# Solving a system of nonlinear equations: show uniqueness or multiplicity of solutions

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)) $$0le d. Then there are two possible choices $$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