proceedings
of the
american mathematical
Volume
104, Number
society
1, September
1988
RADIAL SYMMETRY OF THE FIRST EIGENFUNCTION
FOR THE p-LAPLACIAN IN THE BALL
TILAK BHATTACHARYA
(Communicated
by Walter Littman)
ABSTRACT. We prove the radial symmetry of the eigenfunction corresponding
to the first eigenvalue of the equation: div(|Vu|p_2Vu)
+ A|u|p_2u = 0, when
fî is a ball in Rn and 1 < p < oo.
Introduction.
Let fi be a bounded domain in Rn, n > 2. Consider the problem
ApU + A|u|p_2u = 0
u G W0hp(Q), u^O,
inQ,
XGR, and 1< p < oo,
where Apu = div(|Vu|p-2Vu)
is the p-Laplacian of u.
We say that u is a solution of (1a) if there exists a X, and (1a) holds in the sense
of distributions, i.e.
(l'A)
[ |Vu|p-2Vu • VV = A / |u|p-2uV> W- 6 W01,p(n).
Jn
Jn
Consider the following minimization problem
(2)
inf/(t>),
v G W¿*(Sl),
J(v) = 1,
where I(v) = (1/p) /n \Vv\p and J(v) = (1/p) /n \v\*>-lv.
It can be shown [1, 2] that there exists a smallest Ai > 0 and ua, > 0 that
solves (IaJ- In that case Ai is the infimum in (2). Furthermore u is a solution
(2) iff u solves (IaJ and satisfies J(u) = 1. In the case fi is a ball, the spherically
decreasing rearrangements of positive solutions are also solutions. Furthermore if
qb(r) is a radial solution then all other radial solutions are scalar multiples of <j>(r)
[1]. The radial solution <p(r) can be chosen to be positive.
In [1] the question is raised as to whether or not a solution, corresponding to
Ai in a ball, must be radially symmetric. In the following we prove that if u solves
(IaJ in a ball, then u is radially symmetric (see Theorem 1). Our method is based
on an idea of Pólya and Szegö [4, p. 89].
Radial symmetry.
Let Q be the ball of radius R centered at the origin. Let
u G Wo'^n) solve (IaJ- By a remark at the end of Theorem 4 in [5, p. 264], for
K P < n, ||u||oo,n < C||u||P)n- If P > n, by Theorem 7.17 in [3, p. 163] u G Cß(U)
with ß = 1 —n/p.
Received by the editors August 5, 1987.
1980 Mathematics Subject Classification (1985 Revision). Primary 35P30.
The author thanks
encouragement.
Professors
Allen Weitsman
and Juan Manfredi
©1988
for helpful discussions
American
0002-9939/88
169
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170
TIL AK BH ATTACH ARYA
Thus u G L°°(fi) and by the results in [7], u G CX¿"(Q). Let 4>(r)> 0 be a radial
solution of (IaJ
and let / be such that
(3)
u = fob.
By Theorem 3 in [1], <f>
G Cx(Jï) and hence / G Cx(Vt).
Let vol(S) denote the volume of a region S in Rn and Ai(S) denote the infimum
of (2) in S.
We now prove a few preliminary
results that lead to the main theorem.
We begin
with
PROPOSITION l. Let Oi and Q2 be balls of radius Ri and R2 centered at the
origin with Ri < i?2- Then Ai(fi2) < Ai(fii).
PROOF. Let (¡>ibe radial and minimize (2) in fii.
We take 4>i > 0. Then
I(4>i) = Ai(fii) and J(</>i)= 1.
Set (¡>(r) = c"/p0i(cr) where c = Ri/R2. Then (j>(R2) = 0, J((j>2) = 1, and
I(<i>2)= cpAi(fii). But Ai(f22) is the infimum of (2) in U2 and hence Ai(fi2) <
cpAi(fii)< Ai(nj.
LEMMA l.
Ifu G Wl'p(ü) solves (IaJ,
then either u > 0 or u < 0 in fi.
PROOF. First we show that u does not change sign. Suppose it does. Let
u+ = max{u,0} and fi+ = {x G fi: u > 0}. As u+ is continuous, fi+ is an open
subset of fi with 0 < vol(fi+) < vol(fi). Furthermore, u+ G W01,p(fi+). Scale u+ so
that J(u+) = 1 in fi+. Putting u+ as ip in (1AJ gives I(u+) = Aj(fi) > Ai(fi+).
Let fis be the ball centered at the origin with vol(fis) = vol(fi+).
By Theorem 2 in
[1] (also see [6]), Ai(fi+) > Ai(fis). Proposition 1 then implies Ai(fi) > Ai(fi+) >
Ai(fis) > Ai(fi), a contradiction.
To prove that u never vanishes in fi, we use the strong maximum
principle as
stated in Proposition 3.2.2 in [8].
Note. From now on we will take u > 0. It then follows from (3) that / > 0 in fi.
For notational simplicity we write
IMIp= IMIp.n, l<p<oo.
LEMMA 2. If u G W01,p(Q)solves (IaJ
and <f>(r)> 0 is a radial solution of
(IaJ, and if f is such that u = f<f>,then f G L°°(fi).
PROOF. Now / = u/<f>where 4>(r) is never zero. Let fii be the ball of radius Ri
centered at the origin with 0 < i?i < R. Then
sup/ < Hulloo/inf <t>
<M<
Consider the annulus A = fi\fii.
+00.
Define w(r) = k(R —r) for k > 0 where r = \x\.
Then
Apw = -(n - I)*""1 /r.
Choose k so large such that
(4)
-(n-l)fcp-1/-R<-Ai||u||p0-1
and
k(R - Ri) > H«,.
Then Apw < Apu with w > u on dA. By the weak comparison principle, w > u
in A. Thus u(x) < k(R - \x\) with k as chosen in (4). By Theorem 3 in [1] and
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RADIAL SYMMETRY OF THE FIRST EIGENFUNCTION
171
Proposition 3.2.1 in [8], there is a c > 0 such that <j)'(r) < —c < 0 in Ri < r < R.
Hence, (f>(r) > c(R - r) in A. This implies / = u/cj> < k/c < +oo in A. Thus
/ei°°(fi).
LEMMA 3. We have fp4>G W01,p(fi).
PROOF. It is obvious that fp<f>G Lp(Q). Now
V(/p<A)= /pV0 + p0/p-1V/.
Also
Vu _ _u_ . _ Vu _
V</>
Hence
(5)
¿Vf = Vu - fV4>.
The above yields
\V(fp<¡>)\
< /p|V0| + p/'-^IVul + /|V^|}.
As /, <¡>,
V(/>G L°°(fi) and u, qbG W¿lP{íí), it follows that fp<pG W1*^).
We now
show fp4> is the strong limit in H/1'p(fi) of a sequence of functions belonging to
Cl(ü).
For n = 2,3,4,...,
let Rn = (n - l)R/n and Rn = Rn + R/2n, and fi„ be the
ball of radius Rn centered at the origin. For each n, let 0 < hn (r) < 1 be a radial
function in C¿ (fi) such that
, . .
/int»-)
K'
/ 1
=
\
10
in 0 < r < ñ„,
—
in Rn < r < R,
and |V/in| < 8n/Ä. Then the functions gn = (fp4>)hn belong to C¿(fi). We will
show that the functions gn converge to fp<f>in W1,p(fi).
By Theorem 3 in [1], qbG C1 (fi) and qbvanishes on dfi. Thus there exists a c > 0
such that qb'(r) > —c in R/2 <r<R.
Integrating, we obtain that qb(r) < c(R —r)
in R/2 < r < R. In particular,
it follows that (f>(r) < c(R — Rn) in Rn < r < Rn.
This implies that
(6)
4>(r) < cR/n
in Rn < r < Rn.
We first show that gn —*fp4> in Lp(fi) as n —>oo. We have
/1»„ - r<t>\p
= [ (fp4>r\i- k\v < wñio [
Jn
Jn
Jn\nn
i
< 11/^11^vol(fi\fin).
The result is obtained by noting that vol(fi\fi„)
—►
0 as n —►
oo. Next we show
that Vgn ->■V(fp(f>) in Lp(fi). By Minkowski's inequality
(X |Vff"" V(/W)
<([
P - {L lV(/P^lP|1- /l"|P) P+ (/n(/^)PlV/l«lP)
|V(/P0)|P]
+||/P||oo(/
^|V/i„|p]
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.
172
TILAK BHATTACHARYA
The first term on the right side of the inequality can be made small by the absolute
continuity of the integral. We estimate the second term as follows. Using (6), we
obtain that
qbp\Vhn\p< (cR/n)p(8n/R)p < M < +00 Vn = 2,3,4,....
Hence
/
qbp]Vhn\p< Mvol(Q\n„).
•/n\n„
Since vol(fi\fin) -^ 0 as n -* 00, we obtain that Vgn -* V(/p0) in Lp(fi). This
completes the proof.
LEMMA 4. The following integrals exist:
(i)/i = /n/p|V<P|p<oo,
(ii) 72 = ¡n |/P-2<p(V0 • V/)|V^|p-2|
< 00,
(iii)/3 = /n^|V/|p<oo.
PROOF. We note using (5) that
|4W-V/)|<|V<¿|{|Vu| + /|V0|}
and
02|V/|p<(|Vu|
+ /|V4>|)p.
Thus I1J2, h exist, as /, qb,V<A€ L°°(fi) and u, qbG W0X'P(Ü).
PROPOSITION 2. Let A and B be vectors in Rn, n>2.
(7)
If 1 < p < 00, then
\A + B\p > \A\P + p\A\p~2A ■B + K\B\P
where \V\ denotes the length of a vector V, and
K=\
[
-(,
\A + B\p-\A\p-p\A\p-*A-B\,
mm('-\B\p-J'
BÍ0,
B = 0.
Furthermore, 0<K<1
and K = 0 iff B = 0.
Before proving Proposition
LEMMA 5. Ift>0,a,
(8)
2, we prove the following Lemma.
and s are any real numbers, and if 1 < p < 00, then
{(l + a)2r2 + s2}p/2>(l
+ ap)ip.
Equality holds in (8) iff s = 0 and either a = 0 or t = 0.
PROOF. Observe that for all s,
(9)
((l + a)2í2 + s2)p/2>|l+Q|píp.
To prove (8), it is then sufficient to prove that for all a
(10)
|l-l-a|p
> 1 + ap.
Case 1. If 1 + a > 0, then |1 + q|p = (1 + a)p. Define
F(a) = (l + a)p-1-ap
Va >-1.
Then
F'(a)=p(l
+ a)p-1-p.
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173
RADIAL SYMMETRY OF THE FIRST EIGENFUNCTION
It is easy to see that F'(a) < 0 when —1 < a < 0, and F'(a) > 0 when a > 0.
Noting that F(0) = 0, we obtain (10). It is clear that, in the case 1 + a > 0,
equality in (10) holds, iff a = 0.
Case 2. If 1 + a < 0, then 1 + ap < 0 so (10) follows trivially, and strict inequality
holds.
To prove the 'only if part of the last assertion of the lemma, we note that for
equality in (8), we must have equality in (9). This implies that 3 = 0. If t ^ 0,
then equality must hold in (10), which implies that a = 0.
PROOF OF PROPOSITION 2. If A = 0, the inequality in (7) follows easily. Let
us assume that A ^ 0. Then we can write B = aA + D, for some scalar a and
some vector D orthogonal to A. Such a decomposition is unique. Then
|A + ß|p = ((l + a)2L4|2 + |£>|2)p/2
and
\A\P + p\A\p~2A -B = (l + ap)\A\p.
From Lemma 5
|A + ß|p>|A|p+p|^|p-2A-ß.
where equality holds iff B = 0. Thus, K in (7) is strictly positive whenever B/0.
This finishes the proof.
We now prove the main theorem of this paper.
THEOREM 1. Let fi be a ball in Rn and u G Wq'v(ü) solve (IaJ.
radially symmetric.
Then u is
PROOF. By the note at the end of Lemma 1, u > 0, and hence (2) yields
/ |Vu|p = Xi [ up.
Jn
Jn
Using (3), we get
¡ \V(f4>)\p
= Xi( (f<p)p.
Jn
By Lemma 3, fpqb is a legitimate
replacing u by qb, (l'x ) gives
Ja
test function in (l'x ). Using tp = fpqb, and
/ \Vqb\»-2Vqb
• V(fpqb)= Ai / (fqb)p.
Ja
Ja
Hence,
(11)
[ |V(/0)|p = f \Vqb\p~2Vqb
■V(fpqb).
Jn
Ja
Now,
/"|V(/^)|p= f \fVqb+ qbVf\p.
Jn
Jn
Setting A = fVqb and B = $VF, and using (7), we have
/ iv(/0)r> Jn/ /piv0ip+p/p-viv<^r2(v^.v/)+^p|v/ip,
Jn
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TILAK BHATTACHARYA
174
where K is as defined in Proposition
side of the inequality we obtain
(12)
2. Combining the first two terms on the right
/ |V(/¿)|p > f |V0|P-2V0•V(fpqb)+ Kqbp\Vf\p.
Jn
Jn
Comparing (11) and (12), we get
L
Kqbp\Vf\p< 0.
As qb > 0 in fi, we obtain Ä"|V/|P — 0. From Proposition 2, K and Vf vanish
together. Hence / is a constant and u is radially symmetric.
Note added in proof. The author has been able to extend this method to prove
that A¿ is simple for C2 domains. The details are worked out in his Ph.D. thesis
supervised by A. Weitsman. He has also learned that S. Sakaguchi and A. Anane
have obtained similar results.
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DEPARTMENT
OF MATHEMATICS,
PURDUE UNIVERSITY,
WEST LAFAYETTE,
INDIANA
47907
Current
address:
Department
of Mathematics,
Northwestern
60201
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University,
Evanston,
Illinois