Dynamics of Excitable Media

8/13/2019 Dynamics of Excitable Media

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ChE 360 Mathematical Methods inChemical Engineering

Dynamics of Excitable Media

Fall 2012Serena Xu

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Table of Contents

1. Introduction32. Results and Discussion..4

a. Part 1 Oscillatory Dynamics.4b. Part 2 Excitable Dynamics6c. Part 3 (1)10d. Part 3 (2)11

3. Appendix.13

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Introduction

Neuron behavior may be described by the FitzHugh-Nagumo model,

.These equations describe the behavior of neuron impulses with stimulation (the betaterm). They describe the excitation and propagation of neuron activity that results fromthe flow of sodium and potassium. U allows for self-excitation in the system through apositive feedback, while V represents the slower negative feedback. In theseequations, epsilon, alpha and beta are dimensionless parameters with a strong effect onthe inhibitors and activators.

Numerical integration in Matlab (ode45) was used to calculate the solution of theFitzHugh-Nagumo model through time.

Centered finite differences were used to solve for the steady state behavior of thesecond system in Matlab, by changing the problem from an ordinary differential

equation to a set of algebraic equations.

The Crank Nicolson method, which uses centered differences in space and combinesthe forward Euler and backward Euler method through time. It has the advantage ofbeing more stable than the use of the forward Euler method, while also being moreaccurate than the backward Euler method. It was used to solve the diffusive transportsystem shown above.

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Part 1 Oscil latory Dynamics

Nullclines

Figure 1 Nullclines

A single critical point of (0,0) can be found at the intersection of the nullclines of u and v.Here, the levels of activator and inhibitor are equal and steady.

Figure 2V vs. U

This figure represents the constant oscillations of the activator and inhibitor activity around thesteady state their activity goes through the same cycle endlessly, without excitation.

-1.5 -1 -0.5 0 0.5 1 1.5-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-1.5 -1 -0.5 0 0.5 1 1.5-0.5

0

0.5

1

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Figure 3 U vs. t

The neuron exhibits a constant oscillatory pattern of positive feedback every 50

seconds by releasing activator.

Figure 4 V vs. t

The neuron exhibits a constant oscillatory pattern of negative feedback by releasing

inhibitor.

-50 0 50 100 150 200-1.5

-1

-0.5

0

0.5

1

1.5

0 20 40 60 80 100 120 140 160 180 200-0.5

0

0.5

1

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Part 2, Excitable Dynamics

Figure 5Nullclines

The intersection of the nullclines can be seen at (-.585, -.385). The steady state hasmoved to this point due to the presence of excitation in the system (beta=.2). At thispoint, there is no neuron excitation and the levels of activator and inhibitor are steady.

Figure 6V vs. UThis figure shows the convergence over time of the system to a fixed point of (-.585,.385, as seen above). All three simulations from delta u =.01, .1, .2 all converge tosame point, but from differentdistances. Where delta u is higher, the neuron impulse isstronger.

-1.5 -1 -0.5 0 0.5 1 1.5-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

u

v

-1.5 -1 -0.5 0 0.5 1-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

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Figure 7 U vs. t, delta u=.01

Numerical integration of du/dt from a starting point delta u=.01 from the steady state results in a

series of oscillations of u that slowly converges toward the steady state. This represents

oscillations in the amount of activator released that is greater in the beginning, likely due to

outside stimulus, and slowly returns to the steady state amount.


The greater starting distance from the steady state can be seen in the larger oscillations in the

beginning. The fact that the beginning oscillations in the quantity of activator is likely caused by

a stronger outside stimulus than in the figure 7. For example, a harder pinch would cause a

greater sensation of pain.

0 20 40 60 80 100 120 140 160 180 200

-0.592

-0.59

-0.588

-0.586

-0.584

-0.582

-0.58

-0.578

-0.576

-0.574

-0.572

0 20 40 60 80 100 120 140 160 180 200-0.66

-0.64

-0.62

-0.6

-0.58

-0.56

-0.54

-0.52

-0.5

-0.48

-0.46

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This represents an even stronger stimulus and therefore stronger reaction of the neurons with a

greater release of activator.

Figure 10 V vs. t, delta u=.01

Numerical integration of dv/dt from a starting point delta u=.1 from the steady state results in a

series of oscillations of v that slowly converges toward the steady state quantity of inhibitor. The

larger waves represent the neurons reaction to the increased quantity of activator.

0 20 40 60 80 100 120 140 160 180 200-1.5

-1

-0.5

0

0.5

1

0 20 40 60 80 100 120 140 160 180 200-0.3865

-0.386

-0.3855

-0.385

-0.3845

-0.384

-0.3835

-0.383

-0.3825

-0.382

-0.3815

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Figure 11 V vs. t, delta u=.1Numerical integration of dv/dt from a starting point delta u=.1 from the steady state

results in a series of oscillations of v that slowly converges toward the steady state

quantity of inhibitor. The larger waves represent the neurons reaction to the increased

quantity of activator. The fact that the oscillation of inhibitor in the beginning is larger

than that in figure 10 can be explained by the increased amount of the activator

compared to figure 10 as well.

Figure 12 V vs. t, delta u=.2

The fact that the oscillation of inhibitor in the beginning is larger than that in figure 10

can again be explained by the increased amount of the activator compared to figure 10

as well.

0 20 40 60 80 100 120 140 160 180 200-0.4

-0.39

-0.38

-0.37

-0.36

-0.35

-0.34

0 20 40 60 80 100 120 140 160 180 200-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

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PART 3, 1, Diffusive Transport, steady state

Figure 12Steady state profile of traveling waves

At steady state, the levels of activator and inhibitor converge from the starting point at

x=0 of u=.5 and v=0 to the stable fixed point of u*=-.585 and v*=-.385 due to resistance

to diffusion.

Figure 13 Steady state profiles from different starting points

Convergence to the fixed point of u*,v* occurs no matter the quantity of activator at x=0.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

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PART 3, 2

0 2 4 6 8 10 12 14 16 18 20-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

x

Time = 0.757557

u

v

0 2 4 6 8 10 12 14 16 18 20-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

x

Time = 2.272670

u

v

0 2 4 6 8 10 12 14 16 18 20-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

x

Time = 4.545340

u

v

0 2 4 6 8 10 12 14 16 18 20-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

x

Time = 9.848236

u

v

0 2 4 6 8 10 12 14 16 18 20-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

x

Time = 24.999369

u

v

0 2 4 6 8 10 12 14 16 18 20-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

x

Time = 14.393576

u

v

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The progression of a single wave can be seen in first 8 figures. This represents a single

impulse of the neuron. In the first figure, you can see the flat line, steady state values of

u* and v* (activator and inhibitor). The activator is first released from x=0 at a higher

level than the inhibitor and travels more quickly through the neuron than the inhibitor

(most clearly seen in at times 10 and 15). By time 25, the diffusion of inhibitor has

caught up to the diffusion of activator and a new wave of activator and inhibitor has

begun. The waves of activator and inhibitor continue through time 200, representing a

constant stimulation of neurons (perhaps by a constant pinching of the skin).

0 2 4 6 8 10 12 14 16 18 20

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

x

Time = 199.994950

u

v

0 2 4 6 8 10 12 14 16 18 20-1.5

-1

-0.5

0

0.5

1

x

Time = 49.998737

u

v

0 2 4 6 8 10 12 14 16 18 20-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

x

Time = 149.996212

u

v

0 2 4 6 8 10 12 14 16 18 20

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

x

Time = 99.997475

u

v

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Appendix

Part 1FIGURE !NULLCLINES

u=linspace(-1.5,1.5);v1=u-u.^3;v2=1.11*u;hold onplot(u,v1)plot(u,v2)-------------------------------------------------------------------------------

Function for numerical in tegration

functiondx=osc(t,x)

e=.06;a=.9;b=0;

dx(1,1)= -x(2)+x(1)-x(1).^3;dx(2,1)=e*(x(1)-a*x(2)-b);

return

-------------------------------------------------------------------------------

FIGURE 2PLOT V vs UT=0:0.5:200;

hold onfori=1:100

X0=[0;1];[T,X]=ode45(@(t,x)osc(t,x),T,X0); plot(X(:,1),X(:,2));

end-------------------------------------------------------------------------------

FIGURE 3

PLOT U VS T

hold onfori=1:100

X0=[0;1];[T,X]=ode45(@(t,x)osc(t,x),T,X0); plot(T,X(:,1));

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end-------------------------------------------------------------------------------

FIGURE 4

PLOT V VS T

T=0:0.5:200;

hold onfori=1:100

X0=[0;1];[T,X]=ode45(@(t,x)osc(t,x),T,X0); plot(T,X(:,2));

end-------------------------------------------------------------------------------------------------------------------------------

-------------------------------------------------------------------------------------------------------------------------------

Part 2Figure 5

Nullclines

u=linspace(-1.5,1.5);v1=u-u.^3;v2=(u+.2)/1;

hold onplot(u,v1)

plot(u,v2)

-------------------------------------------------------------------------------

Function for numerical integration

functiondx=exc(t,x)

e=.12;a=1.0;b=-.2;

dx(1,1)= -x(2)+x(1)-x(1).^3;dx(2,1)=e*(x(1)-a*x(2)-b);

return-------------------------------------------------------------------------------

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Figure 6

PLOT V vs U

T=0:0.5:200;fori=1:100

X0=[-.575;-.385];[T,X]=ode45(@(t,x)exc(t,x),T,X0); plot(X(:,1),X(:,2));

end

fori=1:100X0=[-.485;-.385];[T,X]=ode45(@(t,x)exc(t,x),T,X0); plot(X(:,1),X(:,2));

end

fori=1:100X0=[-.385;-.385];[T,X]=ode45(@(t,x)exc(t,x),T,X0); plot(X(:,1),X(:,2));

end-------------------------------------------------------------------------------

PLOT V vs T

Figure 7

T=0:0.5:200;hold onfori=1:100

X0=[-.575;-.385];[T,X]=ode45(@(t,x)exc(t,x),T,X0); plot(T,X(:,1));

end

Figure 8

T=0:0.5:200;hold onfori=1:100


enddeltaU=.2

Figure 9

T=0:0.5:200;hold onfori=1:100


end

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PLOT V vs T

Figure 10

T=0:0.5:200;

hold onfori=1:100X0=[-.575;-.385];[T,X]=ode45(@(t,x)exc(t,x),T,X0); plot(T,X(:,2));

end

-------------------------------------------------------------------------------

Figure 11

T=0: 0. 5: 200;hol d onf or i =1: 100

X0=[ - . 485; - . 385] ; [ T, X] =ode45( @( t , x) exc( t , x) , T, X0) ; pl ot (T, X( : , 2) ) ;

end

-------------------------------------------------------------------------------

Figure 12

T=0:0.5:200;

hold onfori=1:100

X0=[-.385;-.385];[T,X]=ode45(@(t,x)exc(t,x),T,X0);

plot(T,X(:,2));end

-------------------------------------------------------------------------------------------------------------------------------

-------------------------------------------------------------------------------------------------------------------------------

Part 3 (1)

functionf=part3_1(u,N)% Setting up the Discretized equations

% parametersN=300;L=10;dx=L/N;e=.12;a=1.0;b=-.2;F=2.236;

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% Define discretization points and distance between themx=linspace(0,1,N);

% 1st equations at x=0f(1,1)=u(1)-.5;

f(301,1)=u(301);

%Internal equations.fori=2:299

f(i,1)=(u(i+1)-2*u(i)+u(i-1))/(dx^2)-u(300+i)+u(i)-u(i)^3; f(300+i)=(u(i+301)-2*u(300+i)+u(i+299))/(dx^2)+1.2*(u(i)-u(300+i)+.2);

end

% Last equation at x=Lf(N,1)=2*(u(299)-u(300))/(dx^2)-u(600)+u(300)-u(300)^3; f(N+300,1)=2*(u(599)-u(600))/(dx^2)+1.2*(u(300)-u(600)+.2);

return

% Choosing the number of discretization pointsN=600;

% Discretizing spaceR=linspace(0,1,300);

% Choosing an initial guess for the concetration inside the catalystu(1:N,1)=0.5;

% Solving the discretized problem using Newton methodu=fsolve(@(x)part3_1(x,N),u);

%plotting the resultshold onplot(R,u(1:300))plot(R,u(301:600))-------------------------------------------------------------------------------------------------------------------------------

-------------------------------------------------------------------------------------------------------------------------------

Part 3 (2)f uncti on CN

% Numeri cal Par amet ersc = 0. 5; % c = dt / dx 2N = 200; % number of gr i d poi nt sL = 20; % si ze of t he domai nt max = 200; % durati on

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a=1; b=- 0. 2; e=0. 12; d=0. 1; % Space Domai ndx = L/ ( N- 1) ; x = [ 0: dx: L]' ; % Ti me Domai ndt = c*dx 2; Nt = t max/ dt ;

% PDE Paramet ers

% I ni t i al condi t i onsu = - 0. 585*ones( N, 1) ; u( 1)=0. 5; v = - 0. 385*ones( N, 1) ;v(1) =0;%- - - - Mat r i ces f or Cr ank- Ni col son

% Lef t Hand Si deA = zeros( N, 3) ; A( : , 1) = - 1; A( : , 2) = 6- ( dx 2) ; A( : , 3) = - 1; LHSu = spdi ags( A, [ - 1: 1] , N, N) ;

C = zeros( N, 3) ; C( : , 1) = - 1; C( : , 2) = ( 4+( dx 2) *e*a) / d+2; C( : , 3) = - 1; LHSv = spdi ags( C, [ - 1: 1] , N, N) ;

% Neumann Boundar y condi t i on: x=LLHSu( 200, 199) =- 2; LHSu( 200, 200) =6- ( dx 2) ; LHSv(200, 199) =- 2; LHSv(200, 200) =( 4+( dx 2) *e*a) / d+2;

%Di r i chl et

% Ri ght Hand Si deB = zeros( N, 3) ; B( : , 1) = 1; B( : , 2) = 2+dx 2; B( : , 3) = 1; RHSu = spdi ags( B, [ - 1: 1] , N, N) ;

B= zeros( N, 3) ; B( : , 1) = 1; B( : , 2) = ( 4- ( dx 2) *e*a) / d- 2; B( : , 3) = 1; RHSv = spdi ags( B, [ - 1: 1] , N, N) ;

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% Neumann Boundary condi t i on: x=1RHSu( 200, 199) =2; RHSu( 200, 200) =2+dx 2; RHSv( 200, 199) =2; RHSv( 200, 200) =( 4- ( dx 2) *e*a) / d- 2;

%Di r i chl et Boundar y Condi t i onsRHSu( 1, : ) =0;

LHSu( 1, : ) =0; LHSu( 1, 1) =1; RHSv( 1, : ) =0; LHSv( 1, : ) =0; LHSv(1, 1) =1;

f or j = 1: Nt % Nonl i near Ter m

Fu = ( 2*dx 2) *( - v- u. 3) ; Fv= 2*dx 2*e/ d*( u- b) ; Fu( 1)=0. 5; Fv( 1) =0; r hsu = RHSu*u + Fu; r hsv = RHSv*v + Fv;

% Boundar y condi t i ons: Di r i chl et l ef t si deu = LHSu\ r hsu; v = LHSv\ r hsv;

% Pl ot hol d oni f( mod( j , 150) ==0)

pl ot ( x, u, ' b- ' , x, v, ' r . ' ) ; xl abel ( ' x' ) ; yl abel ( ' u' ) ; l egend( ' u' ) ; %axi s([ 0, L, - 1, 1] ) ; t i t l e( s pr i nt f ( ' Ti me = %f ' , dt * j ) ) ; pause( 0. 01) ;

endend

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Dynamics of Excitable Media

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