Mathematics Asked by STF on October 2, 2020
Consider this system of $12$ equations
$$
left{begin{array}{rcrclr}
alpha^{2}p_{i} & + &
left(1 – alpharight)^{2}left(1 – p_{i}right) & = & c_{i}, & forall i =1,2,3,4
\[1mm]
alphaleft(1 – alpharight)p_{i} & + &
left(1 – alpharight)alphaleft(1 – p_{i}right) &
= & d_{i}, & forall i =1,2,3,4
\[1mm]
left(1 – alpharight)^{2}p_{i} & + &
alpha^{2}left(1 – p_{i}right) & = & e_i, &
forall i =1,2,3,4
end{array}right.
$$
where
$alpha in left[0,1right]$
$p_{i} in left[0,1right]$ $forall i = 1, 2, 3, 4$
$c_{i}, d_{i}, e_{i}$ are real numbers $forall i = 1, 2, 3, 4$.
I want to show that this system of equation has ( or does not have ) a unique solution with respect to
$alpha, p_{1}, p_{2}, p_{3}, p_{4}$. Could you help ?.
This is what I have tried and where I’m stacked.
Let $i = 1$. From the second equation, we get
$$
alpha – alpha^{2} = d_{1}
$$
which gives
$$
alpha_{left(1right)} = frac{1 + sqrt{1 – 4d_{1}}}{2},quad alpha_{left(2right)} = frac{1 – sqrt{1 – 4d_{1}}}{2}
$$
From the first equation one can get $p_{1}$. From other equations, I guess one can analogously obtain $p_{2}, p_{3}, p_{4}$.
Is this sufficient to show that the system does not have a unique solution ? Or, is there a way to exclude one between $alpha_{left(1right)},alpha_{left(2right)}$ ?.
As you noticed, second four equations reduce to $alpha-alpha^2=d_i$. So the necessary condition for the system to have a solution is $0le d=d_1=d_2=d_3=d_4le frac 14$. The remaining equations reduce to $$p_i(2alpha-1)=c_i-(alpha-1)^2=alpha^2-e_i.$$ It follows $2alpha^2-2alpha=c_i+e_i-1=-2d_i$. This is an other necessary condition for the system to have a solution. We assume that both groups of necessarily conditions hold. Now the following cases are possible.
1)) $d=tfrac 14$. Then $alpha=tfrac 12$. Then $p_i$ are undetermined by the system, and it has a solution (not unique) iff $e_i=alpha^2=frac 14$ for each $i$
2)) $0le d<frac 14$. Then there are two possible choices $alpha_1$ and $alpha_2$ for $alpha$ and
$$p_i=frac{alpha^2-e_i}{2alpha-1}=frac{alpha-d-e_i}{2alpha-1}=frac 12+frac{1/2-d-e_i}{2alpha-1}.$$
We have $p_iin [0,1]$ iff $|1-2d-2e_i |le |2alpha-1|=|2alpha_j-1|$ for each $i$. If this condition fails for some $i$, then the system has no solutions. Otherwise it has two solutions, one for each $alpha_j$.
Correct answer by Alex Ravsky on October 2, 2020
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