Index exceeds dimensions error

2017 2 13
1 个回答
更新时间：2021 8 20
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0 个投票

I'm trying to code the predictor-corrector method, but I keep getting an "Index exceeds dimensions", and I don't see where the error is.
%function[t y]=abmpc4(f,a,b,y0,h)
f=@(t,y)((y/t)-(y/t)^2);
h=0.1;
a=1;
b=2;
y0=1;
n=(b-a)/h;
t(1)=a;
y(1)=y0;
w=[];
s=[];
s(1)=a;
w(1)=y0;
%RK4 method to get initial values
for i=1:3
k1=h*f(t,y);
k2=h*f(t+(h/2),y+(k1/2));
k3=h*f(t+(h/2),y+(k2/2));
k4=h*f(t+h,y+k3);
y=y+(k1+(2*k2)+(2*k3)+k4)/6;
t=t+h;
w=[w,y];
s=[s,t];
end
%truevals(1,1)=y0;
%Adams-Bashforth 4-Step
% i=4;
% y=y+(55/24)*h*f(s(i,1),w(i,1))-(59/24)*h*f(s(i-1,1),w(i-1,1))+(37/24)*h*f(s(i-2,1),w(i-2,1))-(9/24)*h*f(s(i-3,1),w(i-3,1)); %predictor
% %t=t+h;
% w(i+1,1)=y;
% s(i+1,1)=t;
for i=4:n
%truevals=g;
t=t+h;
y=y+(h/24)*(55*f(t(i),y(i))-59*f(t(i-1),y(i-1))+37*f(t(i-2),y(i-2))-9*f(t(i-3),y(i-3))); %predictor
y=y+(h/24)*(9*f(t(i+1,1),y(i+1,1))+19*f(t(i,1),y(i,1))-5*f(t(i-1,1),y(i-1,1))+f(t(i-2,1),y(i-2,1))); %corrector
w=[w,y];
s=[s,t];
end

1 个评论
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KSSV
KSSV 2017-2-13
You are trying to access t as a vector, but it is a scalar. Check it.

回答（1 个）

Star Strider
Star Strider 2017-2-13

0 个投票

The error:
Index exceeds matrix dimensions.
Error in Elipses (line ###)
y=y+(h/24)*(55*f(t(i),y(i))-59*f(t(i-1),y(i-1))+37*f(t(i-2),y(i-2))-9*f(t(i-3),y(i-3)));
%predictor
The problem is that here ‘t’ is a (1x1) double and ‘i’ is 4. That condition with throw the error you see.
This situation:
q = pi
r = q(4)
will throw the same error.

4 个评论
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cee878
cee878 2017-2-13
How can I fix this?
Star Strider
Star Strider 2017-2-13
It is difficult for me to follow your code. The solution is for ‘t’ and ‘y’ to be vectors, since you are also referring to previous values as well as current values.
It will probably be best to preallocate those vectors prior to the loop. Consider in your initial assignments:
t = [a 0 0 0];
y = [y0 0 0 0];
s = [a 0 0 0];
w = [y0 0 0 0];
Those assignments will create your necessary vectors.
cee878
cee878 2017-2-13
Okay. I fixed it like this:
%function[t y]=abmpc4(f,a,b,y0,h)
f=@(t,y)((y/t)-(y/t)^2);
h=0.1;
a=1;
b=2;
y0=1;
n=(b-a)/h;
t(1,1)=a;
y(1,1)=y0;
w=zeros(n+1,1);
s=zeros(n+1,1);
s(1,1)=a;
w(1,1)=y0;
%RK4 method to get initial values
for i=1:3
k1=h*f(t,y);
k2=h*f(t+(h/2),y+(k1/2));
k3=h*f(t+(h/2),y+(k2/2));
k4=h*f(t+h,y+k3);
y=y+(k1+(2*k2)+(2*k3)+k4)/6;
t=t+h;
w(i+1,1)=y;
s(i+1,1)=t;
end
%truevals(1,1)=y0;
for i=4:n
%truevals=g;
r(i,1)=f(s(i),w(i));
y=y+(h/24)*(55*r(i,1)-59*r(i-1,1)+37*r(i-2,1)-9*r(i-3,1)); %predictor
display(y);
t=t+h;
y=y+(h/24)*(9*f(s(i+1,1),w(i+1,1))+19*r(i,1)-5*r(i-1,1)+r(i-2,1)); %corrector
w(i+1,1)=y;
s(i+1,1)=t;
end
But I don't get an error, but when my code runs, I get NaN values. I don't get what's wrong though.
Star Strider
Star Strider 2017-2-13
You need to update the vectors in your loops by subscripting them.
For example:
for i=4:n-1
. . .
y(n+1) = y(n) + ...
. . .
end
Then reference them appropriately in your other equations that use them, similar to what I outlined here. You will have to experiment.

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cee878
cee878
2017-2-13

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