Solving $x=\frac{\alpha y}{\alpha y+\beta z}$, $y=\frac{\gamma x}{\gamma x+\delta(1-z)}$, $z=\frac{\epsilon(1-x)}{\epsilon(1-x)+\zeta(1-y)}$

Question

Given this set of three rational equations:

$$ x = \frac{\alpha\cdot y}{\alpha\cdot y + \beta\cdot z} $$ $$ y = \frac{\gamma\cdot x}{\gamma\cdot x + \delta\cdot (1-z)} $$ $$ z = \frac{\epsilon\cdot (1-x)}{\epsilon\cdot (1-x) + \zeta\cdot (1-y)} $$

Where all Greek letters are known parameters.

(a) How can it be solved (if indeed possible)?

(b) If not, is there anything non-trivial one can say about the solution?

I tried all sorts of algebraic manipulations but no luck so far, so I think I may have some missing knowledge about systems of this kind.

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edit - equation 2 corrected (x instead of y in the denominator), thanks @Dan for noticing, hopefully the general idea remains the same

Answer 1

A certain number of remarks, with a common geometrical approach.

Let us assume that all constants are non-zero.

I observe that each one of the 3 equations describe a surface belonging to the same family of quadric surfaces, namely Hyperbolic Paraboloids (with reference equation $Z=kXY$, a kind of surface with a "saddle shape" or "manta ray shape") because its equation

$$\alpha y (x-1) = \beta z$$

can be transformed into reference equation $Z=kXY$ by the affine change of variable (which is a mere translation):

$$\begin{cases}X&=&x-1\\Y&=&y\\Z&=&z\end{cases}$$

In the same way, the second equation $y(\gamma y-\delta z+\delta)=\gamma z$

can undergo as well an affine transformation, namely

$$\begin{cases}X&=&\gamma y-\delta z+\delta\\Y&=&y\\Z&=&z\end{cases}$$

and the same kind of transformation for the last one.

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An important remark is that each one of the three surfaces contains a line which is easily found on the initial equations (as given in your text):

• in the first equation (initial version), one sees that the whole line $(x,y,z)=(O,O,z)$ (i.e., all the $z$ axis but value $z=0$) belongs to the surface.

• in the second equation (initial version), one sees that the whole $z$ axis characterized by points $(x,y,z)=(0,0,z)$ (but value $z=1$...) belongs to the surface. Therefore the two first surfaces share a common axis, the $z$ axis (but a point).

• in the third equation (initial version), one sees that a whole horizontal axis characterized by points $(x,y,z)=(1,y,0)$ (but value $y=1$...) belongs to the surface.

This shouldn't come as a surprize because a hyperbolic paraboloid is a (doubly) ruled surface. Indeed equation $Z=kXY$ shows that all lines of the form $Z=kx_0Y$ (for any $x_0$) and $Z=kXy_0$ (for any $y_0$) belong to the surface.

One more comment for the 1st, resp 3rd equation: if one let $y$ tend to $\infty$, $x \to 1$. For the 3rd equation: when $x \to \infty$, $z \to 1$.

Remark: The name "quadric surface" is given to the 3D equivalent of conic curves, which can be described by a second degree polynomial in $x,y,z$ ; they can be either ellipsoids (particular case: spheres), paraboloids, cylinders, cones, hyperboloids (with one or two sheets), cones, or ... hyperbolic paraboloids (the last ones being "ruled" surfaces).

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Addendum (partial conclusion) : the solution set of the restricted system made of the two first equations contains a line ($z$ axis).

Two non-identical Hyp. Paraboloid surfaces sharing a common line could share a second common line, but not a 3rd one (it is known that 3 lines completely determine a unique Hyperbolic Paraboloid, see here and here). Under this assumption, adding the 3rd equation gives in general a transverse intersection to $z$ axis, yielding finally either a single point, or exceptionaly a second point.

This reasoning intends to give a "big picture" of the situation ; there should exist particular cases, with particular values of coefficients where this isn't true, where for example the third surface shares as well a common line with the first two ones.