Electronic Journal of Qualitative Theory of Differential Equations 2005, No. 18, 1-12, http://www.math.u-szeged.hu/ejqtde/

EXISTENCE OF SOLUTIONS TO NONLOCAL AND SINGULAR

VARIATIONAL ELLIPTIC INEQUALITY VIA GALERKIN METHOD

FRANCISCO JULIO S. A. CORRÊA & SILVANO B. DE MENEZES

Abstract. In this article, we study the existence of solutions for nonlocal variational elliptic inequality

−M(∥u∥^{2})Δu ≥ f(x, u) Making use of the penalized method and Galerkin approximations, we establish existence theorems for both cases when M is continuous and when M is discontinuous.

1. Introduction

Let Ω ⊂ ℝ^{ℕ} be a bounded smooth domain with boundary ∂Ω. Let us consider the Sobolev spaces L^{2}(Ω), H_{0}^{1}(Ω) whose inner products and norms will be denoted by (,), |, |, ((,)), ∥, ∥, respectively. We have H_{0}^{1}(Ω) ⊂ L^{2}(Ω) dense and continuous. Then the duals (L^{2}(Ω))′ ⊂ (H_{0}^{1}(Ω))′ also dense and continuous. If we identify L^{2}(Ω) = (L^{2}(Ω))′ we have

H_{0}^{1}(Ω) ⊂ L^{2}(Ω) ⊂ H^{−1}(Ω),

where H^{−1}(Ω) is the dual of H_{0}^{1}(Ω). Throughout this paper, let us represent by K a closed convex set of L^{2}(Ω), with 0 ∈ K, which has the following property:

(H1) There exists a contraction ρ : ℝ ⟶ ℝ (e.g., | ρ(λ_{1}) − ρ(λ_{2}) |≤ | λ_{1} − λ_{2} |) with ρ(0) = 0 such that (P_{K}v)(x) = ρ(v(x)),∀ v ∈ L^{2}(Ω), where P_{K} is the projection operator from L^{2}(Ω) into K.

For example, let us consider K = {v ∈ L^{2}(Ω); a ≤ v(x) ≤ b a.e. in Ω} where −∞ ≤ a ≤ 0 ≤ b ≤ ∞. We define ρ(λ) as follows

ρ(λ) = \{ \begin{matrix}λ & for & a < λ < b \\ b & for & λ ≥ b & (if & b & is & finite) \\ a & for & λ ≤ a & (if & a & is & finite) \end{matrix}

In this paper we study some questions related to the existence of solutions for the nonlocal elliptic variational inequality:

u ∈ K : (−M(∥u∥^{2})Δu, v) ≥ (f, v) , for all v ∈ K ∩ (H_{0}^{1}(Ω) ∩ H^{2}(Ω)) (1.1)

where f ∈ H_{0}^{1}(Ω) and M : ℝ → ℝ is a function whose behavior will be stated later. The main purpose of this work is establishing properties on M under which

EJQTDE, 2005, No. 18, p. 1 problem (1.1), and its nonlinear counterpart, possesses a solution. This inequality has called our attention because the operator

Lu := M(∥u∥^{2})Δu

contains the nonlocal term M(∥u∥^{2}) which poses some interesting mathematical questions. Also the operator L appears in the Kirchhoff equation, which arises in nonlinear vibrations, namely

u_{tt} − M( ∫_{Ω} |∇u|^{2}dx)Δu = f(x, u) in Ω × (0, T),
u = 0 on ∂Ω × (0, T),
u(x, 0) = u_{0}(x), u_{t}(x, 0) = u_{1}(x).

The mathematical aspects of this model were largely investigated. See, for example, Arosio-Spagnolo [2], Hazoya-Yamada [11], Lions [16], Pohozhaev [18]. A survey of the results about the mathematical aspects of Kirchhoff model can be found in Medeiros-Limaco-Menezes [14]. Unilateral problems for nonlinear operators of the Kirchhoff type were initially studied by Kludnev [12], Larkin [13], Medeiros-Milla Miranda [17] among others. Recently, Alves-Corrêa [1] focused their attention on problem related with (1.1) in case M(t) ≥ m_{0} > 0, for all t ≥ 0, where m_{0} is a constant. Among other things they studied the above M-linear problem (1.1) where M, besides the strict positivity mentioned before, satisfies the following assumption:

The function H : ℝ → ℝ with

H(t) = M(t^{2})t

is monotone and H(ℝ) = ℝ. In a previous paper, Menezes-Corrêa [8] proved a similar results by allowing M to

attain negative values and M(t) ≥ m_{0} > 0 only for t large enough. This is possible thanks to a device explored by Alves-de Figueiredo [3], who use Galerkin method to attack a non-variational elliptic system. The technique can be conveniently adapted to problems such as (1.1). In this way we improve substantially the existence result on the above problem mainly because our assumptions on M are weakened. Indeed, we may also consider the case in which M possesses a singularity. The methodology used in our proof consists in transforming, by penalty, the inequality (1.1) into a family of equations depending of a parameter ϵ > 0 and apply Galerkin’s method. In the application of Galerkin’s methods we use the sharp angle lemma(see Lions [15, p.53]). This paper is organized as follows: Section 2 is devoted to the study of (1.1) in the continuous case. In Section 3 the inequality (1.1) is studied in case M possesses a discontinuity. In Section 4 we analyze another type of variational inequality.

2. The M-linear Problem: Continuous Case

In this section we are concerned with the M-linear problem (1.1) where f ∈ H_{0}^{1}(Ω) and M : ℝ → ℝ is a continuous function satisfying

EJQTDE, 2005, No. 18, p. 2 (M1) There exists a positive number m_{0} such that M(t) ≥ m_{0}, for all t ≥ 0.

We have

Theorem 2.1. Assume that and (M1) and (H1) hold. Then for any choice of 0 ≠ f ∈ H_{0}^{1}(Ω) the problem (1.1) admits at least one solution.

The proof of Theorem 2.1 is given by the penalty method. In fact, let us represent by β the operator from L^{2}(Ω) into L^{2}(Ω) defined by β = I − P_{K}, e.g., (βv)(x) = v(x)−ρ(v(x)) for all v ∈ L^{2}(Ω). The operator β is monotone and Lipschitzian. The next result can be found in Haraux [10] p. 58.

Lemma 2.2. Let g : ℝ → ℝ be a Lipschitzian and increasing function with g(0) = 0. Then (g(u), −Δu) ≥ 0 for all u ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω).

We have that g(s) = s − ρ(s) is under the condition of the Lemma 2.2, then

(β(v), −Δv) ≥ 0 for all v ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω).  (2.1)

The penalized problem associated to the problem (1.1), consists in given ϵ > 0, find u_{ϵ} solution in Ω of the problem

\begin{matrix}−M(∥u_{ϵ}∥^{2})Δu_{ϵ} + \frac{1}{ϵ}β(u_{ϵ}) = f in Ω, \\ u_{ϵ} = 0 on ∂Ω, \end{matrix}  (2.2)

where f is given in H_{0}^{1}(Ω). The existence of one solution of the penalized problem (2.2) is give by the

Theorem 2.3. Assume that (M1) hold. Then, for any 0 ≠ f ∈ H_{0}^{1}(Ω) there exists at least one u_{ϵ} ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω), solution of the problem (2.2) .

Proof. We employ the Galerkin Method by using the sharp angle lemma. Let us consider the Hilbertian basis of spectral objects (e_{j})_{j∈ℕ} and (λ_{j})_{j∈ℕ} for the operator −Δ in H_{0}^{1}(Ω), cf. Brezis [5]. We know that the eigenvectors (e_{j})_{j∈ℕ} are orthonormal complete in L^{2}(Ω) and complete in H_{0}^{1}(Ω) ∩ H^{2}(Ω). For each m ∈ ℕ consider

𝕍_{m} = span{e_{1}, …, e_{m}},

the subspace of H_{0}^{1}(Ω) ∩ H^{2}(Ω) generated by m eigenvectors e_{1}, e_{2}, ..., e_{m}. If u_{ϵ_{m}} ∈ V_{m}, then

u_{ϵ_{m}} = ∑_{j=1}^{m} ξ_{j}e_{j} with real ξ_{j}, 1 ≤ j ≤ m.

We will consider u_{m} instead of u_{ϵ_{m}} . The approximate problem consists in finding a solution u_{m} ∈ V_{m} of the system of algebraic equations

M(∥u_{m}∥^{2})((u_{m}, e_{i})) + \frac{1}{ϵ}(β(u_{ϵ}), e_{i}) = (f, e_{i}), i = 1, …, m.  (2.3)

We need to prove that (2.3) has a solution u_{m} ∈ V_{m}. To this end, we will consider the vector η = (η_{i})_{1≤i≤m} of ℝ^{m} defined by

η_{i} = M(∥u_{m}∥^{2})((u_{m}, e_{i})) + \frac{1}{ϵ}(β(u_{m}), e_{i}) − (f, e_{i}), i = 1, …, m.

Let ξ = (ξ_{i})_{1≤i≤m} be the components of the vector u_{m} of V_{m}. The mapping P : ℝ^{m} → ℝ^{m} defined by P ξ = η is continuous. If we prove that ⟨P ξ, ξ⟩ ≥ 0 for ∥ξ∥_{ℝ^{n}} = r, with an appropriate r, it will follow by the sharp angle lemma that there exists a ξ in the ball B_{r}(0) ⊂ ℝ^{m} such that P ξ = 0. This implies the existence of a solution to (2.3). In fact, we have

⟨P ξ, ξ⟩ = ∑_{1}^{m} ξ_{i}η_{i} = M(∥u_{m}∥)((u_{m}, u_{m})) + \frac{1}{ϵ}(β(u_{m}), u_{m}) − (f, u_{m}).

Using (M1), Hölder and Poincaré inequalities and observing that (β(u_{m}), u_{m}) ≥ 0, we get

⟨P ξ, ξ⟩ ≥ m_{0}∥u_{m}∥^{2} − C∥f∥∥u_{m}∥

We can consider r large enough that (P ξ, ξ) ≥ 0 for ∥ξ∥_{ℝ^{m}} = r. Then P ξ = 0 for some ξ ∈ B_{r}(0), which implies that system (2.3) has a solution u_{m} ∈ V_{m} corresponding to this ξ. Thus, there is u_{m} ∈ 𝕍_{m},

∥u_{m}∥ ≤ r,  (2.4)

r does not depend on m and ϵ, such that

M(∥u_{m}∥^{2})((u_{m}, e_{i})) + \frac{1}{ϵ}(β(u_{m}), e_{i}) = (f, e_{i}), i = 1, …, m,

which implies that

M(∥u_{m}∥^{2})((u_{m}, ω)) + \frac{1}{ϵ}(β(u_{m}), ω) = (f, ω), for all ω ∈ 𝕍_{m}.  (2.5)

Because (∥u_{m}∥^{2}) is a bounded real sequence and M is continuous one has

∥u_{m}∥^{2} → t̃_{0},  (2.6)

for some t̃_{0} ≥ 0, and

u_{m} ⇀ u in H_{0}^{1}(Ω), u_{m} → u in L^{2}(Ω), M(∥u_{m}∥^{2}) → M(t̃_{0}),  (2.7)

perhaps for a subsequence.

Take k ≤ m, 𝕍_{k} ⊂ 𝕍_{m}. Fix k and let m → ∞ in equation (2.5) to obtain

M(t̃_{0})((u, ω)) + \frac{1}{ϵ}(β(u), ω) = (f, ω), for all ω ∈ (H_{0}^{1} ∩ H^{2})(Ω).  (2.8)

Since −Δe_{j} = λ_{j}e_{j} we take ω = −Δu_{m} in (2.8). We obtain

M(t̃_{0})(Δu_{m}, Δu_{m}) + \frac{1}{ϵ}(β(u_{m}), −Δu_{m}) = (∇f, ∇u_{m})  (2.9)

By (2.1), we obtain

|Δu_{m}|^{2} ≤ C(|∇f|^{2} + |∇u_{m}|^{2}).  (2.10)

Since f ∈ H_{0}^{1}(Ω) and by (2.6), we have that

|Δu_{m}|^{2} ≤ C  (2.11)

where C does not depend on m and ϵ. By (2.7) and (2.11) and by compact imbedding of (H_{0}^{1}(Ω) ∩ H^{2}(Ω)) ⊂ H_{0}^{1}(Ω) we deduce that

u_{m} → u in H_{0}^{1}(Ω)  (2.12)

EJQTDE, 2005, No. 18, p. 4 The continuity of M and convergence (2.7) and (2.12) permit to pass to the limit in (2.5). We obtain

M(∥u∥^{2})((u, v)) + \frac{1}{ϵ}(β(u), v) = (f, v), for all v ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω). (2.13)

Thus u_{ϵ} is a weak solution of problem (2.2) and the proof of Theorem 2.3 is complete. □

Proof. of Theorem 2.1 Let (ϵ_{n})_{n∈ℕ} be a sequence of real numbers such that

0 < ϵ_{n} < 1 for all n ∈ ℕ and lim_{n→∞} ϵ_{n} = 0.

For each n ∈ ℕ, we get function u_{ϵ_{n}} which satisfies Theorem 2.3. Since the estimates were uniform on ϵ and n, we can see that there exists a subsequence of u_{ϵ_{n}}, again called u_{ϵ_{n}}, and a function u ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω) such that

u_{ϵ_{m}} → u in H_{0}^{1}(Ω).  (2.14)

Consider v in H_{0}^{1}(Ω) ∩ H^{2}(Ω) with v belongs to K. By (2.5), we have

(−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}} − f, v) = (−\frac{1}{ϵ_{n}} β(u_{ϵ_{n}}), v)  (2.15)

and

(−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}} − f, −u_{ϵ_{n}}) = (−\frac{1}{ϵ_{n}} β(u_{ϵ_{n}}), −u_{ϵ_{n}}).  (2.16)

Follows of (2.15) and (2.16) that

\begin{matrix}(−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}} − f, v − u_{ϵ_{n}}) = \frac{1}{ϵ}(−β(u_{ϵ_{n}}), v − u_{ϵ_{n}}) \\ = \frac{1}{ϵ}(β(v) − β(u_{ϵ_{n}}), v − u_{ϵ_{n}}) ≥ 0, \end{matrix}  (2.17)

because v ∈ K and monotonicity of β. Hence

(−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}} − f, v) + (f, u_{ϵ_{n}}) ≥ (−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}}, u_{ϵ_{n}}),  (2.18)

But

\begin{matrix}(−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}}, u_{ϵ_{n}}) = M(∥u_{ϵ_{n}}∥^{2})(∇u_{ϵ_{n}}, ∇u_{ϵ_{n}}) = \\ M(∥u_{ϵ_{n}}∥^{2})(∇(u_{ϵ_{n}} − u), ∇(u_{ϵ_{n}} − u)) + M(∥u_{ϵ_{n}}∥^{2})(∇u_{ϵ_{n}}, ∇u)+ \\ M(∥u_{ϵ_{n}}∥^{2})(∇u, ∇(u_{ϵ_{n}} − u)) ≥ M(∥u_{ϵ_{n}}∥^{2})(∇u_{ϵ_{n}}, ∇u)+ \\ M(∥u_{ϵ_{n}}∥^{2})(∇u, ∇(u_{ϵ_{n}} − u)), \end{matrix}  (2.19)

because M(∥u_{ϵ_{n}}∥^{2})(∇(u_{ϵ_{n}} − u), ∇(u_{ϵ_{n}} − u)) ≥ 0. Then,

lim inf[−M(∥u_{ϵ_{n}}∥^{2})(Δu_{ϵ_{n}}, u_{ϵ_{n}})] ≥ −M(∥u∥^{2})(∇u, u).  (2.20)

By (2.18) and (2.20), we obtain (1.1). In order to prove u ∈ K we observe that from (2.5), with ω = u_{ϵ_{m}}, that

0 ≤ (β(u_{ϵ_{n}}), u_{ϵ_{n}}) ≤ ϵC  (2.21)

therefore

(β(u_{ϵ_{n}}), u_{ϵ_{n}}) converges to zero.  (2.22)

EJQTDE, 2005, No. 18, p. 5 Let v ∈ H_{0}^{1}(Ω) be arbitrarily fixed; then the inequality

(β(u_{ϵ_{n}}) − β(v), u_{ϵ_{n}} − v) ≥ 0

yields (β(v), u − v) ≤ 0, hence (β(u − λω)), ω) ≤ 0 with the choice of v = u − λω with λ > 0 and ω ∈ H_{0}^{1}(Ω). By hemicontinuity of β we can let λ → 0^{+} and obtain (β(u), ω) ≥ 0, hence β(u) = 0 by the arbitrariness of ω. Therefore u ∈ K.

□

3. The M-linear Problem: A Discontinuous Case

In this section we concentrate our atention on problem (1.1) when M possesses a discontinuity. More precisely, we study problem (1.1) with M : ℝ/{θ} → ℝ continuous such that (M2) lim_{t→θ^{+}} M(t) = lim_{t→θ^{−}} M(t) = +∞ (M3) lim sup_{t→+∞} M(t^{2})t = +∞ .

Theorem 3.1. Assume that (M1)–(M3) and (H1) hold. Then for any choice of f ∈ H_{0}^{1}(Ω) the problem (1.1) admits at least one solution solution.

The proof of Theorem 3.1 is given as in the proof of Theorem 2.1. We will formulate the penalized problem, associated with the variational inequality (1.1), as follows. Given ϵ > 0, find a function u_{ϵ} ∈ H_{0}^{1}(Ω) solution of the problem

\begin{matrix}−M(∥u_{ϵ}∥^{2})Δu_{ϵ} + \frac{1}{ϵ}β(u_{ϵ}) = f in Ω, \\ u_{ϵ} = 0 on ∂Ω, \end{matrix}  (3.1)

where f is given in H_{0}^{1}(Ω). The existence of one solution of the penalized problem (3.1) is given by the

Theorem 3.2. Assume that (M1) − (M3) hold. Then, for any 0 ≠ f ∈ H_{0}^{1}(Ω) there exists at least one u_{ϵ} ∈ L^{2}(Ω), solution of the problem (3.1) .

Proof. We first consider the sequence of functions M_{n} : ℝ → ℝ given by

M_{n}(t) = \begin{cases}n, & θ − δ′_{n} ≤ t ≤ θ + δ′′_{n}, \\ M(t), & t ≤ θ − δ′_{n} or t ≥ θ + δ′′_{n}, \end{cases}

for n > m_{0}, where θ − δ′_{n} and θ + δ′′_{n}, δ′_{n}, δ′′_{n} > 0, are, respectively, the points closest to θ, at left and at right, so that

M(θ − δ′_{n}) = M(θ + δ′′_{n}) = n.

We point out that, in this case, δ′_{n}, δ′′_{n} → 0 as n → ∞.

Take n > m_{0} and observe that the horizontal lines y = n cross the graph of M. Hence M_{n} is continuous and satisfies (M1), for each n > m_{0}. In view of this, for each n like above, there is u_{n} ∈ H_{0}^{1}(Ω) ⋂ H^{2}(Ω) satisfying

M_{n}(∥u_{n}∥^{2})((u_{n}, ω)) + \frac{1}{ϵ}(β(u_{n}), ω) = (f, ω),

for all ω ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω).

(3.2)

Taking ω = u_{n} in the above equation one has

M_{n}(∥u_{n}∥^{2})∥u_{n}∥^{2} + \frac{1}{ϵ}(β(u_{n}), u_{n}) = (f, u_{n}),

and so

M_{n}(∥u_{n}∥^{2})∥u_{n}∥ ≤ ∥f∥,

since 0 ≤ (β(u_{n}), u_{n}). Because of (M3) the sequence (∥u_{n}∥) must be bounded. Hence

u_{n} ⇀ u in H_{0}^{1}(Ω),
u_{n} → u in L^{2}(Ω),
∥u_{n}∥^{2} → θ_{0}, for some θ_{0},

(3.3)

perhaps for subsequences. We note that if (M_{n}(∥u_{n}∥^{2}) converges its limit is different of zero. Suppose that ∥u_{n}∥^{2} → θ. If ∥u_{n}∥^{2} > θ +δ′′_{n} or ∥u_{n}∥^{2} < θ −δ′_{n}, for infinitely many n, we would get M_{n}(∥u_{n}∥^{2}) = M(∥u_{n}∥^{2}), for such n, and so

M(∥u_{n}∥^{2})∥u_{n}∥^{2} ≤ (f, u_{n}) ⇒ +∞ ≤ (f, u)

which is a contradiction. On the other hand, if there are infinitely many n so that θ − δ′_{n} ≤ ∥u_{n}∥^{2} ≤ θ + δ′′_{n} ⇒ M_{n}(∥u_{n}∥^{2}) = n and so n∥u_{n}∥^{2} ≤ (f, u_{n}) ⇒ +∞ ≤ (f, u) and we arrive again in a contradiction.

Consequently ∥u_{n}∥^{2} → θ_{0} ≠ θ which implies that for n large enough

∥u_{n}∥^{2} < θ − δ′_{n} or ∥u_{n}∥^{2} > θ + δ′′_{n}

and so M_{n}(∥u_{n}∥^{2}) = M(∥u_{n}∥^{2}) which yields

M(∥u_{n}∥^{2})((u_{n}, ω)) + \frac{1}{ϵ}(β(u_{n}), ω) = (f, ω), ∀ω ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω).  (3.4)

Taking ω = −Δu_{n} in (3.4), using(2.1) and arguments of compactness as in the proof of theorem 2.1, we obtain that

u_{n} → u ∈ H_{0}^{1}(Ω)  (3.5)

The estimates (3.3) and (3.5) are sufficient to pass to the limit in the approximate equation and to obtain

M(∥u∥^{2})((u, ω)) + \frac{1}{ϵ}(β(u), ω) = (f, ω), ∀ω ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω).  (3.6)
□

Proof. of Theorem 3.1 As in the proof of Theorem 2.1, let (ϵ_{n})_{n∈ℕ} be a sequence of real numbers such, that

0 < ϵ_{n} < 1 for all n ∈ ℕ and lim_{n→∞} ϵ_{n} = 0.

Applying Theorem 3.1, for each n ∈ ℕ, we get a function u_{ϵ_{n}} ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω) which satisfies

M(∥u_{ϵ_{n}}∥^{2})((u_{ϵ_{n}}, ω)) + \frac{1}{ϵ_{n}} (β(u_{ϵ_{n}}), ω) = (f, ω), ∀ω ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω). (3.7)

EJQTDE, 2005, No. 18, p. 7 Since the estimates were uniform on ϵ and n, we can see that there exists a subsequence of (u_{ϵ_{n}}), again called (u_{ϵ_{n}}), and a function u ∈ H_{0}^{1}(Ω) ∩ H^{2}(Ω) such that

u_{ϵ_{n}} → u in H_{0}^{1}(Ω).  (3.8)

We note that ∥u_{ϵ_{n}}∥ does not converges to θ. In fact, we assume by contradiction that ∥u_{ϵ_{n}}∥ → θ. By (3.7) and (3.8), with ω = u_{ϵ_{n}}, we get

M(∥u_{ϵ_{n}}∥^{2})∥u_{ϵ_{n}}∥ ≤ C

By hyphotese (M2), letting n → ∞, we have a contradiction. Thus

M(∥u_{ϵ_{n}}∥^{2}) → M(∥u∥^{2}),

since ∥u_{ϵ_{n}}∥ → ∥u∥ and M is continuous in an neighboard of ∥u∥ ≠ θ. Consider v in H_{0}^{1}(Ω) with v belonging to K. By (3.7), we have

(−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}} − f, v) = (−\frac{1}{ϵ}β(u_{ϵ_{n}}), v)  (3.9)

and

(−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}} − f, −u_{ϵ_{n}}) = (−\frac{1}{ϵ}β(u_{ϵ_{n}}), −u_{ϵ_{n}}).  (3.10)

Follows of (3.9) and (3.10) that

\begin{matrix}(−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}} − f, v − u_{ϵ_{n}}) = \frac{1}{ϵ}(−β(u_{ϵ_{n}}), v − u_{ϵ_{n}}) \\ = \frac{1}{ϵ}(β(v) − β(u_{ϵ_{n}}), v − u_{ϵ_{n}}) ≥ 0, \end{matrix}  (3.11)

because v ∈ K and monotonicity of β. Hence

(−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}} − f, v) + (f, u_{ϵ_{n}}) ≥ (−M(∥u_{ϵ_{n}}∥^{2})Δu_{ϵ_{n}}, u_{ϵ_{n}}).  (3.12)

As in the proof of Theorem 2.1, letting n → ∞, we obtain that u satisfies the conditions of Theorem 3.2.

□

4. Another Nonlocal Problem

Next, we make some remarks on a nonlocal problem which is a slight generalization of one studied by Chipot-Lovat [6] and Chipot-Rodrigues [7]. More precisely, the above authors studied the problem

−a( ∫_{Ω} u)Δu = f in Ω,
u = 0 on ∂Ω,

(4.1)

where Ω ⊂ ℝ^{N} is a bounded domain, N ≥ 1, and a : ℝ → (0, +∞) is a given function. Equation (4.1) is the stationary version of the parabolic problem

u_{t} − a( ∫_{Ω} u(x, t)dx)Δu = f in Ω × (0, T),
u = 0 on ∂Ω × (0, T),
u(x, 0) = u_{0}(x), u_{t}(x, 0) = u_{1}(x).

EJQTDE, 2005, No. 18, p. 8 Here T is some arbitrary time and u represents, for instance, the density of a population subject to spreading. See [6, 7] for more details. In particular, [6] studies problem (4.1), with f ∈ H^{−1}(Ω), and proves the following result.

Proposition 4.1. Let a : ℝ → (0, +∞) be a positive function, f ∈ H^{−1}(Ω). Then problem (4.1) has as many solutions μ as the equation

a(μ)μ = ⟨⟨f, φ⟩⟩,

where φ is the function(unique) satisfying

−Δφ = 1 in Ω,
φ = 0 on ∂Ω.

In the present work, we consider the variational inequality

u ∈ K : (−a( ∫_{Ω} |u|^{q})Δu, ω) ≥ (f, ω) for all v ∈ K ∩ (H_{0}^{1}(Ω) ∩ H^{2}(Ω)) (4.2)

where K and f are as before and 1 < q < 2N/(N − 2), N ≥ 3. When q = 2 we have the well known Carrier model.

Theorem 4.2. If t ↦ a(t) is a decreasing and continuous function, for t ≥ 0, lim_{t→+∞} a(t^{q})t = +∞ and t ↦ a(t^{q})t is injective, for t ≥ 0, then, for each 0 ≠ f ∈ H^{−1}(Ω), problem (4.2) possesses a weak solution.

The proof of Theorem 4.2 is given as in the proof of Theorem 2.1. We will formulate the penalized problem, associated with the variational inequality (4.2), as follows: Given ϵ > 0, find a function u_{ϵ} ∈ H_{0}^{1}(Ω) solution of the problem

−a( ∫_{Ω} |u_{ϵ}|^{q})Δu_{ϵ} + \frac{1}{ϵ}β(u_{ϵ}) = f in Ω,

u_{ϵ} = 0 on ∂Ω,

(4.3)

where f is given in H_{0}^{1}(Ω). The existence of one solution of the penalized problem (4.3) is given by the

Theorem 4.3. Under the hypotheses of Theorem 4.2, then for each 0 ≠ f ∈ H^{−1}(Ω) there exists at least one u_{ϵ} ∈ H_{0}^{1}(Ω) solution of the problem (4.3) .

Proof. As in the proof of Theorem 2.1, let ∑ = {e_{1}, …, e_{m}, …} be an orthonormal basis of the Hilbert space H_{0}^{1}(Ω). For each m ∈ ℕ consider the finite dimensional Hilbert space

𝕍_{m} = span{e_{1}, …, e_{m}},

P : ℝ^{m} → ℝ^{m} be the function P (ξ) = (P_{1}(ξ), …, P_{m}(ξ)), where

P_{i}(ξ) = a(∥u∥_{q}^{q})((u, e_{i})) + \frac{1}{ϵ}(β(u), e_{i}) − ⟨⟨f, e_{i}⟩⟩, i = 1, …, m

with u = ∑_{j=1}^{m} ξ_{j}e_{j} and the identifications of ℝ^{m} and 𝕍_{m} mentioned before. So

P_{i}(ξ) = a(∥u∥_{q}^{q})ξ_{i} + \frac{1}{ϵ}(β(u), e_{i}) − ⟨⟨f, e_{i}⟩⟩, i = 1, …, m

and then

< P (ξ), ξ >= a(∥u∥_{q}^{q})∥u∥^{2} + \frac{1}{ϵ}((β(u), u)) − ⟨⟨f, u⟩⟩

EJQTDE, 2005, No. 18, p. 9 We have to show that there is r > 0 so that ⟨P (ξ), ξ⟩ ≥ 0, for all ∥ξ∥_{ℝ^{m}} = r in 𝕍_{m}. Suppose, on the contrary, that for each r > 0 there is u_{r} ∈ 𝕍_{m} such that ∥u_{r}∥ = r and

⟨P (ξ_{r}), ξ_{r}⟩ < 0, ξ_{r} ↔ u_{r}.

Taking r = n ∈ ℕ we obtain a sequence (u_{n}), ∥u_{n}∥ = n, u_{n} ∈ 𝕍_{m} and

⟨P (u_{n}), u_{n}⟩ = a(∥u_{n}∥_{q}^{q})∥u_{n}∥^{2} + \frac{1}{ϵ}(β(u_{n}), u_{n}) − ⟨⟨f, u_{n}⟩⟩ < 0

and so

a(∥u_{n}∥_{q}^{q})∥u_{n}∥ < C∥f∥, ∀n = 1, 2, …,

since(β(u_{n}), u_{n}) > 0. Because of the continuous immersion H_{0}^{1}(Ω) ⊂ L^{q}(Ω) one gets ∥u∥_{q} ≤ C∥u∥ and the monotonicity of a yields a(∥u∥_{q}^{q}) ≥ a(C∥u∥^{q}) and so

a(C∥u_{n}∥^{q})∥u_{n}∥ < C∥f∥.

In view of lim_{t→+∞} a(t^{q})t = +∞ one has that ∥u_{n}∥ ≤ C, ∀n ∈ ℕ, which contradicts ∥u_{n}∥ = n. So, there is r_{m} > 0 such that ⟨F (ξ), ξ⟩ ≥ 0, for all |ξ| = r_{m}. In view of the sharp angle lemma there is u_{m} ∈ 𝕍_{m}, ∥u_{m}∥ ≤ r_{m} such that P_{i}(u_{m}) = 0, i = 1, …, m, that is,

a (∥u_{m}∥_{q}^{q}) ((u_{m}, ω)) + \frac{1}{ϵ}(β(u_{m}), ω) = ⟨⟨f, ω⟩⟩, ∀ω ∈ 𝕍_{m}.  (4.4)

Reasoning as before, by using the facts that t → a(t) is decreasing for t ≥ 0 , lim_{t→+∞} a(t^{q})t = +∞ and ∫_{Ω} β(u)u > 0 we conclude that ∥u_{m}∥ ≤ C, ∀m = 1, 2 … for some constant C that does not depend on m. Hence, u_{m} ⇀ u in H_{0}^{1}(Ω), u_{m} → u in L^{q}(Ω), 1 < q < \frac{2N}{N−2}, and so ∥u_{m}∥_{q} → ∥u∥_{q}. Taking limits on both sides of equation (4.4) we conclude that the function u is a solution of problem (4.3). □

Remark 4.4. The function

a(t) = \frac{1}{t^{2β} + 1},

where β and q are related by 2βq < 1, satisfies the assumptions of Theorem 4.2

Proof. of Theorem 4.2 Let (ϵ_{n})_{n∈ℕ} be a sequence of real numbers such, that

0 < ϵ_{n} < 1 for all n ∈ ℕ and lim_{n→∞} ϵ_{n} = 0.

For each n ∈ ℕ, we get function u_{ϵ_{n}} which satisfies Theorem 4.3. Since the estimates were uniform on ϵ and n, we can see that there exists a subsequence of u_{ϵ_{n}}, again called u_{ϵ_{n}}, and a function u ∈ H_{0}^{1}(Ω) ∩ L^{q}(Ω), 1 ≤ q ≤ \frac{2N}{N−2} such that

u_{ϵ_{n}} ⇀ u in H_{0}^{1}(Ω) ∩ H^{2}(Ω)  (4.5)
u_{ϵ_{n}} → u in L^{q}(Ω), 1 ≤ q ≤ \frac{2N}{N − 2}  (4.6)
∥u_{ϵ_{n}}∥_{q} → ∥u∥_{q}  (4.7)

Consider v in H_{0}^{1}(Ω) with v belongs to K. By (4.3), we have EJQTDE, 2005, No. 18, p. 10

\begin{matrix}(−a(∥u_{ϵ_{n}}∥_{q}^{p})Δu_{ϵ_{n}} − f, v − u_{ϵ_{n}}) = (−\frac{1}{ϵ}β(u_{ϵ_{n}}), v − u_{ϵ_{n}}) = \\ \frac{1}{ϵ}(β(v) − β(u_{ϵ_{n}}), v − u_{ϵ_{n}}) ≥ 0 \end{matrix}  (4.8)

because v ∈ K and monotonicity of β. As in the proof of Theorem 2.1, when ϵ_{n} → 0 we obtain that u satisfies the conditions of Theorem 4.2.

□

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EJQTDE, 2005, No. 18, p. 11

(Received December 12, 2004)

Francisco Julio S. A. Corrêa Departamento de Matemática-CCEN, Universidade Federal do Pará, 66.075-110 Belém Pará Brazil

E-mail address: fjulio@ufpa.br

Silvano D. B. Menezes Departamento de Matemática-CCEN, Universidade Federal do Pará, 66.075-110 Belém Pará Brazil

E-mail address: silvano@ufpa.br