# A Multiplicity Result for Quasilinear Problems with Nonlinear Boundary Conditions in Bounded Domains

Authors: S. Khademloo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We study the following quasilinear problem with nonlinear boundary condition-Δpu-λa(x)u|u|p-2=b(x)u|u|γ-2, in Ω and (1-α)|∇u|p-2(∂u/∂n)+αu|u|p-2=0,
on ∂Ω, where Ω⊆RN is a connected bounded domain with smooth boundary ∂Ω, the outward unit normal to which is denoted by n. Δp is the p-Laplcian operator defined by Δpu=div(|∇u|p-2∇u), the functions a and b are sign changing continuous functions in Ω, 1<p<γ<p*, where p*=Np/(N-p) if N>p and ∞ otherwise. The properties of the first eigenvalue λ1+(α) and the associated eigenvector of the related eigenvalue problem have been studied in (Khademloo, In press). In this paper, it is shown that if λ≤λ1+(α), the original problem admits at least one positive solution, while if λ1+(α)<λ<λ*, for a positive constant λ*, it admits at least two distinct positive solutions. Our approach is variational in character and our results extend those of Afrouzi and Khademloo (2007) in two aspects: the main part of our differential equation is the p-Laplacian, and the boundary condition in this paper also is nonlinear.


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## Body

## 1. Introduction and Results

In this paper, we consider the problem(1.1)-Δpu-λa(x)u|u|p-2=b(x)u|u|γ-2,x∈Ω,(1-α)|∇u|p-2∂u∂n+αu|u|p-2=0,x∈∂Ω,
where Ω⊆RN is a connected bounded domain with a smooth boundary ∂Ω, the outward unit normal to which is denoted by n. Δp is the p-Laplcianian operator defined by Δpu=div(|∇u|p-2∇u). The functions a and b are assumed to be sign changing in Ω. Here, we say a function a(x) changes sign if the measure of the sets {x∈Ω;a(x)>0} and {x∈Ω;a(x)<0} are both positive. λ≥0 is a real parameter and exponent γ is assumed to satisfy the condition 1<p<γ<p*, where p*=Np/(N-p) if N>p and ∞ otherwise.A host of literature exists for this type of problem whenp=2. For the works concerning with problems similar to (1.1) in the case p=2, we refer to [1–3] and references therein.The growing attention in the study of thep-Laplace operator is motivated by the fact that it arises in various applications, for example, non-Newtonian fluids, reaction diffusion problems, flow through porus media, glacial sliding, theory of superconductors, biology, and so forth (see [4, 5] and the references therein).In this paper, we obtain new existence results by using a variational method based on the properties of eigencurves, that is, properties of the mapλ→μ(α,λ), where μ(α,λ) denotes the principal eigenvalue of the problem(1.2)-Δpu-λa(x)u|u|p-2=μu|u|p-2,x∈Ω,(1-α)|∇u|p-2∂u∂n+αu|u|p-2=0,x∈∂Ω.
Similar to [6] our method works provided that the eigenvalue problem(1.3)-Δpu=λa(x)u|u|p-2,x∈Ω,(1-α)|∇u|p-2∂u∂n+αu|u|p-2=0,x∈∂Ω,
has principal eigenvalues and it can be shown that this occurs on an interval [α0,1] where α0≤0. Thus, we are able to obtain existence results for problem (1.2) even in the case of nonlinear Neumann boundary conditions where α is small and negative. Our method depends on using eigencurves to produce an equivalent norm on W1,p(Ω); such an equivalent norm is also introduced in [2]. The results that we obtain in this paper are generalization of the previous results obtained by Pohozaev and Veron [7].It can be shown thatμ(α,λ) has the variational characterization:(1.4)μ(α,λ)=inf⁡{∫Ω|∇u|pdx+α1-α∫∂Ω|u|pdσ-λ∫Ωa|u|pdx;u∈W1,p(Ω),∫Ω|u|pdx=1}
from whence it follows that(i)
λ is a principal eigenvalue of (1.2) if and only if μ(α,λ)=0,(ii)
α→μ(α,λ) is an increasing function,(iii)
λ→μ(α,λ) is a concave function with a unique maximum such that μ(α,λ)→-∞ as λ→±∞ [6].Ifα∈(0,1], then μ(α,λ)>0, and so λ→μ(α,λ) has exactly one negative zero λ1-(α) and one positive zero λ1+(α). Thus, λ1-(α) and λ1+(α) are principal eigenvalues for (1.2).Ifα=0, then μ(0,0)=0. If a(x)>0, then μ(0,λ) is decreasing, and if a(x)<0, then μ(0,λ) is increasing. Assume now that a(x) changes sign in Ω: if ∫Ωa(x)dx<0, there exists a unique λ1+(0)>0 such that μ(0,λ+(0))=0 and μ(0,λ)>0 for λ∈(0,λ1+(0)). If ∫Ωa(x)dx=0, then μ(0,0)=0 and μ(0,λ)<0 for λ≠0. If ∫Ωa(x)dx>0, then there exists a unique λ1-(0)<0 such that μ(0,λ1-(0))=0 and μ(0,λ)>0 for λ∈(λ1-(0),0).Suppose now that∫Ωa(x)dx<0 and that α is small and negative. Then, since α→μ(α,λ) is increasing, it follows that there still exist principal eigenvalues λ1-(α)<λ1+(α) of (2.5), but now both of them are positive.It can be shown that there existsα0<0 such that the above is true for all α∈(α0,0), but for α<α0, μ(α,λ)<0 for all λ so that principal eigenvalues no longer exist.Similar considerations show that when∫Ωa(x)dx>0, there exists α0<0 such that there are principal eigenvalues λ1-(α)<λ1+(α)<0 for α0<α<0 but when ∫Ωa(x)dx=0, there are no principal eigenvalues for α<0 (see [6]).It is easy to see that ifλ1-(α) and λ1+(α) exist, μ(α,λ)>0 for all λ∈(λ1-(α),λ1+(α)).Thus, we assume that the following conditions hold:(𝒜1)
α∈(0,1)orthat∫Ωa(x)dx≠0andα∈(α0,1],(𝒜2)
a(x)∈L∞(Ω),(ℬ1)
b(x)∈L∞(Ω),(ℬ2)
b+≢0,(ℬ3)
∫Ωb(x)(u1+)γdx<0,whereu1+ is the positive principal eigenfunction corresponding to λ1+(α).With these constructions we have the following.Proposition 1.1.
Assume(𝒜1), then for every λ∈(0,λ1+(α)),
(1.5)‖u‖α,λ≔(∫Ω(|∇u|p-λa(x)|u|p)dx+α1-α∫∂Ω|u|pdσ)1/p
defines a norm in W1,p(Ω) which is equivalent to the usual norm of W1,p(Ω), that is,
(1.6)‖u‖=(∫Ω|∇u|pdx+∫Ω|u|pdx)1/p.Proof.
See [2].Now we can state our main results.Theorem 1.2.
Assume(𝒜1), (𝒜2), (ℬ1), and (ℬ2). Then, for every λ∈(0,λ1+(α)), problem (1.2) admits at least one positive solution u∈W1,p(Ω)∩L∞(Ω).Theorem 1.3.
Assume(𝒜1), (𝒜2), (ℬ1), (ℬ2), and (ℬ3). Then,  problem (1.2) has at least one positive solution u∈W1,p(Ω)∩L∞(Ω) for λ=λ1+(α).Theorem 1.4.
Assume(𝒜1), (𝒜2), (ℬ1), (ℬ2), and (ℬ3). Then, there exists λ*>λ1+(α) such that problem (1.2) admits at least two distinct positive weak solutions in W1,p(Ω)∩L∞(Ω), whenever λ1+(α)<λ<λ*.Whenλ=0 and α=0, we have the following.Corollary 1.5.
Assume∫Ωb(x)dx<0. Then, the problem
(1.7)-Δpu=b(x)u|u|γ-2,x∈Ω,|∇u|p-2∂u∂n=0,x∈∂Ω,
has a positive solution.Throughout this paper, c denotes a positive constant. We will use fibrering method in a similar way to those in [8]. A brief description of the method and the proof of Theorem 1.2 are presented in Section 2. We then study the cases λ=λ1+(α) and λ>λ1+(α) in Sections 3 and 4.


## 2. The Case Whenλ<λ1+(α)

In this section, motivated by Pohozaev [8], we will introduce the fibrering map as our framework for the study of problem (1.2).For a (weak) solution of problem (1.2), we mean a function u∈X=W1,p(Ω) such that for every v∈W1,p(Ω), there holds(2.1)∫Ω(|∇u|p-2∇u∇v-λa(x)|u|p-2uv)dx+α1-α∫∂Ω|u|p-2uvdσ=∫Ωb(x)|u|γ-2uvdx.Now let us define the variational functional corresponding to problem (1.2). We set Iλ:X→R as(2.2)Iλ(u)=1p∫Ω(|∇u|p-λa(x)|u|p)dx+αp(1-α)∫∂Ω|u|pdσ-1γ∫Ωb(x)|u|γdx.It is easy to see thatIλ∈C1(X,R), and for v∈X, there holds(2.3)(Iλ′(u),v)=∫Ω(|∇u|p-2∇u∇v-λa(x)|u|p-2uv)dx+α1-α∫∂Ω|u|p-2uvdσ=∫Ωb(x)|u|γ-2uvdx.SinceC0∞(Ω)⊂X, we know that critical points of Iλ are weak solutions of the problem (1.2). Thus, to prove our main theorems, it suffices to show that Iλ admits critical point. We will do this in this paper. Our main tool is the theory of the fibrering maps. First, we will introduce this map for Iλ:X→R.Letϕu(t)=Iλ(tu)(t>0). We refer to such maps as fibrering maps. It is clear that if u is a local minimizer of Iλ, then ϕu has a local minimum at t=1.Lemma 2.1.
Letu∈X-{0} and t>0. Then, (Iλ′(tu),tu)=0 if and only if ϕu′(t)=0.Proof.
The result is an immediate consequence of the fact that(2.4)ϕu′(t)=(Iλ′(tu),u)=1t(Iλ′(tu),tu).Thus critical points ofIλ correspond to stationary points of the maps ϕu that can be given by(2.5)ϕu(t)=Iλ(tu)=tppAλ(u)-tγγB(u),
where(2.6)Aλ(u)=∫Ω(|∇u|p-λa(x)|u|p)dx+α1-α∫∂Ω|u|pdσ,B(u)=∫Ωb(x)|u|γdx.
Hence,(2.7)ϕu′(t)=tp-1Aλ(u)-tγ-1B(u).
Thus, if Aλ(u) and B(u) have the same sign, ϕu has exactly one turning point at(2.8)t(u)=(Aλ(u)B(u))1/(γ-p),
provided that B(u)≠0, and if Aλ(u) and B(u) have opposite signs, ϕu has no turning points. Now substituting (2.8) into (2.5), we get(2.9)Iλ(t(u)u)=1p(Aλ(u)B(u))p/(γ-p)Aλ(u)-1γ(Aλ(u)B(u))γ/(γ-p)B(u)=1pAλ(u)γ/(γ-p)B(u)p/(γ-p)-1γAλ(u)γ/(γ-p)B(u)p/(γ-p)=(1p-1γ)Aλ(u)γ/(γ-p)B(u)p/(γ-p):=Jλ(u).Lemma 2.2.
Suppose thatu0 is a critical point of Jλ, where Aλ(u0) and B(u0) have the same sign. Then, Iλ′(t(u0)u0)=0.Proof.
Letu0∈X, B(u0)≠0, then
(2.10)(Iλ′(t(u0)u0),v)=∫Ω(|∇(t(u0)u0)|p-2∇(t(u0)u0)∇v-λa(x)|t(u0)u0|p-2t(u0)u0v)dx-∫Ωb(x)|t(u0)u0|γ-2t(u0)u0vdx+α1-α∫∂Ω|t(u0)u0|p-2t(u0)u0vdσ=t(u0)p-1(Aλ′(u0),v)-t(u0)γ-1(B′(u0),v)=1t(u0)(Jλ′(u0),v)=0
for all v∈X. This completes the proof.The following lemma shows that the critical points of the functionalIλ can be found by using the conditional variational problem associated with Iλ.Lemma 2.3.
Suppose thatH is a well-defined functional on X and u0 is a minimizer of Jλ on
(2.11)S:={u∈X,H(u)=c,(H′(u),u)≠0}
for some c≠0. Then,Jλ′(u0)=0.Proof.
Ifu0 is a local minimizer of Jλ on S, then u0 is a solution of the optimization problem:
(2.12)minimizeJλ(u)subjecttoγ(u)=0,
where γ(u)=H(u)-c. Hence, by the theory of the Lagrange multipliers, there exists μ∈R such that Jλ′(u0)=μγ′(u0). Thus,
(2.13)(Jλ′(u0),u0)=μ(γ′(u0),u0).
Note that the functional Jλ is 0-homogeneous and the Gateaux derivative of Jλ at the point v∈X, B(v)≠0, in direction v, is zero, that is, (Jλ′(u0),u0)=0. It then follows that μ=0 due to u0∈S. Hence, the proof is complete.The following scheme for the investigation of the solvability of (1.2) is based on previous lemmas. First, we will prove the existence of nonzero critical points of Jλ under the constraint given by suitable functional H. This will be an actual critical point of Jλ and it will generate critical point of the Euler functional Iλ which will coincide with the weak solution of problem (1.2).Proof of Theorem1.2.
Suppose that(𝒜1), (𝒜2), (ℬ1), (ℬ2) satisfy. It follows from variational characterization of μ(α,λ) that Aλ(u)≥0 for all u∈X and 0≤λ<λ1+(α). Moreover, we have
(2.14)(Aλ′(u),u)=pAλ(u),u∈X.
Hence, for using Lemma 2.3, it is sufficient to consider the case Aλ(u)=c≠0, for example c=1. In this case, we have
(2.15)Jλ(u)=(1p-1γ)1B(u)p/(γ-p).
From the necessary condition for the existence of t(u), we have B(u)>0. It follows that we must consider a critical point with B(u)>0.
The functionalJλ(u) is nonnegative and so bounded below, hence we can look for positive local minimizer for Jλ(u) on X.
Let us consider variational problem:(2.16)Mλ=sup⁡{B(u);Aλ(u)=1,B(u)>0}.
Note that for 0≤λ<λ1+(α), this set is not empty, and from Lemma 2.3, the solution of this problem is a minimizer of Jλ(u) on X.
Suppose{un} is the maximizing sequence of this problem. It follows from the equivalent property in Proposition 1.1, {un} is bounded and so we may assume that un⇀u0 in X. Since X may be compactly embedded in Lp(Ω), Lγ(Ω), and Lp(∂Ω), we have un→u0 in Lp(Ω), Lγ(Ω) and Lp(∂Ω). Hence,
(2.17)B(un)⟶B(u0)=Mλ>0.
Moreover, we have
(2.18)Aλ(u0)≤liminf⁡Aλ(un)=1.
Here, the weak lower semicontinuity of the equivalent norm was used.
Assume thatAλ(u0)<1. By using the map:
(2.19)L(t)=Aλ(tu0),
we have L(1)<1 and limt→∞L(t)=∞ and so L(t0)=1 for some t0>1, that is, Aλ(t0u0)=1. So we derive
(2.20)B(t0u0)=t0γB(u0)>Mλ,
which is a contradiction. Hence,Aλ(u0)=1 and so u0 is a maximizer. As Lemma 2.2, the result would follow by considering u1=t(u0)u0∈X. Then, u1 is a weak solution of (1.2) and u1≥0 in Ω. Now, following the bootstrap argument (used, e.g., in [9]), we prove u1∈L∞(Ω). Then, we can apply the Harnack inequality due to Trudinger [5] in order to get u1>0 in Ω (cf. [9]).


## 3. The Case Whenλ=λ1+(α)

Proof of Theorem1.3.
Ifλ=λ1+(α), It is easy to see that the set {u∈X;Aλ(u)=1} is unbounded and we are forced to require an additional assumption (ℬ3). In this case, again we are looking for a maximizing of the problem:
(3.1)Mλ1+(α)=sup⁡{B(u);Aλ1+(α)(u)=1,B(u)>0}.
Suppose {un} is a maximizing sequence of this problem. First, we investigate the case when {un} is unbounded. Then, we may assume without loss of generality that ∥un∥→∞. So, we obtain
(3.2)1=Aλ1+(α)(un)=‖un‖pAλ1+(α)(vn),
where vn=un/∥un∥. Thus, we have
(3.3)Aλ1+(α)(vn)=Aλ1+(α)(un)‖un‖p=1‖un‖p⟶0asn⟶∞.
Since ∥vn∥=1, we may assume that vn⇀v0 in X. Again, since X may be compactly embedded in Lp(Ω), Lγ(Ω) and Lp(∂Ω), we have
(3.4)lim⁡n→∞∫Ωa(x)|vn|pdx=∫Ωa(x)|v0|pdx,lim⁡n→∞∫Ωb(x)|vn|γdx=∫Ωa(x)|v0|γdx,lim⁡n→∞∫∂Ω|vn|pdσ=∫∂Ω|v0|pdσ.
So,
(3.5)∫Ωa(x)|v0|pdx≥1λ1+(α),
And, therefore, v0≢0. Returning to Aλ1+(α)(vn), we have also
(3.6)0≤Aλ1+(α)(v0)≤liminf⁡n→∞Aλ1+(α)(vn)=0.
Due to simplicity of λ1+(α), there exists c≠0 such that v0(x)=cu1+(x). Therefore, we have
(3.7)0<Mλ1+(α)=lim⁡n→∞B(un)=lim⁡n→∞‖un‖γB(vn),
that implies |c|γB(u1+)≥0, which contradicts (ℬ3). Therefore, {un} is bounded and so un⇀u0 in X and un→u0 in Lp(Ω), Lγ(Ω) and Lp(∂Ω). Hence, B(un)→B(u0)=ℳλ1+(α)>0, and so u0≠0.
Moreover, we obtain0≤Aλ1+(α)(u0)≤1. Here, the variational characteristic of λ1+(α) and the weak lower semicontinuity of the norm were used. Let us now consider the case Aλ1+(α)(u0)=0. Again, we obtain u0=cu1+ for some c≠0, and so B(u0)=|c|γB(u1+)=ℳλ1+(α)>0 which contradicts (ℬ3).
Now we prove thatAλ1+(α)(u0)=1. Suppose otherwise, then by using L(t) in (2.20), we obtain some t0>1 such that Aλ1+(α)(t0u0)=1. By direct calculation, we get a contradiction like (4.1). This yields that u0 is a nonnegative solution of problem (1.2). Using the same ideas as the proof of Theorem 1.2, we have proved Theorem 1.3.


## 4. The Case Whenλ>λ1+(α)

As it is proved in [10], we will show that for λ>λ1+(α) but close enough to λ1+(α), we have two distinct positive solutions for problem (1.2). The existence of one of them is obtained by using the following lemma.Lemma 4.1.
Under the assumption of Theorem1.4, there exists a maximizer u1 of the problem:
(4.1)sup⁡{B(u);Aλ(u)=1,B(u)>0},
whenever λ∈(λ1+,λ1+(α)+δ), for some δ>0. Moreover, u1∈X∩L∞(Ω) is a positive weak solution of problem (1.2).Proof.
First note that usingL(t) in (2.20), it is easy to see that u1 is a maximizer of problem (4.2) if and only if u1 is a maximizer of the problem:
(4.2)Mλ=sup⁡{B(u);Aλ(u)≤1,B(u)>0}.
Now suppose that the result is false. Then, there exists a sequence {δk} such that δk→0 and problem (4.1) has no solution for λ1+(α)+δk. For simplicity, we use λk=λ1+(α)+δk. Let {unk} be a maximizing sequence of this problem, that is,
(4.3)Aλk(unk)≤1,B(unk)⟶Mλk>0.
We prove that if {unk} be bounded or unbounded, we arrive at a contradiction and so the lemma is proved.
The first case:{unk} is bounded. Thus, unk⇀u0k in X for some u0k∈X and unk→u0k in Lp(Ω), Lγ(Ω), and Lp(∂Ω). Hence, B(u0k)=ℳλk>0 and the weak lower semicontinuity of the norm gives Aλk(u0k)≤1. Therefore, u0k is a solution of problem (4.5), which is a contradiction.
The case{unk} is unbounded. Then we may assume that ∥unk∥→∞. Let vnk=unk/∥unk∥. Then, vnk⇀v0k in X for some v0k∈X and vnk→v0k in Lp(Ω), Lγ(Ω) and Lp(∂Ω). This implies that
(4.4)B(unk)=‖unk‖γB(vnk)⟶Mλk>0.
Therefore, B(v0k)≥0. Furthermore, ∥unk∥pAλk(vnk)≤1 and so Aλk(vnk)≤1/∥vnk∥p. Thus, we arrive at
(4.5)Aλk(v0k)≤liminf⁡n→∞Aλk(vnk)⟶0,asn⟶∞,
which implies
(4.6)∫Ω|∇v0k|pdx≤1,(4.7)λk∫Ωa(x)|v0k|pdx≥1.
It is a direct consequent of the comactly embedding of X in Lp(Ω), Lγ(Ω) and Lp(∂Ω). Now, we pass to the limit for k→∞. Then, λk→λ1+(α) and since {v0k} is a bounded sequence, we may assume that v0k⇀v0 in X for some v0∈X and v0k→v0 in Lp(Ω), Lγ(Ω), and Lp(∂Ω).
It follows from (4.10) that
(4.8)λ1+(α)∫Ωa(x)|v0|pdx≥1,
and from the variational characteristic of λ1+(α) and (4.6) that
(4.9)0≤Aλ1+(α)(v0)≤0,
which contradicts (ℬ3).
Hence, for someδ>0, problem (4.5) has at least one nonnegative solution u1 for any λ∈(λ1+(α),λ1+(α)+δ).In order to find the second positive solution, we consider the minimizing problem:(4.10)Nλ=inf⁡{Aλ(u);B(u)=-1}.Note that this set is empty forλ<λ1+(α) (because of the variational characterization of λ1+(α)). Hence, this second problem does not have a solution for λ<λ1+(α).Lemma 4.2.
Under the assumption of Theorem1.4, there exists ɛ>0 such that for λ∈(λ1+,λ1+(α)+ɛ), problem (1.1) has a nonnegative solution u2 satisfying Aλ(u2)<0.Proof.
First note that using the auxiliary functionL(t)=|t|γB(u) and the assumption (ℬ3), it is easy to see that B(t1u1+)=-1 for some t1<1. Thus, the set {u∈X;B(u)=-1} is nonempty. Also, for this t1, we have
(4.11)Aλ(t1u1+)=|t1|p(λ1+(α)-λ)∫Ωa(x)|u1+|pdx<0,
for λ>λ1+(α).
Again, we assume that the result is not true. Then, there existsɛk→0 such that for λk=λ1+(α)+ɛk, problem (1.1) has no solution.
Let{unk} be a minimizing sequence of this problem, that is,B(unk)=-1,Aλk(unk)→𝒩λk<0.
Assume that{unk} is bounded. Using similar argument as in the proof of  Lemma 4.1, we find a solution of problem (1.1) which is a contradiction. Let us assume {unk} is bounded. Again, we consider
(4.12)vnk=unk‖unk‖,with‖unk‖⟶∞,{vnk} is bounded, and so vnk⇀v0k in X for some v0k∈X. Thus,B(vnk)=unk∥unk∥γ=-1∥unk∥γ→0.
Lettingn→∞, we arrive at B(v0k)=0. It follows easily that Aλk(unk)=∥unk∥pAλk(vnk)≤0, and so Aλk(vnk)≤0. As might be expected, we arrive at a contradiction with assumption (ℬ3), by using similar steps as in the proof of Theorem 1.3. This completes the proof.Proof of Theorem1.4.
Letη=min{ɛ,δ}, λ*=λ1+(α)+η and w1=t(u1)u1 and w2=t(u2)u2, where t(ui) is defined by (2.16).
It easy to see thatAλ(w1)>0 and Aλ(w2)≤0, and so w1≢w2. Thus, w1 and w2 are two distinct non negative weak solutions to (1.2). Other properties of w1 and w2 follow in the same way as in Section 2.



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