Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 2, Article 38, 2004
BOUNDARY VALUE PROBLEM FOR SECOND-ORDER DIFFERENTIAL
OPERATORS WITH MIXED NONLOCAL BOUNDARY CONDITIONS
M. DENCHE AND A. KOURTA
L ABORATOIRE E QUATIONS D IFFERENTIELLES ,
D EPARTEMENT DE M ATHEMATIQUES ,
FACULTE DES S CIENCES , U NIVERSITE M ENTOURI ,
25000 C ONSTANTINE , A LGERIA .
denech@wissal.dz
Received 15 October, 2003; accepted 16 April, 2004
Communicated by A.G. Ramm
A BSTRACT. In this paper, we study a second order differential operator with mixed nonlocal
boundary conditions combined weighting integral boundary condition with another two point
boundary condition. Under certain conditions on the weighting functions and on the coefficients
in the boundary conditions, called non regular boundary conditions, we prove that the resolvent
decreases with respect to the spectral parameter in Lp (0, 1), but there is no maximal decreasing
at infinity for p ≥ 1. Furthermore, the studied operator generates in Lp (0, 1) an analytic semi
group with singularities for p ≥ 1. The obtained results are then used to show the correct
solvability of a mixed problem for a parabolic partial differential equation with non regular
boundary conditions.
Key words and phrases: Green’s Function, Non Regular Boundary Conditions, Regular Boundary Conditions, Semi group
with Singularities.
2000 Mathematics Subject Classification. 47E05, 35K20.
1. I NTRODUCTION
In the space Lp (0, 1) we consider the boundary value problem

00
 L (u) = u = f (x) ,
0
0
B1 (u) = a0 u (0) + b0 u (0) + c0 u (1) + d0 u (1) = 0,
(1.1)
R
R

0
1
1
B2 (u) = 0 R (t) u (t) dt + 0 S (t) u (t) dt = 0,
where the functions R, S ∈ C([0, 1] , C). We associate to problem (1.1) in space Lp (0, 1) the
operator:
Lp (u) = u00 ,
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
The authors are grateful to the editor and the referees for their valuable comments and helpful suggestions which have much improved the
presentation of the paper.
147-03
2
M. D ENCHE AND A. KOURTA
with domain D(Lp ) = {u ∈ W 2,p (0, 1) : Bi (u) = 0, i = 1, 2} .
Many papers and books give the full spectral theory of Birkhoff regular differential operators
with two point linearly independent boundary conditions, in terms of coefficients of boundary
conditions. The reader should refer to [7, 10, 20, 21, 22, 28, 31, 33] and references therein.
Few works have been devoted to the study of a non regular situation. The case of separated
non regular boundary conditions was studied by W. Eberhard, J.W. Hopkins, D. Jakson, M.V.
Keldysh, A.P. Khromov, G. Seifert, M.H. Stone, L.E. Ward (see S. Yakubov and Y. Yakubov
[33] for exact references). A situation of non regular non-separated boundary conditions was
considered by H. E. Benzinger [2], M. Denche [4], W. Eberhard and G. Freiling [8], M.G.
Gasumov and A.M. Magerramov [12, 13], A.P. Khromov [18], Yu. A. Mamedov [19], A.A.
Shkalikov [24], Yu. T. Silchenko [26], C. Tretter [29], A.I. Vagabov [30], S. Yakubov [32] and
Y. Yakubov [34].
A mathematical model with integral boundary conditions was derived by [9, 23] in the context of optical physics. The importance of this kind of problem has been also pointed out by
Samarskii [27].
In this paper, we study a problem for second order ordinary differential equations with mixed
nonlocal boundary conditions combined with weighted integral boundary conditions and another two point boundary condition. Following the technique in [11, 20, 21, 22], we should
bound the resolvent in the space Lp (0, 1) by means of a suitable formula for Green’s function.
Under certain conditions on the weighting functions and on the coefficients in the boundary
conditions, called non regular boundary conditions, we prove that the resolvent decreases with
respect to the spectral parameter in Lp (0, 1), but there is no maximal decreasing at infinity for
p ≥ 1. In contrast to the regular case this decreasing is maximal for p = 1 as shown in [11]. Furthermore, the studied operator generates in Lp (0, 1) an analytic semi group with singularities
for p ≥ 1. The obtained results are then used to show the correct solvability of a mixed problem
for a parabolic partial differential equation with non regular non local boundary conditions.
2. G REEN ’ S F UNCTION
Let λ ∈ C, u1 (x) = u1 (x, λ ) and u2 (x) = u2 (x, λ ) be a fundamental system of solutions to
the equation
L (u) − λu = 0.
Following [20], the Green’s function of problem (1.1) is given by
(2.1)
G(x, s; λ) =
N (x, s; λ)
,
∆(λ)
where ∆(λ) it the characteristic determinant of the considered problem defined by
(2.2)
∆(λ) =
B1 (u1 ) B1 (u2 )
B2 (u1 ) B2 (u2 )
and
(2.3)
N (x, s; λ) =
u1 (x) u2 (x) g(x, s, λ)
B1 (u1 ) B1 (u2 )
B1 (g)x
B2 (u1 ) B2 (u2 )
B2 (g)x
for x, s ∈ [0, 1].The function g(x, s, λ) is defined as follows
(2.4)
g(x, s; λ) = ±
1 u1 (x)u2 (s) − u1 (s)u2 (x)
,
2 u01 (s)u2 (s) − u1 (s)u02 (s)
where it takes the plus sign for x > s and the minus sign for x < s.
J. Inequal. Pure and Appl. Math., 5(2) Art. 38, 2004
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BVP
Given an arbitrary δ ∈
π
,π
2
FOR
2 ND -O RDER D IFFERENTIAL O PERATORS
3
, we consider the sector
Σδ = {λ ∈ C, |arg λ| ≤ δ, λ 6= 0} .
P
For λ ∈ δ , define ρ as the square root of λ with positive real part. For λ 6= 0, we can consider
a fundamental system of solutions of the equation u00 = λu = ρ2 u given by u1 (t) = e−ρt and
u2 (t) = eρt .
In the following we are going to deduce an adequate formulae for ∆(λ) and G(x, s; λ). First
of all, for j = 1, 2, we have
j
j
B1 (uj ) = a0 + (−1)j b0 ρ + c0 e(−1) ρ + (−1)j d0 ρe(−1) ρ ,
Z 1
Z 1
j
(−1)j ρt
j
B2 (uj ) =
R(t)e
dt + (−1) ρ
S(t)e(−1) ρt dt.
0
0
so we obtain from (2.2)
−ρ
Z
1
(2.5) ∆(λ) = a0 − b0 ρ + c0 e − d0 ρe
(R(t) + ρS(t))e dt
0
Z 1
ρ
ρ
−ρt
− (a0 + b0 ρ + c0 e + d0 ρe )
(R(t) − ρS(t))e dt ,
−ρ
ρt
0
and g(x, s; λ) has the form
g(x, s; λ) =

1 ρ(x−s)


(e
− eρ(s−x) ) if x > s,

 4ρ


1 ρ(s−x)


(e
− eρ(x−s) ) if x < s.
4ρ
Thus we have
B1 (g) = a0 − b0 ρ − c0 e−ρ + d0 ρe−ρ
eρs
4ρ
e−ρs
+ (−a0 − b0 ρ + c0 eρ + d0 ρeρ )
,
4ρ
Z s
Z 1
eρs
−ρt
−ρt
B2 (g) =
(R(t) − ρS(t))e dt +
(−R(t) + ρS(t))e dt
4ρ
0
s
Z s
Z 1
e−ρs
ρt
ρt
+
−
(R(t) + ρS(t))e dt +
(R(t) + ρS(t))e dt .
4ρ
0
s
After a long calculation, formula (2.3) can be written as
N (x, s; λ) = ϕ(x, s; λ) + ϕi (x, s; λ),
(2.6)
where
1
(2.7) ϕ(x, s; λ) =
2ρ
Z
1
(a0 + ρb0 ) (R (t) + ρS (t)) eρt dt
Z s
ρ
ρt
− e (c0 + ρd0 )
(R (t) + ρS (t)) e dt e−ρ(x+s)
0
Z 1
+
(a0 − ρb0 ) (R (t) − ρS (t)) e−ρt dt
s
Z s
ρ
−ρt
ρ(x+s)
− e (c0 − ρd0 )
(R (t) − ρS (t)) e dt e
0
0
J. Inequal. Pure and Appl. Math., 5(2) Art. 38, 2004
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4
M. D ENCHE AND A. KOURTA
and the function ϕi (x, s; λ) is given by
ϕi (x, s; λ) =
(2.8)

 ϕ1 (x, s; λ) if x > s,
 ϕ (x, s; λ) if x < s,
2
with
Z s
eρ(s−x)
(2.9) ϕ1 (x, s; λ) =
(a0 + ρb0 + c0 eρ + ρeρ d0 ) (R (t) − ρS (t)) e−ρt dt
2ρ
0
Z 1
ρt
−
(a0 − ρb0 ) (R (t) + ρS (t)) e dt
s
Z s
eρ(x−s)
+
a0 − ρb0 + c0 e−ρ − ρe−ρ d0 (R (t) + ρS (t)) eρt dt
2ρ
0
Z 1
−ρt
−
(a0 + ρb0 ) (R (t) − ρS (t)) e dt ,
0
and
1
(2.10) ϕ2 (x, s; λ) =
2ρ
Z 1
−
(a0 + ρb0 + c0 eρ + ρeρ d0 ) (R (t) − ρS (t)) e−ρt dt
s
Z 1
−ρ
ρt
+
(c0 − ρd0 ) e (R (t) + ρS (t)) e dt eρ(s−x)
0
Z 1
−
a0 − ρb0 + c0 e−ρ − ρe−ρ d0 (R (t) + ρS (t)) eρt dt
s
Z 1
ρ
−ρt
ρ(x−s)
−
(c0 + ρd0 ) e (R (t) − ρS (t)) e dt e
.
0
3. B OUNDS ON THE R ESOLVENT
Every λ ∈ C such that ∆(λ) 6= 0 belongs to ρ(Lp ), and the associated resolvent operator
R(λ, Lp ) can be expressed as a Hilbert Schmidt operator
Z 1
−1
(3.1)
(λI − Lp ) f = R(λ, Lp )f = −
G(·, s; λ)f (s)ds,
f ∈ Lp (0, 1).
0
Then, for every f ∈ Lp (0, 1) we estimate (3.1)
kR(λ; Lp )f kLp (0,1) ≤
Z
sup
0≤s≤1
and so we need to bound
p1
Z 1
p
=
sup
|G(x, s; λ)| dx
0≤s≤1
0
1
|G(x, s; λ)|p dx
p1
kf kLp (0,1) ,
0
1
|∆(λ)|
Z
p
|N (x, s; λ)| dx
sup
0≤s≤1
1
p1
.
0
3.1. Estimation of N (x, s; λ). We will denote by k·k the supremum norm which is defined by
kRk = sup |R (s)|. Since
0≤s≤1
J. Inequal. Pure and Appl. Math., 5(2) Art. 38, 2004
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BVP
FOR
N (x, s; λ) =
2 ND -O RDER D IFFERENTIAL O PERATORS
5

 ϕ (x, s; λ) + ϕ1 (x, s; λ) if x > s
;
ϕ (x, s; λ) + ϕ2 (x, s; λ) if x < s

then
kN (x, s; λ)kLp ≤ kϕ (x, s; λ)kLp + kϕi (x, s; λ)kLp ,
(3.2)
from (2.8), we have
kϕi (x, s; λ)kLp ≤ 21/p
(3.3)
(Z
s
p
|ϕ2 (x, s; λ)| dx
0
p1
Z
1
p
|ϕ1 (x, s; λ)| dx
+
p1 )
.
s
From (2.7) we have
Z 1
e−(x+s) Re ρ
|ϕ (x, s; λ)| ≤
(kRk + |ρ| kSk) (|a0 | + |ρ| |b0 |)
et Re ρ dt
2 |ρ|
s
Z s
Re ρ
t Re ρ
+ (kRk + |ρ| kSk) (|c0 | + |ρ| |d0 |) e
e
dt
0
Z 1
e(x+s) Re ρ
(kRk + |ρ| kSk) (|a0 | + |ρ| |b0 |)
+
e−t Re ρ dt
2 |ρ|
s
Z s
− Re ρ
−t Re ρ
+e
(kRk + |ρ| kSk) (|c0 | + |ρ| |d0 |)
e
dt
0
then
|ϕ (x, s; λ)| ≤
e−(x+s) Re ρ (kRk + |ρ| kSk) (|a0 | + |ρ| |b0 |) eRe ρ − es Re ρ
2 |ρ| Re ρ
+ (kRk + |ρ| kSk) (|c0 | + |ρ| |d0 |) e(s+1) Re ρ − eRe ρ
e(x+s) Re ρ +
(kRk + |ρ| kSk) (|a0 | + |ρ| |b0 |) e−s Re ρ − e− Re ρ
2 |ρ| Re ρ
+ (kRk + |ρ| kSk) (|c0 | + |ρ| |d0 |) e− Re ρ − e−(s+1) Re ρ
and
Z
0
1
|ϕ (x, s; λ)|p dx
R1
p
2p 0 e−px Re ρ dx h
≤
(kRk + |ρ| kSk)p (|a0 | + |ρ| |b0 |)p e(1−s) Re ρ − 1
p
(|ρ| Re ρ)
p i
+ (kRk + |ρ| kSk)p × (|c0 | + |ρ| |d0 |)p eRe ρ − e(1−s) Re ρ
R1
p
2p 0 epx Re ρ dx
+
[(kRk + |ρ| kSk)p × (|a0 | + |ρ| |b0 |)p 1 − e(s−1) Re ρ
p
(|ρ| Re ρ)
i
p
p
(s−1) Re ρ
−1 Re ρ p
+ (kRk + |ρ| kSk) (|c0 | + |ρ| |d0 |) e
−e
,
J. Inequal. Pure and Appl. Math., 5(2) Art. 38, 2004
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6
M. D ENCHE AND A. KOURTA
after calculation we obtain
Z
p1
1
p
|ϕ (x, s; λ)| dx
0
2
21+ p eRe ρ
≤
1+ p1
|ρ| (Re ρ)
p1/p
−p Re ρ
× 1−e
(kRk + |ρ| kSk) (|a0 | + |ρ| |b0 |) e−s Re ρ − e− Re ρ
p1
−p Re ρ
+ 1−e
p1
× (|c0 | + |ρ| |d0 |)
h
(s−1) Re ρ
1−e
1 − e−p Re ρ
i
p1
+ 1 − e−p Re ρ
p1
+ (kRk + |ρ| kSk)
1 − e−s Re ρ
e(s−1) Re ρ − e− Re ρ
ii
as Re ρ > 0 so
Z
(3.4)
p
|ϕ (x, s; λ)| dx
sup
0≤s≤1
p1
1
0
2
≤
21+ p eRe ρ (kRk + |ρ| kSk) [(|a0 | + |c0 |) + |ρ| (|b0 | + |d0 |)]
1
|ρ| (Re ρ)1+ p p1/p
.
From (2.9) we have
e(s−x) Re ρ |ϕ1 (x, s; λ)| ≤
(|a0 | + |ρ| |b0 |) + eRe ρ (|c0 | + |ρ| |d0 |)
2 |ρ|
Z s
× (kRk + |ρ| kSk)
e−t Re ρ dt
0
Z 1
t Re ρ
+ (|a0 | + |ρ| |b0 |) (kRk + |ρ| kSk)
e
dt
0
e(x−s) Re ρ
+
(kRk + |ρ| kSk) [(|a0 | + |ρ| |b0 |)
2 |ρ|
Z s
Re ρ
+e
(|c0 | + |ρ| |d0 |)]
et Re ρ dt
0
Z 1
−t Re ρ
+ (|a0 | + |ρ| |b0 |) (kRk + |ρ| kSk)
e
dt
0
then
|ϕ1 (x, s; λ)| ≤
e(s−x) Re ρ (|a0 | + |ρ| |b0 |) (kRk + |ρ| kSk) eRe ρ − e−s Re ρ
2 |ρ| Re ρ
+ (|c0 | + |ρ| |d0 |) (kRk + |ρ| kSk) eRe ρ − e(1−s) Re ρ
e(x−s) Re ρ +
(kRk + |ρ| kSk) (|a0 | + |ρ| |b0 |) es Re ρ − e− Re ρ
2 |ρ| Re ρ
+ (|c0 | + |ρ| |d0 |) (kRk + |ρ| kSk) e(s−1) Re ρ − e− Re ρ
J. Inequal. Pure and Appl. Math., 5(2) Art. 38, 2004
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BVP
and
Z
s
FOR
2 ND -O RDER D IFFERENTIAL O PERATORS
7
1
|ϕ1 (x, s; λ)|p dx
R1
p
2p s e−px Re ρ dx h
≤
(|a0 | + |ρ| |b0 |)p (kRk + |ρ| kSk)p e(1+s) Re ρ − 1
p
p
|ρ| (Re ρ)
p i
+ (|c0 | + |ρ| |d0 |)p (kRk + |ρ| kSk)p e(1+s) Re ρ − eRe ρ
R1
p
2p s epx Re ρ dx h
+
(|a0 | + |ρ| |b0 |)p (kRk + |ρ| kSk)p 1 − e−(1+s) Re ρ
p
p
|ρ| (Re ρ)
p
+ (|c0 | + |ρ| |d0 |)p (kRk + |ρ| kSk)p e− Re ρ − e−(1+s) Re ρ ,
this yields
Z 1
|ϕ1 (x, s; λ)|p dx ≤
2p+1 ep Re ρ
p
p
p
p+1 [(|a0 | + |ρ| |b0 |) (kRk + |ρ| kSk)
|ρ| (Re ρ)
p
× 1 − e−(1+s) Re ρ
1 − ep(s−1) Re ρ
p
+ (|c0 | + |ρ| |d0 |)p (kRk + |ρ| kSk)p 1 − e−s Re ρ
1 − ep(s−1) Re ρ
s
as Re ρ > 0 we obtain
Z
(3.5)
p
|ϕ1 (x, s; λ)| dx
sup
0≤s≤1
p1
1
s
2
≤
21+ p eRe ρ (kRk + |ρ| kSk) [(|a0 | + |c0 |) + |ρ| (|b0 | + |d0 |)]
1
|ρ| (Re ρ)1+ p p1/p
.
From (2.10) we have
Z 1
e(x−s) Re ρ
|ϕ2 (x, s; λ)| ≤
(kRk + |ρ| kSk) (|c0 | + |ρ| |d0 |)
e(1−t) Re ρ dt
2 |ρ|
0
Z
− Re ρ
+ (kRk + |ρ| kSk) (|a0 | + |ρ| |b0 |) + e
(|c0 | + |ρ| |d0 |)
1
t Re ρ
e
dt
0
Z 1
e(s−x) Re ρ
+
[(|c0 | + |ρ| |d0 |) (kRk + |ρ| kSk)
e(t−1) Re ρ dt
2 |ρ|
0
Z 1
Re ρ
−t Re ρ
+ (|a0 | + |ρ| |b0 |) + e
(|c0 | + |ρ| |d0 |) (kRk + |ρ| kSk)
e
dt
s
then
|ϕ2 (x, s; λ)| ≤
ex Re ρ (kRk + |ρ| kSk) (|c0 | + |ρ| |d0 |) e(1−s) Re ρ − e− Re ρ
2 |ρ| Re ρ
+ (|a0 | + |ρ| |b0 |) (kRk + |ρ| kSk) e(1−s) Re ρ − 1
e−x Re ρ +
(kRk + |ρ| kSk) (|c0 | + |ρ| |d0 |) eRe ρ − e(s−1) Re ρ
2 |ρ| Re ρ
+ (|a0 | + |ρ| |b0 |) (kRk + |ρ| kSk) 1 − e(s−1) Re ρ .
J. Inequal. Pure and Appl. Math., 5(2) Art. 38, 2004
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8
M. D ENCHE AND A. KOURTA
So
Z
s
2p+1 ep Re ρ h
p
p
(s−2) Re ρ p
(kRk
+
|ρ|
kSk)
(|c
|
+
|ρ|
|d
|)
1
−
e
0
0
p |ρ|p (Re ρ)p+1
× 1 − e−ps Re ρ + (|a0 | + |ρ| |b0 |)p
p i
× (kRk + |ρ| kSk)p 1 − e−ps Re ρ 1 − e(s−1) Re ρ
|ϕ2 (x, s; λ)|p dx ≤
0
as Re ρ > 0 we obtain
Z
(3.6)
p
|ϕ2 (x, s; λ)| dx
sup
0≤s≤1
p1
s
0
2
≤
21+ p eRe ρ (kRk + |ρ| kSk) [(|a0 | + |c0 |) + |ρ| (|b0 | + |d0 |)]
1
|ρ| (Re ρ)1+ p p1/p
.
From (3.4), (3.5) and (3.6), we obtain
2
kR (λ, Lp )k ≤
21+ p eRe ρ (kRk + |ρ| kSk) [(|a0 | + |c0 |) + |ρ| (|b0 | + |d0 |)] 1
|4 (λ)| |ρ| (Re ρ)1+ p p1/p
for ρ ∈ Σ δ = ρ ∈ C : |arg ρ| ≤ 2δ , ρ 6= 0 , we have (Re ρ)−1 <
2
1+ p2
2
1+ p2
2
kR (λ, Lp )k ≤
1
,
|ρ| cos( 2δ )
1+ p2
2
+1 ,
then
+ 1 eRe ρ (kRk + |ρ| kSk) [(|a0 | + |c0 |) + |ρ| (|b0 | + |d0 |)]
.
1+ p1
1
|4 (λ)| |ρ| |ρ|1+ p cos 2δ
p1/p
Finally, we obtain for λ = ρ2 ∈ Σδ
(3.7)
kR (λ, Lp )kLp ≤
c eRe ρ
1
|4 (λ)| |ρ|1+ p
kRk
+ kSk [(|a0 | + |c0 |) + |ρ| (|b0 | + |d0 |)] ,
|ρ|
where
1+ p2
2
c=
p1/p
2
+1
1+ p1 .
cos 2δ
1+ p2
3.2. Estimation of the Characteristic Determinant. The next step is to determine the cases
for which |∆(λ)| remains bounded below. It will then be necessary to bound |∆(λ)| appropriately. However, formula (2.5) is not useful for this purpose. It will be then necessary to make
some additional regularity hypotheses on the functions R and S, and so we assume that the
functions R and S are in C 2 ([0, 1] , C).
Integrating twice by parts in (2.5), we obtain
h
0
0
ρ
(3.8) ∆(λ) = e ρ(d0 S(0) − b0 S(1)) + (d0 S (0) + b0 S (1) + a0 S (1)
1
0
−b0 R (1) − d0 R (0) + c0 S (0)) + (a0 R (1) − c0 R (0) + b0 R (1)
ρ
1
0
0
0
0
0
−a0 S (1) − d0 R (0) + c0 S (0)) + 2 (−a0 R (1) − c0 R (0)) + Φ(ρ) ,
ρ
J. Inequal. Pure and Appl. Math., 5(2) Art. 38, 2004
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BVP
FOR
2 ND -O RDER D IFFERENTIAL O PERATORS
9
where
Z 1 R00 (t) S 00 (t) ρt
−ρ
−2ρ
(3.9) Φ(ρ) = (a0 − ρb0 )e + (c0 − ρd0 )e
−
e dt
ρ2
ρ
0
Z 1 R00 (t) S 00 (t) −ρt
−ρ
+ (a0 + ρb0 )e + (c0 + ρd0 )
−
e dt
ρ2
ρ
0
0
−ρ 1
+e
−b0 R0 (0) + d0 R0 (1) − c0 S 0 (1) + a0 S (0) + c0 R(1) − a0 R(0)
ρ
+ ρ(b0 S(0) − d0 S(1))] + 2e−2ρ [(a0 + ρb0 )(R(1) − ρS(1))
1
0
0
− (c0 − ρd0 )(R(0) + ρS(0)) + 2 (a0 + ρb0 )(R (1) − ρS (1))
ρ
1
0
0
+ 2 (c0 − ρd0 )(R (0) + ρS (0)) .
ρ
After a straightforward calculation, we obtain the following inequality valid for ρ ∈ Σ δ ,with
2
|ρ| sufficiently large,
Re ρ
kR00 k kS 00 k
e
−1
− Re ρ
−2 Re ρ
|Φ(ρ)| ≤ (|a0 | + |ρ| |b0 |) e
+ (|c0 | + |ρ| |d0 |) e
+
|ρ|
Re ρ
|ρ|2
00
kR k kS 00 k
1 − e− Re ρ
− Re ρ
+ (|a0 | + |ρ| |b0 |) e
+ (|c0 | + |ρ| |d0 |)
+
|ρ|
Re ρ
|ρ|2
1
0
0
0
0
+ 2e− Re ρ
−b0 R (0) + d0 R (1) − c0 S (1) + a0 S (0) + c0 R(1) − a0 R(0)
|ρ|
1
−2 Re ρ
+ |ρ| |b0 S(0) − d0 S(1)|] + e
(|a0 | + |ρ| |b0 |) (kRk + |ρ| kSk)
|ρ|
!
0
0
R + |ρ| S
1
+
(|c0 | + |ρ| |d0 |) (kRk + |ρ| kSk) + (|a0 | + |ρ| |b0 |)
|ρ|
|ρ|2
!#
0
0
R + |ρ| S
+ (|c0 | + |ρ| |d0 |)
.
|ρ|2
Then
00
(|a0 | + |c0 |)
2
kR k
00
|Φ(ρ)| ≤
+ (|b0 | + |d0 |)
+ kS k
|ρ|
|ρ|
|ρ| cos 2δ
1
1
0
0
0
0
+
2 −b0 R (0) + d0 R (1) − c0 S (1) + a0 S (0) + c0 R(1) − a0 R(0)
δ
|ρ| cos 2 |ρ|
1
|a0 |
kRk
+ |b0 S(0) − d0 S(1)|] +
+ |b0 |
+ kSk
|ρ|
|ρ|
|ρ| (cos 2δ )2
!
0
0
R
S
|c0 |
kRk
|a0 |
+
+ |d0 |
+ kSk +
+ |b0 |
+
|ρ|
|ρ|
|ρ|
|ρ|
|ρ|2
!#
0
0
R
S
|c0 |
(3.10)
+
+ |d0 |
+
.
|ρ|
|ρ|
|ρ|2
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M. D ENCHE AND A. KOURTA
Where we have used that Re(ρ) ≥ |ρ| cos( 2δ ), (1 − e− Re ρ ) < 1,
e− Re ρ ≤
2
1
ande−2 Re ρ ≤
, e− Re ρ < 1.
2
δ 2
|ρ| (cos 2 )
2 |ρ| (cos 2δ )2
2
There are several cases to analyze depending on the functions R and S.
• Case 1.
Suppose that S 6= 0, d0 6= 0, b0 6= 0, d0 S (0) − b0 S (1) = 0 and
0
0
k1 = d0 S (0) + b0 S (1) + a0 S (1) − b0 R (1) − d0 R (0) + c0 S (0) 6= 0.
From (3.8), we have for |ρ| sufficiently large
h
0
0
|4 (λ)| ≥ eRe ρ d0 S (0) + b0 S (1) + a0 S (1) − b0 R (1) − d0 R (0) + c0 S (0)
−
1
0
0
0
0
a0 R (1) − c0 R (0) + b0 R (1) − a0 S (1) − d0 R (0) + c0 S (0)
|ρ|
1
0
0
− 2 a0 R (1) + c0 R (0) − |Φ(ρ)| .
|ρ|
From (3.10) we deduce for ρ ∈ Σ δ , |ρ| ≥ r0 > 0.
2
|Φ(ρ)| ≤
c(r0 )
.
|ρ|
Then, we have
Re ρ
|4 (λ)| ≥ e
h
0
0
d0 S (0) + b0 S (1) + a0 S (1) − b0 R (1) − d0 R (0) + c0 S (0)
1
0
0
0
0
a0 R (1) − c0 R (0) + b0 R (1) − a0 S (1) − d0 R (0) + c0 S (0)
|ρ|
1
c(r0 )
0
0
− 2 a0 R (1) + c0 R (0) −
.
|ρ|
|ρ|
we can now choose r0 > 0, such that
−
1
0
0
0
0
a0 R (1) − c0 R (0) + b0 R (1) − a0 S (1) − d0 R (0) + c0 S (0)
r0
1
c(r0 )
0
0
+ 2 a0 R (1) + c0 R (0) +
r0
|ρ|
1
0
0
≤ d0 S (0) + b0 S (1) + a0 S (1) − b0 R (1) − d0 R (0) + c0 S (0) ,
2
then, for ρ ∈ Σ δ , |ρ| ≥ r0 > 0, we get
2
eRe ρ
|k1 | .
2
From (3.7), we deduce the following bound, valid for every |arg ρ| ≤ 2δ , ρ 6= 0
|a0 | + |c0 |
kRk
2c
+ (|b0 | + |d0 |)
+ kSk ,
kR (λ, Lp )k ≤
1
|ρ|
|ρ|
|ρ| p |k1 |
then
c1
kR (λ, Lp )k ≤
1 ,
|λ| 2p
|4 (λ)| ≥
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as |λ| −→ +∞, where
c1 =
2c kSk (|b0 | + |d0 |)
.
|k1 |
• Case 2.
If b0 = d0 = 0, S 6= 0 and a0 S (1) + c0 S (0) = 0, with
0
0
k2 = a0 R (1) − c0 R (0) − a0 S (1) + c0 S (0) 6= 0,
we have the following bound, valid for λ ∈ Σδ and |λ| big enough,
2c (|a0 | + |c0 |) kRk
kR (λ, Lp )k ≤
+ kSk ,
1
|ρ|
|ρ| p |k2 |
then
kR (λ, Lp )k ≤
c1
1
,
|λ| 2p
as |λ| −→ +∞, where
c1 =
2c (|a0 | + |c0 |) kSk
.
|k2 |
• Case 3.
If S = 0, b0 6= 0, d0 6= 0 and b0 R(1) + d0 R (0) = 0, with
0
0
k3 = a0 R (1) − c0 R (0) + b0 R (1) − d0 R (0) 6= 0.
Similarly, we get
kR (λ, Lp )k ≤
2c kRk
1
|ρ|1+ p |k3 |
(|a0 | + |c0 |)
+ (|b0 | + |d0 |) ,
|ρ|
then, we have
kR (λ, Lp )k ≤
c1
1
,
|λ| 2p
as |λ| −→ +∞, where
c1 =
2c kRk (|b0 | + |d0 |)
.
|k3 |
• Case 4.
If b0 = d0 = 0, S = 0 and a0 R (1) − c0 R (0) = 0 with
0
0
k4 = a0 R (1) − c0 R (0) 6= 0.
Again in this case, we have
kR (λ, Lp )k ≤
2c (|a0 | + |c0 |) kRk
1
,
|ρ| p |k4 |
then
kR (λ, Lp )k ≤
c1
1
,
|λ| 2p
as |λ| −→ +∞, where
c1 =
J. Inequal. Pure and Appl. Math., 5(2) Art. 38, 2004
c kRk (|a0 | + |c0 |)
.
|k4 |
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12
M. D ENCHE AND A. KOURTA
Definition 3.1. The boundary value conditions in (1.1) are called non regular if the functions
R, S ∈ C 2 ([0, 1] , C) and if and only if one of the following conditions holds
1-d0 S (0) − b0 (1) = 0, b0 6= 0, d0 6= 0, S 6= 0 and
0
0
d0 S (0) + b0 S (1) + a0 S (1) − b0 R (1) − d0 R (0) + c0 S (0) 6= 0
2- b0 = d0 = 0, kSk =
6 0, a0 S (1) + c0 S (0) = 0 with
0
0
a0 R (1) − c0 R (0) − a0 S (1) + c0 S (0) 6= 0
3-S = 0, b0 6= 0, d0 6= 0, b0 R(1) +0 R (0) 6= 0 with
0
0
a0 R (1) − c0 R (0) + b0 R (1) − d0 R (0) 6= 0
4-b0 = d0 = 0, S = 0, a0 R (1) − c0 R (0) = 0 with
0
0
a0 R (1) − c0 R (0) 6= 0.
This proves the following theorem
Theorem 3.1. If the boundary value conditions in (1.1) are non regular, then Σδ ⊂ ρ(Lp ) for
sufficiently large |λ| and there exists c > 0 such that
c
kR (λ, Lp )k ≤
1 .
|λ| 2p
Remark 3.2. From Theorem 3.1 it follows that the operator Lp , for p 6= ∞, generates an
analytic semigroup with singularities [25] of type A (2p − 1, 4p − 1).
3.3. Application. In the following, we apply the above obtained results to the study of a class
of a mixed problem for a parabolic equation with an weighted integral boundary condition
combined with another two point boundary condition of the form

∂u(t, x)
∂ 2 u(t, x)


−
a
= f (t, x)


∂t
∂x2



 L (u) = a u (0, t) + b u0 (0, t) + c u (1, t) + d u0 (1, t) = 0,
1
0
0
0
0
(3.11)
R
R

0
1
1


L2 (u) = 0 R(ξ) u(t, ξ) dξ + 0 S(ξ) u (t, ξ) dξ = 0,





u(0, x) = u0 (x),
where (t, ξ) ∈ [0, T ] × [0, 1]. Boundary value problems for parabolic equations with integral
boundary conditions are studied by [1, 3, 5, 6, 14, 15, 16, 17, 35] using various methods. For
instance, the potential method in [3] and [17], Fourier method in [1, 14, 15, 16] and the energy
inequalities method has been used in [5, 6, 35]. In our case, we apply the method of operator
differential equation. The study of the problem is then reduced to a cauchy problem for a
parabolic abstract differential equation, where the operator coefficients has been previously
studied. For this purpose, let E, E1 , and E2 be Banach spaces. Introduce two Banach spaces
(
)
Cµ ((0, T ] , E) =
f /f ∈ C ((0, T ] , E) , kf k = sup ktµ f (t)k < ∞ , µ ≥ 0,
t∈(0,T ]
(
Cµγ ((0, T ] , E) =
f /f ∈ C ((0, T ] , E) , kf k = sup ktµ f (t)k
t∈(0,T ]
+
sup
−γ µ
kf (t + h) − f (t)k h t < ∞ , µ ≥ 0, γ ∈ (0, 1] ,
0<t<t+h≤T
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13
and a linear space
C 1 ((0, T ] , E1 , E2 ) = f /f ∈ C ((0, T ] , E1 ) ∩ C 1 ((0, T ] , E2 ) , E1 ⊂ E2 ,
where C ((0, T ] , E) and C 1 ((0, T ] , E) are spaces of continuous and continuously differentiable, respectively, vector-function from (0, T ] into E. We denote, for a linear operator A in a
Banach space E, by
n
1 o
E(A) = u/u ∈ D(A), kukE(A) = kAuk2 + kuk2 2 ,
and
n
o
0
C 1 ((0, T ] , E(A), E) = f /f ∈ C ((0, T ] , E(A)) , f ∈ C ((0, T ] , E)) .
Let us derive a theorem which was proved by various methods in [25] and [31, 33]. Consider,
in a Banach space E, the Cauchy problem
( 0
u (t) = Au(t) + f (t), t ∈ [0, T ] ,
(3.12)
u(0) = u0 ,
where A is, generally speaking, unbounded linear operator in E, u0 is a given element of E,
f (t) is a given vector-function and u(t) is an unknown vector-function in E.
Theorem 3.3. Let the following conditions be satisfied:
(1) A is a closed linear operator in a Banach space E and for some β ∈ (0, 1] , α > 0
π
kR (λ, A)k ≤ C |λ|−β ,
|arg λ| ≤ + α, |λ| → ∞;
2
γ
(2) f ∈ Cµ ((0, T ] , E) for some γ ∈ (1 − β, 1] , µ ∈ [0, β) ;
(3) u0 ∈ D(A).
Then the Cauchy problem (3.12) has a unique solution
u ∈ C ([0, T ] , E) ∩ C 1 ((0, T ] , E(A), E)
and for the solution u the following estimates hold
ku(t)k ≤ C kAu0 k + ku0 k + kf kCµ ((0,t],E) , t ∈ (0, T ] ,
0
u (t) + kAu(t)k ≤ C tβ−1 (kAu0 k + ku0 k) + tβ−µ−1 kf kCµγ ((0,t],E) , t ∈ (0, T ] .
As a result of this we get the following theorem
Theorem 3.4. Let the following conditions be satisfied
π
(1) a 6= 0, |arg a| < ,
2
(2) the functions R(t), S(t) ∈ C 2 ([0, 1] , C) and one of the following conditions is satisfied
d0 S (0) − b0 S (1) = 0, b0 6= 0, d0 6= 0, and S 6= 0
0
0
d0 S (0) + b0 S (1) + a0 S (1) − b0 R (1) − d0 R (0) + c0 S (0) 6= 0,
or b0 = 0, d0 = 0, S 6= 0 , a0 S (1) + c0 S (0) = 0 with
0
0
a0 R (1) − c0 R (0) − a0 S (1) + c0 S (0) 6= 0,
or b0 6= 0, d0 6= 0, S = 0, b0 R(1) + d0 R (0) = 0 with
0
0
a0 R (1) − c0 R (0) + b0 R (1) − d0 R (0) 6= 0,
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14
M. D ENCHE AND A. KOURTA
or b0 = d0 = 0, S = 0, a0 R (1) − c0 R (0) = 0 with
0
0
a0 R (1) − c0 R (0) 6= 0.
i
h
1
1
γ
q
(3) f ∈ Cµ ((0, T ] , L (0, 1)) for some γ ∈ 1 − 2q , 1 and some µ ∈ 0, 2q ,
(4) u0 ∈ Wq2 (0, 1) , Li u = 0, i = 1, 2 .
Then the problem (3.11) has a unique solution
u ∈ C ((0, T ] , Lq (0, 1)) ∩ C 1 (0, T ] , Wq2 (0, 1), Lq (0, 1)
and for this solution we have the estimates:
(3.13)
ku(t, ·)kLq (0,1) ≤ c ku0 kWq2 (0,1) + kf kCµ ((0,t],Lq (0,1)) , t ∈ (0, T ] ,
(3.14)
ku00 (t, ·)kLq (0,1) + ku0 (t, ·)kLq (0,1)
1
1
≤ c t 2q −1 ku0 kWq2 (0,1) + t 2q −1−µ kf kCµγ ((0,t],Lq (0,1)) , t ∈ (0, T ] .
Proof. We consider in the space Lq (0, 1) 1 ≤ q < ∞, the operator A defined by
A(u) = au00 (x), D(A) = u ∈ Wq2 (0, 1), Li (u) = 0, i = 1, 2 .
Then problem (3.11) can be written as
( 0
u (t) = Au(t) + f (t),
u(0) = u0 ,
where u(t) = u(t, ·), f (t) = f (t, ·), and u0 = u0 (·) are functions with values in the Banach
1
π
space Lq (0, 1). From Theorem 3.1 we conclude that kR(λ, A)k ≤ c |λ|− 2q , for |arg λ| ≤ +α,
2
as |λ| → ∞.
Then, from Theorem 3.3 the problem (3.11) has a unique solution
u ∈ C ((0, T ] , Lq (0, 1)) ∩ C 1 (0, T ] , Wq2 (0, 1), Lq (0, 1)
and we have the following estimates
(3.15)
ku(t, .)kLq (0,1) ≤ c kAu0 kLq (0,1) + ku0 kLq (0,1) + kf kCµ ((0,t],Lq (0,1)) ,
(3.16)
0
+ kAu(t, ·)kLq (0,1)
1 1
≤ c t 2q −1 kAu0 kLq (0,1) + ku0 kLq (0,1) + t 2q −1−µ kf kCµγ ((0,t],Lq (0,1)) ,
u (t)
Lq (0,1)
where t ∈ [0, T ], from (3.15) we get
00
ku(t, ·)kLq (0,1) ≤ c ku0 kLq (0,1) + ku0 kLq (0,1) + kf kCµ ((0,t],Lq (0,1))
≤ c ku0 kWq2 (0,1) + kf kCµ ((0,t],Lq (0,1)) , t ∈ [0, T ] .
and from (3.16) we get
0
+ ku00 (t, ·)kLq (0,1)
1 1
≤ c t 2q −1 kAu0 kLq (0,1) + ku0 kLq (0,1) + t 2q −1−µ kf kCµγ ((0,t],Lq (0,1))
1
1
−1
−1−µ
2q
2q
ku0 kWq2 (0,1) + t
kf kCµγ ((0,t],Lq (0,1)) , t ∈ [0, T ] .
≤c t
u (t)
Lq (0,1)
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which gives the desired result.
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