An overdetermined problem in Riesz-potential and fractional Laplacian
aa r X i v : . [ m a t h . A P ] F e b An overdetermined problem in Riesz-potential andfractional Laplacian
Guozhen Lu and Jiuyi Zhu
Abstract.
The main purpose of this paper is to address two open questions raisedby W. Reichel in [ R2 ] on characterizations of balls in terms of the Riesz potential andfractional Laplacian. For a bounded C domain Ω ⊂ R N , we consider the Riesz-potential u ( x ) = Z Ω | x − y | N − α dy for 2 ≤ α = N . We show that u = constant on ∂ Ω if and only if Ω is a ball. In the caseof α = N , the similar characterization is established for the logarithmic potential u ( x ) = R Ω log | x − y | dy . We also prove that such a characterization holds for the logarithmicRiesz potential u ( x ) = Z Ω | x − y | α − N log 1 | x − y | dy when the diameter of the domain Ω is less than e N − α in the case when α − N is a non-negative even integer. This provides a characterization for the overdetermined problemof the fractional Laplacian. These results answer two open questions in [ R2 ] to someextent.
1. Introduction
It is well-known that the gravitational potential of a ball of constant mass density isconstant on the surface of the ball. It is shown by Fraenkel [ Fr ] that this property indeedprovides a characterization of balls. In fact, Fraenkel proves the following Theorem A [ Fr ]: Let Ω ⊂ R N be a bounded domain and ω N be the surface measureof the unit sphere in R N . Consider(1.1) u ( x ) = π R Ω log | x − y | dy, N = 2 , N − ω N R Ω 1 | x − y | N − dy, N ≥ . If u ( x ) is constant on ∂ Ω, then Ω is a ball.
Mathematics Subject Classification.
Key words and phrases.
Overdetermined problem, Riesz potential, moving plane method in integralform, fractional Laplacian.Research is partly supported by a US NSF grant
This result has been extended by Reichel [ R2 ] to more general Riesz potential, butunder a more restrictive assumption on the domain Ω, i.e., Ω is assumed to be convex. In[ R2 ], Reichel considers the integral equation(1.2) u ( x ) = R Ω log | x − y | dy, N = α, R Ω 1 | x − y | N − α dy, N = α, and proves the following theorem. Theorem B [ R2 ] : Let Ω ⊂ R N be a bounded convex domain and α >
2, if u ( x ) isconstant on ∂ Ω, then Ω is a ball.This more general Riesz potential is actually closely related to the fractional Lapla-cian ( −△ ) α in R N . Let N be the collection of nonnegative integers. It is known thatthe fundamental solution G ( x, y ) for pseudo-differential operator ( −△ ) α in R N has thefollowing representation(1.3) G ( x, y ) = Γ( N − α )2 α π N Γ( α ) | x − y | α − N , if α − N N , ( − k α − π N Γ( α ) | x − y | α − N log | x − y | , if α − N ∈ N . We note that for the case of α = 2, Fraenkel’s result is under weaker assumption onthe domain Ω, namely, Ω only needs to be bounded and open in R N . The surprising partfor α = 2 is that there is neither regularity nor convexity requirement for Ω. Thus, twoopen problems were raised by Reichel in [ R2 ] Question 1.
Is Theorem B true if we remove the convexity assumption of Ω?
Question 2.
Is there an analogous result as Theorem B for Riesz-Potential of theform(1.4) u ( x ) = Z Ω | x − y | α − N log 1 | x − y | dy ?It is meaningful to study (1.4) because in the case of α − N ∈ N , up to some rescaling, thekernel function in above integral is the fundamental solution of the fractional Laplacian( −△ ) α . Our goal is to address the above two open questions.The first result we establish does remove the convexity assumption in Theorem B.
Theorem 1.
Let Ω ba a C bounded domain. If u in (1.2) is constant on ∂ Ω , then Ω isa ball. As far as Question 2 is concerned, we partially solve it under some additional assump-tion on the diameter of the domain Ω. Since we are only interested in the case when α > N , we will assume this when we address Question 2.
Theorem 2.
Assume α > N . Let Ω be a C bounded domain with diam Ω < e N − α . Thus, Ω is a ball if u ( x ) in (1.4) is constant on ∂ Ω . N OVERDETERMINED PROBLEM IN RIESZ-POTENTIAL 3
Remark 1.1.
In the above two theorems, if the conclusion that Ω is a ball is verified,then we can easily deduce that u ( x ) is radially symmetric with respect to the center of theball. There has been extensive study in the literature about overdetermined problems inelliptic differential equations and integral equations. In his seminal paper [ Se ], Serrinshowed that the overdetermined boundary value determines the geometry of the under-lying set. This is, if Ω is a bounded C domain and u ∈ C ( ¯Ω) satisfies the following(1.5) △ u = − ,u = 0 , ∂u∂n = constant on Ω , then Ω is a ball and u is radially symmetric with respect to its center of the ball. Serrin’sproof is based on what is nowadays called the moving planes method relying on themaximum principle of solutions to the differential equations, which is originally due toAlexandrov, and has been later used to derive further symmetry results for more generalelliptic equations. Important progress as for the moving plane methods since then are theworks of Gidas-Ni-Nirenberg [ GNN ], Caffarelli-Gidas-Spruck [
CGS ], to just name someof the early works in this direction.Immediately after Serrin’s paper, Weinberger [ W ] obtained a very short proof of thesame result, using the maximum principle applied to an auxiliary function. However,compared to Serrin’s approach, Weinberger’s proof relies crucially on the linearity of theLaplace operator.Since the work of [ Se ], many results are obtained about overdetermined problems.The interested reader may refer to [ AB ], [ B ], [ BK ], [ BNST ], [
BNST1 ], [ CS ], [ EP ],[ FG ], [ FGK ], [ FK ], [ FV ], [ G ], [ GL ], [ HPP ], [
Lim ], [
Liu ], [ M ], [ MR ], [ PP ], [ PS ],[ P ], [ Sh ], [ Si ], [ WX ] and references therein, for more general elliptic equations. See also[ R1 ] and reference therein for overdetermined problems in an exterior domain or generaldomain. In [ BNST ], an alternative shorter proof of Serrin’s result, not relying explicitlyon the maximum principle has been given, where they deduce some global informationconcerning the geometry of the solution.Overdetermined problems are important from the point of view of mathematicalphysics. Many models in fluid mechanics, solid mechanics, thermodynamics, and elec-trostatics are relevant to the overdetermined Dirichlet or Newmann boundary problemsof elliptic partial differential equations. We refer the reader to the article [ FG ] for a niceintroduction in that aspect.Instead of a volume potential, single layer potential is also considered in overdeter-mined problems. A single layer potential is given by(1.6) u ( x ) = A R ∂ Ω − π log | x − y | dσ y , N = 2 ,A R ∂ Ω 1( N − ω N | x − y | N − dσ y , N ≥ , where A > u isconstant in ¯Ω, then Ω can be proved to be a ball under different smoothness assumptionon the domain Ω. See [ M ] for the case of n = 2 and [ R1 ] for the case of n ≥
3, and also
GUOZHEN LU AND JIUYI ZHU some related works in [
Lim ] and [ Sh ]. We also refer the reader to the book of C. Kenig[ K ] on this subject of layer potential.Generally speaking, two approaches are widely applied in dealing with overdeterminedproblems. One is the classical moving plane method. In [ Se ], the moving plane methodwith a sophisticated version of Hopf boundary maximum principle plays a very importantrole in the proof. The other way is based on an equality of Rellich type, as well as an inte-rior maximum principle, see [ W ]. Our approach is a new variant of moving plane method- Moving plane in integral forms. It is much different from the traditional methods ofmoving planes used for partial differential equations. Instead of relying on the differentia-bility and maximum principles of the structure, a global integral norm is estimated. Themethod of moving planes in integral forms can be adapted to obtain symmetry and mono-tonicity for solutions. The method of moving planes on integral equations was developedin the work of W. Chen, C. Li and B. Ou [ CLO ], see also Y.Y. Li [ Li ], the book by W.Chen and C. Li [ CL1 ] and an exhaustive list of references therein, where the symmetryof solutions in the entire space was proved. Moving plane method in integral form overbounded domains requires some additional efforts and has been carried out recently insymmetry problems arising from the integral equations over bounded domains, see thework of D. Li, G. Strohmer and L. Wang [
LSW ].We end this introduction with the following remark concerning the characterizationof balls by using the Bessel potential. The Bessel kernel g α in R N with α ≥ g α ( x ) = 1 r ( α ) Z ∞ exp( − πδ | x | ) exp( − δ π ) δ α − N − dδ, where r ( α ) = (4 π ) α Γ( α ) . In the paper [
HLZ ], we consider the Bessel potential type equation:(1.8) u ( x ) = Z Ω g α ( x − y ) dy. Overdetermined problems for Bessel potential over a bounded domain in R N havebeen recently studied in [ HLZ ]. For instance, the following theorem is proved in [
HLZ ],among some other results:
Theorem 3.
Let Ω ba a C bounded domain in R N . If u in (1.8) is constant on ∂ Ω , then Ω is a ball. It is well-known that (1.8) is closely related to the following fractional equation( I − △ ) α u = χ Ω . In the case of α = 2, it turns out to be the ground state of the Schr¨ o dinger equation.The paper is organized as follows. In Section 2, we show Theorem 1. In Section 3,we carry out the proof of Theorem 2. Throughout this paper, the positive constant C isfrequently used in the paper. It may differ from line to line, even within the same line.It also may depends on u in some cases.Finally, we thank Dr. Xiaotao Huang for his comments on our earlier draft of thispaper. N OVERDETERMINED PROBLEM IN RIESZ-POTENTIAL 5
2. Proof of Theorem 1
In this section, we will prove Theorem 1 by adapting the moving plane method inintegral forms, see [
CLO ]. Since we are dealing with the case of bounded domains, wemodify the method accordingly (see also [
LSW ], [ CZ ]).We first introduce some notations. Choose any direction and, rotate coordinate systemif it is necessary such that x -axis is parallel to it. For any λ ∈ R , define T λ = { ( x , ..., x n ) ∈ Ω | x = λ } . Since Ω is bounded, if λ is sufficiently negative, the intersection of T λ and Ω is empty.Then, we move the plane T λ all the way to the right until it intersects Ω. Let λ = min { λ : T λ ∩ ¯Ω = ∅} . For λ > λ , T λ cuts off Ω. We defineΣ λ = { x ∈ Ω | x < λ } . Set x λ = { λ − x , ..., x n } and Σ ′ λ = { x λ ∈ Ω | x ∈ Σ λ } . At the beginning of λ > λ , Σ ′ λ remains within Ω. As the plane keeps moving to theright, Σ ′ λ will still stay in Ω until at least one of the following events occurs:(i)Σ ′ λ is internally tangent to the boundary of Ω at some point P λ not on T λ .(ii) T λ reaches a position where it is orthogonal to the boundary of Ω at some point Q .Let ¯ λ be the first value such that at least one of the above positions is reached.We assert that Ω must be symmetric about T ¯ λ ; i.e.,(2.1) Σ ¯ λ ∪ T ¯ λ ∪ Σ ′ ¯ λ = Ω . If this assertion is verified, for any given direction in R N , there also exists a plane T ¯ λ suchthat Ω is symmetric about T ¯ λ . Moreover, Ω is connected. Then the only domain withthose properties is a ball, see [ Al ].In order to assert (2.1), we introduce u λ ( x ) = u ( x λ ) , Ω λ = Ω \ (Σ λ ∪ Σ ′ λ ) . We first establish some lemmas. Throughout the paper we assume α ≥ Lemma 2.1.
Let l ∈ N with ≤ l < α . Then for any solution in (1.2), u ∈ C l ( R N ) anddifferentiation of order l can be taken under the integral. Proof.
The proof is standard. We refer the reader to [ R2 ]. (cid:3) Lemma 2.2.
For λ < λ < ¯ λ and u ( x ) satisfying (1.2), we have(i) If N ≥ α , u λ ( x ) > u ( x ) for any x ∈ Σ λ . (ii) If N < α , u λ ( x ) < u ( x ) for any x ∈ Σ λ . GUOZHEN LU AND JIUYI ZHU
Proof.
For x ∈ Σ λ , in the case of N = α , we rewrite u ( x ) and u λ ( x ) as u ( x ) = Z Σ λ log 1 | x − y | dy + Z Σ λ log 1 | x λ − y | dy + Z Ω λ log 1 | x − y | dy, and u λ ( x ) = Z Σ λ log 1 | x λ − y | dy + Z Σ λ log 1 | x − y | dy + Z Ω λ log 1 | x λ − y | dy. Then(2.2) u λ ( x ) − u ( x ) = Z Ω λ log | x − y || x λ − y | dy. Since | x − y | > | x λ − y | for x ∈ Σ λ and y ∈ Ω λ , then u λ ( x ) > u ( x ) . While in the case of N = α , u λ ( x ) and u ( x ) have the following representations respectively: u ( x ) = Z Σ λ | x − y | α − N dy + Z Σ λ | x λ − y | α − N dy + Z Ω λ | x − y | α − N dy, and u λ ( x ) = Z Σ λ | x λ − y | α − N dy + Z Σ λ | x − y | α − N dy + Z Ω λ | x λ − y | α − N dy. Thus,(2.3) u λ ( x ) − u ( x ) = Z Ω λ ( | x λ − y | α − N − | x − y | α − N ) dy, Note that | x − y | > | x λ − y | for x ∈ Σ λ and y ∈ Ω λ . Thus, (i) and (ii) are concluded. (cid:3) Lemma 2.3.
Assume that u ( x ) satisfies (1.2) and suppose λ = ¯ λ in the first case; i.e. Σ ′ λ is internally tangent to the boundary of Ω at some point P ¯ λ not on T ¯ λ , then Σ ¯ λ ∪ T ¯ λ ∪ Σ ′ ¯ λ =Ω . Proof.
When N ≥ α , thanks to Lemma 2.1, u ¯ λ ( x ) ≥ u ( x ) for x ∈ Σ ¯ λ . While
N < α , u ¯ λ ( x ) ≤ u ( x ) for x ∈ Σ ¯ λ . We argue by contradiction. Suppose Σ ¯ λ ∪ T ¯ λ ∪ Σ ′ ¯ λ & Ω; thatis, Ω ¯ λ = ∅ . At P ¯ λ , from (2.2) and (2.3), u ( P ¯ λ ) > u ( P ) in the case of N ≥ α . It is acontradiction since P ¯ λ , P ∈ ∂ Ω and u ( P ¯ λ ) = u ( P ) = constant. From the same reason, u ( P ¯ λ ) < u ( P ) when N < α . It also contradicts the fact that u is constant on the boundary.Therefore, the lemma is completed. (cid:3) Lemma 2.4.
Assume that u ( x ) satisfies (1.2) and suppose that the second case occurs:i.e. T ¯ λ reaches a position where is orthogonal to the boundary of Ω at some point Q , then, Σ ¯ λ ∪ T ¯ λ ∪ Σ ′ ¯ λ = Ω . N OVERDETERMINED PROBLEM IN RIESZ-POTENTIAL 7
Proof.
Since u ( x ) is constant on the boundary and Ω ∈ C , ▽ u is parallel to thenormal at Q . As implied in the second case, ∂u∂x | Q = 0. We denote the coordinate of Q by z . Suppose Ω ¯ λ = ∅ , there exits a ball B ⊂⊂ Ω ¯ λ . Choose a sequence { x i } ∞ ∈ Σ ¯ λ \ T ¯ λ such that x i → z as i → ∞ . It is easy to see that x i ¯ λ → z as i → ∞ . Since B ⊂⊂ Ω ¯ λ , wecan also find a δ such that diam Ω > | x i ¯ λ − y | > δ for any y ∈ B and any x i ¯ λ .If N = α , by (2.2), u ( x i ¯ λ ) − u ( x i ) = Z Ω ¯ λ log | x i − y || x i ¯ λ − y | dy. Let e = (1 , , · · · , ∈ R N , then ( x i ¯ λ − x i ) · e is the first component of ( x i ¯ λ − x i ). By theMean Value theorem, u ( x i ¯ λ ) − u ( x )( x i ¯ λ − x i ) · e = Z Ω ¯ λ log | x i − y | − log | x i ¯ λ − y | ( x i ¯ λ − x i ) · e dy = Z Ω ¯ λ ( y − ¯ x i ¯ λ ) · e | y − ¯ x i ¯ λ | dy> C Z B | diam Ω | dy> C, (2.4)where ¯ x i ¯ λ is some point between x i ¯ λ and x i . Nevertheless,lim i →∞ u ( x i ¯ λ ) − u ( x i )( x i ¯ λ − x i ) · e = ∂u∂x | Q = 0 , which contradicts (2.4). Therefore, Ω ¯ λ = ∅ . In the case of
N > α , similarly we have u ( x i ¯ λ ) − u ( x i )( x i ¯ λ − x i ) · e = Z Ω ¯ λ | x i ¯ λ − y | α − N − | x i − y | α − N ( x i ¯ λ − x i ) · e dy = Z Ω ¯ λ ( α − N ) | ¯ x i ¯ λ − y | α − N − (( x i ¯ λ − y ) · e ) dy> Z B ( α − N ) | ¯ x i ¯ λ − y | α − N − (( x i ¯ λ − y ) · e ) dy> C. (2.5)It also contradicts ∂u∂x | Q = 0, thus Ω ¯ λ = ∅ .The same idea can be applied to the case of N < α with minor modification. Inconclusion, Σ ¯ λ ∪ T ¯ λ ∪ Σ ′ ¯ λ = Ω when the second case occurs. (cid:3) Combining Lemma (2.3) and Lemma (2.4), Theorem 1 is implied.
GUOZHEN LU AND JIUYI ZHU
3. Proof of Theorem 2
In this section, we will prove theorem 2 under some restriction on the diameter ofΩ. Since we are mainly interested in the case of α − N ∈ N . This is the case when thefundamental solution of ( −△ ) α has the representation (1.3). Therefore, we will assume α > N in this section. Obviously, u ∈ C ( R N ) in (1.4). We begin with establishingseveral lemmas. Lemma 3.1.
For λ < λ < ¯ λ , assume u ( x ) satisfies (1.4) with diam Ω < e N − α , then u λ ( x ) < u ( x ) for any x ∈ Σ λ . Proof.
Since | x λ − y λ | = | x − y | , and | x λ − y | = | x − y λ | , we write u ( x ) and u λ ( x ) inthe following forms: u ( x ) = Z Σ λ | x − y | α − N log 1 | x − y | dy + Z Σ λ | x λ − y | α − N log 1 | x λ − y | dy + Z Ω λ | x − y | α − N log 1 | x − y | dy, and u λ ( x ) = Z Σ λ | x λ − y | α − N log 1 | x λ − y | dy + Z Σ λ | x − y | α − N log 1 | x − y | dy + Z Ω λ | x λ − y | α − N log 1 | x λ − y | dy. Then,(3.1) u λ ( x ) − u ( x ) = Z Ω λ | x − y | α − N log | x − y | dy − Z Ω λ | x λ − y | α − N log | x λ − y | dy. We consider the function s α − N log s. Note α > N , thus( s α − N log s ) ′ = s α − N − [( α − N ) log s + 1] < , whenever s < e N − α . Since | x − y | > | x λ − y | for x ∈ Σ λ , y ∈ Ω λ , and diam Ω < e N − α , weeasily infer that u λ ( x ) < u ( x ) for any x ∈ Σ λ . (cid:3) Lemma 3.2. u ( x ) satisfies (1.4) and suppose λ = ¯ λ in the first case; i.e. Σ ′ ¯ λ is internallytangent to the boundary of Ω at some point P ¯ λ not on T ¯ λ , then Σ ¯ λ ∪ T ¯ λ ∪ Σ ′ ¯ λ = Ω . Proof.
The proof is essentially the same as that of Lemma (2.3). (cid:3)
Lemma 3.3.
Suppose that u ( x ) satisfies (1.4) with diam Ω < e N − α and that the secondcase occurs: i.e. T ¯ λ reaches a position where is orthogonal to the boundary of Ω at somepoint Q , then, Σ ¯ λ ∪ T ¯ λ ∪ Σ ′ ¯ λ = Ω . Proof.
The argument follows that of the proof of Lemma (2.4). Since u ( x ) is constanton ∂ Ω and Ω ∈ C , ∂u∂x | Q = 0. We denote the coordinate of Q by z . Suppose Ω ¯ λ = ∅ , thereexits a ball B ⊂⊂ Ω ¯ λ . Choosing a sequence { x i } ∞ ∈ Σ ¯ λ \ T ¯ λ such that x i → z as i → ∞ , N OVERDETERMINED PROBLEM IN RIESZ-POTENTIAL 9 then x i ¯ λ → z as i → ∞ . Since B ⊂⊂ Ω ¯ λ , we find a δ such that diam Ω > | x i ¯ λ − y | > δ forany y ∈ B and any x i ¯ λ .From (3.1), by Mean Value Theorem, u ( x i ¯ λ ) − u ( x i )( x i ¯ λ − x i ) · e = Z Ω ¯ λ | x i − y | α − N log | x i − y | − | x i ¯ λ − y | α − N log | x i ¯ λ − y | ( x i ¯ λ − x i ) · e dy = Z Ω ¯ λ −| ¯ x i ¯ λ − y | α − N − (( x i ¯ λ − y ) · e )(( α − N ) log | ¯ x i ¯ λ − y | + 1) dy< Z B −| ¯ x i ¯ λ − y | α − N − (( x i ¯ λ − y ) · e )(( α − N ) log | ¯ x i ¯ λ − y | + 1) dy< − C. (3.2)Where ¯ x i ¯ λ is some point between x i ¯ λ and x i . The assumption diam Ω < e N − α is applied inthe last inequalities. Consequently, (3.2) contradicts ∂u∂x | Q = 0 as i → ∞ . Therefore, thelemma is verified. (cid:3) With the help of the above two lemmas, Theorem 2 is confirmed.
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