How to prove independence of system of quadratic equations

Question

I hope you may be able to help me with my problem. I don't know if it is actually possible to do what I want to do, but maybe I overlooked an answer.

I have a system of quadratic equations

$ \begin {pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = A \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} + B \cdot \begin{pmatrix} x_1^2 \\ x_2^2 \\ x_3^2 \end{pmatrix} + C\cdot \begin{pmatrix} x_1\cdot x_3 \\ x_2\cdot x_3 \\ x_1\cdot x_2 \end{pmatrix} $

The matrices A and B and the y vector are given. I have been solving the system of equations with matlab with a solve() function. I have been varying the values of $y_1, y_2$ and $y_3$, which has been working fine so far. However, I ran into problems when trying to solve the system for a certain value for $y_2 = y_3$. It will work for other values and $y_2 = y_3$.

When I try to solve the equations for $y_3 = 0.99999999 \cdot y_2$ with $y_2$ being the problematic value, it will work just fine. I haven't figured out why this is the case. Therefore I wanted to check if the system is still solvable with $y_2 = y_3$ by checking for independence.

Is there another way to check if it still solvable? Or might it be another problem, maybe the solver just not being able to solve it?

Thank you in advance! Your help is appreciated.

edit:

The values of the vectors and matrices are

$ \begin {pmatrix} 750 \\ 1500 \\ 1500 \end{pmatrix} = \begin {pmatrix} 500 & 0 & 0 \\ 500 & 500 & 0 \\ 0 & 0 & 500 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} + \begin {pmatrix} 0 & 0 & 0 \\ -19.581 & -15.053 & 0 \\ 0 & 0 & -15.053 \end{pmatrix} \cdot \begin{pmatrix} x_1^2 \\ x_2^2 \\ x_3^2 \end{pmatrix} + \begin {pmatrix} 0 & 0 & 0 \\ 11.32 & 11.32 & -34.634 \\ 11.32 & 11.32 & 0 \end{pmatrix} \cdot \begin{pmatrix} x_1\cdot x_3 \\ x_2\cdot x_3 \\ x_1\cdot x_2 \end{pmatrix}$

The problem arose while I was trying to calculate the currents required for a certain power to be converterted at certain points in an asymmetrical electrical dc network. Basically I'm looking at a multiterminal bipolar HVDC network in metallic return mode and an outage at one station.

What might need to be mentioned is that im using the solver of the symbolic toolbox of matlab.

Answer 1

I use Maxima the code and results are below:

E1 : -750 + 500*x1;
E2 : -1500 + 500*x1 + 500*x2 -19.581*x1^2 -15.053*x2^2 + 9.58*x1*x3 + 9.58*x2*x3 -34.634*x1*x2;
E3 : -1500 + 500*x3 -15.053*x3^2 + 9.58*x1*x3 + 9.58*x2*x3;

solve([E1,E2,E3],[x1,x2,x3]);

[x1 = 1.5, x2 = 85.96972563859981, x3 = 87.74766355140187], 
[x1 = 1.5, x2 = 29.27408993576017, x3 = 1.959984609465179], 
[x1 = 1.5, x2 = 1.655006858710562, x3 = 3.102202145680407], 
[x1 = 1.5, x2 = 0.4651535380507343, x3 = 31.2810650887574]

The results are not very accurate? Substituting each result back into the three equations:

           subst(R[1],[E1,E2,E3]);
    (%o13)        [0.0, 0.004048568953294307, - 0.007026502295047976]
    (%i14) subst(R[2],[E1,E2,E3]);
    (%o14)        [0.0, - 7.914470367040849e-4, 2.623238295029751e-6]
    (%i15) subst(R[3],[E1,E2,E3]);
    (%o15)        [0.0, - 1.355784513634717e-5, 1.001358072869607e-4]
    (%i16) subst(R[4],[E1,E2,E3]);
    (%o16)        [0.0, 1.999027597321401e-5, - 0.001041557330609066]

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https://www.wolframalpha.com/input/?i=-750+%2B+500*x1+%3D+0,+-1500+%2B+500*x1+%2B+500*x2+-19.581*x1%5E2+-15.053*x2%5E2+%2B+9.58*x1*x3+%2B+9.58*x2*x3+-34.634*x1*x+%3D+0,+-1500+%2B+500*x3+-15.053*x3%5E2+%2B+9.58*x1*x3+%2B+9.58*x2*x3

Wolfram gives:

x3 (9.58 x1+9.58 x2+500.)-15.053 x3^2-1500.≈-15.053 x3^2+500. x3-1500.

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Using the grobner basis package in maxima:

load(affine);

grobner_basis([-750 + 500*x1,-1500 + 500*x1 + 500*x2 -19.581*x1^2 -15.053*x2^2 + 9.58*x1*x3 + 9.58*x2*x3 -34.634*x1*x2,-1500 + 500*x3 -15.053*x3^2 + 9.58*x1*x3 + 9.58*x2*x3]);

Gives:

$$\left(-576830960000000\right)\,{\it x_3}+\left(- 5411710412472\right)\,{\it x_2}^3+643257403387460\,{\it x_2}^2+ \left(-15004525334998746\right)\,{\it x_2}+24884658413238003 = 0$$

$$ \left(-10823420824944\right)\,{\it x_2}^4+1270279675537504\, {\it x_2}^3+\left(-29892020931835112\right)\,{\it x_2}^2+ 58711593597479768\,{\it x_2}+\left(-20969575034285991\right) = 0$$

$$ \left(-2\right)\,{\it x_1}+3 = 0$$

The third equation gives $x_1 = 1.5$.

The second equation gives a fourth order polynomial in $x_2$.

The first equation is linear in $x_3$ and cubic in $x_2$.

So solve for $x_2$ then substitute into the first equation to find $x_3$.

plot2d(R[2],[x2,-20,100]);

This is what the second equation ($x_2^4$) looks like.

Notice that it rises from zero to approx. $1e16$ very quickly.

You may require higher floating point precision to get accurate answers.