Dani`ele Fournier-Prunaret, Laura Gardini, & Ludwig Reich, Editors
SOME ASPECTS OF THE LOCAL THEORY OF GENERALIZED DHOMBRES
FUNCTIONAL EQUATIONS IN THE COMPLEX DOMAIN
J¨
org Tomaschek
1Abstract. We study the generalized Dhombres functional equation f(zf(z)) =ϕ(f(z)) in the com-plex domain. The functionϕ is given and we are looking for solutionsf with f(0) =w0 and w0 is
a primitive root of unity of orderl≥2. All formal solutions for this case are described in this work, for the situation whereϕcan 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 Dhombresf(zf(z)) =ϕ(f(z)) dans le plan complexe. La fonctionϕest donn´ee et nous cherchons les solutionsf avecf(0) =w0o`uw0est 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 lorsqueϕpeut 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 solutionsf withf(0) =w0 wherew0 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
1Institute for Mathematics and Scientific Computing, Karl-Franzens-Universit¨at Graz, Heinrichstraße 36, A-8010 Graz, Austria;
email: [email protected].
c
EDP Sciences, SMAI 2012
where xis 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 com-plex number w0 which is different from zero. In a first step (1) becomes by puttingf(z) =w0+g(z),g(0) = 0
equivalent to the transformed generalized Dhombres functional equation
g(w0z+zg(z)) = ˜ϕ(g(z)) (2)
where g is a series given by g(z) = ckzk +. . . with ck 6= 0, k ∈ N, and the function ˜ϕ is calculated as ˜
ϕ(ω) =wk
0ω+d2ω2+. . .. By taking an arbitrary but fixed power series solutionT ofT(z)k=g(z) we get
T(w0T−1(z) +T−1(z)zk) =ψ(z)
for the equation above withψ(z)k= ˜ϕ(zk). Finally definingU =T−1 leads to the linear functional equation
(w0+zk)U(z) =U(ψ(z)). (3)
The functionϕin equation (1) is known and so it is possible to computeψ, we getψ(z) =w0z
1 + d2
wk
0k
zk+. . ..
From now on we suppose thatw0is a primitive root of unity of orderl≥2.
1.
Generalized Dhombres functional equation with a root of unity
w
0In this section we want to give a full description of all formal solutions of equation (1) withf(0) =w0andw0
is a primitive root of unity of orderl ≥2. The case wherew0= 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 givenψ andk then there exists a solutionU0 withU0(z)∈zC[[zk]]∩Γ1.
Proof. We writeU(z) =u1zU?(z) withU?(z) = 1 +. . ., then (3) is equivalent to
w0z(1 +w−01zk)
ψ(z) U
?(z) =U?(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
Lnw0z
ψ(z)+ Ln(1 +w −1 0 z
k) =X?(ψ(z))−X?(z) (4)
with X? = LnU? and ord X? > 0. For the function ψ it follows from the assumption ψ(z)k = ˜ϕ(zk) that ψ(z)∈zC[[zk]]∩Γ and hence the left side Lnw0z
ψ(z)+ Ln(1 +w
−1
0 zk) of (4) belongs to C[[zk]]. It is possible to
write
X?(y) = X
ν≥1
ν6≡0(mod k)
ξ?νyν+ X
ν≥1
ν≡0(mod k)
ξ?νyν
forX?. Then we get
X?(ψ(z))−X?(z) = X ν≥1
ν6≡0(mod k)
ξ?ν(ψ(z)ν−zν) + X ν≥1
ν≡0(mod k)
where the first part P
ν≥1
ν6≡0(mod k)
ξ?
ν(ψ(z)ν−zν) does not contain any termzν with ν ≡0 (modk) while the
second part P
ν≥1
ν≡0(mod k)
ξ?
ν(ψ(z)ν−zν) is an element ofC[[zk]]. Now defineY? as Y?(z) := P
ν≥1
ν≡0(mod k)
ξ? νzν
then (4) is equivalent to
Lnw0z
ψ(z)+ Ln(1 +w −1 0 z
k) =Y?(ψ(z))−Y?(z)
and hence (4) has a solutionY?∈
C[[zk]] and after reversing our calculations we obtain a functionU0 with the
desired property.
So we know the structure of a solution U withU(z) =u1U0(z) andU0∈zC[[zk]]∩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
w0z
(1 +w−01zk)
ψ(z) U
?(z) =U?(ψ(z)).
Define ˆU byU?(z)k = ˆU(z), then ˆU is given by ˆU(z) = 1 +. . .and converselyU? is uniquely determined by ˆU.
Taking thek−thpower and usingU?(z)k= ˆU(z) in the equation above leads to
wk
0zk(1 +w
−1 0 zk)k
ψ(z)k U(z) = ˆˆ U(ψ(z))
and again with the formal logarithm this becomes equivalent to
Lnw
k
0z
k(1 +w−1 0 z
k)k
ψ(z)k = ˆX(ψ(z))−Xˆ(z)
with ˆX := Ln ˆU. Now as a consequence of ψ(z)∈zC[[zk]] and some other computations it is possible to write ˆ
X(z) = ˜X(zk) and therefore withy=zk andω
0:=w0k we get
Lnω0y(1 +w −1 0 y)k
˜
ϕ(y) = ˜X( ˜ϕ(y))− ˜ X(y)
and this leads to
Lnω0y ˜
ϕ(y)+kLn(1 +w −1
0 y) = ˜X( ˜ϕ(y))−X(y)˜ (5)
whereω0 is again a root of unity of orderl1 andl1=gcd(lk,l). Now it is necessary to distinguish two cases. The
first case deals with a function ˜ϕwhich is not linearizable and hence the function is conjugated to a non linear semicanonical form. So ˜ϕcan be written as
˜
ϕ(y) = (R−1◦N◦R)(y)
with N(y) = ω0y+δnl1+1y
nl1+1 +. . . and δ
nl1+1 6= 0. The function R
−1 and hence also the function R is
uniquely determined by R−1(y) =y+ρ2y2+. . .and the condition that the coefficients ρµl1+1= 0 for µ≥1.
The numberl1denotes the order ofω0. Then (5) is equivalent to
Lnω0y ˜
ϕ(y)+kLn(1 +w −1
0 y) = ˜X(R
−1(N(R(y))))
−X(y)˜
and substitutingR−1(y) fory and definingX as X:= ˜X◦R−1leads to
Ln ω0R −1(y)
R−1(N(y))+kLn(1 +w
−1 0 R−
Discussing the right hand sideX(N(y))−X(y) of equation (6) in detail leads to the fact that the first coefficient ofX(N(y))−X(y) which is different from zero is the coefficient which belongs to the termy(n+1)l1. So we have
Lemma 1.2. Equation (6)has a solution if and only if the coefficients ofyl1, . . . , ynl1 of the left hand side of
equation (6)are zero.
Proof. We write
ω0R−1(y)
R−1(N(y))=
ω0R−1(y)
ω0y
R−1(N(y))
N(y)
ω0y
N(y)
and therefore
Ln ω0R −1(y)
R−1(N(y)) = Ln
R−1(y)
y −Ln
R−1(N(y))
N(y) −Ln N(y)
ω0y
= LnR −1(y)
y −Ln R−1(u)
u |u=N(y)−Ln N(y)
ω0y
.
Then we obtain
N(y) ω0y
= 1 ω0y
(ω0y+δnl1+1y
nl1+1+. . .) = 1 +ω−1
0 δnl1+1y
nl1+. . .∈
C[[yl1]]
whereω−01δnl1+16= 0 and hence
LnN(y) ω0y
=ω0−1δnl1+1y
nl1+. . . .
Therefore it is only necessary to investigate the coefficients of yl1, . . . , y(n−1)l1 of LnR
−1(y)
y −Ln R−1(u)
u |u=N(y).
Hence we get
LnR −1(y)
y = ˜ρ2y+. . .+ ˜ρl1y
l1−1+ ˜ρ
l1+1y
l1+. . .
with polynomials ˜ρµwhich depend on (ρ2, . . . , ρµ) forµ6≡1 (modl1) and on (ρ2, . . . , ρµ−1) forµ≡1 (modl1).
We write
LnR −1(y)
y =: Θ(y) =θ1y+θ2y
2+. . .
and so we achieve
LnR −1(u)
u |u=N(y)= Θ(N(y)) =θ1(ω0y+δnl1+1y
nl1+1+. . .) +θ
2(ω0y+δnl1+1y
nl1+1+. . .)2+. . .
=θ1ω0y+θ2ω20y
2+. . .+θ
l1−1ω
l1−1
0 y
l1−1+θ
l1y
l1+. . .+θ
2l1y
2l1+. . .+θ
nl1y
nl1+. . .
≡θ1ω0y+θ2ω20y
2+. . . (modord nl 1+ 1) ≡Θ(ω0y) (modord nl1+ 1)
and hence Θ(y)−Θ(N(y))≡Θ(y)−Θ(ω0y) (modord nl1+ 1) and therefore there are no coefficients different
from zero for yl1, . . . , ynl1 in LnR
−1(y)
y −Ln R−1(u)
u |u=N(y). It remains to determine the coefficients of kLn(1 +
w0−1R−1(y)). Computing kLn(1 +w0−1R−1(y)) it is clear that comparing the coefficients ofyl1 leads to
ρl1=Pl1(ρα|α < l1).
with a polynomialPl1. Proceeding in comparing the coefficients ofy
µl1 withµ≥2 leads to
with polynomialsPµl1 for 2≤µ≤(n−1)l1. Fory
nl1 we obtain
−ω0−1δnl1+1+kw
−1
0 ρnl1+Pnl1(ρα|α < nl1) = 0
with a polynomialPnl1 and hence
δnl1+1=kω0w
−1
0 ρnl1+ω0Pnl1(ρα|α < nl1)
Now one obtainsδnl1+16= 0 ifρnl1 is suitably choosen.
It is also possible to determine the numberkifN andR and hence ˜ϕare given. We get
k=δnl1+1−ω0Pnl1(ρα|α < nl1)
ω0w−01ρnl1
. (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
−ω0−1δnl1+1+kw
−1
0 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))) whereN denotes the semicanonical form of ϕ˜. Then the equation
Ln ω0R −1(y)
R−1(N(y))+kLn(1 +w
−1 0 R
−1(y)) =X(N(y))−X(y) (6)
has a solutionX if and only if the polynomial relations (8)and (9)are fulfilled. This solution X is unique.
Proof. We obtain the uniquely determined solution if we compare the coefficients of (6) namley Lnω0R−1(y)
R−1(N(y))+
kLn(1 +w0−1R−1(y)) = X(N(y))−X(y). It is sufficient to look at Lnω0R−1(y)
R−1(N(y)) +kLn(1 +w
−1
0 R−1(y)) =
ˆ
θ1y+ ˆθ2y2+. . .. Let the seriesX be given byX(y) = ˜β1y+ ˜β2y2+. . ., then we obtain ˜βµ= (ω0µ−1)−1θˆµ for
1≤µ≤l1−1. Comparing the coefficients ofyl1 leads to the first polynomial relation of (8) but it is not possible
to determine ˜βl1 from this equation. Proceeding we can compute the coefficients ˜βl1+1, . . . ,β˜2l1−1 and fory
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 = (l1ω
l1−1
0 δnl1+1)
−1θˆ
(n+1)l1+1. Inductively we obtain the uniquely determined
solutionX 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-lution U of (3). In more detail we have ˜X(y) = X(R(y)) and therefore ˆX(y) = ˜X(yk) = X(R(yk). This
leads to ˆU(y) = exp(X(R(yk))). Next we obtain U(y) = u1z
exp(X(R(yk)))
1
k. Then we get T(y) =
u1z
exp(X(R(yk)))1k
[−1]
and finallyg is given by
g(y) =
(
u1z
h
exp(X(R(zk2)))i
1
k
The solutionsf of the generalized Dhombres functional equation are given by
f(z) =w0+
(
u1z
h
exp(X(R(zk2)))i
1
k
[−1])1k
. (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 solutionX is uniquely determined whereas in the situation of Theorem 1.6 the solutionX is depending on some arbitrary coefficients.
But before we prove the next theorem we have to consider the following remark.
Remark 1.5. Letw0be a primitive root of unity of orderl≥1,k∈Nandω0be defined byω0=wk0 where the
order of ω0 equalsl1. Also the seriesN is given byN(y) =ω0y+δnl1+1y
nl1+1+. . .∈y
C[[yl1]] withn≥1 and
the coefficientδnl1+1 is different from zero. Then let the seriesH be given by H(y) =h1y+h2y
2+. . .∈
C[[y]] such that the coefficients hµl1 ofH with 1≤µ≤(n−1)l1are given by the polynomial relations
hµl1 =Pµl1(hα|α < µl1)
where the polynomialsPµl1 are the same as in (8) and such that the coefficientshµl1+1 are zero for µ≥1 and
(H−1◦N◦H)(y) =ω0y+. . .∈C[[y]].
The coeffiecienthnl1 has to be choosen such that
k= δnl1+1−ω0Pnl1(hα|α < nl1)
ω0w−01hnl1
holds. Then for every choice of the coefficients of the seriesN except the coefficientδnl1+1 and for every choice
of the coefficients of the seriesH except these coefficients we get from the polynomial relations and also except the coefficient hnl1 there exists a unique solutionf 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 function ˜R such that ˜ϕ(y) = ˜R−1(ω
0R(y)). But it is also possible to use the notations from the case where a˜
non linear semicanonical formN exists. We may consider the present situation as a limiting case of the previous situation forn→ ∞.
In formula (8)ρµl1 =Pµl1(ρα|α < µl1) exactly (n−1)l1algebraic equations as a consequence of the non linear
semicanonical form are given. If now the semicanonical formN(y) =ω0y+δnl1+1y
nl1+1+. . .withδ
nl1+16= 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)
forµ≥1.
Theorem 1.6. Let k∈Nandϕ˜ be linearizable and given byϕ(y) =˜ R−1(ω
0R(y)). Then the equation
Lnω0R −1(y)
R−1(ω 0y)
+kLn(1 +w0−1R−1(y)) =X(ω0y)−X(y) (6)
Proof. We consider equation (6) namely
Lnω0R −1(y)
R−1(ω 0)
+kLn(1 +w0−1R−1(y)) =X(ω0y)−X(y)
and again it is possible to write Lnω0R−1(y)
R−1(ω0) +kLn(1 +w
−1
0 R−1(y)) = ˜ρ1y+ ˜ρ2y2+. . ..
LetX(y) =β1y+β2y2+. . . ,then
X(ω0y)−X(y) =
X
ν≥1
ν6≡0(mod l1)
βν(ων0−1)yν
and therefore we obtain a special solutionX0which is given by
X0(y) =
X
ν≥1
ν6≡0(mod l1)
˜
ρν(ω0ν−1)
−1yν.
Hence the general solution is given by
X(y) = X
ν≥1
ν6≡0(mod l1)
˜
ρν(ω0ν−1)−
1yν+X
ν≥1
βνl1y
νl1
with arbitraryβνl1 forν ≥1.
Remark 1.7. Another representation of the solution U of (w0+zk)U(z) = U(ψ(z)) where ψ is linearizable
andw0 is a root of unity of orderl≥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 ˜ϕis 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+zk)U(z) =
U(ψ(z)) where the functionψis linearizable. Notice that the function ˜ϕis linearizable if and only if the function ψ is. Letw0 be a root of unity with order l≥2. Then we know that ω0 :=w0k is a root of unity with order
l1= gcd(lk,l).
We iterate equation (3) and so we obtain
(w0+ψ(z)k)U(ψ(z)) =U(ψ2(z))
and hence
(w0+ ˜ϕ(zk))(w0+zk)U(z) =U(ψ2(z)).
Iterating the last equation l1−2 times and usingψ(z)k = ˜ϕ(zk) as well as y = zk and w0l = 1 leads to the
equation
(1 +w0−1ϕ˜l−1(y))(1 +w−01ϕ˜l−2(y)). . .(1 +w0−1ϕ(y))(1 +˜ w−01y) = 1. (12)
Then there exists ap∈Nsuch thatl=l1pand so as a consequence of ˜ϕl1(y) =y we obtain
(1 +w0−1ϕ˜pl1−1(y))(1 +w−1
0 ϕ˜
pl1−2(y)). . .(1 +w−1
0 ϕ(y))(1 +˜ w
−1 0 y) = 1
which becomes
(1 +w0−1ϕ˜l1−1(y))(1 +w−1
0 ϕ˜
l1−2(y)). . .(1 +w−1
0 ϕ(y))(1 +˜ w
Next we want to construct a function ˜ϕsuch that (12) is satisfied. As an Ansatz we use the special M¨ obius-transformation
˜
ϕ(y) = ω0y 1 +βy withβ 6= 0 and ˜ϕl1(y) =y. For 1≤ν ≤l
1−1 we obtain
˜
ϕν(y) = ω
ν
0y
1 +β(1 +ω0+. . .+ω0ν−1)y
.
Forν=l1we obtain ˜ϕl1(y) =y, hence ˜ϕis linearizable as well asψ. Then we get
1 +w−01ϕ˜ν(y) = 1 +w0−1 ω0y
1 +β(1 +ω0+. . .+ων0−1)y
= 1 + [w −1
0 ω0ν+β(1 +ω0+. . .+ων0−1)]y
1 +β(1 +ω0+. . .+ω0ν−1)y
.
Substituting this in equation (12) leads to
1 +w−01y 1 ·
1 + [w−01ω0+β]y
1 +βy ·. . .·
1 + [w0−1ωl1−1
0 +β(1 +ω0+. . .+ω0l1−2)]y
1 +β(1 +ω0+. . . ω0l1−2)y
= 1. (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 isl1while the degree of the denominator isl1−1 which contradicts (13). So it is necessary
that β is one of the numbers
β1=−w0−1ω0
β2=−w0−1ω 2
0(1 +ω0)−1
.. .
βl1−1=−w
−1 0 ω
l1−1
0 (1 +ω0+. . .+ω0l1−2)− 1.
Hence for 1≤µ≤l1−1 we have
βµ=−w0−1(1 +ω0+. . .+ωµ0−1)
−1ωµ
0. (14)
Next we have to show that this is sufficient. Therefore we number the fractions of (13) such that the numerators areZ0 toZl1−2 and the denominators areN1 toNl1−2, for N0 we haveN0 = 1. Then we choose a specialβµ
for 1≤µ≤l1−1. Then it is true thatZν =Nν−µ forν ≥µ and thatZν =Nν+l1−µ forν≤µhold. First we
show that forν ≥µwe getZν=Nν−µ. We obtain
Zν = 1 + [w0−1ω
ν
0+ (−w
−1
0 (1 +ω0+. . .+ωµ0−1)
−1ωµ
0)(1 +ω0+. . .+ων0−1)]y
= 1 + [w0−1ω0ν−w−01ωµ0(1 +ω0+. . .+ω0µ−1)
−1(1 +ω
0+. . .+ων0−1)]y.
ForNν−µ we get
Nν−µ= 1 + (−w0−1(1 +ω0+. . .+ω
µ−1 0 )
−1
ω0µ)(1 +ω0+. . .+ω
ν−µ−1 0 )
So it remains to show that
ων0−ω
µ
0(1 +ω0+. . .+ω0ν−1)
1 +ω0+. . .+ω
µ−1 0
=−ω
µ
0(1 +ω0+. . .+ω0ν−µ−1)
1 +ω0+. . .+ω
µ−1 0
Expanding this equation leads to
ω0ν+ω
ν+1
0 +. . .+ω
ν+µ−1 0 −ω
µ
0 −ω
µ+1
0 −. . .−ω
ν−1 0 −ω
ν
0 −. . .−ω
µ+ν−1 0
for the left hand side and to
−ωµ0 −ω0µ+1−. . .−ων0−1
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 forν ≤µ. Like above we obtain
Zν = 1 + [w−01ω
ν
0−w
−1 0 ω
µ
0(1 +ω0+. . .+ω
µ−1 0 )−
1(1 +ω
0+. . .+ω0ν−1)]y
forZν. The denominatorNν+l1−µ is given by
Nν+l1−µ= 1 + (−w
−1 0 ω
µ
0(1 +ω0+. . .+ω
µ−1 0 )
−1(1 +ω
0+. . .+ω
ν+l1−µ−1
0 ))y.
After cancelling 1 andw0−1 one has to show that
ω0ν−ω
µ
0(1 +ω0+. . .+ω0ν−1)
1 +ω0+. . .+ωµ0−1
=−ω
µ
0(1 +ω0+. . .+ω
ν+l1−µ−1
0 )
1 +ω0+. . . ω0µ−1
is fulfilled. Expanding this equation leads to
ω0ν+ων0+1+. . .+ω0µ+. . .+ω0ν+µ−1−ω0µ−ω0µ+1−. . .−ω0ν+µ−1
for the left hand side and hence only the expression
ω0ν+ων0+1+. . .+ω0µ−1
remains. The right hand side is given by
−ω0µ−ωµ0+1−. . .−ων+l1−1
0 .
Therefore we have to consider the equation
ω0ν(1 +ω0+. . .+ω0µ−ν−1) =−ω
ν
0(ω
µ−ν
0 +ω
µ−ν+1
0 +. . .+ω
l1−1
0 )
which is equivalent to
1 +ω0+. . .+ω0µ−ν−1=−ω
µ−ν
0 −ω
µ−ν+1
0 −. . .−ω
l1−1
0 . (15)
For a root of unityω0of orderl1the equation 1 +ω0+. . .+ωl01−1= 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 solutionU of (3) (w0+zk)U(z) =U(ψ(z)). Remember that our
functions are given by
ψ(z) = w0z (1 +βµzk)
1
k .
for 1≤µ≤l1−1. To show how one can compute a special solution we consider the special caseβ2=−w−01(1 +
ω0)−1ω02 and we try to find a special solution for β2. Therefore the functionψ is given by ψ(z) = w0z (1+β2zk)
1
k .
First we writeU(z) =zU(1)(z) withU(1)(z) = 1 +. . .. Then (3) is equivalent to
w0(1 +w0−1z
and this is equivalent to
w0z(1 +w−01z
k)U(1)(z) =w
0z(1 +β2zk)−
1
kU(1)(ψ(z)).
This leads to
(1 +w0−1zk)(1 +β2zk)
1
kU(1)(z) =U(1)(ψ(z)). (16)
Then we use a new Ansatz, so we considerU(1)(z) = (1 +w0−1zk)α1U(2)(z) and after substituting this expression
in (16) we get
(1 +w−01zk)α1+1(1 +β
2zk)
1
kU(2)(z) = (1 +w−1
0 ψ(z)
k)α1U(2)(ψ(z))
or the equivalent expression
(1 +w−01zk)α1+1(1 +β
2zk)
1
kU(2)(z) =
1 +w−01 ω0z
k
1 +β2zk
α1
U(2)(ψ(z)).
This leads to
(1 +w−01zk)α1+1(1 +β
2zk)
1
k+α1U(2)(z) = (1 + (β
2+w0−1ω0)zk)α1U(2)(ψ(z)). (17)
Then again a new Ansatz is needed and byU(2)(z) = (1 + (β
2+w0−1ω0)zk)α1U(3)(z) we describe this new one.
Substituting this in (17) leads to
(1 +w0−1zk)α1+1(1 +β
2zk)
1
k+α1(1 + (β
2+w−01ω0)zk)α1U(3)(z) =
(1 + (β2+w−01ω0)zk)α1(1 + (β2+w−01ω0)ψ(z)k)α1U(3)(ψ(z)).
This is equivalent to
(1 +w−01zk)α1+1(1 +β
2zk)
1
k+α1U(3)(z) =
1 + (β2+w−01ω0)
ω0zk
1 +β2zk
α1
U(3)(ψ(z)).
Hence we get
(1 +w0−1zk)α1+1(1 +β
2zk)
1
k+α1U(3)(z) =
1 +β
2zk+β2ω0zk+w0−1ω02zk
1 +β2zk
α1
U(3)(ψ(z))
=
1 + [β
2(1 +ω0) +w−01ω02]zk
1 +β2zk
α1
U(3)(ψ(z))
=
1
1 +β2zk
α1
U(3)(ψ(z))
because the condition which the expression β2 has to fulfill is β2(1 +ω0) +w−01ω20 = 0. Hence the following
equation remains
(1 +w0−1zk)α1+1(1 +β
2zk)
1
k+2α1U(3)(z) =U(3)(ψ(z)) (18)
and so we need a last Ansatz. We chooseU(3)(z) = (1 +δzk)α2 and substituting this expression in (18) we get
for the right hand side of (18)
U(3)(ψ(z)) =
1 +δ ω0z
k
1 +β2zk
α2
=
1 + (β
2+δω0)zk
1 +β2zk
α2
Therefore (18) is equivalent to
(1 +w0−1zk)α1+1(1 +β
2zk)
1
k+2α1+α2(1 +δzk)α2 = (1 + (β
2+δω0)zk)α2.
Hence we obtain δ = β2 +δω0 which is equaivalent to δ = (1−ω0)−1β2. Also (1 +w−01zk)α1+1(1 +
β2zk)
1
k+2α1+α2 = 1 has to be satisfied and this leads to α
1 = −1 and α2 = 2− 1k. Then the special
solu-tion forβ2 is given by
U(z) =z(1 +w−01zk)−1(1 + (β2+w−01ω0)zk)−1(1 + (1−ω0)−1β2zk)2−
1
k
which we obtain after putting our results together.
These calculations motivate to state that a special solutionU of equation (3) for aβµ with 1≤µ≤l1−1 has
the representation
U(z) =z
µ−1
Y
ν=0
(1 + [βµ(1 +ω0+. . .+ω0ν−1) +w
−1 0 ω
ν
0]z
k)−1(1 +β
µ(1−ω0)−1zk)µ−
1
k.
To proof this we have to substituteU(z) in (3). Then the left hand side of (3) is
(w0+zk)U(z) =w0(1 +w0−1z
k)z µ−1
Y
ν=0
(1 + [βµ(1 +ω0+. . .+ω0ν−1) +w
−1 0 ω
ν
0]z
k)−1
(1 +βµ(1−ω0)−1zk)µ−
1
k
=w0z(1 +w−01z
k)1−1
µ−1
Y
ν=1
(1 + [βµ(1 +ω0+. . .+ω0ν−1) +w
−1 0 ω
ν
0]z
k)−1
(1 +βµ(1−ω0)−1zk)µ−
1
k.
The right hand side of (3) is given by
U(ψ(z)) =ψ(z)
µ−1
Y
ν=0
1 + [βµ(1 +ω0+. . .+ων0−1) +w
−1 0 ω
ν
0]ψ(z)
k−1
(1 +βµ(1−ω0)−1ψ(z)k)µ−
1
k
=w0z(1 +βµzk)−
1
k
µ−1
Y
ν=0
1 + [βµ(1 +ω0+. . .+ων0−1) +w
−1 0 ω
ν
0]
ω0zk
1 +βµzk −1
1 +βµ(1−ω0)−1
ω0zk
1 +βµzk µ−1k
=w0z(1 +βµzk)−
1
k+µ−µ+k1
µ−1
Y
ν=0
(1 +βµzk+ [βµ(1 +ω0+. . .+ω0ν−1) +w
−1 0 ω
ν
0]
ω0zk)−1(1 +βµzk+βµ(1−ω0)−1ω0zk)µ−
1
k.
The last term of the right hand side leads to
(1 +βµzk+βµ(1−ω0)−1ω0zk)µ−
1
k = (1 +βµ((1 + (1−ω0)−1)ω0)zk)µ−
1
k
= (1 +βµ(1−ω0)−1zk)µ−
1
and hence this terms and the term w0z 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) +w
−1 0 ω
n
0]z
k
)−1
and the (n−1)−thfactor of the right hand side by
(1 +βµzk+ [βµ(1 +ω0+. . .+ω0n−2) +w
−1 0 ω
n−1
0 ]ω0zk)−1.
Now it is easy to see that these expressions are the same. It remains to consider the (µ−1)−thfactor of the right hand side of the equation. We obtain
(1 +βµzk+ [βµ(1 +ω0+. . .+ωµ0−2) +w
−1 0 ω
µ−1
0 ]ω0zk)−1= (1 + [βµ(1 +ω0+. . .+ω0µ−1) +w
−1 0 ω
µ
0]z
k)−1
and this equals 1 becauseβµ has to fulfillβµ(1 +ω0+. . .+ω0µ−1) +w
−1 0 ω
µ
0 = 0. Hence the functionU(z) which
we obtain is a special solution of (3) (w0+zk)U(z) =U(ψ).
In the next step we want to find a function S such that we can linearize the function ψ(z) = w0z
(1+βµzk)
1
k that
means such thatS(ψ(z)) =w0S(z) holds. To construct such aS we can use similar calculations as above. For
1≤µ≤l1−1 we get
S(z) =z(1 +βµ(1−ω0)−1zk)−
1
k.
It is easy to show that the linearization equation holds, we obtain
w0S(z) =w0z(1 +βµ(1−ω0)−1zk)−
1
k
for the right hand side and
S(ψ(z)) =ψ(z)(1 +βµ(1−ω0)−1ψ(z)k)−
1
k
= w0z (1 +βµzk)
1
k
1 +βµ(1−ω0)−1
ω0zk
(1 +βµzk) −1k
=w0z(1 +βµzk)−
1
k+1k(1 +βµzk+βµ(1−ω0)−1ω0zk)−k1
=w0z(1 +βµ(1 + (1−ω0)−1ω0)zk)−
1
k
=w0z(1 +βµ(1−ω0)−1zk)−
1
k
for the left hand side and hence the function S is a linearization of ψ. Now we are interested in the general solutionU of (3) (w0+zk)U(z) =U(ψ(z)) withψ(z) = w0z
(1+βµzk)
1
k
, 1≤µ≤l1−1. The general solution has
the representation U(z) =U0(z)W(z). IfW(z) = U1(z)
U2(z) for two solutionsU1andU2of (3) we get
W(z) =W w0z (1 +βµzk)
1
k
!
⇔W(z) =W(S−1(w0S(z)))
⇔W(S−1(z)) =W(S−1(w0z)).
Defining the function Φ =W◦S−1 we obtain
That means the function Φ is invariant with respect to w0 and hence we have the representation Φ(z) =
1 +φlzl+φ2lz2l+. . .. Introducing the function Φ? such that Φ(z) = Φ?(zl) and Φ?(z) = 1 +. . . holds, leads
toW(z) = Φ?(S(z)l). Finally we receive
U(z) =z
µ−1
Y
ν=0
(1 + [βµ(1 +ω0+. . .+ων0−1) +w
−1 0 ω
ν
0]z
k)−1
(1 +βµ(1−ω0)−1zk)µ−
1
kΦ?
[z(1 +βµ(1−ω0)−1zk)−
1
k]l
as the general solutionU of (3) (w0+zk)U(z) =U(ψ(z)).
Now we summarize this situation. Letψbe linearizable and hence also the function ˜ϕis,ψ(z) =w0z+. . .and
the relationψ(z)k = ˜ϕ(zk), wherew
0 is a primitive root of unity of orderl≥2 holds. We defineω0:=w0k and
hence this is a primitive root of unity of orderl1=gcd(lk,l).
Hence we have proven the following theorem.
Theorem 2.1. Let ϕ˜be given by
˜
ϕ(y) = ω0y 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+zk)U(z) =U(ψ(z))is given by
U(z) =z
µ−1
Y
ν=0
(1 + [βµ(1 +ω0+. . .+ων0−1) +w
−1 0 ω
ν
0]z
k)−1
(1 +βµ(1−ω0)−1zk)µ−
1
kΦ?
[z(1 +βµ(1−ω0)−1zk)−
1
k]l
where the function Φ? is arbitrary and Φ?(z)∈
C[[z]]. Theβµ’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(w0z+zg(z)) = ˜ϕ(g(z)) (2)
has convergent solutionsg 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+zk)U(z) =U(ψ(z)) (3)
withψ(z)k=ϕ(zk) andU(z) =u
1z+. . .. This is the same as
z(w0+zk)
ψ(z) U
withU(z) =u1zU?(z) =u1z(1 +. . .). Hence it is possible to use the formal logarithm which leads to
Lnz(w0+z
k)
ψ(z) +X
?(z) =X?(ψ(z)) (19)
with X? = LnU?. Then it is known that Lnz(w0+zk)
ψ(z) ∈ C[[z
k]] and that Lnz(w0+zk)
ψ(z) is convergent. So (2) is
solvable if and only if (19) has a solutionX?(z) = ˜X(zk). Usingy=zk andA(y) = Lnz(w0+zk)
ψ(z) we obtain
A(y) + ˜X(y) = ˜X( ˜ϕ(y)). (20)
Then ˜ϕ is linearizable and hence, see [5], there exists a convergent R such that ˜ϕ(y) = R−1(ω0R(y)) with
ω0=wk0 holds. WritingA◦R−1=B and ˜X◦R−1=X leads to
B(y) +X(y) =X(ω0y)
with a convergent series B. Let B(y) = P
ν≥1βνyν and X(y) = Pν≥1ξνyν then we get X(ψ(y))−X(y) = P
ν≥1
ν6≡0(mod l1)
ξν(ω0ν−1)yν wherel1denotes the order ofω0. Therefore we obtain
B(y) = X
ν≥1
ν6≡0(mod l1)
ξν(ω0ν−1)y
ν
and hence a special solution of (20) is given by
X0(y) =
X
ν≥1
ν6≡0(mod l1)
βν
(ων
0−1)
yν
and this solution is convergent since there existsr0>0, such that|ων0−1|−1≤r0for allν withν 6≡0 (modl1).
Now we choose the coefficientsξνl1 withν≥1 such that the series
X(y) =X0(y) +
X
ν≥1
ξνl1y
νl1
is also convergent. It is possible to reverse these calculations and hence there exist local analytic solutionsg of the transformed generalized Dhombres functional equation (2).
References
[1] Jean Dhombres,Some Aspects of Functional Equations, Chulalongkorn University Press, Bangkok, 1979.
[2] Harald Fripertinger, Ludwig Reich, On a Linear Functional Equation for Formal Power Series, Sitzungsber. ¨Osterr. Akad. Wiss. Wien, Math.-nat Kl. Abt. II210: 85–134, 2001.
[3] Ludwig Reich, Jaroslav Sm´ıtal, Marta ˇStef´ankov´a,Local Analytic Solutions of the Generalized Dhombres Functional Equations I, Sitzungsber. ¨Osterr. Akad. Wiss. Wien, Math.-nat Kl. Abt. II214: 3–25, 2005.
[4] Ludwig Reich, Jaroslav Sm´ıtal, Marta ˇStef´ankov´a,Local Analytic Solutions of the Generalized Dhombres Functional Equations II, Journal of Mathematical Analysis and Applications355: 821–829, 2009.
