A question about FitzHugh-Nagumo model

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

Answer 1

[Alan Stevens]

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

[shuhao hu]

Thank you for the answer,but there is one thing that I still don't understand.How could the eq2 have the different result with the others,I thought they could all same result.If you know the reason,please tell why.Thank you!

[Alan Stevens]

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.

[shuhao hu]

Thank you very much!I got it,you must be Jesus!