Check, for constructing nice set of linear equations

Nov 2016
24
0
Aus
Hi

Well, I was struggling with solving these linear equations for days. more accurately, It solved but the answers do not check with what I have, so want to find where I have mistaken. It would be appreciated if you can check with me that my understanding is correct for the problem. (do not lose the interest please, it is nice math, promise you)

1- the equations are :

Aij*fj + Bij*gj =S1. S0j -----(1)
Cij*fj + Dij*gj =S2. S0j------(2)

i,j=0,1,2,3.............N

fj,&gj are variables f0,f1,---fN; g0,g1----gN

The constants are: A,B,C,D, (vary for i,&j) --see attached, please.

S1, S2 (are not depending on i&j): they are constants with one value for each

Sij = Kroencker delta (=1, when i=j, & =0:if i≠j )

I used, Sij = zero matrix with diagonal elements = 1
I used, S0j= zero when j≠0, else S0j=1

the details of the constants calculation (depend on i,& j) are in the attached

2-My questions:

-are the equations in a matrix form, or not?

what do I mean, here?

well, if the equations are in a matrix form then the equations will be like [ row (i) from A * column vector of, f, Plus row (i) from B * column vector of g = S1.S0j],

then the number of equations is 2*(N+1) ..(2 for 2 sets of equations, and (+1) to account the N=0,
the number of variables in each equation is 2*(N+1)

But what about (S0j) as each equation, in this case, has all (j) values!. there is a mistake here, right?

So the other option which seems more reasonable based on the presented equations and was tried but no success in the results is

for each (i, j) there is one equation: i.e.; A01*f1+B01*g1=(s1*0; as j≠0, so S0j=0), then, (A01*f1+B01*g1=0)

This will end a number of equations = 2*(N+1)^2
many of these equations are zeros in the matrix of equations
the number of varibales= 2*(N+1)
As the number of equations > a number of variables, lease square method used to minimize the square of error in each equation). This will end up with a number of equation = number of variables can be solved linearly.

-But, in this case, as you clearly can see (trying an example of N=2)
The values of ( fj, & gj =0 for j≠0 !!).
and then the solutions are always for f0, g0 only.
I feel there is a mistake here, as increasing the terms N (then the number of variables) should enhance/convergence the results


Your comment & suggestion would be of great help

Please see the attached

Thanks
 

Attachments