Nonlinear Analysis 72 (2010) 2852–2858
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Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
Generalized 1D Jörgens theorem Przemysław Górka Department of Mathematics and Information Sciences, Warsaw University of Technology, Pl. Politechniki 1, 00-661 Warsaw, Poland
article
abstract
info
Article history: Received 22 June 2009 Accepted 10 November 2009
We study the Cauchy problem for a 1D nonlinear wave equation on R. The nonlinearity can depend on the unknown function and its first order spatial derivative. Using the fixed point theorem we prove the existence of a classical solution. Moreover, the existence of periodic and almost periodic solutions are shown. © 2009 Elsevier Ltd. All rights reserved.
MSC: 35L05 35L15 Keywords: Wave equation Existence of solution Jörgens theorem
1. Introduction In this paper we will study a nonlinear equation of the form utt − uxx = f (t , x, u, ux ) on (0, T ) × R.
(1)
We shall deal with the Cauchy problem for the above equation on (0, T ) × R. It means to find a solution u of Eq. (1) such that u(0, x) = u0 (x),
ut (0, x) = u1 (x) on R.
(2)
The main goal of the present paper is showing the existence and uniqueness of the classical solution to problem (1), (2). Moreover, we shall show the existence of periodic and almost periodic solutions to the periodic and almost periodic data respectively. Our result is a generalization of the Jörgens theorem (see [1]). In fact, Jörgens has worked with the 3D wave equation, where the nonlinearity has depended only on u. The proof of our theorem relies on the application of the d’Alembert formula with Duhamel’s principle. What is more, we use fixed point theorem in a carefully chosen Banach space. 2. Main result We shall show that problem (1) poses a unique local in time classical solution. We distinguish three cases of initial conditions: arbitrary, periodic and almost periodic data. Before we present the main result we give the class of admissible nonlinearities. Definition 1. We shall say that the map f : R4 → R of C 2 -class is admissible if for every compact set K ⊂ R2 the condition sup x∈R2 ,y∈K
|f (x, y)| + |Df (x, y)| + |D2 f (x, y)| < ∞
holds. E-mail address:
[email protected]. 0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.11.028
P. Górka / Nonlinear Analysis 72 (2010) 2852–2858
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Moreover, let us fix k ∈ N. We shall denote by C k (R) the space of k-times differentiable functions such that the norm
kwkC k (R) :=
k X
sup w (n) (x)
n=0 x∈R
is finite. 2.1. Arbitrary data In this subsection we shall work with the problem (1), (2) on the real line. It means: utt − uxx = f (t , x, u, ux ) u(0, x) = u0 (x),
on (0, T ) × R,
ut (0, x) = u1 (x)
(3)
on R.
Now, we can present the main result of this subsection. Theorem 1. Let us assume that f is admissible and u0 ∈ C 2 (R), u1 ∈ C 1 (R). Then there exist T > 0 and a unique solution u ∈ C 2 ([0, T ) × R) of (3). Moreover, if T < ∞, then
max
sup
t ∈[0,T ),x∈R
|u(t , x)|,
sup
t ∈[0,T ),x∈R
|ux (t , x)| = ∞
and T is minimal with this property. Let us mention that the special case of the theorem has been shown in the paper [2]. Proof. First of all, for each T ∈ (0, 1] we introduce the following space: C˜ 2 ([0, T ) × R) = u ∈ C 1 ([0, T ] × R) : uxx , utx , uxt ∈ C 0 ([0, T ) × R),
kukC 1 ([0,T ]×R) , kuxx kC 0 ([0,T )×R) , kutx kC 0 ([0,T )×R) , kuxt kC 0 ([0,T )×R) < ∞ , endowed with the norm:
kukC˜ 2 ([0,T )×R) = kukC 1 ([0,T ]×R) + kuxx kC 0 ([0,T )×R) + kutx kC 0 ([0,T )×R) + kuxt kC 0 ([0,T )×R) . Next, we define the function u˜ 0 as follows u˜ 0 (t , x) =
1 2
(u0 (x + t ) + u0 (x − t )) +
1 2
Z
x +t
u1 (y) dy.
x −t
Subsequently, we define the set
n
o
XT = u ∈ C˜ 2 ([0, T ) × R) : u(0, x) = u0 (x), ut (0, x) = u1 (x), u − u˜ 0 C 1 ([0,T ]×R) ≤ 1, uxx − u˜ 0 xx C 0 ([0,T )×R) ≤ 1 .
Next, we define the map F :
F : XT −→ C 2 ([0, T ) × R), F (v) = u, as follows: for v ∈ XT we define u as unique solution to the linear problem: utt − uxx = f (t , x, v, vx ) u(0, x) = u0 (x),
on (0, T ) × R,
ut (0, x) = u1 (x)
(4)
on R.
The map F is well defined. Indeed, since v ∈ C˜ 2 ([0, T ) × R) we conclude that the function f (t , x, v, vx ) is C 1 and the problem (4) poses unique C 2 solution. Moreover, we can apply d’Alembert formula with the Duhamel’s principle (see [3]) and the map F can be written as follows:
F (v)(t , x) = u˜ 0 (t , x) +
1 2
Z tZ 0
x+t −τ
f (t , y, v, vy ) dy dτ . x−t +τ
Subsequently, let us introduce the notation: xba = x + b − a. Now, we will show that there exists T > 0 such that F : XT → XT . We have to estimate the quantity:
F (u) − u˜ 0
C 1 ([0,T ]×R)
= F (u) − u˜ 0 C 0 ([0,T ]×R) + ∂t F (u) − u˜ 0 C 0 ([0,T ]×R)
+ ∂x F (u) − u˜ 0 C 0 ([0,T ]×R) = I1 + I2 + I3 .
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P. Górka / Nonlinear Analysis 72 (2010) 2852–2858
Let us introduce the quantity:
n
C1 (˜u0 ) = sup |f (x)| : x ∈ R4 , x3 , x4 ∈ −A(˜u0 ), A(˜u0 )
2 o
,
where A(˜u0 ) = 1 + ku0 kC 1 (R) + ku1 kC 0 (R) . Hence, I1 ≤
1 2
T
Z
Z
xTτ
xτT
0
C1 (˜u0 ) dy dτ ≤ T 2 C1 (˜u0 ).
Keeping in mind, that:
∂t
0
∂x
xtτ
Z tZ
xτt
Z tZ 0
xtτ
xτ
! f τ , y, u, uy dy dτ
t
Z 0
! f (τ , y, u, uy ) dy dτ
t
Z
f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) − f (τ , xτt , u(τ , xτt ), uy (τ , xτt )) dτ ,
= 0
t
f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) + f (τ , xτt , u(τ , xτt ), uy (τ , xτt )) dτ ,
=
we obtain: I2 + I3 ≤ 2TC1 (˜u0 ). Next, we need to estimate the norm
F (u) − u˜ 0 . xx C ([0,T )×R) For this purpose we introduce the constants
n
C1i (˜u0 ) = sup ∂xi f (x) : x ∈ R4 , x3 , x4 ∈ −A(˜u0 ), A(˜u0 )
2 o
,
where i = 2, 3, 4. Hence, 1
Z
1
t
∂x 2 Z
2 ∂xx (F (u) − u˜ 0 )(t , x) =
=
2
t
f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) − f (τ , xτt , u(τ , xτt ), uy (τ , xτt )) dτ
0
0
∂x2 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) − ∂x2 f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))
+ ∂x3 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ ))uy (τ , xtτ ) − ∂x3 f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))uy (τ , xτt )
+ ∂x4 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ ))uyy (τ , xtτ ) − ∂x4 f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))uyy (τ , xτt ) dτ . Subsequently, we obtain
0
00
0
F (u) − u˜ 0
u
u
u k k ≤ T C + C 1 + + u + C 1 + + . 12 13 1 14 C R ( ) 0 0 1 C (R) xx C ([0,T )×R) C (R) C (R) Thus, we conclude that if T ≤ T˜ , where
s
T˜ = min 1 +
1 C1 (˜u0 )
1
− 1,
C12 + C13 A(u˜0 ) + C14 1 + u000 C (R) + u01 C (R)
,
then F : XT → XT . Our next goal is to show that the map F is a contraction on the set XT . For this purpose we take two elements u, v ∈ XT and compute the norm of the difference:
kF (u) − F (v)kC˜ 2 ([0,T )×R) = kF (u) − F (v)kC 0 ([0,T )×R) + k∂t (F (u) − F (v))kC 0 ([0,T )×R) + k∂x (F (u) − F (v))kC 0 ([0,T )×R) + k∂xx (F (u) − F (v))kC 0 ([0,T )×R) + k∂tx (F (u) − F (v))kC 0 ([0,T )×R) + k∂xt (F (u) − F (v))kC 0 ([0,T )×R) = II1 + II2 + II3 + II4 + II5 + II6 . Before we turn our attention to the estimating IIi , we introduce the constant:
n
2 o
C2 (˜u0 ) = sup max ∂x3 f (x) , ∂x4 f (x) : x ∈ R4 , x3 , x4 ∈ −A(˜u0 ), A(˜u0 )
.
P. Górka / Nonlinear Analysis 72 (2010) 2852–2858
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We are in position to estimate IIi ,
Z t Z xtτ II1 ≤ sup f (t , y, u, uy ) − f (t , y, v, vy ) dy dτ ≤ T 2 C2 (˜u0 ) ku − vkC 1 ([0,T ]×R) , (t ,x)∈[0,T )×R 2 0 xτt 1
II2 + II3 ≤ 2TC2 (˜u0 ) ku − vkC 1 ([0,T ]×R) . Let us introduce the constants:
n
C3 (˜u0 ) = sup max ∂x23 x2 f (x) , ∂x24 x4 f (x) : x ∈ R4 , x3 , x4 ∈ −A(˜u0 ), A(˜u0 )
2 o
, 2 o
C4 (˜u0 ) = sup max ∂x23 x3 f (x)x4 , ∂x24 x3 f (x)x4 + ∂x3 f (x) : x ∈ R4 , x3 , x4 ∈ −A(˜u0 ), A(˜u0 )
n
2 C5 (˜u0 ) = sup max ∂x23 x4 f (x)z , ∂x24 x4 f (x)z , ∂x4 f (x) : x ∈ R4 , x3 , x4 ∈ −A(˜u0 ), A(˜u0 ) , z ∈ −1 − +ku0 kC 1 (R) − ku1 kC 1 (R) , 1 + ku0 kC 1 (R) + ku1 kC 1 (R) . Now, we turn to the estimating the terms: II4 , II5 and II6 . 2 ∂xx (F (u) − F (v)) (t , x) =
1 2
∂x
t
Z 0
f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) − f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))
− f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ )) + f (τ , xτt , v(τ , xτt ), vy (τ , xτt )) dτ Z 1 t = ∂x2 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) − ∂x2 f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ )) 2
0
+ ∂x2 f (τ , xτt , v(τ , xτt ), vy (τ , xτt )) − ∂x2 f (τ , xτt , u(τ , xτt ), uy (τ , xτt )) + ∂x3 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ ))uy (τ , xtτ ) − ∂x3 f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ ))vy (τ , xtτ ) + ∂x3 f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))vy (τ , xτt ) − ∂x3 f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))uy (τ , xτt ) + ∂x4 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ ))uyy (τ , xtτ ) − ∂x4 f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ ))vyy (τ , xtτ )
+ ∂x4 f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))uyy (τ , xτt ) − ∂x4 f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))vyy (τ , xτt ) dτ . Hence, II4 = k∂xx (F (u) − F (v))kC 0 ([0,T )×R)
≤ TC3 (˜u0 ) ku − vkC 1 ([0,T ]×R) + TC4 (˜u0 ) ku − vkC 1 ([0,T ]×R) + TC5 (˜u0 ) ku − vkC˜ 2 ([0,T )×R) ≤ T C3 (˜u0 ) + C4 (˜u0 ) + C5 (˜u0 ) ku − vkC˜ 2 ([0,T )×R) . In the same manner we can estimate the term II5 . Namely,
∂tx2 (F (u) − F (v)) (t , x) =
1 2
∂t
t
Z 0
f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) − f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))
− f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ )) + f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))dτ Z 1 t = ∂x2 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) + ∂x3 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ ))uy (τ , xtτ ) 2
0
+ ∂x4 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ ))uyy (τ , xtτ ) + ∂x2 f (τ , xτt , u(τ , xτt ), uy (τ , xτt )) + ∂x3 f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))uy (τ , xτt ) + ∂x4 f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))vyy (τ , xτt ) − ∂x2 f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ )) − ∂x3 f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ ))
− ∂x4 f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ ))vyy (τ , xtτ ) − ∂x2 f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))
− ∂x3 f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))uy (τ , xτt ) − ∂x4 f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))vyy (τ , xτt ) dτ . Hence, II5 ≤ T C3 (˜u0 ) + C4 (˜u0 ) + C5 (˜u0 ) ku − vkC˜ 2 ([0,T )×R) .
Finally, we estimate II6 :
∂xt2 (F (u) − F (v)) (t , x) =
1 2
∂x
Z 0
t
f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) + f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))
− f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ )) − f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))dτ
,
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P. Górka / Nonlinear Analysis 72 (2010) 2852–2858
=
t
Z
1 2
0
∂x2 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ )) + ∂x3 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ ))uy (τ , xtτ )
+ ∂x4 f (τ , xtτ , u(τ , xtτ ), uy (τ , xtτ ))uyy (τ , xtτ ) + ∂x2 f (τ , xτt , u(τ , xτt ), uy (τ , xτt )) + ∂x3 f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))uy (τ , xτt ) + ∂x4 f (τ , xτt , u(τ , xτt ), uy (τ , xτt ))vyy (τ , xτt ) − ∂x2 f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ )) − ∂x3 f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ ))
− ∂x4 f (τ , xtτ , v(τ , xtτ ), vy (τ , xtτ ))vyy (τ , xtτ ) − ∂x2 f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))
− ∂x3 f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))uy (τ , xτt ) − ∂x4 f (τ , xτt , v(τ , xτt ), vy (τ , xτt ))vyy (τ , xτt ) dτ . Hence, II6 ≤ T C3 (˜u0 ) + C4 (˜u0 ) + C5 (˜u0 ) ku − vkC˜ 2 ([0,T )×R) .
Finally,
kF (u) − F (v)kC˜ 2 ([0,T )×R) ≤ T (2 + T )C2 (˜u0 ) + T (C3 (˜u0 ) + C4 (˜u0 ) + C5 (˜u0 )) ku − vkC˜ 2 ([0,T )×R) ≤ T 3C2 (˜u0 ) + C3 (˜u0 ) + C4 (˜u0 ) + C5 (˜u0 ) ku − vkC˜ 2 ([0,T )×R) . So, if we take T < min T˜ ,
!
1 2 3C2 (˜u0 ) + C3 (˜u0 ) + C4 (˜u0 ) + C5 (˜u0 )
,
then F : XT → XT has an unique fixed point. Hence we have obtained local solution. Next, the solution u may be extended beyond T , as a C 2 solution. Namely, we can apply the representation for u at the point (t , x), t > T , u(t , x) = u˜ 1 (t , x) +
1 2
xtτ
Z tZ
xτt
T
f (τ , y, u, uy ) dy dτ ,
where u˜ 1 (v)(t , x) = u˜ 0 (t , x) +
1
T
Z
2
Z
0
xtτ xτt
f (τ , y, u, uy ) dy dτ .
Hence, by iteration, we can extend u until a time T , when u or ux become unbounded. By Gronwall inequality one can show that this extension in unique. This finishes proof of the theorem. 2.2. Periodic solution In this section we shall turn our attention into periodic solution. Let us fix α > 0. We shall show existence of α -periodic solution (with respect to spatial variables) for α - periodic data and α -periodic f . Since S1 ≈ R/α Z we can look at our problem as the nonlinear wave equation on S1 . Theorem 2. Let us assume that f is admissible and α -periodic with respect to the spatial variable. If u0 ∈ C 2 (R), u1 ∈ C 1 (R) are α -periodic, then there exist T > 0 and a unique α - periodic solution u ∈ C 2 ([0, T ) × R) of (3). Moreover, if T < ∞, then
max
sup
t ∈[0,T ),x∈R
|u(t , x)|,
sup
t ∈[0,T ),x∈R
|ux (t , x)| = ∞
and T is minimal with this property. Proof. We start with the space of α - periodic functions. Namely, Cper ([0, T ) × R) = {u ∈ C ([0, T ) × R) : u(t , x + α) = u(t , x)} . Subsequently, we define the space 2 C˜ per ([0, T ) × R) = Cper ([0, T ) × R) ∩ C˜ 2 ([0, T ) × R) ,
where C˜ 2 was defined in the proof of the Theorem 1. Now, we can define the set per
XT
n 2 = u ∈ C˜ per ([0, T ) × R) : u(0, x) = u0 (x), ut (0, x) = u1 (x), o
u − u˜ 0 1
uxx − u˜ 0 xx 0 ≤ 1 , ≤ 1 C ([0,T ]×R) C ([0,T )×R)
P. Górka / Nonlinear Analysis 72 (2010) 2852–2858
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and the map Fper : per
−→ C 2 ([0, T ) × R), Z Z 1 t Fper (v)(t , x) = u˜ 0 (t , x) + Fper : XT
2
x+t −τ
f (t , y, v, vy ) dy dτ . x−t +τ
0
It can be checked that Fper (u) − u˜ 0 C 1 ([0,T ]×R) ≤ 1 and Fper (u)
1
Fper (v)(t , x + α) = u˜ 0 (t , x + α) +
= u˜ 0 (t , x) + = u˜ 0 (t , x) +
1
2
0
Z tZ
2
0
xx
− u˜ 0 xx C 0 ([0,T )×R) ≤ 1 for v ∈ XTper . Since
x+α+t −τ
f (τ , y, v, vy ) dy dτ
x−α−t +τ 0 x+t −τ
Z tZ
2 1
Z tZ
f (τ , y + α, v(y + α), vy (y + α)) dy dτ x−t +τ x+t −τ
f (τ , y, v(y), vy (y)) dy dτ , x−t +τ per
per
we obtain that Fper (v) is periodic. Hence, we obtain that Fper : XT −→ XT . per Finally, by the same calculations as in the proof of Theorem 1 we get that Fper is a contraction on the set XT .
2.3. Almost periodic solution In this section we shall show existence of the almost periodic solution to the problem (1), (2) for almost periodic data. First, we recall the notion of almost periodicity. Namely, a function f is almost periodic (see for example [4], [5,6] and [7]) when is continuous and for every sequence {αn }, the corresponding sequence f (x + αn ) contains a subsequence uniformly convergent. This definition is equivalent to the definition based on the condition that the set of almost periods of f is relatively dense in R. Now, we can formulate the result. Theorem 3. Let us assume that f is admissible and almost periodic with respect to the second variable. If u0 ∈ C 2 (R), u1 ∈ C 1 (R) are almost periodic, then there exist T > 0 and a unique solution u ∈ C 2 ([0, T ) × R) of (3) such that u is almost periodic with respect to the spatial variable. Moreover, if T < ∞, then
max
sup
t ∈[0,T ),x∈R
|u(t , x)|,
sup
t ∈[0,T ),x∈R
|ux (t , x)| = ∞
and T is minimal with this property. Proof. We introduce the space of almost periodic functions with respect to the spatial variable. Namely, Caper ([0, T ) × R) = {u ∈ C ([0, T ) × R) : u is almost periodic with respect to x} . Using the Cantor diagonal method one can show that the space Caper ([0, T ) × R) endowed with the sup norm is a Banach space. Next, we define the set 2 C˜ aper ([0, T ) × R) = Caper ([0, T ) × R) ∩ C˜ 2 ([0, T ) × R) ,
where C˜ 2 was defined in the proof of the Theorem 1. Subsequently, we define the set aper
XT
n 2 = u ∈ C˜ aper ([0, T ) × R) : u(0, x) = u0 (x), ut (0, x) = u1 (x), o
u − u˜ 0 1 ≤ 1, uxx − u˜ 0 xx 0 ≤1 . C ([0,T ]×R)
C ([0,T )×R)
Now, we can define the map Faper : aper
−→ C 2 ([0, T ) × R), Z Z 1 t Faper (v)(t , x) = u˜ 0 (t , x) + Faper : XT
2
0
x+t −τ
f (t , y, v, vy ) dy dτ . x−t +τ aper
In the same techniques as in the proof of the Theorem 1 we obtain for v ∈ XT and Faper (v)
xx
− u˜ 0 xx C 0 ([0,T )×R) ≤ 1.
the estimates Faper (v) − u˜ 0 C 1 ([0,T ]×R) ≤ 1
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P. Górka / Nonlinear Analysis 72 (2010) 2852–2858
Next, we shall show that Faper (v) is almost periodic with respect to x. For this purpose we take any sequence {αn } ⊂ R. Then,
Faper (v)(t , x + αn ) = u˜ 0 (t , x + αn ) +
= u˜ 0 (t , x + αn ) +
1 2 1 2
Z tZ 0
Z tZ 0
x+αn +t −τ
f (τ , y, v, vy ) dy dτ
x+αn −t +τ x+t −τ
f (τ , y + αn , v(y + αn ), vy (y + αn )) dy dτ .
(5)
x−t +τ
Since v ∈ Caper ([0, T )×R) and vxx ∈ C 0 ([0, T )×R) we obtain that vx ∈ Caper ([0, T )×R). Hence, keeping in mind assumptions on initial data and on f we obtain from formula (5) the existence of the subsequence of {αnk } such that Faper (v)(t , x + αnk ) aper converges uniformly. Finally, we get that Faper : XT −→ XTaper . aper Arguing in the same way as in the proof of Theorem 1 we have that Faper is a contraction on the set XT . This finishes the proof. Acknowledgements P.G. is very grateful to an anonymous referee for the constructive criticism and comments. This research has been partially supported by the project ‘‘Redes de Anillos R04’’, CONICYT and FONDECYT grant #3100019. References [1] [2] [3] [4]
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