Second order ODE15s going to infinity

Where μ = 1200, t is from 0 to 7000, and ODE15s is to be used.

tfinal = 7000;
delta_t = 0.01;
t = [0:delta_t:tfinal];
initial = [1 1];  % initial conditions
[t,w] = ode15s(@sub2,t,initial);    % line with error
figure
plot(t,w,'LineWidth',2)    



%%%%%%%%%% Subfunction
function d_dt = sub2(t,w)
 
mu = 1200;
% turning d^2y/dt^2 = mu(1-y^2)dydt - y     into     dzdt = mu(1-y)^2 - y
%                                                    dydt = z
z = w(1);  
y = w(2);

d_dt = zeros(2, 1); % initialize the column vector of equations
d_dt(1,1) = mu*z*(1-y)^2 - y;  % first equation
d_dt(2,1) = z;  % second equation

end

I made the code above, but i keep getting the error:

Warning: Failure at t=1.702777e-01.  Unable to meet integration tolerances without reducing the step size below the
smallest value allowed (4.440892e-16) at time t. 
> In ode15s (line 730)
  In sec_ord_diff (line 5) 

I think it has something to do with μ being too large, or maybe the time step is too small/ big. Can anyone help me fix the issue? Thank you

Answer 1

[Star Strider]

Your ‘sub2’ function is not correct.

sub2 = @(t,w,mu) [w(2); w(1)-mu*(w(1)^2-1)*w(2)];
mu = 1200;
tfinal = 7000;
delta_t = 0.1;
t = [0:delta_t:tfinal];
initial = [1 1];  % initial conditions
[t,w] = ode15s(@(t,w)sub2(t,w,mu),t,initial);    % line with error
figure
semilogy(t,w,'LineWidth',2)

[Tarek Chahattou]

Thank you so much! Also, the semilogy was a nice thing i didn't know about.

How did you get the w(1)-mu*(w(1)^2-1)*w(2)? When i try to simplify of of the equations, I keep getting mu*(1-w(1)^2)*w(2)-w(1). It seems to be a sign difference or something. I keep wanting to do:

dy/dt = z

dz/dt - μ(1-y^2)z + y = 0

then

dy/dt = z

dz/dt = μ(1-y^2)z - y

[Star Strider]

As always, my pleasure!

Actually, I let the Symbolic Math Toolbox do the work:

syms t y(t) mu 
Eq = diff(y,2) == mu*(1-y^2)*diff(y)+y;
[VF,Sbs] = odeToVectorField(Eq);

I then couid have used the matlabFunction function, however it was easier to just transcribe the ‘VF’ results.