Electronic Journal of Di erential Equations, Vol. 1995(1995), No. 14, pp. 1-13. ISSN: 1072-6691. URL: http://ejde.math.swt.edu (147.26.103.110) telnet (login: ejde), ftp, and gopher acess: ejde.math.swt.edu or ejde.math.unt.edu
PICONE'S IDENTITY AND THE MOVING PLANE PROCEDURE Walter Allegretto and David Siegel Abstract. Positive solutions of a class of nonlinear elliptic partial di erential equa-
tions are shown to be symmetric by means of the moving plane argument coupled with Spectral Theory results and Picone's Identity. The method adapts easily to situations where the moving plane procedure gives rise to variational problems with positive eigenfunctions.
0. Introduction
Consider the problem:
?u = p(x)g(u)
in
(1) u=0 on @
where is a cylinder in Rn : = (?1; 1) 0 with 0 a domain (= bounded, open, connected set) of Rn?1 . Are all C 2 ( ) positive solutions symmetric in x1 ? If the boundary of 0 is reasonably smooth, then under suitable conditions on p; g the classic moving plane argument of Serrin, [21], as extended by Gidas, Ni, Nirenberg in [14], [15], Berestycki and Nirenberg, [5], Amick and Fraenkel, [3], and elsewhere, [8], [18], [25], applies and the answer is positive. More recently, [6], Berestycki and Nirenberg showed that the result is still true for general nonlinear equations if, for example, u 2 W`2oc;n ( ) \ C ( ) with no regularity assumed on @ ; while Dancer, [11], dealt with the case of u 2 H01;2 ( ) \ L1( ). It is the purpose of this paper to discuss the symmetry of positive solutions under somewhat weaker conditions than those that to the best of our knowledge have been applied earlier. We avoid direct use of pointwise considerations near @
by employing related arguments from Spectral Theory and Picone's Identity. In this way, we are able in particular to bypass various \corner Lemma" and Maximum Principle procedures. Most of the paper is devoted to the problem ?u = f (u) in
(2) u=0 on @
Subject Classi cation: 35B05, 35J60. Key words and phrases: symmetry, positive solutions, nonlinear elliptic, moving plane, Spectral Theory, Picone's Identity.
c 1995 Southwest Texas State University and University of North Texas. Submitted: April 15, 1995. Published: October 6 ,1995. Supported in part by NSERC (Canada)
1
2
W. Allegretto & D. Siegel
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with bounded, Rn (we consider explictly the case n 3); or = Rn itself since these cases illustrate all the ideas. While our approaches are di erent, our results are closest to those obtained by Dancer, [11]. We feel that one possible advantage of our approach is that it adapts easily to variational problems with positive eigenfunctions. This is illustrated explicitly for the case of = Rn ; and at the end we indicate some extensions to more general problems. We were unable to extend our approach to the more general nonlinear equations considered in [6]. Our results in particular show: Theorem 0. Let 0 < p(x) 2 L ( ); > n=2; symmetric in x1 . Assume p is nonincreasing in x1 for x1 0 and: g 2 C`1+ oc ; g( ) > 0 for 0. (a) For small enough, Problem 1 has a bounded (i.e. L1 ) positive solution. (b) All bounded positive solutions to (1) are symmetric with respect to x1 ; and if @u < 0 for 0 < x1 < 1. moreover p 2 Ln+" ( ) then @x 1
By a solution in Theorem 0 { and throughout the paper { we mean at least a function u 2 H01;2 ( ) \ C (K ) for any K ?b ; = (K ); which satis es the problem in the weak sense, i.e. B (u; v) = f (u); v for all v 2 H01;2 ( ) where B denotes the form associated with ?. We do not usually ask that u 2 C ( ). Our approach instead requires a variational linear structure and some higher integrability properties of u. The latter, in at least some cases, may be obtained either from the equation u satis es or from a-priori estimates used in nding u. Heuristically, what we do is related to the sucient condition (ii) of page 4 of [6] and to [7] but without the need for u 2 C [ ] or other speci ed behaviour at @ ; [7], thanks to the variational structure of the linearized problem. The regularity of @ is also irrelevant for most of our procedures, although we assume in most of the arguments that the domains obtained by re ecting about the moving planes are contained in
. In particular, this means that cannot shrink as we move in from the boundary by moving the planes, nor can have any \holes" in the regions involved in the re ection process. If f is not smooth, some assumptions on @ and/or u are added to ensure that now, u 2 C ( ); while for some f we show by the same methods, the symmetry and uniqueness of positive solutions without reference to the moving plane procedure. Some of these results were presented at the UAB-Georgia Tech. International Conference held in Birmingham, Alabama, March 12-17, 1994. 1. Preliminary Considerations
As mentioned in the Introduction, our procedure involves Spectral Theory and Picone's Identity. We begin therefore by considering a family of linear eigenvalue problems. Observe that we deal with functions in H01;2 ( ) in what follows. Without loss of generality we may assume they are de ned in the whole of Rn by means of the trivial extension. Given any function g; we set: g+ = max (g; 0); g? = max (?g; 0). Let f g2[a;b] be a family of bounded open sets in Rn ; their characteristic functions, and assume p : [a; b] ! L ( ). That is: p 2 L (Rn ); for 2 [a; b] and some xed > n=2. We then have:
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Picone's Identity
3
Theorem 1.
(a) For each 2 [a; b]; the map ` : H01;2 ( ) ! L2 ( ) formally given by ` = ? ? pI has a least eigenvalue 0 () to which corresponds a nonnegative eigenfunction u0 (); positive in at least one of the components of . (b) For any K b there exists ? > 0 such that u0 () 2 C (K ). (c) There exists a function ! 2 H 1;2 ( e ) ? H01;2 ( e ) \ C ( e ) with ! 0 and `(!) 0 a.e. e for any component e of i 0 () > 0. (d) 0 () = 0 i there exists at least one ! 2 H 1;2 ( ) \ C ( ); with ! 0 and `(!) 0 a.e., nontrivial in each component of ; and, for any such ! there exists in each component of a constant c 0 with c! = u0 (). (e) If kp kL is bounded, there exists a constant C0 such that if meas ( ) < C0 then 0 () > 0. Proof. (a) If need be we add a positive constant to p and observe that `? 1 de nes a compact self-adjoint map L2 ( ) ! L2 ( ) by Sobolev's Theorem since > n=2; [16]. The existence of the least eigenvalue 0 () then follows. That the associated eigenfunction u0 () has the desired positivity properties is immediate, since by the Courant min.-max. characterization of 0 () we conclude that u0 () 0; [16], and indeed it follows that u0 () > 0 in at least one component by the weak Harnack Inequality, [16]. (b) This is given in [16]. (c) It is here that we employ Picone's Identity. Once again by the weak Harnack Inequality, ! > 0 in since ! is assumed nontrivial in each component. Let ' 2 C01( ); " > 0. We recall Picone's Identity, see eg. [2]: Z
(! + ")2 r[
' ]2 = Z jr'j2 ? p '2 !+"
Z 2 2 r( !'+ " )r(!) ? p '!(+! "+ ") ?
Z Z p"'2 : 2 2 jr'j ? p ' +
! + "
Since j"=(! + ")j 1; and "=(! + ") ! 0 pointwise, we let " ! 0; apply Lebesgue's Dominated Convergence Theorem, followed by letting ' ! u0 () in H01;2 to conclude that either in each component e we have ec! = u0 () for some constant e c 0 or else 0 () > 0. The rst case is impossible since ! 2= H01;2 ( e ) for any e ; and we conclude o () > 0. On the other hand, if 0 () > 0 then ` = p has a solution 2 H01;2 ( ). Set ! = + 1 then ` (!) = 0 and Courant's min.-max. principle shows that if ! is nontrivial in a component then !? = 0 a.e., whence ! > 0 (again by Harnack's inequality). Finally ! 2 H 1;2 ( e ) ? H01;2 ( eR ) for otherwise 1 = ! ? 2 H01;2 ( e ). Since an equivalent norm on H01;2 ( e ) is ( e jruj2 )1=2 then C meas ( e ) k1k2H 1;2 ( e ) = 0 and the result follows. (d) If 0 () = 0 then for any such ! we conclude c! = u0 () by part (c). Observe that we may always construct at least one such ! by using the eigenfunction itself in some components and solving ` (!) = 1 in others. Conversely, since such a ! exists
W. Allegretto & D. Siegel
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then 0 () 0 by part (c). On the other hand, if 0 () > 0 then part (c) shows there exists a !; as desired, with ! 2 H 1;2 ( e ) ? H01;2 ( e ) whence c! 6= u0 () in at least one component. (e) We need only apply Sobolev's Estimate, [16] to obtain: Z
jr'j2 ? p (x)'2
Z
jr'j2 [1 ? K kpkL (meas ( ) )]
where = 2 n?n ; and ' 2 C01 ( ); and observe that kp kL is bounded. As a consequence we have: Corollary 2. If 0 2 C (a; b) with 0 () > 0 for all b ? small enough, and for each there exists a ! 0; dependent on ; in each H 1;2 ( e ) ? H01;2 ( e ) such that `(!) 0; then 0 () > 0 for all 2 (a; b). Otherwise there would exist 0 with 0 (0 ) = 0. The existence of such a ! then contradicts Theorem 1(d). The continuity of 0 { indeed of the entire spectrum { is a classical problem, discussed by Courant and Hilbert, [10], and studied more recently in a variety of papers: [4,13,20,26]. Based on these results we have: Theorem 3. Let = \ fx1 > g 6= ; for 2 [a; b] and let ` = ?u ? P u de ned on H01;2 ( ) where P 2 C ([a; b]; L ( )) for some > n=2; and is a xed bounded domain. Then 0 (); the least eigenvalue of ` in ; is continuous. If meas ( ) is small enough, 0 () > 0. ? ? By P 2 C [a; b]; L ( ) we mean P 2 L (Rn ) and P 2 C [a; b]; L (Rn ) where denotes the characteristic function of . Proof. Observe that these are nested domains. Choose 0 in the interval of interest, and without to a subsequence and suppose rst m # 0 ; with ? loss of generality pass = lim 0 (m ) . Note that 0 (m ) are bounded since > n=2. Let f!m g be associated, normalized in L2 ; eigenfunctions and observe that !m R 0 (!m > 0 if
is connected). By the trivial extension, k!m kL2 ( 0 ) = 1 and Pm !m2 can 0 be estimated. To see this, note that:
j
Z
Z
Pm !m2 j = j Pm !m" !m2?" j K kPm kL k!m k"L2 kr!m kL2?2 "
0
0
where " = 2 ? (n= ). Whence:
kr!mk2L2 K1 + K2 kr!mkL2?2 " : We conclude that k!m kH01;2 ( 0 ) kr!m kL2 ( 0 ) is bounded. Passing to a subsequence, also denoted by f!m g; we may conclude convergence (weakly) in H01;2 ( 0 ) and (strongly) in Lq ( 0 ); for q < (2n)=(n ? 2); to some ! 2 H01;2 ( 0 ). Obviously, ! 0; nontrivial, and `0 (!) = ! in H01;2 ( 0 ). We claim that = 0 (0 ). If 0 is connected, this is immediate by the positivity of ! { and again
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Picone's Identity
5
employing Picone's Identity. If 6= 0 (0 ); then 0 (0 ) < by the min.-max. principle. We conclude that there exists some ' 2 C01 ( 0 ) such that (`0 '; ') < ( ? 2")('; '). But since ' has compact support, then (`m '; ') < ( ? ")('; ') for the de nition of . all large m and some " > 0; i.e. 0 (m ) < ? " contradicting ? Suppose next that m " 0 and again set = lim 0 (m ) . The same procedure as above shows the existence of a subsequence !m and function !. We need to show ! 2 H01;2 ( 0 ). To see this, let g be a cut o function: g 2 C 1(R; R); with ? g() = 0 if < 2; g() = 1 if > 3 and set zm (x) = g (x1 ? 0 )=(0 ? m ) . We then have:
kzm !mk2H 1;2( 0 ) C k!mk2H 1;2( 0 ) + k jrzm j!mk2L2 ( 0 ) : The rst term is clearly bounded. For the second, note: Z 1 2 !m2 k jrzm j!mkL2 ( 0 ) C ( ? )2 0 m
m \fx1 0 if (x1 ; y 1) 2 Z \ K; if K b ; by the positivity of the solution u in the compacta of . Now let be the least number such that we have C = f(x1 ; y1 )j < x1 xe1 g e
. It follows that ( ; y 1 ) 2 @ e and C fxjx 2 C g . By de nition and assumption, ?" for " > 0 small enough, and we conclude that ( ; y 1 ) 2 . I.e. C and thus u > > 0 in C . This contradicts the absolute continuity of the integral and it follows that ! 2= H01;2 ( e 0 ) and thus 0 (0 ) > 0 by Theorem 1-d. Since 0 () > 0 and (u ? v )? 2 H01;2 ( ); then the min.-max. principle again implies (u ? v )? = 0; whence u v . The earlier arguments show that u 6 v in
; and then u > v . This shows part (a). As for part (b), we have u(x) u(x0 ) in 0 by continuity, since u > v on
; > 0 . However is symmetric in this case about x1 = 0 . If we rst perform a re ection?about x1 = 0 ; and then repeat the above procedure we would conclude u(x0 ) u (x0 )0 = u(x) and the result. Finally, for part (c), note that x0 2 and P 2 Ln+" ; imply that ! is, without loss of generality, in C 1+ (B ) for some ball B with x0 2 @B; [16]. Choose a function z 2 C 1+ (B ) such that ?z ? P z = 0 in B; z > 0 in B . Then by considering the equation !=z satis es we conclude, again by [16], that !=z 2 @! (x0 ) = 2 @u (x0 ) < 0. C 2 (B ) \ C 1+ (B ) and @x@ 1 (!=z)(x0 ) < 0; i.e. @x @x1 1 Next assume that f depends on x as well, i.e. f f (x; u). As was the case in the previous references, this situation can also be dealt with if we assume f is monotone in x1 : f (x; ) f (x ; ) for > 0. There is no signi cant change in the proofs. Note that we could thus deal with some cases where f had singularities with respect to x on fx1 = 0 g \ ; or with singularities along planes fxm = cg \ ; m 6= 1; as examples with f (x; u) = p(x)g(u) easily show, as long as the resulting P was in L ( ). One limitation in the applicability of Theorem 5 is given by condition (II). Ob-
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Picone's Identity
7
serve that since kP kL n+2 " is assumed bounded, then it suces that P (x) be pointwise continuous, a.e. in ; which will be immediately the case if f is smooth in u. This follows from the observation that if fgn g is a sequence of functions bounded in L and gn ! g 2 L pointwise then gn ! g in L ?" for any " > 0; and this is because we can nd a constant c {independent of n{ such that gn = gnc + gn0 with gnc = gn if jgn j c; jgnc j = c otherwise, and gn0 of small L ?" norm. We thus only need toR check the boundedness of P . Recall that we may also express P as ? 1 0 P = 0 f tu + (1 ? t)v dt and thus if, for example, u 2 L1 ( ) and f 2 C`1+ oc ; then Theorem 5 applies. Finally, observe that the results apply if f is Lipschitz (locally Lipschitz if u is in L1 ). This follows by setting p = [f (u) ? f (v )]=[u ? v ] if u(x) 6= v (x); p = 0 otherwise, and this example also shows that the continuity of 0 is really only needed from the left: If we let 0 be given by 0 () > 0 if > 0 ; as before, then we repeat the above arguments, in particular Theorem 1-c, and conclude that both 0 (0 ) > 0 and u > v0 . We next observe that P ! 0 P0 pointwise, and that jP j < C for some C; by the Lipschitz assumption on f; as ! (0 )?. We have then the contradiction 0 () > 0 for j ? 0 j small. 3. = Rn
The above approach also works for the case of = Rn and we now consider the modi cations needed to deal with this case. Detailed references of other results for this case may be found in [17]. Speci cally, assume ?u = f (x; u) weakly in Rn ; with 0 < u 2 E (Rn ); u ! 0R at 1. Here E ( ) denotes the closure of C01( ) with respect to the norm kuk2E = jruj2 . Our conditions on f are as follows: (II') Let p be de ned as in condition (II) and assume p : (0; 1] ! L ( ) continuously, with = n=2 and p1 f (x; u)=u; (III) f 2 C`1+ oc and f (x; 0) = 0; (IV) f (x; ) f (x ; ) for: x 2 ; > 0 and > 0; furthermore given any > 0; " > 0 there exists x 2 such that f (x; ) > f (x; ) for some 0 < < ". Observe that in the special case: f (x; ) = p(x) ; then condition (IV) holds if, @p ("; x) < 0 for 0 < " small enough. If u 2 C 1 ; say, for example, @x@p1 0; @x 1 decays fast enough at 1 then u 2 E (Rn ). We give explicit conditions for this to be the case, as well as other convenient results in Theorem 6.
Theorem 6.
(a) If u 2 C 1 and: u 2 L n2?n2 ; uf (x; u) 2 L1 then u 2 E (Rn ). (b) Let = fxjx1 < g; then ! = u ? v 2 E ( ) if u 2 E (Rn ). We recall that it is often possible to show that a solution 0 < u 2 E (Rn ) must, as a consequence, tend to zero at in nity, [1]. Proof of Theorem 6. (a) We observe by direct calculation: Z
Rn
jr('u)j2
=
Z
Rn
u2 jr'j2 +
Z
Rn
'2 uf (x; u)
kuk2L n2?n2 (supp jr'j) k jr'j k2Ln +
Z
Rn
juf (x; u)j
8
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where ' denotes a cut-o function: '(x) = g( jmxj ); 0 g 1; smooth, and g() = 1 if 1; g() = 0 if 2. As m ! 1 we have f'ug bounded in E (Rn ) and thus without loss of generality weakly convergent (to u) in E (Rn ). (b) Since u 2 E (Rn ); then v 2 E (Rn ) and thus ! by de nition. Furthermore if 'm 2 C01 (Rn ); 'm ! u in E (Rn ) then 'm(x) ? 'm (x ) 2 E ( ) since 'm (x) = 'm (x ) on x1 ; and f'm (x) ? 'm (x )g is Cauchy in E ( ); converging to u ? v . Sucient explicit conditions for (II0 ) to hold are: Lemma 7. Assume jfu(x; )j k(x) + h(x)? for 0 with k(x) 2 L n2 \ L1(Rn ); h(x) 2 Ls \ L1(Rn ) with s (2n )= 2n ? (n ? 2) ; then p : (0; 1] ! L ( ) continuously. Note that we require that 0 < 4=(n ? 2) in Lemma 7. This is because we do not postulate any speci c decay conditions at 1 on u; apart from u ! 0 and those implicitly associated with u 2 E . Note that the given bounds on s; reduce exactly to the sucient condition for existence as stated in [1], if e.g. f (x; ) = h(x) +1 . Proof of Lemma 7. We clearly have: Z
1
?
jp (x)j = j fu x; tu + (1 ? t)v dtj 0 C [k(x) + h(x)(juj + jv j )]: The continuity of u and f 2 C`1+ in Rn as ! oc imply p (x) ! p (x) pointwise for 2 (0; 1]. Since p is also uniformly bounded in L1 for 2 (0; 1]; then we immediately have p ! p in L (B ) for any xed ball B Rn . That kp ? p kL (Rn ?B) is small for B large is immediate by the integrability assumption on k; h and the embedding E (Rn ) ,! L n2?n2 (Rn ). We consider the eigenvalue problem: ?z = p z for z 2 E ( ). Observe that (?)?1 (p u) { for xed { can be viewed as a compact map E ( ) ! E ( ); [1], and thus there exist for each an eigenvalue and associated positive eigenfunction given by: (p '; ') : 1 = sup k'k2
'2E( ) '60
E( )
That > 0 follows from the next Lemma. In any case observe that f (x; u) must be positive somewhere, since ?u = f (x; u) and u > 0. Since we also have p ! p1 = f (x; u)=u pointwise, then p cannot be nonpositive for large. I.e. for such at least, > 0.
Lemma 8.
(a) Let ! = u ? v and ; exist then ( ? 1)(p ; !) 0. (b) is continuous in ; 1 and w > 0 for all > 0. Proof. (a) Observe that ` ! = ?! ? p ! = f (x; v ) ? f (x ; v ) = r(x) 0 whence (p ; !) = ( ; ` !) + (p ; !) and the result since > 0. (b) We show rst that 1. We claim that this is true for large enough since otherwise there exists a sequence m ! 1 for which this is not the case.
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Picone's Identity
9
But as m ! 1; pm ! p1 in L ; and thus m > 0 exists. Furthermore: as m ! 1; ! ! u pointwise and thus ! ! u in L (B ) for any large and xed ball B Rn by the uniform boundedness of !; u. Similarily, we note that m is bounded by the min.-max. principle and setting 1 equal to the limit of m ; we may assume that km kE = 1 and thus m ! weakly in E (Rn ) and strongly in 2n?" L n?2 (B ) where denotes a (positive) eigenfunction of ? = 1p1 in E (Rn ); since 6 0; as a consequence of 1 (p1 ; ) = 1. We thus have the existence of two positive eigenfunctions: ; u corresponding to the eigenvalues 1; 1 respectively. However, Picone's Identity also shows the simplicity of the eigenvalue associated with a positive eigenfunction such that (p1 ; ) > 0; and thus 1 = 1 and u. We conclude by part (a) that (p1 u; u) = kuk2E(Rn ) 0 if m < 1; which is a contradiction. Let 0 denote the least such that 1 and ! is nontrivial (and thus positive) for > 0 . The next arguments also show that if 0 = 0 we are done, hence suppose 0 > 0. Assume rst ! is nontrivial in 0 . By the presumed continuity, 0 = 1 since ! is also nontrivial for some < 0 ; and ?! ? p0 ! 0. We conclude (p0 0 ; !) 0 = (p0 0 ; !) 0 + (r; 0 ) 0 ; i.e. r 0 and since ! 2 E ( 0 ); it must be an eigenfunction corresponding to 0 . Indeed, again by Picone's Identity, if ! 6 0 then ! = c0 0 in 0 for some constant c0 . Note that this result applies to all for which = 1 and ! 6 0 in ; and, identically, if > 1 then (?!; !?) (p !; !?) whence (?!?; !? ) (p !?; !? ); i.e. ? ? 1 (kp!?!k2; ! ) E( )
and we have a contradiction unless !? = 0; i.e. once again, if ! 6 0 then ! 0 and ! > 0 by the weak Harnack Inequality. It follows that since for any > 0 we have 1 and ! is nontrivial then ! > 0 in ; ! = 0 on x1 = and @u thus @x1 < 0 ?if x1 > 0 . Now suppose ! is trivial in 0 and then observe ? f x; u(x) f x0 ; u(x) for x 2 0 ; contradicting assumption (IV) on f . Hence if 0 > 0; then 0 = 1 and ! > 0 is its associated eigenfunction. But this is impossible since then r 0 in 0 and again this violates (IV). Finally, the continuity of follows from the properties of p and the min.-max. de nition of . Speci cally, observe rst that if > 0 then p is somewhere positive and thus so is p for j ? j small ( suciently large if = 1). We conclude is bounded above and below, and set equal to any limit point of . The earlier arguments in this proof show that a subsequence of the normalized (in2nE ) positive ?" n eigenfunctions converges to weakly in E (R ) and strongly in L n?2 (B ). We again note that (p ; ) = 1; ? = p in ; and 0; nontrivial. If " then 2 E ( ) and again by Picone's Identity, = and = . If # then we need only show 2 E ( ); since the rest is the same. But this is immediate here by the smoothness of the (plane) boundary. The uniqueness of ; then shows the continuity. We then have under the above assumptions (II0 ); (III), (IV) on f : Theorem 9. If f (x; ) is symmetric (in x1 ) about x1 = 0; then u is symmetric in @u < 0 if x1 6= 0. x1 ; and x1 @x 1
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@u < 0 for Proof. We have by Lemma 8 that 1 and w > 0; i.e. u > v and @x 1 all > 0 and thus u v for = 0. Repeating the procedure for negative we obtain the result.
4. Extensions
We now brie y and heuristically comment on some extensions where the eigenvalue arguments and Picone's Identity still work. Observe that the same procedure can yield some modest results even for problems not involving purely Dirichlet conditions. Consider for example the cylinder (?1; 1) 0 with Dirichlet conditions on f?1g 0 ; f+1g 0 but Neumann elsewhere. The procedure in such a case is identical. Note that a key step involves the fact that the part of the boundary with Neumann conditions does not re ect inside the regions . Assume now that f is not smooth. Suppose rst as in [3], [14] that f = f1 + f2 with f1 smooth and f2 monotone increasing and continuous. Let f2 ( ) 0 if < c for some c > 0; and suppose is bounded. We now also require that u be in C ( ) \ C 1 ( ). This is similar to the requirements of [6]. If f1 (u); f2 (u) are in Lp for some p > n then it suces to assume that satis es an exterior cone condition at every point of @ ; [16]. A more detailed study of the requirements on @ can be found in [23]. We now set p (x) = [?f1 (u) ? f1 (v )]=(u ? v ) in (II) and repeat the earlier procedures. Observe that (u ? v )? ; ` ((u ? v )) = ? f2 (u) ? f2(v ); (u ? v)? . If u v we have an immediate contradiction to ? 0 0 () 0 unless there is a point x such that f2 (u) ? f2 (v ) (u ? v )? j(x0 ) < 0. On the other hand, f2 (u) 0 in a neighborhood of @ by the continuity of u and thus we may keep our arguments away from @ . We conclude in such eventuality that dist (x0 ; (@ ?fx1 = g)) "0 for some "0 > 0. To apply these observations, suppose that for some value of = we have u v in and 0 ( ) > 0; then by the assumed continuity, 0 () > 0 for near . If u v for such then we construct a sequence of points fx0 g with: x0 ! x0 ; dist (x0 ; @ ) "0 ; u(x0 ) < @u (x0 ) 0; contradicting v(x0 ); x0 2 @ . Consequently, x0 2 fx1 = g and @x 1 Theorem 5(c). Observe that the same procedures also work if f2 has a simple jump discontinuity at c; or if f = f1 (x; ) + f2 (x; ); with obvious changes. Suppose next that we consider ?u = f (u) in Rn and assume the existence of ; with f (0) = 0 < u ! 0 at 1; with maximum at the origin, and that f ( ) 2 C`1+ oc f 0(0) = 0. Theorem 6 gives conditions on u and f which suce for u 2 E (Rn ). To apply the above spectral arguments and obtain monotonicity results, we then only further require the continuity of p in Ln=2 (Rnn); and a detailed calculation shows that for this to be the case it suces that u 2 L 2 (Rn ). Note that by following the arguments in [14] we conclude that u ? v 6 0 in ; for > 0; since the maximum of u is at x = 0. We note that we can begin our eigenvalue procedures for any 0 for which we can conclude that (0 ) > 0. If some information is known about the norm of u; then we need not start by considering a very thin domain to ensure (0 ) > 0. We can bypass in this way the requirement that for all relevant ; and still obtain the monotonicity result u > v ; for > 0 say. We note that, as a consequence, we immediately have the observation that if f is Lipschitz, with small
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11
constant, and is symmetric in x1 then all positive solutions must be symmetric, regardless of whether is convex or not in x1 . This result is essentially known, [12]. This approach can also be applied if some extra conditions are imposed on f; as the following arguments in particular indicate. Consider now the situation where f (x; ) is not smooth at = 0. In general we could not obtain results for this case. In special situations, however, we could actually obtain than those obtained earlier. Speci cally, assume now: ? better results f 2 Lip`oc (0; 1) and 0 f (x; )= is nonincreasing in for > 0. The prototype f we have in mind is f (x; ) = p(x) + q(x) with p; q 0; smooth and 1. Our result for these f is as follows (see also [9]): Theorem 10. Suppose is a bounded domain and that u is a positive solution of ?u = f (x; u). Then u is unique up to a constant multiple. If f (x; c) 6 cf (x; ) for any c > 0; > 0 (c 6= 1); and x 2 then u is unique. If both and f (x; ) are symmetric about x1 = 0 then so is u. We observe that we do not require that be convex in x1 nor assume conditions ~ xf . on r Proof. Let ' 2 C01( ). We apply once again Picone's Identity and conclude: Z
jr'j2 =
f (x; u) '2 + Z u2 jr ( ' )j2 : u u
Z
Now let v denote another positive solution and without loss of generality, (u ? v)? 6 0. By our assumptions on f (x; )= we have:
J (') =
Z
Z
jr'j2 ? f (x; uu) ?? fv (x; v) '2
u2 jr ( ' )j2 +
Z
u
f (x; v) ]'2 0 [ f (x;u u) ? f (x; uu) ? ?v
(3)
) f (x;u)?f (x;v) ] = 0 if u = v. since f (x;uu)??fv (x;v) f (x;u u . Note that here we set [ u?v Now let w = (u ? v)? and observe that we may construct a sequence of functions 0 'm (u ? v)? with compact support such that 'm ! (u ? v)? in H 1;2 . We conclude that ? f (x; u) ? f (x; v ) 2 ' f (x; u) ? f (x; v ) [(u ? v )? ]2 m u?v u?v = jf (x; u) ? f (x; v)j(u ? v)? ;
and by integrating and recalling the equation (u ? v) satis es, that J (w) = 0 since: Z
jr'm
j2 !
Z
jrwj2 = ?
Z
r(u ? v) rw:
In the same way, we obtain 0 J (w + "') for given " and any ' 2 C01. We conclude that ?w ? [ f (x;uu)??fv (x;v) ]w = 0 in . But w 0 in and since f (x;uu)??fv (x;v) is
12
W. Allegretto & D. Siegel
EJDE{1995/14
locally in L1 ( ) by the fact that u; v 2 C (K ); Harnack's Inequality, [16], implies that w > 0 or w 0. By assumption, we have w 6 0.? But then (3) yields w = cu for some c > 0; i.e. v = (1 + c)u. It also follows that f x; (1 + c)u = (1 + c)f (x; u) for x 2 from the equations that u; v satisfy, and if this is impossible, we conclude u v. Finally, suppose both and f (x; ) are symmetric in x1 ; and now choose v by v(x; x) = u(?x1 ; x). Clearly 0 < v satis es the same equation in by the assumed symmetry, and we again have w > 0 or w 0. Now w > 0 is impossible since w = 0 on x1 = 0; and it follows that w 0; i.e. u v. In the same way we obtain v u and the result. The proof of Theorem 10 also yields monotonicity results: suppose, for example, f = f (u) and is properly contained in for some > 0 then u > v follows by @u < 0 on \ fx1 = g. choosing v = v in the proof, and we thus have @x 1 As a nal remark, we recall that the classic moving plane argument can be @fi 0. We were extended to systems: ?~u = f~(~u); [22], if we merely assume @u j not able to obtain a similar extension under our conditions on @ ; ~u; f~. 5. Examples
We conclude with some simple examples. We begin with: Proof of Theorem 0. (a) Since p 2 L ; the linear problem: ?ue = p 0; ue 2 H01;2 ( ) has a positive solution ue in L1( ) \ C ( ); [16]. Observe that we thus have ?ue p(x)g(ue) for such that g(ue) 1. Since u = 0 is a subsolution we have the existence of a solution 0 < u 2R L1?( ) \ C ( ) by Schauder's Fixed Point Theorem. (b) In this case p = p(x) 01 g0 tu + (1 ? t)v and kP kL is clearly bounded and pointwise continuous in by the continuity of u; g0 and the continuity n+" in L 2 follows. The symmetry and di erentiability of the positive solutions follow from the comments after Theorem 5. We mention that the uniqueness questions for some of these problems is discussed in [22]. As another example, consider the problem: ?u = q(x)u in Rn ; with: s 1 n 1< < ? (n + 2)=(n ? 2) and q smooth such that 0 < q 2 L \ L (R ) for s = 2n= (n + 2) ? (n ? 2) . The Mountain Pass Lemma, [1], [19] then yields the existence of a decaying positive solution u 2 E (Rn ). If we further assume that q is symmetric with respect to x1 = 0 and x1 @x@q1 < 0 for x1 6= 0 then any such @u < 0 for x1 6= 0. solutions is symmetric in x1 and x1 @x 1 References 1. W. Allegretto and L.S. Yu, Positive Lp -solutions of subcritical nonlinear problems, J. Di . Eq. 87 (1990), 340-352. 2. W. Allegretto, A comparison theorem for nonlinear operators, Ann. Scuola Norm. Sup. Pisa 25 (1971), 41-46. 3. C.J. Amick and L.E. Fraenkel, The uniqueness of Hill's spherical vortex, Arch. Rat. Mech. Anal. 92 (1986), 91-119. 4. I. Babuska, Stabilitat des de nitionsgebietes mit rucksicht auf grundlegende probleme der theorie der partiellen di erential gleichungen auch in zusammenhang mit der elastizitatstheorie, Czechoslovak Math. J. 11 (1961), 165-203. 5. H. Berestycki and L. Nirenberg, Monotonicity, symmetry and antisymmetry of solutions of semilinear elliptic equations, J. Geo. Phys. 5 (1988), 237-275.
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Picone's Identity
13
6. H. Berestycki and L. Nirenberg, On the method of moving planes and the sliding method, Bol. Soc. Brasil Math. (N.S.) 22 (1991), 1-37. 7. H. Berestycki, L. Nirenberg and S.R.S. Varadhan, The principal eigenvalue and maximum principle for second order operators in general domains, Comm. Pure Appl. Math. 47 (1994), 47-92. 8. H. Berestycki and F. Pacella, Symmetry properties for positive solutions of elliptic equations with mixed boundary conditions, J. Funct. Anal. 87 (1989), 177-211. 9. H. Berestycki, Le nombre de solutions de certain problemes semi-lineaires elliptiques, J. Funct. Anal. 40 (1981), 1-29. 10. R. Courant and D. Hilbert, Methods of Mathematical Physics, Vol. 1, Interscience, New York, 1961. 11. E.N. Dancer, Some notes on the method of moving planes, Bull. Austral. Math. Soc. 46 (1992), 425-434. 12. E.N. Dancer, Breaking of symmetries for forced equations, Math. Ann. 262 (1983), 473-486. 13. P. Garabedian and M. Schi er, Convexity of domain functionals, J. d'Analyse Math. 2 (195253), 283-358. 14. B. Gidas, W.M. Ni and L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Phys. 68 (1979), 209-243. 15. B. Gidas, W.M. Ni and L. Nirenberg, Symmetry of positive solutions of nonlinear elliptic equations in Rn , Mathematical Anal. and Applications, Part A, \Advances in Math. Suppl. Studies 7A," L. Nachbin, editor, Academic Press, 1981, 369-402. 16. D. Gilbarg and N.S. Trudinger, Elliptic Partial Di erential Equations of Second Order, Second Edition, Springer-Verlag, New York, 1983. 17. Y. Li and W.M. Ni, Radial symmetry of positive solutions of nonlinear elliptic equation in Rn , Comm. P.D.E. 18 (1993), 1043-1054. 18. P.L. Lions, Two geometrical properties of solutions of semilinear problems, Applicable Anal. 12 (1981), 267-272. 19. P. Rabinowitz, Minimax Methods in Critical Point Theory with Applications to Di erential Equations, American Mathematical Society, Providence, 1986. 20. F. Rellich, Perturbation Theory of Eigenvalue Problems, Gordon and Breach, New York, 1969. 21. J. Serrin, A symmetry problem in potential theory, Arch. Ration. Mech. 43 (1971), 304-318. 22. R. Schaaf, Uniqueness for semilinear elliptic problems { supercritical growth and domain geometry, (preprint). 23. G. Stampacchia, Problemi al contorno ellittici, con dati discontinui, dotati di soluzioni holderiane, Annali di Math. 51-52 (1960), 1-37. 24. G. M. Troianiello, Elliptic Di erential Equations and Obstacle Problems, Plenum Press, New York, 1987. 25. W.C. Troy, Symmetry properties in systems of semilinear elliptic equations, J. Di . Eq. 42 (1981), 400-413. 26. R. Vyborny, Continuous Dependence of Eigenvalues on the Domain, Institute for Fluid Mechanics and Applied Mathematics, University of Maryland, 1964. Department of Mathematics University of Alberta Edmonton, Alberta Canada T6G 2G1
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[email protected] Department of Applied Mathematics University of Waterloo Waterloo, Ontario Canada N2L 3G1
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