IJMMS 30:6 (2002) 327–338
PII. S0161171202004167
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
INITIAL-BOUNDARY VALUE PROBLEM WITH A NONLOCAL
CONDITION FOR A VISCOSITY EQUATION
ABDELFATAH BOUZIANI
Received 9 October 1999
This paper deals with the proof of the existence, uniqueness, and continuous dependence
of a strong solution upon the data, for an initial-boundary value problem which combine
Neumann and integral conditions for a viscosity equation. The proof is based on an energy
inequality and on the density of the range of the linear operator corresponding to the
abstract formulation of the studied problem.
2000 Mathematics Subject Classification: 35L20, 35L82, 35B45, 35D05, 35B30.
1. Formulation of the problem. In this paper, we deal with a class of hyperbolic
equations with time- and space-variable characteristics, with a nonlocal boundary
condition. The precise statement of the problem is as follows: let β > 0, T > 0, and
Q = {(x, t) ∈ R2 : α < x < β, 0 < t < T }. Find a function v(x, t), (x, t) ∈ Q̄, satisfying
ᏸv =
∂2v
∂
∂v
∂2
∂v
−
a(x,
t)
−
b(x,
t)
+ c(x, t)v = f(x, t),
∂t 2 ∂x
∂x
∂t∂x
∂x
(1.1)
the initial condition
1 v = v(x, 0) = Φ(x),
2 v =
x ∈ (α, β),
∂v(x, 0)
= Ψ (x),
∂t
(1.2)
x ∈ (α, β),
the Neumann condition
∂v(α, t)
= µ(t),
∂x
t ∈ (0, T ),
(1.3)
and the integral condition
β
α
v(x, t)dx = E(t),
t ∈ (0, T ),
where Φ, Ψ , µ, E, a, b, c, and f are known functions.
(1.4)
328
ABDELFATAH BOUZIANI
Assumption 1.1. For all (x, t) ∈ Q̄, we assume that
c0 ≤ a(x, t) ≤ c1 ,
∂a(x, t)
≤ c2 ,
∂t
∂a(x, t)
≤ c3 ,
∂x
c4 ≤ b(x, t),
∂b(x, t)
≤ c6 ,
∂t
∂b(x, t)
≤ c7 ,
∂x
∂ 2 b(x, t)
≤ c8 ,
∂t 2
c5 ≤
∂ 2 b(x, t)
≤ c9 ,
∂x∂t
(1.5)
c(x, t) ≤ c10 .
Assumption 1.2. For all (x, t) ∈ Q̄, we assume that
∂ 2 a(x, t)
≤ c11 ,
∂x∂t
∂ 2 b(x, t)
≤ c12 ,
∂x 2
∂ 3 b(x, t)
≤ c13 .
∂x∂t 2
(1.6)
In Assumptions 1.1, 1.2, and in the rest of the paper, we assume that ci , where
i = 0, . . . , 17, are positive constants.
The data satisfies the following compatibility conditions:
dΦ(α)
= µ(0),
dx
dΨ (α)
= µ (0),
dx
β
α
Φ(x)dx = E(0),
β
α
(1.7)
Ψ (x)dx = E (0).
Several authors investigated the initial-boundary value problems in one space variable, which involve an integral over the spatial domain of a function of the desired
solution that may appear in a boundary condition. Along a different line, problems for
parabolic equations which combine classical and integral conditions were considered
by Batten [1], Ionkin [12], Cannon et al. [8, 9, 10, 11], Yurchuk [16], Lin [13], BenouarYurchuk [2], Shi [15], Bouziani et al. [7, 14]. However, most of these papers considered
particular situations like heat equation in the rectangle (0, 1) × (0, T ). Problems with
only boundary integral conditions for a second-order parabolic equation have been
treated in Bouziani-Benouar [5], and for a 2m-parabolic equation in Bouziani [4]. Recently, a problem of this type for second-order pluriparabolic equation is studied
in Bouziani [6].
In this paper, the existence, uniqueness, and continuous dependence of a strong
solution upon the data of problem (1.1), (1.2), (1.3), and (1.4) are demonstrated. We
use a functional analysis method based on an energy inequality and on the density
of the range of the linear operator corresponding to the abstract formulation of the
considered problem.
To this end, we reduce the inhomogeneous boundary conditions (1.3) and (1.4)
to homogeneous conditions, by introducing a new unknown function u defined by
u(x, t) = v(x, t) + K(x, t), where
K(x, t) =
6(x − α)
(x − α)
E(t)
.
(3x − α − 2β)µ(t) −
2(α − β)
(β − α)2
(1.8)
329
INITIAL-BOUNDARY VALUE PROBLEM WITH A NONLOCAL CONDITION . . .
Then, problem (1.1), (1.2), (1.3), and (1.4) becomes
ᏸu = f(x, t) + ᏸK(x, t) = f (x, t),
(1.9)
1 u = u(x, 0) = Φ(x) + 1 K = ϕ(x),
2 u =
(1.10)
∂u(x, 0)
= Ψ (x) + 2 K = (x),
∂t
∂u(α, t)
= 0,
∂x
(1.11)
u(x, t)dx = 0.
(1.12)
β
α
Here we assume that the functions ϕ and satisfy conditions of the form (1.11)
and (1.12), that is,
dϕ(α)
= 0,
dx
β
α
ϕ(x)dx = 0,
d(α)
= 0,
dx
β
α
(x)dx = 0.
(1.13)
Instead of searching for the function v, we search for the function u. So the solution
of problem (1.9), (1.10), (1.11), and (1.12) will be given by v(x, t) = u(x, t) − K(x, t).
2. Energy inequality and its consequences. The solution of problem (1.9), (1.10),
(1.11), and (1.12) can be considered as a solution of the operator equation
Lu = (f , ϕ, ),
(2.1)
where L = (ᏸ, 1 , 2 ). The operator L maps from B to F , where B is the Banach space
x
consisting of functions x u ∈ L2 (Q), where x u = α u(ξ, ·)dξ, having finite norm:
1/2
∂u 2
∂u 2
2
uB = +
u
+
,
x
∂t 2
C(0,T ;L2 (α,β))
∂t C(0,T ;L2 (α,β))
L (Q)
(2.2)
and F is the Hilbert space with the finite norm
1/2
2
2
2
.
LuF = x ᏸuL2 (Q) + 1 uL2 (α,β) + x 2 uL2 (α,β)
(2.3)
The domain D(L) of the operator L is the set of all functions u such that x u ∈
L2 (Q) for which x (∂u/∂t), x (∂ 2 u/∂t 2 ), x (∂ 2 u/∂x 2 ) ∈ L2 (Q), and satisfying (1.11)
and (1.12)
Theorem 2.1. Let Assumption 1.1 be fulfilled. Then the a priori estimate
uB ≤ CLuF
holds for any function u ∈ D(L), where C is a positive constant independent of u.
(2.4)
330
ABDELFATAH BOUZIANI
Proof. Applying operator x to (1.9) by taking into account condition (1.11), multiplying the obtained equality with 2x (∂u/∂t), and integrating over Qτ := (α, β) ×
(0, τ), where 0 ≤ τ ≤ T . Observe that
2
∂u
∂2u
∂u
∂u
dx
dt
−
2
x
dx dt
a(x, t)
x
∂t 2
∂t
∂x
∂t
Qτ
Qτ
∂u
∂
∂u
−2
b(x, t)
x
dx dt
∂x
∂t
Qτ ∂t
∂u
x c(ξ, t)u x
+2
dx dt
τ
∂t
Q
∂u
x f x
=2
dx dt.
∂t
Qτ
x
(2.5)
Integrating by parts the first three integrals on the left-hand side of (2.5), we obtain
2
x
Qτ
∂2u
∂u
x
dx dt =
∂t 2
∂t
−2
Qτ
a
∂u
∂u
x
dx dt =
∂x
∂t
β
α
β
α
Qτ
∂u(ξ, τ)
∂t
2
dx −
β
a(x, τ)u2 (x, τ)dx −
−
−2
x
Qτ
∂a 2
u dx dt + 2
∂t
α
β
Qτ
α
2
x dx,
a(x, 0)ϕ2 dx
∂a
∂u
ux
dx dt,
∂x
∂t
β
∂u 2
∂
∂u
∂b(x, τ) 2
∂u
b
dx dt +
b
x
dx dt = 2
u (x, τ)dx
∂t
∂x
∂t
∂t
∂t
α
Qτ
β
∂b(x, 0) 2
∂2b 2
ϕ dx −
u dx dt
−
2
τ
∂t
α
Q ∂t
∂b ∂u
∂u
∂2b
∂u
x
dx dt + 2
ux
dx dt.
+2
∂t
∂t
Qτ ∂x ∂t
Qτ ∂x∂t
(2.6)
Substituting (2.6) into (2.5), we get
∂u
2
b
∂t
Qτ
2
dx dt +
β α
∂u(ξ, τ) 2
∂b
2
u (x, τ) + x
a+
dx
∂t
∂t
β 2
∂b(x, 0)
a(x, 0) +
dx
ϕ2 + x ∂t
α
Qτ
∂2b
∂a ∂ 2 b
∂a
∂u
+ 2 u2 dx dt − 2
+
ux
dx dt
+
∂t
∂t
∂x ∂x∂t
∂t
Qτ
Qτ
∂u
∂b ∂u
∂u
x
dx dt − 2
dx dt.
−2
x cu x
∂t
∂t
Qτ ∂x ∂t
Qτ
=2
x f x
∂u
dx dt +
∂t
(2.7)
INITIAL-BOUNDARY VALUE PROBLEM WITH A NONLOCAL CONDITION . . .
331
Estimating the first and the three last integrals on the right-hand side of (2.7), by
applying elementary inequalities, we get
∂u 2
dx dt,
x
∂t
Qτ
Qτ
Qτ
2 2 ∂a 2
∂u
∂a
∂ b
∂2b
ux
+
u2 dx dt
+
dx dt ≤ 2
−2
∂x ∂x∂t
∂t
∂x
∂x∂t
Qτ
Qτ
2
x f x
∂u
dx dt ≤
∂t
2
x f dx dt +
+
−2
Qτ
∂u 2
dx dt,
x
∂t
Qτ
∂u 2
dx dt
∂t
Qτ
1
∂u 2
∂b 2
+
dx dt,
x
c4 Qτ ∂x
∂t
∂b ∂u
∂u
x
dx dt ≤ c4
∂x ∂t
∂t
∂u
(β − α)2
dx dt ≤
x (cu)x
−2
∂t
2
Qτ
2
Qτ
2
c u dx dt +
x
Qτ
∂u
∂t
2
dx dt.
(2.8)
Therefore, by formulas (2.7), (2.8), and Assumption 1.1, we obtain
τ
∂u(·, τ) 2
∂u(·, t) 2
u(·, τ)22
x
dt
+
+
2
∂t 2
L (α,β)
∂t
0
L (α,β)
L (α,β)
τ
2
2
x f (·, t)22
≤ c14
L (α,β) dt + ϕL2 (α,β) + x L2 (α,β)
(2.9)
0
+ c15
τ
u(·, t)22
L
0
∂u(·, t) 2
x
+
dt,
(α,β)
∂t
L2 (α,β)
where
max 1, c1 + c6
,
min c4 , c0 + c5 , 1
max c2 + c8 + c32 + c92 , 3 + c72 /c4
c15 =
.
min c4 , c0 + c5 , 1
c14 =
(2.10)
Eliminating the last integral on the right-hand side of inequality (2.9). To this end,
using Gronwall’s lemma, it follows that
τ
2
∂u(·, τ) 2
∂u(·, t) 2
dt + u(·, τ) L2 (α,β) + x
2
∂t 2
∂t
0
L (α,β)
L (α,β)
≤ c16
T
0
where c16 = c14 exp(c15 T ).
x f (·, t)22
2
L (α,β) dt + ϕL2 (α,β)
2
+ x 2
L
(α,β) ,
(2.11)
332
ABDELFATAH BOUZIANI
The right-hand side of (2.11) is independent of τ; hence, replacing the left-hand
1/2
side by the upper bound with respect to τ. Thus inequality (2.4) holds, where C = c16 .
It follows from (2.4) that there is a bounded inverse L−1 on the range R(L) of L.
However, since we have no information concerning R(L) expect that R(L) ⊂ F , we
must extend L (construct its closure L̄) so that (2.4) holds for the extension and its
range is the whole space.
We first show that L : B → F with domain D(L), has a closure, that is, the closure of
the graph G(L) ⊂ B × F of L is a graph G(L̄) = G(L) of a new linear operator L̄, which
we call the closure of L.
Proposition 2.2. The operator L from B into F has a closure.
Proof. Suppose that un ∈ D(L) is a sequence such that
n→∞
un →
0
in B,
n→∞
(f , ϕ, )
Lun →
(2.12)
in F ,
(2.13)
we must prove that f ≡ 0, ϕ ≡ 0, and ≡ 0. Equation (2.12) implies that
n→∞
un →
0
in Ᏸ (Q).
(2.14)
By virtue of the continuity of derivation of Ᏸ (Q) in Ᏸ (Q), we have
n→∞
ᏸun →
0 in Ᏸ (Q).
(2.15)
We see via (2.13) that
n→∞
in L2 (Q),
(2.16)
n→∞
in Ᏸ (Q).
(2.17)
ᏸun →
f
then
ᏸun →
f
By virtue of the uniqueness of the limit in Ᏸ (Q), (2.15) and (2.17) imply that f ≡ 0.
On the other hand, from (2.13) we have
n→∞
ϕ
1 un →
in L2 (α, β).
(2.18)
We see via (2.12) and the obvious inequality
1 un L2 (α,β)
≤ un B ,
∀n,
(2.19)
INITIAL-BOUNDARY VALUE PROBLEM WITH A NONLOCAL CONDITION . . .
333
that
n→∞
0
1 un →
in L2 (α, β).
(2.20)
By virtue of (2.18), (2.20), and the uniqueness of the limit in L2 (α, β), we conclude that
ϕ ≡ 0. The reasoning is similar for proving that ≡ 0.
Definition 2.3. A solution of the equation
L̄u = (f , ϕ, ),
(2.21)
is called a strong solution of problem (1.9), (1.10), (1.11), and (1.12).
Since points of the graph of L̄ are limits of sequences of points of the graph of L,
we extend (2.4) to apply to strong solutions by taking the limits.
Corollary 2.4. Under the conditions of Theorem 2.1, there is a constant C > 0
independent of u such that
uB ≤ CL̄uF ,
∀u ∈ D(L̄).
(2.22)
Corollary 2.4 asserts that, if a strong solution exists, it is unique and depends continuously on (f , ϕ, ), if u is considered in the topology of B and (f , ϕ, ) is considered
in the topology of F .
Corollary 2.5. The range R(L̄) of the operator L̄ equals to the closure R(L) of R(L).
Proof. It follows from the definition of L̄ that R(L̄) ⊆ R(L). It remains to prove the
opposite inclusion. Suppose that w ∈ R(L), then there exists a sequence {wn }∞
n=1 of
elements in R(L) such that limn→∞ wn = w. Consequently, there exists a corresponding sequence un ∈ D(L) such that Lun = wn .
According to Theorem 2.1, we have
um − un ≤ C Lum − Lun B
F
(2.23)
when n and m → ∞. Thus {un } is a fundamental sequence in B which converges to
an element u ∈ B and L̄u = w, then w ∈ R(L̄). This proves Corollary 2.5.
Corollary 2.5 states that, to prove that problem (1.9), (1.10), (1.11), and (1.12) has a
strong solution for arbitrary (f , ϕ, ) ∈ F , it is sufficient to show that R(L) = F .
3. Solvability of the problem
Theorem 3.1. Let Assumptions 1.1 and 1.2 be fulfilled. Then for any x f ∈ L2 (Q),
ϕ ∈ L2 (α, β), and x ∈ L2 (α, β), problem (1.9), (1.10), (1.11), and (1.12) admits a
unique strong solution u = L̄−1 (f , ϕ, ) = L−1 (f , ϕ, ).
334
ABDELFATAH BOUZIANI
Proof. First, we prove that R(L) is dense in F for the special case in which L is
reduced to L0 = (ᏸ0 , 1 , 2 ) with domain D0 (L0 ) = D0 (L), where ᏸ0 is the principal
part of ᏸ, that is,
ᏸ0 u =
∂u
∂2
∂u
∂2u
∂
a(x, t)
−
b(x, t)
,
−
2
∂t
∂x
∂x
∂t∂x
∂x
(3.1)
and D0 (L) = {u/u ∈ D(L) : 1 u = 0 and 2 u = 0}.
Proposition 3.2. Under the conditions of Theorem 3.1. If
x ᏸ0 u · x ωdx dt = 0,
(3.2)
Q
for x ω ∈ L2 (Q) and for all u ∈ D0 (L), then ω vanishes almost everywhere in Q.
Proof of Proposition 3.2. Construct the function x ω. Using the fact that relation (3.2) holds for any function u ∈ D0 (L), we can express x u in a special form.
Let



0 ≤ t ≤ s,
0,
(3.3)
x u = t
2
u
∂

x

 (t − τ)
dτ,
s
≤
t
≤
T
,
∂τ 2
s
and let x (∂ 2 u/∂t 2 ) be a solution of the equation
a(σ , t)x
∂2u
= ∗
t x ω =
∂t 2
T
t
x ωdτ,
(3.4)
where σ is a fixed number in [α, β]. We now have
∂2u
∂2u
∂
a(σ , t)x 2 = −
a(σ , t)x 2 .
x ω = ∗−1
t
∂t
∂t
∂t
(3.5)
Relations (3.3) and (3.4) imply that u is in Ds (L), where Ds (L) is the set of functions
D(L) such that u and ∂u/∂t vanish in the neighborhood of t ≤ s. If we put s = 0, then
u is in D0 (L).
Lemma 3.3. Under the conditions of Proposition 3.2, the function x u, defined by
(3.3) and (3.4), has derivatives with respect to t up to third-order inclusive belonging to
the space L2 (Qs ), where Qs = (α, β) × (s, T ).
The proof of Lemma 3.3 is similar to that of [4, Lemma 1].
Substituting (3.1) and (3.5) into (3.2), we have
−
Qs
x
∂2u ∂
∂2u
a(σ
,
t)
dx dt
x
∂t 2 ∂t
∂t 2
∂2u
∂u ∂
+
a
a(σ , t)x 2 dx dt
∂x ∂t
∂t
Qs
∂
∂2u
∂u ∂
+
b
a(σ , t)x 2 dx dt = 0.
∂x ∂t
∂t
Qs ∂t
(3.6)
INITIAL-BOUNDARY VALUE PROBLEM WITH A NONLOCAL CONDITION . . .
335
Integrating by parts each term of the above equality, we obtain
−
Qs
x
=
Qs
Qs
1
2
∂2u ∂
∂2u
a(σ
,
t)
dx dt
x
∂t 2 ∂t
∂t 2
β
1
∂ 2 u(ξ, s) 2
∂2u 2
a(σ , s) x
dx
−
a
(σ
,
t)
dx dt,
x
∂t 2
2 Qs
∂t 2
α
∂2u
∂u ∂
a(σ , t)x 2 dx dt
∂x ∂t
∂t
∂2u
∂u ∂a
∂a ∂u ∂ 2 a
∂2u
a
=
+
u a(σ , t) 2 dx dt+
+
u a(σ , t)x 2 dx dt,
∂t
∂t
∂t
∂x
∂t
∂t∂x
∂t
Qs
Qs
a
∂u ∂
∂
∂2u
b
a(σ , t)x 2 dx dt
∂t
∂x ∂t
∂t
=
β
2 2
∂ u
∂u(x, T ) 2
∂b
ba(σ , t)
dx
dt
+
dx
a(σ
,
T
)
∂t 2
∂t
α ∂t
Qs
2
2
1
∂b ∂u
∂ b
∂2b
∂2u 2
a
a(σ
,
t)
dx
dt−
a(σ
,
t)+
(σ
,
t)
dx dt
x
2
2 Qs ∂x 2
∂t 2
∂t
∂t
∂t
Qs
∂u
∂2u
∂2b
∂2u
∂2b
a(σ , t)
x 2 dx dt +
a(σ , t)u 2 dx dt
+2
2
∂t
∂t
∂t
Qs ∂x∂t
Qs ∂t
−
+
Qs
∂3b
∂2u
a(σ , t)ux 2 dx dt.
2
∂x∂t
∂t
(3.7)
Substituting (3.7) into (3.6) yields
β
α
∂u(x, T ) 2
∂ 2 u(ξ, s) 2 ∂b
a(σ , s)
x
a(σ , T )
+
dx
2
∂t 2
∂t
∂t
+
=
1
2
2 2
∂ u
ba(σ , t)
dx dt
∂t 2
Qs
Qs
+
Qs
−
a (σ , t) +
∂2u 2
∂2b
a(σ
,
t)
dx dt
x
∂x 2
∂t 2
∂b ∂u 2
∂2b
a
a(σ
,
t)
+
(σ
,
t)
dx dt
∂t 2
∂t
∂t
∂a ∂ 2 b
∂u
∂2u
+
+ 2 u a(σ , t) 2 dx dt
a
∂t
∂t
∂t
∂t
Qs
−
Qs
2
∂ 2 b ∂u
∂ a
∂3b
∂2u
∂a
+2
+
+
u a(σ , t)x 2 dx dt.
2
∂x
∂x∂t ∂t
∂t∂x ∂x∂t
∂t
(3.8)
336
ABDELFATAH BOUZIANI
By virtue of Assumptions 1.1 and 1.2, we obtain
2
2
∂u(·, T ) 2
c0 x ∂ u(·, s) +
c
c
0
5
2
2
∂t 2
∂t
L2 (α,β)
L (α,β)
2
T
2
c2
1
x ∂ u(·, t) c2 + c1 c12 + 1
≤
dt
2
2
2
2
∂t
s
L (α,β)
T ∂u(·, t) 2
c42
+ c32 + 4c92
dt
+ c1 c8 + c2 c6 +
∂t 2
2c0 c4
s
L (α,β)
T
2 2
c c + c82
2
2
u(·, t)22
+ c11
+ c13
+ 1 2
L (α,β) dt.
2c0 c4
s
(3.9)
Using the Friedrichs inequality for the norm of u obtained from the norm of ∂u/∂t.
This yields
∂ 2 u(·, s) 2
x
∂t 2 2
L
∂u(·, T ) 2
+
2
∂t
(α,β)
L (α,β)
T ∂ 2 u(·, t) 2
x
≤ c17
∂t 2 2
s
L
∂u(·, t) 2
+
dt,
∂t 2
(α,β)
L (α,β)
(3.10)
where
2 2
c c + c82
c2
c2 +c1 c12 c12
2
2
+ , c1 c8 + c2 c6 + 4 +c32 + 4c92 +γ 1 2
c17 = max
+c11
+ c13
2
4
2c0 c4
2c0 c4
−1
2
c
× min 0 , c0 c5
2
(3.11)
and γ is the constant of the Friedricks inequality.
Inequality (3.10) is basic in our proof. In order to use it, we introduce a new function
z defined by the formula
z(x, t) = ∗
t
∂2u
=
∂τ 2
T
t
∂2u
dτ.
∂τ 2
(3.12)
Then, ∂u(x, t)/∂t = z(x, s) − z(x, t), ∂u(x, T )/∂t = z(x, s), and we have
T T
∂u(·, t) 2
z(·, t)22
z(·, s)22
dt
≤
2
dt
+
(T
−
s)
∂t 2
L (α,β)
L (α,β) .
s
s
L (α,β)
(3.13)
Consequently, (3.10) becomes
∂ 2 u(·, s) 2
x
∂t 2 2
L (α,β)
2
+ 1 − 2c17 (T − s) z(·, s)L2 (α,β)
T ∂ 2 u(·, t) 2
x
≤ 2c17
∂t 2 2
s
L (α,β)
2
+ z(·, t) 2
L
(α,β) dt.
(3.14)
INITIAL-BOUNDARY VALUE PROBLEM WITH A NONLOCAL CONDITION . . .
337
Hence, if s0 > 0 satisfies 1 − 2c17 (T − s) = 1/2, then inequality (3.14) implies
∂ 2 u(·, s) 2
x
∂t 2 2
L (α,β)
2
+ z(·, s)L2 (α,β)
T ∂ 2 u(·, t) 2
x
≤ 4c17
∂t 2 2
s
L (α,β)
+ z(·, t)2L2 (α,β)
dt
(3.15)
for all s ∈ [T − s0 , T ]. We denote the integral on the right-hand side of (3.15) by y(s).
Hence, we obtain
dy(s)
≤ 4c17 y(s),
ds
(3.16)
d y(s) exp 4c17 s ≤ 0.
ds
(3.17)
−
and, consequently,
−
It follows from (3.17) that y(s) = 0, and thus x ω ≡ 0 almost everywhere in QT −s0 .
Proceeding in this way step by step along a rectangle of side s0 , we prove that x ω ≡ 0,
and thus ω ≡ 0 almost everywhere in Q.
Now, we will prove Theorem 3.1. For this end in view, it is sufficient to prove that
the range R(L) of L is dense in F . Suppose that, for some W = (ω, ω1 , ω2 ) ∈ F be
orthogonal to R(L0 ), so that
Q
x ᏸ0 u · x ω dx dt +
β
α
1 uω1 dx +
β
α
x 2 ux ω2 dx = 0.
(3.18)
We must prove that W ≡ 0. Putting u ∈ D0 (L) in (3.18), we obtain
Q
x ᏸ0 u · x ω dx dt = 0,
u ∈ D0 (L).
(3.19)
Hence Proposition 3.2 implies that ω ≡ 0. Thus (3.18) takes the form
β
α
1 uω1 dx +
β
α
x 2 ux ω2 dx = 0.
(3.20)
Since 1 and 2 are independent and the sets of the operators 1 and 2 are everywhere
β
dense in L2 (α, β) and the space with the norm ( α (x ω2 )2 dx)1/2 , respectively, the
above relation implies that ω1 ≡ 0 and ω2 ≡ 0. Hence W ≡ 0, and thus R(L0 ) = F .
Now consider the general case. If we use the fact that R(L0 ) is dense in F and that
L − L0 = (ᏸ − ᏸ0 , 1 , 2 ) maps continuously B into F , we conclude that we can prove
that R(L) is dense in F by means of the method of continuation along the parameter.
We will not describe the application of this method because it is analogous to the
method used in [3].
Acknowledgment. This work was supported by “Le Centre Universitaire Larbi
Ben M’hidi-Oum El Baouagui.”
338
ABDELFATAH BOUZIANI
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Abdelfatah Bouziani: Département de Mathématiques, Centre Universitaire Larbi
Ben M’hidi-Oum El Baouagui, 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