Sliding mode control of the Hodgkin–Huxley mathematical model

Cecilia Cavaterra; Denis Enăchescu; Gabriela Marinoschi

doi:10.3934/eect.2019043

Article Contents

2019, Volume 8, Issue 4: 883-902. Doi: 10.3934/eect.2019043

This issue Previous Article Existence and extinction in finite time for Stratonovich gradient noise porous media equations Next Article

Sliding mode control of the Hodgkin–Huxley mathematical model

1.
Dipartimento di Matematica "F. Enriques", Università degli Studi di Milano, Via C. Saldini 50, 20133 Milano, Italy
2.
"Gheorghe Mihoc-Caius Iacob" Institute of Mathematical Statistics and Applied Mathematics of the Romanian Academy, Calea 13 Septembrie 13, Bucharest, Romania
3.
Research Group of the Project PN-Ⅲ-P4-ID-PCE-2016-0372, Simion Stoilow Institute of Mathematics of the Romanian Academy, Bucharest, Romania
4.
Istituto di Matematica Applicata e Tecnologie Informatiche "E. Magenes", CNR, Via Ferrata 1, 27100 Pavia, Italy

^* Corresponding author: Gabriela Marinoschi
^* Corresponding author: Gabriela Marinoschi

Received: January 2019

Early access: June 2019

Published: June 2019

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Abstract

In this paper we deal with a feedback control design for the action potential of a neuronal membrane in relation with the non-linear dynamics of the Hodgkin-Huxley mathematical model. More exactly, by using an external current as a control expressed by a relay graph in the equation of the potential, we aim at forcing it to reach a certain manifold in finite time and to slide on it after that. From the mathematical point of view we solve a system involving a parabolic differential inclusion and three nonlinear differential equations via an approximating technique and a fixed point result. The existence of the sliding mode and the determination of the time at which the potential reaches the prescribed manifold are proved by a maximum principle argument. Numerical simulations are presented.

Keywords:

Mathematics Subject Classification: Primary: 35K55, 35K57, 35Q92; Secondary: 93B52, 92C30.

Citation:

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Figure 1. Graphics $ v(t, 0) $ (left), $ v(t, x) $ (center), $ n $, $ m $, $ h $ (right) for $ v_0 = 4.82 $, $ v^* = 0 $, $ \rho = 0 $

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Figure 2. Graphics $ v(t, 0) $ (left), $ v(t, x) $ (center), $ n $, $ m $, $ h $ (right) for $ v_0 = 4.82 $, $ v^* = 0 $, $ \rho = 20 $

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Figure 3. Graphics $ v(t, 0) $ (left), $ v(t, x) $ (center), $ n $, $ m $, $ h $ (right) for $ v_0 = 4.82 $, $ v^* = 0.5\sin(4/\pi *t)+0.6 $, $ \rho = 20 $

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Figure 4. Graphics $ v(t, 0) $ (left), $ v(t, x) $ (center), $ n $, $ m $, $ h $ (right) for $ v_0 = 4.82 $, $ v^* = 0 $, $ \rho = 0 $, $ g_K = 3.8229 $

Download: Full-size image PowerPoint slide

Figure 5. Graphics $ v(t, 0) $ (left), $ v(t, x) $ (center), $ n $, $ m $, $ h $ (right) for $ v_0 = 4.82 $, $ v^* = 0 $, $ \rho = 20 $, $ g_K = 3.8229 $

Download: Full-size image PowerPoint slide

Figure 6. Graphics $ v(t, 0) $ (left), $ v(t, x) $ (center), $ n $, $ m $, $ h $ at $ x = 0.5 $ (right) for $ v_0 = 0.5sin(4\pi x)+0.6 $, $ v^* = 0 $, $ \delta = 50 $, $ g_K = 36 $, $ \rho = 50 $

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