A question about FitzHugh-Nagumo model
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I`ve been trying to built the FitzHugh-Nagumo model with a changeable input,but I failed to make the input changeable and I was unable to figure out why.
I defined three variables which I thought they should be the same,but only the variable 'eq1' and the way I put in the comment worked out,the others ended without graph out.I wonder how could this happened and how should I change the code of 'eq2' and 'eq3' .
syms t
tspan = [0 100];
v0 = 0; w0 = 0;
IC = [v0 w0];
A=0.5;
B=0.05;
Epsilon=0.005;
I=sin(t);
eq1=@(t,vw)fn(t,vw,A,B,Epsilon,sin(t));
eq2=@(t,vw)fn(t,vw,A,B,Epsilon,0)+[sin(t);0];
eq3=@(t,vw)fn(t,vw,A,B,Epsilon,I);
[t, vw] = ode45(eq1, tspan,IC);
%[t, vw] = ode45(@(t,vw)fn(t,vw,A,B,Epsilon,sin(t)), tspan,IC);
v = vw(:,1);
w = vw(:,2);
% results
plot(t,v,'r',t,w,'b'),grid
xlabel('t'),ylabel('v and w')
legend('v','w')
function dvwdt = fn(~,vw,A,B,Epsilon,I)
a = A;
b = B;
epsilon = Epsilon;
i = I;
v = vw(1);
w = vw(2);
dvwdt = [(v*(v-a)*(1-v)-w+i)/epsilon;
v-w-b];
end
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Risposta accettata
Alan Stevens
il 9 Nov 2021
Like this?
tspan = [0 100];
v0 = 0; w0 = 0;
IC = [v0 w0];
A=0.5;
B=0.05;
Epsilon=0.005;
I=@(t)sin(t); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
eq1=@(t,vw)fn(t,vw,A,B,Epsilon,sin(t));
eq2=@(t,vw)fn(t,vw,A,B,Epsilon,0)+[sin(t);0];
eq3=@(t,vw)fn(t,vw,A,B,Epsilon,I(t));
[t1, vw1] = ode45(eq1, tspan,IC);
[t2, vw2] = ode45(eq2, tspan,IC);
[t3, vw3] = ode45(eq3, tspan,IC);
v1 = vw1(:,1); v2 = vw2(:,1); v3 = vw3(:,1);
w1 = vw1(:,2); w2 = vw2(:,2); w3 = vw3(:,2);
% results
figure
plot(t1,v1,'r',t1,w1,'b'),grid
xlabel('t1'),ylabel('v1 and w1')
legend('v1','w1')
figure
plot(t2,v2,'r',t2,w2,'b'),grid
xlabel('t2'),ylabel('v2 and w2')
legend('v2','w2')
figure
plot(t3,v3,'r',t3,w3,'b'),grid
xlabel('t3'),ylabel('v3 and w3')
legend('v3','w3')
function dvwdt = fn(~,vw,A,B,Epsilon,I)
a = A;
b = B;
epsilon = Epsilon;
i = I;
v = vw(1);
w = vw(2);
dvwdt = [(v*(v-a)*(1-v)-w+i)/epsilon;
v-w-b];
end
3 Commenti
Alan Stevens
il 12 Nov 2021
In eq2, the value of I passed to fn is zero. To see the effect this has change eq2 so that it doesn't add the [sin(t); 0] term. You will then see that this last vector just produces oscillations about two very different values.
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