How to use ode solvers inside while/for loops for animation?

It is not lucid from the documnation how to use the solution of an ODE at each step. The examples show how one can pass the entire time span which is good for some cases. However, I need to obtain the solution at each step for animation purposes. I'm using fixed ode solver (i.e. ode4).

classdef System < handle
    properties
        t = 0;
        stepSize=0.01;
        x0 = [0;0];
        %Parameters:
        m = 0.5;       % mass (Kg)
        d = 0.0023e-6; % viscous friction coefficient
        L = 1;         % arm length (m)
        I;
        g = 9.81;      % acceleration due to gravity m/s^2
        % PID tuning
        Kp = 5;
        Kd = 1.9;
        Ki = 0.02;
        % error
        error = 0;
        % control input
        u=0;
        x;
        dx;
        xVec;
        dxVec;
        errorVec;
        idx=1;
    end
    
    methods
        function obj = System()
            obj.I = 1/3*obj.m*obj.L^2; % inertia seen at the rotation axis. (Kg.m^2)
            obj.xVec(obj.idx) = obj.x0(1);
            obj.dxVec(obj.idx) = obj.x0(2);
            obj.errorVec(obj.idx) = 0;
        end
        function move(this)
            tspan = [this.t this.t+this.stepSize];
            y = ode4(@this.ODESolver, tspan, this.x0);
            this.x = y(1);
            this.dx = y(2);
            this.t = this.t + this.stepSize;
        end
        function updateController(this)
            Kp = this.Kp;
            Kd = this.Kd;
            Ki = this.Ki;
            % u: joint torque
            this.u = Kp*(pi/2 - this.x) + Kd*(-this.dx) + Ki*this.error;
            this.error = this.error + (pi/2 - this.x); 
        end
        function dx = ODESolver(this,t,x)
            m=this.m;
            g=this.g;
            I=this.I;
            d=this.d;
            
            dx = zeros(2,1);
            dx(1) = x(2);
            dx(2) = 1/I*(this.u - d*this.dx - m*g*L*sin(this.x));
        end
    end
end

The main script is

clear all
clc

sys = System();

for i=1:100
    sys.updateController;
    sys.move;
end

I get errors at

function move(this)
    tspan = [this.t this.t+this.stepSize]; % <---- I've tried to solve it with this but to no avail. 
    y = ode4(@this.ODESolver, tspan, this.x0); % <--- Error
    this.x = y(1);
    this.dx = y(2);
    this.t = this.t + this.stepSize;
end

Any suggestions? Also, how can I do same scenario with ode45?

Answer 1

[Steven Lord]

If you want to plot the solution of the system of ODEs while it's being solved, use the OutputFcn option in the ODE options structure. See the ballode example for an illustration of how this is used.

If you want to plot the solution after it has been solved, either call the ODE solver with two outputs (as shown in the non-stiff van der Pol equation example on this documentation page) and possibly a tspan input with more than two elements or call it with one output that you can evaluate at specified times using deval.

[Bandar]

Thank you for this answer. But this isn't exactly what I'm looking for. I'm not waiting for any event rather I'm trying to plot the solution from a successful ode45 solution after each iteration. How to implement an event that captures this scenario?

[Steven Lord]

"If you want to plot the solution of the system of ODEs while it's being solved, use the OutputFcn option in the ODE options structure. See the ballode example for an illustration of how this is used."

The description of the OutputFcn option on the documentation page for the odeset function, which creates the ODE options structure, states in part: "The ODE solver calls the output function after each successful time step."

[Bandar]

I've read OutputFcn. It did solve the problem paritally. The problem with OutputFcn is the fact that we don't have full control of the code. I mean if my code will do other stuff, I need to put them inside OutputFcn. For example, sending the command using tcp/ip. I couldn't find a solution where I can put ode45 inside while/for loops similar to Odeint C++ library where the stepper can perform the numerical method once at a time.