Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2014, Article ID 482740, 6 pages http://dx.doi.org/10.1155/2014/482740
Research Article Quasilinear Elliptic Equations with Hardy-Sobolev Critical Exponents: Existence and Multiplicity of Nontrivial Solutions Guanwei Chen School of Mathematics and Statistics, Anyang Normal University, Anyang, Henan 455000, China Correspondence should be addressed to Guanwei Chen;
[email protected] Received 30 November 2013; Accepted 12 January 2014; Published 19 February 2014 Academic Editor: Tai-Ping Chang Copyright Β© 2014 Guanwei Chen. 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. We study the existence of positive solutions and multiplicity of nontrivial solutions for a class of quasilinear elliptic equations by using variational methods. Our obtained results extend some existing ones.
In the case where π = 0 and π = 1 hold, then (1) reduces to the quasilinear elliptic problem:
1. Introduction and Main Results Let us consider the following problem:
βΞπ π’ β |π’|π
β
|π’|π (π )β2 βΞ π π’ β π π’ = π (π₯, π’) , |π₯|π π’ = 0,
π₯ β Ξ© \ {0} ,
(1)
π₯ β πΞ©,
where Ξ π π’ = div(|βπ’|πβ2 βπ’) denotes the π-Laplacian differential operator, Ξ© is an open bounded domain in Rπ (π β₯ 3) with smooth boundary πΞ© and 0 β Ξ©, 0 β€ π < π, πβ (π ) = π(π β π )/(π β π) is the Hardy-Sobolev critical exponent, and πβ = πβ (0) = ππ/(π β π) is the Sobolev critical exponent. Here, we let βπ’β := (β« |βπ’|π ππ₯)
1/π
Ξ©
,
(2)
which is equivalent to the usual norm of Sobolev space 1,π π0 (Ξ©) due to the PoincarΒ΄e inequality. Let π΄ π (Ξ©) :=
inf 1,π
βπ’βπ πβ (π )
π’βπ0 (Ξ©)\{0} (β« (|π’| Ξ©
/|π₯|π )ππ₯)
which is the best Hardy-Sobolev constant.
π/πβ (π )
,
(3)
β
β2
π’ = π (π₯, π’) ,
π’ = 0,
π₯ β πΞ©.
π₯ β Ξ©,
(4)
GoncΒΈalves and Alves [1] have studied (4) in Rπ involving π(π₯, π’) = β(π₯)π’π , π’ β₯ 0 and π’ β‘ΜΈ 0 to obtain existence of positive solutions where 2 β€ π < π, 0 < π < π β 1, or π β 1 < π < πβ β 1 and a suitable β. We should mention that problem (4) with π = 2 has been widely studied since BrΒ΄ezis and Nirenberg; see [2β4] and the references therein. Ghoussoub and Yuan [5] have studied (1) with π(π₯, π’) = π|π’|πβ2 π’, where π < π < πβ . They obtained a positive solution in the case where π > 0, π > 0, and π > [π(π β 1)π + π2 ]/[π + (π β 1)(π β π)] (in particular if π β₯ π2 ) hold. They also obtained a sign-changing solution in the case where π > 0, π > 0, and π > [π(π β 1)π + π]/[1 + (π β 1)(π β π)] (in particular if π > π3 β π2 + π) hold. For other relevant papers, see [6β12] and the references therein. We should mention that the energy functional associated 1,π with (1) is defined on π0 (Ξ©), which is not a Hilbert space for π =ΜΈ 2. Due to the lack of compactness of the embedding β β 1,π 1,π in π0 (Ξ©) σ³¨
β πΏπ (Ξ©) and π0 (Ξ©) σ³¨
β πΏπ (π ) (Ξ©, |π₯|βπ ππ₯), we cannot use the standard variational argument directly. The corresponding energy functional fails to satisfy the classical 1,π Palais-Smale ((PS) for short) condition in π0 (Ξ©). However,
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a local (PS) condition can be established in a suitable range. Then the existence result is obtained via constructing a minimax level within this range and the Mountain Pass Lemma due to Ambrosetti et al. [2] and Rabinowitz [13]. In this paper, we study (1) with a general nonlinearity by using a variational method; besides, we also considerably generalize the results obtained in [5]. In what follows, we always assume that the nonlinearity π satisfies π(π₯, 0) β‘ 0. π‘ Let πΉ(π₯, π‘) := β«0 π(π₯, π )ππ , π₯ β Ξ©. To state our main results, we still need the following assumptions. (π΄ 1 ) π β πΆ(Ξ© Γ R+ , R), limπ‘ β 0+ (π(π₯, π‘)/π‘πβ1 ) = 0, and β limπ‘ β β (π(π₯, π‘)/π‘π β1 ) = 0 uniformly for π₯ β Ξ©. (π΄ 2 ) There exists a constant π with π > π such that 0 < ππΉ (π₯, π‘) β€ π (π₯, π‘) π‘,
βπ₯ β Ξ©, βπ‘ β R+ \ {0} .
The rest of this paper is organized as follows. In Section 2, we give some preliminary lemmas, which are useful in the proofs of our main results. In Section 3, we give the detailed proofs of our main results.
2. Preliminaries In what follows, we let β β
βπ denote the norm in πΏπ (Ξ©). It is obvious that the values of π(π₯, π‘) for π‘ < 0 are irrelevant in Theorem 1, so we may define π (π₯, π‘) β‘ 0
βπ₯ β Ξ©, βπ‘ β R \ {0} .
(5)
(6)
πβ (π )β1
(π’+ ) βΞ π π’ = π |π₯|π
(10)
π₯ β πΞ©.
πβ (π )
π 1 ) , πβ β } π πβ1
(π’+ ) π 1 πΌ (π’) = β« |βπ’|π ππ₯ β β β« π Ξ© π (π ) Ξ© |π₯|π β β« πΉ (π₯, π’) ππ₯, Ξ©
(7)
hold, then (1) has at least one positive solution. Theorem 2. Suppose that π β₯ 3, 0 < π < β, 0 β€ π < π, and 1 < π < π hold. If (π΄ 3 ), (π΄ 4 ), and (7) hold, then (1) has at least two distinct nontrivial solutions. Noting that π > π and π2 β€ π imply that (7) holds, therefore, we have the following corollaries. Corollary 3. Suppose that 0 < π < β, 0 β€ π < π, π2 β€ π, and 1 < π < π. Moreover, (π΄ 1 ) and (π΄ 2 ) hold; then (1) has at least one positive solution. Corollary 4. Suppose that 0 < π < β, 0 β€ π < π, π β€ π, and 1 < π < π. Moreover, (π΄ 3 ) and (π΄ 4 ) hold; then (1) has at least two distinct nontrivial solutions. Remark 5. Theorem 1 generalizes Theorem 1.3 in [5], where the author only studied the special situation that π(π₯, π’) = π|π’|πβ2 π’, π < π < πβ . There are functions π satisfying the assumptions of our Theorem 1 and not satisfying those in [5]. Let (π₯, π‘) β Ξ© Γ R,
(8)
ππ₯ (11)
1,π
π’ β π0 (Ξ©) .
By Hardy-Sobolev inequalities (see [5, 14]) and (π΄ 1 ), we know 1,π πΌ β πΆ1 (π0 (Ξ©), R). Now it is well known that there exists a one-to-one correspondence between the weak solutions of 1,π (10) and the critical points of πΌ on π0 (Ξ©). More precisely 1,π we say that π’ β π0 (Ξ©) is a weak solution of (10), if, for any 1,π V β π0 (Ξ©), there holds σΈ
πβ2
β¨πΌ (π’) , Vβ© = β« |βπ’| Ξ©
πβ (π )β1
(π’+ ) (βπ’, βV) ππ₯ β π β« |π₯|π Ξ©
Vππ₯
β β« π (π₯, π’) Vππ₯ = 0. Ξ©
2
β
π₯ β Ξ© \ {0} ,
The energy functional corresponding to (10) is given by
Theorem 1. Suppose that π β₯ 3, 0 < π < β, 0 β€ π < π, and 1 < π < π hold. If (π΄ 1 ), (π΄ 2 ), and
π (π₯, π‘) := π (π₯) |π‘|πβ2 π‘ + π|π‘|πβ2 π‘,
+ π (π₯, π’) ,
π’ = 0,
Now, our main results read as follows.
π > max {π, πβ (1 β
(9)
We firstly consider the existence of nontrivial solutions to the problem:
(π΄ 3 ) π β πΆ(Ξ© Γ R, R), limπ‘ β 0 (π(π₯, π‘)/|π‘|πβ1 ) = 0, and β lim|π‘| β β (π(π₯, π‘)/|π‘|π β1 ) = 0 uniformly for π₯ β Ξ©. (π΄ 4 ) There exists a constant π with π > π such that 0 < ππΉ (π₯, π‘) β€ π (π₯, π‘) π‘,
for π₯ β Ξ©, π‘ β€ 0.
(12)
Lemma 6 (see [15]). If ππ β π a.e. in Ξ© and βππ βπ β€ πΆ < β for all n and some 0 < π < β, then σ΅© σ΅©π σ΅© σ΅©π σ΅© σ΅©π lim (σ΅©σ΅©σ΅©ππ σ΅©σ΅©σ΅©π β σ΅©σ΅©σ΅©ππ β πσ΅©σ΅©σ΅©π ) = σ΅©σ΅©σ΅©πσ΅©σ΅©σ΅©π .
πββ
(13)
Lemma 7. For any π > 0, 0 β€ π β€ 1, and π β₯ 1, we have (π + π)π β€ ππ + π(π + 1)πβ1 π. Proof. Let
β
where π(π₯) > 0, π β πΏ (Ξ©), π > 0, and π < π < π < π . Obviously, π satisfies all the conditions of Theorem 1 in this paper, while it does not satisfy the conditions of Theorem 1.3 in [5].
β (π₯) := (π + π₯)π β ππ β π(π + 1)πβ1 π₯. Clearly, βσΈ (π₯) < 0, π₯ β (0, 1), so β(π) β€ β(0) = 0.
(14)
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3 πβ
Lemma 8 (see [5]). If 0 β€ π < π hold, then we have (i) π΄ π (Ξ©) is independent of Ξ©, and will henceforth be denoted by π΄ π ;
By the continuity of embedding, we have βπ’π βπβ β€ πΆ1 < β. From [5], going if necessary to a subsequence, one can get that βπ’π σ³¨β βπ’ a.e. in Ξ©, π’π π’ β π₯ π₯
(ii) π΄ π (Ξ©) is attained when Ξ© = Rπ by the functions (πβπ)/π(πβπ )
π β π πβ1 ππ (π₯) = (π (π β π ) ( ) ) πβ1
(15)
weakly in πΏπ (Ξ©) ,
β σ΅¨σ΅¨ σ΅¨σ΅¨πβ (π )β2 π’ σ΅¨π’ σ΅¨ |π’|π (π )β2 π’ Vππ₯, β« σ΅¨ π σ΅¨ π π Vππ₯ σ³¨β β« |π₯| |π₯|π Ξ© Ξ©
1,π
βV β π0 (Ξ©) (20)
(πβπ )/(πβ1) (πβπ)/(πβπ )
Γ (π + |π₯|
)
for some π > 0. Moreover, the functions ππ (π₯) are the only positive radial solutions of β
π’π (π )β1 βΞ π π’ = |π₯|π
(16)
β«
Rπ
σ΅¨σ΅¨ σ΅¨σ΅¨π σ΅¨σ΅¨βππ σ΅¨σ΅¨ ππ₯ = β« Rπ
β 1 σ΅¨ σ΅¨σ΅¨ π|π‘|π σ΅¨σ΅¨π (π₯, π‘) π‘σ΅¨σ΅¨σ΅¨ β€ π (π) + 2πΆ1
for (π₯, π‘) β Ξ© Γ (0, +β) . (21)
Set πΏ := π/2π(π) > 0. When πΈ β Ξ©, meas(πΈ) < πΏ, we get σ΅¨ σ΅¨σ΅¨ β σ΅¨σ΅¨β« π (π₯, π’ ) π’ ππ₯σ΅¨σ΅¨σ΅¨ β€ β« π (π) ππ₯ + π β« σ΅¨σ΅¨σ΅¨π’ σ΅¨σ΅¨σ΅¨π ππ₯ σ΅¨σ΅¨ σ΅¨σ΅¨ π π πσ΅¨ σ΅¨ 2πΆ1 πΈ σ΅¨ σ΅¨ πΈ πΈ
in Rπ, and satisfy σ΅¨σ΅¨ σ΅¨σ΅¨πβ (π ) σ΅¨σ΅¨ππ σ΅¨σ΅¨ (πβπ )/(πβπ ) ππ₯ = (π΄ π (Ξ©)) . |π₯|π
as π β β. By (π΄ 1 ), we know that for any π > 0 there exists π(π) > 0 such that
(22)
< π σ³¨β 0 as meas (πΈ) σ³¨β 0.
(17)
It follows from Vitaliβs theorem that Lemma 9. If (π΄ 1 ), (π΄ 2 ), and 0 < π < ((π β π )/π(π β π ))π΄(πβπ )/(πβπ ) π(πβπ)/(πβπ ) hold, then πΌ satisfies (PS)π condition. π 1,π
Proof. Suppose that {π’π } is a (PS)π sequence in π0 (Ξ©). By (π΄ 2 ), we have
Ξ©
1 σΈ β¨πΌ (π’π ) , π’π β© π
as π σ³¨β β. (23)
Similarly, we can also get
Ξ©
Ξ©
as π σ³¨β β.
(24)
Since πΌσΈ (π’π ) β 0, we have πβ (π )
(π’π+ ) 1 1 1 σ΅© σ΅©π 1 = ( β ) σ΅©σ΅©σ΅©π’π σ΅©σ΅©σ΅© + ( β β ) π β« π π π π (π ) |π₯|π Ξ©
ππ₯
1 β β« (πΉ (π₯, π’π ) β π (π₯, π’π ) π’π ) ππ₯ π Ξ© β₯(
Ξ©
β« πΉ (π₯, π’π ) ππ₯ σ³¨β β« πΉ (π₯, π’) ππ₯
σ΅© σ΅© π + 1 + π (1) σ΅©σ΅©σ΅©π’π σ΅©σ΅©σ΅© β₯ πΌ (π’π ) β
β« π (π₯, π’π ) π’π ππ₯ σ³¨β β« π (π₯, π’) π’ ππ₯
1 1 σ΅©σ΅© σ΅©σ΅©π β ) σ΅©π’ σ΅© , π π σ΅© πσ΅©
β¨πΌσΈ (π’π ) , π’π β© πβ (π )
(π’π+ ) σ΅© σ΅©π = σ΅©σ΅©σ΅©π’π σ΅©σ΅©σ΅© β π β« |π₯|π Ξ©
ππ₯ β β« π (π₯, π’π ) π’π ππ₯ = π (1) . Ξ©
(25)
Let Vπ := π’π β π’, which together with Lemma 6 implies (18)
β
where π = min{π, π (π )}. Hence we conclude that {π’π } is a 1,π 1,π bounded sequence in π0 (Ξ©). So there exists π’ β π0 (Ξ©); going if necessary to a subsequence, we have
(1 < πΎ < πβ ) ,
π’π σ³¨β π’ a.e. in Ξ©, π σ³¨β β.
πβ (π )
(π’+ ) ππ₯ β π β« |π₯|π Ξ©
ππ₯ (26)
β β« π (π₯, π’) π’ ππ₯ = π (1) . Ξ©
From (20), we can obtain
1,π
π’π β π’ in π0 (Ξ©) , π’π σ³¨β π’ in πΏπΎ (Ξ©)
πβ (π )
(Vπ+ ) σ΅©σ΅© σ΅©σ΅©π π σ΅©σ΅©Vπ σ΅©σ΅© + βπ’β β π β« |π₯|π Ξ©
(19)
πβ (π )
(π’+ ) lim β¨πΌ (π’π ) , π’β© = βπ’β β π β« πββ |π₯|π Ξ© σΈ
π
ππ₯
β β« π (π₯, π’) π’ ππ₯ = 0. Ξ©
(27)
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Note that πΌ(π’π ) β π as π β β, which together with Lemma 6 implies πβ (π )
(Vπ+ ) π 1 σ΅© σ΅©π 1 πΌ (π’π ) = σ΅©σ΅©σ΅©Vπ σ΅©σ΅©σ΅© + βπ’βπ β β β« π π π (π ) Ξ© |π₯|π β
β + π (π )
(π’ ) π β« (π ) Ξ© |π₯|π
πβ
β
so that β«Ξ© (|Vπ |π (π ) /|π₯|π ) = 1. Then, by using the argument as used in [5], we can get the following results: σ΅© σ΅©π π΄ π + πΆ2 π(πβπ)/(πβπ ) β€ σ΅©σ΅©σ΅©Vπ σ΅©σ΅©σ΅© β€ π΄ π + πΆ3 π(πβπ)/(πβπ ) ,
ππ₯
ππ₯ β β« πΉ (π₯, π’) ππ₯ + π (1)
σ΅¨ σ΅¨π πΆ4 π(πβπ)π/π(πβπ ) β€ β« σ΅¨σ΅¨σ΅¨Vπ σ΅¨σ΅¨σ΅¨ ππ₯ β€ πΆ5 π(πβπ)π/π(πβπ ) , Ξ©
Ξ©
1 β€ π < πβ (1 β
β
π (π )
= πΌ (π’) +
(Vπ+ ) π 1 σ΅©σ΅© σ΅©σ΅©π β« σ΅©σ΅©Vπ σ΅©σ΅© β β π π (π ) Ξ© |π₯|π
ππ₯ + π (1)
= π + π (1) .
π = πβ (1 β
Therefore, one gets that
1 ), π
πΆ4 π((πβ1)/(πβπ ))(πβπ(πβπ)/π)
πβ (π )
ππ₯ = π + π (1) .
(29)
σ΅¨ σ΅¨π β€ β« σ΅¨σ΅¨σ΅¨Vπ σ΅¨σ΅¨σ΅¨ ππ₯ β€ πΆ5 π((πβ1)/(πβπ ))(πβπ(πβπ)/π) , Ξ©
From (26) and (27), we have
πβ (1 β
πβ (π )
(Vπ+ ) σ΅©σ΅© σ΅©σ΅©π σ΅©σ΅©Vπ σ΅©σ΅© β π β« |π₯|π Ξ©
1 ), π
σ΅¨ σ΅¨π πΆ4 π(πβπ)π/π(πβπ ) |ln π| β€ β« σ΅¨σ΅¨σ΅¨Vπ σ΅¨σ΅¨σ΅¨ ππ₯ β€ πΆ5 π(πβπ)π/π(πβπ ) |ln π| , Ξ©
(28) (Vπ+ ) π 1 σ΅© σ΅©π πΌ (π’) + σ΅©σ΅©σ΅©Vπ σ΅©σ΅©σ΅© β β β« π π (π ) Ξ© |π₯|π
(36)
(30)
ππ₯ = π (1) ;
1 ) < π < πβ . π (37)
π
then βVπ β β 0 as π β β. Otherwise, there exists a subsequence (still denoted by Vπ ) such that πβ (π )
(Vπ+ ) lim π β« πββ |π₯|π Ξ©
σ΅© σ΅©π lim σ΅©σ΅©V σ΅©σ΅© = π, π β βσ΅© π σ΅© By (3), we have σ΅©σ΅© σ΅©σ΅©π σ΅©σ΅©Vπ σ΅©σ΅© β₯ π΄ π (β«
Ξ©
|π₯|
π/πβ (π )
βπ β N;
(32)
π΄(πβπ )/(πβπ ) π(πβπ)/(πβπ ) . π
In the following, we shall give some estimates for the extremal functions. Let π β π πβ1 ) ] πβ1
ππ (π₯) :=
(πβπ)/π(πβπ )
, (34)
ππ (π₯) . πΆπ
Define a function π β πΆ0β (Ξ©) such that π(π₯) = 1 for |π₯| β€ π
, π(π₯) = 0 for |π₯| β₯ 2π
, 0 β€ π(π₯) β€ 1, where π΅2π
(0) β Ξ©. Set π’π (π₯) = π (π₯) ππ (π₯) , Vπ (π₯) =
π’π (π₯)
σ΅¨ σ΅¨πβ (π ) (β«Ξ© (σ΅¨σ΅¨σ΅¨π’π σ΅¨σ΅¨σ΅¨ /|π₯|π ))
for π σ³¨β 0+ .
(38)
Lemma 10. Suppose that 0 β€ π < π. If (π΄ 1 ), (π΄ 2 ) and (7) 1,π hold, then there exists π’0 β π0 (Ξ©), π’0 β‘ΜΈ 0, such that sup πΌ (π‘π’0 )
0. (31)
ππ₯ = π,
π/πβ (π ) πβ (π ) (Vπ+ ) ππ₯) , π
Moreover, by using the Sobolev embedding theorem and (36), one can deduce
π‘β₯0
πβπ π΄(πβπ )/(πβπ ) π(πβπ)/(πβπ ) . π (π β π ) π
(39)
Proof. We consider the functions β
π (π‘) = πΌ (π‘Vπ ) =
π‘π σ΅©σ΅© σ΅©σ΅©π π‘π (π ) β β« πΉ (π₯, π‘Vπ ) ππ₯, σ΅©σ΅©Vπ σ΅©σ΅© β π β π π (π ) Ξ© β
π‘π σ΅© σ΅©π π‘π (π ) π (π‘) = σ΅©σ΅©σ΅©Vπ σ΅©σ΅©σ΅© β π β . π π (π )
(40)
Since limπ‘ β β π(π‘) = ββ, π(0) = 0, and π(π‘) > 0 for π‘ small enough, supπ‘β₯0 π(π‘) is attained for some π‘π > 0. Therefore, we have 0 = πσΈ (π‘π ) β 1 σ΅© σ΅©π = π‘ππβ1 (σ΅©σ΅©σ΅©Vπ σ΅©σ΅©σ΅© β ππ‘ππ (π )βπ β πβ1 β« π (π₯, π‘π Vπ ) Vπ ππ₯) , Ξ© π‘π
(41) and hence
1/πβ (π )
,
(35)
β 1 σ΅©σ΅© σ΅©σ΅©π πβ (π )βπ + πβ1 β« π (π₯, π‘π Vπ ) Vπ ππ₯ β₯ ππ‘ππ (π )βπ . σ΅©σ΅©Vπ σ΅©σ΅© = ππ‘π Ξ© π‘π (42)
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3. Proofs of Main Results
Therefore, we obtain σ΅©σ΅© σ΅©σ΅©π 1/(π σ΅©V σ΅© π‘π β€ ( σ΅© π σ΅© ) π
β
1,π
(π )βπ)
:=
π‘π0 .
(43)
Proof of Theorem 1. Let π := π0 (Ξ©). From the Sobolev and Hardy-Sobolev inequalities, we can easily get β
βπ’βππ β€ πΆβπ’βπ ,
By (π΄ 1 ), we can easily get σ΅¨ σ΅¨σ΅¨ πβ β1 + π (π) π‘πβ1 , σ΅¨σ΅¨π (π₯, π‘)σ΅¨σ΅¨σ΅¨ β€ ππ‘
π (π) > 0.
β«
Ξ©
|π’|π (π ) πβ (π ) , π ππ₯ β€ πΆβπ’β |π₯|
(44)
πβ
Hence, we can get β β σ΅©σ΅© σ΅©σ΅©π πβ (π )βπ + π β« |π‘π |π βπ |Vπ |π ππ₯ + π (π) β« |Vπ |π ππ₯. σ΅©σ΅©Vπ σ΅©σ΅© β€ ππ‘π Ξ© Ξ© (45)
By (36)β(38), when π is small enough, we conclude that β π‘ππ (π )βπ
π΄ β₯ π . 2
σ΅©σ΅© σ΅©σ΅©π(πβπ )/(πβπ ) β€ π΄(πβπ )/(πβπ ) + πΆ7 π(πβπ )/(πβπ ) . σ΅©σ΅©Vπ σ΅©σ΅© π
It follows from (π΄ 1 ) that
On the other hand, the function π(π‘) attains its maximum at π‘π0 and is increasing in the interval [0, π‘π0 ]. Note that (π΄ 2 ) implies πΉ(π₯, π‘) β₯ πΆ8 |π‘|π , which together with (36), (46), and (47) implies that
βπ‘ β (0, πΏ2 )
for all π‘ β R+ and for π₯ β Ξ©. Then one gets
πβπ π΄(πβπ )/(πβπ ) π(πβπ)/(πβπ ) π (π β π ) π
β πΆ8 (
π΄π ) 2
(48)
(π’+ ) π 1 πΌ (π’) = βπ’βπ β β β« π π (π ) Ξ© |π₯|π β₯
β πΆπ σ΅© σ΅©πβ (π ) πΆ 1 β πβπ’βπ β πΆπΆ10 βπ’βπ βπ’βπ β β σ΅©σ΅©σ΅©π’+ σ΅©σ΅©σ΅© π π (π ) π (56)
1,π
π > 0 small enough.
Furthermore, from (7) and (37), we get
(50)
Therefore, by choosing π small enough, we have sup πΌ (π‘Vπ ) = π (π‘π ) < π‘β₯0
(57)
1,π
(49)
By (7), we have π > πβ β π/(π β 1), which implies π (π β π) πβπ πβ1 > (π β ). πβπ πβπ π
Ξ©
πΌ (π’) β₯ π½ βπ’ β ππ΅π = {π’ β π0 (Ξ©) , βπ’β = π} ,
Ξ©
Ξ©
ππ₯ β β« πΉ (π₯, π’) ππ₯
for π small enough. So there exists π½ > 0 such that
β« |Vπ |π ππ₯.
β« |Vπ |π ππ₯ β₯ πΆ4 π((πβ1)/(πβπ ))(πβπ(πβπ)/π) .
(55)
πβ (π )
+ πΆ9 π(πβπ)/(πβπ ) π(πβπ)/(πβπ ) π/(πβ (π )βπ)
β 1 π π|π‘| + πΆ10 |π‘|π π
for all π‘ β R and for π₯ β Ξ©. By (52) and (55) we have
β β« πΉ (π₯, π‘π Vπ ) ππ₯ β€
βπ‘ β [πΏ2 , πΏ1 ]
β 1βπβ σ΅¨ πβ β1 σ΅¨σ΅¨ + ππ‘πβ1 + π β€ ππ‘πβ1 + (1 + ππΏ2 ) π‘π β1 σ΅¨σ΅¨π (π₯, π‘)σ΅¨σ΅¨σ΅¨ β€ π‘ (54)
|πΉ (π₯, π‘)| β€
π β π σ΅©σ΅© σ΅©σ΅©π(πβπ )/(πβπ ) (πβπ)/(πβπ ) π σ΅©V σ΅© π (π β π ) σ΅© π σ΅©
(53)
uniformly for all π₯ β Ξ©. Therefore, we deduce that
Ξ©
Ξ©
σ΅¨ σ΅¨ s.t. σ΅¨σ΅¨σ΅¨π (π‘)σ΅¨σ΅¨σ΅¨ β€ π,
βπ > 0,
π (π‘π ) β€ π (π‘π0 ) β β« πΉ (π₯, π‘π Vπ ) ππ₯ =
β σ΅¨ σ΅¨ s.t. σ΅¨σ΅¨σ΅¨π (π‘)σ΅¨σ΅¨σ΅¨ < π‘π β1 , βπ‘ > πΏ1 ; σ΅¨ σ΅¨ β0 < πΏ2 < πΏ1 , s.t. σ΅¨σ΅¨σ΅¨π (π‘)σ΅¨σ΅¨σ΅¨ < ππ‘πβ1 ,
βπΏ1 > 0,
(46)
(47)
(52)
βπ’ β π.
βπ > 0,
On the one hand, from Lemma 7 and (36), it follows that
β
βπ’βπβ β€ πΆβπ’βπ ,
πβπ π΄(πβπ )/(πβπ ) π(πβπ)/(πβπ ) . π (π β π ) π (51)
Hence, the proof of the lemma is completed by taking π’0 = Vπ .
By Lemma 10, there exists π’0 β π0 (Ξ©) with π’0 β‘ΜΈ 0 such that πβπ π΄(πβπ )/(πβπ ) π(πβπ)/(πβπ ) . sup πΌ (π‘π’0 ) < (58) π (π β π ) π π‘β₯0 It follows from the nonnegativity of πΉ(π₯, π‘) that πβ (π )
β (π’0+ ) π 1 σ΅© σ΅©π πΌ (π‘π’0 ) = π‘π σ΅©σ΅©σ΅©π’0 σ΅©σ΅©σ΅© β β π‘π (π ) β« π π (π ) |π₯|π Ξ©
ππ₯
β β« πΉ (π₯, π‘π’0 ) ππ₯
(59)
Ξ©
πβ (π )
β (π’0+ ) π 1 σ΅© σ΅©π β€ π‘π σ΅©σ΅©σ΅©π’0 σ΅©σ΅©σ΅© β β π‘π (π ) β« π π (π ) |π₯|π Ξ©
ππ₯.
6
Journal of Applied Mathematics
Therefore, limπ‘ β +β πΌ(π‘π’0 ) β ββ, so we can choose π‘0 > 0 such that σ΅© σ΅©σ΅© πΌ (π‘0 π’0 ) β€ 0. (60) σ΅©σ΅©π‘0 π’0 σ΅©σ΅©σ΅© > π, By virtue of the Mountain Pass Lemma in [16], there is a sequence {π’π } β π satisfying πΌσΈ (π’π ) σ³¨β 0,
πΌ (π’π ) σ³¨β π β₯ π½,
(61)
where π = inf max πΌ (β (π‘)) , ββΞ π‘β[0,1]
(62)
Ξ = {β β πΆ ([0, 1] , π) | β (0) = 0, β (1) = π‘0 π’0 } . Note that 0 < π½ β€ π = inf max πΌ (β (π‘)) β€ max πΌ (π‘π‘0 π’0 ) β€ sup πΌ (π‘π’0 ) ββΞ π‘β[0,1]