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## Re: Non-linear ecuation... again...

 From: humberto Nanni Subject: Re: Non-linear ecuation... again... Date: Fri, 06 Oct 2000 12:29:28 -0500

```Yes, l2 is a (very precise) solution, but maybe there are another. Why i

point that?; because now i have the next equations to solve:

function F  =  evalF(x)

g1 = -19475/8314;
g2 = 0;
g3 = 0;
g4 = 20028/8314;
g5 = 395865/8314;
g6 = 92603/8314;
g7 = 0;

Ac = 2;
Ah = 14;
Ao = 3;

F(1) = log(x(1))-log(x(8))+x(9)+4*x(10)-g1;
F(2) = log(x(2))-log(x(8))+2*x(11)-g2;
F(3) = log(x(3))-log(x(8))+2*x(10)-g3;
F(4) = log(x(4))-log(x(8))+x(9)+x(11)-g4;
F(5) = log(x(5))-log(x(8))+x(9)+2*x(11)-g5;
F(6) = log(x(6))-log(x(8))+2*x(10)+x(11)-g6;
F(7) = log(x(7))-log(x(8))+x(9)-g7;
F(8) = x(1)+x(2)+x(3)+x(4)+x(5)+x(6)+x(7)-x(8);
F(9) = x(1)+x(4)+x(5)+x(7)-Ac;
F(10) = 4*x(1)+2*x(3)+2*x(6)-Ah;
F(11) = 2*x(2)+3*x(4)+2*x(5)+x(6)-Ao;

when I solve it, (damped NR), i get the next results:

x =

2.7379e-02
5.9502e-20
6.9452e+00
4.2874e-10
1.5000e+00
3.8942e-05
4.7264e-01
8.9452e+00
2.9405e+00
1.2654e-01
2.3230e+01

F =

-8.8818e-16
0.0000e+00
-1.6653e-16
-1.7764e-15
0.0000e+00
1.7764e-15
0.0000e+00
0.0000e+00
0.0000e+00
0.0000e+00
0.0000e+00

defect =  7.1275e-30

that looks like a solution, but by the physical meaning of the
variables, x(2) and x(4) can´t be zero at the same time (and they are
from a
practical sense). Yeah, the model could be wrong (i hope it doesn´t)....

anyway; somebody have a sugestion to found another solution (real
roots), ( i used a lot
of starting values, from  all x(i)  being 0.0001, to all x(i) equal to
100000, and always get the same), or a suggestion to demostrate that the

non-lin system has only one solution.

well thanks a lot.

H.Nanni

>Hi,
>
>The explanations below were not very satisfactory for me. So I played a

little
>bit with the example. If I did not make any mistake, I got the
following
>solutions starting with the initial guess identical to one:
>
>l2 =
>
>    0.1696
>    0.8485
>    1.5093
>    0.3211
>    5.8123
>    8.6608
>    0.1994
>   25.0683
>    0.7635
>
>They seem to be a solution. I was using a straighforward damped Newton
method
>and a rathet sophisticated affine invariant Newton method. They needed
8 and 7
>Newton steps, respectively, for solving this system.
>
>To be honest, an implementation of a trust-region method and the
back-tracking
>line search along NR did not converge.
>
>Hope that helps,
>
>Michael
>
>
Gerrit Visser wrote on Fre, 24 M<E4>r 2000:
> I've played around with the equations, and I can see why octave can't
> solve it. Some of the variables have a very large influence on the
> equations. At a guess, some of the gradients must be in the 10^9
region
> or higher. To see what I mean, set up a "cost" function which has the
>
> >>> Dirk Laurie <address@hidden> 23/March/2000 05:59pm >>>
> Cederik skryf:
> >
> > I have the next set of nine non-linear ecuations:
> > ln(a)-ln(f)+i+4g+(19720/8314)=0
> > ln(b)-ln(f)+2g+h-(192420/8314)=0
> > ln(c)-ln(f)+h+i-(200240/8314)=0
> > ln(d)-ln(f)+2h+i-(395790/8314)=0
> > ln(e)-ln(f)+2g=0
> > a+c+d-2=0
> > 4a+2b+2e-14=0
> > b+c+2d-3=0
> > a+b+c+d+e-f=0
> >
> > Where:
> > x[1]=a
> > x[2]=b
> > x[3]=c
> > x[4]=d
> > x[5]=e
> > x[6]=f
> > x[7]=g
> > x[8]=h
> > x[9]=i
> >
> > Octave can't solve it... (non-convergent), but in fact i know that
set of
> > ecuations has solution. Because is a Book example.
> > Octave can solve it if i put the initial x's [a,b,c,d,e,f,g,h,i] as
the
> > exactly solutions. But that way isn't usefull for me...
> > Any sugestions?
> >
>
>
>
>
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+---------------------------------------------------------------+
|  Michael Hanke                Royal Institute of Technology   |
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