       least square for 6 variables

• To: mathgroup at smc.vnet.net
• Subject: [mg28183] least square for 6 variables
• From: Stefan.Schenderlein at ferring.de
• Date: Tue, 3 Apr 2001 02:38:01 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```Hi,
using FindMinimum[] with Method->LevenbergMarquardt I did not understand why
I had to expand the equations manually. When I work with the simplified
version the algorithm stops in a local minimum or something without meaning.
For sure I have checked with Mathematica whether the expansion /
simplification was really the same. (There are some parts of the notebook
below. I will send the complete code, if necessary.)
I do not have a problem with that since the expanded version now gives me
the solutions I was looking for. But in a discussion with Visual Numerics
Germany they told me that their IMSL Math Library would rather like the
simplified equations.
I did not find much about correct using of FindMinimum in the manual and
FAQ. My guess is that I run into the trouble due to the extensive use of
subscripts. Perhaps someone could give me a hint.

Thanks.

stefan.schenderlein at ferring.de

Chemical potential for the polymer poor phase: (example for component1)
This is what i would like to use
In:=
\!\(\(\[Mu]\_A1 :=
Log[\[Phi]\_A1] +
1 - \[Phi]\_A1 - \(\[Nu]\_1\/\[Nu]\_2\) \[Phi]\_A2 - \
\(\[Nu]\_1\/\[Nu]\_3\) \[Phi]\_A3 + \((\[Chi]\_12\ \[Phi]\_A2\  +
\[Chi]\_13\ \
\[Phi]\_A3\ )\) \((\[Phi]\_A2 + \[Phi]\_A3)\) - \(\[Nu]\_1\/\[Nu]\_2\)
\[Chi]\
\_23\ \[Phi]\_A2\ \[Phi]\_A3;\)\)

This is what works.
\!\(\(\[Mu]\_A1 :=
Log[\[Phi]\_A1] +
1 - \[Phi]\_A1 - \(\[Nu]\_1\/\[Nu]\_2\) \[Phi]\_A2 - \
\(\[Nu]\_1\/\[Nu]\_3\) \[Phi]\_A3 + \[Chi]\_12\ \[Phi]\_A2\ \[Phi]\_A2 + \
\[Chi]\_12\ \(\[Phi]\_A2\) \[Phi]\_A3 + \[Chi]\_13\ \[Phi]\_A3\ \[Phi]\_A2 +
\
\[Chi]\_13\ \[Phi]\_A3\ \[Phi]\_A3 - \(\[Nu]\_1\/\[Nu]\_2\) \[Chi]\_23\ \
\[Phi]\_A2\ \[Phi]\_A3;\)\)

If the phases are in equilibrium the chemical potential must be equal for
any component in both phases.
In:=
\!\(\(f\_1 := \((\[Mu]\_A1 - \[Mu]\_B1)\);\)\[IndentingNewLine]
\(f\_2 := \((\[Mu]\_A2 - \[Mu]\_B2)\);\)\[IndentingNewLine]
\(f\_3 := \((\[Mu]\_A3 - \[Mu]\_B3)\);\)\)
The material balances have to be fulfilled in each phase.
In:=
\!\(\(f\_4 :=
1 - \((\[Phi]\_A1 + \[Phi]\_A3 + \[Phi]\_A2)\);\)\[IndentingNewLine]
\(f\_5 := 1 - \((\[Phi]\_B1 + \[Phi]\_B3 + \[Phi]\_B2)\);\)\)
The objective function is formulated as as sum of squares.
In:=
\!\(\(F = f\_1\^2 + f\_2\^2 + f\_3\^2 + f\_4\^2 + f\_5\^2;\)\)
In:=
\!\(F := f\_1\^2 + f\_2\^2 + f\_4\^2 + f\_5\^2\)

In:=
\!\(\[Nu]\_1 = 20000; \[Nu]\_2 = 63.6; \[Nu]\_3 =
150000. ; \[Chi]\_12 =  .4; \[Chi]\_13 = 5; \[Chi]\_23 =
0.4; \[Phi]\_B3 = 0; \[Phi]\_A2 =  .01;
init = \ \(solution = \[IndentingNewLine]FindMinimum[
F, \[IndentingNewLine]{\[Phi]\_A1,  .01}, {\[Phi]\_A3,  .8}, \ \ {\
\[Phi]\_B1,  .98}, {\[Phi]\_B2,  .0135},
WorkingPrecision \[Rule] 16, \[IndentingNewLine]Method \[Rule]
LevenbergMarquardt, MaxIterations \[Rule] 350]\)\)

```

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