Multiplicity of Solutions for Neumann Problems for Semilinear Elliptic Equations

and Applied Analysis 3 If not, it is clear from (7) that

studied the semilinear elliptic boundary value problem in any spatial dimension and using variational techniques; they showed that a suitable perturbation will turn the almost resonant situation ( near to   , i.e., near resonance with a nonprincipal eigenvalue) in a situation where the solutions are at least two.In [14], those results were extended to the degenerate elliptic equations in the bounded domain.
Motivated by the above idea, we have the goal in this paper of extending these results in [6,[12][13][14] to some elliptic equations with the Neumann boundary conditions.Here, it is worth pointing out that ( ∞ ) is weaker than ( 1 ) in [13] (or () in [14]).More to the point, there are functions satisfying the assumptions of our main results in Section 2 and not satisfying the assumptions in [13,14].For example, let (, ) = / ln(1 + ||).Then  satisfies the assumptions for our Theorem 5 in Section 2 and does not satisfy ( 1 ) in [13] (or () in [14]).
The rest of our paper is organized as follows.In Section 2 we give some preliminary lemmas and our main results.Section 3 gives the detailed proofs of our main results based on several estimates, whose proof will be presented in Sections 4 and 5.

Preliminaries and Main Results
Let the Sobolev space  =  1 (Ω).Denote (3) to be the norm of  in  and ‖‖  the norm of  in   (Ω).The space  is a Hilbert space.For a discussion about the space setting, we refer to [15] and the references therein.
We denote by   the eigenspace corresponding to an eigenvalue   , and we can decompose  1 (Ω) = ⨁ ∞ =1   .We set The eigenvalues admit the following variational characterizations in terms of the Rayleigh quotient ()/‖‖ 2 2 for all  ∈  1 (Ω): In (7), the infimum is realized on  1 .Also, in (8), both the infimum and the supremum are realized on   .All the eigenspaces have the so-called unique continuation property.The first eigenvalue  1 is simple and it is clear from (7) that the corresponding eigenfunctions do not change sign.Namely, we can suppose that  1 is strictly positive on Ω.We mention that all the other eigenvalues have nodal eigenfunctions.For more properties to the eigenvalue problem ( 4), see [6,7].By the presence of function , weak solutions of (1) must be found in a suitable space.To this purpose, letting  > max{− 1 , 0}, we introduce a new inner product on  1 (Ω) by for , V ∈  1 (Ω) and the associated norm ,  ∈  1 (Ω) . ( Proof.By virtue of H ö lder's inequality, we have where  is the constant of Sobolev imbedding from  1 (Ω) →  2  (Ω),   = /( − 1),  2 = max{1, ,  2 ‖‖  }.On the other hand, if  > max{− 1 , 0}, then there exists If not, it is clear from (7) that Exploiting the 2-homogeneity of  we can find a sequence Passing to a subsequence if necessary, we may assume that The sequential weak lower semicontinuity of  and (15) imply that So () =  1 ‖‖ 2 2 and thus  =  1 , with some  ∈ .

Proof of Theorems
The associated functional of problem (1) is for  ∈  1 (Ω).Under the conditions ( 0 ) and ( ∞ ), it is easy to verify that, for every  ∈ ,  ∈  1 ( 1 (Ω), ) and for , V ∈  1 (Ω).Moreover, critical points of  are exactly weak solutions of problem (1).It follows from ( 0 ) and ( ∞ ) that for every  > 0 there exist M > 0 and for all  ∈  and a.e. ∈ Ω, which implies that for all  ∈  and a.e. ∈ Ω.From this and Hölder's inequality, we have where L = ‖ M ‖ 2 and  is the best embedding constant.
and   ,   , and   , respectively, are their relative boundaries.Theorems 5 and 6 will be a consequence of the geometry in Propositions 9 and 10 stated below, whose proofs will be postponed to Sections 4 and 5.
(The values with index  depend on ; the others may be fixed uniformly.) Proof of Theorem 5. Since the functional  satisfies the (P.S.) condition, we can apply two times the saddle point theorem (see, e.g., [17]); let The first solution, which we denote by  −1 and may be obtained for any  ∈ ( −1 ,   ) with just hypotheses ( 0 ) and ( ∞ ), corresponds to a critical point at the level The criticality of this level is guaranteed by the estimates ( 43) and ( 44), since     and  ⊕  link; that is, the image of any map in Γ −1 intersects  ⊕ .
The second solution, which we denote by   , corresponds to a critical point at the critical level Actually, this is a critical level because of the estimates ( 45) and ( 46), since  1   and  link.
To conclude the proof, we need to show that these two solutions are distinct.
We observe first that by estimate (45) we have that   ≥   , then we observe that we may build a map  0 ∈ Γ −1 in such a way that its image is the union between the annulus { ∈  : ‖‖ ∈ [ 1 ,   ]} and the image of a (−1)-dimensional ball in  1   whose boundary is  1   .By the estimates (46) and (47), we deduce that sup ∈    ( 0 ()) <   , and as a consequence  −1 <   , proving that the two solutions are distinct, for being at different critical levels.
Proof of Theorem 6.Since the functional  satisfies the (P.S.) condition, we can apply the saddle point theorem and Lemma 4.
The first solution, which we denote by   and may be obtained for any  ∈ (  ,  + ) with just hypotheses ( 0 ) and ( ∞ ), is again obtained through the saddle point theorem and corresponds to a critical point at the critical level where now The criticality is guaranteed by estimates (48) and (49), since     and  link.
The second solution, which we denote by  −1 , comes from Lemma 3, where we set  1 =  and  2 =  ⊕ ; actually we have the structure sup and then we have a critical point  −1 at the level  −1 ≤   .
Finally, in order to prove that these two solutions are distinct, we need a sharper estimate for   than that given by (49).For this we use Lemma 3 to guarantee that, for any map  ∈ Γ  , since   >  2 , one has that the image of  either intersects  2   or has a point  ∈  with ‖‖ ≥  2 .This implies that sup by estimates (51) and (52), and then   >   proving that the two solutions are distinct, for being at different critical levels.
Proof of Theorem 7. The proof will be divided into four steps.
Step 4. Three solutions are obtained.

Proof of Estimates
In this section we will prove all the estimates in Propositions 9 and 10.
For  ∈ (  ,  + ), then the same estimate holds but the constant cannot be made independent of , giving (48).
In the same way, let  ∈  ⊕  and set  =   −  > 0; we get Letting  < / 2 ( +   ), it follows that  is bounded below in  ⊕ ; that is, there exists a   such that for all  ∈  ⊕  we have (43), where again the constant   depends on , that is, on .
Finally, (71) with  ∈ (  ,  + ) implies Then, no matter the value of ,  is bounded from below in any bounded subset of  ⊕ , giving (53) for a suitable value of .
Finally, letting  ∈  ⊕  and setting  =  −   > 0, we get Letting  < / 2   , it is clear that (once  is fixed) this goes to −∞ and then we may find the claimed   >  2 such that (49) holds.
Observe that   and  can be chosen uniformly for  ∈ (  ,  + ), while   ,   will in fact depend on .

Estimating the Effect of the Nontrivial Perturbation.
In this section we will prove the remaining inequalities in Propositions 9 and 10, those which rely on the hypothesis ( 1 ) or ( 2 ) or ( 3 ) or ( 4 ), which, roughly speaking, say that the perturbation  is nontrivial in such a way that a new solution arises when  is sufficiently near to the eigenvalue   .The proof is simpler for Theorem 5, since we need to estimate the functional in the compact set   , while for Theorem 6 the same kind of estimate is required in the noncompact set   .
Let  = (+ C|Ω|+ C|Ω| 1/2 ‖ C‖ 2 ) −1 ; one finally obtains It is elementary that is well defined and satisfies the claim.Now we may prove the following.
Proof.Letting  =   + , we see from (70) that property (52) will be satisfied provided that  2 is large enough (say  2 > R); observe that this value can be made independent from  once  is small enough.Now we consider the two sets of hypotheses separately.(i) In case ( 3 ), suppose  ∈   ⊕ ; we can assume that  =  + , with  ∈  and  ∈   .Since   is a finite dimension subspace, all the norms are equivalent, so that there exists  > 0 such that for all  ∈   we have ‖‖ ≤ ‖‖ 1 .In addition, by ( 0 ) and ( 3 ), there exist  2 and  2 ∈  2 (Ω) such that (ii) In case ( 4 ), first we give some conclusions which are similar to Lemma 3 of [18].Under the property of , there for all  ∈ .In fact, for every  ∈ , there exists  ∈  such that