Second order iterative functional equations related to a competition equation

The functional equation related to competition ([2]) fx+y1-xy=fx+fy1+fxfy,x,y∈R,xy≠1,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f\left( \frac{x+y}{1-xy}\right) =\frac{f\left( x\right) +f\left(y\right)} {1+f\left( x\right) f\left( y\right)},\qquad x,y\in\mathbb{R}, xy\neq 1,$$\end{document}for y = cx with a fixed c > 0, leads to the equation f1+cx1-cx2=fx+fcx1+fxfcx,x∈R,x<1c.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f\left( \frac{\left( 1+c\right) x}{1-cx^{2}}\right) =\frac{f\left(x\right) +f\left( cx\right)} {1+f\left( x\right) f\left( cx\right)},\qquad x\in \mathbb{R}, \left\vert x \right\vert <\frac{1}{\sqrt{c}}.$$\end{document}The case c = 1 (a first order iterative functional equation) was treated in [3]. In this paper we consider the case c ≠ 1 (when the equation is of the second order). We show that a functionf:R→R,f0=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${f:\mathbb{R} \rightarrow \mathbb{R},\,f\left( 0\right) =0}$$\end{document}, differentiable at the point 0 satisfies this functional equation iff there is a realpsuch thatf=tanh∘ptan-1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${f=\tanh \circ \left( p\tan ^{-1} \right) }$$\end{document} which extends the main result of [3].


Introduction
The functional equation on a restricted domain (following [1] a conditional functional equation), the so-called competition equation, considered first in [2], was also treated in [3]. If f is a solution then either f (0) = 0, or f (0) = −1 or f (0) = 1. It is shown in [2] (cf. also [3]) that, if f takes value −1 or 1, then it is a constant function (cf. also Remark 1). Therefore only the case f (0) = 0 is interesting. The main result of [3] says that, in this case, a function f : R → R satisfies the equation for all real x, y such that xy < 1, iff f = tanh •α • tan −1 where α : R → R is an additive function. Thus, if f is measurable or continuous at least at one point, then there is a p ∈ R such that f = g p where g p = tanh • p tan −1 .
Taking y = x in (CE) we obtain the diagonalization of the competition equation, which is a first order iterative functional equation (cf. [5,6]). In [2] it is proved that if a function f : R → R , such that f (0) = 0, satisfies this equation for all x ∈ (−1, 1) , and is twice differentiable at the point 0, then f = g p for some real p.
In [4] the following stronger result is presented. A function f : R → R, f (0) = 0, differentiable at the point 0, satisfies this functional equation iff there is a real p such that f = g p . Moreover g p (0) = p (Theorem 1). Applying the theory of iterative functional equations ( [5,6], cf. also [7,8]) it is also shown that in this result the assumption of the differentiability of the solution f at the point 0 cannot be replaced by the continuity of f at 0.
In the present paper we consider a generalization of the above diagonalization problem. Namely, in Sect. 2, for a fixed c > 0 , restricting the competition equation to the straight line y = cx, we obtain the iterative functional equation which is of the second order if c = 1. (For the definition of the nth order of an iterative functional equation see [5], chapter XII, and [6] pp. 237-239.) Our main result says that a function f : R → R, differentiable at the point 0 and such that f (0) = 0, satisfies this equation iff there is a p ∈ R such that f = g p , where g p = tanh • p tan −1 . In Sect. 3 we discuss the case when c ≤ 0. In Sect. 4, recalling the motivation coming from a meteorological phenomenon (hail suppression by competition of small particles via cloud seeding), we discuss the mutual relation between (CE) and a Riccati differential equation.
Proof. It is easy to verify that if f : R → R is differentiable at the point 0, f (0) = 0, and f satisfies equation (1), then the function ϕ : R → R defined by is continuous at 0 and satisfies the functional equation Suppose that the functions ϕ 1 : R → R and ϕ 2 : R → R satisfy this equation, are continuous at the point 0 and Put The function h : is odd, continuous, strictly increasing, convex in (0, ∞) , concave in (−∞, 0), maps I c onto R and It follows that its inverse β Of course, the function γ := cβ has similar properties. Moreover and, consequently, for every t ∈ R, the sequences (β n ) n∈N and (γ n ) n∈N of iterates of the functions β and γ converge uniformly to 0 on compact subsets of R.
With the above solutions ϕ 1 , ϕ 2 of equation (2), we have Hence, putting and subtracting the respective sides of these equalities for j = 1 and j = 2, we get, for all x ∈ I c , Since ϕ 1 and ϕ 2 are continuous at 0 and Moreover, for x ∈ (−δ, δ) , we have (3) and considering the definition of γ, we get Vol. 89 (2015) Competition functional equation 111 where κ (t) = k (β (t)) , μ (t) = m (β (t)), hence and, of course, Of course, σ is continuous, strictly increasing and From inequalities (4) and (5) we obtain, for all t ∈ I, Hence, by induction, where σ n denotes the n-th iterate of σ. Since the decreasing sequence of intervals (σ n (I)) n∈N tends to {0} , this inequality, the continuity of ψ at 0 and ψ (0) = 0 imply that sup |ψ (I)| = 0. Hence, from the definition of ψ we get ϕ 1 (x) = ϕ 2 (x) for all x ∈İ. Now, according to the theory of iterative functional equations (cf. [5], p. 68, Lemma 3.1) we conclude that ϕ 1 = ϕ 2 . Taking into account the definition of the function ϕ, we have proved that, for any p ∈ R, there exists at most one solution f of equation (1) that is differentiable at 0 and such that f (0) = p. Since, in view of Lemma 1, the function f = g p satisfies equation (1), is differentiable at 0 and f (0) = p, the proof is complete.
To justify the assumption f (0) = 0 consider the following

Discussion of the case when c is nonpositive
In this section we consider the functions f : R → R, f (0) = 0, satisfying the competition equation (CE) restricted to the straight line {(x, cx) : x ∈ R} , i.e. the equation where c ≤ 0.  Assume, for instance that −1 < c < 0; setting y = −cx in the competition equation (CE) we get whence, calculating f (x) , we get the functional equation If f : R → R is differentiable at the point 0, f (0) = 0, and f satisfies this equation, then the function ϕ : R → R defined by is continuous at 0 and satisfies the functional equation Suppose that the functions ϕ 1 : R → R and ϕ 2 : R → R satisfy this equation, are continuous at the point 0 and Hence we have Setting and subtracting the respective sides of these equalities for j = 1 and j = 2, performing simple calculations, we get where Since the functions y 1 , y 2 , z 1 , z 2 are continuous at x = 0 and it is easy to check that there is a real number δ > 0 such that Hence, making use of (6), we obtain Now we can argue similarly as in the proof of Theorem 1.

Motivation and remarks on a Riccati-type differential equation and the competition functional equation (CE)
In [2] it was shown that a model of a meteorological phenomenon in cloud physics, interpreted as competition and described with the aid of a Riccati differential equation, can be fully characterized by the functional equation (CE) that does not involve any derivative. The functional equation (CE) reflects symmetry properties of the corresponding model differential equation (and of its solution).

Remark 4. (Differential equation and initial condition)
Some physical models present themselves in this way, with two arbitrary parameters a, b > 0, fixed. (In [2] there is a = b = 1.) This may be reduced to a one-parameter model: (7) can be written in the form Therefore, setting we can write this equation in the form √ a and, consequently, we can write this equation in the form With initial condition f (x 0 ) = f 0 (where x 0 , f 0 ∈ R, fixed), the solution of equation (8) (obtainable by separation of variables) reads as Check: f (x 0 ) = f 0 , and we have . It is to be noted that the last two versions of (9) do not require |f 0 | < 1 (fixed) but admit f 0 ∈ R (fixed). In general, (9) is not an odd function of x. Only for the special initial condition f (0) = 0 (i.e. x 0 = 0, f 0 = 0, called "standard" initial condition), the result (9) reduces to an odd function (called "standard" solution), With this standard solution g p we can represent the general solution f of the differential equation (8) via (9) as a rational (fractional linear) expression in g p .
From (10) we get g −p (x) = g p (−x) = −g p (x), implying that we may relax the condition p > 0 (fixed) to p ∈ R (fixed) for f : R → R. Already from (8) we see that changing the sign of p is equivalent to changing the sign of f [and f 0 in (9)].
Remark 5. If (in a certain physical situation) a slightly more general fourparameter model appears adequate or desirable (e.g. for easy interpretation), with parameters A, B, C, D ∈ R (fixed) and CD = 0, it can be reduced immediately to the two-parameter model (7) by writing abbreviating a := A D , b := BC D 2 , F (X) := D C G (X), we get equation (7). [Further reduction to the one-parameter model (8) may be done as above.] Remark 6. If a differentiable function f : R → R satisfies the functional equation (CE) then it satisfies the Riccati differential equation (8).
Proof. Assume that a differentiable function f : R → R satisfies the functional equation x,y∈ R, xy < 1.