Equality of two variable Cauchy mean values - Springer Link

Aequationes Math. 65 (2003) 61–81 0001-9054/03/010061-21

c Birkh¨ ° auser Verlag, Basel, 2003

Aequationes Mathematicae

Equality of two variable Cauchy mean values ´szlo ´ Losonczi La

Summary. Let I be an interval, f, g : I → R be differentiable functions on I. Then, by the Cauchy mean value theorem for every x, y ∈ I, x 6= y there exists a t between x and y such that f0 f 0 (t) [g(x) − g(y)] = g 0 (t) [f (x) − f (y)] . If g 0 6= 0 and 0 is invertible on I, then t is unique and g ¶ µ 0 ¶ −1 µ f (x) − f (y) f . t= g0 g(x) − g(y) Continuous extension gives t = x if y = x. The number t is called the Cauchy mean value of x, y and is denoted by Df,g (x, y). We solve the equality problem for (two variable) Cauchy means, i.e. we solve the functional equation Df1 ,g1 (x, y) = Df2 ,g2 (x, y)

(x, y ∈ I)

under the basic assumption that the functions f1 , g1 , f2 , g2 are seven times continuously differentiable. First we reduce the number of unknown functions to 3 by a suitable transformation. Then, using the second, fourth and sixth partial derivatives (the first, third and fifth partial derivatives do not give independent equations) of this transformed equation at equal values of the variables we obtain a system of sixth order ordinary differential equations. We show that the solutions of our system can be obtained from the solutions of a Riccati equation ¯ 2 = c (G0 )4/3 , ¯ 0 − 2h 4h where c is an arbitrary constant and G is the solution of the nonlinear equation µ 00 ¶ 3 Gıv G000 G00 G 9 0 − 45 0 + 40 = 0. G G G0 G0 There are 33 families of solutions. The principal solution is f20 (x) = αf10 (x) + βg10 (x) g20 (x) = γf10 (x) + δg10 (x), where α, β, γ, δ are arbitrary constants with αδ −βγ 6= 0. It contains two “arbitrary” functions f10 and g10 (and corresponds to the general solution of the same equality problem for n ≥ 3 variables, see Losonczi [17]). The other 32 families arise only for two variables and they contain only one “arbitrary” function and several arbitrary constants. Mathematics Subject Classification (2000). Primary 39B22. Keywords. Divided differences, Cauchy Mean Value, functional equations. This research has been supported by Kuwait University Grant SM04/99.

62

AEM

L. Losonczi

1. Introduction For a function f : I → R, I being a real interval, the divided differences of f on distinct points xi ∈ I are usually defined inductively by [x1 ]f :=f (x1 ), [x1 , . . . , xn ]f :=

[x1 , . . . , xn−1 ]f − [x2 , . . . , xn ]f x1 − xn

(n = 2, 3, . . . )

(see e.g. Aumann and Haupt [4] §3.17, their expression contains an extra factor n − 1 on the right). This definition must be modified if two or more points of [x1 , . . . , xn ]f coincide: if at most r points xi coincide, the definition is then framed on the assumption that f is (r − 1)-times differentiable on I. A full definition, as the ratio of two determinants, can be found in Schumaker [31]. It is known that [x1 , . . . , xn ]f is independent of the order of its arguments, it is a continuous (analytic) function of x1 , . . . , xn , if f (n−1) is continuous (f is analytic) respectively. Let x1 ≤ · · · ≤ xn , xi ∈ I and assume that f (n−1) , g (n−1) exist, with g (n−1) (u) 6= 0, on I. Then there is a t ∈ [x1 , xn ] (moreover t ∈ (x1 , xn ) if x1 < xn ) such that f (n−1) (t) [x1 , . . . , xn ]f = (n−1) . [x1 , . . . , xn ]g g (t) This is the “Cauchy mean value theorem for divided differences” (see Leach and Sholander [13] and also R¨ atz and Russell [29], P´ ales [24], [25]). Supposing that the function u → f (n−1) (u)/g (n−1) (u) is invertible on I, the value t is unique and ¶ µ (n−1) ¶−1 µ [x1 , . . . , xn ]f f t= [x1 , . . . , xn ]g g (n−1) is a mean value of x1 , . . . , xn which is symmetric in its variables. It is called the Cauchy (or difference) mean of the numbers x1 , . . . , xn and will be denoted by Df,g (x1 , . . . , xn ) or by Df,g (x) where x = (x1 , . . . , xn ). In the special case n = 2,  µ 0 ¶−1 µ ¶ f (x) − f (y)  f , if x 6= y Df,g (x, y) = g0 g(x) − g(y)  x, if x = y. The mean value Df,g was first defined by Leach and Sholander [13] who also found some basic properties of it (they called it extended (f, g)-mean of x1 , . . . , xn ). Some homogeneous two variable Cauchy means (where f, g are power or logarithmic functions) were discovered earlier by Stolarsky [35], [36]. These special means have been studied extensively (see e.g. Alzer [3], Brenner [5], Brenner and Carlson [6], Burk [7], Carlson [8], Dodd [9], Losonczi and P´ ales [18], Lin [14], P´ ales [21],

Vol. 65 (2003)

Equality of two variable Cauchy mean values

63

[22], Pittenger [26], [27], S´ andor [30], Seiffert [32], [33], Sz´ekely [37]). Some of these means have applications in electrostatics [28], in heat conduction and and chemical problems [38], in signal processing theory in connection with time–frequency distributions [10]. Neuman [20] studied multivariable weighted logarithmic means. The systematic study of the (general) Df,g -means started with the paper [17] where the equality problem Df1 ,g1 (x) = Df2 ,g2 (x)

(x ∈ I n , n ≥ 3 is fixed)

was solved. Under suitable differentiability conditions on the functions involved the above equality holds if and only if there exist constants α, β, γ, δ with αδ − βγ 6= 0 such that for all x ∈ I   f (n−1) (x) = αf (n−1) (x) + βg (n−1) (x) 2 1 1  g (n−1) (x) = γf (n−1) (x) + δg (n−1) (x) 2 1 1 holds. For n = 2 not only the above result gets false, but also the method of [17] fails (since one of the differential equations used in the solution becomes an identity). Here we solve the equality problem of two variable Cauchy means. There are a 32 new families of solutions. The result opens the road to finding all homogeneous Cauchy means for any fixed n ≥ 2.

2. Reduction of the number of unknown functions Throughout the paper I is a real interval (open or closed) and En (I) (n ∈ N) denotes the set of all pairs (f, g) of functions f, g : I → R satisfying the following conditions: f, g are n-times continuously differentiable on I, g 0 (u) 6= 0 for u ∈ I, h :=

f0 has non-vanishing (first) derivative on I. g0

We remark that (f, g) ∈ E1 (I) is a sufficient condition for Df,g (x, y) to exist for every possible choice of x, y ∈ I. By the intermediate value property of the derivative (f, g) ∈ E1 (I) implies that the functions g 0 (and h0 ) are either positive or negative on I, therefore h is strictly monotonic and invertible. Further, if (f, g) ∈ E1 (I), then Zx

0

Z1

f (s) ds = (x − y)

f (x) − f (y) = y

0

f 0 (tx + (1 − t)y) dt,

64

AEM

L. Losonczi

hence we have the representation  1 R 0  f (tx + (1 − t)y)dt   −1  0 Df,g (x, y) = h  1   R 0 g (tx + (1 − t)y)dt

(x, y ∈ I).

(1)

0

Our aim is to solve the functional equation Df1 ,g1 (u, v) = Df2 ,g2 (u, v)

(u, v ∈ I).

(2)

We reduce the number of unknown functions in (2) to 3 by Proposition 1. If (f1 , g1 ), (f2 , g2 ) ∈ E1 (I), then (2) holds if and only if Df,g (x, y) = DF,G (x, y) where

(x, y ∈ J)

 −1 −1 h(x) = h1 (h−1  2 (x)), f (x) = f1 (h2 (x)), g(x) = g1 (h2 (x)),     −1 −1 H(x) = h2 (h−1 2 (x)) = x, F (x) = f2 (h2 (x)), G(x) = g2 (h2 (x)),   f 0 (u) f 0 (u)   , h2 (u) = 20  h1 (u) = 10 g1 (u) g2 (u)

(3)

(4)

for x ∈ J = h2 (I) = { h2 (x) : u ∈ I } and u ∈ I. Proof. (2) can be written as ¶ ¶ µ µ f1 (u) − f1 (v) f2 (u) − f2 (v) −1 = h h−1 1 2 g1 (u) − g1 (v) g2 (u) − g2 (v)

(u 6= v, u, v ∈ I).

(5)

If u = v, then both sides of (2) have the common value u which can also be obtained by taking the limit v → u in (5). Applying h2 to both sides and substituting −1 u = h−1 2 (x), v = h2 (y) with x, y ∈ J, equation (5) goes over into ¶ µ F (x) − F (y) f (x) − f (y) −1 = (x 6= y, x, y ∈ J) (6) h g(x) − g(y) G(x) − G(y) and conversely, from (6) we can get (5) by applying h−1 2 to both sides of (6) and substituting x = h2 (u), y = h2 (v). We claim that (6) is of the form (3). To justify this claim we show that (by the formula for the derivative of the inverse h−1 2 ) f10 (h−1 2 (x)) 0 0 f 0 (h−1 f (x) h (h−1 2 (x)) 2 (x)) = 02 −1 = h1 (h−1 = 10 −1 2 (x)) = h(x), 0 g (x) g1 (h2 (x)) g1 (h2 (x)) h02 (h−1 2 (x))

Vol. 65 (2003)

Equality of two variable Cauchy mean values

65

and similarly f20 (h−1 2 (x)) f20 (h−1 F 0 (x) h02 (h−1 2 (x)) 2 (x)) = = h2 (h−1 = 2 (x)) = H(x) = x. −1 −1 0 0 G0 (x) g2 (h2 (x)) g2 (h2 (x)) h02 (h−1 2 (x)) (3) has only three unknown functions: h, g, G as f 0 = hg 0 , F 0 = HG0 , but H, being the identity function, is known.

3. Reduction of (3) to a differential equation A well-known method for solving functional equations is their reduction to differential equations (see e.g. Acz´el [1]). Differentiating the functional equation with respect to some of the variables and substituting suitable values of the variables, one can get usually easily differential equations for the unknown functions. Solving the differential equations and selecting those solutions which also solve the functional equation, completes the solution procedure. There are however some functional equations for which deducing differential equations is not easy at all. (3) is such an equation (for other examples see Losonczi [15], [16], P´ ales [23]). In this section we assume that (f, g), (F, G) ∈ E7 (J). Let us write (3) in the form  R1 0 0 f (tx + (1 − t)y)dt F (tx + (1 − t)y)dt    0 −1  0 =0 h 1 − 1  R 0 R 0 g (tx + (1 − t)y)dt G (tx + (1 − t)y)dt 

R1

0

(x, y ∈ J).

(7)

0

Denote the left-hand side of (7) by E(x, y). We will calculate some partial derivatives of E at the point (x, x). It turns out that only the second, fourth and sixth partial derivatives of our equation E(x, y) = 0 give independent differential equations (at the points (x, x)). Both first partial derivatives give the identity 0 = 0, the third and fifth partial derivatives supply equations which can be obtained from the second and fourth derivatives. To facilitate the calculation of the partial derivatives of E, we write it as E(x, y) = h−1

µ

k(x, y) l(x, y)

¶ −

K(x, y) L(x, y)

66

AEM

L. Losonczi

where Z1

Z1

0

(h · g )(tx + (1 − t)y)dt,

k(x, y) =

l(x, y) =

0

Z1 K(x, y) =

g 0 (tx + (1 − t)y)dt,

0

(i · G0 )(tx + (1 − t)y)dt, L(x, y) =

0

Z1

G0 (tx + (1 − t)y)dt,

0

h · g 0 is the product of the functions h and g 0 and i is the identity function: i(x) = x (x ∈ J). It is easy to check that ∂ m+n k(x, x) = (h · g 0 )(m+n) (x) ∂y m ∂xn ∂

m+n

∂y m

l(x, x) = g (m+n+1) (x) ∂xn

Z1

Z1 tm (1 − t)n dt = 0

tm (1 − t)n dt = 0

(h · g 0 )(m+n) (x) n! m! , (n + m + 1)!

g (m+n+1) (x) n! m! , (n + m + 1)!

and we have analogous expressions for the partial derivatives of K, L too. Using these and the formulae ¡ −1 ¢0 ¡ −1 ¢00 ¡ −1 ¢000 1 h00 h000 (h00 )2 h = , h = − 0 3, h =− 0 4 +3 0 5, 0 h (h ) (h ) (h ) ıv 000 00 00 3 ¡ −1 ¢ıv h h h (h ) h = − 0 5 + 10 0 6 − 15 0 7 , (h ) (h ) (h ) v ıv 00 ¡ −1 ¢v h h h (h000 )2 h000 (h00 )2 (h00 )4 h = − 0 6 + 15 0 7 + 10 0 7 − 105 + 105 , (h ) (h ) (h ) (h0 )8 (h0 )9 ¡ −1 ¢vı hvı hv h00 hıv h000 (hıv (h00 )2 (h000 )2 h00 h = − 0 7 + 21 0 8 + 35 0 8 − 210 − 280 0 9 (h ) (h ) (h ) (h ) (h0 )9 +1260

h000 (h00 )3 (h00 )5 − 945 (h0 )10 (h0 )11

(where the derivatives of h−1 are to be taken at x, the derivatives of h are to be taken at h−1 (x)) the derivatives of E can be expressed using the functions h, g, G and their derivatives. This involves a large amount of elementary calculation which we could do only up to the fourth derivatives. Subsequent derivatives were found and also other long calculations (substitutions of a number of expressions into others) were done using the software package Maple V. Our first equation is · 00 ¸ h00 g 00 G (8) −12Exy = 0 + 2 0 − 2 0 = 0. h g G Here and in the sequel, E with subscripts indicates that the respective partial derivatives are taken at the point (x, x). Furthermore, the derivatives of h, g, G

Vol. 65 (2003)

Equality of two variable Cauchy mean values

67

are taken at the point x. Taking the other possible second derivative, we get the same equation since Exy = −Exx . The second equation we intend to use is ¸ · µ 00 ¶3 µ 00 ¶2 00 hıv h00 g 00 h000 g h h − 30 + 12 + 25 + 20 h0 h0 g 0 h0 h0 h0 g0 " µ 00 ¶2 # 00 " ıv µ 00 ¶3 # h g 000 g g 000 g 00 g g + −8 0 + 20 + 8 0 − 40 0 0 + 40 g g0 h0 g g g g0 " µ 00 ¶3 # G000 G00 Gıv G = 0. − 8 0 − 40 0 0 + 40 G G G G0

−240Exxyy = 7

(9)

Our last equation comes from the sixth derivative: −20160Exxxyyy = 0, in detailed form ¸ · h00 g 00 hv hvı 171 0 + −1197 0 − 486 0 h g h0 "h µ 00 ¶2 µ 00 ¶2 # ıv h h00 g 00 g 000 h000 g h + 3402 0 0 − 1656 0 + 2772 − 2961 0 + 6552 0 0 h g g g h h h0 " ¸ ¶ ¶ · µ µ 3 2 00 2 h00 g 00 h000 g h00 h00 + 11340 0 + 3276 0 + −25690 − 16422 h g h0 h0 h0 g0 # · µ 00 ¶2 ¸ 00 µ 00 ¶3 000 000 00 g g g g ıv h000 h g g + 8064 0 − 14952 0 − 7952 + 7056 0 0 − 864 0 g g h0 g0 g g g h0 #µ ¶ " ¶ µ 00 ¶5 µ 00 ¶4 00 µ 2 3 g 000 g h00 h h g 00 +11865 + 10710 + −6888 + 13860 h0 h0 g0 g0 g0 h0 ¸ µ 00 ¶2 · µ 00 ¶3 000 00 ıv g g g h g + 11760 − 9408 0 0 + 1008 0 g0 g g g h0 (10) # " ¶ ¶ ¶ µ µ µ 2 2 4 h00 g ıv g 00 gv g 00 g 000 g 000 g 00 + 1008 0 0 −9408 0 + 2184 −144 + 7560 g g g g0 g0 g0 g0 h0 " ¸ · µ 00 ¶2 g vı g v g 00 g 000 g ıv g + 144 0 − 1008 0 0 + 6048 − 3024 g g g g0 g0 g0 µ ¶ ¶ ¶5 # µ µ 2 3 000 g 00 g 000 g g 00 g 00 +13440 0 − 30240 + 15120 g g0 g0 g0 g0 " ¶ ¸ · µ 2 Gvı Gv G00 G000 Gıv G00 − 144 0 − 1008 0 0 + 6048 − 3024 G G G G0 G0 G0 µ ¶ ¶ ¶5 # µ µ 2 3 G00 G000 G000 G00 G00 = 0. +13440 0 − 30240 + 15120 G G0 G0 G0 G0

68

AEM

L. Losonczi

At the first glance, trying to solve the system (8), (9), (10) looks hopeless. But it can be solved if we introduce the right new functions. Equation (8) suggests the introduction of new functions for the ratios h00 (u) ¯ , h(u) := 0 h (u)

g¯(u) :=

g 00 (u) , g 0 (u)

G00 (u) ¯ G(u) := 0 G (u)

(u ∈ J).

(11)

h(i) g (i) G(i) , , (i = 2, . . . , 6) can be expressed as h0 g 0 G0 ¯ g¯, G ¯ and their derivatives. Thus our system can be written in polynomials of h, terms of these new functions. This new system is still too complicated. After ¯ g¯ from this system gives a nonlinear several trials we found that eliminating h, equation for G which can be solved. So, we proceed as follows.

It is easy to see that all ratios

(i) ¯ and express g ¯ − 1h as With the new functions (11) write (8) as g¯ = G 2 g0 ¯ and their derivatives: ¯ h polynomials of G,  00 g  ¯ ¯ − 1 h,  =G  0  g 2     µ ¶0    1¯ g 000  ¯ − 1h ¯ 2=G ¯0 + 1 h ¯ 2, ¯ ¯ − 1 h) ¯0 + G ¯2 − G ¯h  = G − + (G h  0  g 2 2 2 4     µ ¶µ ¶0 µ ¶3 1 ¯ 00 1¯ 1¯ g ıv 1¯ (12) ¯ ¯ ¯ ¯ = (G − h) + 3 G − h G− h + G− h  0  g 2 2 2 2        ¯0 + 3 G ¯2 − 3 G ¯ − 3G ¯ ¯0h ¯h ¯h ¯2h ¯ 00 + 3G ¯G ¯0 + G ¯3 − 3 G  =G   2 2 4 2      1 ¯ 00 3 ¯ ¯ 0 1 ¯ 3   + hh − h , − h 2 4 8

and also express the ratios

h(i) (i = 2, 3, 4) as h0

000 ıv h00 ¯ h =h ¯0 + h ¯ 2, h = h ¯ 00 + 3h ¯h ¯0 + h ¯ 3. = h, h0 h0 h0

(13)

Substituting these into (9), we obtain that ¯ 2 ) = 3h ¯ 00 − 3h ¯h ¯ 0. ¯ 0 − 2h ¯ h G(4

(14)

(14) suggests the introduction of the function · 00 ¸2 h000 (x) h (x) 0 2 ¯ ¯ −6 S(x) := 4h (x) − 2h(x) = 4 0 h (x) h0 (x)

(x ∈ J).

(15)

We remark that S/4 is exactly the Schwarzian derivative of h which can also be

Vol. 65 (2003)

Equality of two variable Cauchy mean values



69

 1 00 h   2! 6 . det   1  (h0 )2 1 h00 h000 2! 3! Probably it comes into play here because it is invariant with respect to the transαx + β , see [19], p. 199. formation x → γx + δ Since ¯ 00 − 3h ¯h ¯ 0 = 3 S0, 3h 4 (14) has now the very simple form 4 ¯ (16) S 0 = S G. 3 Writing (8), (15), (16) as written as

h0

4 G00 ¯0 = 1 S + 1 h ¯ 2, S0 = S 0 h 4 2 3 G (i) (i) h g respectively, we can express the ratios 0 , 0 (i = 2, . . . , 6) appearing in (10) h g G00 ¯ as polynomials of 0 , h, S and their derivatives. For example, for i = 3, 4 we find G 000 h ¯0 + h ¯2 = 1 S + 3 h ¯2, =h h0 4 2 ¶0 ¶ µ µ hıv ¯3 ¯ 00 + 3h ¯h ¯0 + h ¯3 = 1 S + 1 h ¯ 1S + 1h ¯ 2 + 3h ¯2 + h = h h0 4 2 4 2 g¯ =

1¯ G00 − h, G0 2

1 G00 ¯ 2 + 3h ¯ 3, S + Sh 3 G0 µ 00 ¶0 µ 00 ¶2 1¯ 1¯ G000 G00 ¯ 1 G G = g¯0 + g¯2 = − + − = 0 − 0h − S, h h 0 0 G 2 G 2 G G 8 =

g 000 g0

g ıv = g¯00 + 3¯ g g¯0 + g¯3 g0 µ 00 µ 00 ¶00 ¶0 µ 00 ¶ µ 00 ¶3 1¯ 1¯ 1¯ 1¯ G G G G = − + 3 − − − h h h + h G0 2 G0 2 G0 2 G0 2 Gıv 3 G000 ¯ 13 G00 1 ¯ − S + S h. h− G0 2 G0 24 G0 16 ¯ disappears and we obtain that Substituting these into (10), h " µ 00 ¶3 # Gıv G000 G00 G S 9 0 − 45 0 0 + 40 = 0. G G G G0 =

(17)

70

AEM

L. Losonczi

With this we have proved Theorem 1. Suppose that (f, g), (F, G) ∈ E7 (J) and F 0 (x)/G0 (x) = x (x ∈ J), the functional equation  1 R 0 R1 0 f (tx + (1 − t)y)dt F (tx + (1 − t)y)dt    0  =0 − h−1  01   R1 0 R 0 g (tx + (1 − t)y)dt G (tx + (1 − t)y)dt 0

h000 (x) −6 S(x) = 4 0 h (x)

µ

h00 (x) h0 (x)

¶2

f 0 (x) is either identically zero or never zero on J. Then either g 0 (x) S(x) = 0

or G satisfies the differential equation 9

(7)

0

is satisfied,

where h(x) =

(x, y ∈ J)

G000 (x) G00 (x) Gıv (x) − 45 0 + 40 0 G (x) G (x) G0 (x)

(x ∈ J) µ

G00 (x) G0 (x)

(18) ¶3 =0

(x ∈ J).

Further, the other unknown functions satisfy the equations  4 G00 (x)   S(x), F 0 (x) = x G0 (x), f 0 (x) = h(x) g 0 (x), S 0 (x) =   3 G0 (x) µ 00 ¶0 µ ¶2   1 h00 (x) G00 (x) 1 h00 (x) 1 g 00 (x) h (x)   S(x) + = − . = , h0 (x) 4 2 h0 (x) g 0 (x) G0 (x) 2 h0 (x)

(19)

(20)

4. The solution of the differential equation (19) Theorem 2. Suppose that (f, g), (F, G) ∈ E7 (J), F 0 (x)/G0 (x) = x (x ∈ J), the functional equation (7) is satisfied and µ 00 ¶2 h (x) h000 (x) ¯ 0 (x) − 2h(x) ¯ 2 = 0 (x ∈ J) −6 = 4h (18) S(x) = 4 0 h (x) h0 (x) holds. Then there exists constants α, β, γ, δ with αδ − βγ 6= 0 such that ( 0 f (x) = αF 0 (x) + βG0 (x) g 0 (x) = γF 0 (x) + δG0 (x)

(21)

holds for all x ∈ J. Conversely, if (21) is valid, then f, g, F, G satisfy the functional equation (7).

Vol. 65 (2003)

Equality of two variable Cauchy mean values

71

¯ whose solutions are h(x) ¯ Proof. (18) is a separable differential equation for h = 2 ¯ and h(x) = 0 where c is an arbitrary constant. From − x+c  − 2 , 00 h (x) 0 0 ¯ = (ln |h (x)|) = h(x) = x+c  h0 (x) 0 we find by integration

(

h(x) =

e +d Ax + B x+c = Cx +D ax + b

(x ∈ J)

(22)

where A, B, C, D are constants satisfying C 2 + D2 > 0 and BC − AD 6= 0 (since h0 6= 0). Rewriting equation (8) in the form  ¯¶ µ ¯ 0 C  00 ¯ G (x) ¯ 0 1 1 (x) h if C = 6 0 − ¯ ¯ = , = h(x) = ln ¯¯ 0 Cx +D ¯ 0  g (x) 2 h (x) 2 0 if C = 0 we get by integration that P G0 (x) = g 0 (x) Cx + D

(x ∈ J)

(23)

where P 6= 0 is a constant. From (22), (23) with h =

f0 F0 , H = i = 0 i(x) = x, 0 g G

we obtain that

Ax + B AF 0 (x) + BG0 (x) f 0 (x) = = , 0 g (x) Cx + D CF 0 (x) + DG0 (x) P P G0 (x) G0 (x) = = . 0 0 g (x) Cx + D CF (x) + DG0 (x)

From the last equation g 0 (x) =

C 0 D F (x) + G0 (x) = γF 0 (x) + δG0 (x) P P

and using this we get f 0 (x) =

A 0 B F (x) + G0 (x) = αF 0 (x) + βG0 (x), P P

proving (21). Here the constants have to satisfy αδ − βγ 6= 0, otherwise

f0 would be a g0

constant and not invertible. The converse statement can be justified by a simple calculation. The solution of (19) is given by

72

AEM

L. Losonczi

Theorem 3. Suppose that G : J → R is four times differentiable, satisfies the differential equation (19) on J and either G00 (x) = 0 (x ∈ J) or G00 (x) 6= 0 (x ∈ J). Then there are constants A, B 6= 0, C4 6= 0 such that for all x ∈ J G0 (x) = C4 , or

(24)

G0 (x) =

C4 , or |x + A|3/2

(25)

G0 (x) =

C4 , or (x + A)3

(26)

G0 (x) = G0 (x) =

C4 [(x + A)2 − B 2 ]

3/2

C4 [(x + A)2 + B 2 ]

3/2

, or

(27)

.

(28)

Proof. If G00 = 0, then G0 is a constant function and we get (24). If G00 6= 0, then G0 is an invertible function, therefore there is a function p on the range of v := G0 , defined by p(z) = G00 ((G0 )−1 (z)) for each z = G0 (x) (x = (G0 )−1 (z)), such that v 0 (x) = G00 (x) = p (G0 (x)) = p (v(x)) . Thus G000 = v 00 = p0 (v)v 0 = p0 (v)p(v), Gıv = v 000 = p00 (v)v 0 p(v) + p0 (v)p0 (v)v 0 = p00 (v)p(v)2 + p0 (v)2 p(v), and from (19) we get 9v 2 (p00 p2 + (p0 )2 p) − 45vp0 p2 + 40p3 = 0. With P = p2 , this equation goes over into the Euler equation 9v 2 P 00 − 45vP 0 + 80P = 0. The solutions are P (v) = C1 v 10/3 + C2 v 8/3 therefore

(C1 , C2 ∈ R, C12 + C22 > 0),

p v 0 = p(v) = ± C1 v 10/3 + C2 v 8/3 .

This is a separable equation for v whose nonconstant solutions are given by Z Z dv p . ± dx = 10/3 C1 v + C2 v 8/3

Vol. 65 (2003)

Equality of two variable Cauchy mean values

73

If C2 6= 0, then C1 can be both positive and negative, while if C2 = 0, then C1 > 0, otherwise the integrand is not defined. Integrating, we obtain  −3 −2/3  if C2 = 0, C1 > 0  √ v  2 C1 ±(x + C3 ) = p    −3 C1 + C2 v −2/3 if C2 6= 0 C2 hence

·

¸ 3 3/2 C4 2C1 0 = if C2 = 0, C1 > 0, G (x) = v(x) = ± |x + C3 |3/2 |x + A|3/2 27 3 C4 |C 2| if C2 6= 0, G0 (x) = v(x) = ± · ¸3/2 = 3/2 [(x + A)2 ± B 2 ] 9C1 2 (x + C3 ) − 2 C2

with A = C3 and suitably chosen constants B, C4 . The first equation gives the solution (25), the second supplies solutions (26), (27), (28) with C1 = 0, C1 > 0, C1 < 0, respectively.

5. The solutions g, G, h of the system (8), (9), (10) In possession of G we first find S by integrating the equation (16): S(x) = C5 G0 (x)4/3

(x ∈ J)

(29)

where C5 is an arbitrary constant. We may suppose that S(x) 6= 0 (x ∈ J), since the case S(x) = 0 (x ∈ J) was settled in Theorem 2. Using G0 given by (24), (25), (26), (27), (28), we obtain from (29) that there are constants A, B 6= 0, C 6= 0 such that for all x ∈ J we have S(x) = ±8C 2 , or

(30)

S(x) =

±8C 2 , or (x + A)2

(31)

S(x) =

±8C 2 , or (x + A)4

(32)

S(x) =

S(x) =

±8C 2 [(x + A)2 − B 2 ] ±8C 2 [(x + A)2 + B 2 ]

2,

2.

or

(33)

(34)

74

AEM

L. Losonczi

¯ corresponding to S using the Riccati differential Next we find the functions h equation ¯ 2 = S. ¯ 0 − 2h (15) 4h The transformation R

d ¯ (35) h(x) = −2 (ln z(x)) dx transforms (15) into the linear homogeneous differential equation 1 (36) z 00 + Sz = 0. 8 The solutions of (36) have been found in [16]. Knowing z, we get from the equation ¢ ¡ h00 ¯ = ln z −2 0 =h (37) 0 h that Z (38) h=K z −2 + L 1

z(x) = e− 2

¯ h(x) dx

(> 0),

where K 6= 0, L are arbitrary constants. G0 is given by (24)–(28), and g 0 can be obtained from (8) or from the third equation of (20) as g 0 = C6 G0 z

(39)

where C6 6= 0 is an arbitrary constant. In the next table we list all solutions of (7) obtained in this way. In addition to the solution given by Theorem 2 we get 32 families of solutions. In our list z and f (for typographical reasons) are not included, but we can easily find them from the equations 1

z(x) = e− 2

R

¯ h(x) dx

= (cons.) · |h0 (x)|−1/2 , f 0 (x) = h(x)g 0 (x).

(40)

For uniformness we also introduced new constants. We remark that the functions ϕ in the table of solutions in [16] are identical to the functions h in the following table. h(x)

g 0 (x)

G0 (x)

K tanh(Cx+D)+L

M cosh(Cx+D)

N

(41)

K coth(Cx+D)+L

M sinh(Cx+D)

N

(42)

K tan(Cx+D)+L

M cos(Cx+D)

N

(43)

K exp(−2Cx)+L

M exp(Cx)

N

(44)

K (ln |x+A|+D)−1 +L

M (ln |x+A|+D) x+A

N |x+A|3/2

(45)

K ln |x+A|+L

M x+A

N |x+A|3/2

(46)

Vol. 65 (2003)

Equality of two variable Cauchy mean values

75

K tanh (C ln |x+A|+D)+L

M cosh (C ln |x+A|+D) x+A

N |x+A|3/2

(47)

K coth (C ln |x+A|+D)+L

M sinh (C ln |x+A|+D) x+A

N |x + A|3/2

(48)

K exp (−2C ln |x + A|) + L

M exp (C ln |x + A|) x+A

N |x + A|3/2

(49)

K tan (C ln |x + A| + D) + L

M cos (C ln |x + A| + D) x+A

N |x + A|3/2

(50)

N (x + A)3

(51)

N (x + A)3

(52)

N (x + A)3

(53)

N (x + A)3

(54)

N (B 2 − (x + A)2 )3/2

(55)

N (B 2 − (x+A)2 )3/2

(56)

N (B 2 −(x+A)2 )3/2

(57)

N (B 2 −(x+A)2 )3/2

(58)

N (B 2 − (x+A)2 )3/2

(59)

N (B 2 − (x+A)2 )3/2

(60)

N ((x+A)2 − B 2 )3/2

(61)

K tanh

K coth

K exp

K tan

K

³

³

³

³

³

C x+A

C x+A

−2C x+A

C x+A

´

´

+D +L

³

K tanh C

+D +L

³

K exp −2C

³

M exp

+L

M cos

+D +L

´

³

C x+A

+D

´

³

C x+A

´

+D

³

C x+A

+D

´

(x + A)2

´ −1

+L

³ M tanh−1

x+A B

+D

´

B 2 − (x + A)2 M B 2 − (x+A)2

A tanh−1 x+ B

x+A +D B

´ −1

³ M cosh C tanh−1

´

x+A +D B

+L ´ +L

+L

´ ³ A +D M sinh C tanh−1 x+ B B 2 − (x+A)2

+L

´

x+A +D B

B 2 − (x+A)2 ³ M coth−1

+L

x+A B

B 2 − (x+A)2 ³ M cos C tanh−1

´

´

x+A +D B

B 2 − (x+A)2

³ M exp C tanh−1

´

A tanh−1 x+ +D B

K coth−1

+D

(x + A)2

´

K coth C tanh−1

C x+A

(x + A)2

A +D tanh−1 x+ B

³

K tan C

M sinh

x+A +L B

K tanh−1

³

(x + A)2

´

tanh−1 x+A B

³

M cosh

x+A +D B

(x+A)2 − B 2

´

´

76

AEM

L. Losonczi

M (x+A)2 − B 2

x+A +L B

K coth−1

³

K tanh C

A coth−1 x+ +D B

³

x+A +D B

K coth C coth−1

³

K exp −2C

A coth−1 x+ B

³

K tan C coth−1

K

³

K tan−1

K tanh C tan−1

K coth C

³

K tan C

(x+A)2 − B 2

+L

x+A +D B

x+A +D B

x+A +D B

A tan−1 x+ +D B

x+A B

³ M cosh C tan−1 +L

+L

A tan−1 x+ +D B

+L

+L

N ((x+A)2 − B 2 )3/2

(65)

N ((x+A)2 − B 2 )3/2

(66)

N ((x+A)2 +B 2 )3/2

(67)

N ((x+A)2 +B 2 )3/2

(68)

N ((x+A)2 +B 2 )3/2

(69)

N ((x+A)2 +B 2 )3/2

(70)

N ((x+A)2 +B 2 )3/2

(71)

N ((x+A)2 +B 2 )3/2

(72)

´

x+A +D B

´

x+A B

´

(x+A)2 +B 2 ³ M cos C tan−1

´

x+A +D B

(x+A)2 +B 2 ³ M exp C tan−1

´

(64)

´

(x+A)2 +B 2 ³ M sinh C tan−1

´

N ((x+A)2 − B 2 )3/2

´

(x+A)2 +B 2

´

(63)

´

(x+A)2 − B 2 ³ M tan−1

+L

x+A B

(x+A)2 − B 2 ³ M cos C coth−1

´

N ((x+A)2 −B 2 )3/2 ´

(x+A)2 − B 2 ³ M exp C coth−1

+L

x+A +D B

M (x+A)2 +B 2

K exp −2C tan−1 ³

+L

x+A +L B

³

³

´ −1

x+A +D B

³ M sinh C coth−1

´

´

x+A +D B

A tan−1 x+ +D B

+L

(62)

´

³ M cosh C coth−1

´

N ((x+A)2 − B 2 )3/2

x+A +D B

(x+A)2 +B 2

´

where A, B 6= 0, C 6= 0, D, K 6= 0, L, M 6= 0, N 6= 0, are arbitrary constants. Theorem 4. Suppose that (i) (f, g), (F, G) ∈ E7 (J) and F 0 (x)/G0 (x) = x (x ∈ J), that is the functions f, g, F, G : J → R are seven times differentiable, the first derivatives of g, G, h := f 0 /g 0 do not vanish and F 0 (x) = x G0 (x) on J,

Vol. 65 (2003)

Equality of two variable Cauchy mean values

(ii) the functional equation  1 R 0 R1 0 F (tx + (1 − t)y)dt  f (tx + (1 − t)y)dt   0  h−1  01 =0 − 1  R 0 R 0 g (tx + (1 − t)y)dt G (tx + (1 − t)y)dt 0

77

(x, y ∈ J)

(7)

0

is satisfied, µ 00 ¶2 h000 (x) h (x) −6 6= 0 (x ∈ J), (iii) S(x) = 4 0 h (x) h0 (x) (iv) either G00 (x) = 0 (x ∈ J) or G00 (x) 6= 0 (x ∈ J). Then the functions h, g 0 , G0 are given by (41)–(72) where A, B 6= 0, C 6= 0, D, K 6= 0, L, M 6= 0, N 6= 0 are arbitrary constants, and for each solution the common domain J of the functions h, g 0 , G0 has to be chosen such that the condition (i) is satisfied.

6. The main result Theorem 5. Suppose that I is a real interval and (i) (f1 , g1 ), (f2 , g2 ) ∈ E7 (I), (ii) J is the range of the function h2 := f20 /g20 on I, µ 00 ¶2 h000 (x) h (x) −6 = 0 or S(x) 6= 0 for x ∈ J, where (iii) either S(x) = 4 0 h (x) h0 (x) µ ¶−1 f 0 f20 h(x) = 10 (x), (x ∈ J) g1 g20 (iv) either G00 (x) = 0 (x ∈ J) or G00 (x) 6= 0 (x ∈ J), where G(x) = g2 (h−1 2 (x)),

(x ∈ J).

Then the functions f1 , g1 , f2 , g2 satisfy the functional equation Df1 ,g1 (u, v) = Df2 ,g2 (u, v)

(u, v ∈ I)

(2)

if and only if either there are constants α, β, γ, δ with αδ − βγ 6= 0 such that f10 (u) = αf20 (u) + βg20 (u),

(73)

g10 (u) = γf20 (u) + δg20 (u), (u ∈ I), where the functions f2 , g2 are arbitrary such that (i) is satisfied or f1 (u) = f (h2 (u)) ,

g1 (u) = g (h2 (u)) ,

f2 (u) = F (h2 (u)) ,

g2 (u) = G (h2 (u)) ,

(u ∈ I),

(74)

holds where the functions f, g, F, G are determined by (41)–(72) (and by the first two equations of (20)), h2 : I → J is an arbitrary, seven times continuously

78

AEM

L. Losonczi

differentiable function with non-vanishing derivative and A, B 6= 0, C 6= 0, D, K 6= 0, L, M 6= 0, N 6= 0 are arbitrary constants. For each solution the common domain J of the functions h, g 0 , G0 has to be chosen such that the condition (i) is satisfied. Remark 1. Conditions (iii) and (iv) are always satisfied on some suitable subintervals of J, since S is either identically zero or not on J. In the latter case there is at least one point x1 ∈ J such that S(x1 ) 6= 0. By the continuity of S also in a neighborhood J1 of x1 the function S does not vanish. Applying now the same argument for G00 on the interval J1 we conclude the existence of a subinterval J2 of J on which (iii) and (iv) hold. Proof. Necessity. If (2) is satisfied, then by Proposition 1 the functions f, g, F, G defined in (4) satisfy (3)≡(6)≡(7). If S(x) = 0, then by Theorem 2 the solutions f, g, F, G of (7) are given by (21). From (21) and (4) we easily obtain (73). If S(x) 6= 0, then by Theorem 4 the solutions f, g, F, G of (7) are given by (41)–(72) and from (4) we obtain the stated form of the solutions. Sufficiency. If S(x) = 0, then by Theorem 2 the functions given by (73) satisfy (2). Thus, on the basis of Proposition 1, it is enough to show that the functions (41)–(72) satisfy (7). If h, f, g, F, G (h = f 0 /g 0 ) is a solution of (7), then it is easy to see that also ˜ = f˜0 /˜ ˜ e = g ◦ h−1 (h g 0 ) is also h = h−1 , f˜ = F ◦ h−1 , g˜ = G ◦ h−1 , Fe = f ◦ h−1 , G 0 0 a solution of (7). Thus if we have one solution triple h, g , G of (7) among (41)– 0 −1 0 −1 ˜ = h−1 , g˜0 = G ◦ h , G e0 = g ◦ h (72), then its “symmetric pair” h is a also h0 ◦ h−1 h0 ◦ h−1 solution (necessarily among (41)–(72)). Thus our 32 solutions can be divided into pairs such that the second solution is the symmetric pair of the first one. One can see that these pairs are (41)&(56), (42)&(62), (43)&(68), (44)&(46), (45)&(53), (47)&(59), (48)&(65), (49)&(49), (50)&(71), (51)&(55), (52)&(61), (54)&(67), (57)&(57), (58)&(63), (60)&(69), (64)&(64), (66)&(70), (72)&(72). It is enough to show that the functions h, g 0 , G0 given by the first member of the above pairs are solutions of the functional equation (7). This gives us 18 solutions to check. We shall not bore the reader with these calculations. We intend to show only that (50) is a solution of (3)=(7) and describe what changes should be made in case of the other 17 solutions. With the notation l(x) = C ln |x + A| + D (50) has the form h(x) = K tan l(x) + L,

, g 0 (x) =

M cos l(x) , x+A

G0 (x) =

N |x + A|3/2

where we have to assume that J is such that for x ∈ J either x + A is positive or

Vol. 65 (2003)

Equality of two variable Cauchy mean values

79

´ ³ π π negative and l(x) ∈ Ik =: − + kπ, + kπ . Hence and from (20) we get 2 2 Z M M cos l(x)l0 (x) dx = sin l(x) + P, g(x) = C C Z M f (x) = h(x)g 0 (x) dx = (−K cos l(x) + L sin l(x)) + Q, C Z −2N sign(x + A) N G(x) = dx = + R, |x + A|3/2 |x + A|3/2 Z 2N A sign(x + A) F (x) = xG0 (x) dx = 2N |x + A|1/2 + + T, |x + A|3/2 where P, Q, R, T are arbitrary constants. In the next calculations we assume that x 6= y, x, y ∈ J. The right-hand side of (6) can be written as U :=

F (x) − F (y) 1/2 = sign(x + A) (|x + A||y + A|) − A. G(x) − G(y)

A simple calculation gives that µ ¶ cos l(x) − cos l(y) l(x) + l(y) f (x) − f (y) =K − + L = K tan +L V := g(x) − g(y) sin l(x) − sin l(y) 2 where we also used the identity tan

cos u − cos v u+v =− 2 sin u − sin v

´ ³ π π u, v ∈ Ik =: − + kπ, + kπ . 2 2

Hence the left-hand side of (6) is ¶ ¶ µ µ l(x) + l(y) −1 −1 −1 V − L −1 h (V ) = l =l = l−1 (l(U )) = U tan K 2 which proves that (6) holds. For the other solutions the calculations are similar, except that we have to use one or more of the identities below (where either u 6= v has to be assumed or the right-hand sides should be defined at u = v by taking the limit as u → v)). tanh

cosh u − cosh v u+v = (u, v ∈ R), 2 sinh u − sinh v

coth

u+v sinh u − sinh v = (uv > 0), 2 cosh u − cosh v

exp

exp u − exp v2 u+v ¡ ¢ ¡ u2¢ =− 2 exp − 2 − exp − v2

(u, v ∈ R),

80

L. Losonczi

AEM

µ ¶ (u2 + 1)−1/2 − (v 2 + 1)−1/2 tan−1 u + tan−1 v = tan−1 − 2 u(u2 + 1)−1/2 − v(v 2 + 1)−1/2 (u, v ∈ R), (1 − u2 )−1/2 − (1 − v 2 )−1/2 tanh−1 u + tanh−1 v = tanh−1 2 u(1 − u2 )−1/2 − v(1 − v 2 )−1/2 (u, v ∈ (−1, 1)), (u2 − 1)−1/2 − (v 2 − 1)−1/2 coth−1 u + coth−1 v = coth−1 2 u(u2 − 1)−1/2 − v(v 2 − 1)−1/2 (u, v > 1 or u, v < −1), µ ¶ |u|1/2 − |v|1/2 ln |u| + ln |v| = ln − −1/2 (uv > 0). 2 |u| − |v|−1/2

References ´l, Lectures on functional equations and their applications, Academic Press, New [1] J. Acze York, 1966. [2] W. H. Adams, Heat transmission, McGraw-Hill, New York, 1954. [3] H. Alzer, Bestm¨ ogliche Absch¨ atzungen f¨ ur spezielle Mittelwerte, Univ. u Novom Sadu Zb. Rad. Prirod.-Mat. Fak. Ser. Mat. 23 (1) (1993), 331–346. [4] G. Aumann and O. Haupt, Einf¨ uhrung in die reelle Analysis, Band II, W. de Gruyter, Berlin–New York, 1979. [5] J. Brenner, A unified treatment and extension of some means of classical analysis I. Comparison theorems, J. Combin. Inform System Sci. 3 (1978), 175–199. [6] J. Brenner and B. C. Carlson, Homogeneous mean values: Weights and asymptotics, J. Math. Anal. Appl. 13 (1987), 265–280. [7] F. Burk, By all means, Amer. Math. Monthly 92 (1985), 50. [8] B. C. Carlson, The logarithmic mean, Amer. Math. Monthly 79 (1972), 615–618. [9] E. L. Dodd, Some generalization of the logarithmic mean and of similar means of two variates which become indeterminate when the two variates are equal, Ann. Math. Statist. 12 (1971), 422–428. `s, Geometry of affine time-frequency distributions, Applied [10] P. Flandrin and P. Gonc ¸ alve and Computational Harmonic Anal. 3 (1996), 10–39. [11] E. Leach and M. Sholander, Extended mean values, Amer. Math. Monthly 85 (1978), 84–90. [12] E. Leach and M. Sholander, Extended mean values II, J. Math. Anal. Appl. 92 (1983), 207–223. [13] E. Leach and M. Sholander, Multi-variable extended mean values, J. Math. Anal. Appl. 104 (1984), 390–407. [14] T. P. Lin, The power mean and the logarithmic mean, Amer. Math. Monthly 81 (1974), 879–883. [15] L. Losonczi, A structure theorem for sum form functional equations, Aequationes Math. 53 (1997), 141–157. [16] L. Losonczi, Equality of two variable weighted means: reduction to differential equations, Aequationes Math. 58 (1999), 223–241. [17] L. Losonczi, Equality of Cauchy mean values, Publ. Math. Debrecen. 57 (2000), 217–230. ´ les, Minkowski’s inequality for two variable difference means, [18] L. Losonczi and Zs. Pa Proc. Amer. Math. Soc. 126 (1998), 779–789. [19] Z. Nehari, Conformal mapping, Dover Publications, New York, 1975.

Vol. 65 (2003)

Equality of two variable Cauchy mean values

81

[20] E. Neuman, The weighted logarithmic mean, J. Math. Anal. Appl. 188 (1994), 885–900. ´ les, Inequalities for differences of powers, J. Math. Anal. Appl. 131 (1988), 271–281. [21] Zs. Pa ´ les, Comparison of two variable homogeneous means, General Inequalities 6. Proc. [22] Zs. Pa 6th Internat. Conf. Math. Res. Inst. Oberwolfach, Birkh¨ auser Verlag Basel, 1992, pp. 59–69. ´ les, On reduction of linear two variable functional equations to differential equations [23] Zs. Pa without substitutions, Aequationes Math. 43 (1992), 236–247. ´ les, A unified form of the classical mean value theorems, in: R. P. Agarwal (ed.), [24] Zs. Pa Inequalities and Applications, World Scientific Publ., Singapore–New Jersey–London–Hong Kong, 1994, 493–500. ´ les, Notes on mean value theorems (manuscript). [25] Zs. Pa [26] A. O. Pittenger, The symmetric, logarithmic, and power means, Univ. Beograd. Publ. Elektrotehn. Fak., Ser. Mat. Fiz. 681 (1980), 19–23. [27] A. O. Pittenger, The logarithmic mean in n variables, Amer. Math. Monthly 92 (1985), 99–104. ´ lya and G. Szego ˝ , Isoperimetric inequalities in mathematical physics, Princeton [28] G. Po Univ. Press, Princeton, N. J., 1951. ¨ tz and D. Russell, An extremal problem related to probability, Aequationes Math. [29] J. Ra 34 (1987), 316–324. ´ ndor, On certain inequalities for means, J. Math. Anal. Appl. 189 (1995), 602–606. [30] J. Sa [31] L. Schumaker, Spline functions, Wiley, New York–Toronto, 1981. [32] H. Seiffert, Ungleichungen f¨ ur einen bestimmten Mittelwert, Nieuw Arch. Wisk. 13 (1995), 195–198. [33] H. Seiffert, Ungleichungen f¨ ur elementare Mittelwerte, Arch. Math. (Basel) 63 (1995), 129–131. [34] J. F. Steffenson, Interpolation, 2nd ed., Chelsea, New York, 1950. [35] K. B. Stolarsky, Generalizations of the logarithmic mean, Math. Mag. 48 (1975), 87–92. [36] K. B. Stolarsky, The power and generalized logarithmic means, Amer. Math. Monthly 87 (1980), 545–548. ´kely, A classification of means, Ann. Univ. Sci. Budapest. E¨ [37] G. Sze otv¨ os Sect. Math. 18 (1975), 129–133. ¨ ´ rka ´ ny, D. Kra ´ lik and R. Somfai, Uber [38] K. Tettamanti, G. Sa die Ann¨ aherung logarithmischer Funktionen durch algebraische Funktionen, Period. Polytech. Chem. Engrg. 14 (1970), 99–111. L. Losonczi Debrecen University Institute of Mathematics and Informatics H–4010 Debrecen Pf. 12 Hungary e-mail: [email protected] Manuscript received: December 12, 2000 and, in final form, December 4, 2001.

To access this journal online: http://www.birkhauser.ch