Radial symmetry and monotonicity for an integral equation

J. Math. Anal. Appl. 342 (2008) 943–949 www.elsevier.com/locate/jmaa

Radial symmetry and monotonicity for an integral equation ✩ Li Ma a,∗ , Dezhong Chen b a Department of Mathematical Sciences, Tsinghua University, Beijing 100084, PR China b Department of Mathematics and Statistics, McMaster University, Hamilton, ON L8S 4K1, Canada

Received 15 April 2007 Available online 28 December 2007 Submitted by W.L. Wendland

Abstract In this paper we study radial symmetry and monotonicity of positive solutions of an integral equation arising from some higherorder semilinear elliptic equations in the whole space Rn . Instead of the usual method of moving planes, we use a new Hardy– Littlewood–Sobolev (HLS) type inequality for the Bessel potentials to establish the radial symmetry and monotonicity results. © 2007 Elsevier Inc. All rights reserved. Keywords: Bessel potential; Radial symmetry; Monotonicity

1. Introduction The radial symmetry and monotonicity of nonnegative solutions of elliptic equations in the unit ball or the whole Euclidean space have been studied by Gidas, Ni and Nirenberg using the technique of moving planes [7,8]. However, their method of moving planes is based on the maximum principle, and hence one is not able to apply it in the absence of the maximum principle. Recently, it was first observed by Chen, Li and Ou [4,5,12] that instead of the maximum principle, one can use a Hardy–Littlewood–Sobolev (HLS) type inequality to obtain the radial symmetry and monotonicity of nonnegative solutions of (systems of) elliptic equations of Yamabe type. Motivated by the works mentioned above, we consider in this paper the positive solutions of the following semilinear partial differential equation in Rn : α

(I − Δ) 2 (u) = uβ , where α > 0, β > 1, and Δ =

(1) n

∂2 i=1 ∂x 2 i

denotes the usual Laplace operator in Rn . We note that under some appropri-

ate decay assumptions of the solutions at infinity, (1) is equivalent to the following integral equation u = gα ∗ uβ , ✩

The research is partially supported by the National Natural Science Foundation of China 10631020 and SRFDP 20060003002.

* Corresponding author.

E-mail addresses: [email protected] (L. Ma), [email protected] (D. Chen). 0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.12.064

(2)

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L. Ma, D. Chen / J. Math. Anal. Appl. 342 (2008) 943–949

where ∗ denotes the convolution of functions and gα is the Bessel kernel whose precise definition will be given in Section 2. An important case of (1) corresponding to α = 2 appears in the study of standing waves of the nonlinear Klein– Gordon equations −Δu + u = u|u|β−1 ,

in Rn .

(3)

Such a kind of equation also arises from the ground states of the Schrödinger equation [2]. Here we just want to mention that it has been shown that under some decay assumptions at the infinity, the smooth positive solutions of (3) are unique and radially symmetric. For more details, we refer to the paper [9] and the references therein. Our main theorem is q n Theorem 1. For any q > max{β, n(β−1) α }, if u ∈ L (R ) is a positive solution of (2), then u must be radially symmetric and monotonely decrease about some point.

The proof of the main theorem is based on the method of moving planes introduced by Chen, Li and Ou. Here the new ingredient is a HLS type inequality for the Bessel potentials, which is established in Theorem 9. 2. Preliminaries on Bessel potentials In this section, we recall some basic properties of the Bessel potentials. For more details, one may refer to [14,17]. Definition 2. The Bessel kernel gα , α  0, is given by 1 gα (x) = γ (α)

∞

    α−n δ π δ 2 −1 dδ, exp − |x|2 exp − δ 4π

0 α 2

where γ (α) = (4π) ( α2 ). There are some elementary facts about the Bessel kernel gα . Proposition 3. (1) The Fourier transform of gα is given by gα (x) =

1 α

(1 + 4π 2 |x|2 ) 2

,

where the Fourier transform of f is defined by  f(x) = f (t) exp(−2πixt) dt. Rn

(2) For each α > 0, gα (x) ∈ L1 (Rn ). (3) The following Bessel composition formula holds gα ∗ gβ = gα+β ,

α, β  0.

Definition 4. The Bessel potentials Bα (f ), f ∈ Lp (Rn ) with 1  p  ∞, is defined by  g ∗ f, α > 0, Bα (f ) = α f, α = 0. For later use, we mention the following two properties of the Bessel potentials.

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Proposition 5. (1) Bα (f )Lp (Rn )  f Lp (Rn ) , 1  p  ∞. (2) Bα ◦ Bβ = Bα+β , α, β  0. Proof. (1) Follows from the Young inequality and the fact that gα L1 (Rn ) = 1. (2) Follows immediately from the Bessel composition formula (4). 2 The most interesting fact concerning Bessel potentials is that they can be employed to characterize the Sobolev spaces W k,p (Rn ). This is expressed in the following theorem where we employ the notation Lα,p (Rn ), α > 0 and 1  p  ∞, to denote all functions u such that u = gα ∗ f for some f ∈ Lp (Rn ). We may define a norm on Lα,p (Rn ) as follows uLα,p (Rn ) = f Lp (Rn ) . With respect to this norm, Lα,p (Rn ) becomes a Banach space. Theorem 6. If k is a nonnegative integer and 1 < p < ∞, then Lk,p (Rn ) = W k,p (Rn ). Moreover, if u ∈ Lk,p (Rn ) with u = gα ∗ f , then C −1 f Lp (Rn )  uW k,p (Rn )  Cf Lp (Rn ) where C = C(α, n, p). α

Before ending this section, we would like to point out that on Sobolev spaces W k,p (Rn ), Bα = (I − Δ)− 2 . 3. A HLS type inequality In this section, we prove a new HLS type inequality which will play an important role in the proof of Theorem 1. To that purpose, we recall the classical Sobolev embedding theorem of the spaces W k,p (Rn ) [1]. Theorem 7 (The Sobolev embedding theorem). For 1  p < ∞, there exist the following embeddings: ⎧ q n np ⎨ L (R ), kp < n and p  q  n−kp ,  k,p n W R → Lq (Rn ), kp = n and p  q < ∞, ⎩ Cb (Rn ), kp > n, where Cb (Rn ) = {u ∈ C(Rn ) | u is bounded on Rn }. The following lemma comes from [1, Theorem 7.63, (d) and (e)]. Lemma 8. np (1) If t  s and 1 < p  q  n−(s−t)p < ∞, then Ls,p (Rn ) → Lt,q (Rn ). s,p (2) If 0  μ  s − n/p < 1, then L (Rn ) → C 0,μ (Rn ).

Now we are ready to prove the following HLS type inequality. q

n q n β Theorem 9. Let q > max{β, n(β−1) α }. If f ∈ L (R ), then Bα (f ) ∈ L (R ). Moreover, we have the estimate Bα (f )Lq (Rn )  Cf  q n , where C = C(α, β, n, q). L β (R )

Proof. By definition, Bα (f ) ∈ L Case 1. α is an integer.

α, βq

q

(Rn ) provided that f ∈ L β (Rn ). We divide the proof into several cases.

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It follows from Theorem 6 that Bα (f ) ∈ W q β

n·

q β

n−α·

α, βq

(Rn ). Note that

n n q ·q > ·q =q > . n(β−1) nβ − αq β nβ − α · α

=

So if α · βq  n, then by the Sobolev embedding theorem, we have Bα (f ) ∈ Lq (Rn ). Now it follows again from Theorem 6 that



Bα (f ) p n  C Bα (f ) α, q  Cf  q n . L (R ) n β

L

L β (R )

(R )

If α · βq > n, then it follows from the Sobolev embedding theorem that Bα (f ) ∈ Cb (Rn ) and Bα (f )Cb (Rn )  CBα (f ) α, q n  Cf  q n . Now a straightforward computation shows that L

β

L β (R )

(R )



Bα (f ) q n = L (R )



  Bα (f )q

1 q

Rn



q

q− q  Bα (f ) β · Bα (f ) β n Cb (R )

 Rn

1 = Bα (f ) β q

L β (Rn )

 f 

1 β q Lβ

= Cf 

(Rn ) q

q

1− β1 · Bα (f ) Cb (R n)

· Cf 

L β (Rn )

1

1− β1 q

L β (Rn )

.

Case 2. α is a fraction. In this case α − [α] > 0, where [α] denotes the integer satisfying [α]  α < [α] + 1. We divide this case into two nβ nβ subcases, i.e. q < α−[α] and q  α−[α] . Case 2.1. q < Let r0 =

nβ α−[α] . n βq

n−(α−[α]) βq α, βq n

. Since [α]  α and

q β

> 1, it follows from Lemma 8(1), that for any r satisfying

we have L (R ) → L[α],r (Rn ). Note that Bα (f ) ∈ L There are two possibilities.

α, βq

q β

 r  r0 ,

(Rn ) and [α] is an integer. Thus Bα (f ) ∈ W [α],r (Rn ).

Case 2.1.1. q  r0 . q [α],q (Rn ). MoreThis is equivalent to q  n(β−1) α−[α] . In this situation, q lies in the interval [ β , r0 ]. Hence Bα (f ) ∈ W over,





Bα (f ) q n  Bα (f ) [α],q n  C Bα (f ) [α],q n  C Bα (f ) α, q = Cf  q n . L (R ) W (R ) L (R ) n L

β

Case 2.1.2. q > r0 . This is equivalent to q < Case 2.1.2.1. q <

nβ α .

n(β−1) α−[α] .

In this situation, we have Bα (f ) ∈ W [α],r0 (Rn ).

(R )

L β (R )

L. Ma, D. Chen / J. Math. Anal. Appl. 342 (2008) 943–949

947

nr0 We have [α]r0 < n. By assumption, q  n(β−1) . Hence q actually lies in the interval [r0 , n−[α]r ]. Now it follows α 0 from the Sobolev embedding theorem that Bα (f ) ∈ Lq (Rn ).

Case 2.1.2.2. q =

nβ α .

We have [α]r0 = n. Again it follows from the Sobolev embedding theorem that Bα (f ) ∈ Lq (Rn ). Case 2.1.2.3. q >

nβ α .

We have [α]r0 > n. Again it follows from the Sobolev embedding theorem that Bα (f ) ∈ Cb (Rn ). A straightforward calculation shows that

1− 1

1

Bα (f ) q n  Bα (f ) β n Bα (f ) β q Cb (R )

L (R )

 C Bα (f )

= C Bα (f ) = Cf  Case 2.2. q 

L β (Rn )

1− β1 q L β (Rn )

1

Bα (f ) β q

L β (Rn )

q

L β (Rn )

q

L β (Rn )

.

nβ α−[α] . α−[α], q

nβ β (Rn ). Note that q  First, we observe that Bα−[α] (f ) ∈ L α−[α] ⇔ 0  μ0 < 1, where μ0 = α − [α] − It follows from Lemma 8(2), that Bα−[α] (f ) ∈ C 0,μ0 (Rn ). Then as before, we obtain the following estimate.



Bα (f ) q n = B[α] ◦ Bα−[α] (f ) q n L (R ) L (R )



 Bα−[α] (f ) Lq (Rn )

1− 1  Bα−[α] (f ) 0,μβ 0 C

(Rn )

1− 1  C Bα−[α] (f ) β

1

Bα−[α] (f ) β q

q α−[α], β

L

 Cf  = Cf  This completes the proof.

1− β1 q L β (Rn ) q

L β (Rn )

f 

L β (Rn )

(Rn )

1

Bα−[α] (f ) β q

L β (Rn )

1 β q

L β (Rn )

.

2

4. Proof of Theorem 1 For a given real number λ, we may define    Σλ = x = (x1 , . . . , xn )  x1  λ . Let x λ = (2λ − x1 , . . . , xn ) and define uλ (x) = u(x λ ). Lemma 10. For any solution u(x) of (2), we have     u(x) − uλ (x) = gα (x − y) − gα x λ − y u(y)β − uλ (y)β dy. Σλ

nβ q .

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L. Ma, D. Chen / J. Math. Anal. Appl. 342 (2008) 943–949

Proof. Let Σλc = {x = (x1 , . . . , xn ) | x1 < λ}. It is easy to see that   β u(x) = gα (x − y)u(y) dy + gα (x − y)u(y)β dy Σλc

Σλ



=



  β gα x − y λ u y λ dy

gα (x − y)u(y) dy + β

Σλ

Σλ





 gα x λ − y uλ (y)β dy.

gα (x − y)u(y) dy +

=

β

Σλ

Σλ

Here we have used the fact that |x − y λ | = |x λ − y|. Substituting x by x λ , we get    λ  λ β u x = gα x − y u(y) dy + gα (x − y)uλ (y)β dy. Σλ

Σλ

Thus  u(x) − u x λ =



 gα (x − y) u(y)β − uλ (y)β dy −

Σλ



  gα x λ − y u(y)β − uλ (y)β dy

Σλ



   gα (x − y) − gα x λ − y u(y)β − uλ (y)β dy.

= Σλ

This completes the proof.

2

We may define Σλ− = {x | x ∈ Σλ , u(x) < uλ (x)}. Lemma 11. For λ 0, Σλ− must be measure zero. Proof. Whenever x, y ∈ Σλ , we have |x − y|  |x λ − y|. Moreover, since λ < 0, |y λ |  |y| for any y ∈ Σλ . Then it follows from Lemma 10 that for any x ∈ Σλ− ,   gα (x − y) uλ (y)β − u(y)β dy uλ (x) − u(x)  Σλ−



β

 β−1  gα (x − y) uλ (uλ − u) (y) dy.

Σλ−

It follows first from the HLS type inequality (Theorem 9) and then the Hölder inequality that for q as in Theorem 1, we have

 β−1 uλ − uLq (Σ − )  β Bα uλ (uλ − u) Lq (Σ − ) λ λ

β−1

q  C uλ (uλ − u) − L β (Σλ )

β−1  Cuλ Lq (Σ − ) uλ λ

− uLq (Σ − )

β−1  Cuλ Lq (Σλ ) uλ

− uLq (Σ − )

β−1 = CuLq (Σ c ) uλ λ

λ

λ

− uLq (Σ − ) . λ

β−1

Since u ∈ Lq (Rn ), we can choose N 0, s.t. for λ  −N , CuLq (Σ c )  12 . Then it follows from the last inequality that uλ − uLq (Σ − ) = 0, and therefore Σλ− must be measure zero. λ

λ

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L. Ma, D. Chen / J. Math. Anal. Appl. 342 (2008) 943–949

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Now we fix an N 0 s.t. for λ  −N , u(x)  uλ (x),

∀x ∈ Σλ .

(5)

We start moving the planes continuously from λ  −N to the right as long as (5) holds. Suppose that at a λ0 < 0, we have u(x)  uλ0 (x) and m({x ∈ Σλ0 | u(x) > uλ0 (x)}) > 0. By Lemma 10, we know that u(x) > uλ0 (x) in the interior of Σλ0 . Let Σλ−0 = {x ∈ Σλ0 | u(x)  uλ0 (x)}. It is easy to see that Σλ−0 has measure zero and limλ→λ0 Σλ− ⊂ Σλ−0 . Let (Σλ− ) be the reflection of Σλ− about the planes x1 = λ. It follows from the above arguments that β−1

uλ − uLq (Σ − )  CuLq (Σ − ) uλ − uLq (Σ − ) . λ

λ

(6)

λ

As before, the integrability condition on u guarantees that there exists an  0, s.t. ∀λ ∈ [λ0 , λ0 + ), β−1 CuLq (Σ − )  12 . Now by (6), we have uλ − uLq (Σ − ) = 0, and hence Σλ− must be measure zero. Therefore λ

λ

the planes can be moved further to the right, i.e., there exist an  depending on n, α, β, q and the solution u, such that u(x)  uλ (x) on Σλ , ∀λ ∈ [λ0 , λ0 + ). The planes either stops at some λ < 0 or can be moved until λ = 0. In the former case, it turns out that uλ (x) = u(x), ∀x ∈ Σλ . In the latter case, we have uλ (x)  u(x), ∀x ∈ Σ0 . However, if we move the planes from the right to the left, then we may see that uλ (x)  u(x), ∀x ∈ Σ0 . Thus uλ (x) = u(x), ∀x ∈ Σ0 . Since we can choose any direction to start the process, we conclude that u is actually radial symmetric and monotone decreases about some point in Rn . This completes the proof of Theorem 1. We point out that there are some interesting works related to ours. See the references [3,6,10,11,13,15,16]. Acknowledgments The authors thank the referee for helpful suggestions. The first named author would like to thank Professor W. Chen for a helpful discussion in the summer of 2004 in Tsinghua University, Beijing.

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