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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.

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  • $\begingroup$ The question has been clarified. Any help or intuition would be very appreciated. $\endgroup$
    – Star
    Commented May 13, 2016 at 18:13

2 Answers 2

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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:

simplex

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 $\vec{Y_3X_{34}}$.

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

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  • $\begingroup$ Thanks for your answer. Can you elaborate more? I have 7 equations. An example in which you show two possible solutions would be enough. My problem is that I can't find such an example. That's why I was wondering whether the fact that I constrained each solution to lie in $(0,1)$ could be enough to pin down one solution. $\endgroup$
    – Star
    Commented May 16, 2016 at 13:36
  • $\begingroup$ Thanks for the example. I don't understand it. It should be $\sum_{i=1}^4 a_i=1$, $\sum_{j=1}^3 d_j=1$. Also, what is $y_{ij}$? $\endgroup$
    – Star
    Commented May 16, 2016 at 14:48
  • $\begingroup$ $y_{ji}$ is explained above. $a_4$ and $d_3$ are useless. $\endgroup$
    – user91684
    Commented May 16, 2016 at 14:51
  • $\begingroup$ Thanks. In your answer, you also have $y_{ij}$. If $a_4$ is useless maybe you should set $i<4$ instead of $i\leq 4$? $\endgroup$
    – Star
    Commented May 16, 2016 at 14:54
  • $\begingroup$ Thanks, I have deleted my questions, I was wrong, apologies. I'm still unsure on accepting your answer because your solutions involve approximations. Could you find 2 solutions without $\approx$? $\endgroup$
    – Star
    Commented May 16, 2016 at 16:49

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