why this equation is not ploting

Tstart = 100;
Tend = 1000.0;
Nt = 2000;
dT = (Tend-Tstart)/Nt;
X0 = 0.3;
Y0 = 0;
Z0 = 0;
N = 20;

k1=1;
k2=2.52;
k3=1;
k4=5;
I=1.5;
s=4;
xr=-1.6;
r=0.01;

T = zeros(Nt+1,1);
X = zeros(Nt+1,1);
Y = zeros(Nt+1,1);
Z = zeros(Nt+1,1);
a = zeros(N+1,1);
b = zeros(N+1,1);
c = zeros(N+1,1);

T(1) = 0.0;
X(1) = X0;
Y(1) = Y0;
Z(1) = Z0;

for j = 2:Nt+1
    
    a(1) = X(j-1);
    b(1) = Y(j-1);
    c(1) = Z(j-1);
    
    for k = 1:N
        
        a(k+1)=(b(k)-k1*a(k).^3+k2*a(k).^2-c(k)+I)/(k+1);
        b(k+1)=(k3-k4*a(k).^2-b(k))/(k+1);
        c(k+1)=r*(s*(a(k)-xr)-c(k))/(k+1);
    end
    
    x = a(1);
    y = b(1);
    z = c(1);
    for k = 2:N+1
        x = x + a(k)*dT^(k-1);
        y = y + b(k)*dT^(k-1);
        z = z + c(k)*dT^(k-1);
        
    end
    
    T(j) = T(j-1) + dT;
    X(j) = x; 
    Y(j) = y; 
    Z(j) = z; 
end

plot(T,X)

It's not showing the plot! How do I resolve this?

[Image Analyst]

It works for us.

[shiv gaur]

why this is not like that pl match with equation very famous hind marsh

https://en.wikipedia.org/wiki/Hindmarsh%E2%80%93Rose_model

Answer 1

[Sam Chak]

Hi @shiv gaur

Simulating the Hindmarsh–Rose model requires solving the differential equations using some numerical ode solvers. You can change the parameters in the code below for your study. If you find my Answer is helpful, please vote and Accept.

function Hindmarsh_Rose
close all
clc  

tspan = [0 1000];                % time span of simulation
y0 = [-1.5 -10 2];               % initial values
[t, y] = ode45(@(t, y) odefcn(t, y), tspan, y0);
plot(t, y(:,1), 'linewidth', 1.5)
grid on
xlabel('Time, t [sec]')
ylabel('x(t)')
title('Hindmarsh–Rose Neuronal Bursting Model')

end

function dydt = odefcn(t, y)    % Ordinary Differential Equations
    dydt = zeros(3,1);
    a = 1;
    b = 3;
    c = 1;
    d = 5;
    I = 2;
    r = 0.0021;
    s = 4;
    xR = -8/5;
    dydt(1) = y(2) - a*y(1)^3 + b*y(1)^2 - y(3) + I;
    dydt(2) = c - d*y(1)^2 - y(2);
    dydt(3) = r*(s*(y(1) - xR) - y(3));
end

Result:

[shiv gaur]

this may be the option

now we have use lorenz system using power series the program is given below so here we are applying power series to hindmarsh equation in same way as lorenz

program is here

Tstart = 0.0;
Tend = 1000.0;
Nt = 20000;
dT = (Tend-Tstart)/Nt;
X0 = 0.0;
Y0 = 1.0;
Z0 = 0.0;
N = 20;
SIGMA = 10;
R = 28;
B = 8/3;

T = zeros(Nt+1,1);
X = zeros(Nt+1,1);
Y = zeros(Nt+1,1);
Z = zeros(Nt+1,1);
a = zeros(N+1,1);
b = zeros(N+1,1);
c = zeros(N+1,1);

T(1) = 0.0;
X(1) = X0;
Y(1) = Y0;
Z(1) = Z0;


for j = 2:Nt+1
    
    a(1) = X(j-1);
    b(1) = Y(j-1);
    c(1) = Z(j-1);
    for k = 1:N
        SB = 0.0;
        SC = 0.0;
        for i= 0:k-1
            SB = SB + a(i+1)*c(k-i);
            SC = SC + a(i+1)*b(k-i);
        end
        a(k+1) = (SIGMA*(b(k) - a(k)))/k; 
        b(k+1) = ((R*a(k) - b(k)) - SB)/k ;
        c(k+1) = (-B*c(k) + SC)/k ;
    end  
    
    
    
    x = a(1);
    y = b(1);
    z = c(1);
    for k = 2:N+1
        x = x + a(k)*dT^(k-1);
        y = y + b(k)*dT^(k-1);
        z = z + c(k)*dT^(k-1);
    end
    
    
    T(j) = T(j-1) + dT;
    X(j) = x; 
    Y(j) = y; 
    Z(j) = z; 
end



plot(T,Y)

[shiv gaur]

from lorenz to apply HM equation

I have use the your value HM equation

the program using power series is right how to match with ode your program with power series program is given below

Tstart = 0;
Tend = 1000.0;
Nt = 2000;
dT = (Tend-Tstart)/Nt;
X0 = -1.5;
Y0 = -10;
Z0 = 2;
N = 200;

k1=1

k1 = 1

k2=3

k2 = 3

k3=1

k3 = 1

k4=5

k4 = 5

I=2

I = 2

s=4;
xr=-1.6;
r=0.0021;

T = zeros(Nt+1,1);
X = zeros(Nt+1,1);
Y = zeros(Nt+1,1);
Z = zeros(Nt+1,1);
a = zeros(N+1,1);
b = zeros(N+1,1);
c = zeros(N+1,1);

T(1) = 0.0;
X(1) = X0;
Y(1) = Y0;
Z(1) = Z0;

for j = 2:Nt+1
    
    a(1) = X(j-1);
    b(1) = Y(j-1);
    c(1) = Z(j-1);
    
    for k = 1:N
        
        a(k+1)=(b(k)-k1*a(k).^3+k2*a(k).^2-c(k)+I)/(k+1);
        b(k+1)=(k3-k4*a(k).^2-b(k))/(k+1);
        c(k+1)=r*(s*(a(k)-xr)-c(k))/(k+1);
    end
    
    x = a(1);
    y = b(1);
    z = c(1);
    for k = 2:N+1
        x = x + a(k)*dT^(k-1);
        y = y + b(k)*dT^(k-1);
        z = z + c(k)*dT^(k-1);
        
    end
    
    T(j) = T(j-1) + dT;
    X(j) = x; 
    Y(j) = y; 
    Z(j) = z; 
end




plot(T,X)

[shiv gaur]

if any one help to match the graph accept the answer

[shiv gaur]

we are matching the ode with power series in lorenz system why in the HMarsh the ode plot is not matching with power series

[Sam Chak]

Ok @shiv gaur, please show all relevant equations and explain how your power series works. Are you researching on a never-seen-before novel mathematical power series based numerical solver? The proposed ode45 solution does not meet your expectation, although it took some time to research on the Hindmarsh–Rose model on a few papers and type out the code.

[shiv gaur]

I have use in lorenz equation just as butterfly using power series so this is to apply in rossler also

why is not applicable to HMarsh equation

[shiv gaur]

I have to match the ode solution with bpower series as in same as in lorenz system

[shiv gaur]

this is the lorenz type fig 4 ,5 paper that program works but in HM equation this is not matching properly

[Torsten]

@shiv gaur

for k = 1:N        
    a(k+1)=(b(k)-k1*a(k).^3+k2*a(k).^2-c(k)+I)/(k+1);
    b(k+1)=(k3-k4*a(k).^2-b(k))/(k+1);
    c(k+1)=r*(s*(a(k)-xr)-c(k))/(k+1);
end

If it were that easy to get the Taylor coefficients for the Hindmarsh-Rose model, the loop for the Lorenz equation would have been

for k = 1:N
    a(k+1) = (SIGMA*(b(k) - a(k)))/k; 
    b(k+1) = ((R*a(k) - b(k)) - a(k)*c(k))/k ;
    c(k+1) = (-B*c(k) + a(k)*b(k))/k ;
end  

instead of

for k = 1:N
    SB = 0.0;
    SC = 0.0;
    for i= 0:k-1
        SB = SB + a(i+1)*c(k-i);
        SC = SC + a(i+1)*b(k-i);
    end
    a(k+1) = (SIGMA*(b(k) - a(k)))/k; 
    b(k+1) = ((R*a(k) - b(k)) - SB)/k ;
    c(k+1) = (-B*c(k) + SC)/k ;
end  

wouldn't it ?

[shiv gaur]

in hmarsh equation we have not use SB and SC so any product of summation is not there so we use same equation

Tstart = 0;

Tend = 1000.0;

Nt = 2000;

dT = (Tend-Tstart)/Nt;

X0 = -1.5;

Y0 = -10;

Z0 = 2;

N = 200;

k1=1

k2=3

k3=1

k4=5

I=2

s=4;

xr=-1.6;

r=0.0021;

T = zeros(Nt+1,1);

X = zeros(Nt+1,1);

Y = zeros(Nt+1,1);

Z = zeros(Nt+1,1);

a = zeros(N+1,1);

b = zeros(N+1,1);

c = zeros(N+1,1);

T(1) = 0.0;

X(1) = X0;

Y(1) = Y0;

Z(1) = Z0;

for j = 2:Nt+1

a(1) = X(j-1);

b(1) = Y(j-1);

c(1) = Z(j-1);

for k = 1:N

a(k+1)=(b(k)-k1*a(k).^3+k2*a(k).^2-c(k)+I)/(k+1);

b(k+1)=(k3-k4*a(k).^2-b(k))/(k+1);

c(k+1)=r*(s*(a(k)-xr)-c(k))/(k+1);

end

x = a(1);

y = b(1);

z = c(1);

for k = 2:N+1

x = x + a(k)*dT^(k-1);

y = y + b(k)*dT^(k-1);

z = z + c(k)*dT^(k-1);

end

T(j) = T(j-1) + dT;

X(j) = x;

Y(j) = y;

Z(j) = z;

end

plot(T,X)

[shiv gaur]

yes in case of lorenz equation is correct in Hmarsh eq is different

https://en.wikipedia.org/wiki/Hindmarsh%E2%80%93Rose_model

[Sam Chak]

@shiv gaur Your Hindmarsh–Rose code generates NaN in a, b, c in the loop "for k = 1:N". The HR ODEs are different from the Lorenz ODEs because they contain power terms. But there is none in the Lorenz system.

When you use .^ operator, the power() function is called, and it executes

.

The NaN was generated most likely because on the 3rd iteration, a(k) becomes negative and log(a(k)) produces a complex number, and your looping code creates a chain of issues. That's why it plots until T = 0.5 sec instead of 1000 sec.

Tstart = 0;

Tend = 1000.0;

Nt = 2000;

dT = (Tend-Tstart)/Nt;

X0 = -1.5;

Y0 = -10;

Z0 = 2;

N = 200;

k1 = 1;

k2 = 3;

k3 = 1;

k4 = 5;

I = 2;

s = 4;

xr = -1.6;

r = 0.0021;

T = zeros(Nt+1,1);

X = zeros(Nt+1,1);

Y = zeros(Nt+1,1);

Z = zeros(Nt+1,1);

a = zeros(N+1,1);

b = zeros(N+1,1);

c = zeros(N+1,1);

T(1) = 0.0;

X(1) = X0;

Y(1) = Y0;

Z(1) = Z0;

for j = 2:Nt+1

a(1) = X(j-1);

b(1) = Y(j-1);

c(1) = Z(j-1);

for k = 1:N % NaN is produced in this loop due to the power terms in Hindmarsh–Rose

a(k + 1) = (b(k) - k1*a(k).^3 + k2*a(k).^2 - c(k) + I)/(k + 1);

b(k + 1) = (k3 - k4*a(k).^2 - b(k))/(k + 1);

c(k + 1) = r*(s*(a(k) - xr) - c(k))/(k + 1);

% Test (no power terms)

% a(k + 1) = (b(k))/(k + 1);

% b(k + 1) = (k3 - k4*a(k) - b(k))/(k + 1);

% c(k + 1) = r*(s*(a(k) - xr) - c(k))/(k + 1);

end

x = a(1);

y = b(1);

z = c(1);

for k = 2:N+1

x = x + a(k)*dT^(k-1);

y = y + b(k)*dT^(k-1);

z = z + c(k)*dT^(k-1);

end

T(j) = T(j-1) + dT;

X(j) = x;

Y(j) = y;

Z(j) = z;

end

plot(T, X)

Try fixing that or you may want to request for help from the Experts like @Torsten and @Image Analyst.

[Torsten]

@shiv gaur

yes in case of lorenz equation is correct in Hmarsh eq is different

No, it's not different. The rules for calculating the coeffcients of the Taylor series for the solutions of systems of ordinary differential equations are not dependent on the differential equation you solve. They are valid in general.

[shiv gaur]

so how to resolve this problem

[Torsten]

What shall I say ?

By correctly calculating the a, b and c arrays.

[shiv gaur]

sir you have all the things equation papers so you are req to resolve the problem

[shiv gaur]

if any one help appriciation for solving hmarsh using power series