Equality of two variable weighted means: reduction to differential equations

Summary. Let F, Y \Phi, \Psi be strictly monotonic continuous functions, F,G be positive functions on an interval I and let n ? \Bbb N \{1} n \in {\Bbb N} \setminus \{1\} . The functional equation¶¶F^-1 ((?_i=1ⁿF(x_i)F(x_i))/(?_i=1ⁿ F(x_i)]) Y^-1 ((?_i=1ⁿY(x_i)G(x_i))/(?_i=1ⁿ G(x_i))])  (x₁,?,x_n ? I) \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 Bajraktarevi' 3] for a fixed n 3 3 n\ge 3 . Assuming that the functions involved are twice differentiable he proved that the above functional equation holds if and only if¶¶Y(x) = (aF(x) + b)/(cF(x) + d)],       G(x) = kF(x)(cF(x) + d) \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(c²+d²)(ad-bc) 1 0 k(c^2+d^2)(ad-bc)\ne 0 . Supposing the functional equation for all n = 2,3,... 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 j = F°Y^-1 \varphi = \Phi\circ\Psi^{-1} . We get 32 new families of solutions.