Howto simpleMatrixLeastSquareFit

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Revision as of 11:11, 19 May 2010 by TonyaJensen (Talk | contribs)

This is a very simple example of how to use the simpleMatrix class to solve a vector-matrix system using the LUsolve functionality

It does not require any dictionaries. I called it clduFoam and I execute it using 'clduFoam -e0 0 -e0 1'.

The values used to fit the function are hardcoded because I wanted to keep it simple

1 Code

Application
    clduFoam

// base functions
// 0 - 1
// 1 - 1.0/T
// 2 - log(T)
// 3 - T^e

Description

\*---------------------------------------------------------------------------*/

#include "fvCFD.H"
#include "simpleMatrix.H"
#include "OFstream.H"

// * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * //
// Main program:

int main(int argc, char *argv[])
{
    argList::validOptions.insert("e0", "scalar");
    argList::validOptions.insert("e1", "scalar");

#   include "setRootCase.H"

    // number of data-points
    label N = 4;

    // number of base functions
    label Nc = 4;

    // number of e-values between e0 and e1
    label Ne = 10;

    simpleMatrix<scalar> A(Nc);
    scalarField coeffs(Nc);
 
    scalar e0 = atof(args.options()["e0"].c_str());
    scalar e1 = atof(args.options()["e1"].c_str());
    scalar de = (e1-e0)/(Ne-1);

    scalarField T(N);
    scalarField pv(N);
    scalarField w(N);

    T[0] = 340.0; pv[0] = 0.133;   w[0] = 5.0;
    T[1] = 368.0; pv[1] = 1.33;    w[1] = 1.0;
    T[2] = 465.0; pv[2] = 1.0e+5;  w[2] = 100.0;
    T[3] = 700.0; pv[3] = 1.0e+8;  w[3] = 1.0e-0;

    scalarField logPv(log(pv));
    scalarField logT(log(T));
    scalarField Tinv(1.0/T);

    for(label it=0; it<Ne; it++)
    {
        scalar e = e0 + it*de;
        scalarField powT(pow(T,e));
   
        A.source()[0] = sum(logPv*w);
        A.source()[1] = sum(logPv*Tinv*w);
        A.source()[2] = sum(logPv*logT*w);
        A.source()[3] = sum(logPv*powT*w);

        A[0][0] = sum(w);
        A[0][1] = sum(Tinv*w);
        A[0][2] = sum(logT*w);
        A[0][3] = sum(powT*w);
   
        A[1][0] = sum(Tinv*w);
        A[1][1] = sum(Tinv*Tinv*w);
        A[1][2] = sum(logT*Tinv*w);
        A[1][3] = sum(powT*Tinv*w);

        A[2][0] = sum(logT*w);
        A[2][1] = sum(Tinv*logT*w);
        A[2][2] = sum(logT*logT*w);
        A[2][3] = sum(powT*logT*w);

        A[3][0] = sum(powT*w);
        A[3][1] = sum(Tinv*powT*w);
        A[3][2] = sum(logT*powT*w);
        A[3][3] = sum(powT*powT*w);

        coeffs = A.LUsolve();
        //Info << coeffs << endl;
        scalar err = 0.0;
        forAll(T,i)
        {
            scalar pc = coeffs[0] + coeffs[1]/T[i] + coeffs[2]*::log(T[i]) + coeffs[3]*::pow(T[i],e);
            err += ::pow(pc - logPv[i], 2);
            pc = ::exp(pc);
       
        }
        cout.precision(12);
        Info << "e = " << e << ", err = " << ::sqrt(err) << ",c = " << coeffs << endl;
    }

    forAll(T,i)
    {
        scalar pc = coeffs[0] + coeffs[1]/T[i] + coeffs[2]*::log(T[i]) + coeffs[3]*::pow(T[i],e1);
        pc = ::exp(pc);
        Info << "T = " << T[i] << ", pv = " << pv[i] << ", pvFit = " << pc << endl;
      
    }

    OFstream dataFile
    (
        "val.dat"
    );
 
    for(scalar Ti=300; Ti <= 710; Ti +=1)
    {
        scalar logpc = coeffs[0] + coeffs[1]/Ti + coeffs[2]*::log(Ti) + coeffs[3]*::pow(Ti,e1);
        scalar pc = ::exp(logpc);
        dataFile << Ti << " " << logpc << " " << pc << endl;
    }

    Info << "End\n" << endl;
    return 0;
}


// ************************************************************************* //

2 Description

I want to do a curve-fit of the function

 y(T) = \exp(c_0 + c_1/T + c_2 \log(T) + c_3 T^e)

to a number of P_v values.

I want to minimize the function

 S_e = \sum_i (y(T_i) - P_{v,i})^2

with respect to the coefficients c_i.

But since the function is an exp, I will minimize the difference in log instead, i.e.

 S = \sum_{T_i} (\log(y(T_i)) - \log(P_{v,i}))^2

Hence

 \frac{\partial S}{\partial c_i} = \frac{\partial}{\partial c_i}\sum_{T_i} [c_0 + c_1/T_i + c_2\log(T_i) + c_3T_i^e - log(P_{v,i})]^2 = 0.

and we get

 \frac{\partial S}{\partial c_0} = 2\sum_{T_i} \{ [c_0 + c_1/T_i + c_2\log(T_i) + c_3T_i^e - log(P_{v,i})] \} = 0

 \frac{\partial S}{\partial c_1} = 2\sum_{T_i} \{ [c_0 + c_1/T_i + c_2\log(T_i) + c_3T_i^e - log(P_{v,i})] (1/T_i)\} = 0

 \frac{\partial S}{\partial c_2} = 2\sum_{T_i} \{ [c_0 + c_1/T_i + c_2\log(T_i) + c_3T_i^e - log(P_{v,i})] \log(T_i)\} = 0

 \frac{\partial S}{\partial c_3} = 2\sum_{T_i} \{ [c_0 + c_1/T_i + c_2\log(T_i) + c_3T_i^e - log(P_{v,i})] T_i^e\} = 0

which we can write as a vector matrix system  A c = b

 (\sum_{T_i} 1)c_0 + (\sum_{T_i} 1/T_i)c_1 + (\sum_{T_i} \log(T_i))c_2 + (\sum_{T_i} T_i^e)c_3 = \sum_{T_i} log(P_{v,i})

 (\sum_{T_i} 1/T_i)c_0 + (\sum_{T_i} (1/T_i)(1/T_i))c_1 + (\sum_{T_i} \log(T_i)/T_i)c_2 + (\sum_{T_i} T_i^e/T_i)c_3 = \sum_{T_i} log(P_{v,i})/T_i

et c. The matrix components should be clear from the code, where weigths have also been added.

--Niklas 07:56, 21 January 2009 (CET)