IJMMS 30:7 (2002) 435–447
PII. S0161171202004507
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
ON THE SOLVABILITY OF A CLASS OF SINGULAR PARABOLIC
EQUATIONS WITH NONLOCAL BOUNDARY CONDITIONS
IN NONCLASSICAL FUNCTION SPACES
ABDELFATAH BOUZIANI
Received 11 January 2000
The aim of this paper is to prove the existence, uniqueness, and continuous dependence
upon the data of a generalized solution for certain singular parabolic equations with initial
and nonlocal boundary conditions. The proof is based on an a priori estimate established in
nonclassical function spaces, and on the density of the range of the operator corresponding
to the abstract formulation of the considered problem.
2000 Mathematics Subject Classification: 35K20, 35B30, 35D05, 46E40, 46E99.
1. Introduction. This paper is devoted to the solvability of a certain singular parabolic problem with a nonlocal boundary condition. It can be a part in the contribution
of the development of the a priori estimates method for solving such problems. The
questions related to these problems are so miscellaneous that the elaboration of a general theory is still premature. Therefore, the investigation of these problems requires
at every time a separate study.
This work can be considered as a continuation of the results of Yurchuk [12], Benuar
and Yurchuk [1], Bouziani [2, 3, 5, 4, 6], Bouziani and Benouar [7, 8], and Mesloub and
Bouziani [9], in so far as, on the one hand, the studied equation is parabolic and, on
the other hand, the boundary condition is of integral type.
The remainder of the paper is divided into four sections. In Section 2, we give the
statement of the problem. Then in Section 3, we first introduce the appropriate function spaces needed in our investigation, the abstract formulation of the problem and
the sense of the generalized solution are presented in Section 3.2, and some properties of special smoothing operators are considered in Section 3.3. The uniqueness and
the continuous dependence upon the data of a solution are established in Section 4.
In Section 5, the existence of the generalized solution is proved.
2. Statement of the problem. In the rectangle Q = (0, b) × (0, T ), we consider the
singular parabolic equation
ᏸz =
∂z a(t) ∂
∂z
−
x
= f (x, t),
∂t
x ∂x
∂x
(2.1)
where b and T are fixed but arbitrary positive numbers, and a(t) is a known function
satisfying the following assumption.
436
ABDELFATAH BOUZIANI
Assumption 2.1. For t ∈ [0, T ], we assume that
c0 ≤ a(t) ≤ c1 ,
a (t) ≤ c2 .
(2.2)
In Assumption 2.1, and throughout, we suppose that ci (where i = 0, . . . , 4) are positive
constants.
We pose the following problem for (2.1): given the data f , Φ, µ, and M, find a function
z = z(x, t) subject to the initial condition
z = z(x, 0) = Φ(x),
for 0 ≤ x ≤ b,
(2.3)
the Dirichlet condition
z(b, t) = µ(t),
for 0 ≤ t ≤ T ,
Φ(b) = µ(0),
(2.4)
and the weighted integral condition
b
b
0
x 2 z(x, t)dx = M(t) for 0 ≤ x ≤ b,
0
x 2 Φ(x)dx = M(0).
(2.5)
We transform problem (2.1), (2.3), (2.4), and (2.5) with inhomogeneous boundary
conditions into a problem with homogeneous boundary conditions. For this, we put
z(x, t) = u(x, t) + ζ(x, t), where
ζ(x, t) =
b3
12
x
µ(t) + 4 (b − x) M(t) −
µ(t) .
b
b
4
(2.6)
Then, problem (2.1), (2.3), (2.4), and (2.5) can be transformed as follows: find a function
u = u(x, t) satisfying
ᏸu =
∂u a(t) ∂
∂u
−
x
= f (x, t) − ᏸζ = f (x, t),
∂t
x ∂x
∂x
u = u(x, 0) = Φ(x) − ζ = ϕ(x),
(2.8)
u(b, t) = 0,
b
0
(2.7)
x 2 u(x, t)dx = 0.
(2.9)
3. Preliminaries
3.1. Function spaces. We first introduce appropriate function spaces. We denote
by C0 (0, b) the vector space of continuous functions with compact support in (0, b).
Since such functions are Lebesgue integrable with respect to dx, we can define on
C0 (0, b) the bilinear form ((·, ·))x given by
(u, w) x =
b
0
∗
∗
x (ξu) · x (ξw)dx,
(3.1)
b
where ∗
x g = x g(ξ, t)dξ. We recall that ((·, ·))x is a scalar product on C0 (0, b) for
which C0 (0, b) is not complete. Thus we are led to introduce its completion.
437
ON THE SOLVABILITY OF A CLASS OF SINGULAR PARABOLIC . . .
1,∗
Definition 3.1. We denote by B2,x (0, b) a completion of C0 (0, b) for the scalar
product defined by (3.1), called the space of square integrable weighted primitive functions on (0, b) (or the weighted Bouziani space).
1,∗
1,∗
Remark 3.2. If x = 1, then the space B2,x (0, b) is identified with the space B2 (0, b),
first introduced in [2, 5].
1
(0, b) a completion of C0 (0, b) for the scalar
Definition 3.3. We denote by B2,ρ
product defined by
T
t ρ(τ)u · t ρ(τ)w dt,
(3.2)
(u, w) ρ =
0
where t g =
t
0
g(x, τ)dτ and ρ(t) = ect .
1,∗
Now, we generalize Definitions 3.1 and 3.3 of weighted Bouziani spaces B2,x (0, b)
1
and B2,ρ
(0, T ).
Definition 3.4. Let (0, b) (resp., (0, T )) be an open interval in R, let σ (x) (resp.,
+
ρ(t)) be a continuous function from (0, b) to R+
∗ (resp., from (0, T ) to R∗ ), let m be
m,∗
m
a non-negative integer and let 1 ≤ p ≤ ∞. Then we define Bp,σ (0, b) (resp., Bp,ρ
(0, T ))
to be the completion of the space C0 (0, b) (resp., C0 (0, T )) for the norm
b
1/p
p
∗m
m,∗
uBp,σ
=
(σ
u)
dx
,
(3.3)
x
(0,b)
0
respectively,
T
m
uBp,ρ
(0,T ) =
0
p
m
t (ρu) dt
1/p
,
(3.4)
and for p = 2, we define a scalar product by
m
∗m
(u, w)B m,∗ (0,b) = ∗
x (σ u), x (σ w) L2 (0,b) ,
(3.5)
m
m
m
(u, w)B2,ρ
(0,T ) = t (ρu), t (ρw) L2 (0,T ) .
(3.6)
2,σ
respectively,
0,∗
0
Remark 3.5. The spaces B2,σ (0, b) and B2,ρ
(0, T ) coincide (with equality of norm
2
2
of graphs) with the spaces Lx (0, b) and Lρ (0, T ), respectively; that is, by the norms
of functions u from L2x (0, b) and L2ρ (0, T ) we understand the nonnegative numbers:
b
T
uL2x (0,b) = { 0 (xu)2 dx}1/2 and uL2ρ (0,T ) = { 0 (ρ(t)u)2 dt}1/2 , respectively.
In this paper, we also use other weighted spaces such as L2σ (0, b), L2s (0, b), and
√
where σ (x) = x 2 , s(x) = x, and r (t) = ρ(t) = ect/2 , which are Hilbert
spaces of (classes of) weighted square integrable functions with finite norms:
b
1/2
uL2σ (0,b) =
(σ u)2 dx
,
L2r (0, T ),
0
b
uL2s (0,b) =
1/2
(su)2 dx
T
uL2r (0,T ) =
,
0
1/2
(r u)2 dt
0
.
(3.7)
438
ABDELFATAH BOUZIANI
Let Hs1 (0, b) = {u/u ∈ L2s (0, b), ∂u/∂x ∈ L2s (0, b),
Hilbert space for the norm
b
0
x 2 u(x, t)dx = 0}, which is the
1/2
∂u 2
.
uHs1 (0,b) = u2L2 (0,b) + s
∂x L2s (0,b)
(3.8)
Let H be a Hilbert space with a norm ·H . We denote by L2 (0, T ; H) (resp., L2r (0, T ; H))
the set of all measurable abstract functions u(·, t) from (0, T ) into H such that
T
uL2 (0,T ;H) =
u(·, t)
2 dt
1/2
H
0
< ∞,
(3.9)
respectively,
T
uL2r (0,T ;H) =
0
ct/2 u(·, t)
2 dt
e
H
1/2
< ∞.
(3.10)
Let C(0, T ; H) be the set of all continuous functions u(·, t) : (0, T ) → H with
uC(0,T ;H) = sup u(·, τ)
H < ∞.
(3.11)
0≤τ≤T
1
(0, T ; H) for the space of functions from (0, T ) into H which are weighted
We write B2,ρ
Bouziani space for the measure dt. It is a Hilbert space for the norm
T
uB 1
2,ρ
(0,T ;H) =
0
2
cτ t e u(·, τ)
H
dt
1/2
.
(3.12)
The following inequalities are well known and are frequently used in this paper. We
list them here for convenience.
Lemma 3.6. For x ∈ (0, b), the following inequalities hold:
u2L2 (0,b) ≤ bu2L2 (0,b) ,
x
u2 1,∗
B2
u2 1,∗
s
(0,b)
B2,x (0,b)
(3.13)
≤ 4u2L2 (0,b) ,
x
≤
2
b
u2L2 (0,b) .
x
2
(3.14)
We are now in a position to give the abstract formulation corresponding to the
problem (2.7), (2.8), and (2.9).
3.2. Abstract formulation. We consider problem (2.7), (2.8), and (2.9) as the solution of the abstract equation
Lu = (f , ϕ),
(3.15)
439
ON THE SOLVABILITY OF A CLASS OF SINGULAR PARABOLIC . . .
where L is the operator which maps u(x, t) to the pair of elements ᏸu and u, so that
Lu = (ᏸu, u).
(3.16)
We consider L as an unbounded operator with domain D(L) consisting of all func1,∗
tions u belonging to L2 (0, T ; B2,x (0, b)) for which ∂u/∂t, (1/x)(∂u/∂x), ∂ 2 u/∂x 2 ∈
1,∗
L2 (0, T ; B2,x (0, b)) and satisfying conditions (2.9). We complete D(L) in the norm
1/2
∂u 2
2
uB = +
u
;
∂t 2
1,∗
C(0,T ;Hs1 (0,b))
L (0,T ;B2,x (0,b))
(3.17)
this yields a Banach space B. The elements of B are continuous functions on [0, T ]
with values in Hs1 (0, b). Hence on B, the following mapping is defined and continuous:
: B u → u = u(x, 0) ∈ Hs1 (0, b).
(3.18)
We write F for the Hilbert space L2 (0, T ; L2s (0, b))×Hs1 (0, b) consisting of all elements
(f , ϕ) for which the norm
1/2
(f , ϕ)
= f 2
+ ϕ2H 1 (0,b)
F
L2 (0,T ;L2 (0,b))
s
s
(3.19)
is finite. We consider the operator L with the above domain as a mapping from B
into F .
Now, we can introduce the concept of a generalized solution of problem (2.7), (2.8),
and (2.9). Let L̄ be the closure of the operator L.
Definition 3.7. A solution of the operator equation
L̄u = (f , ϕ),
(f , ϕ) ∈ F ,
(3.20)
is called a generalized solution of problem (2.7), (2.8), and (2.9).
To prove the solvability of problem (2.7), (2.8), and (2.9) in the sense of Definition 3.7,
we establish the a priori estimate
uB ≤ cLuF ,
u ∈ D(L).
(3.21)
It follows from (3.21) that there is a bounded inverse L−1 on the range R(L) of L.
However, since we have no information concerning R(L) except that R(L) ⊂ F , we
must extend L, so that an a priori estimate like (3.21) holds for the extension. For
this, we prove that L admits a closure. Thus we extend (3.21) to u ∈ D(L̄) by passing
to the limit. It follows that the closure procedure for L reduces to the closure of the
range R(L) in F , so that R(L) = R(L̄), and a bounded inverse L̄−1 exists on R(L̄), so
the uniqueness of a generalized solution. For existence, it remains to prove that R(L)
does not have an orthogonal complement in F .
3.3. Smoothing operators. We consider the operators defined by the relations
−1 1 ρε v = t e(1/ε)(τ−t) v , ε > 0,
ε
−1 ∗
1 (1/ε)(t−τ) v = − ∗
v , ε > 0,
ρε
e
ε t
(3.22)
440
ABDELFATAH BOUZIANI
T
where ∗
t g = t g(x, τ)dτ. These operators, first proposed by Yurchuk in abstract
form in [11], are used as smoothing operators with respect to t [12]. They furnish the
solutions of the problems
−1 ∂ ρε−1 v −1 + ρε v = v,
ρε v(x, 0) = 0,
ε
∂t
(3.23)
−1 ∗
−1 ∗
∂ ρε
v −1 ∗
−ε
v = v,
ρε
v(x, T ) = 0,
+ ρε
∂t
respectively. These operators have, for all v ∈ L2 (0, T , L2 (0, b)), the following
properties:
(P1) the functions (ρε−1 )v and (ρε−1 )∗ v ∈ H 1 (0, T ), with (ρε−1 )v(x, 0) = 0, and
(ρε−1 )∗ v(x, T ) = 0;
(P2) the operators (ρε−1 )∗ are conjugate to (ρε−1 ), that is,
−1 ∗
v · ρε−1 ω dx dt, ∀ω ∈ L2 (0, T );
(3.24)
ρε v · ω dx dt =
Q
(P3)
(P4)
Q
(ρε−1 )(∂v/∂τ) = (∂/∂t)(ρε−1 )v + (1/ε)e−t/ε · v(x, 0);
T
T
T
−1
−1
0 (ρε )vL2 (0,b) dt ≤ 0 vL2 (0,b) dt and 0 (ρε )v
− vL2 (0,b) dt → 0, when
ε → 0;
T
T
T
(P5) 0 (ρε−1 )∗ vL2 (0,b) dt ≤ 0 vL2 (0,b) dt and 0 (ρε−1 )∗ v −vL2 (0,b) dt → 0, when
ε → 0;
(P6) if A(t)v = a(t)(∂/∂x)(x(∂v/∂x)) then
(3.25)
A(t) ρε−1 v = ρε−1 A(τ)v + ε ρε−1 A (τ) ρε−1 v,
where A (t)v = a (t)(∂/∂x)(x(∂u/∂x)).
For the proof of these properties, see, for instance, [4].
4. Uniqueness and continuous dependence. In this section, we first establish an a
priori estimate. The uniqueness and the continuous dependence of the solution upon
the data then are direct corollary of it.
Theorem 4.1. Under Assumption 2.1, the solution of problem (2.7), (2.8), and (2.9)
satisfies the following a priori estimate:
uB ≤ cLuF ,
(4.1)
where c is a positive constant independent of u.
1,∗
Proof. We consider the scalar product in L2 (0, τ; B2,x (0, b) ∩ L2s (0, b)), with 0 ≤
τ ≤ T , of (2.7) and ∂u/∂t, yields
τ
τ
∂u(·, t) 2
∂u(·, t) ∂u(·, t)
a(t) ∂
x
,
dt −
dt
∂t 1,∗
1,∗
x ∂x
∂x
∂t
0
0
B2,x (0,b)
B2,x (0,b)
τ
τ
∂u(·, t) 2
∂u(·, t) ∂u(·, t)
a(t) ∂
x
,
dt
−
dt
∂t 2
x ∂x
∂x
∂t
0
0
Ls (0,b)
L2
s (0,b)
τ
τ
∂u(·, t)
∂u(·, t)
=
dt
+
dt.
f (·, t),
f
(·,
t),
1,∗
∂t
∂t
0
0
B2,x (0,b)
L2
s (0,b)
+
(4.2)
ON THE SOLVABILITY OF A CLASS OF SINGULAR PARABOLIC . . .
441
The standard integration by parts of the second and last terms on the left-hand side
of (4.2) leads to
τ
−
0
τb
a(t) ∂
∂u ∗ ∂u
∂u(·, t) ∂u(·, t)
dt
=
a(t)x
x
,
dx dt,
ξ
1,∗
x ∂x
∂x
∂t
∂x x
∂t
0 0
B2,x (0,b)
τ
−
0
1 b
∂u(x, τ) 2
a(t) ∂
∂u(·, t) ∂u(·, t)
dt =
a(τ)x
dx
x
,
x ∂x
∂x
∂t
2 0
∂x
L2
s (0,b)
1 b
dϕ 2
−
a(0)x
dx
2 0
dx
∂u 2
1 τ b a (t)x
dx dt.
−
2 0 0
∂x
(4.3)
Substituting (4.3) into (4.2), we get
τ
τ
∂u(·, t) 2
∂u(·, t) 2
1 b
∂u(x, τ) 2
dt +
dt +
a(τ)x
dx
∂t 1,∗
∂t 2
2 0
∂x
0
0
B2,x (0,b)
Ls (0,b)
=
τ
τ
1 b
dϕ 2
∂u
∂u(·, t)
dt
+
dt
+
a(0)x
dx
f,
f
(·,
t),
1,∗
∂t B2,x
∂t
2 0
dx
0
0
(0,b)
L2
s (0,b)
+
1
2
τb
0
a (t)x
0
∂u
∂x
τb
2
dx dt −
a(t)x
0
0
∂u ∗ ∂u
dx dt.
x ξ
∂x
∂t
(4.4)
In light of the Cauchy inequality and inequality (3.13), the first two terms and the last
term in the right-hand side of (4.4) are then majorized as follows:
τ
f (·, t),
0
≤
ε1
2
τ
f (·, t),
0
∂u(·, t)
∂t
τ
0
≤
bε3
2
dt
dt +
τ
0
1
2ε1
τ
∂u(·, t) 2
dt,
∂t 1,∗
0
B2,x (0,b)
L2
s (0,b)
dt
f (·, t)2L2 (0,b) dt +
s
0
a(t)x
0
1,∗
B2,x (0,b)
f (·, t)2 1,∗
τb
−
B2,x (0,b)
0
τ
ε2
≤
2
∂u(·, t)
∂t
1
2ε2
τ
∂u(·, t) 2
dt,
∂t 2
0
Ls (0,b)
∂u ∗ ∂u
x ξ
dx dt
∂x
∂t
τ
∂u(·, t) 2
∂u(·, t) 2
1
a2 (t)
dt +
dt.
∂x 2
1,∗
2ε3 0
∂t B2,x
(0,b)
Ls (0,b)
(4.5)
442
ABDELFATAH BOUZIANI
Combining the inequalities (4.5) with (4.4), choosing ε1 = 3/2, ε2 = 3/4, and ε3 = 3/2,
and using Assumption 2.1, we obtain
1
3
2
τ ∂u(·, t) 2
∂u(·, t) 2
c0 ∂u(·, τ) +
dt + ∂t 2
∂t 1,∗
2
∂x L2s (0,b)
0
B2,x (0,b)
Ls (0,b)
≤
3
4
+
τ
0
f (·, t)2 1,∗
B2,x (0,b)
3bc12
4
dt +
3
8
τ
0
f (·, t)2L2 (0,b) dt +
s
2
c1 dϕ 2 dx L2s (0,b)
(4.6)
τ
∂u(·, t) 2
dt.
∂x 2
0
Ls (0,b)
Observing that
1
1
1
u(·, τ)
22
ϕ2L2 (0,b) +
Ls (0,b) ≤
s
3
3
3
τ
0
u(·, t)2L2 (0,b) dt +
s
1
3
τ
∂u(·, t) 2
dt,
∂t 2
0
Ls (0,b)
(4.7)
it follows by using (3.13) and (3.14) that
τ
2
∂u(·, t) 2
dt + u(·, τ)
Hs1 (0,b)
∂t 1,∗
0
B2,x (0,b)
τ
≤ c3
(4.8)
τ
f (·, t)2L2 (0,b) dt + ϕ2H 1 (0,b)
s
s
0
+ c4
u(·, t)2H 1 (0,b) dt,
s
0
where
c3 =
max c1 /2, 3 1 + b3 /8
,
min 1/3, c0 /2
c4 =
max 1/3, 3bc12 /4
.
min 1/3, c0 /2
(4.9)
We eliminate the last term on the right-hand side of (4.8). To do that we use [3,
Lemma 3.1] to obtain
τ
∂u(·, t) 2
2
dt + u(·, τ)
Hs1 (0,b)
∂t 1,∗
0
B2,x (0,b)
T
≤ c3 exp c4 T
−f (·, t)2L2 (0,b) dt + ϕ2H 1 (0,b) .
0
s
(4.10)
s
Since the right-hand side here does not depend on τ; we take the upper bound of the
1/2
left-hand side on τ from 0 to T ; hence (4.1) holds with c = c3 exp(c4 T /2), and this
proves Theorem 4.1.
We show that the operator L admits a closure, that is, the closure of the graph
G(L) ⊂ B × F of L is a graph G(L̄) = G(L) of operator L̄.
Proposition 4.2. The operator L : B → F with domain D(L) has a closure.
Proof. For the proof, the reader should refer to [10].
ON THE SOLVABILITY OF A CLASS OF SINGULAR PARABOLIC . . .
443
We extend the a priori estimate (4.1) to u ∈ D(L̄) by passing to the limit, that is,
uB ≤ c L̄u
F ,
∀u ∈ D L̄ .
(4.11)
From (4.11) we conclude the following corollaries.
Corollary 4.3. Let the conditions of Theorem 4.1 be satisfied. If problem (2.7),
(2.8), and (2.9), has a generalized solution, then this solution is unique and depends
continuously on (f , ϕ).
Corollary 4.4. The range R(L̄) of L̄ is closed in F and R(L̄) = R(L), where R(L) is
the range of L.
5. The existence of the solution. Now we want to prove the solvability of our problem. Our existence theorem reads as follows.
Theorem 5.1. There exists a function u ∈ C(0, T ; Hs1 (0, b)) with ∂u/∂t ∈ L2 (0, T ;
which solves problem (2.7), (2.8), and (2.9) in the sense of Definition 3.7, for
arbitrary f ∈ L2 (0, T ; L2s (0, b)) and ϕ ∈ Hs1 (0, b).
1,∗
B2,x (0, b))
Proof. Corollary 4.4 shows that, to prove that (2.7), (2.8), and (2.9) has a generalized solution for each (f , ϕ) ∈ F , it is sufficient to show that R(L) is dense in F . For
this we need the following proposition.
Proposition 5.2. If
T
0
ᏸu(·, t), ω(·, t) L2s (0,b) dt = 0,
(5.1)
for some function ω ∈ L2 (0, T ; L2s (0, b)) and all u ∈ D0 (L) = {u/u ∈ D(L) : u = 0},
then ω ≡ 0 almost everywhere in Q.
Proof of the proposition. Equation (5.1) may be written in the form
T 0
T
∂u(·, t)
dt =
A(t)u, ω(·, t) L2 (0,b) dt.
, ω(·, t)
2
∂t
0
Ls (0,b)
(5.2)
Substitute in (5.2) u by the smooth function ρε−1 u, hence by property (P3) it follows
that
T
T ∂u
, ω(·, t)
A(t)ρε−1 u, ω(·, t) L2 (0,b) dt.
dt =
(5.3)
ρε−1
2
∂τ
0
0
Ls (0,b)
Applying property (P6) to the right-hand side of (5.3), we get
T 0
ρε−1
T
−1
∂u
, ω(·, t)
dt =
ρε A(t)u + ερε−1 A (t)ρε−1 u, ω(·, t) L2 (0,b) dt.
∂τ
0
L2
(0,b)
s
(5.4)
According to property (P2), it follows that
T 0
∂u(·, t) −1 ∗
ω
, ρε
∂t
T
L2
s (0,b)
dt =
0
∗ A(t)u + εA (t)ρε−1 u, ρε−1 ω L2 (0,b) dt. (5.5)
444
ABDELFATAH BOUZIANI
The standard integration by parts with respect to t of the left-hand side of (5.5) leads to
T u(·, t),
0
∗ T
∗ ∂ ρε−1 ω
dt =
A(t)u + εA (t)ρε−1 u, ρε−1 ω L2 (0,b) dt. (5.6)
∂t
0
L2
s (0,b)
The operator A(t) with boundary conditions (2.9) has, on L2 (0, b), a continuous inverse. Hence, it is easy to see that
T 0
∗ ∂ ρε−1 ω
dt
u(·, t),
∂t
L2
s (0,b)
T
∗ =
A(t)u + εA (t)ρε−1 A−1 (t)A(t)u, ρε−1 ω L2 (0,b) dt
0
T
=
0
T
=
0
(5.7)
∗ A(t)u + εΛε (t)A(t)u, ρε−1 ω L2 (0,b) dt
−1 ∗ ω L2 (0,b) dt.
A(t)u, I + εΛ∗
ρε
ε
The calculations of A−1 (t), Λε (t), and Λ∗
ε (t) are straightforward but somewhat tedious. We only give their definitions
A−1 (t)g =
1
a(t)
b
x
dξ
ξ
Λε (t)g = a (t)ρε−1
−1 ∗
Λ∗
ω=
ε (t) ρε
b
ξ
g(η, t)dη +
x
1
log
a(t)
b
b
g(x, t)dx,
0
1
g(x, τ),
a(τ)
(5.8)
∗
1 −1 ∗ a (τ) ρε−1 ω.
ρ
a(t) ε
The left-hand side of (5.7) shows that the mapping
is a continuous linear functional of u, where
T
0
(A(t)u, Kε (t)(ρε−1 )∗ ω)L2 (0,b) dt
∗
−1 ∗
ω,
Kε (t) ρε−1 ω = I + εΛ∗
ε (t) ρε
(5.9)
if the function Kε has the following properties:
∂Kε
∈ L2 0, T ; L2 (0, b) ,
∂x
∂Kε
∂
x
∈ L2 0, T ; L2 (0, b) ,
∂x
∂x
(5.10)
and satisfies
Kε x=0 = Kε x=b = 0.
Therefore, we conclude from (5.9) and (5.11) that
∗
1 −1 ∗ a (τ) ρε−1 ω|x=0 = 0.
I +ε
ρε
a(t)
For each fixed x ∈ [0, b] and sufficiently small ε, the operator
∗
1 −1 ∗ ρε
a (τ) ρε−1
I +ε
a(t)
(5.11)
(5.12)
(5.13)
ON THE SOLVABILITY OF A CLASS OF SINGULAR PARABOLIC . . .
445
has a continuous inverse operator on L2 (0, T ). Thus from (5.12) we obtain
ω|x=0 = ω|x=b = 0.
(5.14)
2 ω(x, t) = x 3 v − 3∗
x ξ v .
(5.15)
We set
From (5.14) and (5.15), we conclude that
b
0
x 2 v(x, t)dx = 0,
v(b, t) = 0.
(5.16)
If we substitute (5.15) into (5.2), we obtain
T 0
2
∂u(·, t) 4
ξ
v(·,
t)
dt
, x v(·, t) − 3x∗
x
∂t
L2 (0,b)
T
2
=
A(t)u, x 3 v(·, t) − 3∗
x ξ v(·, t) L2 (0,b) dt.
(5.17)
0
Now we put
u = t ecτ v =
t
ecτ v(x, τ)dτ
(5.18)
0
in (5.17), where c is a constant such that
cc0 − c2 − 36b2 c12 ≥ 0,
(5.19)
and integrating by parts by taking into account (5.16), we get
T 0
2
∂u(·, t) 4
, x v(·, t) − 3x∗
x ξ v(·, t)
∂t
T b
=
T
0
0
0
ect x 4 v 2 dx dt +
3
2
T b
0
0
L2 (0,b)
dt
2
ect ∗
x (ξv) dx dt,
2
A(t)u, x 3 v(·, t) − 3∗
x ξ v(·, t) L2 (0,b) dt
∂T ect v 2
e
a(T )x
dx
∂x
0
∂t ecτ v 2
1 T b −ct 4 e x ca(t) − a (t)
dx dt
−
2 0 0
∂x
T b
∂t ecτ v
3
dx dt.
x a(t)v
−6
∂x
0 0
1
=−
2
b
−cT
4
(5.20)
446
ABDELFATAH BOUZIANI
Substituting (5.20) into (5.17), yields
T b
0
0
ect x 4 v 2 dx dt +
3
2
T b
0
0
2
ect ∗
x (ξv) dx dt
∂T ect v 2
e
a(T )x
dx
∂x
0
∂t ecτ v 2
1 T b −ct 4 −
e x ca(t) − a (t)
dx dt
2 0 0
∂x
T b
∂t ecτ v
3
x a(t)v
−6
dx dt.
∂x
0 0
1
=−
2
b
−cT
4
(5.21)
The application of the Cauchy inequality to the last term of the above equality gives
T b
T b
0
0
ect x 4 v 2 dx dt +
≤−
1
2
b
e−cT a(T )x 4
0
T b
−
0
0
0
0
2
ect ∗
x (ξv) dx dt
∂T ect v 2
dx
∂x
(5.22)
∂t ecτ v 2
e−ct x 4 ca(t) − a (t) − 36x 2 a2 (t)
dx dt.
∂x
According to Assumption 2.1 and inequality (5.19), we get
v2L2 (0,T ;L2 (0,b)) + v22
r
σ
1,∗
Lr (0,T ;B2,x (0,b))
≤ − cc0 − c2 − 36b2 c12 v2B 1
2
2,ρ (0,T ;Lσ (0,b))
≤ 0,
(5.23)
and thus v ≡ 0, hence ω ≡ 0 almost everywhere in Q. This proves Proposition 5.2.
Returning to the proof of Theorem 5.1. Since F is a Hilbert space, the density of
R(L) in F is equivalent to the property that orthogonality of a vector W = (ω, ω0 ) ∈ F
to the range R(L), that is, the identity
T
0
ᏸu(·, t), ω(·, t) L2s (0,b) dt + u, ω0 Hs1 (0,b) = 0,
∀u ∈ D(L),
(5.24)
implies W ≡ 0. In particular, if u ∈ D0 (L) then conclude by Proposition 5.2 that ω ≡ 0.
Thus (5.24) implies that
u, ω0 Hs1 (0,b) = 0,
u ∈ D(L).
(5.25)
Now since R() is dense in Hs1 (0, b), it follows that ω0 ≡ 0. Hence R(L) = F . This
completes the proof of Theorem 5.1.
Acknowledgment. This work was supported by “Le Centre Universitaire Larbi
Ben M’hidi de Oum El Bouaghi.”
ON THE SOLVABILITY OF A CLASS OF SINGULAR PARABOLIC . . .
447
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
N. I. Benuar and N. I. Yurchuk, A mixed problem with an integral condition for parabolic
equations with a Bessel operator, Differentsial nye Uravneniya 27 (1991), no. 12,
2094–2098.
A. Bouziani, Mixed problem with boundary integral conditions for a certain parabolic
equation, J. Appl. Math. Stochastic Anal. 9 (1996), no. 3, 323–330.
, Solution forte d’un problème mixte avec condition intégrale pour une classe
d’équations paraboliques, Maghreb Math. Rev. 6 (1997), no. 1, 1–17 (French).
, On a third order parabolic equation with a nonlocal boundary condition, J. Appl.
Math. Stochastic Anal. 13 (2000), no. 2, 181–195.
, On a class of nonclassical hyperbolic equations with nonlocal conditions, to appear
in J. Appl. Math. Stochastic Anal., 2002.
, On the quasi-static flexure of a thermoelastic rod, to appear in Commun. Appl.
Anal., 2002.
A. Bouziani and N. E. Benouar, Problème mixte avec conditions intégrales pour une classe
d’équations paraboliques, C. R. Acad. Sci. Paris Sér. I Math. 321 (1995), no. 9, 1177–
1182 (French).
, Mixed problem with integral conditions for a third order parabolic equation, Kobe
J. Math. 15 (1998), no. 1, 47–58.
S. Mesloub and A. Bouziani, Problème mixte avec conditions aux limites intégrales pour
une classe d’équations paraboliques bidimensionnelles, Acad. Roy. Belg. Bull. Cl. Sci.
(6) 9 (1998), no. 1-6, 61–72.
, On a class of singular hyperbolic equation with a weighted integral condition, Int.
J. Math. Math. Sci. 22 (1999), no. 3, 511–519.
N. I. Yurchuk, Solvability of boundary value problems for certain operator-differential
equations, Differensial’nye Uravneniya 13 (1977), no. 4, 626–636.
, A mixed problem with an integral condition for some parabolic equations, Differentsial nye Uravneniya 22 (1986), no. 12, 2117–2126.
Abdelfatah Bouziani: Département de Mathématiques, Centre Universitaire Larbi
Ben M’hidi-Oum El Bouagui, B.P. 565, 04000, Algeria
Current address: Mathematical Division, The Abdus Salam International Centre for
Theoretical Physics (ICTP), Strada Costiera 11, 34100 Trieste, Italy
E-mail address: bouziani@ictp.trieste.it