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I have a system of non-linear equations

$ a_1=f_0 g_1$

$a_2=f_1 g_1 + f_0 g_2$

$a_3=f_2 g_1 + f_6 g_2 + f_0 g_3 $

$a_4=f_3 g_1 + f_7 g_2 + f_6 g_3 + f_0 g_4 $

$a_5=f_4 g_1 + f_8 g_2 + f_7 g_3 + f_6 g_4 + f_0 g_5$

$a_6=f_5 g_1 + f_9 g_2 + f_8 g_3 + f_7 g4 + f_6 g_5 + f_0 g_6$

$a_7=f_5 g_3 + f_9 g_4 + f_8 g_5 + f_2 g_6 + f_0 g_7 $

$a_8=f_5 g_5 + f_4 g_6 + f_7 g_7 + f_0 g_8$

$a_9=f_5 g_7 + f_8 g_8 + f_0 g_9$

$a_{10}=f_0 g_{10} + f_5 g_9$

here both $f=(f_0,..,f_5,,..f_9)$ and $g=(g_1,…,g_{10})$ are unknown.

My problem is to figure out when this system has no solution, a unique or multiple solutions.

What is known is that this system of equations allows to decompose itself as

$\left(\begin{array}{c} a_1\\ a_2 \\ \vdots \\a_9 \\a_{10} \end{array} \right) = \begin{bmatrix} f_0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ f_1 & f_0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ f_2 & f_6& f_0& 0& 0& 0& 0& 0 &0 & 0 \\ f_3 & f_7 & f_6 & f_0 & 0 & 0 & 0 & 0 & 0& 0 \\ f_4 & f_8 & f_7 & f_6 & f_0& 0& 0& 0& 0& 0 \\ f_5 & f_9 & f_8 & f_7 & f_6& f_0& 0& 0& 0& 0 \\ 0 & 0 & f_5 & f_9 & f_8 & f_2& f_0& 0 & 0& 0 \\ 0 & 0 & 0 & 0 & f_5 & f_4 & f_7 & f_0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & f_5& f_8 & f_0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & f_5 & f_0 \end{bmatrix} \times \left( \begin{array}{c} g_1 \\ g_2 \\ \vdots \\ g_9 \\ g_{10} \end{array} \right) $

Then whenever I take only vector of $f$ as unknowns (and so $g$ are the parameters), or vice versa, there is a unique solution for a generic vector $a$, similarly to how it works for linear systems of equations. While when I am letting one extra element of $f$ or of $g$ be "a parameter", i.e. there are more equations than there are unknown $f$ or unknown $g$, there is no solution.

But I have no idea what happens when both $f$ and $g$ are unknown. I have then 10 equations and 20 unknowns, but it is not clear to me that it automatically should lead to a continuum of solutions. (And I cannot solve this system explicitly..)

If there is a continuum of solutions (for a "generic" $a$), it seems to be an interesting case: a system of $N$ non-linear equations and $M(>N)$ unknowns does not have a solution (for generic vector $a$) whenever the size of the smallest of two sets of unknowns is less than $N$. Or it has a continuum of solutions in the complementary case, but there is no case in-between -- with a unique solution.

Are there any known results for such system of non-linear equations that are "decomposable"?
I would appreciate any opinions on this!

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  • $\begingroup$ This doesn't exploit the "decomposable" structure, but have you looked into Gröbner bases for your system of 10 quadratic equations? en.wikipedia.org/wiki/Gr%C3%B6bner_basis $\endgroup$
    – j.c.
    Commented Sep 22, 2015 at 16:25
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    $\begingroup$ where do this equations come from? If you have a solution you can multiply all f's with a constant and all g's with the inverse of the constant to get a new solution. $\endgroup$
    – user35593
    Commented Sep 22, 2015 at 19:11
  • $\begingroup$ Thank you j.c. for editing, I was trying to find the Gröbner basis, but was unsuccessful (in fact even doing this in Mathematica did not help), so I started to look for more general ideas... $\endgroup$
    – Kass
    Commented Sep 22, 2015 at 19:28
  • $\begingroup$ @user35593: My bad! you are perfectly right, that one can generate easily a continuum of solutions, provided we have at least one. In fact there is less freedom than what I stated. My $g$ sum up to 1. And $f$ belong to [0,1]. As I have been loosing hope to get a solution that satisfies these extra constraints, I started to get rid of them and see for unconstrained solutions. I guess I should edit my question. $\endgroup$
    – Kass
    Commented Sep 22, 2015 at 19:31
  • $\begingroup$ So the system has the form $B(f,g)=a$ for a bilinear map $$B:\mathbb{R}^{10}\times \mathbb{R}^{10} \to \mathbb{R}^{10}$$ and $B(f,\cdot)$ is invertible iff $f_0\neq0$. $\endgroup$
    – Pietro Majer
    Commented Dec 4, 2015 at 8:37

2 Answers 2

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The linear-in-$g$ system is uniquely solvable whenever $f_0\ne0$. therefore the solution set is parametrized by $f$ in the form $$g=f_0^{-10}P(f;a),$$ where $P$ is linear in $a$, polynomial in $f$ and homogenenous of degree $9$. Not only it is a continuum, but it has dimension $10$.

When $f_0=0$, there is no solution in general, say for $a_1\ne0$.

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  • $\begingroup$ So the solution set is a graph over $\mathbb{R}^{10}$, plus (if $a_1=0$) a 9-dimensional vertical component corresponding to $f_0=0$ and $f_1 f_2 f_5 f_6f_7f_8\neq0$; plus possibly other smaller components according to the form of $a$. $\endgroup$
    – Pietro Majer
    Commented Dec 4, 2015 at 8:50
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How about setting $f_0=1$ and $f_1=t$ where $t$ is a parameter and the rest of the $f_i=0?$ Then $g_i=a_i$ for all $i $ except $g_2=a_2-ta_1.$ That seems like a continuum of solutions, even projectively (i.e. up to rescaling.)

It might be interesting to ask if there are always $9$ degrees of freedom (or $10$ with rescaling allowed.)

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