Differentiate matrix wrt vector

3 Mar 2021


      Suppose, I have a 5x5 matrix and I want to differentiate the matrix wrt to
a 5x1 matrix in order to obtain a 3D matrix or tensor. In the following
code, A.diff(uL); cannot compile. Is it possible to differentiate a matrix
with symbolic elements wrt to a vector which also contains symbols?

#include <iostream>
#include <ginac/ginac.h>
using namespace std;
using namespace GiNaC;

int main()
{
    const int DIM = 5;

    symbol g("g");

    symbol uL1("uL1");
    symbol uL2("uL2");
    symbol uL3("uL3");
    symbol uL4("uL4");
    symbol uL5("uL5");

    matrix uL = {
        {uL1},
        {uL2},
        {uL3},
        {uL4},
        {uL5}
    };

    auto qL = matrix(DIM, 1);

    qL(0,0) = sqrt(uL1);
    qL(1,0) = uL2 / qL(0,0);
    qL(2,0) = uL3 / qL(0,0);
    qL(3,0) = uL4 / qL(0,0);
    qL(4,0) = (uL5 + (g - 1) * (uL5 - 0.5 * (pow(uL2,2) + pow(uL3,2) +
pow(uL4,2)) / uL1)) / uL1;

    symbol uR1("uR1");
    symbol uR2("uR2");
    symbol uR3("uR3");
    symbol uR4("uR4");
    symbol uR5("uR5");

    matrix uR = {
        {uR1},
        {uR2},
        {uR3},
        {uR4},
        {uR5}
    };

    auto qR = matrix(DIM, 1);

    qR(0,0) = sqrt(uR1);
    qR(1,0) = uR2 / qR(0,0);
    qR(2,0) = uR3 / qR(0,0);
    qR(3,0) = uR4 / qR(0,0);
    qR(4,0) = (uR5 + (g - 1) * (uR5 - 0.5 * (pow(uR2,2) + pow(uR3,2) +
pow(uR4,2)) / uR1)) / uR1;

    auto q = qL.add(qR);
    q.mul_scalar(0.5);

    matrix B = {
        {2*q(0,0), 0, 0, 0, 0},
        {q(1,0), q(0,0), 0, 0, 0},
        {q(1,0), q(0,0), 0, 0, 0},
        {q(2,0), 0, q(0,0), 0, 0},
        {q(3,0), 0, 0, q(0,0), 0},
        {q(4,0)/g, (g-1)*q(1,0)/g, (g-1)*q(2,0)/g, (g-1)*q(3,0)/g, q(0,0)/g}
    };

    matrix C = {
        {q(1,0), q(0,0), 0, 0, 0},
        {(g-1)*q(4,0)/g, (g+1)*q(1,0)/g, (1-g)*q(2,0)/g, (1-g)*q(3,0)/g,
(g-1)*q(0,0)/g},
        {0, q(2,0), q(1,0), 0, 0},
        {0, q(3,0), 0, q(1,0), 0},
        {0, q(4,0), 0, 0, q(1,0)}
    };

    auto A = B.mul(C.inverse());

    A.diff(uL);

    return 0;
}

Orxan Shibliyev

Vladimir V. Kisil

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