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On a singular two dimensional nonlinear evolution equation with nonlocal conditions
Nonlinear Analysis 68 (2008) 2594–2607 www.elsevier.com/locate/na
On a singular two dimensional nonlinear evolution equation with nonlocal conditions Said Mesloub King Saud University, College of Sciences, Department of Mathematics, P.O. Box 2455, Riyadh 11451, Saudi Arabia Received 21 May 2006; accepted 5 February 2007
Abstract This paper deals with a nonclassical initial boundary value problem for a two dimensional parabolic equation with Bessel operator. We prove the existence and uniqueness of the weak solution of the given nonlinear problem. We start by solving the associated linear problem. After writing this latter in its operator form, we establish an a priori bound from which we deduce the uniqueness of the strong solution. For the solvability of the associated linear problem, we prove that the range of the operator generated by the considered problem is dense. On the basis of the obtained results of the linear problem, we apply an iterative process to establish the existence and uniqueness of the nonlinear problem. c 2007 Elsevier Ltd. All rights reserved.
MSC: 35K; 35R Keywords: Nonlinear evolution equation; Nonlocal condition; A priori estimate; Bessel operator
1. Introduction Extensive research studies have been done for numerous and different nonlocal conditions for the solvability of parabolic, hyperbolic, pseudoparabolic and elliptic equations which were published in different journals. Many problems of modern physics and technology can be effectively described in terms of nonlocal problems for partial differential equations. Some presented articles, Friedman [15] and Kawohl [21], analyze such conditions imposed, and observe that the existence and uniqueness of the solution of parabolic equation is related mainly to the smallness of functions involved in nonlocal conditions. Nonclassical problems formulated for equations of mathematical physics are called nonlocal initial boundary value problems when the role of the boundary (or initial) conditions is played by an expression where the boundary value of an unknown function is expressed by the value(s) of this function at internal point(s). There are two types of nonlocal problems, spatial and in-time nonlocal problems. Nonlocal conditions arise mainly when the data on the boundary cannot be measured directly. Mixed problems with nonlocal conditions such as boundary integral conditions, have many important applications in chemical diffusion, thermoelasticity, heat conduction processes, population dynamics, vibration problems, nuclear reactor dynamics, control theory, medical science, biochemistry and certain biological processes: Cannon [3], Cannon and van der Hoek [2], Capasso and Kunisch [7], Day [9], Kawohl [21], Wang [33], Wang and Lin [34]. Along a different line, mixed problems for second E-mail address: [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.02.006
S. Mesloub / Nonlinear Analysis 68 (2008) 2594–2607
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order parabolic partial differential equations combining boundary local and boundary integral conditions were treated by Cannon and van der Hoek [4,5], Cannon, Esteva and van der Hoek [6], Kartynnik [20], Lin [23], Ionkin [19], Yurchuk [35] and Shi [31]. Problems for elliptic equations with operator nonlocal conditions were considered by Gushin and Mikhailov [18], Skubachevskii and Steblov [32], and Paneiah [27]. Then Gordeziani and Avalishvili [17], devoted a few papers to nonlocal problems for hyperbolic equations. Pulkina [28,29] studied the nonlocal analogue to classical Goursat problem. For mixed problems with purely integral conditions for parabolic, hyperbolic and pseudoparabolic equations, the reader should refer to Mesloub and Bouziani [26,25], Mesloub [24] and Bouziani [1]. We should also mention that many authors researched parabolic equations imposing different nonlocal conditions that have used numerical methods, see for example Chegis [8], Dehghan [10–12], Ekolin [13], Fairweather [14], Goolin and Ionkin [16], and Sapagovas and Chegis [30]. This paper is structured as follows: The position of the proposed studied nonlinear problem and the needed function spaces are described in Section 2. In Section 3, we establish an energy estimate for the solution of the associated linear problem, from which we deduce a uniqueness result and give some of the consequences of the obtained a priori bound. The solvability of the linear associated problem is described in Section 4. Based on the results obtained in previous sections and on an iterative process, Section 5 concludes this paper with the establishment of the well posedness of the nonlinear given problem. Finally, some references are introduced at the end. 2. Position of the nonlinear problem Let Q = Ω × (0, T ) be a bounded domain in I R 3 with Ω = (0, a) × (0, b) where a < ∞, b < ∞ and T < ∞. We consider the problem of finding a solution u, in Q, of the nonlinear partial differential equation Lu = u t −
1 div(x y∇u) = f (x, y, t, u, u x , u y ), xy
(x, y, t) ∈ Q,
(2.1)
subject to the initial condition `u = u(x, y, 0) = ϕ(x, y),
(2.2)
the classical boundary conditions u x (a, y, t) = 0,
u y (x, b, t) = 0,
(2.3)
and the nonlocal boundary integral conditions Z a Z b u(x, y, t)dx = 0, u(x, y, t)dy = 0. 0
(2.4)
0
We suppose that ϕ satisfies the compatibility conditions Z a ϕx (a, y) = 0, ϕ y (x, b) = 0, ϕ(x, y)dx = 0 0
b
Z and
ϕ(x, y)dy = 0.
0
Here ϕ and f are given functions. We assume in addition that the function f is Lipschitz, that is | f (x, y, t, u 1 , v1 , w1 ) − f (x, y, t, u 2 , v2 , w2 )| ≤ δ (|u 1 − u 2 | + |v1 − v2 | + |w1 − w2 |) ,
(2.5)
for all (x, y, t) ∈ Q and δ is a positive constant. We start by giving some results concerning the existence and uniqueness of a strong solution of the following associated linear problem 1 Lu = u t − div(x y∇u) = f (x, y, t), (x, y, t) ∈ Q, xy `u = u(x, y, 0) = ϕ(x, y), (2.6) u x (a, y, t) = 0, u y (x, b, t) = 0, Z Z a b u(x, y, t)dx = 0, u(x, y, t)dy = 0. 0
0
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The proofs are based on an energy inequality and the fact that the range of the operator generated by the problem (2.6) is dense. Let us first introduce appropriate function spaces. Let C(0, T ; X ), L 2 (Ω ) and L 2 (0, T ; X ) be the standard functional spaces, where X is a Hilbert space. Let L 2ρ (Ω ) be the weighted L 2 -space with finite norm Z √ 2 2 k f k L 2 (Ω ) = k x y f k L 2 (Ω ) = x y f 2 dxdy. ρ
Ω
The scalar product in
L 2ρ (Ω )
is defined by
( f, g) L 2ρ (Ω ) = (x y f, g) L 2 (Ω ) . Let L 2 (0, T ; Vρ1 (Ω )), be the subspace of L 2 (0, T ; L 2ρ (Ω )) with finite norm kuk2L 2 (0,T ;V 1 (Ω )) = kuk2L 2 (0,T ;L 2 (Ω )) + ku x k2L 2 (0,T ;L 2 (Ω )) + ku y k2L 2 (0,T ;L 2 (Ω )) . ρ
ρ
Vρ1 (Ω )
is the subspace of
L 2 (Ω )
ρ
ρ
with the finite norm
kuk2V 1 (Ω ) = kuk2L 2 (Ω ) + ku x k2L 2 (Ω ) + ku y k2L 2 (Ω ) . ρ
ρ
ρ
ρ
To problem (2.6), we assign the operator L = (L, `) with domain D(L) consisting of functions u belonging to L 2 (0, T ; L 2ρ (Ω )) for which partial derivatives u t , u x , u y , u t x , u t y , u x x , u yy are in L 2 (0, T ; L 2ρ (Ω )) and satisfying conditions (2.3) and (2.4). The operator L is considered from B to H , where B is the Banach space consisting of functions u ∈ L 2 (0, T ; L 2ρ (Ω )) having finite norm 2 kuk2B = ku t k2L 2 (0,T ;L 2 (Ω )) + kdiv(x y∇u)k2L 2 (0,T ;L 2 (Ω )) + ku(x, y, τ )kC(0,T ;V 1 (Ω )) ρ
ρ
and satisfying conditions (2.3) and (2.4). H is the Hilbert space of vector valued functions ( f, ϕ) ∈ L 2 (0, T ; L 2ρ (Ω ))× Vρ1 (Ω ) with finite norm kFk2F = k f k2L 2 (0,T ;L 2 (Ω )) + kϕk2V 1 (Ω ) , ρ
ρ
F = ( f, ϕ).
The scalar product in H = L 2 (0, T ; L 2ρ (Ω )) × Vρ1 (Ω ) is defined by (F, W ) F = ( f, Ψ ) L 2 (0,T ;L 2ρ (Ω )) + (ϕ, ω0 )Vρ1 (Ω ) , where F = ( f, ϕ) and W = (Ψ , ω0 ). 3. An energy estimate and uniqueness of solution In this section, we prove a uniqueness result for the linear problem (2.6), that is we establish an energy inequality for the operator L and we give some of its consequences. Before stating the uniqueness result, we prove the following Lemma 3.1. For any u ∈ L 2 (0, T ; L 2ρ (Ω )), satisfying conditions (2.4), we have the following inequalities k=x uk L 2 (0,T ;L 2ρ (Ω )) ≤ akuk L 2 (0,T ;L 2ρ (Ω )) ,
(3.1)
k= y uk L 2 (0,T ;L 2ρ (Ω )) ≤ bkuk L 2 (0,T ;L 2ρ (Ω )) ,
(3.2)
k=x y uk L 2 (0,T ;L 2ρ (Ω )) ≤ abkuk L 2 (0,T ;L 2ρ (Ω )) , Rx Ry Rx Ry where =x u = 0 u(ξ, y, t)dξ, = y u = 0 u(x, η, t)dη and =x y u = 0 0 u(ξ, η, t)dηdξ . Proof. It is sufficient to prove (3.1). Observe that Z a Z a x2 2 x=a 2 x (=x u) dx = x 2 u=x udx (=x u) |x=0 − 2 0 0 Z a Z 1 a 1 x 3 u 2 dx + x (=x u)2 dx. ≤ 2 0 2 0
(3.3)
(3.4)
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Thanks to boundary condition (2.4) and Cauchy-ε inequality: ε A2 /2 + B 2 /2ε. It follows from (3.4) that Z a Z a Z a xu 2 dx. x 3 u 2 dx ≤ a 2 x (=x u)2 dx ≤ 0
0
0
An integration with respect to t over [0, T ] of both sides of this last inequality yields k=x uk L 2 (0,T ;L 2ρ (Ω )) ≤ akuk L 2 (0,T ;L 2ρ (Ω )) . Inequalities (3.2) and (3.3) can be proved in the same way. Theorem 3.2. For any function u ∈ D(L), there exists a positive constant λ independent of the function u, such that kuk B ≤ λkLuk H .
(3.5) Rx
Proof. We first introduce the operators =x g = 0 g(ξ, y, t)dξ and Mg = gt + =x y g. And considering the inner product in L 2 (0, τ ; L 2ρ (Ω )) of the differential equation Lu = f (in (2.6)) and the operator Mu, we then have 1 2 − u t , (xu x )x ku t k L 2 (0,τ ;L 2 (Ω )) + (u t , =x y u) 2 L 0,τ ;L 2ρ (Ω ) ρ x L 2 (0,τ ;L 2ρ (Ω )) 1 1 1 − =x y u, (xu x )x − =x y u, (yu y ) y − u t , (yu y ) y x y y L 2 (0,τ ;L 2ρ (Ω )) L 2 (0,τ ;L 2ρ (Ω )) L 2 (0,τ ;L 2ρ (Ω )) = ( f, Mu) L 2 (0,τ ;L 2ρ (Ω )) .
(3.6)
We invoke (2.3) and (2.4), and in (3.6) we carry out appropriate integration by parts which are straightforward but somewhat tedious. We only give their results 1 1 1 (3.7) = ku x (x, y, τ )k2L 2 (Ω ) − kϕx (x, y)k2L 2 (Ω ) , − u t , (xu x )x ρ ρ x 2 2 L 2 (0,τ ;L 2ρ (Ω )) 1 − =x y u, (xu x )x = (u x , = y u) L 2 (0,τ ;L 2ρ (Ω )) , (3.8) x L 2 (0,τ ;L 2ρ (Ω )) 1 1 1 (3.9) = ku y (x, y, τ )k2L 2 (Ω ) − kϕ y (x, y)k2L 2 (Ω ) , − u t , (yu y ) y ρ ρ y 2 2 L 2 (0,τ ;L 2ρ (Ω )) 1 − =x y u, (yu y ) y = (u y , =x u) L 2 (0,τ ;L 2ρ (Ω )) . (3.10) y L 2 (0,τ ;L 2ρ (Ω )) Substitution of (3.7)–(3.10) into (3.6) yields 1 1 ku t k2L 2 (0,τ ;L 2 (Ω )) + ku x (x, y, τ )k2L 2 (Ω ) + ku y (x, y, τ )k2L 2 (Ω ) ρ ρ ρ 2 2 1 = −(u t , =x y u) L 2 (0,τ ;L 2ρ (Ω )) − (u x , = y u) L 2 (0,τ ;L 2ρ (Ω )) − (u y , =x u) L 2 (0,τ ;L 2ρ (Ω )) + kϕx (x, y)k2L 2 (Ω ) ρ 2 1 + kϕ y (x, y)k2L 2 (Ω ) + ( f, u t ) L 2 (0,τ ;L 2ρ (Ω )) + ( f, =x y u) L 2 (0,τ ;L 2ρ (Ω )) . ρ 2 Therefore, using Lemma 3.1 and Young’s inequality, we infer from (3.11),
(3.11)
ku t k2L 2 (0,τ ;L 2 (Ω )) + ku x (x, y, τ )k2L 2 (Ω ) + ku y (x, y, τ )k2L 2 (Ω ) ≤ kϕx (x, y)k2L 2 (Ω ) + kϕ y (x, y)k2L 2 (Ω ) ρ
+ 3k f k2L 2 (0,τ ;L 2 (Ω )) ρ
ρ
ρ
+ ku y k2L 2 (0,τ ;L 2 (Ω )) ρ
+ ku x k2L 2 (0,τ ;L 2 (Ω )) (a 2 ρ
ρ
2
ρ
2 2
+ b + 3a b
)kuk2L 2 (0,τ ;L 2 (Ω )) . ρ
(3.12)
From the differential equation Lu = f (in (2.6)), we have 1 1 1 kdiv(x y∇u)k2L 2 (0,τ ;L 2 (Ω )) ≤ k f k2L 2 (0,τ ;L 2 (Ω )) + ku t k2L 2 (0,τ ;L 2 (Ω )) . ρ ρ 8ab 4 4
(3.13)
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Consider also the obvious inequality, 1 1 1 1 ku(x, y, τ )k2L 2 (0,τ ;L 2 (Ω )) ≤ kϕk2L 2 (Ω ) + kuk2L 2 (0,τ ;L 2 (Ω )) + ku t k2L 2 (0,τ ;L 2 (Ω )) . ρ ρ ρ ρ 4 4 4 4 Combination of (3.12)–(3.14) gives
(3.14)
ku(x, y, τ )k2V 1 (Ω ) + ku t k2L 2 (0,τ ;L 2 (Ω )) + kdiv(x y∇u)k2L 2 (0,τ ;L 2 (Ω )) ρ ρ ρ 2 2 2 ≤ c1 k f k L 2 (Q τ ) + kϕkV 1 (Ω ) + c1 kukV 1 (Q τ ) , ρ
ρ
(3.15)
ρ
where max 26, 2 + 8a 2 + 8b2 + 24a 2 b2 . c1 = 1 min 4, ab Applying Lemma 7.1 in [22] to inequality (3.15), with g1 (t) = ku t k2L 2 (Ω ) + kdiv(x y∇u)k2L 2 (Ω ) , ρ
ρ
g2 (τ ) = ku(x, y, τ )k2V 1 (Ω ) , ρ 2 g3 (t) = c1 k f k L 2 (0,τ ;L 2 (Ω )) + kϕk2V 1 (Ω ) , ρ
ρ
it follows that ku(x, y, τ )k2V 1 (Ω ) + ku t k2L 2 (0,τ ;L 2 (Ω )) + kdiv(x y∇u)k2L 2 (0,τ ;L 2 (Ω )) ≤ c1 ec1 T k f k2L 2 (Q) + kϕk2V 1 (Ω ) . (3.16) ρ
ρ
ρ
ρ
ρ
Observe that the right-hand side of (3.16) is independent of τ , hence replacing the left-hand side by its upper bound √ with respect to τ from 0 to T , thus obtaining the a priori estimate (3.5), where λ = c1 ec1 T /2 . It can be proved in the standard way that the operator L : B → H is closable. Let L be the closure of the operator L. A solution of the equation Lu = F, F ∈ H , is called a strong solution of problem (2.6). We extend the energy estimate (3.5) to u ∈ D(L) by passing to the limit and thus establish uniqueness of a strong solution and closedness of the range R(L) of the operator L. More precisely, as consequences of the extended estimate kuk B ≤ λkLuk H , ∀u ∈ D(L), we have the following results: Corollary 3.3. If a strong solution of problem (2.6) exists, it is unique and depends continuously on F = ( f, ϕ) ∈ H . Corollary 3.4. The range R(L) of the operator L is closed and equal to R(L). 4. Existence of solution For the existence of a strong solution of the linear problem (2.6), we prove that the range R(L) of the operator L is dense in H . Thus we have the following Theorem 4.1. For each f ∈ L 2 (0, T ; L 2ρ (Ω )) and ϕ ∈ Vρ1 (Ω ), there exists a unique strong solution u = L −1 ( f, u 0 ) = L −1 ( f, u 0 ), of problem (2.6) satisfying the a priori bound kuk B ≤ λkLuk H , where λ is a positive constant independent of the solution u. Proof. From the extended estimate kuk B ≤ λkLuk H , it follows that the operator L has an inverse and, from Corollary 3.4, we deduce that the range R(L) is closed. Hence, it suffices to prove that R(L) = H . We first prove the following.
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Theorem 4.2. If for some function θ ∈ L 2 (0, T ; L 2ρ (Ω )) and all u ∈ D(L) with u|t=0 = 0 (Lu, θ ) L 2 (0,T ;L 2ρ (Ω )) = 0,
(4.1)
then θ = 0 almost everywhere in Q. Proof of Theorem 4.2. Define the function v by the relation Z T v(x, y, t) = θ (x, y, τ )dτ. t
Let u be a solution of the equation u t + =x y u = v(x, y, t).
(4.2)
And let u=
Z
T t
u τ dτ,
s≤t ≤T
0,
(4.3)
0 ≤ t ≤ s.
We now have θ (x, y, t) = −u tt − =x y u t .
(4.4)
Lemma 4.3. The function θ defined by the relation (4.4) is in L 2 (0, T ; L 2ρ (Ω )). Proof of Lemma 4.3. From (3.3) and the definition of D(L), we deduce that −=x y u t ∈ L 2 (0, T ; L 2ρ (Ω )). To prove that −u tt ∈ L 2 (0, T ; L 2ρ (Ω )), we use the t-averaging operators T
s−t ω g (x, s) ds, ε t R +∞ where ω ∈ C0∞ (0, T ), ω(t) ≥ 0, −∞ ω(t)dt = 1. Applying the operators ρε and ∂/∂t to Eq. (4.2), we obtain 1 (ρε g) (x, t) = ε
Z
∂ ∂ ∂ ρε u t + ρε =x y u = ρε v. ∂t ∂t ∂t
(4.5)
It follows from (4.5) that
2
∂
ρε u t
∂t
2 L
2
2
∂
2 2 ∂
≤ 2 ρ = u + 2a b ρ v . ε xy ε
∂t ∂t (0,T ;L 2ρ (Ω )) L 2 (0,T ;L 2ρ (Ω )) L 2 (0,T ;L 2ρ (Ω ))
Using properties of operators ρε introduced in [22], yields
2
∂
ρε u t ≤ K k=x y u t k2L 2 (0,T ;L 2 (Ω )) + K
∂t
2 ρ 2 L (0,T ;L ρ (Ω ))
2
∂
ρε v
∂t
2
L (0,T ;L 2ρ (Ω ))
(4.6)
,
where K = 2 max 1, a 2 b2 . Since ρε g → 0 in L 2 (0, T ; L 2ρ (Ω )), and the norm in L 2 (0, T ; L 2ρ (Ω )) of ε→0
θ ∈ L 2 (0, T ; L 2ρ (Ω )). This achieves the proof of Lemma 4.3.
∂ ∂t ρε u t
is bounded, we conclude that
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On going back to proof of Theorem 4.2, we replace the function θ given by (4.4) in (4.1), and we have −(u t , u tt ) L 2 (0,T ;L 2ρ (Ω )) − (u t , =x y u t ) L 2 (0,T ;L 2ρ (Ω )) 1 1 + =x y u t , (xu x )x + u tt , (xu x )x x x L 2 (0,T ;L 2ρ (Ω )) L 2 (0,T ;L 2ρ (Ω )) 1 1 + u tt , (yu y ) y + =x y u t , (yu y ) y = 0. y y 2 2 L (0,T ;L ρ (Ω )) L 2 (0,T ;L 2ρ (Ω ))
(4.7)
Direct evaluation of the first and last four terms on the left-hand side of (4.7), yields 1 −(u t , u tt ) L 2 (0,T ;L 2ρ (Ω )) = ku t (x, y, s)k2L 2 (Ω ) , ρ 2 1 u tt , (xu x )x = ku t x k2L 2 (s,T ;L 2 (Ω )) , ρ x L 2 (0,T ;L 2ρ (Ω )) 1 =x y u t , (xu x )x = −(u x , = y u t ) L 2 (s,T ;L 2ρ (Ω )) , x L 2 (0,T ;L 2ρ (Ω )) 1 = ku t y k2L 2 (s,T ;L 2 (Ω )) , u tt , (yu y ) y ρ y L 2 (0,T ;L 2ρ (Ω )) 1 =x y u t , (yu y ) y = −(u y , =x u t ) L 2 (s,T ;L 2ρ (Ω )) . y L 2 (0,T ;L 2ρ (Ω ))
(4.8) (4.9) (4.10) (4.11) (4.12)
Substitution of (4.8)–(4.12) into (4.7) gives 1 ku t (x, y, s)k2L 2 (Ω ) + ku t x k2L 2 (s,T ;L 2 (Ω )) + ku t y k2L 2 (s,T ;L 2 (Ω )) ρ ρ ρ 2 = (u t , =x y u t ) L 2 (s,T ;L 2ρ (Ω )) + (u x , = y u t ) L 2 (s,T ;L 2ρ (Ω )) + (u y , =x u t ) L 2 (s,T ;L 2ρ (Ω )) , which in turn implies after using Young’s inequality and Lemma 3.1, 1 ku t (x, y, s)k2L 2 (Ω ) + ku t x k2L 2 (s,T ;L 2 (Ω )) + ku t y k2L 2 (s,T ;L 2 (Ω )) ρ ρ ρ 2 2 1 1 + a + b2 + a 2 b2 1 ≤ ku x k2L 2 (s,T ;L 2 (Ω )) + ku y k2L 2 (s,T ;L 2 (Ω )) + ku t k2L 2 (s,T ;L 2 (Ω )) . ρ ρ ρ 2 2 2
(4.13)
It is quite easy to see that ku x (x, y, T )k2L 2 (Ω ) ≤ ku x k2L 2 (s,T ;L 2 (Ω )) + ku t x k2L 2 (s,T ;L 2 (Ω )) ,
(4.14)
ku y (x, y, T )k2L 2 (Ω ) ≤ ku y k2L 2 (s,T ;L 2 (Ω )) + ku t y k2L 2 (s,T ;L 2 (Ω )) .
(4.15)
ρ
ρ ρ
ρ
ρ
ρ
Adding side to side (4.13)–(4.15) to get ku t (x, y, s)k2L 2 (Ω ) + ku x (x, y, T )k2L 2 (Ω ) + ku y (x, y, T )k2L 2 (Ω ) ρ ρ ρ 2 2 2 ≤ γ ku t k L 2 (s,T ;L 2 (Ω )) + ku x k L 2 (s,T ;L 2 (Ω )) + ku y k L 2 (s,T ;L 2 (Ω )) , ρ
where γ = max 3, 1 + a 2 + b2 + a 2 b2 .
ρ
ρ
(4.16)
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Inequality (4.16) is essential in our proof. To use it, we introduce the new function g defined by the formula Z T u τ dτ. g(x, y, t) = t
Then u(x, y, t) = g(x, y, s) − g(x, y, t), u(x, y, T ) = g(x, y, s). Using (4.16) and the above relations, we obtain (1 − 2γ (T − s)) kgx (x, y, s)k2L 2 (Ω ) + kg y (x, y, s)k2L 2 (Ω ) + ku t (x, y, s)k2L 2 (Ω ) ρ ρ ρ ≤ 2γ ku t k2L 2 (s,T ;L 2 (Ω )) + kgx k2L 2 (s,T ;L 2 (Ω )) + kg y k2L 2 (s,T ;L 2 (Ω )) . ρ
ρ
ρ
(4.17)
Hence if s0 > 0 satisfies 1 − 2γ (T − s0 ) = 1/2, then (4.17) implies ku t (x, y, s)k2L 2 (Ω ) + kgx (x, y, s)k2L 2 (Ω ) + kg y (x, y, s)k2L 2 (Ω ) ρ ρ ρ 2 2 2 ≤ 4γ ku t k L 2 (s,T ;L 2 (Ω )) + kgx k L 2 (s,T ;L 2 (Ω )) + kg y k L 2 (s,T ;L 2 (Ω )) ρ
ρ
ρ
(4.18)
for all s ∈ [T − s0 , T ]. The sum of the three norms in the right-hand side of (4.18) is a function of s. We denote it by Z T F(s) = ku t k2L 2 (Ω ) + kgx k2L 2 (Ω ) + kg y k2L 2 (Ω ) dt. s
ρ
ρ
ρ
Then (4.18) takes the form −F 0 (s) ≤ 4γ F (s) .
(4.19)
Integrating (4.19) with respect to t over [s, T ] and taking into account that F(T ) = 0, we obtain F (s) exp(4γ s) ≤ 0. It follows that θ = 0 almost everywhere in Q T −s0 . Proceeding in this way step by step, we prove that θ = 0 almost everywhere in Q. Therefore the proof of Theorem 4.2 is completed. We now complete the proof of Theorem 4.1. Since H is a Hilbert space, density of the set R(L) is equivalent to the property that orthogonality of a vector F = ( f, ϕ) ∈ H to the set R(L), i.e, the equality (Lu, θ ) L 2 (0,T ;L 2ρ (Ω )) + (`u, ϕ)Vρ1 (Ω ) = 0,
∀u ∈ D(L),
(4.20)
implies F = 0. In particular, in (4.20) we set `u = 0. We then conclude by Theorem 4.2 that θ = 0. Thus (4.20) implies that (`u, ϕ)Vρ1 (Ω ) = 0. Now since the range of the trace operator ` is dense in Vρ1 (Ω ), it follows that ϕ = 0. Hence, F = 0. 5. The nonlinear problem We are now in a position to solve the nonlinear problem (2.1)–(2.4). Relying on the results obtained previously, we apply an iterative process to establish the existence and uniqueness of the weak solution of the nonlinear problem. If u is a solution of problem (2.1)–(2.4) and Φ is a solution of the problem 1 div(x y∇Φ) = 0, xy `Φ = Φ(x, y, 0) = ϕ(x, y), Φx (a, y, t) = 0, Φ y (x, b, t) = 0, LΦ = Φt −
(5.1) (5.2) (5.3)
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S. Mesloub / Nonlinear Analysis 68 (2008) 2594–2607 a
Z
Z
b
Φ(x, y, t)dy = 0,
Φ(x, y, t)dx = 0,
(5.4)
0
0
then ξ = u − Φ satisfies 1 div(x y∇ξ ) = F(x, y, t, ξ, ξx , ξ y ), xy ξ(x, y, 0) = 0, Lξ = ξt −
(5.5) (5.6)
ξx (a, y, t) = 0, ξ y (x, b, t) = 0, Z b Z a ξ(x, y, t)dy = 0, ξ(x, y, t)dx = 0,
(5.7) (5.8)
0
0
where F(x, y, t, ξ, ξx , ξ y ) = f (x, y, t, ξ + Φ, ξx + Φx , ξ y + Φ y ), and satisfies the condition |F(x, y, t, u 1 , v1 , w1 ) − F(x, y, t, u 2 , v2 , w2 )| ≤ δ (|u 1 − u 2 | + |v1 − v2 | + |w1 − w2 |) .
(5.9)
Theorem 4.1 shows that problem (5.1)–(5.4), admits a unique solution that depends continuously on the data ϕ ∈ Vρ1 (Ω ). In order to solve problem (5.5)–(5.8), we will show that it has a unique weak solution. The general idea is to construct an iteration sequence (ξ (n) )n∈ N´ which converges to an element ξ ∈ L 2 (0, T ; Vρ1 (Ω )) that will be a solution of the problem under consideration (5.5)–(5.8). √ Theorem 5.1. Suppose that condition (5.9) holds and δ < e−c1 T /2 / 3T , then problem (5.5)–(5.8) admits a weak solution belonging to L 2 (0, T ; Vρ1 (Ω )). Ra Rb Proof. Let us first assume that ξ and υ ∈ C 1 (Q), such that ξ(x, y, 0) = 0, υ(x, y, T ) = 0, 0 ξ dx = 0, 0 ξ dy = Ra Rb 0, 0 υdx = 0, 0 υdy = 0. For υ ∈ C 1 (Q), we have 1 1 (xξx )x , =x y υ − (yξ y ) y , =x y υ ξt , =x y υ L 2 (0,T ;L 2 (Ω )) − ρ x y L 2 (0,T ;L 2ρ (Ω )) L 2 (0,T ;L 2ρ (Ω )) = Lξ, =x y υ L 2 (0,T ;L 2 (Ω )) = F, =x y υ L 2 (0,T ;L 2 (Ω )) . (5.10) ρ
ρ
In light of the above assumptions, we obtain ξt , =x y υ L 2 (0,T ;L 2 (Ω )) = − υt , =x y (ζ ηξ ) L 2 (Q) , ρ 1 − (xξx )x , =x y υ = − xυ, = y (ηξx ) L 2 (Q) , x L 2 (0,T ;L 2ρ (Ω )) 1 − (yξ y ) y , =x y υ = − yυ, =x (ζ ξ y ) L 2 (Q) , y L 2 (0,T ;L 2ρ (Ω )) F, =x y υ L 2 (0,T ;L 2 (Ω )) = υ, =x y (ζ ηF) L 2 (Q) .
(5.14)
Substituting (5.11)–(5.14) into (5.10) yields A(ξ, υ) = υ, =x y (ζ ηF) L 2 (Q) ,
(5.15)
ρ
(5.11) (5.12) (5.13)
where A(ξ, υ) = − υt , =x y (ζ ηξ )
L 2 (Q)
− xυ, = y (ηξx ) L 2 (Q) − yυ, =x (ζ ξ y ) L 2 (Q) .
It should be mentioned here that L 2 (0, T ; L 2 (Ω )) = L 2 (Q). We define the weak solution of problem (5.5)–(5.8) as a function ξ ∈ L 2 (0, T ; Vρ1 (Ω )) satisfying:
(5.16)
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(1) A(ξ, υ) = υ, =x y (ζ ηF) L 2 (Q) . (2) ξx (a, y, t) = 0, ξ y (x, b, t) = 0. As mentioned previously, we define the iteration sequence (ξ (n) )n∈ N´ as follows: Let ξ (0) = 0. Given the element then for n = 1, 2, . . . . solve the iterated problem:
ξ (n−1) ,
(n)
ξt
−
1 div(x y∇ξ (n) ) = F(x, y, t, ξ (n−1) , ξx(n−1) , ξ y(n−1) ), xy
(5.17)
ξ (n) (x, y, 0) = 0, ξx(n) (a, a
Z
y, t) = 0,
(5.18) ξ y(n) (x, b, t)
ξ (n) (x, y, t)dx = 0,
0
b
Z
= 0,
(5.19)
ξ (n) (x, y, t)dy = 0.
(5.20)
0
For fixed n, Theorem 4.1 shows that each problem (5.17)–(5.20) has a unique solution ξ (n) (x, y, t). Now, we set Y (n) (x, y, t) = ξ (n+1) (x, y, t) − ξ (n) (x, y, t), then we obtain the following new problem (n)
Yt
−
1 div(x y∇Y (n) ) = β (n−1) (x, y, t), xy
(5.21)
Y (n) (x, y, 0) = 0, Yx(n) (a, a
Z
y, t) = 0,
(5.22) Y y(n) (x, b, t)
Y (n) (x, y, t)dx = 0,
b
Z
= 0,
(5.23)
Y (n) (x, y, t)dy = 0,
(5.24)
0
0
where β (n−1) (x, y, t) = F(x, y, t, ξ (n) , ξx(n) , ξ y(n) ) − F(x, y, t, ξ (n−1) , ξx(n−1) , ξ y(n−1) ).
Lemma 5.2. Suppose that condition (5.9) holds, then there exists a positive constant C = problem (5.21)–(5.24), we have the a priori bound
√
c2 T ec1 T /2 such that for
kY (n) k L 2 (0,T´ ;V 1 (Ω )) ≤ CkY (n−1) k L 2 (0,T´ ;V 1 (Ω )) ,
(5.25)
ρ
ρ
where c1 = a 2 b2 + a 2 /2 + b2 /2 + 1/2,
c2 = 3δ 2 .
Proof. Taking the inner product in L 2 (0, ν; L 2ρ (Ω )), with 0 ≤ ν ≤ T , of Eq. (5.21) and the integro-differential operator (n)
MY (n) = Yt
+ =x y Y (n) ,
we obtain (n) 1 (n) xY − Y , t x ρ L 2 (0,ν;L 2ρ (Ω )) x L 2 (0,ν;L 2 (Ω )) x ρ 1 (n) 1 xYx(n) − Yt , yY y(n) − =x y Y (n) , x L 2 (0,ν;L 2 (Ω )) y L 2 (0,ν;L 2 (Ω )) x y ρ ρ 1 − =x y Y (n) , yY y(n) y L 2 (0,ν;L 2 (Ω )) y ρ (n) (n−1) (n) = β (n−1) , Yt + β , = Y . x y 2 2 2 2
(n) (n) kYt k2L 2 (0,ν;L 2 (Ω )) + Yt , =x y Y (n)
L (0,ν;L ρ (Ω ))
L (0,ν;L ρ (Ω ))
(5.26)
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S. Mesloub / Nonlinear Analysis 68 (2008) 2594–2607
It is straightforward to see that 1 1 (n) kYt k2L 2 (0,ν;L 2 (Ω )) + kYx(n) (x, y, ν)k2L 2 (Ω ) + kY y(n) (x, y, ν)k2L 2 (Ω ) ρ ρ ρ 2 2 (n) (n−1) (n−1) (n) = β , Yt + β , =x y Y 2 2 2 2 L (0,ν;L ρ (Ω ))
(n)
− Yt , =x y Y (n)
L (0,ν;L ρ (Ω ))
L 2 (0,ν;L 2ρ (Ω ))
− Yx(n) , = y Y (n)
L 2 (0,ν;L 2ρ (Ω ))
− Y y(n) , =x Y (n)
L 2 (0,ν;L 2ρ (Ω ))
.
(5.27)
Estimating terms on the RHS of (5.27), we obtain (n) β (n−1) , Yt
L 2 (0,ν;L 2ρ (Ω ))
1 (n) 2 kY k L 2 (0,ν;L 2 (Ω )) + kβ (n−1) k2L 2 (0,ν;L 2 (Ω )) ρ ρ 4 t 1 (n) 2 3 2 (n−1) 2 ≤ kYt k L 2 (0,ν;L 2 (Ω )) + δ kY k L 2 (0,ν;L 2 (Ω )) ρ ρ 4 2 + kYx(n−1) k2L 2 (0,ν;L 2 (Ω )) + kY y(n−1) k2L 2 (0,ν;L 2 (Ω )) ,
≤
ρ
ρ
β (n−1) , =x y Y (n)
≤
L 2 (0,ν;L 2ρ (Ω ))
a 2 b2 (n) 2 kY k L 2 (0,ν;L 2 (Ω )) ρ 2 3 2 (n−1) 2 + δ kY k L 2 (0,ν;L 2 (Ω )) + kYx(n−1) k2L 2 (0,ν;L 2 (Ω )) ρ 2 ρ (n−1) 2 + kY y k L 2 (0,ν;L 2 (Ω )) , ρ
(n)
− Yt , =x y Y (n)
− Yx(n) , = y Y (n) − Y y(n) , =x Y (n)
L 2 (0,ν;L 2ρ (Ω ))
L 2 (0,ν;L 2ρ (Ω ))
L 2 (0,ν;L 2ρ (Ω ))
≤
1 (n) 2 kY k L 2 (0,ν;L 2 (Ω )) + a 2 b2 kY (n) k2L 2 (0,ν;L 2 (Ω )) , ρ ρ 4 t
(5.28)
(5.29) (5.30)
≤
b2 1 (n) 2 kYx k L 2 (0,ν;L 2 (Ω )) + kY (n) k2L 2 (0,ν;L 2 (Ω )) , ρ ρ 2 2
(5.31)
≤
1 (n) 2 a2 kY y k L 2 (0,ν;L 2 (Ω )) + kY (n) k2L 2 (0,ν;L 2 (Ω )) . ρ ρ 2 2
(5.32)
It is obvious that 1 (n) 1 1 (n) kY (x, y, ν)k2L 2 (Ω ) ≤ kY (n) k2L 2 (0,ν;L 2 (Ω )) + kYt k2L 2 (0,ν;L 2 (Ω )) . ρ ρ ρ 2 2 2
(5.33)
Combination of inequalities (5.28)–(5.33) then yields kY (n) (x, y, ν)k2V 1 (Ω ) ≤ c1 kY (n) k2L 2 (0,ν;V 1 (Ω )) + c2 kY (n−1) k2L 2 (0,ν;V 1 (Ω )) , ρ
ρ
ρ
(5.34)
where c1 = (2a 2 b2 + a 2 + b2 + 1)/2,
c2 = 3δ 2 .
An application of Lemma 7.1 in [22] to (5.34), gives kY (n) (x, y, ν)k2V 1 (Ω ) ≤ c2 ec1 T kY (n−1) k2L 2 (0,ν;V 1 (Ω )) . ρ
ρ
(5.35)
Integrating both sides of (5.35) with respect to ν over the interval [0, T ], we obtain kY (n) k2L 2 (0,T ;V 1 (Ω )) ≤ c2 T ec1 T kY (n−1) k2L 2 (0,T ;V 1 (Ω )) . ρ
ρ
(5.36)
√ P (n) converges if δ < e−c1 T /2 / 3T . Knowing that Y (n) (x, y, t) = It follows from (5.36) that the series ∞ n=1 Y ξ (n+1) (x, y, t) − ξ (n) (x, y, t), then it follows that the sequence defined by
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S. Mesloub / Nonlinear Analysis 68 (2008) 2594–2607
ξ (n) (x, y, t) =
n−1 X
Y (m) (x, y, t) + ξ (0) (x, y, t)
m=0
=
n−1 X
(ξ (m+1) (x, y, t) − ξ (m) (x, y, t)) + ξ (0) (x, y, t),
n = 1, 2, . . .
m=0
converges to an element ξ ∈ L 2 (0, T ; Vρ1 (Ω )) which must be proved that it is a solution of problem (5.5)–(5.8). In other words, ξ must satisfy: (1) A(ξ, υ) = υ, =x y (ζ ηF) L 2 (Q) , (2) ξx (a, y, t) = 0, ξ y (x, b, t) = 0 as mentioned previously. From equalities (5.17)–(5.20), we have (n−1) (n−1) A(ξ (n) , υ) = υ, =x y (ζ ηF(ξ, η, t, ξ (n−1) , ξζ , ξη )) 2 , L (Q)
from which it follows that (n−1) (n−1) A(ξ, υ) + A(ξ (n) − ξ, υ) = υ, =x y (ζ ηF(ξ, η, t, ξ (n−1) , ξζ , ξη )) − =x y (ζ ηF(ξ, η, t, ξ, ξζ , ξη )) L 2 (Q) + (υ, =x y (ζ ηF(ξ, η, t, ξ, ξζ , ξη )))
L 2 (Q)
.
Now, from the partial differential equation (5.17), we have ∂ ∂ ∂ (n) (n) (n) − υ, =x y η A(ξ − ξ, υ) = υ, =x y (ζ η(ξ − ξ )) ζ (ξ − ξ ) ∂t ∂ζ ∂ζ L 2 (Q) L 2 (Q) ∂ ∂ (n) − υ, =x y ζ η (ξ − ξ ) ∂η ∂η L 2 (Q) (n−1) − ξζ , ξη(n−1) − ξη )) 2 . = υ, =x y (ζ ηF(ζ, η, t, ξ (n−1) − ξ, ξζ L (Q)
(5.37)
(5.38)
Performing integration by parts in each term on the RHS of (5.38), and using conditions on υ and ξ to obtain ∂ ∂ (n) υ, =x y (ζ η(ξ (n) − ξ )) + = y υ, (ξ − ξ ) − ∂t ∂x L 2 (Q) L 2ρ (Q) ∂ + =x υ, (ξ (n) − ξ ) = A(ξ (n) − ξ, υ). (5.39) ∂y L 2ρ (Q) Application of Cauchy–Schwarz inequality to terms on the LHS of (5.39) yields
∂υ a 3/2 b32 (n)
∂υ kξ (n) − ξ k L 2ρ (Q) , − , =x y (ζ η(ξ − ξ )) ≤ √
∂t ∂t L 2 (Q) 2 L 2 (Q)
∂ (n)
∂ (n) 1/2 3/2
= y υ, (ξ − ξ ) ≤ a b kυk L 2 (Q) (ξ − ξ ) ,
2 ∂x ∂x L 2ρ (Q) L ρ (Q)
∂ (n)
∂
=x υ, (ξ (n) − ξ ) ≤ a 3/2 b1/2 kυk L 2 (Q) (ξ − ξ ) .
2 ∂y ∂y L 2ρ (Q) L ρ (Q)
(5.40) (5.41) (5.42)
Substitution of (5.40)–(5.42) into (5.39) then gives A(ξ (n) − ξ, υ) ≤ Bkξ (n) − ξ k L 2 (0,T ;Vρ1 (Ω )) kυk L 2 (Q) + kυt k L 2 (Q) , where 3/2 b32 1/2 3/2 3/2 1/2 a . B = max a b , a b , √ 2
(5.43)
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S. Mesloub / Nonlinear Analysis 68 (2008) 2594–2607
On the other side, we have (n−1) (n−1) υ, =x y (ζ ηF(ξ, η, t, ξ (n−1) , ξζ , ξη )) − =x y (ζ ηF(ξ, η, t, ξ, ξζ , ξη )) ≤
a 3/2 b3/2 √
2
L 2 (Q)
δkυk L 2 (Q) kξ (n) − ξ k L 2 (0,T ;Vρ1 (Ω )) .
(5.44)
Invoking (5.43) and (5.44), and taking the limit as n → ∞ in equality (5.37), we obtain A(ξ, υ) = υ, =x y (ζ ηF(ξ, η, t, ξ, ξζ , ξη )) L 2 (Q) . To conclude that problem (5.5)–(5.8) admits a weak solution, we must R t show that conditions (5.7) hold. Rt Since ξ ∈ L 2 (0, T ; Vρ1 (Ω )), then 0 ∂∂ξx (x, y, s)ds ∈ C(Q) and 0 ∂ξ ∂ y (x, y, s)ds ∈ C(Q), and we conclude that ξx (a, y, t) = 0, ξ y (x, b, t) = 0, a.e. It remains now to prove the uniqueness of solution of problem (5.5)–(5.8). Theorem 5.3. If condition (5.9) is satisfied, then the solution of problem (5.5)–(5.8) is unique. Proof. Let ξ1 , ξ2 ∈ L 2 (0, T ; Vρ1 (Ω )) be two solutions of (5.5)–(5.8), then Y = ξ1 − ξ2 ∈ L 2 (0, T ; Vρ1 (Ω )) and satisfies 1 div(x y∇Y ) = β(x, y, t), xy Y (x, y, 0) = 0, Yx (a, y, t) = 0, Y y (x, b, t) = 0, Z b Z a Y (x, y, t)dx = 0, Y (x, y, t)dy = 0,
(5.45)
Yt −
0
(5.46) (5.47) (5.48)
0
where ∂ξ1 ∂ξ1 β(x, y, t) = F x, y, t, ξ, , ∂x ∂y
∂ξ2 ∂ξ2 , − F x, y, t, ξ, ∂x ∂y
.
By following the same procedure used in proving Lemma 5.2, we take the inner product in L 2 (0, T ; Vρ1 (Ω )) of Eq. (5.45) and differential operator DY = Yt + =x y Y , to obtain kY k L 2 (0,T´ ;V 1 (Ω )) ≤ CkY k L 2 (0,T´ ;V 1 (Ω )) , ρ
ρ
(5.49)
where C is the same constant as in Lemma 5.2. Since C < 1, we deduce from (5.49) that (1 − C)kY k L 2 (0,T´ ;V 1 (Ω )) = 0, ρ
which in turns implies that ξ1 − ξ2 = 0, and consequently ξ1 = ξ2 ∈ L 2 (0, T ; Vρ1 (Ω )). The proof of Theorem 5.1 is achieved. Acknowledgment This work has been funded and supported by the Research Center Project No. Math/2006/45 at King Saud University. References [1] A. Bouziani, Solvability of nonlinear pseudoparabolic equation with a nonlocal boundary condition, Nonlinear Anal. 55 (7–8) (2003) 883–904. [2] J.R. Cannon, J. van der Hoek, Diffusion subject to specification of mass, J. Math. Anal. Appl. 115 (1986) 517–529. [3] J.R. Cannon, The one-dimensional heat equation, in: K. Rach (Ed.), Encyclopedia of Mathematics and its Applications, vol. 23, AddisonWesley, Reading, MA, 1984.
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On a singular two dimensional nonlinear evolution equation with nonlocal conditions Said Mesloub King Saud University, College of Sciences, Department of Mathematics, P.O. Box 2455, Riyadh 11451, Saudi Arabia Received 21 May 2006; accepted 5 February 2007
Abstract This paper deals with a nonclassical initial boundary value problem for a two dimensional parabolic equation with Bessel operator. We prove the existence and uniqueness of the weak solution of the given nonlinear problem. We start by solving the associated linear problem. After writing this latter in its operator form, we establish an a priori bound from which we deduce the uniqueness of the strong solution. For the solvability of the associated linear problem, we prove that the range of the operator generated by the considered problem is dense. On the basis of the obtained results of the linear problem, we apply an iterative process to establish the existence and uniqueness of the nonlinear problem. c 2007 Elsevier Ltd. All rights reserved.
MSC: 35K; 35R Keywords: Nonlinear evolution equation; Nonlocal condition; A priori estimate; Bessel operator
1. Introduction Extensive research studies have been done for numerous and different nonlocal conditions for the solvability of parabolic, hyperbolic, pseudoparabolic and elliptic equations which were published in different journals. Many problems of modern physics and technology can be effectively described in terms of nonlocal problems for partial differential equations. Some presented articles, Friedman [15] and Kawohl [21], analyze such conditions imposed, and observe that the existence and uniqueness of the solution of parabolic equation is related mainly to the smallness of functions involved in nonlocal conditions. Nonclassical problems formulated for equations of mathematical physics are called nonlocal initial boundary value problems when the role of the boundary (or initial) conditions is played by an expression where the boundary value of an unknown function is expressed by the value(s) of this function at internal point(s). There are two types of nonlocal problems, spatial and in-time nonlocal problems. Nonlocal conditions arise mainly when the data on the boundary cannot be measured directly. Mixed problems with nonlocal conditions such as boundary integral conditions, have many important applications in chemical diffusion, thermoelasticity, heat conduction processes, population dynamics, vibration problems, nuclear reactor dynamics, control theory, medical science, biochemistry and certain biological processes: Cannon [3], Cannon and van der Hoek [2], Capasso and Kunisch [7], Day [9], Kawohl [21], Wang [33], Wang and Lin [34]. Along a different line, mixed problems for second E-mail address: [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.02.006
S. Mesloub / Nonlinear Analysis 68 (2008) 2594–2607
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order parabolic partial differential equations combining boundary local and boundary integral conditions were treated by Cannon and van der Hoek [4,5], Cannon, Esteva and van der Hoek [6], Kartynnik [20], Lin [23], Ionkin [19], Yurchuk [35] and Shi [31]. Problems for elliptic equations with operator nonlocal conditions were considered by Gushin and Mikhailov [18], Skubachevskii and Steblov [32], and Paneiah [27]. Then Gordeziani and Avalishvili [17], devoted a few papers to nonlocal problems for hyperbolic equations. Pulkina [28,29] studied the nonlocal analogue to classical Goursat problem. For mixed problems with purely integral conditions for parabolic, hyperbolic and pseudoparabolic equations, the reader should refer to Mesloub and Bouziani [26,25], Mesloub [24] and Bouziani [1]. We should also mention that many authors researched parabolic equations imposing different nonlocal conditions that have used numerical methods, see for example Chegis [8], Dehghan [10–12], Ekolin [13], Fairweather [14], Goolin and Ionkin [16], and Sapagovas and Chegis [30]. This paper is structured as follows: The position of the proposed studied nonlinear problem and the needed function spaces are described in Section 2. In Section 3, we establish an energy estimate for the solution of the associated linear problem, from which we deduce a uniqueness result and give some of the consequences of the obtained a priori bound. The solvability of the linear associated problem is described in Section 4. Based on the results obtained in previous sections and on an iterative process, Section 5 concludes this paper with the establishment of the well posedness of the nonlinear given problem. Finally, some references are introduced at the end. 2. Position of the nonlinear problem Let Q = Ω × (0, T ) be a bounded domain in I R 3 with Ω = (0, a) × (0, b) where a < ∞, b < ∞ and T < ∞. We consider the problem of finding a solution u, in Q, of the nonlinear partial differential equation Lu = u t −
1 div(x y∇u) = f (x, y, t, u, u x , u y ), xy
(x, y, t) ∈ Q,
(2.1)
subject to the initial condition `u = u(x, y, 0) = ϕ(x, y),
(2.2)
the classical boundary conditions u x (a, y, t) = 0,
u y (x, b, t) = 0,
(2.3)
and the nonlocal boundary integral conditions Z a Z b u(x, y, t)dx = 0, u(x, y, t)dy = 0. 0
(2.4)
0
We suppose that ϕ satisfies the compatibility conditions Z a ϕx (a, y) = 0, ϕ y (x, b) = 0, ϕ(x, y)dx = 0 0
b
Z and
ϕ(x, y)dy = 0.
0
Here ϕ and f are given functions. We assume in addition that the function f is Lipschitz, that is | f (x, y, t, u 1 , v1 , w1 ) − f (x, y, t, u 2 , v2 , w2 )| ≤ δ (|u 1 − u 2 | + |v1 − v2 | + |w1 − w2 |) ,
(2.5)
for all (x, y, t) ∈ Q and δ is a positive constant. We start by giving some results concerning the existence and uniqueness of a strong solution of the following associated linear problem 1 Lu = u t − div(x y∇u) = f (x, y, t), (x, y, t) ∈ Q, xy `u = u(x, y, 0) = ϕ(x, y), (2.6) u x (a, y, t) = 0, u y (x, b, t) = 0, Z Z a b u(x, y, t)dx = 0, u(x, y, t)dy = 0. 0
0
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The proofs are based on an energy inequality and the fact that the range of the operator generated by the problem (2.6) is dense. Let us first introduce appropriate function spaces. Let C(0, T ; X ), L 2 (Ω ) and L 2 (0, T ; X ) be the standard functional spaces, where X is a Hilbert space. Let L 2ρ (Ω ) be the weighted L 2 -space with finite norm Z √ 2 2 k f k L 2 (Ω ) = k x y f k L 2 (Ω ) = x y f 2 dxdy. ρ
Ω
The scalar product in
L 2ρ (Ω )
is defined by
( f, g) L 2ρ (Ω ) = (x y f, g) L 2 (Ω ) . Let L 2 (0, T ; Vρ1 (Ω )), be the subspace of L 2 (0, T ; L 2ρ (Ω )) with finite norm kuk2L 2 (0,T ;V 1 (Ω )) = kuk2L 2 (0,T ;L 2 (Ω )) + ku x k2L 2 (0,T ;L 2 (Ω )) + ku y k2L 2 (0,T ;L 2 (Ω )) . ρ
ρ
Vρ1 (Ω )
is the subspace of
L 2 (Ω )
ρ
ρ
with the finite norm
kuk2V 1 (Ω ) = kuk2L 2 (Ω ) + ku x k2L 2 (Ω ) + ku y k2L 2 (Ω ) . ρ
ρ
ρ
ρ
To problem (2.6), we assign the operator L = (L, `) with domain D(L) consisting of functions u belonging to L 2 (0, T ; L 2ρ (Ω )) for which partial derivatives u t , u x , u y , u t x , u t y , u x x , u yy are in L 2 (0, T ; L 2ρ (Ω )) and satisfying conditions (2.3) and (2.4). The operator L is considered from B to H , where B is the Banach space consisting of functions u ∈ L 2 (0, T ; L 2ρ (Ω )) having finite norm 2 kuk2B = ku t k2L 2 (0,T ;L 2 (Ω )) + kdiv(x y∇u)k2L 2 (0,T ;L 2 (Ω )) + ku(x, y, τ )kC(0,T ;V 1 (Ω )) ρ
ρ
and satisfying conditions (2.3) and (2.4). H is the Hilbert space of vector valued functions ( f, ϕ) ∈ L 2 (0, T ; L 2ρ (Ω ))× Vρ1 (Ω ) with finite norm kFk2F = k f k2L 2 (0,T ;L 2 (Ω )) + kϕk2V 1 (Ω ) , ρ
ρ
F = ( f, ϕ).
The scalar product in H = L 2 (0, T ; L 2ρ (Ω )) × Vρ1 (Ω ) is defined by (F, W ) F = ( f, Ψ ) L 2 (0,T ;L 2ρ (Ω )) + (ϕ, ω0 )Vρ1 (Ω ) , where F = ( f, ϕ) and W = (Ψ , ω0 ). 3. An energy estimate and uniqueness of solution In this section, we prove a uniqueness result for the linear problem (2.6), that is we establish an energy inequality for the operator L and we give some of its consequences. Before stating the uniqueness result, we prove the following Lemma 3.1. For any u ∈ L 2 (0, T ; L 2ρ (Ω )), satisfying conditions (2.4), we have the following inequalities k=x uk L 2 (0,T ;L 2ρ (Ω )) ≤ akuk L 2 (0,T ;L 2ρ (Ω )) ,
(3.1)
k= y uk L 2 (0,T ;L 2ρ (Ω )) ≤ bkuk L 2 (0,T ;L 2ρ (Ω )) ,
(3.2)
k=x y uk L 2 (0,T ;L 2ρ (Ω )) ≤ abkuk L 2 (0,T ;L 2ρ (Ω )) , Rx Ry Rx Ry where =x u = 0 u(ξ, y, t)dξ, = y u = 0 u(x, η, t)dη and =x y u = 0 0 u(ξ, η, t)dηdξ . Proof. It is sufficient to prove (3.1). Observe that Z a Z a x2 2 x=a 2 x (=x u) dx = x 2 u=x udx (=x u) |x=0 − 2 0 0 Z a Z 1 a 1 x 3 u 2 dx + x (=x u)2 dx. ≤ 2 0 2 0
(3.3)
(3.4)
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Thanks to boundary condition (2.4) and Cauchy-ε inequality: ε A2 /2 + B 2 /2ε. It follows from (3.4) that Z a Z a Z a xu 2 dx. x 3 u 2 dx ≤ a 2 x (=x u)2 dx ≤ 0
0
0
An integration with respect to t over [0, T ] of both sides of this last inequality yields k=x uk L 2 (0,T ;L 2ρ (Ω )) ≤ akuk L 2 (0,T ;L 2ρ (Ω )) . Inequalities (3.2) and (3.3) can be proved in the same way. Theorem 3.2. For any function u ∈ D(L), there exists a positive constant λ independent of the function u, such that kuk B ≤ λkLuk H .
(3.5) Rx
Proof. We first introduce the operators =x g = 0 g(ξ, y, t)dξ and Mg = gt + =x y g. And considering the inner product in L 2 (0, τ ; L 2ρ (Ω )) of the differential equation Lu = f (in (2.6)) and the operator Mu, we then have 1 2 − u t , (xu x )x ku t k L 2 (0,τ ;L 2 (Ω )) + (u t , =x y u) 2 L 0,τ ;L 2ρ (Ω ) ρ x L 2 (0,τ ;L 2ρ (Ω )) 1 1 1 − =x y u, (xu x )x − =x y u, (yu y ) y − u t , (yu y ) y x y y L 2 (0,τ ;L 2ρ (Ω )) L 2 (0,τ ;L 2ρ (Ω )) L 2 (0,τ ;L 2ρ (Ω )) = ( f, Mu) L 2 (0,τ ;L 2ρ (Ω )) .
(3.6)
We invoke (2.3) and (2.4), and in (3.6) we carry out appropriate integration by parts which are straightforward but somewhat tedious. We only give their results 1 1 1 (3.7) = ku x (x, y, τ )k2L 2 (Ω ) − kϕx (x, y)k2L 2 (Ω ) , − u t , (xu x )x ρ ρ x 2 2 L 2 (0,τ ;L 2ρ (Ω )) 1 − =x y u, (xu x )x = (u x , = y u) L 2 (0,τ ;L 2ρ (Ω )) , (3.8) x L 2 (0,τ ;L 2ρ (Ω )) 1 1 1 (3.9) = ku y (x, y, τ )k2L 2 (Ω ) − kϕ y (x, y)k2L 2 (Ω ) , − u t , (yu y ) y ρ ρ y 2 2 L 2 (0,τ ;L 2ρ (Ω )) 1 − =x y u, (yu y ) y = (u y , =x u) L 2 (0,τ ;L 2ρ (Ω )) . (3.10) y L 2 (0,τ ;L 2ρ (Ω )) Substitution of (3.7)–(3.10) into (3.6) yields 1 1 ku t k2L 2 (0,τ ;L 2 (Ω )) + ku x (x, y, τ )k2L 2 (Ω ) + ku y (x, y, τ )k2L 2 (Ω ) ρ ρ ρ 2 2 1 = −(u t , =x y u) L 2 (0,τ ;L 2ρ (Ω )) − (u x , = y u) L 2 (0,τ ;L 2ρ (Ω )) − (u y , =x u) L 2 (0,τ ;L 2ρ (Ω )) + kϕx (x, y)k2L 2 (Ω ) ρ 2 1 + kϕ y (x, y)k2L 2 (Ω ) + ( f, u t ) L 2 (0,τ ;L 2ρ (Ω )) + ( f, =x y u) L 2 (0,τ ;L 2ρ (Ω )) . ρ 2 Therefore, using Lemma 3.1 and Young’s inequality, we infer from (3.11),
(3.11)
ku t k2L 2 (0,τ ;L 2 (Ω )) + ku x (x, y, τ )k2L 2 (Ω ) + ku y (x, y, τ )k2L 2 (Ω ) ≤ kϕx (x, y)k2L 2 (Ω ) + kϕ y (x, y)k2L 2 (Ω ) ρ
+ 3k f k2L 2 (0,τ ;L 2 (Ω )) ρ
ρ
ρ
+ ku y k2L 2 (0,τ ;L 2 (Ω )) ρ
+ ku x k2L 2 (0,τ ;L 2 (Ω )) (a 2 ρ
ρ
2
ρ
2 2
+ b + 3a b
)kuk2L 2 (0,τ ;L 2 (Ω )) . ρ
(3.12)
From the differential equation Lu = f (in (2.6)), we have 1 1 1 kdiv(x y∇u)k2L 2 (0,τ ;L 2 (Ω )) ≤ k f k2L 2 (0,τ ;L 2 (Ω )) + ku t k2L 2 (0,τ ;L 2 (Ω )) . ρ ρ 8ab 4 4
(3.13)
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Consider also the obvious inequality, 1 1 1 1 ku(x, y, τ )k2L 2 (0,τ ;L 2 (Ω )) ≤ kϕk2L 2 (Ω ) + kuk2L 2 (0,τ ;L 2 (Ω )) + ku t k2L 2 (0,τ ;L 2 (Ω )) . ρ ρ ρ ρ 4 4 4 4 Combination of (3.12)–(3.14) gives
(3.14)
ku(x, y, τ )k2V 1 (Ω ) + ku t k2L 2 (0,τ ;L 2 (Ω )) + kdiv(x y∇u)k2L 2 (0,τ ;L 2 (Ω )) ρ ρ ρ 2 2 2 ≤ c1 k f k L 2 (Q τ ) + kϕkV 1 (Ω ) + c1 kukV 1 (Q τ ) , ρ
ρ
(3.15)
ρ
where max 26, 2 + 8a 2 + 8b2 + 24a 2 b2 . c1 = 1 min 4, ab Applying Lemma 7.1 in [22] to inequality (3.15), with g1 (t) = ku t k2L 2 (Ω ) + kdiv(x y∇u)k2L 2 (Ω ) , ρ
ρ
g2 (τ ) = ku(x, y, τ )k2V 1 (Ω ) , ρ 2 g3 (t) = c1 k f k L 2 (0,τ ;L 2 (Ω )) + kϕk2V 1 (Ω ) , ρ
ρ
it follows that ku(x, y, τ )k2V 1 (Ω ) + ku t k2L 2 (0,τ ;L 2 (Ω )) + kdiv(x y∇u)k2L 2 (0,τ ;L 2 (Ω )) ≤ c1 ec1 T k f k2L 2 (Q) + kϕk2V 1 (Ω ) . (3.16) ρ
ρ
ρ
ρ
ρ
Observe that the right-hand side of (3.16) is independent of τ , hence replacing the left-hand side by its upper bound √ with respect to τ from 0 to T , thus obtaining the a priori estimate (3.5), where λ = c1 ec1 T /2 . It can be proved in the standard way that the operator L : B → H is closable. Let L be the closure of the operator L. A solution of the equation Lu = F, F ∈ H , is called a strong solution of problem (2.6). We extend the energy estimate (3.5) to u ∈ D(L) by passing to the limit and thus establish uniqueness of a strong solution and closedness of the range R(L) of the operator L. More precisely, as consequences of the extended estimate kuk B ≤ λkLuk H , ∀u ∈ D(L), we have the following results: Corollary 3.3. If a strong solution of problem (2.6) exists, it is unique and depends continuously on F = ( f, ϕ) ∈ H . Corollary 3.4. The range R(L) of the operator L is closed and equal to R(L). 4. Existence of solution For the existence of a strong solution of the linear problem (2.6), we prove that the range R(L) of the operator L is dense in H . Thus we have the following Theorem 4.1. For each f ∈ L 2 (0, T ; L 2ρ (Ω )) and ϕ ∈ Vρ1 (Ω ), there exists a unique strong solution u = L −1 ( f, u 0 ) = L −1 ( f, u 0 ), of problem (2.6) satisfying the a priori bound kuk B ≤ λkLuk H , where λ is a positive constant independent of the solution u. Proof. From the extended estimate kuk B ≤ λkLuk H , it follows that the operator L has an inverse and, from Corollary 3.4, we deduce that the range R(L) is closed. Hence, it suffices to prove that R(L) = H . We first prove the following.
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Theorem 4.2. If for some function θ ∈ L 2 (0, T ; L 2ρ (Ω )) and all u ∈ D(L) with u|t=0 = 0 (Lu, θ ) L 2 (0,T ;L 2ρ (Ω )) = 0,
(4.1)
then θ = 0 almost everywhere in Q. Proof of Theorem 4.2. Define the function v by the relation Z T v(x, y, t) = θ (x, y, τ )dτ. t
Let u be a solution of the equation u t + =x y u = v(x, y, t).
(4.2)
And let u=
Z
T t
u τ dτ,
s≤t ≤T
0,
(4.3)
0 ≤ t ≤ s.
We now have θ (x, y, t) = −u tt − =x y u t .
(4.4)
Lemma 4.3. The function θ defined by the relation (4.4) is in L 2 (0, T ; L 2ρ (Ω )). Proof of Lemma 4.3. From (3.3) and the definition of D(L), we deduce that −=x y u t ∈ L 2 (0, T ; L 2ρ (Ω )). To prove that −u tt ∈ L 2 (0, T ; L 2ρ (Ω )), we use the t-averaging operators T
s−t ω g (x, s) ds, ε t R +∞ where ω ∈ C0∞ (0, T ), ω(t) ≥ 0, −∞ ω(t)dt = 1. Applying the operators ρε and ∂/∂t to Eq. (4.2), we obtain 1 (ρε g) (x, t) = ε
Z
∂ ∂ ∂ ρε u t + ρε =x y u = ρε v. ∂t ∂t ∂t
(4.5)
It follows from (4.5) that
2
∂
ρε u t
∂t
2 L
2
2
∂
2 2 ∂
≤ 2 ρ = u + 2a b ρ v . ε xy ε
∂t ∂t (0,T ;L 2ρ (Ω )) L 2 (0,T ;L 2ρ (Ω )) L 2 (0,T ;L 2ρ (Ω ))
Using properties of operators ρε introduced in [22], yields
2
∂
ρε u t ≤ K k=x y u t k2L 2 (0,T ;L 2 (Ω )) + K
∂t
2 ρ 2 L (0,T ;L ρ (Ω ))
2
∂
ρε v
∂t
2
L (0,T ;L 2ρ (Ω ))
(4.6)
,
where K = 2 max 1, a 2 b2 . Since ρε g → 0 in L 2 (0, T ; L 2ρ (Ω )), and the norm in L 2 (0, T ; L 2ρ (Ω )) of ε→0
θ ∈ L 2 (0, T ; L 2ρ (Ω )). This achieves the proof of Lemma 4.3.
∂ ∂t ρε u t
is bounded, we conclude that
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On going back to proof of Theorem 4.2, we replace the function θ given by (4.4) in (4.1), and we have −(u t , u tt ) L 2 (0,T ;L 2ρ (Ω )) − (u t , =x y u t ) L 2 (0,T ;L 2ρ (Ω )) 1 1 + =x y u t , (xu x )x + u tt , (xu x )x x x L 2 (0,T ;L 2ρ (Ω )) L 2 (0,T ;L 2ρ (Ω )) 1 1 + u tt , (yu y ) y + =x y u t , (yu y ) y = 0. y y 2 2 L (0,T ;L ρ (Ω )) L 2 (0,T ;L 2ρ (Ω ))
(4.7)
Direct evaluation of the first and last four terms on the left-hand side of (4.7), yields 1 −(u t , u tt ) L 2 (0,T ;L 2ρ (Ω )) = ku t (x, y, s)k2L 2 (Ω ) , ρ 2 1 u tt , (xu x )x = ku t x k2L 2 (s,T ;L 2 (Ω )) , ρ x L 2 (0,T ;L 2ρ (Ω )) 1 =x y u t , (xu x )x = −(u x , = y u t ) L 2 (s,T ;L 2ρ (Ω )) , x L 2 (0,T ;L 2ρ (Ω )) 1 = ku t y k2L 2 (s,T ;L 2 (Ω )) , u tt , (yu y ) y ρ y L 2 (0,T ;L 2ρ (Ω )) 1 =x y u t , (yu y ) y = −(u y , =x u t ) L 2 (s,T ;L 2ρ (Ω )) . y L 2 (0,T ;L 2ρ (Ω ))
(4.8) (4.9) (4.10) (4.11) (4.12)
Substitution of (4.8)–(4.12) into (4.7) gives 1 ku t (x, y, s)k2L 2 (Ω ) + ku t x k2L 2 (s,T ;L 2 (Ω )) + ku t y k2L 2 (s,T ;L 2 (Ω )) ρ ρ ρ 2 = (u t , =x y u t ) L 2 (s,T ;L 2ρ (Ω )) + (u x , = y u t ) L 2 (s,T ;L 2ρ (Ω )) + (u y , =x u t ) L 2 (s,T ;L 2ρ (Ω )) , which in turn implies after using Young’s inequality and Lemma 3.1, 1 ku t (x, y, s)k2L 2 (Ω ) + ku t x k2L 2 (s,T ;L 2 (Ω )) + ku t y k2L 2 (s,T ;L 2 (Ω )) ρ ρ ρ 2 2 1 1 + a + b2 + a 2 b2 1 ≤ ku x k2L 2 (s,T ;L 2 (Ω )) + ku y k2L 2 (s,T ;L 2 (Ω )) + ku t k2L 2 (s,T ;L 2 (Ω )) . ρ ρ ρ 2 2 2
(4.13)
It is quite easy to see that ku x (x, y, T )k2L 2 (Ω ) ≤ ku x k2L 2 (s,T ;L 2 (Ω )) + ku t x k2L 2 (s,T ;L 2 (Ω )) ,
(4.14)
ku y (x, y, T )k2L 2 (Ω ) ≤ ku y k2L 2 (s,T ;L 2 (Ω )) + ku t y k2L 2 (s,T ;L 2 (Ω )) .
(4.15)
ρ
ρ ρ
ρ
ρ
ρ
Adding side to side (4.13)–(4.15) to get ku t (x, y, s)k2L 2 (Ω ) + ku x (x, y, T )k2L 2 (Ω ) + ku y (x, y, T )k2L 2 (Ω ) ρ ρ ρ 2 2 2 ≤ γ ku t k L 2 (s,T ;L 2 (Ω )) + ku x k L 2 (s,T ;L 2 (Ω )) + ku y k L 2 (s,T ;L 2 (Ω )) , ρ
where γ = max 3, 1 + a 2 + b2 + a 2 b2 .
ρ
ρ
(4.16)
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Inequality (4.16) is essential in our proof. To use it, we introduce the new function g defined by the formula Z T u τ dτ. g(x, y, t) = t
Then u(x, y, t) = g(x, y, s) − g(x, y, t), u(x, y, T ) = g(x, y, s). Using (4.16) and the above relations, we obtain (1 − 2γ (T − s)) kgx (x, y, s)k2L 2 (Ω ) + kg y (x, y, s)k2L 2 (Ω ) + ku t (x, y, s)k2L 2 (Ω ) ρ ρ ρ ≤ 2γ ku t k2L 2 (s,T ;L 2 (Ω )) + kgx k2L 2 (s,T ;L 2 (Ω )) + kg y k2L 2 (s,T ;L 2 (Ω )) . ρ
ρ
ρ
(4.17)
Hence if s0 > 0 satisfies 1 − 2γ (T − s0 ) = 1/2, then (4.17) implies ku t (x, y, s)k2L 2 (Ω ) + kgx (x, y, s)k2L 2 (Ω ) + kg y (x, y, s)k2L 2 (Ω ) ρ ρ ρ 2 2 2 ≤ 4γ ku t k L 2 (s,T ;L 2 (Ω )) + kgx k L 2 (s,T ;L 2 (Ω )) + kg y k L 2 (s,T ;L 2 (Ω )) ρ
ρ
ρ
(4.18)
for all s ∈ [T − s0 , T ]. The sum of the three norms in the right-hand side of (4.18) is a function of s. We denote it by Z T F(s) = ku t k2L 2 (Ω ) + kgx k2L 2 (Ω ) + kg y k2L 2 (Ω ) dt. s
ρ
ρ
ρ
Then (4.18) takes the form −F 0 (s) ≤ 4γ F (s) .
(4.19)
Integrating (4.19) with respect to t over [s, T ] and taking into account that F(T ) = 0, we obtain F (s) exp(4γ s) ≤ 0. It follows that θ = 0 almost everywhere in Q T −s0 . Proceeding in this way step by step, we prove that θ = 0 almost everywhere in Q. Therefore the proof of Theorem 4.2 is completed. We now complete the proof of Theorem 4.1. Since H is a Hilbert space, density of the set R(L) is equivalent to the property that orthogonality of a vector F = ( f, ϕ) ∈ H to the set R(L), i.e, the equality (Lu, θ ) L 2 (0,T ;L 2ρ (Ω )) + (`u, ϕ)Vρ1 (Ω ) = 0,
∀u ∈ D(L),
(4.20)
implies F = 0. In particular, in (4.20) we set `u = 0. We then conclude by Theorem 4.2 that θ = 0. Thus (4.20) implies that (`u, ϕ)Vρ1 (Ω ) = 0. Now since the range of the trace operator ` is dense in Vρ1 (Ω ), it follows that ϕ = 0. Hence, F = 0. 5. The nonlinear problem We are now in a position to solve the nonlinear problem (2.1)–(2.4). Relying on the results obtained previously, we apply an iterative process to establish the existence and uniqueness of the weak solution of the nonlinear problem. If u is a solution of problem (2.1)–(2.4) and Φ is a solution of the problem 1 div(x y∇Φ) = 0, xy `Φ = Φ(x, y, 0) = ϕ(x, y), Φx (a, y, t) = 0, Φ y (x, b, t) = 0, LΦ = Φt −
(5.1) (5.2) (5.3)
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Z
Z
b
Φ(x, y, t)dy = 0,
Φ(x, y, t)dx = 0,
(5.4)
0
0
then ξ = u − Φ satisfies 1 div(x y∇ξ ) = F(x, y, t, ξ, ξx , ξ y ), xy ξ(x, y, 0) = 0, Lξ = ξt −
(5.5) (5.6)
ξx (a, y, t) = 0, ξ y (x, b, t) = 0, Z b Z a ξ(x, y, t)dy = 0, ξ(x, y, t)dx = 0,
(5.7) (5.8)
0
0
where F(x, y, t, ξ, ξx , ξ y ) = f (x, y, t, ξ + Φ, ξx + Φx , ξ y + Φ y ), and satisfies the condition |F(x, y, t, u 1 , v1 , w1 ) − F(x, y, t, u 2 , v2 , w2 )| ≤ δ (|u 1 − u 2 | + |v1 − v2 | + |w1 − w2 |) .
(5.9)
Theorem 4.1 shows that problem (5.1)–(5.4), admits a unique solution that depends continuously on the data ϕ ∈ Vρ1 (Ω ). In order to solve problem (5.5)–(5.8), we will show that it has a unique weak solution. The general idea is to construct an iteration sequence (ξ (n) )n∈ N´ which converges to an element ξ ∈ L 2 (0, T ; Vρ1 (Ω )) that will be a solution of the problem under consideration (5.5)–(5.8). √ Theorem 5.1. Suppose that condition (5.9) holds and δ < e−c1 T /2 / 3T , then problem (5.5)–(5.8) admits a weak solution belonging to L 2 (0, T ; Vρ1 (Ω )). Ra Rb Proof. Let us first assume that ξ and υ ∈ C 1 (Q), such that ξ(x, y, 0) = 0, υ(x, y, T ) = 0, 0 ξ dx = 0, 0 ξ dy = Ra Rb 0, 0 υdx = 0, 0 υdy = 0. For υ ∈ C 1 (Q), we have 1 1 (xξx )x , =x y υ − (yξ y ) y , =x y υ ξt , =x y υ L 2 (0,T ;L 2 (Ω )) − ρ x y L 2 (0,T ;L 2ρ (Ω )) L 2 (0,T ;L 2ρ (Ω )) = Lξ, =x y υ L 2 (0,T ;L 2 (Ω )) = F, =x y υ L 2 (0,T ;L 2 (Ω )) . (5.10) ρ
ρ
In light of the above assumptions, we obtain ξt , =x y υ L 2 (0,T ;L 2 (Ω )) = − υt , =x y (ζ ηξ ) L 2 (Q) , ρ 1 − (xξx )x , =x y υ = − xυ, = y (ηξx ) L 2 (Q) , x L 2 (0,T ;L 2ρ (Ω )) 1 − (yξ y ) y , =x y υ = − yυ, =x (ζ ξ y ) L 2 (Q) , y L 2 (0,T ;L 2ρ (Ω )) F, =x y υ L 2 (0,T ;L 2 (Ω )) = υ, =x y (ζ ηF) L 2 (Q) .
(5.14)
Substituting (5.11)–(5.14) into (5.10) yields A(ξ, υ) = υ, =x y (ζ ηF) L 2 (Q) ,
(5.15)
ρ
(5.11) (5.12) (5.13)
where A(ξ, υ) = − υt , =x y (ζ ηξ )
L 2 (Q)
− xυ, = y (ηξx ) L 2 (Q) − yυ, =x (ζ ξ y ) L 2 (Q) .
It should be mentioned here that L 2 (0, T ; L 2 (Ω )) = L 2 (Q). We define the weak solution of problem (5.5)–(5.8) as a function ξ ∈ L 2 (0, T ; Vρ1 (Ω )) satisfying:
(5.16)
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S. Mesloub / Nonlinear Analysis 68 (2008) 2594–2607
(1) A(ξ, υ) = υ, =x y (ζ ηF) L 2 (Q) . (2) ξx (a, y, t) = 0, ξ y (x, b, t) = 0. As mentioned previously, we define the iteration sequence (ξ (n) )n∈ N´ as follows: Let ξ (0) = 0. Given the element then for n = 1, 2, . . . . solve the iterated problem:
ξ (n−1) ,
(n)
ξt
−
1 div(x y∇ξ (n) ) = F(x, y, t, ξ (n−1) , ξx(n−1) , ξ y(n−1) ), xy
(5.17)
ξ (n) (x, y, 0) = 0, ξx(n) (a, a
Z
y, t) = 0,
(5.18) ξ y(n) (x, b, t)
ξ (n) (x, y, t)dx = 0,
0
b
Z
= 0,
(5.19)
ξ (n) (x, y, t)dy = 0.
(5.20)
0
For fixed n, Theorem 4.1 shows that each problem (5.17)–(5.20) has a unique solution ξ (n) (x, y, t). Now, we set Y (n) (x, y, t) = ξ (n+1) (x, y, t) − ξ (n) (x, y, t), then we obtain the following new problem (n)
Yt
−
1 div(x y∇Y (n) ) = β (n−1) (x, y, t), xy
(5.21)
Y (n) (x, y, 0) = 0, Yx(n) (a, a
Z
y, t) = 0,
(5.22) Y y(n) (x, b, t)
Y (n) (x, y, t)dx = 0,
b
Z
= 0,
(5.23)
Y (n) (x, y, t)dy = 0,
(5.24)
0
0
where β (n−1) (x, y, t) = F(x, y, t, ξ (n) , ξx(n) , ξ y(n) ) − F(x, y, t, ξ (n−1) , ξx(n−1) , ξ y(n−1) ).
Lemma 5.2. Suppose that condition (5.9) holds, then there exists a positive constant C = problem (5.21)–(5.24), we have the a priori bound
√
c2 T ec1 T /2 such that for
kY (n) k L 2 (0,T´ ;V 1 (Ω )) ≤ CkY (n−1) k L 2 (0,T´ ;V 1 (Ω )) ,
(5.25)
ρ
ρ
where c1 = a 2 b2 + a 2 /2 + b2 /2 + 1/2,
c2 = 3δ 2 .
Proof. Taking the inner product in L 2 (0, ν; L 2ρ (Ω )), with 0 ≤ ν ≤ T , of Eq. (5.21) and the integro-differential operator (n)
MY (n) = Yt
+ =x y Y (n) ,
we obtain (n) 1 (n) xY − Y , t x ρ L 2 (0,ν;L 2ρ (Ω )) x L 2 (0,ν;L 2 (Ω )) x ρ 1 (n) 1 xYx(n) − Yt , yY y(n) − =x y Y (n) , x L 2 (0,ν;L 2 (Ω )) y L 2 (0,ν;L 2 (Ω )) x y ρ ρ 1 − =x y Y (n) , yY y(n) y L 2 (0,ν;L 2 (Ω )) y ρ (n) (n−1) (n) = β (n−1) , Yt + β , = Y . x y 2 2 2 2
(n) (n) kYt k2L 2 (0,ν;L 2 (Ω )) + Yt , =x y Y (n)
L (0,ν;L ρ (Ω ))
L (0,ν;L ρ (Ω ))
(5.26)
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S. Mesloub / Nonlinear Analysis 68 (2008) 2594–2607
It is straightforward to see that 1 1 (n) kYt k2L 2 (0,ν;L 2 (Ω )) + kYx(n) (x, y, ν)k2L 2 (Ω ) + kY y(n) (x, y, ν)k2L 2 (Ω ) ρ ρ ρ 2 2 (n) (n−1) (n−1) (n) = β , Yt + β , =x y Y 2 2 2 2 L (0,ν;L ρ (Ω ))
(n)
− Yt , =x y Y (n)
L (0,ν;L ρ (Ω ))
L 2 (0,ν;L 2ρ (Ω ))
− Yx(n) , = y Y (n)
L 2 (0,ν;L 2ρ (Ω ))
− Y y(n) , =x Y (n)
L 2 (0,ν;L 2ρ (Ω ))
.
(5.27)
Estimating terms on the RHS of (5.27), we obtain (n) β (n−1) , Yt
L 2 (0,ν;L 2ρ (Ω ))
1 (n) 2 kY k L 2 (0,ν;L 2 (Ω )) + kβ (n−1) k2L 2 (0,ν;L 2 (Ω )) ρ ρ 4 t 1 (n) 2 3 2 (n−1) 2 ≤ kYt k L 2 (0,ν;L 2 (Ω )) + δ kY k L 2 (0,ν;L 2 (Ω )) ρ ρ 4 2 + kYx(n−1) k2L 2 (0,ν;L 2 (Ω )) + kY y(n−1) k2L 2 (0,ν;L 2 (Ω )) ,
≤
ρ
ρ
β (n−1) , =x y Y (n)
≤
L 2 (0,ν;L 2ρ (Ω ))
a 2 b2 (n) 2 kY k L 2 (0,ν;L 2 (Ω )) ρ 2 3 2 (n−1) 2 + δ kY k L 2 (0,ν;L 2 (Ω )) + kYx(n−1) k2L 2 (0,ν;L 2 (Ω )) ρ 2 ρ (n−1) 2 + kY y k L 2 (0,ν;L 2 (Ω )) , ρ
(n)
− Yt , =x y Y (n)
− Yx(n) , = y Y (n) − Y y(n) , =x Y (n)
L 2 (0,ν;L 2ρ (Ω ))
L 2 (0,ν;L 2ρ (Ω ))
L 2 (0,ν;L 2ρ (Ω ))
≤
1 (n) 2 kY k L 2 (0,ν;L 2 (Ω )) + a 2 b2 kY (n) k2L 2 (0,ν;L 2 (Ω )) , ρ ρ 4 t
(5.28)
(5.29) (5.30)
≤
b2 1 (n) 2 kYx k L 2 (0,ν;L 2 (Ω )) + kY (n) k2L 2 (0,ν;L 2 (Ω )) , ρ ρ 2 2
(5.31)
≤
1 (n) 2 a2 kY y k L 2 (0,ν;L 2 (Ω )) + kY (n) k2L 2 (0,ν;L 2 (Ω )) . ρ ρ 2 2
(5.32)
It is obvious that 1 (n) 1 1 (n) kY (x, y, ν)k2L 2 (Ω ) ≤ kY (n) k2L 2 (0,ν;L 2 (Ω )) + kYt k2L 2 (0,ν;L 2 (Ω )) . ρ ρ ρ 2 2 2
(5.33)
Combination of inequalities (5.28)–(5.33) then yields kY (n) (x, y, ν)k2V 1 (Ω ) ≤ c1 kY (n) k2L 2 (0,ν;V 1 (Ω )) + c2 kY (n−1) k2L 2 (0,ν;V 1 (Ω )) , ρ
ρ
ρ
(5.34)
where c1 = (2a 2 b2 + a 2 + b2 + 1)/2,
c2 = 3δ 2 .
An application of Lemma 7.1 in [22] to (5.34), gives kY (n) (x, y, ν)k2V 1 (Ω ) ≤ c2 ec1 T kY (n−1) k2L 2 (0,ν;V 1 (Ω )) . ρ
ρ
(5.35)
Integrating both sides of (5.35) with respect to ν over the interval [0, T ], we obtain kY (n) k2L 2 (0,T ;V 1 (Ω )) ≤ c2 T ec1 T kY (n−1) k2L 2 (0,T ;V 1 (Ω )) . ρ
ρ
(5.36)
√ P (n) converges if δ < e−c1 T /2 / 3T . Knowing that Y (n) (x, y, t) = It follows from (5.36) that the series ∞ n=1 Y ξ (n+1) (x, y, t) − ξ (n) (x, y, t), then it follows that the sequence defined by
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S. Mesloub / Nonlinear Analysis 68 (2008) 2594–2607
ξ (n) (x, y, t) =
n−1 X
Y (m) (x, y, t) + ξ (0) (x, y, t)
m=0
=
n−1 X
(ξ (m+1) (x, y, t) − ξ (m) (x, y, t)) + ξ (0) (x, y, t),
n = 1, 2, . . .
m=0
converges to an element ξ ∈ L 2 (0, T ; Vρ1 (Ω )) which must be proved that it is a solution of problem (5.5)–(5.8). In other words, ξ must satisfy: (1) A(ξ, υ) = υ, =x y (ζ ηF) L 2 (Q) , (2) ξx (a, y, t) = 0, ξ y (x, b, t) = 0 as mentioned previously. From equalities (5.17)–(5.20), we have (n−1) (n−1) A(ξ (n) , υ) = υ, =x y (ζ ηF(ξ, η, t, ξ (n−1) , ξζ , ξη )) 2 , L (Q)
from which it follows that (n−1) (n−1) A(ξ, υ) + A(ξ (n) − ξ, υ) = υ, =x y (ζ ηF(ξ, η, t, ξ (n−1) , ξζ , ξη )) − =x y (ζ ηF(ξ, η, t, ξ, ξζ , ξη )) L 2 (Q) + (υ, =x y (ζ ηF(ξ, η, t, ξ, ξζ , ξη )))
L 2 (Q)
.
Now, from the partial differential equation (5.17), we have ∂ ∂ ∂ (n) (n) (n) − υ, =x y η A(ξ − ξ, υ) = υ, =x y (ζ η(ξ − ξ )) ζ (ξ − ξ ) ∂t ∂ζ ∂ζ L 2 (Q) L 2 (Q) ∂ ∂ (n) − υ, =x y ζ η (ξ − ξ ) ∂η ∂η L 2 (Q) (n−1) − ξζ , ξη(n−1) − ξη )) 2 . = υ, =x y (ζ ηF(ζ, η, t, ξ (n−1) − ξ, ξζ L (Q)
(5.37)
(5.38)
Performing integration by parts in each term on the RHS of (5.38), and using conditions on υ and ξ to obtain ∂ ∂ (n) υ, =x y (ζ η(ξ (n) − ξ )) + = y υ, (ξ − ξ ) − ∂t ∂x L 2 (Q) L 2ρ (Q) ∂ + =x υ, (ξ (n) − ξ ) = A(ξ (n) − ξ, υ). (5.39) ∂y L 2ρ (Q) Application of Cauchy–Schwarz inequality to terms on the LHS of (5.39) yields
∂υ a 3/2 b32 (n)
∂υ kξ (n) − ξ k L 2ρ (Q) , − , =x y (ζ η(ξ − ξ )) ≤ √
∂t ∂t L 2 (Q) 2 L 2 (Q)
∂ (n)
∂ (n) 1/2 3/2
= y υ, (ξ − ξ ) ≤ a b kυk L 2 (Q) (ξ − ξ ) ,
2 ∂x ∂x L 2ρ (Q) L ρ (Q)
∂ (n)
∂
=x υ, (ξ (n) − ξ ) ≤ a 3/2 b1/2 kυk L 2 (Q) (ξ − ξ ) .
2 ∂y ∂y L 2ρ (Q) L ρ (Q)
(5.40) (5.41) (5.42)
Substitution of (5.40)–(5.42) into (5.39) then gives A(ξ (n) − ξ, υ) ≤ Bkξ (n) − ξ k L 2 (0,T ;Vρ1 (Ω )) kυk L 2 (Q) + kυt k L 2 (Q) , where 3/2 b32 1/2 3/2 3/2 1/2 a . B = max a b , a b , √ 2
(5.43)
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On the other side, we have (n−1) (n−1) υ, =x y (ζ ηF(ξ, η, t, ξ (n−1) , ξζ , ξη )) − =x y (ζ ηF(ξ, η, t, ξ, ξζ , ξη )) ≤
a 3/2 b3/2 √
2
L 2 (Q)
δkυk L 2 (Q) kξ (n) − ξ k L 2 (0,T ;Vρ1 (Ω )) .
(5.44)
Invoking (5.43) and (5.44), and taking the limit as n → ∞ in equality (5.37), we obtain A(ξ, υ) = υ, =x y (ζ ηF(ξ, η, t, ξ, ξζ , ξη )) L 2 (Q) . To conclude that problem (5.5)–(5.8) admits a weak solution, we must R t show that conditions (5.7) hold. Rt Since ξ ∈ L 2 (0, T ; Vρ1 (Ω )), then 0 ∂∂ξx (x, y, s)ds ∈ C(Q) and 0 ∂ξ ∂ y (x, y, s)ds ∈ C(Q), and we conclude that ξx (a, y, t) = 0, ξ y (x, b, t) = 0, a.e. It remains now to prove the uniqueness of solution of problem (5.5)–(5.8). Theorem 5.3. If condition (5.9) is satisfied, then the solution of problem (5.5)–(5.8) is unique. Proof. Let ξ1 , ξ2 ∈ L 2 (0, T ; Vρ1 (Ω )) be two solutions of (5.5)–(5.8), then Y = ξ1 − ξ2 ∈ L 2 (0, T ; Vρ1 (Ω )) and satisfies 1 div(x y∇Y ) = β(x, y, t), xy Y (x, y, 0) = 0, Yx (a, y, t) = 0, Y y (x, b, t) = 0, Z b Z a Y (x, y, t)dx = 0, Y (x, y, t)dy = 0,
(5.45)
Yt −
0
(5.46) (5.47) (5.48)
0
where ∂ξ1 ∂ξ1 β(x, y, t) = F x, y, t, ξ, , ∂x ∂y
∂ξ2 ∂ξ2 , − F x, y, t, ξ, ∂x ∂y
.
By following the same procedure used in proving Lemma 5.2, we take the inner product in L 2 (0, T ; Vρ1 (Ω )) of Eq. (5.45) and differential operator DY = Yt + =x y Y , to obtain kY k L 2 (0,T´ ;V 1 (Ω )) ≤ CkY k L 2 (0,T´ ;V 1 (Ω )) , ρ
ρ
(5.49)
where C is the same constant as in Lemma 5.2. Since C < 1, we deduce from (5.49) that (1 − C)kY k L 2 (0,T´ ;V 1 (Ω )) = 0, ρ
which in turns implies that ξ1 − ξ2 = 0, and consequently ξ1 = ξ2 ∈ L 2 (0, T ; Vρ1 (Ω )). The proof of Theorem 5.1 is achieved. Acknowledgment This work has been funded and supported by the Research Center Project No. Math/2006/45 at King Saud University. References [1] A. Bouziani, Solvability of nonlinear pseudoparabolic equation with a nonlocal boundary condition, Nonlinear Anal. 55 (7–8) (2003) 883–904. [2] J.R. Cannon, J. van der Hoek, Diffusion subject to specification of mass, J. Math. Anal. Appl. 115 (1986) 517–529. [3] J.R. Cannon, The one-dimensional heat equation, in: K. Rach (Ed.), Encyclopedia of Mathematics and its Applications, vol. 23, AddisonWesley, Reading, MA, 1984.
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