����������� ������ 53 (2001), 79–89

UDK 517.956.3

���������� ������ ��� research paper

ON A NONLOCAL SINGULAR MIXED EVOLUTION PROBLEM

Said Mesloub and Nadia Lekrine

Abstract. In the present paper, the existence and uniqueness of the strong solution of a mixed problem for a second order plurihyperbolic equation with an integral condition is proved. The proof is essentially based on an a priori bound and on the density of the range of the operator generated by the considered problem. In spite of the apparant simplicity of the problem, the solution requires a delicate set of techniques. It seems very difficult to extend these technics to the considered equation in more than one dimension without imposing complementary conditions.

1. Statement of the problem

In the region Q = (0, a) × (0, T_{1}) × (0, T_{2}), with a < ∞, T_{1} < ∞ and T_{2} < ∞, we consider the one dimensional hyperbolic equation

Lv = v_{t_{1}t_{2}} − \frac{1}{x} (xv_{x})_{x} = F (x, t_{1}, t_{2}),  (1)

The equation (1) is supplemented by boundary and initial conditions

ℓ_{1}v = v(x, 0, t_{2}) = ϕ_{1}(x, t_{2}),  (x, t_{2}) ∈ Q_{2} = (0, a) × (0, T_{2}),  (2)
ℓ_{2}v = v(x, t_{1}, 0) = ϕ_{2}(x, t_{1}),  (x, t_{1}) ∈ Q_{1} = (0, a) × (0, T_{1}),  (3)
v_{x}(a, t_{1}, t_{2}) = Φ(t_{1}, t_{2}),  (t_{1}, t_{2}) ∈ (0, T_{1}) × (0, T_{2}),  (4)
∫_{0}^{a}xv(x, t_{1}, t_{2}) dx = Ψ(t_{1}, t_{2}),  (t_{1}, t_{2}) ∈ (0, T_{1}) × (0, T_{2}).  (5)

where ϕ_{1}(x, t_{2}), ϕ_{2}(x, t_{1}), Φ(t_{1}, t_{2}), Ψ(t_{1}, t_{2}) and F (x, t_{1}, t_{2}) are given functions. The data functions have to satisfy the following compatibility conditions:

\frac{∂ϕ_{1}}{∂x} = Φ(0, t_{2}),  ∫_{0}^{a} xϕ_{1}(x, t_{2}) dx = Ψ(0, t_{2}),

\frac{∂ϕ_{2}}{∂x} = Φ(t_{1}, 0),  ∫_{0}^{a} xϕ_{2}(x, t_{1}) dx = Ψ(t_{1}, 0),

and ϕ_{1}(x, 0) = ϕ_{2}(x, 0).

AMS Subject Classification: 35 L 20, 35 L 67

Key words and phrases: Strong solution, a priori bound, plurihyperbolic equation.

In the early sixties Cannon [6] has proved by means of an integral equation (Potential method), the existence and uniqueness of the solution for a mixed problem combining a classical condition (Dirichlet condition) and an integral one for the homogeneous equation. One year later Kamynin [10] has generalized the results of Cannon by using a system of integral equations (Potential method). The importance of mixed problems with integral conditions has been also pointed out by Samarskii [14]. Problem (1)–(5), can be viewed as a non-local problem for a plurihyperbolic equation (with the Bessel operator). A similar problem for which a homogeneous Dirichlet condition and the linear constraint ∫_{0}^{a} v(x, t) dx = 0 are combined, has been investigated by Benouar and Yurchuk [1]. In their papers [2], [3], [4] and [5], the authors considered hyperbolic and parabolic equations having the operator (α(x, t)v_{x})_{x} instead of the Bessel operator considered in equation (1). For some mixed problems for second order parabolic equations which combine classical and integral conditions the reader should refer to Cannon-van der Hoek [7], [8], Cannon-Esteva-van der Hoek [9], Kartynik [11], Shi [15], Yurchuk [16] and Mesloub-Bouziani [13]. In this paper, the existence and uniqueness of a strong solution of problem (1)–(5) is proved by means of an energy estimate and a density argument.

In point of view of the used method, it is preferable to transform the nonhomogeneous conditions to homogeneous ones. If we set:

u(x, t_{1}, t_{2}) = v(x, t_{1}, t_{2}) − w(x, t_{1}, t_{2}),

where

w(x, t_{1}, t_{2}) = (x − \frac{4(x − a)^{2}}{a}) · Φ(t_{1}, t_{2}) + \frac{12(x − a)^{2}}{a^{4}} · Ψ(t_{1}, t_{2}),

then problem (1)–(5), becomes

Lu = F (x, t_{1}, t_{2}) − Lw = f(x, t_{1}, t_{2}),  (6)
ℓ_{1}u = u(x, 0, t_{2}) = ϕ_{1}(x, t_{2}) − ℓ_{1}w = φ_{1}(x, t_{2}),  (7)
ℓ_{2}u = u(x, t_{1}, 0) = ϕ_{2}(x, t_{1}) − ℓ_{2}w = φ_{2}(x, t_{1}),  (8)
u_{x}(a, t_{1}, t_{2}) = 0,  (9)
∫_{0}^{a} xu(x, t_{1}, t_{2}) dx = 0  (10)

We now introduce the appropriate function spaces needed for the investigation of the posed problem. Let L_{ρ}^{2}(Q) be the weighted L^{2}-space with finite norm

∥u∥_{L_{ρ}^{2}}^{2} = ∫_{Q} xu^{2} dx dt,

t = (t_{1}, t_{2}), dt = dt_{1} dt_{2}. The scalar product in L_{ρ}^{2}(Q) is defined by (u, v)_{L_{ρ}^{2}} = (xu, v)_{L^{2}}. Let V_{ρ}^{1,0}(Q_{i}), V_{ρ}^{1,1}(Q_{1}), and V_{ρ}^{1,1}(Q_{2}), i = 1, 2 be the Hilbert spaces with scalar products respectively

(u, v)_{V_{ρ}^{1,0}(Q_{i})} = (u, v)_{L_{ρ}^{2}(Q_{i})} + (u_{x}, v_{x})_{L_{ρ}^{2}(Q_{i})}, i = 1, 2,
(u, v)_{V_{ρ}^{1,1}(Q_{1})} = (u, v)_{L_{ρ}^{2}(Q_{1})} + (u_{x}, v_{x})_{L_{ρ}^{2}(Q_{1})} + (u_{t_{1}}, v_{t_{1}})_{L_{ρ}^{2}(Q_{1})},
(u, v)_{V_{ρ}^{1,1}(Q_{2})} = (u, v)_{L_{ρ}^{2}(Q_{2})} + (u_{x}, v_{x})_{L_{ρ}^{2}(Q_{2})} + (u_{t_{2}}, v_{t_{2}})_{L_{ρ}^{2}(Q_{2})},

and with associated norms

∥u∥_{V_{ρ}^{1,0}(Q_{i})}^{2} = ∥u∥_{L_{ρ}^{2}(Q_{i})}^{2} + ∥u_{x}∥_{L_{ρ}^{2}(Q_{i})}^{2} , i = 1, 2,
∥u∥_{V_{ρ}^{1,1}(Q_{1})}^{2} = ∥u∥_{L_{ρ}^{2}(Q_{1})}^{2} + ∥u_{x}∥_{L_{ρ}^{2}(Q_{1})}^{2} + ∥u_{t_{1}}∥_{L_{ρ}^{2}(Q_{1})}^{2} ,
∥u∥_{V_{ρ}^{1,1}(Q_{2})}^{2} = ∥u∥_{L_{ρ}^{2}(Q_{2})}^{2} + ∥u_{x}∥_{L_{ρ}^{2}(Q_{2})}^{2} + ∥u_{t_{2}}∥_{L_{ρ}^{2}(Q_{2})}^{2} .

The given problem (6)–(10) can be considered as the resolution of the operator equation

Lu = (Lu, ℓ_{1}u, ℓ_{2}u) = (f, φ_{1}, φ_{2}) = F,

where L is an operator defined on E into F, and E is the Banach space of functions u ∈ L_{ρ}^{2}(Q), satisfying conditions (9) and (10), with the finite norm

∥u∥_{E}^{2} = sup_{0≤τ_{2}≤T_{2}} (∥u(·, ·, τ_{2})∥_{V_{ρ}^{1,1}(Q_{1})}^{2} + ∥ℑ_{x}(ξu_{t_{1}}(·, ·, τ_{2}))∥_{L^{2}(Q_{1})}^{2})

+ sup_{0≤τ_{1}≤T_{1}} (∥u(·, τ_{1}, ·)∥_{V_{ρ}^{1,1}(Q_{2})}^{2} + ∥ℑ_{x}(ξu_{t_{2}}(·, τ_{1}, ·))∥_{L^{2}(Q_{2})}^{2}) ,

where ℑ_{x}(ξu) = ∫_{0}^{a} ξu(ξ, t_{1}, t_{2}) dξ, and F is the Hilbert space L_{ρ}^{2}(Q)× V_{ρ}^{1,1}(Q_{2}) × V_{ρ}^{1,1}(Q_{1}), which consists of elements F = (f, φ_{1}, φ_{2}) with finite norm

∥F∥_{F}^{2} = ∥φ_{1}∥_{V_{ρ}^{1,1}(Q_{2})}^{2} + ∥φ_{2}∥_{V_{ρ}^{1,1}(Q_{1})}^{2} + ∥Lf∥_{L_{ρ}^{2}(Q)}^{2} .

Let D(L) be the set of all functions u ∈ L^{2}(Q) for which u_{t_{1}}, u_{t_{2}}, u_{t_{1}t_{2}}, u_{x}, u_{xx}, u_{xt_{1}}, u_{xt_{2}} ∈ L^{2}(Q) and satisfying conditions (9) and (10).

2. A priori bound and its consequences

Theorem 2.1. For any function u ∈ D(L), there exists a positive constant c independent of the solution u such that

∥u∥_{E} ≤ c ∥Lu∥_{F} .  (11)

Proof. Taking the scalar product in L^{2}(Q^{τ}) of equation (6) and the integrodifferential operator

Mu = x(u_{t_{1}} + u_{t_{2}}) − xℑ_{x}^{2}(ξu_{t_{1}} + ξu_{t_{2}}),

where Q^{τ} = (0, a) × (0, τ_{1}) × (0, τ_{2}) and ℑ_{x}^{2}h = ∫_{0}^{x}∫_{0}^{ξ} h(ζ, t_{1}, t_{2}) dζ dξ, we obtain

(u_{t_{1}t_{2}}, u_{t_{1}} + u_{t_{2}})_{L_{ρ}^{2}(Q^{τ})} − (u_{t_{1}t_{2}}, ℑ_{x}^{2}(ξu_{t_{1}} + ξu_{t_{2}}))_{L_{ρ}^{2}(Q^{τ})}

− (u_{t_{1}} + u_{t_{2}}, (xu_{x})_{x})_{L^{2}(Q^{τ})} + (ℑ_{x}^{2}(ξu_{t_{1}} + ξu_{t_{2}}), (xu_{x})_{x})_{L^{2}(Q^{τ})}

= (Lu, Mu)_{L^{2}(Q^{τ})}. (12)

The successive integration by parts of integrals on the left-hand side of (12) are straightforward but somewhat tedious. We only give their results

(u_{t_{1}t_{2}}, u_{t_{1}} + u_{t_{2}})_{L_{ρ}^{2}(Q^{τ})} =

= \frac{1}{2} ∫_{Q_{1}^{τ_{1}}} x(u_{t_{1}}(x, t_{1}, τ_{2}))^{2} dx dt_{1} − \frac{1}{2} ∫_{Q_{1}^{τ_{1}}} x(\frac{∂φ_{2}}{∂t_{1}} )^{2} dx dt_{1}

+ \frac{1}{2} ∫_{Q_{2}^{τ_{2}}} x(u_{t_{2}}(x, τ_{1}, t_{2}))^{2} dx dt_{1} − \frac{1}{2} ∫_{Q_{2}^{τ_{2}}} x(\frac{∂φ_{1}}{∂t_{2}} )^{2} dx dt_{2}, (13)

− (u_{t_{1}t_{2}}, ℑ_{x}^{2}(ξu_{t_{1}} + ξu_{t_{2}}))_{L_{ρ}^{2}(Q^{τ})} =

= \frac{1}{2} ∫_{Q_{1}^{τ_{1}}} (ℑ_{x}(ξu_{t_{1}}(x, t_{1}, τ_{2}))^{2} dx dt_{1} − \frac{1}{2} ∫_{Q_{1}^{τ_{1}}} (ℑ_{x}(ξ \frac{∂φ_{2}}{∂t_{1}} ))^{2} dx dt_{1}

+ \frac{1}{2} ∫_{Q_{2}^{τ_{2}}} (ℑ_{x}(ξu_{t_{2}}(x, τ_{1}, t_{2}))^{2} dx dt_{2} − \frac{1}{2} ∫_{Q_{2}^{τ_{2}}} (ℑ_{x}(ξ \frac{∂φ_{1}}{∂t_{2}} ))^{2} dx dt_{2}, (14)

− (u_{t_{1}} + u_{t_{2}}, (xu_{x})_{x})_{L^{2}(Q^{τ})} =

= \frac{1}{2} ∫_{Q_{2}^{τ_{2}}} x(u_{x}(x, τ_{1}, t_{2}))^{2} dx dt_{2} − \frac{1}{2} ∫_{Q_{2}^{τ_{2}}} x(\frac{∂φ_{1}}{∂x})^{2} dx dt_{2}

+ \frac{1}{2} ∫_{Q_{1}^{τ_{1}}} x(u_{x}(x, t_{1}, τ_{2}))^{2} dx dt_{1} − \frac{1}{2} ∫_{Q_{1}^{τ_{1}}} x(\frac{∂φ_{2}}{∂x})^{2} dx dt_{1}, (15)

(ℑ_{x}^{2}(ξu_{t_{1}} + ξu_{t_{2}}), (xu_{x})_{x})_{L^{2}(Q^{τ})} =

= − ∫_{Q^{τ}} xu_{x}(ℑ_{x}(ξu_{t_{1}}) + ℑ_{x}(ξu_{t_{2}})) dx dt_{1} dt_{2}. (16)

First observe that

∥ℑ_{x}u∥_{L^{2}(Q^{τ})}^{2} ≤ \frac{a^{2}}{2} ∥u∥_{L^{2}(Q^{τ})}^{2} ,  (17)

then by making use of (13)–(17), the Cauchy ε-inequality αθ ≤ εα^{2}/2 + θ^{2}/2ε, and the identity (12), we obtain

\frac{1}{2} ∥u_{t_{1}}(·, t_{1}, τ_{2})∥_{L_{ρ}^{2}(Q_{1}^{τ_{1}} )}^{2} + \frac{1}{2} ∥u_{t_{2}}(·, τ_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{2}^{τ_{2}} )}^{2} + \frac{1}{2} ∥u_{x}(·, t_{1}, τ_{2})∥_{L_{ρ}^{2}(Q_{1}^{τ_{1}} )}^{2}

+\frac{1}{2} ∥u_{x}(·, τ_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{2}^{τ_{2}} )}^{2}+\frac{1}{2} ∥ℑ_{x}(ξu_{t_{1}}(·, t_{1}, τ_{2}))∥_{L^{2}(Q_{1}^{τ_{1}} )}^{2}+\frac{1}{2} ∥ℑ_{x}(ξu_{t_{2}}(·, τ_{1}, t_{2}))∥_{L^{2}(Q_{2}^{τ_{2}} )}^{2}

≤ (\frac{a^{4}}{4} + \frac{1}{2}) ∥\frac{∂φ_{1}}{∂t_{2}} ∥_{L_{ρ}^{2}(Q_{2})}^{2} + (\frac{a^{4}}{4} + \frac{1}{2}) ∥\frac{∂φ_{2}}{∂t_{1}} ∥_{L_{ρ}^{2}(Q_{1})}^{2} + \frac{1}{2} ∥\frac{∂φ_{2}}{∂x} ∥_{L_{ρ}^{2}(Q_{1})}^{2} + a ∥u_{x}∥_{L_{ρ}^{2}(Q^{τ})}^{2}

+ \frac{1}{2} ∥\frac{∂φ_{1}}{∂x} ∥_{L_{ρ}^{2}(Q_{2})}^{2} + \frac{1}{2} ∥u_{t_{1}}∥_{L_{ρ}^{2}(Q^{τ})}^{2} + \frac{1}{2} ∥u_{t_{2}}∥_{L_{ρ}^{2}(Q^{τ})}^{2} + 2 ∥Lu∥_{L_{ρ}^{2}(Q^{τ})}^{2}

+ (\frac{1}{2} + \frac{a^{3}}{4} ) ∥ℑ_{x}(ξu_{t_{1}})∥_{L^{2}(Q^{τ})}^{2} + (\frac{1}{2} + \frac{a^{3}}{4} ) ∥ℑ_{x}(ξu_{t_{2}})∥_{L^{2}(Q^{τ})}^{2} . (18)

On a nonlocal singular mixed evolution problem 83

Consider the elementary inequalities:

∥u(·, τ_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{2}^{τ_{2}} )}^{2} ≤ ∥u∥_{L_{ρ}^{2}(Q^{τ})}^{2} + ∥u_{t_{1}}∥_{L_{ρ}^{2}(Q^{τ})}^{2} + ∥φ_{1}∥_{L_{ρ}^{2}(Q_{2})}^{2} ,  (19)
∥u(·, t_{1}, τ_{2})∥_{L_{ρ}^{2}(Q_{1}^{τ_{1}} )}^{2} ≤ ∥u∥_{L_{ρ}^{2}(Q^{τ})}^{2} + ∥u_{t_{2}}∥_{L_{ρ}^{2}(Q^{τ})}^{2} + ∥φ_{2}∥_{L_{ρ}^{2}(Q_{1})}^{2} .  (20)

Adding side to side inequalities (18)–(20), we obtain ∥u_{t_{1}}(·, t_{1}, τ_{2})∥_{L_{ρ}^{2}(Q_{1}^{τ_{1}} )}^{2} + ∥u_{t_{2}}(·, τ_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{2}^{τ_{2}} )}^{2} + ∥u_{x}(·, t_{1}, τ_{2})∥_{L_{ρ}^{2}(Q_{1}^{τ_{1}} )}^{2} + ∥u_{x}(·, τ_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{2}^{τ_{2}} )}^{2} + ∥ℑ_{x}(ξu_{t_{1}}(·, t_{1}, τ_{2}))∥_{L^{2}(Q_{1}^{τ_{1}} )}^{2} + ∥ℑ_{x}(ξu_{t_{2}}(·, τ_{1}, t_{2}))∥_{L^{2}(Q_{2}^{τ_{2}} )}^{2}

+ ∥u(·, τ_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{2}^{τ_{2}} )}^{2} + ∥u(·, t_{1}, τ_{2})∥_{L_{ρ}^{2}(Q_{1}^{τ_{1}} )}^{2} ≤ k {∥φ_{1}∥_{V_{ρ}^{1,1}(Q_{2})}^{2} + ∥φ_{2}∥_{V_{ρ}^{1,1}(Q_{1})}^{2} + ∥Lu∥_{L_{ρ}^{2}(Q^{τ})}^{2} + ∥u∥_{L_{ρ}^{2}(Q^{τ})}^{2} + ∥u_{t_{1}}∥_{L_{ρ}^{2}(Q^{τ})}^{2} + ∥u_{t_{2}}∥_{L_{ρ}^{2}(Q^{τ})}^{2} + ∥u_{x}∥_{L_{ρ}^{2}(Q^{τ})}^{2} + ∥ℑ_{x}(ξu_{t_{1}})∥_{L^{2}(Q^{τ})}^{2} + ∥ℑ_{x}(ξu_{t_{2}})∥_{L^{2}(Q^{τ})}^{2}} , (21)

where

k = max \{2a, 4, 1 + \frac{a^{3}}{2} , 1 + \frac{a^{4}}{2} \} .

Now, to eliminate the last six terms on the right-hand side of (21), we use the following lemma which can be proved in the same fashion as in lemma 7.1 from [12].

Lemma 2.2. If f_{1}(τ_{1}, τ_{2}), f_{2}(τ_{1}, τ_{2}) and f_{3}(τ_{1}, τ_{2}) are nonnegative functions on the rectangle (0, T_{1}) × (0, T_{2}), f_{1}(τ_{1}, τ_{2}) and f_{2}(τ_{1}, τ_{2}) are integrable, and f_{3}(τ_{1}, τ_{2}) is nondecreasing in each of its variables separately, then it follows from

∫_{0}^{τ_{1}}∫_{0}^{τ_{2}} f_{1}(τ_{1}, τ_{2}) dt_{1} dt_{2} + f_{2}(τ_{1}, τ_{2})

≤ c ∫_{0}^{τ_{1}} f_{2}(t_{1}, τ_{2}) dt_{1} + c ∫_{0}^{τ_{2}} f_{2}(τ_{1}, t_{2}) dt_{2} + f_{3}(τ_{1}, τ_{2})

that

∫_{0}^{τ_{1}}∫_{0}^{τ_{2}} f_{1}(τ_{1}, τ_{2}) dt_{1} dt_{2} + f_{2}(τ_{1}, τ_{2}) ≤ exp(2c(τ_{1} + τ_{2})) · f_{3}(τ_{1}, τ_{2}).

Then (21) takes the form

∥u(·, t_{1}, τ_{2})∥_{V_{ρ}^{1,1}(Q_{1}^{τ_{1}} )}^{2} + ∥ℑ_{x}(ξu_{t_{1}}(·, t_{1}, τ_{2}))∥_{L^{2}(Q_{1}^{τ_{1}} )}^{2}

+ ∥u(·, τ_{1}, t_{2})∥_{V_{ρ}^{1,1}(Q_{2}^{τ_{2}} )}^{2} + ∥ℑ_{x}(ξu_{t_{2}}(·, τ_{1}, t_{2}))∥_{L^{2}(Q_{2}^{τ_{2}} )}^{2}

≤ ke^{k(T_{1}+T_{2})} {∥φ_{1}∥_{V_{ρ}^{1,1}(Q_{2})}^{2} + ∥φ_{2}∥_{V_{ρ}^{1,1}(Q_{1})}^{2} + ∥Lu∥_{L_{ρ}^{2}(Q^{τ})}^{2}} .

Since the right-hand side of the above inequality is independent of (τ_{1}, τ_{2}), we can take the least upper bound of the left side with respect to (τ_{1}, τ_{2}) from [0, T_{1}) and [0, T_{2}) respectively, we get the desired estimate (11) with c = \sqrt{k}e^{k(T_{1}+T_{2})/2}.

We shall now prove that the operator L admits a closure. For this we must either show that it follows from a well known theorem in the theory of unbouded operators that the operator L^{∗} adjoint to L is defined in a dense set, or else verify directly the following assertion: If u_{n} ∈ D(L) is a sequence such that

u_{n} → 0_{n→∞}  in the norm of E,  (22)

and

Lu_{n} → F_{n→∞} = (f, φ_{1}, φ_{2})  in the norm of F,  (23)

then f = 0, φ_{1} = 0, φ_{2} = 0.

Since (22) holds, then

u_{n} → 0_{n→∞}  in D′(Q),  (24)

where D′(Q) is the space of distributions on Q. By virtue of the continuity of derivation of D′(Q) in D′(Q), (24) implies that

Lu_{n} → 0_{n→∞}  in D′(Q).  (25)

But since

Lu_{n} → f_{n→∞}  in L_{ρ}^{2}(Q),  (26)

then

Lu_{n} → f_{n→∞}  in D′(Q).  (27)

From the uniqueness of the limit in the space D′(Q), we conclude that f = 0. According to (23), we have

ℓ_{1}u_{n} → φ_{1}_{n→∞}  in V_{ρ}^{1,1}(Q_{2}),  (28)

and by the fact that the canonical injection from V_{ρ}^{1,1}(Q_{2}) into D′(Q_{2}) is continuous, (28) implies

ℓ_{1}u_{n} → φ_{1}_{n→∞} in D′(Q_{2}).  (29)

Moreover, since (22) holds and

∥ℓ_{1}u_{n}∥_{V_{ρ}^{1,1}(Q_{2})} ≤ ∥u_{n}∥_{E} ∀n,  (30)

we have

ℓ_{1}u_{n} → 0_{n→∞} in V_{ρ}^{1,1}(Q_{2}).  (31)

Hence

ℓ_{1}u_{n} → 0_{n→∞} in D′(Q_{2}).  (32)

By virtue of the uniqueness of the limit in D′(Q_{2}), we conclude from (29) and (32), that φ_{1} = 0. In the same fashion, we can show that φ_{2} = 0.

On a nonlocal singular mixed evolution problem 85

Let \overline{L} be the closure of the operator L with domain of definition D(\overline{L}).

Definition 2.1. A solution of the operator equation

\overline{L}u = F,

is called the strong solution of the problem (6)–(10).

By passing to the limit, the estimate (11) can be extended to strong solutions, that is we have the inequality

∥u∥_{E} ≤ c ∥\overline{L}u∥_{F} ∀u ∈ D(\overline{L}). (33)

Hence

Corollary 2.3. If a strong solution of (6)–(10) exists, it is unique and depends continuously on elements F = (f, φ_{1}, φ_{2}) ∈ F.

Corollary 2.4. The range R(\overline{L}) of the operator \overline{L} is closed in F and R(\overline{L}) = \overline{R(L)}.

Hence, to prove that a strong solution of problem (6)–(10) exists for any element (f, φ_{1}, φ_{2}) ∈ F, it remains to prove that \overline{R(L)} = F.

3. Solvability of the posed problem Theorem 3.1. If, for some function ω ∈ L^{2}(Q) and for all u ∈ D(L) verifying

ℓ_{1}u = ℓ_{2}u = 0, we have

∫_{Q} xLu · ω dx dt = 0,  (34)

dt = dt_{1} dt_{2}, then ω vanishes almost everywhere in the domain Q.

Proof. Relation (34) holds for any function u in D(L) such that ℓ_{1}u = ℓ_{2}u = 0, so it can be expressed in a particular form. Consider the function g_{ij} defined by

g_{ij}(t_{1}, t_{2}, x) = ∫_{t_{i}}^{T_{i}} ω_{ij} dτ_{i}, i, j = 1, 2.

Let ∂^{2}u/∂t_{i}∂t_{j} be the solution of the equation

∂^{2}u/∂t_{i}∂t_{j} − ∫_{0}^{x}∫_{0}^{ξ} ζ∂^{2}u/∂t_{i}∂t_{j} dζ dξ = g_{ij}(t_{1}, t_{2}, x)  (35)

and let

u = \{ \begin{matrix}0, & 0 ≤ t_{i} ≤ s_{i}, \\ ∫_{s_{1}}^{t_{1}}∫_{s_{2}}^{t_{2}} u_{τ_{1}τ_{2}} dτ_{1} dτ_{2}, & s_{i} ≤ t_{i} ≤ T_{i}, \end{matrix} i = 1, 2. (36)

From the above relations, we have

ω = ∑_{i=1}^{2} ∑_{j=1}^{2} ω_{ij} = − ∑_{i=1}^{2} ∑_{j=1}^{2} (∂^{2}u/∂t_{i}∂t_{j} − ∫_{0}^{x}∫_{0}^{ξ} ζ∂^{2}u/∂t_{i}∂t_{j} dζ dξ)_{t_{i}} .  (37)

86 S. Mesloub, N. Lekrine

Lemma 3.2. The function ω defined by (37) is in L^{2}(Q).

Proof. The proof can be derived as in [2].

To continue the proof of Theorem 3.1, replacing ω in (34) by its representation (37), we have

− (u_{t_{1}t_{2}}, ∑_{j=1}^{2} u_{t_{1}t_{1}t_{j}})_{L_{ρ}^{2}(Q)} + (u_{t_{1}t_{2}}, ∑_{j=1}^{2} ℑ_{x}^{2}(ξu_{t_{1}t_{1}t_{j}}))_{L_{ρ}^{2}(Q)}

+ ((xu_{x})_{x}, ∑_{j=1}^{2} u_{t_{1}t_{1}t_{j}})_{L^{2}(Q)} − ((xu_{x})_{x}, ∑_{j=1}^{2} ℑ_{x}^{2}(ξu_{t_{1}t_{1}t_{j}}))_{L^{2}(Q)}

− (u_{t_{1}t_{2}}, ∑_{j=1}^{2} u_{t_{2}t_{2}t_{j}})_{L_{ρ}^{2}(Q)} + (u_{t_{1}t_{2}}, ∑_{j=1}^{2} ℑ_{x}^{2}(ξu_{t_{2}t_{2}t_{j}}))_{L_{ρ}^{2}(Q)}

+ ((xu_{x})_{x}, ∑_{j=1}^{2} u_{t_{2}t_{2}t_{j}})_{L^{2}(Q)} − ((xu_{x})_{x}, ∑_{j=1}^{2} ℑ_{x}^{2}(ξu_{t_{2}t_{2}t_{j}}))_{L^{2}(Q)} = 0. (38)

Using conditions (9), (10), the particular form of u given by the relations (35), (36) and then integrating by parts each term of (38), we get

−(u_{t_{1}t_{2}}, ∑_{j=1}^{2} u_{t_{1}t_{1}t_{j}})_{L_{ρ}^{2}(Q)} = \frac{1}{2} ∥u_{t_{1}t_{1}}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} ,  (39)

where Q_{s_{1}}^{1} = (0, a) × (s_{1}, T_{1}),

(u_{t_{1}t_{2}}, ∑_{j=1}^{2} ℑ_{x}^{2}(ξu_{t_{1}t_{1}t_{j}}))_{L_{ρ}^{2}(Q)} = \frac{1}{2} ∥ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2}))∥_{L^{2}(Q_{s_{1}}^{1} )}^{2} ,  (40)
((xu_{x})_{x}, ∑_{j=1}^{2} u_{t_{1}t_{1}t_{j}})_{L^{2}(Q)} = \frac{1}{2} ∥u_{xt_{1}}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} ,  (41)
−((xu_{x})_{x}, ∑_{j=1}^{2} ℑ_{x}(ξu_{t_{1}t_{1}t_{j}}))_{L^{2}(Q)} = −(u_{xt_{1}}, ℑ_{x}(ξu_{t_{1}t_{1}}))_{L_{ρ}^{2}(Q_{s})}−
−(u_{xt_{2}}, ℑ_{x}(ξu_{t_{1}t_{1}}))_{L_{ρ}^{2}(Q_{s})} + (xu_{x}(x, t_{1}, T_{2}), ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2})))_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}, (42)

where Q_{s} = (0, a) × (s_{1}, T_{1}) × (s_{2}, T_{2}),

−(u_{t_{1}t_{2}}, ∑_{j=1}^{2} u_{t_{2}t_{2}t_{j}})_{L_{ρ}^{2}(Q)} = \frac{1}{2} ∥u_{t_{2}t_{2}}(x, T_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}^{2} ,  (43)

where Q_{s_{2}}^{2} = (0, a) × (s_{2}, T_{2}),

(u_{t_{1}t_{2}}, ∑_{j=1}^{2} ℑ_{x}^{2}(ξu_{t_{2}t_{2}t_{j}}))_{L_{ρ}^{2}(Q)} = \frac{1}{2} ∥ℑ_{x}(ξu_{t_{2}t_{2}}(x, T_{1}, t_{2}))∥_{L^{2}(Q_{s_{2}}^{2} )}^{2} ,  (44)
((xu_{x})_{x}, ∑_{j=1}^{2} u_{t_{2}t_{2}t_{j}})_{L^{2}(Q)} = \frac{1}{2} ∥u_{xt_{2}}(x, T_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}^{2} ,  (45)
−((xu_{x})_{x}, ∑_{j=1}^{2} ℑ_{x}^{2}(ξu_{t_{2}t_{2}t_{j}}))_{L^{2}(Q)} = −(u_{xt_{2}}, ℑ_{x}(ξu_{t_{2}t_{2}}))_{L_{ρ}^{2}(Q_{s})}−
−(u_{xt_{1}}, ℑ_{x}(ξu_{t_{2}t_{2}}))_{L_{ρ}^{2}(Q_{s})} + (xu_{x}(x, T_{1}, t_{2}), ℑ_{x}(ξu_{t_{2}t_{2}}(x, T_{1}, t_{2})))_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}. (46)

Combining equalities (38)–(46), we get

\frac{1}{2} ∥u_{t_{1}t_{1}}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} + \frac{1}{2} ∥ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2}))∥_{L^{2}(Q_{s_{1}}^{1} )}^{2}

+ \frac{1}{2} ∥u_{xt_{1}}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} + \frac{1}{2} ∥u_{t_{2}t_{2}}(x, T_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}^{2}

+ \frac{1}{2} ∥u_{xt_{2}}(x, T_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}^{2} + \frac{1}{2} ∥ℑ_{x}(ξu_{t_{2}t_{2}}(r, T_{1}, t_{2}))∥_{L^{2}(Q_{s_{2}}^{2} )}^{2}

= (u_{xt_{1}}, ℑ_{x}(ξu_{t_{1}t_{1}}))_{L_{ρ}^{2}(Q_{s})} + (u_{xt_{2}}, ℑ_{x}(ξu_{t_{1}t_{1}}))_{L_{ρ}^{2}(Q_{s})} + (u_{xt_{2}}, ℑ_{x}(ξu_{t_{2}t_{2}}))_{L_{ρ}^{2}(Q_{s})}

+ (u_{xt_{1}}, ℑ_{x}(ξu_{t_{2}t_{2}}))_{L_{ρ}^{2}(Q_{s})} − (xu_{x}(x, T_{1}, t_{2}), ℑ_{x}(ξu_{t_{2}t_{2}}(x, T_{1}, t_{2})))_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}

− (xu_{x}(x, t_{1}, T_{2}), ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2})))_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}. (47)

We now estimate the terms on the right-hand side of (47). We have

(u_{xt_{1}}, ℑ_{x}(ξu_{t_{1}t_{1}}))_{L_{ρ}^{2}(Q_{s})} ≤ \frac{a}{2} ∥u_{xt_{1}}∥_{L_{ρ}^{2}(Q_{s})}^{2} + \frac{1}{2} ∥ℑ_{x}(ξu_{t_{1}t_{1}})∥_{L^{2}(Q_{s})}^{2} , (48)
(u_{xt_{2}}, ℑ_{x}(ξu_{t_{2}t_{2}}))_{L_{ρ}^{2}(Q_{s})} ≤ \frac{a}{2} ∥u_{t_{2}x}∥_{L_{ρ}^{2}(Q_{s})}^{2} + \frac{1}{2} ∥ℑ_{x}(ξu_{t_{2}t_{2}})∥_{L^{2}(Q_{s})}^{2} , (49)
(u_{xt_{2}}, ℑ_{x}(ξu_{t_{1}t_{1}}))_{L_{ρ}^{2}(Q_{s})} ≤ \frac{a}{2} ∥u_{t_{2}x}∥_{L_{ρ}^{2}(Q_{s})}^{2} + \frac{1}{2} ∥ℑ_{x}(ξu_{t_{1}t_{1}})∥_{L^{2}(Q_{s})}^{2} , (50)
(u_{xt_{1}}, ℑ_{x}(ξu_{t_{2}t_{2}}))_{L_{ρ}^{2}(Q_{s})} ≤ \frac{a}{2} ∥u_{t_{1}x}∥_{L_{ρ}^{2}(Q_{s})}^{2} + \frac{1}{2} ∥ℑ_{x}(ξu_{t_{2}t_{2}})∥_{L^{2}(Q_{s})}^{2} , (51)

− (xu_{x}(x, t_{1}, T_{2}), ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2})))_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}

≤ a ∥u_{x}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} + \frac{1}{4} ∥ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2}))∥_{L^{2}(Q_{s_{1}}^{1} )}^{2} . (52)

Consider the elementary inequality

a ∥u_{x}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} ≤ a ∥u_{x}∥_{L_{ρ}^{2}(Q_{s})}^{2} + a ∥u_{xt_{1}}∥_{L_{ρ}^{2}(Q_{s})}^{2} .  (53)

Applying the Poincare-Friedriks inequality to the first term on the right-hand side of (53), then (52) becomes

− (xu_{x}(x, t_{1}, T_{2}), ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2})))_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}

≤ (c_{1}a + a) ∥u_{xt_{1}}∥_{L_{ρ}^{2}(Q_{s})}^{2} + \frac{1}{4} ∥ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2}))∥_{L^{2}(Q_{s_{1}}^{1} )}^{2} . (54)

We also have − (xu_{x}(x, T_{1}, t_{2}), ℑ_{x}(ξu_{t_{2}t_{2}}(x, T_{1}, t_{2})))_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}

≤ (c_{2}a + a) ∥u_{xt_{2}}∥_{L_{ρ}^{2}(Q_{s})}^{2} + \frac{1}{4} ∥ℑ_{x}(ξu_{t_{2}t_{2}}(x, T_{1}, t_{2}))∥_{L^{2}(Q_{s_{2}}^{2} )}^{2} . (55)

Combining the equality (47), the estimates (48)–(51), (54) and (55), we obtain ∥u_{t_{1}t_{1}}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} + ∥ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2}))∥_{L^{2}(Q_{s_{1}}^{1} )}^{2}

+ ∥u_{xt_{1}}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} + ∥u_{t_{2}t_{2}}(x, T_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}^{2}
+ ∥u_{xt_{2}}(x, T_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}^{2} + ∥ℑ_{x}(ξu_{t_{2}t_{2}}(r, T_{1}, t_{2}))∥_{L^{2}(Q_{s_{2}}^{2} )}^{2}

≤ c {∥u_{xt_{1}}∥_{L_{ρ}^{2}(Q_{s})}^{2} + ∥ℑ_{x}(ξu_{t_{1}t_{1}})∥_{L^{2}(Q_{s})}^{2} + ∥u_{t_{2}x}∥_{L_{ρ}^{2}(Q_{s})}^{2} + ∥ℑ_{x}(ξu_{t_{2}t_{2}})∥_{L^{2}(Q_{s})}^{2}} , (56) where

c = max {8a + 4c_{1}a, 8a + 4c_{2}a, 4} .

It results from (56) that

∥u_{t_{1}t_{1}}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} + ∥ℑ_{x}(ξu_{t_{1}t_{1}}(x, t_{1}, T_{2}))∥_{L^{2}(Q_{s_{1}}^{1} )}^{2}

+ ∥u_{xt_{1}}(x, t_{1}, T_{2})∥_{L_{ρ}^{2}(Q_{s_{1}}^{1} )}^{2} + ∥u_{t_{2}t_{2}}(x, T_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}^{2}

+ ∥u_{xt_{2}}(x, T_{1}, t_{2})∥_{L_{ρ}^{2}(Q_{s_{2}}^{2} )}^{2} + ∥ℑ_{x}(ξu_{t_{2}t_{2}}(r, T_{1}, t_{2}))∥_{L^{2}(Q_{s_{2}}^{2} )}^{2} ≤ 0, (57)

thanks to Gronwall’s lemma 2.2. Hence (57) implies that ω = 0 almost everywhere on Q. This achieves the proof of Theorem 3.1.

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

Proof. Suppose that, for some W = (ω, w_{1}, w_{2}) ∈ R(L)^{⊥},

(Lu, ω)_{L_{ρ}^{2}(Q)} + (ℓ_{1}u, w_{1})_{V_{ρ}^{1,0}(Q_{2})} + (ℓ_{2}u, w_{2})_{V_{ρ}^{1,0}(Q_{1})} = 0.  (58)

We must prove that W = 0.

Let

D_{0}(L) = { u ∈ D(L) : ℓ_{1}u = ℓ_{2}u = 0 }

Putting u ∈ D_{0}(L) in (58), we get (Lu, ω)_{L_{ρ}^{2}(Q)} = 0, u ∈ D_{0}(L). Hence, by virtue of Theorem 3.1 it follows that ω = 0. Thus (58) becomes

(ℓ_{1}u, w_{1})_{V_{ρ}^{1,1}(Q_{2})} + (ℓ_{2}u, w_{2})_{V_{ρ}^{1,1}(Q_{1})} = 0.  (59)

ℓ_{1}u, and ℓ_{2}u are independent, and the ranges of the operators ℓ_{1} and ℓ_{2} are everywhere dense in the spaces V_{ρ}^{1,1}(Q_{2}), and V_{ρ}^{1,1}(Q_{1}), respectively. Hence the equality (59) implies that w_{1} = w_{2} = 0. Consequently W = 0. This ends the proof of Theorem 3.3.

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(received 20.01.2001, in revised form 15.12.2001)

Department of Mathematics, University of Tebessa, 12002 Tebessa, Algeria

Department of Mathematics, University of Constantine, 25000 Constantine, Algeria

E-mail: mesloub@hotmail.com