Nonlinear elliptic problems in strip-like domains: symmetry of positive vortex rings

Nonlinear Analysis, Theory, Methodr & Applicarionr, Printed in Great Britain.

Vol. 7, No. 4, pp, 365-379,

1983.

0362-546x/83/040365-15 $03,00/O @ 1983 Pergamon Press Ltd.

NONLINEAR ELLIPTIC PROBLEMS IN STRIP-LIKE DOMAINS: SYMMETRY OF POSITIVE VORTEX RINGS M. J. C.N.R.S.

Laboratoire

d’Analyse

NumCrique,

(Received Keywords:

Semilinear

elliptic problems,

ESTEBAN

UniversitC P. et Marie Curie, France 2

4 Place Jussieu,

75230 Paris Cedex

05,

August 1982)

axial symmetry,

strip, compact

imbedding.

critical

point

INTRODUCTION

paper is divided in three parts. In the first one we study the existence of nontrivial, axially symmetric solutions of the problem: THIS

-Au = f(u) in Q u=OonaQ where Q = 0

x [WP,

(1)

with p > 1 and 0 is a bounded domain in R”, m 2 1. We denote by

N=m+p.

This problem arises when we search certain solitary waves in Schrddinger and Klein-Gordon nonlinear equations. It appears also in other problems of mechanics and physics. This type of problems has been considerably treated when Q is bounded (see [l, 2]), or when C2= RN (see [4] and [16]), but not in strip-like domains (except for [5]). We prove the existence of a positive solution and of infinitely many axially symmetric solutions of (1). Our second section is devoted to the study of the problem: -Au=Af(u) 1

u=O

inQ,L>O on&2

(2)

i.e. we want to solve a nonlinear eigenvalue problem related to (1). The result we give here concerns the existence of infinitely many ‘eigenvalues’ of (2) under optimal assumptions, much more general than the ones in section 1, where (1) is considered. Finally, in section 3 we study the symmetry of positive solutions of the so-called vortex rings problems. This problem is classical in fluid mechanics and has been studied by many authors (see, for example [3, 8, 131). Existence is already known and, for instance, one knows that there exists a positive axially symmetric solution of the vortex rings problem in dimensions 2 and 3. We prove here that in fact, every positive solution of this problem is axially symmetric. This kind of result is to be compared with the general results of Gidas et al [9, lo], and we will use in a fundamental way their techniques. 365

M. J. ESTEBAN

366

Notation.

Throughout

this paper,

and the implicit

summation

1. EXISTENCE

OF

we use the standard

convention

SOLUTIONS

OF

notation

for repeated

for partial

derivatives:

subscripts.

ELLIPTIC SEMI-LINEAR DOMAINS

PROBLEMS

IN STRIP-LIKE

In this section, we are interested in finding solutions of problem (l), and to do so we use some critical point theorems which allow us to prove the existence of axially symmetric solutions of (1). 1.1. Main results Throughout this section that: f(0) where

A1(0)

we suppose = O,f(t)

= g(t) + vt

is the first eigenvalue -_c4

that f and g are two real continuous

of (-A) lim

<

f(‘O lim I’+”

so,, Q 1 ’

G(t)t-‘is

e> Remark.

we

can

v < hi (0)

assume

-kc

that

and g satisfies

f has

the

boundary

conditions. (4)

ifN=2

+~a

for t 3 0,O < i& G(t)t-’ f++X E

(3)

0

ifN>Z,Z<

Vt

Actually

vt, x1 E 0, where Then,

g(t) g(t) -=s*llm -= t Ifi+ t

2,

such

with v< hi(O).

in 0, with Dirichlet

N+2 where I = N-2

non decreasing

Vt,

functions

< + m,

(5)

for some

52.

(6)

more

(4)-(6)

uniformly

(3)-(6),

then,

general

form:

f(t)

= g(t, xl) +

in XI E 0.

we can state the following.

THEOREM

1. Suppose

p 2 2, f and g satisfy -Au

=f(u) u=O

has a solution

u EWf$(Q)

fl HA(Q)

= VU + g(u)

problem

(1):

in Q

(1)

0naQ

(Vq < +w),

which satisfies:

(i) u>Oin 52; (ii) u is axially symmetric, i.e. u has the form u(x,, x’) = u(x1, Ix’(), where x’ E W; moreover u is decreasing in Ix’ 1; (iii) If f is locally Holder-continuous, u E C2 (Q). If we are interested in the existence of infinitely that g is odd. In this case we obtain the following

many solutions result.

.x1 E 0 C

of (l), we have to suppose

Nonlinear elliptic problems in strip-like domains: symmetry of positive vortex rings

367

THEOREM 2. Suppose p 2 2, that f and g satisfy (3)-(6) and that g is odd. Then, problem (1) possesses an infinite sequence of distinct solutions, {u~G)~~N, which are axially symmetric, and (Vk) UkE c(o), if f is locally Holder continuous. We will prove these two theorems in 1.3, but let us now give an auxiliary lemma and an idea of the proof.

1.2. Some auxiliary results If Q = 0 X Iwp, with 0 C R” bounded, then for all u E&(Q),

LEMMA 3.

we have

h lu1Q.r =G11(0)-ij-- IVu12d& where A,(O) is the first eigenvalue of (-A) Proof. Let u E 9(O)

in 0, with Dirichlet boundary conditions.

such that i, JVu12dx G (A,(O) + E) i, ju12dx,

andleta=(ai,...,

a,) be a point in 0. Then, for every 0 < 6 < 1 we define:

u6(x1,. . . , xm, x’, , . . , xJ’> = u(x 1, . . . ,x,,,) p( 8x ‘, . . . , 6xp), where q E ‘9+(RP), q + 0. We have that V6 >O, u6 f 0, ug E 9(Q) In

and (Vu# =JVuj2q2 + u21V~)*(6x’, . . . , 6xP) ,

[VZQ(~ dx G&(o)+&+ /u~12dx

IVVI~(~X)dx I RP I RP

ti(cM dx

dx I IP9J12 I lP12dX. Rp

+ &+ a2

~/l,(O)

w’)

This proves that A =

inf

L&H;(Q) UfO

Jg’

I

a

b12~

c

Al(O) (remark that by the classical Poincare’s

inequality, A > 0). Suppose now that A < Ai. Then, we can find w E 9(Q) such that: IR

jVw12dx

IQ

= p<&(O). (w12dx

Let us now define for every 0 < LYs 1, zf(X1,. . . ,xm,x’)

= w(x1,. . . ,xm, cvx’),

yx1,.

. . ,xd

EO,

VX’E

w.

368

M. J.ESTEBAN

As cr goes to 0 we see that

A,=

I

jVu”I’ dxi, . . . , dx, dx’ +j3($jZ(X

Ia

Iu”J’dx,,

If we consider h E H;(Q)

I,...

,X??l,X’) d.Xl,...

1 dx,,

dx’

’ .

, dx,,

dx’

h(xr . . . .x,)

the function

,x,,x’)l”dx

*lw(* I,..

i

w2(x,,

=

.

I,...

, dx,,

,x,,x’)

.

dx’

, we see that

dx’

!

and that:

I

lVh12 dx,, . .

0

(hi2 dxi, j0

, dx,

.

~(j*(w~~d*,....,dx,,dx'j* c

. . , dx,

lim A, a+ 0

[j- w2 dx,, . . . , dx,, dx '1'

=

8

But VaG 1, A,< p < A, (0). Then this contradicts the fact that Al(O) is the first eigenvalue n of (-A) in 0. This contradiction shows that A = A, (0). is axially Definition. We say that u EHh(t3 x W) U(Xi ) . . . ,&,X’) = u(x1,. . . ;x,, Ix’l), Vx’ E RP. Let us note Hb.,(O X W’) the set of axially symmetric

symmetric functions

if

u

has

the

form

in Hh(O x IV’).

Our results rely on the following compactness lemma due to Esteban & Lions [7], generalizing a lemma due to Strauss [16] (see also Berestycki & Lions [4]). The proof of lemma 4 is given in [7] and it uses strongly the results of Lions [12]. LEMMA 4. (Esteban & Lions [7]). If p 2 2, the Sobolev imbedding of Hb.,(O x [WJ’) in Lq(O x W) is compact for every 2 < q < 2N/N - 2. Furthermore, if {u,} is a bounded sequence in Hh,,(O x Rp), such that {u,,} converges weakly to u and if F is a continuous function such that F(t) = o(t’)

as t+

F(t) = o(jtJ2N”-2)

0 asIt\-+

+a.

Then,

Proof of theorems 1 and 2. As we are first interested of (3), we modify the problem (1) as follows. We define g(t) ‘(‘) Observe

that by the maximum

principle,

= {O

in the existence

of a positive

solution

if t 2 0 ift < 0

the solution

of (3) with g are the positive

solutions

Nonlinear elliptic problems in strip-like domains: symmetry of positive vortex rings

369

of (3) with g. Hence we will consider that whenever g appears, we have replaced it by g, With this modification, g satisfies the stronger conditions: lim

Ig(t)I -<++.

t-0

t

,,,ty_

ltlN+2/N-2

(4bis)

k(t)I =

(5bis)

0.

Under all these assumptions it is easy to prove that S(u) = ;I, jVu12dx - i/e )u12dx - i, G(u) dx is a meaningful C’ functional in E =Hb,,(S2). On the other hand, it is clear that the critical points of S are solutions of (1). Then, in order to solve problem (l), we are going to look for critical points of S in E. To do so we use a critical point result due to Ambrosetti & Rabinowitz (see theorem 3.9 in [l] or [2]). So, the proof of theorem 1 consists of the verification of the conditions needed for the application of this critical point theorem. Since this verification is not difficult, we are only going to check that, for example, S satisfies the Palais-Smale condition, in order to show the main point of the proof. We have to see that if {u,} is a sequence of E satisfying inE’; 0 < CYGS(u,) G M < + 00, S’(u,) -0 ?l-++CC then {u,} has a convergent Let {u,} be such that Vn,

O
subsequence in E. 2 i, IVu,12dx - f

l, /u,,j2dx -

j-- )Vu,12dx - VI, (u,12dx - j--&&n

with JIE&I :

j-c G(u,) dx 6 M dx = (c,, un)

(7) (8)

0, (. , a) the duality bracket between HA and H-‘.

By (6), tg(t) 3 f3G(t), Vt; then from (7) and (8),

So, applying lemma 3 and assumption (3), we deduce, as 0 > 2,

Hence {u,} is bounded in Hi,,(S2). We extract a subsequence, that we still denote by {u,}, which is weakly convergent in E to some u, and we may assume that {u,} converges a.e. in Q to u. Now we can apply lemma 4 to the function (g(s)s)+ and we deduce:

370

M.J.ESTEBAN

On the other

hand we know that

and that u obviously

solves: -Au

In addition,

since un*

u a.e.,

n

u,-

j- (IVu/‘52

n

u in Hi(Q),

the equalities

I

(IVu,(‘-

n

3,

IR

vu.

3 a definite

shows that the preceding

positive

“lim”

quadratic

is actually

form in

“lim” and

we have:

(\Vu,I* - vu,J2 dx _r:

(IVu(’ - vu’) dx.

IR

R

And by lemma

vu;) dx

R

is by lemma

of these inequalities

hold; in particular

we have:

vu*) dx
(indeed remark that Jo (jVul* - vu2) dx Hh( Q)) , and by Fatou’s lemma:

The combination

in H-‘(Q).

= vu + g(u)

vu -vuu

= ((24, u)) is an equivalent

scalar product

on H&Q),

this

implies: u, < u in HA (Q). At this point we have proved theorems 1 and 2 except for the decreasingness of the positive solution that we may obtain either by using the results of Gidas er al. [9, lo] or by the method n of proof of theorem 6 below. Let us consider now the case when p = 1, that is R = 0 x Pi.We suppose as above that f and g are two real continuous functions which satisfy (3), (4), (5) and (6); then we prove by a different method that theorem 1 still holds. THEOREM

6. Under

the above

conditions

theorem

1 still holds.

Proof. We prove this theorem by the following approximation & Rabinowitz [l ,2], there 0 x ] - R, R[; b y th e results of Ambrosetti

i and 0 s j-o (;I

(VuR12 -;j&l’

-AuR

=

UR /dC&

=

VUR 0;

- G(uR) dx)

UR

+

g(UR) >

0

in

method: let QZR= exists uR solution of

QR

in QR

= bR.

In addition, from the explicit formula giving bR in [l, 21, we see that bR is bounded. Therefore as in the proof of theorem 1, uR is bounded in #(QR) and using standard regularity results 6 C (C indpt. of R) for we see that for p 2 1 IIua(ILp(oR)c C (C indpt. of R). Thus ]IuRI/w~.P(~~~) and we may assume that there exists {R,} ---+ RaRo,p<+m, i-m such that u& = n’+Z

Nonlinear elliptic problems in strip-like domains: symmetry of positive vortex rings &I-

n++a

u uniformly

sets of 6 X [w, and u is a solution of

on bounded

-Au=g(u)+ r &j=o,

371

V,U inSZ;uEH&Q) in !2,

u20

u E Wt;f(Q)

(JJ < + m).

Next, by Gidas et al. results [9] we know that UR(X,x’) = u~(x, Ix’/) Vx E 0, Vlx’l E (-R, R) and that uR(x, x’) is nonincreasing for lx’/ E (0, R). By the uniform convergence on bounded sets, u has the same properties, and thus it remains only to prove that u f 0, and consequently, that u > 0 in !2. Since ;ax uR. = max u&(x, 0)) and uR, (X, 0) j u(x, 0) uniformly on 0, it suffices to R

0

prove that ?here exists a constant LY> 0, independei;yf that such a constant

lIuR,IIL” + 0,

OR;,=

R, such thatmp

uR(x, 0) aa:

CCdoes not exist. Then, we can find a sequence RA++

&,

vR;

Suppose

CCsuch that

satisfies [ -AOR;,= (zJ+*)

uR;

inQR;

But since lim -g(s) c 0, and v < &(O), by lemma 3 this is impossible, and this contradiction s-o+ s ends the proof. 2. EXISTENCE

OF SOLUTIONS

OF PROBLEM

(2)

Suppose now that Q = 0 x R’, p 2 1, where 0 is a bounded domain in R”, and that f is a continuous real function such that f(0) = 0, f(t) = 0 Vt c 0, and that f satisfies (4), (5) and

3c> 0

s.t.

F(C) > 0,

(10)

where F(t) = ‘f(s) ds Vt E R . i Let us the: consider for every v > 0 the following minimization =- k,/

Q

Ivvlz d_xover

problem: Minimize J(v)

K,, where K, = (v EH~(Q)~/ F(v) dx ?=q): n

inf J(v). UEK,

(11)

Under the above hypotheses we can state the following. 7. For every q > 0, the problem (11) has a solution u EH~(Q) rl W&Cg ($2) (Vq < +a) which is positive, axially symmetric, i.e. u(x) = u(x~, Ix’/), Vxl E 0, Vx’ E Rp, u is decreasing with respect to Ix’ 1, and u satisfies: F(u) dx = r. In Moreover, there exists a Lagrange multiplier A > 0 such that: THEOREM

Au + /If(u) = 0

inQ;u

= 0

on&?.

M. J. ESTEBAN

372

Remark. This theorem

gives an extension of Pohozaev’s results (see [14]) to the case of a strip-like domain. Similar extensions have already been given for other domains; in the case of G = RN, the result is due to Berestycki & Lions (see 141). Proof of theorem 7 Step 1. Take q > 0. Let us first see that the minimizing set is not empty. 1

,t let QR ={(x, ,x’) E Q(xr E 0, (~‘1< R) , and DR = ((meas O)S,)“p {(x1,x’) E R(x, EO’, lx’1 < R - E}, where 0’ C 0, E> 0, DRC QR. and measDR = meas QR - 1. Let us consider a function WR which satisfies: WR(X) = 0 if x E QR, WR(X) = 5 if x E DR;

For

R >

every

((wR((L”=c;

WR -fh(~).

Thus

I

a

F(wR) dx ?=F(<)(meaS QR - 1) -

Hence, for R large enough,

I

maX IF(S .SE[O, 51

F(wR) > 0.

Now, for every 6 > 0, if we dtnote U’(X) =u(xl, 6x’) , we have uDE I&$(Q) and IR

I

n

F(ub) dx = Kp

F(U) dx.

Thus, after a convenient F(W) dx =n.

choice of R and 6, we find a function w E HA(Q) satisfying:

Let

a

us

now

choose

{u,} CZif&Q)

sequence

with

IQ

F(u,) dx > q,

Vn,

and

; n lvun/* dx \I, I where I =,2: J(u) . ri Step 2. Let U, =(u,t)*, where p* denotes the Steiner symmetrizated function of jpj with respect to the axis {x1 = 0}, and p+ is the positive part of p_ By the properties of the Steiner symmetrization (see [6, ll]), and since F(t) = 0 Vt d 0, we have:

I

R

F(u,) dx =

In

F(u,+) dx =

IR

F(u,) dx

and I

R \VU,-~~ d x G j-- 10~; I2dx s j-- (Vu, I2dx.

So we get a minimizing sequence of nonnegative with respect to Ix’ (. Now, as

axially symmetric functions,

IVii,J2 dx L I3 0 th e sequence {U,} is bounded

t VN, S,,, is the volume

of the unit sphere

of ET”

in H~,JQ)

decreasing

(remark that 0 is

Nonlinear elliptic problems in strip-like domains: symmetry of positive vortex rings

373

bounded and so one may apply Poincare’s inequality in 52). Then, taking a subsequence if necessary, there exists u EZ&(Q) such that ri,,--, u

in H& G?)-weak.

Using lemma 4 or the following compactness lemma 8 when p = 1, due to Esteban & Lions [7], applied to F+(S), where F+(s) (resp. F-(s)) is the positive (resp. negative) part of F(s), we obtain:

I

D

F+(u,) dx + F+(u) dx. n IP

LEMMA 8. Let {u,} be a bounded sequence in H&, (0

X R) with 0 C R” bounded

and let us

assume that {u,} satisfies: u,(xr , . . . ,x,, x’) is nondecreasing

un 2 0, V(x, , . . . , x,) EO, creasing for x1 G 0.

Then (u, ), 3 1 is relatively compact in Lq( Q) , V2
n

for x1 3 0, and nonde-

~4if N = 2) . In addition,

u, a.e., then for each continuous function F satisfying: F(f) = o(t*)

as t+ 0

F(t) = o(ltl 2~N-2)(orO(~t~q),

ifN=2)

q<+m,

asItl++oc

we have: F(u) dx.

On the other hand, since z&f u weakly in HA(Q) ; p IVul* dx G Ii&j-

I

If we prove that u satisfies the constraint

R

)Vu,j* dx = 1.

F(u) ~q,

IQ

we may then conclude.

But by Fatou’s lemma, we have:

I

P

Hence,

IR

F(u) dx >q.

F-(u)

F-(u,) dx 7 lim n--t+mI Q

Actually we have:

IR

dx.

F(u) dx =n.

F(u) dx >q; then necessarily k n (Vv12dx has a local minimum at u = u, I I F(u) dx 3~ > 0. but this implies uQ= 0, which contradicts I F(u) dx = n and u is not only The minimum of J(v) over K, but also that of J(u) Thus I 51

Suppose that

over Kh ={ u E HA(&) i, F(u) dx = n}.

374

M. J. ESTEBAN

Step 3. Let us now see what is the relation (2). If u solyes (ll),

there

exists a Lagrange -Au

between

multiplier

the minimizing

problem

and problem

A such that:

u E H,$2).

= hf(u),

Let us show that A.> 0: first remark that A.# 0, because A = 0 implies impossible. Suppose then that we have A.< 0. Since we have a.e. f(u) # 0 we can find w E 9(Q) such that:

u = 0, which

is

I*f(u)w dx>O. Then

we have for E > 0 small,

since

F(u) dx ECi(H:,(Q)),

JR

I

F(u) dx + E S(u)w dx 2 rj F(u + EW) dx = IR Ia 52 I, (Vu(2 dx + 2&A I,f(z+

(V(u + .sw)12dx=

and we see that we can find v E Hi(Q)

I

dx + E’ 1 (VW1’

satisfying:

F(v) dx >

R

; n IVvl’dx I

IR

F(u) dx = 11

< ;j-- (Vu(‘dx

= I.

But this contradicts the fact that u minimizes J(u) on K, This contradiction shows that A.> 0. Finally, by the strong maximum principle, u > 0 in Q. Remark. As we see above, we can obtain the existence of a positive axially symmetric solution of (2) under much more general conditions than in the case of problem (1). This result suggests that in fact (1) should be solvable under less assumptions than those we made in theorems 1 and 2.

3. SYMMETRY

OF POSITIVE

VORTEX

RINGS

IN DIMENSIONS

2 AND

3

The problem of vortex rings is classical in fluid mechanics and has been studied by many authors (see Fraenkel & Berger [8], Berestycki & Lions [3] and Ni [13]), who proved various results on existence of local and global vortex rings. If w and k are constants, w > 0, k 2 0, and f is a W@(R)--function that satisfies: f(c) = 0, the equations

which model

this problem

Vt s 0,

are the following.

Nonlinear elliptic problems in strip-like domains: symmetry of positive vortex rings

The two-dimensional

375

problem is

-Au = f(u - WI - k)

ifx = (r, z) E II = {x E R*(r > 0)

u/,=0 = 0

(12)

u(x) -I*I--t+mO; u > O and the three-dimensional

problem is given by

(13)

I

u(x)-jx[++m 0; u

> 0.

Under the above conditions on f, and some additional assumptions, existence of solutions and, in particular, of a positive “symmetric” solution, has been proved (see for the most general results, Berestycki & Lions [3]). In addition the solutions satisfy: u - ;r’-

k

Our goal is to prove that every solution of (12) or (13) is necessarily symmetric with respect to a certain axis. Our next results give this symmetry in the two- and the three-dimensional cases. 9. If u is a solution of (12) such that f(u - wr - k) has compact support, exists NE R!such that

THEOREM

U(T, z + cr) = U(T, CX - 2)

there

V(r, 2) E l-I

U(T, z + cu> is decreasing for z 2 0. Before proving it, let us give two lemmas we need. The proof of this theorem basically uses a geometrical technique used by Gidas et al. to prove many symmetry results (see [9] and

[W 10. If u is a solution of (12) and supp f(u - wr - k) is compact in II, there exists a~ R! such that

LEMMA

Vz > (Y 3/z(r) > 0

s.t. r > h(z)

Vz < CY if r > h(2a - z),

implies U&K)< 0;

then U&V)> 0,

where x = (r, z). Proof of lemma 10. It is easy to see that if we apply the Kelvin transform: x-y

lI is mapped into itself, and if we define: U(Y)

= 4X)iY

= (Yl,Y2),

x = k-, 2)

=x lX12’

M. J.ESTEBAN

376 we

get:

-Au(y)

=

fk(y)-wi%-k yy=(y

lY14

1,

’

Y2)En

Next, as supp f(u - wr - k) is compact, we can find 6 > 0 such that Vy E BP(O) II M.

Au(y) = 0

Now, we may extend u oddly to BP(O) n {y(yr < 0}, i.e. we take for every yl < 0, 4Yl>Y2)

= -4-y1,y2).

Therefore, Ay = 0 in BP(O) - (0). Moreover, if I ={y E R*}(yr = 0) fl BP(O), and ulr+ (resp. ulr) when y ~B,(o),yr + O+ (resp. O-),

designates the value of u

ulr- = ulr+ = u&+ = uzlrm = 0; ur/rm = u1Ir+, and u E L”(B,(O)); so, u is harmonic in BP(O). Note that since ~~(0) = u2z(0) = uz2*(0), we have: h(O)

Therefore,

= ~112(0)

= 0.

we can write:

I I

U(Y)

= UlYl +

+

where

Now, we compute Us

9

aijk.../

a12 71y2

YlYi

+

a111 7y1

+ %y:y2

3 +

+ o( (y I”).

ugk...l(O), and al

=

h(Y)

= a1 + Ul2Y2 + 4

u207)

= Ul2Yl +

yyly:

>

0

IYI>

O(lYI>.

in terms of y and u, and we get easily that for z >z

u2(x) < j-$(-C(z)ri

(14)

I

= cr ,

+ o(r3) + O(P)), C(z) >O.

So, if r > h(z), uz(x) < 0. We can prove a similar result for z < LY.In that case, r > h(2~t - z) implies u2(x) < 0. LEMMA 11.For every A > (Y(a is the constant found in lemma lo), there exists R(A) > 0 such that if z < h and 1x1> R(A), we have u(x) >u(x% where x* = (r, 2A - z), i.e. x’ is the symmetric point of x = (r, z) with respect to {z = A}.

Nonlinear elliptic problems in strip-like domains: symmetry of positive vortex rings

377

Proof of lemma 11. First we make a translation in the z-direction in order to put the origin at (0, cu>. Then, for 6(x) = u(r, z + C$ we can write the following problem: -Au(x)

=f(ti(x)

- wr - k)

in H (15)

1 Z.&n= 0. Next, we use again the Kelvin transform,

and we define C(x) =zi x

.

( IA2 >

As, above, we see that li is harmonic in a neighborhood of the origin and that we can expand there a. Moreover, if we compute the value of &z(O), we see that it is 0. When x2 > A/2, the result is easily obtained applying lemma 10. For x2 < h/2, instead of proving that C(x) > fi(x’), we equivalently see that

P($ >p(h). From (14), to verify it is equivalent

to prove that if we write 1 =(I1 ,12) = x

lx12’

h =

(hl, h2) = &

+ 2

(h:h2 - 132) < a~(ll - hl).

But this is easy to prove, showing that as [xl--, +m, h: - r: hIh; - l,R hlh: - 111,’ and II - hI’ II - hI ’ II - hl

h:hz -

/:I2

II- hl

converge to 0. Proof of theorem 9. Using Green’s formula in the half-plane, (12) 7 u, as:

we can write any solution of

Vx”=(ro,zO)EII

u(x”)=&~ln(l+~)f(u-wr-k)dx,

where K = supp f (u - wr - k), s2 = (r - r”)2 + (z - z’)~. Let us define A= max{lz( E lFPj(r, z) E ZQ. Then, for every A 2 1, for every x0 E H with z” < A, we have: ln(l +$J where

>ln(l

+s),

(r”)A = r”, (z”)” = 2A - z”, (S~)~ = (r - r”)2 +

(Z -

(z')')~,

and thus,

Hence, defining AI = inf{A > ~~/u(x) > u(xP),

Vx E II,

x2 < p; Vp > A}

u(x') >

u(x")").

378

M. J. ESTEBAN

we see that (YG Ai s 1. Let us see that hi = cr. If Ai > C_Y, there will be two sequences: {xi} C II, and {Ai}+A, with U(Xi) s u((# ) , and from lemma 11, {xi} is bounded; therefore, {xi} has an accumulation point in I?, X = (?, ~7). By continuity, we have u(Z) c u(@)‘~), and by definition of A, as Z c hi, u(i) 2 u((@). so, u(f) = u((x)*‘). Now, there are two possibilities for f.

(i) J > 0 (ii) P = 0. By the maximum principle, (i) is impossible. Moreover, in case (ii) we apply a lemma due to Gidas et al. (see lemma 5 in [9]), which applies in our case, and gives that u&Z) < 0, and so, u12 < 0 in a neighborhood of f in II, %. Now, from the existence of {xi}, we deduce that there exists another sequence, Cyi}, that converges to f, and such that u2(yi) c 0 Vi. If we consider the yi which are in %, and we integrate u12 in the (-x1)-direction, we shall have u2 > 0 at some points of c?II, but there U2= 0. Remark

13. In the three-dimensional problem, the proof of the axial symmetry theorem is alike. We must only adapt lemmas 10 and 11 to this particular case. First, and using an idea of Ni (see [13]), for every solution, U, of (13) and for every x E 4

R5,

with r2 =,&ix” > 0, we define

u(r, z> where z = x5.

ii(x) = 7,

One may move that zi is a regular solution of -A@)=f /-

[lqx) = 0

(4

ii-zr2-k

tlx

withr>O

(16)

on {r = 0)

where zi > 0, C(X)-+OasIxl++mandsuppf Let us now give the analogous result to lemmas 10 and 11 for the three-dimensional

problem:

LEMMA 14. If U is a solution of (16), there exists a constant CYsuch that for every z > LYwe can find h(z) > 0 which satisfies:

z&(x) < 0

Vx

with r > h(z).

Similarly, z&(x) > 0 if 2 < (Y,r > h(2c~ - z). Furthermore, for every A > (Y,there exists R(h) > 0 such that z < h and 1x1> R(A) imply U(X)> U(x”) where XLis the symmetric of x with respect to {z = A}.

Nonlinear elliptic problems in strip-like domains: symmetry of positive vortex rings

379

Proof of lemma 14. This proof is very similar to the one made for lemmas 10 and 11, and n

we will omit it.

The following theorem, which is equivalent to theorem 9 in the case of the three-dimensional problem, gives us the desired result: THEOREM

15. For every solution of (13) satisfying that the support off

u -f

r* - k

is

compact, there exists (YE R such that u(r, 2) = u(r, 2a - 2)

Vr > 0,

Vz E R;

u(r, z + E) is decreasing for z 2 0. REFERENCES 1. AMBROSETTIA. & RABINOWI~TZ P. H., Variational methods for nonlinear eigenvalue problems, in Eigenualue of NL problems (Edited by G. PRODI), C.I.M.E. Ediz. Cremonese, Roma (1974). 2. AILIBROSETTI A. & RABINOW~TZ P. H., Dual variational methods in critical points theory and applications, J. funct. analysis 14, 349-381 (1973). 3. &RESTY&I H. & LIONS P. ‘L., A minimization approach to the problem of global vortex rings, to appear. 4. BERESTYCKIH. & LIONS P. L.. Existence of solutions of NL equations. I. The ground state. II. Existence of infinitely many bound states. Arch. Rat. Mech. Anal., to appear: 5. BONAJ. L., BOSE D. K. & TURNERR. E. L., Finite-amplitude steady waves in stratified fluids, MRC Technical Summary Report 2401, University of Wisconsin, Madison (1982). 6. BRASCAMPH. J., LIEB E. H. & LU~TINGERJ. M., A general rearrangement inequality for multiple integrals, J. funct. analysis 17, 227-237 (1974). 7. ESTJZBAN M. J. & LIONSP. L., A compactness lemma, Nonlinear Analysis, to appear. 8. FRAENKELL. E. & BERGER M. S., A global theory of steady vortex rings in an idea fluid, Actu Math. 132, 14-51 (1974). 9. GIDASB., NI W. & NIRENBERGL., Symmetry and related properties via the maximum principle, Comm. Math. Physics 68, 209-283

(1979).

10. GIDAS B., NI W. & NIRENBERGL., Symmetry of positive solutions of nonlinear elliptic equations in RN, to appear. 11. LIEB E. H., Existence and uniqueness of the minimizing solution of Choquard’s nonlinear equation, Studies Appl. Math. 57, 93-105 (1977). 12. LIONSP. L., Minimization problems in L’(W3), J. funcf. analysis, to appear. 13. NI W., On the existence of global vortex rings, J. Analyse Math. XXXVII, 208-247 (1980). 14. POHOZAEVS., Eigenfunctions of the equation Au + nf(u) = 0, Sou. Math. Dokludy 5, 1408-1411 (1965). 15. SERRINJ., A symmetry problem in Potentia! theory, Arch. Rat. Mech. 43 304-318 (1971). 16. STRAUSSW. A., Existence of solitary waves in higher dimensions, Comm. Math. Physics 55, 149-162 (1977).