Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 29, Number 3, June 2024
Page(s) 242 - 256
DOI https://doi.org/10.1051/wujns/2024293242
Published online 03 July 2024

© Wuhan University 2024

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

The purpose of this paper is to inquire about asymptotic behavior of radially positive solutions to the kMathematical equation-Hessian equation with a Matukuma-type source:

S k ( D 2 ( - φ ) ) = | x | λ - 2 ( 1 + | x | 2 ) λ / 2 φ q ,     x Ω Mathematical equation(1)

where Sk(D2(-φ))Mathematical equation is the k-Hessian operator, q>k>1,λ>0,n>2k,kNMathematical equation, and ΩMathematical equation is a suitable bounded domain in RnMathematical equation. The operators {Sk: k=1,,n}Mathematical equation are a family of operators including Laplace operator when k=1Mathematical equation and Monge-Ampère operator when k=nMathematical equation. The kMathematical equation-Hessian equation admits several significant applications in fluid mechanic, geometric problem and other applied subjects. For example, the kMathematical equation-Hessian equation is closely related to non-equilibrium phase transitions and statistical physics[1], the problem of prescribing the Gauss curvature of a hypersurface[2] and to the Monge-Ampère equation, which is of interest in complex geometry[3].

When k=1Mathematical equation, equation (1) reduces to the classical Matukuma equation[4]

Δ φ + | x | λ - 2 ( 1 + | x | 2 ) λ / 2 φ q = 0 ,     x R n Mathematical equation(2)

The existence and nonexistence of positive solutions to (2) could be referred to Refs. [5-9]. Batt et al[10] established a comprehensive theory of radially positive solutions to (2) in R3Mathematical equation, and displayed that there exist three different types of solutions: M-solutions (singular at r=0Mathematical equation), E-solutions (regular at r=0Mathematical equation) and F-solutions (whose existence begins away from r=0Mathematical equation). By applying an iterative method[11], Wang et al[12] generalized asymptotic expansions of M-solutions of (2) from n=3Mathematical equation to n>3Mathematical equation. This iterative method also could be used for the Hénon equation -Δpφ=|x|σφqMathematical equation with p=2Mathematical equation, where the accurate asymptotic expansions of M-solutions was systematically derived in Ref. [13]. It is worth noting that the results obtained in Ref. [13] are more precise than those in Refs. [14,15]. Recently, Wang and Zhang[16] extended the work of Ref. [13] from p=2Mathematical equation to 1<p<NMathematical equation. When λ=2Mathematical equation, equation (2) reduces to

Δ φ + 1 ( 1 + | x | 2 ) λ / 2 φ q = 0 ,     x R n Mathematical equation(3)

which was presented by astrophysicist Matukuma[17] for the description of certain stellar globular clusters in a steady state, where q>1,n=3Mathematical equation, and φ>0Mathematical equation is the gravitational potential. Li[18] gave a nearly complete description of the structure of positive radial solutions to (3) when 1<q<5Mathematical equation and proved a symmetry result for general nonlinear elliptic equations. Yanagida[19] established the uniqueness of positive radial entire solution with finite total mass and obtained its explicit structure for n3Mathematical equation and 1<q<(n+2)/(n-2)Mathematical equation. We refer to Refs. [20-22] about the Matukuma equation.

When k>1Mathematical equation, Sánchez and Vergara[23] considered the problem

{ S k ( D 2 φ ) = λ | x | σ ( 1 - φ ) q , x B φ < 0 ,                                   x B φ = 0 ,                                   x B    Mathematical equation(4)

where BMathematical equation is the unit ball in Rn,n>2k,kN,λ>0,q>kMathematical equation, and σ0Mathematical equation. The existence, multiplicity and uniqueness of radially symmetric bounded solutions to (4) were investigated by a dynamical systems approach. Lately, Miyamoto et al[24] extended the problem (4) into

{ S k ( D 2 φ ) = μ | x | λ - 2 ( 1 + | x | 2 ) λ / 2 ( 1 - φ ) q , x B φ < 0 ,                                              x B φ = 0 ,                                              x B    Mathematical equation

where BMathematical equation denotes the unit ball in Rn,n>2k,kN,μ>0,q>kMathematical equation, and λ2Mathematical equation. Combining dynamical-systems tools, the intersection number between a singular and a regular solution and the super/subsolution method, the existence and multiplicity of solutions for the above problem were obtained. The problems with kMathematical equation-Hessian operator have attracted lots of attention, see e.g., Refs. [25-32].

It is known from Refs. [10,12] that the equation (1) with k=1Mathematical equation admits three different types of radially positive solutions: the F-, E- and M-solutions. Furthermore, the E- and F-solutions are regular, and the M-solutions are singular. However, when k>1Mathematical equation, it turns out that the equation (1) only has the E- and M-solutions, see Section 1.1. From the above literatures, the study of M-solutions to (1) is quite scarce. Hence, we shall pay our attention to the existence and asymptotic behavior of the singular solution (i.e., the M-solutions). To this end, let p=n-2+λ.Mathematical equation

When p>[(n-2k)q]/kMathematical equation, i.e., 1<q<[(n-2+λ)k]/(n-2k)Mathematical equation, we firstly give some a priori estimates in Theorem 1. Similar to Refs. [10,13], we find the M-solution admits a splitting form: φ=S+ΘMathematical equation, where SMathematical equation is the singular term and ΘMathematical equation is the regular one. To derive more accurate asymptotic expansions of SMathematical equation and ΘMathematical equation, we introduce a new parameter ω:=p-[(n-2k)q]/kMathematical equation, and choose k0NMathematical equation such that 2k0<(n-2k)/k2(k0+1)Mathematical equation in Theorem 1, 2 and Theorem 3. Furthermore, we separate the range p>[(n-2k)q]/kMathematical equation into three subcases: (i) [(n-2k)q]/k<p<[(n-2k)(q+1)]/kMathematical equation; (ii) p=[(n-2k)(q+1)]/kMathematical equation; (iii) p>[(n-2k)(q+1)]/kMathematical equation in Section 2. It is worth noting that we require more precise ranges of (n-2k)/kMathematical equation and ωMathematical equation for the subcase (i), which is the most complicated and difficult case in these three subcases. Combining a priori estimates with an iterative method of Refs. [10,11], we could obtain the precise asymptotic expansions of SMathematical equation and ΘMathematical equation near the origin.

The case with Laplace operator (i.e., k=1Mathematical equation) and weight term K(r)=rλ-2/(1+r2)λ/2Mathematical equation has been examined in Ref. [12]; the case with pMathematical equation-Laplace operator and weight term K(r)=rσMathematical equation has been discussed in Ref. [16]. These provided us the significant references to solve problems for the case with kMathematical equation-Hessian operator (i.e., k>1Mathematical equation) and weight term K(r)=rλ-2/(1+r2)λ/2Mathematical equation. The schemes we used in current paper are as follows.

First, motivated by Ref. [12], we replace Laplace operator (i.e., k=1Mathematical equation) with kMathematical equation-Hessian operator (i.e., k>1Mathematical equation), which leads to some computational challenges as follows. Since Δφ=r1-n(rn-1φ')'Mathematical equation, we find that the exponent of φ'Mathematical equation is 1Mathematical equation. A straightforward ordinary differential equation (ODE) analysis implies that φ'=-1rn-1[c^+0rsn-1K(s)φq(s)ds],Mathematical equation where K(r)=rλ-2/(1+r2)λ/2Mathematical equation. Based on the Taylor expansion for K(r)Mathematical equation, the asymptotic expansion of φ'Mathematical equation could be obtained. However, since Sk(D2(-φ))=cn,kr1-n(rn-k(-φ')k)',  cn,k=(nk)/n,Mathematical equation we have that the exponent of φ'Mathematical equation is kMathematical equation, and then deduce

φ ' = - 1 r ( n - k ) / k [ c ¯ + 0 r c n , k - 1 s n - 1 K ( s ) φ q ( s ) d s ] 1 / k : = - 1 r ( n - k ) / k K ¯ 1 / k ( r ) Mathematical equation(5)

Not only do we need to use Taylor expansion for K(r)Mathematical equation, but we also need to use Taylor expansion for K¯(r)Mathematical equation. The repeated application of Taylor expansion makes the calculation more complex. Inspired by Ref. [16], we replace weight term K(r)=rσMathematical equation with K(r)=rλ-2/(1+r2)λ/2Mathematical equation. Since Δpφ=r1-n(rn-1|φ'|p-2φ')'Mathematical equation, we have the exponent of φ'Mathematical equation is p-1Mathematical equation, and then derive

φ ' = - 1 r ( n - 1 ) / ( p - 1 ) [ c ˜ + 0 r s n - 1 K ( s ) φ q ( s ) d s ] 1 / ( p - 1 ) : = - 1 r ( n - 1 ) / ( p - 1 ) K ¯ ¯ 1 / ( p - 1 ) ( r ) Mathematical equation(6)

In a similar manner with (5), we also need to use Taylor expansion for K¯¯Mathematical equation corresponding to K¯Mathematical equation in (5), however, in this paper we require to use Taylor expansion for K(r)Mathematical equation once more. Second, we shall state that the precise ranges for (n-2k)/kMathematical equation and ωMathematical equation are necessary in Theorem 1. Wang and Zhang[16] obtained that φMathematical equation is in the form of

φ = c r ( n - p ) / ( p - 1 ) [ 1 + i = 1 n 0 + 1 a ^ i r i μ + o ( r ( n 0 + 1 ) μ ) ] : = c r ( n - p ) / ( p - 1 ) [ 1 + D 1 + o ( r ( n 0 + 1 ) μ ) ]   , Mathematical equation

where μ=N+σ-[(N-p)q]/(p-1)Mathematical equation, σ>pMathematical equation, and n0Mathematical equation is a positive integer.

When n0μ<(n-p)/(p-1)<(n0+1)μMathematical equation, they split the term r-(n-p)/(p-1)D1Mathematical equation into singular term (i.e., i=1n0a^iriμ-(n-p)/(p-1)Mathematical equation) and regular term (i.e., a^n+01r(n+01)μ-(n-p)/(p-1)Mathematical equation). In this paper, we obtain for (n-2k)/k<(n0+1)ωMathematical equation,

φ = c r ( n - 2 k ) / k [ 1 + i = 1 n 0 + 1 a ^ i 0 r i ω + i = 1 n 0 j = 1 η i a ^ i j r i ω + 2 j + o ( r ( n 0 + 1 ) ω ) ] : = c r ( n - 2 k ) / k [ 1 + D 2 + D 3 + o ( r ( n 0 + 1 ) ω ) ]   , Mathematical equation

where ηi=(n0-i+1)k0+n0-iMathematical equation and a^ijMathematical equation are some constants depending upon c,λ,q,k,k0,n,Mathematical equation and n0Mathematical equation. The presence of the term D3Mathematical equation is due to the repeated use of Taylor expansion. In a similar manner with r-(n-p)/(p-1)D1Mathematical equation, we shall split the term r-(n-2k)/kD2Mathematical equation into singular and regular terms when n0ω<(n-2k)/k<(n0+1)ωMathematical equation. But this range is no longer sufficient to divide the term r-(n-2k)/kD3Mathematical equation into singular and regular terms. To solve this problem, we require the following precise range of (n-2k)/kMathematical equation:

2 ( n 0 + 1 ) k 0 + 2 n 0 < ( n - 2 k ) / k < ( n 0 + 1 ) ω . Mathematical equation

On the other hand, when (n-2k)/k=(n0+1)ωMathematical equation, since the fact that the size of the exponents (n-2k)/kMathematical equation and iω+2jMathematical equation could not be determined, we introduce a precise range on ωMathematical equation, i.e., 2k0+2n0/(n0+1)<ω2(k0+1)Mathematical equation.

1 Preliminaries

1.1 Classification of Positive Solutions

In this subsection, we will separate radially positive solutions of a more general problem including (1) into two distinct types. Firstly, we state the definitions of the kMathematical equation-Hessian operator and maximal solution.

Definition 1   Let ϑC2(Ω), 1kn, kNMathematical equation and Λ=(λ1,λ2,,λn)Mathematical equation be the eigenvalues of the Hessian matrix (D2ϑ)Mathematical equation. Then the kMathematical equation-Hessian operator is given by the formula Sk(D2ϑ)=Pk(Λ)=1i1<<ikn λi1λik,Mathematical equation where Pk(Λ)Mathematical equation is the kMathematical equation-th elementary symmetric polynomial in the eigenvalues ΛMathematical equation, see Ref. [33].

Note that the kMathematical equation-Hessian operators are fully nonlinear for k1Mathematical equation. Furthermore, they are not elliptic in general, unless they are restricted to the class

Φ ϑ k ( Ω ) = { ϑ C 2 ( Ω ) C 1 ( Ω ¯ ) : S i ( D 2 ϑ ) 0   i n   Ω , i = 1 , , k } Mathematical equation(7)

Observe that Φϑk(Ω)Mathematical equation belongs to the class of subharmonic functions. Moreover, it follows from the maximum principle[33] that the functions in Φϑk(Ω)Mathematical equation are negative in ΩMathematical equation.

To investigate positive solutions of (1), under the change of variable φ=-ϑMathematical equation, it is not hard to obtain Sk(D2ϑ)=(-1)kSk(D2φ)Mathematical equation by the kMathematical equation-homogeneity of the kMathematical equation-Hessian operator [23,34].

Definition 2   A function φ=-ϑΦϑk(Ω)Mathematical equation is called a supersolution (resp. subsolution) of (1) if

S k ( D 2 ( - φ ) ) ( r e s p . ) | x | λ - 2 ( 1 + | x | 2 ) λ / 2 φ q    i n    Ω Mathematical equation

Observe that the trivial function φ0Mathematical equation is always a subsolution.

Definition 3   We say that a function φMathematical equation is a maximal solution of (1) if φMathematical equation is a solution of (1) and, for every subsolution ψMathematical equation of (1), we have ψφMathematical equation.

Remark 1   Introduction of functional space Φϑk(Ω)Mathematical equation is to ensure that the kMathematical equation-Hessian operators are elliptic. Then the maximum principle and the super/subsolutions method could be applied to investigate existence of the solutions to (1).

Let KMathematical equation be a positive function in C1(R+)Mathematical equation such that r(n-k)/kK(r)Mathematical equation is bounded as r+Mathematical equation. Suppose that φ:(R-,R)(0,+)Mathematical equation is a maximal solution of the problem

c n , k r 1 - n ( r n - k ( - φ ' ) k ) ' = K ( r ) φ q ,    q > k > 1 , n > 2 k Mathematical equation(8)

where 0R-<R<+Mathematical equation. Now we introduce the space of functions ΦkMathematical equation defined on φ=-ϑMathematical equation and Ω=(R-,R)Mathematical equation as in (7), for problem (8):

Φ k = {   φ C 2 ( ( R - , R ) ) C 1 ( [ R - , R ] ) :   ( r n - i ( - φ ' ) i ) ' 0    i n   ( R - , R ) , i = 1 , , k } Mathematical equation

Note that the functions in ΦkMathematical equation are non-negative on [R-,R]Mathematical equation. If (rn-i(-φ')i)'>0Mathematical equation Mathematical equationfor every i=1,,kMathematical equation, then any function in ΦkMathematical equation is positive and strictly decreasing on [R-,R]Mathematical equation. Let r0(R-,R)Mathematical equation and

G ( r ) : = - r 0 n - k ( - φ ' ( r 0 ) ) k - r 0 r c n , k - 1 s n - 1 K ( s ) φ q ( s ) d s   i n   ( R - , R ) Mathematical equation(9)

It follows that G(r)=-rn-k(-φ')k, i.e., (-φ')k=-G(r)rn-kMathematical equation and G'(r)=-cn,k-1rn-1K(r)φq<0  in (R-,R)Mathematical equation. Hence the limit G0:=limrR-G(r)(-,+]Mathematical equation exists.

We claim that G00Mathematical equation. If not, there exists some r*(R-,r0)Mathematical equation such that G(r)>G(r*)>0Mathematical equation in (R-,r*)Mathematical equation. Hence, (-φ')k=-G(r)rn-k<0  in (R-,r*)Mathematical equation, which is impossible.

For G00Mathematical equation, we claim that R-=0Mathematical equation. We argue by contradiction. Suppose that R->0Mathematical equation. Then there exists r*=(R-,r*)Mathematical equation such that G(r*)<0, φ'(r*)<0, G(r)Mathematical equation and φ'(r)Mathematical equation are bounded in (R-,r*)Mathematical equation. Therefore, φ(r)Mathematical equation could be extended beyond R-Mathematical equation, which is a contradiction. Thus R-=0Mathematical equation and φ'(r)<0Mathematical equation in (0,R)Mathematical equation. Therefore, the limit limr0φ(r)(0,+]Mathematical equation exists. In this case, we define R0:=0Mathematical equation and have R0=inf{r(0,R)|φ'(r)<0}Mathematical equation.

The solutions of (8) could be classified as follows:

(i) if limr0φ(r)=+Mathematical equation, then we call φMathematical equation an M-solution;

(ii) if limr0φ(r)<+Mathematical equation, then we call φMathematical equation an E-solution.

Remark 2   It is known Refs. [10,12] that when G0>0R>0R->0Mathematical equation, there exists an F-solution. Because the functions are restricted to ΦkMathematical equation, they are positive and strictly decreasing on [R-,R]Mathematical equation. It follows that G00Mathematical equation. Hence, the kMathematical equation-Hessian equation (8) has no F-solution.

In this paper, we set K(r)=rλ-2/(1+r2)λ/2Mathematical equation with λ>0Mathematical equation, the equation (8) reduces to the radial form of (1) , i.e.,

c n , k r 1 - n ( r n - k ( - φ ' ) k ) ' = r λ - 2 ( 1 + r 2 ) λ / 2 φ q ,    c n , k = ( n k ) / n Mathematical equation(10)

where q>k>1,n>2kMathematical equation.

In the following, we introduce a lemma that will be frequently used to examine the existence and uniqueness of regular term of the M-solutions when p>[(n-2k)q]/kMathematical equation. The proof can be established by similar argument as in Ref. [4].

Lemma 1   Assume that βRMathematical equation and f(r,Θ):(0,)×RRMathematical equation satisfies the following conditions:

(i) f(r,Θ)C((0,)×R)Mathematical equation;

(ii) f(r,β)Lloc1[0,)Mathematical equation;

(iii) there exist a number δ>0Mathematical equation and a function

L β : ( 0 , δ ) [ 0 , ]    w i t h    L β ( r ) L 1 [ 0 , δ ] Mathematical equation

such that for every r(0,δ)Mathematical equation and Θ1,Θ2[β-δ,β+δ]Mathematical equation,

| f ( r , Θ 1 ) - f ( r , Θ 2 ) |   L β ( r ) | Θ 1 - Θ 2 | Mathematical equation

Then the initial value problem:

Θ ' = f ( r , Θ ) ,    Θ ( 0 ) = β Mathematical equation

admits a unique solution ΘMathematical equation on (0,R)Mathematical equation with Θ(0):=limr0Θ=βMathematical equation.

1.2 Transformation to Lotka-Volterra System

In this subsection we discuss the solutions φMathematical equation of (8) when r(0,R)Mathematical equation. Inspired by Refs. [23,24], we adopt a more general transformation

u ( t ) : = r k K ( r ) φ q c n , k ( - φ ' ) k ,    v ( t ) : = r - φ ' φ ,    r : = e t Mathematical equation(11)

where φ'=dφ/drMathematical equation. Set Jφ:=(0,R)Mathematical equation and Iϕ:=lnJφMathematical equation. We find that ϕ:=(u,v):IϕR+×R+Mathematical equation is a maximal solution of a non-autonomous Lotka-Volterra system

{ u ˙ = u [ p ( t ) - u - q v ] , v ˙ = v [ - n - 2 k k + u k + v ] , Mathematical equation(12)

where ""Mathematical equation denotes differentiation with respect to tMathematical equation, R+×R+Mathematical equation is an invariant set of system (12), i.e., the positive uMathematical equation- and vMathematical equation- axes are invariant, and p(t)=n+rK'(r)K(r)Mathematical equation.

Furthermore, the inverse of (11) could be characterized by

φ = [ u ( l n r ) v ( l n r ) k c n , k r 2 k K ( r ) ] 1 / ( q - k ) Mathematical equation(13)

In particular, when K(r)=rλ-2/(1+r2)λ/2Mathematical equation, we have p(t)=n-2+λ-λe2t1+e2tMathematical equation.

Define p:=n-2+λMathematical equation. Thus limt-p(t)=pMathematical equation. Then the limiting system of system (12) as t-Mathematical equation could be written as

{ u ˙ = u [ p - u - q v ]   , v ˙ = v [ - n - 2 k k + u k + v ] . Mathematical equation(14)

2 Singular Solutions

In this section, we mainly study singular solutions φMathematical equation in (0,R)Mathematical equation and their corresponding solutions ϕMathematical equation in (-,T)Mathematical equation, where 0<R<+Mathematical equation and T=lnRMathematical equation. To establish more precisely asymptotic expansions of singular solutions (i.e., the M-solutions) near the origin, we need to divide p>[(n-2k)q]/kMathematical equation into the following three subcases:

(a) [(n-2k)q]/k<p<[(n-2k)(q+1)]/kMathematical equation;

(b) p=[(n-2k)(q+1)]/kMathematical equation;

(c) p>[(n-2k)(q+1)]/kMathematical equation.

For the case p[(n-2k)q]/kMathematical equation one can apply a dynamical system approach to obtain asymptotic expansions of singular solutions, see Refs. [10,12,16]. Firstly, we give a prior estimates of singular solution for p>[(n-2k)q]/kMathematical equation.

Lemma 2   Let p>[(n-2k)q]/kMathematical equation. Then the following statements are equivalent:

(i) φMathematical equation is an M-solution.

(ii) There exists a constant c>0Mathematical equation such that φ=cr(n-2k)/k[1+o(1)],  r0Mathematical equation.

(iii) There exists a constant c>0Mathematical equation such that u(t)=kkcq-k(n-2k)kcn,ke{p-[(n-2k)q]/k}t[1+o(1)], v(t)=n-2kk[1+o(1)],  t-Mathematical equation.

Moreover, φMathematical equation satisfies

r n - k ( - φ ' ) k = c ¯ + 0 r c n , k - 1 s n - 1 K ( s ) φ q d s Mathematical equation(15)

where c¯=((n-2k)ck)kMathematical equation and cMathematical equation are uniquely determined.

Proof   This proof can be established by similar argument as in Refs. [4,10], and is omitted here.

We proceed to prove that the M-solution φMathematical equation has a splitting form φ=S+ΘMathematical equation, where SMathematical equation is the singular term in the form of S=cP/r(n-2k)/kMathematical equation with P=1+o(1)Mathematical equation as r0Mathematical equation, and ΘMathematical equation is the regular term which satisfies

{ Θ ' = - 1 r ( n - k ) / k [ c ¯ + 0 r c n , k - 1 s n - 1 K ( s ) ( Θ + S ) q d s ] 1 / k - S ' , 0 < r < R , Θ ( 0 ) : = l i m r 0 + Θ = β R . Mathematical equation(16)

It follows that

Θ ' = - 1 r ( n - k ) / k [ c ¯ + 0 r c n , k - 1 s n - 1 K ( s ) ( Θ + c P s ( n - 2 k ) / k ) q d s ] 1 / k - ( c P r ( n - 2 k ) / k ) ' = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q c ¯ c n , k 0 r s ω - λ + 1 K ( s ) P q ( 1 + s ( n - 2 k ) / k c P Θ ) q d s ] 1 / k + ( n - 2 k ) c k r ( n - k ) / k P - c r ( n - 2 k ) / k P ' = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q c ¯ c n , k 0 r s ω - λ + 1 K ( s ) P q ( 1 + O   ( s ( n - 2 k ) / k P Θ ) ) d s ] 1 / k + ( n - 2 k ) c k r ( n - k ) / k P - c r ( n - 2 k ) / k P ' = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q c ¯ c n , k 0 r s ω - λ + 1 K ( s ) P q d s + O ( r ω + ( n - 2 k ) / k ) ] 1 / k + ( n - 2 k ) c k r ( n - k ) / k P - c r ( n - 2 k ) / k P ' = : f ( r , Θ ) Mathematical equation(17)

It is clear that if f(r,Θ)Mathematical equation satisfies the assumptions in Lemma 1, then the problem (16) admits a unique solution. In the following, we will discuss three different subcases p=[(n-2k)(q+1)]/kMathematical equation,p>[(n-2k)(q+1)]/kMathematical equation and [(n-2k)q]/k<p<[(n-2k)(q+1)]/kMathematical equation, and establish the expansions of SMathematical equation and ΘMathematical equation near the origin, respectively. In order to do so, we need the following Taylor expansions

( 1 + r ) α = 1 + α r + α ( α - 1 ) 2 ! r 2 + + α ( α - 1 ) ( α - h + 1 ) h ! r h + o ( r h ) , 1 ( 1 + r 2 ) λ / 2 = 1 - λ 2 r 2 + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! r 2 k 0 + O ( r 2 k 0 + 2 ) , Mathematical equation

which are crucial in the following process of proof.

Theorem 1   Let [(n-2k)q]/k<p<[(n-2k)(q+1)]/kMathematical equation. Define ω:=p-[(n-2k)q]/kMathematical equation and choose n0,k0NMathematical equation such that 2(n0+1)k0+2n0<(n-2k)/k(n0+1)ωMathematical equation and 2k0+2n0/(n0+1)<Mathematical equationω2(k0+1)Mathematical equation.Then there exist numbers a^ij(i=0,1,,n0+3;j=0,1,,ηi,Mathematical equation where ηi=(n0-i+1)Mathematical equationk0+n0-i;a^i0:=a^i)Mathematical equation depending on c,λ,q,k,k0,n,Mathematical equation and n0Mathematical equation such that

(i) Every M-solution φMathematical equation has the form φ=S+ΘMathematical equation, where

S = { c r ( n - 2 k ) / k ( 1 + i = 1 n 0 j = 0 η i a ^ i j r i ω + 2 j ) ,                                 ( n - 2 k ) / k < ( n 0 + 1 ) ω , c r ( n 0 + 1 ) ω ( 1 + i = 1 n 0 j = 0 η i a ^ i j r i ω + 2 j + a ^ n 0 + 1 r l ( n 0 + 1 ) ω n r ) ,   ( n - 2 k ) / k = ( n 0 + 1 ) ω , Mathematical equation(18)

and ΘMathematical equation is the solution of (16) with the expansion

Θ = { β + a ^ n 0 + 1 r + ( n 0 + 1 ) ω - ( n - 2 k ) / k o ( r ) ,                                                                         ( n - 2 k ) / k < ( n 0 + 1 ) ω , ( n 0 + 1 ) ω - ( n - 2 k ) / k β + i = 1 n 0 j = η i + 1 η i - 1 a ^ i j r ( i - n 0 - 1 ) ω + 2 j + j = 1 k 0 a ^ n 0 + 1 , j r 2 j + a ^ n 0 + 2 r ω l n r + a ^ n 0 + 3 r ω + o ( r ω ) ,                    ( n - 2 k ) / k = ( n 0 + 1 ) ω , Mathematical equation

for uniquely determined constants c>0,βRMathematical equation.

(ii) Conversely, given any c>0, βRMathematical equation, there exists a unique solution ΘMathematical equation of (16) such that φ=S+ΘMathematical equation is an M-solution, where SMathematical equation is given by (18), and ΘMathematical equation satisfies (17) with

f ( r , Θ ) = { O ( r ( n 0 + 1 ) ω - ( n - k ) / k ) ,        ( n - 2 k ) / k < ( n 0 + 1 ) ω , O ( r n 0 ( 2 k 0 + 2 - ω ) - 1 ) ,          ( n - 2 k ) / k = ( n 0 + 1 ) ω . Mathematical equation

Moreover, the solution (u,v)Mathematical equation possesses the following expansion:

{ u ( t ) = c q c ¯ c n , k e ω t [ 1 - λ 2 e 2 t + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! e 2 k 0 t - [ ω k 2 + ( n - 2 k ) ( q - k ) ] c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k e ω t + o ( e ω t ) ] , v ( t ) = n - 2 k k [ 1 + [ ω k + ( n - 2 k ) ( q - 1 ) ] c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k e ω t + o ( e ω t ) ] ,    t - . Mathematical equation

Proof   (i) Let c, c¯>0Mathematical equation be determined by Lemma 2. We compute

s n - 1 K ( s ) φ q = s p - 1 [ 1 + O ( s 2 ) ]   ( c s ( n - 2 k ) / k [ 1 + o ( 1 ) ] ) q = c q s ω - 1 [ 1 + O ( s 2 ) ] [ 1 + o ( 1 ) ] = c q s ω - 1 [ 1 + o ( 1 ) ] . Mathematical equation

Using (15), we have

( - φ ' ) k = c ¯ r n - k [ 1 + 1 c ¯ c n , k 0 r s n - 1 K ( s ) φ q d s ] = c ¯ r n - k [ 1 + c q ω c ¯ c n , k r ω + o ( r ω ) ] , φ ' = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q ω c ¯ c n , k r ω + o ( r ω ) ] 1 / k = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q ω k c ¯ c n , k r ω + o ( r ω ) ] . Mathematical equation

Thus

φ = c r ( n - 2 k ) / k - ( n - 2 k ) c q + 1 ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k r ω - ( n - 2 k ) / k + o ( r ω - ( n - 2 k ) / k ) + C ^ = c r ( n - 2 k ) / k [ 1 - ( n - 2 k ) c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k r ω + o ( r ω ) ] , Mathematical equation

where C^Mathematical equation is a constant.

Repeating the above iterative process, we deduce

s n - 1 K ( s ) φ q = c q s ω - 1 ( 1 + s 2 ) - λ / 2 [ 1 - ( n - 2 k ) c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k s ω + o ( s ω ) ] q = c q s ω - 1 [ 1 - λ 2 s 2 + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! s 2 k 0 + O ( s 2 k 0 + 2 ) ] [ 1 - q ( n - 2 k ) c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k s ω + o ( s ω ) ] = c q s ω - 1 [ 1 - λ 2 s 2 + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! s 2 k 0 - q ( n - 2 k ) c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k s ω + o ( s ω ) ] . Mathematical equation

Hence,

( - φ ' ) k = c ¯ r n - k [ 1 + 1 c ¯ c n , k 0 r s n - 1 K ( s ) φ q d s ] Mathematical equation

= c ¯ r n - k { 1 + c q c ¯ c n , k 0 r s ω - 1 [ 1 + j = 1 k 0 ( - 1 ) j λ ( λ + 2 ) ( λ + 2 j - 2 ) ( 2 j ) ! ! s 2 j - q ( n - 2 k ) c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k s ω + o ( s ω ) ] d s } = c ¯ r n - k [ 1 + c q ω c ¯ c n , k r ω + j = 1 k 0 ( - 1 ) j λ ( λ + 2 ) ( λ + 2 j - 2 ) c q ( 2 j ) ! ! ( ω + 2 j ) c ¯ c n , k r ω + 2 j - q ( n - 2 k ) c 2 q 2 ω 2 [ ω - ( n - 2 k ) / k ] k 2 ( c ¯ c n , k ) 2 r 2 ω + o ( r 2 ω ) ] . Mathematical equation

Then

φ ' = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q ω k c ¯ c n , k r ω + j = 1 k 0 ( - 1 ) j λ ( λ + 2 ) ( λ + 2 j - 2 ) c q ( 2 j ) ! ! ( ω + 2 j ) k c ¯ c n , k r ω + 2 j - [ q ( n - 2 k ) + ( k - 1 ) ( ω k - n + 2 k ) ] c 2 q 2 ω 2 [ ω - ( n - 2 k ) / k ] k 3 ( c ¯ c n , k ) 2 r 2 ω + o ( r 2 ω ) ] . Mathematical equation

If n0=1Mathematical equation, i.e., 4k0+2<(n-2k)/k2ωMathematical equation, then there exists βRMathematical equation such that when (n-2k)/k=2ωMathematical equation,

φ ' = - 2 ω c r 2 ω + 1 [ 1 + c q ω k c ¯ c n , k r ω + j = 1 k 0 ( - 1 ) j λ ( λ + 2 ) ( λ + 2 j - 2 ) c q ( 2 j ) ! ! ( ω + 2 j ) k c ¯ c n , k r ω + 2 j + ( 2 q - k + 1 ) c 2 q 2 ω 2 k 2 ( c ¯ c n , k ) 2 r 2 ω + o ( r 2 ω ) ] , φ = c r 2 ω [ 1 + 2 c q ω k c ¯ c n , k r ω + j = 1 k 0 ( - 1 ) j 2 ω λ ( λ + 2 ) ( λ + 2 j - 2 ) c q ( 2 j ) ! ! ( ω - 2 j ) ( ω + 2 j ) k c ¯ c n , k r ω + 2 j - ( 2 q - k + 1 ) c 2 q ω k 2 ( c ¯ c n , k ) 2 r 2 ω l n r + o ( r 2 ω l n r ) ] . Mathematical equation

Similarly, we repeat above process again and get

s n - 1 K ( s ) φ q = c q s ω - 1 [ 1 - λ 2 s 2 + + ( - 1 ) 2 k 0 + 1 λ ( λ + 2 ) ( λ + 4 k 0 ) ( 4 k 0 + 2 ) ! ! s 4 k 0 + 2 + 2 q c q ω k c ¯ c n , k s ω + j = 1 k 0 ( - 1 ) j 2 q ω λ ( λ + 2 ) ( λ + 2 j - 2 ) c q ( 2 j ) ! ! ( ω - 2 j ) ( ω + 2 j ) k c ¯ c n , k s ω + 2 j - q ( 2 q - k + 1 ) c 2 q ω k 2 ( c ¯ c n , k ) 2 s 2 ω l n s + o ( s 2 ω l n s ) ] Mathematical equation

and

( - φ ' ) k = c ¯ r n - k [ 1 + c q ω c ¯ c n , k r ω + j = 1 2 k 0 + 1 ( - 1 ) j λ ( λ + 2 ) ( λ + 2 j - 2 ) c q ( 2 j ) ! ! ( ω + 2 j ) c ¯ c n , k r ω + 2 j + q c 2 q ω 2 k ( c ¯ c n , k ) 2 r 2 ω Mathematical equation

+ j = 1 k 0 ( - 1 ) j 2 q ω λ ( λ + 2 ) ( λ + 2 j - 2 ) c 2 q ( 2 j ) ! ! ( ω - 2 j ) ( ω + 2 j ) ( 2 ω + 2 j ) k ( c ¯ c n , k ) 2 r 2 ω + 2 j - q ( 2 q - k + 1 ) c 3 q 3 ω 2 k 2 ( c ¯ c n , k ) 3 r 3 ω l n r + q ( 2 q - k + 1 ) c 3 q 9 ω 3 k 2 ( c ¯ c n , k ) 3 r 3 ω + o ( r 3 ω ) ] . Mathematical equation

Therefore,

φ ' = - 2 ω c r 2 ω + 1 [ 1 + j = 0 2 k 0 + 1 a ˜ 0 , j r ω + 2 j + j = 0 k 0 a ˜ 1 , j r 2 ω + 2 j + a ˜ 2 r 3 ω l n r + a ˜ 3 r 3 ω + o ( r 3 ω ) ] Mathematical equation

with some adequate constants a˜ij(i=1,2,3;j=0,1,,2k0+1)Mathematical equation depending on c,λ,q,k,k0,Mathematical equation and nMathematical equation. Integration yields

φ = c r 2 ω [ 1 + j = 0 k 0 2 ω a 0 , j ω - 2 j r ω + 2 j + 2 ω a 1 r 2 ω l n r ] + β + j = k 0 + 1 2 k 0 + 1 2 ω c a 0 , j ω - 2 j r - ω + 2 j - j = 0 k 0 ω c a 1 , j j r 2 j - 2 c a 2 r ω l n r + 2 c ( a 2 ω - a 3 ) r ω + o ( r ω ) . Mathematical equation

The singular term SMathematical equation and regular term ΘMathematical equation can be acquired. By induction, for the case (n-2k)/k=(n0+1)ωMathematical equation, we may suppose that φ=cr(n0+1)ω[1+i=1n0j=0ηia^ijriω+2j+a^n0+1r(n0+1)ωlnr+o(r(n0+1)ωlnr)],Mathematical equation where a^ij(i=0,1,,n0+1;j=0,1,,ηi)Mathematical equation are some constants depending upon c,λ,q,k,k0,n,Mathematical equation and n0Mathematical equation. In a similar manner, we compute

s n - 1 K ( s ) φ q = c q s ω - 1 ( 1 + s 2 ) - λ / 2 [ 1 + i = 1 n 0 j = 0 η i a ^ i j s i ω + 2 j + a ^ n 0 + 1 s ( n 0 + 1 ) ω l n s + o ( s ( n 0 + 1 ) ω l n s ) ] q = c q s ω - 1 [ 1 - λ 2 s 2 + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! s 2 k 0 + O ( s 2 k 0 + 2 ) ] [ 1 + i = 1 n 0 j = 0 η i a ̆ i j s i ω + 2 j + q a ^ n 0 + 1 s ( n 0 + 1 ) ω l n s + o ( s ( n 0 + 1 ) ω l n s ) ] = c q s ω - 1 [ 1 - λ 2 s 2 + + ( - 1 ) ( n 0 + 1 ) k 0 + n 0 λ ( λ + 2 ) ( λ + 2 ( n 0 + 1 ) k 0 - 2 ( n 0 - 1 ) ) [ 2 ( n 0 + 1 ) k 0 + 2 n 0 ] ! ! s 2 ( n 0 + 1 ) k 0 + 2 n 0      + i = 1 n 0 j = 0 η i a ¯ i j s i ω + 2 j + q a ^ n 0 + 1 s ( n 0 + 1 ) ω l n s + o ( s ( n 0 + 1 ) ω l n s ) ] . Mathematical equation

where ăijMathematical equation and a¯ijMathematical equation are some appropriate constants which depend on c,λ,q,k,k0,n,Mathematical equation and n0Mathematical equation. Hence,

( - φ ' ) k = c ¯ r n - k { 1 + c q c ¯ c n , k 0 r s ω - 1 [ 1 - λ 2 s 2 + + ( - 1 ) ( n 0 + 1 ) k 0 + n 0 λ ( λ + 2 ) ( λ + 2 ( n 0 + 1 ) k 0 - 2 ( n 0 - 1 ) ) [ 2 ( n 0 + 1 ) k 0 + 2 n 0 ] ! ! s 2 ( n 0 + 1 ) k 0 + 2 n 0    + i = 1 n 0 j = 0 η i a ¯ i j s i ω + 2 j + q a ^ n 0 + 1 s ( n 0 + 1 ) ω l n s + o ( s ( n 0 + 1 ) ω l n s ) ] d s } = c ¯ r n - k [ 1 + c q ω c ¯ c n , k r ω - λ c q 2 ( ω + 2 ) c ¯ c n , k r ω + 2 + + ( - 1 ) ( n 0 + 1 ) k 0 + n 0 λ ( λ + 2 ) ( λ + 2 ( n 0 + 1 ) k 0 - 2 ( n 0 - 1 ) ) c q [ 2 ( n 0 + 1 ) k 0 + 2 n 0 ] ! ! ( ω + 2 ( n 0 + 1 ) k 0 + 2 n 0 ) c ¯ c n , k r ω + 2 ( n 0 + 1 ) k 0 + 2 n 0    + i = 1 n 0 j = 0 η i a ¯ i j c q [ ( i + 1 ) ω + 2 j ] c ¯ c n , k r ( i + 1 ) ω + 2 j + q a ^ n 0 + 1 c q ( n 0 + 2 ) ω c ¯ c n , k r ( n 0 + 2 ) ω l n r - q a ^ n 0 + 1 c q ( n 0 + 2 ) 2 ω 2 c ¯ c n , k r ( n 0 + 2 ) ω + o ( r ( n 0 + 2 ) ω ) ] . Mathematical equation

Then

φ ' = - ( n 0 + 1 ) ω c r ( n 0 + 1 ) ω + 1 [ 1 + c q ω k c ¯ c n , k r ω + j = 1 η 0 a ˜ 0 , j r ω + 2 j + i = 1 n 0 j = 0 η i a ˜ i j r ( i + 1 ) ω + 2 j + a ˜ n 0 + 1 r ( n 0 + 2 ) ω l n r - a ˜ n 0 + 2 r ( n 0 + 2 ) ω + o ( r ( n 0 + 2 ) ω ) ] Mathematical equation

with some adequate constants a˜ij(i=0,1,,n0+2;j=0,1,,ηi)Mathematical equation depending upon c,λ,q,k,Mathematical equation and nMathematical equation. By integrating, we derive

φ = c r ( n 0 + 1 ) ω [ 1 + ( n 0 + 1 ) c q n 0 ω k c ¯ c n , k r ω + j = 1 η 0 - k 0 - 1 a ˜ 0 , j ( n 0 + 1 ) ω n 0 ω - 2 j r ω + 2 j + i = 1 n 0 - 1 j = 0 η i - k 0 - 1 a ˜ i j ( n 0 + 1 ) ω ( n 0 + 1 ) ω - ( i + 1 ) ω - 2 j r ( i + 1 ) ω + 2 j - a ˜ n 0 ( n 0 + 1 ) ω r ( n 0 + 1 ) ω l n r ]    + β + c r ( n 0 + 1 ) ω [ j = η 0 - k 0 η 0 a ˜ 0 , j ( n 0 + 1 ) ω n 0 ω - 2 j r ω + 2 j + i = 1 n 0 - 1 j = η i - k 0 η i a ˜ i j ( n 0 + 1 ) ω ( n 0 + 1 ) ω - ( i + 1 ) ω - 2 j r ( i + 1 ) ω + 2 j ] - j = 1 k 0 a ˜ n 0 , j ( n 0 + 1 ) ω c 2 j r 2 j    - a ˜ n 0 + 1 ( n 0 + 1 ) c r ω l n r + ( n 0 + 1 ) ( a ˜ n 0 + 1 ω - a ˜ n 0 + 2 ) c r ω + o ( r ω ) Mathematical equation

= c r ( n 0 + 1 ) ω [ 1 + i = 1 n 0 j = 0 η i a ^ i j r i ω + 2 j + a ^ n 0 + 1 r l ( n 0 + 1 ) ω n r ] + β + i = 1 n 0 j = η i + 1 η i - 1 a ^ i j r ( i - n 0 - 1 ) ω + 2 j + j = 1 k 0 a ^ n 0 + 1 , j r 2 j + a ^ n 0 + 2 r ω l n r + a ^ n 0 + 3 r ω + o ( r ω ) , Mathematical equation

where

{ a ^ 1 = ( n 0 + 1 ) c q n 0 ω k c ¯ c n , k , a ^ i + 1 , j = a ˜ i j ( n 0 + 1 ) ω c ( n 0 + 1 ) ω - ( i + 1 ) ω - 2 j ( i = 0,1 , , n 0 - 1 ; j = 0,1 , , η i ) , a ^ n 0 + 1 = - a ˜ n 0 ( n 0 + 1 ) ω ,      a ^ n 0 + 1 , j = a ˜ n 0 , j ( n 0 + 1 ) ω c 2 j ( j = 0,1 , , η i ) , a ^ n 0 + 2 = - a ˜ n 0 + 1 ( n 0 + 1 ) c ,      a ^ n 0 + 3 = ( n 0 + 1 ) ( a ˜ n 0 + 1 ω - a ˜ n 0 + 2 ) c .    Mathematical equation(19)

Thus the singular term SMathematical equation and regular term ΘMathematical equation could be obtained precisely. For the case 2(n0+1)k0+2n0<(n-2k)/k<(n0+1)ωMathematical equation, we can apply similar arguments to obtain its corresponding conclusion.

(ii) To prove that (16) admits a unique solution ΘMathematical equation, it suffices to verify that f(r,Θ)Mathematical equation fulfills the assumptions in Lemma 1. For the case (n-2k)/k=(n0+1)ωMathematical equation, we find

P = 1 + i = 1 n 0 j = 0 η i a ^ i j r i ω + 2 j + a ^ n 0 + 1 r l ( n 0 + 1 ) ω n r ,   P ' = i = 1 n 0 j = 0 η i a ^ i j ( i ω + 2 j ) r i ω + 2 j - 1 + a ^ n 0 + 1 ( n 0 + 1 ) ω r l ( n 0 + 1 ) ω - 1 n r + a ^ n 0 + 1 r , ( n 0 + 1 ) ω - 1 Mathematical equation

and

f ( r , Θ ) = - ( n 0 + 1 ) ω c r ( n 0 + 1 ) ω + 1 [ 1 + c q c ¯ c n , k 0 r s ω - 1 ( 1 + s 2 ) - λ / 2 ( 1 + i = 1 n 0 j = 0 η i a ^ i j s i ω + 2 j + a ^ n 0 + 1 s l ( n 0 + 1 ) ω n   s ) q d s + O ( r ( n 0 + 2 ) ω ) ] 1 / k + ( n 0 + 1 ) ω c r ( n 0 + 1 ) ω + 1 ( 1 + i = 1 n 0 j = 0 η i a ^ i j r i ω + 2 j + a ^ n 0 + 1 r l ( n 0 + 1 ) ω n r ) - c r ( n 0 + 1 ) ω ( i = 1 n 0 j = 0 η i a ^ i j ( i ω + 2 j ) r i ω + 2 j - 1 + a ^ n 0 + 1 ( n 0 + 1 ) ω r l ( n 0 + 1 ) ω - 1 n r + a ^ n 0 + 1 r ( n 0 + 1 ) ω - 1 ) = - ( n 0 + 1 ) ω c r ( n 0 + 1 ) ω + 1 [ 1 + c q ω c ¯ c n , k r ω - λ c q 2 ( ω + 2 ) c ¯ c n , k r ω + 2 + + ( - 1 ) ( n 0 + 1 ) k 0 + n 0 λ ( λ + 2 ) ( λ + 2 ( n 0 + 1 ) k 0 + 2 ( n 0 - 1 ) ) c q [ 2 ( n 0 + 1 ) k 0 + 2 n 0 ] ! ! ( ω + 2 ( n 0 + 1 ) k 0 + 2 n 0 ) c ¯ c n , k r ω + 2 ( n 0 + 1 ) k 0 + 2 n 0 + i = 1 n 0 j = 0 η i a ¯ i j c q [ ( i + 1 ) ω + 2 j ] c ¯ c n , k r ( i + 1 ) ω + 2 j + q a ^ n 0 + 1 c q ( n 0 + 2 ) ω c ¯ c n , k r ( n 0 + 2 ) ω l n r + o ( r ( n 0 + 2 ) ω l n r ) ] 1 / k Mathematical equation

+ ( n 0 + 1 ) ω c r ( n 0 + 1 ) ω + 1 + i = 1 n 0 j = 1 η i a ^ i j [ ( n 0 - i + 1 ) ω - 2 j ] c r ( i - n 0 - 1 ) ω + 2 j - 1 - a ^ n 0 + 1 c r - 1 = - ( n 0 + 1 ) ω c r ( n 0 + 1 ) ω + 1 [ 1 + c q ω k c ¯ c n , k r ω + j = 1 η 0 a ˜ 0 , j r ω + 2 j + i = 1 n 0 j = 0 η i a ˜ i j r ( i + 1 ) ω + 2 j + a ˜ n 0 + 1 r ( n 0 + 2 ) ω l n r + o ( r ( n 0 + 2 ) ω l n r ) ] + ( n 0 + 1 ) ω c r ( n 0 + 1 ) ω + 1 + i = 1 n 0 j = 0 η i a ^ i j [ ( n 0 - i + 1 ) ω - 2 j ] c r ( i - n 0 - 1 ) ω + 2 j - 1 - a ^ n 0 + 1 c r - 1 = a ^ 1 , n 0 ( k 0 + 1 ) ( 2 k 0 + 2 - ω ) n 0 c r n 0 ( 2 k 0 + 2 - ω ) - 1 + o ( r n 0 ( 2 k 0 + 2 - ω ) - 1 ) Mathematical equation

due to the relations (19) between a^ijMathematical equation and a˜ijMathematical equation. The other case can be similarly handled.

For the case 2(n0+1)k0+2n0<(n-2k)/k<(n0+1)ωMathematical equation, we obtain

u ( t ) = r K ( r ) φ q c n , k ( - φ ' ) k = r λ + k - 2 c n , k ( 1 + r 2 ) λ / 2 φ q ( - φ ' ) k = c q r ω c ¯ c n , k ( 1 + r 2 ) λ / 2 ( 1 + i = 1 n 0 j = 0 η i a ^ i j r i ω + 2 j + o ( r ) n 0 ω + 2 η i ) q [ 1 + i = 1 n 0 j = 0 η i a ^ i j k [ ( n - 2 k ) / k - ( i ω + 2 j ) ] n - 2 k r i ω + 2 j + o ( r ) n 0 ω + 2 η i ] - k = c q c ¯ c n , k r ω [ 1 - λ 2 r 2 + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! r 2 k 0 + O ( r 2 k 0 + 2 ) ] [ 1 - q ( n - 2 k ) c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k r ω + o ( r ω ) ] [ 1 - c q ω c ¯ c n , k r ω + o ( r ω ) ] = c q c ¯ c n , k r ω [ 1 - λ 2 r 2 + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! r 2 k 0 - [ ω k 2 + ( n - 2 k ) ( q - k ) ] c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k r ω + o ( r ω ) ] Mathematical equation

and

v ( t ) = r - φ ' φ = r [ ( n - 2 k ) c k r ( n - k ) / k + i = 1 n 0 j = 0 η i a ^ i j [ ( n - 2 k ) / k - ( i ω + 2 j ) ] c r i ω + 2 j - ( n - k ) / k + o ( r n 0 ω + 2 η i - ( n - k ) / k ) ] [ c r ( n - 2 k ) / k ( 1 + i = 1 n 0 j = 0 η i a ^ i j r i ω + 2 j + o ( r n 0 ω + 2 η i ) ) ] - 1 = n - 2 k k [ 1 + c q ω k c ¯ c n , k r ω + o ( r ω ) ] [ 1 - q ( n - 2 k ) c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k r ω + o ( r ω ) ] = n - 2 k k [ 1 + [ ω k + ( n - 2 k ) ( q - 1 ) ] c q ω [ ω - ( n - 2 k ) / k ] k 2 c ¯ c n , k r ω + o ( r ω ) ] . Mathematical equation

Remark 3   The ranges of the parameters (n,k,q,λ,n0,k0)Mathematical equation are not empty under the assumptions in Theorem 1. Since n>2k, q>k, λ+2k-2>0,Mathematical equation and 0<ω:=p-[(n-2k)q]/k<(n-2k)/kMathematical equation, we can choose n=50,k=2,q=3,Mathematical equation and λ=29Mathematical equation so that (n-2k)/k=23Mathematical equation and ω=8Mathematical equation. By some computations, it is not difficult to find n0=2Mathematical equation and k0=3Mathematical equation such that 2(n0+1)k0+2n0<(n-2k)/kMathematical equation(n0+1)ωMathematical equation and 2k0+2n0/(n0+1)<ω2(k0+1)Mathematical equation.

Theorem 2   Let p=[(n-2k)(q+1)]/kMathematical equation and select k0NMathematical equation such that 2k0<(n-2k)/kMathematical equation2(k0+1)Mathematical equation. Then the following results are valid.

(i) Every M-solution φMathematical equation has the form φ=S+ΘMathematical equation, where

S = c r ( n - 2 k ) / k - c q + 1 k c ¯ c n , k l n r Mathematical equation(20)

and ΘMathematical equation solves the initial value problem (16) with the following expansion

Θ = β - i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( n - 2 k ) c q + 1 2 i [ 2 i + ( n - 2 k ) / k ] ( 2 i ) ! ! k 2 c ¯ c n , k r 2 i + q c 2 q + 1 2 k ( n - 2 k ) ( c ¯ c n , k ) 2 r ( n - 2 k ) / k l n r - q c 2 q + 1 [ 2 ( n - 2 k ) c ¯ c n , k ] 2 r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) Mathematical equation

as r0Mathematical equation for some uniquely determined constants c>0, βRMathematical equation.

(ii) Conversely, for any given c>0Mathematical equation and βRMathematical equation , there exists a unique solution ΘMathematical equation of (16) such that φ=S+ΘMathematical equation is an M-solution, where SMathematical equation is given by (20), and ΘMathematical equation satisfies (17) with f(r,Θ)=O(r)Mathematical equation. In addition, the solution (u,v)Mathematical equation has the following expansion:

{ u ( t ) = c q c ¯ c n , k e ( n - 2 k ) t / k [ 1 + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( 2 i ) ! ! e 2 i t - q c q k c ¯ c n , k t e ( n - 2 k ) t / k + ( q β c - k c q ( n - 2 k ) c ¯ c n , k ) e ( n - 2 k ) t / k + o ( e ( n - 2 k ) t / k ) ] , v ( t ) = n - 2 k k [ 1 + c q k c ¯ c n , k t e ( n - 2 k ) t / k + ( c q ( n - 2 k ) c ¯ c n , k - β c ) e ( n - 2 k ) t / k + o ( e ( n - 2 k ) t / k ) ] ,    t - . Mathematical equation

Proof   (i) Let c,c¯>0Mathematical equation be determined in Lemma 2. By some calculations, we have

s n - 1 K ( s ) φ q = s p - 1 ( 1 + s 2 ) λ / 2 c q s [ ( n - 2 k ) q ] / k ( s ( n - 2 k ) / k c φ ( s ) ) q = c q s ( n - 2 k ) / k - 1 [ 1 + O ( s 2 ) ] [ 1 + o ( 1 ) ] q = c q s ( n - 2 k ) / k - 1 [ 1 + o ( 1 ) ] . Mathematical equation

It follows from (15) that

φ ' = - 1 r ( n - k ) / k [ c ¯ + 0 r c n , k - 1 s n - 1 K ( s ) φ q d s ] 1 / k = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + 1 c ¯ c n , k 0 r s n - 1 K ( s ) φ q d s ] 1 / k = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + 1 k c ¯ c n , k 0 r s n - 1 K ( s ) φ q d s + o ( 0 r s n - 1 K ( s ) φ q d s ) ] = - ( n - 2 k ) c k r ( n - k ) / k - c q + 1 k c ¯ c n , k r + o ( 1 r ) . Mathematical equation

Hence,

φ = c r ( n - 2 k ) / k - c q + 1 k c ¯ c n , k l n r + o ( l n r ) Mathematical equation(21)

It is clear that each term on the expansion of φMathematical equation is still singular. To obtain the regular term of φMathematical equation, we need to repeat the above arguments. Using (21), we find

s n - 1 K ( s ) φ q = c q s ( n - 2 k ) / k - 1 ( 1 + s 2 ) λ / 2 [ 1 - c q k c ¯ c n , k s ( n - 2 k ) / k l n s + o ( s ( n - 2 k ) / k l n s ) ] q = c q s ( n - 2 k ) / k - 1 [ 1 - λ 2 s 2 + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! s 2 k 0 + O ( s 2 k 0 + 2 ) ] [ 1 - q c q k c ¯ c n , k s ( n - 2 k ) / k l n s + o ( s ( n - 2 k ) / k l n s ) ] = c q s ( n - 2 k ) / k - 1 [ 1 + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( 2 i ) ! ! s 2 i - q c q k c ¯ c n , k s ( n - 2 k ) / k l n s + o ( s ( n - 2 k ) / k l n s ) ] . Mathematical equation

Therefore,

φ ' = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + 1 c ¯ c n , k 0 r s n - 1 K ( s ) φ q d s ] 1 / k = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + k c q r ( n - 2 k ) / k ( n - 2 k ) c ¯ c n , k + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) c q r ( n - 2 k ) / k + 2 i ( 2 i + ( n - 2 k ) / k ) ( 2 i ) ! ! c ¯ c n , k ] = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q r ( n - 2 k ) / k ( n - 2 k ) c ¯ c n , k + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) c q r ( n - 2 k ) / k + 2 i ( 2 i + ( n - 2 k ) / k ) ( 2 i ) ! ! k c ¯ c n , k - q c 2 q r 2 ( n - 2 k ) / k l n r 2 k ( n - 2 k ) ( c ¯ c n , k ) 2 + [ q - 2 ( k - 1 ) ] c 2 q r 2 ( n - 2 k ) / k 4 ( n - 2 k ) 2 ( c ¯ c n , k ) 2 + o ( r 2 ( n - 2 k ) / k ) ] . Mathematical equation

Let S'=-(n-2k)ckr(n-k)/k-cq+1kc¯cn,kr, Θ'=φ'-S'.Mathematical equation Note that Θ'Mathematical equation is integrable and Θ(0):=βMathematical equation exists. Therefore,

φ = c r ( n - 2 k ) / k - c q + 1 k c ¯ c n , k l n r + β - i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( n - 2 k ) c q + 1 2 i k ( 2 i k + n - 2 k ) ( 2 i ) ! ! c ¯ c n , k r 2 i + q c 2 q + 1 2 k ( n - 2 k ) ( c ¯ c n , k ) 2 r ( n - 2 k ) / k l n r - [ 3 q - 2 ( k - 1 ) ] c 2 q 4 ( n - 2 k ) 2 ( c ¯ c n , k ) 2 r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) . Mathematical equation

(ii) For any given cMathematical equation and βMathematical equation, we denote S=cr(n-2k)/k-cq+1kc¯cn,klnrMathematical equation. Then

P = 1 - c q k c ¯ c n , k r ( n - 2 k ) / k l n r ,   P ' = - ( n - 2 k ) c q k 2 c ¯ c n , k r ( n - 2 k ) / k - 1 l n r - c q k c ¯ c n , k r ( n - 2 k ) / k - 1 , Mathematical equation

and

f ( r , Θ ) = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q c ¯ c n , k 0 r s ( n - 2 k ) / k - 1 ( 1 + s 2 ) λ / 2 ( 1 - c q k c ¯ c n , k s ( n - 2 k ) / k l n s ) q d s + O ( r 2 ( n - 2 k ) / k ) ] 1 / k + ( n - 2 k ) c k r ( n - k ) / k ( 1 - c q k c ¯ c n , k r ( n - 2 k ) / k l n r ) + c r ( n - 2 k ) / k ( ( n - 2 k ) c q k 2 c ¯ c n , k r ( n - 2 k ) / k - 1 l n r + c q k c ¯ c n , k r ( n - 2 k ) / k - 1 ) = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + k c q r ( n - 2 k ) / k ( n - 2 k ) c ¯ c n , k + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) c q r ( n - 2 k ) / k + 2 i ( 2 i + ( n - 2 k ) / k ) ( 2 i ) ! ! c ¯ c n , k - q c 2 q r 2 ( n - 2 k ) / k l n r 2 ( n - 2 k ) ( c ¯ c n , k ) 2 + o ( r 2 ( n - 2 k ) / k l n r ) ] 1 / k + ( n - 2 k ) c k r ( n - k ) / k + c q + 1 k c ¯ c n , k r = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q r ( n - 2 k ) / k ( n - 2 k ) c ¯ c n , k + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) c q r ( n - 2 k ) / k + 2 i ( 2 i k + n - 2 k ) ( 2 i ) ! ! c ¯ c n , k - q c 2 q r 2 ( n - 2 k ) / k l n r 2 k ( n - 2 k ) ( c ¯ c n , k ) 2 + o ( r 2 ( n - 2 k ) / k l n r ) ] + ( n - 2 k ) c k r ( n - k ) / k + c q + 1 k c ¯ c n , k r = λ ( n - 2 k ) c q + 1 r 2 n k c ¯ c n , k + o ( r ) , Mathematical equation

which satisfies the assumptions of Lemma 1, thus (16) possesses a unique solution ΘMathematical equation. Hence φ=S+ΘMathematical equation is an M-solution. We compute

u ( t ) = r K ( r ) φ q c n , k ( - φ ' ) k = r λ + k - 2 c n , k ( 1 + r 2 ) λ / 2 φ q ( - φ ' ) k = c q r ( n - 2 k ) / k c ¯ c n , k ( 1 + r 2 ) λ / 2 [ 1 - c q k c ¯ c n , k r ( n - 2 k ) / k l n r + r ( n - 2 k ) / k c Θ ] q [ 1 + c q ( n - 2 k ) c ¯ c n , k r ( n - 2 k ) / k - k r ( n - k ) / k ( n - 2 k ) c Θ ' ] k = c q r ( n - 2 k ) / k c ¯ c n , k [ 1 - λ 2 r 2 + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! r 2 k 0 + O ( r 2 k 0 + 2 ) ] [ 1 - q c q k c ¯ c n , k r ( n - 2 k ) / k l n r + q β c r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) ] [ 1 - k c q ( n - 2 k ) c ¯ c n , k r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) ] = c q c ¯ c n , k r ( n - 2 k ) / k [ 1 - λ 2 r 2 + + ( - 1 ) k 0 λ ( λ + 2 ) ( λ + 2 k 0 - 2 ) ( 2 k 0 ) ! ! r 2 k 0 - q c q k c ¯ c n , k r ( n - 2 k ) / k l n r + q β c r ( n - 2 k ) / k - k c q ( n - 2 k ) c ¯ c n , k r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) ] Mathematical equation

and

v ( t ) = r - φ ' φ = r ( n - 2 k ) c k r ( n - k ) / k + c q + 1 k c ¯ c n , k r - Θ ' c r ( n - 2 k ) / k - c q + 1 k c ¯ c n , k l n r + Θ = [ n - 2 k k + c q k c ¯ c n , k r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) ] [ 1 - c q k c ¯ c n , k r ( n - 2 k ) / k l n r + β c r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) ] - 1 Mathematical equation

= n - 2 k k [ 1 + c q k c ¯ c n , k r ( n - 2 k ) / k l n r + ( c q ( n - 2 k ) c ¯ c n , k - β c ) r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) ] . Mathematical equation

Theorem 3   Let p>[(n-2k)(q+1)]/kMathematical equation. Define ω:=p-[(n-2k)q]/kMathematical equation and choose k0NMathematical equation such that 2k0<(n-2k)/k2(k0+1)Mathematical equation. Then the following statements are true.

(i) Every M-solution φMathematical equation has the form φ=S+ΘMathematical equation, where S=cr(n-2k)/kMathematical equation, and ΘMathematical equation solves the initial value problem (16). Moreover,

Θ = β - ( n - 2 k ) c q + 1 k 2 c ¯ c n , k [ 1 ω [ ω - ( n - 2 k ) / k ] r ω - ( n - 2 k ) / k + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( ω + 2 i ) [ ω - ( n - 2 k ) / k + 2 i ] ( 2 i ) ! ! r ω - ( n - 2 k ) / k + 2 i + q β c ω [ ω - ( n - 2 k ) / k ] r ω + o ( r ω ) ] ,      r 0 Mathematical equation

for some uniquely determined constants c>0,βRMathematical equation.

(ii) Conversely, given any c>0Mathematical equation and βRMathematical equation , there exists a unique solution ΘMathematical equation of (16) such that φ=S+ΘMathematical equation is an M-solution with S=c/r(n-2k)/kMathematical equation. In this case, ΘMathematical equation satisfies (17) with f(r,Θ)=O(rω-(n-k)/k)Mathematical equation. In addition, the solution (u,v)Mathematical equation has the following expansion:

{ u ( t ) = c q c ¯ c n , k e ω t [ 1 + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( 2 i ) ! ! e 2 i t + O ( e ( n - 2 k ) t / k ) ] , v ( t ) = n - 2 k k [ 1 - β c e ( n - 2 k ) t / k + c q ( ω k - n + 2 k ) c ¯ c n , k e ω t + o ( e ω t ) ] ,      t - . Mathematical equation

Proof   (i) Let c,c¯>0Mathematical equation be determined by Lemma 2. It follows from (15) that φ'Mathematical equation is integrable near the origin. We compute

φ ' = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + 1 c ¯ 0 r c n , k - 1 s n - 1 K ( s ) φ q d s ] 1 / k = - ( n - 2 k ) c k r ( n - k ) / k - ( n - 2 k ) c k 2 c ¯ c n , k r ( n - k ) / k 0 r s n - 1 K ( s ) φ q d s + o ( 1 r ( n - k ) / k 0 r s n - 1 K ( s ) φ q d s ) . Mathematical equation

Thus, we obtain for certain r0(0,R)Mathematical equation,

ϕ = c r ( n - 2 k ) / k - c r 0 ( n - 2 k ) / k + ϕ ( r 0 ) - r 0 r ( n - 2 k ) c k 2 c ¯ c n , k t ( n - k ) / k 0 t s n - 1 K ( s ) ϕ q d s d t + o ( r ω - ( n - 2 k ) / k ) = : c r ( n - 2 k ) / k + Θ = : S + Θ ,      0 < r < R . Mathematical equation

Clearly, ΘMathematical equation satisfies (16) and (17). Suppose that c1/r(n-2k)/k+Θ1=c2/r(n-2k)/k+Θ2Mathematical equation. It is obvious that c1=c2Mathematical equation and Θ1=Θ2Mathematical equation, which implies that c,βMathematical equation and ΘMathematical equation are unique. It follows from Θ=β+o(1)Mathematical equation that

Θ ' = - ( n - 2 k ) c k 2 c ¯ c n , k r ( n - k ) / k 0 r s n - 1 K ( s ) φ q d s + o ( 1 r ( n - k ) / k 0 r s n - 1 K ( s ) φ q d s ) = - ( n - 2 k ) c q + 1 k 2 c ¯ c n , k r ( n - k ) / k 0 r s ω - 1 ( 1 + s 2 ) - λ / 2 [ 1 + s ( n - 2 k ) / k c Θ ( s ) ] q d s + o ( r ω - 1 ) = - ( n - 2 k ) c q + 1 k 2 c ¯ c n , k r ( n - k ) / k 0 r s ω - 1 [ 1 + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( 2 i ) ! ! s 2 i + O ( s 2 k 0 + 2 ) ] [ 1 + q β c s ( n - 2 k ) / k + o ( s ( n - 2 k ) / k ) ] d s + o ( r ω - 1 ) = - ( n - 2 k ) c q + 1 k 2 c ¯ c n , k r ( n - k ) / k 0 r s ω - 1 [ 1 + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( 2 i ) ! ! s 2 i + q β c s ( n - 2 k ) / k + o ( s ( n - 2 k ) / k ) ] d s + o ( r ω - 1 ) = - ( n - 2 k ) c q + 1 r ω - ( n - k ) / k ω k 2 c ¯ c n , k [ 1 + i = 1 k 0 ( - 1 ) i ω λ ( λ + 2 ) ( λ + 2 i - 2 ) ( ω + 2 i ) ( 2 i ) ! ! r 2 i + ω q β c [ ω - ( n - 2 k ) / k ] r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) ] . Mathematical equation

The expansion for ΘMathematical equation could be obtained directly by integrating.

(ii) For any given cMathematical equation and βMathematical equation, we denote S=cr(n-2k)/kMathematical equation. Then P=1Mathematical equation and

f ( r , Θ ) = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q c ¯ c n , k 0 r s ω - 1 ( 1 + s 2 ) λ / 2 d s + O ( r ω + ( n - 2 k ) / k ) ] 1 / k + ( n - 2 k ) c k r ( n - k ) / k Mathematical equation

= - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q ω c ¯ c n , k r ω + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) c q ( ω + 2 i ) ( 2 i ) ! ! c ¯ c n , k r ω + 2 i + O ( r ω + ( n - 2 k ) / k ) ] 1 / k + ( n - 2 k ) c k r ( n - k ) / k = - ( n - 2 k ) c k r ( n - k ) / k [ 1 + c q ω k c ¯ c n , k r ω + o ( r ω ) ] + ( n - 2 k ) c k r ( n - k ) / k = - ( n - 2 k ) c q + 1 ω k 2 c ¯ c n , k r ω - ( n - k ) / k + o ( r ω - ( n - k ) / k ) . Mathematical equation

Clearly, f(r,Θ)Mathematical equation satisfies the assumptions in Lemma 1. Therefore, the problem (16) has a unique solution ΘMathematical equation. Thus φ=S+ΘMathematical equation is an M-solution. We compute

u ( t ) = r K ( r ) φ q c n , k ( - φ ' ) k = r λ + k - 2 c n , k ( 1 + r 2 ) λ / 2 φ q ( - φ ' ) k = c q r ω c ¯ c n , k ( 1 + r 2 ) λ / 2 [ 1 + r ( n - 2 k ) / k c Θ ] q [ 1 - k ( n - 2 k ) c r ( n - k ) / k Θ ' ] k = c q r ω c ¯ c n , k [ 1 + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( 2 i ) ! ! r 2 i + O ( r 2 k 0 + 2 ) ] [ 1 + q β c r ( n - 2 k ) / k - ( n - 2 k ) q c q ω ( ω k - n + 2 k ) k c ¯ c n , k r ω + o ( r ω ) ] [ 1 - c q ω c ¯ c n , k r ω + o ( r ω ) ] = c q c ¯ c n , k r ω [ 1 + i = 1 k 0 ( - 1 ) i λ ( λ + 2 ) ( λ + 2 i - 2 ) ( 2 i ) ! ! r 2 i + q β c r ( n - 2 k ) / k + o ( r ( n - 2 k ) / k ) ] Mathematical equation

and

v ( t ) = r - φ ' φ = n - 2 k k - r ( n - k ) / k c Θ ' 1 + r ( n - 2 k ) / k c Θ Mathematical equation

= [ n - 2 k k + ( n - 2 k ) c q ω k 2 c ¯ c n , k r ω + o ( r ω ) ] [ 1 - β c r ( n - 2 k ) / k + ( n - 2 k ) c q ω ( ω k - n + 2 k ) k c ¯ c n , k r ω + o ( r ω ) ] = n - 2 k k [ 1 - β c r ( n - 2 k ) / k + c q ( ω k - n + 2 k ) c ¯ c n , k r ω + o ( r ( n - 2 k ) / k ) ] . Mathematical equation

Remark 4   We discuss the expansions of singular solutions in different intervals with p as the index in this section. For each specific subinterval of p, the singular solution is unique, and its corresponding singular and regular terms are also unique. However, for the entire interval, the singular solution is not unique, and its singular and regular terms are not unique either.

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