# Existence of Solution for Double-Phase Problem with Singular Weights

Authors: Zhi-Yuan Chen; Bin Ge; Wen-Shuo Yuan; Xiao-Feng Cao
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

The aim of this paper is to establish the existence of solutions for singular double-phase problems depending on one parameter. This work improves and complements the existing ones in the literature. There seems to be no results on the existence of solutions for singular double-phase problems.


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## Body

## 1. Introduction and Main Results

The study of various mathematical problems involving the double-phase operator has become very attractive in recent decades. The existence and multiplicity of solutions of double-phase Dirichlet problems has been studied by several authors (see, e.g., [1–8]); in particular, for the eigenvalues of the double-phase operator, see [7]. For other double-phase problems with variable exponents, there are the works of Zhang and Radulescu [9], Shi et al. [10], and Cencelj et al. [11].But up to now, to the best of our knowledge, no paper discussing the existence of solutions for singular double-phase problems via critical point theory can be found in the existing literature. In order to fill in this gap, we study double-phase problems from a more extensive viewpoint. More precisely, we are going to prove that problemPλ has at least one solution. To the best of our knowledge, this is one of the first works which combines a singular term and indefinite term in one problem.This paper is concerned with the existence of solutions to the following singular double-phase problem:(1)−div∇up−2∇u+ax∇uq−2∇u=bxu−θ−1u+λfx,u,inΩ,u=0,on∂Ω,Pλ,where Ω is a smooth bounded domain in ℝN, N≥2, 0<θ<1, λ∈ℝ, a:Ω¯↦0,+∞ is Lipschitz continuous, and b is a given measurable function. The precise conditions on the data will be presented later.Problems of the above type arise for instance in nonlinear elasticity. The main reasons are to describe the behavior of Lavrentiev’s phenomenon; we refer to [12–14]. In fact, Zhikov intended to provide models for strongly anisotropic materials in the context of homogenization. In particular, he considered the following functional:
(2)∫Ω∇up+ax∇uqdx,0≤ax≤L,1<p<q,where the modulating coefficient ax dictates the geometry of the composite made of two differential materials, with hardening exponents p and q, respectively. Recently, there is a wide literature on the regularity theory for minimizers of variational problems and solutions of differential equations with the double-phase operator; far from being complete, we refer the readers to [15–21], respectively, and references therein.In the entire paper, we suppose the following assumptions:Ha: a:Ω¯↦0,+∞ is Lipschitz continuous and 1<p<q<N are chosen such that q/p<1+1/N.Hb: b∈Lp/p+θ−1Ω such that bx>0 in Ω.Hf1: f:Ω×ℝ⟶ℝ is a continuous function such that for a.a. x∈Ω, fx,0=0, and
(i)
there exists positive measurable subsetsΩ1⊂Ω and c∈Ls1Ω such that cx≥0 on Ω1, and
(3)limt→0fx,tcxtr1−2t=0uniformlyfora.a.x∈Ω,where 1<r1<p<N<s1;(ii)
there existsd∈Ls2Ω such that dx>0 on Ω, and
(4)limt→+∞fx,tdxtr2−2t=0uniformlyfora.a.x∈Ω,where 1<r2<p<N<s2;(iii)
there existsM>0 such that
(5)∫ΩFx,tdx>0,∀∣t∣>M,where Fx,t=∫0tfx,sds.Example 1.
The following function satisfies hypothesesHf1:
(6)fx,t=cxtp−2t−tq−2t,if∣t∣≤1,dxtα−2t−t∣t∣,if∣t∣≥1,with 1<α<r2.We are now in the position to state our main results. Firstly, problemPλ has a solution when λ≤0.Theorem 1.
Assume thatHa, Hb, and Hf1 hold. Then for all λ≤0, problem Pλ has at least one nontrivial weak solution with negative energy.Moreover, we also show that problemPλ has a solution when λ>0. In order to do this task, the following conditions are needed:Hf2: f:Ω×ℝ⟶ℝ is a Carathéodory function such that for a.a. x∈Ω, fx,0=0, and
(i)
there existsC>0 and d∈Ls2Ω such that
(7)fx,t≤Cdxtr2−2t,∀x,t∈Ω×ℝ,where 1<r2<p<s2;(ii)
there exists a positive measurable subsetΩ1⊂Ω such that
(8)fx,t,dx>0,∀x∈Ω1,∀t>0.Example 2.
The following functions satisfy hypothesesHf2:
(9)f1x,t=Cdxtr2−2t,f2x,t=Cdxtp−1,ift∈0,1,Cdxtr2−1,ift∈1,+∞,−Cdx−tr2−1,ift∈−1,0,−Cdx−tp−1,ift∈−∞,−1.Theorem 2.
Assume thatHa, Hb, and Hf2 hold. Then for all λ≥0, problem Pλ has at least one nontrivial weak solution with negative energy.The rest of this paper is organized as follows. In Section2, we present some necessary preliminary knowledge on space W01,HΩ. In Section 3, the proof of the main results is given.


## 2. Preliminaries

In order to discuss problemP, we need some facts on space W01,HΩ which are called Musielak-Orlicz-Sobolev spaces. For this reason, we will recall some properties involving the Musielak-Orlicz spaces, which can be found in [7, 22–24] and references therein.Denote byNΩ the set of all generalized N-function. For 1<p<q and 0≤a·∈L1Ω, we define Hx,t=tp+axtq,∀x,t∈Ω×0,+∞. It is clear that H∈NΩ is a locally integrable function and Hx,2t≤2qHx,t,∀x,t∈Ω×0,+∞ which is called condition Δ2.The Musielak-Orlicz spaceLHΩ is defined by
(10)LHΩ=u:Ω⟶ℝmeasurable:∫ΩHx,udx<+∞,endowed with the Luxemburg norm uH=infλ>0:∫ΩHx,u/λdx≤1. The Musielak-Orlicz-Sobolev space W1,HΩ is defined by
(11)W1,HΩ=u∈LHΩ:∇u∈LHΩ,and it is equipped with the norm u=uH+∇uH. We denote by W01,HΩ the completion of C0∞Ω in W1,HΩ. With these norms, the spaces LHΩ, W01,HΩ, and W1,HΩ are separable reflexive Banach spaces (see [7] for the details).Proposition 3.
([1], Proposition 2.1).
SetρHu=∫Ωup+axuqdx. For u∈LHΩ, we have
(i)
Foru≠0,uH=λ⇔ρHu/λ=1(ii)
uH<1=1;>1⇔ρHu<1=1;>1(iii)
IfuH≥1, then uHp≤ρHu≤uHq(iv)
IfuH≤1, then uHq≤ρHu≤uHpProposition 4.
([7], Proposition 2.15, Proposition 2.18).
(1)
If1≤s≤Np/N−p, then the embedding from W01,HΩ to LsΩ is continuous. In particular, if s∈1,Np/N−p, then the embedding W01,HΩ↪LsΩ is compact(2)
Assume thatHa holds. Then, Poincare’s inequality holds; that is, there exists a positive constant C0 such that uH≤C0u,∀u∈W01,HΩBy the above Proposition, there existscτ>0 such that uτ≤cτu,∀u∈W0HΩ, where uτ denotes the usual norm in LτΩ for all 1≤τ<Np/N−p. It follows from (2) of Proposition 4 that ∇uH is an equivalent norm in W01,HΩ. We will use the equivalent norm in the following discussion and write u=∇uH for simplicity.In order to discuss the problemP, we need to define a functional in W01,HΩ:
(12)Ju=∫Ω1p∇up+axq∇uqdx.We know that (see [25], P63, example) J∈C1W01,HΩ,ℝ and the double-phase operator
(13)−div∇up−2∇u+ax∇uq−2∇u,is the derivative operator of J in the weak sense. Moreover, similar to the proof of Theorem 3.1 in [25], we know that the energy functional J is sequentially weakly lower semicontinuous.


## 3. Variational Setting and Proof of the Main Results

For anyλ∈ℝ and each u∈E, we define
(14)φλu=Ju−∫Ωbx1−θu1−θdx−λ∫ΩFx,udx,where Ju=∫Ω1/p∇up+ax/q∇uqdx. By using Hf1, we get r1<s1r1/s1−1<Np/N−p and r2<s2r2/s2−1<Np/N−p. Also, by Proposition 4 (1) we deduce that embeddings E↪Ls1r1/s1−1Ω and E↪Ls2r2/s2−1Ω are compact and continuous. Furthermore, there exists a constant C0>0 such that
(15)maxus1r1/s1−1,up,us2r2/s2−1≤C0u,∀u∈E.Now, we are ready to prove Theorem1.Proof of Theorem 1.
To complete the proof of the main result, we need to consider the following three steps.Step 1.
We first show that for everyλ≤0, the functional φλ is coercive on E.
Letλ≤0 be fixed. Put ΩM=Ω¯×−M,M. Clearly, from the continuity of f, there exists KM>0 such that
(16)KM=maxx,t∈ΩMfx,t.
Thus, we deduce that for anyx∈Ω and ∣t∣≤M,
(17)Fx,t=∫0tfx,sds≤∫0tfx,sds≤MKM.
By virtue of assumptionHf1 (iii), (15), (17), and Proposition 3, one has for any u∈E with u>1(18)φλu≥1qup−11−θbp/p+θ−1u1−θp/1−θ−λ∫ux≤MFx,udx−λ∫ux>MFx,udx≥1qup−11−θbp/p+θ−1up1−θ+λMKMΩ≥1qup−11−θbp/p+θ−1C01−θu1−θ+λMKMΩ.
Since0<θ<1 and 1−θ<p, so this implies φλu⟶+∞ as u⟶+∞. The proof of Step 1 is now completed.Step 2.
We show that there existsξ∈E with ξ>0, φλtξ<0 for t>0 small enough.
Letξ∈C0∞Ω such that suppξ⊂Ω1⊂Ω, ξ=1 in a subset Ω0⊂suppξ, and 0≤ξ≤1 in Ω1. Thus, by condition Hf1 (i), it follows that there exists t0∈0,1 such that
(19)Fx,tξx≤cxr1tξxr1≤cxtξxr1,∀t∈0,t0,x∈Ω.
Hence, for anyt∈0,t0, from Hb and Hf1 (i), we deduce that
(20)φλtξ≤tppmaxξp,ξq−t1−θ1−θ∫Ωbxξ1−θdx−λ∫ΩFx,tξdx≤tppmaxξp,ξq−11−θ∫Ωbxtξ1−θdx−λtr1∫Ω1cxξr1dx≤tr1maxξp,ξq−λ∫Ω1cxξr1dx−t1−θ1−θ∫Ωbxξ1−θdx.
Sincer1>1−θ, we have φλtξ<0 for t<t1 with
(21)0<t1<mint0,1/1−θ∫Ωbxξ1−θdxmaxξp,ξq−λ∫Ω1cxξr1dx1/r1+θ−1.
The proof of Step2 is now complete.Step 3.
We show that there existsuλ∈E such that φλuλ=infu∈Eφλu for any λ≤0.
Letuλn⊂E be a minimizing sequence of φλ. Then, using Step 1, we get that uλn is a bounded sequence. So, there exists uλ∈E such that, up to a subsequence,
(22)uλn⇀uλinE,uλn⟶uλinLpΩ,uλnx⇀uλxa.e.inΩ.
Recall thatJ is sequentially weakly lower semicontinuous, and so we deduce that
(23)Juλ≤liminfn→+∞Juλn.
Now, using Hölder’s inequality, we get that, asn⟶+∞,
(24)∫Ωbxuλn1−θdx≤∫Ωbxuλ1−θdx+∫Ωbxuλn−uλ1−θdx≤∫Ωbxuλ1−θdx+bp/p+θ−1uλn−uλ1−θp/1−θ=∫Ωbxuλ1−θdx+bp/p+θ−1uλn−uλp1−θ=∫Ωbxuλ1−θdx+on1.
Analogously,(25)∫Ωbxuλ1−θdx≤∫Ωbxuλn1−θdx+∫Ωbxuλ−uλn1−θdx≤∫Ωbxuλn1−θdx+bp/p+θ−1uλn−uλ1−θp/1−θ=∫Ωbxuλn1−θdx+bp/p+θ−1uλn−uλp1−θ=∫Ωbxuλn1−θdx+on1.
Hence, by (24) and (25), one yields
(26)limn→+∞∫Ωbxuλn1−θdx=∫Ωbxuλ1−θdx.
Moreover, using assumptionsHf1 (i) and Hf1 (ii), for all ε>0, there exists Cε>0 such that
(27)Fx,uλ≤εcxuλr1+Cεdxuλr2.
The above information and Hölder’s inequality imply(28)∫ΩFx,uλdx≤εCs1uλr1s1/s1−1+Cεds2uλr2s2/s2−1d=εCs1uλs1r1/s1−1r1+Cεds2uλs2r2/s2−1r2.
Again, by Proposition4 (1), we deduce that
(29)E↪Ls1r1/s1−1Ω,uλn⟶uinLs1r1/s1−1Ω,E↪Ls2r2/s2−1Ω,uλn⟶uinLs2r2/s2−1Ω,uλn⟶ufora.a.x∈Ω,Fx,uλnx⟶Fx,uλxfora.a.x∈Ω.
Thus, using the fact thatuλn is bounded in E and the dominated convergence theorem, we can infer that
(30)limn→+∞∫ΩFx,uλnx=∫ΩFx,uλxdx.
Hence, for everyλ<0, by (26) and (30), one yields
(31)liminfn→∞φλuλn=liminfn→∞Juλn−∫Ωbx1−θuλn1−θdx−λ∫ΩFx,uλndx=liminfn→∞Juλn−limn→∞∫Ωbx1−θuλn1−θdx−λlimn→∞∫ΩFx,uλndx≥Juλ−∫Ωbx1−θuλ1−θdx−λ∫ΩFx,uλdx=Juλ−Φuλ−λΨuλ=φλuλ,which implies that φλ is weakly lower semicontinuous, and consequently,
(32)φλuλ≤liminfu∈Eφλuλn=infv∈Eφλv≤φλuλ,which implies that
(33)φuλ=infu∈Eφλuλn.
So, we complete Step3.Therefore, combining the above Steps2 and 3, we deduce that uλ is the required nontrivial solution of problem Pλ. Therefore, we complete the Proof of Theorem 1.Now, we are ready to prove Theorem2.Proof of Theorem 2.
To complete the proof of the main result, we need to consider the following three steps.Step 1.
We first show that for everyλ≥0, the functional φλ is coercive on E.
Firstly, due to conditionHf2 (i), one has
(34)Fx,t≤Cdxr2tr2,∀x,t∈Ω×ℝ.
Again, using the conditionHf2 (i), Hölder’s inequality, Proposition 3, and relation (34), we deduce that for any u∈E with u>1, the following inequality holds true:
(35)φλu≥1qup−11−θbp/p+θ−1u1−θp/1−θ−λCr2∫Ωdxur2dx≥1qup−11−θbp/p+θ−1up1−θ−λCr2ds2ur2s2/s2−1=1qup−11−θbp/p+θ−1up1−θ−λCC0r2r2ds2ur2.
Since1−θ<1<r2<p, we infer that φλu⟶+∞ as ∥u∥⟶+∞. The proof of Step 1 is now complete.Step 2.
We show that there existsξ∈E with ξ>0, φλtξ<0 for t>0 small enough.
Letξ∈C0∞Ω such that suppξ⊂Ω1⊂Ω, ξ=1 in a subset Ω0⊂suppξ, and 0≤ξ≤1 in Ω1. Thus, by condition Hf2 (ii), it follows that
(36)Fx,tξx≥0,∀t∈0,1,x∈Ω.
Hence, for anyt∈0,1, from Hb and Hf2 (ii), we deduce that
(37)φλtξ=∫Ω1p∇up+axq∇uqdx−∫Ωbx1−θtξ1−θdx−λ∫ΩFx,tξdx≤tppmaxξp,ξq−t1−θ1−θ∫Ω1bxξ1−θdx.
Sincep>1>1−θ, we have φλtξ<0 for t<t2 with
(38)0<t2<min1,p∫Ωbxξ1−θdx1−θmax∥ξ∥p,∥ξ∥q1/p+θ−1.
The proof of Step2 is now complete.Step 3.
We show that there existsuλ∈E such that φλuλ=infu∈Eφλu for any λ≥0.
Letuλn⊂E be a minimizing sequence of φλ. Then, using Step 1, we get that uλn is a bounded sequence. So, there exists uλ∈E such that, up to a subsequence,
(39)uλn⇀uλinE,uλn⟶uλinLpΩ,uλnx⇀uλxa.e.inΩ.
Thus, as the proof of Step3 in Theorem 1, we also obtain that
(40)limn→+∞∫Ωbxuλn1−θdx=∫Ωbxuλ1−θdx,limn→+∞∫ΩFx,uλnx=∫ΩFx,uλxdx,and φλ is weakly lower semicontinuous, and consequently,
(41)φλuλ≤liminfu∈Eφλuλn=infv∈Eφλv≤φλuλ,which implies that
(42)φuλ=infu∈Eφλuλn.
The proof of Step3 is complete.Therefore, combining the above Steps2 and 3, we deduce that uλ is the required nontrivial solution of problem Pλ. Thus, we complete the Proof of Theorem 2.



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