Symmetry of Solutions to Semilinear Equations Involving the Fractional Laplacian on R n and R n +
aa r X i v : . [ m a t h . A P ] O c t Symmetry and Nonexistence of Solutions to SemilinearEquations Involving the Fractional Laplacian on R n and R n + ✩ Lizhi Zhang, Yongzhong Wang ∗ Department of Applied Mathematics, Northwestern Polytechnical University, Xi’an, Shaanxi, 710129, P. R.China
Abstract
In this paper, we investigate the following semilinear equations involving the fractional Laplacian( −△ ) α/ u ( x ) = f ( u ) , on R n and R n + respectively, 0 < α <
2. Applying a direct method of moving planes for thefractional Laplacian, we prove symmetry and nonexistence of positive solutions on R n and R n + under mild conditions on f . Keywords: the fractional Laplacian, radial symmetry, nonexistence, semilinear equation, adirect method of moving planes.MSC(2010): 35S15, 35B06, 35J61.
1. Introduction
Symmetry and nonexistence properties are very important and useful in studying semilinearelliptic equations. For example, these properties played an essential role in deriving a prioribounds for solutions in [2], [27], [32], [33], and they were used to obtain uniqueness of solutionsin [9], [35], [37], [38], [40]. There are many other applications.Gidas and Spruck [31] proved that there is no nontrivial C solution to the problem −△ u ( x ) = u p , in R n ,u ( x ) ≥ , in R n , (1.1)if n ≥ < p < n +2 n − . Gidas, Ni and Nirenberg in [30] pointed out that in the critical case p = n +2 n − the only solutions of (1.1) with some decay at infinity are of the form u ( x ) = [ n ( n − λ ] n − ( λ + | x − x | ) n − for some λ > x ∈ R n . This elegant result was also established by Caffarelli, Gidas andSpruck [14] without any condition at infinity. Later Chen and Li [15] found a simpler proof ofthe result based on the Kelvin transform and the Alexandrov-Serrin moving plane method. ✩ This work is supported by the National Natural Science Foundation of China (No. 11271299) and the NaturalScience Basic Research Plan in Shaanxi Province of China (No. 2016JM1023). ∗ Corresponding author. Tel.: +86 029 88431660.
Email addresses: [email protected] (Lizhi Zhang), [email protected] (Yongzhong Wang)
Preprint submitted to Journal of Mathematical Analysis and Applications August 7, 2018 n the upper half Euclidean space R n + = { x = ( x , · · · , x n ) : x n > } , Gidas and Spruck [31] showed that there are no nontrivial solutions to the Dirichlet problem −△ u ( x ) = u p , in R n + ,u ( x ) ≥ , in R n + ,u ( x ) = 0 , on ∂ R n + , (1.2)where n ≥ < p ≤ n +2 n − .These results were generalized to the problems with more general nonlinearity g ( u ). Bianchi[4] concluded that if u ∈ C ( R n ) is a positive solution to △ u + g ( u ) = 0 , in R n , (1.3)where g ( t ) ≥ t > g ( t ) /t n +2 n − is non-increasing, then either u is aconstant or u ( x ) = k ( | x − x | + h ) ( n − / (1.4)for some positive constants k, h and some point x ∈ R n . Damascelli and Gladiali [25] investigatedthe weak solutions u ∈ W , loc ( R n ) ∩ C ( R n ) of problem (1.3) under the slightly different conditionsand obtained the same conclusion.Using the same technique, Damascelli and Gladiali [25] also considered the weak solutions tothe corresponding Dirichlet problem in R n + −△ u ( x ) = g ( u ) , in R n + ,u ( x ) ≥ , in R n + ,u ( x ) = 0 , on ∂ R n + , (1.5)where g : [0 , + ∞ ) → R is a continuous function satisfying( i ) g ( t ) /t n +2 n − is nonincreasing in (0 , + ∞ );( ii ) g + ( t ) /t is bounded for t → iii ) g ( s ) > for every s >
0, lim inf s →∞ g ( s ) > . They derived that the solution u depends only on x n under ( i ) and ( ii ), and u is trivial under( i ) , ( ii ) and ( iii ).In this paper, we are concern with symmetry and nonexistence of positive solutions for semi-linear elliptic equations involving the fractional Laplacian on R n and R n + . Let us begin with thedefinition of the fractional Laplacian.The fractional Laplacian in R n is a nonlocal pseudo-differential operator with the form( −△ ) α/ u ( x ) = C n,α P V Z R n u ( x ) − u ( z ) | x − z | n + α dz, (1.6)where 0 < α < P V stands for the Cauchy principle value.In recent years, the fractional Laplacian has been frequently used to model diverse physicalphenomena, such as the turbulence, water waves, the anomalous diffusion, quasi-geostrophicflows, relativistic quantum mechanics of stars and molecular dynamics (see [6] [8] [23] [42] andthe references therein). It also has various applications in probability and finance (see [1] [3] [22]).2n particular, the fractional Laplacian can be understood as the infinitestmal generator of a stableL´evy process [3]. We refer the readers to Di Nezza, Palatucci and Valdinoci’s survey paper [39].The operator in (1.6) is well defined in the Schwartz space S of rapidly decreasing C ∞ functionson R n . In the space S , it can be equivalently defined by the Fourier transform \ ( −△ ) α/ u ( ξ ) = | ξ | α ˆ u ( ξ ) , where ˆ u is the Fourier transform of u . One can also extend this operator to a wider space offunctions L α = { u | Z R n | u ( x ) | | x | n + α dx < ∞} by < ( −△ ) α/ u, φ > = Z R n u ( −△ ) α/ φdx, for all φ ∈ C ∞ ( R n ) . It is easy to verify that for u ∈ L α ∩ C , loc , the integral on the right hand side of (1.6) is welldefined. Here, we consider the fractional Laplacian ( −△ ) α/ in this setting.A great deal of results about symmetry and nonexistence of the solutions to elliptic equationsinvolving △ were appeared (see [10], [12], [48] and the references therein). However, there aremuch fewer articles studying similar properties of solutions to equations involving ( −△ ) α/ . Themain difficulty is caused by the non-locality of the fractional Laplacian. To circumvent it, Caf-farelli and Silvestre [21] introduced the extension method which reduced this nonlocal probleminto a local one in higher dimensions. Chen, Li and Ou [19] provided the method of moving planesin integral forms, and treated the partial differential equations involving the fractional Laplacianby studying the equivalent integral equations. With these methods, a series of fruitful resultshave been obtained (see [5], [13], [16], [17], [24], [26], [44] and the references therein). For morearticles concerning nonlocal equations, please see [11], [20], [28], [29], [34], [36], [43], [45], [46],[47] and the references therein.Either by the extension method or by integral equations, one may need to impose extraconditions on the solutions. But these extra conditions will not be necessary if we consider thepseudo differential equations, including equations involving ( −△ ) α/ , directly.Recently, a direct approach to carry on the method of moving planes for nonlocal problemson bounded or unbounded domains was developed by Chen, Li, and Li [18], which enables onesto systematically investigate equations involving the fractional Laplacian. In this approach, the narrow region principle and the decay at infinity for anti-symmetric functions will be repeatedlyused. For readers’ convenience, we will collect them in Section 2. Based on the approach, Chen,Li, and Li [18] proved that if u ∈ L α ∩ C , loc is a nonnegative solution of( −△ ) α/ u = u p ( x ) , x ∈ R n , (1.7)then( i ) in the critical case p = n + αn − α , u is radially symmetric and monotone decreasing about somepoint;( ii ) in the subcritical case 1 < p < n + αn − α , u ≡ . To the Dirichlet problem on R n + ( −△ ) α/ u ( x ) = u p ( x ) , x ∈ R n + ,u ( x ) ≡ , x / ∈ R n + , (1.8)3he authors of [18] derived that if 1 < p ≤ n + αn − α and u ∈ L α ∩ C , loc is a nonnegative solution of(1.8), then u ≡ −△ ) α/ with more general nonlinearity.Throughout this paper, we suppose u ∈ L α ∩ C , loc .Let us state the main results. Theorem 1.1.
Suppose that f ( t ) > is strictly increasing in t > , and f ( t ) /t n + αn − α is non-increasing. Let u be a positive solution of ( −△ ) α/ u ( x ) = f ( u ) , x ∈ R n , (1.9) then u ( x ) = k ( | x − x o | + h ) ( n − α ) / (1.10) for some constants k, h > and some points x o ∈ R n , and f ( t ) = ct n + αn − α with c > for any t ∈ (0 , max R n u ] . Remark 1.1. (i) We know from Theorem 1.1 that u is bounded on R n , and actually, only on(0 , max R n ] can we derive that f ( t ) = ct n + αn − α with some c > f ( t ) /t n + αn − α is strictly decreasing, then it contradict theconclusion that f ( t ) = ct n + αn − α and hence (1.9) possesses no positive solution. Theorem 1.2.
Suppose that f ( t ) > is strictly increasing in t > and f ( t ) /t n + αn − α is non-increasing. Then ( ( −△ ) α/ u ( x ) = f ( u ) , x ∈ R n + ,u ( x ) ≡ , x / ∈ R n + , (1.11) possesses no positive solution. To prove Theorems 1.1 and 1.2, some new ideas are involved. Here we take the proof ofTheorem 1.1 as an example to illustrate them. In the proof of Theorem 1.1, to apply the methodof moving planes, we need to make a proper Kelvin transform centered at x ∈ R n ( x is arbitrarilychosen): v ( x ) = 1 | x − x | n − α u ( x + x − x | x − x | ) , x ∈ R n \ { x } , then v ( x ) satisfies( −△ ) α/ v ( x ) = 1 | x − x | n + α f ( | x − x | n − α v ( x )) , x ∈ R n \ { x } . (1.12)Through the moving plane method, we derive two possible cases: (i) v is symmetric about somepoint Q = x , (ii) v is symmetric about x . Here the new idea we want to underline is that incase (i), using the known fact that u is bounded on R n , we derive by (1.12) and the symmetry of v that f ( t ) = ct n + αn − α on (0 , max R n ] for some c >
0, and reduce (1.9) into( −△ ) α/ u ( x ) = Cu n + αn − α ( x ) , x ∈ R n , which leads to (1.10) by the well known result [19]. To (ii), it immediately follows that there isno positive solution. Remark 1.2.
When f ( t ) ≡
0, we already derived the Liouville Theorem for α -harmonic functionson R n + in our previous paper [47]. The same result on R n was established by Bogdan, Kulczyckiand Nowak [7]. 4 emark 1.3. ( i ) Theorems 1.1 and 1.2 are generalizations of the elegant results in Bianchi[4] and Damascelli and Gladiali [25] from equations involving Laplacian to the correspondingequations involving the fractional Laplacian. It is worth noting that we drop the condition that f is locally Lipschitz continuous. More specifically, we do not need the continuity of f at all.( ii ) In addition, we extend the results in [18] to the problems with more general nonlinearity f ( u ) on R n and R n + , and furthermore derive the exact representation of solution u on R n .The organization of the paper is as follows. In Section 2, we list two important maximumprinciples for anti-symmetric functions which were given in [18]. In Section 3, we prove Theorem1.1. Section 4 is devoted to the proof of Theorem 1.2. In Section 5, we prove two claims whichare used in Section 3.
2. Two Known Maximum Principles
Let T be a hyperplane in R n with the form T = { x = ( x , x ′ ) ∈ R n | x = λ f or some λ ∈ R } , where x ′ = ( x , x , · · · , x n ). Denote the reflection of x about the plane T by˜ x = (2 λ − x , x , · · · , x n )and let H = { x ∈ R n | x < λ } and ˜ H = { x | ˜ x ∈ H } . We describe two key ingredients for the direct method of moving planes for the fractionalLaplacian introduced in [18].
Theorem 2.1. ([18], Narrow Region Principle) Let Ω ⊂ H be a bounded narrow region containedin { x | λ − l < x < λ } with small l > . Suppose that c ( x ) is bounded from below in Ω , u ∈L α ∩ C , loc (Ω) is lower semi-continuous on ¯Ω and satisfying ( −△ ) α/ u ( x ) + c ( x ) u ( x ) ≥ , in Ω ,u ( x ) ≥ , in H \ Ω ,u (˜ x ) = − u ( x ) , in H, (2.1) then we have (i) u ( x ) ≥ in Ω; (2.2) (ii) furthermore, if u = 0 at some point in Ω , then u ( x ) = 0 almost everywhere in R n ; (2.3) (iii) conclusions (i) and (ii) hold for the unbounded region Ω if we further assume that lim | x | → ∞ u ( x ) ≥ . heorem 2.2. ([18], Decay at Infinity) Let Ω be an unbounded region in H . Assume that u ∈L α ∩ C , loc (Ω) is a solution of ( −△ ) α/ u ( x ) + c ( x ) u ( x ) ≥ , in Ω ,u ( x ) ≥ , in H \ Ω ,u (˜ x ) = − u ( x ) , in H, (2.4) with lim | x |→∞ | x | α c ( x ) ≥ , (2.5) then there exists a constant R > (depending on c ( x ) but independent of u ) such that if u ( x ) = min Ω u ( x ) < , (2.6) then | x | ≤ R . Remark 2.1.
In Theorems 2.1 and 2.2, the inequality( −△ ) α/ u ( x ) + c ( x ) u ( x ) ≥ , in Ω , and the condition (2.5) are only required at points where u is negative.
3. The proof of Theorem 1.1
In this section, we prove Theorem 1.1.
Proof of Theorem 1.1.
Since there is no decay condition on u near infinity, we are notable to carry the method of moving planes on u directly. To circumvent this difficulty, we makea Kelvin transform. For any point x ∈ R n , let v ( x ) = 1 | x − x | n − α u ( x + x − x | x − x | ) , x ∈ R n \ { x } , be the Kelvin transform of u centered at x . Then( −△ ) α/ v ( x ) = 1 | x − x | n + α f ( | x − x | n − α v ( x )) , x ∈ R n \ { x } . (3.1)The function v is positive, decays to 0 at infinity as | x − x | α − n and may have a singularity at x . Next we prove that v is radially symmetric, which implies the desired property of u .For x = ( x , x , · · · , x n ), choose any direction to be the x direction. To prove that v isradially symmetric, it suffices to show that v is symmetric in x .Write x = ( x , x , · · · , x n ) and let that for λ ∈ R and λ < x , T λ = { x ∈ R n | x = λ } , x λ = (2 λ − x , x , · · · , x n ) , Σ λ = { x ∈ R n | x < λ } , v λ ( x ) := v ( x λ ) , and w λ ( x ) = v λ ( x ) − v ( x ) , Σ − λ = { x ∈ Σ λ | w λ ( x ) < } . w λ ( x ) satisfies( −△ ) α/ w λ ( x )= f ( | x λ − x | n − α v λ ( x )) | x λ − x | n + α − f ( | x − x | n − α v ( x )) | x − x | n + α , x ∈ Σ λ \ { ( x ) λ } , (3.2)Noting lim | x |→∞ w λ ( x ) = 0 , we have that if w λ is negative somewhere in Σ λ , then the negative minima of w λ are attained inthe interior of Σ λ . Since f ( t ) > t > f ( t ) /t n + αn − α is non-increasing, it follows from (3.2)that at points x ∈ Σ − λ \ B ǫ (( x ) λ ) ( ǫ small),( −△ ) α/ w λ ( x )= f ( | x λ − x | n − α v λ ( x )) | x λ − x | n + α − f ( | x − x | n − α v ( x )) | x − x | n + α = f ( | x λ − x | n − α v λ ( x ))( | x λ − x | n − α v λ ( x )) n + αn − α v n + αn − α λ ( x ) − f ( | x − x | n − α v ( x ))( | x − x | n − α v ( x )) n + αn − α v n + αn − α ( x ) ≥ f ( | x − x | n − α v ( x ))( | x − x | n − α v ( x )) n + αn − α (cid:18) v n + αn − α λ ( x ) − v n + αn − α ( x ) (cid:19) = n + αn − α · f ( | x − x | n − α v ( x ))( | x − x | n − α v ( x )) n + αn − α φ αn − α ( x ) w λ ( x ) ≥ C f ( | x − x | n − α v ( x ))( | x − x | n − α v ( x )) n + αn − α v αn − α ( x ) w λ ( x ) , (3.3)where v λ ( x ) < φ ( x ) < v ( x ). Here and below C denotes a positive constant whose value may bedifferent at different lines.Denote t = | x − x | n − α v ( x ), we know by the definition of v ( x ) that t → u ( x ) > | x | → ∞ .Hence for any x ∈ Σ − λ \ B ǫ (( x ) λ ), there exists a constant c > t ≥ c . Using that f ( t ) /t n + αn − α is non-increasing, it yields f ( t ) t n + αn − α ≤ f ( c ) c n + αn − α , and by (3.3), ( −△ ) α/ w λ ( x ) ≥ C f ( | x − x | n − α v ( x ))( | x − x | n − α v ( x )) n + αn − α v αn − α ( x ) w λ ( x ) (3.4) ≥ Cv αn − α ( x ) w λ ( x ) . (3.5)Therefore, ( −△ ) α/ w λ ( x ) + c ( x ) w λ ( x ) ≥ , x ∈ Σ − λ \ B ǫ (( x ) λ ) (3.6)with c ( x ) = − Cv αn − α ( x ) . (3.7) Step 1.
We prove that for λ < | λ | large enough, w λ ≥ λ \ { ( x ) λ } (3.8)by using Theorem 2.2. 7et us first admit: Claim 3.1.
For λ sufficiently negative, there exists ǫ > and c λ > , such that w λ ( x ) ≥ c λ , x ∈ B ǫ (( x ) λ ) \ { ( x ) λ } . (3.9)Its proof will be given in Section 5. From (3.9), one can see that Σ − λ has no intersection with B ǫ (( x ) λ ). Using (3.7), it is easy to verify that for | x | sufficiently large, c ( x ) ∼ | x | α , (3.10)hence c ( x ) satisfies (2.5). Applying Theorem 2.2 to w λ with H = Σ λ and Ω = Σ − λ , we conclude that there exists R > λ ), such that if ¯ x is a negative minimum of w λ in Σ λ , then | ¯ x | ≤ R . (3.11)Now for λ ≤ − R , we have (3.8). Step 2.
Step 1 provides a starting point such that we can now move the plane T λ to theright as long as (3.8) holds to its limiting position. Denoting λ = sup { λ < x | w µ ( x ) ≥ , x ∈ Σ µ \ { ( x ) µ } , µ ≤ λ } , we show w λ ( x ) ≡ , x ∈ Σ λ \ { ( x ) λ } . (3.12)To do so, let us consider two possibilities: (i) λ < x ; and (ii) λ = x . (i). For λ < x , we deduce that if w λ ( x ) , x ∈ Σ λ \ { ( x ) λ } , (3.13)then the plane T λ can be moved further to the right, to be more rigorous, there exists some ε > λ ∈ ( λ , λ + ε ), w λ ( x ) ≥ , x ∈ Σ λ \ { ( x ) λ } . (3.14)This is a contradiction with the definition of λ . Hence we must have (3.12). (i) . Now we prove (3.14) by combining Theorems 2.1 and 2.2. Let us admit: Claim 3.2.
There exists c > such that for sufficiently small η , w λ ( x ) ≥ c , x ∈ B η (( x ) λ ) \ { ( x ) λ } . (3.15)We put its proof in Section 5. From (3.15), one knows that Σ − λ does not intersect with B η (( x ) λ ), hence for λ sufficiently close to λ , Σ − λ has no intersection with B η (( x ) λ ) either. Tosee this, for any x ∈ B η (( x ) λ ) \ { ( x ) λ } , we find ¯ x ∈ B η (( x ) λ ) \ { ( x ) λ } such that x λ = ¯ x λ .Then, for λ sufficiently close to λ , since v ( x ) is continuous in Σ λ , it follows w λ ( x ) = [ v λ ( x ) − v λ (¯ x )] + [ v λ (¯ x ) − v (¯ x )] + [ v (¯ x ) − v ( x )] ≥ c + [ v (¯ x ) − v ( x )] ≥ , x ∈ B η (( x ) λ ) \ { ( x ) λ } , (3.16)8hich verifies that Σ − λ does not intersect with B η (( x ) λ ).By (3.11), the negative minimum of w λ connot be attained outside of B R (0). Next we willprove that it can neither be attained inside of B R (0), i.e., for λ sufficiently close to λ , w λ ( x ) ≥ , x ∈ (Σ λ ∩ B R (0)) \ { ( x ) λ } . (3.17)Actually, using Theorem 2.1, there is a small δ >
0, such that for λ ∈ [ λ , λ + δ ), if w λ ( x ) ≥ , x ∈ Σ λ − δ \ { ( x ) λ } , (3.18)then w λ ( x ) ≥ , x ∈ (Σ λ \ Σ λ − δ ) \ { ( x ) λ } . (3.19)To derive (3.19) from (3.18), we use Theorem 2.1 with H = Σ λ and the narrow region Ω = (Σ − λ \ Σ λ − δ ) , while the lower bound of c ( x ) can be seen from (3.7) and (3.10).Now if (3.18) holds, then one can immediately get (3.17) from (3.18) and (3.19), so what leftis to show (3.18), and by (3.11) and (3.16) we only need to prove w λ ( x ) ≥ , x ∈ (Σ λ − δ ∩ B R (0)) \ B η (( x ) λ ) . (3.20)In fact, when λ < x , we have w λ ( x ) > , x ∈ Σ λ \ { ( x ) λ } . (3.21)If not, then there exists some ˆ x such that w λ (ˆ x ) = min Σ λ w λ ( x ) = 0and it implies ( −△ ) α/ w λ (ˆ x )= C n,α P V Z R n − w λ ( y ) | ˆ x − y | n + α dy = C n,α P V Z Σ λ − w λ ( y ) | ˆ x − y | n + α dy + Z R n \ Σ λ − w λ ( y ) | ˆ x − y | n + α dy = C n,α P V Z Σ λ − w λ ( y ) | ˆ x − y | n + α dy + Z Σ λ w λ ( y ) | ˆ x − y λ | n + α dy = C n,α P V Z Σ λ (cid:18) | ˆ x − y λ | n + α − | ˆ x − y | n + α (cid:19) w λ ( y ) dy ≤ . (3.22)On the other hand, we have by the monotonicity of f ( t ) /t n + αn − α that( −△ ) α/ w λ (ˆ x )= f ( | ˆ x λ − x | n − α v λ (ˆ x )) | ˆ x λ − x | n + α − f ( | ˆ x − x | n − α v (ˆ x )) | ˆ x − x | n + α = f ( | ˆ x λ − x | n − α v λ (ˆ x ))( | ˆ x λ − x | n − α v λ (ˆ x )) n + αn − α v n + αn − α λ (ˆ x ) − f ( | ˆ x − x | n − α v (ˆ x ))( | ˆ x − x | n − α v (ˆ x )) n + αn − α v n + αn − α (ˆ x )= f ( | ˆ x λ − x | n − α v (ˆ x ))( | ˆ x λ − x | n − α v (ˆ x )) n + αn − α − f ( | ˆ x − x | n − α v (ˆ x ))( | ˆ x − x | n − α v (ˆ x )) n + αn − α ! v n + αn − α (ˆ x ) ≥ . (3.23)9rom (3.22) and (3.23), one derives that w λ ( x ) ≡ λ \ { ( x ) λ } , which is a contradictionwith (3.13). This proves (3.21).It follows from (3.21) that there exists a constant c >
0, such that w λ ( x ) ≥ c , x ∈ (Σ λ − δ ∩ B R (0)) \ B η (( x ) λ ) . Since w λ depends continuously on λ , there exists ε > ε < δ , such that for all λ ∈ ( λ , λ + ε ),we have w λ ( x ) ≥ , x ∈ (Σ λ − δ ∩ B R (0)) \ B η (( x ) λ ) . (3.24)This verifies (3.20) and hence (3.18) is also proved.Combining (3.11), (3.16), (3.19) and (3.24), we conclude (3.14) for all λ ∈ ( λ , λ + ε ), whichis a contradiction with the definition of λ . Therefore, we must have (3.12). (i) . Now we know by (3.12) that v ( x ) is radially symmetric about some point Q differentfrom x and v ( x ) c is bounded near x , thus u ( x ) c is bounded in R n too. And also,( −△ ) α/ v ( x ) is radially symmetric about the same point Q , which can be easily proved throughelementary computation with the help of (1.6). Thus the right hand side of the following equation( −△ ) α/ v ( x ) = 1 | x − x | n + α f ( | x − x | n − α v ( x )) , x ∈ R n \ { x } (3.25)should have the same symmetry. From this, we are able to prove that f ( | x − x | n − α v ( x )) = C (cid:2) | x − x | n − α v ( x ) (cid:3) n + αn − α , x ∈ R n \ { x } , (3.26)for some positive constant C ( C cannot be 0, otherwise v ( x ) ≡ c > x , refer to [11]). Yet the proof of (3.26) is sophisticated, and we postpone the proof for a while.Now set t ( x ) = | x − x | n − α v ( x ) , x ∈ R n \ { x } . Since u is bounded on R n , it infers by | x − x | n − α v ( x ) = u ( x + x − x | x − x | )that t ∈ (0 , max R n u ]. Noting (3.26), we get f ( t ) = Ct n + αn − α for any t ∈ (0 , max R n u ] and somepositive constant C and (1.9) becomes( −△ ) α/ u ( x ) = Cu n + αn − α ( x ) , x ∈ R n , which gives by [19] that u ( x ) = k ( | x − x o | + h ) ( n − α ) / (3.27)for some constants k, h > x o ∈ R n .( i ) . Now let us prove (3.26). We turn (3.25) into the following form[( −△ ) α/ v ( x )] v n − αn + α = f ( | x − x | n − α v ( x ))[ | x − x | n − α v ( x )] n + αn − α , x ∈ R n \ { x } . (3.28)Since the left hand side of (3.28) is radially symmetric about Q , it implies that f ( | x − x | n − α v ( x ))[ | x − x | n − α v ( x )] n + αn − α = g ( | x − Q | ) , x ∈ R n \ { x } ,
10s also radially symmetric about Q for some proper function g . For simplicity, we denote F ( t ) := f ( t ) t n + αn − α , t = t ( x ) = | x − x | n − α v ( x ) . Let l Qx be the line passing through points Q and x , and B R ( Q ) the sphere centered at Q with radius R , then for any x ∈ B R ( Q ), one can easily see that the maximum point and theminimum point of t ( x ) are located on the line l Qx . Suppose that the maximum point of t ( x ) on R n is ˜ P , then ˜ P ∈ l Qx ( ˜ P = Q and ˜ P = x ), and ˜ P , x are at the different sides of Q on l Qx .Define ˜ x := 2 Q − x , then the symmetry point of ˜ P about Q is P := 2 Q − ˜ P . Let s x P be theline segment between points x and P , P ∈ s x P , but x s x P . Then for any A ∈ s x P , wehave t ( A ) < t ( ˜ A ) and t ( ˜ A ) − t ( A ) ≥ c for some positive constant c . Without loss of generality,we may assume t ( A ) = m and t ( ˜ A ) = M . Then, because g ( A ) = g ( ˜ A ), so F ( m ) = F ( M ). Usingthat F ( t ) is non-increasing in t on (0 , + ∞ ), we derive F ( t ) = C > , on [ m, M ] with M − m ≥ c > . (3.29)Concretely, when A → x , one has t ( A ) → A → ˜ x , t ( ˜ A ) → t ( ˜ x ) := a , a > t ( x ) and the above technique, we have F ( t ) = C > , on (0 , a ) . (3.30)When A = P , suppose t ( P ) = b , we already know t ( ˜ P ) = max R n u . Then repeat the aboveprocess, we have F ( t ) = C > , on [ b, max R n u ] . (3.31)If a ≥ b , then by (3.30), (3.31) and the continuity of F ( t ) for t >
0, we already proved that F ( t ) = C > , on (0 , max R n u ] . (3.32)If a < b , we can also show that (3.32) is still true. In fact, let us seek a finite sequence ofpoints { A i } ⊂ s x P , i = 1 , , · · · , k , such that t ( A i +1 ) = t ( ˜ A i ) := m i , i = 1 , , · · · , k − ,t ( A ) = a := m , t ( ˜ A k ) := m k and m k − ≤ b ≤ m k . Since f ( t ) ∈ (0 , b ] is continuous on s x P , and m i − m i − ≥ c which can be seen from (3.29), i = 1 , , · · · , k , we can find such { A i } indeed, and we also know that k ≤ ⌊ b − ac ⌉ + 1, where ⌊ b − ac ⌉ denotes the integer part of b − ac . It follows from (3.29) that F ( t ) = c i > , on [ m i , m i +1 ] , i = 0 , , , · · · , k. (3.33)According to(0 , a ) ∪ [ a, m ] ∪ [ m , m ] ∪ · · · ∪ [ m k − , m k ] ∪ [ b, max R n u ] = (0 , max R n u ] , then by (3.30), (3.31), (3.33) and the continuity of F ( t ) on t >
0, we can also derive (3.32) for a < b .Hence (3.26) is proved.Till now, we have proved Theorem 1.1 under Possibility ( i ).11 ii). From the definition of λ , one knows that λ = x and w λ ≥ , x ∈ Σ λ \ { ( x ) λ } . (3.34)Now we move the plane T λ from + ∞ to the left and similarly derive that either λ > x and w λ ≡ , x ∈ Σ λ . (3.35)or λ = x and w λ ≤ , x ∈ Σ λ . (3.36)The case described by (3 .
35) can be handled with the same way as ( i ). Now from (3.34) and(3.36), we have λ = x and w λ ≡ , x ∈ Σ λ . So far, we have proved that v is symmetric about the plane T x . Since the x direction canbe chosen arbitrarily, we have actually shown that v is radially symmetric about x .For any two points X i ∈ R n , i = 1 ,
2, choose x = X + X . Since v is radially symmetricabout x , so is u , hence u ( X ) = u ( X ). This implies that u is a constant, without loss ofgenerality, we may suppose u ( x ) = a >
0. Now (1.9) holds if and only if f ( a ) = 0, this is acontradiction with the condition “ f ( t ) > t >
4. The Proof of Theorem 1.2
We first indicate two propositions.
Proposition 4.1.
Assume that u ∈ L α ∩ C , loc ( R n + ) is a locally bounded positive solution to theproblem ( −△ ) α/ u ( x ) = f ( u ) , x ∈ R n + ,u ( x ) ≡ , x / ∈ R n + , (4.1)then it is also a solution to the integral equation u ( x ) = Z R n + G ∞ ( x, y ) f ( u ( y )) dy ; (4.2)and vice versa. Here G ∞ ( x, y ) is the Green function to (4.1): G ∞ ( x, y ) = A n,α s n − α " − B t + s ) n − Z st ( s − tb ) n − b α/ (1 + b ) db , where A n,α , B are positive constants, t = 4 x n y n and s = | x − y | .The proof of Proposition 4.1 is similar to Theorem 4.1 in [13]. Since the condition “ f ( t ) > t >
0” guarantees that f ( t ) is locally bounded on (0 , + ∞ ) and f ( t ) ≥ C for t > R , where R > C is a positive constant, we only need to substitute f ( u ) for u p in the proof of [13]. We omit the details.12 roposition 4.2. ([13]) If ts is sufficiently small, then for any x = ( x ′ , x n ), y = ( y ′ , y n ) ∈ R n + , c n,α s ( n − α ) / t α/ s α/ ≤ G ∞ ( x, y ) ≤ C n,α s ( n − α ) / t α/ s α/ , (4.3)that is G ∞ ( x, y ) ∼ t α/ s n/ , (4.4)where t = 4 x n y n , s = | x − y | , c n,α and C n,α stand for different positive constants and onlydepend on n and α .Now let us prove Theorem 1.2. Proof of Theorem 1.2.
Let us make a Kelvin transform again: v ( x ) = 1 | x − x | n − α u ( x + x − x | x − x | ) , x ∈ R n + , where the center x is on the boundary ∂ R n + in order that R n + is invariant under the transform.Obviously, v satisfies( −△ ) α/ v ( x ) = 1 | x − x | n + α f ( | x − x | n − α v ( x )) , x ∈ R n + , (4.5)and is positive, decays to 0 at infinity as | x − x | α − n and may have a singularity at x . For x = ( x , x , · · · , x n ), choose any direction in R n − (the boundary of R n + ) to be x direction. Wefirst prove v = v ( x n ) , (4.6)that is, v is rotational symmetric about the x n axis. Next we derive a contradiction by (4.6) andprove that any positive solution of (1.11) does not exist.To prove (4.6), it suffices to show that v is axisymmetric about any line parallel to x n axis.Write x = ( x , x , · · · , x n ), for λ < x , λ ∈ R , and for any x ∈ R n + , denote T λ = { x ∈ R n | x = λ } , x λ = (2 λ − x , x , · · · , x n ) , Σ λ = { x ∈ R n + | x < λ } , v λ ( x ) := v ( x λ ) , and w λ ( x ) = v λ ( x ) − v ( x ) , Σ − λ = { x ∈ Σ λ | w λ ( x ) < } . Then w λ ( x ) satisfies( −△ ) α/ w λ ( x ) = f ( | x λ − x | n − α v λ ( x )) | x λ − x | n + α − f ( | x − x | n − α v ( x )) | x − x | n + α , x ∈ Σ λ . (4.7)Since f ( t ) > t >
0, and f ( t ) /t n + αn − α is non-increasing, similarly to (3.6) and (3.7) in theprevious section, we have by (4.7) that for ǫ small,( −△ ) α/ w λ ( x ) + c ( x ) w λ ( x ) ≥ , x ∈ Σ − λ \ B ǫ (( x ) λ ) , (4.8)where c ( x ) = − Cv αn − α ( x ) . (4.9)Hence the lower bound of c ( x ) in Σ − λ \ B ǫ (( x ) λ ) can be obtained from (4.9), and it follows bythe asymptotic behavior v ( x ) ∼ | x | n − α , for | x | large13hat lim | x |→∞ w λ ( x ) = 0 and c ( x ) ∼ | x | α for | x | large . These guarantee that we can apply Theorems 2.1 and 2.2 to show the following:
Step 1.
For λ sufficiently negative, w λ ( x ) ≥ , x ∈ Σ λ . Step 2.
Denoting λ = sup { λ | w µ ( x ) ≥ , x ∈ Σ µ ; µ ≤ λ < x } , we have w λ ( x ) ≡ , x ∈ Σ λ . (4.10)The proofs of two steps are in general similar to the proof of Theorem 1.1, here we only statethe differences.To prove (4.10), we also need to consider two possibilities: (i) λ < x ; and (ii) λ = x . (i). A similar proof obtaining (3.12) leads to λ < x , w λ ≡ , x ∈ Σ λ . (4.11)By l Q we denote the line parallel to x n axis passing through the point Q ∈ ∂ R n + . For any point P = ( p , · · · , p n ), denote p ′ = ( p , · · · , p n − ).Since one can choose the x direction arbitrarily, it easily derives by (4.11) that v ( x ) isaxisymmetric about the line l Q different from l x . Thus similarly to (3.25), the right hand sideof the following equation( −△ ) α/ v ( x ) = 1 | x − x | n + α f ( | x − x | n − α v ( x )) , x ∈ R n + , should also be axisymmetric about line l Q . Since l Q = l x , similarly to (3.26), we have f ( | x − x | n − α v ( x )) = C (cid:2) | x − x | n − α v ( x ) (cid:3) n + αn − α , for some positive constants C (note C = 0, otherwise v ( x ) = v ( x n ) is axisymmetric about l x ,refer to [47]).Till now, we have shown that v ( x ) is bounded near x and f ( t ) = Ct n + αn − α for some positiveconstants C . Hence u ( x ) ∼ | x | n − α near infinity (4.12)and (1.11) becomes ( −△ ) α/ u ( x ) = Cu n + αn − α ( x ) , x ∈ R n + ,u ( x ) ≡ , x / ∈ R n + . (4.13)The result of section 3.2.1 in [18] yields that (1.11) possesses no positive solution. (ii). From the definition of λ , we know λ = x and w λ ≥ , x ∈ Σ λ . (4.14)Now let us move the plane T λ from + ∞ to the left. Similarly, we can derive that either λ > x and w λ ≡ , x ∈ Σ λ (4.15)14r λ = x and w λ ≤ , x ∈ Σ λ . (4.16)The case (4.15) can be handled with the same way as in ( i ). Now from (4.14) and (4.16), weget λ = x and w λ ≡ , x ∈ Σ λ . So far, we have proved that v is symmetric about the plane T x . Noting that the x direction canbe chosen arbitrarily, we have actually shown that v is axially symmetric about the line parallelto x n axis and passing through x . Because x is any point on ∂ R n + , we already deduce that theoriginal solution u is independent of the first n − u = u ( x n ).Next, we will show that this will contradict with the finiteness of u . By Proposition 4.1, weonly need to show that u = u ( x n ) contradicts the finiteness of the integral Z R n + G ∞ ( x, y ) f ( u ( y )) dy. In fact, by Proposition 4.2, if u ( x ) = u ( x n ) is a solution of (4.2), then for each fixed x ∈ R n + and R large enough, we have by (4.3),+ ∞ > u ( x ) = Z ∞ f ( u ) Z R n − G ∞ ( x, y ) dy ′ dy n ≥ C Z ∞ R f ( u ( y n )) y α/ n Z R n − \ B R (0) | x − y | n dy ′ dy n . (4.17)When y n → + ∞ , the behavior of u ( y n ) may have two cases: (ii) u ( y n ) → + ∞ ; (ii) there is a positive constant c such that u ( y n ) ≤ c .However in both cases, we will be able to derive contradictions in the sequel. (ii) . It is easy to see that u ( y n ) ≥ C for y n > R and f ( u ( y n )) ≥ C (since f ( t ) > t > x = ( x ′ , x n ) , y = ( y ′ , y n ) ∈ R n + , r = | x ′ − y ′ | and a = | x n − y n | ,then we have form (4.17),+ ∞ > u ( x ) ≥ C Z ∞ R f ( u ( y n )) y α/ n Z R n − \ B R (0) | x − y | n dy ′ dy n ≥ C Z ∞ R y α/ n Z ∞ R r n − ( r + a ) n drdy n ≥ C Z ∞ R y α/ n | x n − y n | Z ∞ R/a τ n − ( τ + 1) n dτ dy n (4.18) ≥ C Z ∞ R y α/ − n dy n = ∞ , (4.19)which is a contradiction. We derive (4.18) by letting τ = ra . (ii) . Since f ( t ) /t n + αn − α is non-increasing about t in (0 , + ∞ ) and u ( y n ) ≤ c for y n > R , itfollows that for y n > R , f ( u ( y n )) u n + αn − α ( y n ) ≥ f ( c ) c n + αn − α := C , ∞ > u ( x ) ≥ C Z ∞ R f ( u ( y n )) y α/ n Z R n − \ B R (0) | x − y | n dy ′ dy n ≥ C Z ∞ R f ( u ( y n )) u n + αn − α ( y n ) u n + αn − α ( y n ) y α/ n Z R n − \ B R (0) | x − y | n dy ′ dy n ≥ C Z ∞ R u n + αn − α ( y n ) y α/ n Z R n − \ B R (0) | x − y | n dy ′ dy n ≥ C Z ∞ R u n + αn − α ( y n ) y α/ n Z ∞ R r n − ( r + a ) n drdy n ≥ C Z ∞ R u n + αn − α ( y n ) y α/ n | x n − y n | Z ∞ R/a τ n − ( τ + 1) n dτ dy n ≥ C Z ∞ R u n + αn − α ( y n ) y α/ − n dy n , (4.20)which implies that there exists a sequence { y in } → ∞ as i → ∞ , such that u n + αn − α ( y in )( y in ) α/ → . (4.21)The rest proof is completely similar to the last part of section 3.2 in [13], here we only need tosubstitute u n + αn − α for u p , and we can finally derive a contradiction with (4.21).At present, we have proved that u = u ( x n ) contradicts the finiteness of u . Hence (1.11)possesses no positive solution.
5. Proofs of Two Claims
Let us prove Claims 3.1 and 3.2 which are used in Section 3. Without loss of generality, welet x = 0. Proof of Claim 3.1 . Since v ( x ) → | x | → ∞ , and B ǫ (0 λ ) ⊂ Σ λ when λ is sufficientlynegative, we only need to prove v λ ( x ) ≥ c λ in B ǫ (0 λ ) \ { λ } , i.e., v ( x ) ≥ c λ in B ǫ (0) \ { } . Using u ( x ) = 1 | x | n − α v ( x | x | ) , it suffices to prove u ( x ) ≥ c λ | x | n − α , when | x | sufficiently large.Denote ϕ ( x ) = c n, − α Z R n η ( y ) f ( u ( y )) | x − y | n − α dy, where η ( y ) ∈ C ∞ ( R n ) is a cutoff function and η ( y ) = , y ∈ B (0) , , y / ∈ B (0) . (5.1)Then ( −△ ) α/ ϕ ( x ) = c n, − α Z R n ( −△ ) α ( 1 | x − y | n − α ) η ( y ) f ( u ( y )) dy = η ( x ) f ( u ( x )) (5.2)16nd ( −△ ) α/ ( u − ϕ ) = f ( u ) − ηf ( u )= f ( u )(1 − η ) ≥ , in R n . (5.3)For any x ∈ B cR (0), we have c n, − α Z B (0) f ( u ( y )) | x − y | n − α dy ≤ ϕ ( x ) ≤ c n, − α Z B (0) f ( u ( y )) | x − y | n − α dy. Because u ∈ C , loc ( R n ), f > t >
0, it yields that for any x ∈ B cR (0) and R large enough, there exist two positive constants C and C such that C | x | n − α ≤ ϕ ( x ) ≤ C | x | n − α ≤ C R n − α . (5.4)Combining (5.3) and (5.4), we get ( −△ ) α/ ( u − ϕ + C R n − α ) ≥ , in B R (0) ,u − ϕ + C R n − α ≥ , in B cR (0) (5.5)and u − ϕ + C R n − α ≥ , in B R (0) , by the maximum principle in [41]. Letting R → ∞ , it derives u ≥ ϕ, x ∈ R n , and by (5.4), u ( x ) ≥ c λ | x | n − α , for | x | sufficiently large . This complete the proof of Claim 3.1.Before proving Claim 3.2, we first narrate
Proposition 5.1.
Assume that u ∈ L α ∩ C , loc ( R n ) is a locally bounded positive solution to( −△ ) α/ u ( x ) = f ( u ) , x ∈ R n , (5.6)then it is also a solution to u ( x ) = Z R n C | x − y | n − α f ( u ( y )) dy. (5.7)And vice versa.The proof of Proposition 5.1 is entirely similar to the proof of Theorem 2.1 in [44] and we onlyneed to guarantee that: ( i ) for any positive constant c , f ( c ) >
0; ( ii ) f ( u ( x )) is locally boundedon R n . These can be easily acquired by the condition on f ( t ). Here we omit the details. Proof of Claim 3.2. If w λ ( x ) , x ∈ Σ λ \ { λ } , then there exists a point x ∈ Σ λ \ { λ } such that w λ ( x ) > , δ such that w λ ( x ) > , x ∈ B δ ( x ) , (5.8)and w λ ( x ) ≥ c , x ∈ B δ/ ( x ) . (5.9)From Proposition 5.1 and (3.1) (please note that we already let x = 0 for simplicity at thebeginning of this section), we have( −△ ) α/ v ( x ) = Z R n C | x − y | n − α | y | n + α f ( | y | n − α v ( y )) dy, x ∈ R n \ { } . (5.10)Through an elementary computation, we have that for any x ∈ R n \ { λ } , w λ ( x ) = C Z Σ λ ( 1 | x − y | n − α − | x − y λ | n − α )( f ( | y λ | n − α v ( y λ )) | y λ | n + α − f ( | y | n − α v ( y )) | y | n + α ) dy. (5.11)It follows from the monotonicities of f ( t ) and f ( t ) /t n + αn − α that for any x ∈ B ǫ (0 λ ) \ { λ } , w λ ( x ) = C Z Σ λ ( 1 | x − y | n − α − | x − y λ | n − α )( f ( | y λ | n − α v ( y λ )) | y λ | n + α − f ( | y | n − α v ( y )) | y | n + α ) dy ≥ C Z Σ λ ( 1 | x − y | n − α − | x − y λ | n − α )( f ( | y | n − α v ( y λ )) − f ( | y | n − α v ( y )) | y | n + α ) dy ≥ C Z B δ/ ( x ) ( 1 | x − y | n − α − | x − y λ | n − α )( f ( | y | n − α v ( y λ )) − f ( | y | n − α v ( y )) | y | n + α ) dy. (5.12)For any y ∈ B δ/ ( x ), | y | n − α v ( y λ ) − | y | n − α v ( y ) = | y | n − α w λ ( y ) , one knows by the continuity of v and (5.9) that c ≤ | y | n − α v ( y λ ) − | y | n − α v ( y ) ≤ c for some constants c , c > c < c . Since f is strictly increasing, there exists c , c > c < c such that for any x ∈ B δ/ ( x ), c ≤ f ( | y | n − α v ( y λ )) − f ( | y | n − α v ( y )) ≤ c . Hence f ( | y | n − α v ( y λ )) − f ( | y | n − α v ( y )) ≥ C ( | y | n − α v ( y λ ) − | y | n − α v ( y ))= C ( | y | n − α w λ ( y )) , for some C >
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