A Fokker-Planck Model in for Two Interacting Populations of Neurons - PowerPoint PPT Presentation

A Fokker-Planck Model in for Two Interacting Populations of Neurons J.A. Carrillo 1 , S. Cordier 2 , S.Mancini 2 1 ICREA, Universitat Aut` onoma de Barcelona 2 MAPMO, F´ ed´ eration Denis Poisson, Universit´ e d’Orl´ eans Stochastic Models in Neuroscience CIRM, 18 January 2010

Outline • The Wilson-Cowan and Deco-Mart ` ı model • The stochastic ODE system and its Fokker-Planck equation • Stationnary solution (equilibrium) • Time dependent problem and Generalized relative entropy • Numerical simulations • Slow-fast behavior

Wilson-Cowan model ⊲ Two neurons populations/cells ⊲ Neglect spatial interactions ⊲ Time dynamics E ( t ) = excitatory cells prop. activ/unit time at time t I ( t ) = inhibitory cells prop. activ/unit time at time t Model : The value of E ( t + τ ) and I ( t + τ ) is proportional to number of cells which are sensitive (not refractory) and which also receive at least threshold, θ , excitation at time t. [WC] H.R. Wilson, J.D. Cowan, Biophysical Journal (1972)

Wilson-Cowan, the model ⊲ Excit./Inhibit. sensitives prop. of cells ( r refractory time): � t � t t − r E ( t ′ ) dt ′ , t − r I ( t ′ ) dt ′ 1 − 1 − ⊲ Sub-pop. prop. receiving at least θ excitation/time, as a fonction of the mean excitation x ( t ) = > response fonction (sigmoide): � x S ( x ) = θ D ( θ ) dθ , D ( θ ) = thresholds distribution ⊲ α ( t ) decreasing rate of stimuli effect, c i coeff.connexion, I e ( t ) , I i ( t ) stimuli; � t �� t � � � t − r E ( t ′ ) dt ′ −∞ α ( t − t ′ )[ c 1 E ( t ′ ) − c 2 I ( t ′ ) + I e ( t ′ )] dt ′ 1 − S e E ( t + τ ) = � t �� t � � � t − r I ( t ′ ) dt ′ −∞ α ( t − t ′ )[ c 3 E ( t ′ ) − c 4 I ( t ′ ) + I i ( t ′ )] dt ′ I ( t + τ ) = 1 − S i � �� � x ( t )

Wilson-Cowan, simplified Time averaging ⇒ : � t � t t − r E ( t ′ ) dt ′ → rE ( t ) , −∞ α ( t − t ′ ) E ( t ′ ) dt ′ → kE ( t ) First order expansion in τ = 0 ⇒ τ dE dt = − E + (1 − rE ) S e ( kc 1 E − kc 2 I + kI e ( t )) τ ′ dI dt = − I + (1 − rI ) S i ( k ′ c 3 E − k ′ c 4 I + k ′ I i ( t )) ⊲ Hysteresis multiple loops and limit cycles. ⊲ Results are independent of the choice of the sigmoide.

The Deco-Mart ` ı model w + excitation coeff. between neu- rons of the same population w − excitation coeff. between neu- rons of different populations w I inhibition coeff. between all neu- ronal populations Synaptic force between population i and j :  w + − w I i = j  w ij = ⇒ w 11 = w 22 , w 12 = w 21 w − − w I i � = j  ⊲ Unbiased : λ 2 = λ 1 ⊲ Biased : λ 2 = λ 1 + 0 . 1 [DM] G.Deco, D.Marti, Biological Cybernetics (2007)

The associated stochastic system ⊲ ν 1 = ν 1 ( t ) , ν 2 = ν 2 ( t ) firing rates of the 2 sub-populations: ⊲ ξ = ξ ( t ) white noise of amplitude β (brownian motion with variance β 2 / 2).  � � λ 1 + � τ ˙ ν 1 = − ν 1 + φ + ξ j =1 , 2 w 1 j ν j  � � λ 2 + � τ ˙ ν 2 = − ν 2 + φ + ξ j =1 , 2 w 2 j ν j  where φ ( x ) is the sigmoide (response fct.), strictly monotone and bounded: ν c φ ( x ) = α, ν c ∈ R 1 + exp( − α ( x/ν c − 1)) ,

The Fokker-Planck equation Let dx t = f ( x t ) dt + g ( x t ) dB t a ods syst. and L = � f i ( x ) ∂ x i + (1 / 2) � d ij ∂ x i j the generator of T , then u ( x, t ) = T t ϕ ( x ) satisfies ∂ t u = Lu . def = R 2 ⊲ f ( t, ν 1 , ν 2 ) probability distribution, t ≥ 0 and ν = ( ν 1 , ν 2 ) ∈ Ω + ⊲ ∂ t f + ∇ · ( F f ) − β 2 2 ∆ f = 0 ( FP ) Ff = ( − ν + Φ(Λ + W · ν )) f � � Ff − β 2 2 ∇ f · n = 0 ⊲ flux incoming F · n ≤ 0 ( H 1) � ⊲ normalization Ω f dν = 1 ( H 2)

No explicit equilibrium solution Remark : The flux F is not the gradient of a potential A : ∂ ν 2 F 1 � = ∂ ν 1 F 2 ⊲ No explicit solution of the stationnary problem associated to ( FP ). ⊲ If one proves the existence, uniqueness and positivity of a solution f ∞ of the stationnary problem, then it is possible to split F in the sum of a gradient and non-gradient terms: let A = − log f ∞ (ie. f ∞ = e − A ),   � � � � Fe − A − β 2 F + β 2   2 ∇ ( e − A ) e − A   ∇ · = 0 , hence ∇ · 2 ∇ A = 0       � �� � G F = − β 2 and so: 2 ∇ A + G Remark : G is s.t. ∇· ( Ge − A ) = 0, but we do not have an explicit form for G [ACJ] A. Arnold, E. Carlen, Q. Ju, Communication in Stochastic Analysis (2008)

Stationnary problem We first consider the stationnary problem associated to ( FP ) � � A f = − β 2 Ff − β 2 2 ∆ f + ∇ · ( Ff ) = 0 , 2 ∇ f · n = 0 ( S ) Theorem: If ( H 1) and ( H 2) hold, then there exists an unique positive solution f ∞ ( ν ) to ( S ) . Proof: Based on the Krein-Rutman theorem : • T : L 2 (Ω) → L 2 (Ω), s.t. ∀ g ∈ L 2 (Ω), Tg = f , with f the unique solution of : ( Ff − β 2 A f + ρf = g in Ω , 2 ∇ f ) · n = 0 on ∂ Ω • T : H 2 → H 2 is a compact operator, and T : K → K str. pos., with K = W 2 , 2 + (Ω). • KR th. = ⇒ r ( T ) > 0 and ∃ g > 0 s.t. Tg = r ( T ) g. We have, 1 A f + ρf = λf, f = r ( T ) g > 0 , λ = and r ( T ) � A f = ( λ − ρ ) f ⇒ ( λ − ρ ) ⇒ ρ = λ ⇒ A f = 0 . f dx = 0 � Ω

Time dependent problem We consider here the parabolic problem: � � Ff − β 2 ∂ t f + A f = 0 , 2 ∇ f · n = 0 ( P ) with the initial condition: f 0 ( · ) ∈ L 2 (Ω) Theorem: ( P ) admits an unique solution f ( t, ν 1 , ν 2 ) . Proof: Let a ( t, f, g ) be the bi-linear form associated to A : � � β 2 ∀ f, g ∈ H 1 (Ω) , a ( t, f, g ) = 2 ∇ f · ∇ g dν − fF · ∇ g dν , ( a ) Ω Ω • a ( t, f, g ) is continuous, • a ( t, f, g ) + ρ < f, g > is coercive for ρ ∈ R large enough. � Remark : The maximum principle does not apply ! ( F has negative divergence ∇ · F ≤ 0)

Generalized relative entropy Theorem: Let f 1 , f 2 > 0 solutions to ( P ) , and g > 0 a solution of the dual problem:  ∂ t g = − F · ∇ g − β 2 in Ω × [0 , T ] ,  2 ∆ g, ∂g ∂n = 0 on ∂ Ω  Then, ∀ H convex � � 2 = − β 2 � � d Ω gf 1 H ( f 2 Ω gf 1 H ′′ ( f 2 � ∇ ( f 2 � � � � ≤ 0 , dν ) ) dν ( GRE ) � � dt f 1 2 f 1 f 1 � � �� � � �� � H g ( f 2 | f 1 ) D g ( f 2 | f 1 ) � � f 1 �� ∂t [ gf 1 H ] = −∇ · [ Fgf 1 H ] + β 2 2 gf 1 H ′′ |∇ ( f 2 /f 1 ) | 2 − β 2 g 2 ∇ ∂ 2 ∇ · Proof: g H � � �� � =0 integr. by part � � Lemme: The solution f to ( P ) satisfies: Ω f ( t, ν ) dν = Ω f 0 ( ν ) dν. Proof: Integration of d/dt � Ω fg � [MMP] P. Michel, S.Mischler, B.Perthame, J.Math. Pures Appl. (2005).

Corollaries Corollary 1: If f 0 ( ν ) > 0 , then f ( t, ν ) > 0 ∀ t . Proof: We choose in ( GRE ): g = 1, f 1 = f ∞ , f 2 solution of ( P ) with f 2 (0 , ν ) > 0, and H ( f 2 /f 1 ) = ε ( f 2 /f 1 ) − . Then: � d f ∞ ( f 2 /f 1 ) − dν ≤ 0 . dt Ω � �� � h ( t ) h (0) = 0, h ( t ) is decreasing and h ( t ) is positive ⇒ h ( t ) = 0 ∀ t � Corollary 2: If ( H 1) holds and f is a solution to ( P ) with initial con- dition f 0 , then � Ω | f ( t, ν ) − f ∞ ( ν ) | 2 dν = 0 lim t →∞ Proof: We choose in ( GRE ): g = 1, f 1 = f ∞ , f 2 = f and H ( s ) = s 2 / 2; and applying the Aubin-Lions theorem �

Numerical approximation - finite difference method Let f k ( i, j ) = f ( k ∆ t, n i , n j ) with n i = ( i + 1 2 )∆ N 1 , i = 0 ...N 1 − 1 and n j = ( j + 1 2 )∆ N 2 , j = 0 ...N 2 − 1. Then, the Fokker-Planck equation is discretised by : f k +1 ( i, j ) f k ( i, j ) = � � F k ( i + 1 / 2 , j ) − F k ( i − 1 / 2 , j ) + ∆ t / ∆ N 1 � � G k ( i, j + 1 / 2) − G k ( i, j − 1 / 2) + ∆ t / ∆ N 2 , where F k ( i + 1 2 , j ), G k ( i, j + 1 2 ) are the flux at the interfaces : � � F k ( i + 1 / 2 , j ) f k ( i + 1 / 2 , j ) = − n i +1 / 2 + Φ( λ + w 11 n i +1 / 2 + w 12 n j ) β 2 � � f k ( i + 1 , j ) − f k ( i, j ) − , 2∆ N 1 � � G k ( i, j + 1 / 2) f k ( i, j + 1 / 2) = − n j +1 / 2 + Φ( λ + w 21 n i + w 22 n j +1 / 2 ) β 2 � � f k ( i, j + 1) − f k ( i, j ) − . 2∆ N 2 and we choose linear interpolation for f at the interfaces: f k ( i + 1 / 2 , j ) = f k ( i + 1 , j ) + f k ( i, j ) f k ( i, j + 1 / 2) = f k ( i, j + 1) + f k ( i, j ) , . 2 2 Remark : Adaptatif ∆ t (gain factor 100) ⇒ for i, j s.t. f k ( i, j ) � = 0 and F k ( i, j ) � = 0: f k ( i, j ) ∆ t = min 2 |F k ( i, j ) | i,j

Computed quantities Marginales of f ( t, ν 1 , ν 2 ) with respect to ν 2 and ν 1 : � ν M � ν M N 1 ( t, ν 1 ) = f ( t, ν 1 , ν 2 ) dν 2 , N 2 ( t, ν 2 ) = f ( t, ν 1 , ν 2 ) dν 1 . 0 0 First order moments: � � µ i ( t ) = Ω ν i f ( ν 1 , ν 2 , t ) dν 1 dν 2 , i = 1 , 2 Second order moments: � � γ ij ( t ) = Ω ν i ν j f ( ν 1 , ν 2 , t ) dν 1 dν 2 , i, j = 1 , 2 . Probabilities ρ i ( t ) to belong to the domains Ω i , i = 1 , 2 , 3 : � � ρ i ( t ) = f ( ν 1 , ν 2 , t ) dν 1 dν 2 . Ω i We have N 1 = N 2 = 200 discretisation points, and stop computations when we get a 10 − 10 difference between to successive iterations. β = 0 . 1, α = 4, ν c = 20, λ 1 = 15, r = 0 . 3, w + = 2 . 35, w I = 1 . 9, τ = 10 − 2 .