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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco

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On sign-changing solutions for (p,q)-Laplace equations with two parameters

Vladimir BobkovORCID iD: https://orcid.org/0000-0002-4425-0218
  • Corresponding author
  • Institute of Mathematics, Ufa Scientific Center, Russian Academy of Sciences, Chernyshevsky str. 112, Ufa 450008, Russia; and Department of Mathematics and NTIS, Faculty of Applied Sciences, University of West Bohemia, Univerzitní 8, Plzeň 306 14, Czech Republic
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/ Mieko Tanaka
Published Online: 2016-12-20 | DOI: https://doi.org/10.1515/anona-2016-0172


We investigate the existence of nodal (sign-changing) solutions to the Dirichlet problem for a two-parametric family of partially homogeneous (p,q)-Laplace equations -Δpu-Δqu=α|u|p-2u+β|u|q-2u where pq. By virtue of the Nehari manifolds, the linking theorem, and descending flow, we explicitly characterize subsets of the (α,β)-plane which correspond to the existence of nodal solutions. In each subset the obtained solutions have prescribed signs of energy and, in some cases, exactly two nodal domains. The nonexistence of nodal solutions is also studied. Additionally, we explore several relations between eigenvalues and eigenfunctions of the p- and q-Laplacians in one dimension.

Keywords: eigenvalue problem; first eigenvalue; second eigenvalue; nodal solutions; sign-changing solutions; Nehari manifold; linking theorem; descending flow

MSC 2010: 35J62; 35J20; 35P30

1 Introduction

In this article, we study the existence and nonexistence of sign-changing solutions for the problem

{-Δpu-Δqu=α|u|p-2u+β|u|q-2uin Ω,u=0on Ω,(${\mathrm{GEV};\alpha,\beta}$)

where ΩN, N1, is a bounded domain with a sufficiently smooth boundary Ω, and α,β are parameters. The operator Δru:=div(|u|r-2u) is the classical r-Laplacian, r={q,p}>1, and without loss of generality we assume that q<p.

Boundary value problems with a combination of several differential operators of different nature (in particular, as in ((${\mathrm{GEV};\alpha,\beta}$))) arise mainly as mathematical models of physical processes and phenomena, and have been extensively studied in the last two decades; see, e.g., [15, 30, 19, 13] and the references below. Among the historically first examples one can mention the Cahn–Hilliard equation [12] describing the process of separation of binary alloys, and the Zakharov equation [33, (1.8)] which describes the behavior of plasma oscillations. Elliptic equations with the (2,6)- and (2,p)-Laplacians were considered explicitly in [8, 7] with the aim of obtaining soliton-type solutions (in particular, as a model for elementary particles).

The considered problem ((${\mathrm{GEV};\alpha,\beta}$)) attracts special attention due to its symmetric and partially homogeneous structure; cf. [31, 32, 28, 20, 34, 10, 4]. By developing the results of [31, 28, 20], the authors of the present article obtained in [10] a reasonably complete description of the subsets of the (α,β)-plane which correspond to the existence/nonexistence of positive solutions to problem ((${\mathrm{GEV};\alpha,\beta}$)). At the same time, to the best of our knowledge, analogous results for sign-changing solutions have not been obtained circumstantially so far, although a particular information on the existence can be extracted from [24, 32, 1]. The main reason for this is a crucial dependence of the structure of the solution set to problem ((${\mathrm{GEV};\alpha,\beta}$)) on parameters α and β. As a consequence, the existence can not be treated by a unique approach, and various tools have to be used for different parts of the (α,β)-plane.

The aim of the present article is to allocate and characterize the sets of parameters α and β for which problem ((${\mathrm{GEV};\alpha,\beta}$)) possesses or does not possess sign-changing solutions (see Figure 1). In this sense, this work can be seen as the second part of the article [10].

1.1 Notations and preliminaries

Before formulating the main results, we introduce several notations. In what follows, Lr(Ω) with r(1,+) and L(Ω) stand for the Lebesgue spaces with the norms


respectively, and W01,r:=W01,r(Ω) denotes the Sobolev space with the norm ur. For uW01,r we define u±:=max{±u,0}. Note that u±W01,r and u=u+-u-.

By a (weak) solution of ((${\mathrm{GEV};\alpha,\beta}$)) we mean function uW01,p which satisfies


for all φW01,p. If u is a solution of ((${\mathrm{GEV};\alpha,\beta}$)) and u±0 (a.e. in Ω), then u is called nodal or sign-changing solution. It is not hard to see that any solution of ((${\mathrm{GEV};\alpha,\beta}$)) is a critical point of the energy functional Eα,βC1(W01,p,) defined by




Notice that the supports of u+ and u- are disjoint for any uW01,p. This fact, together with evenness of the functionals Hα and Gβ, easily implies that


Remark 1.

Any solution uW01,p of problem ((${\mathrm{GEV};\alpha,\beta}$)) belongs to C01,γ(Ω¯) for some γ(0,1). In fact, uL(Ω) by the Moser iteration process; cf. [25, Appendix A]. Furthermore, the regularity up to the boundary in [21, Theorem 1] and [22, p. 320] provides uC01,γ(Ω¯), γ(0,1).

Next, we recall several facts related to the eigenvalue problem for the Dirichlet r-Laplacian, r>1. We say that λ is an eigenvalue of -Δr, if the problem

{-Δru=λ|u|r-2uin Ω,u=0on Ω(${\mathrm{EV};r,\lambda}$)

has a nontrivial (weak) solution. Analogously to the linear case, the set of all eigenvalues of ((${\mathrm{EV};r,\lambda}$)) will be denoted as σ(-Δr). It is well known that the lowest positive eigenvalue λ1(r) can be obtained through the nonlinear Rayleigh quotient as (cf. [2])


The eigenvalue λ1(r) is simple and isolated, and the corresponding eigenfunction φrW01,p (defined up to an arbitrary multiplier) is strictly positive (or strictly negative) in Ω. Moreover, λ1(r) is the unique eigenvalue with a corresponding sign-constant eigenfunction [2]. Note also that any eigenfunction φ of -Δr belongs to C01,γ(Ω¯) for some γ(0,1).

The following lemma directly follows from the definition of λ1(r) and its simplicity.

Lemma 2.

Assume that uW01,p{0}. Then we have the following results:

  • (i)

    Let αλ1(p) . Then Hα(u)0 , and Hα(u)=0 if and only if α=λ1(p) and u=tφp for some t{0}.

  • (ii)

    Let βλ1(q) . Then Gβ(u)0 , and Gβ(u)=0 if and only if β=λ1(q) and u=tφq for some t{0}.

Although the structure of σ(-Δr) is not completely known except for the case r=2 or N=1 (see, e.g., [17, Theorem 3.1]), several unbounded sequences of eigenvalues can be introduced by virtue of minimax variational principles. In what follows, by {λk(r)}k we denote a sequence of eigenvalues for ((${\mathrm{EV};r,\lambda}$)) introduced in [18]. It can be described variationally as


where Sk-1 is the unit sphere in k and

k(r):={hC(Sk-1,S(r)):h is odd},(1.4)S(r):={uW01,r:ur=1}.

It is known [18] that λk(r)+ as k+. Moreover, λ2(r) coincides with the second eigenvalue of -Δr, i.e.,


and it can be alternatively characterized as in [16]:


where the first eigenfunction φr is normalized such that φrS(r). We denote any eigenfunction corresponding to λ2(r) as φ2,r. Notice that λ2(r)>λ1(r). Furthermore, in the one-dimensional case the sequence (1.3) describes the whole σ(-Δr) (cf. [17, Theorem 4.1], where this result is proved for the Krasnosel’skii-type eigenvalues).

Finally, we introduce the notation for the eigenspace of -Δr at λ:

ES(r;λ):={vW01,r:v is a solution of (EV; rλ)}.(1.6)

It is clear that ES(r;λ){0} if and only if λσ(-Δr).

1.2 Main results

Let us state the main results of this article. We begin with the nonexistence of nodal solutions for ((${\mathrm{GEV};\alpha,\beta}$)).

Theorem 3.

Assume that


Then ((${\mathrm{GEV};\alpha,\beta}$)) has no nodal solutions.

In the one-dimensional case Theorem 3 can be refined as follows.

Theorem 4.

Let N=1. If (α,β)(-,λ2(p)]×(-,λ2(q)], then ((${\mathrm{GEV};\alpha,\beta}$)) has no nodal solutions.

In the case of general dimensions an additional information on hypothetical nodal solutions to ((${\mathrm{GEV};\alpha,\beta}$)) for α(λ1(p),λ2(p)] and β(λ1(q),λ2(q)] is given in Lemma 5 below.

The case λ2⁢(q)<λ3⁢(q){\lambda_{2}(q)<\lambda_{3}(q)}, λ2⁢(p)<λ3⁢(p){\lambda_{2}(p)<\lambda_{3}(p)}, and (λ2⁢(p),λ3⁢(p))∩σ⁢(-Δp)=∅{(\lambda_{2}(p),\lambda_{3}(p))\cap\sigma(-\Delta_{p})=\emptyset}. Existence (light gray, solid lines),nonexistence (dark gray, zigzag lines), unknown (white, dashed lines).
Figure 1

The case λ2(q)<λ3(q), λ2(p)<λ3(p), and (λ2(p),λ3(p))σ(-Δp)=. Existence (light gray, solid lines),nonexistence (dark gray, zigzag lines), unknown (white, dashed lines).

Now we formulate the existence result for nodal solutions with a positive energy. Let us define the following “lower” critical value depending on α:




and put β(α)=+ whenever the admissible set (α) is empty.

Theorem 5.

Let α>λ2(p). Then for all β<βL(α) problem ((${\mathrm{GEV};\alpha,\beta}$)) has a nodal solution u with Eα,β(u)>0 and precisely two nodal domains.

Several main properties of the function β(α) are collected in Lemma 11 below. Let us remark that the parametrization by α in (1.7) is different from the parametrization by s of the form (α,β)=(λ+s,λ) which was used in [10] in order to construct a critical curve for the existence of positive solutions. In the context of the present article, the parametrization by α makes problem ((${\mathrm{GEV};\alpha,\beta}$)) easier to analyze. We also note that (1.7) is conceptually similar to the characterization of the first nontrivial curve of the Fučík spectrum given in [26]. In Section 2 below, we introduce and study several other critical points besides (1.7), which although are not directly used in the proofs of the main results, increase the understanding of the construction of the (α,β)-plane, and could be employed in further investigations.

Next, we state the existence of negative energy nodal solutions for ((${\mathrm{GEV};\alpha,\beta}$)). Consider the “upper” critical value


where α, and set β𝒰*(α)=- provided ασ(-Δp). Several lower and upper bounds for β𝒰*(α) are given in Lemmas 6 and 7 below. Define kα:=min{k:α<λk+1(p)} and notice that λkα+1(q)λ2(q) for all α.

Theorem 6.

Let αRσ(-Δp)¯. Then for all β>max{βU*(α),λkα+1(q)} problem ((${\mathrm{GEV};\alpha,\beta}$)) has a nodal solution u satisfying Eα,β(u)<0.

Evidently, if σ(-Δp) is a discreet set (as it is for p=2 or N=1), then σ(-Δp)¯=. Moreover, λ1(p) and λ2(p) belong to σ(-Δp)¯ for all p>1 and N1 since λ1(p) is isolated and there are no eigenvalues between λ1(p) and λ2(p) (see Section 1.1).

One of the main ingredients for the proof of Theorem 6 is the result on the existence of three nontrivial solutions (positive, negative and sign-changing) to the problem with the (p,q)-Laplacian and a nonlinearity in the general form given by Theorem 13 below. This result is of independent interest.

Theorem 6 can be refined as follows.

Theorem 7.

Assume that


Then ((${\mathrm{GEV};\alpha,\beta}$)) has a nodal solution u satisfying Eα,β(u)<0.

Remark 8.

In the one-dimensional case we have φpqq/φpqq<λ2(q) (see Lemma 2 in Appendix A), and hence the assertion of Theorem 7 holds for all (α,β)(-,λ2(p))×(λ2(q),+).

Let us note that unlike the case of positive solutions, the structure of the set of nodal solutions for problem ((${\mathrm{GEV};\alpha,\beta}$)) is more complicated, and we are not aware of the maximality of the regions obtained in Theorems 5 and 6.

The article is organized as follows: In Section 2, we apply the method of the Nehari manifold in order to prove Theorem 5. In Section 3, by means of linking arguments and the descending flow method, we provide two general existence results which yield, in particular, Theorems 6 and 7. For the convenience of the reader we collect the proofs of the main theorems in Section 4. In Appendix A, we prove several additional facts on the relation between eigenvalues and eigenfunctions of the p- and q-Laplacians in the one-dimensional case. Finally, in Appendix B, we give a sketch of the proof of Theorem 13.

2 Nodal solutions with positive energy

The classical Nehari manifold for problem ((${\mathrm{GEV};\alpha,\beta}$)) is defined by


It can be readily seen that 𝒩α,β contains all nontrivial solutions of ((${\mathrm{GEV};\alpha,\beta}$)). On the other hand, if uW01,p is a sign-changing solution of ((${\mathrm{GEV};\alpha,\beta}$)), then


These equalities bring us to the definition of the so-called nodal Nehari set for ((${\mathrm{GEV};\alpha,\beta}$)):


By construction, α,β contains all sign-changing solutions of ((${\mathrm{GEV};\alpha,\beta}$)), and hence α,β𝒩α,β.

Let us divide α,β into the following three subsets:


Evidently, α,β=α,β1α,β2α,β3 and all α,βi are mutually disjoint. The main aim of this section is to prove the existence of nodal solutions for ((${\mathrm{GEV};\alpha,\beta}$)) through minimization of Eα,β over α,β1 in an appropriate subset of the (α,β)-plane.

2.1 Preliminary analysis

In this subsection, we mainly study the properties of the sets α,β1, α,β2, and α,β3. First of all, we give the following auxiliary lemma, which is in fact analogous to [10, Proposition 6] and can be proved in the same manner.

Lemma 1.

Let uW01,p. If Hα(u)Gβ(u)<0, then there exists a unique critical point t(u)>0 of Eα,β(tu) with respect to t>0 and t(u)uNα,β. In particular, if


then t(u) is the unique maximum point of Eα,β(tu) with respect to t>0 and Eα,β(t(u)u)>0.

We start our consideration of the sets α,βi with several simple facts.

Lemma 2.

Let α,βR. The following hold:

  • (i)

    If βλ1(q) , then α,β1=α,β and, consequently, α,β2,α,β3=.

  • (ii)

    If αλ1(p) , then α,β2=α,β and, consequently, α,β1,α,β3=.


Let us first prove assertion (i). Assume that βλ1(q) and wα,β. Then Lemma 2 implies that Gβ(w±)0 and in fact Gβ(w±)>0, since otherwise w±=φq, which is impossible in view of the strict positivity of φq in Ω. Thus, the Nehari constraints Hα(w±)+Gβ(w±)=0 yield Hα(w±)<0, whence wα,β1. Assertion (ii) can be shown by the same arguments. ∎

Let us introduce the following sets:


Obviously, α,β11(α) and α,β22(α). Moreover, we have the following result.

Lemma 3.

Let α,βR. The following hold:

  • (i)

    If αλ2(p) , then 1(α)= and, consequently, α,β1=.

  • (ii)

    If βλ2(q) , then 2(α)= and, consequently, α,β2=.


We give the proof of assertion (i). The second part can be proved analogously. Suppose, by contradiction, that αλ2(p) and there exists w1(α). These assumptions read as


On the other hand, it is shown in [9, Proposition 4.2] that the second eigenvalue λ2(r), r>1, can be characterized as follows:


Comparing (2.4) and (2.5) (with r=p), we obtain a contradiction. ∎

Lemmas 2 and 3 readily entail the following information about the emptiness of α,β and, consequently, the nonexistence of nodal solutions for ((${\mathrm{GEV};\alpha,\beta}$)).

Lemma 4.

If αλ2(p) and βλ1(q), or αλ1(p) and βλ2(q), then Mα,β=.

Lemma 5.

Let αλ2(p) and βλ2(q). If u is a nodal solution of ((${\mathrm{GEV};\alpha,\beta}$)), then


Let us now subsequently treat the emptiness and nonemptiness of α,β1 and α,β2. First we consider α,β1. Introduce the critical value


for each α, where the admissible set 1(α) is defined by (2.2), or, equivalently,


We assume that β1(α)=- whenever 1(α) is empty. Consider also


where ES(p,λ2(p)) is the eigenspace of the second eigenvalue λ2(p) defined by (1.6).

The main properties of β1(α) are collected in the following lemma.

Lemma 6.

The following assertions hold:

  • (i)

    β1(α)=- for any αλ2(p) , and β1(α)[β1*,+) for all α>λ2(p).

  • (ii)

    β1(α) is nondecreasing for α(λ2(p),+).

  • (iii)

    β1(α) is left-continuous for α(λ2(p),+).

  • (iv)

    β1(α)+ as α+.

  • (v)

    α,β1 if and only if α>λ2(p) and β<β1(α).


(i) If αλ2(p), then 1(α)= in view of Lemma 3, and hence β1(α)=-. On the other hand, if α>λ2(p), then any second eigenfunction φ2,p satisfies Hα(φ2,p±)<0 and, in consequence, it belongs to 1(α). This implies that ES(p,λ2(p)){0}1(α) and β1(α)β1*.

Consider the set


It is known that for any α there exists C(α)>0 such that vpC(α)vq for all vX(α); see [31, Lemma 9]. Therefore, since u±X(α) for any u1(α), the Hölder inequality yields the existence of a constant C1>0 such that

C1u±qu±pC(α)u±qfor all u1(α),

which gives the boundedness of β1(α) from above.

(ii) If λ2(p)<α1α2, then 1(α1)1(α2), which implies the desired monotonicity.

(iii) Let us fix an arbitrary α0>λ2(p). Since assertion (ii) readily leads to limαα0-0β1(α)β1(α0), it is enough to show that limαα0-0β1(α)β1(α0). By the definition of β1(α0), for any ε>0 there exists uε1(α0) such that


Recalling that Hα0(uε±)<0, we can find δ=δ(ε)>0 such that Hα(uε±)<0 for any α(α0-δ,α0]. Therefore, uε1(α), and for all α(α0-δ,α0] the definition of β1(α) leads to


Combining (2.8) and (2.9), we obtain the inequality limαα0-0β1(α)β1(α0), since ε>0 is arbitrary.

(iv) Let L>λ1(q) be an arbitrary positive constant. Recalling that for the variational eigenvalues λk(q) there holds λk(q)+ as k+, we can find kL2 such that λkL(q)>L. Take an eigenfunction φ corresponding to λkL(q). Since φC01,γ(Ω¯) and φ changes its sign in Ω (see Section 1.1), there exists αL satisfying


Therefore, φ1(αL), and from the definition of β1(αL) and its monotonicity it follows that


provided ααL. Since L can be chosen arbitrary large, we conclude that limα+β1(α)=+.

(v) If α>λ2(p) and β<β1(α), then, by the definition of β1(α), there exists u1(α) such that


This means that Hα(u±)<0 and Gβ(u±)>0. Hence, by Lemma 1 we obtain t±>0 such that t±u±𝒩α,β, whence t+u+-t-u-α,β1.

Suppose now that there exists uα,β1 for some α,β. Lemma 3 implies that α>λ2(p). On the other hand, uα,β11(α). Hence, from the Nehari constraints it follows that Gβ(u±)>0, and we arrive to (2.10). ∎

Consider now the set α,β2. The corresponding critical value, parametrized again by α, appears to be the following:


where the admissible set 2(α) is defined by (2.3).

The main properties of β2(α) are similar to those for β1(α) and collected in the following lemma.

Lemma 7.

The following assertions hold:

  • (i)

    β2(α)[λ2(q),+) for any α.

  • (ii)

    β2(α) is nondecreasing for α , and β2(α)=β2(λ1(p))=λ2(q) for αλ1(p).

  • (iii)

    β2(α) is right-continuous for α.

  • (iv)

    α,β2 if and only if α and β>β2(α).


(i) It is easy to see that for any α the admissible set 2(α) is nonempty. For example, any eigenfunction corresponding to λσ(-Δp) belongs to 2(α) provided λ>max{α,λ1(p)}. Hence, β2(α)<+. On the other hand, the definition of β2(α) and characterization (2.5) with r=q directly imply that β2(α)λ2(q) for any α since 2(α)W01,pW01,q.

(ii) If α1α2, then 2(α2)2(α1), which leads to the desired monotonicity. Since any sign-changing function wW01,p satisfies Hλ1(p)(w±)>0 (see Lemma 2), we get 2(α)=2(λ1(p))={uW01,p:u±0} for all αλ1(p), and hence β2(α)=β2(λ1(p)) for all αλ1(p). In order to show that β2(λ1(p))=λ2(q), let us recall that any eigenfunction φ2,q corresponding to λ2(q) belongs to C01,γ(Ω¯) (see Section 1.1). Hence, φ2,q2(λ1(p)) and, consequently,


where the equality follows from (2.5) with r=q, and the last inequality is given by assertion (i).

Assertions (iii) and (iv) can be proved in much the same way as in Lemma 6. ∎

For the further proof of the existence of nodal solutions to ((${\mathrm{GEV};\alpha,\beta}$)) in α,β1, let us study the properties of the critical value (1.7) defined as


where the admissible set (α) is given by (1.8), or, equivalently,


We put β(α)=+ whenever (α)=. Arguing as in the proof of Lemma 3, it can be shown that (α)= if and only if α<λ2(p). Note that α,β11(α)(α).

First we give two auxiliary results.

Lemma 8.

Let α>0, βR, and {un}nN be an arbitrary sequence in BL(α) (or in Mα,β1). Denote by {vn}nN a sequence normalized as follows:


Then the following assertions hold:

  • (i)

    vn(α) (or vnα,β1 ) for all n.

  • (ii)

    vn converges, up to a subsequence, to some v0W01,p weakly in W01,p and strongly in Lp(Ω).

  • (iii)

    v0±0 and Hα(v0±)0 , that is, v0(α).


Obviously, vn±=un±/un±p, and hence assertion (i) follows from the p-homogeneity of Hα. Assertion (ii) is a consequence of the boundedness of {vn}n in W01,p. Since Hα(vn±)0 for all n, we get vn±pp1/α, whence v0±0 a.e. in Ω, due to the strong convergence of vn in Lp(Ω). Moreover, using the weak lower semicontinuity of the W01,p-norm, we conclude that Hα(v0±)lim infn+Hα(vn±)0. This is assertion (iii). ∎

Proposition 9.

For any αλ2(p) there exists a minimizer uαBL(α) of βL(α).


If αλ2(p), then (α) is nonempty, since Hα(φ2,p±)0 for any second eigenfunction φ2,p corresponding to λ2(p). Thus, there exists a minimizing sequence {un}n(α) for β(α). Consider the corresponding normalized sequence {vn}n(α) given by (2.11). Lemma 8 implies that the limit point v0(α), and hence

β(α)min{Ω|v0+|q𝑑xΩ|v0+|q𝑑x,Ω|v0-|q𝑑xΩ|v0-|q𝑑x}lim infn+min{Ω|vn+|q𝑑xΩ|vn+|q𝑑x,Ω|vn-|q𝑑xΩ|vn-|q𝑑x}=β(α),

which means that v0 is a minimizer of β(α). ∎

Remark 10.

The definition (1.7) of β(α) is equivalent to


This can be seen by testing β(α) either with the corresponding minimizer uα or with -uα.

Consider now the critical value


The following lemma contains the main properties of β(α).

Lemma 11.

The following assertions hold:

  • (i)

    β(α)=+ for any α<λ2(p) , and β(α)(λ1(q),β*] for any αλ2(p).

  • (ii)

    β(α) is nonincreasing for α[λ2(p),+).

  • (iii)

    β(α) is right-continuous for α[λ2(p),+).

  • (iv)

    𝒦α,β if and only if αλ2(p) and ββ(α) , where 𝒦α,β is defined by



(i) As stated in the proof of Lemma 3, we easily see that (α)= for all α<λ2(p), and hence β(α)=+. If αλ2(p), then ES(p,λ2(p)){0}(α), and using (2.12), we obtain that β(α)β*. Since any sign-changing function wW01,p satisfies w±qq>λ1(q)w±qq (see Lemma 2), taking a minimizer uα of β(α) (see Proposition 9), we conclude that

β(α)=uα+qq/uα+qq>λ1(q)for all αλ2(p).

Assertion (ii) can be proved as in Lemma 6.

(iii) Due to assertion (ii), it is sufficient to show that β(α0)limαα0+0β(α) for all α0λ2(p). Since β(α) is monotone and bounded in a right neighborhood of α0, for any decreasing sequence {αn}n such that αnα0+0 as n+ there holds


According to Proposition 9, for each n there exists a minimizer un(αn) of β(αn), and we can assume that un±p=1. Thus, passing to an appropriate subsequence, un converges to some u0W01,p weakly in W01,p and strongly in Lp(Ω). Moreover, u0±0 in Ω since Hαn(un±)0 implies that un±pp1/αn. Furthermore, due to the weak lower semicontinuity of the W01,p-norm, we have Hα0(u0±)0, and hence u0(α0). Consequently, using (2.12), we conclude that

β(α0)Ω|u0+|q𝑑xΩ|u0+|q𝑑xlim infn+Ω|un+|q𝑑xΩ|un+|q𝑑x=lim infn+β(αn)=limαα0+0β(α).

(iv) Assume that αλ2(p) and ββ(α). Let u(α) be a minimizer of β(α). Then Hα(u±)0 and, in view of (2.12), we may suppose that Gβ(α)(u+)=0. Therefore, Gβ(u+)Gβ(α)(u+)=0 and hence u𝒦α,β.

Suppose now that there exists u𝒦α,β for some α,β. Since 𝒦α,β(α), assertion (i) implies that αλ2(p). Moreover, since Gβ(u+)0, the definition of β(α) leads to


which completes the proof. ∎

In the sequel, it will be convenient to use the notation


Remark 12.

Due to Lemmas 6 and 11, the definitions of β1* and β* (see (2.6) and (2.13)) imply that β(α)β*β1*β1(α) for all α>λ2(p), and hence α,β1 for any (α,β)Σ.

Remark 13.

In the one-dimensional case we have


Indeed, if Ω=(0,T), then the second eigenfunction φ2,p is given explicitly through the first eigenfunction φp by φ2,p(x)=φp(2x) for x(0,T/2], and φ2,p(x)=-φp(2x-T) for x(T/2,T) (see Appendix A). Hence, Lemma 2 in Appendix A implies that


and, consequently, (2.16) holds.

2.2 Existence of positive energy nodal solutions

In this subsection, we prove the existence of nodal solutions in the set Σ defined by (2.15). To this end, we consider the minimization of the energy functional Eα,β over the set α,β1.

First, we prepare the following auxiliary lemma.

Lemma 14.

Let {un}nN be an arbitrary sequence in Mα,β1 and let {vn}nNMα,β1 be a corresponding normalized sequence given by (2.11) in Lemma 8. If un+p+ as n+, and {Eα,β(un+)}nN is bounded from above, then Gβ(v0+)0. Consequently, v0Kα,β.


Assume that {Eα,β(un+)}n is bounded from above. Recalling that -Gβ(un±)=Hα(un±)<0 by unα,β1 and noting that the equalities


hold for all u𝒩α,β, we get the boundedness of Gβ(un+):


Consequently, the weak lower semicontinuity and the assumption that un+p+ as n+ imply

Gβ(v0+)lim infn+Gβ(vn+)=lim infn+Gβ(un+)un+pq=0.

Combining this inequality with the fact that v0(α) (see Lemma 8), we conclude that v0𝒦α,β. ∎

From Remark 12 we know that α,β1 for any (α,β)Σ. Hence, there exists a minimizing sequence for Eα,β over α,β1. Moreover, this minimizing sequence, in fact, converges.

Theorem 15.

Let (α,β)ΣL. Then there exists a minimizer uMα,β1 of Eα,β over Mα,β1.


Assume {un}nα,β1 to be a minimizing sequence for Eα,β over α,β1. Equalities (2.17) imply that Eα,β(un±)>0, and hence {Eα,β(un)}n and {Eα,β(un±)}n are bounded. Applying Lemma 14, we conclude that if un+p+ as n+, then the limit v0 of a normalized sequence (2.11) belongs to the set 𝒦α,β defined by (2.14). However, 𝒦α,β= for all (α,β)Σ, due to Lemma 11 (iv). This is a contradiction. Thus, {un+}n is bounded in W01,p. Since {-un}n is also a minimizing sequence for Eα,β over α,β1, we apply the same arguments to derive that (-un)+un- is bounded in W01,p, which finally yields the boundedness of the whole sequence {un}n.

Let us now show that un+p and un-p do not converge to zero. Applying assertions (ii) and (iii) of Lemma 8 to the corresponding normalized sequence {vn}n given by (2.11), we see that its limit point v0 belongs to (α). Suppose, by contradiction, that un+p0 as n+. Then, using the Nehari constraints, we get

0<Gβ(vn+)=-un+pp-qHα(vn+)0as n+

since Hα is bounded on a bounded set and vn+p=1. Consequently, Gβ(v0+)lim infn+Gβ(vn±)=0 and Hα(v0±)lim infn+Hα(vn±)0, i.e., v0𝒦α,β, and we obtain a contradiction as above. In the case un-p0, we consider -un instead of un, and again obtain a contradiction. As a result, there holds


Now, choosing an appropriate subsequence, we get unu0 weakly in W01,p and unu0 strongly in Lp(Ω), where u0W01,p. Inequalities (2.18) together with Hα(un±)<0 imply that un±ppδ±/α for all n, and hence u0±0. At the same time, the weak lower semicontinuity yields

Hα(u0±)lim infn+Hα(un±)0.(2.19)

Let us show that


Indeed, since 𝒦α,β is empty for (α,β)Σ, we see that u0+-u0-𝒦α,β and u0--u0+𝒦α,β. This leads to Gβ(u0±)>0 since Hα(u0±)0 by (2.19). Finally, from the Nehari constraints and the weak lower semicontinuity we derive that

Hα(u0±)+Gβ(u0±)lim infn+(Hα(un±)+Gβ(un±))=0.

This means that Hα(u0±)-Gβ(u0±)<0, and hence (2.20) is shown.

According to (2.20), Lemma 1 implies the existence of unique maximum points t0+>0 of Eα,β(tu0+) and t0->0 of Eα,β(tu0-) with respect to t>0, and t0±u0±𝒩α,β. Accordingly, we conclude from (2.20) that t0+u0+-t0-u0-α,β1. Therefore,

infα,β1Eα,βEα,β(t0+u0+-t0-u0-)lim infn+Eα,β(t0+un+-t0-un-)=lim infn+(Eα,β(t0+un+)+Eα,β(t0-un-))lim infn+(Eα,β(un+)+Eα,β(un-))=lim infn+Eα,β(un)=infα,β1Eα,β.

The last inequality in this formula is due to the fact that maxt>0Eα,β(tun±)=Eα,β(un±); see Lemma 1. Consequently, t0+u0+-t0-u0-α,β1 is the minimizer of Eα,β over α,β1. ∎

Lemma 16.

Let (α,β)ΣL. If uMα,β1 is a minimizer of Eα,β over Mα,β1, then u is a critical point of Eα,β on W01,p.


The proof can be handled in much the same way as the proof of [9, Lemma 3.2], where a variant of the deformation lemma was used in a framework of the problem with indefinite nonlinearities; see also [6, Proposition 3.1]. ∎

2.3 Qualitative properties

In this subsection, we show that any minimizer u of Eα,β over α,β1 for (α,β)Σ has exactly two nodal domains (that is, connected components of Ωu-1(0)).

Lemma 17.

Let (α,β)ΣL and let uMα,β1 be a minimizer of Eα,β over Mα,β1. Then u has exactly two nodal domains.


Suppose, contrary to our claim, that there exists a minimizer uα,β1 of Eα,β over α,β1 with (at least) three nodal domains. We decompose u such that u=u1+u2+u3, where ui0 for i=1,2,3, and each ui is of a constant sign on its support. Note that each ui𝒩α,β. Indeed, uiW01,p (cf. [16, Lemma 5.6]), and since u is a solution of ((${\mathrm{GEV};\alpha,\beta}$)) by Lemma 16, we obtain

0=Eα,β(u),ui=Hα(ui)+Gβ(ui) for i=1,2,3.(2.21)

Assume, without loss of generality, that u+=u1+u2 and u-=-u3. Since uα,β1, we have


Moreover, we may assume that Hα(u2)Hα(u1), whence Hα(u2)<0. This assumption splits into the following four cases:

  • (i)


  • (ii)


  • (iii)

    Hα(u2)<0<Hα(u1) and Hα(u1)+Hα(u3)0.

  • (iv)

    Hα(u2)<0<Hα(u1) and Hα(u1)+Hα(u3)<0.

Now we will subsequently show a contradiction for each case.

(i) It is easy to see that u1+u3α,β1. Since Hα(u2)<0 leads to Eα,β(u2)>0, we have a contradiction by the following inequality:


(ii) Since Hα(u1)=0, we can derive from (2.21) that Gβ(u1)=0. Recalling that Hα(u2)<0, we get u1-u2𝒦α,β, which contradicts assertion (iv) of Lemma 11.

(iii) Recall Hα(u3)<0 and set


Since u1,u3𝒩α,β, we obtain


On the other hand, since Gβ(u3)>0, t01, and p>q, we have


Consequently, Hα(u1-t0u3)=0 and Gβ(u1-t0u3)0. Considering a function w=u1-t0u3-u2, we get w+=u1-t0u3 and w-=u2, which implies that w𝒦α,β. This is again a contradiction to the emptiness of 𝒦α,β.

(iv) Consider a function w=u1-u3-u2. Then w+=u1-u3 and w-=u2. By the assumptions, we have Hα(w±)<0. Therefore, wα,β1 and


that is, w is also a minimizer of Eα,β over α,β1 and hence a weak solution of ((${\mathrm{GEV};\alpha,\beta}$)) in view of Lemma 16. This implies that for any ξW01,p there holds


On the other hand, since u=u1+u2+u3 is also a weak solution of ((${\mathrm{GEV};\alpha,\beta}$)), we obtain


for all ξW01,p. Summarizing (2.22) and (2.23) and noting that |u||w| and |u||w|, we get


for each ξW01,p, since the supports of ui are mutually disjoint. This means that u1 is a nonnegative solution of ((${\mathrm{GEV};\alpha,\beta}$)) in Ω. However, the strong maximum principle implies that u1>0 in Ω; cf. [10, Remark 1, p. 3284]. Hence, u20 and u30, which is a contradiction. ∎

3 Nodal solutions with negative energy

In this section, we provide the main ingredients for the proofs of Theorems 6 and 7.

3.1 Auxiliary results

Consider the set


where λ0. Hereinafter, by S+k we denote the closed unit upper hemisphere in k+1 with the boundary Sk-1. We begin with the following linking lemma.

Lemma 1.

Let kN. Then h(S+k)Y(λk+1(p)) for any hC(S+k,W01,p) provided h|Sk-1 is odd.


Fix any hC(S+k,W01,p) such that h|Sk-1 is odd. If up=0 for some uh(S+k), then, obviously, uY(λk+1(p)). Thus, we may assume that up>0 for every uh(S+k). Define the map

h~(z):={h(z)/h(z)pif zS+k,-h(-z)/h(-z)pif zS-k.

It is not hard to see that h~k+1(p), where k+1(p) is the set given by (1.4) with r=p. By the definition (1.3) of λk+1(p), there exists z0Sk such that h~(z0)ppλk+1(p). Since h~(z0)S(p), we have h~(z0)ppλk+1(p)h~(z0)pp. Moreover, since h~ is odd, we may suppose that z0S+k. Consequently, we obtain h(z0)Y(λk+1(p)). ∎

Lemma 2.

Let α,βR and let λ>max{α,0}. Then Eα,β is bounded from below on Y(λ).


Assume that uY(λ) with λ>max{α,0}. Using the Hölder inequality, we obtain


which implies the desired conclusion since q<p. ∎

Lemma 3.

Assume α,βR and let {un}nN be a sequence in W01,p which satisfies unp+ and Eα,β(un)/unpp-10 in (W01,p)* as n+. Then vn:=un/unp has a subsequence strongly convergent in W01,p to some v0ES(p;α){0}, that is, ασ(-Δp).


Since vnp=1 for any n, passing to an appropriate subsequence, we may assume that vn converges to some v0 weakly in W01,p and strongly in Lp(Ω). In particular, Hα(v0),vnHα(v0),v0 as n+. Moreover,


as n+, by the assumption. Using these facts, we get


where the last inequality is obtained by Hölder’s inequality. Hence, vnpv0p=1 as n+, and the uniform convexity of W01,p implies that vn converges to v0 strongly in W01,p.

On the other hand, for any ξW01,p the following equality holds:


Therefore, passing to the limit as n+, we derive


for all ξW01,p, that is, v0ES(p;α){0}. ∎

Lemma 4.

If ασ(-Δp), then Eα,β satisfies the Palais–Smale condition.


Let {un}nW01,p be a Palais–Smale sequence for Eα,β, that is,


as n+, where c is a constant. Due to the (S+)-property for the operator -Δp-Δq (see Remark 5 below), it is sufficient to show that {un}n is bounded in W01,p. If we suppose, by contradiction, that unp+ as n+, then Lemma 3 implies that ασ(-Δp), which contradicts the assumption of the lemma. ∎

Remark 5.

For the reader’s convenience we show that the operator -Δp-Δq has the (S+)-property, namely, any sequence {un}nW01,p converging to some u0 weakly in W01,p and satisfying

lim supn+-Δpun-Δqun,un-u00(3.1)

converges strongly in W01,p. Let unu0 in W01,p as n+, and let (3.1) hold. Then the Hölder inequality yields


which implies that unpu0p and unqu0q as n+. Due to the uniform convexity of W01,p, we conclude that un converges to u0 strongly in W01,p.

Recall the definition (1.9):


Lemma 6.

If ασ(-Δp), then λ1(q)βU*(α)<+.


Let ασ(-Δp). Recall that [31, Lemma 9] implies the existence of a constant C(α)>0 such that upC(α)uq for any uX(α), where X(α) is defined by (2.7). Thus, applying the Hölder inequality, we get


for any uX(α). Therefore, β𝒰*(α)<+ since ES(p;α)X(α). On the other hand, it is clear that β𝒰*(α)λ1(q) provided ES(p;α){0}. ∎

In the one-dimensional case we can clarify the bounds for β𝒰*(α) as follows.

Lemma 7.

Let N=1 and α=λk(p), kN. Then



Let Ω=(0,T), T>0, and α=λk(p) for some k. It is known that λk(r)=(r-1)(kπrT)p for any r>1 and k (cf. Appendix A), and hence the first and third equalities in (3.3) are satisfied.

Note that the eigenspace ES(p;λk(p)) is one-dimensional, as it follows from [17, Proposition 2.1]. Denoting the corresponding eigenfunction as φk, we directly get β𝒰*(λk(p))=φkqq/φkqq. On the other hand, φk has exactly k nodal domains of equivalent length (see Appendix A), and hence the standard scaling yields β𝒰*(λk(p))=kqφpqq/φpqq, where φp is the first eigenfunction of -Δp. The inequalities in (3.3) follow from Lemma 2 below. ∎

The following lemma ensues readily from the definition (3.2).

Lemma 8.

Let ασ(-Δp) and β>βU*(α). Then Gβ(φ)<0 for all φES(p;α){0}.

Lemma 9.

Let αR and kN. If β>λk+1(q), then there exist an odd map h0C(Sk,W01,p) and t0>0 such that



Let β>λk+1(q) and choose ε satisfying


By the definition of λk+1(q), there exists a map h1k+1(q) such that


Note that by taking t>0 small enough it is easy to get maxzSkEα,β(th1(z))<0. However, h1C(Sk,S(q)), and we do not know a priori that h1C(Sk,W01,p). Hence the arguments below are needed.

Since C0(Ω) is a dense subset of W01,q and h1 is odd, for any zSk we can find uzC0(Ω) such that


By the continuity of h1, for any zSk there exists δ(z)(0,1) such that

h1(z)-h1(y)q<εfor all ySk with |z-y|<δ(z).(3.7)

Considering min{δ(z),δ(-z)} instead of δ(z), we may assume that δ is even. Note that (3.6) and (3.7) lead to

uz-h1(y)q<2εfor all ySk such that |z-y|<δ(z).(3.8)

Due to the compactness of Sk, we may choose a finite number of points ziSk, i=1,2,,m, such that


where B(zi,δ(zi))k+1 is a ball of radius δ(zi) centered at the point zi. Now, for each i=1,2,,m we take a function ρiC0(k+1) such that

suppρi=B(zi,δ(zi))¯andρi>0 in B(zi,δ(zi)).

Note that B(zi,δ(zi))B(-zi,δ(-zi))= for all i=1,2,,m since δ(zi)=δ(-zi)<1. Thus, ρi(-z)=0 whenever ρi(z)>0. Define

ρ~i(z):=ρi(z)j=1m(ρj(z)+ρj(-z))for zSk.

Since {B(zi,δ(zi))B(-zi,δ(-zi))}i=1m is an open covering of Sk, it is easy to see that ρ~iC(Sk) for all i=1,2,,m. Moreover,

0ρ~i1,ρ~i(-z)=0provided ρ~i(z)>0 and j=1m(ρ~j(z)+ρ~j(-z))=1(3.9)

for all zSk and i=1,2,,m. That is, {ρ~i}i=1m forms a partition of unity of Sk. Set

h0(z):=i=1m(ρ~i(z)uzi+ρ~i(-z)u-zi)i=1muzi(ρ~i(z)-ρ~i(-z))for zSk.

Evidently, h0 is odd, and the continuity of ρ~i implies that h0C(Sk,W01,p).

Let us show that maxzSkEα,β(th0(z))<0 for sufficiently small t>0. First, for all zSk there holds


where we used that uziqq<λk+1(q)+2ε, by virtue of (3.6) and (3.5). Moreover, h0(z)0 for all zSk. Indeed, using the convexity of qq, the oddness of h1, (3.9) and (3.8), we derive


since ρ~i(-z)>0 if and only if -zB(zi,δ(zi)). Hence, h0(z)qh1(z)q-2ε=1-2ε>0 for every zSk. Now using (3.10) and (3.4), we get


for all zSk. Thus, for sufficiently small t>0 and any zSk we obtain


since q<p. This is the desired conclusion. ∎

In the sequel, we will also need the following variant of the deformation lemma. We refer the reader to [14, Theorem 3.2] for the proof.

Lemma 10.

Let Ψ be a C1-functional on a Banach space W, let Ψ satisfy the Palais–Smale condition at any level c[a,b] and let Ψ have no critical values in (a,b). Assume that either Ka:={uW:Ψ(u)=0,Ψ(u)=a} consists only of isolated points, or Ka=. Define Ψc:={uW:Ψ(u)c}. Then there exists ηC([0,1]×W,W) such that the following hold:

  • (i)

    Ψ(η(s,u)) is nonincreasing in s for every uW.

  • (ii)

    η(s,u)=u for any uΨa, s[0,1].

  • (iii)

    η(0,u)=u and η(1,u)Ψa for any ΨbKb.

  • (iv)

    If Ψ is even, then η(s,) is odd for all s[0,1].

That is, Ψa is a strong deformation retract of ΨbKb.

3.2 General existence result via minimax arguments

In this subsection, we prove a result on the existence of an abstract nontrivial solution to ((${\mathrm{GEV};\alpha,\beta}$)). Let us emphasize that this result does not guarantee that the obtained solution is sign-changing. (However, it is shown in [10] that for sufficiently large α and β problem ((${\mathrm{GEV};\alpha,\beta}$)) has no sign-constant solutions).

Recall that we denote kα:=min{k:α<λk+1(p)}.

Theorem 11.

Assume that αRσ(-Δp)¯. Then for any β>max{βU*(α),λkα+1(q)} problem ((${\mathrm{GEV};\alpha,\beta}$)) has a nontrivial solution u with Eα,β(u)<0, where βU*(α) is defined by (1.9).


Since ασ(-Δp)¯, we need to investigate two cases:

  • (i)


  • (ii)

    ασ(-Δp) and there exists a sequence {αn}nσ(-Δp) such that limn+αn=α.

Case (i): Let β>λkα+1(q)=max{β𝒰*(α),λkα+1(q)}. Then Lemma 9 guarantees the existence of an odd h0C(Skα,W01,p) and of t0>0 such that


Moreover, by the definition of kα we have α<λkα+1(p), and hence Lemma 2 implies that Eα,β is bounded from below on Y(λkα+1(p)), that is,


Since t0h0() is odd and Eα,β is even, Lemma 1 justifies that Eα,β(t0h0(z0))δ0 for some z0S+kα, and hence δ0ρ. We are going to show that Eα,β has at least one critical value in [δ0-1,ρ]. Suppose, by contradiction, that Eα,β has no critical values in [δ0-1,ρ]. Recall that Eα,β satisfies the Palais–Smale condition by Lemma 4 because we are assuming that ασ(-Δp). Then, due to Lemma 10, there exists ηC([0,1]×W01,p,W01,p) such that η(s,) is odd for every s[0,1] and

Eα,β(η(1,t0h0(z)))δ0-1for all zSkα.(3.12)

On the other hand, noting that η(1,t0h0())|S+kαC(S+kα,W01,p) and η(1,t0h0())|Skα-1 is odd, Lemma 1 guarantees the existence of a point z1S+kα such that


whence δ0Eα,β(η(1,t0h0(z1))) by the definition of δ0 (see (3.11)). However, this contradicts (3.12).

Case (ii): Let β>max{β𝒰*(α),λkα+1(q)}. As in the former case, according to Lemma 9, there exist an odd map h0C(Skα,W01,p) and t0>0 such that


Recalling that α<λkα+1(p) and discarding, if necessary, a finite number of terms of the sequence {αn}n, we may suppose that αn<λkα+1(p) and


for all n. Since αnσ(-Δp), we apply the proof of case (i) to each αn<λkα+1(p) and β>λkα+1(q), and hence obtain a sequence of critical values cn of Eαn,β such that


Let unW01,p be a critical point of Eαn,β corresponding to the level cn, i.e., Eαn,β(un)=cn. We proceed to show that {un}n is bounded in W01,p. Suppose, by contradiction, that unp+ as n+. Set vn:=un/unp and note that


as n+. Thus, due to Lemma 3, we have that vn converges strongly in W01,p, up to a subsequence, to some v0ES(p,α){0}. Let us prove that Gβ(v0)=0. By (3.14), we have


To obtain a converse estimate, we show that δn is bounded from below. Since limn+αn=α<λkα+1(p), we can choose α0 such that αn<α0<λkα+1(p) for all sufficiently large n. Thus, Lemma 2 implies that Eα0,β is bounded from below on Y(λkα+1(p)). Noting that Eαn,β(u)Eα0,β(u) for any uW01,p, we get δninf{Eα0,β(u):uY(λkα+1(p))}>- for all n large enough, which is the desired boundedness. Using this fact, the two equalities in (3.17), and (3.15), we derive that


as n+, which leads to Gβ(v0)=0, because vnv0 strongly in W01,p. On the other hand, since ασ(-Δp) and β>β𝒰*(α), we get

Gβ(φ)=φqq-βφqq0for all φES(p;α){0};(3.18)

see Lemma 8. Hence, we obtain a contradiction since Gβ(v0)=0 and v0ES(p,α){0}. Thus, from (3.16) it follows that {un}n is a bounded Palais–Smale sequence for Eα,β. Then the (S+)-property of the operator -Δp-Δq (see Remark 5) implies that un converges strongly in W01,p, up to a subsequence, to some critical point u0 of Eα,β. Furthermore, u0 is nontrivial and its energy is negative since

Eα,β(u0)=lim supn+Eαn,β(un)=lim supn+cnlim supn+ρnρ+o(1)<0

by (3.13) and (3.14). ∎

Remark 12.

Note that the proof of case (ii) gives more. Namely, if ασ(-Δp) and limn+αn=α for some sequence {αn}nσ(-Δp), and β>λkα+1(q) is such that (3.18) holds, then there exists a nontrivial solution to ((${\mathrm{GEV};\alpha,\beta}$)).

3.3 General existence result via the descending flow

In the last part of this section, we use the descending flow method to provide an existence result for (p,q)-Laplace equations with a nonlinearity in the general form.

Suppose that h:Ω× is a Carathéodory function satisfying h(x,0)=0 for a.e. xΩ and there exists C>0 such that

|h(x,s)|C(1+|s|p-1)for every s and a.e. xΩ.(3.19)

Under (3.19), we define a C1-functional J on W01,p by


For simplicity, we denote the positive cone in C01(Ω¯) by

P:={uC01(Ω¯):u(x)>0 for all xΩ}.(3.21)

The following result can be proved by the same arguments as [27, Theorem 11]. For the reader’s convenience, we give a sketch of the proof in Appendix B.

Theorem 13.

Assume that the following conditions hold:

  • (A1)

    There exists λ0>0 such that

    h(x,u)u+λ0(|u|q+|u|p)0for every u and a.e. xΩ.

  • (A2)

    There exists γC([0,1],C01(Ω¯)) such that γ(0)P, γ(1)-P and maxs[0,1]J(γ(s))<0.

If, moreover, J is coercive on W01,p, then J has at least three critical points w1intP, w2-intP, and w3C01(Ω¯)(P-P) such that J(wi)maxs[0,1]J(γ(s))<0 for i=1,2,3. Here

intP:={uP:u(x)/ν<0 for all xΩ},

and ν denotes the unit outer normal vector to Ω.

We say that vW01,p is a (weak) super-solution of ((${\mathrm{GEV};\alpha,\beta}$)) whenever for all nonnegative φW01,p there holds


Applying Theorem 13 to a truncated functional corresponding to Eα,β, we show the following result on the existence of nodal solutions to ((${\mathrm{GEV};\alpha,\beta}$)) with a negative energy.

Proposition 14.

Let αR and β>λ2(q). If there exists a super-solution of ((${\mathrm{GEV};\alpha,\beta}$)) which belongs to intP, then ((${\mathrm{GEV};\alpha,\beta}$)) has a nodal solution u such that Eα,β(u)<0.


Let vintP be a super-solution of ((${\mathrm{GEV};\alpha,\beta}$)) with α and β>λ2(q). Note that -v becomes a negative sub-solution of ((${\mathrm{GEV};\alpha,\beta}$)). Using v, we truncate the right-hand side of ((${\mathrm{GEV};\alpha,\beta}$)) as follows:

f(x,s):={αv(x)p-1+βv(x)q-1if s>v(x),α|s|p-2s+β|s|q-2sif -v(x)sv(x),-αv(x)p-1-βv(x)q-1if s<-v(x).

It is easy to see that f is the Carathéodory function and f(x,0)=0 for all xΩ. Moreover, f satisfies (3.19) and, taking λ0=max{|α|,|β|}, it satisfies assumption (A1) of Theorem 13.

Define a corresponding truncated C1-functional I on W01,p by


Note that the boundedness of v in Ω implies the boundedness of f, and therefore I is coercive on W01,p. To apply Theorem 13 it remains to show that (A2) holds. To this end, let us construct an appropriate path γ0. Choose ε>0 satisfying λ2(q)+2ε<β. By the characterization (1.5) of λ2(q), there exists γC([0,1],S(q)) such that


Using the density arguments (as in the proof of Lemma 9), we can obtain a path γ~C([0,1],C01(Ω¯){0}) such that γ~(0)P, γ~(1)-P, and


for every s[0,1]. Since vintP and γ~C([0,1],C01(Ω¯){0}), we get for any t>0 small enough, s[0,1] and xΩ that


and hence




for sufficiently small t>0 since


Thus, for such a small t>0 the path γ0(s):=tγ~(s) satisfies assumption (A2) of Theorem 13.

As a result, according to Theorem 13, we obtain a sign-changing critical point uC01(Ω¯)(P-P) of I satisfying I(u)maxt[0,1]I(γ0(t))<0. By the standard argument, we can show that -vuv in Ω. In fact, recalling that v is a super-solution of ((${\mathrm{GEV};\alpha,\beta}$)) and taking (u-v)+W01,p as a test function for I(u)-Eα,β(v), we obtain


which implies that (u-v)+0 and hence uv in Ω. Similarly, taking -(u-(-v))- as a test function, we get u-v. Therefore, u is a nodal solution of ((${\mathrm{GEV};\alpha,\beta}$)) and Eα,β(u)=I(u)maxs[0,1]I(γ0(s))<0. ∎

4 Proofs of the main results

In this section, we collect the proofs of our main results stated in Section 1.2.

Proof of Theorem 3.

Recall that any sign-changing solution of ((${\mathrm{GEV};\alpha,\beta}$)) belongs to the nodal Nehari set α,β defined by (2.1). At the same time, α,β is empty under the assumptions of the theorem, as is shown in Lemma 4, which completes the proof. ∎

Proof of Theorem 5.

The desired conclusion follows directly from the combination of Theorem 15 and Lemmas 16 and 17. ∎

Proof of Theorem 6.

Note that problem ((${\mathrm{GEV};\alpha,\beta}$)) possesses an abstract nontrivial solution uW01,p with Eα,β(u)<0 for any


by Theorem 11. If u is a nodal solution, then we are done. If u is a nontrivial nonnegative solution, then uintP (see, e.g., [10, Remark 1, p. 3284]), and hence Proposition 14 guarantees the existence of a nodal solution v of ((${\mathrm{GEV};\alpha,\beta}$)) such that Eα,β(v)<0. ∎

Proof of Theorem 7.

If α<λ1(p) or λ1(p)<α<λ2(p), then for all β>λ2(q) there exists a nodal solution, as follows from Theorem 6. If α=λ1(p), then, as noted in Remark 12, Theorem 11 implies the existence of an abstract nontrivial negative energy solution of ((${\mathrm{GEV};\alpha,\beta}$)) for any β>λ2(q) such that Gβ(φp)0. Since the first eigenfunction φp of -Δp is unique, up to a multiplier, we derive the existence under the assumption βφpqq/φpqq. If the obtained solution changes its sign, then we are done. Otherwise, we apply Proposition 14 and obtain the existence of a nodal solution with a negative energy. ∎

Finally, we will prove the nonexistence result in the one-dimensional case.

Proof of Theorem 4.

Let N=1 and Ω=(0,T), T>0. We temporarily denote by λk(r,S) the kth eigenvalue of -Δr on (0,S) subject to zero Dirichlet boundary conditions, r>1, S>0 (see Appendix A). Suppose, by contradiction, that αλ2(p,T) and βλ2(q,T), but there exists a nodal solution u for ((${\mathrm{GEV};\alpha,\beta}$)). Evidently, there is at least one nodal domain of u which length S is less than or equal to T/2. Using, if necessary, the translation of the coordinate axis, we may assume that u is a constant-sign solution of ((${\mathrm{GEV};\alpha,\beta}$)) on interval (0,S). Define v:=u on (0,S) and v=0 on [S,T/2]. Clearly, vW01,p(0,S)W01,p(0,T/2). Moreover, it is not hard to see that


for any r>1. Thus, (1.2) and the assumption ST/2 lead to the inequalities


Taking now v as a test function for (1.1), we arrive at


and hence we have equalities in (4.1). On the other hand, the simplicity of λ1(r,S) implies that v is the first eigenfunction corresponding to λ1(p,S) and λ1(q,S), simultaneously. However, this is a contradiction, since φp and φq are linearly independent for N=1 (see [20, Lemma 4.3] or Lemma 1 below). ∎

A Appendix A

In this section, we show some relations between eigenvalues and eigenfunctions of the p- and q-Laplacians in the one-dimensional case. Consider the eigenvalue problem

{-(|u|r-2u)=λ|u|r-2uin (0,T),u(0)=u(T)=0,

where r>1 and T>0. It is known (cf. [17, Theorem 3.1]) that σ(-Δr) is exhausted by eigenvalues

λk(r)=(r-1)(kπrT)r,where πr=2πrsin(π/r).

(It is not hard to see that πr is a decreasing function of r>1.) The corresponding eigenfunctions are denoted by sinr(kπrtT), where sinr(t) is the inverse function of 0x(1-sr)-1/r𝑑s, x[0,1], extended periodically and anti-periodically from [0,πr/2] to the whole (see also [11]). By construction, sinr(kπrtT) has exactly k nodal domains of the length T/k on (0,T). As usual, we denote the first eigenfunction sinr(πrtT) as φr.

For the convenience of the reader we briefly prove that the first eigenfunctions φp and φq are linearly independent; see also [20, Lemma 4.3] for a different proof.

Lemma 1.

Let N=1 and qp. Then φp and φq are linearly independent.


Suppose, by contradiction, that φp(t)=φq(t) for all t[0,T]. In particular, we have


for all t[0,T/2]. By the definitions of sinp and sinq, we obtain

1πp0x(1-sp)-1/p𝑑s=1πq0x(1-sq)-1/q𝑑sfor all x[0,1].

Using a Taylor series, we get (1-sp)-1/p=1+O(sp) and (1-sq)-1/q=1+O(sq) in a neighborhood of s=0. Thus,


for sufficiently small x>0, which implies that πp=πq since p,q>1. However, this contradicts the monotonicity of πr with respect to r>1. ∎

Next, we prove the main result of the section.

Lemma 2.

Let N=1 and 1<q<p<+. Then



The first inequality is trivial because the first eigenvalue λ1(q) is simple and φpφq (see [20] or Lemma 1). Let us prove by direct calculations that

φpqqφpqq<λ2(q)for q<p.

Note that


Using the formulas


from [11, Proposition 3.1], where B(x,y):=01tx-1(1-t)y-1𝑑t is the beta function with real x,y>0, it becomes sufficient to prove that


We will subsequently simplify (A.2), to obtain an easier sufficient condition. Note that, by definition,


Note that


where Γ(y) is the gamma function; cf. [3, Theorem 1.1.4]. Hence, combining the Euler reflection formula Γ(y)Γ(1-y)=πsinπy (see, e.g., [3, p. 9]) with the identity xΓ(x)=Γ(x+1), we obtain


Applying (A.3) to B(q+1p,p-1p) with x=q/p and y=1/p, we get


Therefore, using the estimate


we arrive at


Thus, comparing the right-hand sides of (A.2) and (A.4), we get the following sufficient condition for the assertion of the lemma:


To prove this inequality, we first obtain an appropriate upper bound for its left-hand side. From [3, p. 8] we know that


since for all n there holds


Next, we will get a suitable lower bound for the right-hand side of (A.5). Since πr is a decreasing function of r>1 (in fact dπr/dr<0), we have πq/πp>1 for q<p. Hence,


since sinx<x for all x>0.

Let us consider three cases. Assume first that 1<q<p2. By a direct analysis, the minimum value of the right-hand side 2q/q of (A.7) is greater than 16/9. Since the right-hand side


of (A.6) is strictly increasing with respect to p>1, it is easy to see that

2p(p+2)(p+1)2169for all 1<p2.

Combining these facts, we prove that (A.5) holds for 1<q<p2.

Secondly, assume that 2q<p. Noting that 2q/q is, in fact, strictly increasing for q2, we obtain


for all q2 and p>1. Thus, (A.6) and (A.7) yield (A.5) for 2q<p.

Finally, we assume that 1<q<2p. Since πr is decreasing, p2 implies that πpπ, and we refine inequality (A.7) in the following way:


It is not hard to check that

22q-1qq>2>2p(p+2)(p+1)2for all q(1,2),

which again implies (A.5).

Therefore, (A.5) holds for all 1<q<p<+, which completes the proof. ∎

If we swap p and q in Lemma 2, then an opposite situation occurs.

Lemma 3.

Let N=1. Then for any kN there exist 1<q0<p0 such that

φqppφqpp>λk(p)for all 1<q<q0 and p>p0.


The case k=1 is obvious. Let k2. Similar to (A.1) and (A.2), it is sufficient to show that


Note first that


and for q<p there holds


Therefore, (A.8) can be simplified as


Note that sin(πq)=sin(πq(q-1))<πq(q-1). Hence, using the estimates


we arrive at the following sufficient inequality:


At the same time, (q-1)p-1p(q-1)12 for 1<q<2<p, and, choosing p1>2 large enough, we obtain 2p2/p for any pp1. Therefore, to prove (A.8) it is sufficient to show that


However, (A.9) is obviously satisfied for any q<1+1(4k)2 and sufficiently large p>p1. ∎

B Appendix B: Sketch of the proof of Theorem 13

Let us consider a map Tλ:W01,p(W01,p)* defined for λ>0 by


for u,vW01,p. The following properties of Tλ can be proved in much the same way as in the proof of [27, Propositions 9, 10].

Lemma 1.

Tλ is invertible and Tλ-1:(W01,p)*W01,p is continuous. Moreover, if 1<pN and r>N/p, then there exists a constant D0>0 such that for all uLr(Ω) we have


Let us define ψ(u):=|u|p-2u+|u|q-2u and a map Bλ:W01,pW01,p by


for uW01,p and λ>0. According to Lemma 1 and assumption (3.19), we see that Bλ is well-defined and continuous. Moreover, critical points of the energy function J given by (3.20) correspond to fixed points of Bλ, see [27, Remark 12]. Throughout this section, K:={uW01,p:J(u)=0} is the set of critical points of J, and, to shorten notation, we write u instead of up for uW01,p.

By the standard calculations, we have the following facts (cf. [5, Lemmas 3.7 and 3.8] for details).

Lemma 2.

Let λ>0. Then there exist constants di=di(λ)>0, i=1,2,,6 such that for all uW01,p the following assertions hold:

  • (i)

    J(u),u-Bλ(u)d1u-Bλ(u)2((u+Bλ(u))p-2+(u+Bλ(u))q-2) for 1<q<p2.

  • (ii)

    J(u),u-Bλ(u)d2(u-Bλ(u)p+u-Bλ(u)q) for 2q<p.

  • (iii)

    J(u),u-Bλ(u)d3u-Bλ(u)2(u+Bλ(u))q-2+d3u-Bλ(u)p for 1<q2p.

  • (iv)

    J(u)(W01,p)*d4(u-Bλ(u)p-1+u-Bλ(u)q-1) for 1<q<p2.

  • (v)

    J(u)(W01,p)*d5u-Bλ(u)((u+Bλ(u))p-2+(u+Bλ(u))q-2) for 2q<p.

  • (vi)

    J(u)(W01,p)*d6u-Bλ(u)(u+Bλ(u))p-2+d6u-Bλ(u)q-1 for 1<q2p.

Then similar arguments as in [27, Lemma 17] (see also [5, Lemma 4.1]) can be applied to prove the following result on the existence of a locally Lipschitz continuous pseudo-gradient vector field in order to produce an invariant descending flow with respect to the positive and negative cones ±P defined by (3.21).

Lemma 3.

Let λ>λ0, where λ0>0 is given by assumption (A1) of Theorem 13. Then, there exists a locally Lipschitz continuous operator Vλ:C01(Ω¯)KC01(Ω¯) such that the following hold:

  • (i)

    For any uC01(Ω¯)K we have

    J(u),u-Vλ(u)d12u-Bλ(u)2{(u+Bλ(u))p-2+(u+Bλ(u))q-2}for 1<q<p2,J(u),u-Vλ(u)d22(u-Bλ(u)p+u-Bλ(u)q)for 2q<p,J(u),u-Vλ(u)d32u-Bλ(u)2(u+Bλ(u))q-2+d32u-Bλ(u)pfor 1<q2p,12u-Bλ(u)u-Vλ(u)2u-Bλ(u).

    Here d1, d2 , and d3 are the positive constants from Lemma 2.

  • (ii)

    Vλ(u)±intP for every u±PK , respectively.

  • (iii)

    Let p*:=NpN-p for N>p , and p*:=p+1 otherwise. Set r0:=p* and define a sequence {rn}n inductively as follows:


    Then for any n there exists a constant Cn*>0 such that

    Vλ(u)rn+1Cn+1*(2+|Ω|+urn)for all uC01(Ω¯)K.

  • (iv)

    If Np and r>max{N/p,1/(p-1)} , then there exists a constant D1>0 such that

    Vλ(u)D1(ur(p-1)+2+|Ω|)for all uC01(Ω¯)K.

  • (v)

    There exists a constant D2>0 such that

    Vλ(u)D2(2+u)for all uC01(Ω¯)K.

  • (vi)

    For every R>0 there exist γ(0,1) and M>0 such that

    Vλ(u)C01,γ(Ω¯)Mfor all uC01(Ω¯)K with uR.

Now, we will give the proof of Theorem 13.

Proof of Theorem 13.

Note first that the boundary of ±P in C01(Ω¯) does not intersect with K{0} since any nonnegative (resp. nonpositive) and nontrivial solution of corresponding equation is strictly positive (resp. negative) in Ω and u/ν<0 (resp. >0) on Ω under assumption (A1) of the theorem, due to the strong maximum principle and boundary point lemma (see [29, Theorem 5.3.1 and Theorem 5.5.1]).

Take λ>λ0 and let Vλ be a locally Lipschitz continuous operator given by Lemma 3. Consider the following initial value problem in C01(Ω¯):


Denote by η(t,u)C01(Ω¯) its unique solution on the right maximal interval [0,τ(u)). According to assertion (ii) of Lemma 3, η(t,u) is the invariant descending flow with respect to the positive cone P and the negative cone -P, namely, η(t,u)±intP for all 0<t<τ(u) provided u±PK (see [23, Lemma 3.2]). Define the sets

Q±:={uC01(Ω¯)K:η(t,u)±intP for some t[0,τ(u))}(±intP).

It is known that Q± are open subsets of C01(Ω¯) invariant for the descending flow η, and Q± are closed subsets of C01(Ω¯) invariant for η; see [23, Lemma 2.3].

Choose a constant c satisfying maxs[0,1]J(γ(s))<c<0, where γ is the continuous path given by assumption (A2) of the theorem. Since γ(0)Q+, γ(1)Q-, and Q± are open in C01(Ω¯), there exist 0<s+s-<1 such that γ(s+)Q+ and γ(s-)Q-. Put u1:=γ(0), u2:=γ(1), and u3:=γ(s+). Due to assertion (i) of Lemma 3, we know that


which implies that

-<infW01,pJJ(η(t,ui))c<0for every t[0,τ(ui)).

Hence, the coercivity of J guarantees the existence of R>0 such that for all t[0,τ(ui)) we have


Therefore, if τ(ui)< for i=1,2,3, then for every 0<t1<t2<τ(ui)< we have


by assertion (i) of Lemma 3 and (B.1). Thus, η(t,ui) converges to some wi in W01,p as tτ(ui)-0 whenever τ(ui)<. On account of Lemma 3 and [27, Lemma 18 (ii)], it is not hard to prove that wiK and η(t,ui) converges to wi in C01(Ω¯) as tτ(ui)-0. Recalling now that Q± and Q± are invariant, we see that J(wi)J(ui)c<0, i=1,2,3, and w1intP, w2-intP, w3Q+. Since Q+(±P{0})= (note that ±P{0}Q±), our conclusion is proved provided τ(ui)< for i=1,2,3.

Assume that τ(ui)= for some i{1,2,3}. In this case, we can prove the existence of a sequence {tn}n+ such that

tn+andJ(η(tn,ui))0in (W01,p)* as n+.(B.2)

Note that there exists a sequence {tn}n+ such that tn+ and ddtJ(η(tn,ui))0 as n+ since

-<infW01,pJJ(η(t,ui))cfor all t0

and J(η(t,ui)) is nondecreasing in t. Let us show that this sequence satisfies (B.2). If 1<q<p2, then Lemma 2 (iv), Lemma 3 (i), and (B.1) imply


for all t>0. Hence,

J(η(tn,ui))(W01,p)*0as n+.

The cases 2q<p and 1<q2p can be handled in a similar way using the estimates of Lemma 2 and Lemma 3 (i).

Combining now (B.2) with (B.1), we conclude that {η(tn,ui)}n is a bounded Palais–Smale sequence to J. At the same time, it is not hard to show that J satisfies the Palais–Smale condition because the coercivity of J implies the boundedness of any Palais–Smale sequence (see Lemma 4). Thus, there exists wiW01,pK such that limn+η(tn,ui)=wi in W01,p, up to an appropriate subsequence. Furthermore, arguing as in the proof of [27, Lemma 18 (iii)], using Lemma 3 (iii)–(vi) and (B.1), we see that {η(t,ui):t0} is bounded in C01,ν(Ω¯) for some ν(0,1). Thus, the compactness of C01,ν(Ω¯)C01(Ω¯) and limn+η(tn,ui)=wi in W01,p imply that limn+η(tn,ui)=wi in C01(Ω¯). Therefore, w1intP, w2-intP and w3C01(Ω¯)(P-P). ∎


The first author wishes to thank the Tokyo University of Science, where the main constructions and results of the article were obtained, for the invitation and hospitality.


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About the article

Received: 2016-08-02

Accepted: 2016-09-27

Published Online: 2016-12-20

Funding Source: Japan Society for the Promotion of Science

Award identifier / Grant number: 15K17577

Funding Source: Ministerstvo Školství, Mládeže a Tělovýchovy

Award identifier / Grant number: LO1506

This work was supported by JSPS KAKENHI grant number 15K17577. The work of the first author was also supported by the project LO1506 of the Czech Ministry of Education, Youth and Sports.

Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 101–129, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2016-0172.

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