Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 89, pp. 1–19.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

FRACTIONAL-POWER APPROACH FOR THE STUDY OF

ELLIPTIC SECOND-ORDER BOUNDARY-VALUE PROBLEMS

WITH VARIABLE-OPERATOR COEFFICIENTS IN AN

UNBOUNDED DOMAIN

FATIHA BOUTAOUS

Communicated by Jerome A. Goldstein

Abstract. In this article, we give new results on the study of elliptic complete
abstract second order differential equations with variable operator coefficients

under Dirichlet boundary conditions, and set in R+. In the framework of

Hölderian spaces and under some compatibility conditions, we prove the main
results on the existence, uniqueness and maximal regularity of the classical

solution of this kind of problems which have not been studied in variable coef-

ficients case. We use semigroups theory, fractional powers of linear operators,
Dunford’s functional calculus and interpolation theory. In this work, we con-

sider some differentiability assumptions on the resolvents of square roots of
linear operators.

1. Introduction

In a Complex Banach space X, we consider the complete second-order differential
equation with variable operator coefficients,

u′′(x) +B(x)u′(x) +A(x)u(x)− λu(x) = f(x), x ∈ (0,+∞), (1.1)

under the Dirichlet nonhomogeneous boundary conditions

u(0) = ϕ, u(+∞) = 0. (1.2)

Here λ is a positive real number, ϕ is a given element in X, f ∈ Cθ∞([0,∞);X),
0 < θ < 1, where Cθ∞([0,∞);X) is the space of bounded and θ-Hölder continuous-
vector valued functions φ : [0,∞)→ X such that

sup
x≥0
‖φ(x)‖X < +∞,

∃C > 0 : ∀x, s ≥ 0, ‖φ(x)− φ(s)‖X ≤ C|x− s|θ,
with lim

x→+∞
φ(x) = 0,

2010 Mathematics Subject Classification. 34G10, 34K10, 34K30, 35J25, 44A45, 47D06.
Key words and phrases. Fractional powers of linear operators; analytic semigroups;

classical solution; Dunford’s functional calculus.
c©2020 Texas State University.

Submitted January 12, 2018. Published September 1, 2020.

1



2 F. BOUTAOUS EJDE-2020/89

endowed with the norm

‖φ‖Cθ∞([0,∞);X) := sup
x≥0
‖φ(x)‖X + sup

x 6=s

‖φ(x)− φ(s)‖X
|x− s|θ

. (1.3)

(B(x))x≥0 is a family of bounded linear operators, and (A(x))x≥0 is a family of
closed linear operators in X, with domains D(A(x)) not necessarily dense in X.
Set

Aλ(x) = A(x)− λI, λ > 0.

We seek for a classical solution u(·) to Problem (1.1)-(1.2), that is

u ∈ C2
∞([0,∞), X), u(x) ∈ D(Aλ(x)) for every x ≥ 0;

x 7→ Aλ(x)u(x) ∈ C∞([0,∞), X);

and u satisfies Problem (1.1)-(1.2).

(1.4)

We suppose that the family of operators (B(x))x≥0 satisfies

∃C > 0 : ∀x ∈ [0,∞), ‖B(x)‖L(X) ≤ C. (1.5)

The term B(x)u′(x) is considered as a “perturbation” in some sense.
We consider Problem (1.1)-(1.2) in an elliptic situation: the family of linear

closed operators (Aλ(x))x≥0 satisfies

∃C > 0 : ∀x ≥ 0,∀z ≥ 0, ∃(Aλ(x)− zI)−1 ∈ L(X)

and ‖(Aλ(x)− zI)−1‖L(X) ≤ C/(1 + z),
(1.6)

which holds in some sector Πθ0,r0 ⊂ ρ(Aλ(x)) (the resolvent set of Aλ(x))

Πθ0,r0 = {z ∈ C\{0} : | arg(z)| ≤ θ0} ∪ {z ∈ C : |z| ≤ r0},
where θ0 and r0 are small positive numbers. According to Assumption (1.6), for
every x ≥ 0 and every λ > 0, the square roots

Kλ(x) = −(−Aλ(x))1/2

are well defined and generate analytic semigroups (eyKλ(x))y>0 not necessarily
strongly continuous in 0 (see Balakrishnan [2] for dense domains and Martinez-
Sanz [16] for nondense domains).

Equation (1.1) has been studied by several authors via various approaches:

• The constant case of operators A(x) = A and B(x) = B. For the bounded
interval, when B = 0, Krein [13] provided an approach based on the frac-
tional powers of linear operators, while in [7] the authors provided another
approach based on the sum theory of linear operators. Moreover, a study
for the general case when B(x) = B, can be found in [9]. Now, for the
interval [0,∞), when B(x) = 0, Berroug has used in [3] a method based on
direct calculus using Dunford’s operational calculus, while in [4] he stud-
ied Equation (1.1) by the Krein’s method in the frame work of Hölderian
spaces. For the general case, a study in the frame work of Lp spaces can
be found in [8].

• In the variable case of operators A(x) and B(x). For the bounded interval,
when B(x) = 0, a direct method based on Dunford’s operational calculus
has been used in [14] under some Hypotheses on differentiability of resol-
vents of operators Aλ(x). Moreover, the case of bounded operators B(x)
has been studied in [10] by using the previous method. However, in [5]
the authors used the Krein’s approach, under some natural differentiability



EJDE-2020/89 ELLIPTIC SECOND-ORDER BOUNDARY-VALUE PROBLEMS 3

Assumptions on the resolvents of the square roots Kλ(x) combining those
of Yagi [20], Da Prato-Grisvard [7] and Acquistapace-Terreni [1].

This article extends and improves the studies done in [5] and [6] by considering
the interval [0,∞), where we study the existence, the uniqueness and the maximal
regularity of the classical solution of Problem (1.1)-(1.2). In particular, we give
necessary and sufficient conditions to obtain a unique classical solution of Problem
(1.1)-(1.2) satisfying maximal regularity.

The remainder of this article is structured as follows: In Section 2, the As-
sumptions of this work are given and the representation of the solution of Problem
(1.1)-(1.2) is built using the analytic semigroups and the Dunford’s operational
calculus. Section 3 is devoted to the study of the regularity of the classical solu-
tion, and Section 4 includes the equation verified by the solution and its resolution.
In section 5, our main results on the existence, the uniqueness and the maximal
regularity of the classical solution are proved. Finally, in section 6, we provide an
example to which our abstract results apply.

2. Assumptions and construction of the solution

2.1. Assumptions. We begin by recalling that, from Hypothesis (1.6) it is well
known that there exists a sector S

θ1+π/2,r1
⊂ ρ(Kλ(x)) defined by

S
θ1+π/2,r1

= {z ∈ C\{0} : | arg(z)| ≤ θ1 + π/2} ∪ {z ∈ C : |z| ≤ r1},

with a small θ1 > 0 and r1 > 0. Let us consider, the curve

Γ = {z = ρe±i(θ1+π/2) : ρ ≥ r1} ∪ {z ∈ C : |z| = r1, | arg(z)| ≥ θ1 + π/2}

oriented from ∞e−i(θ1+π/2) to ∞ei(θ1+π/2). Therefore, for all x ≥ 0, y > 0 and
positive integer ω, we have

eyKλ(x) = − 1

2iπ

∫
Γ

eyz(Kλ(x)− zI)−1dz,

(Kλ(x))ωeyKλ(x) = − 1

2iπ

∫
Γ

zωeyz(Kλ(x)− zI)−1dz,

and for all z ≥ 0, x ≥ 0, we have

(Kλ(x)− zI)−1 =
1

π

∫ ∞
0

√
s(Aλ(x)− sI)−1

s+ z2
ds.

This last equality has an analytic continuation (in z) in the sector S
θ1+π/2,r1

. (see

[19, (2.32) p. 37]). Before giving the remaining assumptions of this work, we recall
the following basic result that will be useful in the sequel.

Lemma 2.1. Under Hypothesis (1.6), there exists a constant C > 0, such that

(1) ∀z ∈ S
θ1+π/2,r1

, ∀x ≥ 0, ‖(Kλ(x)− zI)−1‖L(x) ≤ C
|z| .

(2) ∃δ > 0 : ∀x ≥ 0,∀y > 0, ‖eyKλ(x)‖L(X) ≤ Ce−δy.

(3) ∃δ > 0 : ∀α ∈ N,∀x ≥ 0,∀y > 0, ‖(Kλ(x))αeyKλ(x)‖L(X) ≤ Cy−αe−δy.

The proof of statement (1) is based on analytic semigroup’s properties. While
the proof of (2) and (3) can be found in [17, Theorem 6.13, p. 74].

In addition to Assumptions (1.5)-(1.6), we assume that for all z ∈ S
θ1+π/2,r1

, the

mapping x 7→ (Kλ(x)− zI)−1, defined on [0,∞), is in C2([0,∞), L(X)) and there



4 F. BOUTAOUS EJDE-2020/89

exist C > 0, ρ ∈ (1/2, 1] and α ∈ (0, 1) such that for all z ∈ S
θ1+π/2,r1

and all

x, s ∈ [0,+∞),

‖ ∂
∂x

(Kλ(x)− zI)−1‖L(X) ≤
C

|z|ρ
, (2.1)

‖ ∂
∂x

(Kλ(x)− zI)−1 − ∂

∂s
(Kλ(s)− zI)−1‖L(X) ≤

C|x− s|α

|z|ρ
with α+ ρ− 1 > 0 ,

(2.2)

‖ ∂
2

∂x2
(Kλ(x)− zI)−1‖L(X) ≤ C|z|1−ρ, (2.3)

‖ d
2

dx2
(Kλ(x))−1 − d2

ds2
(Kλ(s))−1‖L(X) ≤ C|x− s|α. (2.4)

B(0)(X) ⊂ D(Kλ(0)) = D(A(0)), (2.5)

d

dx
(Kλ(x))−1

∣∣
x=0

(D(Kλ(0))) ⊂ D(Kλ(0)) = D(A(0)) , (2.6)

To study the regularity of the solution we give an essential result.

Lemma 2.2. Under Hypotheses (1.6) and (2.1), we have:

(1) The map x 7→ exKλ(x)ϕ belongs to the space C([0,∞);X) if and only if

ϕ ∈ D(Kλ(0)) = D(A(0)), in this case limx→0 e
xKλ(x)ϕ = ϕ.

(2) The map x 7→ exKλ(0)ϕ belongs to the space Cθ([0,∞);X) if and only if
ϕ ∈ DKλ(0)(θ,+∞) = DA(0)(θ/2,+∞), where

DKλ(0)(θ,+∞) = {φ ∈ X : sup
r>0
‖rθKλ(0)(Kλ(0)− rI)−1φ‖X < +∞}

is the known real interpolation space defined, for instance, in [11].

Proof. We just give the sketches of the proof.
(1) For the sake of the proof, we will write

exKλ(x)ϕ = exKλ(0)ϕ+ (exKλ(x) − exKλ(0))ϕ.

Thanks to Sinestrari [18, Proposition 1.2-(i), p. 20], we obtain

x 7→ exKλ(0)ϕ ∈ C([0,∞);X) if and only if ϕ ∈ D(Kλ(0)) = D(A(0)).

This last equality follows from the properties of fractional powers of sectorial oper-
ators, see Haase [12]. Moreover, from (2.1) we obtain

‖(exKλ(x) − exKλ(0))ϕ‖X = ‖
∫ x

0

∂

∂ξ
exKλ(ξ)ϕdξ‖X

≤ Cxρ‖ϕ‖X → 0, as x→ 0.

Hence limx→0 e
xKλ(x)ϕ = ϕ.

Item (2) follows by a similar proof as the one in [1, Proposition 3.4-(iii), p. 26]
and in [18, Theorem 3.1-(b) and (f), p. 39]. Finally, the equality DKλ(0)(θ,+∞) =
DA(0)(θ/2,+∞) follows from the Lions-Peetre reiteration interpolation property
given in [15]. �

Remark 2.3. All the constants given above are independent of x and always, we
have α + ρ − 1 < α and α + ρ − 1 < ρ. Moreover, we can replace z by

√
λ + z in

Assumptions (2.1), (2.2) and (2.3).



EJDE-2020/89 ELLIPTIC SECOND-ORDER BOUNDARY-VALUE PROBLEMS 5

2.2. Construction of the solution. First, we recall briefly that in the case B ≡ 0,
A(x) = A is a complex scalar z ∈ C\R+, such that

√
−z is the analytic determi-

nation defined by Re(
√
−z) > 0. By using the method based on the variation of

constant and Green’s Kernel, the solution of Problem (1.1)-(1.2) is

u(x) = e−x
√
−zϕ−

∫ +∞

0

k√−z(x, s)f(s) ds,

where

k√−z(x, s) =


e−
√
−zx sinh

√
−zs√

−z , if 0 6 s 6 x,

e−
√
−zs sinh

√
−zx√

−z , if x 6 s.

The later formula can be also written as

k√−z(x, s) =


e−
√
−z(x−s)

2
√
−z (1− e−2

√
−zs), if 0 6 s 6 x,

e−
√
−z(s−x)

2
√
−z (1− e−2

√
−zx), if x 6 s.

According to the Dunford’s functional calculus and the definition of analytic semi-
groups generated by the square roots K = −(−A)1/2, the solution of Problem
(1.1)-(1.2) can be written as

u(x) = − 1

2πi

∫
Γ

e−
√
−zx(K − zI)−1ϕdz

+
1

4πi

∫
Γ

∫ x

0

e−
√
−z(x−s)
√
−z

(1− e−2
√
−zs)(K − zI)−1f(s) ds dz

+
1

4πi

∫
Γ

∫ +∞

x

e−
√
−z(s−x)

√
−z

(1− e−2
√
−zx)(K − zI)−1f(s)dsdz

= exKϕ− 1

2

∫ +∞

0

e(x+s)KK−1f(s)ds+
1

2

∫ x

0

e(x−s)KK−1f(s)ds

+
1

2

∫ +∞

x

e(s−x)KK−1f(s)ds.

It is worth noting that so far the operators A(x) were considered constant. From
now on, operators A(x) will be variable and satisfy the natural ellipticity Hypothesis
(1.6). We now consider the following representation of the solution u (with Kλ(x) =
−(−Aλ(x))1/2),

u(x) = exKλ(x)ϕ∗ − 1

2

∫ +∞

0

e(x+s)Kλ(x)(Kλ(x))−1g∗(s) ds

+
1

2

∫ x

0

e(x−s)Kλ(x)(Kλ(x))−1g∗(s) ds

+
1

2

∫ ∞
x

e(s−x)Kλ(x)(Kλ(x))−1g∗(s) ds,

(2.7)

where ϕ∗ and g∗ are unknown elements to be determined in an adequate space
(g∗ ∈ Cβ∞([0,∞);X), (0 < β < 1)) see ((1.3), in order to obtain a classical solution
u of Problem (1.1)-(1.2), see (1.4). A formal calculus gives

u(0) = ϕ∗ = ϕ and u(+∞) = 0.



6 F. BOUTAOUS EJDE-2020/89

For this last condition, we can have∫ x

0

e(x−s)Kλ(x)(Kλ(x))−1g∗(s) ds

=

∫ x/2

0

e(x−s)Kλ(x)(Kλ(x))−1g∗(s)ds+

∫ x

x/2

e(x−s)Kλ(x)(Kλ(x))−1g∗(s) ds

= (I) + (II).

Therefore, by Lemma 2.1-(2), we obtain

‖(I)‖X ≤ C‖g∗‖Cβ∞([0,∞);X)

(∫ x

x/2

e−δ(x−s)ds
)

≤ C‖g∗‖Cβ∞([0,∞);X)

(
e−

δx
2 − e−δx

)
→ 0, as x→ +∞.

Since g∗ ∈ Cβ∞([0,∞);X) we have limx→+∞ sups∈[ x2 ,x] ‖g∗(s)‖X = 0, which leads
to

‖(II)‖X ≤ C sup
s∈[ x2 ,x]

‖g∗(s)‖X
(∫ x

x/2

e−δ(x−s)ds
)

≤ C sup
s∈[ x2 ,x]

‖g∗(s)‖X
(

1− e−δx/2
)
→ 0, as x→ +∞.

Now, it suffices to seek g∗ in an appropriate space such that the following repre-
sentation

u(x) = exKλ(x)ϕ− 1

2

∫ +∞

0

e(x+s)Kλ(x)(Kλ(x))−1g∗(s) ds

+
1

2

∫ x

0

e(x−s)Kλ(x)(Kλ(x))−1g∗(s) ds

+
1

2

∫ +∞

x

e(s−x)Kλ(x)(Kλ(x))−1g∗(s) ds

= u0(x) +m0(x, g∗) + w(x, g∗),

(2.8)

where w(x, g∗) is defined by the last two integrals, gives a classical solution for
Problem (1.1)-(1.2). By Lemma 2.1, all these integrals are absolutely convergent.
Indeed, for the last term in (2.8), one can write (the other terms will be treated
likewise)

‖1

2

∫ +∞

x

e(s−x)Kλ(x)(Kλ(x))−1g∗(s) ds‖X

≤ C
∫ +∞

x

e−(s−x)δ‖(Kλ(x))−1g∗(s)‖Xds

≤ C‖g∗‖C∞([0,∞);X) < +∞.

3. Regularity of the solution

From the previous representation of the solution (see (2.8)), let us start this
section by considering the operator defined by

Op(u)(x) = u′′(x) +B(x)u′(x) +Aλ(x)u(x), (3.1)

for all x > 0. Now, we shall analyze the behavior of (3.1) near 0.



EJDE-2020/89 ELLIPTIC SECOND-ORDER BOUNDARY-VALUE PROBLEMS 7

3.1. Regularity of operator Op(u0). To study the regularity of the function
x 7→ Op(u0(x)), where

u0(x) = exKλ(x)ϕ = − 1

2iπ

∫
Γ

exz(Kλ(x)− zI)−1ϕdz,

we first study the behavior of semigroups exKλ(x)ϕ and their derivatives d
dx (exKλ(x))ϕ

and d2

dx2 (exKλ(x))ϕ near 0. For this purpose, we need the use of Lemmas 2.1, 2.2
and the basic results proved in [5, Section 3, p. 7-15].

Proposition 3.1. Let ϕ ∈ D(A(0)). Under Assumptions (1.5)–(2.6), the function
x 7→ Op(u0(·))(x) belongs to Cmin(α,ρ)([0,+∞[;X).

Proof. For x > 0, we have

Aλ(x)u0(x) = −(Kλ(x))2exKλ(x)ϕ =
1

2iπ

∫
Γ

z2exz(Kλ(x)− zI)−1ϕdz,

and

u′0(x) =
d

dx
exKλ(x)ϕ

= − 1

2iπ

∫
Γ

zexz(Kλ(x)− zI)−1ϕdz − 1

2iπ

∫
Γ

exz
∂

∂x
(Kλ(x)− zI)−1ϕdz,

which leads to

B(x)u′0(x) = B(x)
d

dx
exKλ(x)ϕ

= B(x)Kλ(x)exKλ(x)ϕ− B(x)

2iπ

∫
Γ

exz
∂

∂x
(Kλ(x)− zI)−1ϕdz.

By [5, Lemma 3.5 p. 10] we can prove that

x 7→ B(x)u′0(x) ∈ Cmin(α,ρ)([0,∞);X), B(x)u′0(x)→ B(0)Kλ(0)ϕ,

as x→ 0. Moreover,

u′′0(x) =
d2

dx2

(
exKλ(x)ϕ

)
= − 1

2iπ

∫
Γ

z2exz(Kλ(x)− zI)−1ϕdz − 1

iπ

∫
Γ

zexz
∂

∂x
(Kλ(x)− zI)−1ϕdz

− 1

2iπ

∫
Γ

exz
∂2

∂x2
(Kλ(x)− zI)−1ϕdz.

By [5, Lemmas 3.6, 3.7 and 3.8], all these integrals are absolutely convergent.
Therefore,

Op(u0(x)) = − 1

iπ

∫
Γ

zexz
∂

∂x
(Kλ(x)− zI)−1ϕdz

− 1

2iπ

∫
Γ

exz
∂2

∂x2
(Kλ(x)− zI)−1ϕdz

+B(x)exKλ(x)ϕ− B(x)

2iπ

∫
Γ

exz
∂

∂x
(Kλ(x)− zI)−1ϕdz

= Gλ(x)ϕ =

4∑
i=1

Gλi(x)ϕ.

(3.2)



8 F. BOUTAOUS EJDE-2020/89

Now, by [5, Lemma 3.7 p. 13]) we can show that

x 7→ Gλ1(x)ϕ ∈ Cmin(α,ρ)([0,∞);X), Gλ1(x)ϕ→ 0, as x→ 0.

According to [5, Lemma 3.8 p. 14]) we obtain that, for all x > 0,

Gλ,2(x)ϕ = exKλ(0)(− d2

dx2
(Kλ(x))−1

∣∣
x=0

(Kλ(0)ϕ)) +G(x)ϕ,

where the function x 7→ G(x)ϕ belongs to Cmin(α,ρ)([0,∞);X). Moreover, from
Lemma 2.2-(1) we obtain

Gλ,2(x)ϕ→ − d2

dx2
(Kλ(x))−1

∣∣
x=0

(Kλ(0))ϕ, as x→ 0

if and only if

− d2

dx2

(
Kλ(x)

)−1∣∣
x=0

(Kλ(0))ϕ ∈ D(Kλ(0)) = D(A(0)).

On the other hand, the function x 7→ Gλ3(x)ϕ belongs to the space C([0,∞);X)
because of ϕ ∈ D(A(0)), see Lemma 2.2-(1). Finally, as in [5, Lemma 3.5 p. 10]) it
follows that Gλ4(x)ϕ→ 0 as x→ 0. Consequently,

Op(u0) ∈ Cmin(α,ρ)([0,∞);X).

�

3.2. Regularity of operator Op(m0). Observe that by using Dunford functional
calculus, one can write for all x > 0,

m0(x, g∗) = −1

2

∫ +∞

0

e(x+s)Kλ(x)(Kλ(x))−1g∗(s) ds

=
1

4iπ

∫
Γ

∫ +∞

0

e(x+s)z(Kλ(x)− zI)−1(Kλ(x))−1g∗(s)dsdz.

In the sequel, we will treat very carefully the convergence of all integrals obtained
with respect to the variables z on the curve Γ and s at +∞.

Proposition 3.2. Assume (1.5)–(2.6). Then the function x 7→ Op(m0(·, g∗))(x)
belongs to Cα+ρ−1

∞ ([0,∞);X).

Proof. For g∗ ∈ Cβ∞([0,∞);X), we have

Aλ(x)m0(x, g∗) =
1

2

∫ +∞

0

Kλ(x)e(x+s)Kλ(x)g∗(s) ds

= − 1

4iπ

∫
Γ

∫ +∞

0

ze(x+s)z(Kλ(x)− zI)−1g∗(s) ds dz.

This term is well defined. Indeed, for z ∈ Γ and |z| sufficiently large, we have

|e(x+s)z| ≤ e−(x+s) Re(z) = e−C0(x+s)|z|, where C0 = sin θ1.

It follows that

‖Aλ(x)m0(x, g∗)‖X ≤ C
∫

Γ

(∫ +∞

0

e−C0(x+s)|z|ds
)
‖g∗‖C∞([0,∞);X)d|z|

≤ C
(∫

Γ

e−C0x|z|

|z|
d|z|

)
‖g∗‖C∞([0,∞);X) < +∞.



EJDE-2020/89 ELLIPTIC SECOND-ORDER BOUNDARY-VALUE PROBLEMS 9

Now, since

(Kλ(x)− zI)−1(Kλ(x))−1 =
1

z
[(Kλ(x)− zI)−1 − (Kλ(x))−1],

we can write

m0(x, g∗) =
1

4iπ

∫
Γ

∫ +∞

0

e(x+s)z

z
(Kλ(x)− zI)−1g∗(s) ds dz.

Therefore,

m′0(x, g∗) =
1

4iπ

∫
Γ

∫ +∞

0

e(x+s)z(Kλ(x)− zI)−1g∗(s) ds dz,

+
1

4iπ

∫
Γ

∫ +∞

0

e(x+s)z

z

∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz,

and

B(x)m′0(x, g∗) =
B(x)

4iπ

∫
Γ

∫ +∞

0

e(x+s)z(Kλ(x)− zI)−1g∗(s) ds dz,

+
B(x)

4iπ

∫
Γ

∫ +∞

0

e(x+s)z

z

∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz.

On the other hand, we have

m′′0(x, g∗) =
1

4iπ

∫
Γ

∫ +∞

0

ze(x+s)z(Kλ(x)− zI)−1g∗(s) ds dz

+
1

2iπ

∫
Γ

∫ +∞

0

e(x+s)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

+
1

4iπ

∫
Γ

∫ +∞

0

e(x+s)z

z

∂2

∂x2
(Kλ(x)− zI)−1g∗(s) ds dz.

Hence

Op(m0(x, g∗)) =
1

2iπ

∫
Γ

∫ +∞

0

e(x+s)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

+
1

4iπ

∫
Γ

∫ +∞

0

e(x+s)z

z

∂2

∂x2
(Kλ(x)− zI)−1g∗(s) ds dz

+
B(x)

4iπ

∫
Γ

∫ +∞

0

e(x+s)z(Kλ(x)− zI)−1g∗(s) ds dz

+
B(x)

4iπ

∫
Γ

∫ +∞

0

e(x+s)z

z

∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

= Tλ(g∗)(x) =

4∑
i=1

Tλi(g
∗)(x).

(3.3)

Now, by Hypotheses (2.1)-(2.3), all these integrals are absolutely convergent. In
fact, for the first and second terms (similarly we treat the other terms), by Hypoth-
esis ((2.2), we have

Tλ1(g∗)(x)

=
1

2iπ

∫
Γ

∫ 1

0

e(x+s)z
( ∂
∂x

(Kλ(x)− zI)−1 − ∂

∂x
(Kλ(x)− zI)−1

∣∣
x=0

)
ds dz



10 F. BOUTAOUS EJDE-2020/89

+
1

2iπ

∫
Γ

∫ 1

0

e(x+s)z ∂

∂x
(Kλ(x)− zI)−1

∣∣
x=0

ds dz

+
1

2iπ

∫
Γ

∫ +∞

1

e(x+s)z
( ∂
∂x

(Kλ(x)− zI)−1 − ∂

∂x
(Kλ(x)− zI)−1

∣∣
x=0

)
ds dz

+
1

2iπ

∫
Γ

∫ +∞

1

e(x+s)z ∂

∂x
(Kλ(x)− zI)−1

∣∣
x=0

, ds dz

= (a1) + (a2) + (a3) + (a4).

Therefore,

‖(a1)‖X ≤ C
∫ 1

0

∫
Γ

e−C0(x+s)|z| x
α

|z|ρ
‖g∗‖C∞([0,∞);X)d|z|ds

≤ C
∫ 1

0

∫
Γ

e−C0σ

σρ
dσ

x+ s
(x+ s)ρ‖g∗‖C∞([0,∞);X)ds

≤ Cxα((x+ 1)ρ − xρ)‖g∗‖C∞([0,∞);X)

≤ Cxα+ρ−1‖g∗‖C∞([0,∞);X).

Similarly, it follows that

‖(a2)‖X ≤ C((x+ 1)ρ − xρ)‖g∗‖C∞([0,∞);X) < +∞,

‖(a3)‖X ≤ Cxα
(∫

Γ

e−C0x|z|

|z|ρ+1
d|z|

)
‖g∗‖C∞([0,∞);X) < +∞,

‖(a4)‖X ≤ C
(∫

Γ

e−C0x|z|

|z|ρ+1
d|z|

)
‖g∗‖C∞([0,∞);X) < +∞ .

From (2.3), it follows that

‖Tλ2(g∗)(x)‖X ≤ C
(∫

Γ

e−C0x|z|

|z|ρ+1
d|z|

)
‖g∗‖C∞([0,∞);X) < +∞.

Finally, by Remark 2.3, it results that

Op(m0) ∈ Cmin(α,ρ,α+ρ−1)
∞ ([0,∞);X) = Cα+ρ−1

∞ ([0,∞);X).

�

3.3. Regularity of operator Op(w).

Proposition 3.3. Assume (1.5)–(2.6). Then the function x 7→ Op(w(·, g∗))(x)

belongs to C
min(β,α+ρ−1)
∞ ([0,∞);X).

Proof. Recall that for x > 0, we have

w(x, g∗) =
1

2

∫ x

0

e(x−s)Kλ(x)(Kλ(x))−1g∗(s) ds

+
1

2

∫ +∞

x

e(s−x)Kλ(x)(Kλ(x))−1g∗(s) ds,



EJDE-2020/89 ELLIPTIC SECOND-ORDER BOUNDARY-VALUE PROBLEMS 11

where g∗ ∈ Cβ∞([0,∞);X) (β will be specified, 0 < β < 1). Thus

Aλ(x)w(x, g∗) = −1

2

∫ x

0

Kλ(x)e(x−s)Kλ(x)(g∗(s)− g∗(x)) ds

− 1

2

∫ +∞

x

Kλ(x)e(s−x)Kλ(x)(g∗(s)− g∗(x)) ds

+ g∗(x)− 1

2
exKλ(x)g∗(x).

(3.4)

From the properties of analytic semigroups and g∗ being Hölder continuous, the
first two integrals are absolutely convergent. For instance, for the second integral,
we have

‖ − 1

4πi

∫
Γ

∫ +∞

x

ze(s−x)z(Kλ(x)− zI)−1(g∗(s)− g∗(x)) ds dz‖X

≤ C
∫

Γ

(∫ +∞

x

e−C0(s−x)|z|(s− x)βds
)
‖g∗‖Cβ∞([0,∞);X)d|z|.

Now, by using the Hölder inequality, we obtain∫ +∞

x

e−C0(s−x)|z|(s− x)βds

≤
(∫ +∞

x

e−C0(s−x)|z|ds
)1−β(∫ +∞

x

e−C0(s−x)|z|(s− x) ds
)β

≤
(C1

|z|

)1−β( C2

|z|2
)β

≤ C

|z|1+β
.

Hence

‖ − 1

4πi

∫
Γ

∫ +∞

x

ze(s−x)z(Kλ(x)− zI)−1(g∗(s)− g∗(x)) ds dz‖X

≤
(
C

∫
Γ

d|z|
|z|1+β

)
‖g∗‖Cβ∞([0,∞);X) < +∞.

Moreover,

w′(x, g∗) =
1

2

∫ x

0

e(x−s)Kλ(x)g∗(s) ds− 1

2

∫ +∞

x

e(s−x)Kλ(x)g∗(s) ds

+
1

2

∫ x

0

e(x−s)Kλ(x) d

dx
(Kλ(x))−1g∗(s) ds

+
1

2

∫ +∞

x

e(s−x)Kλ(x) d

dx
(Kλ(x))−1g∗(s) ds

− 1

4iπ

∫
Γ

∫ x

0

e(x−s)z ∂

∂x
(Kλ(x)− zI)−1(Kλ(x))−1g∗(s) ds dz

− 1

4iπ

∫
Γ

∫ +∞

x

e(s−x)z ∂

∂x
(Kλ(x)− zI)−1(Kλ(x))−1g∗(s) ds dz

=

6∑
i=1

wi(x, g
∗).



12 F. BOUTAOUS EJDE-2020/89

All these integrals are absolutely convergent (see Lemma 2.1-(2) and Assumption
(2.1)). On the other hand, it is very important to note here, that the calculation of
the term w′′(x, g∗) is not easy to justify. For this purpose, we need to simplify the
terms (w3(x, g∗) + w5(x, g∗)) and (w4(x, g∗) + w6(x, g∗)) in order to justify their
derivatives. Indeed, by using Dunford’s calculus and similar calculus as in [5, p.
23, 24] we obtain

(w3(x, g∗) + w5(x, g∗)) + (w4(x, g∗) + w6(x, g∗))

= − 1

4iπ

∫
Γ

∫ x

0

e(x−s)z

z

∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

− 1

4iπ

∫
Γ

∫ +∞

x

e(s−x)z

z

∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

= Sλ(g∗)(x).

(3.5)

Therefore,

w′(x, g∗) = w1(x, g∗) + w2(x, g∗)

− 1

4iπ

∫
Γ

∫ x

0

e(x−s)z

z

∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

− 1

4iπ

∫
Γ

∫ +∞

x

e(s−x)z

z

∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz,

and

B(x)w′(x, g∗) = B(x)w1(x, g∗) +B(x)w2(x, g∗)

− B(x)

4iπ

∫
Γ

∫ x

0

e(x−s)z

z

∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

− B(x)

4iπ

∫
Γ

∫ +∞

x

e(s−x)z

z

∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

= Nλ(g∗)(x).

(3.6)

All these terms are well defined because of Lemma 2.1-(2) and Assumptions (1.5)
and (2.1). Now, to differentiate w′(x, g∗), we also need to justify the term w′1(·, g∗)+
w′2(·, g∗). For this end, we use the method presented, for instance, in [19, Theorem
3.3.4, p. 70]. For 0 < ε ≤ x, let us introduce the function

w′ε(x, g
∗) = wε1(x, g∗) + wε2(x, g∗) + Sλ(g∗)(x),

where

wε1(x, g∗) + wε2(x, g∗) =
1

2

∫ x−ε

0

e(x−s)Kλ(x)g∗(s) ds

− 1

2

∫ +∞

x+ε

e(s−x)Kλ(x)g∗(s) ds.

It is a simple matter to see that these integrals are absolutely convergent and

lim
ε→0

w′ε(x, g
∗) = w′(x, g∗).

The calculus of the term (wε1)′(x, g∗) + (wε2)′(x, g∗) gives

(wε1)′(x, g∗) + (wε2)′(x, g∗)

=
1

2
eεKλ(x)(g∗(x− ε) + g∗(x+ ε))− eεKλ(x)g∗(x) +

1

2
exKλ(x)g∗(x)



EJDE-2020/89 ELLIPTIC SECOND-ORDER BOUNDARY-VALUE PROBLEMS 13

+
1

2

∫ x−ε

0

Kλ(x)e(x−s)Kλ(x)(g∗(s)− g∗(x)) ds

+
1

2

∫ +∞

x+ε

Kλ(x)e(s−x)Kλ(x)(g∗(s)− g∗(x)) ds

− 1

4iπ

∫
Γ

∫ x−ε

0

e(x−s)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

+
1

4iπ

∫
Γ

∫ +∞

x+ε

e(s−x)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz.

Hence

lim
ε→0

[(wε1)′(x, g∗) + (wε2)′(x, g∗)]

=
1

2
exKλ(x)g∗(x) +

1

2

∫ x

0

Kλ(x)e(x−s)Kλ(x)(g∗(s)− g∗(x)) ds

+
1

2

∫ +∞

x

Kλ(x)e(s−x)Kλ(x)(g∗(s)− g∗(x)) ds

− 1

4iπ

∫
Γ

∫ x

0

e(x−s)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

+
1

4iπ

∫
Γ

∫ +∞

x

e(s−x)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

= w′1(x, g∗) + w′2(x, g∗).

(3.7)

On the other hand, from (3.5) we obtain

(Sλ(g∗))′(x) = − 1

4iπ

∫
Γ

∫ x

0

e(x−s)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

− 1

4iπ

∫
Γ

∫ x

0

e(x−s)z

z

∂2

∂x2
(Kλ(x)− zI)−1g∗(s) ds dz

+
1

4iπ

∫
Γ

∫ +∞

x

e(s−x)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

− 1

4iπ

∫
Γ

∫ +∞

x

e(s−x)z

z

∂2

∂x2
(Kλ(x)− zI)−1g∗(s) ds dz.

(3.8)

Now, since

w′′ε (x, g∗) = (wε1)′(x, g∗) + (wε2)′(x, g∗) + (Sλ(g∗))′(x),

by summing (3.7) and (3.8), we obtain

lim
ε→0

w′′ε (x, g∗) = w′′(x, g∗). (3.9)

Finally, by summing (3.4), (3.6) and (3.9), it follows that

Op(w(x, g∗)) = g∗(x) + Vλ(g∗)(x) +Nλ(g∗)(x), (3.10)



14 F. BOUTAOUS EJDE-2020/89

where

Vλ(g∗)(x) = − 1

2iπ

∫
Γ

∫ x

0

e(x−s)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

+
1

2iπ

∫
Γ

∫ +∞

x

e(s−x)z ∂

∂x
(Kλ(x)− zI)−1g∗(s) ds dz

− 1

4iπ

∫
Γ

∫ x

0

e(x−s)z

z

∂2

∂x2
(Kλ(x)− zI)−1g∗(s) ds dz

− 1

4iπ

∫
Γ

∫ +∞

x

e(s−x)z

z

∂2

∂x2
(Kλ(x)− zI)−1g∗(s) ds dz

=

4∑
i=1

Vλi(g
∗)(x).

(3.11)

These integrals are absolutely convergent because of (2.1), (2.2) and (2.3). To see
this, let us for instance treat the convergence of Vλ1(g∗)(x).

Vλ1(g∗)(x)

= − 1

2iπ

∫ x

0

∫
Γ

e(x−s)z
( ∂
∂x

(Kλ(x)− zI)−1 − ∂

∂x
(Kλ(x)− zI)−1

∣∣
x=0

)
g∗(s) dz ds

− 1

2iπ

∫ x

0

∫
Γ

e(x−s)z ∂

∂x
(Kλ(x)− zI)−1

∣∣
x=0

g∗(s) dz ds

= φ1(x) + φ2(x).

From (2.2), we obtain

‖(φ1(x))‖X ≤ C‖g∗‖C∞([0,∞);X)

∫ x

0

∫
Γ

e−C0(x−s)|z| x
α

|z|ρ
d|z|ds

≤ C‖g∗‖C∞([0,∞);X)

∫ x

0

∫
Γ

e−C0σxα
(x− s)ρ

σρ
dσ

(x− s)
ds

≤ Cxα+ρ‖g∗‖C∞([0,∞);X).

Assumption (2.1) leads to

‖φ2(x)‖X ≤ Cxρ‖g∗‖C∞([0,∞);X).

By Remark 2.3, it follows that Vλ(g∗) + Nλ(g∗) ∈ Cα+ρ−1
∞ ([0,∞);X). Finally,

the function x 7→ Op(w(x, g∗)) belongs to C
min(β,α+ρ−1)
∞ ([0,∞);X), because g∗ ∈

Cβ∞([0,∞);X). �

Remark 3.4. Note that all the Hölderianities studied above were done near 0
which allow us to deduce the Hölderianities in [0,∞).

4. Equation satisfied by the solution and its resolution

The previous calculus prove that the representation given in (2.8) satisfies the
abstract equation

u′′(x) +B(x)u′(x) +Q(x)u(x)

= Op(u(x, g∗))

= g∗(x) +Gλ(x)ϕ+ Tλ(g∗)(x) + Vλ(g∗)(x) +Nλ(g∗)(x)

= f(x), for x ∈ [0,∞).

(4.1)



EJDE-2020/89 ELLIPTIC SECOND-ORDER BOUNDARY-VALUE PROBLEMS 15

To determine the unknown function g∗, we need to prove the following result.

Proposition 4.1. Let ϕ ∈ D(A(0)) and f ∈ Cθ∞([0,∞);X), where 0 < θ < 1.
Under Hypotheses (1.5)–( (2.6), suppose that u given in (2.8) is a classical solution
of Problem (1.1)-(1.2). Then in the space C∞([0,∞);X), the function g∗ satisfies

(I + Pλ)(g∗)(x) = f(x)−Gλ(x)ϕ, (4.2)

where
Pλ(g∗)(·) = Tλ(g∗)(x) + Vλ(g∗)(x) +Nλ(g∗)(x). (4.3)

Moreover, there exists λ∗ > 0 such that for every λ ≥ λ∗, operator I + Pλ is

invertible in the space C
min(α,ρ)
∞ ([0,∞);X) and

g∗(·) = (I + Pλ)−1f(·). (4.4)

Proof. To solve (4.2) in C∞([0,∞);X), we should estimate ‖Pλ‖L(C∞([0,∞);X)) (see
(4.3)), for a large λ > 0. Let us, for instance, estimate some terms contained in
Vλ(g∗)(x) (see formula (3.11)). By applying the same arguments used in [5, p. 27,
28], it follows that

‖Vλ1(g∗)(x)‖X ≤
C

λ(1−α)/2
‖g∗‖C∞([0,∞);X),

‖Vλ3(g∗)(x)‖X ≤ C
( 1

λ1/4
+

1

λ(ρ+α−1)/2

)
‖g∗‖C∞([0,∞);X).

A similar analysis for the remaining terms proves the existence of some λ∗ such
that for all λ ≥ λ∗, we have ‖Pλ‖L(C([0,∞);X)) ≤ 1

2 . Therefore operator I + Pλ is
invertible for λ ≥ λ∗ in C∞([0,∞);X) and thus (4.4) follows. �

In the sequel, we need the following result concerning g∗(0).

Proposition 4.2. Let ϕ ∈ D(A(0)) and f ∈ Cθ∞([0,∞);X), where 0 < θ < 1.
Under Hypotheses (1.5)–(2.6), suppose that u given in (2.8) is a classical solution
of Problem (1.1)-(1.2). Then

g∗(0) = f(0) +
d2

dx2
(Kλ(x))−1

∣∣
x=0

Kλ(0)ϕ+ Ψ∗0(ϕ) + s0(g∗),

where Ψ∗0(ϕ), s0(g∗) ∈ D(Kλ(0)) = D(A(0)). Moreover g∗ ∈ Cβ∞([0,∞);X),
where β = min(α+ ρ− 1, θ).

Proof. We have

g∗(0) = f(0)− [Gλ(0)ϕ+ Tλ(g∗)(0) + Vλ(g∗)(0) +Nλ(g∗)(0)],

and the term
s0(g∗) = Tλ(g∗)(0) + Vλ(g∗)(0) +Nλ(g∗)(0)

is in D(Kλ(0)) = D(A(0)). From (3.2), it follows that

Gλ(0)ϕ = Ψ∗0(ϕ)− d2

dx2
(Kλ(x))−1

∣∣
x=0

Kλ(0)ϕ,

where Ψ∗0(ϕ) ∈ D(Kλ(0)) (see (2.5)). Therefore

g∗(0) = f(0) +
d2

dx2
(Kλ(x))−1

∣∣
x=0

Kλ(0)ϕ+ Ψ∗0(ϕ) + s0(g∗),

where Ψ∗0(ϕ), s0(g∗) ∈ D(Kλ(0)). On the other hand, the function

x 7→ Gλ(x)ϕ+ Tλ(g∗)(x) + Vλ(g∗)(x) +Nλ(g∗)x)



16 F. BOUTAOUS EJDE-2020/89

belongs to Cα+ρ−1
∞ ([0,∞);X), and thus we deduce that if g∗ exists, then it belongs

necessarily to C
min(θ,α+ρ−1)
∞ ([0,∞);X). Therefore β = min(θ, α+ ρ− 1). �

5. Main results

According to the above study on the regularity of the solution, we are ready
to state our first main result on the existence and the uniqueness of the classical
solution of Problem (1.1)-(1.2).

Theorem 5.1. Let ϕ ∈ D(A(0)) and f ∈ Cθ∞([0,∞);X), where 0 < θ < 1.
Then, under Hypotheses (1.5)–(2.6), there exists λ∗ > 0 such that for all λ ≥ λ∗,
the function u given in the representation (2.7) is the unique classical solution of
Problem (1.1)-(1.2) if and only if

(−A(0))ϕ+ f(0) +
d2

dx2
(λI −A(x))−1/2

∣∣
x=0

(λI −A(0))1/2ϕ ∈ D(A(0)). (5.1)

Proof. It suffices to prove that

x 7→ Aλ(x)u(x) = −(Kλ(x))2u(x) ∈ C∞([0,∞), X).

We have

(Kλ(x))2u(x) = (Kλ(x))2[u0(x)ϕ+m0(x, g∗) + w(x, g∗)]

= (Kλ(x))2exK(x)ϕ− g∗(x) + exK(x)g∗(x)

− 1

2

∫ +∞

0

Kλ(x)e(x+s)Kλ(x)(g∗(s)− g∗(0)) ds

+
1

2

∫ x

0

Kλ(x)e(x−s)Kλ(x)(g∗(s)− g∗(x)) ds

+
1

2

∫ +∞

x

Kλ(x)e(s−x)Kλ(x)(g∗(s)− g∗(x)) ds.

We treat only the regularity of the following term (the other terms are regular near
0),

R(x) = (Kλ(x))2exKλ(x)ϕ+ exKλ(x)g∗(0)

= [(Kλ(x))2 − (Kλ(0))2]exKλ(x)ϕ+ (Kλ(0))2[exKλ(x) − exKλ(0)]ϕ

+ [exKλ(x) − exKλ(0)]g∗(0) + (Kλ(0))2exKλ(0)ϕ+ exKλ(0)g∗(0)

=

5∑
i=1

Ri(x).

Hence R1(x) and R2(x) tend to 0, as x→ 0. Moreover, (2.1) leads to

‖R3(x)‖X = ‖ − 1

2iπ

∫
Γ

exz
(∫ x

0

∂

∂r
(Kλ(r)− zI)−1dr

)
g∗(0)dz‖X

≤ Cxρ‖g∗‖C∞([0,∞);X) → 0, as x→ 0.

Now, by Proposition 4.2 and Lemma 2.2-(1), the term

R4(x) +R5(x) = exKλ(0)[(Kλ(0))2ϕ+ f(0) +
d2

dx2
(Kλ(x))−1

∣∣
x=0

Kλ(0)ϕ]

+ exKλ(0)[Φ∗0(ϕ) + s0(g∗)]



EJDE-2020/89 ELLIPTIC SECOND-ORDER BOUNDARY-VALUE PROBLEMS 17

is in C([0,∞);X) if and only if

(Kλ(0))2ϕ+ f(0) +
d2

dx2
(Kλ(x))−1

∣∣
x=0

Kλ(0)ϕ ∈ D(Kλ(0)) = D(A(0))

and Φ∗0(ϕ) + s0(g∗) ∈ D(Kλ(0)) = D(A(0)). This completes the proof. �

Before giving our second main result on the maximal regularity of the classical
solution of problem (1.1)-(1.2), we recall the interpolation spaces that play a crucial
role in our proof. For this reason, consider the interpolation spaces

D(−A(0))(θ/2,+∞) = DKλ(0)(θ,+∞) ⊂ D(Kλ(0)) = D(A(0)).

We replace Assumptions (2.5)-(2.6) by the following Hypotheses:

B(0)(X) ⊂ DKλ(0)(θ,+∞), (5.2)

d

dx
(Kλ(x))−1

∣∣
x=0

(D(Kλ(0))) ⊂ DKλ(0)(θ,+∞). (5.3)

By using Lemma 2.2-(2) and similar arguments to those applied in the proof of
Theorem 5.1, we can prove the following result.

Theorem 5.2. Let ϕ ∈ D(A(0)) and f ∈ Cθ∞([0,∞);X), where 0 < θ < 1. Then,
under Hypotheses (1.5)–(2.4), (5.2) and (5.3), there exists λ∗ > 0 such that for
all λ ≥ λ∗, the function u given in the representation (2.7) is the unique classical
solution of Problem (1.1)-(1.2) satisfying

u′′(·), B(·)u′(·), Aλ(·)u(·) ∈ Cβ∞([0,∞);X), β ∈ min(θ, α+ ρ− 1)

if and only if

(−A(0))ϕ+ f(0) +
d2

dx2
(λI −A(x))−1/2

∣∣
x=0

(λI −A(0))1/2ϕ ∈ DA(0)(θ/2,+∞).

(5.4)

6. Applications

Consider the Banach space X = C([0, 1]) with its usual supremum norm and
define the family of closed linear operators (−A(x))1/2 for all x ≥ 0 by

D(−(−A(x))1/2) = {ϕ ∈ C2([0, 1]) : ϕ(0)− a(x)ϕ′(0) = 0;ϕ(1) = 0}

(−(−A(x))1/2ϕ)(y) = ϕ′′(y), y ∈ [0, 1].

Then

D(A(x)) =
{
ϕ ∈ C4([0, 1]) : ϕ(0)− a(x)ϕ′(0) = 0, ϕ(1) = 0,

ϕ′′(0)− a(x)ϕ′′′(0) = 0, ϕ′′(1) = 0
}

(A(x)ϕ)(y) = −ϕ(iv)(y), y ∈ [0, 1].

We assume that a ∈ C2,κ[0,∞), a(x) ≥ 0 and minx≥0 a(x) > 0. The family of
bounded linear operators (B(x))x≥0 is defined by

D(B(x)) = X, (B(x)ϕ)(y) = ω(x)ϕ(y).

Then, all our results apply to the following model on a concrete quasi-elliptic
boundary-value problem, for a large positive λ,

∂2u

∂x2
(x, y) + ω(x)

∂u

∂x
(x, y)− ∂4u

∂y4
(x, y)− λu(x, y) = f(x, y), x > 0, y ∈ (0, 1),



18 F. BOUTAOUS EJDE-2020/89

u(x, 0)− a(x)
∂u

∂y
(x, 0) = 0, x ≥ 0,

∂2u

∂y2
(x, 0)− a(x)

∂3u

∂y3
(x, 0) = 0, x ≥ 0,

u(x, 1) = 0 =
∂2u

∂y2
(x, 1), x ≥ 0,

u(0, y) = ϕ(y), u(+∞, y) = 0, y ∈ [0, 1], f ∈ Cθ∞([0,∞);X).

Conclusions. In this article, we obtained interesting results, namely Theorems 5.1
and 5.2, on the existence, the uniqueness and the maximal regularity of the classical
solution of Problem (1.1)-(1.2) by using semigroups theory, the fractional powers
of linear operators, the Dunford’s functional calculus and the interpolation spaces.
Moreover, we established necessary and sufficient conditions of compatibility (see
(5.4)) to obtain the solution. For future work, we think it is interesting to gener-
alize the study of this problem to more complicated situations, where the variable
operators (A(x))x≥0 and (B(x))x≥0 are closed (unbounded), in the two functional
frames Hölderian and Lp.

Acknowledgements. The author is grateful to Professor Rabah Labbas for his
valuable advice and suggestions to improve this work. The author also thanks the
referee for her/his helpful comments.

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Fatiha Boutaous

Department of Mathematics, Faculty of Sciences, Saâd Dahlab University, 09000, Blida,

Algeria.
Lab. NLPDE and Hist. of Maths, Ecole Normale Supérieure, 16050, Kouba, Algeria

Email address: boutaous.fatiha2009@hotmail.com


	1. Introduction
	2. Assumptions and construction of the solution
	2.1. Assumptions
	2.2. Construction of the solution

	3. Regularity of the solution
	3.1. Regularity of operator Op(u0)
	3.2. Regularity of operator Op(m0)
	3.3. Regularity of operator Op(w)

	4. Equation satisfied by the solution and its resolution
	5. Main results
	6. Applications
	Conclusions
	Acknowledgements

	References