RK4 method to solve a 2nd ODE.

I am trying to solve a 2nd ODE using the RK4 method, however the increase in time step doesn't seem to be working correctly?

%% Initial Conditions

u0 = 0.719622752231628; % Initial Displacement

z0 = 0; % Initial Velocity

%%Defining the equations

f1 = @(u,z,t) z0; % Initial f1

f2 = @(u,z,t) A*z0(t)+B*u0(t); % Initial f2

% RK4 Loop

for i = 0:1:4001

% Update time

t(i+1) = t(i)+dt;

% Update u and z

K1 = f(x(i),y(i));

L1 = f(x(i),y(i));

K2 = f(x(i)+0.5*h,y(i)+h*0.5*k1);

L2 = f(x(i)+0.5*h,y(i)+h*0.5*k1);

K3 = f(x(i)+0.5*h,y(i)+h*0.5*k2);

L3 = f(x(i)+0.5*h,y(i)+h*0.5*k2);

K4 = f(x(i)+ h,y(i)+h *k2);

L4 = f(x(i)+ h,y(i)+h *k2);

u(i+1) = y(i)+h/6*(K1+2*K2+2*K3+K4);

z(i+1) = u(i)+h/6*(L1+2*L2+2*L3+L4);

end

Error message:

Array indices must be positive integers or logical values.

Error in RK4 (line 26)

t(i+1) = t(i)+dt;

Answer 1

[Torsten]

1. The definition of the function handles f1 and f2 is wrong:

     f1 = @(u,z,t) z0; % Initial f1
     f2 = @(u,z,t) A*z0(t)+B*u0(t); % Initial f2

You don't reference u, z and t in the definition of the functions

Further z0 and u0 are scalars, not functions. So z0(t) and u0(t) don't exist.

2. Array indices start with 1. You address t(0) in t(i+1) = t(i) + dt.

3. f is undefined. You defined f1 and f2, but no f.

4. dt and h are identical. So give them the same name.

5. dt and h are undefined.

6. k1 and k2 are undefined. You work with K1, K2, K3 K4,L1,L2,L3 and L4.

7. The updates

     K4 = f(x(i)+    h,y(i)+h    *k2);
     L4 = f(x(i)+    h,y(i)+h    *k2);

are wrong.