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International Journal of Differential Equations
Volume 2011 (2011), Article ID 801706, 13 pages
http://dx.doi.org/10.1155/2011/801706
Research Article

On the Weak Solution of a Semilinear Boundary Value Problem without the Landesman-Lazer Condition

Department of Mathematics and Statistics, University of Uyo, Akwa Ibom, 520003 Uyo, Nigeria

Received 4 August 2011; Accepted 15 October 2011

Academic Editor: Yuji Liu

Copyright © 2011 Sikiru Adigun Sanni. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We prove the existence of weak solution to a semilinear boundary value problem without the Landesman-Lazer condition.

1. Introduction

We consider the nonlinear boundary value problem Δ𝑢+𝜆𝑘𝑢+𝑔(𝑢)=(𝑥)inΩ,(1.1)𝑢=0on𝜕Ω,(1.2) where Ω𝑛 is open and bounded, 𝐿2(Ω), 𝜆𝑘 is a simple eigenvalue of Δ corresponding to the eigenvector 𝜙𝑘, and the nonlinearity 𝑔 satisfies the following conditions: ||||𝑔(𝑢)𝑔(𝑣)𝐿|𝑢𝑣|(Lipschitzcontinuity)forsomeconstant𝐿>0.(H)

Landesman and Lazer [1] considered the problem (1.1)-(1.2) with continuous function 𝑔 satisfying 𝑔()<𝑔(𝜉)<𝑔(), where 𝑔(±)=lim𝑠±𝑔(𝑠) exist and are finite. The authors showed that if 𝜙𝑘 is an eigenfunction corresponding to 𝜆𝑘, Ω+={𝑥Ω𝜙𝑘>0} and Ω={𝑥Ω𝜙𝑘<0}, then the necessary and sufficient condition for the existence of weak solution of (1.1)-(1.2) is that𝑔()Ω+𝜙𝑘𝑑𝑥+𝑔()Ω𝜙𝑘𝑑𝑥<Ω𝜙𝑘𝑑𝑥<𝑔()Ω+𝜙𝑘𝑑𝑥+𝑔()Ω𝜙𝑘𝑑𝑥.(1.3) The condition (1.3) is the well-known Landesman-Lazer condition, named after the authors. The result of the paper [1] has since been generalized by a number of authors which include [29], to mention a few.

We mention, briefly, few works without the assumption of the Landesman-Lazer condition. The perturbation of a second order linear elliptic problems by nonlinearity without Landesman-Lazer condition was investigated in [10]. The function 𝑔(𝑢) was assumed to be a bounded continuous function satisfying 𝑔(𝑡)𝑡0,𝑡.(1.4) The nonhomogeneous term was assumed to be an 𝐿-function orthogonal to an eigenfunction 𝜙 in 𝐿2, which corresponds to a simple eigenvalue 𝜆1. Ha [11] considered the solvability of an operator equation without the Landesman-Lazer condition. The author used a nonlinear Carathéodory function 𝑔(𝑥,𝑢) which satisfies the conditions ||||𝑔(𝑥,𝑢)𝑏(𝑥),𝑢𝑔(𝑥,𝑢)0,(1.5) for almost all 𝑥Ω and all 𝑢, where 𝑏𝐿2(Ω). The solvability of the operator equation is proved under some hypotheses on 𝑔(𝑥,𝑢). The nonhomogeneous term was assumed to be an 𝐿2-function. Iannacci and Nkashama proved existence of solutions to a class of semilinear two-point eigenvalue boundary value problems at resonance without the Landesman-Lazer condition, by imposing the same conditions as in [11] in conjunction with some other hypotheses on 𝑔 and . Furthermore, the existence of solution was proved only for the eigenvalue 𝜆=1. Assuming a Carathéodory function 𝑓(𝑥,𝑢) with some growth restriction and assuming an 𝐿2-function , Santanilla [12] proved existence of solution to a nonlinear eigenvalue boundary value problem (for eigenvalue 𝜆=1) without Landesman-Lazer condition. Du [13] proved the existence of solution for nonlinear second-order two-point boundary value problems, by allowing the eigenvalue 𝜆 of the problem to change near the eigenvalues of 𝑚2𝜋2 of the problem 𝑦+𝑚2𝜋2𝑦=0,𝑦(0)=𝑦(1)=0. The author did not use the Landesman-Lazer condition and imposed weaker conditions on 𝑔(𝑢) than in [12]. Recently, Sanni [14] proved the existence of solution to the same problem considered by Du [13] with 𝜆=𝑚2𝜋2 exactly, without assuming the Landesman-Lazer condition. The author assumed that |𝑔(𝑢)|𝐶=constant and 𝐿2(0,1). Other works without the assumption of Landesman-Lazer condition include [1521]. We mention that most of the papers on this topic use the methods in [22] and [12]. The method of upper and lower solutions is used in [14]. For several other related resonance problems, we refer the reader to the book of Rădulescu [23].

The current work constitutes further deductions on the problem considered by Landesman and Lazer [1] and is motivated by previous works and by asking if it is possible to obtain a weak solution of (1.1)-(1.2) by setting 𝑢=𝜙𝑘𝑣(𝑥). The answer is in the affirmative. The substitution gives rise to a degenerate semilinear elliptic equation. Consequently, we prove the existence of weak solution to the degenerate semilinear elliptic equation in a 𝜙2𝑘-weight Sobolev’s space, by using the Schaefer’s fixed point theorem. For information on weighted Sobolev’s spaces, the reader is referred to [24, 25]. The current work is significant in that the condition H enables a relaxation of the Landesman-Lazer condition (1.3), and the solution 𝑢 to (1.1)-(1.2) is constructed using the eigenfunctions 𝜙𝑘. Furthermore, the current analysis takes care of the situation where 𝑔()=𝑔()=0.

The remaining part of this paper is organized as follows: the weighted Sobolev’s spaces used are defined in Section 2. In addition, we use the substitution 𝑢=𝜙𝑘𝑣 to get the degenerate semilinear elliptic equation in 𝑣, from which we give a definition of a weak solution. Furthermore, we state two theorems used in the proof of the existence result. In Section 3, we prove the existence and uniqueness of solution to an auxiliary linear problem. In Section 4, we prove a necessary condition for the existence of solution to (1.1)-(1.2) before proving the existence of solution to (1.1)-(1.2). At the end of Section 4, we prove that 𝑢=𝜙𝑘𝑣 is in 𝐻10(Ω), provided that 𝑣𝑋. Finally, we give an illustrative example in Section 5 for which our result applies.

2. Preliminaries

We define the following weighted Sobolev’s spaces used in this paper:𝐿2Ω,𝜙2𝑘=𝑤Ωsuchthat𝑤𝐿2(Ω,𝜙2𝑘)<,(2.1) where 𝑤𝐿2(Ω,𝜙2𝑘)=Ω𝜙2𝑘𝑤2𝑑𝑥.𝐻1Ω,𝜙2𝑘=𝑤Ωsuchthat𝑤𝐻1(Ω,𝜙2𝑘)<,(2.2) where 𝑤𝐻1(Ω,𝜙2𝑘)=Ω𝜙2𝑘𝑤2𝑑𝑥+Ω𝜙2𝑘|𝑤|2𝑑𝑥.

For brevity, we set 𝑋=𝐻1(Ω,𝜙2𝑘).

Set 𝑢=𝜙𝑘𝑣(𝑥) in (1.1) to deduce Δ𝜙𝑘+𝜆𝑘𝜙𝑘𝑣𝜙𝑘Δ𝑣2𝜙𝑘𝜙𝑣=𝑔𝑘𝑣(𝑥)inΩ.(2.3) Note that the first term on the left of (2.3) vanishes, multiply (2.3) by 𝜙𝑘 and use (1.2) to deduce 𝜙2𝑘𝑣=𝜙𝑘𝑔𝜙𝑘𝑣𝜙𝑘𝜙(𝑥)inΩ,𝑘𝑣=0on𝜕Ω.(2.4) Thus, if we can prove the existence of solution to (2.4), then 𝑢=𝜙𝑘𝑣 solves (1.1)-(1.2). Indeed, we will prove that the solution 𝑢 belongs to the Sobolev space 𝐻10(Ω).

Definition 2.1. We say that 𝑣𝑋 is a weak solution of the problem (2.4) provided Ω𝜙2𝑘𝑣𝜁𝑑𝑥=Ω𝜙𝑘𝜙𝜁𝑔𝑘𝑣𝑑𝑥Ω𝜙𝑘𝜁𝑑𝑥,(2.5) for each 𝜁𝑋.

Definition 2.2. Let 𝑋 be a Banach space and 𝐴𝑋𝑋 a nonlinear mapping. 𝐴 is called compact provided for each bounded sequence {𝑢𝑘}𝑘=1 the sequence {𝐴[𝑢𝑘]}𝑘=1 is precompact; that is, there exists a subsequence {𝑢𝑘𝑗}𝑗=1 such that {𝐴[𝑢𝑘𝑗]}𝑗=1 converges in 𝑋 (see [26]).
The following theorems are applied in this paper.

Theorem 2.3 (Bolzano-Weierstrass). Every bounded sequence of real numbers has a convergent subsequence (see [27]).

Theorem 2.4 (Schaefer’s Fixed Point Theorem). Let 𝑋 be a Banach space and 𝐴𝑋𝑋(2.6) a continuous and compact mapping. Suppose further that the set [𝑢]{𝑢𝑋𝑢=𝜏𝐴forsome0𝜏1}(2.7) is bounded. Then 𝐴 has a fixed point (see [26]).

3. Auxiliary Linear Problem

Consider the linear boundary value problem: 𝜙𝐿𝑣=2𝑘𝑣+𝜇𝜙2𝑘𝑣=𝜇𝜙2𝑘𝑠+𝜙𝑘𝑔𝜙𝑘𝑠𝜙𝑘𝜙inΩ,(3.1)𝑘𝑣=0on𝜕Ω,(3.2) where 𝜇 is a strictly positive constant; 𝑠𝐿2(Ω,𝜙2𝑘), 𝑔(𝜙𝑘𝑠), and are functions of 𝑥 only.

Theorem 3.1 (a priori estimates). Let 𝑣 be a solution of (3.1)-(3.2). Then 𝑣𝑋 and we have the estimate 𝑣2𝑋𝐶𝑠2𝐿2(Ω,𝜙2𝑘)+2𝐿2(Ω)+1<,(3.3) for some appropriate constant 𝐶>0.

Proof. Multiply (3.1) by 𝑣, integrate by parts and apply (3.2) to get Ω𝜙2𝑘||||𝑣2𝑑𝑥+𝜇Ω𝜙2𝑘𝑣2𝑑𝑥=𝜇Ω𝜙2𝑘𝑣𝑠𝑑𝑥+Ω𝑣𝜙𝑘𝑔𝜙𝑘𝑠𝑑𝑥Ω𝑣𝜙𝑘𝑑𝑥𝜇Ω𝜙2𝑘𝑣2𝑑𝑥1/2Ω𝜙2𝑘𝑠2𝑑𝑥1/2+Ω𝜙2𝑘𝑣2𝑑𝑥1/2Ω||𝑔𝜙𝑘𝑠||2𝑑𝑥1/2+Ω𝜙2𝑘𝑣2𝑑𝑥1/2Ω2𝑑𝑥1/2̈(byHolder'sinequality)3𝜖Ω𝜙2𝑘𝑣21𝑑𝑥+𝜇4𝜖2Ω𝜙2𝑘𝑠2𝑑𝑥+Ω||𝑔𝜙𝑘𝑠||2𝑑𝑥+Ω2𝑑𝑥(byCauchy'sinequalitywith𝜖).(3.4) Using 𝐻, the second term in the bracket on the right side of (3.4) may be estimated as ||𝑔𝜙𝑘𝑠||𝑔(0)2𝐿2||𝜙𝑘𝑠||2||𝑔𝜙or𝑘𝑠||2||||𝑔(0)2||𝑔𝜙+2𝑘𝑠||||||𝑔(0)+𝐿2||𝜙𝑘𝑠||2||||𝑔(0)2+12||𝑔𝜙𝑘𝑠||2||||+2𝑔(0)2+𝐿2||𝜙𝑘𝑠||2(byYoung'sinequality).(3.5) Simplifying (3.5), we deduce ||𝑔𝜙𝑘𝑠||||𝜙𝐶1+𝑘𝑠||,(3.6) (see [26]) for some constant 𝐶=𝐶(𝐿,|𝑔(0)|). Notice that (3.6) implies that Ω||𝑔𝜙𝑘𝑠||2𝑑𝑥𝐶1+𝑠𝐿2(Ω,𝜙2𝑘)2<,(3.7) so that 𝑔(𝜙𝑘𝑠)𝐿2(Ω).
Using (3.7) and choosing 𝜖>0 sufficiently small in (3.4) and simplifying, we deduce (3.3).

Definition 3.2. (i) The bilinear form 𝐵[,] associated with the elliptic operator 𝐿 defined by (3.1) is 𝐵[]𝑣,𝜁=Ω𝜙2𝑘𝑣𝜁𝑑𝑥+𝜇Ω𝜙2𝑘𝑣𝜁𝑑𝑥,(3.8) for 𝑣,𝜁𝑋,
(ii) 𝑣𝑋 is called a weak solution of the boundary value problem (3.1)-(3.2) provided 𝐵[]=𝑢,𝜁𝜇𝜙2𝑘𝑠+𝜙𝑘𝑔𝜙𝑘𝑠𝜙𝑘,,𝜁(3.9) for all 𝜁𝑋, where (,) denotes the inner product in 𝐿2(Ω).

Theorem 3.3. 𝐵[𝑢,𝑣] satisfies the hypotheses of the Lax-Milgram theorem precisely. In other words, there exists constants 𝛼,𝛽 such that(i)|𝐵[𝑣,𝜁]|𝛼𝑣𝑋𝜁𝑋, (ii)𝛽𝑣2𝑋𝐵[𝑣,𝑣], for all 𝑣,𝜁𝑋.

Proof. We have ||𝐵[]||=||||𝑣,𝜁Ω𝜙2𝑘𝑣𝜁𝑑𝑥+𝜇Ω𝜙2𝑘||||𝑣𝜁𝑑𝑥𝜇Ω𝜙2𝑘𝑣2𝑑𝑥1/2Ω𝜙2𝑘𝜁2𝑑𝑥1/2+Ω𝜙2𝑘||||𝑣2𝑑𝑥1/2Ω𝜙2𝑘||||𝜁2𝑑𝑥1/2̈byHoldersinequality𝛼𝑣𝑋𝜁𝑋,(3.10) for appropriate constant 𝛼>0. This proves (i).
We now proof (ii). We readily check that 𝛽𝑣2𝑋Ω𝜙2𝑘||||𝑣2𝑑𝑥+𝜇Ω𝜙2𝑘𝑣2[],𝑑𝑥=𝐵𝑣,𝑣(3.11) for some constant 𝛽>0. We can for example take 𝛽=min{1,𝜇}.

Theorem 3.4. There exists unique weak solution to the degenerate linear boundary value problem (3.1)-(3.2).

Proof. The hypothesis on and (3.7) imply that 𝑔(𝜙𝑘𝑠)𝐿2(Ω). For fixed 𝑔(𝜙𝑘𝑠), set 𝜇𝜙2𝑘𝑠+𝜙𝑘𝑔(𝜙𝑘𝑠)𝜙𝑘,𝜁=(𝜇𝜙2𝑘𝑠+𝜙𝑘𝑔(𝜙𝑘𝑠)𝜙𝑘,𝜁)𝐿2(Ω) for all 𝜁𝑋 (where ,, denotes the pairing of 𝑋 with its dual). This is a bounded linear functional on 𝐿2(Ω) and thus on 𝑋. Lax-Milgram theorem (see, e.g., [26]) can be applied to find a unique function 𝑣𝑋 satisfying 𝐵[]=𝑣,𝜁𝜇𝜙2𝑘𝑠+𝜙𝑘𝑔𝜙𝑘𝑠𝜙𝑘,,𝜁(3.12) for all 𝜁𝑋. Consequently, 𝑣 is the unique weak solution of the problem (3.1)-(3.2).

4. Main Results

Theorem 4.1. The necessary condition that 𝑢𝐻10(Ω) be a weak solution to (1.1)-(1.2) is that Ω𝑔(𝑢)𝜙𝑘𝑑𝑥=Ω𝜙𝑘𝑑𝑥.(4.1)

Proof. Suppose 𝑢𝐻10(Ω) is a weak solution of (1.1)-(1.2). For a test function 𝜙𝑘, using integration by parts, we have: ΩΔ𝑢𝜙𝑘𝑑𝑥+𝜆𝑘Ω𝑢𝜙𝑘𝑑𝑥+Ω𝑔(𝑢)𝜙𝑘𝑑𝑥=Ω𝑢𝜙𝑘+𝜆𝑘Ω𝑢𝜙𝑘𝑑𝑥+Ω𝑔(𝑢)𝜙𝑘=𝑑𝑥Ω𝑢Δ𝜙𝑘+𝜆𝑘𝜙𝑘𝑑𝑥+Ω𝑔(𝑢)𝜙𝑘𝑑𝑥=Ω𝜙𝑘𝑑𝑥,(4.2) from which (4.1) follows, since Δ𝜙𝑘+𝜆𝑘𝜙𝑘=0.

Theorem 4.2. Let the condition (4.1) of Theorem 4.1 holds. Then there exists a weak solution to the problem (2.4).

Proof. The proof is split in seven steps.
Step 1. A fixed point argument to (2.4) is 𝜙2𝑘𝑤+𝜇𝜙2𝑘𝑤=𝜇𝜙2𝑘𝑣+𝜙𝑘𝑔𝜙𝑘𝑣𝜙𝑘𝜙(𝑥)inΩ,𝑘𝑤=0on𝜕Ω.(4.3) Define a mapping 𝐴𝑋𝑋(4.4) by setting 𝐴[𝑣]=𝑤 whenever 𝑤 is derived from 𝑣 via (4.3). We claim that 𝐴 is a continuous and compact mapping. Our claim is proved in the next two steps.Step 2. Choose ̃𝑣,𝑣𝑋, and define ̃𝑤𝐴[𝑣]=𝑤,𝐴[𝑣]=. For two solutions 𝑤,𝑤𝑋 of (4.3), we have 𝜙2𝑘𝑤𝑤+𝜇𝜙2𝑘𝑤𝑤=𝜇𝜙2𝑘̃(𝑣𝑣)+𝜙𝑘𝑔𝜙𝑘𝑣𝜙𝑘𝑔𝜙𝑘̃𝑣𝜙inΩ,𝑘𝑤𝑤=0on𝜕Ω.(4.5) Using (4.5), we obtain an analogous estimate to (3.4), namely: Ω𝜙𝑘||𝑤||𝑤2𝑑𝑥+𝜇Ω𝜙2𝑘||𝑤||𝑤23𝜖Ω𝜙2𝑘||𝑤||𝑤2+1𝑑𝑥𝜇4𝜖2Ω𝜙𝑘||̃𝑣||𝑣2𝐿2(Ω,𝜙2𝑘)+Ω||𝑔𝜙𝑘𝑣𝜙𝑔𝑘̃𝑣||2.𝑑𝑥(4.6) Now Ω||𝑔𝜙𝑘𝑣𝜙𝑔𝑘̃𝑣||2𝑑𝑥Ω𝜙2𝑘𝐿2||̃𝑣||𝑣2𝑑𝑥,(4.7) using the condition (H). We may now use (4.7) in (4.6) and simplify to deduce [𝑣][̃𝑣]𝐴𝐴𝑋=𝑤𝑤𝑋̃𝐶𝑣𝑣𝐿2(Ω,𝜙2𝑘)̃𝐶𝑣𝑣𝑋,(4.8) for some constant 𝐶>0. Thus, the mapping 𝐴 is Lipschitz continuous, and hence continuous.Step 3. Let {𝑣𝑘}𝑘=1 be a bounded sequence in 𝑋. By Bolzano-Weierstrass theorem, it has a convergent subsequence, say {𝑣𝑘𝑗}𝑗=1. Define 𝑣=lim𝑘𝑗𝑣𝑘𝑗.(4.9) Using (4.8)-(4.9), we deduce lim𝑘𝑗𝐴𝑣𝑘𝑗[𝑣]𝐴𝑋lim𝑘𝑗𝐶𝑣𝑘𝑗𝑣𝑋=0.(4.10) Thus, 𝐴[𝑣𝑘𝑗]𝐴[𝑣] in 𝑋. Therefore, 𝐴 is compact.Step 4. Define a set 𝐾={𝑝𝑋𝑝=𝜏𝐴[𝑝]forsome0𝜏1}. We will show that 𝐾 is a bounded set. Let 𝑣𝐾. Then 𝑣=𝜏𝐴[𝑣] for some 𝜏[0,1]. Thus, we have 𝑣/𝜏=𝐴[𝑣]. By the definition of the mapping 𝐴, 𝑤=𝑣/𝜏 is the solution of the problem 𝜙2𝑘𝑣𝜏+𝜇𝜙2𝑘𝑣𝜏=𝜇𝜙2𝑘𝑣+𝜙𝑘𝑔𝜙𝑘𝑣𝜙𝑘𝜙(𝑥)inΩ,𝑘𝑣𝜏=0on𝜕Ω.(4.11) Now, (4.11) are equivalent to 𝜙2𝑘𝑣+𝜇𝜙2𝑘𝑣=𝜇𝜏𝜙2𝑘𝑣+𝜏𝜙𝑘𝑔𝜙𝑘𝑣𝜏𝜙𝑘𝜙(𝑥)inΩ,𝑘𝑣=0on𝜕Ω.(4.12) Using (4.12) we have an analogous estimate to (3.3) of Theorem 3.1, namely: 𝑣2𝑋𝜏𝐶𝑣2𝐿2(Ω,𝜙2𝑘)+2𝐿2(Ω).+1(4.13) Choosing 𝜏[0,1] sufficiently small in (4.13) and simplifying, we conclude that 𝑣𝑋𝐶2𝐿2(Ω)+1<(4.14) for some constant 𝐶>0. Equation (4.14) implies that the set 𝐾 is bounded, since 𝑣 was arbitrarily chosen.
Since the mapping 𝐴 is continuous and compact and the set 𝐾 is bounded, by Schaefer’s fixed point theorem (see, e.g., [26]), the mapping 𝐴 has a fixed point in 𝑋.
Step 5. Write 𝜙𝑘𝑣0=𝜙𝑘𝑣|𝜕Ω=0. For 𝑚=0,1,2,, inductively define 𝑣𝑚+1𝑋 to be the unique weak solution of the linear boundary value problem 𝜙2𝑘𝑣𝑚+1+𝜇𝜙2𝑘𝑣𝑚+1=𝜇𝜙2𝑘𝑣𝑚+𝜙𝑘𝑔𝜙𝑘𝑣𝑚𝜙𝑘𝜙(𝑥)inΩ,(4.15)𝑘𝑣𝑚+1=0on𝜕Ω.(4.16) Clearly, our definition of 𝑣𝑚+1𝑋 as the unique weak solution of (4.15)-(4.16) is justified by Theorem 3.4. Hence, by the definition of the mapping 𝐴, we have for 𝑚=0,1,2,𝑣𝑚+1𝑣=𝐴𝑚.(4.17) Since 𝐴 has a fixed point in 𝑋, there exists 𝑣𝑋 such that lim𝑚𝑣𝑚+1=lim𝑚𝐴𝑣𝑚[𝑣]=𝐴=𝑣.(4.18)Step 6. Using (4.15)-(4.16), we obtain an analogous estimate to (3.3), namely: 𝑣𝑚+12𝑋𝑣𝐶𝑚2𝐿2(Ω,𝜙2𝑘)+2𝐿2(Ω)𝑣+1𝐶𝑚2𝑋+2𝐿2(Ω)+1(4.19) for some appropriate constant 𝐶>0. Using (4.18), we take the limit on the right side of (4.19) to deduce that sup𝑚𝑣𝑚𝑋<.(4.20) Equation (4.20) implies the existence of a subsequence {𝑣𝑚𝑗}𝑗=1 converging weakly in 𝑋 to 𝑣𝑋.
Furthermore, using (3.7), we deduce Ω||𝑔𝜙𝑘𝑣𝑚||2𝑣𝑑𝑥𝐶1+𝑚2𝐿2(Ω,𝜙2𝑘)2.(4.21) Again, we use (4.18) to obtain the limit on the right side of (4.21) to deduce that sup𝑚𝑔𝜙𝑘𝑣𝑚𝐿2(Ω)<.(4.22) Equation (4.22) implies the existence of a subsequence {𝑔(𝜙𝑘𝑣𝑚𝑗)}𝑗=1 converging weakly in 𝐿2(Ω) to 𝑔(𝜙𝑘𝑣) in 𝐿2(Ω).
Step 7. Finally, we verify that 𝑣 is a weak solution of (2.4). For brevity, we take the subsequences of the last step as {𝑣𝑚}𝑚=1 and {𝑔(𝜙𝑘𝑣𝑚)}𝑚=1. Fix 𝜁𝑋. Multiply (4.15) by 𝜁, integrate by parts and apply (4.16) to get Ω𝜙2𝑘𝑣𝑚+1𝜁𝑑𝑥+𝜇Ω𝜙2𝑘𝑣𝑚+1𝜁𝑑𝑥=𝜇Ω𝜙2𝑘𝑣𝑚𝜁𝑑𝑥+Ω𝜁𝜙𝑘𝑔𝜙𝑘𝑣𝑚𝑑𝑥Ω𝜁𝜙𝑘𝑑𝑥.(4.23) Using the deductions of the last step, we let 𝑚 in (4.23) to obtain Ω𝜙𝑘𝑣𝜁𝑑𝑥+𝜇Ω𝜙2𝑘𝑣𝜁𝑑𝑥=𝜇Ω𝜙2𝑘𝑣𝜁𝑑𝑥+Ω𝜁𝜙𝑘𝑔𝜙𝑘𝑣𝑑𝑥Ω𝜁𝜙𝑘𝑑𝑥,(4.24) from which canceling the terms in 𝜇, we obtain (2.5) as desired.

Theorem 4.3. Let 𝑣𝑋 be the solution of (3.1)-(3.2). Then, the solution 𝑢=𝜙𝑘𝑣 of (1.1)-(1.2) belongs to 𝐻10(Ω), and we have the estimate 𝑢𝐻10(Ω)𝐶𝑣𝑋,(4.25) for some constant 𝐶>0.

Proof. We split the proof in two steps.
Step 1. Recall that 𝜙𝑘 satisfies the equations: Δ𝜙𝑘+𝜆𝑘𝜙𝑘=0inΩ𝑛,𝜙(4.26)𝑘=0on𝜕Ω.(4.27) Multiplying (4.26) by 𝑣2𝜙𝑘, integrating by parts and applying (4.27) we compute Ω𝑣2𝜙𝑘Δ𝜙𝑘𝑑𝑥+𝜆𝑘Ω𝑣2𝜙2𝑘𝑑𝑥=0orΩ𝑣2𝜙𝑘𝜙𝑘𝑑𝑥=𝜆𝑘Ω𝑣2𝜙2𝑘𝑑𝑥orΩ||𝜙𝑘||2𝑣2𝑑𝑥=𝜆𝑘Ω𝑣2𝜙2𝑘𝑑𝑥2Ω𝜙𝑘𝑣𝑣𝜙𝑘𝑑𝑥(4.28)𝜆𝑘Ω𝑣2𝜙2𝑘𝑑𝑥+𝜖Ω||𝜙𝑘||2𝑣21𝑑𝑥+𝜖Ω𝜙2𝑘||||𝑣2𝑑𝑥,(4.29) by Cauchy’s inequality with 𝜖. Choosing 𝜖>0 sufficiently small in (4.29) and simplifying, we deduce Ω||𝜙𝑘||2𝑣2𝑑𝑥𝐶𝑣2𝑋,(4.30) for some constant 𝐶>0.Step 2. We have Ω𝑢2𝑑𝑥=Ω𝜙2𝑘𝑣2𝑑𝑥,Ω||||𝑢2𝑑𝑥=Ω||𝜙𝑘𝑣||2𝑑𝑥=Ω||𝜙𝑘𝑣+𝜙𝑘||𝑣2𝑑𝑥2Ω||𝜙𝑘||2𝑣2𝑑𝑥+2Ω𝜙2𝑘||||𝑣2𝑑𝑥𝐶𝑣2𝑋,(using(4.30))(4.31) for some constant 𝐶>0. Thus, 𝑢𝐻1(Ω). Hence, by a Sobolev’s embedding theorem (see [26, page 269]), we have that 𝑢𝐻10(Ω), since 𝑢|𝜕Ω=0.

5. Illustrative Example

Consider the following special case for 𝑛=1: 𝑢+𝑢2𝑢=1in(0,𝜋),𝑢(0)=𝑢(𝜋)=0.(5.1) In this case, the eigenfunction 𝜙𝑘=sin𝑥, 𝑔(𝑢)=2𝑢, and =1. Clearly 𝑔(𝑢) is Lipschitz continuous and 𝐿2(Ω). Provided the necessary condition2𝜋0𝑢sin𝑥𝑑𝑥=𝜋0sin𝑥𝑑𝑥(5.2) is satisfied; Theorems 4.2 and 4.3 ensure the existence of a solution 𝑢=𝜙𝑘𝑣(𝑥)𝐻10(Ω) of the problem (5.1). Now, the problem (5.1) admits the solution𝑢=sinh(𝜋𝑥)+sinh𝑥sinh𝜋1.(5.3) Using (5.3) in (5.2), it is not difficult to verify that the necessary condition 2𝜋0sinh(𝜋𝑥)+sinh𝑥sinh𝜋1sin𝑥𝑑𝑥=𝜋0sin𝑥𝑑𝑥=2(5.4) is satisfied.

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