Saddle point in a non-convex objective function

Question

I am investigating the following minimization problem as follows. It is the summation of two fractional functions that mirror each other around $(y_c,y_n)=(\frac{1}{2},\frac{1}{2})$:

$Min\ z(y_c,y_n)=\frac{{\beta }_c{y_c}^2+{\beta }_my_cy_n+{\beta }_n{y_n}^2}{h-r_cy_c-r_ny_n}+\frac{{\beta }_c{(1-y_c)}^2+{\beta }_m(1-y_c)(1-y_n)+{\beta }_n{(1-y_n)}^2}{h-r_c(1-y_c)-r_n(1-y_n)}$

Subject to

$0\le y_c\le 1$

$0\le y_n\le 1$

Note that parameters are practically defined in a way that $h-r_cy_c-r_ny_n>0$ is always valid, and all parameters $r_c$, $r_n$,\textit{ }${\beta }_c$, ${\beta }_m$, ${\beta }_n$\textit{ }and\textit{ }$h$\textit{ }are positive real numbers. During the investigation, I checked the roots of derivatives of the objective function:

$\frac{df(y_c,y_n)}{dy_c}=\frac{2{\beta }_cy_c+{\beta }_my_n}{h-r_cy_c-r_ny_n}+r_c\frac{{\beta }_c{y_c}^2+{\beta }_my_cy_n+{\beta }_n{y_n}^2}{{\ (h-r_cy_c-r_ny_n)}^2}-\frac{2{\beta }_c(1-y_c)+{\beta }_m(1-y_n)}{h-r_c(1-y_c)-r_n(1-y_n)}-r_c\frac{{\beta }_c{(1-y_c)}^2+{\beta }_m(1-y_c)(1-y_n)+{\beta }_n{(1-y_n)}^2}{{(h-r_c(1-y_c)-r_n(1-y_n))}^2}=0$

$\frac{df(y_c,y_n)}{dy_n}=\frac{2{\beta }_ny_n+{\beta }_my_c}{h-r_cy_c-r_ny_n}+r_n\frac{{\beta }_c{y_c}^2+{\beta }_my_cy_n+{\beta }_n{y_n}^2}{{(h-r_cy_c-r_ny_n)}^2}-\frac{2{\beta }_n(1-y_n)+{\beta }_m(1-y_c)}{h-r_c(1-y_c)-r_n(1-y_n)}-r_n\frac{{\beta }_c{(1-y_c)}^2+{\beta }_m(1-y_c)(1-y_n)+{\beta }_n{(1-y_n)}^2}{{(h-r_c(1-y_c)-r_n(1-y_n))}^2}=0$

While I see similar terms in each equation, I could not solve this system of two equations with two variables by hand. But Maple software solved it easily and gave me the answer for roots as $(y_c,y_n)=(\frac{1}{2},\frac{1}{2})$. Via plots, I see that this point is a saddle point.

Is there any property, theorem, etc., to show why this point is a root for such a system of equations? Or how it can be obtained and whether it is the only root? Thank you!

Answer 1

Wanting to understand through a simple example, to solve a system of rational equations (A):

$$ \frac{(x-1)(y-2)}{(x-3)} = 0 \quad \land \quad \frac{(y-4)(x-3)}{(y-2)} = 0 $$

the general idea is to transform it into an equivalent system of polynomial equations (B):

$$ \small (x-1)(y-2) = 0 \quad \land \quad (y-4)(x-3) = 0 \quad \land \quad (x-3)u-1 = 0 \quad \land \quad (y-2)v-1 = 0 $$

The advantage is that we can now compute a Gröbner basis of these polynomials (C):

$$ x-1 = 0 \quad \land \quad y-4 = 0 \quad \land \quad 2u+1 = 0 \quad \land \quad 2v-1 = 0 $$

which, in general, will be a triangular system of polynomial equations, i.e. the first equation will be in the variable $x$ only, the second in the variables $x,y$, the third in the variables $x,y,u$ and the fourth in the variables $x,y,u,v$. It's the generalization of the Gaussian elimination method.

So, systems (C), (B) are true for $\small (x,y,u,v)=(1,4,-1/2,1/2)$, i.e. system (A) for $(x,y)=(1,4)$.

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In particular, if we want to consider a generalization of the system of equations subject of the topic:

$$ \small w(x,y) := \frac{ax^2+bxy+cy^2+dx+ey+f}{gx+hy+i}, \quad \quad z(x,y) := w(x-2j,y-2j) + w(2k-x,2k-y) $$

$$ \nabla z(x,y) = (0,0) \quad \quad \Leftrightarrow \quad \quad (x,y) = (j+k,j+k) $$

in Mathematica we can calculate it almost instantaneously by writing:

w[x_, y_] := (a x^2 + b x y + c y^2 + d x + e y + f) / (g x + h y + i)
z[x_, y_] := w[x - 2 j, y - 2 j] + w[2 k - x, 2 k - y]
{zx, zy} = Grad[z[x, y], {x, y}];
Solve[{zx == 0, zy == 0}, {x, y}]

{{x -> j + k, y -> j + k}}

or, wanting to implement the strategy just described:

{nzx, dzx} = NumeratorDenominator[Together[zx]];
{nzy, dzy} = NumeratorDenominator[Together[zy]];
polys = {nzx, nzy, dzx u - 1, dzy v - 1};
Simplify[GroebnerBasis[polys, {v, u, y, x}, CoefficientDomain->RationalFunctions]]

{-j - k + x, -j - k + y, -1 + (i - (g + h) (j - k))^4 u, -1 + (i - (g + h) (j - k))^4 v}

where it's clear that the solution found with Solve[] is the only one allowed, as we wanted to prove. On the other hand, I have no idea how to do it by hand, as I find it a very daunting task!