How to solve first-order nonlinear differential equation where the solution is coupled with an integral?

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Califfo 2019-7-3

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评论： David Goodmanson 2019-7-8

采纳的回答： David Goodmanson

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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)

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Answer 1

David Goodmanson 2019-7-4

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https://ww2.mathworks.cn/matlabcentral/answers/470110-how-to-solve-first-order-nonlinear-differential-equation-where-the-solution-is-coupled-with-an-integ#answer_381885

编辑：David Goodmanson 2019-7-4

在 MATLAB Online 中打开

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