Attempted to access X(3); index out of bounds because numel(X)=1.

fourLumpData=[...

0, 6.593, 7.912, 4.222, 0

60, 5.149, 6.468, 5.667, 1.444

120, 3.371, 4.690, 7.444, 3.222

180, 3.149, 4.468, 7.667, 3.444

240, 2.704, 4.023, 8.111, 3.889

300, 2.037, 3.356, 8.778, 4.556

];

time=fourLumpData(2:6,1)'

tf=max(time)

tspan=(0:1:tf)'

X0(1:4,1)=fourLumpData(1,2:5)'

[t,Xcal]=ode45(@modelEquation,tspan,X0);

subfunction： function dxdt = modelEquation(X,K)

f1=K(2)*X(3)*X(4)- K(1)*X(1)*X(2);

f2= K(2)*X(3)*X(4)-K(1)*X(1)*X(2);

f3= K(1)*X(1)*X(2)-K(2)*X(3)*X(4);

f4=K(1)*X(1)*X(2)-K(2)*X(3)*X(4);

dxdt = [f1;f2;f3;f4]; ??? Attempted to access X(3); index out of bounds because numel(X)=1.

Error in ==> modelEquation at 2

f1=K(2)*X(3)*X(4)- K(1)*X(1)*X(2);

I dont't know why，and when I add the sentence disp(X(3)) after X0(1:4,1)=fourLumpData(1,2:5)'，

it displays 4.222 which is the right value. Then what dose the information mean?

Answer 1

[the cyclist]

I am not an expert in using ode45, but looking at the documentation for it, it is expecting the user-defined function to be f(t,y), which in your notation translates to modelEquation(t,X). So, your function seems to have two problems, I think:

• You defined your input arguments in the incorrect order. X should be the second argument. (Notice that if you breakpoint inside of modelEquation, K is taking the values you expected for X.)
• You are trying to use a second vector argument "K", but MATLAB is expecting a scalar argument "t" there instead (which it is going to fill in using your variable "tspan".)