A Note on Singular Solutions of the Matukuma Equation in Higher Dimensional Space

WANG Biaoand ZHANG Zhengce

School of Mathematics and Statistics,Xi’anJiaotongUniversity,Xi’an710049,China.

A Note on Singular Solutions of the Matukuma Equation in Higher Dimensional Space

WANG Biao?and ZHANG Zhengce

School of Mathematics and Statistics,Xi’anJiaotongUniversity,Xi’an710049,China.

.We revise one case of the M-solutions of B.Wang,et al.,J.Diff.Eqs.253 (2012),pp.3232-3265andobtain morepreciseform ofthe singular termandthe regular term.

Matukuma equation;singular solutions;asymptotic expansion.

1 Introduction

In this paper,we review one case of the M-solutions of positive radial solutions φ=φ(r) of the differential equation

In[1],in order to apply a theorem on asymptotically autonomous systems by Thieme[2], Emden-Fowler equation was extended into the following generalized one

By using the substitutionone could transfer the solutions of(1.2)into the solutions ?=(u,v)of the Lotka-Volterra system

Applying another substitution(see[1]for more details)

to solutions of(1.1),we get

Here the coefflcient

is time-dependent,but the limits

exist,and we deflne q:=N?2+λ.Then(1.4)is asymptotically autonomous with respect to(1.3)for t→?∞and to(EFSp,N?2,N)for t→+∞,and this fact makes this system an asymptotically autonomous one.

Since satisfles the assumptions of[1,Lemma 2.1],it sufflces to investigate whether h2(r)satisfles those assumptions.For the case(N?2)p＜q＜(N?2)(p+1),Wang et al.[1,Theorem 3.4]derived an implicit form of S and Θ,they did not give exact form of S and Θ.Hence,in this paper we would like to improve this result.In this paper,we shall separate two cases to state the main theorem.For k0=1, [μ,2μ)withμ=:q?(N?2)p is a bigger interval than that of N=3,ifμ/=2m,we need to divide this interval into m+1 cases,i.e.,

such that the singular term S and the regular term Θ can be exactly read off.Ifμ=2m,it sufflces to classify m cases:

However,when k0≥2,the story will change dramatically.Compared with the case k0=1, [k0μ,(k0+1)μ)withμ/=2m needs to be further partitioned into k0m+1 cases such that we could obtain more precise form of S and Θ,and this makes the case k0≥2 more complicated than that of k0=1.

2 Preliminaries

2.1 Classiflcation of positive solutions

Let K be a positive function in C1(R+)with r2K(r)bounded away from zero for r→∞, and p＞1.Assume φ:(R?,R)→(0,∞)is a radial solution of

Then 0≤R?＜R≤∞.Let r0∈(R?,R)and deflne

If H0＞0,we can prove that R?＞0 and there exists R0∈(R?,R)with H(R0)=0=φ′(R0); If H0≤0,we could obtain that R?=0.In the latter case we deflne R0=0 and have

for all solutions.We can classify the solutions as follows:

2.2 Linearization of(1.4)

Clearly,the system(1.3)with p＞1 and q＞N?2 has the following stationary points:

for q＜(N?2)p,

For P1we have eigenvalues λ1=q,λ2=?N+2 and eigenvectors ξ1=(1,0),ξ2=(0,1), and P1is a saddle.

For P2we have λ1=q?(N?2)p,λ2=N?2 and ξ1=(q?(N?2)(p+1),N?2),ξ2=(0,1). when q＞(N?2)p,P2is an unstable improper node;when q＞(N?2)(p+1),P2is a 2-tangential improper node;when q=(N?2)(p+1),P2is a 1-tangential node;when (N?2)p＜q＜(N?2)(p+1),P2is a 2-tangential improper node;when q=(N?2)p,P2is a unstable 2-tangential node;when q＜(N?2)p,P2is a saddle.

If q＜(N?2)p,

P4is a center.Because

3 The main theorem

Since 2m≤μ＜2m+2,then 2k0m≤k0μ＜2k0m+2k0,which can be classifled into 2(k0?1)(k0≥2)cases:

Theorem 3.1.Let(N?2)p＜q＜(N?2)(p+1).Letμ:=q?(N?2)p∈(0,N?2)and k0,m∈N such that k0μ≤N?2＜(k0+1)μ,2m≤μ＜2(m+1).If 2k0m≤k0μ＜2k0m+2,then

(i)Every M-solution φ has the form φ=S+Θ with S and Θ possessing the following expansions.Moreover Θ is a solution of(1.5).

(a)For k0=1,ifμ/=2m,we have

Case 1.Ifμ≤N?2＜μ+2,

Case 2.Ifμ+2≤N?2＜μ+4,

······

Case m+1.Ifμ+2m≤N?2＜2μ,for uniquely determined constants c＞0,β∈R.Ifμ=2m,there are m cases,we also have the similar conclusion.

(b)For k0≥2,ifμ/=2m,then there exist constants d1i,d2i,···,dk0,i,dk0+1(i=0,1,···,m) depending on p,q,N such that

······

······

······

Case k0m+1.If k0μ+2m≤N?2＜(k0+1)μ,there exists constant dk0(m+1)+1depending on

p,q,N such that

On the other hand,ifμ=2m,we only have m cases:k0μ≤N?2＜k0μ+2,k0μ+2≤N?2＜k0μ+4,···,k0μ+2(m?1)≤N?2＜(k0+1)μ.

(ii)Conversely,given any c＞0,β∈R,there exists a unique solution Θ of(1.5)with S given in(i)and φ:=S+Θ is an M-solution(as long as φ is positive).

Ifμ=2m,h2(r)has the similar forms.

For k0≥2,ifμ/=2m,then

with i=0,1,···,m?1;j=1,2,···,k0?1.Ifμ=2m,

Proof.(i)Let c＞0 be determined by[1,Theorem 3.1],which also gives

where C is some constant.Sinceμ＜N?2,we continue to apply this iterative process and have

Again we have

Becauseμ≤N?2＜2μwith k0=1,there exists some β∈R such that

Sinceμ∈(0,N?2),[μ,2μ)should be a bigger interval than that of N=3.Hence we need to classify m+1 cases withμ/=2m,i.e.,μ≤N?2＜μ+2,μ+2≤N?2＜μ+4,···,μ+2m≤N?2＜2μsuch that the singular term S and the regular term Θ can be exactly read off. Similarly,ifμ=2m,[μ,2μ)can be separated into m cases.

If k0≥2,we need to continue the above process.For the next step of iteration,we have to classify two cases.

a)2m≤μ＜2m+1.

with λi=[λ(λ+2)···(λ+2i?2)](i=0,1,···,m).

Similarly,we get

whereμi,j=[2μ+2(i+j)?N+2][2μ+2(i+j)](μ+2i?N+2)(μ+2i)(2i)!!(2j)!!.

b)2m+1≤μ＜2m+2.

If we denote the flrst step of iteration by

then for kth step of iteration,there are 2(k?1)(k≥2)cases:

with appropriate constants dj,idepending on p,q,N.

Repeating above iterative process,we obtain

then(3.1)holds for k+1.By induction,we can repeat the process until the biggest k with kμ≤N?2＜(k+1)μ,which is k0.Note that ifμ/=2m,we have the following fact:

thus we need to further divide the interval[k0μ,(k0+1)μ)into k0m+1 cases such that the singular term S and the regular term Θ can be more precisely read off.Ifμ=2m,it sufflces to partition the interval[k0μ,(k0+1)μ)into m cases:k0μ≤N?2＜k0μ+2,k0μ+2≤N?2＜k0μ+4,···,k0μ+2(m?1)≤N?2＜(k0+1)μ.

The rest can be treated by similar arguments.

For k0≥2,an easy computations shows for k0μ≤N?2＜μ+2(k0?1)m+2,

The rest can be studied by the similar arguments.It is easy to verify that h2satisfles the assumptions of[1,Lemma 2.1],thus(1.5)admits a unique solution Θ.

Acknowledgements

The authors would thank ProfessorLi Yi very much for helpful discussion.Zhang is supported by the National Natural Science Foundation of China(No.11371286,11401458) and by the Scientiflc Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry.

[1]Wang B.,Zhang Z.C.,Li Y.,The radial positive solutions of the Matukuma equation in higher dimensional space:Singular solution.J.Diff.Eqs.,253(2012),3232-3265.

[2]Thieme H.R.,Asymptotically autonomous differentialequations in the plane.Rocky Mountain J.Math.,24(1994),351-380.

[3]Batt J.,Li Y.,The positive solutions of the Matukuma equation and the problem of flnite radius and flnite mass.Arch.Rational Mech.Anal.,198(2010),613-675.

[4]Li Y.,Asymptotic behavior of positive solutions of equation?u+K(|x|)|u|p=0 in Rn.J. Diff.Eqs.,95(1992),304-330.

[5]Li Y.,Ni W.M.,On conformal scalar curvature equations in Rn.Duck Math.J.,57(1988), 895-924.

?Correspondingauthor.Email addresses:wang.biao@stu.xjtu.edu.cn(B.Wang),zhangzc@mail.xjtu.edu. cn(Z.C.Zhang)

Received 7 May 2015;Accepted 29 August 2015

AMS Subject Classiflcations:35J15,35J61,35B40

Chinese Library Classiflcations:O175.25,O175.29