Harald Fripertinger¹,       Ludwig Reich

On covariant embeddings of a linear functional equation with respect to an analytic iteration group

Institut für Mathematik
Karl Franzens Universität Graz
Heinrichstr. 36/4
A-8010 Graz, AUSTRIA
Email: harald.fripertinger@kfunigraz.ac.at
Email: ludwig.reich@kfunigraz.ac.at

Abstract

j(p(x)) = a(x)j(x)+b (x),

(L)

(for the unknown series j(x) Î C [[x]]) with respect to (p(s,x))_{s Î C} we understand families (a(s,x))_{s Î C} and (b(s,x))_{s Î C} with entire coefficient functions in s, such that the system of functional equations and boundary conditions

j(p(s,x)) = a(s,x)j(x)+b(s,x)

(Ls)

a(t+s,x) = a(s,x)a(t,p(s,x))

(Co1)

b(t+s,x) = b(s,x)a(t,p(s,x)) +b(t,p(s,x))

(Co2)

a(0,x) = 1        b(0,x) = 0

(B1)

a(1,x) = a(x)        b(1,x) = b(x)

(B2)

holds for all solutions j(x) of (L) and s,t Î C. In this paper we solve the system ((Co1),(Co2)) (of so called cocycle equations) completely, describe when and how the boundary conditions (B1) and (B2) can be satisfied and present a large class of equations (L) together with iteration groups (p(s,x))_{s Î C} for which there exist covariant embeddings of (L) with respect to (p(s,x))_{s Î C}.

1  Introduction

j(p(x)) = a(x)j(x)+b (x),

(L)

where p(x),a(x),b(x) Î C [[x]] are given formal power series and j(x) Î C [[x]] should be determined by the functional equation. We always assume that

p(x) = rx+c2x2+c3x3+... = rx+ å n ³ 2  cnxn

with multiplier r ¹ 0 and

a(x) = a0+a1x+a2x2+... = å n ³ 0  anxn

with a₀ ¹ 0. For a foundation of the basic calculations with formal power series we refer the reader to [Henrici, 1977] and to [Cartan, 1963] or [Cartan, 1966]. If y(x) Î C [[x]] is of the form y(x) = å_{n ³ k}d_kx^k with d_k ¹ 0, then k is the order of y, which will be indicated as ord y(x) = k. Hence ord p(x) = 1 and ord a(x) = 0. The set of all formal power series of order 1 is indicated by G, which is a group with respect to the substitution in C [[x]]. In addition to this let G₀ indicate the set of all formal power series of the form x+d₂x²+... Î G.

Furthermore, the notion of congruence modulo order r will be useful. We write j º y mod ord r for formal power series j(x),y(x) Î C [[x]] if the difference j(x)-y(x) is a series of order greater than or equal to r.

A formal power series j(x) can be substituted into the series y(x) = å_{n ³ 0}d_nxⁿ Î C [[x]], i.e. the series y(j(x)) = å_{n ³ 0}d_nj(x)ⁿ can be computed, if and only if ord j(x) ³ 1.

The exponential series is given as

exp(x) = å n ³ 0  xn n! ,

and the formal logarithm is the series defined by

ln(1+x) = å n ³ 1  (-1)nxn n .

A family p: = (p(s,·))_{s Î C} in G is called an iteration group, (see e.g. [Scheinberg, 1970]), or a one-parameter group in G, if the translation equation

p(t+s,x) = p(t,p(s,x))

(T)

holds for all t,s,x Î C. Hence p(0,x) = x and p(-1,x) = p^-1(1,x), the inverse of p with respect to substitution. If we express p(s,x) in the form å_{n ³ 1}p_n(s)xⁿ, then p is called an analytic iteration group, if all the coefficient functions p_n(x) are entire functions.

The formal power series p(x) is called (analytically) iterable, or embeddable, if there exists an (analytic) iteration group p in G, such that p(1,x) = p(x) for all x Î C.

There exist only three different types of analytic iteration groups in G.

1. p(s,x) = x for all s Î C.
2. p(s,x) = S^-1(e^lsS(x)) for all s Î C, where l Î C\{ 0} and S(x) = x+s₂x²+... belongs to G₀. These iteration groups are called iteration groups of the first type. Each iteration group of this type is simultaneously conjugate to the iteration group (e^lsx)_{s Î C}.
3. p(s,x) = x+c_ksx^k+P_k+1^(k)(s)x^k+1+... for all s Î C, where c_k ¹ 0, k ³ 2 and P_r^(k)(s) are certain polynomials in s for r > k. These iteration groups are called iteration groups of the second type.

If r is a complex root of 1 and r ¹ 1, the series p(x) need not have an analytic embedding. But if such a p(x) has an analytic embedding, then it is of the first type. In this situation, however, the embedding need not be unique.

If p(x) = x+c_kx^k+..., with c_k ¹ 0 and k ³ 2, then there exists exactly one analytic embedding of p(x) in an iteration group of second type. (These facts about analytic iteration groups in C [[x]] can also be deduced as special cases of the results in [Reich & Schwaiger, 1977].)

Assume that a(x) and p(x) are formal power series given as above. For n Î Z we form the natural iterates of p(x) defined by

pn(x): = ì ï í ï î x, n = 0 p(pn-1(x)), n > 0 (p-1)-n(x), n < 0.

Furthermore for n ³ 0 we define

a(n,x): = n-1 Õ r = 0  a(pr(x))

and

b(n,x): = an(x) n-1 å r = 0  b(pr(x)) r Õ j = 0  a(pj(x)) .

Then the conditions

a(0,x) = 1        b(0,x) = 0

(B1)

a(1,x) = a(x)        b(1,x) = b(x)

(B2)

are clearly satisfied.

Lemma 1 The two families (a(n,x))_{n Î N0} and (b(n,x))_{n Î N0} satisfy

a(n+m,x) = a(m,x)a(n,pm(x))

(C1)

b(n+m,x) = b(m,x)a(n,pm(x)) +b(n,pm(x))

(C2)

for all n,m ³ 0.

We leave the proof by induction to the reader.

If for n < 0 we define

a(n,x): = 1 a(-n,pn(x)) =

1 -n-1 Õ r = 0  a(pr+n(x)) = 1 -1 Õ r = n  a(pr(x))

and

b(n,x): = -b(-n,pn(x)) a(-n,pn(x)) = -a(n,x)b(-n,pn(x)),

then Lemma 1.1 holds for all n,m Î Z.

Lemma 2 If j(x) satisfies (L), then it also satisfies

j(pn(x)) = a(n,x)j(x)+b(n,x)

(Ln)

for all n Î Z.

Proof. Obvious from Lemma 1.1 and its generalization for all n Î Z.       ^[¯]

Motivated by (Ln), (C1) and (C2) for natural iterates L. Reich introduced in [Reich, 1998] the following notion.

The linear functional equation (L) has a covariant embedding with respect to the analytic iteration group (p(s,x))_{s Î C} of p(x), if there exist families (a(s,x))_{s Î C} and (b(s,x))_{s Î C} of formal power series with entire coefficient functions a_n(s) and b_n(s) for all n ³ 0, such that

j(p(s,x)) = a(s,x)j(x)+b(s,x)

(Ls)

holds for all s Î C and for all solutions j(x) of (L) in C [[x]]. Moreover it is assumed that a and b satisfy both the boundary conditions (B1) and (B2) and the cocycle equations

a(t+s,x) = a(s,x)a(t,p(s,x))

(Co1)

b(t+s,x) = b(s,x)a(t,p(s,x)) +b(t,p(s,x))

(Co2)

for all s,t Î C.

Such embeddings were studied in a much more general setting by Z. Moszner in [Moszner, 1999] and for real-valued functions by G. Guzik in [Guzik, 1999], [Guzik, 2000] and [Guzik, 2001]. For the theory of linear functional equations we refer the reader to [Kuczma et al., 1990] and to [Kuczma, 1968]. We will treat this problem in the ring of formal power series C [[x]]. In section 2 we solve the underlying functional equations (Co1) and (Co2) completely. Then in section 3 we show how to adjust these solutions to given boundary conditions. And finally, in the last section we describe how to embed the linear functional equation (L) in the generic cases.

When dealing with analytic iteration groups (p(s,x))_{s Î C} of the first type, it is enough to consider p(s,x) = e^lsx. This is explained in the next

Theorem 1 Let p(s,x) = S^-1(e^lsS(x)) for l ¹ 0 and S(x) Î G₀ be an embedding of p(x).

1. The formal power series j(x) is a solution of (L) if and only if [(j)\tilde]: = j°S^-1 satisfies

   
   

   ~ j   (elsy) = ~ a   (y) ~ j   (y)+ ~ b   (y)

   (

   ~ L

   )

   for all s,y Î C, where [a\tilde]: = a°S^-1 and [b\tilde]: = b°S^-1.

2. The system (Ls), (Co1), (Co2), (B1) and (B2) is equivalent to the system

   
   

   ~ j   (elsy) = ~ a   (s,y) ~ j   (y)+ ~ b   (s,y)

   (

   ~ L

   s)

   
   

   ~ a   (t+s,y) = ~ a   (s,y) ~ a   (t,elsy)

   (

   ~ C

   o1)

   
   

   ~ b   (t+s,y) = ~ b   (s,y) ~ a   (t,elsy) + ~ b   (t,elsy)

   (

   ~ C

   o2)

   
   

   ~ a   (0,y) = 1        ~ b   (0,y) = 0

   (

   ~ B

   1)

   
   

   ~ a   (1,y) = ~ a   (y)        ~ b   (1,y) = ~ b   (y),

   (

   ~ B

   2)

   where [(a)\tilde](s,y) = a(s,S^-1(y)) and [(b)\tilde](s,y) = b(s,S^-1(y)).

Proof. The formal series j(x) satisfies (L) if and only if

j(S-1(elsS(x))) = a(x)j(x) +b(x)Û

(j°S-1)(elsS(x)) =

(a°S-1)(S(x))(j°S-1)(S(x))+(b°S-1)(S(x)),

which is equal to ([L\tilde]) after replacing S(x) by y.

Assuming that (Ls) holds we deduce

j(S-1(elsS(x))) = a(s,x)j(x) +b(s,x)Þ

(j°S-1)(elsS(x)) =

a(s,S-1(S(x)))(j°S-1)(S(x))+b(s,S-1(S(x))),

which is equal to ([L\tilde]s) after replacing S(x) by y. The boundary conditions ([B\tilde]1) and ([B\tilde]2) are naturally equivalent to (B1) and (B2). Finally,

~ a   (t+s,y) = a(t+s,S-1(y)) =

a(s,S-1(y)) a(t,p(s,S-1(y))) =

~ a   (s,y)a(t,S-1(elsS(S-1y))) =

~ a   (s,y)a(t,S-1(elsy)) = ~ a   (s,y) ~ a   (t,elsy),

hence ([C\tilde]o1) is satisfied. Using similar methods it is possible to show that ([C\tilde]o2) is a consequence of (Co2).

Since (p(s,x))_{s Î C} and (e^lsx)_{s Î C} are conjugate via the formal power series S(x) it is clear how to prove the implications into the converse direction.       ^[¯]

2  Solutions of the cocycle equations

Lemma 3 Let E(x): = e₀+e₁x+... Î C [[x]], e₀ ¹ 0 and let m Î C. Then

a(s,x): = ems E(p(s,x)) E(x)

is a solution of (Co1).

Proof. Since p satisfies the translation equation (T) it is clear that

a(t+s,x) = em(t+s) E(p(t+s,x)) E(x) =

emtems E(p(t,p(s,x))) E(x) =

emt E(p(t,p(s,x))) E(p(s,x)) ems E(p(s,x)) E(x) =

a(t,p(s,x))a(s,x).

      ^[¯]

Lemma 2.1 also holds, when e^ms is replaced by a generalized exponential function.

If we express a(s,x) in the form

a(s,x) = ¥ å n = 0  an(s)xn,

then it follows from the cocycle equation (Co1) that a₀(t+s) = a₀(s)a₀(t). Hence, taking into account the regularity conditions for the coefficients of a and the fact that a₀(s) ¹ 0, it is clear that a₀(s) = e^ms for some m Î C. Consequently a(s,x) = e^ms [^(a)](s,x) and [^(a)](s,x) = 1+[^(a)]₁(s)x+... . Using the formal logarithm there exists exactly one [(a)\tilde](s,x) Î C [[x]], such that ord _x [(a)\tilde](s,x) ³ 1 for all s Î C and [^(a)](s,x) = exp([(a)\tilde](s,x)). The coefficient functions of [(a)\tilde] are analytic if and only if the coefficient functions of [^(a)] are analytic, which is equivalent to the fact that the coefficient functions of a are analytic. Furthermore, [^(a)] is a solution of (Co1) if and only if [(a)\tilde] satisfies

~ a   (t+s,x) = ~ a   (s,x)+ ~ a   (t,p(s,x)).

(Co1¢)

Theorem 2 The family [(a)\tilde] of formal power series is a solution of (Co1¢) and [(a)\tilde](0,x) = 0 if and only if there exists a formal power series K(y) Î C [[y]], ord K(y) ³ 1, such that

~ a   (s,x) = ó õ s 0  K(p(s,x))ds,

where integration is taken coefficientwise.

Proof. First assume that [(a)\tilde] is a solution of (Co1¢) with [(a)\tilde](0,x) = 0. Coefficientwise differentiation of (Co1¢) with respect to the variable t and the chain rule for this differentiation yields

~ a   ¢(t+s,x) = ~ a   ¢(t,p(s,x)).

For t = 0 we get [(a)\tilde]¢(s,x) = [(a)\tilde]¢(0,p(s,x)). Since ord _x [(a)\tilde](s,x) ³ 1, also ord _x [(a)\tilde]¢(s,x) ³ 1. Putting K(y): = [(a)\tilde]¢(0,y), we obtain ord K(y) ³ 1 and [(a)\tilde]¢(s,x) = K(p(s,x)). By coefficientwise integration it follows that

~ a   (s,x) = ó õ s 0  K(p(s,x))ds.

Conversely, assume that [(a)\tilde](s,x) is given as the integral above. We prove that [(a)\tilde] satisfies (Co1¢):

~ a   (t+s,x) = ó õ t+s 0  K(p(s,x))ds =

ó õ s 0  K(p(s,x))ds+ ó õ t+s s  K(p(s,x))ds =

~ a   (s,x)+ ó õ t 0  K(p(t+s,x))dt =

~ a   (s,x)+ ó õ t 0  K(p(t,p(s,x)))dt =

~ a   (s,x)+ ~ a   (t,p(s,x)),

by applying (T). From the definition of [(a)\tilde] it is obvious that [(a)\tilde](0,x) = 0.       ^[¯]

Corollary 1 Using the notation from above, we have:

1. The family ([^(a)](s,x))_{s Î C} is a solution of (Co1) if and only if there exists K(y) Î C [[y]], ord K(y) ³ 1, such that

   
   

   ^ a   (s,x) = exp ó õ s 0  K(p(s,x))ds.

2. The family (a(s,x))_{s Î C} is a solution of (Co1) if and only if there exist m Î C and K(y) Î C [[y]], ord K(y) ³ 1, such that

   
   

   a(s,x) = emsexp ó õ s 0  K(p(s,x))ds.

Now we assume that a satisfies (Co1). Since ord _xa(s,x) = 0, it is possible to define g(s,x) Î C [[x]] by

g(s,x): = b(s,x) a(s,x)        "s Î C.

The coefficient functions of g are analytic if and only if the coefficient functions of b are analytic.

Lemma 4 The families a and b satisfy the system ((Co1),(Co2)) if and only if a satisfies (Co1) and g is a solution of

g(t+s,x) = g(s,x)+ g(t,p(s,x)) a(s,x) .

(Co2¢)

Proof. Assume first that a and b satisfy (Co1) and (Co2). Then

g(t+s,x) = b(t+s,x) a(t+s,x) =

b(s,x)a(t,p(s,x)) +b(t,p(s,x)) a(s,x)a(t,p(s,x)) =

b(s,x) a(s,x) + b(t,p(s,x)) a(s,x)a(t,p(s,x)) =

g(s,x)+ g(t,p(s,x)) a(s,x) .

Assuming conversely that g is a solution of (Co2¢) we get

b(t+s,x) = a(t+s,x)g(t+s,x) =

a(s,x)a(t,p(s,x))[g(s,x)+ g(t,p(s,x)) a(s,x) ] =

a(s,x)g(s,x)a(t,p(s,x))+

a(t,p(s,x))g(t,p(s,x)) =

b(s,x)a(t,p(s,x))+ b(t,p(s,x)) .

      ^[¯]

Theorem 3 Assume that a satisfies the cocycle equation (Co1). Then a and b are a solution of (Co2) if and only if there exists a series L(y) Î C [[y]], such that

b(s,x) = a(s,x) ó õ s 0  L(p(s,x)) a(s,x) ds,

where integration is taken coefficientwise.

Proof. First assume that a and b satisfy (Co2). Then Lemma 2.4 implies that a and g satisfy (Co2¢). Coefficientwise differentiation of (Co2¢) with respect to the variable t yields

g¢(t+s,x) = g¢(t,p(s,x)) a(s,x) .

For t = 0 we get g¢(s,x) = g¢(0,p(s,x))/a(s,x). Putting L(y): = g¢(0,y), we obtain g¢(s,x) = L(p(s,x))/a(s,x). By coefficientwise integration it follows that

g(s,x) = ó õ s 0  L(p(s,x)) a(s,x) ds

and

b(s,x) = a(s,x) ó õ s 0  L(p(s,x)) a(s,x) ds.

Conversely, if b is given by that formula, then

g(t+s,x) = ó õ t+s 0  L(p(s,x)) a(s,x) ds =

ó õ s 0  L(p(s,x)) a(s,x) ds+ ó õ t+s s  L(p(s,x)) a(s,x) ds =

g(s,x)+ ó õ t 0  L(p(t+s,x)) a(t+s,x) dt =

g(s,x)+ ó õ t 0  L(p(t,p(s,x))) a(s,x)a(t,p(s,x)) dt =

g(s,x)+ g(t,p(s,x)) a(s,x) .

In other words, a and g satisfy (Co2¢), hence by Lemma 2.4 a and b satisfy (Co2).       ^[¯]

Now we will describe a different form of representing the general solution of (Co1), and of the system ((Co1),(Co2)), involving as few integrals as possible. In Lemma 2.1 we already derived solutions a of (Co1) which could be represented without integrals at all. Their form is a motivation for the representation of the general solution of (Co1) we have in mind here. In the first part of Lemma 2.7 we will, similarly, present a class of solutions of ((Co1),(Co2)), which are free of integrals. This motivates the representation of the general solution of ((Co1),(Co2)) and will be applied in the proof of the form of the general solution. In this context it will be necessary and helpful to distinguish between the different types of iteration groups (p(s,x))_{s Î C}, and also to consider certain special cases of m and of (m,l), if iteration groups of the first type are used. In particular, we will investigate under which conditions the solutions can be expressed without integrals. Theorem 2.6 summarizes our results concerning (Co1), Theorem 2.8 the results concerning the system ((Co1),(Co2)). The above mentioned form of the general solutions will be useful in solving the boundary conditions.

Theorem 4

1. Let p(s,x) = e^lsx for l ¹ 0. Then a is a solution of (Co1) if and only if there exist m Î C and a formal power series E(x) = 1+e₁x+... Î C [[x]], such that

   
   

   a(s,x) = ems E(elsx) E(x) .

   The series E(x) is uniquely determined by a.

2. Let p(s,x) = x+c_ksx^k+... Î C [[x]] with c_k ¹ 0 and k ³ 2. If a(s,x) = e^ms(1+[^(a)]_k(s)x^k+...) º e^ms mod ord _x k, then a is a solution of (Co1) if and only if there exist m Î C and a series E(x) = 1+e₁x+... Î C [[x]], such that

   
   

   a(s,x) = ems E(p(s,x)) E(x) .

   The series E(x) is uniquely determined by a.

3. The general solution a of (Co1) for iteration groups (p(s,x))_{s Î C} of second type is

   
   

   a(s,x) =

   
   

   ems k-1 Õ n = 1  æ è exp ó õ s 0  p(s,x)nds ö ø kn   E(p(s,x)) E(x) ,

   with arbitrary k_n Î C.

Proof. In Lemma 2.1 we described solutions a of (Co1) which could be expressed without integrals. In Corollary 2.3 all solutions of this equation in integral form were determined. Combining these two results we investigate when

emsexp ó õ s 0  K(p(s,x))ds = ems E(p(s,x)) E(x)

(1)

holds, where E(x) º 1 mod ord 1. After applying the formal logarithm we have to check when

ó õ s 0  K(p(s,x))ds = ~ E   (p(s,x))- ~ E   (x)

is true, for [E\tilde](x): = lnE(x). Coefficientwise differentiation of the last equation with respect to the variable s yields

K(p(s,x)) = d ~ E   dy ê ê y = p(s,x)  p¢(s,x),

(2)

where we used the ``mixed'' chain rule for this derivation. If (p(s,x))_{s Î C} is an iteration group of the first type, this means

K(elsx) = d ~ E   dy ê ê y = elsx  lelsx .

In this formula e^lsx can be replaced by the indeterminate y, hence we get K(y) = ly[d[E\tilde](y)/dy]. Since ord K(y) ³ 1 and l ¹ 0, it is possible to divide by ly and we end up with a differential equation

K(y) ly = : ~ K   (y) = d ~ E   (y) dy .

(3)

Assume that [K\tilde](y) = å_{n ³ 0} [(k)\tilde]_nyⁿ, then the Ansatz [E\tilde](y) = å_{n ³ 1} [e\tilde]_nyⁿ leads to [e\tilde]_n+1 = [(k)\tilde]_n/(n+1) for all n ³ 0. Hence, all the coefficients e_n of E(x) for n ³ 1 and n = 0 are uniquely determined.

So far we proved that the series E is uniquely defined by K. Next we show that each solution [E\tilde] of the differential equation (3) with [E\tilde](0) = 0 leads to a solution E of (1) by setting E(y) = exp[E\tilde](y). Since [E\tilde] is a solution of the differential equation, it is clear that

K(elsx) = lelsx d ~ E   dy ê ê y = elsx  .

The right hand side of this equation is [(¶)/(¶s)][E\tilde](e^lsx), hence

ó õ s 0  K(elsx) ds = ~ E   (elsx) - ~ E   (el0 x)

and

emsexp ó õ s 0  K(elsx) ds =

emsexp( ~ E   (elsx) - ~ E   (x)) =

ems exp ~ E   (elsx) exp ~ E   (x) = ems E(elsx) E(x) .

Moreover, the coefficient e₀ of E(x) is equal to 1, since E(x) = exp[E\tilde](X) and ord [E\tilde](x) ³ 1, which finishes the proof for iteration groups p of the first type.

If (p(s,x))_{s Î C} is an analytic iteration group of the second type, then from iteration theory (cf. [Scheinberg, 1970] or [Reich & Schwaiger, 1977]) it follows that p¢(s,x) = H(p(s,x)), where H(y): = p¢(s,y)|_{s = 0} is the infinitesimal generator of p. In the present situation H(y) = c_ky^k+..., hence ord H(y) = k, and (2) means

K(p(s,x)) = d ~ E   dy ê ê y = p(s,x)  H(p(s,x)).

After replacing p(s,x) by the indeterminate y we realize that ord K(y) ³ k, since K(y) = H(y) [d[E\tilde](y)/dy]. (This however is equivalent to a(s,x) º e^ms mod ord _x k.) Hence we end up with the differential equation

K(y) H(y) = : ~ K   (y) = d ~ E   (y) dy ,

(4)

which, similar as in the first part of the proof, has exactly one solution [E\tilde](y) = å_{n ³ 1} [e\tilde]_nyⁿ.

Finally it remains to prove that each solution [E\tilde] of this differential equation with [E\tilde](0) = 0 yields a solution E of (1). Let [E\tilde] be a solution of (4) with [E\tilde](0) = 0, then

K(p(s,x)) = d ~ E   dy ê ê y = p(s,x)  H(p(s,x)) =

d ~ E   dy ê ê y = p(s,x)  p¢(s,x) = ¶ ¶s ~ E   (p(s,x))

and

ó õ s 0  K(p(s,x)) ds = ~ E   (p(s,x))- ~ E   (p(0,x)) =

~ E   (p(s,x))- ~ E   (x).

Substitution into the exponential series and multiplication by e^ms yields

emsexp ó õ s 0  K(p(s,x)) ds =

emsexp( ~ E   (p(s,x))- ~ E   (x)) = ems E(p(s,x)) E(x) .

Hence E(x) satisfies (1) and E(x) = exp[E\tilde](x) º 1 mod ord 1.

From Corollary 2.3 we deduce that the general solution a of (Co1) is given by

a(s,x) =

emsexp ó õ s 0  æ è k-1 å n = 1  knp(s,x)n+ å n ³ k  knp(s,x)n ö ø ds =

emsexp æ è k-1 å n = 1  kn ó õ s 0  p(s,x)nds ö ø ·

·exp ó õ s 0  ^ K   (p(s,x))ds

with [^K](y) = å_{n ³ k}k_nyⁿ. By Corollary 2.3

exp ó õ s 0  ^ K   (p(s,x))ds

is a solution of (Co1), and it is of the form 1+[^(a)]_k(s)x^k+..., since ord [^K](y) ³ k. Hence, by the second part of the present theorem there exists a unique series E(x) = 1+e₁x+..., such that

exp ó õ s 0  ^ K   (p(s,x))ds = E(p(s,x)) E(x) .

Summarizing, we found

a(s,x) =

ems k-1 Õ n = 1  æ è exp ó õ s 0  p(s,x)nds ö ø kn   E(p(s,x)) E(x) .

      ^[¯]

Lemma 5 Let E(x): = e₀+e₁x+... Î C [[x]], e₀ ¹ 0, and assume that F(x) Î C [[x]] and m Î C.

1. The series

   
   

   b(s,x): =

   
   

   emsE(p(s,x))[F(x)-e-msF(p(s,x))]

   together with a given in Lemma 2.1 satisfies (Co2) for any analytic iteration group p.

2. Assume that p(s,x) = x+c_ksx^k+... Î C [[x]] with k ³ 2 and c_k ¹ 0 is an analytic iteration group of the second type, and let P(s,x) denote the series

   
   

   P(s,x): = k-1 Õ n = 1  æ è exp ó õ s 0  p(s,x)nds ö ø kn   .

   Then b defined by

   
   

   b(s,x): =

   
   

   emsP(s,x)E(p(s,x))[F(x)-e-ms F(p(s,x)) P(s,x) ]

   together with a given in the third part of Theorem 2.6 satisfies (Co2).

Proof. The families a and b satisfy (Co2) if and only if

b(t+s,x)-b(t,p(s,x)) = b(s,x)a(t,p(s,x))

for all s,t Î C. If we express b and a by E, F, p, this is

em(t+s) E(p(t+s,x))[F(x)-e-m(t+s)F(p(t+s,x))]-

emt E(p(t,p(s,x))) é ë F(p(s,x))-

e-mtF(p(t,p(s,x))) ù û =

emsE(p(s,x))[F(x)-e-msF(p(s,x))]

emt E(p(t,p(s,x))) E(p(s,x)) .

Application of (T) together with simplification of both sides yields

em(t+s) E(p(t+s,x))F(x)

-emt E(p(t,p(s,x)))F(p(s,x)) =

emsF(x)emtE(p(t,p(s,x)))-

F(p(s,x))emt E(p(t,p(s,x))),

which is always true since p satisfies (T).

The proof of the second part is similar to the proof above, the reader only has to take into account that (P(s,x))_{s Î C} is a solution of (Co1).       ^[¯]

Theorem 5 Let a be a solution of (Co1).

1. Assume that p(s,x) = e^lsx for l ¹ 0 and that a is given as in the first part of Theorem 2.6.

   If m-nl ¹ 0 for all n Î N₀, then (a,b) is a solution of (Co2) if and only if there exists a formal power series F(x) Î C [[x]], such that

   
   

   b(s,x) = emsE(elsx)[F(x)-e-msF(elsx)].

   The series F(x) is uniquely determined by a and b.

   If m = n₀l for n₀ Î N₀, then (a,b) is a solution of (Co2) if and only if there exist a formal power series F(x) Î C [[x]] and l_n0 Î C, such that

   
   

   b(s,x) =

   
   

   emsE(elsx)[ln0sxn0+F(x)-e-msF(elsx)].

2. Let p(s,x) = x+c_ksx^k+... Î C [[x]] with c_k ¹ 0 and k ³ 2.

   Assume that a can be expressed as in the second part of Theorem 2.6. If m ¹ 0, then (a,b) is a solution of (Co2) if and only if there exists a formal power series F(x) Î C [[x]], such that

   
   

   b(s,x) =

   
   

   emsE(p(s,x))[F(x)-e-msF(p(s,x))].

   The series F(x) is uniquely determined by a and b.

   If m = 0, then (a,b) is a solution of (Co2) if and only if there exists a series F(x) Î C [[x]], such that

   
   

   b(s,x) = E(p(s,x))[F(x)-F(p(s,x))].

   Furthermore ord _xb(s,x) ³ k.

   Let a be the general solution of (Co1) given in the third part of Theorem 2.6, and assume that

   
   

   P(s,x): = k-1 Õ n = 1  æ è exp ó õ s 0  p(s,x)nds ö ø kn

   actually occurs. Then let n₀ be the minimum of { n | 1 £ n £ k-1,  k_n ¹ 0} . If m ¹ 0 or n₀ ¹ k-1 or k_k-1-m c_k ¹ 0 for all m Î N, then (a,b) is a solution of (Co2) if and only if there exists a formal power series F(x) Î C [[x]], such that

   
   

   b(s,x) =

   
   

   emsP(s,x)E(p(s,x))[F(x)-e-ms F(p(s,x)) P(s,x) ].

   The series F(x) is uniquely determined by a and b, and furthermore ord _xb(s,x) ³ n₀.

   If m = 0, n₀ = k-1 and k_k-1 = n₁ c_k for n₁ Î N, then (a,b) is a solution of (Co2) if and only if there exists a series F(x) Î C [[x]] and l_n1+n0¢¢ Î C, such that b(s,x) equals

   
   

   E(p(s,x))P(s,x)·[ F(x)-

   
   

   F(p(s,x)) P(s,x) +ln1+n0¢¢ ó õ s 0  p(s,x)n1+n0 P(s,x) ds ù ú û .

   Also in this situation ord _xb(s,x) ³ k-1.

Proof. We apply similar ideas and arguments as in the proof of Theorem 2.6. In Lemma 2.7 we described special solutions, in Theorem 2.5 all solutions of (Co2) in integral form. If a can be expressed without any integrals, then we check when

a(s,x) ó õ s 0  L(p(s,x)) a(s,x) ds = emsE(p(s,x))[F(x)-e-msF(p(s,x))]

(5)

holds. This is equivalent to

ó õ s 0  L(p(s,x)) a(s,x) ds = E(x)[F(x)-e-msF(p(s,x))].

Coefficientwise differentiation of the last equation with respect to the variable s yields

L(p(s ,x)) a(s,x) =

E(x)[me-msF(p(s,x))-e-ms d F d y ê ê y = p(s,x)  p¢(s,x)].

If p is an iteration group of the first type this means

L(elsx) =

a(s,x)E(x)[me-msF(elsx)-e-ms d F d y ê ê y = elsx  lelsx].

Using the special form of a from the first part of Theorem 2.6 and replacing e^lsx by the indeterminate y gives

~ L   (y): = L(y) E(y) = mF(y)-ly dF(y) dy .

(6)

Assume that [L\tilde](y) = å_{n ³ 0} l_nyⁿ, then the Ansatz F(y) = å_{n ³ 0} f_nyⁿ leads to

å n ³ 0  lnyn = å n ³ 0  (m-nl) fnyn.

If m-nl ¹ 0 for all n ³ 0, then F(y) is uniquely given by

fn = ln m-nl        "n ³ 0.

So far we proved that in this situation the series F is uniquely defined by L. Next we show that each solution F of the differential equation (6) is a solution of (5). Since F is a solution of (6) it is clear that

L(y) = E(y)[mF(y)-ly dF(y) dy ].

After replacing y by e^lsx and using the special form of a we derive that

L(elsx) a(s,x) =

E(x)[me-msF(elsx)-e-mslelsx dF dy ê ê y = elsx  ] =

E(x) ¶ ¶s (-e-msF(elsx)).

Coefficientwise integration finally yields the desired result.

Still we are dealing with analytic iteration groups (p(s,x))_{s Î C} of the first type. But now we assume that m = n₀l. In this situation comparing the coefficients of y^n₀ yields the condition 0 = (m-n₀l)f_n0 = l_n0. If l_n0 ¹ 0, then we split L(y) in the form

å [(n ³ 0) || (n ¹ n0)]  ln yn +ln0 yn0.

From Theorem 2.6 we know that b is given as

b(s,x) = a(s,x) ó õ s 0  L(elsx) a(elsx) ds =

ems E(elsx) E(x) ó õ s 0  L(elsx)E(x) emsE(elsx) ds =

emsE(elsx) ó õ s 0  e-ms å n ³ 0  lnenlsxnds =

emsE(elsx) æ è ó õ s 0  ln0ds  xn0 +

+ ó õ s 0  å [(n ³ 0) || (n ¹ n0)]  e(nl-m)s ln xn ds ö ø =

emsE(elsx)[ln0sxn0+F(x)-e-msF(elsx)].

For n ¹ n₀ the coefficients f_n of F(x) are uniquely given by

fn = ln m-nl ,

whereas f_n0 can be arbitrarily chosen in C.

In the next part of the proof we assume that (p(s,x))_{s Î C} is an iteration group of the second type. We investigate when (5) holds. For doing this we assume that a is given as in the second part of Theorem 2.6. Inserting the special form of a, coefficientwise differentiation with respect to s and expressing p¢(s,x) as H(p(s,x)), where H is the infinitesimal generator of p, yields the equation

L(p(s,x)) =

E(p(s,x))[mF(p(s,x))- dF dy ê ê y = p(s,x)  H(p(s,x))].

After replacing p(s,x) by y we end up with the differential equation

~ L   (y): = L(y) E(y) = mF(y)- dF(y) dy H(y).

(7)

Assume that [L\tilde](y) = å_{n ³ 0} l_nyⁿ and H(y) = å_{n ³ k}h_nyⁿ, where h_k = c_k ¹ 0, then the Ansatz F(y) = å_{n ³ 0} f_nyⁿ leads to

å n ³ 0  lnyn =

m k-1 å n = 0  fn yn+ å n ³ k  æ è mfn - å [(r+s = n) || (r ³ k)]  (r+1)fr+1hs ö ø yn.

Comparing coefficients yields

ln = mfn, n < k ln = mfn - n-k+1 å r = 1  r fr hn-r+1,    n ³ k.

If m ¹ 0, then F is uniquely determined by

fn = ln m , n < k fn = 1 m æ è ln+ n-k+1 å r = 1  r fr hn-r+1 ö ø ,     n ³ k.

Assuming conversely that F is a solution of (7), then it is left to the reader to prove that F satisfies (5). (The proof is similar to that given in the first part of this proof.)

If m = 0, then (7) reduces to

~ L   (y) = - dF(y) dy H(y)

(8)

or in more details

å n ³ 0  lnyn = - å n ³ k  æ è n-k+1 å r = 1  r fr hn-r+1 ö ø yn.

Comparing coefficients yields a necessary condition for the coefficients of [L\tilde], namely

ln = 0,        n < k

and a formula to determine recursively the values of f_n by

fn = - lk+n-1+ n-1 å r = 1  r fr hk+n-r n ck ,       n ³ 1,

since h_k = c_k. The value f₀ can be arbitrarily chosen in C. Hence ord [L\tilde](y) ³ k, which implies that ord L(y) ³ k and finally ord _x b(s,x) ³ k.

Assuming conversely that F is a solution of (8), then it is left to the reader to prove that F satisfies (5) for m = 0.

Finally, let a be the general solution of (Co1). Then

b(s,x) =

emsP(s,x) E(p(s,x)) E(x) ó õ s 0  L(p(s,x))E(x) emsP(s,x)E(p(s,x)) ds =

emsP(s,x)E(p(s,x)) ó õ s 0  e-ms ~ L   (p(s,x)) P(s,x) ds.

Now we have to check when

emsP(s,x)E(p(s,x)) ó õ s 0  e-ms ~ L   (p(s,x)) P(s,x) ds = emsP(s,x)E(p(s,x))[F(x)-e-ms F(p(s,x)) P(s,x) ]

(9)

holds. This is obviously equivalent to

ó õ s 0  e-ms ~ L   (p(s,x)) P(s,x) ds = F(x)-e-ms F(p(s,x)) P(s,x) .

Coefficientwise differentiation with respect to the variable s gives

e-ms ~ L   (p(s,x)) P(s,x) = - ¶ ¶s e-ms F(p(s,x)) P(s,x) =

me-ms F(p(s,x)) P(s,x) -e-ms æ ç è ¶ ¶s 1 P(s,x) ö ÷ ø F(p(s,x))-

e-ms 1 P(s,x) dF dy ê ê y = p(s,x)  H(p(s,x)),

where H is the infinitesimal generator of p. Since

¶ ¶s 1 P(s,x) = k-1 å n = 1  (-kn)p(s,x)n P(s,x)

we end up with the following differential equation after replacing p(s,x) by y,

~ L   (y) = æ è m+ k-1 å n = 1  kn yn ö ø F(y)- dF(y) dy H(y).

(10)

The usual Ansatz leads to

å n ³ 0  ln = m å n ³ 0  fn yn+ å n ³ 1  æ è min{ k-1,n} å r = 1  kr fn-r ö ø yn-

å n ³ k  æ è n-k+1 å r = 1  r fr hn-r+1 ö ø yn.

Hence the coefficients satisfy

l0 = mf0,

ln = mfn+ n å r = 1  kr fn-r

for 1 £ n < k, and

ln = mfn+ k-1 å r = 1  kr fn-r - n-k+1 å r = 1  r fr hn-r+1

for n ³ k. If m ¹ 0, then F is uniquely given by

f0 = l0 m ,

fn = 1 m æ è ln - n å r = 1  kr fn-r ö ø

for 1 £ n < k , and

fn = 1 m æ è ln - k-1 å r = 1  kr fn-r + n-k+1 å r = 1  r fr hn-r+1 ö ø

for n ³ k.

Again it is left to the reader to prove that each of these solutions F satisfies (9).

What happens in the case m = 0? If n₀ denotes min{ n | 1 £ n £ k-1,  k_n ¹ 0} , then (10) reduces to

~ L   (y) = æ è k-1 å n = n0  kn yn ö ø F(y)- dF(y) dy H(y).

Since the right hand side is a power series of order ³ n₀, the coefficients of [L\tilde](y) satisfy

ln = 0

for 0 £ n < n₀,

ln = n å r = n0  kr fn-r

for n₀ £ n < k and

ln = k-1 å r = n0  kr fn-r - n-k+1 å r = 1  r fr hn-r+1

for n ³ k.

Consequently ord [L\tilde](y) ³ n₀ and ord _xb(s,x) ³ n₀. Hence for n₀ £ n < k the coefficients f_n-n0 are uniquely determined by

fn-n0 = 1 kn0 æ è ln- n å r = n0+1  kr fn-r ö ø .

(11)

For n ³ k we still have to consider different cases. If n₀ < k-1, then n-n₀ > n-k+1 and f_n-n0 are uniquely given by the recursive formula

fn-n0 = 1 kn0 æ è ln- k-1 å r = n0+1  kr fn-r+ n-k+1 å r = 1  r fr hn-r+1 ö ø .

If n₀ = k-1, then for n ³ k

ln = kk-1fn-k+1- n-k+1 å r = 1  r frhn-r+1 = (kk-1- (n-k+1)hk)fn-k+1- n-k å r = 1  r fr hn-r+1.

(12)

Hence, if k_k-1-mc_k ¹ 0 for all m Î N, then

fn-n0 = fn-k+1 = ln + n-k å r = 1  r fr hn-r+1 kk-1- (n-k+1)ck

(13)

for n ³ k.

Finally we have to consider the case that m = 0, n₀ = k-1 and there exists n₁ Î N, such that k_k-1 = n₁c_k. Then (12) means

ln = (kk-1-(n-k+1)ck) fn-k+1 - n-k å r = 1  r fr hn-r+1

for n ³ k. If n-k+1 = n₁, which is equivalent to n = n₁+k-1, we have

ln1+k-1 = (kk-1-n1ck) fn1 - n1-1 å r = 1  r fr hn1+k-r =

- n1-1 å r = 1  r fr hn1+k-r.

This is a necessary condition for writing b(s,x) in the above form. In general let

ln1+k-1¢: = - n1-1 å r = 1  r fr hn1+k-r,

then

ó õ s 0  ~ L   (p(s,x)) P(s,x) ds = ó õ s 0  1 P(s,x) å n ³ k  lnp(s,x)nds =

ó õ s 0  æ ç ç ç è å [(n ³ k) || (n ¹ n1+k-1)]  lnp(s,x)n+ln1+k-1¢p(s,x)n1+k-1 P(s,x) +

(ln1+k-1-ln1+k-1¢)p(s,x)n1+k-1 P(s,x) ö ÷ ø ds =

F(x)- F(p(s,x)) P(s,x) +ln1+k-1¢¢ ó õ s 0  p(s,x)n1+k-1 P(s,x) ds,

where l_n1+k-1¢¢ = l_n1+k-1-l_n1+k-1¢. For n ¹ n₁+k-1 the coefficients f_n-k+1 of the series F(x) are uniquely given by the two formulae (11) and (13), and f_n1 can be arbitrarily chosen in C.

Also in the last cases it is left to the reader to prove that each solution F of the differential equation also satisfies (9) for m = 0.       ^[¯]

3  Solutions which satisfy the boundary conditions

a(x) = å n ³ 0  anxn,  a0 ¹ 0 and b(x) = å n ³ 0  bnxn.

From the results of the previous section it is obvious that (B1) is always satisfied. We only have to consider (B2) for further investigations.

First we will deal with analytic iteration groups of the first type, i.e. we consider p(s,x) = e^lsx for l ¹ 0. Before describing the solutions a which satisfy the boundary conditions we need a preliminary result. If J = J(l) denotes the set { n Î N | nl Î 2Zpi} , then the following lemma holds.

Lemma 6 Assume that J is not empty and let j₀ be the minimum of J. Then J = Nj₀.

Proof. Since j₀ Î J, there exists z₀ Î Z, such that j₀l = 2z₀pi. Then nj₀l = 2nz₀pi Î 2Zpi for all n Î N. Hence Nj₀ is a subset of J. Conversely, assume that n Î J, then by division we deduce that n = qj₀+r with uniquely determined r, such that 0 £ r < j₀. From 2Zpi ' nl = (qj₀+r)l = qj₀l+rl = 2qz₀pi+rl it follows that rl Î 2Zpi and consequently r = 0. Hence n Î Nj₀, which finishes the proof.       ^[¯]

In the first part of Theorem 2.6 the general solution a of (Co1) for analytic iteration groups (p(s,x))_{s Î C} of the first type was described. We want to analyze how to adjust it to the condition a(1,x) = a(x).

Theorem 6 Assume that a(x) is a given formal power series of order 0.

If J = Æ, then there exists exactly one formal power series E(x), such that

a(s,x) = ems E(elsx) E(x)

is a solution of (Co1) and satisfies a(1,x) = a(x).

If J ¹ Æ, then there exist formal power series E(x), such that a(s,x) of the above form satisfies both (Co1) and the boundary condition if and only if for n Î J the coefficients a_n satisfy

an = - n-1 å r = 1  aren-r,

where e_n are the coefficients of E(x).

Proof. Writing a as indicated above, the assumption a(1,x) = a(x) is equivalent to

em å n ³ 0  enenl xn = å n ³ 0  n å r = 0  aren-r.

Comparing coefficients yields for n = 0 that e^m = a₀, since e₀ = 1. Hence m is a logarithm lna₀. For n ³ 1 we get

emenlen = a0en+ n-1 å r = 1  aren-r+an,

which implies

a0(enl -1)en = n-1 å r = 1  aren-r+an.

For n Ï J the coefficient e_n is uniquely determined by

en = n å r = 1  aren-r a0(enl-1) .

However, for n Î J the coefficient e_n can be chosen arbitrarily in C and a_n must satisfy the condition above. We will not analyze these conditions further in this paper.       ^[¯]

Before we adjust b to the condition b(1,x) = b(x) we need another preliminary result. Let K = K(m,l) denote the set { n Î N₀ | m-nl Î 2Zpi} . Then the following lemma holds.

Lemma 7 Assume that the cardinality of K is greater than 1. Then J is not empty and

K = { k0+nj0 | n Î N} =

{ n Î N | n º k0 mod j0} ,

where k₀: = minK and j₀: = minJ. If | K| = 1, then J = Æ.

Proof. First we prove that when n₁ and n₂ are two different elements of K such that n₁ > n₂, then n₁-n₂ belongs to J. Since n₁,n₂ Î K, there exist z₁,z₂ Î Z, such that m-n₁l = 2z₁pi and m-n₂l = 2z₂pi. Then (n₁-n₂)l = (m-n₂l)-(m-n₁l) = 2(z₂-z₁)pi Î 2Zpi and n₁-n₂ Î N. Hence n₁-n₂ Î J.

Since k₀ Î K and j₀ Î J, there exist z₀,z₁ Î Z, such that m-k₀l = 2z₀pi and j₀l = 2z₁pi. Let n Î N₀, then m-(k₀+nj₀)l = m-k₀l-nj₀l = 2z₀pi-2nz₁pi = 2(z₀-nz₁)pi Î 2Zpi and consequently k₀+nj₀ Î K. Thus k₀+N₀j₀ Í K.

In the next step we prove that k₀ < j₀. (Then it is clear that k₀+N₀j₀ is the set of all positive integers congruent k₀ modulo j₀.) If we assume that k₀-j₀ ³ 0, then k₀-j₀ Î K since m- (k₀-j₀)l = 2(z₀+z₁)pi Î 2Zpi. Moreover k₀-j₀ < k₀ which is a contradiction to the construction of k₀.

Finally we have to prove that K Í k₀+N₀j₀. Let n Î K. If n ¹ k₀, then n > k₀ and then there exists z Î Z, such that m-nl = 2zpi. Moreover (n-k₀)l = 2(z₀-z)pi Î 2Zpi, thus n-k₀ Î J = Nj₀ by Lemma 3.1. Hence n Î k₀+Nj₀.

If | K| = 1, then J = Æ. If we assume that J ¹ Æ and k₀ Î K, then J = Nj₀. Hence k₀+J Ì K, which is a contradiction to | K| = 1.       ^[¯]

Let a be a solution of (Co1) where p is an analytic iteration group of the first type. The general form of b, which satisfies together with a the cocycle equation (Co2), was given in the first part of Theorem 2.8.

Theorem 7 Let p be an analytic iteration group of the first type. Assume that b(x) is a given formal power series and a is a solution of (Co1) given by

a(s,x) = ems E(elsx) E(x) .

1. If K = Æ, then there exists exactly one formal power series F(x), such that

   
   

   b(s,x) = emsE(elsx)[F(x)-e-msF(elsx)]

   together with a is a solution of (Co2) satisfying b(1,x) = b(x).

2. If K ¹ Æ and m-nl ¹ 0 for all n Î N₀, then there exist formal power series F(x), such that

   
   

   b(s,x) = emsE(elsx)[F(x)-e-msF(elsx)]

   together with a is a solution of (Co2) satisfying the boundary condition if and only if for n Î K the coefficients b_n satisfy

   
   

   bn = n-1 å r = 0  en-rfre(n-r)l(em-erl),

   (14)

   where e_n and f_n are the coefficients of E(x) and F(x).

3. If K ¹ Æ and m-n₀l = 0, then there exist formal power series F(x) and l_n0 Î C, such that

   
   

   b(s,x) =

   
   

   emsE(elsx)[ln0sxn0+F(x)-e-msF(elsx)]

   together with a is a solution of (Co2) satisfying the boundary condition if and only if for n Î K\{ n₀} the coefficients b_n satisfy (14) for n < n₀, and b_n equals

   
   

   n-1 å [r = 0 || (r ¹ n0)]  en-rfre(n-r)l(em-erl) +ln0en-n0enl

   for n > n₀.

Proof. Writing b in the form

b(s,x) = emsE(elsx)[F(x)-e-msF(elsx)],

and assuming that there is no n₀ Î N such that m = n₀l, then the assumption b(x) = b(1,x) is equivalent to

å n ³ 0  bnxn = em å n ³ 0  æ è n å r = 0  en-rfre(n-r)l ö ø xn- å n ³ 0  æ è n å r = 0  en-rfr ö ø enlxn,

which yields for all n ³ 0 that b_n equals