why range of x axis of time is not showing

Tstart = 0.0;
Tend = 100.0;
Nt = 2000;
dT = (Tend-Tstart)/Nt;
X0 = 1;
Y0 = 2;
Z0 = 3;
N = 20;

k1=1
k2=3
k3=1
k4=5
I=0.5
s=4;
xr=-1.6;
r=0.0001;
% Initialize coefficient arrays
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




figure(1)
plot(T,X,'Color','red')
hold on
figure(2)
plot(T,Y,'Color','red')
hold on
figure(3)
plot(T,Z,'Color','red')
hold on  
figure(4)
plot3(X,Y,Z,'Color','red')
hold on

the time range is coming 1.5 in
place of 100

[Rik]

Serious question: why do you refuse to use formatting?

Answer 1

[Walter Roberson]

Because after that your values go to infinity or nan. Automatic selection of plot axes range is based only on the finite coordinates of what is drawn.

[shiv gaur]

why other type is showing time value on x axis

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;
%
% Initialize coefficient arrays
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
    % Compute Taylor coefficients      
    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
    
    % Prepare for T = T + dT
    T(j) = T(j-1) + dT;
    X(j) = x; 
    Y(j) = y; 
    Z(j) = z; 
end
figure(1)
plot(T,X,'Color','red')
hold on
figure(2)
plot(T,Y,'Color','red')
hold on
figure(3)
plot(T,Z,'Color','red')
hold on 

[shiv gaur]

pl resolve this problem for time rannge

Answer 2

[Davide Masiello]

That's happening because all the values of X,Y, and Z after T = 1.5 are NaN, which will not be plotted by MatLab.

I guess there might be a mistake in your equations that causes them to predict your variables to be NaN.

[shiv gaur]

in ploting other software like scilab the graph is going upto 100

[Davide Masiello]

yes, but as you can see, there's no line after T=1.5, because the y-values are NaNs.

You can obtain the same result as shown in the plot above by adjusting the plot axis in MatLab.

You could write

figure(1)
plot(T,X,'Color','red')
axis([0 100 -inf +inf])
hold on
figure(2)
plot(T,Y,'Color','red')
axis([0 100 -inf +inf])
hold on
figure(3)
plot(T,Z,'Color','red')
axis([0 100 -inf +inf])
hold on  
figure(4)
plot3(X,Y,Z,'Color','red')
axis([0 100 -inf +inf])
hold on

[shiv gaur]

in which parameter is the problem for inf or nan

[Walter Roberson]

Y0.

I showed you the mathematical analysis before using the Symbolic Toolbox. If your initial conditions are 1 then your formulas explode.

[shiv gaur]

what value of y0 we have to take so in fullrange

[Torsten]

@shiv gaur

Show us the ODEs behind your Taylor expansion.

[shiv gaur]

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

these are the equation

[Torsten]

And from where did you get the Taylor series expansion of the solution to compute the Taylor coefficient vectors a,b and c ?

[shiv gaur]

from your solution lorenz convert in to marsh model no multiplication summation i so we have not use taylor

we use only simple but matching is not there

[Walter Roberson]

I modified the code to use symbolic Y0, and added some optimizations, and I cut down on the loop limits a lot.

The result is still insanely slow.

I am not even sure what step is so slow: I asked it about an hour ago to pause and it has not done so yet, so it must be deep inside the symbolic engine calculating something

In the part that I debugged out earlier, after a very small number of steps, the calculations for a, b, c, involved polynomials with coefficients on the order of 10^26 and degree 5. At least at that stage, the calculations would have exploded for Y0 > 1e-5

[Walter Roberson]

I had my program continue calculatign with symbolic Y0.

Before the end of the 7th j loop, some of the expressions involve Y0 to the power of over 320000, and coefficients over 10 to the million.

The arithmetic round-off involved in the calculations is horrible.

If your Y0 is anything with absolute value greater than 1 then for certain your values would explode beyond the range of double precision; if your Y0 is anything with absolute value less than 1, then most of the terms are going to underflow.

[Walter Roberson]

I substituted in 10^-10 for Y0 in c(6) . The result was less than -2 times 10 to the one million.

There is just no possibility of getting useful numbers out of the equations you have given.