aequationes mathematicae

, Volume 58, Issue 3, pp 223–241

# Equality of two variable weighted means: reduction to differential equations

• L. Losonczi
Article

DOI: 10.1007/s000100050110

Losonczi, L. Aequ. math. (1999) 58: 223. doi:10.1007/s000100050110

## Summary.

Let $$\Phi, \Psi$$ be strictly monotonic continuous functions, F,G be positive functions on an interval I and let $$n \in {\Bbb N} \setminus \{1\}$$. The functional equation¶¶$$\Phi^{-1}\,\left({\sum\limits_{i=1}^{n}\Phi(x_{i})F(x_{i})\over\sum\limits_{i=1}^{n} F(x_{i}}\right) \Psi^{-1}\,\left({\sum\limits_{i=1}^{n}\Psi(x_{i})G(x_{i})\over\sum\limits_{i=1}^{n} G(x_{i})}\right)\,\,(x_{1},\ldots,x_{n} \in I)$$¶was solved by Bajraktarević [3] for a fixed$$n\ge 3$$. Assuming that the functions involved are twice differentiable he proved that the above functional equation holds if and only if¶¶$$\Psi(x) = {a\Phi(x)\,+\,b\over c\Phi(x)\,+\,d},\qquad G(x) = kF(x)(c\Phi(x) + d)$$¶where a,b,c,d,k are arbitrary constants with $$k(c^2+d^2)(ad-bc)\ne 0$$. Supposing the functional equation for all$$n = 2,3,\dots$$ Aczél and Daróczy [2] obtained the same result without differentiability conditions.¶The case of fixed n = 2 is, as in many similar problems, much more difficult and allows considerably more solutions. Here we assume only that the same functional equation is satisfied for n = 2 and solve it under the supposition that the functions involved are six times differentiable. Our main tool is the deduction of a sixth order differential equation for the function $$\varphi = \Phi\circ\Psi^{-1}$$. We get 32 new families of solutions.

Keywords. Mean value, functional equation.