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

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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?

采纳的回答

the cyclist
the cyclist 2011-8-2
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".)

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