Unrecognized function or variable 'ode4'.

I am trying to use ode4 to validate with the Runge-Kutta method, but keep getiing this error when I call ode4 "Unrecognized function or variable 'ode4'."

h=0.1;                                             % step size
x = 0:h:60;                                        
%y(1) = [-0.5;0.3;0.2];
y(1) = -0.5; 
F_xy = @(t) -3*cos(t/2) + 4*sin(t/2) + 1;                    
for i=1:(length(x)-1)                              % calculation loop
    k_1 = F_xy(x(i));
    k_2 = F_xy(x(i)+ 0.5*h*k_1);
    k_3 = F_xy((x(i)+0.5*h*k_2));
    k_4 = F_xy((x(i)+k_3*h));
    y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;  % main equation
  
end
% validate using a decent ODE integrator  
y0 = -0.5;
yx = ode4(F_xy, 0,h,60, y0)
plot(x,y,'o-', x, yx, '--')

Answer 1

[Walter Roberson]

https://www.mathworks.com/matlabcentral/answers/98293-is-there-a-fixed-step-ordinary-differential-equation-ode-solver-in-matlab-8-0-r2012b contains the code.

[Khang Nguyen]

I still get the same error.

h=0.1;                                             % step size
x = 0:h:60;
F_xy = @(t) -3*cos(t/2) + 4*sin(t/2) + 1;
y0 = -0.5;
L = length(x)
tspan = linspace(0,60,L)
yx = ode4(F_xy, tspan, y0)

[Walter Roberson]

Did you download the .zip file from that Question, and unzip it and place the directory on your MATLAB path ?

Answer 2

[Torsten]

Look below at how k_1,...,k_4 must be computed in the case that your ODE function only depends on t, not y.

The classical Runge-Kutta boils down to the usual Simpson's rule.

h=0.1;                                             % step size
x = 0:h:60;                                        
%y(1) = [-0.5;0.3;0.2];
y(1) = -0.5; 
F_xy = @(t) -3*cos(t/2) + 4*sin(t/2) + 1;                    
for i=1:(length(x)-1)                              % calculation loop
    k_1 = F_xy(x(i));
    k_2 = F_xy(x(i)+0.5*h);
    k_3 = F_xy(x(i)+0.5*h);
    k_4 = F_xy(x(i)+h);
    y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;  % main equation
    
end
% validate using a decent ODE integrator  
y0 = -0.5;
yx = ode4(@(t,y)F_xy(t), 0,h,60, y0);
plot(x,y,'o-', x, yx, '--')

function yout = ode4(F,t0,h,tfinal,y0)
% ODE4  Classical Runge-Kutta ODE solver.
%   yout = ODE4(F,t0,h,tfinal,y0) uses the classical
%   Runge-Kutta method with fixed step size h on the interval
%      t0 <= t <= tfinal
%   to solve
%      dy/dt = F(t,y)
%   with y(t0) = y0.

%   Copyright 2014 - 2015 The MathWorks, Inc.

y = y0;
yout = y;
for t = t0 : h : tfinal-h
    s1 = F(t,y);
    s2 = F(t+h/2, y+h*s1/2);
    s3 = F(t+h/2, y+h*s2/2);
    s4 = F(t+h, y+h*s3);
    y = y + h*(s1 + 2*s2 + 2*s3 + s4)/6;
    yout = [yout; y]; %#ok<AGROW>
end
end

[Khang Nguyen]

Thank you. How do you find the error between these two value? I tried to subtract the two data points.

f = y - yx'
plot(x,f)

[Walter Roberson]

The error between what two values? You cannot calculate the error unless you have an analytic solution.

h=0.1;                                             % step size
x = 0:h:60;                                        
y(1) = -0.5; 
F_xy = @(t) -3*cos(t/2) + 4*sin(t/2) + 1;
syms t C
intF = int(F_xy(t), t)

intF = 

eqn = subs(intF, t, 0) + C == y(1)

eqn = 

sol = solve(eqn)

sol = 

intF = subs(intF + C, C, sol)

intF = 

... that should be the analytic solution.

[Torsten]

@Khang Nguyen

If you compare your code and ode4, you'll see that they are identical. So there should be no difference in the results.