Ground state solutions for a semilinear problem with
critical exponent
Andrzej Szulkin∗
Tobias Weth
Michel Willem†
To Patrick Habets and to Jean
Mawhin with admiration and
friendship.
Abstract
This work is devoted to the existence and to qualitative properties of ground
state solutions of the Dirchlet problem for the semilinear equation −∆u − λu =
∗
|u|2 −2 u in a bounded domain. Here 2∗ is the critical Sobolev exponent, and the
term ground state refers to minimizers of the corresponding energy within the set
of nontrivial solutions. We focus on the indefinite case where λ is larger than the
first Dirichlet eigenvalue of the Laplacian, and we present a particularly simple
approach to the study of ground states.
Key words. Semilinear Dirichlet problem, critical exponent, ground state, Morse
index, radial solutions.
1.
Introduction
We consider the existence of solutions of the problem
∗
−∆u − λu = |u|2 −2 u in Ω,
(Pλ )
u=0
on ∂Ω,
where Ω is a smooth bounded domain in RN , N ≥ 3, and 2∗ = 2N/(N − 2) denotes the
critical Sobolev exponent. Let
0 < λ1 < λ2 ≤ λ3 . . .
∗
†
Supported in part by the Swedish Research Council.
Supported by Crédit aux chercheurs FNRS 1.5.211.08.
1
be the eigenvalues of the problem
−∆u = λu in Ω,
u=0
on ∂Ω.
When 0 < λ < λ1 , the existence of (positive) ground state solutions was proved by
Brezis and Nirenberg in the classical paper [BN]. When λ ≥ λ1 , the existence of nontrivial solutions was considered in [CFP] (see also [GR],[CW],[Z]). This work is devoted
to the existence of ground state solutions when λ ≥ λ1 .
In section 2, we give a minimax characterization of the ground states and we define
the corresponding Nehari manifold. We use the manifold introduced by Pankov in [P]
and we adapt to the critical case the method of [SW]. Other reduction methods for
variational problems are used in [CL], [ES], [RT], [RY].
In section 3, we prove, under some assumptions, the existence of ground states when
λ ≥ λ1 using the minimax characterization and the direct method of the calculus of
variations. As observed in [SW], this analytical minimax principle is much simpler than
the usual topological minimax theorems (see e.g. [W0]). Moreover, it is not necessary
to use the deformation lemma or the Ekeland variational principle.
Section 4 is devoted to the Morse index and to nodal properties of ground states.
In sections 5 and 6, we consider symmetry properties of ground states and radial
ground states.
As has been pointed out above, our results on the existence of nontrivial solutions
are not new. The novelty (besides the method) is that we can prove there exist ground
states and characterize them by a minimax principle. Moreover, if Ω is a ball, then
we show that for λ between λ1 and λ2 there are two solutions which respectively are a
(nonradial) ground state and a radial ground state. These solutions have exactly two
nodal domains.
2.
Characterization of ground states
We consider the Hilbert space X = H01 (Ω) with scalar product
Z
hu, vi =
∇u · ∇v dx
Ω
and the induced norm ||·||. We fix a positive integer m and assume that λm ≤ λ < λm+1 .
Moreover, we let e1 , . . . , em be normalized eigenfunctions corresponding to λ1 , . . . , λm ,
and we consider the subspaces Z := span(e1 , . . . , em ) and Y := Z ⊥ of X.
The solutions of (Pλ ) are the critical points of the C 1 -functional
Z
Z
1
1
∗
2
2
|u|2 dx
ϕ(u) =
|(∇u| − λu ) dx − ∗
2 Ω
2 Ω
2
defined on X. Let P (resp. Q) be the orthogonal projector onto Y (resp. Z) in X.
Define the map
F : X \ {0} → R × Z ' Rm+1 ,
F (u) = (h∇ϕ(u), ui, Q∇ϕ(u))
It is clear that if u is a nontrivial critical point of ϕ, then F (u) = 0, and if u ∈ Z \ {0},
then h∇ϕ(u), ui < 0.
Lemma 2.1. Let u ∈ X \ {0} be such F (u) = 0. Then F 0 (u) is onto.
Proof. It suffices to prove that for every (t, z) ∈ R × Z, (t, z) 6= 0,
[F 0 (u)(tu + z)] · (t, z) 6= 0.
Here and in the following, we consider the scalar product on R × Z defined by
(t1 , z1 ) · (t2 , z2 ) = t1 t2 + hz1 , z2 i
for t1 , t2 ∈ R, z1 , z2 ∈ Z.
We modify an argument in [P]. Let (t, z) 6= 0. Since h∇ϕ(u), ui = h∇ϕ(u), zi = 0, we
obtain
[F 0 (u)(tu + z)] · (t, z)
= tϕ00 (u)(tu + z, u) + th∇ϕ(u), tu + zi + ϕ00 (u)(tu + z, z)
= ϕ00 (u)(tu + z, tu + z) − th∇ϕ(u), tu + 2zi
Z
Z
∗
2
2
=
|∇z| − λz dx − [(2∗ − 1)(tu + z)2 − tu(tu + 2z)]|u|2 −2 dx
Z
Z
∗
2
2
=
|∇z| − λz dx − [(2∗ − 2)t2 u2 + 2(2∗ − 2)tzu + (2∗ − 1)z 2 ]|u|2 −2 dx.
The last integrand is positive definite in t and z(x) whenever u(x) 6= 0. If
Z
|∇z|2 − λz 2 dx < 0,
the proof is complete. Assume that
Z
|∇z|2 − λz 2 dx = 0.
Then either λ = λm and z is an eigenfunction, or z = 0. If z 6= 0, then z 6= 0 a.e. on Ω
and if z = 0, then t 6= 0. In both cases it is easy to see that [F 0 (u)(tu + z)] · (t, z) < 0.
By the preceding lemma,
M := {u ∈ X \ {0} : F (u) = 0}
is a C 1 -submanifold of X with codimension m + 1.
3
Theorem 2.2. Let u ∈ X. Then u is a nontrivial critical point of ϕ if and only if
(2.1)
u ∈ M, ϕ0 (u)
= 0.
Tu M
Proof. The necessary condition is clear. Assume that u satisfies (2.1). By definition,
ϕ0 (u) vanishes on the subspace Ru ⊕ Z and on Tu M. Since, by the proof of Lemma
2.1, Ru ⊕ Z is transverse to Tu M, ϕ0 (u) vanishes on X.
By Proposition 2.3 and Lemma 2.8 in [SW], for every v ∈ Y \{0}, there exists a
unique (f (v), g(v)) ∈ ]0, +∞[ ×Z, continuously depending on v and such that
F (f (v)v + g(v)) = 0.
Moreover
ϕ(f (v)v + g(v)) = max
ϕ(tv + w).
t>0
w∈Z
We define the value
c = c(λ) = inf ϕ = inf ϕ(f (v)v + g(v))
M
v∈Y
v6=0
= inf max
ϕ(tv + w).
t>0
v∈Y
v6=0
w∈Z
Since M is a C -submanifold of X, any u ∈ M with ϕ(u) = c satisfies ϕ (u)
0
1
Tu M
= 0,
therefore it is a critical point of ϕ by Theorem 2.2. We call such a function u a ground
state solution, since Theorem 2.2 implies that u minimizes ϕ within the set of nontrivial
solutions of (Pλ ).
3.
Existence of ground states
Let us recall that the best Sobolev constant
R
S=
inf1
u∈H0 (Ω)
u6=0
|∇u|2 dx
2/2∗
R
|u|2∗ dx
Ω
Ω
is independent of Ω.
We shall use the following elementary lemma.
Lemma 3.1. Let A > 0, B > 0. Then
2
N/2
∗
t
1
t2
A
max A − B ∗ =
.
t>0
2
2
N B 2/2∗
Let us also recall that λm ≤ λ < λm+1 .
4
Lemma 3.2. Suppose
c < S N/2 /N.
(3.1)
Then there exists v ∈ Y \{0} such that
max
ϕ(tv + w) = ϕ(f (v)v + g(v)) = c.
t>0
w∈Z
Proof. Let (vn ) ⊂ Y be such that ||vn || = 1 and
max
ϕ(tvn + w) → c.
t>0
(3.2)
w∈Z
We can assume that
vn * v in Y,
vn → v in L2 (Ω),
vn → v a.e. on Ω.
We define
Z
|∇(vn − v)| dx,
A = lim
n→∞
Z
2
B = lim
n→∞
Ω
∗
|vn − v|2 dx.
Ω
Using the Brezis-Lieb lemma (see [BL] or [W1]), we obtain, from (3.2), for every t > 0
and for every w ∈ Z,
∗
t2
t2
ϕ(tv + w) + A − B ∗ ≤ c.
2
2
(3.3)
Assume that v = 0 and B = 0. Since ||vn || = 1, it follows that A = 1 and, for every
t > 0,
t2
≤ c,
2
a contradiction.
Assume now that B 6= 0. We obtain from the Sobolev inequality that
(3.4)
S N/2
1
≤
N
N
A
B 2/2∗
N/2
2
∗
t
t2
= max A − B ∗ .
t>0
2
2
If v = 0, it follows from (3.1), (3.3) and (3.4) that
S N/2
S N/2
≤c<
,
N
N
a contradiction. Thus v 6= 0.
5
Let us define h = (f (v))−1 g(v). Using the definition of c, we have
(3.5)
c ≤ ϕ(f (v)(v + h)) = max ϕ(t(v + h))
t>0
R
N/2
[|∇v|2 + |∇h|2 − λ(v 2 + h2 )]dx
1
Ω
R
=
.
N
( Ω |v + h|2∗ dx)2/2∗
Inequality (3.3) implies that
∗
t2
t2
max ϕ(t(v + h)) + A − B ∗
t>0
2
2
R
N/2
1 A + Ω [|∇v|2 + |∇h|2 − λ(v 2 + h2 )]dx
R
=
≤ c.
N
(B + Ω |v + h|2∗ dx)2/2∗
(3.6)
It follows from (3.1) (3.4) (3.5) and (3.6) that
(N c)2/N B +
Z
"
2/2∗
Z
2/2∗ #
∗
∗
|v + h| dx
< (N c)2/N B 2/2 +
|v + h|2 dx
2∗
Ω
Ω
Z
< A+
[|∇v|2 + |∇h|2 − λ(v 2 + h2 )]dx
Ω
2/N
≤ (N c)
2/2∗
Z
2∗
B + |v + h| dx
,
Ω
a contradiction.
Thus B = 0 and we conclude from (3.3) that, by the definition of c,
c ≤ ϕ(f (v)v + g(v)) ≤ c.
In order to prove inequality (3.1), we need some preliminary results.
Lemma 3.3. Let w ∈ Z be such that w = 0 on a nonempty open subset ω of Ω. Then
w = 0 on Ω.
Proof. Suppose w 6= 0. We can assume that w = w1 + . . . + wk , where wj is an
eigenfunction corresponding to µj ∈ {λ1 , . . . , λm }. We can also assume that µ1 < . . . <
µk . Since, on ω,
0 = µk w + ∆w = (µk − µ1 )w1 + . . . + (µk − µk−1 )wk−1
we can replace wk by 0. After k − 1 steps, we obtain an eigenfunction (corresponding
to µ1 ) vanishing on ω. This is impossible.
6
We may assume that 0 ∈ Ω. Let ψ ∈ D(Ω) be a radial test function such that
0 ≤ ψ ≤ 1 and ψ ≡ 1 in a neighbourhood of 0. Let us recall that the optimal function
for the Sobolev inequality in D1,2 (RN ) is the instanton
Uε (x) = [N (N − 2)](N −2)/4
ε(N −2)/2
.
(ε2 + |x|2 )(N −2)/2
Lemma 3.4. The truncated instanton
uε (x) = ψ(x)Uε (x)
satisfies the following estimates :
Z
a)
|∇uε |2 dx = S N/2 + O(εN −2 ),
Ω
Z
∗
u2ε dx = S N/2 + O(εN ),
b)
Ω
Z
N −2 Z
N −2 Z
N −2 2∗ −1
uε dx = O ε 2 , uε dx = O ε 2 , |∇uε |dx = O ε 2 ,
c)
Ω
Ω
Ω
 2
2
Z
 dε | ln ε| + O(ε ), if N = 4,
2
uε dx ≥
d)
 2
Ω
dε + O(εN −2 ),
if N ≥ 5,
where d is a positive constant depending on N .
Proof. See [BN] and [W0] pp. 35 and 52.
Lemma 3.5. Let N = 4 and λm < λ < λm+1 or let N ≥ 5 and λm ≤ λ < λm+1 .
Then c(λ) < S N/2 /N .
Proof. It suffices to prove that, for ε > 0 small enough,
(3.7)
max
ϕ(tuε + w) < S N/2 /N.
t>0
w∈Z
Let ω = Ω\supp ψ. By Lemma 3.3, ||w||L2∗ (ω) defines a norm on Z. Since
dim Z = m,
all the norms are equivalent on Z. We obtain, by convexity, for every t > 0 and for
every w ∈ Z,
Z
Z
Z
∗
2∗
2∗
|tuε + w| dx =
|tuε + w| dx + |w|2 dx
Ω
Ω\ω
Z
Zω
∗
∗
∗
∗
∗
≥ t2
u2ε dx + 2∗ t2 −1 u2ε −1 wdx + 2∗ c1 ||w||2 .
Ω
Ω
7
It follows that
Z
ϕ(tuε + w) ≤ ϕ(tuε ) + t ∇uε · ∇w − λuε wdx
Ω
Z
Z
1
∗
∗
2
2
2∗ −1
u2ε −1 wdx − c1 ||w||2 .
+
|∇w| − λw dx − t
2 Ω
Ω
(3.8)
In particular, if 0 < ε < 1,
ϕ(tuε + w) ≤ A(t2 + t||w|| + t2
∗ −1
∗
∗
||w||) − B(t2 + ||w||2 )
for some A, B > 0. Hence there exists R > 0 such that, for 0 < ε < 1, t > R and
w ∈ Z, ϕ(tuε + w) ≤ 0. Inequality (3.8) implies that, for t ≤ R,
N −2 N −2 ∗
ϕ(tuε + w) ≤ ϕ(tuε ) + O ε 2 ||w|| − c1 ||w||2 ≤ ϕ(tuε ) + O εN N +2 .
Indeed, for any p > 1, by Young’s inequality,
max(rs − p1 sp ) =
s>0
p
p−1 p−1
r .
p
Since N (N − 2)/(N + 2) > 2 whenever N ≥ 5, we obtain, from Lemmas 3.1 and 3.4,
for ε > 0 small enough,
R
|∇uε |2 − λu2ε dx
Ω
2/2∗
R
u2∗ dx
Ω ε
1
max
ϕ(tu
ε + w) ≤
t>0
N
w∈Z
!N/2
N −2 + O εN N +2
N −2 S N/2
1
2 (2−N )/2
N −2 N/2
S − λdε S
+ O(ε
)
+ O εN N +2 <
.
≤
N
N
Assume now that N = 4 and λm < λ < λm+1 . Inequality (3.8) implies that, for
t ≤ R,
ϕ(tuε + w) ≤ ϕ(tuε ) + O(ε)||w|| − c2 ||w||2 ≤ ϕ(tuε ) + O(ε2 ).
We obtain, from Lemma 3.4, for ε > 0 small enough,
max
ϕ(tuε + w) ≤
t>0
w∈Z
=
1
4
R
|∇uε |2 − λu2ε dx
Ω
∗
R
2∗ dx 2/2
u
ε
Ω
!2
+ O(ε2 )
2
S2
1
S − λdε2 | ln ε|S −1 + O(ε2 ) + O(ε2 ) <
.
4
4
Theorem 3.6. Let N = 4 and λm < λ < λm+1 or N ≥ 5 and λm ≤ λ < λm+1 . Then
there exists a solution u of (Pλ ) such that ϕ(u) = c.
8
Proof. Lemma 3.2 and Lemma 3.5 imply the existence of u ∈ M such that ϕ(u) = c.
In particular ϕ0 (u)
= 0. It follows from Theorem 2.2 that ϕ0 (u) = 0.
Tu M
Remarks. a) When λm < λ < λm+1 , the inequality (3.7) is contained in [CFP]. When
λm = λ, this inequality seems new; however, see [GR] where a similar result may be
found for a maximum defined in a somewhat different way (the subspace Z is modified
in order to make it orthogonal to uε ).
b) Under the assumptions of Theorem 3.6, the existence of a nontrivial solution of
(Pλ ) was proved in [CFP] (see also [Z] and [GR]).
c) A nontrivial solution also exists if N = 4 and λ = λ1 . This is a direct consequence
of a more general multiplicity result in [CW]. However, the existence of a ground state
solution remains an open question in this case.
4.
Nodal properties of the ground states
We need the following elementary lemma.
Lemma 4.1. Let M be a C 1 -submanifold of the Hilbert space X, ϕ ∈ C 2 (X, R) and
u ∈ M be such that ϕ0 (u) = 0 and ϕ(u) = inf ϕ. Then, for every h ∈ Tu M,
M
ϕ00 (u)(h, h) ≥ 0.
Proof. Let γ : ] − 1, 1[ → M be a C 1 -curve such that γ(0) = u and γ̇(0) = h. Since
d
ϕ ◦ γ(t) = h∇ϕ(γ(t)), γ̇(t)i,
dt
we obtain
d2
1
ϕ ◦ γ(0) = lim
h∇ϕ(γ(t)), γ̇(t)i
2
t→0
dt
t
t6=0
= lim
h
t→0
t6=0
∇ϕ(γ(t)) − ∇ϕ(γ(0))
, γ̇(t)i
t
= ϕ00 (γ(0))(γ̇(0), γ̇(0))
and
d2
ϕ ◦ γ(0) = ϕ00 (u)(h, h).
dt2
The Morse index of a critical point u of ϕ will be denoted by M (u).
0≤
Theorem 4.2. Let λm ≤ λ < λm+1 and let u ∈ M be such that ϕ(u) = c. Then
ϕ0 (u) = 0 and M (u) = m + 1.
9
Proof. Since c = inf ϕ, it is clear that ϕ (u)
0
M
Tu M
= 0. By Theorem 2.2, u is a critical
point of ϕ. The preceding lemma implies that, for every h ∈ Tu M = (F 0 (u))−1 (0),
ϕ00 (u)(h, h) ≥ 0.
We conclude that
M (u) ≤ m + 1.
It follows from the proof of Lemma 2.1 that
M (u) ≥ m + 1.
The number of nodal regions of a (classical) solution u of (Pλ ) is denoted by nod(u).
The following result is due to Benci and Fortunato [BF]. We recall the proof for the
sake of completeness.
Theorem 4.3. Let u be a nontrivial critical point of ϕ. Then nod(u) ≤ M (u).
Proof. By elliptic regularity theory, u is a classical solution of (Pλ ). Let Ω1 , ..., Ωn
be the nodal domains of u and define uj = χΩj u, j = 1, ..., n. Theorem IX.17 and
Remarque 20 in [B] imply that uj ∈ H01 (Ω). We obtain
Z
Z
∗
00
2
2
∗
|∇uj | − λuj dx − (2 − 1) |u|2 −2 |uj |2 dx
ϕ (u)(uj , uj ) =
Ω
ZΩ
Z
∗
<
|∇uj |2 − λu2j dx − |uj |2 dx
Ω
Ω
= hϕ0 (uj ), uj i = 0,
and therefore n ≤ M (u).
Corollary 4.4. Let λm ≤ λ < λm+1 and let u ∈ M be such that ϕ(u) = c. Then
nod(u) ≤ m + 1. If moreover m = 1, then nod(u) = 2.
Proof. The above theorems imply that
nod(u) ≤ M (u) = m + 1.
If m = 1, then λ1 ≤ λ < λ2 , and it is easy to verify that 2 ≤ nod(u).
Remark. It was proved in [BF] that, under the assumptions of Theorem 3.6, there
exists a non trivial solution of (Pλ ) such that
nod(u) ≤ M (u) ≤ m + 1.
10
5.
Symmetry properties
In this section we assume that Ω is a radial bounded domain, so it is a ball or an annulus
in RN centered at the origin. We focus on the case λ1 ≤ λ < λ2 , and we discuss the
shape of ground state solutions of (Pλ ). We recall that a function u defined on a radial
N
domain is said to be foliated Schwarz symmetric if there is a unit
vector p ∈ R , |p| = 1
x
such that u(x) only depends on r = |x| and θ = arccos |x|
· p and u is nonincreasing
in θ.
Theorem 5.1.
Let λ1 ≤ λ < λ2 and let u ∈ M be such that ϕ(u) = c = inf ϕ. Then
M
u is foliated Schwarz symmetric. Moreover, if the underlying domain Ω is a ball, then
u is nonradial.
Proof. Let, as before, e1 denote the first eigenfunction of the Dirichlet Laplacian in Ω.
In the present case we have
Z
Z
o
n
∗
2
2
2∗
M = u ∈ X \ {0} : (|∇u| − λu − |u| ) dx = 0 = (∇u·∇e1 − λue1 − |u|2 −2 ue1 ) dx
Ω
Z
ZΩ
o
n
∗
∗
(|∇u|2 − λu2 − |u|2 ) dx = 0 = [(λ1 − λ)ue1 − |u|2 −2 ue1 ] dx
= u ∈ X \ {0} :
Ω
Ω
Now let u ∈ M be such that ϕ(u) = c. We pick x0 ∈ int(Ω), x0 6= 0, with
u(x0 ) = max{u(x) : x ∈ Ω, |x| = |x0 |}
and put p := |xx00 | . Moreover, we let Hp denote the family of all closed halfspaces H in
RN such that 0 lies in the hyperplane ∂H and p is contained in the interior of H. For
H ∈ Hp we consider the reflection σH : RN → RN with respect to the boundary of H.
We claim that, for every H ∈ Hp ,
u(x) ≥ u(σH (x))
(5.1)
for all x ∈ H ∩ Ω.
To prove this, we fix H ∈ Hp and recall a simple rearrangement, namely the polarization
of u with respect to H defined by
(
max{u(x), u(σH (x))},
x∈Ω∩H
uH (x) =
min{u(x), u(σH (x))},
x ∈ Ω \ H.
It is well known and fairly easy to prove that
Z
Z
Z
Z
2
2
p
|∇uH | dx =
|∇u| dx and
|uH | dx =
|u|p dx
Ω
Ω
Ω
for every p ∈ [1, ∞].
Ω
Moreover, since e1 is a radial function, we also have
Z
Z
Z
Z
p−2
uH e1 dx =
ue1 dx and
|uH | uH e1 dx =
|u|p−2 ue1 dx for every p ∈ [1, ∞]
Ω
Ω
Ω
Ω
11
(for details on these invariance properties, see e.g. [BWW, Section 2]). Consequently,
uH ∈ M and ϕ(uH ) = ϕ(u) = c, which implies that both u and uH are critical points
of ϕ and hence classical solutions of (Pλ ). Therefore w := uH − u is a nonnegative
function in Ω ∩ H which solves the Dirichlet problem
−∆w = q(x)w
in int(H ∩ Ω),
w = 0 on ∂Ω ∩ H,
where
∗
Z
q(x) = λ + (2 − 1)
1
|(1 − t)u(x) + tuH (x)|2
∗ −2
dt
for x ∈ Ω ∩ H.
0
Since q ∈ L∞ (Ω ∩ H), the strong maximum principle implies that either w ≡ 0 or
w > 0 in int(H ∩ Ω). The latter case is ruled out since x0 ∈ int(H ∩ Ω) and w(x0 ) =
uH (x0 ) − u(x0 ) = 0 by the choice of x0 . We therefore obtain w ≡ 0, hence u = uH and
(5.1) holds.
Now from (5.1) it follows by continuity that u is symmetric with respect to every
hyperplane containing p, so it is axially
symmetric
with respect to p. Hence u(x) only
x
depends on r = |x| and θ = arccos |x| · p . Moreover, it also follows from (5.1) that u
is nonincreasing in θ. We conclude that u is foliated Schwarz symmetric.
Finally, if Ω is a ball, a result of Aftalion and Pacella (see [AP]) implies that every
radial solution of (Pλ ) has Morse index greater than or equal to N + 1, whereas u has
Morse index 2 by Theorem 4.2. Hence u is nonradial in this case.
Remarks. a) In the radial setting considered here, the Dirichlet eigenvalues of the
Laplacian satisfy λ1 < λ2 = · · · = λN +1 < λN +2 . The proof of Theorem 5.1 does not
extend to the case λ ≥ λN +1 , since in this case the set M is not invariant with respect
to polarization.
b) In dimensions 3 ≤ N ≤ 6 where the critical nonlinearity in (Pλ ) has a convex
derivative, Theorem 5.1 can also be deduced from Theorem 4.2 above and Theorem
1.1 in [PW], the latter being a rather general result relating Morse index estimates to
foliated Schwarz symmetry. However, the proof given here is simpler and works for
every N ≥ 3.
6.
Radial ground states
As in the preceding section, we assume Ω is a ball or an annulus centered at the origin.
1
Let Xr = H0,r
(Ω) be the subspace of H01 (Ω) consisting of radial functions and let
ϕr := ϕ|Xr . Since the group O(N ) of orthogonal transformations acts isometrically
on X by means of the formula gu(x) = u(g −1 x) (where g ∈ O(N )) and Xr is the
space of fixed points of this action, it follows from the principle of symmetric criticality
[W0, Theorem 1.28] that critical points of ϕr are radial solutions of (Pλ ). Denote the
eigenvalues of −∆ to which there correspond eigenfunctions in Xr by λri . It is well
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known that all λri are simple (in Xr ), λ1 = λr1 and λ2 = ... = λN +1 < λr2 . An easy
inspection shows that the results of Sections 2-4 remain valid in Xr (with λm replaced
by λrm ). Let Mr := M ∩ Xr and
cr := inf ϕr .
Mr
Theorem 6.1. (i) Let N = 4 and λrm < λ < λrm+1 or N ≥ 5 and λrm ≤ λ < λrm+1 .
Then there exists ur such that ϕr (ur ) = cr . Moreover, each ur ∈ Mr such that ϕr (ur ) =
cr is a radial solution of (Pλ ).
(ii) Let λrm ≤ λ < λrm+1 and let ur ∈ Mr be such that ϕr (ur ) = cr . Then nod(ur ) =
m + 1.
Proof. We only have to prove that nod(ur ) ≥ m+1. The function ur is an eigenfunction
of

 −∆v − V (x)v = µn v in Ω,
v is a radial function,

v=0
on ∂Ω,
∗
where V = |ur |2 −2 is a radially symmetric function and µn = λ ≥ λrm is the n-th
eigenvalue of the operator −∆ − V in Xr . Since V ≥ 0 and V 6≡ 0, we have that
µn < λrn . It follows that m < n. By the Sturm-Liouville theory, nod(ur ) = n ≥ m + 1.
If Ω is a ball and λ1 ≤ λ < λ2 , then by Theorem 5.1, any u ∈ M such that ϕ(u) = c
is nonradial. Hence we obtain the following
Corollary 6.2. Suppose Ω is a ball and λ1 < λ < λ2 if N = 4, λ1 ≤ λ < λ2 if N ≥ 5.
Then the ground state solution u obtained in Theorem 3.6 and the radial ground state
solution ur obtained in Theorem 6.1 are distinct. Moreover, c < cr and both u and ur
have exactly two nodal domains.
Remarks. a) If Ω is a ball, existence of nontrivial solutions of (Pλ ) for each λ > 0
and N ≥ 4 has been shown in [FJ]. However, the solutions obtained there, although
symmetric, are not radial and need not be ground states.
b) Let Ω be the unit ball in RN , N = 4 or 5. According to [ABP0] and [ABP1], the
bifurcation branch arising from λr2 tends asymptotically to λ1 (= λr1 ) as the L∞ -norm
of the solution tends to infinity. It is shown in [AGGS], by a computer assisted proof,
that this branch does not cross λ1 when N = 4. This corresponds to our assumption
λ1 < λ < λr2 , when N = 4. On the other hand, when N = 5, this branch crosses λ1
(see [GG]).
c) Let Ω be a ball in R3 . By Theorem 1 in [C], for every λ ≥ λ1 there exists a nodal
solution of (Pλ ). Again, these solutions, although symmetric, are not radial and need
not be ground states.
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