1
$\begingroup$

I have a set of 6 variables that satisfy 4 linear equations and 2 non-linear equations. In particular, my first four equations can be written as $Ax = 0$, where $x \in \mathbb{R}^6$. This constraint implies that $x \in ker(A)$.

What I think this means is that if I can find the null space of $A$, I can reduce the number of free parameters from 6 to 2 for the non-linear portion of the solution. Where I'm getting stuck is how to use that parameterization to solve the non-linear system.

Concretely, the two non-linear constraints look like this (with known constants $C_1$ and $C_2$):

$${x_1}^2 + {x_2}^2 = C_1$$ $$(x_4 - x_3 - x_1)^2 + (x_6 + x_5 + x_2)^2 = C_2$$

Do I just need to use a non-linear solver here or is there some other trick I can pursue to try to simplify this further and perhaps find an analytical solution?

$\endgroup$

2 Answers 2

2
$\begingroup$

After expressing everything in terms of two parameters (let's say $s$ and $t$) from the linear portion, you have two equations of the form

$$ \eqalign{(a_1 + b_1 s + c_1 t)^2 + (a_2 + b_2 s + c_2 t)^2 &= C_1 \cr ( a_3 + b_3 s + c_3 t)^2 + (a_4 + b_4 s + c_4 t)^2 &= C_2 \cr}$$ Presumably $C_1, C_2 > 0$ so this is nontrivial.

The resultant of these with respect to one of $s$ and $t$ will be a quartic polynomial. Each real root of that should give you a solution.

$\endgroup$
4
  • $\begingroup$ If the resultant is quartic, I'm guessing that means an analytic solution is probably going to be rather ugly. Is there a good reason to prefer converting the system to this form and using a polynomial root solver over just applying a non-linear solver like Levenberg-Marquardt directly to the original constraints? $\endgroup$
    – Dan Bryant
    Sep 27, 2016 at 21:22
  • 1
    $\begingroup$ In terms of numerical methods, a direct use of Levenberg-Marquardt might well be faster and/or more stable. If you have the coefficients as exact rational numbers, one advantage of the quartic (even if you don't solve it by radicals) is that you can determine the number of real roots, and isolate them in intervals, using Sturm's theorem. $\endgroup$ Sep 28, 2016 at 0:44
  • $\begingroup$ Looking at this again, if $x \in ker(A)$, doesn't that imply that all $a_i = 0$ here? I also just realized that I have the ability to force $C_1 = C_2$ in the way I construct my measurement setup, which might help here. $\endgroup$
    – Dan Bryant
    Sep 28, 2016 at 14:51
  • $\begingroup$ Researching this a bit more, it sounds like I need to find a good book on algebraic geometry. This is unfortunately a topic I never got a chance to study and a lot of the terminology I'm running into is new for me. $\endgroup$
    – Dan Bryant
    Sep 28, 2016 at 14:55
1
$\begingroup$

If I well understand you can find $x_3,x_4,x_5,x_6$ as linear functions of $x_1$ and $x_2$ so your system become somethig as:

$$ \begin {cases} x_1^2+x_2^2=C_1\\ (ax_1+bx_2)^2+(cx_1+dx_2)^2=C_2 \end{cases} $$ In general this can be reduced to a quartic equation which can be solved analytically but with a lot of work (so it is better to search a numerical solution).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .