Index in position 2 exceed array bounds

Here's my code and im tryna figure out why i always get "index in position 2 exceed arra bounds"

What do i have to change?

function [t, y] = Euler_implicite(fonction, J ,t0,y0, h,e, Nt)

y(1,:) = y0;

t=zeros(Nt);

t(1) = t0;

for i = 1:Nt

g= y(:,i+1)-y(:,i)- h*fonction;

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

y = newton_systeme(g,J,SP, niteration, precision);

end

newton_systeme(g,J,y0,e,Nt);

end

[Bob Thompson]

Which line gives the error?

[Maureen Domtche]

this one " g= y(:,i+1)-y(:,i)- h*fonction; "

[Bob Thompson]

Are you intentionally recalculating y inside of your loop? Is it supposed to be the same y as y(1,:) = y0? Does the new calculation of y have the same size as the original?

[Maureen Domtche]

No the size is not supposed to remain de same.Its supposed to grow along with the iterations.i wrote y(1,:)=y0 , just as an a starting point like itteration 0.What can do plz to not have that error anymore?While making sure y0 stay de first element.

[Adam Danz]

Could you provide us with the values of the inputs?

fonction, J ,t0,y0, h,e, Nt

[Bob Thompson]

So y0 is a single value?

If that is true then the first iteration of y will also be a single values, so the size of y is 1x1. It is impossible then to gather y(1,2) (this comes from y(1,i+1) where i = 1 during the first iteration), which is causing the error.

To work around this you either need to add some kind of conditional statement for calculating g when i = 1, or you need to change the loop or equation to not look for values that don't exist yet.

If y0 is not a single value, then yes, please provide sample inputs for the function, as Adam requested.

[Maureen Domtche]

N = 100;

b = 1;

u = 0.05;r = 0.5;

nmax = 100;

precision = 1e-2;

h=0.5;

t0=0;

y0=10

sp= @(S,I) u*N-u*S-b/N*S*I;

ip = @(S,I) (b/N)*S*I-r*I-u*I;

fonction = {sp, ip};

J = {@(S,I)-u-b/N*I, @(S,I)-b/N*S;

@(S,I)(b/N)*I, @(S,I)(b/N)*S-r-u};

[Adam Danz]

What about Nt (and e)?

[Maureen Domtche]

Nt = 200 e = 0.05

Answer 1

[Adam Danz]

Just before the i-loop, you set y equal to 10.

%from your code...
y0 = 10; 
y(1,:) = y0;  % same as y = 10

Then on the first iteration of the i-loop, i=1 so when you execute this line

g= y(:,i+1)-y(:,i)- h*fonction;
% ----------

you're trying to access the second value of y but y only has 1 value (y=10)!

Even if that didn't cause an error this section below would (also) cause an error.

g= y(:,i+1)-y(:,i)- h*fonction;
%                   -----------

K>> h*fonction
Undefined operator '*' for input arguments of type 'cell'. 

So there's at least a couple areas of your code you need to re-think. In other words, your errors are conceptual errors. Your code isn't written properly and that's something you'll need to work on in the context of your project.

I suggest going through your code line-by-line to confirm that it's doing what it's supposed to be doing.

[Maureen Domtche]

I tried going line by line to figure out whats wrong with my code and it turns out that the problematic line is this one "g= y(:,n+1)-y(:,n)- h*f(y(:,n),t(n+1)); " So if any of you guys have ideas on how to fix that let me know please

[Adam Danz]

We don't have enough information to fix it (or confirm that it's broken). It's just a bunch of letters to us.

[Bob Thompson]

What you need to do to fix your code is to figure out how to get more than one element in y. As has been mentioned already the calculation of g is looking for two elements of y when only one exists. We are not particularly able to help you fix this particular problem because it is a very contextual solution, not something to do with the code. All we can tell you is that this is why the code isn't working.

Answer 2

[jahanzaib ahmad]

there is nothing in y(:,i+1)

[jahanzaib ahmad]

u want to minus before and after values but when u reach last value it gives error .

for for i = 1:Nt -1

the last value will not change .

[Maureen Domtche]

i tried it didn't work , i get the same erroe message