research article existence of multiple solutions for a...

Hindawi Publishing CorporationDiscrete Dynamics in Nature and SocietyVolume 2013 Article ID 809262 5 pageshttpdxdoiorg1011552013809262

Research ArticleExistence of Multiple Solutions for a Class ofBiharmonic Equations

Chunhan Liu and Jianguo Wang

Department of Mathematics Qilu Normal University Jinan 250013 China

Correspondence should be addressed to Chunhan Liu chh-liu1981163com

Received 17 July 2013 Revised 28 October 2013 Accepted 11 November 2013

Academic Editor Cengiz Cinar

Copyright copy 2013 C Liu and J Wang This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

By a symmetric Mountain Pass Theorem a class of biharmonic equations with Navier type boundary value at the resonant andnonresonant case are discussed and infinitely many solutions of the equations are obtained

1 Introduction and Main Results

In this paper we study the following fourth-order ellipticequation

Δ2119906 + 119888Δ119906 = 120583ℎ (119909) |119906|

119901minus2119906 + 119891 (119909 119906) 119909 isin Ω

119906 = Δ119906 = 0 119909 isin 120597Ω

(1)

where Δ2 is the biharmonic operator 119888 is a constantΩ sub 119877

119873

is a bounded smooth domain 1 lt 119901 lt 2 120583 ge 0 is a parameterℎ isin 119871

infin(Ω) ℎ(119909) ge 0 ℎ(119909) equiv 0 119891(119909 119904) is a continuous

function onΩ times 119877This fourth-order semilinear elliptic problem can be con-

sidered as an analogue of a class of second-order problemswhich have been studied by many authors A main tool ofseeking solutions of the problem is the Mountain Pass The-orem (see [1ndash3]) In [4] Pei studied the following problem

Δ2119906 + 119888Δ119906 = 119891 (119909 119906) 119909 isin Ω

119906 = Δ119906 = 0 119909 isin 120597Ω

(2)

and obtained at least three nontrivial solutions by using theminimax method and Morse theory

Problem (1) has been studied extensively in recent yearswe refer the reader to [5ndash11] and the references therein In [12]the author showed that the problem (1) admits at least three(or four or five) nontrivial solutions by using the minimaxmethod and Mountain Pass Theory

However to the best of authorrsquos knowledge there havebeen very few results dealingwith (1) using a symmetricMou-ntain PassTheoremThis paper will make some contributionin the research field In this paper we study the problem (1)by a symmetric Mountain Pass Theorem at the resonant andnonresonant case and obtain infinitely many solutions of theequation

Consider eigenvalue problem

minusΔ119906 = 120582119906 119909 isin Ω

119906 = 0 119909 isin 120597Ω

(3)

Let us denote that 120582119896(119896 isin 119873) are the eigenvalues and 120593

119896(119896 isin

119873) are the corresponding eigenfunctions of the eigenvalueproblem (3) It is well known that 0 lt 120582

1lt 1205822le 1205823le sdot sdot sdot le

120582119896

rarr +infin and the first eigenfunction 1205931gt 0 119909 isin Ω

It is easy to see that 120582119896(120582119896minus119888) 119896 = 1 2 are eigenvalues

of the problem

Δ2119906 + 119888Δ119906 = Λ119906 119909 isin Ω

119906 = Δ119906 = 0 119909 isin 120597Ω

(4)

and 120593119896(119896 = 1 2 ) are still the corresponding eigenfunc-

tionsLet 119867 = 119867

2(Ω) cap 119867

1

0(Ω) be the Hilbert space equipped

with the inner product

⟨119906 V⟩119867

= intΩ

(Δ119906ΔV + nabla119906nablaV) d119909 (5)

2 Discrete Dynamics in Nature and Society

and the deduced norm

119906119867 = [intΩ

(|Δ119906|2+ |nabla119906|

2) d119909]12

(6)

Suppose that 119888 lt 1205821 Let us define a norm of the space 119867

as follows

119906 = [intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909]12

(7)

It is easy to verify that the norm sdot is equivalent to thenorm sdot

119867on 119867 and for all 119906 isin 119867 the following Poincare

inequality holds

1199062ge 1205821(1205821minus 119888) |119906|

2

2 (8)

where |119906|2

2= intΩ|119906|2119889119909

Throughout this paper the weak solutions of (1) are thecritical points of the associated functional

Φ120583(119906)

=1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus120583

119901intΩ

ℎ (119909) |119906|119901d119909 minus int

Ω

119865 (119909 119906) d119909 forall 119906 isin 119867

(9)

where 119865(119909 119906) = int119906

0119891(119909 119905) 119889119905

ObviouslyΦ120583isin 1198621(119867 119877) and for all 119906 120593 isin 119867

⟨Φ1015840

120583(119906) 120593⟩

= intΩ

(Δ119906Δ120593 minus 119888nabla119906nabla120593 d119909)

minus 120583intΩ

ℎ (119909) |119906|119901minus2

119906120593 d119909 minus intΩ

119891 (119909 119906) 120593 d119909

(10)

In order to establish solutions for problem (1) we make thefollowing assumptions

(1198650) ℎ isin 119871

infin(Ω) ℎ(119909) ge 0

(1198651) 119891(119909 0) = 0 119891(119909 minus119905) = minus119891(119909 119905) for all 119909 isin Ω

119905 isin 119877(1198652) lim|119905|rarr0

119891(119909 119905)119905 = 120572 lim|119905|rarrinfin

119891(119909 119905)119905 = 120573

uniformly for ae 119909 isin Ω where 0 le 120572 lt 120582119896(120582119896minus 119888) lt

120573 or 120573 = 120582119896(120582119896minus 119888)

(1198653) lim|119905|rarrinfin

(119891(119909 119905)119905minus2119865(119909 119905)) = minusinfin uniformly in119909 isin Ω

Definition 1 (see [13]) Let Φ isin 1198621(119883 119877) we say that Φ sati-

sfies the Cerami condition at the level 119888 isin 119877 if any sequence119906119899 sub 119883 with

Φ(119906119899) 997888rarr 119888 (1 +

10038171003817100381710038171199061198991003817100381710038171003817)Φ1015840(119906119899) 997888rarr 0 as 119899 997888rarr infin

(11)

possesses a convergent subsequence Φ satisfies the (119862)

condition ifΦ satisfies (119862)119888for all 119888 isin 119877

Lemma 2 Let119883119896= span120593

1 1205932 120593

119896119867 = 119883

119896oplus119883perp

119896 then

1199062le 120582119896(120582119896minus 119888) |119906|

2

2 forall 119906 isin 119883

119896

1199062ge 120582119896+1

(120582119896+1

minus 119888) |119906|2

2 forall 119906 isin 119883

perp

119896 119896 ge 2

(12)

Proof It is similar to the proof of Lemma 25 in [13]

Theorem 3 (see [14] a symmetric Mountain Pass Theorem)Suppose that Φ isin 119862

1(119864 119877) is evenΦ(120579) = 0 and

(i) there exist 120588 120572 gt 0 and a finite dimensional linearsubspace 119885 such that

Φ|119885perpcap120597119861120588(120579)

ge 120572 (13)

(ii) there exist a sequence of linear subspace 119885119898

dim(119885119898) = 119898 and 119877

119898gt 0 such that

Φ (119909) le 0 forall 119909 isin 119885119898

119861119877119898

119898 = 1 2 (14)

If Φ satisfies (119875119878) condition then Φ possesses infinitely manydistinct critical points corresponding to positive critical values

Remark 4 If Φ satisfies the (119862) condition Theorem 3 stillholds

The main results of this paper are as follows

Theorem 5 Assume that (1198650)minus(1198652) hold and 119888 lt 120582

1 120582119896(120582119896minus

119888) lt 120573 120573 is not an eigenvalue of (4) then there exist 120583lowast gt 0

such that for 120583 isin (0 120583lowast) (1) has infinitely many solutions

Theorem 6 Assume that (1198650)minus (1198653) hold and 119888 lt 120582

1 120573 =

120582119896(120582119896minus 119888) then there exist 120583lowast gt 0 such that for 120583 isin (0 120583

lowast) (1)

has infinitely many solutions

2 Proofs of Theorems

Proof of Theorem 5 (i) Assume that 119906119899 sub 119867 is a (119862) seq-

uence that is

(1 +1003817100381710038171003817119906119899

1003817100381710038171003817)Φ1015840

120583(119906119899) 997888rarr 0 Φ

120583(119906119899) 997888rarr 119888 (15)

We claim that 119906119899 is boundedAssume as a contradiction that

|119906119899|2

rarr infin Setting V119899= 119906119899|119906119899|2 then |V

119899|2= 1 Without

loss of generality we assume

V119899 V in 119867 V

119899997888rarr V in 119871

2(Ω)

V119899997888rarr V ae 119909 isin Ω

(16)

From (15) we know that100381610038161003816100381610038161003816100381610038161003816

intΩ

(ΔV119899Δ120593 minus 119888nablaV

119899nabla120593) d119909

minus120583intΩ

ℎ (119909)

10038161003816100381610038161199061198991003816100381610038161003816

119901minus2

119906119899

100381610038161003816100381611990611989910038161003816100381610038162

120593 d119909 minus intΩ

119891 (119909 119906119899)

100381610038161003816100381611990611989910038161003816100381610038162

120593 d1199091003816100381610038161003816100381610038161003816100381610038161003816

Discrete Dynamics in Nature and Society 3

=1

100381610038161003816100381611990611989910038161003816100381610038162

10038161003816100381610038161003816⟨Φ1015840

120583(119906119899) 120593⟩

10038161003816100381610038161003816

le1

100381610038161003816100381611990611989910038161003816100381610038162

10038171003817100381710038171003817Φ1015840

120583(119906119899)10038171003817100381710038171003817

10038171003817100381710038171205931003817100381710038171003817 997888rarr 0 forall 120593 isin 119867

(17)

Next we consider the two possible cases (a)V = 0 (b)V = 0In case (a) from (119865

2) we derive

10038161003816100381610038161199061198991003816100381610038161003816 =

1003816100381610038161003816V11989910038161003816100381610038161003816100381610038161003816119906119899

10038161003816100381610038162997888rarr infin lim

119899rarrinfin

119891 (119909 119906119899)

119906119899

= 120573 (18)

For V119899

rarr V in 1198712(Ω) we have

lim119899rarrinfin

119891 (119909 119906119899)

100381610038161003816100381611990611989910038161003816100381610038162

= lim119899rarrinfin

119891 (119909 119906119899)

119906119899

V119899= 120573V ae 119909 isin Ω

(19)

In case (b) we have

119891 (119909 119906119899)

100381610038161003816100381611990611989910038161003816100381610038162

le 1198881003816100381610038161003816V119899

1003816100381610038161003816 997888rarr 0 (20)

Then

lim119899rarrinfin

intΩ

119891 (119909 119906119899) 120593

100381610038161003816100381611990611989910038161003816100381610038162

d119909 = lim119899rarrinfin

intΩ

119891 (119909 119906119899) V119899

119906119899

120593 d119909

= intΩ

120573V120593 d119909 forall 120593 isin 119867

(21)

By (17) and lim119899rarrinfin

intΩℎ(119909)(|119906

119899|119901minus2

119906119899|119906119899|2)120593119889119909 = 0 we

have

intΩ

(ΔVΔ120593 minus 119888nablaVnabla120593) d119909

= intΩ

120573V120593d119909 forall 120593 isin 119867

(22)

We can easily see that V equiv 0 In fact if V equiv 0 then|V|2= 0 which contradicts lim

119899rarrinfin|V119899|2= |V|2= 1 Hence

120573 is an eigenvalue of the problem (4) this contradicts ourassumption Then 119906

119899 is bounded there exist a subsequence

of 119906119899(we can also denote by 119906

119899) and 119906 isin 119867 such that

119906119899 119906 in119867 By Lemma 2 and (15) we have

1003817100381710038171003817119906119899 minus 119906119898

1003817100381710038171003817 = 120583intΩ

ℎ (119909)1003816100381610038161003816119906119899 minus 119906

119898

1003816100381610038161003816

119901d119909

+ intΩ

119891 (119909 119906119899minus 119906119898) (119906119899minus 119906119898) d119909

+ 119900 (1)1003817100381710038171003817119906119899 minus 119906

119898

1003817100381710038171003817

le 120583|ℎ|infin

intΩ

1003816100381610038161003816119906119899 minus 119906119898

1003816100381610038161003816

119901d119909

+ intΩ

119891 (119909 119906119899minus 119906119898) (119906119899minus 119906119898) d119909

+ 119900 (1)1003817100381710038171003817119906119899 minus 119906

119898

1003817100381710038171003817

(23)

where 119900(1) rarr 0 and

1003816100381610038161003816100381610038161003816intΩ

119891 (119909 119906119899minus 119906119898) (119906119899minus 119906119898) d119909

1003816100381610038161003816100381610038161003816

le1003816100381610038161003816119891 (119909 119906

119899minus 119906119898)10038161003816100381610038162

1003816100381610038161003816119906119899 minus 119906119898

10038161003816100381610038162

(24)

as 119899119898 rarr infin Hence 119906119899

rarr 119906 in119867 We verifyΦ120583(119906) satisfies

(119862) condition(ii) There exists some 120588 120574 gt 0 such that Φ

120583|119883perp

119896cap120597119861120588(120579)

ge

120574 gt 0 where 119861120588(120579) = 119906 isin 119867 119906 le 120588

By (1198651) and (119865

2) taking 120590 isin (2 2

lowast) for any given 120576 gt 0

there exists 1198621gt 0 such that

119865 (119909 119906) le1

2(120572 + 120576) |119906|

2+ 1198621|119906|120590 forall 119909 isin Ω (25)

where

2lowast=

2119873

119873 minus 2 119873 gt 2

+infin 119873 le 2

(26)

Taking 120576 gt 0 such that 120572 + 120576 lt 120582119896(120582119896minus 119888) combining

Lemma 2 Poincare inequality and Sobolev embedding wehave

Φ120583(119906) ge

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus120583|ℎ|infin

119901intΩ

|119906|119901d119909 minus int

Ω

119865 (119909 119906) d119909

ge1

21199062minus

120572 + 120576

2|119906|2

2minus

120583|ℎ|infin

119901|119906|119901

119901minus 1198621|119906|120590

120590

ge1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888)) 1199062minus 1205831198622119906119901minus 1198623119906120590

= (1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888))

minus1205831198622119906119901minus2

minus 1198623119906120590minus2

) 1199062

(27)

where 1198622 1198623are constant

Let 119892(119905) = 1205831198622119905119901minus2

+ 1198623119905120590minus2 we claim that there exists 119905

0

such that

119892 (1199050) lt

1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888)) (28)


It is easy to see that 119892(119905) has a minimum at 1199050

= ((1205831198622(2 minus

119901))(1198623(120590 minus 2)))

1(120590minus119901) substituting 1199050in 119892(119905) we have

119892 (1199050) = 120583(120590minus2)(120590minus119901)

[(1198622(1198622(2 minus 119901)

1198623(120590 minus 2)

))

(119901minus2)(120590minus119901)

+(1198623(1198622(2 minus 119901)

1198623(120590 minus 2)

))

(120590minus2)(120590minus119901)

]

lt1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888))

(29)

where 0 lt 120583 lt 120583lowast

= [12(1 minus ((120572 + 120576)(120582119896+1

(120582119896+1

minus 119888))))

((1198622(1198622(2 minus 119901)119862

3(120590 minus 2)))

(119901minus2)(120590minus119901)+ (1198623(1198622(2 minus 119901)119862

3(120590 minus

2)))(120590minus2)(120590minus119901)

)](2minus120590)(120590minus119901)

We take 119906 = 120588 = 1199050gt 0 then there exists 120574 gt 0 such

thatΦ120583|119883perp

119896cap120597119861120588(120579)

ge 120574 gt 0 where 119861120588(120579) = 119906 isin 119867 119906 le 120588

(iii) There exists 119877119896gt 120588 such that

Φ120583(119906) le 0 forall 119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (30)

For 120573 gt 120582119896(120582119896minus119888) (119865

2) implies that for any 120576 gt 0 there exists


119865 (119909 119906) ge1

2(120573 minus 120576) |119906|

2minus 1198624 (31)

Taking 120576 gt 0 such that 120573 minus 120576 gt 120582119896(120582119896minus 119888) By Lemma 2 and

(120583119901) intΩℎ(119909)|119906|

119901d119909 ge 0 we have

Φ120583(119906) =

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus120583

119901intΩ

ℎ (119909) |119906|119901 d119909 minus int

Ω

119865 (119909 119906) d119909

le1

21199062minus

120573 minus 120576

2|119906|2

2+ 1198624 |Ω|

le1

2(1 minus

120573 minus 120576

120582119896(120582119896minus 119888)

) 1199062

+ 1198624 |Ω| 997888rarr minusinfin forall 119906 isin 119883

119896 as 119906 997888rarr +infin

(32)

Hence there exists 119877119896gt 120588 such that

Φ120583(119906) le 0 forall119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (33)

Summing up the above proofsΦ120583satisfies all the conditions

of Theorem 3 and Remark 4 Then the problem (1) has inf-initely many solutions

Proof of Theorem 6 Similar to the proof of Theorem 5(i) wehave

intΩ


= intΩ

120582119896V120593 d119909 forall 120593 isin 119867

(34)

We can easily see that V equiv 0 In fact if V equiv 0 then |V|2

=

0 which contradicts lim119899rarrinfin

|V119899|2

= |V|2

= 1 Then V isa corresponding eigenfunction of the eigenvalue 120582

119896 hence

|119906119899| rarr infin ae 119909 isin ΩBy (1198653) we have

lim119899rarrinfin

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899))

= minusinfin uniformly in 119909 isin Ω

(35)

It follows from Fatous Lemma that

lim119899rarrinfin

intΩ

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899)) d119909 = minusinfin (36)

On the other hand (15) implies that

2119888 larr997888 2Φ120583(119906119899) minus ⟨Φ

1015840

120583(119906119899) 119906119899⟩

= intΩ

(1003816100381610038161003816Δ119906119899

1003816100381610038161003816

2

minus 1198881003816100381610038161003816nabla119906119899

1003816100381610038161003816

2

) d119909

minus2120583

119901intΩ

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816


2119865 (119909 119906119899) d119909

minus intΩ

(1003816100381610038161003816Δ119906119899

1003816100381610038161003816

2

minus 1198881003816100381610038161003816nabla119906119899

1003816100381610038161003816

2

) d119909

+ 120583intΩ

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816

119901

119889119909 + intΩ

119891 (119909 119906119899) 119906119899d119909

= intΩ

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899)) d119909

minus (2

119901minus 1)120583int

Ω

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816

119901 d119909

(37)

where lim119899rarrinfin

(2119901 minus 1)120583 intΩℎ(119909)|119906

119899|119901119889119909 = infin which con-

tradicts (36) hence 119906119899 is bounded Similar to the proof of

Theorem 5(i) we have 119906119899 rarr 119906 in119867 Hencewe verifyΦ

120583(119906)

satisfies (119862) conditionSimilar to [15] let 119867(119909 119905) = 119865(119909 119905) minus (12)120582

119896(120582119896minus 119888)1199052

and 119891(119909 119905) = 120582119896(120582119896minus 119888)119905 + ℎ(119909 119905) then

lim|119905|rarrinfin

2119867 (119909 119905)

1199052= 0


(ℎ (119909 119905) 119905 minus 2119867 (119909 119905)) = minusinfin

(38)

It follows that for every119873 gt 0 there exists 119877119873

gt 0 such that

ℎ (119909 119905) 119905 minus 2119867 (119909 119905) le minus119873 forall 119905 isin 119877

|119905| ge 119877119873 ae 119909 isin Ω

(39)

For 119905 gt 0 we have

119889

119889119905[119867 (119909 119905)

1199052] =

ℎ (119909 119905) 119905 minus 2119867 (119909 119905)

1199053 (40)

over the interval [119905 119904] sub [119879 +infin) we have

119867(119909 119904)

1199042minus

119867 (119909 119905)

1199052le

119873

2(1

1199042minus

1

1199052) (41)


Letting 119904 rarr +infin we see that119867(119909 119905) ge (1198732) 119905 isin 119877 119905 ge 119877119873

ae 119909 isin Ω In a similar way we have 119867(119909 119905) ge (1198732) 119905 isin 119877119905 le minus119877

119873 ae 119909 isin Ω Hence


119867(119909 119905) = +infin ae 119909 isin Ω (42)

By Lemma 2 and ℎ(119909) ge 0 we have

Φ120583(119906) =

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus1

2120582119896(120582119896minus 119888) int

Ω

1199062 d119909

minus120583

119901intΩ

ℎ (119909) |119906|119901 d119909 minus int

Ω

119867(119909 119906) d119909

le minus intΩ

119867(119909 119906) d119909 997888rarr minusinfin forall 119906 isin 119883119896

as 119906 997888rarr +infin

(43)


Φ120583(119906) le 0 forall 119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (44)

Summing up the above proofs andTheorem 5 (ii)Φ120583satisfies

all the conditions of Theorem 3 and Remark 4 then theproblem (1) has infinitely many solutions

Acknowledgments

This work was supported by the National Nature ScienceFoundation of China (10971179) and the Project of ShandongProvince Higher Educational Science and Technology Pro-gram (J09LA55 J12LI53)

References

[1] A Ambrosetti and P H Rabinowitz ldquoDual variational methodsin critical point theory and applicationsrdquo Journal of FunctionalAnalysis vol 14 no 4 pp 349ndash381 1973

[2] H Brezis and L Nirenberg ldquoPositive solutions of nonlinearelliptic equation involving critical Sobolev exponentsrdquoCommu-nications on Pure and Applied Mathematics vol 36 no 4 pp437ndash477 1983

[3] P H Rabinowitz Minimax Methods in Critical Point Theorywith Applications to Differential Equations vol 65 of CBMsRegional Conference Series in Mathematics American Mathe-matical Society Providence RI USA 1986

[4] R Pei ldquoMultiple solutions for a fourth-order nonlinear ellipticproblemrdquo Boundary Value Problem vol 2012 no 39 pp 1ndash82012

[5] A C Lazer and P J Mckenna ldquoLarge-amplitude periodic osci-llations in suspension bridges Some new connections withnonlinear analysisrdquo SIAM Review vol 32 no 4 pp 537ndash5781990

[6] G Tarantello ldquoA note on a semilinear elliptic problemrdquo Differ-ential Integral Equations vol 5 no 3 pp 561ndash565 1992

[7] C V Pao ldquoOn fourth-order elliptic boundary value problemsrdquoProceedings of the American Mathematical Society vol 128 no4 pp 1023ndash1030 2000

[8] A M Micheletti and A Pistoia ldquoMultiplicity results for afourth-order semilinear elliptic problemrdquo Nonlinear AnalysisTheory Methods and Applications vol 31 no 7 pp 895ndash9081998

[9] G Xu and J Zhang ldquoExistence results for some fourth-ordernonlinear elliptic problems of local superlinearity and subli-nearityrdquo Journal of Mathematical Analysis and Applications vol281 no 2 pp 633ndash640 2003

[10] Z Jihui ldquoExistence results for some fourth-order nonlinearelliptic problemsrdquo Nonlinear Analysis Theory Methods andApplications vol 45 no 1 pp 29ndash36 2001

[11] J Zhou and X Wu ldquoSign-changing solutions for some fourth-order nonlinear elliptic problemsrdquo Journal of MathematicalAnalysis and Applications vol 342 no 1 pp 542ndash558 2008

[12] YWei ldquoMultiplicity results for some fourth-order elliptic equa-tionsrdquo Journal of Mathematical Analysis and Applications vol385 no 2 pp 797ndash807 2012

[13] BMarino and S Enrico Semmilinear Elliptic Equations for Begi-nners Springer London UK 2011

[14] K Chang Infinite Dimensional MorseTheory andMultipli Solu-tions Problems Birkhaauser Boston Mass USA 1993

[15] J Liu and J Su ldquoRemarks on multiple nontrivial solutions forQuasi-linear resonant problemsrdquo Journal of Mathematical Ana-lysis and Applications vol 258 no 1 pp 209ndash222 2001

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


and the deduced norm

119906119867 = [intΩ

(|Δ119906|2+ |nabla119906|

2) d119909]12

(6)

Suppose that 119888 lt 1205821 Let us define a norm of the space 119867

as follows

119906 = [intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909]12

(7)

It is easy to verify that the norm sdot is equivalent to thenorm sdot

119867on 119867 and for all 119906 isin 119867 the following Poincare

inequality holds

1199062ge 1205821(1205821minus 119888) |119906|

2

2 (8)

where |119906|2

2= intΩ|119906|2119889119909

Throughout this paper the weak solutions of (1) are thecritical points of the associated functional

Φ120583(119906)

=1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus120583

119901intΩ

ℎ (119909) |119906|119901d119909 minus int

Ω

119865 (119909 119906) d119909 forall 119906 isin 119867

(9)

where 119865(119909 119906) = int119906

0119891(119909 119905) 119889119905

ObviouslyΦ120583isin 1198621(119867 119877) and for all 119906 120593 isin 119867

⟨Φ1015840

120583(119906) 120593⟩

= intΩ

(Δ119906Δ120593 minus 119888nabla119906nabla120593 d119909)

minus 120583intΩ

ℎ (119909) |119906|119901minus2

119906120593 d119909 minus intΩ

119891 (119909 119906) 120593 d119909

(10)

In order to establish solutions for problem (1) we make thefollowing assumptions

(1198650) ℎ isin 119871

infin(Ω) ℎ(119909) ge 0

(1198651) 119891(119909 0) = 0 119891(119909 minus119905) = minus119891(119909 119905) for all 119909 isin Ω

119905 isin 119877(1198652) lim|119905|rarr0

119891(119909 119905)119905 = 120572 lim|119905|rarrinfin

119891(119909 119905)119905 = 120573

uniformly for ae 119909 isin Ω where 0 le 120572 lt 120582119896(120582119896minus 119888) lt

120573 or 120573 = 120582119896(120582119896minus 119888)

(1198653) lim|119905|rarrinfin

(119891(119909 119905)119905minus2119865(119909 119905)) = minusinfin uniformly in119909 isin Ω

Definition 1 (see [13]) Let Φ isin 1198621(119883 119877) we say that Φ sati-

sfies the Cerami condition at the level 119888 isin 119877 if any sequence119906119899 sub 119883 with

Φ(119906119899) 997888rarr 119888 (1 +

10038171003817100381710038171199061198991003817100381710038171003817)Φ1015840(119906119899) 997888rarr 0 as 119899 997888rarr infin

(11)

possesses a convergent subsequence Φ satisfies the (119862)

condition ifΦ satisfies (119862)119888for all 119888 isin 119877

Lemma 2 Let119883119896= span120593

1 1205932 120593

119896119867 = 119883

119896oplus119883perp

119896 then

1199062le 120582119896(120582119896minus 119888) |119906|

2

2 forall 119906 isin 119883

119896

1199062ge 120582119896+1

(120582119896+1

minus 119888) |119906|2

2 forall 119906 isin 119883

perp

119896 119896 ge 2

(12)

Proof It is similar to the proof of Lemma 25 in [13]

Theorem 3 (see [14] a symmetric Mountain Pass Theorem)Suppose that Φ isin 119862

1(119864 119877) is evenΦ(120579) = 0 and

(i) there exist 120588 120572 gt 0 and a finite dimensional linearsubspace 119885 such that

Φ|119885perpcap120597119861120588(120579)

ge 120572 (13)

(ii) there exist a sequence of linear subspace 119885119898

dim(119885119898) = 119898 and 119877


Φ (119909) le 0 forall 119909 isin 119885119898

119861119877119898

119898 = 1 2 (14)

If Φ satisfies (119875119878) condition then Φ possesses infinitely manydistinct critical points corresponding to positive critical values

Remark 4 If Φ satisfies the (119862) condition Theorem 3 stillholds

The main results of this paper are as follows

Theorem 5 Assume that (1198650)minus(1198652) hold and 119888 lt 120582

1 120582119896(120582119896minus

119888) lt 120573 120573 is not an eigenvalue of (4) then there exist 120583lowast gt 0

such that for 120583 isin (0 120583lowast) (1) has infinitely many solutions

Theorem 6 Assume that (1198650)minus (1198653) hold and 119888 lt 120582

1 120573 =

120582119896(120582119896minus 119888) then there exist 120583lowast gt 0 such that for 120583 isin (0 120583

lowast) (1)

has infinitely many solutions

2 Proofs of Theorems

Proof of Theorem 5 (i) Assume that 119906119899 sub 119867 is a (119862) seq-

uence that is

(1 +1003817100381710038171003817119906119899

1003817100381710038171003817)Φ1015840

120583(119906119899) 997888rarr 0 Φ

120583(119906119899) 997888rarr 119888 (15)

We claim that 119906119899 is boundedAssume as a contradiction that

|119906119899|2

rarr infin Setting V119899= 119906119899|119906119899|2 then |V

119899|2= 1 Without

loss of generality we assume

V119899 V in 119867 V

119899997888rarr V in 119871

2(Ω)

V119899997888rarr V ae 119909 isin Ω

(16)

From (15) we know that100381610038161003816100381610038161003816100381610038161003816

intΩ

(ΔV119899Δ120593 minus 119888nablaV

119899nabla120593) d119909

minus120583intΩ

ℎ (119909)

10038161003816100381610038161199061198991003816100381610038161003816

119901minus2

119906119899

100381610038161003816100381611990611989910038161003816100381610038162


119891 (119909 119906119899)

100381610038161003816100381611990611989910038161003816100381610038162

120593 d1199091003816100381610038161003816100381610038161003816100381610038161003816


=1

100381610038161003816100381611990611989910038161003816100381610038162

10038161003816100381610038161003816⟨Φ1015840

120583(119906119899) 120593⟩

10038161003816100381610038161003816

le1

100381610038161003816100381611990611989910038161003816100381610038162

10038171003817100381710038171003817Φ1015840

120583(119906119899)10038171003817100381710038171003817

10038171003817100381710038171205931003817100381710038171003817 997888rarr 0 forall 120593 isin 119867

(17)


2) we derive

10038161003816100381610038161199061198991003816100381610038161003816 =

1003816100381610038161003816V11989910038161003816100381610038161003816100381610038161003816119906119899

10038161003816100381610038162997888rarr infin lim

119899rarrinfin

119891 (119909 119906119899)

119906119899

= 120573 (18)

For V119899


lim119899rarrinfin

119891 (119909 119906119899)

100381610038161003816100381611990611989910038161003816100381610038162


119891 (119909 119906119899)

119906119899

V119899= 120573V ae 119909 isin Ω

(19)

In case (b) we have

119891 (119909 119906119899)

100381610038161003816100381611990611989910038161003816100381610038162

le 1198881003816100381610038161003816V119899

1003816100381610038161003816 997888rarr 0 (20)

Then

lim119899rarrinfin

intΩ

119891 (119909 119906119899) 120593

100381610038161003816100381611990611989910038161003816100381610038162


intΩ

119891 (119909 119906119899) V119899

119906119899

120593 d119909

= intΩ

120573V120593 d119909 forall 120593 isin 119867

(21)


intΩℎ(119909)(|119906

119899|119901minus2

119906119899|119906119899|2)120593119889119909 = 0 we

have

intΩ


= intΩ

120573V120593d119909 forall 120593 isin 119867

(22)








1003817100381710038171003817119906119899 minus 119906119898

1003817100381710038171003817 = 120583intΩ

ℎ (119909)1003816100381610038161003816119906119899 minus 119906

119898

1003816100381610038161003816

119901d119909

+ intΩ

119891 (119909 119906119899minus 119906119898) (119906119899minus 119906119898) d119909

+ 119900 (1)1003817100381710038171003817119906119899 minus 119906

119898

1003817100381710038171003817

le 120583|ℎ|infin

intΩ

1003816100381610038161003816119906119899 minus 119906119898

1003816100381610038161003816

119901d119909

+ intΩ

119891 (119909 119906119899minus 119906119898) (119906119899minus 119906119898) d119909

+ 119900 (1)1003817100381710038171003817119906119899 minus 119906

119898

1003817100381710038171003817

(23)


1003816100381610038161003816100381610038161003816intΩ

119891 (119909 119906119899minus 119906119898) (119906119899minus 119906119898) d119909

1003816100381610038161003816100381610038161003816

le1003816100381610038161003816119891 (119909 119906

119899minus 119906119898)10038161003816100381610038162

1003816100381610038161003816119906119899 minus 119906119898

10038161003816100381610038162

(24)




120583|119883perp

119896cap120597119861120588(120579)

ge

120574 gt 0 where 119861120588(120579) = 119906 isin 119867 119906 le 120588

By (1198651) and (119865

2) taking 120590 isin (2 2



119865 (119909 119906) le1

2(120572 + 120576) |119906|

2+ 1198621|119906|120590 forall 119909 isin Ω (25)

where

2lowast=

2119873

119873 minus 2 119873 gt 2

+infin 119873 le 2

(26)



Φ120583(119906) ge

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)


119901intΩ

|119906|119901d119909 minus int

Ω

119865 (119909 119906) d119909

ge1

21199062minus

120572 + 120576

2|119906|2

2minus

120583|ℎ|infin

119901|119906|119901

119901minus 1198621|119906|120590

120590

ge1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1


= (1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888))

minus1205831198622119906119901minus2

minus 1198623119906120590minus2

) 1199062

(27)


Let 119892(119905) = 1205831198622119905119901minus2


0

such that

119892 (1199050) lt

1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888)) (28)



= ((1205831198622(2 minus

119901))(1198623(120590 minus 2)))


119892 (1199050) = 120583(120590minus2)(120590minus119901)

[(1198622(1198622(2 minus 119901)

1198623(120590 minus 2)

))

(119901minus2)(120590minus119901)

+(1198623(1198622(2 minus 119901)

1198623(120590 minus 2)

))

(120590minus2)(120590minus119901)

]

lt1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888))

(29)


= [12(1 minus ((120572 + 120576)(120582119896+1

(120582119896+1

minus 119888))))

((1198622(1198622(2 minus 119901)119862

3(120590 minus 2)))


3(120590 minus

2)))(120590minus2)(120590minus119901)

)](2minus120590)(120590minus119901)



119896cap120597119861120588(120579)



Φ120583(119906) le 0 forall 119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (30)

For 120573 gt 120582119896(120582119896minus119888) (119865



119865 (119909 119906) ge1

2(120573 minus 120576) |119906|

2minus 1198624 (31)


(120583119901) intΩℎ(119909)|119906|

119901d119909 ge 0 we have

Φ120583(119906) =

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus120583

119901intΩ

ℎ (119909) |119906|119901 d119909 minus int

Ω

119865 (119909 119906) d119909

le1

21199062minus

120573 minus 120576

2|119906|2

2+ 1198624 |Ω|

le1

2(1 minus

120573 minus 120576

120582119896(120582119896minus 119888)

) 1199062


119896 as 119906 997888rarr +infin

(32)


Φ120583(119906) le 0 forall119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (33)




intΩ


= intΩ

120582119896V120593 d119909 forall 120593 isin 119867

(34)


=


|V119899|2

= |V|2


119896 hence


lim119899rarrinfin

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899))


(35)


lim119899rarrinfin

intΩ

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899)) d119909 = minusinfin (36)


2119888 larr997888 2Φ120583(119906119899) minus ⟨Φ

1015840

120583(119906119899) 119906119899⟩

= intΩ

(1003816100381610038161003816Δ119906119899

1003816100381610038161003816

2

minus 1198881003816100381610038161003816nabla119906119899

1003816100381610038161003816

2

) d119909

minus2120583

119901intΩ

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816


2119865 (119909 119906119899) d119909

minus intΩ

(1003816100381610038161003816Δ119906119899

1003816100381610038161003816

2

minus 1198881003816100381610038161003816nabla119906119899

1003816100381610038161003816

2

) d119909

+ 120583intΩ

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816

119901

119889119909 + intΩ

119891 (119909 119906119899) 119906119899d119909

= intΩ

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899)) d119909

minus (2

119901minus 1)120583int

Ω

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816

119901 d119909

(37)


(2119901 minus 1)120583 intΩℎ(119909)|119906

119899|119901119889119909 = infin which con-



120583(119906)


119896(120582119896minus 119888)1199052

and 119891(119909 119905) = 120582119896(120582119896minus 119888)119905 + ℎ(119909 119905) then


2119867 (119909 119905)

1199052= 0



(38)


gt 0 such that


|119905| ge 119877119873 ae 119909 isin Ω

(39)


119889

119889119905[119867 (119909 119905)

1199052] =

ℎ (119909 119905) 119905 minus 2119867 (119909 119905)

1199053 (40)


119867(119909 119904)

1199042minus

119867 (119909 119905)

1199052le

119873

2(1

1199042minus

1

1199052) (41)






119867(119909 119905) = +infin ae 119909 isin Ω (42)


Φ120583(119906) =

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus1

2120582119896(120582119896minus 119888) int

Ω

1199062 d119909

minus120583

119901intΩ

ℎ (119909) |119906|119901 d119909 minus int

Ω

119867(119909 119906) d119909

le minus intΩ



(43)


Φ120583(119906) le 0 forall 119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (44)



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










=1

100381610038161003816100381611990611989910038161003816100381610038162

10038161003816100381610038161003816⟨Φ1015840

120583(119906119899) 120593⟩

10038161003816100381610038161003816

le1

100381610038161003816100381611990611989910038161003816100381610038162

10038171003817100381710038171003817Φ1015840

120583(119906119899)10038171003817100381710038171003817

10038171003817100381710038171205931003817100381710038171003817 997888rarr 0 forall 120593 isin 119867

(17)


2) we derive

10038161003816100381610038161199061198991003816100381610038161003816 =

1003816100381610038161003816V11989910038161003816100381610038161003816100381610038161003816119906119899

10038161003816100381610038162997888rarr infin lim

119899rarrinfin

119891 (119909 119906119899)

119906119899

= 120573 (18)

For V119899


lim119899rarrinfin

119891 (119909 119906119899)

100381610038161003816100381611990611989910038161003816100381610038162


119891 (119909 119906119899)

119906119899

V119899= 120573V ae 119909 isin Ω

(19)

In case (b) we have

119891 (119909 119906119899)

100381610038161003816100381611990611989910038161003816100381610038162

le 1198881003816100381610038161003816V119899

1003816100381610038161003816 997888rarr 0 (20)

Then

lim119899rarrinfin

intΩ

119891 (119909 119906119899) 120593

100381610038161003816100381611990611989910038161003816100381610038162


intΩ

119891 (119909 119906119899) V119899

119906119899

120593 d119909

= intΩ

120573V120593 d119909 forall 120593 isin 119867

(21)


intΩℎ(119909)(|119906

119899|119901minus2

119906119899|119906119899|2)120593119889119909 = 0 we

have

intΩ


= intΩ

120573V120593d119909 forall 120593 isin 119867

(22)








1003817100381710038171003817119906119899 minus 119906119898

1003817100381710038171003817 = 120583intΩ

ℎ (119909)1003816100381610038161003816119906119899 minus 119906

119898

1003816100381610038161003816

119901d119909

+ intΩ

119891 (119909 119906119899minus 119906119898) (119906119899minus 119906119898) d119909

+ 119900 (1)1003817100381710038171003817119906119899 minus 119906

119898

1003817100381710038171003817

le 120583|ℎ|infin

intΩ

1003816100381610038161003816119906119899 minus 119906119898

1003816100381610038161003816

119901d119909

+ intΩ

119891 (119909 119906119899minus 119906119898) (119906119899minus 119906119898) d119909

+ 119900 (1)1003817100381710038171003817119906119899 minus 119906

119898

1003817100381710038171003817

(23)


1003816100381610038161003816100381610038161003816intΩ

119891 (119909 119906119899minus 119906119898) (119906119899minus 119906119898) d119909

1003816100381610038161003816100381610038161003816

le1003816100381610038161003816119891 (119909 119906

119899minus 119906119898)10038161003816100381610038162

1003816100381610038161003816119906119899 minus 119906119898

10038161003816100381610038162

(24)




120583|119883perp

119896cap120597119861120588(120579)

ge

120574 gt 0 where 119861120588(120579) = 119906 isin 119867 119906 le 120588

By (1198651) and (119865

2) taking 120590 isin (2 2



119865 (119909 119906) le1

2(120572 + 120576) |119906|

2+ 1198621|119906|120590 forall 119909 isin Ω (25)

where

2lowast=

2119873

119873 minus 2 119873 gt 2

+infin 119873 le 2

(26)



Φ120583(119906) ge

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)


119901intΩ

|119906|119901d119909 minus int

Ω

119865 (119909 119906) d119909

ge1

21199062minus

120572 + 120576

2|119906|2

2minus

120583|ℎ|infin

119901|119906|119901

119901minus 1198621|119906|120590

120590

ge1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1


= (1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888))

minus1205831198622119906119901minus2

minus 1198623119906120590minus2

) 1199062

(27)


Let 119892(119905) = 1205831198622119905119901minus2


0

such that

119892 (1199050) lt

1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888)) (28)



= ((1205831198622(2 minus

119901))(1198623(120590 minus 2)))


119892 (1199050) = 120583(120590minus2)(120590minus119901)

[(1198622(1198622(2 minus 119901)

1198623(120590 minus 2)

))

(119901minus2)(120590minus119901)

+(1198623(1198622(2 minus 119901)

1198623(120590 minus 2)

))

(120590minus2)(120590minus119901)

]

lt1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888))

(29)


= [12(1 minus ((120572 + 120576)(120582119896+1

(120582119896+1

minus 119888))))

((1198622(1198622(2 minus 119901)119862

3(120590 minus 2)))


3(120590 minus

2)))(120590minus2)(120590minus119901)

)](2minus120590)(120590minus119901)



119896cap120597119861120588(120579)



Φ120583(119906) le 0 forall 119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (30)

For 120573 gt 120582119896(120582119896minus119888) (119865



119865 (119909 119906) ge1

2(120573 minus 120576) |119906|

2minus 1198624 (31)


(120583119901) intΩℎ(119909)|119906|

119901d119909 ge 0 we have

Φ120583(119906) =

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus120583

119901intΩ

ℎ (119909) |119906|119901 d119909 minus int

Ω

119865 (119909 119906) d119909

le1

21199062minus

120573 minus 120576

2|119906|2

2+ 1198624 |Ω|

le1

2(1 minus

120573 minus 120576

120582119896(120582119896minus 119888)

) 1199062


119896 as 119906 997888rarr +infin

(32)


Φ120583(119906) le 0 forall119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (33)




intΩ


= intΩ

120582119896V120593 d119909 forall 120593 isin 119867

(34)


=


|V119899|2

= |V|2


119896 hence


lim119899rarrinfin

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899))


(35)


lim119899rarrinfin

intΩ

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899)) d119909 = minusinfin (36)


2119888 larr997888 2Φ120583(119906119899) minus ⟨Φ

1015840

120583(119906119899) 119906119899⟩

= intΩ

(1003816100381610038161003816Δ119906119899

1003816100381610038161003816

2

minus 1198881003816100381610038161003816nabla119906119899

1003816100381610038161003816

2

) d119909

minus2120583

119901intΩ

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816


2119865 (119909 119906119899) d119909

minus intΩ

(1003816100381610038161003816Δ119906119899

1003816100381610038161003816

2

minus 1198881003816100381610038161003816nabla119906119899

1003816100381610038161003816

2

) d119909

+ 120583intΩ

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816

119901

119889119909 + intΩ

119891 (119909 119906119899) 119906119899d119909

= intΩ

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899)) d119909

minus (2

119901minus 1)120583int

Ω

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816

119901 d119909

(37)


(2119901 minus 1)120583 intΩℎ(119909)|119906

119899|119901119889119909 = infin which con-



120583(119906)


119896(120582119896minus 119888)1199052

and 119891(119909 119905) = 120582119896(120582119896minus 119888)119905 + ℎ(119909 119905) then


2119867 (119909 119905)

1199052= 0



(38)


gt 0 such that


|119905| ge 119877119873 ae 119909 isin Ω

(39)


119889

119889119905[119867 (119909 119905)

1199052] =

ℎ (119909 119905) 119905 minus 2119867 (119909 119905)

1199053 (40)


119867(119909 119904)

1199042minus

119867 (119909 119905)

1199052le

119873

2(1

1199042minus

1

1199052) (41)






119867(119909 119905) = +infin ae 119909 isin Ω (42)


Φ120583(119906) =

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus1

2120582119896(120582119896minus 119888) int

Ω

1199062 d119909

minus120583

119901intΩ

ℎ (119909) |119906|119901 d119909 minus int

Ω

119867(119909 119906) d119909

le minus intΩ



(43)


Φ120583(119906) le 0 forall 119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (44)



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











= ((1205831198622(2 minus

119901))(1198623(120590 minus 2)))


119892 (1199050) = 120583(120590minus2)(120590minus119901)

[(1198622(1198622(2 minus 119901)

1198623(120590 minus 2)

))

(119901minus2)(120590minus119901)

+(1198623(1198622(2 minus 119901)

1198623(120590 minus 2)

))

(120590minus2)(120590minus119901)

]

lt1

2(1 minus

120572 + 120576

120582119896+1

(120582119896+1

minus 119888))

(29)


= [12(1 minus ((120572 + 120576)(120582119896+1

(120582119896+1

minus 119888))))

((1198622(1198622(2 minus 119901)119862

3(120590 minus 2)))


3(120590 minus

2)))(120590minus2)(120590minus119901)

)](2minus120590)(120590minus119901)



119896cap120597119861120588(120579)



Φ120583(119906) le 0 forall 119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (30)

For 120573 gt 120582119896(120582119896minus119888) (119865



119865 (119909 119906) ge1

2(120573 minus 120576) |119906|

2minus 1198624 (31)


(120583119901) intΩℎ(119909)|119906|

119901d119909 ge 0 we have

Φ120583(119906) =

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus120583

119901intΩ

ℎ (119909) |119906|119901 d119909 minus int

Ω

119865 (119909 119906) d119909

le1

21199062minus

120573 minus 120576

2|119906|2

2+ 1198624 |Ω|

le1

2(1 minus

120573 minus 120576

120582119896(120582119896minus 119888)

) 1199062


119896 as 119906 997888rarr +infin

(32)


Φ120583(119906) le 0 forall119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (33)




intΩ


= intΩ

120582119896V120593 d119909 forall 120593 isin 119867

(34)


=


|V119899|2

= |V|2


119896 hence


lim119899rarrinfin

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899))


(35)


lim119899rarrinfin

intΩ

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899)) d119909 = minusinfin (36)


2119888 larr997888 2Φ120583(119906119899) minus ⟨Φ

1015840

120583(119906119899) 119906119899⟩

= intΩ

(1003816100381610038161003816Δ119906119899

1003816100381610038161003816

2

minus 1198881003816100381610038161003816nabla119906119899

1003816100381610038161003816

2

) d119909

minus2120583

119901intΩ

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816


2119865 (119909 119906119899) d119909

minus intΩ

(1003816100381610038161003816Δ119906119899

1003816100381610038161003816

2

minus 1198881003816100381610038161003816nabla119906119899

1003816100381610038161003816

2

) d119909

+ 120583intΩ

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816

119901

119889119909 + intΩ

119891 (119909 119906119899) 119906119899d119909

= intΩ

(119891 (119909 119906119899) 119906119899minus 2119865 (119909 119906

119899)) d119909

minus (2

119901minus 1)120583int

Ω

ℎ (119909)1003816100381610038161003816119906119899

1003816100381610038161003816

119901 d119909

(37)


(2119901 minus 1)120583 intΩℎ(119909)|119906

119899|119901119889119909 = infin which con-



120583(119906)


119896(120582119896minus 119888)1199052

and 119891(119909 119905) = 120582119896(120582119896minus 119888)119905 + ℎ(119909 119905) then


2119867 (119909 119905)

1199052= 0



(38)


gt 0 such that


|119905| ge 119877119873 ae 119909 isin Ω

(39)


119889

119889119905[119867 (119909 119905)

1199052] =

ℎ (119909 119905) 119905 minus 2119867 (119909 119905)

1199053 (40)


119867(119909 119904)

1199042minus

119867 (119909 119905)

1199052le

119873

2(1

1199042minus

1

1199052) (41)






119867(119909 119905) = +infin ae 119909 isin Ω (42)


Φ120583(119906) =

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus1

2120582119896(120582119896minus 119888) int

Ω

1199062 d119909

minus120583

119901intΩ

ℎ (119909) |119906|119901 d119909 minus int

Ω

119867(119909 119906) d119909

le minus intΩ



(43)


Φ120583(119906) le 0 forall 119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (44)



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra














119867(119909 119905) = +infin ae 119909 isin Ω (42)


Φ120583(119906) =

1

2(intΩ

(|Δ119906|2minus 119888|nabla119906|

2) d119909)

minus1

2120582119896(120582119896minus 119888) int

Ω

1199062 d119909

minus120583

119901intΩ

ℎ (119909) |119906|119901 d119909 minus int

Ω

119867(119909 119906) d119909

le minus intΩ



(43)


Φ120583(119906) le 0 forall 119906 isin 119883

119896 119861119877119896

(120579) 119896 = 1 2 (44)



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article existence of multiple solutions for a...

Documents