Sang et al. Boundary Value Problems (2018) 2018:97 https://doi.org/10.1186/s13661-018-1018-7

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Two positive solutions for quasilinear elliptic equations with singularity and critical exponents Yanbin Sang1* , Xiaorong Luo1 and Zongyuan Zhu2,3 *

Correspondence: [email protected] 1 Department of Mathematics, School of Science, North University of China, Taiyuan, China Full list of author information is available at the end of the article

Abstract In this paper, we consider the quasilinear elliptic equation with singularity and critical exponents ⎧ ∗ |u|p–2 u |u|p (t)–2 u –s ⎪ ⎨–p u – μ |x|p = Q(x) |x|t + λu , u > 0, ⎪ ⎩ u = 0,

in , in , on ∂,

where p = div(|∇u|p–2 ∇u) is a p-Laplace operator with 1 < p < N. p∗ (t) := p(N–t) is a N–p critical Sobolev–Hardy exponent. We deal with the existence of multiple solutions for the above problem by means of variational and perturbation methods. Keywords: Quasilinear; Singularity; Critical; Sobolev–Hardy exponent

1 Introduction and preliminaries The main goal of this paper is to consider the following singular boundary value problem: ⎧ ∗ |u|p–2 u |u|p (t)–2 u ⎪ u – μ = Q(x) + λu–s , in , – p p t ⎪ |x| |x| ⎨ u > 0, in , ⎪ ⎪ ⎩ u = 0, on ∂,

(1.1)

where  is a bounded domain in RN , p = div(|∇u|p–2 ∇u) is a p-Laplace operator with )p . p∗ (t) := p(N–t) is a critical 1 < p < N . λ > 0, 0 < s < 1, 0 ≤ t < p, and 0 ≤ μ < μ¯ := ( N–p p N–p Sobolev–Hardy exponent, Q(x) ∈ C() and Q(x) is positive on . In recent years, the elliptic boundary value problems with critical exponents and singular potentials have been extensively studied [2, 6, 7, 10–23, 25, 26, 28, 30–34]. In [19], Han considered the following quasilinear elliptic problem with Hardy term and critical exponent: ⎧ ⎨– u – μ |u|p–2 u = Q(x)|u|p∗ –2 u + λ|u|p–2 u, in , p |x|p ⎩u = 0, on ∂,

(1.2)

© The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

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where 1 < p < N . The existence of multiple positive solutions for (1.2) was established. Furthermore, Hsu [21] studied the following quasilinear equation: ⎧ ⎨– u – μ |u|p–2 u = Q(x)|u|p∗ –2 u + λf (x)|u|q–2 u, in , p |x|p ⎩u = 0, on ∂,

(1.3)

where 1 < q < p < N . We should point out that the authors of [19, 21] both investigated the effect of Q(x). If p = 2, μ = 0, and t = 0, Liao et al. [27] proved the existence of two solutions for problem (1.1) by the constrained minimizer and perturbation methods. Compared with [2, 4, 8, 12, 19, 21, 22, 29], problem (1.1) contains the singular term 1,p λu–s . Thus, the functional corresponding to (1.1) is not differentiable on W0 (). We will remove the singularity by the perturbation method. Our idea comes from [24, 27]. 1,p

Definition 1.1 A function u ∈ W0 () is a weak solution of problem (1.1) if, for every 1,p ϕ ∈ W0 (), there holds   |∇u|

p–2



    ∗ + –s |u|p–2 uϕ Q(x)(u+ )p (t)–1 ϕ dx = ∇u∇ϕ – μ +λ u ϕ dx. |x|p |x|t 

The energy functional corresponding to (1.1) is defined by

Iλ,μ (u) =

1 p

  |∇u|p – μ 

   ∗ + 1–s |u|p (u+ )p (t) λ 1 u Q(x) dx – dx. dx – p ∗ t |x| p (t)  |x| 1–s 

Throughout this paper, Q satisfies (Q1 ) Q(0) = QM = maxx∈ Q(x) and there exists β ≥ p(b(μ) –

Q(x) – Q(0) = o |x|β ,

N–p ) p

such that

as x → 0,

where b(μ) is given in Sect. 1. In this paper, we use the following notations:  p 1,p (i) up =  (|∇u|p – μ |u| ) dx is the norm in W0 (), and the norm in Lp () is |x|p denoted by | · |p ; (ii) C, C1 , C2 , C3 , . . . denote various positive constants; (iii) u+n (x) = max{un , 0}, u–n (x) = max{0, –un }; (iv) We define

1,p ∂Br = u ∈ W0 () : u = r ,



1,p Br = u ∈ W0 () : u ≤ r .

Let S be the best Sobolev–Hardy constant  S :=

inf

1,p

u∈W0 ()\{0}

p  (|∇u|

(



∗

p

– μ |u| ) dx |x|p

|u|p (t)  |x|t

p

dx) p∗ (t)

Our main result is the following theorem.

.

(1.4)

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Theorem 1.1 Suppose that (Q1 ) is satisfied. Then there exists  > 0 such that, for every λ ∈ (0, ), problem (1.1) has at least two positive solutions. The following well-known Brézis–Lieb lemma and maximum principle will play fundamental roles in the proof of our main result. Proposition 1.1 ([3]) Suppose that un is a bounded sequence in Lp () (1 ≤ p < ∞), and un (x) → u(x) a.e. x ∈ , where  ⊂ RN is an open set. Then 

 |un |p dx –

lim

n→∞



  |un – u|p dx = |u|p dx.





Proposition 1.2 ([23]) Assume that  ⊂ RN is a bounded domain with smooth boundary, 0 ∈ , u ∈ C 1 (\{0}), u ≥ 0, u ≡ 0, and –p u ≥ 0 in . Then u > 0 in . By [22, 23], we assume that 1 < p < N , 0 ≤ t < p, and 0 ≤ μ < μ. Then the limiting problem ⎧ ⎨– u – μ up–1 = up∗ (t)–1 , in RN \{0}, p |x|p |x|t ⎩u > 0, in RN \{0}, u ∈ D1,p (RN ) has positive radial ground states V (x) =

p–N p

    p–N x |x| p = ∀ > 0 Up,μ Up,μ

that satisfy   

   ∗ p   N–t |V (x)|p (t) ∇V (x)p – μ |V (x)| dx = dx = S p–t , t |x|p |x| 

where the function Up,μ (x) = Up,μ (|x|) is the unique radial solution of the above limiting problem with  Up,μ (1) =

(N – t)(μ – μ) N –p



1 p∗ (t)–p

,

and lim ra(μ) Up,μ (r) = c1 > 0,

r→0+

lim rb(μ) Up,μ (r) = c2 > 0,

r→+∞

   lim ra(μ)+1 Up,μ (r) = c1 a(μ) ≥ 0,

r→0+

   lim rb(μ)+1 Up,μ (r) = c2 b(μ) ≥ 0,

r→+∞

a(μ) b(μ) ν c3 ≤ Up,μ (r) r ν + r ν ≤ c4 ,

ν :=

N –p , p

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where ci (i = 1, 2, 3, 4) are positive constants depending on N , μ, and p, and a(μ) and b(μ) are the zeros of the function t ≥ 0,

h(t) = (p – 1)t p – (N – p)t p–1 + μ,

satisfying 0 ≤ a(μ) < ν < b(μ) ≤ N–p . p–1 Take ρ > 0 small enough such that B(0, ρ) ⊂ , and define the function u (x) = η(x)V (x) =

p–N p

 η(x)Up,μ

 |x| ,

where η ∈ C0∞ () is a cutoff function ⎧ ⎨1, |x| ≤ ρ , 2 η(x) = ⎩0, |x| > ρ. The following estimates hold when −→ 0:

N–t u p = S p–t + O b(μ)p+p–N ,  ∗

N–t |u |p (t) ∗ dx = S p–t + O b(μ)p (t)–N+t . t |x| 

2 Existence of the first solution of problem (1.1) 1,p In this section, we will get the first solution which is a local minimizer in W0 () for (1.1). Lemma 2.1 There exist λ0 > 0, R, ρ > 0 such that, for every λ ∈ (0, λ0 ), we have Iλ,μ (u)|u∈∂BR ≥ ρ,

inf Iλ,μ (u) < 0.

u∈BR

Proof We can deduce from Hölder’s inequality that p∗ (t) 1 1 λ ∗ C0 u1–s Iλ,μ (u) ≥ up – ∗ QM S– p up (t) – p p (t) 1–s   p∗ (t) 1 λ – p 1–s 1 –1+s+p –1+s+p∗ (t) u C0 , = u – ∗ QM S u – p p (t) 1–s

where C0 is a positive constant. Put f (x) = p1 x–1+s+p – there is a constant R = [ (1–s)f (R) , C0

p∗ (t)S pQM

p∗ (t) p (–1+s+p) ∗ 1 p (t)–p (–1+s+p∗ (t))

]

1

p∗ (t)

QM S –

p∗ (t) p

∗ (t)

x–1+s+p

, we find that

> 0 such that f (R) = maxx>0 f (x) > 0. Letting

λ0 = we have that there is a constant ρ > 0 such that Iλ,μ (u)|u∈∂BR ≥ ρ for every λ ∈ (0, λ0 ). For given R, choosing u ∈ BR with u+ = 0, we have 1–s  1 p r up – λr1–s  (u+ )1–s dx – Iλ,μ (ru) p = lim lim r→0 r 1–s r→0 r1–s 

λ 1–s u+ =– dx < 0, 1–s 

∗

rp (t) p∗ (t)



 Q(x)

∗

(u+ )p (t) |x|t

dx

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since p∗ (t) > p > 1 > s > 0 for 0 ≤ t < p. For all u+ = 0 such that Iλ,μ (ru) < 0 as r → 0, that is, u sufficiently small, we have

= inf Iλ,μ (u) < 0.

(2.1)

u∈BR



The proof of Lemma 2.1 is completed. 1.p

Theorem 2.2 Problem (1.1) has a positive solution u1 ∈ W0 () with Iλ,μ (u1 ) < 0 for λ ∈ (0, λ0 ), where λ0 is defined in Lemma 2.1. Proof By Lemma 2.1, we have 1 1 up – ∗ p p (t) 1 1 up – ∗ p p (t)



∗

Q(x) 



(u+ )p (t) dx ≥ ρ, |x|t

∀u ∈ ∂BR , (2.2)

∗

(u+ )p (t) Q(x) dx ≥ 0, |x|t 

∀u ∈ BR .

From (2.1) we guarantee that there exists a minimizing sequence {un } ⊂ BR such that limn→∞ Iλ,μ (un ) = < 0. Obviously, the minimizing sequence is a closed convex set in BR . 1,p Going if necessary to a sequence still called {un }, there exists u1 ∈ W0 () such that ⎧ ⎪ ⎪ ⎨un  u1 , ⎪ ⎪ ⎩

1.p

in W0 (), 

un −→ u1 ,

1 ≤ p < p∗ (t),

in Lp (, |x|–t ),

(2.3)

un (x) −→ u1 (x), a.e. in ,

and ⎧ ⎪ ∇un (x) −→ ∇u1 (x), ⎪ ⎪ ⎨ p–2 p–2 u |un | un 1| 1  |u|x| p–1 p–1 , |x| ⎪ ⎪ ⎪ ⎩ |un |p∗ (t)–2 un v dx −→    |x|t

a.e. in , p

in L p–1 (), ∗

|u1 |p (t)–2 u1 v dx, |x|t

1,p

∀v ∈ W0 ().

For s ∈ (0, 1), applying Hölder’s inequality, we obtain that 





1–s dx – u+n

 

+ 1–s dx ≤ u1

 

 + 1–s + 1–s   u  dx – u

≤ 

1

n



 +  u – u+ 1–s dx n 1

1–s  1+s ≤ u+ – u+  || p , n

1 p

thus,  



1–s u+n dx =

 

+ 1–s u1 dx + o(1).

(2.4)

Let ωn = un – u1 , by the Brézis–Lieb lemma, one has 

 |∇un |p dx = 

 |∇ωn |p dx +



|∇u1 |p dx + o(1), 

(2.5)

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∗

Q(x) 

Page 6 of 19

(u+n )p (t) dx = |x|t

∗

Q(x) 

(ωn+ )p (t) dx + |x|t

p



∗

Q(x) 

(u+1 )p (t) dx + o(1). |x|t

(2.6)

p

Noting that u1 p = |∇u1 |p – μ|u1 /x|p , we have that

lim un p – ωn p = u1 p .

n→∞

If u1 = 0, then ωn = un , it follows that ωn ∈ BR . If u1 = 0, from (2.2), we derive that 1 1 ωn p – p p ∗ (t)



∗

Q(x) 

(ωn+ )p (t) dx ≥ 0. |x|t

(2.7)

By (2.3)–(2.7), we have

= Iλ,μ (un ) + o(1)  ∗ + 1–s (u+n )p (t) λ u dx – dx + o(1) t |x| 1–s  n    ∗ + 1–s 1 1 (ω+ )p (t) λ ω Q(x) n t dx – dx + o(1) = Iλ,μ (u1 ) + ωn p – ∗ p p (t)  |x| 1–s  n

1 1 = un p – ∗ p p (t)



Q(x)

≥ Iλ,μ (u1 ) + o(1). Consequently, ≥ Iλ,μ (u1 ) as n → ∞. Since BR is convex and closed, so u1 ∈ BR . We get that Iλ,μ (u1 ) = < 0 from (2.1) and u1 ≡ 0. It means that u1 is a local minimizer of Iλ,μ . Now, we claim that u1 is a solution of (1.1) and u1 > 0. Letting r > 0 small enough, and 1.p for every ϕ ∈ W0 (), ϕ ≥ 0 such that (u1 + rϕ) ∈ BR , one has 0 < Iλ,μ (u1 + rϕ) – Iλ,μ (u1 )   ∗

1–s 1 ((u1 + rϕ)+ )p (t) λ 1 (u1 + rϕ)+ Q(x) dx – dx = u1 + rϕp – ∗ t p p (t)  |x| 1–s    ∗ + 1–s 1 1 (u+ )p (t) λ – u1 p + ∗ u Q(x) 1 t dx + dx p p (t)  |x| 1–s  1 1 1 ≤ u1 + rϕp – u1 p . p p

(2.8)

Next we prove that u1 is a solution of (1.1). According to (2.8), we have  

1–s + 1–s  λ (u1 + rϕ)+ dx – u1 1–s   ∗ ∗  1 1 [((u1 + rϕ)+ )p (t) – (u+1 )p (t) ] ≤ u1 + rϕp – u1 p – ∗ Q(x) dx. p p (t)  |x|t Dividing by r > 0 and taking limit as r → 0+ , we have  λ ((u1 + rϕ)+ )1–s – (u+1 )1–s lim inf dx 1 – s r→0+  t    |u1 |p–2 u1 ϕ |∇u1 |p–2 ∇u1 ∇ϕ – μ dx ≤ |x|p   ∗ (u+ )p (t)–1 ϕ – Q(x) 1 t dx. |x| 

(2.9)

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However, λ ((u1 + rϕ)+ )1–s – (u+1 )1–s =λ 1–s t





(u1 + ξ rϕ)+

–s

ϕ dx,



where ξ −→ 0+ and limr→0+ ((u1 + ξ rϕ)+ )–s ϕ = (u+1 )–s ϕ (ξ → 0+ ) a.e. x ∈ . Since ((u1 + ξ rϕ)+ )–s ϕ ≥ 0. By Fatou’s lemma, we obtain that 

+ –s λ u1 ϕ dx ≤ lim inf λ 1 – s r→0+ 



((u1 + rϕ)+ )1–s – (u+1 )1–s dx. t



Hence, from (2.9), we obtain that     + –s |u1 |p–2 u1 ϕ |∇u1 |p–2 ∇u1 ∇ϕ – μ dx – λ u1 ϕ dx p |x|    ∗ (u+ )p (t)–1 ϕ – Q(x) 1 t dx ≥ 0 |x| 

(2.10)

for ϕ ≥ 0. Since Iλ,μ (u1 ) < 0, combining with Lemma 2.1, we can derive that u1 ∈/ ∂BR , thus u1  < R. There exists δ1 ∈ (0, 1) such that (1 + θ )u1 ∈ BR (|θ | ≤ δ1 ). Let h(θ ) = Iλ,μ ((1 + θ )u1 ). Apparently, h(θ ) attains its minimum at θ = 0. Note that h (θ ) =

d Iλ,μ (1 + θ )u1 dθ ∗ (t)–1



= (1 + θ )p–1 u1 p – (1 + θ )p

∗

Q(x) 

(u+1 )p (t) dx – λ(1 + θ )–s |x|t

 



u+1

1–s

dx.

Furthermore, 

∗

(u+ )p (t) h (θ )|θ=0 = u1  – Q(x) 1 t dx – λ |x|  





p



u+1

1–s

dx = 0.

1,p

Define  ∈ W0 () by

+  = u+1 + εψ ,

1,p

for every ψ ∈ W0 () and ε > 0,

where (u+1 + tψ)+ = max{u+1 + tψ, 0}. We deduce from (2.10) and (2.11) that     ∗ |u1 |p–2 u1  (u+1 )p (t)–1  p–2 |∇u1 | ∇u1 ∇ – μ dx – Q(x) dx 0≤ |x|p |x|t    + –s u1  dx –λ 







|u1 |p–2 u1 (u+1 + εψ) |∇u1 |p–2 ∇u1 ∇ u+1 + εψ – μ |x|p {x|u+ 1 +εψ>0}  ∗ + –s +

(u+ )p (t)–1 (u+1 + εψ) u – Q(x) 1 – λ u + εψ dx 1 1 |x|t   

|u1 |p–2 u1 (u+1 + εψ) |∇u1 |p–2 ∇u1 ∇ u+1 + εψ – μ – dx = |x|p  {x|u+ 1 +εψ≤0}

=

(2.11)

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 + –s +

(u+1 + εψ) u – λ u + εψ dx 1 1 |x|t     ∗ + 1–s (u+ )p (t) u1 dx + ε |∇u1 |p–2 ∇u1 ∇ψ ≤ u1 p – Q(x) 1 t dx – λ |x|     ∗ + –s |u1 |p–2 u1 ψ (u+1 )p (t)–1 ψ –μ – Q(x) – λ u ψ dx 1 |x|p |x|t    +

|u1 |p–2 u1 (u+1 + εψ) p–2 |∇u1 | ∇u1 ∇ u1 + εψ – μ dx – |x|p {x|u+ 1 +εψ≤0}    ∗ + –s +

(u+ )p (t)–1 (u+1 + εψ) u + + λ u + εψ dx Q(x) 1 1 1 |x|t {x|u+ 1 +εψ≤0}    ∗ + –s |u1 |p–2 u1 ψ (u+1 )p (t)–1 ψ p–2 – Q(x) – λ u1 ψ dx ≤ε |∇u1 | ∇u1 ∇ψ – μ |x|p |x|t     |u1 |p–1 u1 ψ |∇u1 |p–2 ∇u1 ∇ψ – μ dx. (2.12) –ε |x|p {x|u+ 1 +εψ≤0} ∗ (t)–1

– Q(x)

(u+1 )p

Since the measure of {x | u+1 + εψ ≤ 0} → 0 as ε → 0, we have  lim

ε→0

  |u1 |p–2 u1 ψ |∇u1 |p–2 ∇u1 ∇ψ – μ dx = 0. |x|p {x|u+ 1 +εψ≤0}

Dividing by ε and letting ε → 0+ in (2.12), we deduce that   |∇u1 |p–2 ∇u1 ∇ψ – μ 

 ∗ + –s |u1 |p–2 u1 ψ (u+1 )p (t)–1 – Q(x) ψ – λ u ψ dx ≥ 0. 1 |x|p |x|t

1.p

Since ψ ∈ W0 () is arbitrary, replacing ψ with –ψ, we have  

|u1 |p–2 u1 ψ |x|p   ∗ + –s (u+ )p (t)–1 ψ – Q(x) 1 – λ u ψ dx = 0, 1 |x|t |∇u1 |p–2 ∇u1 ∇ψ – μ

1.p

∀ψ ∈ W0 (),

(2.13)

which implies that u1 is a weak solution of problem (1.1). Putting the test function ψ = u–1 in (2.13), we obtain that u1 ≥ 0. Noting that Iλ,μ (u1 ) = < 0, then u1 ≡ 0. In terms of the maximum principle, we have that u1 > 0, a.e. x ∈ . The proof of Theorem 2.2 is completed.



3 Existence of a solution of the perturbation problem In order to find another solution, we consider the following problem: ⎧ ∗ ⎨– u – μ |u|p–2 u = Q(x) (u+ )p (t)–1 + λ(u+ + γ )–s , in , p |x|p |x|t ⎩u = 0, on ∂,

(3.1)

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where γ > 0 is small. The solution of (3.1) is equivalent to the critical point of the following 1,p C 1 -functional on W0 (): 1 1 Iγ (u) = up – ∗ p p (t)



∗

(u+ )p (t) λ Q(x) dx – t |x| 1 –s 





u+ + γ

1–s

 – γ 1–s dx.



1,p

1,p

For every ϕ ∈ W0 (), the definition of weak solution u ∈ W0 () gives that   |∇u|

p–2



|u|p–2 uϕ ∇u∇ϕ – μ |x|p







–λ

+

u +γ

–s





∗

(u+ )p (t)–1 ϕ ϕ – Q(x) = 0. (3.2) |x|t 

Lemma 3.1 For R, ρ > 0, suppose that λ < λ0 , then Iγ satisfies the following properties: (i) Iγ (u) ≥ ρ > 0 for u ∈ ∂BR ; 1,p (ii) There exists u2 ∈ W0 () such that u2  > R and Iγ (u2 ) < ρ, where R, ρ, and λ0 are given in Lemma 2.1. Proof (i) By the subadditivity of t 1–s , we have

u+ + γ

1–s

1–s – γ 1–s ≤ u+ ,

1,p

∀u ∈ W0 (),

(3.3)

which leads to 1,p

Iγ (u) ≥ Iλ,μ (u),

∀u ∈ W0 ().

Hence, if λ < λ0 for ρ, λ0 > 0, we can obtain the conclusion from Lemma 2.1. 1.p (ii) ∀u+ ∈ W0 (), u+ = 0 and r > 0, which yields 

∗

(u+ )p (t) λ Q(x) dx – t |x| 1 –s   ∗ rp (u+ )p (t) ∗ ≤ up – rp (t) Q(x) dx p |x|t 

rp ∗ Iγ (ru) = up – rp (t) p

→ –∞





ru+ + γ

1–s

 – γ 1–s dx



(r → +∞).

Therefore, there exists u2 such that u2  > R and Iγ (u2 ) < ρ. This completes the proof of Lemma 3.1.



Lemma 3.2 Assume that 0 < γ < 1. Then Iγ satisfies the (PS)c condition with c < N–t

(p–t) S p–t p(N–t) N–p p–t QM

p

– Dλ p+s–1 , where

p+s–1 D= p



p    s–1  p+s–1 p 1 N –p p + C2 . 1 – s p(N – t) (N – t)(1 – s)

1,p

Proof Choose {τn } ⊂ W0 () satisfying Iγ (τn ) → c,

and

Iγ (τn ) → 0 (n → ∞).

(3.4)

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1,p

We assert that {τn } is bounded in W0 (). Otherwise, we assume that limn→∞ τn  → ∞. By (3.4), we have c = Iγ (τn ) –

 1  Iγ (τn ), τn + o(1)

p∗ (t)

1 1 = τn p – ∗ p p (t)



∗

Q(x) 



(τn+ )p (t) λ dx – t |x| 1–s

 

 + 

1–s τn + γ – γ 1–s dx

 +

–s 1 1 λ p τn + γ τn dx + o(1) τ  + dx + n ∗ ∗ ∗ p (t) p (t)  p (t)    

1–s  +  1 λ 1 – τn p – – γ –s dx = τ +γ p p∗ (t) 1–s  n  +

–s λ τn + γ τn dx + o(1) + ∗ p (t)    1 1 p–t p |τn |1–s dx + o(1) τn  – λ + ≥ p(N – t) 1 – s p∗ (t)    1 p–t 1 τn p – λ + ∗ C1 τn 1–s + o(1). ≥ p(N – t) 1 – s p (t) ∗ (τ + )p (t)–1 τn Q(x) n |x|t

–

1,p

The last inequality is absurd thanks to 0 < 1 – s < 1. That is, {τn } is bounded in W0 (). Hence, up to a sequence, there exists a subsequence, still called {τn }. We assume that there 1,p exists {τ1 } ∈ W0 () such that ⎧ ⎪ τn  τ 1 , ⎪ ⎪ ⎪ ⎪ ⎨τ −→ τ , n 1 ⎪ ⎪ τn (x) −→ τ1 (x), ⎪ ⎪ ⎪ ⎩ |τn (x)| ≤ h(x),

1,p

in W0 (), in Lp (, |x|–t ),

1 ≤ p < p∗ (t),

a.e. in , a.e. in  for all n with h(x) ∈ L1 ().

Since 



(τn – τ1 ) τ + + γ –s  ≤ γ –s h + |τ1 | , n it follows from the dominated convergence theorem that  lim

n→∞ 



–s (τn – τ1 ) τn+ + γ dx = 0.

Furthermore, by |τ1 |(τn+ + γ )–s ≤ |τ1 |γ –s , and applying the dominated convergence theorem again, we have  lim



n→∞ 

τn+

+γ

–s





τ1 dx = 

τ1+ + γ

–s

τ1 dx.

Thus, we deduce that  lim

n→∞ 



τn+ + γ

–s

 τn dx = 



τ1+ + γ

–s

τ1 dx.

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 Now we prove that τn → τ1 strongly in W0 (). Set ωn = τn – τ1 . Since Iλ,μ (τn ) → 0 in 1,p ∗ (W0 ()) , we have 1,p



∗

τn p –

Q(x) 

(τn+ )p (t)–1 τn dx – λ |x|t

 

+

–s τn + γ τn dx = o(1).

According to the Brézis–Lieb lemma, together with (3.4), we have 

∗

(ω+ )p (t)–1 ωn ωn  + τ1  – Q(x) n dx – |x|t   +

–s τ1 + γ τ1 dx = o(1), –λ p

p



∗

(τ + )p (t)–1 τ1 Q(x) 1 dx |x|t 



and   lim Iγ (τn ), τ1 = τ1 p –



∗

Q(x)

n→∞



(τ1+ )p (t)–1 τ1 dx – λ |x|t

 



τ1+ + γ

–s

τ1 dx = 0.

Thus 

∗

(ωn+ )p (t)–1 ωn dx = l, n→∞  |x|t   ∗ ∗ Q(x) |ωn |p (t) Q(x) (ωn+ )p (t)–1 ωn dx ≥ dx ≥ dx. |x|t |x|t  QM  QM

lim ωn p = lim

n→∞



∗



|ωn |p (t) |x|t

Q(x)

Sobolev’s inequality implies that 

∗

ωn  ≥ S p



|ωn |p (t) dx |x|t



p p∗ (t)

.

p

Consequently, l ≥ S( QlM ) p∗ (t) . We guarantee that l = 0. Otherwise, we suppose that N–t

l≥

S p–t

N–p p–t

.

QM

It follows that c = Iγ (τn ) – =

 1  Iγ (τn ), τn + o(1)

p∗ (t)

λ (p – t) τn p – p(N – t) 1–s

 



τn+ + γ

1–s

 – γ –s dx +

λ p∗ (t) 

 



τn+ + γ

–s

τn dx + o(1)

N–t  1 p–t (p – t) S p–t 1 p τ + ∗ +  – λ |τn |1–s dx + o(1) ≥ 1 N–p p(N – t) p–t p(N – t) 1 – s p (t)  QM N–t   1 p–t 1 (p – t) S p–t p τ + C2 τ1 1–s + o(1) +  – λ ≥ 1 ∗ (t) p(N – t) N–p p(N – t) 1 – s p QMp–t

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N–t

p (p – t) S p–t ≥ – Dλ p+s–1 , N–p p(N – t) p–t QM

which contradicts the condition of Lemma 3.2. Hence l = 0. Therefore τn → τ1 . This proof of Lemma 3.2 is finished.



1,p

Lemma 3.3 For 0 < s < 1 and λ > 0 small enough, there exists u2 ∈ W0 () such that N–t

p (p – t) S p–t – Dλ p–1+s , sup Iλ,μ (tu2 ) ≤ N–p p(N – t) t≥0 p–t QM

(3.5)

where D is defined in Lemma 3.2. Proof For every r ≥ 0, we have ∗

Iγ (ru ) =

rp rp (t) u p – ∗ p p (t)



∗

Q(x) 

(u+ )p (t) λ dx – |x|t 1–s





ru+ + γ

1–s

 – γ 1–s dx,



which implies that there exists a positive constant 0 such that lim Iγ (ru ) = 0,

r→0

∀ ∈ (0, 0 ),

and ∀ ∈ (0, 0 ),

lim Iγ (ru ) = –∞,

r→+∞

where u is defined in Sect. 1. Let  ∗ ∗ rp rp (t) (u+ )p (t) u p – ∗ Q(x) t dx; p p (t)  |x|   + 

1–s 1 B (r) = – ru + γ – γ 1–s dx, 1–s  A (r) =

because of limr→∞ A (r) = –∞, A (0) = 0, and limr→0+ A (r) > 0, so A (r) attains its maximum at some positive number. In fact, we let ∗ (t)–1

A (r) = rp–1 u p – rp



∗

Q(x) 

(u+ )p (t) dx = 0, |x|t

therefore  r=





u p

+ p∗ (t)

(u )  Q(x) |x|t

1 p∗ (t)–p

:= T . dx

Noting that A (r) > 0 for every 0 < r < T and A (r) < 0 for every r > T , our claim is proved. Thus, the properties of Iγ (ru ) at r = 0 and r = +∞ tell us that supr≥0 Iγ (ru ) is attained for some r > 0.

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From condition (Q1 ), we have ∗     p∗ (t) p∗ (t)    p (t) 

 Q(x) u dx – QM u dx ≤ Q(x) – Q(0) u dx = O β .   t t t |x| |x| |x|   

It follows that 

p∗ (t)

Q(x) 



N–t u ∗ dx = Q(0)S p–t + O b(μ)p (t)–N+t + O β . t |x|

(3.6)

By (3.6), we deduce that

A (T ) =

 1 p 



u p

p p∗ (t)–p

p∗ (t)

u  Q(x) |x|t

 1 – ∗ p (t) 

dx 

u p

p∗ (t) p∗ (t)–p



p∗ (t)

 p–t = p(N – t) 

p∗ (t)

Q(x)

u  Q(x) |x|t

≤

u p



dx 

u p

p p∗ (t)–p

p∗ (t)

u  Q(x) |x|t

u dx |x|t

u p

dx

N–t p–t



S p–t + O b(μ)p+p–N + O β . N–p p(N – t) (Q(0)) p–t

(3.7)

Next, we will estimate B . Here, we use the following inequality from [24, 27]: 1–s 4

x1–s – (x + y)1–s ≤ –(1 – s)y

x

3(1–s) 4

,

0 < x < y.

(3.8)

Observe from (3.8) that   1–s  1 – (r u + γ )1–s dx 1–s γ 1 – s {x||x|≤ 2p }  1–s 4 dx ≤ –C3 1–s (r u )

B (r ) ≤

{x||x|≤ 2p }



 ≤ –C3

1–s

{x||x|≤ 2p }∩{η(x)=1}

 ≤ –C4



1–s–2p 2p



–

N–p p

r –

Up,μ (y)

N–p p

 1–s 4

 Up,μ

|x|

 1–s 4 dx

yN–1 N dy

0

≤ –C5

–

(N–p)(1–s) +N 4p





1–s–2p 2p

y–b(μ)p+N–1 dy 0

⎧ (N–p)(1–s) – +N ⎪ 4p ⎪ , b(μ) > Np , ⎪ ⎨ (N–p)(1–s) ≤ –C5 – 4p +N | ln |, b(μ) = Np , ⎪ ⎪ (N–p)(1–s) (1–s–2p)(–b(μ)p+N) ⎪ ⎩ – 4p +N+ 2p , b(μ) < Np .

(3.9)

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From (3.7) and (3.9), we find that there exists a positive constant  λ0 such that, for every λ ∈ (0, λ0 ), one has Iγ (r u ) = A (r ) + λB (r ) N–t



p – t S p–t + O b(μ)p–N+p + O β ≤ N–p p(N – p) p–t QM ⎧ (N–p)(1–s) – +N ⎪ 4p ⎪ , b(μ) > Np , ⎪ ⎨ (N–p)(1–s) – C5 – 4p +N | ln |, b(μ) = Np , ⎪ ⎪ (N–p)(1–s) (1–s–2p)(–b(μ)p+N) ⎪ ⎩ – 4p +N+ 2p , b(μ) < N , p

N–t p–t

<

p p–t S – Dλ p+s–1 . N–p p(N – p) p–t QM

This completes the proof of Lemma 3.3.



Theorem 3.4 For 0 < γ < 1, there is λ∗ > 0 such that λ ∈ (0, λ∗ ), problem (3.1) admits a 1,p positive solution τγ ∈ W0 () satisfying Iγ (τγ ) > ρ, where ρ is given in Lemma 2.1. Proof Let λ∗ = min{λ0 , λ0 }, then Lemmas 3.1–3.3 hold for 0 < λ < λ∗ . Based on Lemma 3.1, we know that Iγ satisfies the geometry of the mountain pass lemma [1]. Therefore, there 1,p is a sequence {τn } ⊂ W0 () such that Iγ (τn ) → cγ > ρ > 0,

Iγ (τn ) → 0,

(3.10)

where

cγ = inf max Iγ φ(r) , φ∈ r∈[0,1]



1,p  = φ ∈ C [0, 1], W0 () : φ(0) = 0, φ(1) = u2 . So, according to Lemmas 3.1 and 3.3, one has 0 < ρ < cγ ≤ max Iγ (ru2 ) ≤ sup Iγ (ru2 ) r∈[0,1]

r≥0

N–t p–t

<

p p–t S – Dλ p+s–1 . N–p p(N – p) p–t QM

(3.11)

From Lemma 3.2, note that {τn } has a convergent subsequence, still denoted by {τn } ({τn } ⊂ 1,p 1,p W0 ()). Assume that limn→∞ τn = τγ in W0 (). Hence, combining (3.10) and (3.11), we have Iγ (τγ ) = lim Iγ (τn ) = cγ > ρ > 0, n→∞

which implies that τγ ≡ 0. By the continuity of Iγ , we know that τγ is a solution of (3.1). Furthermore, τγ ≥ 0. Hence, applying the strong maximum principle, we obtain that τγ is a positive solution of (3.1). 

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4 Existence of the second solution of problem (1.1) Theorem 4.1 For λ ∈ (0, λ∗ ), problem (1.1) possesses a positive solution τ1 satisfying Iλ,μ (τ1 ) > 0, where λ∗ is given in Theorem 3.4. Proof Let {τγ } be a family of positive solutions of (1.1), we will show that {τγ } has a uniform lower bound. Indeed, we denote ∗ (t)–1

d(r) = rp

+

λ ; (r + p – 1)s d(r) ≥

case (i) 0 < r < 1, r ≥ 1,

case (ii)

λ λ = ; (1 + p – 1)s ps

d(r) ≥ 1.

Therefore, for every γ ∈ (0, 1), r ≥ 0, we get r

p∗ (t)–1

  λ λ λ p∗ (t)–1 + ≥r + ≥ min 1, s . (r + γ )s (r + p – 1)s p

Recall that e is a weak solution of the following problem: ⎧ ⎨– u – μ |u|p–2 u = 1, p |x|p ⎩u = 0,

in , on ∂,

so e(x) > 0 in . According to the comparison principle, we have   λ τγ ≥ min{1, Qm } min 1, s e > 0, p

(4.1)

where Qm = minx∈Q Q(x) > 0. Since {τγ } are solutions of problem (3.1), one has 

p∗ (t)

τγ τγ  – Q(x) dx – λ |x|t 

 (τγ + γ )–s τγ dx = 0.

p



Combining with (3.3), (4.2), and Theorem 3.4, we have N–t

p p – t S p–t – Dλ p+s–1 N–p p(N – p) p–t QM

 1  I (τγ ), τγ p∗ (t) γ     λ λ p–t (τγ + γ )1–s – γ 1–s dx τγ p + ∗ (τγ + γ )–s τγ dx – = p(N – t) p (t)  1–s     p–t λ ≥ (τγ + γ )1–s – γ 1–s dx τγ p – p(N – t) 1–s  > Iγ (τγ ) –

=

λC6 p–t τγ p – τγ 1–s , p(N – t) 1–s

(4.2)

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1,p

since s ∈ (0, 1), so {τγ } is bounded in W0 (). Going if necessary to a subsequence, also 1,p called {τγ }, there exists τ1 ∈ W0 () such that ⎧ 1.p ⎪ in W0 (), ⎪ ⎨ τγ  τ1 ,  τγ −→ τ1 , in Lp (, |x|–t ), ⎪ ⎪ ⎩ τγ (x) −→ τ1 (x), a.e. in .

1 ≤ p < p∗ (t),

(4.3)

1.p

Now, we show that τγ → τ1 in W0 () as γ → 0. Set wγ = τγ – τ1 , then wγ  → 0 as γ → 0; otherwise, there exists a subsequence (still denoted by wγ ) such that limγ →0 wγ  = l > τγ 0. Since 0 ≤ (τγ +γ ≤ τγ1–s , applying Hölder’s inequality and (4.3), we have )s 





τγ (τγ + γ ) dx ≤



τγ1–s dx ≤

–s





|wγ |

1–s

= |wγ |1–s p ||

1+s p



 |τ1 |1–s dx

+ 

 ≤

|τ1 |1–s dx

dx +



|τ1 |1–s dx + o(1). 

Similarly, 

 |τ1 |

1–s

dx ≤



τγ (τγ + γ )–s dx + o(1). 

Therefore 

 –s

lim

γ →0 

τγ (τγ + γ ) dx = 

τ11–s dx.

It follows from Iγ (τγ ), τγ  = 0 and the Brézis–Lieb lemma that 

p∗ (t)

wγ dx – wγ  + τ1  – Q(x) |x|t  p

p



p∗ (t)

τ Q(x) 1 t dx – λ |x| 

 

τ11–s dx = o(1).

(4.4)

1,p

Note that τγ  τ1 as γ → 0+ . Choose the test function ϕ = φ ∈ W0 () ∩ C0 () in (3.2). Letting γ → 0+ and using (4.1), we deduce that τ1 ≥ min{1, Qm } min{1, pλs }e > 0, and   |∇τ1 |p–2 ∇τ1 ∇φ – μ 

   p∗ (t)–1 |τ1 |p–2 τ1 φ τ1 dx = Q(x) φ dx + λ τ1–s φ dx. |x|p |x|t   1,p

(4.5)

1,p

We show that (4.5) holds for every φ ∈ W0 (). In fact, since W0 () ∩ C0 () is dense in 1,p 1,p 1,p W0 (), then for every φ ∈ W0 (), there exists a sequence {φn } ⊂ W0 () ∩ C0 () such that limn→∞ φn = φ. For m, n ∈ N+ large enough, replacing φ with φn – φm in (4.5) yields    |τ1 |p–2 τ1 |φn – φm | |∇τ1 |p–2 ∇τ1 ∇(φn – φm ) – μ dx |x|p    p∗ (t) τ = Q(x) 1 t |φn – φm | dx + λ τ1–s |φn – φm | dx. |x|  

(4.6)

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On the one hand, using φn → φ and (4.6), we have that { φτ1n } is a Cauchy sequence in Lp (), hence there exists ν ∈ Lp () such that limn→∞ φτ ns = ν, which implies that limn→∞ φτ ns = ν 0

0

in measure. By Riesz’s theorem, without loss of generality, choose a subsequence of { φτ ns }, 0

still denoted by { φτ ns }, such that 0

lim

n→∞

φn = ν(x), τ0s

a.e. x ∈ .

(4.7)

On the other hand, from (4.7), we have that ν =  lim

n→∞ 



φn (x) dx = τ0s



φ , τ0s

which leads to

φ(x) dx. τ0s 1,p

Therefore, we deduce that (4.5) holds for φ ∈ W0 (). Setting φ = τ1 in (4.5), we have 

p∗ (t)

τ τ1  – Q(x) 1 t dx – λ |x| 



p



τ11–s dx = 0.

(4.8)

Together with (4.4), we obtain that 

p∗ (t)

wγ dx = o(1). wγ  – Q(x) |x|t  p

(4.9)

Hence 

p∗ (t)



Q(x) |wγ |p (t) dx ≥ QM |x|t

wγ lim wγ  = lim+ Q(x) dx = l > 0. γ →0+ γ →0 |x|t  p

Since  

∗

|wγ |p (t) dx ≥ |x|t



∗

 

∗ (t)

Q(x) (w+γ )p QM |x|t

dx.

N–t

Then l ≥

S p–t

N–p p–t

. By (4.8), we have

QM

 p∗ (t) τ1 λ Q(x) dx – τ 1–s dx |x|t 1–s  1    p–t 1 1 p = τ1  – λ – τ 1–s dx p(N – t) 1 – s p∗ (t)  1   1 1 p–t τ1 p – λ + ∗ C2 τ1 1–s ≥ p(N – t) 1 – s p (t)

1 1 Iλ,μ (τ1 ) = τ1 p – ∗ p p (t)



p

> –Dλ p+s–1 . At the same time, it follows from (4.4) and (4.9) that Iλ,μ (τ1 ) = Iγ (τγ ) –

p–t wγ p + o(1) p(N – t)

(4.10)

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p–t < p(N – t) ≤ –Dλ



p p–1+s

Page 18 of 19



N–t

S p–t

N–p p–t

p

– l – Dλ p–1+s

QM ,

which contradicts (4.10). Therefore, we deduce that Iλ,μ (τ1 ) = lim Iγ (τγ ) > ρ > 0. γ →0

Consequently, problem (1.1) has two different solutions u1 and τ1 . Furthermore, τ1 ≡ 0, together with the maximum principle, we conclude that τ1 > 0 a.e. x ∈ . That is, τ1 is a positive solution of problem (1.1). The proof of Theorem 4.1 is completed.  Remark 4.1 In order to apply the Brézis–Lieb lemma, we need to establish the convergence results for the sequences with gradient terms [5, 9]. Furthermore, the strong maximum principle for a p-Laplace operator is also used.

Acknowledgements We would like to thank the referees for their valuable comments and suggestions to improve our paper. Funding This project is supported by the Natural Science Foundation of Shanxi Province (201601D011003), NSFC (11401583) and the Fundamental Research Funds for the Central Universities (16CX02051A). Availability of data and materials Data sharing not applicable to this article as no data sets were generated or analyzed during the current study. Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally to this work. All authors read and approved the final manuscript. Author details 1 Department of Mathematics, School of Science, North University of China, Taiyuan, China. 2 School of Data Sciences, Zhejiang University of Finance and Economics, Hangzhou, China. 3 China Academy of Financial Research, Zhejiang University of Finance and Economics, Hangzhou, China.

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