Georgian Mathematical Journal Volume 9 (2002), Number 1, 149–159

A STRONG SOLUTION OF AN EVOLUTION PROBLEM WITH INTEGRAL CONDITIONS

S. MESLOUB, A. BOUZIANI, AND N. KECHKAR

Abstract. The paper is devoted to proving the existence and uniqueness of a strong solution of a mixed problem with integral boundary conditions for a certain singular parabolic equation. A functional analysis method is used. The proof is based on an energy inequality and on the density of the range of the operator generated by the studied problem.

2000 Mathematics Subject Classification: 35K20. Key words and phrases: Parabolic equation, integral boundary conditions, strong solution, energy inequality.

1. Posing of the Problem

In the domain Q = Ω × (0, T), with Ω = (0, a) × (0, b), where a < ∞, b < ∞ and T < ∞. We shall determine a solution u, in Q , of the differential equation

Lu = u_{t} − \frac{1}{x}(xu_{x})_{x} − \frac{1}{x^{2}} u_{yy} = f(x, y, t), (x, y, t) ∈ Q,  (1)

satisfying the initial condition

ℓu = u(x, y, 0) = φ(x, y), 0 < x < a, 0 < y < b,  (2)

the classical conditions

u(a, y, t) = 0, 0 < t < T, 0 < y < b,  (3)
u_{y}(x, b, t) = 0, 0 < t < T, 0 < x < a,  (4)

and the integral conditions

∫_{0}^{a} xu(x, y, t)dx = 0, ∫_{0}^{b} u(x, y, t)dy = 0.  (5)

For consistency, we have

φ(a, y) = 0, φ_{y}(x, b) = 0,

∫_{0}^{a} xφ(x, y)dx = 0, ∫_{0}^{b} φ(x, y)dy = 0.

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

Many methods were used to investigate the existence and uniqueness of the solution of mixed problems which combine classical and integral conditions. J. R. Cannon [5] used the potential method, combining a Dirichlet and an integral condition for an equation of the parabolic type. L. A. Mouravey and V. Philinovoski [12] used the maximum principal, combining a Neumann and an integral condition for the heat equation. Ionkin [10] used the Fourier method for the same purpose.

Mixed problems for one-dimensional second order parabolic equations, for which a local and an integral condition are combined, can be found in the papers by Cannon, Estiva, and van der Hoeck [6], Cannon and Van der hoeck [7]–[8], Kamynin [11], Yurchuk [16], Bouziani [2], Peter Shi [15], Mesloub and Bouziani [13]. Problems with purely integral conditions are studied by Bouziani [3], and Benouar and Bouziani [4], Mesloub and Bouziani [14]. In this paper, we prove the existence and uniqueness of a strong solution for the problem (1)–(5). The result and the method used here are a further elaboration of those from the paper by Benouar and Yurchuk [1].

We introduce appropriate function spaces. Let L^{2}(Q) be the Hilbert space of square integrable functions having the norm and scalar product denoted respectively by ∥·∥_{L^{2}(Q)} and (·, ·)_{L^{2}(Q)}. Let V^{1,0}(Q) be a subspace of L^{2}(Q) with the finite norm

∥u∥_{V^{1,0}(Q)}^{2} = ∥u∥_{L^{2}(Q)}^{2} + ∥u_{x}∥_{L^{2}(Q)}^{2} ,

having the scalar product defined by

(u, v)_{V^{1,0}(Q)} = (u, v)_{L^{2}(Q)} + (u_{x}, v_{x})_{L^{2}(Q)}.

In general, a function in the space V^{k,m}(Q), with k, m nonnegative integers, possesses x-derivatives up to kth order in L^{2}(Q), and t-derivatives up to mth order in L^{2}(Q). To problem (1)–(5) we associate the operator L = (L, ℓ) with the domain of definition

D(L) = {u ∈ L^{2}(Q)| u_{t}, u_{x}, u_{y}, u_{xx}, u_{yy}, u_{xt} ∈ L^{2}(Q)}

satisfying (3)–(5). The operator L is considered from E to F, where E is the Banach space consisting of functions u ∈ L^{2}(Q) satisfying the boundary conditions (3)–(5) and having the finite norm

∥u∥_{E}^{2} = ∫_{Q} (x^{3}(ℑ_{y}u_{t})^{2} + x(ℑ_{xy}(ξu_{t}))^{2} + x^{3}u_{x}^{2} + xu_{y}^{2}) dxdydt

+ sup_{0≤τ≤T} ∫_{Ω} ((x + x^{3}) u^{2}(·, ·, τ) + x^{3}(ℑ_{y}u_{x}(·, ·, τ))^{2}) dxdy,

where ℑ_{x}u = ∫_{0}^{x} u(ξ, y, t)dξ, ℑ_{y}u = ∫_{0}^{y} u(x, η, t)dη, ℑ_{xy}u = ℑ_{x}(ℑ_{y}u) (below we

will use also the notation ℑ_{yy}u = ℑ_{y}^{2}u = ℑ_{y}(ℑ_{y}u)), ℑ_{xyy}u = ℑ_{x}(ℑ_{yy}u)) and F is the Hilbert space of vector-valued functions F = (f, φ) having the norm

∥F∥_{F}^{2} = ∥f∥_{L^{2}(Q)}^{2} + ∥φ∥_{V^{1,0}(Ω)}^{2} .

Remark 1.1. The weights appearing in this paper arise because of singular coefficients and for the annihilation of inconvenient terms during integration by parts.

2. A Priori Estimate and Its Consequences

Theorem 2.1. For any function u ∈ D(L), we have the following a priori estimate:

∥u∥_{E} ≤ c ∥Lu∥_{F} ,  (6)

where c is a positive constant independent of the solution u.

Proof. In we take the inner product in L^{2}(Q^{τ}) of equation (1) and the operator

Mu = −x^{3}ℑ_{y}^{2}u_{t} + 2x^{2}ℑ_{y}^{2}u_{x} + x^{3}ℑ_{xyy}(ξu_{t}) + x^{3}ℑ_{y}u_{y},

where Q^{τ} = Ω × (0, τ), then, in light of the initial condition (2), the boundary conditions (3)–(5), and a standard integration by parts, we get

∫_{Q^{τ}} x^{3}(ℑ_{y}u_{t})^{2}dxdydt + \frac{1}{2} ∫_{Ω} x^{3}(ℑ_{y}u_{x}(·, ·, τ))^{2}dxdy

+ \frac{1}{2} ∫_{Ω} xu^{2}(·, ·, τ)dxdy + a^{2} ∫_{0}^{b} ∫_{0}^{τ}(ℑ_{y}u_{x}(a, y, t))^{2}dtdy

+ ∫_{0}^{b} ∫_{0}^{τ} u^{2}(0, y, t)dtdy + ∫_{Q^{τ}} x(ℑ_{xy}(ξu_{t}))^{2}dxdydt

+ \frac{1}{2} ∫_{Ω} x^{3}u^{2}(·, ·, τ)dxdy + ∫_{Q^{τ}} x^{3}u_{x}^{2}dxdydt + ∫_{Q^{τ}} xu_{y}^{2}dxdydt

= \frac{1}{2} ∫_{Ω} x^{3}(ℑ_{y}φ_{x})^{2}dxdy + \frac{1}{2} ∫_{Ω} x^{3}φ^{2}dxdy + \frac{1}{2} ∫_{Ω} xφ^{2}dxdy

+ ∫_{Q^{τ}} x^{4}ℑ_{y}u_{x} · ℑ_{y}u_{t}dxdydt + 2 ∫_{Q^{τ}} x^{2}ℑ_{y}u_{x} · ℑ_{xy}(ξu_{t})dxdydt

− ∫_{Q^{τ}} xu_{y}ℑ_{xy}(ξu_{t})dxdydt + 2 ∫_{Q^{τ}} x^{2}u_{y}ℑ_{y}u_{x}dxdydt

− ∫_{Q^{τ}} x^{3}Luℑ_{y}^{2}u_{t}dxdydt + 2 ∫_{Q^{τ}} x^{2}Luℑ_{y}^{2}u_{x}dxdydt

+ ∫_{Q^{τ}} x^{3}Luℑ_{y}u_{y}dxdydt + ∫_{Q^{τ}} x^{3}Luℑ_{xyy}(ξu_{t})dxdydt.  (7)

152 S. MESLOUB, A. BOUZIANI, AND N. KECHKAR

By virtue of the elementary inequality

∫_{0}^{a}(ℑ_{x}u)^{2}dx ≤ \frac{a^{2}}{2} ∫_{0}^{a} u^{2}dx  (8)

(see [3]) and the Cauchy’s ε-inequality

αβ ≤ \frac{ε}{2}α^{2} + \frac{1}{2ε}β^{2},  (9)

we can estimate the terms on the right-hand side of (7) as follows:

∫_{Q^{τ}} x^{4}ℑ_{y}u_{x} · ℑ_{y}u_{t}dxdydt ≤ \frac{ε_{1}a^{2}}{2} ∫_{Q^{τ}} x^{3}(ℑ_{y}u_{x})^{2}dxdydt
+ \frac{1}{2ε_{1}} ∫_{Q^{τ}} x^{3}(ℑ_{y}u_{t})^{2}dxdydt,  (10)
2 ∫_{Q^{τ}} x^{2}ℑ_{y}u_{x} · ℑ_{xy}(ξu_{t})dxdydt ≤ ε_{2} ∫_{Q^{τ}} x^{3}(ℑ_{y}u_{x})^{2}dxdydt
+ \frac{1}{ε_{2}} ∫_{Q^{τ}} x(ℑ_{xy}(ξu_{t})^{2}dxdydt,  (11)
− ∫_{Q^{τ}} xu_{y}ℑ_{xy}(ξu_{t})dxdydt ≤ \frac{ε_{3}}{2} ∫_{Q^{τ}} xu_{y}^{2}dxdydt
+ \frac{1}{2ε_{3}} ∫_{Q^{τ}} x(ℑ_{xy}(ξu_{t}))^{2}dxdydt,  (12)
2 ∫_{Q^{τ}} x^{2}u_{y}ℑ_{y}u_{x}dxdydt ≤ ε_{4} ∫_{Q^{τ}} xu_{y}^{2}dxdydt
+ \frac{1}{ε_{4}} ∫_{Q^{τ}} x^{3}(ℑ_{y}u_{x})^{2}dxdydt,  (13)
− ∫_{Q^{τ}} x^{3}Luℑ_{y}^{2}u_{t}dxdydt ≤ \frac{a^{3}}{2ε_{5}} ∫_{Q^{τ}} f^{2}dxdydt
+ \frac{ε_{5}b^{2}}{4} ∫_{Q^{τ}} x^{3}(ℑ_{y}u_{t})^{2}dxdydt,  (14)
2 ∫_{Q^{τ}} x^{2}Luℑ_{y}^{2}u_{x}dxdydt ≤ \frac{a}{ε_{6}} ∫_{Q^{τ}} f^{2}dxdydt
+ \frac{ε_{6}b^{2}}{2} ∫_{Q^{τ}} x^{3}(ℑ_{y}u_{x})^{2}dxdydt,  (15)

∫_{Q^{τ}} x^{3}Luℑ_{y}u_{y}dxdydt ≤ \frac{a^{5}}{2ε_{7}} ∫_{Q^{τ}} f^{2}dxdydt

+ \frac{ε_{7}b^{2}}{4} ∫_{Q^{τ}} xu_{y}^{2}dxdydt,  (16)

∫_{Q^{τ}} x^{3}Luℑ_{xyy}(ξu_{t})dxdydt ≤ \frac{a^{5}}{2ε_{8}} ∫_{Q^{τ}} f^{2}dxdydt

+ \frac{ε_{8}b^{2}}{4} ∫_{Q^{τ}} x(ℑ_{xy}(ξu_{t}))^{2}dxdydt,  (17)

\frac{1}{2} ∫_{Ω} x^{3}(ℑ_{y}φ_{x})^{2}dxdy ≤ \frac{a^{3}b^{2}}{4} ∫_{Ω} φ_{x}^{2}dxdy,  (18)

then taking ε_{1} = 2, ε_{2} = 8, ε_{3} = 1, ε_{4} = \frac{1}{8}, ε_{5} = \frac{1}{b^{2}} , ε_{6} = \frac{1}{b^{2}} , ε_{7} = \frac{1}{2b^{2}} , ε_{8} = \frac{1}{2b^{2}} .

Substituting (10)–(18) into (7), and taking into account that the fourth and fifth terms in (7) are positive, we get

∫_{Q^{τ}} (x^{3}(ℑ_{y}u_{t})^{2} + x(ℑ_{xy}(ξu_{t}))^{2} + x^{3}u_{x}^{2} + xu_{y}^{2}) dxdydt

+ ∫_{Ω} (x^{3}(ℑ_{y}u_{x}(·, ·, τ))^{2} + (x + x^{3})u^{2}(·, ·, τ)) dxdy

≤ k {∫_{Q^{τ}} x^{3}(ℑ_{y}u_{x})^{2}dxdydt + ∫_{Q^{τ}} f^{2}dxdydt + ∫_{Ω} (φ^{2} + φ_{x}^{2})dxdy} , (19)

where

k = max(2a^{3} + 2a, 8a^{5}b^{2} + 4ab^{2} + 2a^{3}b^{2}, 4a^{2} + 66).

We now conclude from (19) and Gronwall’s lemma that

∫_{Q^{τ}} (x^{3}(ℑ_{y}u_{t})^{2} + x(ℑ_{xy}(ξu_{t}))^{2} + x^{3}u_{x}^{2} + xu_{y}^{2}) dxdydt

+ ∫_{Ω} (x^{3}(ℑ_{y}u_{x}(·, ·, τ))^{2} + (x + x^{3})u^{2}(·, ·, τ)) dxdy

≤ ke^{kT} {∫_{Q^{τ}} f^{2}dxdydt + ∫_{Ω} (φ^{2} + φ_{x}^{2})dxdy} .  (20)

Since the right-hand side of (20) does not depend on τ, we take the least upper bound in its left-hand side with respect to τ from 0 to T, thus obtaining (6), where c = \sqrt{ke^{kT}} .

It can be proved in a standard way that the operator L : E → F is closable. Let \overline{L} be the closure of this operator, with the domain of definition D(\overline{L}).

Definition 2.1. A solution of the operator equation

\overline{L}u = F

is called a strong solution of problem (1)–(5).

The a priori estimate (6) can be extended to strong solutions, i.e., we have the estimate

∥u∥_{E} ≤ c ∥\overline{L}u∥_{F} , ∀u ∈ D(\overline{L}).  (21)

Inequality (21) implies the following corollaries.

Corollary 2.1. A strong solution of (1)–(5) is unique and depends continuously on F = (f, φ).

Corollary 2.2. The range R(\overline{L}) of \overline{L} is closed in F and \overline{R(L)} = R(\overline{L}).

The latter corollary shows that to prove that problem (1)–(5) has a strong solution for arbitrary F = (f, φ), it suffices to prove that the set R(L) is dense in F.

3. Solvability of the Problem

Theorem 3.1. If, for some function ω ∈ L^{2}(Q) and for all elements u ∈ D_{0}(L) = {u|u ∈ D(L) : ℓu = 0} , we have

∫_{Q} Lu · ωdxdydt = 0,  (22)

then ω vanishes almost everywhere in Q.

Proof. Using the fact that relation (22) holds for any function u ∈ D_{0}(L), we can express it in a special form. First define the function h by the relation

h(x, y, t) + ∫_{t}^{T} (x^{6}ℑ_{xyy}(ξu_{τ}) − 10x^{3}ℑ_{y}^{2}u_{τ} + 20x^{2}ℑ_{y}^{2}u_{x}) dτ

= ∫_{t}^{T} ωdτ.  (23)

Let u_{t} be a solution of the equation

−x^{5}ℑ_{y}^{2}u_{t} = h,  (24)

and let the function u to be given by

u = \begin{cases}0, & 0 ≤ t ≤ s, \\ ∫_{s}^{t} u_{τ}dτ, & s ≤ t ≤ T. \end{cases}  (25)

From the above relations we have

ω(x, y, t) = x^{5}ℑ_{y}^{2}u_{tt} + x^{6}ℑ_{xyy}(ξu_{t}) − 10x^{3}ℑ_{y}^{2}u_{t} + 20x^{2}ℑ_{y}^{2}u_{x}.  (26)

Lemma 3.2. The function ω represented by (26) belongs to L^{2}(Q).

Proof. Using a Poincaré type inequality of form (8), we easily prove that the last three terms of (26) are in L^{2}(Q). To show that the term x^{5}ℑ_{y}^{2}u_{tt} is in L^{2}(Q), we use t-averaging operators ρ_{ε} of the form

(ρ_{ε}g)(x, t) = \frac{1}{ε} ∫_{0}^{T} w (\frac{ν − t}{ε}) g(x, t)dν,

where w ∈ C_{0}^{∞}(0, T), w ≥ 0, ∫_{−∞}^{+∞}w(t)dt = 1.

Applying the operators ρ_{ε} and ∂/∂t to equation (24), and then estimating, we obtain

∫_{Q} (x^{5} \frac{∂}{∂t}ρ_{ε}ℑ_{y}^{2}u_{t})^{2} dxdydt ≤ 2 ∫_{Q} (\frac{∂}{∂t}ρ_{ε}h)^{2} dxdydt

+ 2 ∫_{Q} [\frac{∂}{∂t} (ρ_{ε}x^{5}ℑ_{y}^{2}u_{t} − x^{5}ρ_{ε}ℑ_{y}^{2}u_{t})]^{2} dxdydt.

Using the properties of ρ_{ε} introduced in [9], it follows that

∫_{Q} (x^{5} \frac{∂}{∂t}ρ_{ε}ℑ_{y}^{2}u_{t})^{2} dxdydt ≤ 2 ∫_{Q} (\frac{∂}{∂t}ρ_{ε}h)^{2} dxdydt.

Since ρ_{ε}g → g_{ε→0} in L^{2}(Q), and ∫_{Q} (x^{5} \frac{∂}{∂t}ρ_{ε}ℑ_{y}^{2}u_{t})^{2} dxdydt is bounded, we conclude

that ω ∈ L^{2}(Q).

We now return to the proof of Theorem 3.1. Replacing ω in relation (22) by its representation (26), invoking the special form of u given by (24) and (25) and the boundary conditions (3)–(5), and then carrying out appropriate integrations by parts, we obtain

\frac{1}{2} ∫_{Ω} x^{5}(ℑ_{y}u_{t}(·, ·, s))^{2}dxdy + \frac{3}{2} ∫_{Ω} x^{2}(ℑ_{x}(ξu(·, ·, T)))^{2}dxdy

+ 5 ∫_{Ω} x^{3}(ℑ_{y}u_{x}(·, ·, T))^{2}dxdy + 5 ∫_{Ω} xu^{2}(·, ·, T)dxdy

+ ∫_{Q_{s}} x^{5}(ℑ_{y}u_{tx})^{2}dxdydt + 2 ∫_{Q_{s}} x^{3}(ℑ_{y}u_{t})^{2}dxdydt

+ \frac{5}{2} ∫_{Q_{s}} x^{4}(ℑ_{xy}(ξu_{t}))^{2}dxdydt + ∫_{Q_{s}} x^{3}u_{t}^{2}dxdydt

+ 10a^{2} ∫_{s}^{T} ∫_{0}^{b}(ℑ_{y}u_{x}(a, y, t))^{2}dydt + 10 ∫_{s}^{T} ∫_{0}^{b} u^{2}(0, y, t)dydt

= − ∫_{Q_{s}} x^{7}ℑ_{y}uℑ_{y}u_{tx}dxdydt − 25 ∫_{Q_{s}} x^{4}ℑ_{y}uℑ_{xy}(ξu_{t})dxdydt

− ∫_{Q_{s}} x^{4}u_{t}ℑ_{x}(ξu)dxdydt − 12 ∫_{Q_{s}} x^{6}ℑ_{y}uℑ_{y}u_{t}dxdydt,  (27)

where Q_{s} = Ω × [s, T].

We now estimate each term on the right-hand side of (27) by using inequalities (8) and (9) and, taking into account that the last two terms on the left-hand side are positive, we get

∫_{Ω} x^{5}(ℑ_{y}u_{t}(·, ·, s))^{2}dxdy + ∫_{Ω} x^{2}(ℑ_{x}(ξu(·, ·, T)))^{2}dxdy

+ ∫_{Ω} x^{3}(ℑ_{y}u_{x}(·, ·, T))^{2}dxdy + ∫_{Ω} xu^{2}(·, ·, T)dxdy

+ ∫_{Q_{s}} x^{5}(ℑ_{y}u_{tx})^{2}dxdydt + ∫_{Q_{s}} x^{3}(ℑ_{y}u_{t})^{2}dxdydt

+ ∫_{Q_{s}} x^{4}(ℑ_{xy}(ξu_{t}))^{2}dxdydt + ∫_{Q_{s}} x^{3}u_{t}^{2}dxdydt

≤ c {∫_{Q_{s}} x^{3}(ℑ_{y}u_{x})^{2}dxdydt + ∫_{Q_{s}} x^{5}(ℑ_{y}u_{t})^{2}dxdydt

+ ∫_{Q_{s}} x^{2}(ℑ_{x}(ξu))^{2}dxdydt + ∫_{Q_{s}} xu^{2}dxdydt} ,  (28)

where

c = max \{12 + 12a^{2} + a^{4}, 12a^{4} + a^{6}, a^{3}, \frac{5^{4}a^{3}b^{2}}{8} \} .

To use the essential inequality (28), we note that the constant c is independent of s. However, the function u in (28) does depend on s. To avoid this difficulty, we introduce a new function by the formula

η(x, y, t) = ∫_{t}^{T} u_{τ}(x, y, τ)dτ.

Then

u(x, y, t) = η(x, y, s) − η(x, y, t)

AN EVOLUTION PROBLEM WITH INTEGRAL CONDITIONS 157

and we have

∫_{Q_{s}} (x^{5}(ℑ_{y}u_{tx})^{2} + x^{3}(ℑ_{y}u_{t})^{2} + x^{4}(ℑ_{xy}(ξu_{t}))^{2} + x^{3}u_{t}^{2}) dxdydt

+ ∫_{Ω} x^{5}(ℑ_{y}u_{t}(·, ·, s))^{2}dxdy + (1 − 2c(T − s)) {∫_{Ω} xη^{2}(·, ·, s)dxdy

+ ∫_{Ω} x^{2}(ℑ_{x}(ξη(·, ·, s)))^{2}dxdy + ∫_{Ω} x^{3}(ℑ_{y}η_{x}(·, ·, s))^{2}dxdy}

≤ 2c {∫_{Q_{s}} (x^{5}(ℑ_{y}η_{t})^{2} + x^{3}(ℑ_{y}η_{x})^{2} + x^{2}(ℑ_{x}(ξη))^{2} + xη^{2}) dxdydt} . (29)

If we choose s_{0} > 0 such that 1 − 2c(T − s_{0}) = 1/2, then (29) implies

∫_{Q_{s}} (x^{5}(ℑ_{y}u_{tx})^{2} + x^{3}(ℑ_{y}u_{t})^{2} + x^{4}(ℑ_{xy}(ξu_{t}))^{2} + x^{3}u_{t}^{2}) dxdydt

+ (∫_{Ω} x^{5}(ℑ_{y}u_{t}(·, ·, s))^{2} + ∫_{Ω} xη^{2}(·, ·, s)

+ ∫_{Ω} x^{2}(ℑ_{x}(ξη(·, ·, s))^{2} + ∫_{Ω} x^{3}(ℑ_{y}η_{x}(·, ·, s))^{2}) dxdy

≤ 4c{ ∫_{Q_{s}} (x^{5}(ℑ_{y}η_{t}(x, y, t))^{2} + x^{3}(ℑ_{y}η_{x}(x, y, t))^{2}

+ x^{2}(ℑ_{x}(ξη(x, y, t)))^{2} + xη^{2}(x, y, t)) dxdydt}  (30)

for all s ∈ [T − s_{0}, T].

If we denote the sum of the four integral terms on the right-hand side of (30) by α(s), we obtain

∫_{Q_{s}} (x^{5}(ℑ_{y}u_{tx})^{2} + x^{3}(ℑ_{y}u_{t})^{2}) dxdydt

+ ∫_{Q_{s}} (x^{4}(ℑ_{xy}(ξu_{t}))^{2} + x^{3}u_{t}^{2}) dxdydt − \frac{dα(s)}{ds}

≤ 4cα(s).

Consequently,

−\frac{d}{ds} (α(s)e^{4cs}) ≤ 0.  (31)

Taking into account that α(T) = 0, (31) gives

α(s)e^{4cs} ≤ 0.  (32)

It follows from (32) that ω = 0 almost everywhere in Q_{T −s_{0}} = Ω × [T − s_{0}, T]. Proceeding in this way step by step along the cylinders of heigth s_{0}, we prove that ω = 0 almost everywhere in Q. This completes the proof of Theorem 3.1.

Now to conclude, we have to prove

Theorem 3.3. The range of the operator L coincides with F.

Proof. Since F is a Hilbert space, R(L) = F is equivalent to the orthogonality of the vector W = (ω, ω_{0}) ∈ F to the set R(L), i.e., if and only if the relation

∫_{Q} Lu.ωdxdydt + ∫_{Ω} (ℓu · ω_{0} + \frac{dℓu}{dx})dxdy = 0,  (33)

where u runs over E and W = (ω, ω_{0}) ∈ F, implies that W = 0.

Putting u ∈ D(L_{0}) in (33), we get

∫_{Q} Lu · ωdxdydt = 0.

Hence Theorem 3.1 implies that ω = 0. Thus (33) becomes

∫_{Ω} (ℓu · ω_{0} + \frac{dℓu}{dx})dxdy = 0, ∀u ∈ D(L).  (34)

Since the range R(ℓ) of the trace operator ℓ is everywhere dense in V^{1,0}(Ω), then it follows from (34) that ω_{0} = 0. Hence W = 0. This completes the proof of Theorem 3.3.

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(Received 20.11.2000; revised 10.07.2001)

Authors’ addresses:

S. Mesloub Department of Mathematics University of Tebessa Tebessa 12002 Algeria E-mail: mesloub@hotmail.com

A. Bouziani Department of Mathematics University of Oum el bouaghi, 04000, Algeria E-mail: af bouziani@hotmail.com

N. Kechkar Department of Mathematics University of Constantine, 25000 Algeria