c
Applied Mathematics E-Notes, 10(2010), 285-293 Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
A Study Of Boundary Value Problem For An Elliptic
Equation In Hölder Spaces∗
Tarik Berroug†
Received 10 January 2010
Abstract
We give in this work some new results about the existence, uniqueness and
optimal regularity for the strict solution of an abstract second-order differential
equation set in an unbounded interval. We use similar techniques with those of
Labbas [9], when the right-hand term is Hölder continuous function.
1
Introduction
The aim of this paper is to study the following second order abstract differential equation
u00 (t) + Au(t) = f(t), t ∈ (0, +∞),
(1)
under the non-homogeneous boundary conditions
u(0) = ϕ, u(+∞) = 0,
(2)
where A is a closed linear operator with dense domain D(A) in a complex Banach space
E and ϕ is a given element of D(A). The vector-valued function f is continuous on
[0, +∞[ into E and verifies
lim kf(t)kE = 0.
(3)
t→+∞
Throughout this work we assume that there exists K > 0 such that for all λ > 0,
k(A − λI)−1 kL(E) 6
K
.
1+λ
(4)
We recall that for m ∈ N, BU C m ([0, +∞[; E) denotes the space of vector-valued
functions with uniformly continuous and bounded derivatives up to the order m in
[0, +∞[.
For σ ∈]0, 1[, the Banach space C σ ([0, +∞[; E) denotes the space of the bounded
and σ-Hölder continuous functions f : [0, +∞[−→ E, such that
supt∈[0,+∞[ kf(t)kE < ∞,
∃C > 0 : ∀t, τ ∈ [0, +∞[, kf(t) − f(τ )kE 6 C|t − τ |σ ,
∗ Mathematics
Subject Classifications: 12H20, 35J25, 47D06.
of Mathematics and Applications, Université Sultan Moulay Slimane, B.P. 523, BéniMellal, Morocco
† Laboratory
285
286
BVP for an Elliptic Equation
endowed with the norm
kfkC σ ([0,+∞[;E) =
sup
kf(t)kE + sup
t∈[0,+∞[
t6=τ
kf(t) − f(τ )kE
= kfk∞ + [f]C σ ([0,+∞[;E) .
|t − τ |σ
For simplicity, we shall write C σ (E) instead of C σ ([0, +∞[; E).
We say that u ∈ BU C([0, +∞[; E) is a strict solution of (1)-(2) if
u ∈ BU C 2 ([0, +∞[; E) ∩ BU C([0, +∞[; D(A)),
and u satisfies (1) and (2).
Observe that equation (1) has been studied by many authors, in different situations,
but on a bounded domain. See, for example Krein [7], Sobolevskii [13], Kuyazyuk
[8], Da Prato-Grisvard [3], Labbas [9], Favini-Labbas-Lemrabet-Sadallah [4], FaviniLabbas-Tanabe-Yagi [5].
In the present study, the principal goal is to give an alternative approach to that
used in Berroug-Labbas-Sadallah [2]. The techniques we use are essentially based on the
theory of fractional powers of linear operators in Banach spaces and on the semigroups
estimates generated by them as in Krein [7] and in Sinestrari [12]. We make use of the
real Banach interpolation space DA (θ, +∞), between D(A) and E. It is characterized
in [6], by
DA (θ, +∞) = {ξ ∈ E : sup ktθ A(A − tI)−1 ξkE < ∞}.
t>0
We will prove the following main results.
THEOREM 1 (Existence and uniqueness). Let 0 < θ < 1/2, ϕ ∈ D(A) and
f ∈ C 2θ(E) with assumption (3). Then problem (1)-(2) admits a unique strict solution.
THEOREM 2 (Regularity). Let 0 < θ < 1/2, ϕ ∈ D(A) and f ∈ C 2θ (E) with
assumption (3). If f(0) − Aϕ ∈ DA (θ, +∞), then the unique strict solution of (1)-(2)
satisfies the property of maximal regularity Au(.), u00 (.) ∈ C 2θ (E).
2
Proof of Theorem 1
We start by some recall of the theory of fractional powers of linear operators as developed in Balakrishnan [1], Krein [7] and Pazy [11]. It is well known that assumption (4)
implies that (−(−A)1/2 ) is the infinitesimal generator of an analytic semigroup {V (t)},
t > 0 (for details, see [1]). Moreover we have the practical well known results
PROPOSITION 1.
(1) ∃M, δ > 0 such that kV (t)kL(E) 6 M e−δt ,
(2) there exists C > 0 such that for t > 0, k(−A)1/2 V (t)kL(E) 6 Ct−1 e−δt .
PROPOSITION
2. For all x ∈ E we have
Rt
(1) 0 V (s)xds = (−A)−1/2 (x − V (t)x),
R +∞
(2) t
V (s)xds = (−A)−1/2 V (t)x.
287
T. Berroug
Let 0 < θ < 1/2 and f ∈ X = C 2θ (E) with assumption (3). First, we seek for a
particular solution v(.) to equation (1). Let us set for t ∈ [0, +∞[
1
v(t) = −
2
Z
t
−1/2
V (t − s)(−A)
0
1
f(s)ds −
2
Z
+∞
V (s − t)(−A)−1/2 f(s)ds.
t
Notice that the second integral is convergent. Indeed, Proposition 1 implies
Z +∞
Z +∞
−1/2
0
V (s − t)(−A)
f(s)ds 6 M
e−δ(s−t)dskfkX .
t
t
E
We can see that the derivative v0 (t) exists and
v0 (t) =
1
2
Z
t
1
2
V (t − s)f(s)ds −
0
Z
+∞
V (s − t)f(s)ds.
t
To show that v(t) ∈ D(A) and Av(.) is continuous we write (thanks to Proposition
2)
Av(t)
=
=
Z t
1
−1/2
− A(−A)
V (t − s)(f(s) − f(t))ds
2
0
Z t
1
−1/2
V (t − s)ds f(t)
A(−A)
−
2
0
Z +∞
1
+ A(−A)−1/2
V (s − t)(f(t) − f(s))ds
2
t
Z +∞
1
−1/2
−
A(−A)
V (s − t)ds f(t)
2
t
1
1
(U (t) + S(t)) + f(t) − V (t)f(t).
2
2
where
U (t) =
Z
0
S(t) =
Z
+∞
t
t
∂V
(t − s)(f(s) − f(t))ds =
∂s
∂V
(s − t)(f(t) − f(s))ds =
∂s
Z
t
(−A)1/2 V (t − s)(f(s) − f(t))ds,
0
Z
+∞
−(−A)1/2 V (s − t)(f(t) − f(s))ds.
t
Furthermore, as f is Hölder-continuous, v0 (.) is differentiable with
00
v (t)
=
=
Z t
1/2
1
1/2
f(t) − (−A)
e−(t−s)(−A) f(s)ds
2
0
Z +∞
1/2
1
− (−A)1/2
e(t−s)(−A) f(s)ds
2
t
Z t
1/2
1
e−(t−s)(−A) (f(s) − f(t))ds
f(t) − (−A)1/2
2
0
288
BVP for an Elliptic Equation
1
−
2
=
=
Z t
1/2
−(t−s)(−A)1/2
(−A)
e
ds f(t)
0
Z +∞
1/2
1
e(t−s)(−A) (f(s) − f(t))ds
− (−A)1/2
2
t
Z +∞
1/2
1
(−A)1/2
e(t−s)(−A) ds f(t)
−
2
t
1
1
1
1
f(t) − U (t) − (I − V (t))f(t) − S(t) − f(t)
2
2
2
2
1
1
1
− U (t) − S(t) + V (t)f(t),
2
2
2
so
v00 (t) + Av(t) = f(t).
Hence v is a strict solution to (1) satisfying the boundary conditions
Z
1 +∞
V (s)(−A)−1/2 f(s)ds, v(+∞) = 0,
v(0) = −
2 0
for the last condition we use the estimate
Z
t
−1/2
V (t − s)(−A)
f(s)ds
6
t/2
E
6
M max
kf(r)kE
t
r∈[ 2 ,t]
Z
t
−δ(t−s)
e
t/2
!
ds
M
− δ2 t
max
1
−
e
kf(r)k
.
E
δ r∈[ 2t ,t]
On the other hand we have
(−A)v(0)
=
=
Z +∞
Z +∞
1
1
1/2
1/2
− (−A)
V (s)(f(s) − f(0))ds − (−A)
V (s)f(0)ds
2
2
0
0
Z
1
1 +∞ ∂V
(s)(f(s) − f(0))ds − f(0),
2 0
∂s
2
from Proposition 1, we conclude since f is Hölder-continuous, that v(0) ∈ D(A).
We will also use the following lemma
LEMMA 1. Assume (4) and let ξ ∈ D(A). Then the homogeneous Problem
00
u (t) + Au(t) = 0, t ∈ [0, +∞[,
u(0) = ξ, u(+∞) = 0,
admits a unique strict solution u(.).
PROOF. Let us set u(t) = V (t)ξ. Since ξ ∈ D(A) we can easily see that
u0 (t) = −V (t)(−A)1/2 ξ,
and
u00 (t) = V (t)(−A)ξ,
(5)
289
T. Berroug
then
u00 (t) = (−A)u(t),
u(0) = ξ, u(+∞) = 0.
Let us return to the proof of Theorem 1. The Problem
 00
 u (t) + Au(t) = 0, t ∈ [0, +∞[,
u(0) = x0 ,

u(+∞) = 0,
with
x0 = ϕ − v(0),
admits a unique strict solution u. Indeed, we know that v(0) ∈ D(A) and ϕ ∈ D(A),
thus Lemma 1 applies. Therefore,
u(.) = v(.) + u(.),
is the unique strict solution to problem (1)-(2).
3
Proof of Theorem 2
Let 0 < θ < 1/2, ϕ ∈ D(A) and f ∈ X = C 2θ (E) with assumption (3). Let us suppose
that f(0) − Aϕ ∈ DA (θ, +∞). It is enough to do it for Au(.), for this purpose we write
Au(t) =
=
=
Av(t) + V (t)(Aϕ − Av(0))
1
1
(U (t) + S(t)) + f(t) − V (t)f(t)
2
2
Z
1
1 +∞ ∂V
(s)(f(s) − f(0))ds − f(0)
+V (t) Aϕ +
2 0
∂s
2
1
(U (t) + S(t)) + f(t) + K(t),
2
where

Rt
1/2
U (t) = 0 (−A)1/2 e−(−A) (t−s)(f(s) − f(t))ds


R +∞

1/2 −(−A)1/2 (s−t)
S(t) = t
−(−A)
e
(f(t) − f(s))ds
R

1
∂V
1
1
+∞
 K(t) = V (t) Aϕ +

(s)(f(s) − f(0))ds − f(0) − V (t)f(t).
2 0
∂s
2
2
Let us show the holderianity of U (.), S(.) and K(.). For 0 6 r < t, we get
Z t
1/2
U (t) − U (r) =
(−A)1/2 e−(−A) (t−s)(f(s) − f(t))ds
r
Z r
1/2
+
(−A)1/2 e−(−A) (t−s)(f(s) − f(t))ds
Z0 r
1/2
−
(−A)1/2 e−(−A) (r−s) (f(s) − f(r))ds
0
290
BVP for an Elliptic Equation
= a + b − c.
We have
kakE 6 C
Z
r
t
(t − s)2θ
dskfkX 6 C(t − r)2θ kfkX ,
(t − s)
on the other hand we can see that
Z r
1/2
1/2
b−c =
(−A)1/2 e−(−A) (t−s) − e−(−A) (r−s) (f(s) − f(r))ds
0
Z r
1/2
+
(−A)1/2 e−(−A) (t−s)(f(r) − f(t))ds
0
Z r Z t−s
1/2
=
−((−A)1/2 )2 e−(−A) σ (f(s) − f(r))dσds
0
r−s
Z r
1/2
+
(−A)1/2 e−(−A) (t−s)(f(r) − f(t))ds
0
Z r Z t−s
1/2
=
−((−A)1/2 )2 e−(−A) σ (f(s) − f(r))dσds
0
r−s
i
h
1/2
−(−A)1/2 t
+ e
− e−(−A) (t−r) (f(t) − f(r))
= b 1 + c1 ,
and
kb1 kE
Z
6
r
0
C
6
Z
Z
t−s r−s
r
C
6
E
(r − s)2θ
0
Z
1/2
−((−A)1/2 )2 e−(−A) σ (f(s) − f(r)) dσds
r
0
Z
t−s
r−s
2θ−1
1
dσdskfkX
σ2
(r − s)
(t − r)
dskfkX .
(t − r + r − s)
Now, by making the change of variable (r − s) = (t − r)ξ, it follows
Z
0
r
(r − s)2θ−1 (t − r)
ds 6 (t − r)2θ
(t − r + r − s)
+∞
Z
0
ξ 2θ−1
dξ 6 C(t − r)2θ .
1+ξ
Holderianity of c1 is obvious.
For S(.), one has
S(r) − S(t)
=
Z
+∞
−(−A)1/2 e−(−A)
1/2
(s−t)
(f(s) − f(t))ds
t
+
Z
t
(−A)1/2 e−(−A)
1/2
(s−r)
(f(s) − f(r))ds
r
+
Z
t
+∞
(−A)1/2 e−(−A)
1/2
(s−r)
(f(s) − f(r))ds
291
T. Berroug
=
Z
t
1/2
(−A)1/2 e−(−A) (s−r) (f(s) − f(r))ds
r
Z +∞
1/2
+
(−A)1/2 e−(−A) (s−r) (f(t) − f(r))ds
t
+
+∞
Z
t
=
1/2
1/2
(−A)1/2 e−(−A) (s−r) − e−(−A) (s−t) (f(s) − f(t))ds
ã + b̃ + c̃,
thus
kãkE 6 C
Z
t
r
(s − r)2θ
dskfkX 6 C(t − r)2θ kfkX .
(s − r)
It is easy to check the result for b̃. Finally
Z +∞ Z s−r h
i2
1/2
−(−A)1/2 σ
kc̃kE = − (−A)
e
(f(s) − f(t))dσds
t
6 C
s−t
+∞
Z
(s − t)2θ
t
6 C
Z
Z
s−r
s−t
+∞
(s − t)2θ−1
t
E
dσ
dskfkX
σ2
(t − r)
dskfkX ,
(s − t + t − r)
setting (s − t) = ξ(t − r) in this last inequality we obtain
Z +∞ 2θ−1
ξ
(t − r)2θ
kc̃kE 6 C
dξkfkX 6 C(t − r)2θ kfkX .
(1 + ξ)
0
For K(.), we note that
K(t) − K(r)
=
=
Z
1 +∞ ∂V
(V (t) − V (r)) Aϕ − f(0) +
(s)(f(s) − f(0))ds
2 0
∂s
1
1
− (V (t) − V (r)) (f(r) − f(0)) − V (t)(f(t) − f(r))
2
2
k1 + k2 + k3 ,
we then have the estimate
kk2 kE
6
6
6
Z t
1/2
(−A)1/2 e−(−A) s (f(r) − f(0)) ds
E
r
Z t
C
s−1 r 2θ dskfkX
r
Z t
C
s2θ−1 dskfkX
C
r
6
moreover
C(t − r)2θ kfkX ,
2θ
kk3 kE 6 C (t − r)
kfkX .
292
BVP for an Elliptic Equation
Now, for k1 we use the following result proved in Sinestrari [12]
LEMMA 2. Setting for x ∈ E and t > 0
v(t) = V (t)x = e−(−A)
1/2
t
x,
if x ∈ D(−A)1/2 (2θ, +∞) then v ∈ C 2θ (E).
Thanks to the reiteration theorem in interpolation theory (see [10]) we have the
equality
D(−A)1/2 (2θ, +∞) = DA (θ, +∞).
(6)
Therefore, it suffices to show that (see Sinestrari [12, p.24])
Z +∞ 1−2θ 1/2
1/2
V (r)
−(−A)
V (s)(f(s) − f(0))ds
sup r
−(−A)
6 K.
r>0
0
E
Let r > 0, we have
Z +∞ 1−2θ 1/2
1/2
r
−(−A)
V (r)
−(−A)
V (s)(f(s) − f(0))ds
0
E
Z +∞ 2
1−2θ
1/2
= −(−A)
V (s + r)(f(s) − f(0))ds
r
6
r 1−2θ
Z
0
0
+∞
E
s2θ
dskfkX ,
(s + r)2
by making the change of variable s = rξ, we obtain
Z +∞
Z +∞
s2θ
ξ 2θ
ds
=
dξ.
r 1−2θ
2
(s + r)
(1 + ξ)2
0
0
Consequently
Z
1 +∞ ∂V
(s)(f(s) − f(0))ds ∈ C 2θ (E).
V (.) Aϕ − f(0) +
2 0
∂s
Hence Au(.) ∈ C 2θ (E). This ends the proof of Theorem 2.
EXAMPLE. We present now a simple example to illustrate equations (1)-(2). Consider, for instance, in E = L2 (R) the operator A defined by
D(A) = H 4 (R), Au = au(4) − bu,
with a < 0, b > 0,
for more details concerning A, see [5]. All previous abstract results can be applied to
the following problem

 ∂2u
∂4u
+ a 4 − bu = f(t, x), (t, x) ∈ Σ,
2
∂t
∂x
 u(0,
x) = u0 (x), u(+∞, x) = 0, x ∈ R,
T. Berroug
293
where Σ = (0, +∞) × R, u0 ∈ H 4 (R) and f ∈ C 2θ ([0, +∞[; L2(R)).
Acknowledgment. The author would like to thank Professors Rabah Labbas
(Université du Havre, France), Stéphane Maingot (Université du Havre, France) and
Boubaker-Khaled Sadallah (Ecole Normale Supérieure d’Alger, Algeria) for their precious help concerning the realization of this work.
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