theta )\epsilon }{(\alpha \sigma +\theta )^2\gamma }\bar{i}^2+G_{31}\bar{s}\bar{i}+G_{32}\bar{v}\bar{i},\\ \end{array}\right. } \end{aligned}$$

(13)

the linear part here is the Jordan canonical form of the matrix K, and

$$\begin{aligned} \begin{aligned} G_{11}&=\frac{\left\{ [(\sigma -\epsilon )\eta -\gamma \epsilon ]\alpha +[(\epsilon -1)\eta +\gamma \epsilon ](\gamma -\theta )\right\} (\alpha +\theta )\eta }{\left( \alpha \sigma +\theta \right) \gamma \left( \alpha -\gamma +\theta \right) }, G_{12}=\frac{(\sigma -1)(\alpha +\theta )\alpha \eta ^2}{(\alpha \sigma +\theta )\gamma (\alpha -\gamma +\theta )}, \\ G_{21}&=\frac{-\eta }{(\alpha -\gamma +\theta )(\alpha \sigma +\theta )^2}\left\{ {(\sigma -1)\theta ^3+[\alpha \sigma ^2+((\epsilon -1)\eta -\gamma )\sigma +(2-\epsilon )\gamma -\alpha +(1-\epsilon )\eta ]\theta ^2}\right. \\&\quad \left. {+[-\alpha (\eta -\alpha +\gamma )\sigma ^2+(((2-\epsilon )\alpha -2(\epsilon -1)\eta )\gamma +\alpha (-\alpha +\eta (\epsilon +1)))\sigma +(\epsilon -1)\gamma ^2}\right. \\&\quad \left. {+(\epsilon -1)(\eta -\alpha )\gamma -\alpha \eta \epsilon ]\theta +\sigma \gamma (\eta \alpha \sigma +(\epsilon -1)(\eta +\alpha )\gamma +(1-\epsilon )\alpha ^2-\alpha \eta \epsilon )\gamma }\right\} , \\ G_{22}&=\frac{-\eta }{(\alpha -\gamma +\theta )(\alpha \sigma +\theta )^2}\left\{ {\sigma \alpha ^3+[(\sigma ^2+\sigma +1)\theta -\gamma \sigma ]\alpha ^2+[(\sigma ^2+\sigma +1)\theta ^2}\right. \\&\quad \left. {+((-\eta -\gamma )\sigma ^2+2\eta \sigma -\eta -\gamma )\theta +\eta \gamma \sigma (\sigma -1)]\alpha -\theta ^2\sigma (\gamma -\theta )}\right\} , \\ G_{31}&=\frac{(\alpha +\theta )\left[ (\sigma -\epsilon )\alpha +(\epsilon -1) (\gamma -\theta )\right] \eta }{\alpha (\alpha \sigma +\theta )}, G_{32}=\frac{(\sigma -1)(\alpha +\theta )\eta }{\alpha \sigma +\theta }. \end{aligned} \end{aligned}$$

Let $R_*=\frac{[(\alpha \sigma +\theta ) (\gamma +\eta )+\gamma \eta \sigma ](\alpha \sigma +\theta )}{\eta (\alpha +\theta ) [(\alpha \sigma +\theta )\epsilon +\gamma \sigma ]}$. Then, if $R_0\ne R_*$,there exists a one-dimensional central manifold

$$\begin{aligned} \bar{s}&=\frac{\eta ^2\alpha \left\{ \eta (1-R_*)(\alpha +\theta )\left[ (\alpha \sigma +\theta )\epsilon +\gamma \sigma \right] +\gamma \theta \epsilon (\alpha \sigma +\alpha +\theta ) \right\} }{(\alpha -\gamma +\theta )(\alpha \sigma +\theta )^2\gamma ^3} \bar{i}^2+O\left( |\bar{i}|^3\right) , \\ \bar{v}&=\left\{ \frac{\left[ \theta (\sigma -1)(\alpha +\theta -\gamma )+\eta \theta (1-\sigma )+\gamma \eta \sigma \right] (R_*-1)\left[ (\alpha +\theta )\sigma +\gamma \sigma \right] }{(\alpha \sigma +\theta )^3(\alpha +\theta )(\alpha +\theta -\gamma )\gamma }+\frac{\eta \sigma ^2}{(\alpha \sigma +\theta )^3}\right\} \bar{i}^2\\&\quad +\frac{\eta \epsilon (\sigma \theta -\gamma \sigma -\theta )}{(\alpha \sigma +\theta )^2(\alpha +\theta -\gamma )}\bar{i}^2 +O\left( |\bar{i}|^3\right) , \end{aligned}$$

Therefore, system (2) is simplified to

$$\begin{aligned} \begin{aligned} \frac{d\bar{i}}{dt}&=-\frac{\eta \left[ (\eta +\gamma )(\alpha \sigma +\theta )^2+\eta \alpha \gamma (\sigma ^2-\sigma )-\eta (\alpha +\theta )(\alpha \sigma +\theta )\epsilon \right] }{(\alpha \sigma +\theta )^2\gamma }\bar{i}^2+O\left( |\bar{i}|^3\right) \\&=\frac{\eta ^2(1-R_*)(\alpha +\theta )\left[ (\alpha \sigma +\theta )\epsilon +\gamma \sigma \right] }{(\alpha \sigma +\theta )^2\gamma } \bar{i}^2+O\left( |\bar{i}|^3\right) . \end{aligned} \end{aligned}$$

It is to see that disease-free equilibrium $P_0$ is a saddle node. When $R_0=R_*$, according to the center manifold theorem, the center manifold of system (2) at the far point can be expressed as

$$\begin{aligned} \begin{aligned} \frac{d\bar{i}}{dt}&=-\frac{\eta }{\gamma (\alpha \sigma +\theta )^4}\left\{ \left[ \eta (\alpha +\theta ) (\alpha \sigma +\theta )\epsilon +\eta \gamma \sigma ^2\alpha \right] (\alpha \sigma +\theta )^2 +\gamma \alpha \eta ^2\sigma ^2(1-\sigma )(\alpha +\theta )\right\} \bar{i}^3\\&\quad +O\left( |\bar{i}|^4\right) .\\ \end{aligned} \end{aligned}$$

Since $\sigma , \epsilon \in [0,1]$, the disease-free equilibrium $P_0$ is a stable node, when $R_0=1$ and $R_0=R_*$. $\square$

The proof of Theorem 2

Proof

The Jacobian matrix of system (2) at the equilibrium $P_2$ is

$$\begin{aligned} K(P_2) = \left( \begin{array}{ccc} -\beta i_2-\alpha -\gamma & \theta -\gamma & -\frac{\beta (\beta i_{2} \sigma +\theta )\gamma (1-i_{2})}{\beta ^2i_{2}^2\sigma +\beta (\alpha \sigma +\gamma \sigma +\theta )i_{2}+\gamma (\alpha +\theta )}-\gamma \\ \alpha & -\sigma \beta i_2-\theta & -\frac{ \beta \sigma \alpha \gamma (1-i_{2})}{\beta ^2i_{2}^2\sigma +\beta (\alpha \sigma +\gamma \sigma +\theta )i_{2}+\gamma (\alpha +\theta )}\\ \beta (1-\epsilon )i_2 & \beta (\sigma -\epsilon )i_2 & \frac{\beta \left[ (1-\epsilon )(\beta i_{2} \sigma +\theta )+(\sigma -\epsilon )\alpha \right] \gamma (1-i_{2})}{\beta ^2i_{2}^2\sigma +\beta (\alpha \sigma +\gamma \sigma +\theta )i_{2}+\gamma (\alpha +\theta )}+\beta \epsilon (1-2i_2)-\eta \\ \end{array} \right) . \end{aligned}$$

The characteristic equation of $K(P_2)$ is

$$\begin{aligned} \Lambda ^3+d_2\Lambda ^2+d_1\Lambda +d_0=0, \end{aligned}$$

(14)

where

$$\begin{aligned} d_2&=\frac{1}{h}\left\{ {b_3 i_2^3+\beta ^3\sigma i_2^3(\sigma +1)+\beta ^2 i_2^2\left( \sigma ^2\alpha +\gamma \sigma \epsilon +\sigma ^2\gamma +2\gamma \sigma +\theta \right) +2\beta ^2 i_2^2\sigma (\alpha +\theta )+\beta ^2i_2^2\epsilon (\alpha \sigma +\theta )}\right. \\&\quad \left. {+\gamma \beta i_2(\alpha \sigma +\theta )+i_2\beta \left[ \alpha \sigma +\theta +\gamma \left( 2\sigma +\epsilon +1\right) \right] (\alpha +\theta )+\beta \gamma ^2\sigma i_2+\gamma \left( \alpha +\theta \right) \left( \alpha +\gamma +\theta \right) }\right\} \\&\quad +f(i_2), \\ d_1&=\frac{1}{h}\big \{\big [(1+\sigma )b_3+\beta ^3\sigma ^2\big ]\beta i_2^4+\beta ^2 i_2^3\big [(\epsilon +2)(\alpha +\gamma )\sigma ^2+(2\alpha +\gamma +2\theta ) \epsilon \sigma +(2\sigma +\epsilon )\theta \big ]\\&\quad +\beta ^2i_2^2\big [\epsilon (\alpha +\theta )(\alpha \sigma +2\gamma \sigma +\gamma +\theta )+\theta ^2+2\sigma (\alpha +2\gamma )\theta +3\alpha \gamma \sigma +(\alpha ^2+\alpha \gamma +\gamma ^2)\\&\quad \sigma ^2+\gamma \sigma \beta (1-\epsilon )\big ]+\beta i_2\gamma \big [\beta \theta (1-\epsilon )-\alpha \beta \epsilon \sigma +\alpha \beta \sigma ^2+\alpha ^2\epsilon +2\alpha ^2\sigma +2\alpha \epsilon \theta +2\alpha \gamma \sigma \\&\quad +2\alpha \sigma \theta +\theta ^2\epsilon +2\gamma \sigma \theta +\alpha \gamma +2\alpha \theta +\gamma \theta +2\theta ^2\big ] +\gamma ^2(\alpha +\theta )^2\big \}, \\ d_0&=\frac{1}{h}\left[ \left( \alpha \gamma +\gamma \theta -\beta ^2\sigma i_2\right) ^2f(i_2)+\left( \beta \theta i_2^2+\alpha \gamma i_2+\gamma \theta i_2+\beta \gamma \sigma i_2^2+\alpha \beta \sigma i_2^2+\beta ^2\sigma i_2^3\right) g(i_2)\right] , \\ h_2&=\beta ^2 i_2^2 \sigma +\beta (\alpha \sigma +\gamma \sigma +\theta )i_2+\gamma (\alpha +\theta ). \end{aligned}$$

Obviously, $d_2>0$, $f(i_2)=0$. According to $R_0<1$ and $R_*<R_0$, we have that $g(i_2)<0$, $d_0<0$. Let the three roots of Eq. (14) be $\Lambda _1$, $\Lambda _2$ and $\Lambda _3$, respectively. According to Vieta’s theorem, it has

$$\begin{aligned}&\Lambda _1\Lambda _2\Lambda _3=-d_0>0, \end{aligned}$$

(15)

$$\begin{aligned}&\Lambda _1+\Lambda _2+\Lambda _3=-d_2<0. \end{aligned}$$

(16)

Therefore, according to expression (15), the roots of Eq. (14) include the following three cases: three positive roots, one positive root and a pair of conjugate complex roots, one positive root and two negative roots. According to expression (16), Eq. (14) has a pair of conjugate complex roots or negative roots with negative real parts. In summary, there are two cases of the roots of Eq. (14): one positive root and a pair of conjugate complex roots, one positive root and two negative roots. Therefore, the endemic equilibrium $P_2$ is unstable. $\square$

The proof of Theorem 3

Proof

The Jacobian matrix of system (2) at the endemic equilibrium $P_3$ is

$$\begin{aligned} K(P_3) = \left( \begin{array}{ccc} -\beta i_3-\alpha -\gamma & \theta -\gamma & -\frac{\gamma \left[ \gamma (\alpha +\theta )+\beta \theta \right] }{\beta \theta i_3+\gamma (\alpha +\theta )} \\ \alpha & -\theta & 0\\ \beta (1-\epsilon )i_3 & -\beta \epsilon i_3 & \frac{-2\beta ^2\epsilon \theta i_3^2-\left[ (\gamma -\beta )\epsilon \theta +(\eta +\gamma ) \theta +\alpha \gamma \epsilon \right] \beta i_3-\gamma \left[ \alpha \eta +(\eta -\beta )\theta \right] }{(\beta i_3+\gamma )\theta +\alpha \gamma }\\ \end{array} \right) , \end{aligned}$$

The characteristic equation of $K(P_3)$ is

$$\begin{aligned} \Lambda ^3+e_2\Lambda ^2+e_1\Lambda +e_0=0, \end{aligned}$$

(17)

where

$$\begin{aligned} e_2&=\frac{1}{h_3}\big \{\beta ^2\theta i_3^2\left( 1+\epsilon \right) +\beta i_3\left[ (\epsilon +2)\theta \gamma +\alpha (1+\epsilon )\gamma +\theta (\alpha +\theta )\right] +\gamma \left( \alpha +\theta \right) (\alpha +\gamma +\theta )\big \}\\&\quad +F(i_3), \\ e_1&=\frac{1}{h_3}\big \{\beta ^2 i_3^2\left[ \left( \alpha \gamma +\alpha \theta +\theta ^2\right) \epsilon +\theta ^2\right] +\beta \gamma i_3 \big [\alpha ^2\epsilon +2(1+\epsilon )\theta \alpha +(\eta +\gamma ) \alpha +\theta ^2(2+\epsilon )\big ]\\&\quad +\gamma ^2(\alpha +\theta )^2\big \}+(\alpha +\gamma +\theta )F(i_3) \left( \beta i_3^2+\gamma \i _3\right) G(i_3),\\ \\ e_0&=\frac{1}{h_3}\left[ \left( \alpha \gamma +\gamma \theta \right) ^2F(i_3)+\left( \beta \theta i_2^2+\alpha \gamma i_3+\gamma \theta i_3\right) G(i_3)\right] , h_3=\beta \theta i_3+\gamma (\alpha +\theta ). \end{aligned}$$

Obviously, $e_2>0$, $e_1>0$, $F(i_3)=0$. According to $R_{**}<R_{0}$, $R_*<R_0$ and $R_*<1$, we obtain that $G(i_3)>0$, and then $e_0>0$. Therefore,

$$\begin{aligned} H_1&=e_2>0, \\ H_2&=\left| \begin{array}{cc} e_2 & e_0 \\ 1 & e_1\\ \end{array}\right| \\&=\big (2\alpha \beta ^4\epsilon ^2\gamma \theta +\alpha \beta ^4\epsilon ^2\theta ^2+ \beta ^4\epsilon ^2\theta ^3 +\alpha \beta ^4\epsilon \gamma \theta +\alpha \beta ^4\epsilon \theta ^2 +3\alpha \beta ^3\epsilon ^2\gamma ^2\theta +4\alpha \beta ^3\epsilon ^2\gamma \theta ^2 \\&\quad +2\beta ^4\epsilon \theta ^3 +\beta ^4\theta ^3\big )i_3^4 +\big [2\gamma ^2\alpha \beta ^3\epsilon \theta \left( 1-\epsilon \right) +\alpha ^2\beta ^3\epsilon ^2\gamma ^2 +2\alpha ^2\beta ^3\epsilon ^2\gamma \theta +2\beta ^3\epsilon ^2\gamma \theta ^3 \\&\quad +\alpha ^2\beta ^3\epsilon \gamma ^2 +3\alpha ^2\beta ^3\epsilon \gamma \theta +\alpha ^2\beta ^3\epsilon \theta ^2 +\alpha \beta ^3\eta \gamma \theta +4\alpha \beta ^3\epsilon \gamma ^2\theta +9\alpha \beta ^3\epsilon \gamma \theta ^2 +2\alpha \beta ^3\epsilon \theta ^3 \\&\quad +6\beta ^3\epsilon \gamma \theta ^3 +\beta ^3\epsilon \theta ^4 +\alpha \beta ^3\eta \gamma \theta +\alpha \beta ^3\gamma ^2\theta +3\alpha \beta ^3\gamma \theta ^2 +\alpha \beta ^3\theta ^3 +4\beta ^3\gamma \theta ^3 +\beta ^3\theta ^4\big ]i_3^3 \\&\quad +\big (\alpha ^3\beta ^2\epsilon ^2\gamma ^2 +3\alpha ^2\beta ^2\epsilon ^2\gamma ^2\theta +3\alpha \beta ^2\epsilon ^2\gamma ^2\theta ^2 +\beta ^2\epsilon ^2\gamma ^2\theta ^3 +2\alpha ^3\beta ^2\epsilon \gamma ^2 +2\alpha ^3\beta ^2\epsilon \gamma \theta \\&\quad +2\alpha ^2\beta ^2\epsilon \gamma ^3 +10\alpha ^2\beta ^2\epsilon \gamma ^2\theta +6\alpha ^2\beta ^2\epsilon \gamma \theta ^2 +2\alpha \beta ^2\epsilon \gamma ^3\theta +14\alpha \beta ^2\epsilon \gamma ^2\theta ^2 +6\alpha \beta ^2\epsilon \gamma \theta ^3\\&\quad +6\beta ^2\epsilon \gamma ^2\theta ^3 +2\beta ^2\epsilon \gamma \theta ^4 \alpha ^2\beta ^2\eta \gamma ^2 +\alpha ^2\beta ^2\eta \gamma \theta +\alpha ^2\beta ^2\gamma ^3 +4\alpha ^2\beta ^2\gamma ^2\theta +3\alpha ^2\beta ^2\gamma \theta ^2 \\&\quad +3\alpha \beta ^2\eta \gamma ^2\theta +\alpha \beta ^2\eta \gamma \theta ^2 +2\alpha \beta ^2\gamma ^3\theta +10\alpha \beta ^2\gamma ^2\theta ^2 +6\alpha \beta ^2\gamma \theta ^3 +6\beta ^2\gamma ^2\theta ^3 +3\beta ^2\gamma \theta ^4\big )i_3^2 \\&\quad +\big (\alpha ^2\beta \epsilon \gamma ^2 +2\alpha ^3\beta \epsilon \gamma ^3 +4\alpha ^3\beta \epsilon \gamma ^2\theta +6\alpha ^2\beta \epsilon \gamma ^3\theta +6\alpha ^2\beta \epsilon \gamma ^2\theta ^2 +\alpha \beta ^2\gamma ^3\theta +6\alpha \beta \epsilon \gamma ^3\theta ^2 \\&\quad +4\alpha \beta \epsilon \gamma ^2\theta ^3 +2\beta \epsilon \gamma ^3\theta ^3 +\beta \epsilon \gamma ^2\theta ^4 +\alpha ^3\beta \eta \gamma ^2 +2\alpha ^3\beta \gamma ^3 +3\alpha ^3\beta \gamma ^2\theta +2\alpha ^2\beta \eta \gamma ^2\theta \\&\quad +\alpha ^2\beta \gamma ^4 +8\alpha ^2\beta \gamma ^3\theta +9\alpha ^2\beta \gamma ^2\theta ^2 +\alpha \beta ^2\gamma ^3\theta +\alpha \beta \eta \gamma ^2\theta ^2 +\alpha \beta \gamma ^4\theta +10\alpha \beta \gamma ^3\theta ^2 \\&\quad +9\alpha \beta \gamma ^2\theta ^3\big )i_3 +4\beta \gamma ^3\theta ^3 +3\beta \gamma ^2\theta ^4 +\gamma ^3(\alpha +\theta )^3(\alpha +\gamma +\theta ) +\left( \beta i_3\epsilon +\gamma \right) \alpha \gamma \beta i_3f(i_3) \\&\quad +\big [\beta ^3\theta (\epsilon +1)i_3^4 +\beta ^2\big (2\epsilon \theta \gamma +3\theta \gamma +\alpha \gamma +\alpha ^2+\alpha \theta \big )i_3^3 +\big (\epsilon \theta \gamma +3\theta \gamma +\alpha \epsilon \gamma +\alpha ^2\\&\quad +\alpha \theta \big )\beta \gamma i_3^2+\left( \alpha +\theta \right) i_3\big ]g(i_3)>0, \\ H_3&=\left| \begin{array}{ccc} e_2 & e_0 & 0 \\ 1 & e_1 & 0\\ 0 & e_2 & e_0\\ \end{array}\right| =e_0H_2>0. \end{aligned}$$

According to the Hurwitz criterion, all the roots of Eq. (17) have negative real parts. Therefore, the endemic equilibrium $P_3$ is locally asymptotically stable. $\square$

The proof of Lemma 2

Proof

Let

$$\begin{aligned} \vec {f_*}(t, \vec {x})&=\left( \begin{array}{c} -\beta si+\theta v+\gamma (1-s-v-i)\\ -\sigma \beta vi-\theta v\\ \beta \left[ s+\sigma v+\epsilon (1-s-v-i)\right] i-\eta i \end{array} \right) , \\ \vec {g_*}(t,\vec {x},\vec {u})&=\left( \begin{array}{ccc} \beta si & -s & 0 \\ (1-u_3)\sigma \beta vi & s & (1-u_1)\sigma \beta vi\\ -\beta \left[ s+\left( 1-u_3\right) \sigma v+\epsilon (1-s-v-i)\right] i & 0 & -\left( 1-u_1\right) \beta \sigma v i\\ \end{array} \right) \end{aligned}$$

Then, system (7) can be rewritten as $\vec {x}'(t)=\vec {f_*}(t,\vec {x})+\vec {g_*}(t,\vec {x},\vec {u})\vec {u}$. We prove that the following four conditions are satisfied:

(i)
$\vec {f_*}(t,\vec {x})+\vec {g_*}(t,\vec {x},\vec {u})u$ is first-order continuously differentiable, and there exists a constant C such that
$$\begin{aligned} |\vec {f_*}(t,0,0)|\le C, |\vec {f_*}(\vec {x})+\vec {g_*}(\vec {x},\vec {u})\vec {u}|\le C(1+|\vec {u}|), |\vec {g_*}(\vec {x},\vec {u})|\le C. \end{aligned}$$
(ii)
The solution set of system (7) corresponding to the control parameters in the control set U is non-empty.
(iii)
The dominating set U is a closed convex compact set.
(iv)
The integrand in the objective function J is convex in U.

Firstly, it is easy to see that $\vec {f_*}(t,\vec {x})+\vec {g_*}(t,\vec {x},\vec {u})u$ is first-order continuously differentiable, and $|\vec {f_*}(t,0,0)|=0$. Since s, v, i are non-negative and bounded, there exists a constant C such that $|\vec {f_*}(t,0,0)|\le C, |\vec {f_*}(\vec {x})+\vec {g_*}(\vec {x},\vec {u})\vec {u}|\le C(1+|\vec {u}|), |\vec {g_*}(\vec {x},\vec {u})|\le C.$ This shows that condition (i) is true, It can be seen that the system (7) has a unique solution, which means that condition (ii) is established.

The control set U is a closed convex compact set and the integrand of the objective function is a constant function. So conditions (iii) and (iv) are valid. In summary, the time optimal control problem has an optimal solution. $\square$

The proof of Theorem 4

Proof

In order to find the minimum disease eradication time and the time-dependent control variable $\vec {u}^*(t)$, it is equivalent to the problem of minimizing the Hamiltonian system. Let the Hamiltonian system be

$$\begin{aligned} \begin{aligned} H(\vec {x},\Lambda _1,\Lambda _2,\Lambda _3,t)&=1+\Lambda _1 \frac{ds}{dt}+\Lambda _2 \frac{dv}{dt}+\Lambda _3(t) \frac{di}{dt}\\&=1+\Lambda _1(t)\left[ -\left( 1-u_1\right) \beta si-u_2 s+\theta v+\gamma (1-s-v-i)\right] \\&\quad +\Lambda _2(t) \left[ -\left( 1-u_1\right) \left( 1-u_3\right) \sigma \beta vi-\theta v+u_2 s\right] \\&\quad +\Lambda _3(t) \left\{ \left( 1-u_1\right) \beta \left[ s+\left( 1-u_3\right) \sigma v+\epsilon (1-s-v-i)\right] i-\eta i\right\} , \end{aligned} \end{aligned}$$

(18)

where $\Lambda _1, \Lambda _2, \Lambda _3$ are the adjoint variables associated with the state variables s, v, i. Using the Pontryagin maximum principle³⁶, we can see that the relationship between adjoint control and optimal control is as follows

$$\begin{aligned} \frac{d\Lambda _1}{dt}=-\frac{\partial H}{\partial s}, \frac{d\Lambda _2}{dt}=-\frac{\partial H}{\partial v}, \frac{d\Lambda _3}{dt}=-\frac{\partial H}{\partial i}, \end{aligned}$$

(19)

and the transversality condition is $\Lambda _1(T)=\Lambda _2(T)=\Lambda _3(T)=0$.

Therefore, the adjoint system is

$$\begin{aligned} \begin{aligned}&\frac{d\Lambda _1}{dt}=\Lambda _1 \left[ \left( 1-u_1\right) \beta i+u_2 +\gamma \right] -\Lambda _2 u_2+\Lambda _3 \left( 1-u_1\right) \beta (-1+\epsilon )i,\\&\frac{d\Lambda _2}{dt}=\Lambda _1(-\theta +\gamma )+\Lambda _2(t) \left[ \left( 1-u_1\right) \left( 1-u_3\right) \sigma \beta i+\theta \right] +\Lambda _3 \left( 1-u_1\right) \beta \left[ -(1-u_3)\sigma +\epsilon \right] i,\\&\frac{d\Lambda _3}{dt}=\Lambda _1(t)\left[ \left( 1-u_1\right) \beta s+\gamma \right] +\Lambda _2 \left( 1-u_1\right) \left( 1-u_3\right) \sigma \beta v-\Lambda _3\left( 1-u_1\right) \beta [s+\left( 1-u_3\right) \sigma v\\&\quad \quad +\epsilon (1-s-v-2i)]i+\Lambda _3\eta . \end{aligned} \end{aligned}$$

(20)

The corresponding switching functions and their time derivatives are

$$\begin{aligned} & \Phi (\vec {x},\Lambda _1,\Lambda _2,\Lambda _3,t)=\left( \begin{array}{c} (\Lambda _1-\Lambda _3) \beta s i+(\Lambda _2-\Lambda _3) (1-u_3)\sigma \beta v i+\Lambda _3 \beta \epsilon (1-s-v-i)i\\ (\Lambda _2-\Lambda _1) s\\ (\Lambda _2-\Lambda _3) (1-u_1)\sigma \beta v i\\ \end{array} \right) , \end{aligned}$$

(21)

$$\begin{aligned} & \frac{d\Phi }{dt}(\vec {x},\Lambda _1,\Lambda _2,\Lambda _3,t)=\left( \begin{array}{c} \phi _1\\ (\Lambda _2-\Lambda _1) \frac{ds}{dt}+(\frac{d\Lambda _2}{dt}-\frac{d\Lambda _1}{dt}) s \\ \phi _3 \\ \end{array} \right) , \end{aligned}$$

(22)

where

$$\begin{aligned}&\phi _1=(\frac{d\Lambda _1}{dt}-\frac{d\Lambda _3}{dt}) \beta s i+(\Lambda _1-\Lambda _3)\beta \left( \frac{ds}{dt}i+\frac{di}{dt}s\right) +\left( \frac{d\Lambda _2}{dt}-\frac{d\Lambda _3}{dt}\right) (1-u_3)\sigma \beta v i+(\Lambda _2-\Lambda _3)\\&\quad \quad \left( 1-\frac{d u_3}{dt}\right) \sigma \beta v i+(\Lambda _2-\Lambda _3) (1-u_3)\sigma \beta \left( \frac{dv}{dt} i+v \frac{d i}{dt}\right) +\frac{d\Lambda _3}{dt} \beta \epsilon (1-s-v-i)i\\&\quad \quad +\Lambda _3 \beta \epsilon (1-s-v-i)\frac{d i}{dt}-\Lambda _3 \beta \epsilon \left( \frac{d s}{dt}+\frac{d v}{dt}+\frac{d i}{dt}\right) i, \\ \phi _3&=\left( \frac{d\Lambda _2}{dt}-\frac{d\Lambda _3}{dt}\right) (1-u_1)\sigma \beta v i+(\Lambda _2-\Lambda _3) \left( 1-\frac{d u_1}{dt}\right) \sigma \beta v i+(\Lambda _2-\Lambda _3) (1-u_1)\sigma \\&\quad \beta \left( \frac{dv}{dt} i+v \frac{d i}{dt}\right) , \end{aligned}$$

Next, we prove that the switching function $\Phi$ disappears only at isolated points. Assuming that $\Phi (\vec {x},\Lambda _1,\Lambda _2,\Lambda _3,t)$ disappears in an open set L, then in the neighborhood L, $\Phi =\frac{d\Phi }{dt}=0$, so there is $\Lambda _1=\Lambda _2=\Lambda _3=0$. Then, the Hamiltonian system calculated along the optimal solution can be $H(\vec {x},\Lambda _1,\Lambda _2,\Lambda _3,t)=1\ne 0$, which creates a contradiction. This implies that $\vec {u}^*(t)$ is a Bang-Bang control, that is a piecewise function. $\square$

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Chen, Y., Jing, W., Zhang, J. et al. Bang-bang control optimization in infectious disease model with incorporating breakthrough and reinfection. Sci Rep (2026). https://doi.org/10.1038/s41598-026-44921-7

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Received: 11 December 2025
Accepted: 16 March 2026
Published: 31 March 2026
DOI: https://doi.org/10.1038/s41598-026-44921-7