Brain Rhythms Sue Ann Campbell Department of Applied Mathematics University of Waterloo October 14, 2017
Outline Biological Background 1 Mathematical Background 2 Modelling Rhythms 3 Summary 4 Sue Ann Campbell (Waterloo) FILOMACS 2 / 30
What are brain rhythms? The electrical activity of the brain exhibits characteristic wave forms
What are brain rhythms? The electrical activity of the brain exhibits characteristic wave forms These vary depending on the brain state and are characterized by their frequency
How do brain rhythms arise? Need to understand a bit of physiology
How do brain rhythms arise? A magnified slice of the brain
How do brain rhythms arise? An individual neuron
How do brain rhythms arise? Behaviour of individual neurons
How do brain rhythms arise? Behaviour of brain - network of neurons
What is an oscillator? Anything that varies periodically in time
What is an oscillator? Anything that varies periodically in time Pendulum
What is an oscillator? Anything that varies periodically in time Metronome
What is an oscillator? Anything that varies periodically in time Firefly
What is an oscillator? Anything that varies periodically in time 3 2 θ 1 0 2 4 6 8 t –1 –2 –3
Neurons are oscillators
What happens when oscillators are connected? Movies Brain slice: https://www.youtube.com/watch?v=t3TaMU_qXMc Fireflies: https://www.youtube.com/watch?v=ZGvtnE1Wy6U 32 metronomes: https://www.youtube.com/watch?v=5v5eBf2KwF8 2 metronomes: https://www.youtube.com/watch?v=yysnkY4WHyM
What happens when oscillators are connected? Synchronization - oscillators all reach maximum at same time
What happens when oscillators are connected? Synchronization - oscillators all reach maximum at same time Phase-locking - oscillators have fixed phase difference
Some Mathematics - Recursively Defined Sequences Recall the Fibonacci Sequence: 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . .
Some Mathematics - Recursively Defined Sequences Recall the Fibonacci Sequence: 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . Can write a general formula/algorithm for generating the terms of this sequence x 0 = 1 = 1 x 1 x n = x n − 1 + x n − 2 , n = 2 , 3 , 4 , . . .
Some Mathematics - Recursively Defined Sequences Recall the Fibonacci Sequence: 1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . Can write a general formula/algorithm for generating the terms of this sequence x 0 = 1 = 1 x 1 x n = x n − 1 + x n − 2 , n = 2 , 3 , 4 , . . . A recursively defined sequence
Modelling With Recursively Defined Sequences Let x n represent the value of some variable at time n Basic Idea: current value = (previous value) + (change)
Modelling With Recursively Defined Sequences Let x n represent the value of some variable at time n Basic Idea: current value = (previous value) + (change) x n + 1 = x n + ∆
Modelling With Recursively Defined Sequences Let x n represent the value of some variable at time n Basic Idea: current value = (previous value) + (change) x n + 1 = x n + ∆ Assumption: change in current value only depends on previous value
Modelling With Recursively Defined Sequences Let x n represent the value of some variable at time n Basic Idea: current value = (previous value) + (change) x n + 1 = x n + ∆ Assumption: change in current value only depends on previous value x n + 1 = x n + g ( x n )
Modelling With Recursively Defined Sequences Let x n represent the value of some variable at time n Basic Idea: current value = (previous value) + (change) x n + 1 = x n + ∆ Assumption: change in current value only depends on previous value x n + 1 = x n + g ( x n ) Recursive formula for an unknown sequence x 0 , x 1 , x 2 , x 3 , ...
Simple Model Assumption: Change is proportional to current value g ( x n ) = ax n where a is a constant.
Simple Model Assumption: Change is proportional to current value g ( x n ) = ax n where a is a constant. Model: x n + 1 = x n + ax n = ( 1 + a ) x n
Simple Model Solving the Model Let starting value (time 0) be arbitrary: x 0
Simple Model Solving the Model Let starting value (time 0) be arbitrary: x 0 Day 1: x 1 = ( 1 + a ) x 0 = ( 1 + a ) x 0
Simple Model Solving the Model Let starting value (time 0) be arbitrary: x 0 Day 1: x 1 = ( 1 + a ) x 0 = ( 1 + a ) x 0 Day 2: x 2 = ( 1 + a ) x 1 ( 1 + a ) 2 x 0 =
Simple Model Solving the Model Let starting value (time 0) be arbitrary: x 0 Day 1: x 1 = ( 1 + a ) x 0 = ( 1 + a ) x 0 Day 2: x 2 = ( 1 + a ) x 1 ( 1 + a ) 2 x 0 = ( 1 + a ) n x 0 Day n : x n =
Simple Model Solving the Model Let starting value (time 0) be arbitrary: x 0 Day 1: x 1 = ( 1 + a ) x 0 = ( 1 + a ) x 0 Day 2: x 2 = ( 1 + a ) x 1 ( 1 + a ) 2 x 0 = ( 1 + a ) n x 0 Day n : x n = A geometric sequence
Simple Model Solving the Model Let starting value (time 0) be arbitrary: x 0 Day 1: x 1 = ( 1 + a ) x 0 = ( 1 + a ) x 0 Day 2: x 2 = ( 1 + a ) x 1 ( 1 + a ) 2 x 0 = ( 1 + a ) n x 0 Day n : x n = A geometric sequence What happens to population if a = 0? a > 0? − 1 < a < 0?
Plotting the Solution a = 0 , x 0 = 5 a = 0 : Constant solution
Plotting the Solution a > 0 , x 0 = 1 a > 0 : Exponential Growth
Plotting the Solution − 1 < a < 0 , x 0 = 10 a < 0 : Exponential Decay
Modelling Oscillators Define the phase of the oscillator Amplitude t (time) t 0 t+ ∆ t t +T 0 0 ∆ t/T 0 1 θ (phase) 2π 2π∆ t/T 0
Modelling Oscillators with Recursive Sequences Think of oscillator as angle of point moving on circle with fixed radius y θ x Change in θ in a small time interval θ n + 1 = θ n + Ω
Two Coupled Oscillators θ 1 , n + 1 = θ 1 n + Ω + A sin( θ 2 n − θ 1 n ) = θ 2 n + Ω + A sin( θ 1 n − θ 2 n ) θ 2 , n + 1 where A is a small number ( | A | << 1 ).
Two Coupled Oscillators θ 1 , n + 1 = θ 1 n + Ω + A sin( θ 2 n − θ 1 n ) = θ 2 n + Ω + A sin( θ 1 n − θ 2 n ) θ 2 , n + 1 where A is a small number ( | A | << 1 ). Define phase difference φ n = θ 2 n − θ 1 n φ n + 1 = φ n − 2 A sin( φ n )
Two Coupled Oscillators Simulations of model with A = 0 . 1 φ 0 = 6 , 5 , 4 , 3 , 2 , 1 , 0 φ n + 1 = φ n − 2 A sin( φ n )
Two Coupled Oscillators Simulations of model with A = 0 . 1 φ 0 = 6 , 5 , 4 , 3 , 2 , 1 , 0 φ n + 1 = φ n − 2 A sin( φ n )
Two Coupled Oscillators Simulations of model with A = 0 . 1 φ 0 = 6 , 5 , 4 , 3 , 2 , 1 , 0 φ n + 1 = φ n − 2 A sin( φ n )
Two Coupled Oscillators Simulations of model with A = 0 . 1 φ 0 = 6 , 5 , 4 , 3 , 2 , 1 , 0 φ n + 1 = φ n − 2 A sin( φ n )
Two Coupled Oscillators Simulations of model with A = 0 . 1 φ 0 = 6 , 5 , 4 , 3 , 2 , 1 , 0 φ n + 1 = φ n − 2 A sin( φ n )
Two Coupled Oscillators Simulations of model with A = 0 . 1 φ 0 = 6 , 5 , 4 , 3 , 2 , 1 , 0 φ n + 1 = φ n − 2 A sin( φ n )
Two Coupled Oscillators Simulations of model with A = 0 . 1 φ 0 = 6 , 5 , 4 , 3 , 2 , 1 , 0 φ n + 1 = φ n − 2 A sin( φ n )
Two Coupled Oscillators Simulations of model with A = 0 . 1 φ 0 = 6 , 5 , 4 , 3 , 2 , 1 , 0 φ n + 1 = φ n − 2 A sin( φ n )
Two Coupled Oscillators φ n + 1 = φ n − 2 A sin( φ n ) Special constant solutions: φ n = φ ∗ , n = 1 , 2 , . . . (equilibrium solutions) Correspond to state the phase difference between the two oscillators is fixed (phase locking).
Two Coupled Oscillators φ n + 1 = φ n − 2 A sin( φ n ) Special constant solutions: φ n = φ ∗ , n = 1 , 2 , . . . (equilibrium solutions) Correspond to state the phase difference between the two oscillators is fixed (phase locking). Occur when sin( φ ∗ ) = 0
Two Coupled Oscillators φ n + 1 = φ n − 2 A sin( φ n ) Two possibilities φ = 0 ( 2 π ) : periodic solutions with two oscillators in-phase φ = π : periodic solutions with two oscillators anti-phase (one half period out of phase).
Two Coupled Oscillators
Two Coupled Oscillators φ n + 1 = φ n − 2 A sin( φ n ) How the equilibrium solutions affect the behaviour depends on the sign of A .
Two Coupled Oscillators φ n + 1 = φ n − 2 A sin( φ n ) How the equilibrium solutions affect the behaviour depends on the sign of A . A > 0
Two Coupled Oscillators φ n + 1 = φ n − 2 A sin( φ n ) How the equilibrium solutions affect the behaviour depends on the sign of A . A < 0
Two Coupled Oscillators φ n + 1 = φ n − 2 A sin( φ n ) How the equilibrium solutions affect the behaviour depends on the sign of A . A < 0
Two Coupled Oscillators - Different Frequencies θ 1 , n + 1 = Ω 1 + A sin( θ 2 n − θ 1 n ) θ 2 , n + 1 = Ω 2 + A sin( θ 1 n − θ 2 n ) ⇓ φ n + 1 = ω + φ n − 2 A sin( φ n )
Two Coupled Oscillators - Different Frequencies θ 1 , n + 1 = Ω 1 + A sin( θ 2 n − θ 1 n ) = Ω 2 + A sin( θ 1 n − θ 2 n ) θ 2 , n + 1 ⇓ = ω + φ n − 2 A sin( φ n ) φ n + 1 Equilibrium solutions: φ ∗ such that sin( φ ∗ ) = ω .
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