On singular solutions of Lane-Emden equation on the Heisenberg group

Juncheng Wei; Ke Wu

doi:10.1515/ans-2023-0084

Article Open Access

On singular solutions of Lane-Emden equation on the Heisenberg group

Juncheng Wei and Ke Wu

Published/Copyright: November 20, 2023

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From the journal Advanced Nonlinear Studies Volume 23 Issue 1

Abstract

By applying the gluing method, we construct infinitely many axial symmetric singular positive solutions to the Lane-Emden equation:

Δ H u + u p = 0 , in H n \ { 0 }

on the Heisenberg group H n , where n > 1 , Q ⁄ ( Q − 4 ) < p < p J L ( Q − 2 ) , and Q = 2 n + 2 is the homogeneous dimension of H n .

Keywords: singular solutions; Heisenberg group; Lane-Emden equation; supercritical exponent; gluing method

MSC 2010: Primary: 35A21, 35B09; Secondary: 35J60

1 Introduction

Let n > 1 and H n be the Heisenberg group ( R 2 n + 1 , ∘ ) equipped with the group action

ξ 0 ∘ ξ = x + x 0 , y + y 0 , t + t 0 + 2 ∑ i = 1 n ( x i y 0 i − y i x 0 i )

for

ξ = ( x 1 , x 2 , … , x n , y 1 , y 2 , … , y n , t ) ≔ ( x , y , t ) ∈ R 2 n + 1 .

Let Δ H be the subelliptic Laplacian defined by

Δ H = ∑ i = 1 n ( X i 2 + Y i 2 ) ,

with

X i = ∂ ∂ x i + 2 y i ∂ ∂ t , Y i = ∂ ∂ y i − 2 x i ∂ ∂ t .

A direct calculation shows that

Δ H = ∑ i = 1 n ∂ 2 ∂ x i 2 + ∂ 2 ∂ y i 2 + 4 y i ∂ 2 ∂ x i ∂ t − 4 x i ∂ 2 ∂ y i ∂ t + 4 ( x i 2 + y i 2 ) ∂ 2 ∂ t 2 .

Let Q = 2 n + 2 denote the homogeneous dimension of H n . In a seminal work, Jerison and Lee [9] proved the following celebrated classification result.

Theorem 1.1

All positive solutions of the following equation:

(1.1) Δ H u + u Q + 2 Q − 2 = 0 , in H n

satisfying the integrability condition

(1.2) ∫ H n u 2 Q Q − 2 < + ∞

can be written as ω λ , ξ for some λ > 0 and ξ ∈ H n , where

ω λ , ξ = λ 2 − Q 2 ω ∘ δ λ − 1 ∘ τ ξ − 1

and

(1.3) ω ( x , y , t ) = c 0 1 ( t 2 + ( 1 + ∣ x ∣ 2 + ∣ y ∣ 2 ) 2 ) Q − 2 4 ,

with c 0 being a suitable positive constant. (Here τ ξ ( ξ ′ ) = ξ ∘ ξ ′ is the left translation on H n and δ λ ( ξ ) = ( λ x , λ y , λ 2 t ) is the natural dilation.)

The work of Jerison-Lee completely solved the so-called CR-Yamabe problem on H n and it opened door in the study of more general Lane-Emden equations on H n :

(1.4) Δ H u + u p = 0 , in H n .

In [12], Malchiodi and Uguzzoni proved that the positive solution ω in (1.3) is nondegenerate in the sense that ψ ∈ S 0 1 ( H n ) is a solution of the linearized equation

(1.5) Δ H ψ + Q + 2 Q − 2 ω 4 Q − 2 ψ = 0

if and only if there exist coefficients μ , ν 1 , ν 2 , … , ν 2 n , ν 2 n + 1 such that

(1.6) ψ = μ ∂ ω λ , ξ ∂ λ ( λ , ξ ) = ( 1 , 0 ) + ∑ ν = 1 2 n + 1 ν i ∂ ω λ , ξ ∂ ξ i ( λ , ξ ) = ( 1 , 0 ) ,

where S 0 1 ( H n ) is the Folland-Stein Sobolev space (see [12] for the details of the definition).

In the subcritical case 1 < p < ( Q + 2 ) ⁄ ( Q − 2 ) , equation (1.4) was first considered by Birindelli et al. in [2]. It is proved in [2] that if 1 < p ≤ Q ⁄ ( Q − 2 ) and if u is a nonnegative solution of (1.4), then u ≡ 0 . In [10], Lu and Wei considered Lane-Emden equations in more general stratified groups and the existence and non-existence of solutions were obtained. By applying the moving plane method, Birindelli and Prajapat [3] proved that if 1 < p < ( Q + 2 ) ⁄ ( Q − 2 ) and if u is a nonnegative solution of equation (1.4) such that u ( x , y , t ) = u ( r , t ) with r = ∣ x ∣ 2 + ∣ y ∣ 2 , then u ≡ 0 . In [15], Yu generalized the method in [3] to some semilinear elliptic equations with general nonlinearities. In [14], Xu improved the result in [2] to the range n > 1 , 1 < p < ( Q ( Q + 2 ) ) ⁄ ( Q − 1 ) 2 . Since the proof of Xu [14] is based on integration by part, it is not necessary to assume that solutions satisfy any symmetry. In a recent interesting article, Ma and Ou [11] gave a complete classification of nonnegative solutions to equation (1.4) when p is subcritical. The proof in [11] is based on a generalized Obata-type formula found by Jerison and Lee [9].

In this article, we consider positive solutions of the following Lane-Emden equation on H n

(1.7) Δ H u + u p = 0 , in H n \ { 0 } .

Compared with equation (1.4), the results concerning (1.7) are less known. In [11], a pointwise estimate for positive solutions of (1.7) near the isolated singularity was proved when p is subcritical. In [1], Afeltra constructed a family of positive singular solutions to the following equation:

(1.8) Δ H u + u Q + 2 Q − 2 = 0 , in H n \ { 0 } .

Similar to the Fowler solutions of the Yamabe problem on R n , the positive singular solutions constructed in [1] satisfy the homogeneity property

(1.9) u ∘ δ T = T − Q − 2 2 u

for some T large enough. It will be an interesting problem to prove whether any positive singular solution of (1.8) satisfies the homogeneity property (1.9).

In this article, we apply the gluing method in [5] to construct positive singular solutions to equation (1.7) in the supercritical case p > Q ⁄ ( Q − 4 ) . In order to give the statement of our main result, we introduce the Joseph-Lundgren exponent:

(1.10) p J L ( n ) = ∞ , if 3 ≤ n ≤ 10 , ( n − 2 ) 2 − 4 n + 8 n − 1 ( n − 2 ) ( n − 10 ) , if n ≥ 11 .

The exponent (1.10) is related to the classification of stable solutions of the Lane-Emden equation

(1.11) Δ u + u p = 0 , in R n .

Here, a solution of (1.11) is called stable if

∫ R n ∣ ∇ ψ ∣ 2 d x − p ∫ R n u p − 1 ψ 2 d x ≥ 0 , ∀ ψ ∈ C 0 ∞ ( R n ) .

Indeed, it was proved by Farina [6] that if u ∈ C 2 ( R n ) is a stable solution of (1.11) with 1 < p < p J L ( n ) , then u ≡ 0 . Moreover, (1.11) admits a smooth positive, bounded, stable, and radial solution for n ≥ 11 , p > p J L ( n ) .

The main result in this article is the following.

Theorem 1.2

Assume that n > 1 and

(1.12) Q Q − 4 < p < p J L ( Q − 2 ) ,

then equation (1.7) admits infinitely many singular solutions.

Remark 1.3

Since Q = 2 n + 2 , then Q ⁄ ( Q − 4 ) = ( n + 1 ) ⁄ ( n − 1 ) is the critical exponent of the Hardy equation:

Δ u + ∣ x ∣ − 1 u p = 0 , in R n + 1 .

Remark 1.4

If p > Q ⁄ ( Q − 4 ) , then

(1.13) n − 1 − 1 p − 1 > n − 1 2 .

Moreover, since p < p J L ( Q − 2 ) , then

(1.14) 4 n − 1 − 1 p − 1 − n − 1 − 2 p − 1 2 > 0 .

Indeed, by the properties of the Joseph-Lundgren exponent, we can check that if p < p J L ( N ) , then

2 p p − 1 N − 2 − 2 p − 1 > ( N − 2 ) 2 4 .

Therefore, if p < p J L ( Q − 2 ) = p J L ( 2 n ) , then

2 p p − 1 2 n − 2 − 2 p − 1 > ( 2 n − 2 ) 2 4 .

But this is exactly (1.14).

The content of this article will be organized as follows. In Section 2, we present some preliminary results. In Section 3, we construct inner solutions by studying an initial value problem. In Section 4, we study the asymptotic behavior of the outer problem. In Section 5, we match the inner solutions and the outer solutions to obtain solutions of equation (1.7).

2 Preliminaries

A function u is called cylindrical if for all ( x , y , t ) ∈ H n , u ( x , y , t ) = u ( r , t ) . If u is a cylindrical solution of equation (1.7), then

(2.1) u r r + 2 n − 1 r u r + 4 r 2 u t t + u p = 0 .

Let us consider the transform

(2.2) ρ 2 = r 4 + t 2 , θ = arccot t r 2 , θ ∈ ( 0 , π ) .

By applying these new coordinates, u satisfies the following equation:

(2.3) r 2 ρ 2 u ρ ρ + 4 r 2 ρ 4 u θ θ + ( 2 n + 1 ) r 2 ρ 3 u ρ + 4 n t ρ 4 u θ + u p = 0 .

We want to find a solution of the form

u ( ρ , θ ) = ρ − 2 p − 1 Φ ( θ ) .

After some computations, we can check that Φ satisfies the following equation:

(2.4) 4 sin θ Φ θ θ + 4 n cos θ Φ θ − β sin θ Φ + Φ p = 0 ,

where

(2.5) β = 4 p − 1 n − 1 p − 1 .

Φ ( θ ) = Φ ( π − θ ) , for 0 ≤ θ < π 2 ,

then Φ satisfies the following equation:

(2.6) 4 sin θ Φ θ θ + 4 n cos θ Φ θ − β sin θ Φ + Φ p = 0 , in 0 , π 2 , Φ ( θ ) > 0 , in 0 , π 2 , Φ ′ ( 0 ) exists , Φ ′ π 2 = 0 .

Remark 2.1

It is important to observe that equation (2.4) has an explicit singular solution. Indeed, the function

(2.7) Φ * ( θ ) = 4 p − 1 n − 1 − 1 p − 1 1 p − 1 [ sin θ ] − 1 p − 1

is a singular solution of (2.4) with two singular points θ = 0 and θ = π .

3 Inner solutions

In this section, we study solutions Φ of (2.6) with Φ ( 0 ) = Λ and analyze their behaviors near θ = 0 , where Λ is a sufficiently large constant. Since Λ is sufficiently large, it is convenient to set

Λ = ε − α , α = 1 p − 1 ,

with ε sufficiently small. Let

(3.1) Φ ( θ ) = ε − α v ( s ) , s = θ ε ,

we obtain from (2.6) that v satisfies the initial value problem

(3.2) v ″ ( s ) + n ε cot ( ε s ) v ′ ( s ) − β 4 ε 2 v ( s ) + ε 4 sin ( ε s ) v p ( s ) = 0 , v ( 0 ) = 1 .

Since for ε > 0 sufficiently small,

cot ( ε s ) = cos ( ε s ) sin ( ε s ) = 1 ε s − 1 3 ( ε s ) + ∑ k = 1 ∞ α k ( ε s ) 2 k + 1 , csc ( ε s ) = 1 sin ( ε s ) = 1 ε s + 1 6 ( ε s ) + ∑ k = 1 ∞ β k ( ε s ) 2 k + 1 ,

we have

(3.3) v ″ ( s ) + n s v ′ ( s ) − n 3 ε 2 s v ′ ( s ) + n ∑ k = 1 ∞ α k ε 2 ( k + 1 ) s 2 k + 1 v ′ ( s ) − β 4 ε 2 v ( s ) + 1 4 s v p ( s ) + 1 24 ε 2 s v p ( s ) + 1 4 ∑ k = 1 ∞ β k ε 2 ( k + 1 ) s 2 k + 1 v p ( s ) = 0 , v ( 0 ) = 1 .

The first approximation to the solution of (3.3) is the radial solution of the Hardy equation

(3.4) Δ v + 1 4 ∣ x ∣ v p = 0 , in R n + 1 , v ( 0 ) = 1 .

Since p > Q ⁄ ( Q − 4 ) = ( n + 1 ) ⁄ ( n − 1 ) , it is proved in [4] and [8] (see also [13]) that equation (3.4) has a unique positive radial solution.

Our first objective in this section is to characterize the asymptotic behavior of the unique positive radial solution of (3.4). More precisely, we have the following result.

Lemma 3.1

Let Q ⁄ ( Q − 4 ) < p < p J L ( Q − 2 ) , then there exist constants a 0 , b 0 , and S 0 such that for s ≥ S 0 , the unique positive radial solution v 0 of (3.4) satisfies

(3.5) v 0 ( s ) = A p s − α + a 0 cos ( ω ln s ) + b 0 sin ( ω ln s ) s n − 1 2 + O s − n − 1 − 1 p − 1 ,

where

(3.6) A p p − 1 = 4 p − 1 n − 1 − 1 p − 1

and

(3.7) ω = 1 2 4 n − 1 − 1 p − 1 − n − 1 − 2 p − 1 2 .

Moreover, in (3.5), we have a 0 2 + b 0 2 ≠ 0 .

Proof

Let

v 0 ( s ) = s − 1 p − 1 w ( τ ) , τ = ln s .

It follows from (3.4) that w satisfies the following equation:

(3.8) w ″ ( τ ) + n − 1 − 2 p − 1 w ′ ( τ ) − 1 p − 1 n − 1 − 1 p − 1 w ( τ ) + 1 4 w p ( τ ) = 0 .

By Lemma 4.1 in [4], we know that

(3.9) lim ∣ x ∣ → ∞ ∣ x ∣ 1 p − 1 v 0 ( x ) = 4 p − 1 n − 1 − 1 p − 1 1 p − 1 .

(3.9) is equivalent to lim τ → ∞ w ( τ ) = A p , where A p is defined in (3.6). Let

w ( τ ) = A p + V ( τ ) ,

then V satisfies the following equation:

(3.10) V ″ ( τ ) + n − 1 − 2 p − 1 V ′ ( τ ) + p − 1 4 A p p − 1 V ( τ ) + g ( V ( τ ) ) = 0 ,

with

g ( V ( τ ) ) = 1 4 [ ( A p + V ( τ ) ) p − A p p − p A p p − 1 V ( τ ) ] .

We note that

p − 1 4 A p p − 1 = n − 1 − 1 p − 1 ,

then (3.10) can also be written as:

(3.11) V ″ ( τ ) + n − 1 − 2 p − 1 V ′ ( τ ) + n − 1 − 1 p − 1 V ( τ ) + g ( V ( τ ) ) = 0 .

Step 1: For each ε > 0 sufficiently small, there exists T > 0 such that if τ > T , then

∣ V ( τ ) ∣ + ∣ V ′ ( τ ) ∣ ≤ ε .

Indeed, it follows from lim τ → ∞ w ( τ ) = A p that lim τ → ∞ V ( τ ) = 0 . Therefore, there exists T 1 > 0 such that if τ > T 1 , then ∣ V ( τ ) ∣ ≤ ε ⁄ 2 . Fix a constant T 1 satisfying this condition, by the method of variation of constants, we have

(3.12) V ( τ ) = e σ τ [ α 0 cos ( ω τ ) + α 1 sin ( ω τ ) ] + 1 ω ∫ T 1 τ e σ ( τ − τ ′ ) sin ( ω ( τ − τ ′ ) ) g ( V ( τ ′ ) ) d τ ′ ,

where

σ = − 1 2 n − 1 − 2 p − 1

and ω is given by (3.7). In (3.12), α 0 and α 1 depend only on T 1 . By the definition of g and (3.12), we know that there exists a constant c 0 independent of τ for τ ≥ T 1 such that

(3.13) V ′ ( τ ) = σ e σ τ [ α 0 cos ( ω τ ) + α 1 sin ( ω τ ) ] + e σ τ [ − ω α 0 sin ( ω τ ) + ω α 1 cos ( ω τ ) ] + σ ω ∫ T 1 τ e σ ( τ − τ ′ ) sin ( ω ( τ − τ ′ ) ) g ( V ( τ ′ ) ) d τ ′ + ∫ T 1 τ e σ ( τ − τ ′ ) cos ( ω ( τ − τ ′ ) ) g ( V ( τ ′ ) ) d τ ′ ≤ ( ω + ∣ σ ∣ ) ( ∣ α 0 ∣ + ∣ α 1 ∣ ) e σ τ + c 0 ε 2 .

By choosing ε small, and T 2 > T 1 large, we have ∣ V ′ ( τ ) ∣ ≤ ε ⁄ 2 provided τ ≥ T 2 . Hence the proof of Step 1 is completed.

Step 2: Fix a constant T 0 > T , where T is a constant satisfying Step 1. By the method of variation of constants, we have

(3.14) V ( τ ) = e σ τ [ α T 0 cos ( ω τ ) + β T 0 sin ( ω τ ) ] + 1 ω ∫ T 0 τ e σ ( τ − τ ′ ) sin ( ω ( τ − τ ′ ) ) g ( V ( τ ′ ) ) d τ ′ .

We claim that α T 0 and β T 0 satisfy

∣ α T 0 ∣ + ∣ β T 0 ∣ ≤ c 1 ε e − σ T 0 ,

where c 1 is a positive constant independent of T 0 . Indeed, in (3.14), α T 0 and β T 0 are chosen so that

V ( T 0 ) = e σ T 0 [ α T 0 cos ( ω T 0 ) + β T 0 sin ( ω T 0 ) ] , V ′ ( T 0 ) = e σ T 0 [ ( α T 0 σ + β T 0 ω ) cos ( ω T 0 ) + ( β T 0 σ − α T 0 ω ) sin ( ω T 0 ) ] .

Then Step 2 follows from Step 1 immediately.

Step 3: Set V ˜ ( τ ) = e − σ τ V ( τ ) with T 0 = ln S 0 be a sufficiently large constant, then

V ˜ ( τ ) = α T 0 cos ( ω τ ) + β T 0 sin ( ω τ ) + 1 ω ∫ T 0 τ e − σ τ ′ sin ( ω ( τ − τ ′ ) ) g ( e σ τ ′ V ˜ ( τ ′ ) ) d τ ′ .

Let

ℬ = { V ˜ ∈ C [ T 0 , ∞ ) : ‖ V ˜ ‖ 0 = sup T 0 < τ < ∞ ∣ V ˜ ( τ ) ∣ ≤ 2 C 1 } ,

where C 1 is a positive constant. Consider

N V ˜ ( τ ) = α T 0 cos ( ω τ ) + β T 0 sin ( ω τ ) + 1 ω ∫ T 0 τ e − σ τ ′ sin ( ω ( τ − τ ′ ) ) g ( e σ τ ′ V ˜ ( τ ′ ) ) d τ ′

as a map on ℬ . If V ˜ ∈ ℬ , then

(3.15) ∣ g ( e σ τ V ˜ ( τ ) ) ∣ ≤ c 2 e 2 σ τ C 1 2

and

(3.16) ‖ N V ˜ − ( α T 0 cos ( ω τ ) + β T 0 sin ( ω τ ) ) ‖ 0 ≤ C ′ e σ T 0 C 1 2

for some positive constants c 2 , C ′ independent of T 0 . Since σ < 0 , we can choose T 0 > 1 suitably large so that C ′ e σ T 0 C 1 = C ′ ⁄ ( 2 C ′ + 10 ) , then

(3.17) ‖ N V ˜ − ( α T 0 cos ( ω τ ) + β T 0 sin ( ω τ ) ) ‖ 0 ≤ C 1 2 .

If it is necessary, we can also choose T 0 large enough and ε small so that

∣ α T 0 ∣ + ∣ β T 0 ∣ ≤ c 1 ε e − σ T 0 = ( 2 C ′ + 10 ) c 1 ε C 1 ≤ C 1 ,

then ‖ N V ˜ ‖ 0 ≤ 2 C 1 . In particular, N V ˜ is a map from ℬ into itself. Similarly, we can prove that

(3.18) ‖ N V ˜ 1 − N V ˜ 2 ‖ 0 ≤ C 1 e σ T 0 ‖ V ˜ 1 − V ˜ 2 ‖ 0 ≤ 1 10 ‖ V ˜ 1 − V ˜ 2 ‖ 0 .

Therefore, N V ˜ is a contraction mapping from ℬ into itself. The contraction mapping theorem ensures that N V ˜ has a fixed point W in ℬ .

Step 4: We have V ˜ = W .

Indeed, if W is a fixed point, then W ˜ ( τ ) = e σ τ W ( τ ) satisfies

(3.19) W ˜ ″ ( τ ) + n − 1 − 2 p − 1 W ˜ ′ ( τ ) + n − 1 − 1 p − 1 W ˜ ( τ ) + g ( W ˜ ( τ ) ) = 0 .

We have chose α T 0 , β T 0 so that

V ( T 0 ) = e σ T 0 [ α T 0 cos ( ω T 0 ) + β T 0 sin ( ω T 0 ) ] , V ′ ( T 0 ) = e σ T 0 [ ( α T 0 σ + β T 0 ω ) cos ( ω T 0 ) + ( β T 0 σ − α T 0 ω ) sin ( ω T 0 ) ] .

Then W ˜ also satisfies

W ˜ ( T 0 ) = e σ T 0 [ α T 0 cos ( ω T 0 ) + β T 0 sin ( ω T 0 ) ] , W ˜ ′ ( T 0 ) = e σ T 0 [ ( α T 0 σ + β T 0 ω ) cos ( ω T 0 ) + ( β T 0 σ − α T 0 ω ) sin ( ω T 0 ) ] .

Since both W ˜ and V satisfy (3.19) with the same initial values, by the uniqueness of the solution of the ordinary differential equation, we conclude that W ˜ = V .

Step 5: By the aforementioned analysis, we conclude that

(3.20) V ( τ ) = e σ τ O ( 1 ) .

By (3.20), we know that there exist two constants a 1 ′ and a 2 ′ such that

(3.21) 1 ω ∫ T τ e σ ( τ − τ ′ ) sin ( ω ( τ − τ ′ ) ) g ( V ( τ ′ ) ) d τ ′ = a 1 ′ e σ τ sin ( ω τ ) − 1 ω e σ τ sin ( ω τ ) ∫ τ ∞ e − σ τ ′ cos ( ω τ ′ ) g ( V ( τ ′ ) ) d τ ′ + b 1 ′ e σ τ cos ( ω τ ) + 1 ω e σ τ cos ( ω τ ) ∫ τ ∞ e − σ τ ′ sin ( ω τ ′ ) g ( V ( τ ′ ) ) d τ ′ = a 1 ′ e σ τ sin ( ω τ ) + b 1 ′ e σ τ cos ( ω τ ) + O ( e 2 σ τ ) .

By (3.12), (3.21), and the definition of w , we have

(3.22) V ( τ ) = e σ τ [ ( α T 0 + a 1 ′ ) sin ( ω τ ) + ( β T 0 + b 1 ′ ) cos ( ω τ ) ] + O ( e 2 σ τ ) .

Take

b 0 = α T 0 + a 1 ′ , a 0 = β T 0 + b 1 ′ ,

then (3.22) implies that for s ∈ ( S 0 , ∞ ) ,

(3.23) v 0 ( s ) = A p s − α + a 0 cos ( ω ln s ) + b 0 sin ( ω ln s ) s n − 1 2 + O s − n − 1 − 1 p − 1 .

(3.23) is exactly (3.5). Next, we show that a 0 2 + b 0 2 ≠ 0 . If it is false, then (3.12) and (3.21) imply

(3.24) V ( τ ) = − 1 ω e σ τ sin ( ω τ ) ∫ τ ∞ e − σ τ ′ cos ( ω τ ′ ) g ( V ( τ ′ ) ) d τ ′ + 1 ω e σ τ cos ( ω τ ) ∫ τ ∞ e − σ τ ′ sin ( ω τ ′ ) g ( V ( τ ′ ) ) d τ ′ .

Similar to the previous arguments, (3.24) can define a contraction mapping on ℬ . It is clear that 0 is a fixed point of the new contraction mapping, hence V ≡ 0 . Since we have assumed that v 0 ≠ A p s − α , this is a contradiction. Hence, we have proved that a 0 2 + b 0 2 ≠ 0 . The proof of the lemma is completed.□

Lemma 3.2

Let Q ⁄ ( Q − 4 ) < p < p J L ( Q − 2 ) , v 1 be the unique solution of the initial value problem

(3.25) v 1 ′ ′ ( s ) + n s v 1 ′ ( s ) + p 4 s v 0 p − 1 ( s ) v 1 ( s ) − β 4 v 0 ( s ) − n 3 s v ′ ( s ) + 1 24 s v 0 p ( s ) = 0 , v 1 ( 0 ) = 0 , v 1 ′ ( 0 ) = 0 .

Then for s ∈ [ S 0 , ∞ ) ,

(3.26) v 1 ( s ) = C p s 2 − α + s 2 − n − 1 2 ( a 1 cos ( ω ln s ) + b 1 sin ( ω ln s ) ) + o s 2 − n − 1 2 ,

with C p satisfying

(3.27) ( 2 − α ) ( n + 1 − α ) + p 4 A p p − 1 C p = A p β 4 − n 3 ( p − 1 ) − 1 24 A p p − 1 .

Moreover, ( a 1 , b 1 ) is the solution of

(3.28) D 1 a 1 + 4 ω b 1 = β a 0 + n 3 b 0 ω − n ( n − 1 ) 6 a 0 − E 1 a 0 , − 4 ω a 1 + D 1 b 1 = β b 0 − n 3 a 0 ω − n ( n − 1 ) 6 b 0 − E 1 b 0 ,

where

D 1 = ( 5 − n ) ( n + 3 ) 4 − ω 2 + p A p p − 1 , E 1 = p 4 ( p − 1 ) A p p − 2 C p − 1 24 p A p p − 1 ,

a 0 , b 0 , and ω are given by Lemma 3.1.

Proof

The existence and the uniqueness of solutions of (3.25) follow from standard ordinary differential equation theory. Analyzing the terms which contain v 0 and using the Taylor expansion, we can find that the leading terms are of the forms

s − α , s − n − 1 2 cos ( ω ln s ) , s − n − 1 2 sin ( ω ln s ) .

We also note that

O s − n − 1 − 1 n − 1 = o s − n − 1 2 ,

provided that p > Q ⁄ ( Q − 4 ) . Hence, it is natural to assume that v 1 can be written as

v 1 ( s ) = C p s 2 − α + s 2 − n − 1 2 ( a 1 cos ( ω ln s ) + b 1 sin ( ω ln s ) ) + o s 2 − n − 1 2 .

With the help of this explicit form, (3.27) and (3.28) can be derived by direct calculation.□

Remark 3.3

Since Q ⁄ ( Q − 4 ) < p < p J L ( Q − 2 ) , then ω ≠ 0 . Therefore, the matrix

J = D 1 4 ω − 4 ω D 1

is invertible. In particular, (3.28) is solvable. Moreover, since

b 0 β a 0 + n 3 b 0 ω − n ( n − 1 ) 6 a 0 − E 1 a 0 − a 0 β b 0 − n 3 a 0 ω − n ( n − 1 ) 6 b 0 − E 1 b 0 ≠ 0 ,

we conclude that a 1 2 + b 1 2 ≠ 0 .

Lemma 3.4

Let Q ⁄ ( Q − 4 ) < p < p J L ( Q − 2 ) , then for ε > 0 sufficiently small, equation (3.2) has a solution v such that

(3.29) v ( s ) = v 0 ( s ) + ∑ k = 1 ∞ ε 2 k v k ( s ) .

Moreover, for s ∈ [ S 0 , ∞ ) ,

(3.30) v k ( s ) = ∑ j = 1 k d j k s 2 j − α + ∑ j = 1 k e j k s 2 j − n − 1 2 sin ( ω ln s + E j k ) + o s 2 k − n − 1 2 ,

where d j k , e j k , E j k ( j = 1 , 2 , … , k ) are constants. Moreover,

(3.31) d 1 1 = C p , e 1 1 = a 1 2 + b 1 2 , sin E 1 1 = a 1 e 1 1 , cos E 1 1 = b 1 e 1 1 .

Proof

We take (3.29) into (3.3), and we expand (3.3) according to the order of ε . By calculation, we can check that for k ≥ 2 , v k satisfies the following equation:

(3.32) v k ″ ( s ) + n 3 v ′ ( s ) − n 3 s v ′ ( s ) + ∑ j = 1 k − 1 n α j v ′ ( s ) − β 4 v k − 1 ( s ) + 1 4 s d k d t k ∑ l = 0 k t l v l p ∣ t = 0 ( s ) + 1 24 s d k − 1 d t k − 1 ∑ l = 0 k − 1 t l v l p ∣ t = 0 ( s ) + 1 4 ∑ j = 1 k − 1 β j s 2 j + 1 d k − j − 1 d t k − j − 1 ∑ l = 0 k − j − 1 t l v l p ∣ t = 0 ( s ) , v k ( 0 ) = 0 , v ′ ( 0 ) = 0 .

Similar to the proof of Lemma 3.2, we can find that the leading order of the terms involves only v 0 , v 1 , … , v k − 1 . Since we have obtained the expansion of v 0 and v 1 , then the expansion of v k can be derived by using the Taylor expansion of v p and the induction argument.□

By Lemma 3.4 and the definition of the function v , we can obtain the following proposition.

Proposition 3.5

Let Q ⁄ ( Q − 4 ) < p < p J L ( Q − 2 ) , and let Φ ε inn be an inner solution of (2.6) with Φ ε inn ( 0 ) = ε − α . Then for any sufficiently small ε > 0 and θ > S 0 ε but θ is also sufficiently small,

(3.33) Φ ε inn ( θ ) = A p θ α + C p θ α − 2 + ∑ k = 2 ∞ ∑ j = 1 k d j k ε 2 ( k − j ) θ 2 j − α + ε n − 1 2 − α a 0 cos ω ln θ ε + b 0 cos ω ln θ ε θ n − 1 2 + ε n − 1 2 − α a 1 cos ω ln θ ε + b 1 cos ω ln θ ε θ n − 1 2 − 2 + ε n − 1 2 − α ∑ k = 2 ∞ ∑ j = 1 k e j k ε 2 ( k − j ) θ 2 j − n − 1 2 sin ω ln θ ε + E j k + ε n − 1 2 − α ε n − 1 2 − α O θ σ − n − 1 2 + ∑ k = 1 ∞ o θ 2 k − n − 1 2 .

Proof

Since

Φ ε inn ( θ ) = ε − α v θ ε = ε − α v 0 θ ε + ∑ k = 1 ∞ ε 2 k v k θ ε ,

then (3.33) is a direct consequence of Lemma 3.4.□

The results obtained above can be summarized as the following theorem.

Theorem 3.6

Let Q ⁄ ( Q − 4 ) < p < p J L ( Q − 2 ) , and let Φ Λ inn be an inner solution of equation (2.6) with Φ Λ ( 0 ) = Λ . Then for any sufficiently large Λ > 0 ,

(3.34) Φ Λ inn ( θ ) = A p θ α + C p θ α − 2 + ∑ k = 2 ∞ ∑ j = 1 k d j k Λ − 2 ( p − 1 ) ( k − j ) θ 2 j − α + Λ σ α a 0 cos ( ω ln ( Λ p − 1 θ ) ) + b 0 sin ( ω ln ( Λ p − 1 θ ) ) θ n − 1 2 + Λ σ α a 1 cos ( ω ln ( Λ p − 1 θ ) ) + b 1 sin ( ω ln ( Λ p − 1 θ ) ) θ n − 1 2 − 2 + Λ σ α ∑ k = 2 ∞ ∑ j = 1 k e j k Λ − 2 ( p − 1 ) ( k − j ) θ 2 j − n − 1 2 sin ( ω ln ( Λ p − 1 θ ) + E j k ) + Λ σ α Λ σ α O θ σ − n − 1 2 + ∑ k = 1 ∞ o θ 2 k − n − 1 2

provided that θ = ∣ O Λ σ ( 2 − σ ) α ∣ .

Finally, we prove two lemmas, which will be useful in the proof of the main theorem.

Lemma 3.7

Let Q ⁄ ( Q − 4 ) < p < p J L ( Q − 2 ) , and let v 0 be the unique positive radial solution of equation (3.4). We define

(3.35) v ( Λ , θ ) = Λ v 0 ( Λ p − 1 θ ) ,

then for Λ p − 1 θ ≥ S 0 , v ( Λ , θ ) satisfies the following.

For k = 0 , 1 , 2 ,
∂ k ∂ Λ k ( v ( Λ , θ ) ) = ∂ k ∂ Λ k A p θ α + C ∂ k ∂ Λ k θ − n − 1 2 Λ − ( n − 1 ) ( p − 1 ) 2 − 1 sin ( ω ln ( Λ p − 1 θ ) + D ) + Λ − k − [ ( p − 1 ) n − 1 − 1 p − 1 − 1 ] O θ − n − 1 − 1 p − 1 .
For k = 0 , 1 , 2 ,
∂ k ∂ Λ k ( v θ ( Λ , θ ) ) = − α ∂ k ∂ Λ k A p θ α + 1 + C ∂ k + 1 ∂ Λ k ∂ θ θ − n − 1 2 Λ − ( n − 1 ) ( p − 1 ) 2 − 1 sin ( ω ln ( Λ p − 1 θ ) + D ) + Λ − k − n − 1 − 1 p − 1 ( p − 1 ) − 1 O θ − n − 1 p − 1 ,

where

(3.36) C = a 0 2 + b 0 2 , D = tan − 1 b 0 a 0 .

Proof

We know from Lemma 3.1 that

v 0 ( s ) = A p s − α + a 0 cos ( ω ln s ) + b 0 sin ( ω ln s ) s n − 1 2 + O s − n − 1 − 1 p − 1 = A p s − α + C s − n − 1 2 sin ( ω ln s + D ) + O s − n − 1 − 1 p − 1 ,

where C and D are given by (3.36). Then

(3.37) v ( Λ , θ ) = A p θ α + C Λ − ( n − 1 ) ( p − 1 ) 2 − 1 θ − n − 1 2 sin ( ω ln ( Λ p − 1 θ ) + D ) + Λ − n − 1 − 1 p − 1 ( p − 1 ) − 1 O θ − n − 1 − 1 p − 1 .

With the help of (3.37), (i) and (ii) can be proved directly.□

Lemma 3.8

In the region θ = ∣ O Λ σ ( 2 − σ ) α ∣ , the solution Φ ( Λ , θ ) of (2.6) with

Φ ( Λ , 0 ) = Λ , Φ θ ( Λ , 0 ) = 0

satisfies

∂ Φ ∂ Λ ( Λ , θ ) − ∂ v ∂ Λ ( Λ , θ ) = Λ − ( p − 1 ) ( n − 1 ) 2 o θ − n − 1 2 ;
∂ Φ θ ∂ Λ ( Λ , θ ) − ∂ v θ ∂ Λ ( Λ , θ ) = Λ − ( p − 1 ) ( n − 1 ) 2 o θ − n + 1 2 ;
∂ 2 Φ ∂ Λ 2 ( Λ , θ ) − ∂ 2 v ∂ Λ 2 ( Λ , θ ) = Λ − ( p − 1 ) ( n − 1 ) 2 + 1 θ − n − 1 2 ;
∂ 2 Φ θ ∂ Λ 2 ( Λ , θ ) − ∂ 2 v θ ∂ Λ 2 ( Λ , θ ) = Λ − ( p − 1 ) ( n − 1 ) 2 + 1 ∣ o θ − n + 1 2 ∣ .

Proof

By (3.1), we deduce that

Φ ( Λ , θ ) = Λ v ( Λ p − 1 θ ) = Λ ( v 0 ( Λ p − 1 θ ) + ∑ k = 1 ∞ Λ − 2 k α v k ( Λ p − 1 θ ) ) = v ( Λ , θ ) + ∑ k = 1 ∞ Λ 1 − 2 k α v k ( Λ p − 1 θ ) .

Since θ = ∣ O Λ σ ( 2 − σ ) α ∣ , then

Λ p − 1 θ = ∣ O ( Λ 2 ( p − 1 ) 2 − σ ) ∣ > S 0 ,

provided that Λ is sufficiently large. Note that

ε = Λ − 1 α , − σ α = ( p − 1 ) ( n − 1 ) 2 − 1 .

Then this lemma can be obtained from Lemma 3.7 and Proposition 3.5.□

4 Outer solutions

In this section, we study the asymptotic behavior of solutions of (2.6) far from θ = 0 . Let Φ * be the singular solution in Remark 2.1, we first obtain the following lemma.

Lemma 4.1

The ordinary differential equation

(4.1) 4 sin θ ϕ ″ ( θ ) + 4 n cos θ ϕ ′ ( θ ) − β sin θ ϕ ( θ ) + p A p p − 1 [ sin θ ] − 1 ϕ ( θ ) = 0

admits two fundamental solutions ϕ 1 and ϕ 2 such that any solution ϕ of (4.1) can be written as:

ϕ ( θ ) = c 1 ϕ 1 ( θ ) + c 2 ϕ 2 ( θ ) ,

where c 1 and c 2 are two constants. Moreover, as θ → 0 , there exist two constants c 1 ′ and c 2 ′ such that

(4.2) ϕ ( θ ) = θ − n − 1 2 c 1 ′ cos ω ln θ 2 + c 2 ′ sin ω ln θ 2 + O θ 2 − n − 1 2 .

If ϕ ≠ 0 , then c 1 ′ 2 + c 2 ′ 2 ≠ 0 .

Proof

Let

ϕ ( θ ) = [ sin θ ] − 1 p − 1 ϕ ˜ ( θ ) .

We know from (4.1) that ϕ ˜ satisfies

(4.3) 4 sin 2 θ ϕ ˜ ″ ( θ ) + 4 n − 2 p − 1 sin θ cos θ ϕ ˜ ′ ( θ ) + ( p − 1 ) A p p − 1 ϕ ˜ ( θ ) = 0 .

Under the Emden-Fowler transformations:

ψ ( τ ) = ϕ ˜ ( θ ) , τ = ln tan θ 2 ,

we obtain that ψ satisfies

(4.4) ψ ″ ( τ ) + n − 1 − 2 p − 1 1 − 2 e 2 τ 1 + e 2 τ ψ ′ ( τ ) + n − 1 − 1 p − 1 ψ ( τ ) = 0 .

By the ordinary differential equation theories, we know that for every a , (4.4) has a unique solution such that ψ ( 0 ) = a , ψ ′ ( 0 ) = 0 . Moreover, (4.4) admits two fundamental solutions ψ 1 , ψ 2 ∈ C 2 ( − ∞ , 0 ) such that any solution ψ of (4.4) can be written as:

ψ ( τ ) = c 1 ψ 1 ( τ ) + c 2 ψ 2 ( τ ) .

By the method of variation of constant, we have

(4.5) ψ ( τ ) = e σ τ [ ℓ 3 cos ( ω τ ) + ℓ 4 sin ( ω τ ) ] + 1 ω ∫ T τ e σ ( τ − τ ′ ) sin ( ω ( τ − τ ′ ) ) j ( ψ ) ( τ ′ ) d τ ′ ,

where T ∈ ( − ∞ , 0 ) and

j ( ψ ) ( τ ′ ) = − n − 1 − 2 p − 1 2 e 2 τ ′ 1 + e 2 τ ′ ψ ′ ( τ ′ ) .

Let

ψ ˆ ( τ ) = e − σ τ ψ ( τ ) ,

then

(4.6) ψ ˆ ( τ ) = [ ℓ 3 cos ( ω τ ) + ℓ 4 sin ( ω τ ) ] + 1 ω ∫ T τ sin ( ω ( τ − τ ′ ) ) j ( ψ ˆ ) ( τ ′ ) d τ ′ ,

with

(4.7) j ( ψ ˆ ) ( τ ′ ) = − n − 1 − 2 p − 1 2 e 2 τ ′ 1 + e 2 τ ′ ( σ ψ ˆ ( τ ′ ) + ψ ˆ ′ ( τ ′ ) ) .

We claim that by choosing ∣ T ∣ suitably large, there exists a constant c that depends only on p , n , T , c 1 , c 2 such that

(4.8) ‖ ψ ˆ ‖ 0 ≤ c , ‖ ψ ˆ ′ ‖ 0 ≤ c ,

where ‖ ψ ˆ ‖ 0 = sup τ < τ ′ < T ∣ ψ ˆ ( τ ′ ) ∣ and ‖ ψ ˆ ′ ‖ 0 = sup τ < τ ′ < T ∣ ψ ˆ ′ ( τ ′ ) ∣ . Indeed, it follows from (4.6) and (4.7) that

(4.9) ‖ ψ ˆ − [ ℓ 3 cos ( ω τ ) + ℓ 4 sin ( ω τ ) ] ‖ 0 ≤ c 0 e 2 T ( ∣ σ ∣ ‖ ψ ˆ ‖ 0 + ‖ ψ ˆ ′ ‖ 0 ) ,

where c 0 is a positive constant independent of T . On the other hand, we can check that z ( τ ) ≔ ψ ′ ( τ ) satisfies

(4.10) z ″ ( τ ) + ( n − 1 − 2 α ) z ′ ( τ ) + ( n − 1 − α ) z ( τ ) + h ( τ , ψ ( τ ) , ψ ′ ( τ ) ) = 0 ,

where

(4.11) h ( τ , ψ ( τ ) , ψ ′ ( τ ) ) = ( n − 1 − 2 α ) 2 2 e 2 τ 1 + e 2 τ 1 − 2 e 2 τ 1 + e 2 τ ψ ′ ( τ ) + 2 ( n − 1 − α ) ( n − 1 − 2 α ) 2 e 2 τ 1 + e 2 τ ψ ( τ ) − 2 ( n − 1 − 2 α ) 2 e 2 τ ( 1 + e 2 τ ) 2 ψ ′ ( τ ) .

Therefore,

e − σ τ ψ ′ ( τ ) = [ ℓ 5 cos ( ω τ ) + ℓ 6 sin ( ω τ ) ] + 1 ω ∫ T τ sin ( ω ( τ − τ ′ ) ) h ( τ ′ , ψ ˆ ( τ ′ ) , ψ ˆ ′ ( τ ′ ) ) d τ ′ ,

with

(4.12) h ( τ , ψ ˆ ( τ ) , ψ ˆ ′ ( τ ) ) = ( n − 1 − 2 α ) 2 2 e 2 τ 1 + e 2 τ 1 − 2 e 2 τ 1 + e 2 τ ( σ ψ ˆ ( τ ) + ψ ˆ ′ ( τ ) ) − 2 ( n − 1 − 2 α ) 2 e 2 τ ( 1 + e 2 τ ) 2 ( σ ψ ˆ ( τ ) + ψ ˆ ′ ( τ ) ) + 2 ( n − 1 − α ) ( n − 1 − 2 α ) 2 e 2 τ 1 + e 2 τ ψ ˆ ( τ ) .

Similar to (4.9), we can obtain that

(4.13) ‖ e − σ τ ψ ′ ( τ ) − [ ℓ 5 cos ( ω τ ) + ℓ 6 sin ( ω τ ) ] ‖ 0 ≤ c 0 e 2 T ( ∣ σ ∣ ‖ ψ ˆ ‖ 0 + ‖ ψ ˆ ′ ‖ 0 ) .

Since

ψ ˆ ′ ( τ ) = e − σ τ ψ ′ ( τ ) − σ ψ ˆ ( τ ) ,

then we can obtain (4.8) by combining (4.9) and (4.13). Equations (4.6), (4.7), and (4.8) imply that there exist two constants ℓ 3 ′ and ℓ 4 ′ such that

(4.14) ψ ˆ ( τ ) = ℓ 3 ′ cos ( ω τ ) + ℓ 4 ′ sin ( ω τ ) + 1 ω ∫ − ∞ τ sin ( ω ( τ − τ ′ ) ) j ( ψ ˆ ) ( τ ′ ) d τ ′ = ℓ 3 ′ cos ( ω τ ) + ℓ 4 ′ sin ( ω τ ) + O ( e 2 τ ) .

Therefore, as τ → ∞ ,

(4.15) ψ ( τ ) = e σ τ [ ℓ 3 ′ cos ( ω τ ) + ℓ 4 ′ sin ( ω τ ) + O ( e 2 τ ) ] .

This implies that as θ → 0 ,

(4.16) ϕ ( θ ) = [ sin θ ] − α tan θ 2 σ ℓ 3 ′ cos ω ln tan θ 2 + ℓ 4 ′ sin ω ln tan θ 2 + O tan θ 2 2 .

Since

[ sin θ ] − α = 1 θ α + 1 6 ( p − 1 ) 1 θ α − 2 + O 1 θ α − 4 , tan θ 2 σ = θ 2 σ + σ 3 θ 2 σ + 2 + O ( θ σ + 4 ) ,

then (4.2) follows from (4.16).

Finally, we prove that if ϕ ≠ 0 , then ℓ 3 ′ 2 + ℓ 4 ′ 2 ≠ 0 . If it is false, we obtain from (4.5) that

(4.17) ψ ( τ ) = 1 ω ∫ − ∞ τ e σ ( τ − τ ′ ) sin ( ω ( τ − τ ′ ) ) j ( ψ ) ( τ ′ ) d τ ′ = O ( e ( σ + 2 ) τ ) .

Taking the derivative with respect to (4.17), we can obtain that

(4.18) ψ ′ ( τ ) = O ( e ( σ + 2 ) τ ) .

We take (4.18) into (4.17), then

(4.19) ψ ( τ ) = O ( e ( σ + 4 ) τ ) .

By repeating the aforementioned arguments, we can obtain that ψ ≡ 0 , this is a contradiction. Hence, we have finished the proof of Lemma 4.1.□

Remark 4.2

By the proof of Lemma 4.1, we can obtain that for any δ > 0 , if c 1 and c 2 in (4.2) satisfy

c 1 = c ˜ 1 δ , c 2 = c ˜ 2 δ ,

where c ˜ 1 and c ˜ 2 are constants, then as θ → 0 ,

(4.20) ϕ ( θ ) ≔ ϕ δ ( θ ) = δ θ − n − 1 2 c ˜ 1 ′ cos ω ln θ 2 + c ˜ 2 ′ sin ω ln θ 2 + δ O θ 2 − n − 1 2 ,

where c ˜ 1 ′ and c ˜ 2 ′ are two constants independent of δ .

For any δ > 0 sufficiently small, if Φ ∈ C 2 ( 0 , 2 π ) is a solution of equation (2.6) such that

Φ ( θ ) = Φ * ( θ ) + δ ϕ δ ( θ ) + δ 2 ψ δ ( θ ) ,

where

ϕ δ ( θ ) = c ˜ 1 δ ϕ 1 ( θ ) + c ˜ 2 δ ϕ 2 ( θ )

is a solution of (4.1) with

c 1 = c ˜ 1 δ , c 2 = c ˜ 1 δ .

Then ψ δ satisfies the following equation:

(4.21) 4 sin θ ψ ″ ( θ ) + 4 n cos θ ψ ′ ( θ ) − β sin θ ψ ( θ ) + p Φ * p − 1 ( θ ) ψ ( θ ) + δ − 2 H ( θ ) = 0 , in 0 , π 2 , ψ ′ π 2 = − c ˜ 1 ϕ 1 ′ π 2 + c ˜ 2 ϕ 2 ′ π 2 ,

where

H ( θ ) = ( Φ * ( θ ) + δ ϕ δ ( θ ) + δ 2 ψ ( θ ) ) p − Φ * p ( θ ) − p δ Φ * p − 1 ( θ ) ϕ δ ( θ ) − p δ 2 Φ * p − 1 ( θ ) ψ ( θ ) .

For equation (4.21), we have the following result.

Lemma 4.3

For any δ > 0 sufficiently small and each fixed pair ( c ˜ 1 , c ˜ 2 ) , equation (4.21) admits a solution ψ δ ∈ C 2 0 , π 2 .

Proof

We set the initial value conditions of (4.21) at θ = π 2 : ψ π 2 = 1 , provided

ψ ′ π 2 = − c ˜ 1 ϕ 1 ′ π 2 + c ˜ 2 ϕ 2 ′ π 2 = 0 ;

ψ π 2 = 0 , provided

ψ ′ π 2 = − c ˜ 1 ϕ 1 ′ π 2 + c ˜ 2 ϕ 2 ′ π 2 ≠ 0 .

Then the shooting argument implies that (4.21) admits a unique nontrivial solution ψ δ in C 2 0 , π 2 .□

Proposition 4.4

Let δ be a sufficiently small constant and let ψ δ be the function given by Lemma 4.3, then for θ = O δ 2 2 − σ ,

(4.22) ψ δ ( θ ) = θ − n − 1 2 d ˜ 1 cos ω ln θ 2 + d ˜ 2 sin ω ln θ 2 + O θ 2 − n − 1 2 ,

where d ˜ 1 and d ˜ 2 are constants depending on c ˜ 1 and c ˜ 2 but independent of δ .

Proof

We set

ψ δ ( θ ) = [ sin θ ] − α ψ ˜ δ ( θ ) ,

then ψ ˜ δ satisfies the following equation:

(4.23) 4 sin 2 θ ψ ˜ ″ ( θ ) + 4 n − 2 p − 1 sin θ cos θ ψ ˜ ′ ( θ ) + ( p − 1 ) A p p − 1 ψ ˜ ( θ ) + G ( ψ ˜ ( θ ) ) = 0 ,

where

G ( ψ ˜ ( θ ) ) = δ − 2 [ sin θ ] 1 + α [ ( Φ * ( θ ) + δ ϕ δ ( θ ) + δ 2 [ sin θ ] − α ψ ˜ ( θ ) ) p − Φ * p ( θ ) − p δ Φ * p − 1 ( θ ) ϕ δ ( θ ) − p δ 2 Φ * p − 1 ( θ ) [ sin θ ] − α ψ ˜ ( θ ) ] .

Consider the Emden-Fowler transformations

z ( τ ) = ψ ˜ ( θ ) , τ = ln tan θ 2 ,

then for τ ∈ ( − ∞ , 0 ) , z ( τ ) satisfies the following equation:

(4.24) z ″ ( τ ) + n − 1 − 2 p − 1 1 − 2 e 2 τ 1 + e 2 τ z ′ ( τ ) + n − 1 − 1 p − 1 z ( τ ) + G ( z ( τ ) ) = 0 .

Let

ϕ ˜ 1 ( τ ) = [ sin θ ] α ϕ 1 ( θ ) , ϕ ˜ 2 ( τ ) = [ sin θ ] α ϕ 2 ( θ ) .

By the method of variation of constants, we know that for T ∈ ( − ∞ , 0 ) and ∣ T ∣ suitably large,

(4.25) z ( τ ) = ϑ 1 ϕ ˜ 1 ( τ ) + ϑ 2 ϕ ˜ 2 ( τ ) + ∫ T τ − ϕ ˜ 1 ( τ ) ϕ ˜ 2 ( τ ′ ) + ϕ ˜ 2 ( τ ) ϕ ˜ 1 ( τ ′ ) ϕ ˜ 1 ( τ ′ ) ϕ ˜ 2 ′ ( τ ′ ) − ϕ ˜ 1 ′ ( τ ′ ) ϕ ˜ 2 ( τ ′ ) G ( z ( τ ′ ) ) d τ ′ = e σ τ [ ϑ 1 cos ( ω τ ) + ϑ 2 sin ( ω τ ) ] + O ( e ( σ + 2 ) τ ) + p ( p − 1 ) 2 ω ∫ T τ e σ τ sin ( ω ( τ − τ ′ ) ) [ e σ τ ′ δ 2 ] [ ρ ( τ ′ ) ] 2 d τ ′ + 1 ω ∫ T τ e σ τ sin ( ω ( τ − τ ′ ) ) O ( [ e σ τ ′ δ 2 ] 2 [ ρ ( τ ′ ) ] 3 ) d τ ′ + 1 ω ∫ T τ e σ τ sin ( ω ( τ − τ ′ ) ) O ( e 2 τ ′ ) [ e σ τ ′ δ 2 ] [ ρ ( τ ′ ) ] 2 d τ ′ + 1 ω ∫ T τ e σ τ sin ( ω ( τ − τ ′ ) ) O ( e 2 τ ′ ) O ( [ e σ τ ′ δ 2 ] 2 [ ρ ( τ ′ ) ] 3 ) d τ ′ ,

where

ρ ( τ ′ ) = c ˜ 1 cos ( ω τ ′ ) + c ˜ 2 sin ( ω τ ′ ) + e − σ τ ′ z ( τ ′ ) .

Let

z ˆ ( τ ) = e − σ τ z ( τ ) .

Similar to the proof of Lemma 4.1, we know that there exists a positive constant M ≔ M ( n , p , T ) but independent of δ such that

(4.26) ‖ z ˆ − ( ϑ 1 cos ( ω τ ) + ϑ 2 sin ( ω τ ) ) ‖ 0 ≤ M ,

provided that for τ ∈ [ 10 T , 2 T ] ,

(4.27) δ 2 = ∣ O ( e ( 2 − σ ) τ ) ∣ .

Therefore,

z ( τ ) = e σ τ [ ϑ 1 cos ( ω τ ) + ϑ 2 sin ( ω τ ) ] + O ( e ( σ + 2 ) τ ) ,

provided that (4.27) holds. It follows that Proposition 4.4 holds.□

Theorem 4.5

For any δ > 0 sufficiently small, equation (2.6) admits an outer solution Φ δ out ∈ C 2 0 , π 2 such that

(4.28) Φ δ out ( θ ) = Φ * ( θ ) + δ ϕ δ ( θ ) + δ 2 ψ δ ( θ ) , in 0 , π 2 , ( Φ δ out ) ′ π 2 = 0 .

Moreover, if

(4.29) θ = ∣ O ( δ 2 2 − σ ) ∣ ,

then

(4.30) Φ δ out ( θ ) = A p θ α + A p 6 ( p − 1 ) 1 θ α − 2 + δ 2 ϑ 3 cos ω ln θ 2 + ϑ 4 sin ω ln θ 2 θ n − 1 2 + δ 2 O 1 θ n − 1 2 − 2 ,

where ϑ 3 and ϑ 4 are constants, which are independent of δ . In particular, if

(4.31) δ 2 = ∣ O ( θ 2 − σ ) ∣ ,

then Φ δ out can be written as:

(4.32) Φ δ out ( θ ) = A p θ α + A p 6 ( p − 1 ) 1 θ α − 1 + δ 2 θ − n − 1 2 ϑ 3 cos ω ln θ 2 + ϑ 4 sin ω ln θ 2 + δ 4 O 1 θ n − 1 2 − σ .

Proof

It follows from the expression of Φ * , Lemma 4.1, and Proposition 4.4 that

Φ δ out ( θ ) = Φ * ( θ ) + δ 2 ( c ˜ 1 ϕ 1 ( θ ) + c ˜ 2 ϕ 2 ( θ ) ) + δ 2 ψ δ ( θ ) = A p [ sin θ ] − 1 p − 1 + δ 2 θ n − 1 2 c ˜ 1 ′ cos ω ln θ 2 + c ˜ 2 ′ sin ω ln θ 2 + O θ 2 − n − 1 2 + δ 2 θ n − 1 2 d ˜ 1 cos ω ln θ 2 + d ˜ 2 sin ω ln θ 2 + O θ 2 − n − 1 2 .

Since for δ > 0 sufficiently small and θ = ∣ O ( δ 2 2 − σ ) ∣ ,

(4.33) O ( θ 4 − α ) = δ 2 O θ 2 − n − 1 2 ,

then (4.33) follows from the Taylor expansion of sin θ . If

δ 2 = ∣ O ( θ 2 − σ ) ∣ ,

we have

δ 2 O 1 θ n − 1 2 − 2 = δ 4 O 1 θ n − 1 2 − σ .

Then (4.32) follows from (4.30).□

Remark 4.6

Similar to the proof of Lemma 4.1 and Proposition 4.4, we can prove that ϑ 3 2 + ϑ 4 2 ≠ 0 . This fact will also be used in the proof of Theorem 1.2. Without loss of generality, we will assume that θ 3 ≠ 0 .

5 The proof of the main theorem

In this section, we will construct infinitely many regular solutions of the following equation:

(5.1) 4 sin θ Φ θ θ + 4 n cos θ Φ θ − β sin θ Φ + Φ p = 0 , in 0 , π 2 , Φ ( θ ) > 0 , in 0 , π 2 , Φ ( 0 ) = Λ , Φ ′ π 2 = 0

by matching the inner and outer solutions obtained in Theorems 3.6 and 4.5. For this purpose, we will find Θ ∈ 0 , π 2 such that the following matching conditions hold:

(5.2) Θ = O Λ σ ( 2 − σ ) α ,

(5.3) ( Φ Λ inn ( θ ) − Φ δ out ( θ ) ) ∣ θ = Θ = 0 ,

(5.4) ( Φ Λ inn ( θ ) − Φ δ out ( θ ) ) ′ ∣ θ = Θ = 0 .

First, we have the following identity.

Lemma 5.1

A p and C p satisfy

(5.5) A p 6 ( p − 1 ) = C p .

Proof

It is easy to check that

(5.6) 2 − 1 p − 1 n + 1 − 1 p − 1 + p 4 A p p − 1 = 2 − 1 p − 1 n + 1 − 1 p − 1 + p p − 1 n − 1 − 1 p − 1 = 3 n + 1 − 5 p − 1 .

On the other hand, we have

(5.7) β 4 − n 3 ( p − 1 ) − 1 24 A p p − 1 = 1 p − 1 n − 1 p − 1 − n 3 ( p − 1 ) − 1 6 ( p − 1 ) n − 1 − 1 p − 1 = 1 6 ( p − 1 ) 3 n + 1 − 5 p − 1 .

By (3.27), (5.6), and (5.7), we can obtain (5.5).□

It follows from Lemma 5.1 that the first two terms of Φ Λ inn and Φ δ out can be matched. Moreover, we have

ϑ 3 cos ω ln θ 2 + ϑ 4 sin ω ln θ 2 = E sin ω ln θ + ω ln 1 2 + η , and

a 0 cos ( ω ln ( Λ p − 1 θ ) ) + b 0 sin ( ω ln ( Λ p − 1 θ ) ) = C sin ( ω ln θ + ω ln Λ p − 1 + D ) ,

where

(5.8) C = a 0 2 + b 0 2 , E = ϑ 3 2 + ϑ 4 2 , D = tan − 1 b 0 a 0 , η = tan − 1 ϑ 4 ϑ 3 .

In order to match the next term, we choose Λ * and δ * 2 such that

(5.9) δ * 2 = a 0 2 + b 0 2 ϑ 3 2 + ϑ 4 2 Λ * σ α ,

(5.10) ω ln Λ * p − 1 + D = ω ln 1 2 + η + 2 m π ,

where m is a large positive integer. Consider small perturbations of Λ * and δ * defined in (5.9) and (5.10), i.e.,

(5.11) Λ = Λ * 1 + O Λ * 2 σ ( 2 − σ ) α ,

(5.12) δ 2 = δ * 2 1 + O Λ * 2 σ ( 2 − σ ) α .

We will see that the parameters Λ and δ required to satisfy the matching conditions (5.2), (5.3), and (5.4) can be obtained as the aforementioned small perturbations. To show this, we define

(5.13) F ( Λ , δ 2 ) = Θ n − 1 2 ( Φ Λ inn ( Θ ) − Φ δ out ( Θ ) ) Θ ω [ θ n − 1 2 ( Φ Λ inn ( θ ) − Φ δ out ( θ ) ) ] ′ ∣ θ = Θ ,

where we treat δ 2 as a new variable. Taking Λ = Λ * and δ 2 = δ * 2 in (5.13), then Theorems 3.6 and 4.5 imply

(5.14) ∣ Θ − n − 1 2 F ( Λ * , δ * 2 ) ∣ ≤ M δ * 4 Θ σ − n − 1 2 + small terms .

As in [5] and [7], we evaluate the Jacobian of F at ( Λ * , δ * 2 ) . By Lemmas 3.7, 3.8, Theorems 3.6 and 4.5, we can obtain that

(5.15) ∂ F ( Λ , δ 2 ) ∂ ( Λ , δ 2 ) = C σ α sin τ + ω ( p − 1 ) cos τ Λ * σ α − 1 , − E sin τ C σ α cos τ − ω ( p − 1 ) sin τ Λ * σ α − 1 , − E cos τ + small terms ,

where

τ = ω ln Θ + ω ln Λ * p − 1 + D = ω ln Θ + ω ln 1 2 + η + 2 m π .

To simplify this expression, we define

G ( x , y ) = F Λ * + x Λ * 1 − σ α , δ * 2 + y .

By (5.14), Theorem 4.5, and Lemma 3.7, Lemma 3.8, we can express G in the following form:

(5.16) G ( x , y ) = C + ( L + small terms ) x y + E ( x 2 ( δ * 2 ) − 1 + y 2 Θ σ ) ,

where C is a constant vector which is bounded by M δ * 4 Θ σ and L is given by

L = C σ α sin τ + ω ( p − 1 ) cos τ , − E sin τ C σ α cos τ − ω ( p − 1 ) sin τ , − E cos τ .

Also ∣ E ∣ is bounded independent of x , y , Λ , and δ . Thus,

(5.17) G ( x , y ) = C + L x y + T ( x , y ) .

By Lemma 3.1 and Remark 4.6, we have C ≠ 0 , E ≠ 0 . It follows that L is invertible. Moreover,

∣ L − 1 ∣ ≤ 1 ( p − 1 ) ω C E .

Let J be the operator defined by

J ( x , y ) = − ( L − 1 C + L − 1 T ( x , y ) ) ,

and let

B = ( x , y ) : ( x 2 + y 2 ) 1 2 ≤ 4 M δ * 4 Θ σ ( p − 1 ) ω C E .

Since ∣ C ∣ is bounded by M δ * 4 Θ σ and ∣ E ∣ is bounded independent of x , y , Λ , δ , it is easy to see that J maps the ball B into itself. By the Brouwer fixed point theorem, we conclude that J has a fixed point in B . This point ( x , y ) satisfies G ( x , y ) = 0 and

( x 2 + y 2 ) 1 2 ≤ A δ * 4 Θ σ ,

where A is a constant independent of δ * , Λ * and Θ . By substituting for Λ and δ , and then taking Θ to have the upper limiting value of Λ * σ ( 2 − σ ) α , we obtain (5.11) and (5.12).

The aforementioned arguments yield the following result.

Theorem 5.2

For m ≫ 1 large and Λ and δ given in (5.11) and (5.12), Problem (5.1) admits a C 2 solution Φ Λ , δ . Moreover, there is Θ = ∣ O Λ σ ( 2 − σ ) α ∣ such that (5.3) and (5.4) hold. As a consequence, equation (2.6) admits infinitely many nonconstant positive solutions.

Dedicated to David Jerison on the occasion of his 70th birthday, with admiration.

Acknowledgments

The authors are grateful to the referees for their careful reading of the manuscript and for their helpful remarks.

Funding information: The research of J. Wei is partially supported by NSERC of Canada. The research of K. Wu is supported by the China Postdoctoral Science Foundation (2023M732712).
Conflict of interest: Authors state no conflict of interest.
Data availability statement: Data availability is not applicable to this article as no new data were created or analyzed in this study.

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Received: 2022-11-19

Revised: 2023-06-13

Accepted: 2023-06-29

Published Online: 2023-11-20

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/ans-2023-0084

Keywords for this article

singular solutions; Heisenberg group; Lane-Emden equation; supercritical exponent; gluing method

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