Modelling of Mammalian Lungs with Hysteresis operators Denis Flynn MURPHYS-HSFS 2016 WORKSHOP
Overview Mammalian lungs Hysteresis in the Pressure-Volume Rela9onship Mathema9cal Modelling Results Further work
Physiology of the lungs Source: Wikicommons: http://upload.wikimedia.org/wikipedia/commons/d/db/Illu_bronchi_lungs.jpg
Hysteresis: Volume-Pressure curves of the respiratory system Adapted from: Harris R.S. Pressure-Volume curves of the respiratory system, Respiratory care 50 (2005), 78-79
Motivations for studying the PV relationship • Help with Diagnosis • Acute Lung Injury • Acute Respiratory Distress Syndrome • Set parameters for ventilators • Aid understanding physiology of the lungs
Possible mechanisms giving rise to Hysteresis in Lungs • Opening and closing of alveoli and/or airways • Non elastic behavior of the lung tissue • Surfactant lining the alveoli • Disease states: • Hysteresis is more pronounced in certain diseases.
Surfactant lining the alveoli Young Laplace − 2 T P : = a b r recoil R 2 R 1 Without Surfactant With Surfactant R 0.1mm, R 0.05mm R 0.1mm, R 0.05mm = = = = 1 2 1 2 T 0.72N/m, T 0.72N/m T 0.02N/m, T 0.01N/m = = = = 1 2 1 2 P 1440N/m, P 2880N/m P 400N/m, P 400N/m = = = = 1 2 1 2 Reference: http://www.medicine.mcgill.ca/physio/resp-web/TEXT3.htm
Evidence that surfactant plays the major role in hysteresis Source: Harris R.S. Pressure-Volume curves of the respiratory system, Respiratory care 50 (2005), 78-79
Mathema9cal Modelling • Venegas’ four parameter model: b V ( p ) = a + (1 + e ( p − c ) /d ) Parameters: a,b,c and d. V is volume and p is pressure.
Preisach Model Density f α '( ) ( ) ρ α = β − α max • Can easily recover f( α ): x β max = ∫ ∫ f x ( ) ( ) d d ρ α β α 0 α f (0) 0 • Provided =
Preisach Model β β = α ( α , β ) max max α • Exhalation curve : α max β max V max = f ( α max ) = ρ ( α ) d β d α ∫ ∫ 0 α
Preisach Model β β = α ( α , β ) max max x(t) α • Following the exhalation curve f(x) x β max = ∫ ∫ f x ( ) ( ) d d ρ α β α 0 α
Preisach Model β β = α ( α , β ) max max x(t) ( α , β ) 1 1 α • On switching direction, we can calculate inhalation curve : x x + ∫∫ f ( ) x f ( ) ( ) d d = α ρ α β α inhalation 1 α α 1
Restricted Preisach Model w ( α ) x β max f '( α ) f ( x ) = ρ ( α ) d β d α ∫ ∫ ρ ( α ) = β max − w ( α ) 0 w ( α )
Restricted Preisach Model x β max f '( α ) f ( x ) = ρ ( α ) d β d α ∫ ∫ ρ ( α ) = β max − w ( α ) 0 w ( α ) w ( α ) = α + e • Chose simple w ( α ):
Model 1 b f ( x ) = a + Exhaling: c − x 1 + e d f '( x ) ρ ( x ) = Preisach density: ( β max − x ) bSech 2 ( c − x 2 d ) Where : ρ ( x ) = 4 d ( β max − x )
Model 2 (restricted Preisach) b f ( x ) = a + Exhaling: c − x 1 + e d f '( x ) ρ ( x ) = Preisach density: ( β max − ( x + e )) b Sech 2 � c − x � Where : 2 d ρ ( x ) = 4 d ( x + e − β max)
Model 1 Model 2
Results SSE Error = V Max Cannine Lungs Murnine Lungs Model 1: 0.47 0.58 Model 2 0.32 0.45
Further work • Modelling rate dependent hysteresis • Development of alterna9ve operators
Rate Dependence: Time series Source: Harris R.S. Pressure-Volume curves of the respiratory system, Respiratory care 50 (2005), 78-79
Rate dependence: PV Loops Source: Harris R.S. Pressure-Volume curves of the respiratory system, Respiratory care 50 (2005), 78-79
Rate Dependence • Speed of input changes PV loops • Previous work by Bates & Irvin [2] • Approach using Dynamic Preisach Model[1] [1] Mayergoyz, Isaak D. Mathematical models of hysteresis and their applications. Academic Press, 2003. [2] Bates, Jason HT, and Charles G. Irvin. "Time dependence of recruitment and derecruitment in the lung: a theoretical model." Journal of Applied Physiology 93.2 (2002): 705-713.
Dynamic Preisach Model ρ ( α , β , dV • Density: dt ) • Series expansion: ρ ( α , β , p ( t ) , ˙ V ( t )) = ρ 0 ( α , β , p ( t )) + ˙ V ( t ) ρ 1 ( α , β , p ( t )) + · · · V ( t ) = P 0 [ p ( t )] + ˙ V P 1 [ p ( t )] •
Other Hysteresis Operators
Piecewise defini9on of operator B V ( x ) = { A + η = 1 e − Kp +1 , otherwise V,
Discrete Model: Rate Independent 50 hysterons • Forcing func9on • Forcing func9on 20(0.9 + Sin(0.2 t - π /2)) 20(0.9 + Sin(3t - π /2))
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