Uniqueness of the ground state solutions of quasilinear Schrödinger equations

Nonlinear Analysis 75 (2012) 819–833

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Uniqueness of the ground state solutions of quasilinear Schrödinger equations Shinji Adachi a,∗ , Tatsuya Watanabe b a

Division of Basic Engineering, Faculty of Engineering, Shizuoka University, 3-5-1 Johoku, Naka-ku, Hamamatsu, 432-8561, Japan

b

Department of Mathematics, Faculty of Science, Kyoto Sangyo University, Motoyama, Kamigamo, Kita-ku, Kyoto-City, 603-8555, Japan

article

info

Article history: Received 11 April 2011 Accepted 9 September 2011 Communicated by S. Carl MSC: 35J62 35A02

abstract We are concerned with the uniqueness result of positive solutions for a class of quasilinear elliptic equation arising from plasma physics. We convert a quasilinear elliptic equation into a semilinear one and show the unique existence of positive radial solution for original equation under the suitable conditions on the power of nonlinearity and quasilinearity. We also investigate the non-degeneracy of a positive radial solution for a converted semilinear elliptic equation. © 2011 Elsevier Ltd. All rights reserved.

Keywords: Quasilinear elliptic equation Uniqueness of positive radial solution Non-degeneracy

1. Introduction In this paper, we consider the following quasilinear Schrödinger equation: i

∂z = −1z − |z |p−1 z − κ ∆(|z |α )|z |α−2 z , ∂t

(t , x) ∈ (0, ∞) × RN ,

(1.1)

where κ > 0, α > 1, N ≥ 1 and p > 1. Eq. (1.1) with α = 2 derives from a superfluid film equation in plasma physics, which was introduced in [1,2]. We are interested in the standing wave solution of the form: z (t , x) = u(x)eiλt , λ > 0, where u : RN → C is a complex valued function. Then we obtain the following quasilinear elliptic problem:

− 1u + λu − κ ∆(|u|α )|u|α−2 u = |u|p−1 u in RN . Here we call u a (weak) solution of (1.2) if

∫ ℜ RN

(∇ u · ∇φ + λuφ + κα(α − 2)|∇|u ‖2 |u|2α−4 uφ + κα∇|u| · ∇(uφ)|u|2α−3 − |u|p−1 uφ)dx = 0

for all φ ∈ C0∞ (RN , C), where ℜ(z ) is the real part of z ∈ C and z is the complex conjugate of z ∈ C.

∗

Corresponding author. E-mail address: [email protected] (S. Adachi).

0362-546X/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2011.09.015

(1.2)

820

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

Eq. (1.2) has a variational structure, that is, one can obtain solutions of (1.2) as critical points of the associated functional J defined by

∫ ∫ κ 1 |∇|u |α |2 dx − |u|p+1 dx 2 RN 2α RN p + 1 RN ∫ ∫ ∫ 1 ακ 1 = (|∇ u|2 + λ|u|2 )dx + |∇|u ‖2 |u|2α−2 d − |u|p+1 dx. (1.3) 2 RN 2 RN p + 1 RN  We remark that nonlinear functional RN |∇|u ‖2 |u|2α−2 dx is not defined on all H 1 (RN , C) except for N = 1. Thus the natural function space for N ≥ 2 is given by   ∫ XC := u ∈ H 1 (RN , C); |∇|u ‖2 |u|2α−2 dx < ∞ . (1.4) J (u) =

1

∫

(|∇ u|2 + λ|u|2 )dx +

RN

The existence of a solution of (1.2) has been studied in [3–7]. The purpose of this paper is to study the uniqueness of the ground state solution of (1.2). We believe it is important for applications, for example, the stability of the standing wave solution. To state the existence of a ground state solution, we use the following notation. We define the ground state energy level and the set of ground states by m := inf{J (u); J ′ (u) = 0, u ∈ XC \ {0}},

G := {u ∈ XC \ {0}; J (u) = m, J ′ (u) = 0}. In [3,8], they showed the set G has at least one element which is real valued, positive, radially symmetric, decreasing with respect to r = |x| and has the exponential decay. To prove the uniqueness of ground states, we need more precise properties on ground states. Actually we have the following result which was obtained by Colin et al. [9] for the case α = 2. Theorem 1.1. Let λ > 0, κ > 0, α > 1 and 1 < p < α ∗ := iθ

(2α−1)N +2 N −2

for N ≥ 3, 1 < p < ∞ for N = 1, 2. Then G ̸= ∅

and any u ∈ G is of the form u(x) = e |u(x)| for some θ ∈ R. Moreover any real non-negative ground state w ∈ G satisfies the following properties: (i) (ii) (iii) (iv)

w ∈ C 2 (RN ). w(x) > 0 for all x ∈ RN . w is radially symmetric: w(x) = w(|x|) and decreases with respect to r = |x|. There exist c , c ′ > 0 such that √ N −1 ∂w lim e λr (r + 1) 2 = −c ′ . r →∞ ∂r

√

N −1 lim e λ|x| (|x| + 1) 2 w(x) = c ,

|x|→∞

The derivation of α ∗ is rather clear by (1.4). Indeed (1.4) implies ‖∇|u|α ‖L2 < ∞. By the Sobolev embedding, it follows 2α N

2N

|u|α ∈ L N −2 (RN ), which is equivalent to u ∈ L N −2 (RN ) and α ∗ = N2α−N2 − 1. We can also see in Theorem 2.11 below that p = α ∗ is the critical exponent for (1.2). When N = 1, the uniqueness of ground states was already studied in [9]. The main purpose of this paper is to prove the uniqueness for N ≥ 2. Now Theorem 1.1 implies that any u ∈ G is, up to a phase shift (gauge transformation), real valued and non-negative. Thus we show that the set

{u ∈ X ∩ C 2 (RN ); u is a positive radial solution of (1.2)} consists of just one element if we assume additional conditions on κ, λ, α and p, where X denotes the restriction of XC to the real function:

 X =

u ∈ H 1 (RN , R);

∫ RN

 |∇ u|2 |u|2α−2 dx < ∞ .

Our main result is the following. Theorem 1.2. Suppose N ≥ 3, α > 1 and



1 < p < α∗ α − 1 ≤ p < α∗

if 1 < α ≤ 2, if α > 2. 2α−2

There exists c0 = c0 (p, α) > 0 such that if κλ p−1 ≥ c0 , then (1.2) has at most one real positive radial solution u and hence the ground state solution of (1.2) is unique up to translation and phase shift, that is, any element of G is of the form eiθ u(x + y) for θ ∈ R and y ∈ RN .

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

821

2 We emphasize that the uniqueness in Theorem 1.2 holds even if p is H 1 -supercritical because α ∗ > NN + whenever α > 1. −2 As we will see later, we need a stronger assumption when N = 2. We obtain the following sufficient conditions for the uniqueness of positive radial solutions when N = 2. 2α−2

Theorem 1.3. Suppose N = 2, α > 1 and 2α − 1 ≤ p < ∞. There exists c1 = c1 (p, α) > 0 such that if κλ p−1 ≥ c1 , then (1.2) has at most one real positive radial solution u and hence the ground state solution of (1.2) is unique up to translation and phase shift, that is, any element of G is of the form eiθ u(x + y) for θ ∈ R and y ∈ R2 . 1

1

Remark 1.4. For a solution u of (1.2), we rescale u˜ (x) as u(x) = λ p−1 u˜ (λ 2 x). Then we can see that (1.2) is reduced to

−1u˜ + u˜ − κλ

2α−2 p−1

∆(|˜u|α )|˜u|α−2 u˜ = |˜u|p−1 u˜ in RN .

Thus it seem to be natural to describe the condition for the uniqueness in terms of κλ

2α−2 p−1

.

To prove Theorems 1.1–1.3, we adapt the dual approach as in [3,4]. More precisely, we convert our quasilinear equation into a semilinear equation by using a suitable translation f . We will see that the set of ground states G has one-to-one correspondence to that of the semilinear problem. This enables us to apply the uniqueness result [10–13] for semilinear elliptic equations. This paper is organized as follows. In Section 2, we introduce the dual approach and prove Theorem 1.1. In Sections 3 and 4, we study the uniqueness for N ≥ 3 and N = 2 respectively. Finally in Section 5, we study the non-degeneracy of the ground state solution of the corresponding semilinear problem. Notation. Throughout this paper, c > 0 denote various positive constants which are not essential to our problem and may change from line to line. 2. Dual approach and proof of Theorem 1.1 2.1. Properties of the unique solution of the ODE related to (1.2) In this subsection, we study detailed properties of the unique solution of the ODE related to (1.2). As we will see later, this unique solution gives one-to-one correspondence between (1.2) and a semilinear elliptic problem. Let f˜ be a function defined by f˜ (s) :=

∫ s

1 + κα t 2α−2 dt .

(2.1)

0

Then f˜ is positive, monotone, convex and C ∞ on (0, ∞). For s < 0, we put f˜ (s) = −f˜ (−s). Since f˜ is monotone, we can define the inverse function f . Then f satisfies the following ODE: 1 f ′ (s) =  1 + ακ f (s)2α−2

on s ∈ [0, ∞),

f (0) = 0.

(2.2)

From (2.2), we can observe that f ′′ (s) = −κα(α − 1)f 2α−3 (f ′ )4 = (α − 1) 1

f ′′′ (s) =

f2

(f ′ )4 f

− (α − 1)

Lemma 2.1. f (s) satisfies the following properties: (i) f (s) ≤ s, 0 < f ′ (s) ≤ 1, f ′′ (s) ≤ 0 for all s ≥ 0. (ii) α1 f (s) ≤ sf ′ (s) ≤ f (s) for all s ≥ 0. f (s)

(iii) ( sf ′ (s) )′ > 0 for all s > 0. Lemma 2.2. It follows

(ii) (iii)

f (s) 1

1

= ( ακ ) 2α , lims→0

sα f ′ (s) lims→∞ 1−α s α f (s) lims→∞ sf ′ (s)

f

,

{4(α − 1)2 (f ′ )7 − (6α − 5)(α − 1)(f ′ )5 + (α − 1)(2α − 1)(f ′ )3 }.

The next two lemmas play important roles in Sections 3–5.

(i) lims→∞

(f ′ )2

1 2α

= α1 ( ακ ) . = α.

f (s) s

= 1.

(2.3) (2.4)

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S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

For the proof of (i) and (ii) of Lemmas 2.1 and 2.2, we refer to [3]. Here we give the proof of (iii) of Lemma 2.1. Proof. By (2.3) and (ii) of Lemma 2.1, we have



′

f sf ′

=

1

(sf ′ )2

(s(f ′ )2 − ff ′ − sff ′′ )

1

(s(f ′ )2 − s(f ′ )2 − sff ′′ ) (sf ′ )2 −ff ′′ 1 − (f ′ )2 = = (α − 1 ) . s(f ′ )2 s ≥

Since 0 < f ′ < 1 for all s > 0, we obtain the desired inequality.



Next we study further properties of f . The next lemma follows easily from (ii) of Lemma 2.2. Lemma 2.3. Suppose α, m > 1. Then it follows

lim sf ′ (s)m =

s→∞

  0     

if α >

 m+2 1

m−1  m     +∞

m−1 2

if α =



c

(ii) There exists c > 0 such that lims→∞ ψ(s) = +∞

ψ′ = α −

1 f ′2

f , f′

, , m

m−1

.

Then ψ(s) ≥ 0 and it satisfies:

(i) ψ ′ (s) = (α − 1)f ′ (s)2 .

Proof. Since ψ = α s −

m−1

if 1 < α <

α sf ′ (s)−f (s) . f ′ (s)

Lemma 2.4. Let ψ(s) =

κ−

m m−1 m

if α > 2, if 1 < α ≤ 2.

we have from (2.3) that 1

(f ′2 − ff ′′ ) = α −

f ′2

(f ′2 − (α − 1)f ′4 + (α − 1)f ′2 ) = (α − 1)f ′2 .

Next we show (ii). From (i), we have

ψ(s) = (α − 1)

s

∫

f ′ (t )2 dt . 0

First we suppose α > 2. By (ii) of Lemma 2.2, there exists L > 0 such that for s > L, s

∫

f ′ (t )2 dt ≤ c

s

∫

t

2−2α

α

dt

L

L

∞ for some c > 0. If α > 2, then 2−α2α < −1. This implies 0 f ′ (t )2 dt < ∞. Thus we have lim ψ(s) = (α − 1)

s→∞

∞

∫

f ′ (t )2 dt ∈ (0, ∞). 0

Next we suppose 1 < α ≤ 2. By (ii) of Lemma 2.2, there exists L > 0 such that for s > L, s

∫

f ′ (t )2 dt ≥ c

s

∫

L

t

2−2α

α

 dt =

L

for some c > 0. This implies lims→∞

2−α

cs α − c c log s − c

s 0

if 1 < α < 2, if α = 2

f ′ (t )2 dt = ∞ and hence lims→∞ ψ(s) = ∞.



Finally we show the following asymptotic behavior which also plays an important role in Sections 3–5. Lemma 2.5. It follows lim

s→∞

sf

′3

α sf ′ − f

=

 0 

2−α

α(α − 1)

if α ≥ 2, if 1 < α < 2.

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833 sf ′3 α sf ′ −f

sf ′2

823 sf ′2

Proof. First we observe that = ψ . By Lemma 2.3 and (ii) of Lemma 2.4, we have lims→∞ ψ = 0 if α ≥ 2. In the case 1 < α < 2, we apply l’Hôpital’s rule. Then by Lemma (i) of Lemma 2.4 and from (2.3), we have lim

s→∞

sf ′2

f ′2 + 2sf ′ f ′′ = lim s→∞ s→∞ (α − 1)f ′2 ψ ψ′    ′ 1 2sf ′′ 1 sf ′ ′2 sf = lim 1 + = lim 1 − 2(α − 1) + 2(α − 1)f . α − 1 s→∞ f′ α − 1 s→∞ f f

= lim

f ′2 + 2sf ′ f ′′

By (ii) and (iii) of Lemma 2.2, we obtain lim

s→∞

sf ′2

=

ψ



1

1−

α−1

2(α − 1)



α

and the proof of Lemma 2.3 is complete.

=

2−α α(α − 1)



2.2. Dual variational formulation In this subsection, we introduce a dual variational formulation of (1.2) and prove Theorem 1.1. Let f (s) be a function defined in (2.2). Using the function f , we consider the following semilinear problem:

− 1v + λf (v)f ′ (v) = |f (v)|p−1 f (v)f ′ (v) in RN .

(2.5)

The functional associated to (2.5) is defined by I (v) =

1 2

∫ RN

(|∇v|2 + λf (v)2 ) dx −

1

∫

p+1

|f (v)|p+1 dx.

RN

From Lemma 2.2, we can see that I is well-defined on H 1 (RN ) = H 1 (RN , R) (see [3,4] for the proof). Moreover we have the following relation between (1.2), (2.5), which was already shown in [3,4]. For the sake of completeness, we give the proof. Lemma 2.6. Suppose v is a nontrivial critical point of I and v > 0. Then u = f (v) is a real positive solution of (1.2). Proof. We can easily see that if v ∈ H 1 (RN ) is a nontrivial critical point of I (v), then v is a solution of (2.5). By standard elliptic regularity theory, we see that v ∈ C 2 (RN ). Moreover v > 0 implies u > 0. Since f ∈ C ∞ (0, ∞), we also have u ∈ C 2 (RN ). For v = f˜ (u), we have

∇v = f˜ ′ (u)∇ u,

1v = f˜ ′′ (u)|∇ u|2 + f˜ ′ (u)1u.

From (2.2), it follows f˜ ′ (s) =

1 f ′ (f˜ (s))

 =

1 + κα|f (f˜ (s))|2α−2 =

κα(α − 1)|s|2α−4 s

f˜ ′′ (s) = 

1 + κα|s|2α−2



1 + κα|s|2α−2 ,

.

Thus we have

 κα(α − 1)|u|2α−4 u 1v =  |∇ u|2 + 1 + κα|u|2α−2 1u. 1 + κα|u|2α−2 From (2.5), we can observe that u satisfies

− 1u − κα|u|2α−2 1u − κα(α − 1)|u|2α−4 u|∇ u|2 + λu = |u|p−1 u. α

(2.6)

Now u > 0 implies |u| ∈ C . Then it follows from 2

∆(|u|α ) = div (α|u|α−2 u∇ u) = α|u|α−2 u1u + ∇ u · ∇(α|u|α−2 u) = α|u|α−2 u1u + α(α − 1)|u|α−2 |∇ u|2 that

∆(|u|α )|u|α−2 u = α|u|2α−2 1u + α(α − 1)|u|2α−4 u|∇ u|2 .

(2.7)

Thus from (2.6) and (2.7), we see that if v is a nontrivial critical point of I and v > 0, then u = f (v) is a positive solution of (1.2). 

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S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

Remark 2.7. If α ≥ 2, then f ∈ C 2 [0, ∞) and |f (v)|α ∈ C 2 for any nontrivial critical point v of I. Thus Lemma 2.6 holds for any nontrivial critical point (possibly sign-changing) of I if α ≥ 2. Lemma 2.6 tells us that we have only to show the existence of a positive solution of (2.5) in order to find a real positive solution of (1.2). However to show the existence of a ground state solution, we need more information on the relation between (1.2) and (2.5). Actually we have the following relations. Lemma 2.8. It follows (i) X = f (H 1 (RN )), that is, X = {f (v); v ∈ H 1 (RN )} =: Y . (ii) For any v ∈ H 1 (RN ), we put u = f (v). Then it follows J (u) = I (v). Proof. (i) First we show Y ⊂ X . For v ∈ H 1 (RN ), we put u = f (v). Then we have 1

|∇ f (v)|2 = |f ′ (v)|2 |∇v|2 =

1 + κα|f (v)|2α−2

|∇v|2 .

By (i) of Lemmas 2.1 and 2.2, we obtain

∫ RN

(|∇ u|2 + u2 )dx + κα

∫

|∇ u|2 |u|2α−2 dx =

RN

∫ RN

(|∇v|2 + f (v)2 )dx ≤ C ‖v‖2H 1 < ∞.

Thus it follows Y ⊂ X . To show X ⊂ Y , it suffices to show f˜ (u) ∈ H 1 (RN ) for all u ∈ X . For u ∈ X , we put v = f˜ (u). Then it follows

∫

∫

2

|(f˜ )′ (u)|2 |∇ u|2 dx =

|∇v| dx = RN

RN

∫ RN

(1 + κα|u|2α−2 )|∇ u|2 dx < ∞.

Next by (i) of Lemma 2.2, it follows lim

s→0

f˜ (s) s

= 1,

lim

f˜ (s) sα

s→∞

=c

for some c > 0. Thus there exist constants C1 , C2 > 0 such that

|f˜ (s)| ≤ C1 χ|s|≤1 |s| + C2 χ|s|≥1 |s|α for all s ∈ R. Then we have 2N α

|v|2 ≤ C1 χ|u|≤1 |u|2 + C2 χ|u|≥1 |u|2α ≤ C1 |u|2 + C2 |u| N −2 . By Sobolev’s inequality, we obtain

∫

∫

2

2

|v| dx ≤ C1

∫

2N α

|u| dx + C2

RN

|u| N −2 dx

RN

∫ ≤ C1 RN

RN

|u|2 dx + C2′

∫ RN

 NN−2 < ∞. α 2 |∇ u|2 |u|2α−2 dx

Thus it follows X ⊂ Y and hence X = Y . (ii) We substitute u = f (v) into J (u). Then from (2.2), it follows 1

∫

κα

∫ 1 |∇ f (v)|2 |f (v)|2α−2 dx − |f (v)|p+1 dx 2 RN 2 RN p + 1 RN ∫ ∫ 1 1 = (|∇v|2 (1 + κα|f (v)|2α−2 )|f ′ (v)|2 + λf (v)2 )dx − |f (v)|p+1 dx 2 RN p + 1 RN ∫ ∫ 1 1 = (|∇v|2 + λf (v)2 )dx − |f (v)|p+1 dx 2 RN p + 1 RN = I (v).

J (u) =

and we obtain (ii).

(|∇ f (v)|2 + λf (v)2 )dx +

∫



Finally we prepare the following Pohozaev-type identity. Lemma 2.9. Let u ∈ XC be a solution of (1.2). Then u satisfies the following identity: P (u) :=

N −2 2N

= 0.

∫ RN

(|∇ u|2 + κα|∇|u ‖2 |u|2α−2 )dx +

λ 2

∫

|u|2 dx − RN

1 p+1

∫

|u|p+1 dx RN

(2.8)

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

825

Proof. For t > 0, we put ut (x) = u( xt ). Then we have

∫

∫ ∫ |∇ u|2 dx, |ut |2 dx = t N |u|2 dx, RN RN RN ∫ ∫ ∫ |∇|ut ‖2 |ut |2α−2 dx = t N −2 |∇|u ‖2 |u|2α−2 dx, |ut |p+1 dx = t N |∇ ut |2 dx = t N −2

∫R

N

RN

∫

RN

If u is a solution of (1.2), then

d J dt

RN

(ut )|t =1 = 0. From this equality, we obtain (2.8).

RN

|u|p+1 dx.



Now we are ready to prove Theorem 1.1. Proof of Theorem 1.1. We argue as in [9]. By the results due to [14–16], there exists w ˜ ∈ H 1 (RN ) such that w ˜ > 0, radial and

˜ > 0. I (w) ˜ = inf{I (v); I ′ (v) = 0, v ∈ H 1 (RN ) \ {0}} =: m By Lemma 2.6, w = f (w) ˜ satisfies J ′ (w) = 0. We claim that w ∈ G. Indeed we define N −2 P˜ (v) := 2N

∫

2

|∇v| dx + RN

λ

∫

2

RN

f (v) dx −

∫

1

2

p+1

RN

|f (v)|p+1 dx,

v ∈ H 1 (RN ).

Then it follows P (u) = P˜ (v) for u = f (v) where P (u) was defined in (2.8). Moreover w ˜ can be characterized by

w ˜ ∈ A := {v ∈ H 1 (RN ); P˜ (v) = 0},

˜ = I (w) m ˜ = inf I (v).

(2.9)

v∈A

Now let u ∈ XC be a nontrivial critical point of J. Then from (2.8), we have J (u) =

1

∫

N

RN

(|∇ u|2 + ακ|∇|u ‖2 |u|2α−2 )dx =

1 N

∫

|∇ u|2 dx + RN

κ αN

∫ RN

|∇|u |α |2 dx.

By the pointwise inequality |∇|u(x) ‖≤ |∇ u(x)| a.e. x ∈ RN , it follows J (u) ≥ J (|u|) = I (f˜ (|u|)). If N = 2, then (2.8) implies P (|u|) = 0 and hence P˜ (f˜ (|u|)) = 0. Then we obtain J (u) ≥ I (f˜ (|u|)) ≥ I (w) ˜ = J (w). If N ≥ 3, we distinguish cases P (|u|) = 0 and P (|u|) < 0. If P (|u|) = 0, then we have J (u) ≥ J (w) as in the case N = 2. Suppose P (|u|) = P˜ (f˜ (|u|)) < 0. We put v˜ = f˜ (|u|) and define v˜ θ (x) = v˜ ( θx ) for θ ∈ (0, 1). We can see that there exists

θ0 ∈ (0, 1) such that P˜ (˜vθ0 ) = 0, that is, v˜ θ0 ∈ A. Then we have ∫ ∫ θ N −2 θ N −2 κ I (˜vθ0 ) = 0 |∇ v˜ |2 dx = 0 (|∇|u ‖2 + |∇|u |α |2 )dx N N N N α R R ∫ θ0N −2 κ ≤ (|∇ u|2 + |∇|u |α |2 )dx = θ0N −2 J (u) < J (u). N α RN

(2.10)

Since v˜ θ0 ∈ A, we obtain J (u) > I (˜vθ0 ) ≥ I (w) ˜ = J (w). This implies w ∈ G and hence G ̸= ∅. Moreover by (ii) of Lemma 2.8, we have

˜ = I (w) m ˜ = J (w) = m. Next we show that if u ∈ G, then |u| ∈ G. Indeed if P (|u|) < 0, then we get from (2.9) and (2.10)

˜ = m. m = J (u) > I (˜vθ0 ) ≥ m This is a contradiction. Thus it follows P (|u|) = 0. Arguing as above, we obtain J (u) = J (|u|). This implies |u| ∈ G. Now suppose u ∈ G satisfies LN ({x ∈ RN ; |∇|u(x) ‖< |∇ u(x)|}) > 0, where LN (S ) is the N-dimensional Lebesgue measure of S. We immediately get a contradiction, because we obtain m = J (|u|) < J (u) = m. Thus we have for u ∈ G

|∇|u(x) ‖= |∇ u(x)| a.e. x ∈ RN .

(2.11) iθ(x)

Using the polar form, we put u(x) = r (x)e

|∇|u(x) ‖ = |∇ r (x)| , 2

2

. Then a direct calculation yields

|∇ u(x)| = |∇ r (x)|2 + r (x)2 |∇θ (x)|2 . 2

826

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

Thus (2.11) implies |∇θ (x)| = 0 a.e. x ∈ RN and this leads a conclusion that any u ∈ G is of the form u(x) = eiθ |u(x)| for some θ ∈ R. Next we show properties (i)–(iv) for any real non-negative ground state w ∈ G. N 2 N (i) We can see if w ∈ G, then w ∈ L∞ loc (R ). By the elliptic regularity theory, it follows w ∈ C (R ).

(ii) Since |w| ∈ G for any w ∈ G, it follows w ≥ 0. We put w ˜ = f˜ (w). By the maximum principle, we have w ˜ > 0. This implies w > 0. (iii) We observe that if w ∈ G, then w ˜ = f˜ (w) is a ground state solution of (2.5). In fact for any nontrivial real valued critical point u of J, v = f˜ (u) is a nontrivial critical point of I. Then by Lemma 2.8, we have I (v) = J (u) ≥ m = J (w) = I (w). ˜ This implies w ˜ is a ground state solution of (2.5). By the result of [17], any ground state solution of (2.5) is radially symmetric and decreasing with respect to r = |x|. We can easily see that w(x) = w(|x|) if and only if w( ˜ x) = f˜ (w(x)) satisfies w( ˜ x) = w(| ˜ x|). Thus claim (iii) holds. (iv) Let w ∈ G and w ˜ = f˜ (w). From (ii), w ˜ is a positive radial solution of (2.5). Then by the standard comparison principle, it follows

|Dk w( ˜ x)| ≤ ce−δ|x| for all δ ∈ (0,

√ λ), x ∈ RN ,

|k| ≤ 2 and some c > 0.

√ λ), x ∈ RN ,

|k| ≤ 2 and some c > 0.

By Lemma 2.1 (i), we can see that

|Dk w(x)| ≤ ce−δ|x| for all δ ∈ (0, Thus for |x| ≫ 1, we have

−1w + λw ≤ ce−pδ|x| + ce−(2α−1)δ|x| . Since 2α − 1 > 1 and we can take δ arbitrarily close to

√ λ, it follows

−1w + λw = o(G(x)) as |x| → ∞, where G is the fundamental solution of −∆ + λI. Then by Gidas–Ni–Nirenberg’s asymptotic result [18], we obtain (iv).



Next we give a relation on the sets of positive radial solutions of (1.2) and (2.5). Lemma 2.10. It follows

{u ∈ X ∩ C 2 (RN ); J ′ (u) = 0, u > 0, u(x) = u(|x|)} = f ({v ∈ H 1 ∩ C 2 (RN ); I ′ (v) = 0, v > 0, v(x) = v(|x|)}). Proof. By Lemma 2.6, we know that

{u ∈ X ∩ C 2 (RN ); J ′ (u) = 0, u > 0} ⊇ {f (v); I ′ (v) = 0, v > 0, v ∈ H 1 ∩ C 2 (RN )}. Suppose the equality does not hold. Then there exists u0 ∈ X such that u0 is a positive solution of (1.2) but u0 ̸= f (v) for any positive solution v ∈ H 1 (RN ) of (2.5). On the other hand by (i) of Lemma 2.8, we know if u0 ∈ X , then there exists v0 such that u0 = f (v0 ). Since u0 is a positive solution of (1.2), we can see that v0 is a positive solution of (2.5). This is a contradiction. Thus we have

{u ∈ X ∩ C 2 (RN ); J ′ (u) = 0, u > 0} = {f (v); I ′ (v) = 0, v > 0, v ∈ H 1 ∩ C 2 (RN )}. Finally we can easily see that u(x) = u(|x|) if and only if v(x) = f˜ (u(x)) satisfies v(x) = v(|x|).



It follows from the proof of Theorem 1.1 that if u1 = eiθ1 |u1 |, u2 = eiθ2 |u2 | ∈ G, then |u1 |, |u2 | ∈ G and both |u1 | and |u2 | are positive radial solutions. This implies if (1.2) has at most one positive radial solution, then |u1 | = |u2 | and hence we obtain the desired uniqueness results up to gauge transformation. On the other hand, Lemma 2.10 tells us that if (2.5) has at most one positive radial solution v , then (1.2) also has at most one positive radial solution u = f (v). Thus we have only to study the uniqueness of the positive solution of semilinear problem (2.5). (2α−1)N +2 Finally in this section, we give the non-existence result for p ≥ α ∗ = , which is an easy consequence of the N −2 Pohozaev identity. Theorem 2.11. Suppose N ≥ 3 and p ≥ α ∗ . Then (1.2) has no nontrivial solution u ∈ XC .

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

827

Proof. Suppose u ∈ XC is a nontrivial solution of (1.2) and p ≥ α ∗ . From J ′ (u)u = 0, we have

∫ RN

(|∇ u|2 + λ|u|2 + κα 2 |∇|u ‖2 |u|2α−2 − |u|p+1 )dx = 0.

On the other hand, u satisfies (2.8). Then we obtain

(α − 1)(N − 2) 2α

∫

|∇ u|2 dx +

RN

(α − 1)N + 2 2α

∫ RN

λ|u|2 dx =



N p+1

−

N −2 2α

∫ RN

|u|p+1 dx.

Since the left hand side is positive, it follows N p+1

−

N −2 2α

> 0,

This is a contradiction.

that is, p <

(2α − 1)N + 2 . N −2



3. Uniqueness of positive radial solutions for N ≥ 3 In this section, we study the uniqueness of the positive radial solution of (2.5). We put g (s) := f (s)p f ′ (s) − λf (s)f ′ (s) for s ≥ 0

Kg (s) :=

and

sg ′ (s) g (s)

.

(3.1)

We apply the following uniqueness result due to Serrin and Tang [13]. Proposition 3.1 ([13]). Suppose that there exists b > 0 such that (i) g is continuous on (0, ∞), g (s) ≤ 0 on (0, b] and g (s) > 0 for s > b. (ii) g ∈ C 1 (b, ∞) and Kg′ (s) ≤ 0 on (b, ∞). Then the semilinear problem:

−1u = g (u) in RN , u > 0, u → 0 as |x| → ∞,

u(0) = max u(x)

has at most one positive radial solution. Now we can see that g defined in (3.1) is of the class C 1 [0, ∞) and 1 g (s) = 0 ⇐⇒ f p−1 (s) = λ ⇐⇒ s = f −1 (λ p−1 ). 1

We put b := f −1 (λ p−1 ). Since (s − b)g (s) = (s − b)ff ′ (f p−1 − λ), we can see (i) of Proposition 3.1 holds. From (2.2), we can also observe that 1

f ′ (b) = 

1 + ακλ

2α−2 p−1

.

Since f ′ (s) → 0 as s → ∞, this implies b → ∞ if and only if κλ

2α−2 p−1

→ ∞.

(3.2)

Lemma 3.2. Suppose α > 1 and



1 < p < α∗ α − 1 ≤ p < α∗

if 1 < α ≤ 2, if α > 2.

Then there exists c0 = c0 (p, α) > 0 such that if κλ

2α−2 p−1

≥ c0 , then g satisfies (ii) of Proposition 3.1.

We notice that α − 1 < α ∗ for any α > 1 and N ≥ 3. Proof. We observe that Kg′ (s) =

1 g (s)2

(g ′′ (s)g (s)s + g ′ (s)g (s) − g ′ (s)2 s).

Thus we have only to show that sg ′′ g + g ′ g − sg ′2 < 0 for s > b.

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S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

Firstly we compute g ′ and g ′′ . It follows g (s) = f p f ′ − λff ′ = ff ′ (f p−1 − λ), g (s) = (f ′

g (s) = (f ′′

(3.3)

− λ)(ff + f ) + (p − 1)f

p−1

′′

p−1

− λ)(ff

′2

′′′

f ,

p−1 ′2

+ 3f f ) + (p − 1)f ′ ′′

p−2

(3.4)

(3ff f + pf ). ′ ′′

′3

(3.5)

Now we use

(f p−1 − λ)f p−1 = (f p−1 − λ)2 + λf p−1 − λ2 = (f p−1 − λ)2 + λ(f p−1 − λ), f 2p−2 = (f p−1 − λ)2 + 2λf p−1 − λ2 = (f p−1 − λ)2 + 2λ(f p−1 − λ) + λ2 . We regard sg ′′ g + gg ′ − sg ′2 as a polynomial of f p−1 − λ. By direct computations, we have sg ′′ g + g ′ g − sg ′2 = (f p−1 − λ)2 (sf 2 f ′ f ′′′ + psff ′2 f ′′ + f 2 f ′ f ′′ − sf 2 f ′′2 − psf ′4 + pff ′3 )

+ λ(p − 1)(f p−1 − λ)(sff ′2 f ′′ − psf ′4 + ff ′3 ) − λ2 (p − 1)2 sf ′4 Next from (2.3) and (2.4), we express f ′′ and f ′′′ in terms of f and f ′ . Then we have sf 2 f ′ f ′′′ + psff ′2 f ′′ + f 2 f ′ f ′′ − sf 2 f ′′2 − psf ′4 + pff ′3 = 3(α − 1)2 sf ′8 + (α − 1)(p − 4α + 3)sf ′6 + (α − 1)ff ′5

− α(p + 1 − α)sf ′4 + (p + 1 − α)ff ′3 , sff ′2 f ′′ − psf ′4 + ff ′3 = (α − 1)sf ′6 − (p + α − 1)sf ′4 + ff ′3 . Thus we obtain sg ′′ g + g ′ g − sg ′2 = (f p−1 − λ)2 (3(α − 1)2 sf ′8 + (α − 1)(p − 4α + 3)sf ′6 + (α − 1)ff ′5

− α(p + 1 − α)sf ′4 + (p + 1 − α)ff ′3 ) + λ(p − 1)(f p−1 − λ)((α − 1)sf ′6 − (p + α − 1)sf ′4 + ff ′3 ) − λ2 (p − 1)2 sf ′4 =: (f p−1 − λ)2 H1 (s) + λ(p − 1)(f p−1 − λ)H2 (s) − λ2 (p − 1)2 sf ′4 .

(3.6)

For s > b, it follows f (s)p−1 − λ > 0. Thus it suffices to show that H1 (s) < 0, H2 (s) < 0 for s > b in order to prove sgg ′′ + g ′ g − sg ′2 < 0. First we study the sign of H2 (s). It follows H2 (s) = sf

′4



−(p + α − 1) +

f sf ′

+ (α − 1)f

′2



.

By (ii) and (iii) of Lemma 2.2, it follows

 lim

s→∞

−(p + α − 1) +

f sf ′

+ (α − 1)f ′2



= −(p − 1) < 0.

Thus for sufficiently large b, we obtain H2 (s) < 0 for s > b. Next we investigate the sign of H1 (s). Firstly we suppose p = α − 1. Then from f < α sf ′ , it follows H1 (s) = 3(α − 1)2 sf ′8 − (α − 1)(3α − 2)sf ′6 + (α − 1)ff ′5

< 3(α − 1)2 sf ′8 − 2(α − 1)2 sf ′6 = (α − 1)2 sf ′6 (3f ′2 − 2). Since f ′ (s) → 0 as s → ∞, we have H1 (s) < 0 for s > b if b is sufficiently large. Next we suppose p > α − 1. From f < α sf ′ , we have H1 (s) < 3(α − 1)2 sf ′8 + (α − 1)(p − 3α + 3)sf ′6 − (p + 1 − α)f ′3 (α sf ′ − f )



= f (α sf − f ) 3(α − 1) f ′3

′

2 ′2

sf ′3

α sf ′ − f

+ (α − 1)(p − 3α + 3)

sf ′3

α sf ′ − f

 − (p + 1 − α)

=: f (α sf − f )h1 (s). ′3

′

Since α sf ′ − f > 0, it suffices to show that h1 (s) < 0 for s > b. If α ≥ 2, then we have by Lemma 2.5 that lim h1 (s) = −(p + 1 − α) < 0.

s→∞

If 1 < α < 2, then we also have lim h1 (s) =

s→∞

(p − 3α + 3)(2 − α) − (p + 1 − α). α

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

829

Now we observe that

(p − 3α + 3)(2 − α) − (p + 1 − α) < 0 ⇐⇒ p > 2α − 3. α In the case 1 < α < 2, it follows 1 > α − 1 > 2α − 3. Thus for p > 1, we obtain lim h1 (s) =

s→∞

(p − 3α + 3)(2 − α) − (p + 1 − α) < 0. α

Therefore for sufficiently large b, it follows H1 (s) < 0 for s > b. From (3.2) and (3.6), there exists c0 = c0 (p, α) > 0 such that if κλ

2α−2 p−1

≥ c0 , then we obtain sg ′′ + gg ′ − sg ′2 < 0 for s > b.



By Lemma 3.2, we can apply Proposition 3.1. Hence we obtain the uniqueness of positive radial solutions of (1.2) when N ≥ 3 and the proof of Theorem 1.2 is complete. 4. Uniqueness of positive radial solutions for N = 2 In this section, we study the uniqueness of positive radial solutions of (2.5) for N = 2. Under same notation as in Section Section 3, we apply the following uniqueness result due to Mcleod and Serrin [12]. Proposition 4.1. [12] Suppose that there exist b > 0 and τ ≥ 1 such that (i) g ∈ C 1 [0, ∞), g (0) = 0, g ′ (0) < 0. (ii) g (s) < 0 for s ∈ (0, b), g (s) > 0 for s ∈ (b, ∞). (iii) g ′ (b) > 0. (iv) (

g (s) ′ 0 sτ g (s) ′ ′ s sτ

) > for s > 0, s ̸= b. ) ) < 0 for s > b.

(v) ( (

Then the semilinear problem:

−1u = g (u) in R2 , u > 0, u → 0 as |x| → ∞,

u(0) = max u(x)

has at most one positive radial solution. As mentioned in [11], (iv) and (v) of Proposition 4.1 imply (ii) of Proposition 3.1. This means that we need a stronger condition on the nonlinearity to show the uniqueness for N = 2. Thus we have to restrict λ, κ , α and p more strongly to show (iv) and (v). Now we can see g defined in (3.1) satisfies (i)–(iii) of Proposition 4.1. Lemma 4.2. Assume 2α − 1 ≤ p < ∞. Then g defined in (3.1) satisfies (iv) of Proposition 4.1 with τ =

p+1−α

α

.

Proof. A direct calculation yields that sg ′ (s) − τ g (s) = (p + 1 − α)sf p−1 f ′2 + (α − 1)sf p−1 f ′4 + (α − 2)λsf ′2 − (α − 1)λsf ′4 − τ f p f ′ + τ λff ′

≥ (p + 1 − α − ατ )sf p−1 f ′2 + (τ − 1)λsf ′2 + (α − 1)sf p−1 f ′4 . Here we used Lemma 2.1. Choosing τ =

p+1−α

α

, we obtain τ ≥ 1 and sg ′ − τ g > 0 for s > 0, s ̸= b.



Lemma 4.3. Suppose 2α − 1 ≤ p < ∞. Then there exists c1 = c1 (p, α) > 0 such that if κλ satisfies (v) of Proposition 4.1 with b = f

−1

(λ

1 p−1

2α−2 p−1

≥ c1 , then g

).

Proof. By a direct computation, it follows

  ′ ′ g s

sτ

=

1 s τ +1

(s2 g ′′ + (1 − 2τ )sg ′ + τ 2 g ).

Thus it suffices to show that s2 g ′′ + (1 − 2τ )sg ′ + τ 2 g < 0 for s > b. From (3.3)–(3.5), we have s2 g ′′ + (1 − 2τ )sg ′ + τ 2 g = f p−2 (s2 f 2 f ′′′ + 3ps2 ff ′ f ′′ − (2τ − 1)sf 2 f ′′ + p(p − 1)s2 f ′3 − p(2τ − 1)sff ′2 + τ 2 f 2 f ′ )

− λ(s2 ff ′′′ + 3s2 f ′ f ′′ − (2τ − 1)sff ′′ − (2τ − 1)sf ′2 + τ 2 ff ′ ).

830

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

From (2.3) and (2.4), we express f ′′ and f ′′′ in terms of f and f ′ . Then we have s2 g ′′ + (1 − 2τ )sg ′ + τ 2 g = f p−2 (4(α − 1)2 s2 f ′7 + (α − 1)(3p − 6α + 5)s2 f ′5 − (2τ − 1)(α − 1)sff ′4

+ (p + 1 − α)(p + 1 − 2α)s2 f ′3 − (2τ − 1)(p + 1 − α)sff ′2 + τ 2 f 2 f ′ ) λ − (4(α − 1)2 s2 f ′7 − 2(3α − 4)(α − 1)s2 f ′5 − (2τ − 1)(α − 1)sff ′4 f

+ 2(α − 1)(α − 2)s2 f ′3 + (2τ − 1)(α − 2)sff ′2 + τ 2 f 2 f ′ ) 1

=:

f

(f p−1 K1 (s) − λK2 (s)).

(4.1)

We claim that for sufficiently large b, K1 ( s ) < 0

K1 (s) ≤ K2 (s) for s > b.

and

(4.2)

If (4.2) is fulfilled, then it follows f p−1 K1 − λK2 ≤ K1 (f p−1 − λ). Since f p−1 − λ > 0, f > 0 and K1 < 0 for s > b, we have from (4.1) that s2 g ′′ + (1 − 2τ )g ′ + τ 2 g ≤

K1 f

(f p−1 − λ) < 0 for s > b.

Firstly we show that K1 − K2 ≤ 0 for s > b. From (4.1), we have K1 − K2 = 3(α − 1)(p − 1)s2 f ′5 + (p − 3α + 3)(p − 1)s2 f ′3 − (2τ − 1)(p − 1)sff ′2

  f = (p − 1)s2 f ′3 3(α − 1)f ′2 − (2τ − 1) ′ + p − 3α + 3 . sf

By (ii) and (iii) of Lemma 2.2 and from τ =

 lim

s→∞

3(α − 1)f

′2

− (2τ − 1)

p+1−α

α

, it follows that



f sf ′

+ p − 3α + 3 = −(p − 1) < 0.

Thus for sufficiently large b, we obtain K1 (s) − K2 (s) ≤ 0 for s > b. Next we show that K1 (s) < 0 for s > b. First we suppose p > 2α − 1. We put K1 (s) = s2 f ′3 (4(α − 1)2 f ′4 − (2τ − 1)(α − 1)f ′2

f sf ′

+ (α − 1)(3p − 6α + 5)f ′2 + L1 (s)),

L1 (s) := (p + 1 − α)(p + 1 − 2α) − (2τ − 1)(p + 1 − α)

f sf ′

We claim that L1 (s) < 0 for s > b if b is large. Indeed we put φ(s) =

+ τ2



f (s) . sf ′ (s)

f sf ′

2

.

Then by (ii) of Lemma 2.1 and (iii) of Lemma 2.2,

φ(s) is increasing and 1 ≤ φ(s) < α for all s > 0. We also put p1 = p + 1 − α . Then τ = L1 (s) = p1 (p1 − α) − (2p1 − α)

p1

α

φ(s) +

p21

α2

p1

α

and

φ(s)2 .

Finally we define for 1 ≤ t ≤ α , L2 (t ) : = p1 t 2 − (2p1 − α)α t + α 2 (p1 − α)

= (t − α)(p1 t − α(p1 − α)). Then it follows L1 (s) = α12 L2 (φ(s)) and p

L2 ( t ) < 0

 for t ∈

α(p1 − α) p1



,α .

By (iii) of Lemma 2.1, φ(s) ↗ α as s → ∞. Thus we have

α(p1 − α) p1

< φ(s) < α for s > b

if b is sufficiently large. Thus we have L1 (s) < 0 for s > b and taking b larger if necessary, we obtain K1 (s) < 0 for s > b.

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

831

Next we suppose p = 2α − 1. Then it follows τ = 1, p + 1 − α = α , 3p − 6α + 5 = 2 and hence K1 (s) = 4(α − 1)2 s2 f ′7 + 2(α − 1)s2 f ′5 − (α − 1)sff ′4 − α sff ′2 + f 2 f ′

  s2 f ′6 s2 f ′4 = −ff ′ −4(α − 1)2 − 2(α − 1) + (α − 1)sf ′3 + α sf ′ − f f f   ′ ′3 sf sf sf ′ sf ′3 sf ′3 ′2 = −ff ′ (α sf ′ − f ) −4(α − 1)2 f − 2 (α − 1 ) + (α − 1 ) + 1 f α sf ′ − f f α sf ′ − f α sf ′ − f ′ ′ =: −ff (α sf − f )L3 (s). It follows from f ′ → 0 as s → ∞, (iii) of Lemmas 2.2 and 2.5 that if α ≥ 2,

1 4(α − 1)

 lim L3 (s) =

s→∞

if 1 < α < 2.

α2

In the both cases, we see that L3 (s) > 0 for sufficiently large s and hence we obtain K1 (s) < 0 for s > b provided b is large. From (3.2), there exists c1 = c1 (p, α) > 0 such that if κλ s2 g ′′ + (1 − 2τ )sg ′ + τ 2 g < 0 for s > b. 

2α−2 p−1

≥ c1 , then (4.2) is satisfied. Then we obtain

By Lemmas 4.2 and 4.3, we can apply Proposition 4.1. Hence we obtain the uniqueness of positive radial solutions of (1.2) for N = 2 and the proof of Theorem 1.3 is complete. 5. Non-degeneracy of the ground state solution As we have shown in Sections 3 and 4, problem (1.2) and (2.5) has at most one positive radial solution. In this section, we study spectral properties of the linearized operator. We will show that the unique positive radial solution of (2.5) is 1 non-degenerate in Hrad (RN ) for N ≥ 3. To this aim, we apply the result due to Bates and Shi [19]. We consider the following semilinear problem:

− 1u = g (u) in RN , u > 0,

u(x) = u(|x|), u → 0 as |x| → ∞,

u(0) = max u(x).

(5.1)

Proposition 5.1 ([19]). Suppose N ≥ 3 and g ∈ C 1 ([0, ∞)) satisfies the following properties. (g1) (g2) (g3) (g4) (g5)

There exists b > 0 such that g (0) = g (b) = 0, g (s) < 0 for s ∈ (0, b), g ′ (0) < 0 and g ′ (b) > 0. θ There exists θ > b such that 0 g (s) ds = 0. g (s) > 0 for all s > b. 2 ) as s → ∞. Kg (s) (defined in (3.1)) is non-increasing in [θ , ∞) and Kg (s) → K∞ ∈ [1, NN + −2 Kg (s) ≥ Kg (θ ) for s ∈ (b, θ] and Kg (s) ≤ K∞ for s ∈ (0, b).

Under assumptions (g1)–(g5) , let u be the unique positive radial solution of (5.1) and L0 := −∆ − g ′ (0). Then (1) σ (L0 ) = σp (L0 ) ∪ σe (L0 ). (2) σe (L0 ) = [−g ′ (0), ∞), σp (L0 ) ⊂ (−∞, −g ′ (0)). (3) If µ ∈ σp (L0 ), then the corresponding eigenfunction φ(x) satisfies

|φ(x)| ≤ Cϵ e−



−g ′ (0)−µ+ϵ 2

| x|

for x ∈ RN

for any small ϵ > 0 and some Cϵ > 0. (4) If µ ∈ σp (L0 ) ∩ (−∞, 0), then the corresponding eigenfunction is radially symmetric. (5) The principal eigenvalue µ1 (L0 ) < 0 is simple and the corresponding eigenfunction φ1 can be chosen to be positive. (6) µ2 (L0 ) = 0 and the eigenspace corresponding to the eigenvalue µ = 0 is spanned by



 ∂u ; i = 1, . . . , N . ∂ xi

We show that (g1)–(g5) hold for g (s) defined in (3.1). We can easily see that (g1) and (g3) hold. Next we put θ = 1

p+1 f˜ (( λ) p−1 ). Then we have

θ

2

on p, then (g4) and (g5) hold.

0

g (s) ds = 0, which implies (g2). Next we show that if we suppose one additional condition

Lemma 5.2. Assume 2α − 1 ≤ p < α ∗ . Then g(s) defined in (3.1) satisfies lim Kg (s) =

s→∞

p+1−α

α

[  N +2 =: K∞ ∈ 1, . N −2

832

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

Proof. By direct computation, we have Kg (s) =

g ′ (s)s g (s)

(p − 1)sf p−1 f ′2 + (2 − α)(f p−1 − λ)sf ′2 + (α − 1)(f p−1 − λ)sf ′4 (f p−1 − λ)ff ′  ′  p−1 sf (p − 1)f ′3 = + ( 2 − α) + (α − 1 ) f . f f p−1 − λ =

Since lims→∞

sf ′ f

1

=

lim Kg (s) =

α

and lims→∞ f ′ = 0, we obtain

p−1

s→∞

From 2α − 1 ≤ p <

α

+

2−α

α

(2α−1)N +2 , N −2

=

p+1−α

it follows

α p+1−α

α

.

+2 ∈ [1, NN − ). Thus we show the claim of Lemma 5.2. 2

Lemma 5.3. Suppose α > 1 and 2α − 1 ≤ p < α ∗ . If κλ in (3.1) satisfies (g4) and (g5) in Proposition 5.1.

2α−2 p−1



≥ c0 (c0 is given in Lemma 3.2), then g (s) defined

Proof. By Lemmas Lemmas 3.2 and 5.2, we can see (g4) and the first inequality in (g5) hold. Next we show that Kg (s) ≤ 1 for s ∈ (0, b). Now from (2.3), (3.3) and (3.4), we have sg ′ − g = (f p−1 − λ) (α − 1)sf ′4 − (α − 2)sf ′2 − ff ′ + (p − 1)sf p−1 f ′2 .





By the definition of b, we see f p−1 − λ < 0 for s ∈ (0, b). On the other hand by (ii) of Lemma 2.1, we have sf ′2 ≤ ff ′ . Using f ′ ≤ 1, we obtain

(α − 1)sf ′4 − (α − 2)sf ′2 − ff ′ ≤ (α − 1)sf ′4 − (α − 1)sf ′2 = (α − 1)sf ′2 (f ′2 − 1) ≤ 0. Thus we have sg ′ (s) − g (s) ≥ (p − 1)sf p−1 f ′2 > 0 for all s ∈ (0, b). Since g (s) < 0 for s ∈ (0, b), it follows Kg (s) =

sg ′ (s) g (s)

<

g (s) g (s)

= 1 for s ∈ (0, b).

Finally by Lemma 5.2, we have Kg (s) ≤ 1 ≤ K∞ =

p+1−α

α

.

Thus the second inequality in (g5) holds.



By Lemmas 5.2 and 5.3, we obtain the following. 2α−2

Theorem 5.4. Suppose N ≥ 3, α > 1 and 2α − 1 ≤ p < α ∗ . Assume further κλ p−1 ≥ c0 where c0 is given in Lemma 3.2. Let v be the unique positive solution of (2.5) and L := −∆ − g ′ (v); H 2 (RN ) → L2 (RN ). Then (1) σ (L) = σp (L) ∪ σe (L). (2) σe (L) = [λ, ∞), σp (L) ⊂ (−∞, λ). (3) If µ ∈ σp (L), then the corresponding eigenfunction φ(x) satisfies

|φ(x)| ≤ Cϵ e

−



λ−µ+ϵ 2

|x|

for x ∈ RN

for any small ϵ > 0 and some Cϵ > 0. (4) If µ ∈ σp (L) ∩ (−∞, 0), then the corresponding eigenfunction is radially symmetric. (5) The principal eigenvalue µ1 (L) < 0 is simple and the corresponding eigenfunction φ1 can be chosen to be positive. (6) µ2 (L) = 0 and the eigenspace corresponding to the eigenvalue µ = 0 is spanned by



 ∂v ; i = 1, . . . , N . ∂ xi

1 From (6) of Theorem 5.4, we obtain the non-degeneracy of the ground state solution of (2.5) in Hrad (RN ).

Corollary 5.5. Under the assumptions of Theorem 5.4, let v be the unique positive solution of (2.5). Then v is non-degenerate in 1 Hrad (RN ), that is, if 1 −1φ − g ′ (v)φ = 0 in RN and φ ∈ Hrad (RN ),

then φ ≡ 0.

S. Adachi, T. Watanabe / Nonlinear Analysis 75 (2012) 819–833

833

Remark 5.6. We have shown the non-degeneracy of the ground state solution of semilinear problem (2.5). However we do not know if this implies the non-degeneracy for the ground state solution of quasilinear problem (1.2). Acknowledgments The authors thank the unknown referee for carefully reading the manuscript and for useful suggestions. The first Author is supported in part by Grant-in-Aid for Young Scientists (B) of Japan Society for the Promotion of Science. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19]

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