Unrecognized function or variable 'ode4'.

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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, '--')

Accepted Answer

Walter Roberson
Walter Roberson on 11 Oct 2022
Edited: Walter Roberson on 11 Oct 2022
  2 Comments
Khang Nguyen
Khang Nguyen on 11 Oct 2022
Edited: Khang Nguyen on 11 Oct 2022
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
Walter Roberson on 11 Oct 2022
Did you download the .zip file from that Question, and unzip it and place the directory on your MATLAB path ?

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More Answers (1)

Torsten
Torsten on 11 Oct 2022
Edited: Torsten on 11 Oct 2022
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
  3 Comments
Walter Roberson
Walter Roberson on 11 Oct 2022
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
Torsten on 11 Oct 2022
If you compare your code and ode4, you'll see that they are identical. So there should be no difference in the results.

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