ESAIM: PROCEEDINGS, April 2012, Vol. 36, p. 1-14 Dani` ele Fournier-Prunaret, Laura Gardini, & Ludwig Reich, Editors
SOME ASPECTS OF THE LOCAL THEORY OF GENERALIZED DHOMBRES FUNCTIONAL EQUATIONS IN THE COMPLEX DOMAIN
¨ rg Tomaschek 1 Jo Abstract. We study the generalized Dhombres functional equation f (zf (z)) = ϕ(f (z)) in the complex domain. The function ϕ is given and we are looking for solutions f with f (0) = w0 and w0 is a primitive root of unity of order l ≥ 2. All formal solutions for this case are described in this work, for the situation where ϕ can be transformed into a function which is linearizable and local analytic in a neighbourhood of zero we also show that we obtain local analytic solutions. We also discuss an example where it is possible to use other methods than we use in the general case. AMS (2010) subject classification. 39B12, 39B32, 30D05. Keywords. Generalized Dhombres equation, Iterative functional equation.
R´ esum´ e. Nous ´etudions la fonctionnelle de Dhombres f (zf (z)) = ϕ(f (z)) dans le plan complexe. La fonction ϕ est donn´ee et nous cherchons les solutions f avec f (0) = w0 o` u w0 est une racine primitive de l’unit´e d’ordre l ≥ 2. Nous d´ecrivons dans ce travail toutes les solutions formelles dans ce cas, et lorsque ϕ peut ˆetre transform´ee en une fonction lin´earisable et localement analytique au voisinage de z´ero nous montrons ´egalement comment obtenir des solutions analytiques locales. Nous discutons enfin un exemple o` u il est possible d’utiliser des m´ethodes diff´erentes de celles que nous mettons en œuvre dans le cas g´en´eral. Mots clefs. Equation de Dhombres g´en´eralis´ee, ´equation fonctionnelle it´erative.
Introduction The generalized Dhombres functional equation in the complex domain is given by f (zf (z)) = ϕ(f (z))
(1)
where the function ϕ is a given power series. We are looking for local analytic or formal solutions f of (1) with f (0) = w0 where w0 is a primitve root of unity of order l ≥ 2. The generalized Dhombres functional equation was first studied in the complex domain in the year 2005 in [3]. In this paper solutions f with f (0) = 0 were considered. In 2009 this equation was studied again. This time the interest was focussed on solutions f with f (0) = w0 where w0 is a complex number which is different from zero and no root of unity. The original Dhombres equation is given by f (xf (x)) = f (x)2 1
Institute for Mathematics and Scientific Computing, Karl-Franzens-Universit¨ at Graz, Heinrichstraße 36, A-8010 Graz, Austria; email:
[email protected]. c EDP Sciences, SMAI 2012
Article published online by EDP Sciences and available at http://www.esaim-proc.org or http://dx.doi.org/10.1051/proc/201236001
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where x is a real number. This original equation arised from a population model which was studied by Jean Dhombres (see [1], chapter 6). The following transformations are part of [4] and these transformations can be done for an arbitrary complex number w0 which is different from zero. In a first step (1) becomes by putting f (z) = w0 + g(z), g(0) = 0 equivalent to the transformed generalized Dhombres functional equation g(w0 z + zg(z)) = ϕ(g(z)) ˜
(2)
where g is a series given by g(z) = ck z k + . . . with ck 6= 0, k ∈ N, and the function ϕ˜ is calculated as ϕ(ω) ˜ = w0k ω + d2 ω 2 + . . .. By taking an arbitrary but fixed power series solution T of T (z)k = g(z) we get T (w0 T −1 (z) + T −1 (z)z k ) = ψ(z) for the equation above with ψ(z)k = ϕ(z ˜ k ). Finally defining U = T −1 leads to the linear functional equation (w0 + z k )U (z) = U (ψ(z)).
(3)
The function ϕ in equation (1) is known and so it is possible to compute ψ, we get ψ(z) = w0 z 1 + From now on we suppose that w0 is a primitive root of unity of order l ≥ 2.
d2 k z w0k k
+ ... .
1. Generalized Dhombres functional equation with a root of unity w0 In this section we want to give a full description of all formal solutions of equation (1) with f (0) = w0 and w0 is a primitive root of unity of order l ≥ 2. The case where w0 = 1 has to be treated seperately. We start with a first Lemma but before we note that equation (3) is called solvable, if it has a solution U which is different from zero. Lemma 1.1. If (3) is solvable for given ψ and k then there exists a solution U0 with U0 (z) ∈ zC[[z k ]] ∩ Γ1 . Proof. We write U (z) = u1 zU ? (z) with U ? (z) = 1 + . . ., then (3) is equivalent to w0 z(1 + w0−1 z k ) ? U (z) = U ? (ψ(z)). ψ(z) The series above on the left hand side starts with 1 and hence one can introduce the formal logarithm. Then using the functional equation for the formal logarithm we get Ln
w0 z + Ln(1 + w0−1 z k ) = X ? (ψ(z)) − X ? (z) ψ(z)
(4)
with X ? = LnU ? and ord X ? > 0. For the function ψ it follows from the assumption ψ(z)k = ϕ(z ˜ k ) that −1 k w0 z k k ψ(z) ∈ zC[[z ]] ∩ Γ and hence the left side Ln ψ(z) + Ln(1 + w0 z ) of (4) belongs to C[[z ]]. It is possible to write X X X ? (y) = ξν? y ν + ξν? y ν ν≥1
ν≥1
ν6≡0(mod k)
ν≡0(mod k)
for X ? . Then we get X ? (ψ(z)) − X ? (z) =
X
ξν? (ψ(z)ν − z ν ) +
X
ν≥1
ν≥1
ν6≡0(mod k)
ν≡0(mod k)
ξν? (ψ(z)ν − z ν )
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P where the first part ξν? (ψ(z)ν − z ν ) does not contain any term z ν with ν ≡ 0 (mod k) while the ν≥1 ν6≡0(mod k) P P second part ξν? (ψ(z)ν − z ν ) is an element of C[[z k ]]. Now define Y ? as Y ? (z) := ξν? z ν ν≥1 ν≥1 ν≡0(mod k)
ν≡0(mod k)
then (4) is equivalent to
w0 z + Ln(1 + w0−1 z k ) = Y ? (ψ(z)) − Y ? (z) ψ(z) and hence (4) has a solution Y ? ∈ C[[z k ]] and after reversing our calculations we obtain a function U0 with the desired property. Ln
So we know the structure of a solution U with U (z) = u1 U0 (z) and U0 ∈ zC[[z k ]] ∩ Γ1 of (3). Next we have to show that there exists a solution. Therefore we have to do some transformations. Like in the proof above we have the equation (1 + w0−1 z k ) ? w0 z U (z) = U ? (ψ(z)). ψ(z) ˆ by U ? (z)k = U ˆ (z), then U ˆ is given by U ˆ (z) = 1 + . . . and conversely U ? is uniquely determined by U ˆ. Define U ? k ˆ (z) in the equation above leads to Taking the k − th power and using U (z) = U w0k z k (1 + w0−1 z k )k ˆ ˆ (ψ(z)) U (z) = U ψ(z)k and again with the formal logarithm this becomes equivalent to Ln
w0k z k (1 + w0−1 z k )k ˆ ˆ = X(ψ(z)) − X(z) ψ(z)k
ˆ := LnU ˆ . Now as a consequence of ψ(z) ∈ zC[[z k ]] and some other computations it is possible to write with X k ˆ ˜ X(z) = X(z ) and therefore with y = z k and ω0 := w0k we get Ln
ω0 y(1 + w0−1 y)k ˜ ϕ(y)) ˜ = X( ˜ − X(y) ϕ(y) ˜
and this leads to
ω0 y ˜ ϕ(y)) ˜ + kLn(1 + w0−1 y) = X( ˜ − X(y) (5) ϕ(y) ˜ l where ω0 is again a root of unity of order l1 and l1 = gcd(k,l) . Now it is necessary to distinguish two cases. The first case deals with a function ϕ˜ which is not linearizable and hence the function is conjugated to a non linear semicanonical form. So ϕ˜ can be written as Ln
ϕ(y) ˜ = (R−1 ◦ N ◦ R)(y) with N (y) = ω0 y + δnl1 +1 y nl1 +1 + . . . and δnl1 +1 6= 0. The function R−1 and hence also the function R is uniquely determined by R−1 (y) = y + ρ2 y 2 + . . . and the condition that the coefficients ρµl1 +1 = 0 for µ ≥ 1. The number l1 denotes the order of ω0 . Then (5) is equivalent to Ln
ω0 y ˜ −1 (N (R(y)))) − X(y) ˜ + kLn(1 + w0−1 y) = X(R ϕ(y) ˜
˜ ◦ R−1 leads to and substituting R−1 (y) for y and defining X as X := X Ln
ω0 R−1 (y) + kLn(1 + w0−1 R−1 (y)) = X(N (y)) − X(y). R−1 (N (y))
(6)
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Discussing the right hand side X(N (y)) − X(y) of equation (6) in detail leads to the fact that the first coefficient of X(N (y)) − X(y) which is different from zero is the coefficient which belongs to the term y (n+1)l1 . So we have Lemma 1.2. Equation (6) has a solution if and only if the coefficients of y l1 , . . . , y nl1 of the left hand side of equation (6) are zero. Proof. We write ω0 R−1 (y) = R−1 (N (y))
ω0 R−1 (y) ω0 y R−1 (N (y)) N (y)
ω0 y N (y)
and therefore Ln
ω0 R−1 (y) R−1 (y) R−1 (N (y)) N (y) = Ln − Ln − Ln R−1 (N (y)) y N (y) ω0 y −1 −1 R (y) R (u) N (y) = Ln − Ln |u=N (y) − Ln . y u ω0 y
Then we obtain N (y) 1 = (ω0 y + δnl1 +1 y nl1 +1 + . . .) = 1 + ω0−1 δnl1 +1 y nl1 + . . . ∈ C[[y l1 ]] ω0 y ω0 y where ω0−1 δnl1 +1 6= 0 and hence Ln
N (y) = ω0−1 δnl1 +1 y nl1 + . . . . ω0 y −1
−1
Therefore it is only necessary to investigate the coefficients of y l1 , . . . , y (n−1)l1 of Ln R y (y) − Ln R u(u) |u=N (y) . Hence we get R−1 (y) = ρ˜2 y + . . . + ρ˜l1 y l1 −1 + ρ˜l1 +1 y l1 + . . . Ln y with polynomials ρ˜µ which depend on (ρ2 , . . . , ρµ ) for µ 6≡ 1 (mod l1 ) and on (ρ2 , . . . , ρµ−1 ) for µ ≡ 1 (mod l1 ). We write R−1 (y) Ln =: Θ(y) = θ1 y + θ2 y 2 + . . . y and so we achieve Ln
R−1 (u) |u=N (y) = Θ(N (y)) = θ1 (ω0 y + δnl1 +1 y nl1 +1 + . . .) + θ2 (ω0 y + δnl1 +1 y nl1 +1 + . . .)2 + . . . u = θ1 ω0 y + θ2 ω02 y 2 + . . . + θl1 −1 ω0l1 −1 y l1 −1 + θl1 y l1 + . . . + θ2l1 y 2l1 + . . . + θnl1 y nl1 + . . . ≡ θ1 ω0 y + θ2 ω02 y 2 + . . . ≡ Θ(ω0 y)
(mod ord nl1 + 1)
(mod ord nl1 + 1)
and hence Θ(y) − Θ(N (y)) ≡ Θ(y) − Θ(ω0 y) (mod ord nl1 + 1) and therefore there are no coefficients different −1 −1 from zero for y l1 , . . . , y nl1 in Ln R y (y) − Ln R u(u) |u=N (y) . It remains to determine the coefficients of kLn(1 + w0−1 R−1 (y)). Computing kLn(1 + w0−1 R−1 (y)) it is clear that comparing the coefficients of y l1 leads to ρl1 = Pl1 (ρα |α < l1 ). with a polynomial Pl1 . Proceeding in comparing the coefficients of y µl1 with µ ≥ 2 leads to ρµl1 = Pµl1 (ρα |α < µl1 )
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with polynomials Pµl1 for 2 ≤ µ ≤ (n − 1)l1 . For y nl1 we obtain −ω0−1 δnl1 +1 + kw0−1 ρnl1 + Pnl1 (ρα |α < nl1 ) = 0 with a polynomial Pnl1 and hence δnl1 +1 = kω0 w0−1 ρnl1 + ω0 Pnl1 (ρα |α < nl1 ) Now one obtains δnl1 +1 6= 0 if ρnl1 is suitably choosen.
It is also possible to determine the number k if N and R and hence ϕ˜ are given. We get k=
δnl1 +1 − ω0 Pnl1 (ρα |α < nl1 ) . ω0 w0−1 ρnl1
(7)
In summary we achieve (n − 1)l1 algebraic equations ρµl1 = Pµl1 (ρα |α < µl1 )
(8)
for 1 ≤ µ ≤ (n − 1)l1 and one additional equation −ω0−1 δnl1 +1 + kw0−1 ρnl1 + Pnl1 (ρα |α < nl1 ) = 0.
(9)
Theorem 1.3. Let k ∈ N be given by (7) and let ϕ˜ be not linearizable and be given by ϕ(y) ˜ = R−1 (N (R(y))) where N denotes the semicanonical form of ϕ. ˜ Then the equation Ln
ω0 R−1 (y) + kLn(1 + w0−1 R−1 (y)) = X(N (y)) − X(y) R−1 (N (y))
(6)
has a solution X if and only if the polynomial relations (8) and (9) are fulfilled. This solution X is unique. −1
(y) 0R Proof. We obtain the uniquely determined solution if we compare the coefficients of (6) namley Ln Rω−1 (N (y)) + −1
(y) −1 −1 0R kLn(1 + w0−1 R−1 (y)) = X(N (y)) − X(y). It is sufficient to look at Ln Rω−1 (y)) = (N (y)) + kLn(1 + w0 R µ 2 2 −1 θˆ1 y + θˆ2 y + . . .. Let the series X be given by X(y) = β˜1 y + β˜2 y + . . ., then we obtain β˜µ = (ω0 − 1) θˆµ for 1 ≤ µ ≤ l1 −1. Comparing the coefficients of y l1 leads to the first polynomial relation of (8) but it is not possible to determine β˜l1 from this equation. Proceeding we can compute the coefficients β˜l1 +1 , . . . , β˜2l1 −1 and for y 2l1 we obtain the second polynomial relation of (8). These steps can be done till we compute the coefficient of y (n+1)l1 because there we get β˜l1 = (l1 ω0l1 −1 δnl1 +1 )−1 θˆ(n+1)l1 +1 . Inductively we obtain the uniquely determined solution X of (6).
Remark 1.4. It is possible to go back the way from X to the function U and hence we get also the so˜ ˆ ˜ k ) = X(R(y k ). This lution U of (3). In more detail we have X(y) = X(R(y)) and therefore X(y) = X(y k1 k k ˆ (y) = exp(X(R(y ))). Next we obtain U (y) = u1 z exp(X(R(y ))) . Then we get T (y) = leads to U 1 [−1] u1 z exp(X(R(y k ))) k and finally g is given by ( g(y) =
h i k1 [−1] 2 u1 z exp(X(R(z k )))
) k1 .
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The solutions f of the generalized Dhombres functional equation are given by ( f (z) = w0 +
i k1 [−1] k u1 z exp(X(R(z ))) h
) k1
2
.
(10)
This remark also holds for the following theorem, namely Theorem 1.6. That means also in 1.6 we can reverse the calculations like we do in 1.4 and finally we get solutions of the form (10). The difference is that in the case of Theorem 1.3 the solution X is uniquely determined whereas in the situation of Theorem 1.6 the solution X is depending on some arbitrary coefficients. But before we prove the next theorem we have to consider the following remark. Remark 1.5. Let w0 be a primitive root of unity of order l ≥ 1, k ∈ N and ω0 be defined by ω0 = w0k where the order of ω0 equals l1 . Also the series N is given by N (y) = ω0 y + δnl1 +1 y nl1 +1 + . . . ∈ yC[[y l1 ]] with n ≥ 1 and the coefficient δnl1 +1 is different from zero. Then let the series H be given by H(y) = h1 y + h2 y 2 + . . . ∈ C[[y]] such that the coefficients hµl1 of H with 1 ≤ µ ≤ (n − 1)l1 are given by the polynomial relations hµl1 = Pµl1 (hα |α < µl1 ) where the polynomials Pµl1 are the same as in (8) and such that the coefficients hµl1 +1 are zero for µ ≥ 1 and (H −1 ◦ N ◦ H)(y) = ω0 y + . . . ∈ C[[y]]. The coeffiecient hnl1 has to be choosen such that k=
δnl1 +1 − ω0 Pnl1 (hα |α < nl1 ) ω0 w0−1 hnl1
holds. Then for every choice of the coefficients of the series N except the coefficient δnl1 +1 and for every choice of the coefficients of the series H except these coefficients we get from the polynomial relations and also except the coefficient hnl1 there exists a unique solution f of the generalized Dhombres functional equation f (zf (z)) = w0 + H −1 (N (H(f (z)))). Now we have to consider the second situation, this case deals with a linearizable ϕ. ˜ Then there exists a ˜ such that ϕ(y) ˜ −1 (ω0 R(y)). ˜ function R ˜ =R But it is also possible to use the notations from the case where a non linear semicanonical form N exists. We may consider the present situation as a limiting case of the previous situation for n → ∞. In formula (8) ρµl1 = Pµl1 (ρα |α < µl1 ) exactly (n−1)l1 algebraic equations as a consequence of the non linear semicanonical form are given. If now the semicanonical form N (y) = ω0 y + δnl1 +1 y nl1 +1 + . . . with δnl1 +1 6= 0 becomes linear that means n → ∞ then we obtain infinitely many algebraic equations instead of finitely many ones. Therefore we get the algebraic relations ρµl1 = Pµl1 (ρα |α < µl1 )
(11)
for µ ≥ 1. Theorem 1.6. Let k ∈ N and ϕ˜ be linearizable and given by ϕ(y) ˜ = R−1 (ω0 R(y)). Then the equation Ln
ω0 R−1 (y) + kLn(1 + w0−1 R−1 (y)) = X(ω0 y) − X(y) R−1 (ω0 y)
has infinitely many solutions X if the polynomial relations (11) are fulfilled.
(6)
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Proof. We consider equation (6) namely Ln
ω0 R−1 (y) + kLn(1 + w0−1 R−1 (y)) = X(ω0 y) − X(y) R−1 (ω0 ) −1
R (y) −1 −1 (y)) = ρ˜1 y + ρ˜2 y 2 + . . .. and again it is possible to write Ln ωR0−1 (ω0 ) + kLn(1 + w0 R 2 Let X(y) = β1 y + β2 y + . . . , then
X
X(ω0 y) − X(y) =
βν (ω0ν − 1)y ν
ν≥1
ν6≡0(mod l1 )
and therefore we obtain a special solution X0 which is given by X ρ˜ν (ω0ν − 1)−1 y ν . X0 (y) = ν≥1
ν6≡0(mod l1 )
Hence the general solution is given by X(y) =
X ν≥1
ρ˜ν (ω0ν − 1)−1 y ν +
X
βνl1 y νl1
ν≥1
ν6≡0(mod l1 )
with arbitrary βνl1 for ν ≥ 1.
Remark 1.7. Another representation of the solution U of (w0 + z k )U (z) = U (ψ(z)) where ψ is linearizable and w0 is a root of unity of order l ≥ 2 can be found in [2] Theorem 5 p. 90.
2. A class of examples In this section we want to consider an example of a solution of the generalized Dhombres functional equation where the function ϕ˜ is linearizable. This example is very interesting because we compute it in contrast to the previous section without the use of the formal logarithm. Therefore again one considers (3) (w0 + z k )U (z) = U (ψ(z)) where the function ψ is linearizable. Notice that the function ϕ˜ is linearizable if and only if the function ψ is. Let w0 be a root of unity with order l ≥ 2. Then we know that ω0 := w0k is a root of unity with order l . l1 = gcd(k,l) We iterate equation (3) and so we obtain (w0 + ψ(z)k )U (ψ(z)) = U (ψ 2 (z)) and hence (w0 + ϕ(z ˜ k ))(w0 + z k )U (z) = U (ψ 2 (z)). Iterating the last equation l1 − 2 times and using ψ(z)k = ϕ(z ˜ k ) as well as y = z k and w0l = 1 leads to the equation (1 + w0−1 ϕ˜l−1 (y))(1 + w0−1 ϕ˜l−2 (y)) . . . (1 + w0−1 ϕ(y))(1 ˜ + w0−1 y) = 1. (12) l1 Then there exists a p ∈ N such that l = l1 p and so as a consequence of ϕ˜ (y) = y we obtain (1 + w0−1 ϕ˜pl1 −1 (y))(1 + w0−1 ϕ˜pl1 −2 (y)) . . . (1 + w0−1 ϕ(y))(1 ˜ + w0−1 y) = 1 which becomes (1 + w0−1 ϕ˜l1 −1 (y))(1 + w0−1 ϕ˜l1 −2 (y)) . . . (1 + w0−1 ϕ(y))(1 ˜ + w0−1 y) = 1.
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Next we want to construct a function ϕ˜ such that (12) is satisfied. As an Ansatz we use the special M¨obiustransformation ω0 y ϕ(y) ˜ = 1 + βy with β 6= 0 and ϕ˜l1 (y) = y. For 1 ≤ ν ≤ l1 − 1 we obtain ϕ˜ν (y) =
ω0ν y . 1 + β(1 + ω0 + . . . + ω0ν−1 )y
For ν = l1 we obtain ϕ˜l1 (y) = y, hence ϕ˜ is linearizable as well as ψ. Then we get 1 + w0−1 ϕ˜ν (y) = 1 + w0−1
1 + [w0−1 ω0ν + β(1 + ω0 + . . . + ω0ν−1 )]y ω0 y = . 1 + β(1 + ω0 + . . . + ω0ν−1 )y 1 + β(1 + ω0 + . . . + ω0ν−1 )y
Substituting this in equation (12) leads to 1 + [w0−1 ω0l1 −1 + β(1 + ω0 + . . . + ω0l1 −2 )]y 1 + w0−1 y 1 + [w0−1 ω0 + β]y · · ... · = 1. 1 1 + βy 1 + β(1 + ω0 + . . . ω0l1 −2 )y
(13)
If in each linear factor in the numerator of the left hand side of (13) the coefficient of y is not zero, then the degree of the product is l1 while the degree of the denominator is l1 − 1 which contradicts (13). So it is necessary that β is one of the numbers β1 = −w0−1 ω0 β2 = −w0−1 ω02 (1 + ω0 )−1 .. . βl1 −1 = −w0−1 ω0l1 −1 (1 + ω0 + . . . + ω0l1 −2 )−1 . Hence for 1 ≤ µ ≤ l1 − 1 we have βµ = −w0−1 (1 + ω0 + . . . + ω0µ−1 )−1 ω0µ .
(14)
Next we have to show that this is sufficient. Therefore we number the fractions of (13) such that the numerators are Z0 to Zl1 −2 and the denominators are N1 to Nl1 −2 , for N0 we have N0 = 1. Then we choose a special βµ for 1 ≤ µ ≤ l1 − 1. Then it is true that Zν = Nν−µ for ν ≥ µ and that Zν = Nν+l1 −µ for ν ≤ µ hold. First we show that for ν ≥ µ we get Zν = Nν−µ . We obtain Zν = 1 + [w0−1 ω0ν + (−w0−1 (1 + ω0 + . . . + ω0µ−1 )−1 ω0µ )(1 + ω0 + . . . + ω0ν−1 )]y = 1 + [w0−1 ω0ν − w0−1 ω0µ (1 + ω0 + . . . + ω0µ−1 )−1 (1 + ω0 + . . . + ω0ν−1 )]y. For Nν−µ we get Nν−µ = 1 + (−w0−1 (1 + ω0 + . . . + ω0µ−1 )−1 ω0µ )(1 + ω0 + . . . + ω0ν−µ−1 ) So it remains to show that ω0ν −
ω0µ (1 + ω0 + . . . + ω0ν−1 ) ω0µ (1 + ω0 + . . . + ω0ν−µ−1 ) = − . 1 + ω0 + . . . + ω0µ−1 1 + ω0 + . . . + ω0µ−1
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Expanding this equation leads to ω0ν + ω0ν+1 + . . . + ω0ν+µ−1 − ω0µ − ω0µ+1 − . . . − ω0ν−1 − ω0ν − . . . − ω0µ+ν−1 for the left hand side and to −ω0µ − ω0µ+1 − . . . − ω0ν−1 for the right hand side. So everything cancels and hence we get Zν = Nν−µ for ν ≥ µ. Next we show that Zν = Nν+l1 −µ for ν ≤ µ. Like above we obtain Zν = 1 + [w0−1 ω0ν − w0−1 ω0µ (1 + ω0 + . . . + ω0µ−1 )−1 (1 + ω0 + . . . + ω0ν−1 )]y for Zν . The denominator Nν+l1 −µ is given by Nν+l1 −µ = 1 + (−w0−1 ω0µ (1 + ω0 + . . . + ω0µ−1 )−1 (1 + ω0 + . . . + ω0ν+l1 −µ−1 ))y. After cancelling 1 and w0−1 one has to show that ω0ν −
ω0µ (1 + ω0 + . . . + ω0ν+l1 −µ−1 ) ω0µ (1 + ω0 + . . . + ω0ν−1 ) = − 1 + ω0 + . . . + ω0µ−1 1 + ω0 + . . . ω0µ−1
is fulfilled. Expanding this equation leads to ω0ν + ω0ν+1 + . . . + ω0µ + . . . + ω0ν+µ−1 − ω0µ − ω0µ+1 − . . . − ω0ν+µ−1 for the left hand side and hence only the expression ω0ν + ω0ν+1 + . . . + ω0µ−1 remains. The right hand side is given by −ω0µ − ω0µ+1 − . . . − ω0ν+l1 −1 . Therefore we have to consider the equation ω0ν (1 + ω0 + . . . + ω0µ−ν−1 ) = −ω0ν (ω0µ−ν + ω0µ−ν+1 + . . . + ω0l1 −1 ) which is equivalent to 1 + ω0 + . . . + ω0µ−ν−1 = −ω0µ−ν − ω0µ−ν+1 − . . . − ω0l1 −1 .
(15)
1 + ω0 + . . . + ω0l1 −1
For a root of unity ω0 of order l1 the equation = 0 holds and hence (15) is fulfilled. Therefore our claim is valid which means that for every βµ with 1 ≤ µ ≤ l1 − 1 the corresponding solution is sufficient to solve (13). In the next step we are looking for a special solution U of (3) (w0 + z k )U (z) = U (ψ(z)). Remember that our functions are given by w0 z ψ(z) = 1 . (1 + βµ z k ) k for 1 ≤ µ ≤ l1 − 1. To show how one can compute a special solution we consider the special case β2 = −w0−1 (1 + w0 z ω0 )−1 ω02 and we try to find a special solution for β2 . Therefore the function ψ is given by ψ(z) = 1 . k (1+β2 z ) k
First we write U (z) = zU
(1)
(z) with U
(1)
(z) = 1 + . . .. Then (3) is equivalent to
w0 (1 + w0−1 z k )zU (1) (z) = ψ(z)U (1) (ψ(z))
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and this is equivalent to 1
w0 z(1 + w0−1 z k )U (1) (z) = w0 z(1 + β2 z k )− k U (1) (ψ(z)). This leads to 1
(1 + w0−1 z k )(1 + β2 z k ) k U (1) (z) = U (1) (ψ(z)). Then we use a new Ansatz, so we consider U in (16) we get
(1)
(z) =
(1+w0−1 z k )α1 U (2) (z)
(16)
and after substituting this expression
1
(1 + w0−1 z k )α1 +1 (1 + β2 z k ) k U (2) (z) = (1 + w0−1 ψ(z)k )α1 U (2) (ψ(z)) or the equivalent expression (1 +
w0−1 z k )α1 +1 (1
k
1 k
+ β2 z ) U
(2)
(z) = 1 + w0−1
ω0 z k 1 + β2 z k
α1
U (2) (ψ(z)).
This leads to 1
(1 + w0−1 z k )α1 +1 (1 + β2 z k ) k +α1 U (2) (z) = (1 + (β2 + w0−1 ω0 )z k )α1 U (2) (ψ(z)).
(17)
Then again a new Ansatz is needed and by U (2) (z) = (1 + (β2 + w0−1 ω0 )z k )α1 U (3) (z) we describe this new one. Substituting this in (17) leads to 1
(1 + w0−1 z k )α1 +1 (1 + β2 z k ) k +α1 (1 + (β2 + w0−1 ω0 )z k )α1 U (3) (z) = (1 + (β2 + w0−1 ω0 )z k )α1 (1 + (β2 + w0−1 ω0 )ψ(z)k )α1 U (3) (ψ(z)). This is equivalent to 1
(1 + w0−1 z k )α1 +1 (1 + β2 z k ) k +α1 U (3) (z) =
α1 ω0 z k U (3) (ψ(z)). 1 + (β2 + w0−1 ω0 ) 1 + β2 z k
Hence we get α1 1 + β2 z k + β2 ω0 z k + w0−1 ω02 z k U (3) (ψ(z)) 1 + β2 z k α1 1 + [β2 (1 + ω0 ) + w0−1 ω02 ]z k = U (3) (ψ(z)) 1 + β2 z k α1 1 = U (3) (ψ(z)) 1 + β2 z k
1
(1 + w0−1 z k )α1 +1 (1 + β2 z k ) k +α1 U (3) (z) =
because the condition which the expression β2 has to fulfill is β2 (1 + ω0 ) + w0−1 ω02 = 0. Hence the following equation remains 1
(1 + w0−1 z k )α1 +1 (1 + β2 z k ) k +2α1 U (3) (z) = U (3) (ψ(z)) and so we need a last Ansatz. We choose U for the right hand side of (18) U (3) (ψ(z)) =
(3)
1+δ
k α2
(z) = (1 + δz )
ω0 z k 1 + β2 z k
α2
=
(18)
and substituting this expression in (18) we get
1 + (β2 + δω0 )z k 1 + β2 z k
α2 .
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Therefore (18) is equivalent to 1
(1 + w0−1 z k )α1 +1 (1 + β2 z k ) k +2α1 +α2 (1 + δz k )α2 = (1 + (β2 + δω0 )z k )α2 . Hence we obtain δ = β2 + δω0 which is equaivalent to δ = (1 − ω0 )−1 β2 . Also (1 + w0−1 z k )α1 +1 (1 + 1 β2 z k ) k +2α1 +α2 = 1 has to be satisfied and this leads to α1 = −1 and α2 = 2 − k1 . Then the special solution for β2 is given by 1
U (z) = z(1 + w0−1 z k )−1 (1 + (β2 + w0−1 ω0 )z k )−1 (1 + (1 − ω0 )−1 β2 z k )2− k which we obtain after putting our results together. These calculations motivate to state that a special solution U of equation (3) for a βµ with 1 ≤ µ ≤ l1 − 1 has the representation U (z) = z
µ−1 Y
1
(1 + [βµ (1 + ω0 + . . . + ω0ν−1 ) + w0−1 ω0ν ]z k )−1 (1 + βµ (1 − ω0 )−1 z k )µ− k .
ν=0
To proof this we have to substitute U (z) in (3). Then the left hand side of (3) is (w0 + z k )U (z) = w0 (1 + w0−1 z k )z
µ−1 Y
(1 + [βµ (1 + ω0 + . . . + ω0ν−1 ) + w0−1 ω0ν ]z k )−1
ν=0 1
(1 + βµ (1 − ω0 )−1 z k )µ− k = w0 z(1 + w0−1 z k )1−1
µ−1 Y
(1 + [βµ (1 + ω0 + . . . + ω0ν−1 ) + w0−1 ω0ν ]z k )−1
ν=1
(1 + βµ (1 − ω0 )
1 −1 k µ− k
z )
.
The right hand side of (3) is given by U (ψ(z)) = ψ(z)
µ−1 Y
1 + [βµ (1 + ω0 + . . . + ω0ν−1 ) + w0−1 ω0ν ]ψ(z)k
−1
ν=0 1
(1 + βµ (1 − ω0 )−1 ψ(z)k )µ− k −1 µ−1 Y 1 ω0 z k ν−1 −1 ν k −k = w0 z(1 + βµ z ) 1 + [βµ (1 + ω0 + . . . + ω0 ) + w0 ω0 ] 1 + βµ z k ν=0 µ− k1 k −1 ω0 z 1 + βµ (1 − ω0 ) 1 + βµ z k 1 1 k −k +µ−µ+ k
= w0 z(1 + βµ z )
µ−1 Y
(1 + βµ z k + [βµ (1 + ω0 + . . . + ω0ν−1 ) + w0−1 ω0ν ]
ν=0 1
ω0 z k )−1 (1 + βµ z k + βµ (1 − ω0 )−1 ω0 z k )µ− k . The last term of the right hand side leads to 1
1
(1 + βµ z k + βµ (1 − ω0 )−1 ω0 z k )µ− k = (1 + βµ ((1 + (1 − ω0 )−1 )ω0 )z k )µ− k 1
= (1 + βµ (1 − ω0 )−1 z k )µ− k
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ESAIM: PROCEEDINGS
and hence this terms and the term w0 z cancel in the equation. For 1 ≤ n ≤ µ − 1 the n − th factor of the product of the left hand side of the equation is given by (1 + [βµ (1 + ω0 + . . . + ω0n−1 ) + w0−1 ω0n ]z k )−1 and the (n − 1) − th factor of the right hand side by (1 + βµ z k + [βµ (1 + ω0 + . . . + ω0n−2 ) + w0−1 ω0n−1 ]ω0 z k )−1 . Now it is easy to see that these expressions are the same. It remains to consider the (µ − 1) − th factor of the right hand side of the equation. We obtain (1 + βµ z k + [βµ (1 + ω0 + . . . + ω0µ−2 ) + w0−1 ω0µ−1 ]ω0 z k )−1 = (1 + [βµ (1 + ω0 + . . . + ω0µ−1 ) + w0−1 ω0µ ]z k )−1 and this equals 1 because βµ has to fulfill βµ (1 + ω0 + . . . + ω0µ−1 ) + w0−1 ω0µ = 0. Hence the function U (z) which we obtain is a special solution of (3) (w0 + z k )U (z) = U (ψ). w0 z In the next step we want to find a function S such that we can linearize the function ψ(z) = 1 that k (1+βµ z ) k
means such that S(ψ(z)) = w0 S(z) holds. To construct such a S we can use similar calculations as above. For 1 ≤ µ ≤ l1 − 1 we get 1
S(z) = z(1 + βµ (1 − ω0 )−1 z k )− k . It is easy to show that the linearization equation holds, we obtain 1
w0 S(z) = w0 z(1 + βµ (1 − ω0 )−1 z k )− k for the right hand side and 1
S(ψ(z)) = ψ(z)(1 + βµ (1 − ω0 )−1 ψ(z)k )− k − k1 w0 z ω0 z k −1 = 1 + βµ (1 − ω0 ) 1 (1 + βµ z k ) (1 + βµ z k ) k 1
1
1
= w0 z(1 + βµ z k )− k + k (1 + βµ z k + βµ (1 − ω0 )−1 ω0 z k )− k 1
= w0 z(1 + βµ (1 + (1 − ω0 )−1 ω0 )z k )− k 1
= w0 z(1 + βµ (1 − ω0 )−1 z k )− k for the left hand side and hence the function S is a linearization of ψ. Now we are interested in the general w0 z solution U of (3) (w0 + z k )U (z) = U (ψ(z)) with ψ(z) = 1 , 1 ≤ µ ≤ l1 − 1. The general solution has k (1+βµ z ) k
0
the representation U (z) = U (z)W (z). If W (z) =
W (z) = W
U1 (z) U2 (z)
for two solutions U1 and U2 of (3) we get
!
w0 z 1
(1 + βµ z k ) k
⇔ W (z) = W (S −1 (w0 S(z))) ⇔ W (S −1 (z)) = W (S −1 (w0 z)).
Defining the function Φ = W ◦ S −1 we obtain Φ(z) = Φ(w0 z).
ESAIM: PROCEEDINGS
13
That means the function Φ is invariant with respect to w0 and hence we have the representation Φ(z) = 1 + φl z l + φ2l z 2l + . . .. Introducing the function Φ? such that Φ(z) = Φ? (z l ) and Φ? (z) = 1 + . . . holds, leads to W (z) = Φ? (S(z)l ). Finally we receive
U (z) = z
µ−1 Y
(1 + [βµ (1 + ω0 + . . . + ω0ν−1 ) + w0−1 ω0ν ]z k )−1
ν=0
1 1 (1 + βµ (1 − ω0 )−1 z k )µ− k Φ? [z(1 + βµ (1 − ω0 )−1 z k )− k ]l as the general solution U of (3) (w0 + z k )U (z) = U (ψ(z)). Now we summarize this situation. Let ψ be linearizable and hence also the function ϕ˜ is, ψ(z) = w0 z + . . . and the relation ψ(z)k = ϕ(z ˜ k ), where w0 is a primitive root of unity of order l ≥ 2 holds. We define ω0 := w0k and l hence this is a primitive root of unity of order l1 = gcd(k,l) . Hence we have proven the following theorem. Theorem 2.1. Let ϕ˜ be given by ω0 y 1 + βy where β is one of the complex numbers βµ given by (14). Then the general solution U of the linear functional equation (3) (w0 + z k )U (z) = U (ψ(z)) is given by ϕ(y) ˜ =
U (z) = z
µ−1 Y
(1 + [βµ (1 + ω0 + . . . + ω0ν−1 ) + w0−1 ω0ν ]z k )−1
ν=0
1 1 (1 + βµ (1 − ω0 )−1 z k )µ− k Φ? [z(1 + βµ (1 − ω0 )−1 z k )− k ]l where the function Φ? is arbitrary and Φ? (z) ∈ C[[z]]. The βµ ’s with 1 ≤ µ ≤ l1 − 1 are given by (14).
3. Local analytic solutions Theorem 3.1. Let ϕ˜ be linearizable and convergent in a sufficiently small neighbourhood of zero. Then the equation g(w0 z + zg(z)) = ϕ(g(z)) ˜ (2) has convergent solutions g and hence also the generalized Dhombres functional equation f (zf (z)) = ϕ(f (z))
(1)
has non constant convergent solutions f with f (0) = w0 . Proof. It is known that (2) can be written as linear functional equation (w0 + z k )U (z) = U (ψ(z)) with ψ(z)k = ϕ(z k ) and U (z) = u1 z + . . .. This is the same as z(w0 + z k ) ? U (z) = U ? (ψ(z)) ψ(z)
(3)
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ESAIM: PROCEEDINGS
with U (z) = u1 zU ? (z) = u1 z(1 + . . .). Hence it is possible to use the formal logarithm which leads to Ln
z(w0 + z k ) + X ? (z) = X ? (ψ(z)) ψ(z)
0 +z with X ? = LnU ? . Then it is known that Ln z(wψ(z)
k
)
0 +z ∈ C[[z k ]] and that Ln z(wψ(z)
(19) k
)
is convergent. So (2) is
˜ k ). Using y = z k and A(y) = Ln z(w0 +z solvable if and only if (19) has a solution X (z) = X(z ψ(z) ?
˜ ˜ ϕ(y)). A(y) + X(y) = X( ˜
k
)
we obtain (20)
Then ϕ˜ is linearizable and hence, see [5], there exists a convergent R such that ϕ(y) ˜ = R−1 (ω0 R(y)) with ˜ ◦ R−1 = X leads to ω0 = w0k holds. Writing A ◦ R−1 = B and X B(y) + X(y) = X(ω0 y) P P with a convergent series B. Let B(y) = ν≥1 βν y ν and X(y) = ν≥1 ξν y ν then we get X(ψ(y)) − X(y) = P ξν (ω0ν − 1)y ν where l1 denotes the order of ω0 . Therefore we obtain ν≥1 ν6≡0(mod l1 )
X
B(y) =
ξν (ω0ν − 1)y ν
ν≥1
ν6≡0(mod l1 )
and hence a special solution of (20) is given by X0 (y) =
βν yν (ω0ν − 1)
X ν≥1
ν6≡0(mod l1 )
and this solution is convergent since there exists r0 > 0, such that |ω0ν − 1|−1 ≤ r0 for all ν with ν 6≡ 0 (mod l1 ). Now we choose the coefficients ξνl1 with ν ≥ 1 such that the series X(y) = X0 (y) +
X
ξνl1 y νl1
ν≥1
is also convergent. It is possible to reverse these calculations and hence there exist local analytic solutions g of the transformed generalized Dhombres functional equation (2).
References [1] [2] [3] [4] [5]
Jean Dhombres, Some Aspects of Functional Equations, Chulalongkorn University Press, Bangkok, 1979. ¨ Harald Fripertinger, Ludwig Reich, On a Linear Functional Equation for Formal Power Series, Sitzungsber. Osterr. Akad. Wiss. Wien, Math.-nat Kl. Abt. II 210: 85–134, 2001. ˇ ankov´ Ludwig Reich, Jaroslav Sm´ıtal, Marta Stef´ a, Local Analytic Solutions of the Generalized Dhombres Functional Equations ¨ I, Sitzungsber. Osterr. Akad. Wiss. Wien, Math.-nat Kl. Abt. II 214: 3–25, 2005. ˇ ankov´ Ludwig Reich, Jaroslav Sm´ıtal, Marta Stef´ a, Local Analytic Solutions of the Generalized Dhombres Functional Equations II, Journal of Mathematical Analysis and Applications 355: 821–829, 2009. Ludwig Reich, A remark on Schr¨ oder’s equation in the complex domain, in preparation.
