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Non—existence of positive solutions to semilinearelliptic equations on r” or r” through the method ofmoving planesBianchi Gabriele aa Dipartimento di Matematica, Universitá di Ferrara, Via Machiavelli 35, Ferrara,I–44100, ItalyVersion of record first published: 18 May 2010.

To cite this article: Bianchi Gabriele (1997): Non—existence of positive solutions to semilinear elliptic equations on r” orr” through the method of moving planes, Communications in Partial Differential Equations, 22:9-10, 1671-1690

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COMMUN. IN PARTIAL DIFFERENTIAL EQUATIONS, 22(9&10), 1671-1690 (1997)

NON-EXISTENCE OF POSITIVE SOLUTIONS TO SEMILINEAR ELLIPTIC EQUATIONS ON Rn OR R;

THROUGH THE METHOD OF MOVING PLANES

Gabriele Bianchi*

Dipartinlento di Matematica, Universitk di Ferrara, Via Machiavelli 35, 1-44100 Ferrara, Italy

0. INTRODUCTION

In [6] Ding and Ni proved the following result for the problem

Theorem A. Let p = and K( r ) E COIO, +oo) be non decreasing and not identically constant. The11 problem (0.1) has no radially symmetric solution.

This result was later generalized to any p > 1 by T. Kusano and M. Naito [15] who proved that when -& ( r ( n+2 -p (n -2 ) ) / 2~ ( r ) ) > 0, $ 0 problem (0.1) has no radially symmetric solution.

A natural question, asked by Li and Ni in [lo], is whether under the hypotheses of Theorem A, (0.1) has non-radial solutions. We are able to give a negative answer to this question.

Theorem 1. Let K(T) be non negative and C1(O, +m) and let o = n + 2 - p(n - 2) . If either p > and -& ( r °K(r ) ) 2 0, or 1 < p 5 and & ( r 0 / 2 ~ i ( r ) ) > 0, $ 0 then problem (0.1) has no solution.

- -

"E-mail address: bia@unife.it

Copyright O 1997 by Marcel Dekker, Inc.

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When 0 < p < 2, -& ( r u I i ( r ) ) 2 0, $ 0 and as r + 0 IC(r) = ~ ( r - ~ ) , then any solution to (0.1) is radially symmetric about the origin.

It follows thus, for instance, that there is no nontrivial entire positive solution for Au + I x J ~ u P = 0 when p > s, y 2 -a and also when 1 < p 5 s, 7 > - L T / ~ . Theorem 1 is a consequence of the following symmetry result.

Proposition 6. Let u be apositive solution ofAu+g(lxl, u ) = 0 in R'"{O}, which belongs to CO(!Rn) n C2(R" \ (0)). Let us assume that

(i) g ( r , t ) 2 0 is locally Lipschitz in t > 0 and continuous in r > 0, and, for any compact set H c (0, m), s ~ ~ ~ , ~ E H , ~ + ~ M r , t ) - g( r , s)) l(t - s ) = o(p.-2) as r -+ 0:

(ii) g(r , t ) is non decreasing in t for any r > 0, (iii) g ( l / r , r , n + 2 t ) / rn+2 is non increasing in r for any t > 0,

Then Ir /2-'Lu(x/ /xi2) is radially symmetric about some point. Moreover if / x / " % ( ~ / l x / ~ ) is not coiltinuous a t the origin then u is radially s;vmmetric about the origin. This last conclusion holds also if there exists F 2 0 such that for any t > 0 the function in (iii) is not constant in any neighborhood of 1;.

It is known that a solution u to (0.1) is radially symmetric if I i ( r ) is decreasing and u decays to 0 at infinity fast enough, see for instance Ni and Li [ l l ] . The novelty of proposition 6 is that the nonlinearity g is not required to be monotone decreasing with respect to r. and that 110 decay condition is required on u . The generalized Matukuma equation

( A > 0) proposed as a mathematical model in astrophysics. see [12], is an example of an equation where, for some X and p, the coefficient of UP is not monotone decreasing and proposition 6 applies. This equation also shows that the non existence part of theorem 1 for 1 < p < ( n + 2)/(72 - 2) is optimal: ru t2 K ( r ) is 11011 decreasing if and only if p 5 1 and it is known that for X = 2, p > 1 there are solutions. Another symmetry result in the same spirit is theorem 2.

Theorem 2. Let u be a positive solution of Au + K(lxl)uP = 0, p > 0, in R" \ {0), in the class C2(Rn \ (0)).

Let us assume that K ( r ) is non negative and rontinuous in (0, m) and that there exists f > 0 such that

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is non-increasing in ( 0 , r] and non decreasing in [ F , co). If the function in (0.2) is not identically constant then u is radially sym-

metric about the origin. If the function in (0.2) is identically constant and if u , or its Kelvin trans-

form: is not continuous in the origin the same conclusion holds.

Note that when p E (1, s] any I< non-increasing in (O,r] and non de- creasing in [F, w) satisfies (0.2) , that this result applies also to solutions to (0.1) which are singular in the origin, and that it holds for any p > 0 . The- orem 2 generalizes a result proved by the present author in [ I ] , in the case p = 5 and u ( x ) = 0(1~1~-") at infinity.

The method of proof of theorem 2 and of proposition 6 is relatively sim- ple: we use the moving plane method to prove symmetry of a suitable Kelvin transform of a solution. This Kelvin transform might be singular in some point but when the nonlinearity in the equation is increasing in u , this prob- lem ran be handled without much difficulty.

For the equation Au + g(u) = 0, where

g is locally Lipschitz in t > 0 , g is non negative, and (0.3) 232

g( t ) / t n - 2 is non increasing,

the following result holds.

Theorem 3. Let u be a C2 positive solution of A u + g ( u ) = 0 in Wn, where g satisfies assumption (0.3).

Then either u is constant and g vanishes on that constant or

for some positive constants k, h and some point xo E Rn and g ( t ) is a suitable 232 positive multiple o f t n-2 for any t E (0, mazRn u ] .

Note that in theorem 3 g is not required to be increasing. In [3] Caffarelli, Gidas and Spruck proved the radial symmetry of all entire

positive solutions of Au + g ( u ) = 0 when g satisfies (0.3) and other extra assumptions. They could then prove, via ODE arguments, that all positive solutions of Au + U P = 0 , p = ( n + 2 ) / ( n - 2) . have the form (0.4) . Gidas and Spruck had already proved in [17] that the same equation does not have any positive entire solution for 1 5 p < ( n + 2 ) / ( n - 2) . A simpler proof of all this was later given by W. Chen and C. Li [4]. Theorem 3 generalizes and further simplifies these results.

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The proof is the following: - use the techniques introduced for proposition 6 to prove that any (i.e.

centered at any point) Kelvin transform of a solution is radially symmetric, - apply the following lemma (we state it in a simplified form):

Lemma 7. If a functzon u , defined zn Rn, has the property that two suztable Kelvzn transforms of u are radzally symmetrzc, then either u zs constant o r (0.4) holds.

The precise statement of Lemma 7 is in sectio113 and its proof is absolutely elementary.

After we essentially completed our work we learned that C. C. Chen and C. S. Lin [5] proved theorem 3 under the extra assumptions g increasing,

ze g ( t ) = O(tn-2) as t -+ m.

Under some extra assumptions theorem 3 can he generalized to allow g to depend also on 1x1.

Theorem 4. Let u be a C2 positive solution of Au + g((x1, u) = 0 in R", where g ( r , t ) satisfies the assumptions ofproposition 6 and it is non decreasing in 7. Then either u is constant and g r 0 or

for some real constants k > 0, h and some point xo E Rn and g(r , t ) is a &

suitable positive multiple o f t n-2 for any t E (0, n z a x ~ n u ] .

In the case g ( r , u) = I<(r)uP p 2 this result coincides with Theorem

1, while it is weaker for 1 < p < s, in that it requires I< be non decreasing. WP remark that also the proof of this theorem, as well as that of theorem 3. does not depend on any ODE argument or on any Pohozaev identity and is based solely on repeated application of the nloving plane method and on the previously mentioned Lemma 7. This provides a completely different proof of Theorem A.

These ideas, combined with some technical lemmas in [13] and in [16], ran be used to prove a result similar to Theorem 3 for the corresponding nonlinear Neumann problem on the half space EX; = {x, > 0)

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Theorem 5. Let u be a C 2 ( R T ) n C 1 ( i Q positive solution ofproblem (0.5). Let us assume that g and h are locally Lipschitz in ( 0 , fm), g 2 0 is non decreasing, g( t ) / t ( n+2) l (n -2 ) and -h ( t ) / t n l (n -2 ) are non increasing.

Then one of the following assertions is true: - there are some constants a 2 0, b > 0 such that

u ( x ) = a x , + b , g ~ 0 , and h ( b ) = a ;

- there are some constants k > 0, d < 0 and some point xo E { x , < 0 ) such that

and h ( t ) = dt V t E ( 0 , max u ] ; aiWIf

moreover dk2 / (n -2 ) = ( 2 - n) ) ( zO) , l , where ( z o ) , is the n-th component of xo;

- there are some constants k > 0 , p > 0 , c > 0 , d and some point zo E R n such that

k u ( 5 ) = n-2 ,

( I " - "012 + p 2 ) , 3%

g ( t ) = ct "-2 V t E ( 0 , max u ] , h ( t ) = dt* Vt 6 (0 , max u ] ; "7 arw;

moreover ~ k ~ / ( ~ - ~ ) = ( n - 2)nP2 and bk2/ (n-2) = ( n - 2) ( zo )n , where ( x o ) , is the n-th component of xo.

When h ( u ) = and either g is identically zero or or g ( u ) = cup, 1 < p 5 ( n + 2)/(7% - 2 ) this result is already known, it has been proved in part by B. Hu in [7] and in part by Y. Y. Li and M. Zhu in [13]. Here the method of proof is new, although similar in spirit.

After completing the manuscript the author was informed by Yan Yan Li that Y. Lou and M. Zhu proved in [14] that all non-negative solutions of Au = 0 in Rn+ with boundary condition duldz, = u4 are linear functions of x , when q > 1. This is contained in theorem 5 only if q 2 n / ( n - 2). In their proof the result for q E (1, n / ( n - 2 ) ) follows from the analogous result for q E [ n / ( n - 2 ) , co); when q is sub-critical they put the problem in an higher dimensional setting, making the power q super-critical in that setting (if u denotes a solution and m > n is a positive integer the func- tion u ( y 1 . . . . , y(,-,), xl , . . . , z,) := u(rc1, . . . , x,) solves the same problem in Rm+) . This idea implies that theorem 5 holds also if the hypothesis 1 'g( t ) / t (n+2) l (n-2) and - h ( t ) / t n l ( n - 2 ) are non increasing "is substituted by "for some m > 7a g ( t ) / t (m+2) l (m-2 ) and -h(t) / tml(n"2) are non increasing ,I

Finally we mention that (21 contains some non existence results for the problem nu + ~ < ( z ) u ( ~ + 2 ) / ( n - 2 ) = 0 in { x n > O), du ldx , + h ( ~ ) u ~ l ( ~ - ' ) = 0 on {x, = 0).

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1. PROOFS OF THEOREMS 1 A N D 2 AND OF PROPOSITION 6

Proof of Theorem 2. To prove that u is radial it suffices to show that u is axisy~nmetric about any axis through the origin. Let 1 be such an axis: without loss of generality we can assume that 1 coincides with the x, axis.

Let 5 = (0 , . . . ,0 , F) and let us consider the following conformal transfor- mation of Rn:

The associated Kelvin transform of u

is a solution of ~ . u + l ; ' v ~ = O in IR7"{3,0)

The function u is positive, C2(IW'' \ {2,0)) (0 denotes the origin), decays to 0 at infinity as j ~ 1 ~ - ' h n d may have a singularity at 3: and 0. Proving that u is axisymrnetric with respect to the x, axis is equivalent to prove that v is axisymmetric with respect to the z, axis.

Let us write z' = ( z l , . . . , z , - ~ ) . The function K depends only on lz'l and 2,: let us prove that it is non increasing as a function of lz'l It suffices (by approximation) to prove it under the assumption that A' E C1. The inequality ar;'/a\z1\ 5 0 is equivalent to

Writing r. := T ~ z I / I z - 31 the previous inequality becomes

~2 - r 2 K t ( r ) - + ( ( n - 2 ) p - 11 - 2)Iz'(r) 2 0

r

for any r > 0. and this is equivalent to the assumption on K ( r ) made in the state~nent .

To prove that 1, depends only on \ztl and z, we will use the moving plane method. This method is almost standard now, the only new difficulty here is due to the fact that v may be singular in some point. This part of the proof follows quite closely that of Theorem 4 in [8]. D

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For A E R and for any z E Rn let zX = (2X - i l , 2 2 , . . . , zn), C(X) = {Z E R n : zl < A) \ {sX, 0'). Finally let w(z, A) = v(zX) - v(z). The function w satisfies

where t belongs to the interval whose endpoints are v(z) and v(zX). Let g(2) = I z I - ~ , CY E (0,n - 2), and U(Z, A) = W(Z, A)/g(z). The function satisfies

(1.2) v g

A + 2 - ( ) + ( ( + ( z ) ) ( z ) 0 in 6(A). 9

First step. We prove that w 2 0 in C(A) for X < 0, \ A \ large enough. Suppose the contrary, then infc(X,) w(z, A , ) < 0 on a sequence A, 4 -m.

If z is large this infimum can not be attained by a sequence converging to 5': or to 0". In fact, for instance, liminf,,,~ w(z, A) = liminf,,, v(z) - ~ ( 5 ' ) > 0 for A < 0, / X I large, since v is non negative and superharmonic and v(z) - 0 as 2 -+ m.

Also W ( Z , A) -+ 0 as z + OC, and thus, for each z, the previous infi- mum is attained at a point q, E C(X,). In these points Aw(q,, A , ) > 0 and Vwjq,, A,) = 0: in order to prove that (1.2) is violated in q, and get a contradiction, it suffices to prove that the coefficient of G(., A,) in (1.2) is negative.

The sequence (q;') is not bounded (otherwise ~(q:', A,) would be even- tually positive) and thus the number t appearing in (1.2), which belongs to [v(4,X1),v(q,)], is O(lq,)2-n). Therefore

Second step. Let p = sup(p 5 0 : w(z, A ) < 0 in C(A) for any X 5 p } . Let us prove that if fi < 0 then w(z,p) E 0 (that is v is symnletric with respect to the hyperplane 21 = fi) . Suppose that fi < 0, then there exists a sequence of X i in (fi, 0) converging to P such that for any i infc(X, ) ~ ( z , A,) < 0.

The function w(z, p ) is non negative in C ( j i ) , therefore, by (1.1), it is su- perharmonic. If it is not identically 0, liminf,,j.c. w(z,p) > c > 0 and thus also liminf,,,~, w(z , A,) > 0 for i large enough. Similarly liminf,,,~, w(z, A,) > 0 for i large enough.

The infimum of w(z, A,) in C(Xi) is thus attained at a point qi. Repeating the arguments used in the first step one proves that the sequence (q;) stays

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bounded. A subsequence, that we still denote by (q,), converges to some point q E C(p) where w(q,p) = 0, Vw(q,p) = 0. This and the fact that w(z, 0 ) 2 0 imply, via the strong maximum principle or the Hopf bou~ldary lemma, that w(z, p ) z 0.

Conclusion. Let us prove that

(a) if I<(r)l? - F' l(n+2-(n-2)p)/2 is not identically constant,or (b) if U(Z), or 1~1 ' -~u(x / ( jx1~) ) , are not continuous in the origin,

then ,ii = 0. Suppose the contrary, then v(z) is symnletric about the hyper- plane {zl = p) and thus k = -Au/uP has the same symmetry. It is easy to see that, if ji # 0, this happens only if K(r)1r2 - is identi- cally constant. This proves the assertion in case (a). Similarly if (b) holds 11 is not continuous at the origin or at 3. and thus it can not be symmetric with respect to 21 = p when ,ii f 0.

Therefore p = 0 and v(z l ,zz , . . . ,z,) 5 v(-z l ,zz , . . . ,z,) for any z with 21 < 0. Applying the same reasoning to the function u(-zl, 22,. . . ,z,) one concludes that v is symmetric with respect to zl = 0. Similarly one ran prove that v is symmetric with respect to any hyperplane containing the z, axis. 0

Proof of proposzteon 6. This proof is similar to the previous one: in this case applying the moving plane method to the function v defined as the Kelvin transform of u about the origin one proves that u is radially symmetric about some point Q. We omit the details. We just remark that the assumption about the growth rate of sup,l,12EH (g(r, t l ) - g(r, t z ) ) /( t l - t2) as r -+ 0 is necessary to prove that in the equation which corresponds to (1.2) the coefficient of tc is negative when z is large.

Clearly if Q is not the origin then v is continuous at the origin and more- over, for any s > 0, the function g ( l / r , rn-'t)/rn+', where t is the value of 21 on the sphere {jx - Ql = A), is constant in [IIQI - s I , + S] .

Proof of theorem 1. Let g ( r , t ) = K(r)tP and let 1~ be a solution to (0.1). In this case g ( l / r , rn -2 t ) / tn+2 = K(l / r ) r -" . Moreover the assumption about the behaviour of r u I i ( r ) or of rO/'K(r) implies implicitly that, when p > 1, I<(T) = o(r-') as r -t 0. Thus proposition 6 applies and u is radially symmetric about the origin, except possibly when p > n + 2, h'(r)ru is constant and the Kelvin transform of u is continuous at the origin. Except for this case the non-existence part of theorem 1 follows from the results proved by Ding and Ni and by Kusano and Naito quoted in the introduction.

To deal with the case which is still open, let u denote the Kelvin transform of u: by proposition 6 v is radially symmetric about some point Q. More- over (1 is continuous a t the origin and it satisfies problem (0.1) with h'(lx1)

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substituted by K ( l / l x ( ) l ~ l - ~ , which is identically constant. This problem is thus radially symmetric about Q and the result of Kusano and Naito proves that it has no solution.

2. PROOFS OF THEOREMS 3 A N D 4

We start with two lemmas.

Lemma 6. Let u be a positive solution of Au +g(u) = 0 in Rn \ {0), which belongs to CO(Rn) n C2(JRn \ (0)). Let us assume that g satisfies assumption (0.3). Then 1x12-nu(x/lx(2) is radially symmetric about some point.

Proof. This proof also is similar to that of Theorem 2. Since g( t ) is not assumed to be increasing in t, one new ingredient is needed to prove that, when @ < 0, either w(x,p) r 0 or liminf,,ori w(x,p) > c > 0, (we use the notations introduced in the proof of theorem 2). We only show how to prove this and we omit all the other details. The function w ( x , p) := v(xP) - v(x) satisfies

and is non negative in C(ji). Suppose that w(x,p) 0 and lim i n f , , ~ ~ w(x, 0) = 0. Then w(x, @) is not superharmonic and therefore the right hand side of (2.1) is positive on a sequence xk 4 O P . Since v(xP) 2 v(x)

3% in C(@) and g(t)/ t n - 2 is non increasing, it should be

that is v -+ m on the sequence x: -+ 0. A result proved by C. Li (Theorem 2 in [ 9 ] ) implies that a non negative solution of Au(x) +g(JxJ'L-2u(x))/ Ixln+2 = 0 in (0 < 1x1 < 1) is asymptotically radial, i.e. u(x) = ~ ( l x J ) ( l + o(1)) as x -+ 0, where a ( r ) is the average of u on (1x1 = r } . Thus, b y (2.2), u(x) 4 m

as x -t 0: this contradicts the assumption liminf,,o~ w(z, p ) = 0. 0

Lemma 7. Let u be a function defined on Rn and let us suppose that either

(i) u is radially symmetric about some point P and also its Kelvin trans- form centered in a point different from P is radially symmetric about some point Q or

(ii) the Kelvin transform of u centered in the origin is radia1l.y symmetric about some point P and also the Kelvin transform of u centered in a point T , T # P , llTl1 = 1, is radially symmetric about some point Q

then either u is constant or u(5) = k / ( j x - x 0 1 ~ + h ~ ) ( ~ - ~ ) / ~ for ~ o m e constants k , h E R and some point xo E Rn.

Proof. Let us prove the result under assumption (i) .

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Let vo denote the Kelvin transform mentioned in the statement. To sim- ~ l i f y the notations let us assume that the distance from P to the center of inversion of uo is 1, in the general case the proof works similarly. After a suitable rotation and translation we may suppose that P = (0 , . . . ,0 , I ) and that the center of inversion of uo is the origin. The symmetry about P implies that u ( x ) = f (1s - P I ) for a suitable function f . The confor- mal transformation z = x/jx12 maps the sphere Ix - PI2 = r2 in the sphere S ( r ) = { / z - ( 0 , . . . .0 ,1/ (1 - r 2 ) ) I 2 = ? / ( I - r 2 ) ' ) , when r # 1 , and in the hyperplane S(1) = { z , = 1/2), when r = 1. Thus, for any r 2 0,

(2.3) u,(z) = f ( r ) ) z i2 -" for any z E S ( r ) .

This relation implies that vo is axisymmetric about the x , axis and that the center of symmetry Q of vo lies on the z,, axis unless u E 0. Let Q = ( 0 , . . . , 0 , q ) . If q > 1 or q < 0 then, for a suitable value of v, the sphere S ( r ) is centered in Q and (2.3) contradicts the radial symmetry of uo about Q. Therefore q E [O,l].

The constant value of vo on a sphere centered in Q is determined, through (2.3). by the value of vo in the intersection of that sphere with S ( r ) . More precisely for a given r let I ( r ) be the interval {c > 0 : B(Q, c) n S ( r ) # 0) (B(Q, c ) denotes the sphere centered in Q and with radius c ) . Then for any 2 such that 1: - Q J E I ( r ) we have

In particular this representation implies that if the two intervals I ( r l ) and I ( Q ) intersect for rl # r2 then the value of f ( r ) ( l - q + p 2 ) * at rl and 1 ~ 2 must coincide. On the other hand we claim that given m y two positive numbers rl < 7-2 it is always possible to find a finite number of intermediate value S, = T I < ~2 < . . . < S , = 7.2 such that I ( s , ) n I ( S , + ~ ) f 0 (or, equivalently, such that for each t t h e is a sphere in the family S ( r ) that intersect both B ( Q , s,) and B ( Q , s,+l)). This is an elementary fact which is due to the relative positions of the spheres centered in Q and of those in the family S ( r ) .

Then f ( r ) ( 1 - q + q r 2 ) does not depend on r and

for some constants k and q E [O, 11

Let us now prove the lemma under assumption (ii). Dow

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We may suppose that T = (0, . . . ,0,1). Let vo be as before and v~ be the Kelvin transform of u centered in T. The function v~ is also the Kelvin transform, centered in the origin, of uo(xl, . . . , x,-l, l -I,). This last func- tion is centrally symmetric about some point different from the origin (more precisely about the point (pl , . . . , p,-l , l -p,), where P = (pl , . . . , p,)). One can then apply part (i) of the Lemma to UO. Writing u in terms of uo gives the desired result.

Proof of Theorem 3. By Lemma 6 any Kelvin transform of u, centered at any point, is radially symmetric about some point. Then lemma 7(ii) implies the thesis. 0

Proof of theorem 4 . Let us argue by contradiction. Let u be an entire positive solution of Au+g(Ix(, u) = 0 and let v(s) be a Kelvin transformof u centered at the origin. By proposition 6 v is radially symmetric about some point Q. Let us prove that v satisfies the assumptions of lemma 7(i). This will conclude the proof since the conclusions of Lemma 7 are invariant under Kelvin transformations and thus not only v but also u satisfies the conclusion of Lemma 7.

We may suppose that Q belongs to the non negative x, semiaxis. Let us make a Kelvin transform w of v choosing a point T on the positive s, semiaxis as the center of inversion. w is axisymmetric w.r.t. the x, axis , to prove that it is radially symmetric it suffices then to prove that it is also symmetric w.r.t. some hyperplane whose normal 0 forms an angle with the x, axis which is an irrational multiple of n-. To prove this we use the moving plane method.

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1682 BIANCHI

Let us introduce some notation. The conformal map z ( x ) associated to the Kelvin transform centered in T maps the origin (where v may be singular) in H := T - T / I T I ; it maps Q (the center of symmetry of v ) in S := T + ( Q - T ) / I Q - T I Z , moreover Ix - QI = d l z ( x ) - S I / I z ( x ) - TI, (d := IQ - T I ) .

The radial symmetry of v is thus equivalent to

It is important to remark that thv point S lies in between T and H, see fig. 1. The function w ( z ) is a C2(Rn \ { H ) ) solution of

Let us suppose also that 6' points towards { z , > 0). For X E R and for any z E R" let z X = z + 20(X - 2 . 6 ' ) be the symmetric

point of z with respect to the hyperplane z 6' = A; let C ( X ) = { r E Rn : 2 . e L: A) \ { H A ) .

Let us start the moving plane procedure proving that w ( z ) 5 w ( z X ) in C(X) when X = X 1 := 8 . T . Let now z E C ( X I ) : the relative position of the points S and T with respect to the hyperplane z .6 = X I implies that Iz- S l /d l z -TI < jzAl - SJ /d l zA l - TI . The function v ( r ) is superharmonic and consequently non increasing: therefore v( l z - S l /d l z - T I ) 5 v(IzX1 - Sl/dlzxl - TI ) . Since Iz - TI = / zAl - TI (2.4) concludes the proof of the claim.

Let us now zncrease X from X 1 up to X2, where X2 = H.8 . Let j!i = SUP{^ E [X,,X2) : w ( z A ) 2 W ( Z ) 2n C ( X ) for any X E [XI,p)) and le t us prove that P < S . 6 < X2. To prove this we compute, using (2.4),

azo lim [ z - TIn- = (2 - n ) ( X - T . 6')v(d) - ( S - T ) . 0dv1 (d ) . r.8=X,lzl-tw do

The function r."*v(l/r) is superharmonic and bounded from below, there- fore either it is constant or its derivative is negative for r large. In the first case v already satisfies the conclusion of Lemma 7 (and there is nothing to prove), in the second case let us choose T in such a way that the previous de- rivative is negative in d = I H - TI . This implies that (2 - n ) v ( d ) - d v l ( d ) < 0 and therefore the previous limit is negative for any X < S . 8, X close to S . 8. Thus f?$ < 0 in some point on the hyperplane z . 8 = X and for those values of X w ( z X ) < w ( z ) in some point of C(X) . This proves that 1 < S .6'.

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METHOD OF MOVING PLANES 1683

Let now X E [ A ] , Xz] and let z E S(X). Since (z - HI 2 IzX - H I , lz - T J 5 /zX - TI the assumptions on f imply that

for any s > 0. Via this inequality one can prove. as in the step 2 of the proof of Theorem 2, that v is symmetric with respect to the hyperplane z . O = D . 0

The next lemma is the equivalent of lemma 7 in an half space. By lQ we denote the line parallel to the x, axis passing through a point Q E a R 3 . Moreover for any point P = ( P I , . . . ,p,) let P' denote ( p l , . . . ,p,-1).

Lemma 8. Let u be a function bounded and positive on and let us suppose that either

(i) u is axisymmetric about some line l p and also its Kelvin transform centered in a point which belongs to dR; and which is different from P is axisymmetric about some line lQ or

(ii) the Kelvin transform ofu centered in the origin is axisymmetric about some line 1 and also the Kelvin transfornl of u centered in a point T E dR;, T # P. JJTII = 1, is axisymmetric about some line lQ

then either

- u depends only on z, or - there exists a point so E dR;, a constant X > 0 and a fami1.y of

disjoint spheres S ( d ) := {Jz ' - (zo)'I2 + (x,, - d)2 = d2 - X2) (d E ( A , +m)) whose union is W;, such that on each sphere S ( d )

for some constants k = k(d) > 0 (see fig.2).

Remark. The last assertion of the previous lemma is equivalent to the radial symmetry about some point of a suitable Kelvin transform of u (see the proof of theorem 5, second case, for the details).

Proof. Let us prove the result under assumption (i).

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BI ANCHI

FIG 2: the spheres S(d)

Let vo denote the Kelvin transforni in the statement. To simplify the notations let us assume that the distance from P to the center of inversion of vo is 1, in the general case the proof works similarly. After a suitable rotation and translation we may suppose that P = (1 ,0 , . . . , 0 ) and that the center of inversion of 110 is the origin.

The conformal transformation z = x/ lxI2 maps the ( n - 2)-dimensional sphere {jx' - P'j = a , x , = b ) ( a , b non-negative constants) in a (n - 2)- dinlensional sphere, which we denote by C ( a , b).

The symmetry of u implies that u ( x ) = f ( ( x l - P1l,x , ) for a suitable function f . Thus, for any a, b > 0,

(3.2) v o ( z ) = f ( a , b)lzj2-" for any z E C ( a , b)

By assumption vo is axisyn~metric about lQ . The set C ( a , b ) is symmetric about the hyperplane containing the zl and the z, axes. Thus (3.2) implies that Q lies on the zl axis. Let Q = ( q , 0, . . . ,0) . If q > 1 or q < 0 then, for a suitable value of a , the sphere C ( a , 0) lies on the hyperplane ( x , = 0) and it is centered in Q: (3.2) would then contradict the radial symmetry of v(zl, 0) about Q. Therefore q E [0, I].

The symmetry of vo implies that its value on each set (12'-Q'I = e, z , = p ) ( e , p non negative constants) is constant: if such a set intersect C ( a , b ) the value of vo on such a set it is determined, through (3.2), by the value of vo in the intersection of that set with C ( a , b). If for a given ( a , b) we denote by

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METHOD OF MOVING PLANES 1685

I ( a , b) the set of ( e , p ) such that {lz' - Q'I = e,zfL = p ) intersect C(a, b) one easily computes that for any z such that (lz' - Q'), z,) E I (a , 6) it is

In particular this representation implies that if the two sets I ( a l , bl) and

I(a2, b 2 ) intersect for ( a ] , bl) # (az , b2) then f(a1, bl)b: n-2)/2 - - (11-2)/2

f(az,bz)b, . For r > (1 - q)/q if q > 0, r > 0 otherwise, let B ( r ) denote the set

{ ( a , b) : 2br = 1 - q(1- a2 - b2)}. We claim that if ( a l , bl), (a,, b2) E B(r ) it is possible to find a finite number of couples (cy,,P,), 2 = 1 , . . . , n , such that (.I,&) = ( a ] , b ~ ) , ( ~ n , P n ) = (a2, b2) and I ( Q , , ~ , ) n I (a ,+ l , &+I) # 0 (or, equivalently, such that for each z there is a set of the form (1s' - & ' I = e, 2, = p ) that intersect both C(cy2, P,) and C ( U , + ~ , P,+,)). This is an elementary fact which is due to the relative positions of the (n-2)-spheres C(a, b) and of the sets of the form {lz' - &'I = e, z, = p}.

As a consequence

(3.3) f (a, b)b(,-')I2 is constant on B(r) .

We recall that ~ ( x ) = f(lxl - P1l,x,). When q = 0 (3.3) implies that u

depends only on x,. When q E (0,1] (3.3) implies that on each sphere {I.' - P'I2 + ( 2 , - rlqI2 = (r/q)2 - (1 - q)/q)

for some constants k = k(r) . Finally let d = r/q and X2 = (1 - q)/q: to conclude the proof of lemma 8(i) it suffices to observe that on each of those spheres x, is a multiple of X 2 + jx - PI2.

Lemma S(ii) follows from lemma 8(i) in the same way that lemma 7(ii) follows from lemnla 7(i). 0

Lemma 9. Let PL, g and lz be as in theorem 5. Then l s ( 2 - n u ( ~ / l x ) 2 ) is axisymmetric about some line parallel to the x, axis.

Proof. The function u(z) = Iz/2-nu(z/lz)" is a solution of the following problem

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1686 BIANCHI

which decays to 0 at infinity as and which might be singular in the origin. Theorem 1.1 in [16] implies that a solution to problem (3.4) whach zs regular at t h p orzgzn is axisymnletric about some line parallel to the z,, axis. This is proved through the moving plane method. Lemma 9 can be proved as Theorem 1 1 in [16] with the help of the following lemma. We omit the details.

For X > 0 and for any 2 E PS; let zX = (2X - zl, 2 2 , . . . . zn), C(X) = {z E IW; : zl 5 A). Finally let W ( Z , A ) = u(zX) - z)(r).

Lemma 10.

(i) The function v is larger than a positive constant in son~e neighbor- hood of the origin.

(ii) Let us suppose that for some X < 0 the function w(., A ) is non negative in C(X). Then either ro 5 0 or w is larger than a positive constant in some neighborhood of 0'.

Proof of Lemma 10. This proof i\ based on ideas presented in the proof of Lemma 2.1 in [13].

Let us prove the first claim. First let us remark that the assumptions on h imply that h(0) < 0. The constant 6 := n~inl, i , l , , ,>~ v is positive, since v is positive in IW; and v can not vanish on dR; \ {O), due to the Hopf boundary Lernma and to the sign of h(0).

Let c = m a x { h ( ~ / 2 ) / ( ~ / ? ) , 0) and, for 6 E (0. I ) , let

Let us prove that v 2 4 in (6 5 /zl < 1, z,, >_ 0). Since v - 4 is superharmonic it suffices to prove that v 2 0 on the boundary of that set. It can be easily checked that this is true on {lzl = 1, z,, 2 0) (where v 2 E 2 4) and also on {I:/ = 6, zn 2 0) (where q!~ 5 0, if b is chosen small enough). Let us assume, by contradiction, that mins<l,l<l,,,,O(v - 4) is negative. Let z0 be a point where that minimum is attained. Then

On the other hand the derivative of (v - 4) with respect to z, should be non negative in zo, that is h(lzoln-2v(zo))/)zoln 2 ~ / 2 . Since h(t)/t"l('L-2) is non decreasing this implies

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METHOD OF MOVING PLANES 1687

Due to the definition of c this inequality contradicts (3.5). We have thus proved that v 2 4 in ( 6 _< ( z ( 5 I , z , > 0) . Letting 6 + 0 we obtain

which concludes the proof of the first claim. The second claim can be proved similarly. Since v ( z X ) 2 v ( z ) in C ( X )

and g is non-decreasing the function w ( z ) := v ( z X ) - v ( z ) is superharmonic. - Moreover it satisfies the following boundary condition on C ( X ) n ( 2 , = 0 ) \ {OX)

Since h(t) / tnl("- '1 is non decreasing, (3.6) implies that

at any point z E C(X)n { z , = 0) \ (0 ') where v ( z X ) > v ( z ) , and

at any point z E C(X)n { z , = 0 ) \ {OX) where v ( z X ) = v ( z ) To simplify the notations let us assume that X < -1: in this case the

distance from 0' to the hyperplane { z l = A) is larger than 1 and B + ( o X , 1) := {0 < jz - 0'1 < 1 , z , 2. 0 ) is contained in C(X) .

Let E := m i n l z - o ~ I = l r Z n ~ O w: if w + 0 then E > 0. This is true since w can not vanish in any point in the interior of C ( X ) due to the strong maximum principle and w can not vanish on dR; \ {OX) , due to the Hopf boundary Lemma and to (3.7).

We have already observed that v can not vanish on dlW; \ ( 0 ) : let m = rning+~ox, l )n~,n,o~ v . The function h is locally lipschitz in (0 , +m) and there- fore the supremum of

h ( l ~ ( ~ - ' t ) - h ( l z I n - ' ~ ) Iz17"t - S )

where t , s E [m, m + &/2], t > s, r E B + ( o X , 1) n { z , = 0 ) is bounded. Let c be that supremum and let 4 be defined as above (with 121 replaced by 12 - 0 ' 1 ) . The proof of the second claim of lemma 10 can now be concluded exactly as that of the first one.

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Proof of Theorem 5. Let u be a solution to problem (0.5). Due to Lemma 9, u satisfies the assumptions of Lemma 8(ii). We can thus distinguish two cases: u depends only on x,, or u satisfies (3.1).

First case If u = u(x,) then it satisfies

If g 0 then u ( . r R ) = a x , + b with a 2 0, 6 > 0 and a = h(b). If g $ 0 the previous problem has no solution. This is true since u is

concave and to be positivr on ( 0 : +m) it should he non decreasing. If this happens g ( u ) is larger than a positive constant (g is non decreasing) and the equation implies that d2u/dz :L is less t,llan a negative constant. Therefore u can not he no11 decreasing in [O, +m).

Second case In this case let us consider the conformal transformation

where T = ( P ' . -1). This transfornlation maps IW; in the ball lz - &I 5 2X, where Q = (P', A ) , and it maps each sphere S ( d ) in a sphere centered in Q (see (3.1) for the definition of S(d) ) . Moreover the identities U(Z) = k ( d ) / ( l x - Pi2 + X 2 ) ( n - 2 ) / 2 are equivalent to the radial symmetry about Q of the function

The function 1 ) satisfies the following nonlinear Neumann problem:

( I / denote the unit exterior normal to the boundary of / z - &I 5 2X). Since 1) is radially symmetric about Q also the nonlinear terms in the equation and in t,he boundary condition should be radially symmetric. Since T # & and the value of v 011 the boundary is different from zero (since h(0) 5 0) it should be

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METHOD OF MOVING PLANES 1689

for any t in the range { ( l z - T1/2X)n-2u(z) : lz - QI = 2X) that is for any t E (0, maxaq u]. Similarly

for some c > 0, and for any t E (0 ,max~; u]. If c = 0 then v is constant and

for some k > 0. A posteriori one computes that this satisfies the boundary condition of problem (0.5) if and only if d l ~ ' / ( ~ - ' ) = ( 2 - n ) IT,, 1, where Tn is the n-th component of T. Note that this forces d to be negative.

If c > 0 then, by the uniqueness for the ODE problem, u is a multiple of (12 - QI2 f o ' ) - ( ~ - ~ ) / ' for some a > 0. Expressing u in terms of v, using the relation 12 - Q J = 2Xlx - Q l / [ x - TI and by a simple computation, one gets

u ( x ) = k

( I " - x 0 l 2 + , 2 2 ) q

for some positive constants k and p and some point xo E !Rn. A posteriori one computes that this satisfies the equation of problem (0.5) if and only if k 4 / i n - 2 ) c = ( n -2)np2 and the boundary condition if and only if b k 2 / ( ' L - 2 ) = (n - 2 ) ( x 0 ) , , where ( x o ) , is the n-th component of " 0 . U

G . Bianchi, Non-existence and s y m m e t r y of solulions to the scalar curvature equalion, Comm. P.D.E. 21 (1996), 229-234. G . Bianchi, X. B. Pan, Yamabe equations o n half spaces (preprint). L. A. Caffarelli. B. Gidas, J . Spruck, Asympto t i c s y m m e t r y and local behavior of semtltnear elltptic equations wi th crztical Sobolev growth, Comm. Pure Appl. Math. 42 (1989), 271-297. W. Chen, C. Li, Classification of solutions to some nonlinear equations, Duke Math. J . 63 (1991), 615-622. C. C. Chen, C. S. Lin, Local behavior of singular posztive solutions of s emiknear elliptic equations w i th Sobolev ezponent , Duke Math. J . 78 (1995), 315-334. W. Y. Ding, W. M. Ni, O n the elliptic equation A u + ~ u ( " + ~ ) l ( ~ - ~ ) = 0 and related topzcs, Duke Math. J . 52 (1985), 485-506. B. Hu, Nonextstence of a posative solutaon of the Laplace equation wi th a nonlznear boundary condi t ton , Differential Integral Equations 7 (1994), 301-313. C . Li, Monotonic i ty and s y m m e t r y of solutions of fully nonlmear elliptac equations o n unbounded domains , Comm. P.D.E. 16 (1991), 585-615. -, Local asymptot ic s y m m e t r y of singular solutions to nonlinear elliptic equa- t ions . . Invent. Math 123 (1996), 221-231.

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[lo] Y Li, w. M . N i , O n c o n f o r m a l scalar curvature equat ion m Wn, Duke Math. J . 57 (1988). 895-924.

[ l l ] . O n t h e asymptotzc behavzor and radzal s y m m e t r y of positive so lu t ions of sernzlznear ellzptzc equataons i n Rn 11. Radial s y m m e t r y , Arch. Rational Mech. Anal. 118 (1992), 223-243.

1121 ___. 0 7 1 t h e exzstence and s y m m e t r y propertaes o f f ina te to ta l m a s s so lu t ions of t h e M a t u k u m a equataon, t h e Eddzngton equatzon, and the ir generalzrataons, Arch. Rat. Mech ,411. 108 (1989). 175-194.

[13] Y . Y. Li. M . Zhu, Unaqueness theorems through t h e method of m o v i n g spheres, Duke Math. J . 80 (1995), 383-417.

1141 Y . Lou, hI . Zhu, Classifacatxons of non-negat ive aolutaons t o s o m e ellaptic problem, . - Differential Integral Equations (to appear) .

[15] T. Kusano, h I . Naito. Oscillatzon theory of entzre solulzons of second ordersuper ianear ellzptac equatzons. Funkcial. Ekvac. 30 (1987), 269-28'2.

[16] S . 'I'erraciri S y m m e t r y propertzes of posztzue s o / u t i o m to s o m e slhptrc equatzons w i t h nonlzncar boundary condifzon,s. Differential Integral Equations 8 (1995). 1911-1922.

[ I T ] 8 . Gidas. J . Spruck. Global a n d local behaoaour of posilzve so lu t tons of nonlrnear e lkp tzc equatrons. Comrn. Pure Appl. Math. 34 (1981). 525-598.

Received January 1996 Revised June 1997

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