Liouville theorems for quasi-harmonic functions

Nonlinear Analysis 73 (2010) 2890–2896

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Liouville theorems for quasi-harmonic functions Zhu Xiangrong, Wang Meng ∗ Department of Mathematics, Zhejiang University, Hangzhou, 310027, PR China

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Article history: Received 3 March 2009 Accepted 12 June 2010

abstract Let N be a compact Riemannian manifold. A self-similar solution for the heat flow is a harmonic map from (Rn , e−|x| /2(n−2) ds20 ) to N (n ≥ 3), which was also called a quasiharmonic sphere (cf. Lin and Wang (1999) [1]). (Here ds20 is the Euclidean metric in Rn .) It arises from the blow-up analysis of the heat flow at a singular point. When N = R and without the energy constraint, we call this a quasi-harmonic function. In this paper, we prove that there is neither a nonconstant positive quasi-harmonic function nor a non2

Keywords: Liouville theorem Quasi-harmonic function

2 n ) quasi-harmonic function. However, for all 1 ≤ constant Lp (Rn , e−|x| /2(n−2) ds20 )(p > n− 2 2 p ≤ n/(n − 2), there exists a nonconstant quasi-harmonic function in Lp (Rn , e−|x| /2(n−2) 2 ds0 ). © 2010 Elsevier Ltd. All rights reserved.

1. Introduction Let M and N be two compact Riemannian manifolds with dim M = n. By Nash’s embedding theorem we may assume that N ⊂ Rm , isometrically for some m. A nonconstant smooth map ω : Rl → N is called a quasi-harmonic sphere S l if ω satisfies the equation 1

τ (w) =

x · ∇w 2 with the property that

Z

2 /4

|∇w|2 e−|x| Rn

(1.1)

dx < ∞,

(1.2)

where τ (w) = ∆ω + A(ω)(dω, dω) is the tension field of w . In other words, ω is a critical point of the Dirichlet integral

R

Rl

|∇v|2 e−

|y|2 4

dy. In [1,2], it is proved that quasi-harmonic spheres and harmonic spheres are the only two reasons for blow-





up. Note that, for such a quasi-harmonic sphere ω, if one lets u(x, t ) = ω √x−t , then u is a self-similar solution of a heat flow from Rl × R− into N. It is interesting to show the existence or nonexistence of such quasi-harmonic spheres. Fan [3] provided the first examples of quasi-harmonic spheres for N = S m (3 ≤ m ≤ 6), and Gastel [4] gave more examples with N = S m , for all m ≥ 3. The solutions that they found are equivariant. In [5], it is proved that the nonconstant equivariant quasi-harmonic sphere must be discontinuous at infinity. Even if N = R, that is, w is a function, Eq. (1.1) is new to us. In this case the equation reduces to a linear equation in Rn :

∆(w) = ∗

1 2

x · ∇w

Corresponding author. E-mail addresses: [email protected] (X. Zhu), [email protected] (M. Wang).

0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.06.045

(1.3)

X. Zhu, M. Wang / Nonlinear Analysis 73 (2010) 2890–2896

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which can be rewritten as div(F (x)∇ u) = 0, where F (x) = e−

|x|2 4

(1.4)

. We call a function which satisfies (1.3) a quasi-harmonic function. We can think of it as a harmonic

function on Rn with metric ds2 = e−|x| /2(n−2) k=1 dx2k . By [5], we note that Rn with the metric ds2 is a noncompact Riemannian manifold with a single singularity at ∞. The existence of the solution of (1.4) was also found in [5]. However, we will give the following proposition and prove it by the method of ODE in the Appendix for completeness of this paper: 2

Pn

Proposition 1. For ∀ω ∈ C ∞ (S n−1 ), the following elliptic system:

   2 div e− |x4| ∇ u(x) = 0, 

(1.5)

u|S n−1 = ω(θ )

has a unique smooth solution u on Rn . There has been much interest in Liouville type theorems for harmonic maps. In [6], a nonexistence theorem for quasiharmonic spheres which implies the global existence of heat flow was proved. For a detailed survey and progress in this direction, see works by Hildebrandt [7], Eells and Lemaire [8], Cheng [9], Hildebrandt [7], Hildebrandt, Jost and Widman [10], Hildebrandt and Kaul [11], Li and Wang [12] and Yau [13]. In this paper, we are interested in the Liouville theorems for the quasi-harmonic functions. It is proved in [6] that there is no nonconstant quasi-harmonic function with polynomial growth. It is easy to see that there are no nonconstant quasiharmonic functions with finite energy, from [14]. It was proved by Yau in [13] that there is no nonconstant positive harmonic function on a complete noncompact manifold with nonnegative Ricci curvature. In [15] Yau also obtained that there are no nonconstant nonnegative Lp (p > 1) subharmonic functions on a complete Riemannian manifold without boundary, which implies that there are no nonconstant Lp (p > 1) harmonic functions on a complete manifold without boundary. When p = 1, we know from [16] that there exists a complete two-dimensional manifold with a nonconstant L1 harmonic function. In this note we obtain the corresponding results for quasi-harmonic functions. We think of the quasi-harmonic function as a harmonic function with the metric ds2 = e−|x| /2(n−2) k=1 dx2k . We will show that there is neither a nonconstant positive quasi-harmonic function nor a nonconstant Lp (Rn , ds2 ) (p > n/(n − 2) n ≥ 3) quasi-harmonic functions. However, for all 1 ≤ p ≤ n/n − 2, there exists a nonconstant quasi-harmonic function in Lp (Rn , ds2 ) (n ≥ 3). Note that if we think of the 2

Pn

quasi-harmonic function as a harmonic function, the metric ds2 = e−|x| /2(n−2) k=1 dx2k is quite singular at infinity, it is not complete, the volume of this manifold is finite, and the curvature on the manifold isn’t always nonpositive or nonnegative. So the problem is different from that of [13,15]. The following theorems are our main results. 2

Pn

Theorem 2. For n > 1, suppose u is a smooth positive solution to the elliptic system div(F (x)∇ u) = 0,

x ∈ Rn

(1.6)

2

|x| where F (x) = e− 4 , then u is a constant.

n Theorem 3. For n ≥ 3, suppose that u is a smooth solution to the elliptic system (1.6). If there exists a p > n− such that 2 2

Z

|u|p e

− 4n(n|x−| 2)

Rn

dx < ∞,

(1.7)

then u is also a constant. Theorem 4. For n ≥ 3 and all 1 ≤ p ≤ n/n − 2, there exists a nonconstant smooth solution to the elliptic system (1.6) satisfying 2

R

Rn

|u|p e

− 4n(n|x−| 2)

dx < ∞.

We will give the proof in the next section. Remark 5. We use the method of cut-off functions to prove Theorem 2. However, we could not obtain Theorem 3 by the same method. Then we use the decomposition of u w.r.t. the spectrum of ∆θ on S n to prove Theorem 3. Although the proof of Theorem 2 can be obtained by the proof of Theorem 3, the proof that we now give is more straightforward. Pn 2 In fact, (1.7) is the Lp norm of u on Rn equipped with the metric ds2 = e−|x| /2(n−2) k=1 dx2k , i.e. 2

Z k uk



|x|2



− Lp Rn ,e 2(n−2) dx

|u|p e

= Rn

− 4n(n|x−| 2)

dx.

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X. Zhu, M. Wang / Nonlinear Analysis 73 (2010) 2890–2896

2. Proof of the main theorems Proof of Theorem 2. Let u be a smooth positive solution of (1.6). Set v(r ) =



d

r n − 1 e−

dr

d

r2 4

dr

R

S n−1

 v(r ) = 0.

u(r , θ )dθ . We have (2.1)

Then there exists a constant C such that d

r2

dr

v(r ) = Cr 1−n e 4 .

(2.2)

Note that for n > 1, v(r ) satisfies

Z

1

v (r )dr < ∞, 0

0

so the constant C in (2.2) must be zero. Thus we obtain that v(r ) is a constant, i.e.

v(r ) =

Z S n−1

u(r , θ )dθ = v(0).

Set Ωr = x : r − 1r < |x| < r + get that for any θ ∈ S n−1 ,



1 r

(2.3)



when r > 2. For in Ωr , using the standard theory of uniform elliptic systems we can

C

u(r , θ ) ≤  x : |x − (r , θ )| < 1 r Z < Cr n u(x)dx

Z |x−(r ,θ)|< 1r

u(x)dx

Ωr

≤ Cr 2n−1

Z

r + 1r

r − 1r

v(t )dt

≤ Cr 2n−2 v(0) ≤ Cr 2n−2

(2.4)

where C is a positive constant. On the other hand, from (1.6) we have

Z

e−

|x|2 4

∇ u · ∇ϕ = 0 for any function ϕ ∈ C01 (Rn ).

(2.5)

A standard argument shows that for any r > 0 it holds that

Z

|∇ u|2 e−

|x|2 4

dx ≤

|x|
C

Z

r2

u2 e −

|x|2 4

r <|x|<2r

dx.

(2.6)

Now from (2.4) and (2.6) we can obtain that

Z

|∇ u|2 e−

|x|2 4

dx ≤

|x|
≤

C

Z

r2 C

u2 e − 2r

Z

r2

for any r big enough. So we have

Rn

i.e. u is a constant.

|x|2 4

dx = 0, 

dx

r 4(n−1) r n−1 e−

r r2

|∇ u|2 e−

4

r <|x|<2r

≤ Cr 5n−6 e− 4

Z

|x|2

r2 4

ds

X. Zhu, M. Wang / Nonlinear Analysis 73 (2010) 2890–2896

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Proof of Theorem 3. The proof will be divided into three steps. Step 1: Let ϕl be the lth-order homogeneous harmonic polynomial restriction on S n−1 . It is known that ∆θ ϕl = −λl ϕl , λl = l(n+l−1) ∼ l2 and the dimension of the lth-order homogeneous harmonic polynomial is Cll+n−1 −Cll+−n2−3 ∼ ln−2 . Set fk (r ) = hu(r ), ϕk i. In the L2 sense it can be shown that ∞ X

u( r , θ ) =

X

fk (r )ϕk .

(2.7)

l=0 k,λk =l(n+l−1)

For convenience we omit the index l below. Using polar coordinates we can rewrite the equation as

 urr +

n−1 r



Fr

+

ur +

F

∆θ u

= 0.

r2

(2.8)

Use h, i to denote the L2 inner product on S n−1 . It is easy to get that

hur , ϕk i = fk0 ,

hurr , ϕk i = fk00 .

(2.9)

Multiplying by ϕk in (2.8) and integrating on S 1 , from (2.9) we can get the following ODE: fk00 +



n−1 r

F0

+



F

fk0 =

λk fk

(2.10)

r2

which can be written as

(r n−1 F (r )fk0 (r ))0 = λk r n−3 F (r )fk (r ).

(2.11)

In particular, when k = 0(λ0 = 0), we have (2.2).  Step 2: We estimate ku(r , ∗)k∞ . Consider the elliptic system in Ωr = x : r − F (y) e

x, y ∈ Ωr it holds that 0 that it holds that

1 r

< |x| < r +

1 r



. Note that when

< F (x) < eF (y); using the standard theory of uniform elliptic equations we have for any p > C

|u(r , θ )|p <  x : |x − (r , θ )| < 1 r Z < Cr n |u(x)|p dx

Z |x−(r ,θ)|< 1r

|u(x)|p dx

Ωr

≤ Cr

2n−1

Z

r + 1r r − 1r

Z S n−1

|u(t , ϕ)|p dϕ dt

(2.12)

where C = C (p, n) is a positive constant. Step 3: Finally we estimate ku(r , ∗)k2 . As u isn’t a constant, there must be a k > 0 such that fk (1) > 0 (or <0). As limr →0 r n−1 F (r )(fk )0 (r ) = 0, from (2.11) we can get that

(fk )0 (r ) =

λk

Rr 0

t n−3 F (t )fk (t )dt r n−1 F (r )

.

(2.13)

From the maximal principle it is easy to see that fk is increasing on [0, ∞). (Note that fk (0) = 0 and fk (1) > 0.) When r > 2 it can be obtained that

(fk ) (r ) > 0

> =

λk

Rr

λk

R2

1

t n−3 F (t )fk (t )dt r n −1 F ( r )

1

t n−3 F (t )dtfk (1) r n−1 F (r )

c r n −1 F ( r )

where c is a positive constant which doesn’t depend on r. So when r > 3 we have fk ( r ) >

c r n −1 F ( r )

.

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X. Zhu, M. Wang / Nonlinear Analysis 73 (2010) 2890–2896

Now from the definition of fk (r ) we can get that c

ku(r , ∗)k2 >

r n−1 F (r )

,

r > 3,

(2.14)

which implies that

ku(r , ∗)k∞ >

c

r2

= cr 1−n e 4 ,

r n−1 F (r )

r > 3.

(2.15)

On the other hand from (2.12) we have

|u(r , θ )|p ≤ Cr 2n−1

r + 1r

Z

Z

r − 1r

nr 2

≤ Cr n e 4(n−2)

r + 1r

Z

Z

r − 1r n

= Cr e

nr 2 4(n−2)

|u(t , ϕ)|p dϕ dt

S n−1

S n−1

|u(t , ϕ)|p t n−1 e

2

− 4(nr n−2)

dϕ dt

2

Z

|u|p e

− 4n(n|x−| 2)

dϕ dt

Ωr nr 2

≤ Cr n e 4(n−2) .

(2.16)

(2.15) and (2.16) imply that for any r > 3 it holds that

 r

1 −n

e

r2 4

p

nr 2

< Cr n e 4(n−2) .

n Because p > n− , this is a contradiction. So u must be a constant on Rn . 2



Proof of Theorem 4. By Proposition 1, we know that the following system:

   2 div e− |x4| ∇ u(x) = 0, 

u|S n−1 = ϕk (θ ).

has a unique solution uk (r , θ ), where ϕk is the kth-order homogeneous harmonic polynomial restriction on S n−1 . By (2.7), fk (0) = 0, and the maximum principle, we know that uk (r , θ ) = fk (r )ϕk (θ ), and fk is increasing on [0, ∞)(Note that fk (1) = 1). By (2.13), for s large enough, fk (s) − fk (1) =

s

Z

fk0 (r )dr 1

= λk = λk

s

Z

r 1 −n e

Z0 s

r

Z

r2 4

t n − 3 e−

t2 4

fk (t )dtdr

0

t n − 3 e−

t2 4

fk (t )

Z

0

≤ 3λk s e −n

s2 4

s

Z

t n − 3 e−

s

r 1 −n e

r2 4

drdt

max{t ,1} t2 4

fk (t )dt .

(2.17)

0 2

Rs

In the last inequality, we have used lims→+∞ s2

sn e− 4 (fk (s) − fk (1)) ≤ 3λk

Z

s

1−n e 1 r

− r4

dr

s2 s−n e 4

t n − 3 e−

t2 4

t n − 3 e−

t2 4

= 2. Then

fk (t )dt

0

≤ 3λk

1

Z

fk (t )dt + 3λk fk (1)

0

≤ 3λk

t n − 3 e− 0

Therefore, by Grönwall’s lemma, 2

s fk (s) ≤ Cn,k e 4 s−n + 1,

t n−3 e−

t2 4

dt + 3λk

t2 4

dt + 3λk

s

Z

t n − 3 e− 1

s

Z

1

∞

Z

s

Z

t n − 3 e−

t2 4

(fk (t ) − fk (1)) dt

1 t2 4

(fk (t ) − fk (1)) dt .

(2.18)

X. Zhu, M. Wang / Nonlinear Analysis 73 (2010) 2890–2896

2895

for s large enough. Noting that fk (s) is increasing on [0, ∞) and fk (0) = 0, we have n|x|2 p − 4(n−2) dx e

Z

|uk (x)|

Rn

2

Z = Rn

for 1 ≤ p ≤ n/n − 2.

|fk (|x|)ϕk (x0 )|p e

− 4n(n|x−| 2)

dx < ∞



Acknowledgement This work was supported by PDSFC (20060400336), NSFC (10571156, 10601046, 10701064) and ZJNSF (RC97017). Appendix Proof of Proposition 1. For all k ≥ 1, we consider the ODE (2.10) fk00 +



n−1

+

r

F0



F

f0 =

λk fk r2

.

By the theory of ODE, we know that (2.10) has the following solution in 0 < |r | < ∞: fk ( r ) = r α

∞ X

ai r i ,

(A.1)

i=0



where a0 6= 0, and α = −(n − 2) +

p

ai [(i + α)2 + n(i + α) − λk ] =

 (n − 2 + 2k)2 + 4k /2 > k is a constant. Furthermore, we can get a2i+1 = 0 and

ai−2 (i − 2 + α) 2

for i ≥ 2. Let uk (r , θ ) = fk (r )ϕk (θ ) for r > 0, and uk (0) = 0. It is easy to see that uk ∈ C ∞ (Rn \ {0}). As α > k ≥ 1, uk is continuous and differentiable on Rn . Moreover, it satisfies (2.8) for all r > 0, which means that



div e−

|x|2 4

 ∇ uk

= 0 on Rn \ {0}.

(A.2)

By the divergence theorem, (A.2) and (A.1), for all φ ∈ C01 (Rn ),

Z

e−

|x|2 4

Rn

∇ uk · ∇φ = 0,

which means that uk is a weak solution of (1.6) on Rn . For all R > 0, uk ∈ W 1,2 (BR ) is a weak solution of (1.6) in BR ; then also uk ∈ C ∞ (BR ). R P∞ ∀ω ∈ C ∞ (S n−1 ), ω(θ ) = ω(θ )ϕk (θ )dθ . We can solve (2.10) with fk (1) = Ck ; then k=0 Ck ϕk (θ ), where Ck = S n−1 P∞ u= k=0 uk (r , θ ) solves (1.5) (the case of k = 0 is trivial). Moreover, we can show that the solution of (1.5) is unique. If u is the solution of (1.5) for ω = 0, by the classical theory for elliptic equations (see [17]), we have that e u = 0 on B1 is the only solution of the following Dirichlet problem:

 

div e−



u|S n−1 = 0.



|x|2 4

 ∇ u(x) = 0,

on B1

(A.3)

So, u = 0 on B1 . Assume that u = k=0 fk (r )ϕk (θ ) where fk (r ) = S n−1 u(r , θ )ϕk (θ )dθ satisfies (2.10). Then fk (r ) = 0 for all 0 ≤ r ≤ 1 and k ≥ 0. Therefore fk (r ) = 0 for r ≥ 1 is the unique solution for (2.10) with fk (1) = 0, f 0 (1) = 0. Then u ≡ 0, from which the uniqueness is proved. 

P∞

R

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