A priori bounds and a Liouville theorem on a half-space for higher order elliptic Dirichlet problems
aa r X i v : . [ m a t h . A P ] S e p A PRIORI BOUNDS AND A LIOUVILLE THEOREM ON AHALF-SPACE FOR HIGHER-ORDER ELLIPTIC DIRICHLET PROBLEMS
WOFGANG REICHEL AND TOBIAS WETH
Abstract.
We consider the 2 m -th order elliptic boundary value problem Lu = f ( x, u ) ona bounded smooth domain Ω ⊂ R N with Dirichlet boundary conditions u = ∂∂ν u = . . . =( ∂∂ν ) m − u = 0 on ∂ Ω. The operator L is a uniformly elliptic operator of order 2 m given by L = (cid:0) − P Ni,j =1 a ij ( x ) ∂ ∂x i ∂x j (cid:1) m + P | α |≤ m − b α ( x ) D α . For the nonlinearity we assume thatlim s →∞ f ( x,s ) s q = h ( x ), lim s →−∞ f ( x,s ) | s | q = k ( x ) where h, k ∈ C (Ω) are positive functions and q > N ≤ m , 1 < q < N +2 mN − m if N > m . We prove a priori bounds, i.e, we show that k u k L ∞ (Ω) ≤ C for every solution u , where C > u is a classical, bounded, non-negative solutionof ( − ∆) m u = u q in R N + with Dirichlet boundary conditions on ∂ R N + and q > N ≤ m ,1 < q ≤ N +2 mN − m if N > m then u ≡ Introduction
A priori bounds for solutions of elliptic boundary value problems have been of majorimportance at least as far back as Schauder’s work in the 1930s. In this paper we prove apriori estimates on bounded, smooth domains Ω ⊂ R N for solutions of higher order boundaryvalue problems of the form(1.1) Lu = f ( x, u ) in Ω , u = ∂∂ν u = . . . = (cid:18) ∂∂ν (cid:19) m − u = 0 on ∂ Ω . Here ν is the unit exterior normal on ∂ Ω and L = (cid:16) − N X i,j =1 a ij ( x ) ∂ ∂x i ∂x j (cid:17) m + X | α |≤ m − b α ( x ) D α is a uniformly elliptic operator with coefficients b α ∈ L ∞ (Ω) and a ij ∈ C m − (Ω) such thatthere exists a constant λ > λ − | ξ | ≤ P Ni,j =1 a ij ( x ) ξ i ξ j ≤ λ | ξ | for all ξ ∈ R N , x ∈ Ω.Our main result is the following:
Theorem 1.
Suppose Ω ⊂ R N is a bounded domain with ∂ Ω ∈ C m . Let m ∈ N and assumethat q > if N ≤ m and < q < N +2 mN − m if N > m . Suppose further that there exist Date : October 25, 2018.2000
Mathematics Subject Classification.
Primary: 35J40; Secondary: 35B45.
Key words and phrases.
Higher order equation, a priori bounds, Liouville theorems, moving plane method. positive, continuous functions k, h : Ω → (0 , ∞ ) such that (1.2) lim s → + ∞ f ( x, s ) s q = h ( x ) , lim s →−∞ f ( x, s ) | s | q = k ( x ) uniformly with respect to x ∈ Ω . Then there exists a constant C > depending only on thedata a ij , b α , Ω , N, q, h, k such that k u k ∞ ≤ C for every solution u of (1.1) . Remark 2.
Suppose the nonlinearity depends on a real parameter λ , i.e, f λ : Ω × R → R and lim s → + ∞ f λ ( x, s ) λs q = h ( x ) , lim s →−∞ f λ ( x, s ) λ | s | q = k ( x ) uniformly with respect to x ∈ Ω and λ ∈ [ λ , ∞ ) where λ > . Then the a priori bound ofTheorem 1 depends additionally on λ but not on λ . This is important in the study of globalsolution branches of a parameter dependent version of (1.1) , which we will pursue in futurework. We are focusing on the case of superlinear nonlinearities f ( x, u ) with subcritical growth.A model nonlinearity is f ( x, s ) = | s | q . Our results hold with no restriction on the shape ofthe domain Ω and for general, possibly sign-changing solutions. This is important since thelack of the maximum principle for higher order equations does not allow to restrict attentionto positive solutions only.In the second-order case m = 1 a priori bounds for positive solutions have been establishedfor subcritical, superlinear nonlinearities via different methods by Brezis, Turner [7], Gidas,Spruck [13], DeFigueiredo, Lions and Nussbaum [10] and recently by Quittner, Souplet [20]and McKenna, Reichel [19]. In the higher-order case m ≥ L = ( − ∆) m . For Navier boundary conditions the picture is morecomplete. Let L = ( − L ) m where L = a ij ∂ ∂x i ∂x j + b α ∂∂x α is a second order operator andsuppose the boundary conditions are of Navier-type:(1.3) u = ( − L ) u = . . . = ( − L ) m − u = 0 on ∂ Ω . Soranzo [23] proved a priori bounds for positive solutions if L = ∆ and Ω is a boundedsmooth convex domain. Recently, Sirakov [22] improved this result to general operators L = ( − L ) m and general bounded smooth domains. Both authors strongly use the fact thatthe boundary conditions (1.3) allow to write the problem as a coupled system of second orderequations, where each equation is complemented with Dirichlet boundary conditions. In thiscase maximum principles are available. In contrast, the higher order Dirichlet problem cannot be rewritten as a system and therefore requires different techniques.In our approach we extend the so-called “scaling argument” of Gidas and Spruck [13],which they used to deal with the second order case m = 1 and positive solutions. Let usgive a brief sketch of their method. Gidas and Spruck assume that there exists a sequenceof positive solutions with L ∞ -norm tending to + ∞ . After rescaling the solutions to norm1 and blowing-up the coordinates one can take a limit of the rescaled solutions and obtains PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 3 a nontrivial positive solution of a limit boundary value problem − ∆ u = u q on either thefull-space R N or the half-space R N + = { x ∈ R N : x > } together with Dirichlet boundaryconditions. Then a contradiction is reached provided that a Liouville-type result is available,i.e., a result which shows that the non-negative solutions of the limit problem must beidentically zero. For subcritical q Gidas and Spruck [13], [14] proved both the full-space andthe half-space Liouville theorem for − ∆ u = u q via the method of moving planes.In order to deal with the higher order Dirichlet problem (1.1) and solutions which maychange sign, the blow up procedure has to be modified. Indeed, even under assumption(1.2), there seems to be no direct argument to exclude the case of negative blow up (i.e., theexistence of a sequence of solutions which is not uniformly bounded from below). Instead, itis excluded a posteriori after passing to the limit equation. Once this is done, we still needLiouville theorems for nonnegative solutions of the higher order problems on R N , R N + . Thefull-space Liouville theorem stated next is already known; it was proved by Lin [18] if m = 2and for general m ≥ Theorem 3 (Wei, Xu) . Let m ∈ N and assume that q > if N ≤ m and < q < N +2 mN − m if N > m . If u is a classical non-negative solution of ( − ∆) m u = u q in R N , then u ≡ . Even in the case of the Navier boundary conditions, the corresponding Liouville theoremfor the polyharmonic problem in the half-space is harder to achieve and has only recently beenproved by Sirakov [22]. Due to the lack of a (local) maximum principle, the correspondingDirichlet problem is even more difficult to deal with. Here we show the following new LiouvilleTheorem for the half-space which complements Theorem 3.
Theorem 4.
Let m ∈ N and assume that q > if N ≤ m and < q ≤ N +2 mN − m if N > m .If u is a classical non-negative bounded solution of (1.4) ( − ∆) m u = u q in R N + , u = ∂∂x u = . . . = ∂ m − ∂x m − u = 0 on ∂ R N + then u ≡ . We point out that the critical case q = N +2 mN − m is allowed in Theorem 4. Note also thatTheorem 4 holds in the class of bounded solutions. It remains an open problem to extendthe result to the class of all (possibly unbounded) classical positive solutions.Let us outline the proof of Theorem 4 and point out the main difficulties. Following Gidasand Spruck, we transform the half-space problem via a Kelvin inversion into a problem in theunit ball, where the point at infinity is mapped onto the boundary point P = ( − , , . . . , − ∆) m v = 2 m | x − P | − α v q pointwise in B (0) ,v = ∂∂ν v = . . . = (cid:18) ∂∂ν (cid:19) m − v = 0 on ∂B (0) \ { P } , WOFGANG REICHEL AND TOBIAS WETH where α = N +2 m − q ( N − m ) ≥
0. The key step is to show that v is axially symmetric aroundthe x -axis. In the second order case m = 1 this is proved with the classical moving planemethod, which is a local method based on the maximum principle. The same local approachfails for the higher order case m ≥ v of(1.5) may have a singularity at P ∈ ∂B (0). To overcome this problem, a large part ofthis work is devoted to show that every solution v of (1.5) which corresponds to a boundedsolution of (1.4) can be of represented via the Green function. In this step we also use Greenfunction estimates of Grunau and Sweers [15]. Then we apply a moving plane argument,using the Green function representation and the Hardy-Littlewood-Sobolev inequality, to getthe desired symmetry result. Comparing this variant of the moving plane method with theone in [4], we point out that Berchio, Gazzola and Weth allow more general (non-Lipschitz)nonlinearities, but their argument relies on Green function representations for directionalderivatives of the solution which in our situation might not exist.Once the symmetry result for v is established, we readily conclude – following Gidas andSpruck [13], [14] again – that the corresponding solution u of (1.4) is axially symmetricaround any axis parallel to the x -axis. Consequently, u is a function of x only and hencesolves an ordinary differential equation. It is then easy to conclude that u ≡ ∞ or −∞ . The rest of the paper is devoted to the proof of Theo-rem 4. In Section 3 we prove a Green-representation formula on half-spaces (Theorem 9) byapproximating the half-space by a family of growing balls. Based on the Green-representationfor balls and by a careful estimate of the boundary integrals and the monotone convergencetheorem we obtain a Green-representation for the half-space. Finally, in Section 4 we proveTheorem 4. 2. Proof of Theorem 1 – the blow-up argument
In this section we give the details of the blow-up argument for the proof of Theorem 1under the assumption of the validity of the Liouville-type result of Theorem 4. The proof
PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 5 uses standard linear L p - W m,p estimates for linear problems Lu = g ( x ) in Ω , (2.1) u = ∂∂ν u = . . . = (cid:18) ∂∂ν (cid:19) m − u = 0 on ∂ Ω . (2.2)Recall the following basic estimate of Agmon, Douglis, Nirenberg [2]. Theorem 5 (Agmon, Douglis, Nirenberg) . Let Ω ⊂ R N be a bounded domain with ∂ Ω ∈ C m , m ∈ N . Let a ij ∈ C m − (Ω) , b α ∈ L ∞ (Ω) , g ∈ L p (Ω) for some p ∈ (1 , ∞ ) . Suppose u ∈ W m,p (Ω) ∩ W m,p (Ω) satisfies (2.1) . Then there exists a constant C > depending onlyon k a ij k C m − , k b α k ∞ , λ, Ω , N, p, m and the modulus of continuity of a ij such that k u k W m,p (Ω) ≤ C ( k g k L p (Ω) + k u k L p (Ω) ) . We will also be using the following local analogue of this result. Though the proof may bestandard we give it for the reader’s convenience.
Corollary 6.
Let Ω be a ball { x ∈ R N : | x | < R } or a half-ball { x ∈ R N : | x | < R, x > } .Let m ∈ N , a ij ∈ C m − (Ω) , b α ∈ L ∞ (Ω) , g ∈ L p (Ω) for some p ∈ (1 , ∞ ) . Suppose u ∈ W m,p (Ω) satisfies (2.1)(i) either on the ball (ii) or on the half-ball together with the boundary conditions u = ∂∂x u = . . . = ∂ m − ∂x m − u = 0 on { x ∈ R N : | x | < R, x = 0 } .Then there exists a constant C > depending only on k a ij k C m − , k b α k ∞ , λ, Ω , N, p, m , themodulus of continuity of a ij and R such that for any σ ∈ (0 , k u k W m,p (Ω ∩ B σR ) ≤ C (1 − σ ) m ( k g k L p (Ω) + k u k L p (Ω) ) . Proof.
It is sufficient to prove the result for R = 1. For σ ∈ (0 ,
1) let η ∈ C m ( B ) bea cut-off function with 0 ≤ η ≤ η ≡ B σ , η ≡ | x | ≥ σ ′ where σ ′ = σ and | D γ η | ≤ (cid:16) − σ (cid:17) | γ | for | γ | ≤ m . Then L ( uη ) = gη + X | β |≤ m − | γ |≤ m −| β | c β,γ ( x ) D β uD γ η in Ω , WOFGANG REICHEL AND TOBIAS WETH where c β,γ are bounded functions with k c β,γ k ∞ ≤ C and C = C ( m ) max {k b α k ∞ , k a ij k C m − } .By Theorem 5 k∇ m u k L p (Ω ∩ B σ ) ≤ C (cid:16) k g k L p (Ω) + X ≤ k ≤ m − ≤ l ≤ m − k k∇ k u k L p (Ω ∩ B σ ′ ) (1 − σ ) − l (cid:17) ≤ C (cid:16) k g k L p (Ω) + m − X k =0 k∇ k u k L p (Ω ∩ B σ ′ ) (1 − σ ) k − m (cid:17) . If we introduce for k ∈ N the weighted norm Φ k = sup <σ< (1 − σ ) k k∇ k u k L p (Ω ∩ B σ ) then thelast inequality implies(2.3) Φ m ≤ C (cid:16) k g k L p (Ω) + m − X k =0 Φ k (cid:17) . Recall the standard interpolation inequality, see Adams, Fournier [1], for 0 ≤ k ≤ m − k∇ k u k L p (Ω ∩ B σ ) ≤ ǫ k∇ m u k L p (Ω ∩ B σ ) + Cǫ − k m − k k u k L p (Ω ∩ B σ ) , where C is homothety invariant and hence independent of σ . Using this we find that forevery fixed δ > σ ( δ ) ∈ (0 ,
1) such thatΦ k ≤ (1 − σ ) k k∇ k u k L p (Ω ∩ B σ ) + δ ≤ (1 − σ ) k (cid:16) (1 − σ ) m − k ǫ k∇ m u k L p (Ω ∩ B σ ) + Cǫ − k m − k (1 − σ ) − k k u k L p (Ω ∩ B σ ) (cid:17) + δ = ǫ (1 − σ ) m k∇ m u k L p (Ω ∩ B σ ) + Cǫ − k m − k k u k L p (Ω ∩ B σ ) + δ. Since δ > k ≤ ǫ Φ m + C ǫ Φ . Hence it follows from (2.3) thatΦ m ≤ C ( k g k L p (Ω) + k u k L p (Ω) ), i.e., k∇ m u k L p (Ω ∩ B σ ) ≤ C (1 − σ ) m ( k g k L p (Ω) + k u k L p (Ω) ) . Using the interpolation inequality again we obtain the claim. (cid:3)
Proof of Theorem 1.
It is convenient to rewrite the operator L in the form L = ( − m X | α | =2 m a α ( x ) D α + X | α |≤ m − c α ( x ) D α . Here a α ( x ) = P I ∈M α a i i ( x ) · a i i ( x ) · · · a i m − i m ( x ), where M α is the set of all vectors I =( i , . . . , i m ) ∈ { , . . . , N } m satisfying { j : i j = l } = α l for l = 1 , . . . , N . Hence a α iscontinuous on Ω and c α ∈ L ∞ (Ω). Assume for contradiction that there exists a sequence u k ofsolutions of (1.1) with M k := k u k k ∞ → ∞ as k → ∞ . By considering a suitable subsequencewe can assume that there exists x k ∈ Ω such that either M k = u k ( x k ) for all k ∈ N (positiveblow-up) or M k = − u k ( x k ) for all k ∈ N (negative blow-up). Define v k ( y ) := 1 M k u k ( M − q m k y + x k ) . PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 7
Then k v k k ∞ = 1 and either v k (0) = 1 for all k ∈ N (positive blow-up) or v k (0) = − k ∈ N (negative blow-up). We may also assume that x k → ¯ x ∈ Ω.Case 1: ¯ x ∈ Ω. In this case v k is well-defined on the sequence of balls B ρ k (0) with ρ k := M q − m k dist( x k , ∂ Ω) → ∞ as k → ∞ . Note that D α v k ( y ) = M − q m | α |− k ( D α u k )( M − q m k y + x k ) . For y ∈ B ρ k (0) let(2.4) ¯ a kα ( y ) := a α ( M − q m k y + x k ) , ¯ c kα ( y ) := M ( q − | α | m − k c α ( M − q m k y + x k )and define the operator(2.5) ¯ L k := ( − m X | α | =2 m ¯ a kα ( y ) D α + X | α |≤ m − ¯ c kα ( y ) D α . The function v k satisfies(2.6) ¯ L k v k ( y ) = f k ( y ) in B ρ k (0) , where f k ( y ) := 1 M qk f ( M − q m k y + x k , M k v k ( y )) . By our assumption on the nonlinearity f ( x, s ) we have that k f k k L ∞ ( B ρk (0)) is bounded in k .Note that while the ellipticity constant and the L ∞ -norm of the coefficients of ¯ L k are the sameas for L , the modulus of continuity of ¯ a kα is smaller than that of a α . By applying Corollary 6on the ball B R (0) for any R > p ≥ C p,R > k v k k W m,p ( B R (0)) ≤ C p,R uniformly in k. For large enough p we may extract a subsequence (again denoted v k ) such that v k → v in C m − ,α ( B R (0)) as k → ∞ for every R >
0, where v ∈ C m − ,αloc ( R N ) is bounded with k v k ∞ = 1 = ± v (0). Taking yet another subsequence we may assume that f k ∗ ⇀ F in L ∞ ( K )as k → ∞ for every compact set K ⊂ R N . Also we see that(2.7) F ( y ) = (cid:26) h (¯ x ) v ( y ) q if v ( y ) > ,k (¯ x ) | v ( y ) | q if v ( y ) < , because, e.g., if v ( y ) > k such that v k ( y ) > k ≥ k and hence M k v k ( y ) → ∞ as k → ∞ . Therefore the assumption on f ( x, s ) implies that f k ( y ) → h (¯ x ) v ( y ) q as k → ∞ , and a similar pointwise convergence holds at points where v ( y ) < f k on the set Z + = { y ∈ R N : v ( y ) > } and Z − = { y ∈ R N : v ( y ) < } determine due to the dominated convergence theorem theweak ∗ -limit F of f k on the set Z + ∪ Z − . Since ¯ b kα ( y ) → a kα ( y ) → a α (¯ x ) as k → ∞ and since we may assume that v k → v in W m,ploc ( R N ) we find that v is a bounded, weak W m,ploc ( R N )-solution of(2.8) L v = F in R N , where L = ( − m X | α | =2 m a α (¯ x ) D α = (cid:16) − N X i,j =1 a ij (¯ x ) ∂ ∂y i ∂y j (cid:17) m . WOFGANG REICHEL AND TOBIAS WETH
Since F ∈ L ∞ ( R N ) we get that v ∈ W m,ploc ( R N ) ∩ C m − ,αloc ( R N ) is a bounded, strong solutionof (2.8). Because D m v = 0 a.e. on the set { y ∈ R N : v ( y ) = 0 } we see that v is a strongsolution of L v = h (¯ x ) v ( y ) q if v ( y ) > , v ( y ) = 0 ,k (¯ x ) | v ( y ) | q if v ( y ) < R N . Notice that the right-hand side of the equation is C ( R N ). Hence v is a classical C m,αloc ( R N ) solution. By a linear change of variables we may assume that v solves(2.9) ( − ∆) m v = g ( v ) in R N , where g ( s ) = (cid:26) h (¯ x ) s q if s ≥ ,k (¯ x ) | s | q if s ≤ . By Lemma 15 of the Appendix we find that v ≥
0. This already excludes negative blow-upand implies that g ( v ( y )) = h (¯ x ) v ( y ) q , v (0) = 1. Theorem 3 tells us that this is impossible.This finishes the contradiction argument in the first case.Case 2: ¯ x ∈ ∂ Ω. By flattening the boundary through a local change of coordinates we mayassume that near ¯ x = 0 the boundary is contained in the hyperplane x = 0, and that x > ∂ Ω is locally a C m -manifold, this change of coordinatestransforms the operator L into a similar operator which satisfies the same hypotheses as L .For simplicity we call the transformed variables x and the transformed operator L . Now thefunction v k is well-defined on the set B ρ k (0) ∩ { y ∈ R N : y > − M q − m k x k, } . Since1 = | v k (0) | {z } = ± − v k ( − M q − m k x k, , , . . . , | {z } =0 | ≤ M q − m k x k, k∇ v k k ∞ we see that either M q − m k x k, is unbounded and we can conclude as in Case 1, or (by extractinga subsequence) τ k := M q − m k x k, → τ > k → ∞ . In this case we make a further change ofcoordinates and define w k ( z ) := v k ( z − τ k , z , . . . , z N ) , ˜ a kα ( z ) := ¯ a kα ( z − τ k , z , . . . , z N ) , ˜ c kα ( z ) := ¯ c kα ( z − τ k , z , . . . , z N )and likewise the operator ˜ L k . Note that w k ( τ k , , . . . ,
0) = ±
1. Let R N + = { z ∈ R N : z > } and B + R = B R (0) ∩ R N + for R >
0. For k sufficiently large the coefficients ˜ a kα , ˜ c kα and theoperator ˜ L k are well-defined in B + R . As before w k satisfies˜ L k w k ( z ) = ˜ f k ( z ) in B + R , where ˜ f k ( z ) := 1 M qk f ( M − q m k z + (0 , x k, , . . . , x k,n ) , M k w k ( z )) . together with Dirichlet-boundary conditions on { z ∈ R N : | z | < R, z = 0 } . Hence we mayapply Corollary 6 on the half-ball B + R for any R > p ≥ PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 9 a constant C p,R > k w k k W m,p ( B + R ) ≤ C p,R uniformly in k. As in Case 1 we can extract convergent subsequences w k → w in C m − ,αloc ( R N + ) and f k ∗ ⇀ F in L ∞ ( R N + ) as k → ∞ , where F ≥ , w is a bounded, strong W m,ploc ( R N + ) ∩ C m − ,αloc ( R N + )-solution of L w = F in R N + , ∂∂z w = . . . = ∂ m − ∂z m − w = 0 on ∂ R N + with L as in (2.8). By a linear change of variables we may assume that w solves(2.10) ( − ∆) m w = g ( w ) in R N + , ∂∂z w = . . . = ∂ m − ∂z m − w = 0 on ∂ R N + , where g is defined as in (2.9) of Case 1. The representation formula of Theorem 9 shows that w is positive and that g ( w ( z )) = h (¯ x ) w ( z ) q . Therefore w is a positive, bounded and classicalsolution C m -solution of ( − ∆) m w = h (¯ x ) w q in R N + with Dirichlet boundary conditions and w (0) = 1. A contradiction is reached by Theorem 4. (cid:3) Proof of Remark 2.
Take sequences of solutions ( u k , λ k ) such that M k := k u k k ∞ → ∞ as k → ∞ , λ k ≥ λ > v k ( y ) := 1 M k u k (cid:18)(cid:16) M − qk λ k (cid:17) / m y + x k (cid:19) . Due to the assumption λ k ≥ λ > M − qk /λ k → k → ∞ . Define further¯ a kα ( y ) := a α (cid:18)(cid:16) M − qk λ k (cid:17) / m y + x k (cid:19) , ¯ c kα ( y ) := M ( q − | α | m − k λ | α | m − k c α (cid:18)(cid:16) M − qk λ k (cid:17) / m y + x k (cid:19) with the corresponding operator ¯ L k . Then v k satisfies¯ L k v k ( y ) = f k ( y ) where f k ( y ) := 1 M qk λ k f (cid:18)(cid:16) M − qk λ k (cid:17) / m y + x k , M k v k ( y ) (cid:19) . Note that lim k →∞ f k ( y ) = h (¯ x ) v ( y ) q on Z + and similarly on Z − . The rest of the proof is asbefore. (cid:3) Green representation
The main result of this section is Theorem 9. There we state conditions on a function u on the half-space R N + under which the Green representation formula u ( x ) = Z R N + G + ∞ ( x, y )( − ∆) m u ( y ) dy for all x ∈ R N + holds. Here G + ∞ is the half-space Green function, see (3.1) below. In the next section, in theproof of Theorem 4, this representation formula will be applied to solutions of (1.4).Let us fix some notation. We recall Boggio’s celebrated formula [6] for the Green functionof the operator ( − ∆) m with Dirichlet boundary conditions on the unit ball B = { x ∈ R N : | x | < } : G ( x, y ) = k mN | x − y | m − N Z ( ψ ( x,y )+1) / ( z − m − z N − dz = k mN | x − y | m − N Z ψ ( x,y )0 z m − ( z + 1) N/ dz with ψ ( x, y ) = (1 − | x | )(1 − | y | ) | x − y | for x, y ∈ B . Here k mN is a suitable normalization constant. By dilation we find the Greenfunction for the ball B R = { x ∈ R N : | x | < R } as follows G R ( x, y ) = R m − N G (cid:16) xR , yR (cid:17) = k mN | x − y | m − N Z ψ R ( x,y )0 z m − ( z + 1) N/ dz with ψ R ( x, y ) = ( R − | x | )( R − | y | ) R | x − y | . Next we set P R := ( R, , . . . , ∈ R N + and we denote by B + R := { x ∈ R N : | x − P R | < R } theball of radius R shifted by P R . If we let G + R denote the Green function on B + R with respectto Dirichlet boundary conditions then we find the explicit formula G + R ( x, y ) = R m − N G (cid:18) x − P R R , y − P R R (cid:19) = k mN | x − y | m − N Z ψ + R ( x,y )0 z m − ( z + 1) N/ dz with ψ + R ( x, y ) = ( R − | x − P R | )( R − | y − P R | ) R | x − y | , x, y ∈ B R . Finally, if we let G + ∞ denote the Green function of the operator ( − ∆) m on the half-space R N + subject to Dirichlet boundary conditions then(3.1) G + ∞ ( x, y ) = k mN | x − y | m − N Z ψ ∞ ( x,y )0 z m − ( z + 1) N/ dz with ψ ∞ ( x, y ) = 4 x y | x − y | for x, y ∈ R N + . Lemma 7.
The Green function G + R on B + R converges pointwise and monotonically to theGreen function G + ∞ on R N + .Proof. The pointwise convergence is easily checked. Let x, y ∈ B + R . The monotonicity of G + R ( x, y ) with respect to R is equivalent to the monotonicity of ψ + R ( x, y ) with respect to R .Thus ddR ( R − | x − P R | )( R − | y − P R | ) R = − R ( R − | x − P R | )( R − | y − P R | ) + 2 x R ( R − | y − P R | ) + 2 y R ( R − | x − P R | ) . PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 11
Setting a := ( x − P R ) /R , b := ( y − P R ) /R we have | a | , | b | ≤ ddR ( R − | x − P R | )( R − | y − P R | ) R = 2 R (cid:16) − (1 − | a | )(1 − | b | ) + (1 + a )(1 − | b | ) + (1 + b )(1 − | a | ) (cid:17) = R (cid:16) ( a + 12 + 12 | a | | {z } ( | a | +2 a +1) )(1 − | b | ) + ( b + 12 + 12 | b | | {z } ( | b | +2 b +1) )(1 − | a | ) (cid:17) , and clearly | a | + 2 a + 1 = | ( a + 1 , a , . . . , a N ) | ≥
0. This establishes the proof. (cid:3)
In [15], Lemma 3.4, Grunau and Sweers proved the following estimates for the polyharmonicGreen function G on the unit ball if | k | ≥ m and x ∈ B , y ∈ ∂ B :(3.2) | D ky G ( x, y ) | ≤ C k,N,m | x − y | m − N −| k | (1 − | x | ) m for some constant C k,N,m >
0. For the Green function G R on B R and G + R on B + R the estimate(3.2) transforms as follows:(3.3) | D ky G R ( x, y ) | ≤ C k,N,m | x − y | m − N −| k | ( R − | x | ) m if x ∈ B R , y ∈ ∂B R . Likewise,(3.4) | D ky G + R ( x, y ) | ≤ C k,N,m | x − y | m − N −| k | | x | m if x = ( x , , . . . , ∈ B + R with x ∈ (0 , R ) , y ∈ ∂B + R . Lemma 8.
Let G be the Green function of ( − ∆) m with Dirichlet boundary condition on anarbitrary ball B ⊂ R n with exterior unit normal ν on ∂B . For any function v ∈ C m − ( B ) ∩ W m,p ( B ) with p > N m one has the following Poisson-Green representation for x ∈ B : for m even v ( x ) = m/ X i =1 I ∂B (cid:16) ∆ i − v ( y ) ∂ ν y ∆ m − iy G ( x, y ) − ∆ m − iy G ( x, y ) ∂ ν y ∆ i − v ( y ) (cid:17) ds y (3.5) + Z B G ( x, y )( − ∆) m v ( y ) dy. and for m odd v ( x ) = − ( m − / X i =1 I ∂B (cid:16) ∆ i − v ( y ) ∂ ν y ∆ m − iy G ( x, y ) − ∆ m − iy G ( x, y ) ∂ ν y ∆ i − v ( y ) (cid:17) ds y (3.6) − I ∂B ∆ ( m − / v ( y ) ∂ ν y ∆ ( m − / y G ( x, y ) ds y + Z B G ( x, y )( − ∆) m v ( y ) dy. Proof.
First assume v ∈ C m ( B ). Consider the identity m X i =1 div (cid:0) ∆ i − v ∇ ∆ m − i G − ∆ m − i G ∇ ∆ i − v (cid:1) = m X i =1 (cid:0) ∆ i − v ∆ m − i +1 G − ∆ m − i G ∆ i v (cid:1) = v ∆ m G − G ∆ m v in B. If we integrate this identity over B and take into account that D αy G ( x, y ) = 0 for | α | ≤ m − x ∈ B, y ∈ ∂B then we obtain the claim. For v ∈ C m − ( B ) ∩ W m,p ( B ) we can argue byapproximation and Lebesgue’s dominated convergence theorem if we take into account that R B G ( x, y ) | h ( y ) | dy ≤ const . k h k L p ( B ) provided h ∈ L p ( B ) and p > N m . (cid:3) Theorem 9.
Suppose that u ∈ C m − ( R N + ) ∩ W m,ploc ( R N + ) , p > N m is a function with thefollowing properties: (i) u and all partial derivatives of u of order less than or equal to m − are bounded, (ii) u satisfies Dirichlet boundary conditions on ∂ R N + , (iii) ( − ∆) m u ∈ L ploc ( R N + ) is non-negative in R N + .Then (3.7) u ( x ) = Z R N + G + ∞ ( x, y )( − ∆) m u ( y ) dy for every x ∈ R N + .Proof. Let us first consider the case where m is even. It clearly suffices to prove (3.7) for x = ( x , , . . . , ∈ R N + with x > R > x . Then x ∈ B + R , x ∈ (0 , R ), and (3.3) yields for y ∈ ∂B + R and i ≤ m the following estimates: | ∆ m − iy G + R ( x, y ) | ≤ C i,N,m | x − y | − m − N +2 i | x | m , | ∂ ν y ∆ m − iy G + R ( x, y ) | ≤ C i,N,m | x − y | − m − N +2 i − | x | m . Combining this with (3.5), we get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z B + R G + R ( x, y )( − ∆) m u ( y ) dy − u ( x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C | x | m m/ X i =1 I ∂B + R (cid:16) | ∆ i − u ( y ) || x − y | − m − N +2 i − + | x − y | − m − N +2 i | ∂ ν y ∆ i − u ( y ) | (cid:17) ds y . Since | x − y | ≥ | x | for y ∈ ∂B + R , we conclude that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z B + R G + R ( x, y )( − ∆) m u ( y ) dy − u ( x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C x m/ X i =1 I ∂B + R | x − y | − N (cid:0) | ∆ i − u ( y ) | + | ∂ ν y ∆ i − u ( y ) | (cid:1) ds y = C x m/ − X i =0 I ∂B + R | x − y | − N (cid:16) | ∆ i u ( y ) | + | ∂ ν y ∆ i u ( y ) | (cid:17) ds y . (3.8) PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 13
We claim that, for every x ∈ R N + ,(3.9) m/ − X i =0 I ∂B + R | x − y | − N (cid:16) | ∆ i u ( y ) | + | ∂ ν y ∆ i u ( y ) | (cid:17) ds y → R → ∞ .For N = 1 this is obvious since I ∂B + R | x − y | − N (cid:16) | ∆ i u ( y ) | + | ∂ ν y ∆ i u ( y ) | (cid:17) ds y = | x − R | − (cid:16) | u (2 i ) (2 R ) | + | u (2 i +1) (2 R ) | (cid:17) for i ≤ m − N ≥ ≤ a ≤ b ≤ R let us consider the set B ( a, b ) := { y ∈ ∂B + R : a ≤ y ≤ b } . On B ( a, b ) we have y = ( y , y ′ )with | y ′ | = p Ry − y . For N ≥ B ( a, b ) by the map F : ( ( a, b ) × S N − → B ( a, b ) , ( y , ϕ ) (cid:16) y , p Ry − y θ ( ϕ ) (cid:17) , where ϕ = ( ϕ , . . . , ϕ N − ) and θ ( ϕ ) = cos ϕ sin ϕ sin ϕ . . . sin ϕ N − sin ϕ sin ϕ sin ϕ . . . sin ϕ N − cos ϕ sin ϕ . . . sin ϕ N − ...cos ϕ N − sin ϕ N − cos ϕ N − . Let DF = ( b | b | · · · | b N − ) be the Jacobian Matrix of the map F and gr( DF ) = det( DF T · DF ) = det( b i · b j ) i,j =1 ,...,N − be the Gram determinant of DF . Since b = (cid:0) , R − y √ Ry − y θ (cid:1) T and b i = (cid:0) , p Ry − y ∂θ∂ϕ i (cid:1) T for i = 2 , . . . , N − b · b = R Ry − y , b · b i = 0 for i = 2 , . . . , N −
1. Therefore p gr( DF ) = p b · b · (2 Ry − y ) N − | det( Dθ ) | = R (2 Ry − y ) N − · | det( Dθ ) | . Since det( Dθ ) = ( − N − sin ϕ (sin ϕ ) . . . (sin ϕ N − ) N − we obtain finally p gr( DF ) ≤ R (2 Ry − y ) N − . Therefore we can write the surface integral as follows (if N = 2 the line integral is parame-terized by ( y , ± p Ry − y ) T ) I B ( a,b ) | x − y | − N ds y = Z ba Z S N − | x − F ( y , ϕ ) | − N p gr DF dϕ dy if N ≥ , Z ba R (2 Ry − y ) − / x + 2 Ry − x y dy if N = 2 . Since | x − F ( y , ϕ ) | = x + 2 Ry − x y and b ≤ R , we can now estimate as follows: I B ( a,b ) | x − y | − N ds y ≤ c Z Ra R (2 Ry − y ) N − ( x + 2 Ry − x y ) N dy = c R Z Ra (2 y R − ( y R ) ) N − ( x R + 2 y R (1 − x R )) N dy = c R Z aR (2 t − t ) N − ( x R + 2 t (1 − x R )) N dt ≤ c R Z aR t N − (2 − t ) N − ( x R + t ) N dt = c R Z aR t − (2 − t ) N − x R + t (cid:18) t x R + t (cid:19) N − dt ≤ c R Z aR t − (2 − t ) − x R + t dt with c = 2 N − c . Here we have also used that R ≥ x and N ≥
2. From now on we assume a ≤ R and split the remaining integral as follows:(3.10) I B ( a,b ) | x − y | − N ds y ≤ c R Z aR t − / (2 − t ) − / x R + t dt | {z } =: I + c R Z t − / (2 − t ) − / x R + t dt | {z } =: I . In the first integral I we have (2 − t ) ≥ I ≤ c R Z aR t − / x R + t dt. If a >
0, we conclude that(3.11) I ≤ c R Z aR t − dt ≤ c √ aR , while for a = 0 the substitution z = R x t yields(3.12) I ≤ c R Z t − / x R + t dt = c x Z R x √ z (1 + z ) dz ≤ c x Z ∞ √ z (1 + z ) dz = c x . For I we have(3.13) I ≤ c R Z (2 − t ) − / dt = c R .
Collecting the inequalities (3.10), (3.11), (3.12) and recalling that R ≥ x , we obtain(3.14) I B ( a,b ) | x − y | − N ds y ≤ c ( /x for a = 0,1 / √ aR for a > ε >
0. By the standard mean-value theorem using the Dirichlet boundary conditionsand the boundedness of the derivatives of orders up to m , there exists δ > m/ − X i =0 (cid:16) | ∆ i u ( y ) | + | ∂ ν y ∆ i u ( y ) | (cid:17) ≤ ε on { y ∈ R N + : | y | ≤ δ } . PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 15
Hence we apply (3.14) and obtain m/ − X i =0 I ∂B + R | x − y | − N (cid:16) | ∆ i u ( y ) | + | ∂ ν y ∆ i u ( y ) | (cid:17) ds y ≤ ε I B (0 ,δ ) | x − y | − N ds y + c I B ( δ, R ) | x − y | − N ds y ≤ ε c x + c c √ δR → ε c x as R → ∞ .Since ε was chosen arbitrarily, we conclude that (3.9) holds.Using (3.8), (3.9) and Lemma 7 together with the monotone convergence theorem we get u ( x ) = lim R →∞ Z B + R G + R ( x, y )( − ∆) m u ( y ) dy = Z R N + G + ∞ ( x, y )( − ∆) m u ( y ) dy. This finishes the proof in the case where m is an even integer. In the case where m is odd onlyminor modifications are needed. We use (3.6) instead of (3.5) together with the estimatesarising from (3.4): for x ∈ B + R , y ∈ ∂B + R and i ≤ m − , | ∆ m − iy G + R ( x, y ) | ≤ C i,N,m | x − y | − m − N +2 i | x | m and for i ≤ m +12 , | ∂ ν y ∆ m − iy G + R ( x, y ) | ≤ C i,N,m | x − y | − m − N +2 i − | x | m . By essentially the same estimates as before, we now obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z B + R G + R ( x, y )( − ∆) m u ( y ) dy − u ( x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C x I ∂B + R | x − y | − N (cid:16) ( m − / X i =0 | ∆ i u ( y ) | + ( m − / X i =0 | ∂ ν y ∆ i u ( y ) | (cid:17) ds y . Again, as a consequence of the boundary conditions, for every ε > δ > ( m − / X i =0 | ∆ i u ( y ) | + ( m − / X i =0 | ∂ ν y ∆ i u ( y ) | ≤ ε on { y ∈ R N + : | y | ≤ δ } .We therefore may conclude as in the case where m is even that I ∂B + R | x − y | − N (cid:16) ( m − / X i =0 | ∆ i u ( y ) | + ( m − / X i =0 | ∂ ν y ∆ i u ( y )) | (cid:17) ds y → R → ∞ .Using again the monotone convergence theorem, we conclude that u ( x ) = lim R →∞ Z B + R G + R ( x, y )( − ∆) m u ( y ) dy = Z R N + G + ∞ ( x, y )( − ∆) m u ( y ) dy. (cid:3) Proof of the Liouville Theorem in the half-space
This section is devoted to the proof of Theorem 4. Let m ∈ N and assume that q > N ≤ m and 1 < q < N +2 mN − m if N > m . Let u be a classical non-negative bounded solutionof ( − ∆) m u = u q in R N + , u = ∂u∂x = . . . = ∂ m − u∂x m − = 0 on ∂ R N + . We need to show that u ≡
0. From Theorem 9 we know that(4.1) u ( x ) = Z R N + G + ∞ ( x, y ) u q ( y ) dy for every x ∈ R N + .We consider the conformal diffeomorphism ϕ : B → R N + , ϕ ( y ) = 2 y + e | y + e | − e , where e = (1 , , . . . ,
0) is the first coordinate vector. The following formula shows how G + ∞ is related to the Green function G on the unit ball. Lemma 10. G + ∞ ( ϕ ( x ) , ϕ ( y )) = (cid:16) | x + e || y + e | (cid:17) m − N G ( x, y ) for all x, y ∈ B .Proof. An easy calculation yields(4.2) | ϕ ( x ) − ϕ ( y ) | = 2 | x − y || x + e || y + e | for x, y ∈ B .Considering the functions ψ ( x, y ) = (1 −| x | )(1 −| y | ) | x − y | and ψ ∞ ( x, y ) = x y | x − y | as in Section 3, weobtain ψ ∞ ( ϕ ( x ) , ϕ ( y )) = | x + e | ϕ ( x ) | y + e | ϕ ( y ) | x − y | = (2 x + 2 − | x + e | )(2 y + 2 − | y + e | ) | x − y | = (1 − | x | )(1 − | y | ) | x − y | = ψ ( x, y ) for x, y ∈ B .We conclude that G + ∞ ( ϕ ( x ) , ϕ ( y )) = k mN | ϕ ( x ) − ϕ ( y ) | m − N Z ψ ∞ ( ϕ ( x ) ,ϕ ( y ))0 z m − ( z + 1) N/ dz = (cid:16) | x + e || y + e | (cid:17) m − N k mN | x − y | m − N Z ψ ( x,y )0 z m − ( z + 1) N/ dz = (cid:16) | x + e || y + e | (cid:17) m − N G ( x, y ) for all x, y ∈ B . (cid:3) Corollary 11.
Define the function v : B → R by v ( x ) := | x + e | m − N u ( ϕ ( x )) PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 17 and the function h : B × [0 , ∞ ) → [0 , ∞ ) by h ( x, t ) := 2 m | x + e | − α t q where α := N + 2 m − q ( N − m ) ≥ by assumption on q . Then v satisfies (4.3) v ( x ) = Z B G ( x, y ) h ( y, v ( y )) dy for all x ∈ B .Proof. The Jacobian determinant of ϕ satisfies | J ϕ ( y ) | = N | y + e | N for y ∈ R N + . Therefore wehave u ( ϕ ( x )) = Z R N + G + ∞ ( ϕ ( x ) , y ) u q ( y ) dy = Z B G + ∞ ( ϕ ( x ) , ϕ ( y )) u q ( ϕ ( y )) | J ϕ ( y ) | dy = Z B (cid:18) | x + e || y + e | (cid:19) m − N G ( x, y ) (cid:16) | y + e | N − m v ( y ) (cid:17) q N | y + e | N dy = 1 | x + e | m − N Z B G ( x, y ) h ( y, v ( y )) dy for all x ∈ B , so that v ( x ) = | x + e | m − N u ( ϕ ( x )) = Z B G ( x, y ) h ( y, v ( y )) dy. (cid:3) Proposition 12.
The function v : B → R is axially symmetric with respect to the x -axis.Proof. We assume that N ≥ v
0, since otherwise the statement is trivial. Theintegral representation (4.3) implies that v is strictly positive in B . Note that for every x ∈ ∂ B \ {− e } we have thatlim x → x , x ∈ B v ( x ) = lim x → x , x ∈ B | x + e | m − N u ( ϕ ( x )) = | x + e | m − N u (cid:16) x + e | x + e | − e (cid:17) = 0since 2 x + e | x + e | − e ∈ ∂ R N + . Hence the function v – extended trivially on ∂ B \ {− e } – iscontinuous in B \ {− e } . We fix a unit vector e ∈ R N perpendicular to e (i.e., | e | = 1 and e · e = 0), and we show that v is symmetric with respect to the hyperplane T := { x ∈ R N : x · e = 0 } . For this we apply a moving plane argument based on the integral representation(4.3) and reflection inequalities derived in [4, 11] for G . We need some notation. For λ ≥ H λ = { x ∈ R N : x · e > λ } and the reflection x x λ := x − x · e − λ ) e at the hyperplane ∂H λ . We also consider the set J λ := { x ∈ B : x · e < λ and x λ B } which has nonempty interior if λ >
0. With these definitions, the inequalities stated in [4,Lemma 4] (see also [11, Lemma 3] for the biharmonic case) translate into the followingreflection inequalities:(4.4) G ( x λ , y λ ) > G ( x, y λ ) and G ( x λ , y λ ) − G ( x, y ) > G ( x, y λ ) − G ( x λ , y ) ) for all x, y ∈ H λ ∩ B and(4.5) G ( x λ , y ) − G ( x, y ) > x ∈ H λ ∩ B , y ∈ J λ .Now (4.5) and the strict positivity of v in B imply that(4.6) Z J λ [ G ( x λ , y ) − G ( x, y )] h ( y, v ( y )) dy > λ > x ∈ H λ ∩ B .We claim that the following reflection inequality holds for every λ > C λ ) v ( x ) ≤ v ( x λ ) for all x ∈ H λ ∩ B .We put λ ∗ := inf { λ > C λ ′ ) holds for all λ ′ ≥ λ } . Then λ ∗ ≤
1. Using the continuity of v in B , it is easy to see that ( C λ ∗ ) holds. We supposefor contradiction that λ ∗ >
0. Since 0 < | x + e | − α ≤ | x λ ∗ + e | − α for x ∈ H λ ∗ ∩ B and v ispositive in B , ( C λ ∗ ) yields(4.7) h ( x, v ( x )) ≤ h ( x λ ∗ , v ( x λ ∗ )) for x ∈ H λ ∗ ∩ B .We claim that(4.8) v ( x ) < v ( x λ ∗ ) for all x ∈ H λ ∗ ∩ B .Indeed, by using (4.4), (4.6) and (4.7) we have v ( x λ ∗ ) − v ( x ) = Z B [ G ( x λ ∗ , y ) − G ( x, y )] h ( y, v ( y )) dy = Z H λ ∗ ∩ B . . . dy + Z B \ H λ ∗ . . . dy = Z H λ ∗ ∩ B (cid:16) [ G ( x λ ∗ , y ) − G ( x, y )] h ( y, v ( y )) + [ G ( x λ ∗ , y λ ∗ ) − G ( x, y λ ∗ )] h ( y λ ∗ , v ( y λ ∗ )) (cid:17) dy + Z J λ ∗ [ G ( x λ ∗ , y ) − G ( x, y )] h ( y, v ( y )) dy> Z H λ ∗ ∩ B [ G ( x λ ∗ , y λ ∗ ) − G ( x, y λ ∗ )] [ h ( y λ ∗ , v ( y λ ∗ )) − h ( y, v ( y ))] dy ≥ x ∈ H λ ∗ ∩ B .Hence (4.8) is true. For 0 < µ ≤ λ ∗ we now consider the difference function w µ : H µ ∩ B → R , w µ ( x ) = v ( x µ ) − v ( x )and the set W µ := { x ∈ H µ ∩ B : w µ ( x ) < } . We note that W λ ∗ = ∅ , and we claim that(4.9) | W µ | → µ → λ ∗ ,where | · | denotes the Lebesgue measure. Indeed, let ε ∈ (0 , λ ∗ ), and consider the compactset K := { x ∈ B : x · e ≥ λ ∗ + ε, | x | ≤ − ε } ⊂ H λ ∗ ∩ B PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 19
Then inf x ∈ K w λ ∗ ( x ) > v in B implies that there exists λ ∈ ( λ ∗ − ε, λ ∗ ) such that inf x ∈ K w µ ( x ) > λ ≤ µ ≤ λ ∗ .Hence W µ ⊂ ( H µ ∩ B ) \ K ⊂ { x ∈ B : | x − λ ∗ | ≤ ε or | x | ≥ − ε } and therefore | W µ | ≤ N ε + εω N − for λ ≤ µ ≤ λ ∗ ,where ω N − denotes the area of the N − ε was chosenarbitrarily small, (4.9) follows. Next we note that, for µ ≥ λ ∗ and x ∈ W µ , h ( x µ , v ( x µ )) − h ( x, v ( x )) = | x µ + e | − α v q ( x µ ) − | x + e | − α v q ( x ) ≥ | x + e | − α ( v q ( x µ ) − v q ( x )) ≥ (cid:16) λ ∗ (cid:17) − α [ v q ( x µ ) − v q ( x )] ≥ (cid:16) λ ∗ (cid:17) − α qv q − ( x )[ v ( x µ ) − v ( x )] ≥ c ( λ ∗ ) w µ ( x )(4.10)with c ( λ ∗ ) = (cid:16) λ ∗ (cid:17) − α q (cid:16) sup x ∈ H λ ∗ / ∩ B v ( x ) (cid:17) q − < ∞ . We also note that(4.11) h ( x µ , v ( x µ )) − h ( x, v ( x )) ≥ x ∈ H µ \ W µ .From now on we assume that λ ∗ ≤ µ ≤ λ ∗ . For x ∈ W µ we use (4.4), (4.6), (4.10) and (4.11)to estimate0 > w µ ( x ) = Z B [ G ( x µ , y ) − G ( x, y )] h ( y, v ( y )) dy> Z H µ ∩ B (cid:16) [ G ( x µ , y ) − G ( x, y )] h ( y, v ( y )) + [ G ( x µ , y µ ) − G ( x, y µ )] h ( y µ , v ( y µ )) (cid:17) dy ≥ Z H µ ∩ B [ G ( x µ , y µ ) − G ( x, y µ )] [ h ( y µ , v ( y µ )) − h ( y, v ( y ))] dy ≥ Z W µ [ G ( x µ , y µ ) − G ( x, y µ )] [ h ( y µ , v ( y µ )) − h ( y, v ( y ))] dy ≥ c ( λ ∗ ) Z W µ [ G ( x µ , y µ ) − G ( x, y µ )] w µ ( y ) dy ≥ c N,m c ( λ ∗ ) Z W µ | x − y | − N w µ ( y ) dy, (4.12)where in the last step we use the estimate0 < G ( x, y ) ≤ c N,m | x − y | − N for x, y ∈ B , x = y with some c N,m >
0, which is easily deduced from the integral representation of G inSection 3. Next we pick s > < q := N + s < s . Then theHardy-Littlewood-Sobolev inequality (see e.g. [17, Section 4.3]) implies that (cid:18)Z R N (cid:12)(cid:12)(cid:12)Z W µ | x − y | − N w µ ( y ) dy (cid:12)(cid:12)(cid:12) s dx (cid:19) s = (cid:13)(cid:13)(cid:13) | · | − N ∗ (1 W µ w µ ) (cid:13)(cid:13)(cid:13) L s ( R N ) ≤ c s,q k w µ k L q ( W µ ) with a constant c s,q >
0. Combining this inequality with (4.12) and H¨older’s inequality, weobtain(4.13) k w µ k L s ( W µ ) ≤ c k w µ k L q ( W µ ) ≤ c | W µ | s − qsq k w µ k L s ( W µ ) with c := c N,m c ( λ ∗ ) c s,q .Now (4.9) and (4.13) imply that k w µ k L s ( W µ ) = 0 if µ < λ ∗ is close enough to λ ∗ . Henceproperty ( C µ ) holds if µ < λ ∗ is close enough to λ ∗ , which contradicts the definition of λ ∗ . Itfollows that λ ∗ = 0, thus ( C λ ) holds for all λ >
0, as claimed. By continuity, we now deducethat v ( x ) ≤ v ( x ) for x ∈ H ∩ B .Repeating the moving plane procedure for − e in place of e , we get v ( x ) ≥ v ( x ) for x ∈ H ∩ B .Hence equality holds, i.e., v is symmetric with respect to the hyperplane T = { x ∈ R N : x · e = 0 } as claimed. Since e was chosen arbitrarily with | e | = 1 and e · e = 0 we concludethat v is axially symmetric with respect to the x -axis. (cid:3) Corollary 13.
The function u only depends on the x -variable.Proof. Since v is axially symmetric with respect to the x -axis, the same is true for u . Let z ∈ R N − be arbitrary, and consider the function U z : R N + → R , U z ( x , x ′ ) = u ( x , x ′ − z ) for ( x , x ′ ) ∈ [0 , ∞ ) × R N − .Then U z satisfies the same assumptions as u , so it is also axially symmetric with respect tothe x -axis. This readily implies that u only depends on the x -variable. (cid:3) Theorem 14.
Let f : [0 , ∞ ) → [0 , ∞ ) be a continuous function with f (0) = 0 and f ( s ) > for s > . If u is a classical non-negative bounded solution of the one-dimensional problem ( − m u (2 m ) = f ( u ) in (0 , ∞ ) , u (0) = u ′ (0) = . . . = u ( m − (0) = 0 then u ≡ .Proof. The differential equation admits a first integral given by H = m − X i =1 ( − i u ( i ) u (2 m − i ) + ( − m (cid:16)
12 ( u ( m ) ) + F ( u ) (cid:17) , where F ( s ) := R s f ( s ) ds . Indeed, we calculate that dHdt = m − X i =1 ( − i (cid:16) u ( i +1) u (2 m − i ) + u ( i ) u (2 m − ( i − (cid:17) + ( − m (cid:16) u ( m ) u ( m +1) + f ( u ) u ′ (cid:17) = − u ′ u (2 m ) + ( − m − u ( m ) u ( m +1) + ( − m (cid:16) u ( m ) u ( m +1) + f ( u ) u ′ (cid:17) = u ′ (cid:16) ( − m f ( u ) − u (2 m ) (cid:17) = 0 in (0 , ∞ ).Suppose for contradiction that u
0. Since u has a Green function representation byTheorem 9, we infer that u is strictly positive in (0 , ∞ ), so that u (2 m ) = ( − ( m ) f ( u ) has no PRIORI BOUNDS FOR HIGHER-ORDER ELLIPTIC PROBLEMS 21 zero in (0 , ∞ ). By the mean value theorem, this implies that u ( j ) has at most 2 m − j zeros in(0 , ∞ ) for j = 0 , . . . , m . Hence every u ( j ) is eventually monotone and has a limit as t → ∞ since it is bounded. From this it clearly follows that u ( j ) ( t ) → t → ∞ for j = 1 , . . . , m, but then also u ( t ) → f again. Since F (0) = 0,we thus find that H ≡ , ∞ ). In particular, the boundary conditions yield0 = H (0) = ( − m u ( m ) (0)] , so that u ( m ) (0) = 0. If m = 1 we conclude u (0) = u ′ (0) = 0, so that u ≡ m >
1, we set v = − u ′′ and find that v is a bounded solution of( − m − v (2( m − = f ( u ) > , ∞ ), v (0) = v ′ (0) = . . . = v ( m − (0) = 0 . So v also has a Green function representation by Theorem 9, and thus v > , ∞ ). Insum, we have u (0) = 0 and u > , u ′′ < , ∞ ),which forces u ′ (0) > (cid:3) Proof of Theorem 4 (completed) . By Corollary 13, we have u ( x ) = ˆ u ( x ) with somefunction ˆ u : [0 , ∞ ) → R . Since ˆ u satisfies the assumptions of Theorem 14, we conclude thatˆ u ≡ u ≡ (cid:3) Appendix
Lemma 15.
Let v be a classical bounded solution of ( − ∆) m v = g ( v ) in R N . If g : R → [0 , ∞ ) is convex and non-negative with g ( s ) > for s < then v ≥ .Proof. Notice that the boundedness of v implies the boundedness of D j v for j = 1 , . . . , m .First we show that ( − ∆) l v ≥ R N for l = 1 , . . . , m −
1. Assume that there exists l ∈{ , . . . , m − } and x ∈ R N with ( − ∆) l v ( x ) < − ∆) j v ≥ R N for j = l + 1 , . . . , m .We may assume w.l.o.g. that x = 0. Let v l := ( − ∆) l v for l = 1 , . . . , m − v = v .Then we have − ∆ v = v , − ∆ v = v , . . . − ∆ v m − = g ( v ) in R N . If we define spherical averages ¯ w ( x ) = r N − ω N H ∂B r (0) w ( y ) dσ y , r = | x | then the radial func-tions ¯ v , ¯ v , . . . , ¯ v m − satisfy − ∆¯ v = ¯ v , − ∆¯ v = ¯ v , . . . − ∆¯ v m − ≥ g (¯ v ) in R N , where we have used Jensen’s inequality and the convexity of g . Since v l (0) < v l (0) <
0. Moreover ( r N − ¯ v ′ l ) ′ = − r N − ¯ v l +1 or ≤ − r N − g (¯ v ), which in both cases is non-positive. Since ¯ v ′ l (0) = 0 we see that ¯ v ′ l ( r ) ≤ r >
0. In particular ¯ v l ( r ) ≤ ¯ v l (0) < Next we integrate the equation ∆¯ v l − = − ¯ v l ≥ − ¯ v l (0) > r N − ¯ v ′ l − ( r ) ≥ − r N N ¯ v l (0), i.e, ¯ v ′ l − ( r ) ≥ − rN ¯ v l (0). The unboundedness of ¯ v ′ l − yields a contradiction.Finally, we need to show that v = v ≥
0. Assume that v ( x ) < x = 0.Since ∆¯ v = − ¯ v ≤ v ′ ( r ) ≤ α := lim r →∞ ¯ v ( r ) <
0. Thenlim r →∞ ∆¯ v m − ( r ) = − g ( α ) <
0, i.e., ∆¯ v m − ( r ) ≤ − g ( α ) < r ≥ r . By integration thisleads to lim r →∞ ¯ v ′ m − ( r ) = −∞ , which contradicts the boundedness of ¯ v ′ m − . This finishesthe proof of the lemma. (cid:3) Acknowledgement.
The authors thank Hans-Christoph Grunau (Magdeburg) for helpfulcomments and remarks.
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W. ReichelMathematisches Institut, Universit¨at GiessenArndtstr. 2, D-35392 Giessen, Germany
E-mail address : [email protected] T. WethMathematisches Institut, Universit¨at GiessenArndtstr. 2, D-35392 Giessen, Germany
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