Nonlinear elliptic–parabolic problem involving p-Dirichlet-to-Neumann operator with critical exponent

Yanhua Deng; Zhong Tan; Minghong Xie

doi:10.1515/anona-2022-0306

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Nonlinear elliptic–parabolic problem involving p-Dirichlet-to-Neumann operator with critical exponent

Yanhua Deng , Zhong Tan und Minghong Xie

Veröffentlicht/Copyright: 27. März 2023

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Aus der Zeitschrift Advances in Nonlinear Analysis Band 12 Heft 1

Abstract

We consider the nonlinear elliptic–parabolic boundary value problem involving the Dirichlet-to-Neumann operator of p-Laplace type at the critical Sobolev exponent. We first obtain the existence and asymptotic estimates of the global solution, and the sufficient conditions of finite time blowup of the solution by using the energy method. Second, we improve the regularity of solution by Moser-type iteration. Finally, we analyze the long-time asymptotic behavior of the global solution. Moreover, with the help of the concentration compactness principle, we present a precise description of the concentration phenomenon of the solution in the forward time infinity.

Keywords: Dirichlet-to-Neumann operator; critical exponent; blowup; concentration phenomenon

MSC 2010: 35K55; 35B33; 35B40; 35B44

1 Introduction

Let Ω be a bounded smooth domain in R N , N ≥ 2 , and V = Ω × ( 0 , ∞ ) with the lateral boundary ∂ L V ≔ ∂ Ω × [ 0 , ∞ ) . This article is concerned with the nonlinear elliptic–parabolic boundary value problem with critical growth of the following form:

(1.1) − Δ p v ( x , y , t ) = 0 , for t ∈ R + , ( x , y ) ∈ V , v ( x , y , t ) = 0 , for t ∈ R + , ( x , y ) ∈ ∂ L V , ∂ v ( x , 0 , t ) ∂ t = − Λ p ( v ( x , 0 ) ) + ∣ v ∣ p ∗ − 2 v , for t ∈ R + , ( x , 0 ) ∈ Ω × { 0 } , v ( x , y , 0 ) = v 0 , for ( x , y ) ∈ V ,

where Δ p v = ∇ ⋅ ( ∣ ∇ v ∣ p − 2 ∇ v ) is the p-Laplace operator, 2 ( N + 1 ) N + 2 < p < N + 1 , and p ∗ = p N N + 1 − p is the critical exponent for Sobolev imbedding into trace spaces. Let n denote the unit outer normal on Ω × { 0 } . The condition imposed on the bottom boundary is the so-called dynamical boundary condition, where Λ p ( v ( x , 0 ) ) = ∣ ∇ v ∣ p − 2 ∂ v ( x , 0 , t ) ∂ n is the p-normal derivative known as the p-Dirichlet-to-Neumann operator (hereafter p-DN operator, see the Appendix in [29] for the precise definition and more details) related to an inverse problem for the nonlinear p-Laplace type equation

(1.2) ∇ ⋅ ( γ ∣ ∇ v ∣ p − 2 ∇ v ) = 0 ,

which arises as the Euler-Lagrange equation for minimizing the weighted p-Dirichlet energy functional

E ( v ) = ∫ D γ ∣ ∇ v ∣ p d x d y .

If γ ≡ 1 , equation (1.2) reduces to the well-known p-Laplace equation and v is called a p-harmonic function, see, e.g., [17] for details.

Recall that DN operator occurring in the Calderón’s problem (or the inverse conductivity problem), which is a mathematical model for electrical impedance tomography (EIT), is to recover the unknown conductivity distribution from voltage and current measurements taken at its surface, thereby determining the shape and location of unknown inclusions. It is well known that EIT is an inverse method proposed for medical and industrial imaging. We refer to [33,34] and references therein for a review of EIT.

We now describe more precisely the Calderón’s problem. Let D be a bounded open domain with electrical conductivity γ ∈ L ∞ ( D ) . By the linear Ohm’s law, the current density J is given by J = γ ∇ v , where v denotes the electric potential (voltage). Then, the Kirchhoff’s law entails the linear conductivity equation ∇ ⋅ ( γ ∇ v ) = 0 . The DN operator, or the voltage-to-current map, is defined as follow:

Λ γ ( f ) = γ ∂ v ∂ ϱ ∂ D ,

where ϱ is the unit outer normal on ∂ D and v solves uniquely the Dirichlet problem as follows:

∇ ⋅ ( γ ∇ v ) = 0 , in D , v = f , on ∂ D .

Compared to the linear case, less is known about the variants of the Calderón’s problem for nonlinear elliptic equations. This work gives an insight into p-DN operator coming from the weighted p-harmonic model in which the linear Ohm’s law is replaced by a non-Ohmic law: J = γ ∣ ∇ v ∣ p − 2 ∇ v , such nonlinear law appears naturally in various physical and chemical phenomena. In glaciology, it can describe the stationary motion of a glacier, where v denotes the horizontal velocity component of the ice and γ the function resulting from a constitutive law for the ice [13]. It can also model the Hele-Shaw flows driven by injection of a highly shear-thinning power-law fluid in the absence of surface tension, where J denotes the velocity of the fluid and v the pressure [21]. In analogy with the linear case, the Kirchhoff’s law implies that if J is divergence-free, we reach the equation (1.2). The nonlinear p-DN operator is formally defined as follows:

Λ γ p ( f ) = γ ∣ ∇ v ∣ p − 2 ∂ v ∂ ϱ ∂ D ,

where v is the unique solution to equation (1.2) coupled with the Dirichlet boundary condition f . Especially, if the diffusivity coefficient γ = 1 , then we have that Λ γ p = Λ p .

Let v ( x , 0 , t ) = f ( x , t ) . It is worth noting that the elliptic–parabolic boundary problem (1.1) is in fact equivalent to the following initial value problem for parabolic equation involving the p-DN operator:

(1.3) ∂ f ( x , t ) ∂ t = − Λ p ( f ( x , t ) ) + ∣ f ∣ p ∗ − 2 f ( x , t ) , for t ∈ R + , x ∈ Ω , f ( x , t ) = 0 , for t ∈ R + , x ∈ ∂ Ω , f ( x , 0 ) = f 0 , for x ∈ Ω .

Parabolic and elliptic problems associated with the p-DN operator have attracted much interests in recent years. For the linear case of p = 2 , fruitful results have been achieved, e.g., [1,9,30]. For the general case 1 < p < ∞ , to the best of our knowledge, most of the results are about boundary determination for the Calderón’s problem associated with the p-Laplace equation, which was first introduced in [29], where the authors recovered the conductivity on the boundary of the domain. Brander improved the result to gradient of a scalar conductivity [3]. Further progress on the p-Calderón problem was obtained by Kar et al. [4]. As for the study of the parabolic problems with the p-DN operator, in [14], Hauer established the well-posedness of initial value problem for homogeneous equation ∂ f ∂ t = − Λ p ( f ) on ∂ Ω × ( 0 , + ∞ ) , where Ω is a bounded domain and their result also implied existence and uniqueness of the inhomogeneous equation ∂ f ∂ t = − Λ p ( f ) + g for every g ∈ L 2 ( 0 , T ; L 2 ( ∂ Ω ) ) . In [8], Chill et al. developed a general theory of j -elliptic energy functionals and proved the well-posedness of the elliptic–parabolic system with free force on the boundary of the bounded Lipschitz domain.

With the consideration of the non-locality of p-DN operator and motivated by the ideas of equivalent transformation in [8,14] to study problem (1.3), we treat problem (1.1) equivalently, namely coupling heat flow to the Dirichlet problem on a semi-cylindrical infinite domain. Thus, an initial value problem of an elliptic equation with Dirichlet and Neumann boundary conditions is obtained, also known as elliptic–parabolic boundary value problem. Recently, the DN operator associated with the p-Laplace-Beltrami operator and lower-order terms on a compact Riemannian manifold with a Lipschitz boundary was treated by Hauer [15]. Hauer and Mazón [16] presented first insights about the DN operator in L 1 associated with the 1-Laplace operator or total variational flow operator, which is equivalent to a singular elliptic–parabolic boundary value problem similar to the problem (1.1). The special case of p = 2 of problem (1.1) has been studied by Fang and Tan in [11]. In this work, we give a further study of the range 2 ( N + 1 ) N + 2 < p < N + 1 . Our work is a complement to the existing literature on elliptic and parabolic problems involving the p-DN operator.

To state the main idea, we first give some useful definitions and notations. The energy functional corresponding to (1.1) is defined as follows:

(1.4) E ( v ( t ) ) = 1 p ∫ V ∣ ∇ v ( t ) ∣ p d x d y − 1 p ∗ ∫ Ω × { 0 } ∣ v ( t ) ∣ p ∗ d x , v ∈ W 0 , L 1 , p ( V ) ,

where

(1.5) W 0 , L 1 , p ( V ) = { v ∈ W 1 , p ( V ) : v = 0 a.e. on ∂ L V = ∂ Ω × [ 0 , ∞ ) }

endowed with the norm ‖ ∇ v ‖ p = ∫ V ∣ ∇ v ∣ p d x d y 1 p . Hereafter, denote Ω 0 = Ω × { 0 } for convenience. In this article, we work with solution of (1.1) in the following sense.

Definition 1.1

We say that function v is a weak solution to the problem (1.1) in V × ( 0 , T ) if

v ∈ L ∞ ( 0 , T , W 0 , L 1 , p ( V ) ) , v t ∈ L 2 ( 0 , T ; L 2 ( Ω 0 ) )

and satisfies problem (1.1) in the distribution sense, that is

(1.6) − ∫ Ω 0 ∣ ∇ v ∣ p − 2 ∂ v ( x , 0 , t ) ∂ n ϕ d x + ∫ V ∣ ∇ v ∣ p − 2 ∇ v ∇ ϕ d x d y = 0 , for all ϕ ∈ W 0 , L 1 , p ( V ) , a.e. t ∈ [ 0 , T ] .

Introduce a homogeneous Sobolev space

W ˙ 1 , p ( R + N + 1 ) = { f ∈ L p ∗ ( R + N + 1 ) , s.t. ∇ f ∈ L p ( R + N + 1 ) } ,

where R + N + 1 = { ( x , y ) ∈ R N + 1 : y > 0 } . From [24], we know that for all f ∈ W ˙ 1 , p ( R + N + 1 ) ,

‖ f ‖ L p ∗ ( ∂ R + N + 1 ) ≤ C N ( p ) ‖ ∇ f ‖ L p ( R + N + 1 ) ,

where C N ( p ) is the best constant for the inequality to hold.

For f ∈ W 0 , L 1 , p ( V ) , its extension by zero in R + N + 1 ⧹ V can be approximated by functions with compact support in R + N + 1 ¯ . Therefore, we have

‖ f ‖ L p ∗ ( Ω 0 ) ≤ C N ( p ) ‖ ∇ f ‖ p .

By Hölder’s inequality, the fact that Ω is bounded implies that

(1.7) ‖ f ‖ L q ( Ω 0 ) ≤ C N ( p , ∣ Ω ∣ ) ‖ ∇ f ‖ p , 1 ≤ q ≤ p ∗ .

We also introduce the concept of lower (high)-energy initial value.

Definition 1.2

We say that function v 0 possesses lower energy if v 0 satisfies

E ( v 0 ) < p − 1 N p S N / ( p − 1 ) .

Otherwise, we say that v 0 possesses high energy. Here, S is the best Sobolev constant, see [24] for details.

Regarding the best Sobolev constant S , we recall some remarks.

Remark 1.1

Set S be the minimizer for the Sobolev quotient, i.e.,
(1.8) S = inf ‖ ∇ w ‖ p p ( ‖ w ( x , 0 ) ‖ L p ∗ ( Ω 0 ) ) p : w ∈ W 0 , L 1 , p ( V ) .
Then, S is independent of Ω and depends only on N , p .
The infimum S is never achieved when Ω is a bounded domain in R N because of the constant functions.
When Ω = R N , then S is achieved by the extremal functions
(1.9) U x 0 , ε ( x , y ) = ε p / 2 ∣ x − x 0 ∣ 2 + ( y + ε ) 2 N + 1 − p 2 ( p − 1 ) = ε − N + 1 − p p U x − x 0 ε , y ε
for some ε > 0 and x 0 ∈ R N , with
(1.10) U ( x , y ) = 1 ∣ x ∣ 2 + ( y + 1 ) 2 N + 1 − p 2 ( p − 1 ) .

With the above preparation, we now describe the main results.

Theorem 1.1

Let E ( v 0 ) ≤ 0 , 2 ( N + 1 ) N + 2 < p < N + 1 , then v ( x , y , t ; v 0 ) blows up in finite time.

Theorem 1.2

Let v 0 ( ≢ 0 ) be a lower-energy initial value with E ( v 0 ) > 0 .

If ∫ Ω 0 ∣ v 0 ∣ p ∗ d x < S N / ( p − 1 ) , then problem (1.1) has a global solution v ( x , y , t ; v 0 ) .
Moreover,
1. if p = 2 , there exists α > 0 , so that
  ‖ ∇ v ( t ) ‖ 2 2 = O ( e − α t ) , a s t → ∞ ,
2. if 2 < p < N + 1 , we have
  ‖ ∇ v ( t ) ‖ p p = O ( t − 2 / ( p − 2 ) ) , a s t → ∞
If ∫ Ω 0 ∣ v 0 ∣ p ∗ d x ≥ S N / ( p − 1 ) , then the solution to problem (1.1) blows up in finite time.

Theorem 1.3

Let v ( x , y , t ; v 0 ) be a global positive solution with lower-energy initial value v 0 ( x ) . Then, v ∈ L q ( Ω 0 × [ t 0 , ∞ ) ) for all q ( 2 ≤ q < ∞ ) and t 0 > 0 , and

‖ v ‖ L q ( Ω 0 × [ t 0 , T ] ) ≤ C ,

where constant C depends only on N , q , and t 0 .

Theorem 1.4

If v ( x , y , t ; v 0 ) is a global solution to problem (1.1) and uniformly bounded in W 0 , L 1 , p ( V ) with respect to t, then, for all subsequence t n → ∞ , there exists a stationary solution ω such that v ( x , y , t n ; v 0 ) ⇀ w in W 0 , L 1 , p ( V ) .

Theorem 1.5

If v ( x , y , t ; v 0 ) is a global solution to problem (1.1), then the ω -limit set of v contains a stationary solution w. The ω -limit set is defined as follows:

ω ( v 0 ) = { ω ∈ W 0 , L 1 , p ( V ) : there e x i s t s t n → + ∞ , so t h a t v ( x , y , t n , v 0 ) ⇀ w i n W 0 , L 1 , p ( V ) } .

We collect preliminary facts as follows for the description of last theorem. Let

C 0 , L ∞ ( V ¯ ) = { ϕ ∈ C ∞ ( V ) : ϕ = 0 on ∂ L V , supp ( ϕ ) is compact in V ¯ } ,

and

C 0 ∞ ( R + N + 1 ¯ ) = { ϕ ∈ C ∞ ( R + N + 1 ¯ ) : supp ( ϕ ) is compact in R + N + 1 ¯ } .

Let D 1 , p ( R + N + 1 ) denote the completion of the space C 0 ∞ ( R + N + 1 ¯ ) with respect to the norm

‖ w ‖ D 1 , p ( R + N + 1 ) = ∫ R + n + 1 ∣ ∇ w ∣ p d x d y 1 p .

Obviously, D 1 , p ( R + N + 1 ) is a Banach space.

Theorem 1.6

Let v be a positive global solution to problem (1.1). Assume that the ω -limit set ω ( v 0 ) equipped with the strong topology of W 0 , L 1 , p ( V ) is empty. Then, there exist { t n } , t n → ∞ and v ∞ , a stationary solution of (1.1), such that v ( t n ) ⇀ v ∞ in W 0 , L 1 , p ( V ) . Moreover, there exist a nonnegative integer k , x n j ∈ Ω , R n j ∈ R + and U j ∈ D 1 , p ( R + N + 1 ) such that

(1.11) v n − v ∞ − ∑ j = 1 k ( R n j ) N + 1 − p p U j ( R n j ( x − x n j ) , R n j y ) W 0 , L 1 , p ( V ) = o ( 1 ) ,

(1.12) E ( v n ) = E ( v ∞ ) + ∑ j = 1 k E 0 ( U j , R + N + 1 ) + o ( 1 ) ,

(1.13) R n j d i s t ( x n j , ∂ Ω ) → + ∞ ,

where U j is a positive solution of

(1.14) − Δ p v = 0 , i n R + N + 1 = R N × ( 0 , ∞ ) , ∣ ∇ v ∣ p − 2 ∂ v ∂ n = ∣ v ∣ p ∗ − 2 v , o n R N × { 0 }

and

(1.15) E 0 ( v , R N ) = 1 p ∫ R + N + 1 ∣ ∇ v ∣ p d x d y − 1 p ∗ ∫ R N × { 0 } ∣ v ∣ p ∗ d x .

Remark 1.2

In [10,31], semilinear parabolic equations have the similar bubbling description, which is derived for the harmonic maps on surface [22,26,27].

The rest of this article is organized as follows: in Section 2, we give the blowup analysis and decay estimates of the solution; in Section 3, by using Moser iteration, we obtain the L q regularity; and in Section 4, the asymptotic behaviors are obtained by employing the concentration compactness principle.

2 Blow-up analysis and decay estimates

In this section, we consider the existence of finite time blowup and decay estimates of solution to problem (1.1).

First, we prove the following result.

Theorem 2.1

If there exists some t 0 ≥ 0 such that E ( v ( t 0 ) ) ≤ 0 , then v ( x , y , t ; v 0 ) blows up in finite time.

As a by-product, Theorem 1.1 is immediately obtained.

Proof of Theorem 2.1

We shall use the classical concavity method [7,12,18,25,31] to prove the conclusion. Let f ( t ) = 1 2 ∫ t 0 t ∫ Ω 0 v 2 d x d s . We can calculate

(2.1) f ′ ( t ) = 1 2 ∫ Ω 0 v 2 d x = 1 2 ∫ Ω 0 ∣ v ( t 0 ) ∣ 2 d x − ∫ t 0 t ∫ V ∣ ∇ v ∣ p d x d y d s + ∫ t 0 t ∫ Ω 0 ∣ v ∣ p ∗ d x d s ,

(2.2) f ″ ( t ) = − ∫ V ∣ ∇ v ∣ p d x d y + ∫ Ω 0 ∣ v ∣ p ∗ d x .

On the other hand, multiply both sides of the p-Laplacian equation by v t and integral in V × ( t 0 , t ) , we have an energy identity as follows:

(2.3) ∫ t 0 t ∫ Ω 0 v s 2 d x d s + 1 p ∫ V ∣ ∇ v ∣ p d x d y − 1 p ∗ ∫ Ω 0 ∣ v ∣ p ∗ d x = E ( v ( t 0 ) ) .

By (2.2) and (2.3), it follows that

(2.4) f ″ ( t ) = p ∗ p − 1 ∫ V ∣ ∇ v ∣ p d x d y − p ∗ E ( v ( t 0 ) ) + p ∗ ∫ t 0 t ∫ Ω 0 v s 2 d x d s .

Then, it can be easily seen that the assumption E ( v ( t 0 ) ) ≤ 0 indicates

(2.5) p ∗ p − 1 ∫ V ∣ ∇ v ∣ p d x d y − p ∗ E ( v ( t 0 ) ) ≥ 0 for all t ≥ t 0 ,

where 1 < 2 ( N + 1 ) N + 2 < p < N + 1 ensures p ∗ p − 1 > 0 .

If we had T = ∞ , then

f ″ ( t ) ≥ p ∗ ∫ t 0 t ∫ Ω 0 v s 2 d x d s

would imply that

lim t → ∞ f ′ ( t ) = lim t → ∞ f ( t ) = ∞

and

f ( t ) f ″ ( t ) ≥ p ∗ 2 ∫ t 0 t ∫ Ω 0 v 2 d x d s ⋅ ∫ t 0 t ∫ Ω 0 v s 2 d x d s ≥ p ∗ 2 ∫ t 0 t ∫ Ω 0 v v s d x d s 2 = p ∗ 2 ( f ′ ( t ) − f ′ ( t 0 ) ) 2 .

As t → ∞ , we have

f ( t ) f ″ ( t ) ≥ ( 1 + α ) ( f ′ ( t ) ) 2

for some α = p ∗ 2 − 1 > 0 and for all t ≥ t 0 . Hence, f − α ( t ) is concave on [ t 0 , ∞ ] . But f − α ( t ) > 0 , and lim t → ∞ f − α ( t ) = 0 . This contradiction proves that T < ∞ , which completes the proof of Theorem 2.1.□

Proof of Theorem 1.2 (1)

We will take three steps to prove Theorem 1.2 (1). Here, we just give the proof of (b) in detail. (a) is a special case of (b), and one can find more details in [11].

Step 1: For convenience, let

Σ = v : v ∈ W 0 , L 1 , p ( V ) , v ≥ 0 , v ≢ 0 , E ( v ) < p − 1 p N S N / ( p − 1 ) , ∫ Ω 0 ∣ v ∣ p ∗ < S N / ( p − 1 ) .

Obviously, v 0 ∈ Σ .

From [14], we know that the p-DN operator is m-completely accretive for 1 ≤ p ≤ ∞ , due to the celebrated Crandall-Liggett theorem [20], which ensures the existence and uniqueness of the local solution, continuous with respect to time t , to equation (1.3) with given initial value, which is equivalent to (1.1).

As regards global existence, multiplying the equation by v t and integrating, one has

(2.6) ∫ 0 t ∫ Ω 0 v s 2 d x d s + E ( v ( t ) ) = E ( v 0 ) < p − 1 p N S N / ( p − 1 ) .

Thus,

E ( v ( x , y , t ) ) < p − 1 p N S N / ( p − 1 )

for all t > 0 . Note that if

∫ Ω 0 ∣ v ∣ p ∗ d x < S N / ( p − 1 ) ,

then

∫ V ∣ ∇ v ( x , y , t ) ∣ p d x d y > ∫ Ω 0 ∣ v ( x , y , t ) ∣ p ∗ d x .

Now, we show that v ( x , y , t ) ∈ Σ for all t > 0 . If not, there would exist a constant t ∗ such that v ( x , y , t ∗ ) ∈ ∂ Σ , then we have

E ( v ( x , y , t ∗ ) ) = p − 1 p N S N / ( p − 1 )

∫ V ∣ ∇ v ( x , y , t ∗ ) ∣ p d x d y = ∫ Ω 0 ∣ v ( x , y , t ∗ ) ∣ p ∗ d x

both of which imply that S is achieved in Ω 0 , which is a contradiction. Hence, from (2.6), we obtain

∫ V ∣ ∇ v ∣ p d x d y < S N / ( p − 1 ) ,

∫ 0 t ∫ Ω 0 v s 2 d x d s < p − 1 p N S N / ( p − 1 ) , for all t > 0 .

Therefore, v ( x , y , t ) is a global solution to problem (1.1).

Step 2: We adopt the similar argument as in [19]. Let

H ( v ( t ) ) = ∫ V ∣ ∇ v ∣ p d x d y − ∫ Ω 0 ∣ v ∣ p ∗ d x .

Note that if ∫ Ω 0 ∣ v ∣ p ∗ < S N / ( p − 1 ) , then ∫ V ∣ ∇ v ∣ p d x d y > ∫ Ω 0 ∣ v ∣ p ∗ d x , and hence, H ( v ( t ) ) > 0 for all t ≥ 0 . By Sobolev trace inequality,

∫ Ω 0 ∣ v ∣ p ∗ d x < 1 / S p ∗ p ∫ V ∣ ∇ v ∣ p d x d y p ∗ p

and

(2.7) E ( v 0 ) ≥ E ( v ( t ) ) = 1 p ∫ V ∣ ∇ v ( t ) ∣ p d x d y − 1 p ∗ ∫ Ω 0 ∣ v ( t ) ∣ p ∗ d x = 1 p ∫ V ∣ ∇ v ( t ) ∣ p d x d y + 1 p ∗ H ( v ( t ) ) − ∫ V ∣ ∇ v ( t ) ∣ p d x d y = p − 1 p N ∫ V ∣ ∇ v ( t ) ∣ p d x d y + 1 p ∗ H ( v ( t ) ) ≥ p − 1 p N ∫ V ∣ ∇ v ( t ) ∣ p d x d y , on [ t 0 , ∞ ) .

We obtain

(2.8) ∫ Ω 0 ∣ v ∣ p ∗ d x < 1 / S p ∗ p p N p − 1 E ( v 0 ) p ∗ p − 1 ∫ V ∣ ∇ v ∣ p d x d y .

For convenience, denote 1 − 1 / S p ∗ p p N p − 1 E ( v 0 ) p ∗ p − 1 = γ , and from 0 < E ( v 0 ) < p − 1 p N S N / ( p − 1 ) , it easily follows that 0 < γ < 1 ). Then, we have

(2.9) ∫ Ω 0 ∣ v ( t ) ∣ p ∗ d x < ( 1 − γ ) ∫ V ∣ ∇ v ( t ) ∣ p d x d y .

Let T > t 0 be a fixed number, then from

1 2 d d t ∫ Ω 0 ∣ v ( t ) ∣ 2 d x = − H ( v ( t ) )

and Sobolev trace inequality, we have

(2.10) ∫ t T H ( v ( s ) ) d s = 1 2 ∫ Ω 0 ( ∣ v ( t ) ∣ 2 ) d x − 1 2 ∫ Ω 0 ( ∣ v ( T ) ∣ 2 ) d x ≤ 1 2 ∫ Ω 0 ( ∣ v ( t ) ∣ 2 ) d x ≤ C ∫ V ( ∣ ∇ v ( t ) ∣ p ) d x d y 2 p .

Therefore, combining (2.7) and (2.10), we obtain

(2.11) ∫ t T H ( v ( s ) ) d s ≤ C ( Ω ) ( E ( v ( t ) ) ) 2 p on [ t 0 , T ] .

On the other hand, using (2.9), we obtain

(2.12) γ ∫ V ∣ ∇ v ( t ) ∣ p d x d y ≤ H ( v ( t ) ) on [ t 0 , ∞ ) .

According to (2.7) and (2.12), one has

(2.13) E ( v ( t ) ) ≤ p − 1 p N γ + 1 p ∗ H ( v ( t ) ) .

Furthermore, (2.11) and (2.13) provide

C 1 ∫ t T E ( v ( s ) ) d s ≤ ( E ( v ( t ) ) ) 2 p , on [ t 0 , T ] ,

where

C 1 = C ( Ω ) p − 1 p N γ + 1 p ∗ − 1 .

Then, there exists T 0 large enough so that

(2.14) ∫ t ∞ E ( v ( s ) ) d s ≤ T 0 ( E ( v ( t ) ) ) 2 p on [ t 0 , ∞ ) .

Set y ( t ) = ∫ t ∞ E ( v ( s ) ) d s . It follows from (2.14) that

y ( t ) ≤ C 2 t − 2 / ( p − 2 )

for some C 2 = C 2 ( T 0 , p ) > 0 . Thus, we obtain

(2.15) T 0 E ( v ( T 0 + t ) ) ≤ ∫ t T 0 + t E ( v ( s ) ) d s ≤ ∫ t ∞ E ( v ( s ) ) d s ≤ C 2 t − 2 / ( p − 2 ) .

By using (2.7), we have

p − 1 p N ∫ V ∣ ∇ v ( T 0 + t ) ∣ p d x d y ≤ E ( v ( T 0 + t ) ) ≤ C 3 t − 2 / ( p − 2 )

with some constant C 3 > 0 for enough large t > T 0 . Hence,

∫ V ∣ ∇ v ( t ) ∣ p d x d y = O ( t − 2 / ( p − 2 ) ) , as t → ∞ ,

which completes the proof of Theorem 1.2 (1).□

Remark 2.1

If 1 < 2 ( N + 1 ) N + 2 < p < 2 , there is no corresponding decay estimate.

Proof of Theorem 1.2 (2)

The similar proof was provided in [31]. But for the facility of the readers, we provide it in some details.

(1) First of all, we define a set

O = v 0 : E ( v 0 ) < p − 1 p N S N / ( p − 1 ) , ∫ Ω 0 ∣ v 0 ∣ p ∗ d x = S N / ( p − 1 ) .

We will see that the set O is an empty set.

Actually, let v 0 ∈ O . If v 0 satisfies

∫ V ∣ ∇ v 0 ∣ p d x d y ≤ ∫ Ω 0 ∣ v 0 ∣ p ∗ d x ,

then

S N / ( P − 1 ) = ∫ Ω 0 ∣ v 0 ∣ p ∗ d x ≥ ∫ V ∣ ∇ v 0 ∣ p d x d y ≥ S ∫ Ω 0 ∣ v 0 ∣ p ∗ d x p p ∗ = S N / ( p − 1 ) .

Hence,

∫ V ∣ ∇ v 0 ∣ p d x d y = ∫ Ω 0 ∣ v 0 ∣ p ∗ d x = S N / ( p − 1 ) ,

E ( v 0 ) = 1 p ∫ V ∣ ∇ v 0 ∣ p d x d y − 1 p ∗ ∫ Ω 0 ∣ v 0 ∣ p ∗ d x = p − 1 p N S N / ( p − 1 ) ,

which contradicts the conditions of the set O .

If v 0 satisfies

∫ V ∣ ∇ v 0 ∣ p d x d y > ∫ Ω 0 ∣ v 0 ∣ p ∗ d x ,

by E ( v 0 ) < p − 1 p N S N / ( p − 1 ) , we know that

p − 1 p N S N / ( p − 1 ) > E ( v 0 ) = 1 p ∫ V ∣ ∇ v 0 ∣ p d x d y − 1 p ∗ ∫ Ω 0 v 0 p ∗ d x > p − 1 p N ∫ Ω 0 ∣ v 0 ∣ p ∗ d x ,

which implies that

∫ Ω 0 ∣ v 0 ∣ p ∗ d x < S N / ( p − 1 ) .

Again, this contradicts the conditions of the set O . Therefore, the set O is an empty set.

(2) Based on the above analysis, we only need to contemplate the following case:

(2.16) E ( v 0 ) < p − 1 p N S N / ( p − 1 ) , ∫ Ω 0 ∣ v 0 ∣ p ∗ d x > S N / ( p − 1 ) .

Clearly, in this case, we have

S N / ( p − 1 ) < ∫ V ∣ ∇ v 0 ∣ p d x d y < ∫ Ω 0 ∣ v 0 ∣ p ∗ d x .

To complete the proof of the Theorem 1.2 (2), we need to prove the following claims.

Claim 1: For all t ∈ [ 0 , T ] , the following inequalities hold:

(2.17) S N / ( p − 1 ) < ∫ V ∣ ∇ v ( x , y , t ) ∣ p d x d y < ∫ Ω 0 ∣ v ( x , 0 , t ) ∣ p ∗ d x .

Indeed, if there exists t ∗ ∈ [ 0 , T ) such that

∫ V ∣ ∇ v ( x , y , t ∗ ) ∣ p d x d y = ∫ Ω 0 ∣ v ( x , 0 , t ∗ ) ∣ p ∗ d x ,

then we have

S N / ( p − 1 ) ≤ ∫ V ∣ ∇ v ( x , y , t ∗ ) ∣ p d x d y .

But

p − 1 p N S N / ( p − 1 ) > E ( v 0 ) ≥ E ( v ( t ∗ ) ) = p − 1 p N ∫ V ∣ ∇ v ( x , y , t ∗ ) ∣ p d x d y ,

which is a contradiction.

Claim 2: If v 0 satisfies (2.16), then there exists a sufficiently small constant η > 0 , which is independent of t and relies on v 0 such that

∫ Ω 0 ∣ v ( x , 0 , t ) ∣ p ∗ d x ≥ ( 1 + η ) ∫ V ∣ ∇ v ( x , y , t ) ∣ p d x d y for all t ∈ [ 0 , T ) .

This claim can be proved by a standard argument from Ishii [18], we omit here.

(3) Next, we complete the proof of Theorem 1.2 (2) by employing the same argument as in the proof of Theorem 1.1. Suppose that T = ∞ and denote f ( t ) = 1 2 ∫ t 0 t ∫ Ω 0 v 2 d x d s . Using the claims above and equation (2.2), we obtain

(2.18) f ″ ( t ) ≥ − ∫ V ∣ ∇ v ( x , y , t ) ∣ p d x d y + ( 1 + η ) ∫ V ∣ ∇ v ( x , y , t ) ∣ p d x d y = η ∫ V ∣ ∇ v ( x , y , t ) ∣ p d x d y .

If we have T = ∞ , then (2.18) leads to

lim t → ∞ f ′ ( t ) = lim t → ∞ f ( t ) = ∞ .

On the other hand, combining the above identities (2.1)–(2.3), we obtain

(2.19) f ″ ( t ) ≥ p ∗ p − 1 ∫ V ∣ ∇ v ∣ p d x d y − p ∗ E ( v ( t 0 ) ) + p ∗ ∫ t 0 t ∫ Ω 0 v s 2 d x d s .

From (2.17), it easily follows that

p ∗ p − 1 ∫ V ∣ ∇ v ∣ p d x d y − p ∗ E ( v ( t 0 ) ) > 0 .

So, we have

f ″ ( t ) > p ∗ ∫ t 0 t ∫ Ω 0 v s 2 d x d s

and

f ( t ) f ″ ( t ) ≥ p ∗ 2 ( f ′ ( t ) − f ′ ( t 0 ) ) 2 .

The remainder of the proof is the same as that in [7].□

3 Regularity

In this section, we will apply Moser-type iteration to derive an L q -estimates for the solution with initial value v 0 ∈ ∑ to improve the regularity of solution with respect to spatial variables. Our argument is an adaptation of Brezis and Kato [2] for elliptic equation. In fact, the parabolic Moser-iteration and other methods to obtain L q or even L ∞ -regularity (in time and space) for semigroups generated by − A satisfying nonlinear evolution problems of the form u t + A u + F ( u ) = 0 is discussed in detail in the monograph, where A could be the p-Laplace operator or the DN operator associated with the p-Laplace operator, or the doubly nonlinear operators, and F is a Lipschitz-continuous nonlinearity [6]. Since the trace Sobolev inequality (1.7) is available, the regularity result of this section is already included in their results. Nevertheless, what we are using here is a concrete test function technique as opposed to their abstract discussion.

Proof of Theorem 1.3

For all fixed t 0 > 0 and T > 0 , choosing η ∈ C ∞ ( 0 , T ) with 0 ≤ η ≤ 1 in [ 0 , T ] , η = 1 in [ t 0 , T ] , η = 0 in [ 0 , t 0 2 ] , and ∣ η t ∣ ≤ 1 t 0 . Let ϕ = v p s + 1 η p (for s > 0 to be chosen later) be the test function for (1.6), we obtain

(3.1) ∫ 0 T ∫ Ω 0 [ ϕ v t − ϕ v p ∗ − 2 v ] d x + ∫ V ∣ ∇ v ∣ p − 2 ∇ v ∇ ϕ d x d y d t = 0 .

Suppose v ∈ L p s + 2 ( V × [ 0 , T ] ) . For the first term in (3.1), we have

(3.2) ∫ 0 T ∫ Ω 0 ϕ v t d x d t = ∫ 0 T ∫ Ω 0 v t v p s + 1 η p d x d t = 1 p s + 2 ∫ 0 T ∫ Ω 0 ( v p s + 2 ) t η p d x d t = 1 p s + 2 ∫ 0 T ∫ Ω 0 ( v p s + 2 η p ) t d x d t − p p s + 2 ∫ 0 T ∫ Ω 0 v p s + 2 η p − 1 η t d x d t = 1 p s + 2 ∫ Ω 0 v ( T ) p s + 2 η ( T ) p d x − p p s + 2 ∫ 0 T ∫ Ω 0 v p s + 2 η p − 1 η t d x d t .

For the third term in (3.1), we have

(3.3) ∫ 0 T ∫ V ∣ ∇ v ∣ p − 2 ∇ v ∇ ϕ d x d y d t = p s + 1 ( s + 1 ) p ∫ 0 T ∫ V ∣ ∇ v s + 1 ∣ p η p d x d y d t .

For the second term in (3.1), using Hölder’s inequality, we can obtain

(3.4) ∫ 0 T ∫ Ω 0 ϕ v p ∗ − 2 v d x d t = ∫ 0 T ∫ Ω 0 η p v p s + 2 v p ∗ − 2 d x d t ≤ M ∫ 0 T ∫ Ω 0 η p v p s + 2 d x d t + ∫ 0 T η p ∫ { v p ∗ − 2 ≥ M } η p v p s + p v p ∗ − p d x d t ≤ M ∫ 0 T ∫ Ω 0 η p v p s + 2 d x d t + ∫ 0 T η p ∫ { v p ∗ − 2 ≥ M } v p ∗ d x 1 − p p ∗ ⋅ ∫ Ω 0 ( v s + 1 ) p ∗ d x p p ∗ d t ≤ M ∫ 0 T ∫ Ω 0 η p v p s + 2 d x d t + 1 S sup t ∫ { v p ∗ − 2 ≥ M } v p ∗ d x p − 1 N ⋅ ∫ 0 T η p ∫ V ∣ ∇ v s + 1 ∣ p d x d y d t .

From (3.1)–(3.4), we obtain

(3.5) 1 p s + 2 sup t ∫ Ω 0 v p s + 2 η p d x + p s + 1 ( s + 1 ) p ∫ 0 T ∫ V ∣ ∇ v s + 1 ∣ p η p d x d y d t ≤ p p s + 2 ∫ 0 T ∫ Ω 0 η ∣ η t ∣ v p s + 2 d x d t + M ∫ 0 T ∫ Ω 0 η p v p s + 2 d x d t + 1 S sup t ∫ { v p ∗ − 2 ≥ M } v p ∗ d x p − 1 N ⋅ ∫ 0 T ∫ V ∣ ∇ v s + 1 ∣ p η p d x d y d t .

Now we prove a claim as below.

Claim: Let

ε ( M ) = 1 S sup t ∫ { v p ∗ − 2 ≥ M } v p ∗ d x p − 1 N ,

then ε ( M ) → 0 as M → ∞ .

Proof of Claim: Denote

g ( t ) ≔ ∫ Ω 0 v p ∗ d x

and

g M ( t ) ≔ ∫ { v p ∗ − 2 ≥ M } v p ∗ d x .

We have to prove that for all ε > 0 , there exists a positive number M 0 ∈ R + such that

g M ( t ) < ε , as M > M 0 , t ∈ ( 0 , ∞ ) .

Employing Theorem 1.2 (1) and Sobolev trace inequality, we can choose a T 0 such that

g M ( t ) ≤ g ( t ) ≤ ε for t > T 0 .

So we will complete the proof if we prove that

g M ( t ) < ε as M > M 0 for t ∈ [ 0 , T 0 ] .

In fact, by v ∈ C [ 0 , T 0 ] W 0 , L 1 , p ( V ) , we know that g ( t ) is continuous function on [ 0 , T 0 ] . So, it is uniformly continuous function on [ 0 , T 0 ] . The rest is obvious, and this completes the proof of the claim.

Thus, we may conclude that

(3.6) 1 p s + 2 sup t ∫ Ω 0 v p s + 2 η p d x + p s + 1 ( s + 1 ) p − ε ( M ) ∫ t 0 / 2 T ∫ V ∣ ∇ v s + 1 ∣ p η p d x d y d t ≤ p p s + 2 ∫ t 0 / 2 T ∫ Ω 0 η ∣ η t ∣ v p s + 2 d x d t + M ∫ t 0 / 2 T ∫ Ω 0 η p v p s + 2 d x d t ≤ p t 0 ( p s + 2 ) + M ∫ t 0 / 2 T ∫ Ω 0 v p s + 2 d x d t

for M large enough.

Now we can complete the proof of L q -estimate. Let s 0 = 0 , p s i + 2 = p ( s i − 1 + 1 ) + ( p s i − 1 + 2 ) ( p − 1 ) N , for i ≥ 1 . For all given q ( 2 < q < ∞ ) , there exists i 0 such that p s i 0 − 1 + 2 < q ≤ p s i 0 + 2 . Combining Hölder’s inequality, Sobolev trace inequality, and (3.6), we have

(3.7) ∫ t 0 T ∫ Ω 0 v p ( s + 1 ) + ( p s + 2 ) ( p − 1 ) N d x d t ≤ ∫ t 0 / 2 T ∫ Ω 0 v p ( s + 1 ) + ( p s + 2 ) ( p − 1 ) N η p ( 1 + p − 1 N ) d x d t ≤ 1 S sup t ∫ Ω 0 v p s + 2 η p d x p − 1 N ⋅ ∫ t 0 / 2 T ∫ V ∣ ∇ v s + 1 ∣ p η p d x d y d t ≤ C ∫ t 0 / 2 T ∫ Ω 0 v p s + 2 d x d t 1 + p − 1 N ,

where C depends on N , s , q , t 0 . Let s 0 = 0 , by Theorem 1.2 (1), iterating equations (3.6) and (3.7) until s = s i 0 , we can complete the proof.□

4 Asymptotic behaviors

In this section, we focus on the asymptotic behavior of any global solutions with high-energy initial value. We show that there exists a subsequence { t n } such that the asymptotic behavior of v ( t n ) as t n → ∞ is the Palais-Smale (PS) sequence to the stationary equation of (1.1). Therefore, in order to describe exactly the asymptotic profile of the solutions v ( t n ) , we analyze the (PS) sequence by means of the concentration compactness principle [23].

Proof of Theorem 1.4

For all t n → ∞ , let v n = v ( x , y , t n ; u 0 ) . Since { v n } is bounded in W 0 , L 1 , p ( V ) , it follows that there exists a subsequence which we still denote { v n } and a function w such that

v n ⇀ w in W 0 , L 1 , p ( V ) ,

v n p ∗ − 1 ⇀ w p ∗ − 1 in ( L p ∗ ( Ω × { 0 } ) ) ∗ .

Now, we choose a test function

(4.1) φ ( x , y , t ) = ρ ( t − t n ) ψ ( x , y ) , for t > t n , x ∈ Ω ¯ , y ∈ ( 0 , ∞ ) , 0 , for 0 ≤ t ≤ t n , x ∈ Ω ¯ , y ∈ ( 0 , ∞ ) ,

where ψ ∈ W 0 , L 1 , p ( V ) , ρ ∈ C 0 2 ( 0 , 1 ) , ρ ≥ 0 , and ∫ 0 1 ρ ( s ) d s = 1 .

From Definition 1.1, we obtain

∫ t n t n + 1 ∫ V ρ ( t − t n ) ∣ ∇ v ∣ p − 2 ∇ v ∇ ψ d x d y − ∫ Ω 0 ( v ρ ′ ( t − t n ) ψ + ∣ v ∣ p ∗ − 2 v ρ ( t − t n ) ψ ) d x d t = 0 .

Performing the change of variables s = t − t n , it follows that

(4.2) ∫ 0 1 ∫ V ρ ( s ) ∣ ∇ v ∣ p − 2 ∇ v ( t n + s ) ∇ ψ d x d y − ∫ Ω 0 ( v ρ ′ ( s ) ψ + ∣ v ∣ p ∗ − 2 v ( t n + s ) ρ ( s ) ψ ) d x d s = 0 .

Since v ( t n + s ) is uniformly bounded in W 0 , L 1 , p ( V ) for 0 ≤ s ≤ 1 . Thus, we can choose the same subsequence of { t n } and functions w s and w such that

v ( t n + s ) → w s , strongly in L q ( Ω 0 ) ( p ≤ q < p ∗ ) ,

and

v ( t n ) → w , strongly in L q ( Ω 0 ) ( p ≤ q < p ∗ ) .

Then, we show that w s = w a.e. in Ω 0 . In fact, by the energy identity

(4.3) ∫ 0 t ∫ Ω 0 v s 2 d x d s + E ( v ( t ) ) = E ( v 0 ) ,

we have

∫ Ω 0 ∣ v ( t n + s ) − v ( t n ) ∣ 2 d x ≤ s ∫ t n t n + s ∫ Ω 0 ∣ v τ ∣ 2 d x d τ → 0 , as t n → ∞ , for 0 ≤ s ≤ 1 .

Thus, we have

v ( t n + s ) − v ( t n ) → 0 , strongly in L 2 ( Ω 0 ) , as t n → ∞ , for 0 ≤ s ≤ 1 .

Hence, w s = w a.e. in Ω 0 .

We rewrite (4.2) as follows:

∫ 0 1 ∫ V − ρ ( s ) ∣ ∇ v ∣ p − 2 ∇ v ( t n ) ∇ ψ d x d y + ∫ Ω 0 ( v ( t n ) ρ ′ ( s ) ψ + ∣ v ∣ p ∗ − 2 v ( t n ) ρ ( s ) ψ ) d x d s + ∫ 0 1 ∫ Ω 0 [ v ( t n + s ) − v ( t n ) ] ρ ′ ( s ) ψ d x d s − ∫ 0 1 ∫ V [ ∣ ∇ v ( t n + s ) ∣ p − 2 ∇ v ( t n + s ) − ∣ ∇ v ( t n ) ∣ p − 2 ∇ v ( t n ) ] ρ ( s ) ∇ ψ d x d y d s + ∫ 0 1 ∫ Ω 0 [ ∣ v ∣ p ∗ − 2 v ( t n + s ) − v p ∗ − 2 v ( t n ) ] ρ ( s ) ψ d x d s = 0 .

By the dominated convergence theorem and the choice of ρ and v ( t n ) → w strongly in L 2 ( Ω 0 ) , we have

∫ 0 1 ∫ Ω 0 ρ ( s ) ∣ ∇ v ∣ p − 2 ∇ v ( t n ) ∇ ψ d x d y − ∫ Ω 0 ∣ v ∣ p ∗ − 2 v ( t n ) ρ ( s ) ψ d x d s = o ( 1 ) , as n → ∞ .

From the choice of ρ , we obtain

∫ V ∣ ∇ v ∣ p − 2 ∇ v ( t n ) ∇ ψ d x d y − ∫ Ω 0 ∣ v ∣ P ∗ − 2 v ( t n ) ψ d x = o ( 1 ) , as n → ∞ ,

which completes the proof of Theorem 1.4.□

Proof of Theorem 1.5

Denote v ( x , y , t ; v 0 ) and v ( x , y , t n ; v 0 ) by v and v n , respectively, for convenience. Theorem 1.1 implies that E ( v ( t ) ) > 0 for all t ≥ 0 . So, by energy inequality, we have

(4.4) 0 < E ( v ( t ) ) ≤ E ( v 0 )

and

(4.5) ∫ 0 ∞ ∫ Ω 0 v s 2 d x d s ≤ C < ∞ .

Then, there exists a sequence { t n } satisfying t n → ∞ as n → ∞ such that

(4.6) ∫ Ω 0 ∣ v t ( x , y , t n ; u 0 ) ∣ 2 d x → 0 , as n → ∞ .

Using the energy inequality and equations (4.4)–(4.6), we know that the sequence { v n : t n → ∞ } is a (PS) sequence of the corresponding stationary problem of (1.1). Such a situation has been well investigated in the theory of nonlinear elliptic equations [5,24,35]. It is not difficult to prove that there exists a constant C < + ∞ such that

∫ V ∣ ∇ v n ∣ p d x d y ≤ C .

Thus, there exist a subsequence (not relabeled) and a function w such that

v n ⇀ w , in W 0 , L 1 , p ( V ) ,

v n → w , in L q ( Ω 0 ) ( p ≤ q < p ∗ ) .

Thus, with the theory of elliptic equation, we can access that w is a stationary solution, which completes the proof of Theorem 1.5.□

The rest of the article is dedicated to the proof of Theorem 1.6. We need the following lemmas.

Lemma 4.1

([23], Lemma 2.3) Let { u n } be a bounded sequence in W 0 , L 1 , p ( V ) . We may assume that u n converges a.e. u ∈ W 0 , L 1 , p ( V ) , and ∣ ∇ u n ∣ p , u n p ∗ ( x , 0 ) ⊗ δ ( y ) converges weakly to some bounded, nonnegative measures on R N + 1 μ , ν and s u p p ( ν ) ⊂ { y = 0 } .

Then, for some at most countable set J , distinct points { x j } j ∈ J in R N × { 0 } and { ν j } j ∈ J in ( 0 , ∞ ) , we have
(4.7) ν = ∣ u ∣ p ∗ ( x , 0 ) ⊗ δ 0 ( y ) + ∑ j ∈ J ν j δ x j ,

(4.8) μ ≥ ∣ ∇ u ∣ p + ∑ j ∈ J S ν j p p ∗ δ x j ,
where δ x j are Dirac measures assigned to x j .
If u ≡ 0 and μ ( R N + 1 ) ≤ S ν ( R n + 1 ) p / p ∗ , then J = { x 0 } for some x 0 ∈ R N × { 0 } and ν = c 0 δ x 0 , as well as μ = S c 0 p / p ∗ δ x 0 for some c 0 > 0 .

Lemma 4.2

If p ≥ 2 , we have

[ ∣ ∇ u 1 ∣ p − 2 ∇ u 1 − ∣ ∇ u 2 ∣ p − 2 ∇ u 2 ] [ ∇ u 1 − ∇ u 2 ] ≥ c ∣ ∇ u 1 − ∇ u 2 ∣ p .

If 1 < p < 2 , we have

[ ∣ ∇ u 1 ∣ p − 2 ∇ u 1 − ∣ ∇ u 2 ∣ p − 2 ∇ u 2 ] [ ∇ u 1 − ∇ u 2 ] ≥ c ∣ ∇ u 1 − ∇ u 2 ∣ p ( ∣ ∇ u 1 ∣ + ∣ ∇ u 2 ∣ ) 2 − p

for some constant c > 0 .

Lemma 4.3

Assume that { u n } is bounded in W 0 , L 1 , p ( V ) solving the following problem:

(4.9) − Δ p u n = 0 , i n V , u n = 0 , o n ∂ L V , ∣ ∇ u n ∣ p − 2 ∂ u n ∂ n = ∣ u n ∣ p ∗ − 2 u n + f n , o n Ω 0 .

There are functions u and f such that

(4.10) u n ⇀ u i n W 0 , L 1 , p ( V ) a n d f n ⇀ f i n L p ( Ω 0 ) ,

where q = p / ( p − 1 ) is a conjugate of p . Then, there exist at most finitely many points x 1 , x 2 , … x l ∈ Ω 0 such that

u n → u in W 0 , L , loc 1 , p ( V ⧹ { x 1 , x 2 , … x l } ) .

Proof of Lemma 4.3

Since { ∇ u n } is bounded in L p ( V ) , there exists a T ∈ L p ( V ) such that

∇ u n ⇀ T in L p ( V ) .

Apparently, T satisfies

(4.11) ∫ V T ∇ ψ d x d y = ∫ Ω 0 ( ∣ u ∣ p ∗ − 1 ψ + f ψ ) d x , for all ψ ∈ W 0 , L 1 , p ( V ) .

By Lemma 4.1, suppose that there exists a countable set { x 1 , x 2 , ⋯ , x i , ⋯ ∣ x i ∈ Ω 0 } , ν j > 0 such that

(4.12) ∣ u n ∣ p ∗ ⇀ ν = ∣ u ∣ p ∗ ( x , 0 ) ⊗ δ ( y ) + ∑ j ∈ J ν j δ x j ,

(4.13) ∣ ∇ u n ∣ p ⇀ μ ≥ ∣ ∇ u ∣ p + ∑ j ∈ J S ν j p p ∗ δ x j .

We show that X ≔ { x 1 , x 2 , … } is a finite set. Let j ∈ J be fixed, select a monotonically nonincreasing cutoff function ϕ ∈ C 0 ∞ ( R + N + 1 ) to satisfy

(4.14) ϕ ( x , y ) = 1 , ( x , y ) ∈ B 1 + ( x j ) ≔ B 1 ( x j ) × [ 0 , ∞ ) , 0 , ( x , y ) ∈ B 2 + ( x j ) c ≔ B 2 ( x j ) c × [ 0 , ∞ ) ,

where 0 ≤ ϕ ( x , y ) ≤ 1 . Let ψ n = ϕ ( x − x j n , y n ) u n concentrate on x i . Multiplying the equation by ψ n , concentrate on x j , and integrating and letting n → + ∞ , from (4.11), we have

lim n → + ∞ ∫ Ω 0 ϕ d ν = lim n → + ∞ ∫ V ϕ d μ − ∫ V ϕ T i D i u d x d y + ∫ Ω 0 ∣ u ∣ p ∗ ϕ d x ,

then μ ( { x i } ) = ν ( { x i } ) . Moreover, from the relation μ ( { x j } ) ≥ S ( ν ( { x j } ) ) p / p ∗ , if ν j = ν ( { x j } ) > 0 , then ν j ≥ S N / ( p − 1 ) . Note that ν ( Ω ¯ ) < ∞ , which implies that X is a finite set.

Choose a function φ ∈ C 0 , L ∞ ( V ¯ ) , φ ≥ 0 , φ ≡ 0 on ∂ L V , and φ ( x j ) = 0 for all x j ∈ X . We obtain

∫ Ω 0 ∣ φ u n ∣ p ∗ d x → ∫ Ω 0 ∣ φ u ∣ p ∗ d x + ∑ j ν j φ p ∗ ( x j ) = ∫ Ω 0 ∣ φ u ∣ p ∗ d x .

So that φ u n → φ u in L p ∗ ( Ω 0 ) . Furthermore, combining with Lemma 4.2, we easily obtain

(4.15) ∫ V φ ∣ ∇ u n − ∇ u ∣ p d x d y ≲ ∫ V φ [ ∣ ∇ u n ∣ p − 2 ∇ u n − ∣ ∇ u ∣ p − 2 ∇ u ] [ ∇ u n − ∇ u ] d x d y = ∫ Ω 0 ( ∣ u n ∣ p ∗ − 1 − ∣ u ∣ p ∗ − 1 ) ( u n − u ) φ d x + ∫ Ω 0 φ ( f n − f ) ( u n − u ) d x − ∫ V ( u n − u ) [ ∣ ∇ u n ∣ p − 2 ∇ u n − ∣ ∇ u ∣ p − 2 ∇ u ] ∇ φ d x d y → 0 , as n → ∞ .

This implies that

□ u n → u in W 0 , L , loc 1 , p ( V ⧹ { x 1 , x 2 , … x l } ) .

Lemma 4.4

Suppose that u m ∈ W 0 , L 1 , p ( V ) satisfy (4.9), and (4.10), u m ⇀ 0 in W 0 , L 1 , p ( V ) , and f = 0 . Then, there exist three sequences { x m } ⊂ Ω , R m → ∞ , and w m ∈ W 0 , L 1 , p ( V ) such that

w m ( x , y ) = u m − R m ( N + 1 − p ) / p U 0 ( R m ( x − x m ) , R m y ) + o ( 1 ) ,

∣ E ( u m ) − E ( w m ) − E 0 ( U 0 , R N ) ∣ → 0 ,

E ′ ( w m ) → 0 ,

R m d i s t ( x m , ∂ Ω ) → + ∞ ,

where o ( 1 ) → 0 in D 1 , p ( R + N + 1 ) and U 0 is a solution to

(4.16) − Δ p u = 0 , i n R + N + 1 , ∣ ∇ u ∣ p − 2 ∂ u ∂ n = u p ∗ − 1 , o n R N × { 0 } .

Proof of Lemma 4.4

Assume that

(4.17) ∣ u m ∣ p ∗ → ∑ ν j δ x j , x j ∈ Ω 0 ¯ .

Then, there exists at least one ν j ≠ 0 . Otherwise, { u m } → 0 in L p ∗ ( Ω 0 ) . By f = 0 , we have { u m } → 0 in W 0 , L 1 , p ( V ) .

Set concentration function

(4.18) Q m ( r ) = sup x ∈ Ω ¯ ∫ B r ( x ) × ( 0 , ∞ ) ∣ ∇ u m ∣ p d x d y + ∫ B r ( x ) × { 0 } ∣ u m ∣ p ∗ d x .

For sufficiently small τ ∈ ( 0 , S N / ( p − 1 ) ) , choose R m = R m ( τ ) > 0 and x m ∈ Ω ¯ such that

(4.19) ∫ B 1 / R m ( x m ) × ( 0 , ∞ ) ∣ ∇ u m ∣ p d x d y + ∫ B 1 / R m ( x m ) × { 0 } ∣ u m ∣ p ∗ d x = Q m 1 R m = τ .

Set

(4.20) u ˜ m ( x ) = R m ( p − 1 − N ) / p u m x R m + x m , y R m , x ∈ Ω m = x : x R m + x m ∈ Ω .

Then, we have

(4.21) Q ˜ m ( r ) = sup x ∈ R N ∫ B r ( x ) × ( 0 , ∞ ) ∣ ∇ u ˜ m ∣ p d x d y + ∫ B r ( x ) × { 0 } ∣ u ˜ m ∣ p ∗ d x = sup x ∈ R N ∫ B r / R m ( x ) × ( 0 , ∞ ) ∣ ∇ u m ∣ p d x d y + ∫ B r / R m ( x ) × { 0 } ∣ u m ∣ p ∗ d x = Q m r R m .

Hence,

(4.22) ∫ B 1 ( 0 ) × ( 0 , ∞ ) ∣ ∇ u ˜ m ∣ p d x d y + ∫ B 1 ( 0 ) × { 0 } ∣ u ˜ m ∣ p ∗ d x = Q ˜ m ( 1 ) = τ .

In the following, we prove that there exists τ ∈ ( 0 , S N / ( p − 1 ) ) small enough such that R m ( τ ) → + ∞ , as m → + ∞ . If not, for each ε > 0 , there would be a constant M ε > 0 such that R m ( ε ) ≤ M ε , which yields

(4.23) ∫ B 1 / M ε ( x ) × { 0 } ∣ u m ∣ p ∗ d x ≤ sup x ∈ Ω 0 ¯ ∫ B 1 / R m ( x m ) × ( 0 , ∞ ) ∣ ∇ u m ∣ p d x d y + ∫ B 1 / R m ( x m ) × { 0 } ∣ u m ∣ p ∗ d x = Q m 1 R m = ε for all x ∈ Ω ¯ .

Futhermore,

ν j ≤ ∫ B 1 / M ε ( x ) × { 0 } ∣ u m ∣ p ∗ d x + o ( 1 ) ≤ ε + o ( 1 ) for all ε > 0 .

A contradiction!

Now, we distinguish two cases as follows:

R m dist ( x m , ∂ Ω ) → + ∞ , in this case Ω m → Ω ∞ = R N .
R m dist ( x m , ∂ Ω ) → M < + ∞ , uniformly. After an orthogonal transformation,
Ω m → Ω ∞ = R + N = { x = ( x 1 , … , x N ) , x 1 > 0 } .

We define u ˜ m = 0 for ( x , y ) ∉ Ω m × ( 0 , ∞ ) . Since

∫ R N × ( 0 , ∞ ) ∣ ∇ u ˜ m ∣ p d x d y = ∫ Ω m × ( 0 , ∞ ) ∣ ∇ u ˜ m ∣ p d x d y = ∫ Ω × ( 0 , ∞ ) ∣ ∇ u m ∣ p d x d y ,

we can assume that u ˜ m ⇀ U 0 in W 0 , L 1 , p ( V ) .

Since, in each case, for all φ ∈ C 0 , L ∞ ( Ω ∞ × ( 0 , + ∞ ) ¯ ) , we have that φ ∈ C 0 , L ∞ ( Ω m × ( 0 , + ∞ ) ¯ ) , for large m , there holds

(4.24) ∫ R N × ( 0 , ∞ ) ∣ ∇ u ˜ m ∣ p − 2 ∇ u ˜ m ∇ φ d x d y − ∫ R N × { 0 } ∣ u ˜ m ∣ p ∗ − 1 φ d x − ∫ R N × { 0 } f ˜ m φ d x = ∫ R N × ( 0 , ∞ ) ∣ ∇ u m ∣ p − 2 ∇ u m ∇ φ m ∗ d x d y − ∫ R N × { 0 } ∣ u m ∣ p ∗ − 1 φ m ∗ d x − ∫ R N × { 0 } f m φ m ∗ d x → 0 as m → + ∞ ,

where φ m ∗ ( x ) = R m ( N + 1 − p ) / p φ ( R m ( x − x m ) , R m y ) ∈ W 0 , L 1 , p ( V ) . We are going to prove that

(4.25) u ˜ m → U 0 in D loc 1 , p ( R N × ( 0 , ∞ ) ) .

To have this done, by (4.24), we simply prove that

u ˜ m → U 0 in L loc p ∗ ( R N × { 0 } ) .

In fact, choose a smooth cutoff function η ∈ W 0 , L 1 , p ( V ) such that 0 ≤ η ≤ 1 , η = 1 in B 1 , and η = 0 outside B 2 , where B ρ = { ( x , y ) : ∣ ( x , y ) ∣ ≤ ρ , and y > 0 } . Assume that

(4.26) ∣ η u ˜ m ∣ p ∗ ⇀ ν = ∣ η U 0 ∣ p ∗ ( x , 0 ) ⊗ δ 0 ( y ) + ∑ j ∈ J ν j δ x j ,

(4.27) ∣ ∇ ( η u ˜ m ) ∣ p ⇀ μ ≥ ∣ ∇ ( η U 0 ) ∣ p + ∑ j ∈ J S ν j p p ∗ δ x j .

Similar to Lemma 4.1, we can easily show that if ν j = ν ( { x j } ) > 0 , then ν j ≥ S N / ( p − 1 ) , where x j ∈ B k . Moreover

S N / ( p − 1 ) > τ = Q ˜ m ( 1 ) ≥ ∫ B 1 ( x j ) × ( 0 , ∞ ) ∣ ∇ u ˜ m ∣ p d x d y ≥ μ ( { x j } ) ≥ S N / ( p − 1 ) .

Contradiction. So, ν j = ν ( { x j } ) = 0 and (4.25) fulfills.

We then demonstrate that case (ii) is not true. Indeed, using (4.24) and (4.25), we have

(4.28) − Δ p U 0 = 0 , in R + N × ( 0 , ∞ ) , U 0 = 0 , on ∂ R + N × [ 0 , ∞ ) , ∣ ∇ U 0 ∣ p − 2 ∂ U 0 ∂ n = U 0 p ∗ − 1 on R + N × { 0 } .

Analogous to the proof in Theorem 1.1 in [24], we can prove that U 0 = 0 , thanks to the Pohozaev identity and strong maximum principle (see [5]). This is in contradiction with the following relation:

(4.29) 0 = ∫ B 1 ( 0 ) × ( 0 , ∞ ) ∣ ∇ U 0 ∣ p d x d y + ∫ B 1 ( 0 ) × { 0 } ∣ U 0 ∣ p ∗ d x = lim m → ∞ ∫ B 1 ( 0 ) × ( 0 , ∞ ) ∣ ∇ u ˜ m ∣ p d x d y + ∫ B 1 ( 0 ) × { 0 } ∣ u ˜ m ∣ p ∗ d x = τ > 0 .

Hence, R m dist ( x m , ∂ Ω 0 ) → + ∞ .

Let α ∈ C ( B 2 ) satisfying 0 ≤ α ≤ 1 , α = 1 in B 1 , and α = 0 outside B 2 . Set

w m ( x , y ) = u m ( x , y ) − R m ( N + 1 − p ) / p U 0 ( R m ( x − x m ) , R m y ) α ( R ¯ m ( x − x m ) , R ¯ m y ) ∈ W 0 , L 1 , p ( V ) ,

where the sequence R ¯ m is chosen such that R ¯ m dist ( 0 , ∂ Ω ) and R ˜ m ≔ R m R ¯ m → ∞ , then we have

w ˜ m ( x , y ) = u ˜ m ( x , y ) − U 0 ( x , y ) α x R ˜ m , y R ˜ m .

Resembling [28], it is easy to prove that

(4.30) w m ( x , y ) = u m − R m ( N + 1 − p ) / p U 0 ( R m ( x − x m ) , R m y ) + o ( 1 ) ,

where o ( 1 ) → 0 in D 1 , p ( R + N + 1 ) ,

(4.31) w ˜ m = u ˜ m − U 0 + o ( 1 ) .

Thus, by (4.25), we have

(4.32) ∫ V ∣ ∇ u m ∣ p d x d y − ∫ V ∣ ∇ w m ∣ p d x d y = ∫ V ∣ ∇ u m ∣ p d x d y − ∫ V ∣ ∇ u m − R m ( N + 1 − p ) / p ∇ U 0 ( R m ( x − x m ) , R m y ) ∣ p d x d y + o ( 1 ) = ∫ R N × ( 0 , ∞ ) ∣ ∇ u ˜ m ∣ p d x d y − ∫ R N × ( 0 , ∞ ) ∣ ∇ u ˜ m − ∇ U 0 ∣ p d x d y → ∫ R N × ( 0 , ∞ ) ∣ ∇ U 0 ∣ p d x d y .

By continuing as (4.32), we have

∫ Ω 0 ∣ u m ∣ p ∗ d x − ∫ Ω 0 ∣ w m ∣ p ∗ d x → ∫ R N × { 0 } ∣ U 0 ∣ p ∗ d x .

□ ∣ E ( u m ) − E ( w m ) − E 0 ( U 0 , R N ) ∣ → 0 .

Lemma 4.5

Suppose that u m ∈ W 0 , L 1 , p ( V ) satisfy (4.9) and (4.10), and u m ⇀ u in W 0 , L 1 , p ( V ) . Then, there exist a nonnegative integer k , x m j ∈ Ω , R m j ∈ R + , and U j ∈ D 1 , p ( R + N + 1 ) such that

(4.33) u m − u − ∑ j = 1 k ( R m j ) N + 1 − p p U j ( R m j ( x − x m j ) , y ) W 0 , L 1 , p ( V ) = o ( 1 ) ,

(4.34) E ( u m ) = E ( u ) + ∑ j = 1 k E 0 ( U j , R + N + 1 ) + o ( 1 ) ,

(4.35) R m j d i s t ( x m j , ∂ Ω ) → + ∞ ,

where U j is a positive solution of

(4.36) − Δ p u = 0 , i n R + N + 1 = R N × ( 0 , ∞ ) , ∣ ∇ u ∣ p − 2 ∂ u ∂ n = u p ∗ − 1 , o n R N × { 0 }

and

(4.37) E 0 ( u , R N ) = 1 p ∫ R + N + 1 ∣ ∇ u ∣ p d x d y − 1 p ∗ ∫ R N × { 0 } ∣ u ∣ p ∗ d x .

Proof of Lemma 4.5

Applying Lemma 4.4 to the sequences,

v m 1 = u m − u ,

v m j = u m − u − ∑ i = 1 j − 1 U m i = v m j − 1 − U m j − 1 , j > 1 ,

where U m i ( x ) = ( R m i ) N + 1 − p p U i ( R m i ( x − x n i ) , R m i y ) .

By induction,

E 0 ( v m j ) = E ( u m ) − E ( u ) − ∑ i = 1 j − 1 E 0 ( U j ) ≤ E ( u m ) − ( j − 1 ) β ∗ ,

where β ∗ = p − 1 N p S N / ( p − 1 ) .

Due to the fact that the latter will be negative for large j , the iteration must stop after finite steps; moreover, for this index, we have

v m k + 1 = u m − u − ∑ j = 1 k U m j → 0

and

E ( u m ) − E ( u ) − ∑ j = 1 k E 0 ( U j ) → 0 .

The proof is complete.□

Proof of Theorem 1.6

We can easily see that one can always extract a sequence of t n → ∞ such that { v n } satisfies (4.9) and (4.10), where f n = v t ( t n ) actually converges to zero in L p ( Ω × { 0 } ) . Consequently, Theorem 1.6 follows from the above discussion.□

Funding information: This article is supported by the National Natural Science Foundation of China (No. 12071391)
Conflict of interest: The authors declare that they have no conflict of interest.

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Received: 2021-09-30

Revised: 2023-01-29

Accepted: 2023-01-29

Published Online: 2023-03-27

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https://doi.org/10.1515/anona-2022-0306

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Dirichlet-to-Neumann operator; critical exponent; blowup; concentration phenomenon

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