Liouville Type Theorems for Two Mixed Boundary Value Problems with General Nonlinearities
aa r X i v : . [ m a t h . A P ] O c t LIOUVILLE TYPE THEOREMS FOR TWO MIXED BOUNDARYVALUE PROBLEMS WITH GENERAL NONLINEARITIES
XIAOHUI YU ∗ Abstract:
In this paper, we study the nonexistence of positive solutions for the following two mixed boundaryvalue problems. The first problem is the mixed nonlinear-Neumann boundary value problem − ∆ u = f ( u ) in R N + , ∂u∂ν = g ( u ) on Γ , ∂u∂ν = 0 on Γ and the second is the nonlinear-Dirichlet boundary value problem − ∆ u = f ( u ) in R N + , ∂u∂ν = g ( u ) on Γ ,u = 0 on Γ , where R N + = { x ∈ R N : x N > } , Γ = { x ∈ R N : x N = 0 , x < } and Γ = { x ∈ R N : x N = 0 , x > } . We willprove that these problems possess no positive solution under some assumptions on the nonlinear terms. The maintechnique we use is the moving plane method in an integral form. keywords: Liouville type theorem, Moving plane method, Maximum principle, Mixed boundary value. Introduction
In this paper, we study the nonexistence of positive solutions for the following twomixed boundary value problems with general nonlinearities. The first one is the nonlinear-Neumann boundary value problem − ∆ u = f ( u ) in R N + , ∂u∂ν = g ( u ) on Γ , ∂u∂ν = 0 on Γ (1.1) *The Center for China’s Overseas Interests, Shenzhen University, Shenzhen Guangdong, 518060, ThePeople’s Republic of China(yuxiao [email protected]) Mathematics Subject Classification (2010) : 35J60, 35J57, 35J15. ∗ and the second one is the nonlinear-Dirichlet problem − ∆ u = f ( u ) in R N + , ∂u∂ν = g ( u ) on Γ ,u = 0 on Γ , (1.2)where R N + = { x ∈ R N : x N > } , N ≥
3, Γ = { x ∈ R N : x N = 0 , x < } andΓ = { x ∈ R N : x N = 0 , x > } . In the following, we assume that f, g are continuousfunctions.Liouville type theorems are close related to the existence results and prior estimates forelliptic equations. More precisely, after blowing up, elliptic equation in bounded domainturns to be an equation in R N or R N + . Using the respective Liouville theorem, we get acontradiction, so the prior estimate is proved. Then we can use the topological method toprove the existence results. For more details, we refer to [9][11] and etc.In the past few decades, there are plenty of works on the Liouville type theorems forelliptic equations and elliptic systems. The first result is [10], in which the authors studiedthe nonexistence results for the following elliptic equation − ∆ u = u p in R N , u ≥ . (1.3)The authors proved, among other things, that the only solution for problem (1.3) is u ≡ < p < N +2 N − . This result is optimal in the sense that for any p ≥ N +2 N − , there areinfinitely many positive solutions to (1.3). Thus the Sobolev exponent N +2 N − is the dividingexponent between existence and nonexistence of positive solutions. Nonexistence result forproblem in half space was also obtained in [11]. Later, W.Chen and C.Li proved similarresults by using the moving plane method in [4]. By using the moving plane method, theauthors proved the solution is symmetric in every direction and with respect every point,hence the solution must be u ≡
0. Recently, in an interesting paper [6], L.Damascelli andF.Gladiali studied the nonexistence result of positive solution for the following nonlinearproblem with general nonlinearity − ∆ u = f ( u ) in R N , u ≥ . (1.4)If f is assumed to be increasing and subcritical, then the authors proved the only solutionfor problem (1.4) is u ≡
0. The main tool they used is the method of moving planes.We note that f is only assumed to be continuous in this paper. So we can not concludethat the weak solutions of problem (1.4) are of C class. In developing the method ofmoving plane, the authors in [6] used the technique based on integral inequalities, an ideaoriginally due to S.Terracini’s work [22] and [23]. After the work of [6], a lot of worksconcern the Liouville type theorems for elliptic equation with general nonlinearity, see[12][13][14][15][24][25][26][27]. For Liouville theorem on nonlinear elliptic systems, we referto [8][16][18][19][20][21] and etc..In this paper, we are concerned with the nonexistence of positive solution for the twomixed boundary value problems. Mixed boundary problems were widely studied in the IOUVILLE TYPE THEOREMS FOR TWO MIXED BOUNDARY VALUE PROBLEMS WITH GENERAL NONLINEARITIES3 past few decades. For example, E.Colorado and I.Peral studied the existence results in[5]. The concentration behaviors of singularly perturbed mixed boundary problems werestudied in [1][2][7] and the references therein. Nonexistence results for Neumann-Dirichletmixed boundary problem have also been obtained in the past, see [3] and [6]. However,up to our knowledge, it seems to no result for nonlinear-Neumann and nonlinear-Dirichletmixed boundary value problems and this is the main purpose of this paper.First, we need to define weak solution for problem (1.1) and problem (1.2). Let E = W , loc ( R N + ) ∩ C ( R N + ), we say that u is a weak solution of problem (1.1), if u ∈ E and satisfies Z R N ∇ u ∇ ϕ dx = Z R N f ( u ) ϕ dx + Z Γ g ( u ) ϕ dx ′ for any ϕ ∈ C c ( R N + ). Similarly, we denote by W = { ϕ ∈ C c ( R N + ) , supt { ϕ } ⊂ A } with A = R N + ∪ Γ , then we say that u is a weak solution of problem (1.2), if u ∈ E and satisfies Z R N ∇ u ∇ ϕ dx = Z R N f ( u ) ϕ dx + Z Γ g ( u ) ϕ dx ′ for any ϕ ∈ W .Our first result concerns the nonexistence of positive solution for nonlinear-Neumannmixed boundary value problem, i.e., problem (1.1). Our main result is the following Theorem 1.1.
Let u ∈ E be a bounded nonnegative solution of problem (1.1) , where f, g : [0 , + ∞ ) → [0 , + ∞ ) are continuous functions with the properties(i) f ( t ) , g ( t ) are nondecreasing in (0 , + ∞ ) .(ii) h ( t ) = f ( t ) t N +2 N − , k ( t ) = g ( t ) t NN − are nonincreasing in (0 , + ∞ ) .Then u ≡ c with f ( c ) = g ( c ) = 0 . Next, we study the mixed nonlinear-Dirichlet mixed boundary value problem (1.2). Oursecond conclusion is the following
Theorem 1.2.
Let u ∈ E be a bounded nonnegative solution of problem (1.2) , where f, g : [0 , + ∞ ) → [0 , + ∞ ) are continuous functions with the properties(i) f ( t ) , g ( t ) are nondecreasing in (0 , + ∞ ) .(ii) h ( t ) = f ( t ) t N +2 N − , k ( t ) = g ( t ) t NN − are nonincreasing in (0 , + ∞ ) .Then u ≡ .Remark . By the Doubling Lemma in [17], the boundedness assumptions in Theorem1.1 and Theorem 1.2 can be dropped.Here we should emphasize that we only need that the nonlinearities are continuous inthe above theorems, we don’t need they are Lipschitz continuous. Since no Lipschitzassumption was made on the nonlinearities, the usual maximum principle does not work.We resorted to some integral inequality, which was first introduced by S.Terracini in [22][23]and then was widely used in [6][12][13] and [14] respectively. By the same sprit of this,X.Yu studied the nonlinear Liouville type theorems for other equations in [24][25][26][27]
XIAOHUI YU ∗ and [28]. The main idea of this paper is the same as the above works, we use integralinequality to substitute the usual maximum principle.The rest of this paper is devoted the proof of Theorem 1.1 and Theorem 1.2. We dividethe proof of these theorems into the following steps. First, we show that the nonnegativesolutions of problems (1.1) and (1.2) are nondecreasing as x decreases. Next, we show thatthe nonnegative solutions either depend only on x and x N or are regular at infinity. Finally,if u depends only on x and x N , then the limit function w ( x N ) = lim x →−∞ u ( x , x N )exists and w ( x N ) satisfies a proper equation and it can only be the trivial solution underthe assumptions in Theorem 1.1 and Theorem 1.2. On the other hand, if the solution isregular at infinity, the we get a contradiction directly from the monotonicity of u in x direction. In the rest of this paper, we denote C by a positive constant, which may varyfrom line to line. 2. Proof of Theorem 1.1
The key step is to prove the monotonicity of the nonnegative solutions for problem (1.1)in the x direction. We use the moving plane method to prove our result. The first step ofmoving plane is to show that this procedure can be started at some point. Since we don’tknow the decay behaviors of u , it seems difficult to use this method directly on u . So wemake use of the Kelvin transformation v of u . More precisely, for any µ ∈ R , we denote p µ = ( µ, , ..., ∈ ∂ R N + and define the Kelvin transformation v µ of u at p µ as v µ ( x ) = 1 | x − p µ | N − u ( x − p µ | x − p µ | + p µ ) . (2.1)Then we deduce from the definition of Kelvin transformation that v µ ( x ) decays at the rateof | x − p µ | − N as | x | → ∞ . In particular, we have v µ ( x ) ∈ L ∗ ∩ L ∞ ( R N + \ B r ( p µ )) (2.2)for any r >
0, where 2 ∗ = NN − is the usual Sobolev critical exponent. Moreover, a directcalculation shows that v µ satisfies the following equation − ∆ v µ = f ( | x − p µ | N − v µ ( x ))[ | x − p µ | N − v µ ( x )] N +2 N − v µ ( x ) N +2 N − , x ∈ R N + , ∂v µ ∂ν = g ( | x − p µ | N − v µ ( x ))[ | x − p µ | N − v µ ( x )] NN − v µ ( x ) NN − , x ∈ { x ∈ ∂ R N + : µ − µ < x < µ } , ∂v µ ∂ν = 0 , x ∈ { x ∈ ∂ R N + : x > µ or x < µ − µ } (2.3) IOUVILLE TYPE THEOREMS FOR TWO MIXED BOUNDARY VALUE PROBLEMS WITH GENERAL NONLINEARITIES5 for µ ≥
0, where we interpret µ = + ∞ for µ = 0. However, for µ < v µ satisfies − ∆ v µ = f ( | x − p µ | N − v µ ( x ))[ | x − p µ | N − v µ ( x )] N +2 N − v µ ( x ) N +2 N − , x ∈ R N + , ∂v µ ∂ν = g ( | x − p µ | N − v µ ( x ))[ | x − p µ | N − v µ ( x )] NN − v µ ( x ) NN − , x ∈ { x ∈ ∂ R N + : x < µ or x > µ − µ } , ∂v µ ∂ν = 0 , x ∈ { x ∈ ∂ R N + : µ < x < µ − µ } . (2.4)Moreover, we infer from the definitions of h and k that v µ satisfies − ∆ v µ = h ( | x − p µ | N − v µ ( x )) v µ ( x ) N +2 N − , x ∈ R N + , ∂v µ ∂ν = k ( | x − p µ | N − v µ ( x )) v µ ( x ) NN − , x ∈ { x ∈ ∂ R N + : µ − µ < x < µ } , ∂v µ ∂ν = 0 , x ∈ { x ∈ ∂ R N + : x > µ or x < µ − µ } (2.5)for µ ≥ v µ satisfies − ∆ v µ = h ( | x − p µ | N − v µ ( x )) v µ ( x ) N +2 N − , x ∈ R N + , ∂v µ ∂ν = k ( | x − p µ | N − v µ ( x )) v µ ( x ) NN − , x ∈ { x ∈ ∂ R N + : x < µ or x > µ − µ } , ∂v µ ∂ν = 0 , x ∈ { x ∈ ∂ R N + : µ < x < µ − µ } (2.6)for µ < µ >
0. We will show that the moving plane procedure can becarried out from ∞ to λ = µ . For this purpose, we defineΣ λ = { x ∈ R N + : x > λ } , T λ = { x ∈ R N + : x = λ } and ∂ Σ λ = { x ∈ ∂ Σ λ : x N = 0 } . Moreover, for any x ∈ Σ λ , the reflection of x with respect to T λ is x λ = { λ − x , x , ..., x N } . In the following, we denote by u λ ( x ) = u ( x λ ) and p λµ = { λ − µ, , ..., } . With the abovenotations, we have the following key lemma. Lemma 2.1.
For any fixed λ > µ ≥ , the functions v µ and ( v µ − v λµ ) + belong to L ∗ (Σ λ ) ∩ L ∞ (Σ λ ) with ∗ = NN − . Further more, if we denote A λµ = { x ∈ Σ λ | v µ > v λµ } and B λµ = { x ∈ ∂ Σ λ | v µ ( x ) > v λµ ( x ) } , then there exists C λ > , which is nonincreasing in λ , such that Z Σ λ |∇ ( v µ − v λµ ) + | dx ≤ C λ ( Z A λµ | x − p µ | N dx ) N · ( Z Σ λ |∇ ( v µ − v λµ ) + | dx ) . (2.7) XIAOHUI YU ∗ Proof.
Since λ > µ , there exists r > λ ⊂ R N + \ B r ( p µ ). Hence, we deduce fromequation (2.2) and the decay behavior of v µ that v µ , ( v µ − v λµ ) + ∈ L ∗ (Σ λ ) ∩ L ∞ (Σ λ ) . Now we choose a cut off function η = η ε ∈ C ( R N + , [0 , η ( x ) = (cid:26) ε ≤ | x − p λµ | ≤ ε , | x − p λµ | < ε or | x − p λµ | > ε , |∇ η | ≤ ε for ε < | x − p λµ | < ε and |∇ η | ≤ ε for ε < | x − p λµ | < ε . Moreover, if we define ϕ = ϕ ε = η ε ( v µ − v λµ ) + and ψ = ψ ε = η ε ( v µ − v λµ ) + , then a direct calculation shows that |∇ ψ | = ∇ ( v µ − v λµ ) ∇ ϕ + [( v µ − v λµ ) + ] |∇ η | . On the other hand, it is easy to see that equation − ∆( v µ ( x ) − v λµ ( x )) = h ( | x − p µ | N − v µ ( x )) v µ ( x ) N +2 N − − h ( | x λ − p µ | N − v µ ( x λ )) v µ ( x λ ) N +2 N − (2.8)holds in Σ λ \ { p λµ } . So if we multiply equation (2.8) by ϕ , then we get Z Σ λ ∩{ ε ≤| ξ − p λ |≤ ε } |∇ ( v µ − v λµ ) + | dx ≤ Z Σ λ |∇ ψ | dx = Z Σ λ ∇ ( v µ − v λµ ) ∇ ϕ dx + Z Σ λ [( v µ − v λµ ) + ] |∇ η ε | dx = Z A λµ − ∆( v µ − v λµ ) ϕ dx + Z ∂ Σ λ ∂ ( v µ − v λµ ) ∂ν ϕ dx ′ + I ε ≤ Z A λµ [ h ( | x − p µ | N − v µ ( x )) v N +2 N − µ − h ( | x λ − p µ | N − v λµ )( v λµ ) N +2 N − ] ϕ dx + I ε , (2.9)where I ε = R Σ λ [( v − v λ ) + ] |∇ η ε | dx , and the last inequality follows from that ∂ ( v − v λ ) ∂ν ≤ ∂ Σ λ . Since h is nonincreasing, the above equation implies Z Σ λ ∩{ ε ≤| ξ − p λ |≤ ε } |∇ ( v µ − v λµ ) + | dx ≤ Z A λµ h ( | x − p µ | N − v µ ( x ))[ v N +2 N − µ − ( v λµ ) N +2 N − ] ϕ dx + I ε ≤ C Z A λµ h ( | x − p µ | N − v µ ( x )) v N − µ [ v µ − v λµ ] ϕ dx + I ε . (2.10) IOUVILLE TYPE THEOREMS FOR TWO MIXED BOUNDARY VALUE PROBLEMS WITH GENERAL NONLINEARITIES7
Moreover, since Σ λ ′ ⊂ Σ λ for λ ′ > λ and | x − p µ | N − v µ ( x ) is bounded in Σ λ , then thereexists a constant C λ , which is nonincreasing in λ , such that Z Σ λ ∩{ ε ≤| ξ − p λ |≤ ε } |∇ ( v µ − v λµ ) + | dx ≤ C λ Z A λµ v N − µ [ v µ − v λµ ] + ϕ dx + I ε . (2.11)On the other hand, we deduce from the decay property of v µ that Z Σ λ ∩{ ε ≤| ξ − p λ |≤ ε } |∇ ( v µ − v λµ ) + | dx ≤ C λ Z A λµ | x − p µ | [( v µ − v λµ ) + ] η dx + I ε ≤ C λ ( Z A λµ | x − p µ | N dx ) N ( Z Σ λ [( v µ − v λµ ) + ] NN − dx ) N − N + I ε . (2.12)We claim that I ε → ε →
0. In fact, if we denote D ε = { x ∈ Σ λ | ε < | x − p λµ | < ε } and D ε = { x ∈ Σ λ | ε < | x − p λµ | < ε } , then we get Z D ε |∇ η | N dx ≤ C ε N · ε N = C. Similarly, we also have Z D ε |∇ η | N dx ≤ Cε N · ε N = C. Hence, it follows from the Holder inequality that I ε ≤ ( Z D ε ∪ D ε [( v µ − v λµ ) + ] ∗ dx ) ∗ · ( Z Σ |∇ η | N dx ) N → ε → v − v λ ) + ∈ L ∗ (Σ λ ).Finally, letting ε → Z Σ λ |∇ ( v µ − v λµ ) + | dx ≤ C λ ( Z A λµ | x − p µ | N dx ) N Z Σ λ |∇ ( v µ − v λµ ) + | dx, (2.13)which completes the proof of this lemma. (cid:3) Before we continue the proof of Theorem 1.1, we give some comments on Lemma 2.1.Since we don’t require the nonlinear terms f and g are Lipschitz continuous, the usualmaximum principle does not work. Thanks to equation (2.7), it can play the same role asthe maximum principle. In fact, if we can prove C λ ( Z A λµ | x − p µ | N dx ) N < , XIAOHUI YU ∗ then we get v µ ≤ v λµ , the same conclusion as the maximum principle implies.The next lemma shows that we can start the moving plane from some place. Lemma 2.2.
Under assumptions of Theorem 1.1, there exists λ > µ , such that for all λ ≥ λ , we have v µ ≤ v λµ in Σ λ .Proof. The conclusion of this lemma is a direct corollary of Lemma 2.1. In fact, by thedecay speed of v µ , we can choose λ large enough such that C λ ( Z Σ λ | x − p µ | N dx ) N < λ ≥ λ , then equation (2.7) implies that Z Σ λ |∇ ( v µ − v λµ ) + | dx = 0 . The assertion follows. (cid:3)
Now we can move the plane from the right to the left such that v µ ≤ v λµ in Σ λ andsuppose this process stops at some λ . More precisely, we define λ = inf { λ | v µ ≤ v λµ in Σ λ } , (2.14)then we have the following Lemma 2.3. If u , then λ ≤ µ .Proof. We prove the conclusion by contradiction. Suppose on the contrary that λ > µ ,then we claim that v µ ≡ v λ µ .We prove the claim by contradiction. Suppose v µ v λ µ , then we show that the planecan be moved to the left a little. That is, we will show that there exists δ >
0, such that v µ ( x ) ≤ v λµ ( x ) in Σ λ for all λ ∈ [ λ − δ, λ ]. This contradicts the choice of λ .To prove this claim, we first infer from the continuity that v µ ( x ) ≤ v λ µ ( x ). Moreover,we have h ( | x − p µ | N − v µ ) v N +2 N − µ = f ( | x − p µ | N − v µ ) | x − p µ | N +2 ≤ f ( | x − p µ | N − v λ µ ) | x − p µ | N +2 = f ( | x − p µ | N − v λ µ )[ | x − p µ | N − v λ µ ] N +2 N − ( v λ µ ) N +2 N − ≤ f ( | x λ − p µ | N − v λ µ )[ | x λ − p µ | N − v λ µ ] N +2 N − ( v λ µ ) N +2 N − = h ( | x λ − p µ | N − v λ µ )( v λ µ ) N +2 N − , where the first inequality holds since f is nondecreasing, the second inequality is a conse-quence of (ii) in Theorem 1.1. IOUVILLE TYPE THEOREMS FOR TWO MIXED BOUNDARY VALUE PROBLEMS WITH GENERAL NONLINEARITIES9
This equation implies − ∆ v µ ≤ − ∆ v λ µ , then we infer from the maximum principle that v µ < v λ µ in Σ λ since v µ v λ µ . Moreover,since | x − p µ | N χ A λµ → λ → λ and | x − p µ | N χ A λµ ≤ | x − p µ | N χ A λ − δµ for λ ∈ [ λ − δ, λ ]and some δ >
0, then the dominated convergence theorem implies Z A λµ | x − p µ | N dx → λ → λ . That is, there exists δ >
0, such that C λ ( Z A λµ | x − p µ | N dx ) N < λ ∈ [ λ − δ, λ ]. Then we infer from Lemma 2.1 that v µ ≤ v λµ for all λ ∈ [ λ − δ, λ ].This contradicts the definition of λ . So we prove the claim.Next, we show that v µ ≡ v λ µ implies v µ ≡ u ≡
0. In fact, if v µ ≡ v λ µ , thenwe deduce from the Neumann boundary value condition and the nonlinear boundary valuecondition that v µ ( x ) = 0 for some x ∈ ∂ R N + , this contradicts the Hopf Theorem unless v µ ≡ (cid:3) The above two lemmas implies the following
Corollary 2.4.
For any µ ≥ , we have v µ ( x ) ≤ v µ ( x µ ) for x ∈ Σ µ . Next, we consider the case µ <
0. In this case, it seems difficult to start the movingplane method as µ > v µ . Instead of choosing µ arbitrary for µ ≥
0, we move µ from 0 to −∞ gradually. We first need some integralinequality similar to equation (2.7) to substitute the maximum principle. In this case, weneed to evaluate the term R ∂ Σ λ ∂ ( v − v λ ) ∂ν ϕ dx ′ in equation (2.9) more carefully since it is notnonpositive any longer. Similar to Lemma 2.1, we have the following result. Lemma 2.5.
Suppose that µ < , then for any fixed µ < λ ≤ µ − µ , the functions v µ and ( v µ − v λµ ) + belong to L ∗ (Σ λ ) ∩ L ∞ (Σ λ ) with ∗ = NN − . Further more, if we denote A λµ = { x ∈ Σ λ | v µ > v λµ } and B λµ = { x ∈ ∂ Σ λ | v µ ( x ) > v λµ ( x ) } , then there exists C λ > ,which is nonincreasing in λ , such that Z Σ λ |∇ ( v µ − v λµ ) + | dx ≤ C λ [( Z A λµ | x − p µ | N dx ) N + ( Z B λµ | x − p µ | N − dx ′ ) N − ] · ( Z Σ λ |∇ ( v µ − v λµ ) + | dx ) . (2.15) Proof.
The proof of this lemma is similar to the proof of Lemma 2.1. We sketch it. Wedenote η, ϕ, ψ as the proof of Lemma 2.1, then we have the following inequality similar to ∗ equation (2.9). It reads Z Σ λ ∩{ ε ≤| ξ − p λ |≤ ε } |∇ ( v µ − v λµ ) + | dx ≤ Z A λµ − ∆( v µ − v λµ ) ϕ dx + Z ∂ Σ λ ∂ ( v µ − v λµ ) ∂ν ϕ dx ′ + I ε ≤ Z A λµ [ h ( | x − p µ | N − v µ ( x )) v N +2 N − µ − h ( | x λ − p µ | N − v λµ )( v λµ ) N +2 N − ] ϕ dx + Z B λµ [ k ( | x − p µ | N − v µ ( x )) v NN − µ − k ( | x λ − p µ | N − v λµ )( v λµ ) NN − ] ϕ dx ′ + I ε , (2.16)where I ε = R Σ λ [( v µ − v λµ ) + ] |∇ η ε | dx . Since h and k are nonincreasing, the above equationimplies Z Σ λ ∩{ ε ≤| ξ − p λ |≤ ε } |∇ ( v µ − v λµ ) + | dx ≤ Z A λµ h ( | x − p µ | N − v µ ( x ))[ v N +2 N − µ − ( v λµ ) N +2 N − ] ϕ dx + Z B λµ k ( | x − p µ | N − v µ ( x ))[ v µ ( x ) NN − − ( v λµ ) NN − ] ϕ dx ′ + I ε ≤ C Z A λµ h ( | x − p µ | N − v µ ( x )) v N − µ [ v µ − v λµ ] ϕ dx + C Z B λµ k ( | x − p µ | N − v µ ( x )) v µ ( x ) N − [ v µ ( x ) − v µ ( x λ )] ϕ dx ′ + I ε . ≤ C λ Z A λµ v N − µ [ v µ − v λµ ] + ϕ dx + C λ Z B λµ v µ ( x ) N − [ v µ ( x ) − v µ ( x λ )] ϕ dx ′ + I ε ≤ C λ ( Z A λµ | x − p µ | N dx ) N ( Z Σ λ [( v µ − v λµ ) + ] NN − dx ) N − N + C λ ( Z B λµ | x − p µ | N − dx ′ ) N − ( Z ∂ Σ λ [( v µ − v λµ ) + ] N − N − dx ′ ) N − N − + I ε ≤ C λ [( Z A λµ | x − p µ | N dx ) N + ( Z B λµ | x − p µ | N − dx ′ ) N − ] · ( Z Σ λ |∇ ( v µ − v λµ ) + | dx ) + I ε . (2.17)Letting ε → (cid:3) IOUVILLE TYPE THEOREMS FOR TWO MIXED BOUNDARY VALUE PROBLEMS WITH GENERAL NONLINEARITIES11
With the above preparations, we can start the moving plane procedure for µ < µ <
0, it seems difficult to move the plane T λ because of the boundaryvalue conditions of v µ . Instead of choosing µ arbitrary for µ ≥
0, we move the plane µ from0 to −∞ step by step. In the first step, we will show that the parameter µ can be movedfrom 0 to µ = − √ . This procedure is composed of the following two lemmas. First, weshow that this procedure can be started for µ small enough. More precisely, we have thefollowing result. Lemma 2.6.
There exists µ < , such that for all µ ≤ µ ≤ , we have v µ ≤ v µ for all x ∈ Σ .Proof. By Corollary 2.4, we know that v ≤ v for all x ∈ Σ , hence the functions | x − p µ | N χ A µ → | x − p µ | N − χ B µ → µ → − . We deduce from the dominateconvergence theorem that there exists µ < µ ≤ µ ≤ C λ [( Z A µ | x − p µ | N dx ) N + ( Z B µ | x − p µ | N − dx ′ ) N − ] ≤ . Then we infer from Lemma 2.5 that R A µ |∇ ( v µ − v µ ) + | dx = 0, that is v µ ≤ v µ in Σ . (cid:3) Next, we will show that the plane can be moved down to − √ . More precisely, we define µ = inf { µ < v µ ≤ v µ in Σ } , then we have Lemma 2.7. If u , then we must have µ ≤ − √ .Proof. We prove the conclusion by contradiction. Suppose on the contrary that µ > − √ ,then we claim that v µ ≡ v µ .We prove the claim by contradiction. Suppose v µ v µ , then we show that the planecan be moved to the left a little. That is, we will show that there exists δ >
0, such that v µ ( x ) ≤ v µ ( x ) in Σ for all µ ∈ [ µ − δ, µ ]. This contradicts the choice of µ . ∗ To prove this, we first infer from the continuity that v µ ( x ) ≤ v µ ( x ) for x ∈ Σ . More-over, we have h ( | x − p µ | N − v µ ) v N +2 N − µ = f ( | x − p µ | N − v µ ) | x − p µ | N +2 ≤ f ( | x − p µ | N − v µ ) | x − p µ | N +2 = f ( | x − p µ | N − v µ )[ | x − p µ | N − v µ ] N +2 N − ( v µ ) N +2 N − ≤ f ( | x − p µ | N − v µ )[ | x − p µ | N − v µ ] N +2 N − ( v µ ) N +2 N − = h ( | x − p µ | N − v µ )( v µ ) N +2 N − , where the first inequality holds since f is nondecreasing, the second inequality is a conse-quence of (ii) in Theorem 1.1.This equation implies − ∆ v µ ≤ − ∆ v µ , then we infer from the maximum principle that v µ < v µ in Σ since v µ v µ . Since | x − p µ | N χ A µ → , | x − p µ | N − χ B µ → µ → µ , moreover, as | x − p µ | N χ A µ ≤ | x − p µ | N χ A µ − δ and | x − p µ | N − χ B µ ≤ | x − p µ | N − χ B µ − δ for µ ∈ [ µ − δ, µ ] and some δ > Z A µ | x − p µ | N dx + Z B µ | x − p µ | N − dx ′ → µ → µ . That is, there exists δ >
0, such that C λ [( Z A µ | x − p µ | N dx ) N + ( Z B µ | x − p µ | N − dx ′ ) N − ] < µ ∈ [ µ − δ, µ ]. Then we infer from Lemma 2.5 that v µ ≤ v µ for all µ ∈ [ µ − δ, µ ].This contradicts the definition of µ . This proves the claim.Similar to Lemma 2.3, we can show that v µ ( x ) ≡ v µ implies v µ ≡
0. In fact, if v µ ≡ v µ , then we deduce from the Neumann boundary value condition and the nonlinearboundary value condition that v µ ( x ) = 0 for some x ∈ ∂ R N + , this contradicts the HopfTheorem unless v µ ≡
0. If v µ ≡
0, then the proof of Theorem 1.1 is complete. So if u
0, then we must have µ ≤ − √ . (cid:3) Now, let µ ∈ [ µ ,
0) fixed, then the same procedure as above shows that we move canthe plane T λ from λ = 0 to λ = µ , in particular, we have v µ ( x ) ≤ v µ ( x µ ) for x ∈ Σ µ and µ ∈ [ µ , v µ ( x ) ≤ v µ ( x µ ) for x ∈ Σ µ , we can movethe plane µ from µ = − √ to µ = − √ , then for µ ∈ [ µ , µ ), we can move the plane T λ from µ to µ . In particular, we have v µ ( x ) ≤ v µ ( x µ ) for x ∈ Σ µ and µ ∈ [ µ , µ ). IOUVILLE TYPE THEOREMS FOR TWO MIXED BOUNDARY VALUE PROBLEMS WITH GENERAL NONLINEARITIES13
Continuing the above procedure, in the ( k + 1)th step, we can show that the parameter µ can be moved from µ k to µ k +1 = µ k − √ µ k +22 . A direct calculation shows that µ k → −∞ as k → ∞ . Hence, we have proved Proposition 2.8.
For any µ < , we have v µ ( x ) ≤ v µ ( x µ ) for x ∈ Σ µ . Proposition 2.9.
For fixed ( x , x , · · · , x N ) ∈ R N − , u ( x , x , x , · · · , x N ) is nondecreas-ing as x decreases.Proof. By Corollary 2.4 and Proposition 2.8, we have v µ ≤ v µµ for any µ ∈ R . In particular,we have 1 | x − p µ | N − v ( p µ + x − p µ | x − p µ | ) ≤ | x µ − p µ | N − v ( p µ + x µ − p µ | x µ − p µ | ) . Since | x − p µ | = | x µ − p µ | and x µ − µ = µ − x , then we have u ( x , x , x , · · · , x N ) ≤ u (2 µ − x , x , x , · · · , x N )for x ≥ µ . Because µ is arbitrary, then we conclude that u is nondecreasing as x decreases. (cid:3) Proof of Theorem 1.1:
To prove Theorem 1.1, we first start the moving plane method inthe x direction. For this purpose, we define the Kelvin transformation v µ of u at p µ =(0 , µ, , · · · ,
0) as v µ ( x ) = 1 | x − p µ | N − u ( p µ + x − p µ | x − p µ | ) . Moreover, we define Σ λ = { x ∈ R N + | x > λ } , T λ = { x ∈ R N + | x = λ } and x λ be thereflection of x with respect to T λ as before.If we denote v λµ = v µ ( x λ ), then by the same reason as Lemma 2.1 and Lemma 2.5, wehave the following inequality Z Σ λ |∇ ( v − v λ ) + | dx ≤ C λ [( Z A λµ | x − p µ | N dx ) N + ( Z B λµ | x − p µ | N − dx ′ ) N − ] · ( Z Σ λ |∇ ( v − v λ ) + | dx ) , (2.18)where C λ is a nonincreasing constant, A λµ = { x ∈ Σ λ | v µ ( x ) > v λµ } and B λµ = { x ∈ ∂ Σ λ | v µ ( x ) > v λµ , x N = 0 } . With the above inequality, we can start the movingplane method for λ large enough. Now, move the plane T λ from the right to the left andsuppose the process stops at some λ . If λ > µ , then we can prove that v µ is symmetricwith respect to T λ as before. This means that v µ is regular at p µ . In particular, we have u ∈ L NN − ( R N + ). Otherwise, we must have λ ≤ µ , which implies v µ ≤ v µµ . At the sametime, we can start moving plane method from the left. Similarly, if this process stops atsome λ < µ , then u ∈ L NN − ( R N + ), otherwise, we have λ = µ and v µµ ≤ v µ .We have proved the following fact: if the moving plane procedure stops at some λ = µ ,then u ∈ L NN − ( R N + ), otherwise, we must have v µ ≡ v µµ and hence u ≡ u µ . ∗ With the above preparations, we can prove Theorem 1.1 now. We distinguish two cases.Case 1: During the moving plane method process in the x , · · · , x N − directions, thereexists some µ and some direction, such that λ = µ , i.e., the moving plane method stopsat some λ = µ . Then we conclude that u is regular at infinity and hence u ∈ L NN − ( R N + ).This is impossible unless u ≡ u is nondecreasing as x decreases by Proposition 2.9.Case 2: For all µ and all directions x , · · · , x N − , the moving plane method stops at λ = µ . In this case, the function u is independent of x , · · · , x N − since µ is arbi-trary. So we have u = u ( x , x N ). Moreover, Proposition 2.9 implies that u ( x , x n ) isnondecreasing as x decreases. On the other hand, since u is bounded, then the limit w ( x N ) = lim x →−∞ u ( x , x N ) exists and w ( x N ) satisfies w ′′ ( x N ) = − f ( w ) for x N > , ∂w∂x N = − g ( u ) for x N = 0 . (2.19)If w c with f ( c ) = g ( c ) = 0, then the first equation implies that w is a concave function,while the second equation implies w ′ (0) <
0. Hence there exists r > w ( x N ) < x N > r , this contradicts that u is a nonnegative solution to problem (1.1).Finally, since w ≡ c with f ( c ) = g ( c ) = 0 and f, g are nondecreasing, then equation(1.1) turns to be − ∆ u = 0 in R N + , ∂u∂ν = 0 on ∂ R N + . Since u is bounded, then the classical Liouville theorem implies u ≡ c with f ( c ) = g ( c ) = 0.This finishes the proof of Theorem 1.1. (cid:3) Proof of Theorem 1.2
The sprit of the proof of Theorem 1.2 is the same as the proof of Theorem 1.1. We onlygive the difference between the two problems.As before, for any µ ∈ R , we define the Kelvin transformation v µ of u at p µ = ( µ, , ..., v µ ( x ) = 1 | x − p µ | N − u ( p µ + x − p µ | x − p µ | ) , then a direct calculations shows that v µ satisfies − ∆ v µ = f ( | x − p µ | N − v µ ( x ))[ | x − p µ | N − v µ ( x )] N +2 N − v µ ( x ) N +2 N − , x ∈ R N + , ∂v µ ∂ν = g ( | x − p µ | N − v µ ( x ))[ | x − p µ | N − v µ ( x )] NN − v µ ( x ) NN − , x ∈ { x ∈ ∂ R N + : µ − µ < x < µ } ,v µ = 0 , x ∈ { x ∈ ∂ R N + : x > µ or x < µ − µ } (3.1) IOUVILLE TYPE THEOREMS FOR TWO MIXED BOUNDARY VALUE PROBLEMS WITH GENERAL NONLINEARITIES15 for µ ≥
0, where we interpret µ = + ∞ for µ = 0. However, for µ < v µ satisfies − ∆ v µ = f ( | x − p µ | N − v µ ( x ))[ | x − p µ | N − v µ ( x )] N +2 N − v µ ( x ) N +2 N − , x ∈ R N + , ∂v µ ∂ν = g ( | x − p µ | N − v µ ( x ))[ | x − p µ | N − v µ ( x )] NN − v µ ( x ) NN − , x ∈ { x ∈ ∂ R N + : x < µ or x > µ − µ } ,v µ = 0 , x ∈ { x ∈ ∂ R N + : µ < x < µ − µ } . (3.2)Moreover, we infer from the definitions of h and k that v µ satisfies − ∆ v µ = h ( | x − p µ | N − v µ ( x )) v µ ( x ) N +2 N − , x ∈ R N + , ∂v µ ∂ν = k ( | x − p µ | N − v µ ( x )) v µ ( x ) NN − , x ∈ { x ∈ ∂ R N + : µ − µ < x < µ } ,v µ = 0 , x ∈ { x ∈ ∂ R N + : x > µ or x < µ − µ } (3.3)for µ ≥ v µ satisfies − ∆ v µ = h ( | x − p µ | N − v µ ( x )) v µ ( x ) N +2 N − , x ∈ R N + , ∂v µ ∂ν = k ( | x − p µ | N − v µ ( x )) v µ ( x ) NN − , x ∈ { x ∈ ∂ R N + : x < µ or x > µ − µ } ,v µ = 0 , x ∈ { x ∈ ∂ R N + : µ < x < µ − µ } (3.4)for µ < x direction for µ ≥
0. Forproblem (3.4), we have the same inequality as Lemma 1.2.
Lemma 3.1.
For any fixed λ > µ ≥ , the functions v µ and ( v µ − v λµ ) + belong to L ∗ (Σ λ ) ∩ L ∞ (Σ λ ) with ∗ = NN − . Further more, if we denote A λµ = { x ∈ Σ λ | v µ > v λµ } and B λµ = { x ∈ ∂ Σ λ | v µ ( x ) > v λµ ( x ) } , then there exists C λ > , which is nonincreasing in λ , such that Z Σ λ |∇ ( v µ − v λµ ) + | dx ≤ C λ ( Z A λµ | x − p µ | N dx ) N · ( Z Σ λ |∇ ( v µ − v λµ ) + | dx ) . (3.5) Proof.
We observe from the boundary value condition of v µ that ∂ ( v µ − v λµ ) ∂ν ϕ ( x ) = 0for any x ∈ ∂ Σ λ . Then the rest of the proof is the same as the proof of Lemma 2.1. (cid:3) With the above inequality, we can start the moving plane procedure now. Similar to theproof of Theorem 1.1, we have
Proposition 3.2.
For any µ ≥ , we have v µ ( x ) ≤ v µµ ( x ) for x ∈ Σ µ . ∗ Next, we consider the case µ <
0. Similar to the result for problem (1.1), we have thefollowing result.
Lemma 3.3.
Suppose that µ < , then for any fixed µ < λ < µ − µ , the functions v µ and ( v µ − v λµ ) + belong to L ∗ (Σ λ ) ∩ L ∞ (Σ λ ) with ∗ = NN − . Further more, if we denote A λµ = { x ∈ Σ λ | v µ > v λµ } and B λµ = { x ∈ ∂ Σ λ | v µ ( x ) > v λµ ( x ) } , then there exists C λ > ,which is nonincreasing in λ , such that Z Σ λ |∇ ( v µ − v λµ ) + | dx ≤ C λ [( Z A λµ | x − p µ | N dx ) N + ( Z B λµ | x − p µ | N − dx ′ ) N − ] · ( Z Σ λ |∇ ( v µ − v λµ ) + | dx ) . (3.6) Proof.
We only need to note that Z ∂ Σ λ ∂ ( v µ − v λµ ) ∂ν ϕ ( x ) dx ′ ≤ Z B λµ [ k ( | x − p µ | N − v µ ( x )) v NN − µ − k ( | x λ − p µ | N − v λµ )( v λµ ) NN − ] ϕ dx ′ . The rest of the proof is the same the proof of Lemma 2.5, we omit it. (cid:3)
With the above inequality, we can start the moving plane procedure for µ <
0. First,we can move µ from µ = 0 to µ = − √ , then for µ ∈ [ µ , T λ from λ = 0 to λ = µ . That is, we have v µ ( x ) ≤ v µ ( x µ ) for any µ ∈ [ µ ,
0) and x ∈ Σ µ .Next, we can move the plane µ from µ to µ = − √ . Similarly, for any µ ∈ [ µ , µ ),we can move the plane T λ from λ = µ to λ = µ . That is, we have v µ ( x ) ≤ v µ ( x µ ) forany µ ∈ [ µ , µ ) and x ∈ Σ µ . Continue the above procedure, in the ( k + 1)th step, theplane µ can be moved from µ k from µ k +1 = µ k − √ µ k +22 and T λ can be moved from µ k to µ ∈ [ µ k +1 , µ k ). A direct calculation shows that µ k → −∞ as k → ∞ . So we have provedthe following result. Proposition 3.4.
For any µ < , we have v µ ( x ) ≤ v µµ ( x ) for x ∈ Σ µ . As a direct consequence of Proposition 3.2 and Proposition 3.4, we have the followingresult.
Proposition 3.5.
For fixed ( x , x , · · · , x N ) ∈ R N − , u ( x , x , x , · · · , x N ) is nondecreas-ing as x decreases. With the above results, we are in the position of proving Theorem 1.2 now.
Proof of Theorem 1.2.
As the proof of Theorem 1.1, we first start the moving plane methodin the x direction. For this purpose, we define the Kelvin transformation v µ of u at p µ = (0 , µ, , · · · ,
0) as v µ ( x ) = 1 | x − p µ | N − u ( p µ + x − p µ | x − p µ | ) . IOUVILLE TYPE THEOREMS FOR TWO MIXED BOUNDARY VALUE PROBLEMS WITH GENERAL NONLINEARITIES17 and Σ λ , T λ , x λ as before.If we denote v λµ = v µ ( x λ ), then by the same reason as Lemma 3.1 and Lemma 3.3, wehave the following inequality Z Σ λ |∇ ( v − v λ ) + | dx ≤ C λ [( Z A λµ | x − p µ | N dx ) N + ( Z B λµ | x − p µ | N − dx ′ ) N − ] · ( Z Σ λ |∇ ( v − v λ ) + | dx ) (3.7)with C λ , A λ and B λ as before. With this inequality, we can start the moving plane methodfor λ large enough. Now, moving the plane T λ from the right to the left and supposethat the process stops at some λ . If λ > µ , then we can prove that v µ is symmetricwith respect to T λ as before. This means that v µ is regular at p µ . In particular, we have u ∈ L NN − ( R N + ). Otherwise, we must have λ ≤ µ , which implies v µ ≤ v µµ . At the sametime, we can start the moving plane method from the left. Similarly, if this process stopsat some λ < µ , then u ∈ L NN − ( R N + ), otherwise, we have λ = µ and v µµ ≤ v µ .So during the process of moving plane in the x , · · · , x N − directions, two cases mayoccur.Case 1: There exists some direction and some µ , such that λ = µ , i.e., the moving planemethod stops at some λ = µ . Then we conclude that u is regular at infinity and hence u ∈ L NN − ( R N + ). This is impossible unless u ≡ u is nondecreasing as x decrease byProposition 3.5.Case 2: For all µ and all directions x , · · · , x N − , the moving plane method stops at λ = µ . In this case, the function u is independent of x , · · · , x N − since µ is arbitrary. Sowe have u = u ( x , x N ). Moreover, Proposition 3.5 implies that u ( x , x n ) is nondecreasingas x decreases. On the other hand, since u is bounded, then the limit function w ( x N ) =lim x →−∞ u ( x , x N ) exists and w ( x N ) satisfies w ′′ ( x N ) = − f ( w ) for x N > , ∂w∂x N = − g ( u ) for x N = 0 . (3.8)If w c with f ( c ) = g ( c ) = 0, then the first equation implies that w is a concave function,while the second equation implies w ′ (0) <
0. Hence there exists r > w ( x N ) < x N > r , this contradicts that u is a nonnegative solution to problem (1.2). So we have w ≡ c and the Dirichlet boundary value condition implies u ≡
0. This finishes the proofof Theorem 1.2. (cid:3)
Acknowledgement : This work is supported by NSFC, No.11101291.
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