Asked by Califfo
on 3 Jul 2019

I'm trying to solve this nonlinear ODE

- where q is a nonlinear function, solution of ODE;
- represents the velocity and it is equal to: ;
- tis the time:
- the over dot denotes the derivative with respect to time;
- the initial condition is

λ is a degradation parameter of function q and it is equal to:

The integral depends to the solution of ODE.

So I have written this code, but the solution is bad because there isn't degradation of q function

clc

clear

close all

tspan = [0 pi*5];

q0 = 0;

x=@(t)t.*sin(t);

xdot=@(t)t.*cos(t)+sin(t);

lambda = @(t,q) 1+0.01*integral(@(t)q*xdot(t),0,t,'ArrayValued',true);

qdot = @(t,q) xdot(t)*(1-(abs(q)*lambda(t,q)*(0.5+0.5*sign(xdot(t)*q))));

[t,q] = ode45(qdot, tspan, q0);

plot(x(t),q,'LineWidth',2)

Answer by David Goodmanson
on 4 Jul 2019

Edited by David Goodmanson
on 4 Jul 2019

Accepted Answer

Hi Califfo;

This may be in line with what you want. At least it's changing size It's based on the idea that you know not only qdot, but also lambdadot. That quanity is simply the integrand, .01*q*xdot, and you know that lambda has a starting value of 1. You can make a vector from [q, lambda], which I arbitrarily called z, and then use ode45..

tspan = [0 pi*10];

g0 = 0;

lam0 = 1;

z0 = [g0; lam0];

[t,z] = ode45(@fun, tspan, z0);

x = t.*sin(t);

plot(x,z(:,1))

function zdot = fun(t,z)

xdot = t*cos(t)+sin(t);

q = z(1);

lam = z(2);

qdot = xdot*(1-(abs(q))*(lam/2)*(1+ sign(xdot*q)));

lamdot = 0.01*q*xdot;

zdot = [qdot; lamdot]

end

Califfo
on 4 Jul 2019

David it's perfect, thank you very much!!!

Probably I think I will write you if something is not quite clear to me.

David Goodmanson
on 8 Jul 2019

Yes, let me and the website know if there is anything that needs clarification.

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