ON THREE-POINT BOUNDARY VALUE PROBLEM
WITH A WEIGHTED INTEGRAL CONDITION FOR
A CLASS OF SINGULAR PARABOLIC EQUATIONS
ABDELFATAH BOUZIANI
Received 10 May 2002
We deal with a three-point boundary value problem for a class of singular parabolic equations with a weighted integral condition in place of one of standard boundary conditions. First we establish an a priori estimate in weighted
spaces. Then, we prove the existence, uniqueness, and continuous dependence
of a strong solution.
1. Introduction
In Q = {(x,t) ∈ R2 : 0 < x < b, 0 < t < T }, we consider the following problem:
given functions g, v0 , M, m, µ, µ1 , and µ2 , find the function v = v(x,t) as a solution of
ᏸv =
a
0
∂v
∂v α(t) ∂
−
x
= f (x,t), (x,t) ∈ Q,
∂t
x ∂x ∂x
0 v = v(x,0) = v0 (x), 0 < x < b,
xv(x,t)dx = m(t),
(1.1)
0 < a < b, 0 < t < T,
and, on x = b, one of the following conditions:
v(b,t) = µ(t),
∂v(b,t)
= µ1 (t),
∂x
∂v(b,t)
+ βv(b,t) = µ2 (t),
∂x
0 < t < T,
(1.2)
0 < t < T,
(1.3)
0 < t < T, β ∈ R∗+ ,
(1.4)
Copyright © 2002 Hindawi Publishing Corporation
Abstract and Applied Analysis 7:10 (2002) 517–530
2000 Mathematics Subject Classification: 35K20, 35D05, 35B45, 35B30
URL: http://dx.doi.org/10.1155/S1085337502206041
518
Singular parabolic equation with integral condition
together with the compatibility conditions
a
0
xv0 (x)dx = m(0),
v0 (b) = µ(0),
(1.5)
v0 (b) = µ1 (0), v0 (b) + βv0 (b) = µ2 (0), resp. .
(1.6)
We describe the complete investigation for problem (1.1) and (1.2). The investigation is similar for problems (1.1), (1.3) and (1.1), (1.4).
We note that problem (1.1) and (1.2) has not been studied previously. It
arises from some physical problems. For instance, if u denotes temperature in
a heat conduction problem, then m(t) represents the heat moment in the region
0 < x < a at time t. Problems for second-order singular parabolic equations with
two-point boundary values were considered in [1, 6, 11]. In [1], it is treated
a problem with homogeneous Dirichlet condition and the integral condition
b
0 v(x,t)dx = 0. As for [6, 11], are investigated problems which combine Dirichb
let condition with the integral condition 0 x2 v(x,t)dx = m(t), and Neumann
b
condition with the integral condition 0 xv(x,t)dx = m(t), respectively. However, in [6, 11], we cannot replace Dirichlet condition by Neumann or Robin
conditions and conversely, owing to operators of multiplication constructed to
establish a priori estimates for the considered problems. For other equations
with integral boundary condition(s), we refer the reader to [2, 3, 4, 5, 8, 7, 9],
and the references therein.
In this paper, we prove that problem (1.1) and (1.2) admits a unique strong
solution that depends continuously upon the data. The proof is based on an
energy inequality and on the density of the range of the operator associated to
the abstract formulation of the stated problem.
The paper is divided as follows. Section 2 is devoted to some preliminaries
needed in throughout. In Section 3, we prove the uniqueness and continuous
dependence of the solution. Then, in Section 4 we establish the existence of the
solution.
2. Preliminaries
We start by reducing (1.1) and (1.2) to an equivalent problem with homogeneous boundary conditions. For this purpose, we introduce a new unknown
function u which represents the deviation of the function v(x,t) from the function
U(x,t) =
6(x − b)
2a − 3x
m(t).
µ(t) + 3
2a − 3b
2a − 3a2 b
(2.1)
Therefore, problem (1.1) and (1.2) becomes: find a function u = u(x,t) solution
of
Abdelfatah Bouziani 519
ᏸu =
a
0
∂u
∂u α(t) ∂
−
x
= f (x,t),
∂t
x ∂x ∂x
0 u = u(x,0) = u0 (x),
xu(x,t)dx = 0,
(2.2)
(2.3)
u(b,t) = 0,
(2.4)
u0 (b) = 0.
(2.5)
α (t) ≤ c2 .
(2.6)
with
a
0
xu0 (x)dx = 0,
Assumption 2.1. For all t ∈ [0,T], we assume
0 < c0 ≤ α(t) ≤ c1 ,
Assumption 2.2. For all t ∈ [0,T], we assume
α (t) ≤ c4 .
0 < c3 ≤ α (t),
(2.7)
Throughout the paper, we use the following notation:
Qτ = (0,b) × (0,τ),
a
∗
u(ξ, ·)dξ,
xu=
x

1, x ∈ [0,a],
δ(x) = 
0, x ∈ [a,b],
0 ≤ τ ≤ T;
∗
x u=
2
aa
x
ξ
u(η, ·)dη dξ,

x2 ,
x ∈ [0,a],
x ∈ [a,b],
w(x) =  2
a,
(2.8)
∂u
w (x) ∗ ∂u
x ξ
+ w(x) .
Mu =
x
∂t
∂t
It is easy to observe, for x ∈ [0,a], that
∗
∗
∗
0 (xu) = a (xu) = a (xu) = 0,
a
a
∗ 2
x u dx ≤ 4 x2 u2 dx.
2
0
(2.9)
(2.10)
0
We introduce function spaces needed in our investigation. We denote by
L2σ (0,b) the weighted Lebesgue space that consists of all measurable functions
u equipped with the finite norm
uL2σ (0,b) :=
b
0
2
σ(x)u(x, ·) dx
1/2
.
(2.11)
By Hσ1 (0,b) we denote the weighted Sobolev space defined as the space of all
functions u ∈ L2σ (0,b) such that ∂u/∂x belongs to L2σ (0,b). The corresponding
520
Singular parabolic equation with integral condition
norm is
uHσ1 (0,b) :=
b
0
1/2
2 ∂u(x, ·) 2
dx
σ(x) u(x, ·) + .
∂x (2.12)
If σ(x) = 1, L2σ (0,b) and Hσ1 (0,b) are identified with the standard spaces L2 (0,b)
and H 1 (0,b), respectively. By H 1 (0,b;x2 δ,x), we denote the weighted Sobolev
space with the finite norm
b
uH 1 (0,b;x2 δ,x) :=
0
1/2
2
∂u(x, ·) 2
dx
x δ(x) u(x, ·) + x
.
∂x 2
(2.13)
1,∗
We denote by B2,x
(0,a) the weighted Bouziani space, first introduced in [5, 6],
with finite norm
1,∗
uB2,x
(0,a) :=
a
0
∗ (ξu)2 dx
x
1/2
.
(2.14)
Moreover, we use C(0,T;H) and L2 (0,T;H) for the sets of continuous and
L2 -integrable mappings (0,T) → H, respectively.
We, now, write problem (2.2), (2.3), and (2.4) as an operator equation
Lu = f ,u0 ,
(2.15)
where L = (ᏸ,) is an unbounded operator, with domain D(L), acting from
B to F, where D(L) is the set of functions u belonging to L2 (0,T;L2x (0,b)) for
which ∂u/∂t,(1/x)(∂u/∂x),∂2 u/∂x2 ,∂2 u/∂t∂x ∈ L2 (0,T;L2x (0,b)) and u satisfying
conditions (2.4), B is the Banach space obtained by completing the set D(L) with
respect to the norm
uB :=
2
∂u u2C(0,T;Hxw
1 (0,b)) + ∂t
1/2
,
(2.16)
L2 (0,T;L2xw (0,b))
1 (0,b) consisting of all elements
and F is the Hilbert space L2 (0,T;L2x (0,b)) × Hxw
( f ,u0 ) for which the following norm is finite
1/2
2
f ,u0 = f 2 2
u 0 1
+
.
2
L (0,T;Lx (0,b))
Hxw (0,b)
F
(2.17)
Abdelfatah Bouziani 521
3. Uniqueness and continuous dependence
First, we establish the following energy inequality.
Theorem 3.1. Under Assumption 2.1, the following estimate holds for any function
u ∈ D(L):
uB ≤ cLuF ,
(3.1)
where c is a positive constant independent of u.
Proof. Taking the scalar product in L2x (0,b) of (2.2) and Mu, and integrating the
result over (0,τ), we have
τ
0
f (·,t),Mu ·,t)
=
L2x (0,b) dt
τ
τ
∂u(·,t) 2
∂u(·,t) ∗ ∂u(·,t)
dt
+
ξ
dt
,w
∂t 2
x
∂t
∂t
0
0
L2 (0,b)
Lxw (0,b)
τ
(3.2)
∂
∂u(·,t) ∂u(·,t)
−
α(t)
x
,
dt
0
−
∂x
∂x
∂t
L2w (0,b)
∂
∂u(·,t) w ∗ ∂u(·,t)
α(t)
x
, x ξ
dt.
∂x
∂x
x
∂t
L2 (0,b)
τ
0
Integrating by parts the last three terms on the right-hand side of (3.2), we obtain
τ
0
∂u(·,t) ∗ ∂u(·,t)
dt
,w x ξ
∂t
∂t
L2 (0,b)
τ
∂u(·,t) ∗ ∂u(·,t)
=2
x
dt
, x ξ
∂t
∂t
0
L2 (0,a)
a
τ
∂u 2 ∗
= − x ξ
dt = 0,
∂t
0
−
τ
0
0
∂
∂u(·,t) ∂u(·,t)
α(t)
x
,
∂x
∂x
∂t
τ
dt
L2w (0,b)
a
τ
b
∂u ∂u dt − xa2 α(t) ∂u ∂u dt
∂x ∂t 0
∂x ∂t a
0
0
τ
∂u(·,t) ∂u(·,t)
α(t)
dt
,
+2
∂x
∂t
0
L2x2 (0,a)
=−
x3 α(t)
τ
+
0
=2
τ
−
0
α(t)
∂u(·,t) ∂2 u(·,t)
,
∂x
∂t∂x
∂u(·,t) ∂u(·,t)
,
α(t)
∂x
∂t
α(0) 1
u 2 2
0 Lxw (0,b) −
2
2
τ
0
dt
L2xw (0,b)
2
α(τ) ∂u(·,τ) dt +
2
∂x L2xw (0,b)
L2x2 (0,a)
∂u(·,t) 2
∂x 2
α (t)
Lxw (0,b)
dt,
522
Singular parabolic equation with integral condition
−
τ
0
∂
∂u(·,t) w ∗ ∂u(·,t)
x
, x ξ
dt
∂x
∂x
x
∂t
L2 (0,b)
τ
a
∂u(·,t) ∗ ∂u dt
= −2 α(t)x
x ξ
∂x
∂t 0
0
τ
∂u(·,t) ∂u(·,t)
−2
α(t)
dt.
,
∂x
∂t
0
L2x2 (0,a)
α(t)
(3.3)
Substituting (3.3) into (3.2), yields
τ
∂u(·,t) 2
∂u(·,τ) 2
dt + α(τ)
2 ∂t L2xw (0,b)
∂x L2xw (0,b)
0
τ
=2
f (·,t),Mu(·,t) L2x (0,b) dt
0
2
+ α(0)u0 L2xw (0,b) +
τ
0
(3.4)
∂u(·,t) 2
∂x 2
α (t)
dt.
Lxw (0,b)
According to Cauchy inequality and (2.10), the first term on the right-hand side
of (3.4) is estimated as follows:
τ
2
0
f (·,t),Mu(·,t)
≤ 2ε1
τ
0
L2x (0,b) dt
Lx (0,a) dt + ε2
2
∂u(·,t) 2 τ
+
∂t 1,∗
ε
1
τ
f (·,t)2 2
0
≤ 2ε1 a + ε2 a2
τ
0
B2,x (0,a)
0
f (·,t)2 2
Lxw (0,b) dt
2
1 τ
∂u(·,t) dt +
∂t 2
ε
0
2
f (·,t)2 2
Lx (0,b) dt +
dt
Lxw (0,b)
8a 1
+
ε1 ε2
τ ∂u(·,t) 2
∂t 2
0
dt.
Lxw (0,b)
(3.5)
Inserting (3.5) into (3.4), and choosing ε1 and ε2 so that the last integral in the
right-hand side of (3.5) will be absorbed in the left-hand side of (3.4), we get by
virtue of Assumption 2.1 that
3
2
τ
∂u(·,t) 2
∂t 2
0
≤ 68a2
τ
0
Lxw (0,b)
∂u(·,τ) 2
∂x 2
dt + c0 2
f ·,t 2
Lx (0,b)
Lxw (0,b)
2
dt + c1 u 2
0
Lxw (0,b) + c2
τ
∂u(·,t) 2
∂x 2
0
Lxw (0,b)
dt.
(3.6)
Abdelfatah Bouziani 523
Adding the obvious inequality
u(·,τ)2 2
Lxw (0,b)
τ
2
≤ u 0 2
Lxw (0,b) +
0
∂u(·,t) 2
Lxw (0,b) + ∂t 2
u(·,t)2 2
dt
Lxw (0,b)
(3.7)
to (3.6), we conclude that
τ
2
∂u(·,t) 2
dt + u(·,τ)Hxw
1 (0,b)
∂t 2
0
Lxw (0,b)
(3.8)
τ
τ
2
2
f (·,t)2 2
u 0 1
≤ c5
dt
+
+
c
u(
·
,t)
dt,
1 (0,b)
6
Lx (0,b)
Hxw (0,b)
Hxw
0
0
where
c5 =
max 68a2 ,c1 ,1
,
min 1/2,c0
c6 =
max c2 ,1
.
min 1/2,c0
(3.9)
Hence, application of Gronwall’s type inequality [3, Lemma 1] gives
τ
∂u(·,t) 2
∂t 2
0
Lxw (0,b)
2
dt + u(·,τ)Hxw
1 (0,b)
2
≤ c5 exp c6 T f 2L2 (0,T;L2x (0,b)) dt + u0 Hxw
1 (0,b) .
(3.10)
As the right-hand side of (3.10) is independent of τ, we take the upper bound of
the left-hand side. Hence, we obtain the required estimate.
Since we have no information concerning LB except that LB ⊂ F, we extend L,
so that inequality (3.1) holds for the extension and LB is the whole space. For
this purpose, we state the following proposition.
Proposition 3.2. Under the hypotheses of Theorem 3.1, the operator L : B → F
possesses a closure.
Proof. The proof is similar to that in [8].
Let L be the closure of the operator L, and D(L) its domain.
Definition 3.3. The solution of equation
Lu = f ,u0
is called strong solution of problem (2.2), (2.3), and (2.4).
(3.11)
524
Singular parabolic equation with integral condition
Passing to the limit in inequality (3.1), we obtain
uB ≤ cLuF ,
(3.12)
from which, we have the following corollaries.
Corollary 3.4. Under the hypotheses of Theorem 3.1. If, there exists a strong solution of problem (2.2), (2.3), and (2.4), then it is unique and depends continuously
upon ( f ,u0 ).
Corollary 3.5. The range LB is closed in F, and LB = LB.
4. Existence of the solution
Theorem 4.1. Under Assumptions 2.1 and 2.2, problem (2.2), (2.3), and (2.4)
admits a unique solution in the sense of Definition 3.3, for arbitrary f ∈ L2 (0,T;
1 (0,b), such that u ∈ C(0,T;H 1 (0,b)) and ∂u/∂t ∈ L2 (0,T;
L2x (0,b)) and u0 ∈ Hxw
xw
2
Lxw (0,b)).
Proof. Section 3 implies that L is injective. Therefore, to show the existence of
the solution, it suffices to prove that L is surjective. This can be fulfilled if we
establish the density of LB in F. To this end, we state the following result which
we need below.
Proposition 4.2. Under the hypotheses of Theorem 4.1, if
ᏸu,ω
L2 (0,T;L2x (0,b))
=0
(4.1)
for arbitrary u ∈ D0 (L) = {u/u ∈ D(L) : u = 0} and some ω ∈ L2 (0,T;L2x (0,b)),
then ω vanishes almost everywhere in Q.
Suppose that for the moment that Proposition 4.2 has been proved, and turning to the proof of Theorem 4.1. Let the element ( f ,u0 ) ∈ F be orthogonal to LB,
that is, let
Lu, f ,u0
F
= ᏸu, f L2 (0,T;L2x (0,b)) + u,u0 Hxw
1 (0,b) = 0,
u ∈ D(L).
(4.2)
Assuming in (4.2) that u is replaced by any element of D0 (L). It follows from
Proposition 4.2 that f = 0. Hence
u,u0
1 (0,b)
Hxw
= 0,
u ∈ D(L).
(4.3)
1 (0,b). The above relation implies that
But the set B is everywhere dense in Hxw
u0 = 0. Consequently LB = F. To complete the proof of Theorem 4.1, it remains
to prove Proposition 4.2.
Abdelfatah Bouziani 525
Proof of Proposition 4.2. We start by constructing the function ω. Since equality
(4.1) holds for arbitrary element of D0 (L), we express it in the following form:



0,
v =  t ∂u

dτ,

0 ≤ t ≤ s,
∂τ
s
(4.4)
s ≤ t ≤ T,
and ∂v/∂t is solution of the equation
−α(t)δ(x)∗
x
2
ξ
∂v
∂v
+ α(t) =
∂t
∂t
T
t
ω(x,τ)dτ = g,
(4.5)
where s is an arbitrary fixed number in [0,T].
It is easy to see from (4.4) that v ∈ Ds (L) = {u/u ∈ D(L) : u = 0 for t ≤ s} ⊆
D0 (L), and from (4.5) that ∂v/∂t |t=T = 0. Differentiating (4.5) with respect to t,
it yields
ω(x,t) =
∂v
∂v
∂
2
α(t)δ(x)∗x ξ
.
− α(t)
∂t
∂t
∂t
(4.6)
Lemma 4.3. Under the hypotheses of Theorem 4.1, the function ω defined by (4.6)
belongs to L2 (s,T;L2x (0,b)).
Proof of Lemma 4.3. Set y(x,t) := δ(x)∗x (ξ(∂v/∂t)) − ∂v/∂t, then ω(x,t) =
α (t)y + α(t)(∂y/∂t). We first prove that α (t)y is bounded in L2 (s,T;L2x (0,b)).
Indeed, we have according to Assumption 2.1
2
2
α y 2
L (s,T;L2x (0,b))
≤ c22 y 2L2 (s,T;L2x (0,b))
Tb ∂v
∂v 2
2
≤ c22
x δ(x)∗x ξ
dx dt
−
s
∂t
0
∂t
2
Tb
T a ∗2 ∂v 2
∂v 2
≤ 2c2
xx ξ
x(t)
dx dt .
dx dt +
s
∂t
0
s
0
∂t
(4.7)
By virtue of inequality (2.10), it yields
2
α y 2
L (s,T;L2x (0,b))
2
∂v ≤ 32c22 a6 ∂t
≤ 2c22 max
L2 (s,T;L2x (0,a))
∂v 2
6
2
∂v ∂t 2
+ 2c22 L (s,T;L2x (0,b))
(4.8)
16a ,1 .
∂t L2 (s,T;L2x (0,b))
It remains to prove that α(t)(∂y/∂t) belongs to L2 (s,T;L2x (0,b)). To this end, we
must use the following lemma, in which we summarize some of the properties
of the averaging operator ε defined by
1
ε y (·,t) :=
ε
T s
ϕ
t − t y ·,t dt ,
ε
(4.9)
526
Singular parabolic equation with integral condition
where ϕ ∈ C ∞ (R), ϕ = 0 in the neighborhood of t = s and t = T and outside the
interval (s,T), and such that
1
ε
ϕ
R
t − t
ε
dt =
1
ε
R
t − t
ε
ϕ
dt = 1.
(4.10)
Lemma 4.4. (i) If y ∈ L2 (s,T;H), the function ε y ∈ C ∞ (s,T;H);
(ii) if ε y L2 (s,T;H) ≤ y L2 (s,T;H) , and ε y − y L2 (s,T;H) → 0 as ε → 0;
(iii) (d/dt)ε y = ε (d/dt)y, u ∈ Ds (L);
(iv) if y ∈ L2 (s,T;H), then (∂/∂t)(αε y − ε αy)L2 (s,T;H) → 0 as ε → 0.
Proof. Proof of this lemma is similar to that of [10, Lemma 9.1].
We apply operators ε and ∂/∂t to (4.5), it follows
α(t)
∂ ∂
∂
ε y =
α(t)ε y − ε α(t)y − α (t)ε y,
ε g +
∂t
∂t
∂t
(4.11)
from which, we obtain, using (4.8), and properties (ii), (iii) of Lemma 4.4, that
2
α(t) ∂ ε y 2
∂t
L (s,T;L2x (0,b))
∂
2
≤3 ε g 2
∂t
L (s,T;L2x (0,b))
2
∂
+
∂t
αε y − ε αy 2
∂g ≤ 3
∂t
L2 (s,T;L2x (0,b))
L2 (s,T;L2x (0,b))
∂
+ 3
∂t
2
+ α ε y L2 (s,T;L2x (0,b))
(4.12)
2
αε y − ε αy L2 (s,T;L2x (0,b))
∂v 2
+ 6c22 max 16a6 ,1 ∂t 2
.
L (s,T;L2x (0,b))
Since the norm of α(t)(∂/∂t)ε y is bounded in L2 (s,T;L2x (0,b)), then we pass
to the limit in the above inequality, by taking into account property (iv) in
Lemma 4.4, we conclude that
∂ 2
α y ∂t 2
L (s,T;L2x (0,b))
≤ max
3,96c22 a6 ,6c22
∂g 2
∂t 2
L (s,T;L2x (0,b))
2
∂v +
∂t 2
(4.13)
.
L (s,T;L2x (0,b))
It then follows from (4.8) and (4.13) that ω ∈ L2 (s,T;L2x (0,b)).
Abdelfatah Bouziani 527
Substituting (2.2) and (4.6) into (4.1), we get
x
∂v ∂
∂v
∂v ∂
∂v
2
α∗x ξ
− x ,
α
,
∂t ∂t
∂t
∂t ∂t ∂t
L2 (s,T;L2 (0,a))
∂
∂v ∂
∂v
2
− α
x
,
α∗x ξ
∂x ∂x ∂t
∂t
L2 (s,T;L2 (0,a))
∂
∂v ∂
∂v
+ α
x
,
α
= 0.
∂x ∂x ∂t ∂t L2 (s,T;L2 (0,b))
L2 (s,T;L2 (0,b))
(4.14)
Integrating by parts each term of (4.14), by taking into account (4.4) and
(4.5), we obtain
x
∂v ∂
∂v
2
α∗x ξ
,
∂t ∂t
∂t
=
− x
0
1
2
b
0
− α
L2 (s,T;L2 (0,a))
Ta
∗ ∂v(ξ,s) 2
∗ ∂v 2
ξ
dx − 1
dx,
α(s)
ξ
α
(t)
x
x
∂v ∂
∂v
,
α
∂t ∂t ∂t
=
1
2
a
∂t
2
s
0
∂t
L2 (s,T;L2 (0,b))
2
∂v(x,s) 2
1 T b ∂v xα(s)
xα (t)
∂t dx − 2
∂t dx dt,
∂
∂v ∂
∂v
2
x
,
α∗x ξ
∂x ∂x ∂t
∂t
s
0
L2 (s,T;L2 (0,a))
2
∂v 2
1 a 2
2 2
=
x α (t) dx dt +
x α(T)α (T)v(x,T) dx
∂t
2 0
s
0
1 T a 2 2
−
x α (t) + α(t)α (t) |v|2 dx dt
Ta
2
−
− α
s
0
T a
s
α(t)∗x
0
∂
∂v ∂
∂v
x
,
α
∂x ∂x ∂t ∂t
∂v
∂v
+ α (t)∗x v x dx dt,
∂t
∂t
L2 (s,T;L2 (0,b))
2 2
∂ v ∂v(x,T) 2
1 b
=
α2 (t)x
xα(T)α (T)
∂x∂t dx dt + 2
∂x dx
s
0
0
∂v 2
1 T b 2
−
α (t) + α(t)α (t) x
∂x dx dt.
2
Tb
s
0
It then follows from (4.15) that
Ta
2
s
0
2
2 2
Tb
∂v ∂ v dx dt + 2
xα2 (t)
∂x∂t dx dt
∂t
x2 α2 (t)
s
0
(4.15)
528
Singular parabolic equation with integral condition
2
b
2
∂v(ξ,s) dx + xα(s) ∂v(x,s) dx
∂t
∂t
0
0
a
b
∂v(x,T) 2
2
2
dx
+ x α(T)α (T) v(x,T) dx + xα(T)α (T)
∂x 0
0
2
Ta
Tb
∗ ∂v 2
∂v =
α (t)x ξ
dx +
xα (t)
∂t dx dt
∂t s
0
s
0
a
+
Ta
+
s
∗
α(s)
x ξ
0
Tb
x2 α (t) + α(t)α (t) |v|2 dx dt
2
2
∂v 2
α (t) + α(t)α (t) x
∂x dx dt
s
0
T a
∂v
∂v
−2
α(t)∗x
+ α (t)∗x v x dx dt.
∂t
∂t
s
0
+
(4.16)
Applying the Cauchy inequality and inequality (2.10) to the last term on the
right-hand side of (4.16), we obtain
−2
T a
∂v
∂v
+ α (t)∗x v x dx dt
∂t
∂t
s
0
T
a
T a 2
∗ ∂v 2
∂v 2
≤
x2 dx
dt
+
2
α
dt
∂t x ∂t dx
s
0
s
0
α(t)∗x
T
+2
s
α dt
2
a
0
∗ 2
v dx
x
T a 2
T
a 2
∂v ∂v ≤a
x
α2 dt x
∂t dx dt + 8a
∂t dx
s
0
s
0
T
a
2
+ 8 α dt x2 |v|2 dx
s
(4.17)
0
b 2
∂v ≤
a + 8aα dt x
∂t dx
s
0
2 T
b
∂v 2
2
2
x δ(x)|v| + x
+ 8 α dt
dx.
T
2
s
∂x
0
Inserting (4.17) into (4.16) and employing Assumptions 2.1 and 2.2, we conclude that
2
∂v ∂t 2
L (s,T;H 1 (0,b;x2 δ,x))
∂v(·,s) 2
∂t 1,∗
+
T ∂v(·,t) 2
≤ c7
∂t 1,∗
s
B2,x (0,a)
B2,x (0,a)
∂v(·,s) 2
∂t 2
+
∂v(·,t) 2
+
∂t 2
Lx (0,b)
Lx (0,b)
2
+ v(·,T)H 1 (0,b;x2 δ,x)
2
+ v(·,t) 1
H (0,b;x2 δ,x)
dt,
(4.18)
Abdelfatah Bouziani 529
where
max c1 ,c1 + 8ac12 + a,c22 + c1 c4 + 8c12
c7 =
.
min 2c02 ,c0 ,c0 c3
(4.19)
To continue, we introduce a new function θ defined by
θ(x,t) :=
T
∂v
dτ,
∂τ
t
(4.20)
from which we deduce that θ(x,s) = v(x,T) and v(x,t) = θ(x,s) − θ(x,t). Hence,
inequality (4.18) becomes
2
∂v ∂t 2
∂v(·,s)2
∂t 1,∗
+
L (0,T;H 1 (0,b,x2 δ,x))
≤ 2c7
2
+ 1 − 2c7 (T − s) θ(·,s)H 1 (0,b;x2 δ,x)
B2,x (0,a)∩L2x (0,b)
T ∂v(·,t) 2
∂t 1,∗
B2,x (0,a)∩L2x (0,b)
s
2
+ θ(·,t)H 1 (0,b;x2 δ,x) dt.
(4.21)
If s0 verifies 1 − 2c7 (T − s0 ) = 1/2, then the above inequality implies
2
∂v ∂t 2
L (0,T;H 1 (0,b,x2 δ,x))
∂v(·,s) 2
∂t 1,∗
+
B2,x (0,a)∩L2x (0,b)
T ∂v(·,t) 2
≤ 4c7
∂t 1,∗
s
B2,x (0,a)∩L2x (0,b)
2
+ θ(·,s)H 1 (0,b;x2 δ,x)
(4.22)
2
+ θ(·,t) 1
H (0,b;x2 δ,x)
dt,
for all s ∈ [s0 ,T]. By virtue of [3, Lemma 1] for the interval (s,T), (4.22) implies
that
2
∂v ∂t 2
L (0,T;H 1 (0,b,x2 δ,x))
2
+ θ(·,s) 1
H
∂v(·,s) 2
(0,b;x2 δ,x) + ∂t 1,∗
B2,x (0,a)∩L2x (0,b)
≤ 0,
(4.23)
from which we conclude that ∂v/∂t ≡ 0 and thus ω ≡ 0 almost everywhere in
(s0 ,T) × (0,b). As s is an arbitrary fixed point in [0,T], using the same reasoning
again, from which, step by step, we deduce that ω ≡ 0 almost everywhere in Q.
This achieves the proof of Theorem 4.1.
References
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[2]
N.-E. Benouar and N. I. Yurchuk, A mixed problem with an integral condition for
parabolic equations with a Bessel operator, Differentsial’nye Uravneniya 27 (1991),
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530
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Singular parabolic equation with integral condition
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Abdelfatah Bouziani: Département de Mathématiques, Centre Universitaire
Larbi Ben M’hidi, Oum El Bouagui, 04000, Algeria
Current address: Mathematical Division, The Abdus Salam International Centre
for Theoretical Physics (ICTP), Strada Costiera 11, 34100 Trieste, Italy
E-mail address: af bouziani@hotmail.com, bouziani@ictp.trieste.it