Can some inequalities help to pin down an unique solution in a linear system of equations with infinite solutions?

Question

I need to discuss the number of solutions of the following system of equations. Any help would be very appreciated.

Consider the known parameters $a_1,...,a_4;d_1,d_2,d_3$ such that

$0< a_i< 1$ $\forall i \in \{1,...,4\}$ and $\sum_{i=1}^4 a_i=1$

$0< d_i< 1$ $\forall i \in \{1,2,3\}$ and $\sum_{i=1}^3 d_i=1$

The system, with unknowns $y^i_j$ $\forall i \in \{1,...,4\}$, $\forall j \in \{1,2,3\}$, is $$ \begin{cases} d_1=a_1y_1^1+a_2y_1^2+...+a_4y_1^4\\ d_2=a_1y_2^1+a_2y_2^2+...+a_4y_2^4\\ d_3=a_1y_3^1+a_2y_3^2+...+a_4y_3^4\\ y_1^i+y_2^i+y_3^i=1 \text{ $\forall i \in \{1,...,4\}$}\\ 0<y^i_j<1 \text{ $\forall i \in \{1,...,4\}$, $\forall j \in \{1,2,3\}$} \end{cases} $$

Without the finalinequalities I believe the system would have an infinite amount of solutions. I'm wondering whether the inequalities can help to pin down one solution and why.

Answer 1

STF,

Your notation of using superscript to differentiate between variables is confusing - I originally thought they are exponents. Below I will add parentheses to all superscripts.

Your problem has an infinite number of solutions as loup blanc pointed out. There is actually a nice geometric interpretation of your system and you would be able to write down a parametrized form for all solutions.

Consider the simplex $\Delta_2$: $X_1+X_2+X_3=1$ in ${\mathbb R}^3$, where $X_i>0$ are the coordinates. Given a point $D\in\Delta_2$ and weights $\sum_{j=1}^{4}a_j=1$, we need to find four points $Y^{(j)}\in\Delta_2$ so that the weighted vector sum satisfies $\sum_{j=1}^{4}a_jY^{(j)}=D$.

Algebraically this translates exactly to your system, where the coordinates of $D$ and $Y^{{j}}$ are denoted by $d_i$ and $y_i^{(j)}$. Geometrically $D$ is the centroid (barycenter / center of mass) with weight $a_j$ at each point $Y^{(j)}$. We can illustrate the configuration by the graph below:

where $\left|Y_1X_{12}\right|:\left|Y_2X_{12}\right|=a_2:a_1$, $\left|Y_3X_{34}\right|:\left|Y_4X_{34}\right|=a_4:a_3$ and $\left|DX_{12}\right|:\left|DX_{34}\right|=(a_3+a_4):(a_2+a_1)$.

It's easy to see as long as $D$ is within the simplex, $X$'s can be chosen sufficiently close to $D$ that satisfy the constraints above; similarly for $Y$'s. We can then stretch/rotate the quadrilateral in arbitrary ways (plus the permutation of the vertices) as long as it fits inside the simplex. We can also write down explicitly all the solutions by parametrizing e.g., the vectors $\vec{DX_{12}}$, $\vec{Y_1X_{12}}$ and $\vec{Y_3X_{34}}$.

Answer 2

Since $d_1+d_2+d_3=a_1+a_2+a_3+a_4=1$, you have only $6$ equations in the $12$ unknows $(y_{ij})$ and some inequalities. a priori, I do not see how you can expect a unique solution. Here $y_j^i$ is denoted by $y_{ji}$.

EDIT. That follows is an example with at least $2$ solutions (then with an infinity of solutions).

$a_i=0.1$ witjh $i\leq 3hklwuy$ and $d_j=0.45$ with $j\leq 2$ and the $2$ solutions in $y_{ji}$ with $j\leq 2,i\leq 4$ are:

i) $y_{ij}=0.495$ except $y_{24}=y_{14}\approx 0.4307$.

ii) $y_{11}=0.4,y_{12}\approx 0.4974,y_{13}=0.49with_{14}\approx 0.4439,y_{21}=0.5,y_{22}\approx 0.4926,y_{23}=0.495,y_{24}\approx 0.4303$