Either

the equations of the nomogram scales in Fig. 1. We assume that the functions$f,g$And$h$are continuous and monotonic in the narrow sense. Between the sides$x,y,z$points$P,Q,R$, located on the same line, the relation takes place

Or$F(x)=\frac{d_{1}}{d_{1}+d_{2}}f(x),G(y)=\frac{d_{2}}{d_{1}+d_{2}}g(y)$,$H(z)=h(z)$The function$z=f(x,y)$The function defined by (1) is continuous and montone (with respect to each of the variables). In the following considerations, all functions of one and two variables will be assumed to be continuous and monotone. The scales of the nomograms do not allow for broader assumptions.

The problem arises of characterizing the functions$z=f(x,y)$, which can be put in the form (1).

If we limit ourselves to the functions$z=f(x,y)$admitting partial derivatives up to the third order, then the necessary and sufficient condition for that$f(x,y)$either of the form (1) is expressed by the condition of SAINT-ROBERT [1]:

J. ACZÉL characterized, under the conditions specified above, the particular case of the class of functions

by the functional equation
(3)

the so-called equation of bisymmetry [2], [3], [4]. This author arrived at this result by first imposing two additional conditions on the solution of equation (3). Independently of this, he also demonstrated [5] that the so-called equation of associativity

admits as a solution the functions (2), with$a=b=1,c=0$J.
ACZÉL formulated the problem of finding a functional equation that characterizes functions (1) in general [6]. M. hosszú found the following result [7]: the solutions of the functional equation with three unknown functions
(5)

functions that are montone in the restricted sense and admit first-order partial derivatives are

Besides the differentiability assumption, this result has the drawback that to decide whether a given function is of the form (1) or not, besides this function there are still two unknown functions, which makes the application very difficult.

In this work we will give various necessary and sufficient conditions for the function$z=f(x,y)$either of the form (1), The function$f$being continuous and montone in the restricted sense, the equation$z=f(x,y)$can be resolved in relation to$x$And$y:x=\bar{f}(y,z),y=\tilde{f}(z,x)$One of the given conditions can be written as follows:

which is a special case of equation (5). The differentiability assumption would not come into play there.

A geometric interpretation of equation (6) is known in the theory of hexagonal tissues [8]. Another leads to the following result: nomograms with three scales on the same cubic represent an equation of the form (1), and there are no other type of collinear point nomograms for equation (1). The other conditions that will be given can also be interpreted geometrically in two different ways.

Functional equations (3) and (4) and other similar equations can be easily solved using condition (6). Thus, a new method is established for studying a certain class of functional equations.

The study of the function's value domain$H(x)$in (2), which is required for different values ​​of$ABC$is also done in this work.

If$z=/(x,y)$verifies (6) and can therefore be put in the form (1), the problem of determining the functions remains$F,G,H$that is, to determine the scales, which leads to studying the functional equation

to three unknown functions$\varphi=F^{-1},\psi=G^{-1},\chi=H^{-1}$of a single variable. It was considered by J. ACZÉL for$\varphi=\psi=\chi[4]$We will reduce the general case to this.

We have linked to the equation of associativity (4) many facts which until now have appeared unrelated to each other, in particular: the equation of bisymmetry, the properties of hexagonal tissues, nomograms whose scales are located on the same cubic, properties characteristic of cubics.

Let us consider the function of two variables

where the functions$F,G,H$Functions of a single variable are continuous and monotonic in the narrow sense. Function (1) satisfies the
condition$T$ :$f\left(x_{1},y_{2}\right)=f\left(x_{2},y_{1}\right),f\left(x_{1},y_{3}\right)=f\left(x_{3},y_{1}\right)\rightarrow f\left(x_{2},y_{3}\right)=f\left(x_{3},y_{2}\right)$Indeed
, by virtue of the monotonicity of the function$H$relationships

train

from which we obtain by subtraction

Or

The condition$T$obviously leads to the

The geometric interpretation of this condition is as follows: the level lines of function (1) and the parallels to the axes form a fabric. Let$S\left(x_{2},y_{2}\right)$any point on the plane and$L\left(x_{2},y_{3}\right)$a point belonging to the line$x=x$ : (fig. 2); the line parallel to the axis$Ox$led by$L$meets the contour line by$S$in$M$the parallel to the axis$Oy$by$M$cuts the line parallel to$Ox\operatorname{par}S$in$N$, the contour line by$N$and the line parallel to$Oy$by$S$intersect at the point$P$, the parallel to$Ox$by$P$and the contour line by$S$intersect at$Q$Finally, the lines parallel to the axes through Q and$S$respectively meet in$R$ the condition$B$exemplify the property of points$L$And$R$to be located on the same level. In other words: the curvilinear hexagon whose sides and diagonals are level lines and straight lines parallel to the coordinate axes closes. This curvilinear hexagon is called the Brianchon figure. Therefore, the condition$B$expresses the fact that all the Brianchon figures are closing.

Conversely, in the theory of hexagonal tissues it is demonstrated that for all functions$f(x,y)$continuous and montone in the restricted sense, the closure of Brianchon figures has the consequence that$f(x,y)$is of the form (1) [6]. We present here a more direct proof of this theorem.

We admit that the function$z=f(x,y)$, defined in a domain$D$is continuous and monotonic in the restricted sense with respect to each of${}^{\text{ses }}$variables and satisfies the condition$B$for all points of the domain$D$We can also admit that the domain$D$contains within it the origin of the axes and that the function$f(x,y)$is increasing, because by applying
a suitable linear transformation to the independent variables, we can arrive at this case.

Consider a level line that intersects the coordinate axes at the points$A_{1}$And$B_{1}$, so that the domain$OA_{1}B_{1}$either in the first quadrant and inside of$D$(fig. 3). The level line drawn through the point

intersection$B_{2}$parallels to the axes by$A_{1}$And$B_{1}$cuts the axes in$A_{2}$And$C_{2}$respectively. By virtue of the condition$B$The Brianchon figures close, therefore the points of intersection$B_{3}$And$C_{3}$parallels to the axes by$A_{2}$And$B_{1},A_{1}$And$C_{2}$respectively are located on the same level line, which intersects the axes at the points$A_{3}$And$D_{3}$The points$B_{4},C_{4}$And$D_{4}$The parallels obtained in a similar manner are also on the same level line. This operation is continued until all the points of intersection of the parallels already considered are located outside the part situated in the first quadrant of the domain.$D$Obviously, the same construction can be used for all dials; that is$C_{1}$for example the point of intersection of the contour line by$B_{1}$with the line parallel to the axis$Ox$by$C_{2};B_{0}$And$A_{-1}$respectively, the intersection of the line parallel to the axis$Oy$by$C_{1}$with the parallel to$Ox$by$B_{1}$And$O$respectively; the points$B_{0}$And$O$will be on the same level, etc. We will designate by$x_{i,1}$the common x-coordinate of the points$A_{i},B_{i+1},C_{i+2},\ldots$, by$y_{1,1}$the common ordinate of the points$B_{k}$, by$y_{2,1}$that of the points$C_{k}$, etc., and by$z_{i,1}$the common values ​​of the function$z=f(x,y)$at the points$A_{i},B_{i}$,$C_{l},\ldots$

Let's define the functions$F(x),G(y)$And$H(z)$, for the moment on$\mathrm{l}_{\mathrm{tg}}$discrete points$x_{i,1},y_{i,1}$And$z_{i,1}$, Thus :

Then the given function$z=f(x,y)$satisfied at the tops of the network built on the relationship

By virtue of continuity and montony in the restricted sense of the function$f(x,y)$the system of equations$f(\alpha,\beta)=z_{1,1},f(\alpha,0)=f(0,\beta$admits a single solution$\alpha,\beta$And$0<\alpha<x_{1},0<\beta<y_{1}$ that is to say, it exists on the level line$A_{1}B_{1}$a single point$B_{2}^{\prime}$whose projections to the axes$A_{1}^{\prime}$And$B_{1}^{\prime}$are on the same level. Let's construct a rectangle with sides parallel to the axes and opposite vertices$B_{2}$And$B_{2}^{\prime}$ ; then the other two vertices$B_{3}^{\prime}$And$C_{3}^{\prime}$must be located on the same level line. Similarly$B_{4}^{\prime}$being the intersection of the parallel to$Ox$by$B_{3}^{\prime}$with the contour line by$B_{2}$, And$A_{3}^{\prime}$the projection of$B_{4}^{\prime}$on the axis$Ox$, the points$B_{3}^{\prime}$And$A_{3}^{\prime}$must be located on the same level line. Continuing this operation leads to a refinement of the network obtained by the first step of the construction. By designating the abscissas and ordinates of the new parallels by$x_{i,2}$And$y_{i,2}$respectively, we have$x_{2k,2}=x_{k,1},y_{2k,2}=y_{k,1}$Let$z_{i,2}$the values ​​of the function$f(x,y)$on the new contour lines; then$z_{2k,2}=z_{k,1}$By defining

we have at the vertices of the original network the values ​​defined above, and we can observe that relation (7) is verified for the vertices of the refined network.

We will continue this network refinement operation indefinitely.$n$-th refinement we define the sets$\left\{x_{i,n}\right\},\left\{y_{i,n}\right\}$,$\left\{z_{i,n}\right\}$and the values ​​of the functions$F,G$And$H$in these points by posing

This does not change the values ​​of these functions at the points of the ($n-1$)th network and relation (7) is verified step by step.
are positive and form a decreasing sequence, which therefore has a limit$x^{*}\geqslant 0$We have

It follows from the second equality, by virtue of continuity and montony all

• —

  the second

  Report issue for preceding element

narrow sense of$f(x,y)$that$y_{1,n}$also has a limit, that is$y^{*}$If in the first equality$n\rightarrow\infty$, we obtain$f\left(x^{*},y^{*}\right)=f\left(x^{*},0\right)$, that's to say$y^{*}=0$The second equality entails$f\left(x^{*},0\right)=f(0,0)$, SO$x^{*}=0$.

Let's consider a domain$OAB$, interior to$D$, limited by the coordinate axes and by the contour line$AB,A$being a network apex. So the whole$\left\{x_{i,n}\right\}$is dense on the segment$OA$, When$i$And$n$vary. Indeed, suppose that there exists an interval$(\gamma,\delta)\in OA$which does not contain points$x_{i,n}$. For each$n$, we determine the number$k=k(n)$so that the interval ($\gamma,\delta$) is between$A_{k}^{(n)}$And$A_{k+1}^{(n)}$The contour line by$A_{k}^{(n)}$cut the axle$Oy$in$Q_{k}^{(n)}$, the parallel to$Ox$at this point meets the contour line by$A_{k+1}^{(n)}$in$Q_{k+1}^{(n)}$(fig. 4). The projection of the point$Q_{k+1}^{(n)}$on$Ox$that's precisely the point$A_{1}^{(n)}$Taking into account that for$n\rightarrow\infty$,$\overline{Q_{k}^{(n)}Q_{k+1}^{(n)}}=\overline{OA_{1}^{(n)}}=x_{1,n}\rightarrow 0$we have$\overline{A_{k}^{(n)}A_{k+1}^{(n)}}\rightarrow 0$, contradicting our hypothesis. It follows that the network points form a dense set on the domain$OAB$.

Let us consider a domain$OMM^{*}$, limited by the axes and by the contour line$MM^{*}$, assuming that it no longer passes through network points (fig. 5). We will demonstrate that the network points form a dense set on$OMM^{*}$It suffices to demonstrate that$M$is an accumulation point of the whole$\mathcal{E}$formed by the points$x_{i,n}$located on the segment$OM$Let's suppose the opposite, that is to say that$Q=\sup\mathcal{E}$either located to the left of$M$We know that$\mathcal{E}$is dense on$OQ$So we can choose the points$Q^{\prime}$And$Q^{\prime\prime}$belonging to$\mathscr{E}$in such a way that the rectangle$Q^{\prime}Q^{\prime\prime}RS$have the peaks$S$And$Q^{\prime\prime}$on the same level line and the summit$R$either in the field$QMM^{*}Q^{*}$It follows that$R$And$R^{\prime}$are network points, that is to say that$R^{\prime}{}_{\epsilon}S$which contradicts the hypothesis$Q=\sup\mathcal{E}$.