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Nonlinear fractional stochastic delay modeling and computational analysis of herpes simplex virus type II dynamics
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  • Published: 02 February 2026

Nonlinear fractional stochastic delay modeling and computational analysis of herpes simplex virus type II dynamics

  • Ali Raza1,3,
  • Mansoor Alsulami2,
  • Marek Lampart1 &
  • …
  • Umar Shafique1 

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

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We are providing an unedited version of this manuscript to give early access to its findings. Before final publication, the manuscript will undergo further editing. Please note there may be errors present which affect the content, and all legal disclaimers apply.

Subjects

  • Computational biology and bioinformatics
  • Diseases
  • Mathematics and computing

Abstract

Herpes simplex virus (HSV) is a widespread infection responsible for painful blisters and ulcers. According to the World Health Organization, approximately 519.5 million people aged 15–49 years (13.3%) worldwide are infected with herpes simplex virus type II (HSV-II), the primary cause of genital herpes. In this study, we develop a nonlinear stochastic fractional delay differential equation (SFDDE) model to describe the transmission dynamics of HSV-II in a human population. The population is divided into susceptible \(\:S\left(t\right)\), exposed \(\:E\left(t\right)\), asymptomatic \(\:A\left(t\right)\), symptomatic \(\:I\left(t\right)\), HSV-infected \(\:H\left(t\right)\), and recovered \(\:R\left(t\right)\)compartments. The model’s fundamental properties, including existence, uniqueness, positivity, and boundedness of solutions, are established. Local and global stability analyses are conducted around the HSV-free and HSV-present equilibrium points, and the basic reproduction number is derived using the next-generation matrix method along with sensitivity analysis. Numerical simulations based on a stochastic nonstandard finite difference (NSFD) scheme confirm the theoretical results and demonstrate the stability of the proposed model. These findings highlight the importance of nonlinear fractional stochastic modeling in understanding and controlling HSV-II transmission dynamics.

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Data availability

The datasets analyzed during the current study are available from the corresponding author upon reasonable request.

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Acknowledgements

and Funding section: This work was supported by the Ministry of Education, Youth and Sports of the Czech Republic through the e-INFRA CZ (ID:90254), with the financial support of the European Union under the REFRESH – Research Excellence For Region Sustainability and High-tech Industries project number CZ.10.03.01/00/22_003/0000048 via the Operational Programme Just Transition.

Author information

Authors and Affiliations

  1. IT4Innovations, VSB-Technical University of Ostrava, 17 listopadu 2172/15, Ostrava, 708 33, Czech Republic

    Ali Raza, Marek Lampart & Umar Shafique

  2. Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia

    Mansoor Alsulami

  3. Jadara University Research Center , Jadara University, Irbid, Jordan

    Ali Raza

Authors
  1. Ali Raza
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  2. Mansoor Alsulami
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  3. Marek Lampart
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  4. Umar Shafique
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Contributions

Ali Raza contributed to the conceptualization, methodology, software, validation, formal analysis, investigation, resources, data curation, writing of the original draft, review and editing, visualization, supervision, project administration, and funding acquisition. Mansoor Alsulami contributed to methodology, Visualization, validation, and writing – review and editing. Marek Lampart contributed to resources, supervision, validation,writing – review and editing,and project administration. Umar Shafique contributed to data curation, software, methodology,validation and writing – review and editing. All authors reviewed and approved the final version of the manuscript. All authors read and approved a manuscript with the given study.

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Correspondence to Ali Raza.

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Appendices

Appendix A

Proof: First, we analyze the Lipschitz’s condition for the function. For this, we take help with the following and :

$$\parallel {\hbar _1}\left( {t,\;S} \right) - {\hbar _1}\left( {t,\;{S_1}} \right)\parallel = \parallel \left( {{{{\Lambda }}^\alpha } - {\beta ^\alpha }S\left( {I + {q^\alpha }A} \right){e^{ - {\mu ^\alpha }\tau }} - {\mu ^\alpha }S + {\omega ^\alpha }R} \right) - \left( {{{{\Lambda }}^\alpha } - {\beta ^\alpha }{S_1}\left( {I + {q^\alpha }A} \right){e^{ - {\mu ^\alpha }\tau }} - {\mu ^\alpha }{S_1} + {\omega ^\alpha }R} \right)\parallel$$
$$\:\parallel{{\hslash\:}}_{1}\left(t,\:S\right)-{{\hslash\:}}_{1}\left(t,\:{S}_{1}\right)\parallel=\parallel\left({\beta\:}^{\alpha\:}\left(S-{S}_{1}\right)\left(I+{q}^{\alpha\:}A\right){e}^{-{\mu\:}^{\alpha\:}\tau\:}+{\mu\:}^{\alpha\:}\left(S-{S}_{1}\right)\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{1}\left(t,\:S\right)-{{\hslash\:}}_{1}\left(t,\:{S}_{1}\right)\parallel\le\:\parallel{\beta\:}^{\alpha\:}\left(S-{S}_{1}\right)\left(I+{q}^{\alpha\:}A\right){e}^{-{\mu\:}^{\alpha\:}\tau\:}\parallel+\parallel{\mu\:}^{\alpha\:}\left(S-{S}_{1}\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{1}\left(t,\:S\right)-{{\hslash\:}}_{1}\left(t,\:{S}_{1}\right)\parallel\le\:\left({\beta\:}^{\alpha\:}{e}^{-{\mu\:}^{\alpha\:}\tau\:}{\parallel}I+{q}^{\alpha\:}A\parallel+{\mu\:}^{\alpha\:}\right){\parallel}S-{S}_{1}\parallel$$
$$\:\parallel{{\hslash\:}}_{1}\left(t,\:S\right)-{{\hslash\:}}_{1}\left(t,\:{S}_{1}\right)\parallel\le\:\left({\beta\:}^{\alpha\:}{e}^{-{\mu\:}^{\alpha\:}\tau\:}\left({{\epsilon}}_{4}+{q}^{\alpha\:}{{\epsilon}}_{3}\right)+{\mu\:}^{\alpha\:}\right){\parallel}S-{S}_{1}\parallel$$
$$\:\parallel{{\hslash\:}}_{1}\left(t,\:S\right)-{{\hslash\:}}_{1}\left(t,\:{S}_{1}\right)\parallel\le\:{\xi\:}_{1}{\parallel}S-{S}_{1}\parallel$$

Since, \(\:{\xi\:}_{1}=\left({\beta\:}^{\alpha\:}{e}^{-{\mu\:}^{\alpha\:}\tau\:}\left({{\epsilon}}_{4}+{q}^{\alpha\:}{{\epsilon}}_{3}\right)+{\mu\:}^{\alpha\:}\right)\). Lipschitz’s condition is satisfied. Next, for \(\:{{\hslash\:}}_{2}\left(t,\:E\right)\) consider \(\:E\) and \(\:{E}_{1}\).

$$\:\parallel{{\hslash\:}}_{2}\left(t,\:E\right)-{{\hslash\:}}_{2}\left(t,\:{E}_{1}\right)\parallel=\parallel\left({\beta\:}^{\alpha\:}S\left(I+{q}^{\alpha\:}A\right){e}^{-{\mu\:}^{\alpha\:}\tau\:}-\left({\eta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)E\right)-\left({\beta\:}^{\alpha\:}S\left(I+{q}^{\alpha\:}A\right){e}^{-{\mu\:}^{\alpha\:}\tau\:}-\left({\eta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){E}_{1}\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{2}\left(t,\:E\right)-{{\hslash\:}}_{2}\left(t,\:{E}_{1}\right)\parallel=\parallel\left(\left({\eta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\left(E-{E}_{1}\right)\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{2}\left(t,\:E\right)-{{\hslash\:}}_{2}\left(t,\:{E}_{1}\right)\parallel\le\:\left({\eta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){\parallel}E-{E}_{1}\parallel$$
$$\:\parallel{{\hslash\:}}_{2}\left(t,\:E\right)-{{\hslash\:}}_{2}\left(t,\:{E}_{1}\right)\parallel\le\:{\xi\:}_{2}{\parallel}E-{E}_{1}\parallel$$

For, \(\:{\xi\:}_{2}=\left({\eta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\). Lipschitz condition is satisfied.

Next, for \(\:{\hslash\:}_{3}\left(t,\:A\right)\) analyzing for \(\:A\) and \(\:{A}_{1}\).

$$\:\parallel{{\hslash\:}}_{3}\left(t,\:A\right)-{{\hslash\:}}_{3}\left(t,\:{A}_{1}\right)\parallel=\parallel\left(\left(1-{\rho\:}^{\alpha\:}\right){\eta\:}^{\alpha\:}E-\left({\phi\:}^{\alpha\:}+{\gamma\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)A\right)-\left(\left(1-{\rho\:}^{\alpha\:}\right){\eta\:}^{\alpha\:}E-\left({\phi\:}^{\alpha\:}+{\gamma\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){A}_{1}\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{3}\left(t,\:A\right)-{{\hslash\:}}_{3}\left(t,\:{A}_{1}\right)\parallel=\parallel\left({\phi\:}^{\alpha\:}+{\gamma\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\left(A-{A}_{1}\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{3}\left(t,\:A\right)-{{\hslash\:}}_{3}\left(t,\:{A}_{1}\right)\parallel\le\:\left({\phi\:}^{\alpha\:}+{\gamma\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){\parallel}A-{A}_{1}\parallel$$
$$\:\parallel{{\hslash\:}}_{3}\left(t,\:A\right)-{{\hslash\:}}_{3}\left(t,\:{A}_{1}\right)\parallel\le\:{\xi\:}_{3}{\parallel}A-{A}_{1}\parallel$$

For, \(\:{\xi\:}_{3}=\left({\phi\:}^{\alpha\:}+{\gamma\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\). Lipschitz condition is satisfied.

Next, for \(\:{\hslash\:}_{4}\left(t,\:I\right)\) analyzing for \(\:I\) and \(\:{I}_{1}\).

$$\:\parallel{{\hslash\:}}_{4}\left(t,\:I\right)-{{\hslash\:}}_{4}\left(t,\:{I}_{1}\right)\parallel=\parallel\left({\rho\:}^{\alpha\:}{\eta\:}^{\alpha\:}E-\left({\varphi\:}^{\alpha\:}+{\theta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)I\right)-\left({\rho\:}^{\alpha\:}{\eta\:}^{\alpha\:}E-\left({\varphi\:}^{\alpha\:}+{\theta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){I}_{1}\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{4}\left(t,\:I\right)-{{\hslash\:}}_{4}\left(t,\:{I}_{1}\right)\parallel=\parallel\left({\varphi\:}^{\alpha\:}+{\theta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\left(I-{I}_{1}\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{4}\left(t,\:I\right)-{{\hslash\:}}_{4}\left(t,\:{I}_{1}\right)\parallel\le\:\left({\varphi\:}^{\alpha\:}+{\theta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){\parallel}I-{I}_{1}\parallel$$
$$\:\parallel{{\hslash\:}}_{4}\left(t,\:I\right)-{{\hslash\:}}_{4}\left(t,\:{I}_{1}\right)\parallel\le\:\left({\varphi\:}^{\alpha\:}+{\theta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){\parallel}I-{I}_{1}\parallel$$

For, \(\:{\xi\:}_{4}=\left({\varphi\:}^{\alpha\:}+{\theta\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\). Lipschitz condition is satisfied.

Next, for \(\:{\hslash\:}_{5}\left(t,\:H\right)\) analyzing for \(\:H\) and \(\:{H}_{1}\).

$$\:\parallel{{\hslash\:}}_{5}\left(t,\:H\right)-{{\hslash\:}}_{5}\left(t,\:{H}_{1}\right)\parallel=\parallel\left({\phi\:}^{\alpha\:}A+{\varphi\:}^{\alpha\:}I-\left({\delta\:}^{\alpha\:}+{\xi\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)H\right)-\left({\phi\:}^{\alpha\:}A+{\varphi\:}^{\alpha\:}I-\left({\delta\:}^{\alpha\:}+{\xi\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){H}_{1}\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{5}\left(t,\:H\right)-{{\hslash\:}}_{5}\left(t,\:{H}_{1}\right)\parallel=\parallel\left(\left({\delta\:}^{\alpha\:}+{\xi\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\left(H-{H}_{1}\right)\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{5}\left(t,\:H\right)-{{\hslash\:}}_{5}\left(t,\:{H}_{1}\right)\parallel\le\:\left({\delta\:}^{\alpha\:}+{\xi\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){\parallel}H-{H}_{1}\parallel$$

For, \(\:{\xi\:}_{5}=\left({\delta\:}^{\alpha\:}+{\xi\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\). Lipschitz condition is satisfied.

Next, for \(\:{{\hslash\:}}_{6}\left(t,\:R\right)\) analyzing for \(\:R\) and \(\:{R}_{1}\).

$$\:\parallel{{\hslash\:}}_{6}\left(t,\:R\right)-{{\hslash\:}}_{6}\left(t,\:{R}_{1}\right)\parallel=\parallel\left({\gamma\:}^{\alpha\:}A+{\theta\:}^{\alpha\:}I+{\delta\:}^{\alpha\:}H-\left({\omega\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)R\right)-\left({\gamma\:}^{\alpha\:}A+{\theta\:}^{\alpha\:}I+{\delta\:}^{\alpha\:}H-\left({\omega\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){R}_{1}\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{6}\left(t,\:R\right)-{{\hslash\:}}_{6}\left(t,\:{R}_{1}\right)\parallel=\parallel\left(\left({\omega\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\left(R-{R}_{1}\right)\right)\parallel$$
$$\:\parallel{{\hslash\:}}_{6}\left(t,\:R\right)-{{\hslash\:}}_{6}\left(t,\:{R}_{1}\right)\parallel\le\:\left({\omega\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right){\parallel}R-{R}_{1}\parallel$$
$$\:\parallel{{\hslash\:}}_{6}\left(t,\:R\right)-{{\hslash\:}}_{6}\left(t,\:{R}_{1}\right)\parallel\le\:{\xi\:}_{6}{\parallel}R-{R}_{1}\parallel$$

For, \(\:{\xi\:}_{6}=\left({\omega\:}^{\alpha\:}+{\mu\:}^{\alpha\:}\right)\). Lipschitz condition is satisfied.

Next, there is constant in (13–18).

$$\:{S}_{n}\left(t\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\left(t-s\right)}^{\alpha\:-1}{{\hslash\:}}_{1}\left(s,\:{S}_{n-1}\right)ds$$
(19)
$$\:{E}_{n}\left(t\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\left(t-s\right)}^{\alpha\:-1}{{\hslash\:}}_{2}\left(s,\:{E}_{n-1}\right)ds$$
(20)
$$\:{A}_{n}\left(t\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\left(t-s\right)}^{\alpha\:-1}{{\hslash\:}}_{3}\left(s,\:{A}_{n-1}\right)ds$$
(21)
$$\:{I}_{n}\left(t\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\left(t-s\right)}^{\alpha\:-1}{{\hslash\:}}_{4}\left(s,\:{I}_{n-1}\right)ds$$
(22)
$$\:{H}_{n}\left(t\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\left(t-s\right)}^{\alpha\:-1}{{\hslash\:}}_{5}\left(s,\:{H}_{n-1}\right)ds$$
(23)
$$\:{R}_{n}\left(t\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\left(t-s\right)}^{\alpha\:-1}{{\hslash\:}}_{6}\left(s,\:{R}_{n-1}\right)ds$$
(24)

Remaining variation is as follows:

$$\:{\psi\:}_{n-1}\left(t\right)=\left({S}_{n}\left(t\right)-{S}_{n-1}\left(t\right)\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}\left({{\hslash\:}}_{1}\left(s,\:{S}_{n-1}\right)-{{\hslash\:}}_{1}\left(s,\:{S}_{n-2}\right)\right)ds$$
(25)
$$\:{\phi\:}_{n-1}\left(t\right)=\left({E}_{n}\left(t\right)-{E}_{n-1}\left(t\right)\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}\left({{\hslash\:}}_{2}\left(s,\:{E}_{n-1}\right)-{{\hslash\:}}_{2}\left(s,\:{E}_{n-2}\right)\right)ds$$
(26)
$$\:{\vartheta\:}_{n-1}\left(t\right)=\left({A}_{n}\left(t\right)-{A}_{n-1}\left(t\right)\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}\left({{\hslash\:}}_{3}\left(s,\:{A}_{n-1}\right)-{{\hslash\:}}_{3}\left(s,\:{A}_{n-2}\right)\right)ds$$
(27)
$$\:{\varpi\:}_{n-1}\left(t\right)=\left({I}_{n}\left(t\right)-{I}_{n-1}\left(t\right)\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}\left({{\hslash\:}}_{4}\left(s,\:{I}_{n-1}\right)-{{\hslash\:}}_{4}\left(s,\:{I}_{n-2}\right)\right)ds$$
(28)
$$\:{{\Psi\:}}_{n-1}\left(t\right)=\left({H}_{n}\left(t\right)-{H}_{n-1}\left(t\right)\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}\left({{\hslash\:}}_{5}\left(s,\:{H}_{n-1}\right)-{{\hslash\:}}_{5}\left(s,\:{H}_{n-2}\right)\right)ds$$
(29)
$$\:{\zeta\:}_{n-1}\left(t\right)=\left({R}_{n}\left(t\right)-{R}_{n-1}\left(t\right)\right)=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}\left({{\hslash\:}}_{6}\left(s,\:{R}_{n-1}\right)-{{\hslash\:}}_{6}\left(s,\:{R}_{n-2}\right)\right)ds$$
(30)

Therefore, we have

$$\:{S}_{n}\left(t\right)=\sum\:_{i=0}^{n}{\psi\:}_{i}\left(t\right)$$
(31)
$$\:{E}_{n}\left(t\right)=\sum\:_{i=0}^{n}{\phi\:}_{i}\left(t\right)$$
(32)
$$\:{I}_{n}\left(t\right)=\sum\:_{i=0}^{n}{\vartheta\:}_{i}\left(t\right)$$
(33)
$$\:{A}_{n}\left(t\right)=\sum\:_{i=0}^{n}{\varpi\:}_{i}\left(t\right)$$
(34)
$$\:{H}_{n}\left(t\right)=\sum\:_{i=0}^{n}{{\Psi\:}}_{i}\left(t\right)$$
(35)
$$\:{R}_{n}\left(t\right)=\sum\:_{i=0}^{n}{\zeta\:}_{i}\left(t\right)$$
(36)

Let,

$$\:\parallel{\psi\:}_{n}\left(t\right)\parallel=\parallel{S}_{n}\left(t\right)-{S}_{n-1}\left(t\right)\parallel$$
$$\:\parallel{\psi\:}_{n}\left(t\right)\parallel=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}\left({{\hslash\:}}_{1}\left(s,\:{S}_{n-1}\right)-{{\hslash\:}}_{1}\left(s,\:{S}_{n-2}\right)\right)ds$$
$$\:\parallel{\psi\:}_{n}\left(t\right)\parallel=\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\parallel}{S}_{n}\left(t\right)-{S}_{n-1}\left(t\right){\parallel}ds$$
$$\:\parallel{\psi\:}_{n}\left(t\right)\parallel=\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\psi\:}_{n-1}\left(t\right)ds$$
(37)

Similarly,

$$\:\parallel{\phi\:}_{n}\left(t\right)\parallel=\frac{{\xi\:}_{2}}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\phi\:}_{n-1}\left(t\right)ds$$
(38)
$$\:\parallel{\vartheta\:}_{n}\left(t\right)\parallel=\frac{{\xi\:}_{3}}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\vartheta\:}_{n-1}\left(t\right)ds$$
(39)
$$\:\parallel{\varpi\:}_{n}\left(t\right)\parallel=\frac{{\xi\:}_{4}}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\varpi\:}_{n-1}\left(t\right)ds$$
(40)
$$\:\parallel{{\Psi\:}}_{n}\left(t\right)\parallel=\frac{{\xi\:}_{5}}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{{\Psi\:}}_{n-1}\left(t\right)ds$$
(41)
$$\:\parallel{\zeta\:}_{n}\left(t\right)\parallel=\frac{{\xi\:}_{6}}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}{\zeta\:}_{n-1}\left(t\right)ds$$
(42)

As required.

Appendix B

Proof

Consider the \(\:S\left(t\right),\:E\left(t\right),A\left(t\right),I\left(t\right),H\left(t\right)\:\text{a}\text{n}\text{d}\:R\left(t\right)\) are bounded. Then.

$$\:\parallel{\psi\:}_{n}\left(t\right)\parallel\le\:{\parallel}S\left(0\right)\parallel{\parallel\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}\left(t\right)\parallel}^{n}$$
(43)
$$\:\parallel{\vartheta\:}_{n}\left(t\right)\parallel\le\:{\parallel}E\left(0\right)\parallel{\parallel\frac{{\xi\:}_{2}}{{\Gamma\:}\left(\alpha\:\right)}\left(t\right)\parallel}^{n}$$
(44)
$$\:\parallel{\psi\:}_{n}\left(t\right)\parallel\le\:{\parallel}A\left(0\right)\parallel{\parallel\frac{{\xi\:}_{3}}{{\Gamma\:}\left(\alpha\:\right)}\left(t\right)\parallel}^{n}$$
(45)
$$\:\parallel{\varpi\:}_{n}\left(t\right)\parallel\le\:{\parallel}I\left(0\right)\parallel{\parallel\frac{{\xi\:}_{4}}{{\Gamma\:}\left(\alpha\:\right)}\left(t\right)\parallel}^{n}$$
(46)
$$\:\parallel{{\Psi\:}}_{n}\left(t\right)\parallel\le\:{\parallel}H\left(0\right)\parallel{\parallel\frac{{\xi\:}_{5}}{{\Gamma\:}\left(\alpha\:\right)}\left(t\right)\parallel}^{n}$$
(47)
$$\:\parallel{\zeta\:}_{n}\left(t\right)\parallel\le\:{\parallel}R\left(0\right)\parallel{\parallel\frac{{\xi\:}_{6}}{{\Gamma\:}\left(\alpha\:\right)}\left(t\right)\parallel}^{n}$$
(48)

Since, \(\:S\left(t\right),\:E\left(t\right),A\left(t\right),I\left(t\right),H\left(t\right)\:\text{a}\text{n}\text{d}\:R\left(t\right)\) will converge because the system (31–36) exists and consistent. For this, consider \(\:n\) changes as \(\:{A}_{n}\left(t\right),\:{B}_{n}\left(t\right),\:{C}_{n}\left(t\right),\:{D}_{n}\left(t\right),\:{X}_{n}\left(t\right)\:\text{a}\text{n}\text{d}\:{Y}_{n}\left(t\right)\). Thus,

$$\:S\left(t\right)-S\left(0\right)={S}_{n}\left(t\right)-{A}_{n}\left(t\right)$$
(49)
$$\:E\left(t\right)-E\left(0\right)={E}_{n}\left(t\right)-{B}_{n}\left(t\right)$$
(50)
$$\:A\left(t\right)-A\left(0\right)={A}_{n}\left(t\right)-{C}_{n}\left(t\right)$$
(51)
$$\:I\left(t\right)-I\left(0\right)={I}_{n}\left(t\right)-{D}_{n}\left(t\right)$$
(52)
$$\:H\left(t\right)-H\left(0\right)={H}_{n}\left(t\right)-{X}_{n}\left(t\right)$$
(53)
$$\:R\left(t\right)-R\left(0\right)={R}_{n}\left(t\right)-{Y}_{n}\left(t\right)$$
(54)

The result of Lipschitz condition for \(\:\left({\xi\:}_{1}\right)\) and the triangle inequality, and \(\:{\hslash\:}_{1}\) for \(\:i=\text{1,2},\text{3,4},\text{5,6}\), fulfills the Lipschitz condition.

$$\:\parallel{A}_{n}\left(t\right)\parallel=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}\left({{\hslash\:}}_{1}\left(s,\:{S}_{n-1}\right)-{{\hslash\:}}_{1}\left(s,\:{S}_{n-2}\right)\right)ds$$
$$\:\parallel{A}_{n}\left(t\right)\parallel\le\:\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}\parallel{S}_{n}\left(t\right)-{S}_{n-1}\left(t\right)\parallel$$
(55)

with repetition of (55),

$$\:\parallel{A}_{n}\left(t\right)\parallel\le\:{\parallel\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}\left(t\right)\parallel}^{n+1}{{\epsilon}}_{1}$$
(56)

Next, at \(\:{t}_{\text{*}}\), one acquires

$$\:\parallel{A}_{n}\left(t\right)\parallel\le\:{\parallel\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}\left({t}_{\text{*}}\right)\parallel}^{n+1}{{\epsilon}}_{1}$$
(57)

Assuming \(\:n\:\to\:\:{\infty\:}\) as the limit.

$$\:\underset{n\to\:{\infty\:}}{\text{lim}}\parallel{A}_{n}\left(t\right)\parallel\le\:\underset{n\to\:{\infty\:}}{\text{lim}}{\parallel\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}\left({t}_{\text{*}}\right)\parallel}^{n+1}{{\epsilon}}_{1}$$
(58)

By applying the hypothesis \(\:\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}\left({t}_{\text{*}}\right)<1\), we get

$$\:\underset{n\to\:{\infty\:}}{\text{lim}}\parallel{A}_{n}\left(t\right)\parallel=0$$
(59)

Similarly,

$$\:\parallel{B}_{n}\left(t\right)\parallel\to\:0$$
(60)
$$\:\parallel{C}_{n}\left(t\right)\parallel\to\:0$$
(61)
$$\:\parallel{D}_{n}\left(t\right)\parallel\to\:0$$
(62)
$$\:\parallel{X}_{n}\left(t\right)\parallel\to\:0$$
(63)
$$\:\parallel{Y}_{n}\left(t\right)\parallel\to\:0$$
(64)

As desired.

Appendix c

Proof

Examine how the sets \(\:{S}_{1},{E}_{1},{A}_{1},{I}_{1},{H}_{1},\:\text{a}\text{n}\text{d}\:{R}_{1}\) represent the solutions to (1–6).

$$\:{\parallel}S\left(t\right)-{S}_{1}\left(t\right)\parallel=\frac{1}{{\Gamma\:}\left(\alpha\:\right)}{\int\:}_{0}^{t}\left({{\hslash\:}}_{1}\left(s,\:S\right)-{{\hslash\:}}_{1}\left(s,\:{S}_{1}\right)\right)ds$$

.

$$\:{\parallel}S\left(t\right)-{S}_{1}\left(t\right)\parallel\le\:\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}{\parallel}S\left(t\right)-{S}_{1}\left(t\right)\parallel$$
(65)

After simplifying,

$$\:\left(1-\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}\left(t\right)\right){\parallel}S\left(t\right)-{S}_{1}\left(t\right)\parallel\le\:0$$
(66)

By applying the hypothesis \(\:\left(1-\frac{{\xi\:}_{1}}{{\Gamma\:}\left(\alpha\:\right)}\left(t\right)\right)>0\), we have from (66) yield.

$$\:{\parallel}S\left(t\right)-{S}_{1}\left(t\right)\parallel=0$$
(67)

Similarly,

$$E(t) = {E_1}(t)$$
(68)
$$A(t) = {A_1}(t)$$
(69)
$$I(t) = {I_1}(t)$$
(70)
$$H(t) = {H_1}(t)$$
(71)
$$R(t) = {R_1}(t)$$
(72)

Hence proved.

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Raza, A., Alsulami, M., Lampart, M. et al. Nonlinear fractional stochastic delay modeling and computational analysis of herpes simplex virus type II dynamics. Sci Rep (2026). https://doi.org/10.1038/s41598-026-37658-w

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  • Received: 12 November 2025

  • Accepted: 23 January 2026

  • Published: 02 February 2026

  • DOI: https://doi.org/10.1038/s41598-026-37658-w

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Keywords

  • Nonlinear model
  • Dynamical analysis
  • Stability and sensitivity analysis
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