on the plane R

POLONICI MATHEMATICI LVIII.1 (1993)

On homeomorphic and diffeomorphic solutions of the Abel equation on the plane

by Zbigniew Le´ sniak (Krak´ow)

Abstract. We consider the Abel equation ϕ[f (x)] = ϕ(x) + a

on the plane R², where f is a free mapping (i.e. f is an orientation preserving homeomor- phism of the plane onto itself with no fixed points). We find all its homeomorphic and diffeomorphic solutions ϕ having positive Jacobian. Moreover, we give some conditions which are equivalent to f being conjugate to a translation.

The aim of this paper is to find all homeomorphic and diffeomorphic solutions with positive Jacobian of the Abel equation

(1) ϕ[f (x)] = ϕ(x) + a

on the plane R

. We assume that a 6= (0, 0) and f is an orientation pre- serving homeomorphism of the plane onto itself with no fixed points (such a homeomorphism will be called a free mapping). By a curve is meant a homeomorphic image of a straight line. A curve is said to be a line (an open line in [4]) if it is a closed set.

1. We note that the existence of homeomorphic solutions ϕ of (1) is equivalent to f being conjugate to the translation T (x) = x + a (i.e. f = ϕ

◦ T ◦ ϕ, where ϕ is a homeomorphism). S. Andrea [1] has proved that a free mapping f is conjugate to a translation if and only if

(H) for all x, y ∈ R

there exists a curve segment C with endpoints x, y such that f

[C] → ∞ as n → ∓∞, where f

is the nth iterate of f . In the present paper we give some other conditions equivalent to (H).

We introduce the following conditions:

1991 Mathematics Subject Classification: Primary 39B10; Secondary 54H20, 26A18.

Key words and phrases: functional Abel equation, free mapping.

(A) There exists a homeomorphism ϕ of the plane onto itself satisfy- ing (1).

(A

) There exists a homeomorphism ϕ of the plane into itself satisfy- ing (1).

(B) There exists a line K such that

(2) K ∩ f [K] = ∅,

(3) U

∩ f [U

] = ∅,

(4) [

n∈Z

f

[U

] = R

,

where U

:= M

∪ f [K] and M

is the strip bounded by K and f [K]. (See Fig. 1.)

Fig. 1

Fig. 2

(C) There exist a family of curves {C

: α ∈ I} and a line K such that

(5) f [C

] = C

for α ∈ I,

(6) C

∩ C

= ∅ for α, β ∈ I, α 6= β , (7) card(K ∩ C

) = 1 for α ∈ I ,

(8) [

α∈I

C

= R

. (See Fig. 2.) We shall show

Theorem 1. If f is a free mapping, then conditions (A), (A

), (B) and (C) are equivalent.

2. First note the following

Lemma 1. Let a = (a

, a

) ∈ R

\ {(0, 0)} and T (x) := x + a for x ∈ R

. Then there exists a homeomorphism ψ of the plane onto itself such that (9) T (x) = ψ

[ψ(x) + (1, 0)] .

P r o o f. Set

ψ(x

, x

) =  1 a

x

, − a

a

x

+ x



if a

6= 0 , ψ(x

, x

) =  1

a

x

, x



if a

= 0 . By Lemma 1, from now on we may assume that a = (1, 0).

Lemma 2. If f is a free mapping, then (A

) implies (B).

P r o o f. Since ϕ is a homeomorphism, ϕ[R

] is a region. Moreover, ϕ[R

]

= ϕ[R

] + (1, 0). Put T (x

, x

) := (x

+ 1, x

) and T

(x

, x

) := (x

, x

) for (x

, x

) ∈ ϕ[R

]. Write L := {(x

, x

) ∈ ϕ[R

] : x

= 0}. Since T

[L] = {(x

, x

) ∈ ϕ[R

] : x

= n} for n ∈ Z, we have

(10) T

[L] ∩ T

[L] = ∅ for n, m ∈ Z, n 6= m .

Put L

:= T

[L], N

:= {(x

, x

) ∈ ϕ[R

] : x

∈ (n, n + 1)} and W

:=

N

∪ L

for n ∈ Z. Note that

N

= T

[N

] for n ∈ Z , W

∩ W

= ∅ for n ∈ Z \ {0}, [

n∈Z

W

= ϕ[R

] .

Put K := ϕ

[L] and K

:= f

[K] for n ∈ Z. By (A

), ϕ ◦ f

= T

◦ ϕ, whence

(11) f

◦ ϕ

= ϕ

◦ T

.

Hence

(12) K

= f

[ϕ

[L]] = ϕ

[T

[L]] = ϕ

[L

] .

Therefore, by (10), K

∩ K

= ∅ for n, m ∈ Z, n 6= m.

Since K = ϕ

[L] and L is closed in ϕ[R

], the curve K is a line, and so is K

= f

[K] for every n ∈ Z.

For each n ∈ Z, denote by M

the strip bounded by K

and K

. Since ϕ is a homeomorphism,

(13) M

= ϕ

[N

] for n ∈ Z . Hence by (11) we have

(14) f

[M

] = f

[ϕ

[N

]] = ϕ

[T

[N

]] = ϕ

[N

] = M

for n ∈ Z.

Put U

:= M

∪ K

for n ∈ Z. Then by (12) and (13), U

= ϕ

[N

∪ L

] = ϕ

[W

] and by (14),

U

= f

[M

∪ f [K]] = f

[U

] . Hence

[

n∈Z

f

[U

] = [

n∈Z

U

= ϕ

h [

n∈Z

W

i

= R

and

f [U

] ∩ U

= U

∩ U

= ϕ

[W

∩ W

] = ∅ . This completes the proof.

Theorem 2. Let f be a free mapping of the plane onto itself and let a = (a

, a

) ∈ R

\ {(0, 0)}. Assume that condition (B) is satisfied. Let ϕ

: U

∪ K → R

be continuous and suppose

(15) ϕ

[f (x)] = ϕ

(x) + a for x ∈ K . Then:

(a) There exists a unique solution ϕ of (1) such that (16) ϕ(x) = ϕ

(x) for x ∈ U

∪ K . This solution ϕ is continuous.

(b) If ϕ

is one-to-one and ϕ

[U

] ∩ (ϕ

[U

] + ka) = ∅ for all k ∈ Z \ {0}

then ϕ is a homeomorphism.

(c) If ϕ

is one-to-one, ϕ

[K] is a line and ϕ

[K]∩ D

6= ∅ for all γ ∈ R, where D

= {(x

, x

) ∈ R

: a

x

− a

x

= γ}, then ϕ is a homeomorphism.

(d) If ϕ

is one-to-one, ϕ

[K] is a line, ϕ

[K] ∩ D

6= ∅ for all γ ∈ R, and ϕ

[int U

] = N

, where N

is the strip bounded by ϕ

[K] and ϕ

[K]+a, then ϕ is a homeomorphism of R

onto itself. (See Fig. 3.)

P r o o f. Since K ∩ f [K] = ∅,

f

[K] ∩ f

[K] = ∅ for n ∈ Z .

Fig. 3

Moreover, for each n ∈ Z, the curve f

[K] is a line, since so is K. Denote by M

the strip bounded by f

[K] and f

[K]. Let U

:= M

∪ f

[K]

for n ∈ Z. Since f is a homeomorphism,

U

= f

[U

] for n ∈ Z .

Furthermore, for every n ∈ Z, f

[K] lies in the strip between f

[K] and f

[K], f

[K] ∩ M

= ∅ and f

[K] ∩ M

= ∅. Otherwise we would have f

[U

] ∩ f

[U

] 6= ∅, which contradicts (3).

Define ϕ : R

→ R

by setting

(17) ϕ(x) = ϕ

[f

(x)] + ka, x ∈ U

, k ∈ Z .

It is clear that ϕ is a unique solution of (1) satisfying (16) and that ϕ is continuous in S

n∈Z

int U

.

Take any x

∈ K. We now show that ϕ is continuous at x

. Let P be a closed disc with centre at x

such that P ∩ f

[K] 6= ∅ and P ∩ f [K] 6= ∅.

Then P ∩ f

[K] and P ∩ f [K] are compact. Let R be an open disc with centre at x

and radius smaller than min{̺(x

, P ∩f

[K]), ̺(x

, P ∩f [K])}, where ̺ is the Euclidean metric on the plane. Then we have

(18) R ∩ f

[K] = ∅ and R ∩ f [K] = ∅ .

Put R

:= R ∩ int U

, R

:= R ∩ int U

, R

:= R ∩ K. Then ϕ(x) =

ϕ

(x) for x ∈ R

, and ϕ(x) = ϕ

[f (x)] − a for x ∈ R

∪ R

. As x

∈ R

we

hence get ϕ(x

) = ϕ

(x

) by (15).

Let x

→ x

, x

∈ R. If x

∈ R

∪ R

, then

k→∞

lim ϕ(x

) = lim

k→∞

ϕ

(x

) = ϕ

(x

) = ϕ(x

) ,

because ϕ

is continuous in R

∪ R

⊂ K ∪ U

. If x

∈ R

, then f (x

) ∈ U

and f (x

) → f (x

) ∈ U

. Thus

k→∞

lim ϕ(x

) = lim

k→∞

(ϕ

[f (x

)] − a) = ϕ

[f (x

)] − a = ϕ

(x

) = ϕ(x

) , because ϕ

is continuous in U

and x

∈ K. Consequently, ϕ is continuous at x

∈ K.

Let x

∈ R

\ S

k∈Z

int U

. There exists an m ∈ Z such that x

∈ U

\ int U

= f

[K]. Let V be a neighbourhood of x

such that V ∩f

[K] = ∅ and V ∩f

[K] = ∅ (proceed as in the proof of the existence of R satisfying (18)). Note that V ⊂ U

∪ int U

, thus f

[V ] ⊂ U

∪ int U

and f

[V ] is a neighbourhood of f

(x

) ∈ K.

Take any sequence {x

} in V such that x

→ x

. Then f

(x

) → f

(x

). Since ϕ is continuous on K and f

(x

) ∈ K, we have ϕ[f

(x

)] → ϕ[f

(x

)]. From (17) we have

ϕ(x

) = ϕ

[f

(x

)] + (m + 1)a and

ϕ(x

) = ϕ

[f

(x

)] + (m + 1)a for k ∈ N . Thus ϕ(x

) → ϕ(x

). Consequently, ϕ is continuous on the plane.

Now assume, in addition, that ϕ is one-to-one in U

∪K and ϕ

(x)+na 6∈

ϕ

[U

] for all x ∈ U

and n ∈ Z \ {0}. We show that ϕ is one-to-one in the plane. Suppose x, y ∈ R

and ϕ(x) = ϕ(y). By (4), x ∈ U

and y ∈ U

for some k, l ∈ Z. From (17) it follows that

ϕ(x) = ϕ

[f

(x)] + ka, ϕ(y) = ϕ

[f

(y)] + la . Therefore

ϕ

[f

(x)] = ϕ

[f

(y)] + (l − k)a .

Suppose that l − k 6= 0. Then ϕ

[f

(y)] + (l − k)a 6∈ ϕ

[U

], since f

(y) ∈ U

. Hence ϕ

[f

(x)] 6∈ ϕ

[U

], which is a contradiction, since f

(x) ∈ U

. Thus l = k, and consequently x = y. Note that ϕ, being a continuous one- to-one mapping of R

into R

, is a homeomorphism (see [3], p. 186).

Now we show (c). Note that ϕ[K] + a = ϕ[f [K]]. Moreover, as D

∩ ϕ

[K] 6= ∅, we have D

∩ ϕ

[f [K]] 6= ∅. Since ϕ|

is continuous and one- to-one, it is a homeomorphism and ϕ[int U

] is a region. Moreover, ϕ(x) 6∈

ϕ[int U

], for every x ∈ K ∪f [K], since ϕ is one-to-one in U

∪K. Hence each

y ∈ ϕ[K]∪ϕ[f [K]] is a boundary point of ϕ[int U

], since ϕ

is continuous on

U

∪ K. Therefore ϕ[int U

] ⊂ N

, because ϕ

[K] ∩ D

6= ∅ and ϕ

[f [K]] ∩

D

6= ∅ for γ ∈ R and ϕ

[K] is a line.

Let x, y ∈ R

. Then x ∈ U

and y ∈ U

for some k, l ∈ Z. Assume that ϕ(x) = ϕ(y). Then

ϕ

[f

(x)] = ϕ

[f

(y)] + (l − k)a .

Since ϕ

[U

] ⊂ N

∪ ϕ

[f [K]], we have l − k = 0, whence x = y. Thus ϕ is continuous and one-to-one, and hence a homeomorphism.

Assume, in addition, that ϕ

[int U

] = N

. Put W

:= N

∪ (ϕ[K] + a).

Then W

= ϕ[U

]. Let y ∈ R

. Then there exists an n ∈ Z such that y − na ∈ W

. Take an x ∈ U

such that ϕ(x) = y − na. Then by (1),

ϕ[f

(x)] = ϕ(x) + na = y .

Thus ϕ[R

] = R

. Consequently, ϕ is a homeomorphism of R

onto itself satisfying (1). This completes the proof.

Obviously, we also have the following

R e m a r k 1. Let f be a free mapping of the plane onto itself and let a = (a

, a

) ∈ R

\ {(0, 0)}. Assume that condition (B) is satisfied. Let ϕ be any homeomorphic solution of equation (1). Let ϕ

:= ϕ|

. Then

(a) ϕ

is one-to-one and ϕ

[U

] ∩ (ϕ

[U

] + ka) = ∅ for k ∈ Z \ {0};

(b) if ϕ is a homeomorphism of R

onto itself, then ϕ

is one-to-one, ϕ

[K] is a line, ϕ

[K] ∩ D

6= ∅ for γ ∈ R, and ϕ

[int U

] = N

, where D

and N

are as in the statement of Theorem 2.

From part (d) of Theorem 2 we have

Corollary 1. If f is a free mapping, then (B) implies (A).

From Lemma 2, Corollary 1 and the fact that (A) implies (A

) we have Corollary 2. Let f be a free mapping. Then conditions (A), (A

) and (B) are equivalent.

Now we are going to prove

Lemma 3. Let f be a free mapping. Then (A) implies (C).

P r o o f. Put L := {(x

, x

) ∈ R

: x

= 0}, and D

:= {(x

, x

) ∈ R

: x

= α} for α ∈ R. Let K := ϕ

[L] and C

:= ϕ

[D

] for α ∈ R, where ϕ is a homeomorphism satisfying ϕ ◦ f = T ◦ ϕ with T (x

, x

) = (x

+ 1, x

) for x

, x

∈ R. Let I := R. Since f ◦ ϕ

= ϕ

◦ T and T [D

] = D

for α ∈ R, we have

f [C

] = ϕ

[T [D

]] = ϕ

[D

] = C

for α ∈ R . Moreover, note that

C

∩ C

= ϕ

[D

∩ D

] = ∅ for α, β ∈ R, α 6= β , [

α∈R

C

= ϕ

h [

α∈R

D

i

= ϕ

[R

] = R

and

card(K ∩ C

) = card ϕ

[L ∩ D

] = card(L ∩ D

) = 1 for α ∈ R. This completes the proof.

Theorem 3. Let f be a free mapping. Then condition (C) implies (B).

P r o o f. Suppose that (C) holds.

1. First, we show that for the line K which appears in (C), K ∩ f [K]=∅.

Suppose x

∈ K ∩ f [K]. On account of (8), x

∈ C

for some α ∈ I.

By (5) we get f

(x

) ∈ C

, and clearly f

(x

) ∈ K. Since card(K∩C

) = 1, x

= f

(x

). Hence x

is a fixed point of f , a contradiction.

2. Now we prove that

card(f

[K] ∩ C

) = 1 for α ∈ I and n ∈ Z . Fix any α ∈ I. Let n ∈ Z \ {0}. Take x

∈ K ∩ C

. By (5),

f

(x

) ∈ f

[K] ∩ C

.

Suppose there exist y

, y

∈ f

[K] ∩ C

such that y

6= y

. Then f

(y

), f

(y

) ∈ K ∩ C

and f

(y

) 6= f

(y

), which contradicts (7).

3. Let x ∈ C

. We now prove that, for every n ∈ Z, f

(x) lies between f

(x) and f

(x) on the curve C

. For any x, y ∈ C

denote by hx, yi the segment of C

with endpoints x, y. Let (x, y) := hx, yi \ {x, y}.

Let n ∈ Z. If f

(x) ∈ (f

(x), f

(x)) ⊂ C

, then

f (hf

(x), f

(x)i) = hf

(x), f

(x)i ⊂ hf

(x), f

(x)i . Hence by Brouwer’s Theorem f has a fixed point, which is impossible. Simi- larly, if f

(x) ∈ (f

(x), f

(x)) ⊂ C

, then

f

(hf

(x), f

(x)i) = hf

(x), f

(x)i ⊂ hf

(x), f

(x)i . Hence f

has a fixed point, contradiction again. Thus

(19) f

(x) ∈ (f

(x), f

(x)) .

4. Now we show that (3) holds. Since, f

[K] is a line for all n ∈ Z, R

\ f

[K] consists of two unbounded regions, called the side domains of f

[K]. Since K ∩ f [K] = ∅, we have f

[K] ∩ f

[K] = ∅ for all n ∈ Z. For each n ∈ Z, denote by M

the strip between the lines f

[K] and f

[K].

Let M

be the side domain of f

[K] which does not contain the line f

[K], and M

the side domain of f

[K] which does not contain f

[K].

Then

M

∪ f

[K] ∪ M

∪ f

[K] ∪ M

= R

for n ∈ Z .

Now we show that f

[K] ⊂ M

for n ∈ Z. Suppose otherwise. Then for some n ∈ Z,

(20) f

[K] ⊂ M

∪ f

[K] ∪ M

,

since f

[K] ∩ f

[K] = ∅. Take any x

∈ K. By (8), x

∈ C

for some α ∈ I. By (5), f

(x

) ∈ C

for all n ∈ Z. From (20) we obtain f

(x

) ∈ M

∪ f

[K] ∪ M

. Thus by (19),

C

⊂ M

∪ f

[K] ∪ M

∪ {f

(x

)} , since

f

(x

), f

(x

) ∈ M

∪ f

[K] ∪ M

, f

[K] ∩ C

= {f

(x

)}

and C

has no self-intersections. Consequently, we have shown that for each α ∈ I,

C

⊂ M

∪ f

[K] ∪ M

∪ f

[K] = R

\ M

,

which contradicts (8). Thus f

[K] ⊂ M

for all n ∈ Z. Hence M

, n ∈ Z, are mutually disjoint and f

[K] ∩ K = ∅ for n ∈ Z \ {0}. Since f is a homeomorphism, we have

(21) f

[M

] = M

for n ∈ Z .

Thus f

[M

] ∩ M

= ∅ for n ∈ Z \ {0}. Moreover, for every n ∈ Z \ {0}, f

[M

∪ f [K]] ∩ (M

∪ f [K]) = (f

[M

] ∪ f

[K]) ∩ (M

∪ f [K]) = ∅ . Thus, for all n ∈ Z \ {0}, f

[U

] ∩ U

= ∅, where U

= M

∪ f [K].

5. To complete the proof we show that [

n∈Z

f

[U

] = R

.

For each α ∈ I let K ∩ C

=: {x

} and C

= (x

, f (x

)) ⊂ C

. First, we show that S

α∈I

C

= M

.

Suppose that x

∈ C

and x

6∈ M

. Then C

has either a common point with K different from x

or a common point with f [K] different from f (x

), which is impossible.

For each α ∈ I denote by C

the set of all x ∈ C

such that f (x

) ∈ (x

, x) ⊂ C

, and by C

the set of all x ∈ C

such that x

∈ (x, f (x

))

⊂ C

.

Take any x

∈ M

. Then x

∈ C

for some α ∈ I. Suppose that x

∈ C

. Since card(C

∩ f [K]) = 1 and f (x

) ∈ C

∩ f [K], we have C

∩ f [K] = ∅. Hence C

is contained either in M

or in M

∪ K ∪ M

. Since f

(x

) ∈ C

∩ M

, we have C

⊂ M

, whence x

∈ M

, but this is impossible, since x

∈ M

.

Now suppose x

∈ C

. Since card(C

∩ K) = 1 and x

∈ C

∩ K, we have C

∩ K = ∅. Hence C

is contained either in M

or in M

∪ f [K] ∪ M

. Since f

(x

) ∈ C

∩ M

, we have C

⊂ M

, whence x

∈ M

, and this is also impossible. Consequently,

(22) [

α∈I

C

= M

.

For every α ∈ I and every n ∈ Z, let C

:= (f

(x

), f

(x

)) ⊂ C

. Since f is a homeomorphism, we have by (5),

(23) C

= f

[C

] for α ∈ I and n ∈ Z . Hence, for all n ∈ Z, we get by (21) and (22),

M

= f

[M

] = [

α∈I

f

[C

] = [

α∈I

C

.

Let x

∈ R

. If there exists an n ∈ Z such that x

∈ f

[K], then x

∈ f

[U

]. Now assume that x

∈ R

\ S

n∈Z

f

[K]. Then x

∈ C

for some α ∈ I. Since f

(x

) → ∞ as n → ∓∞ (see [1], Prop. 1.2), there is an n ∈ Z such that x

∈ C

. Hence by (22) and (23),

x

∈ f

[C

] ⊂ f

[M

] ⊂ f

[U

] . Consequently, R

= S

n∈Z

f

[U

]. This completes the proof.

Note that Theorem 1 is a consequence of Corollary 2, Lemma 3 and Theorem 3.

Moreover, from the proof of Theorem 3 we have the following

Corollary 3. Let f be a free mapping. Let K be a line on the plane. If K satisfies condition (C), then it also satisfies (B).

3. In this section we study diffeomorphic solutions of equation (1). First we quote the following

Lemma 4 (see [5]). If the functions f and ϕ are of class C

(p > 0) in a region U ⊂ R

such that f [U ] ⊂ U , then for x ∈ U ,

∂

∂x

. . . ∂x

ϕ[f (x)] =

q

X

k=1 n

X

j1,...,jk=1

b

(x)ϕ

[f (x)] ,

q = 1, . . . , p, where

ϕ

(x) = ∂

∂x

. . . ∂

ϕ(x) ,

b

(x) may be expressed by means of sums and products of a

(x), . . . . . . , a

q−k+1

(x), a

(x) =

i1...∂x_ik

f

(x) , k = 1, . . . , p, and f = (f

, . . . , f

). Consequently, b

are of class C

. In particular,

b

(x) = a

(x) · . . . · a

(x) .

Now let f be a free mapping. Assume that condition (B) is satisfied.

Definition (see [5]) . Let ψ be a continuous function defined in U

∪ K, p times continuously differentiable in int U

. We write

ψ

(x

) = lim

x→x0

x∈int U⁰

∂

∂x

. . . ∂x

ψ(x), k = 1, . . . , p ,

for x

∈ K ∪ f [K] (provided this limit exists). The function ψ is said to be of class C

in U

∪ K if all the functions ψ, ψ

, . . . , ψ

are continuous in U

∪ K.

All diffeomorphic solutions of equation (1) having positive Jacobian can be obtained from the following

Theorem 4. Let f be a free C

mapping of the plane having positive Jacobian at every x ∈ R

and let a = (a

, a

) ∈ R

\ {(0, 0)}. Assume that condition (B) is satisfied. Let ψ be a C

function from U

∪K into the plane which satisfies

ψ[f (x)] = ψ(x) + a for x ∈ K ,

q

X

k=1 2

X

j1,...,jk=1

b

(x)ψ

[f (x)] = ψ

(x)

for x ∈ K, q = 1, . . . , p, i

, . . . , i

= 1, 2, where the functions b

are those occurring in Lemma 4. Then there exists a unique solution ϕ of equation (1) such that

ϕ(x) = ψ(x) for x ∈ U

∪ K .

This solution is of class C

in the plane. Moreover, if ψ is one-to-one, the Jacobian, jac

ψ, is positive for x ∈ int U

, and det[ψ

(x), ψ

(x)] > 0 for x ∈ K ∪ f [K], and either

ψ[U

] ∩ (ψ[U

] + ka) = ∅ for k ∈ Z \ {0}

or

ψ[K] ∩ D

6= ∅ for γ ∈ R and ψ[K] is a line,

where D

= {(x

, x

) ∈ R

: a

x

− a

x

= γ}, then ϕ is an orientation preserving diffeomorphism of class C

having positive Jacobian.

P r o o f. Define ϕ by setting

(24) ϕ(x) = ψ[f

(x)] + ka, x ∈ U

, k ∈ Z,

where U

= f

[U

]. For p = 0 we have Theorem 1. Let p > 0. From (24) it follows that ϕ is of class C

in S

k∈Z

int U

.

Let x

∈ K. Then there exists an open disc R with centre at x

such

that R ∩ f

[K] = ∅ and R ∩ f [K] = ∅ (see the proof of Theorem 2). The

proof of ϕ being C

in R runs in the same way as that of Theorem 3.1 in

[5], part 2.

Let x

∈ R

\ S

k∈Z

int U

. There is an m ∈ Z such that f

(x

) ∈ K.

We have already proved that ϕ is C

in a neighbourhood R of f

(x

).

The function f

is a C

map of R onto a neighbourhood f

[R] of x

. Since ϕ is a solution of (1), we have

ϕ(x

) = ϕ[f

(x

)] + (m + 1)a . Hence ϕ is C

in f

[R].

Now assume, in addition, that ψ is one-to-one, ψ(x) + ka 6∈ ψ[U

] for x ∈ U

and k ∈ Z \ {0} [or ψ[K] ∩ D

6= ∅ for all γ ∈ R and ψ[K] is a line], jac

ψ > 0 for x ∈ int U

and det[ψ

(x), ψ

(x)] > 0 for x ∈ K ∪ f [K]. On account of Theorem 2, ϕ is a homeomorphism.

If x ∈ U

, then jac

ϕ = jac

ψ > 0. If x ∈ f [K], then jac

ϕ = det[ψ

(x), ψ

(x)] > 0, since (∂ϕ/∂x

)(x) = ψ

(x) and (∂ϕ/∂x

)(x) = ψ

(x) for x ∈ f [K]. Thus jac

ϕ > 0 for x ∈ U

.

Let x ∈ R

. Then x ∈ f

[U

] for some n ∈ Z. Since ϕ(x) = ϕ[f

(x)] + na, we have

jac

ϕ = jac

ϕ · jac

f

.

Hence jac

ϕ > 0, since f

(x) ∈ U

and jac

f

> 0. Thus ϕ preserves orientation. Since ϕ is invertible and of class C

, and jac

ϕ 6= 0 for x ∈ R

, ϕ

is C

(see e.g. [6], p. 205). This completes the proof.

References

[1] S. A. A n d r e a, On homeomorphisms of the plane which have no fixed points, Abh.

Math. Sem. Hamburg 30 (1967), 61–74.

[2] —, The plane is not compactly generated by a free mapping, Trans. Amer. Math.

Soc. 151 (1970), 481–498.

[3] R. E n g e l k i n g and K. S i e k l u c k i, Topology. A Geometric Approach, Sigma Ser.

Pure Math. 4, Heldermann, Berlin 1992.

[4] T. H o m m a and H. T e r a s a k a, On the structure of the plane translation of Brouwer , Osaka Math. J. 5 (1953), 233–266.

[5] M. K u c z m a, On the Schr¨oder equation, Rozprawy Mat. 34 (1963).

[6] R. S i k o r s k i, Advanced Calculus. Functions of Several Variables, Monograf. Mat.

52, PWN, Warszawa 1969.

[7] M. C. Z d u n, On continuous iteration groups of fixed-point free mapping in R²space, in: Proc. European Conference on Iteration Theory, Batschuns 1989, World Scientific, Singapore 1991, 362–368.

INSTITUTE OF MATHEMATICS

PEDAGOGICAL UNIVERSITY OF CRACOW PODCHORA¸ ˙ZYCH 2

30-084 KRAK ´OW, POLAND

Re¸cu par la R´edaction le 1.8.1990 R´evis´e le 16.3.1992 et 10.5.1992