A class of self-similar solutions to a singular and degenerate diffusion equation

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www.elsevier.com/locate/na

A class of self-similar solutions to a singular and degenerate diffusion equation

Chunpeng Wang

^a

, Tong Yang

^b^,∗

, Jingxue Yin

^a

aDepartment of Mathematics, Jilin University, Changchun, Jilin 130012, PR China bDepartment of Mathematics, City University of Hong Kong, Hong Kong

Received10 September 2003; accepted23 September 2004

Abstract

In this paper we study self-similar solutions to the singular and degenerate diffusion equation

u_t=(|(p(u))_x|⁻²(p(u))_x)_x, −∞< x <+ ∞, t >0,

where 1<<2. The existence and uniqueness for the solutions are established. In addition, the asymptotic behavior is investigated.

䉷

Keywords: Singular and degenerate diffusion equation; Existence; Uniqueness and asymptotic behavior

1. Introduction

Consider the singular and degenerate diffusion equation

u_t=(|(p(u))_x|⁻²(p(u))_x)_x, −∞< x <+ ∞, t >0, (1.1) where 1<<2,p(s) ∈ C¹([0,+∞)),p(s) >0 for s >0 and p(s)is a monotone increasing function in(0,+∞). The equation may have singularity or degeneracy. In fact, the equation is singular at the points wherep(u)_x=0, while degenerate at the points where u=0 ifp(0)=0. This type of equation has a wide range of applications in physics and engineering sciences, see[9,15,17,18,20]andreferences therein.

∗Corresponding author. Tel.: +852 27889819; fax: +852 27888561.

E-mail address:[email protected](T. Yang).

doi:10.1016/j.na.2004.09.051

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During the past decades, equations of the form (1.1) have been paid much attention by many mathematicians. In particular, the investigation for the important case=2 has been the subject of intensive study, see[1,3–8,14,16,19,21,22]. When =2 andp(s)= s^m(m¹), (1.1) is calledthe non-Newtonian polytropic ﬁltration equation which has also been studied extensively, see[12]for the existence, uniqueness andregularity of solutions in one-dimensional case, and[11]andreferences therein for multi-dimensional case, see also[2,10,13].

Another way to view this equation and the degeneracy comes from the nonlinear damping problem for Euler equations with vacuum. Consider the following system for isentropic ﬂow:

_t+(u)_x=0,

(u)_t+(u²+P ())_x= −(u),

where^,uandP ()are density, velocity and pressure, respectively, while>0 is the damping coefﬁcient, and>0 is the power in the nonlinear damping term. It is known that usually when timet tends to inﬁnity, the convection term in the second equation, i.e., (u)_t +(u²)_x, decays faster than the other terms. This makes it possible to combine the above two equations to have the nonlinear porous media equation in the form of

_t =1

^{((P (})_x)¹^/)_x.

If the pressure function satisﬁesP (0)=0, then it is clear that the above equation is degenerate at=0, i.e., at vacuum states. We believe that the study in this paper will be useful to the study of the large time behavior of solutions to the above system.

In this paper, we will consider Eq. (1.1) with general functionp(u)under some assumption to obtain some existence andregularity results. Moreover, the necessary andsufﬁcient condition on the solutions with compact support is given with detailed behavior of the solutions at the interface. For this, we consider the self-similar solutions of the form

u(x, t)=w(), =x(t+1)⁻¹^/.

A direct calculation shows thatw=w()satisﬁes the following equation:

−1

w =(|(p(w))|⁻²(p(w))). (1.2)

We will investigate the inﬁnite two-point boundary value problem of (1.2) with

w(−∞)=w−, w(+∞)=w+, (1.3)

wherew_±^0.

Since Eq. (1.2) is singular at the points where(p(w)) =0 andmay be degenerate at the points whereu=0, the classical solution may not exist. For this, the solution to (1.2) and (1.3) is deﬁned as follows.

Deﬁnition 1. A non-negative functionw()∈C(−∞,+∞)is a solution of Eq. (1.2), if p(w)∈ C¹(−∞,+∞),wand(p(w)) are absolutely continuous in(−∞,+∞)so that

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(1.2) holds almost everywhere. Moreover, if

→−∞lim w()=w−, lim

→+∞w()=w+,

w()is calleda solution of the inﬁnite two-point boundary value problem (1.2) and(1.3).

Ifw()is a solution of the problem (1.2) and(1.3), thenw(˜ )=w(−)is also a solution of Eq. (1.2) with

w(−∞)=w₊, w(+∞)=w₋.

Therefore, we can assume 0w₋w₊without loss of generality.

The main results of this paper are the following:

Theorem 1. There exists a unique solution of the inﬁnite two-point boundary value problem (1.2) and (1.3).

Theorem 2. Letw()be the solution of the inﬁnite two-point boundary value problem (1.2) and (1.3) with 0< w₋< w₊. Thenw()is strictly increasing and satisﬁes

→−∞lim ||^/(²⁻⁾(w()−w−)= 2− p(w₋)

2(−1)p(w₋) 2−

1/(2−)

,

→+∞lim ^/(²⁻⁾(w₊−w())= 2− p(w₊)

2(−1)p(w₊) 2−

1/(2−)

.

Theorem 3. Letw()be the solution of the inﬁnite two-point boundary value problem (1.2) and (1.3) withw₋=0 andw₊>0. Set

ˆ=inf{∈(−∞,+∞):w() >0}.

Forˆ, we have the following two cases:

(i) If₁

0 p(s)s⁻¹^/(−¹⁾ds <+ ∞, then−∞<ˆ<0 and the solution satisﬁes d

d

_w()

0

p(s)s⁻¹^/(−¹⁾ds

=ˆ⁺

= −ˆ

1/(−1)

>0.

Moreover,w()is strictly increasing on[ˆ,+∞), whilew()≡0 on(−∞,ˆ]. (ii) If₁

0 p(s)s⁻¹^/(−¹⁾ds= +∞, thenˆ= −∞and the solution satisﬁes

→−∞lim ||w()=0, lim

→−∞||^/(²⁻⁾p(w())=0. Andw()is strictly increasing in(−∞,+∞).

Furthermore, for both cases, we have

→+∞lim ^/(²⁻⁾(w+−w())= 2− p(w₊)

2(−1)p(w₊) 2−

1/(2−)

.

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Remark 1. For p(s)=s^m(m¹), Theorem 3 yields the following properties on the solutions:

(i) Ifm >1/(−1), then−∞<ˆ<0 and d

d^(w^m−¹^/(−¹⁾⁽))|_=ˆ⁺= 1 m

m− 1

−1

−ˆ

₁_/(−₁₎

>0.

(ii) If 1m¹/(−1), thenˆ= −∞.

Remark 2. Consider the Euler equations for polytropic gas. WhenP ()=²^with¹ being the adiabatic constant andbeing a constant, the explanation on the vacuum behavior from Theorem 3 can be statedas follows. When>, the gas can connect to vacuum in finite distance at any time with the physical vacuum boundary condition being(⁻)_x =0 andboundedat the vacuum interface. This is consistent with the work done on the Euler equations with linear damping when=1. On the other hand, when 1^{, the gas} canonically does not connect to vacuum in finite distance for any time, but at infinity.

Theorem 4. There exists one solution to the Cauchy problem of Eq. (1.1) with the initial value

u(x,0)=

u−, x <0,

u₊, x >0, (1.4)

whereu_±^0.

2. Preliminaries

Letv=p(w). Then Eq. (1.2) is transformedto

−1

(q(v)) =(|v|⁻²v), (2.1)

i.e.

v= − 1

(−1)q (v)|v|²⁻v, (2.2)

whereq(s)=p⁻¹(s)is the inverse function ofp. Andthe boundary value (1.3) is transformed to

v(−∞)=v₋, v(+∞)=v₊, (2.3)

wherev₋=p(w₋)andv₊=p(w₊). It is easy to see thatq(s)∈C(R(p))∩C¹(R₊(p)), q(s) >0 fors∈R+(p)andq (s)is a monotone decreasing function inR+(p), where

R(p)= {p(s):s∈ [0,+∞)}, R+(p)= {p(s):s∈(0,+∞)}.

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According to Deﬁnition 1, we see that

Deﬁnition 2. A functionv()∈ C(I)is calleda solution of Eq. (2.2) in an intervalI, if v(I)⊂ R(p),v ∈ C¹(I),q(v)andv are absolutely continuous inI so that (2.2) holds almost everywhere. Additionally, ifI=(−∞,+∞)and

→−∞lim v()=v₋, lim

→+∞v()=v₊,

we sayv()is a solution of the inﬁnite two-point boundary value problem (2.2) and (2.3).

From Eq. (2.2) andDeﬁnition 2, ifvis a solution of (2.2) in an intervalI, thenvis a classical solution of (2.2) in{∈I :v()∈R₊(p)}. Noticing 2−>0, it is evident that for any solutionvof Eq. (2.2), ifv(0)∈R₊(p)andv(0)=0 for some0∈(−∞,+∞), thenv ≡0. That is,

Proposition 2.1. Assume I is an interval,v() (∈I)is a solution of Eq. (2.2) andv(I)⊂ R₊(p).

(i) If there exists0∈Isuch thatv(0) >0, then v() >0, ∀∈I.

(ii) If there exists0∈Isuch thatv(0)=0, then v()=0, ∀∈I.

(iii) If there exists0∈Isuch thatv(0) <0, then v() <0, ∀∈I.

Lemma 2.1. Assume I is an interval,v() (∈I)is a strictly increasing solution of Eq.

(2.2), 0∈I,v(I)⊂R+(p),b∈R+(p)and v()b, ∀∈I.

Then

v()(v⁻²(0)+C²)¹^/(−²⁾, ∀∈I, where

C=(2−)q(b) 2(−1) >0.

Proof. Sincevis strictly increasing, (2.2) can be rewritten as v= − 1

(−1)q (v)v³⁻,

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i.e.

(v⁻²) = 2−

(−1)q(v).

Thus

v⁻²()=v⁻²(0)+ 2− (−1)

0

sq(v(s))ds, ∀∈I.

Sinceq(s) >0 fors ∈R₊(p)andq(s)is a monotone decreasing function inR₊(p), we have

0< q(b)q (v(s)), ∀s∈I.

Noticing that 1<<2 andv(0) >0, we see that 0< v⁻²(0)+(2−)q(b)

2(−1) ²v⁻²(), ∀∈I.

Thus

v()

v⁻²(0)+(2−)q(b) 2(−1) ²

1/(−2)

, ∀∈I.

The proof is complete.

Lemma 2.2. Letv1()andv2()be two strictly increasing solutions of Eq. (2.2) in an intervalI andv1(I), v2(I)⊂R₊(p).

(i) If there exists0∈ [0,+∞)∩Isuch that v1(0)v2(0), v₁(0) > v₂(0), then

v1() > v2(), ∀∈ [0,+∞)∩I.

(ii) If there exists0∈(−∞,0)∩I such that v1(0)=v2(0), v1(0) > v2(0), then

v₁() > v₂(), ∀∈ [0,+∞)∩I.

Proof. (i) Assume the conclusion was invalid. Let ^∗=inf{∈ [0,+∞)∩I :v1()v2()}.

Then 00<^∗^,v₁(^∗)=v₂(^∗)andv₁() > v₂()for0<<^∗^{. Since}v1(0)v2(0), v1(^∗) > v2(^∗). Eq. (2.2) impliesv₁(^∗) > v₂(^∗), which contradicts thatv₁(^∗)=v₂(^∗) andv₁() > v₂()for0<<^∗^.

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(ii) From (i), we needto only prove that v1() > v2(), ∀∈ [0,0] ∩I.

Assume the conclusion was invalid. Let _∗=inf{x∈ [0,0] ∩I :v₁()v₂()}.

Then0<_∗^0,v₁(_∗)=v₂(_∗)andv₁() > v₂()for0<<_∗^{. Since}v1(0)=v2(0), v1() > v2()for0<_∗. Integrating (2.1) from0to_∗^{, we get}

v⁻¹^∗

0

= −1

_∗

0

(q(v)) d=1

_∗

0

q(v)d−1 q(v)

∗

0

.

Therefore,

v⁻2 ¹(0)−v⁻1 ¹(0)=1

_∗

0

(q(v1)−q(v2))d−1

_∗(q(v1(_∗))−q(v2(_∗))).

Since_∗^0,q(v1(_∗))−q(v2(_∗)) >0,_∗

0 (q(v1)−q(v2))d>0, we get that v⁻2 ¹(0) > v⁻1 ¹(0),

which contradictsv₁(0) > v₂(0). The proof is complete.

3. The case without degeneracy

In this section, we consider the case without degeneracy, i.e.v_±∈R₊(p). By Proposition 2.1, problems (2.2) and(2.3) have a unique solution forv₋=v₊. Therefore, we just need to investigate the casev₋< v₊.

Lemma 3.1. Assume0 ∈(−∞,+∞)anda ∈ R₊(p). For sufﬁciently small>0, Eq.

(2.2) with the initial value v(0)=a,

v(0)= ^(3.1)

can be extended to+∞.

Proof. We ﬁrst prove the conclusion for0=0. Sincep∈C¹(0,+∞),R₊(p)is an open interval. Thus, there existsb ∈ R₊(p)such thatb > a. Choose0>0 such that for any 0<<0,

_+∞

0

(⁻²+C²)¹^/(−²⁾d< b−a,

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where

C=(2−)q(b) 2(−1) >0.

By Lemma 2.1, it is easy to see for any 0<<0, the solution of (2.2) and(3.1) with0=0 can be extended to+∞.

If0>0, we can obtain the conclusion directly by the result of0=0 andLemma 2.2.

If0<0, by the result of0=0 andthe continuity of solution to initial, we needonly prove that the solutionvof (2.2) and(3.1) can be extendedto 0 andsatisﬁesv(0)∈ R₊(p)for sufﬁciently small>0. By Proposition 2.1,v >0. Therefore, Eq. (2.2) can be transformed to

(v⁻²) = 2−

(−1)q(v).

Integrating from0to^{, we get} v⁻²()=⁻²+ 2−

(−1)

0

sq(v(s))ds,

which impliesvcan be extendedto 0 andsatisﬁesv(0)∈R₊(p)when>0 is sufﬁciently small. The proof is complete.

Lemma 3.2. Assume1<2,a < banda, b∈R+(p). There exists a solution of Eq. (2.2) with the boundary value

v(1)=a, v(2)=b. (3.2)

Proof. We denotev(;)the solution of Eq. (2.2) with the initial value v(1)=a,

v(1)=,

where>0. From Lemma 3.1, we know thatv(;)is existent on[1,+∞)for sufﬁciently small>0. By the continuity of solution to initial,

→lim0⁺ v(2;)=a.

Thus the set

E= {>0:v(;)is existent on[1,2]andv(2;)b}

is not empty. We declare thatEis bounded additionally.

(i) The case10. Eq. (2.2) can be transformedto (v⁻²) = 2−

(−1)q(v).

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Integrating from1to^{, we get} v⁻²()=⁻²+ 2−

(−1)

1

sq(v(s))ds.

Therefore, for sufﬁciently large>0, ifv(;)is existent on[1,2], thenv(2;) > b. This impliesEis bounded.

(ii) The case1<0. Sincev(;) >0 for<0 andv(;)is strictly increasing, b−a

min{2,0} −1

,+∞

∩E= ∅, which impliesEis bounded too.

Let

0=supE.

Then 0<0<+ ∞. By the continuity of solution to initial,0∈E. Assumev(2;0) < b. By Lemma 2.2 andthe continuity of solution to initial,v(;0+ )is existent on[1,2] andv(2;0+ ) < bfor sufﬁciently small >0, which contradicts the deﬁnition of0. Hencev(2;0)=b. This implies thatv(;0)is a solution of the problem (2.2) and(3.2).

The proof is complete.

Now we can study existence, uniqueness and asymptotic behavior of solutions of problems (2.2) and(2.3) for the case without degeneracy.

Proposition 3.1. Assumev₋< v₊. There exists at least one solution of problems (2.2) and (2.3).

Proof. We denotev0()the solution of Eq. (2.2) with the boundary value v(0)=v₋, v(1)=v₊,

andv_n()the solution of Eq. (2.2) with the boundary value v(−n)=v−, v(n)=v+,

wheren is a positive integer. By Lemma 3.2, v0 and v_n are existent. We declare that v_n(0)v₀(0)for eachn. Otherwise, for somen^1,

v_n(0) > v0(0).

Sincev_n(0) > a=v0(0), Lemma 2.2 implies that v_n() > v0(), ∀∈ [0,1]

which contradicts thatv_n(0) > a=v0(0)andv_n(1)b=v0(1). By Lemma 2.1, v_n()(v_n⁻²(0)+C²)¹^/(−²⁾(v0−2

(0)+C²)¹^/(−²⁾,

∀∈ [−n,+n], (3.3)

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whereC >0 is a constant independent ofn. This implies{v_n},{v_n}and{v_n}are all uniformly bounded. Therefore,{v_n}and{v_n}are both uniformly boundedandequi-continuous on any compact subset of(−∞,+∞). By Arzela–Ascoli’s theorem, there exists a subsequence, denoted by{v_n_k}, anda functionv ∈ C¹(−∞,+∞)such that{v_n_k}and{v_n_k}converge uniformly tovandv on any compact subset of(−∞,+∞), respectively. Sincev_n_k satisfy Eq. (2.2),v∈C²(−∞,+∞)andsatisﬁes Eq. (2.2). From (3.3) andthe boundary value of v_n, it is easy to see that

→−∞lim v()=v−, lim

→+∞v()=v+.

Thus,vis a solution of problems (2.2) and(2.3). The proof is complete.

Proposition 3.2. Assumev₋< v₊. Problems (2.2) and (2.3) admit at most one solution.

Proof. We argue by contradiction. Assumev1andv2are two solutions of problems (2.2) and(2.3), andthere exists0∈(−∞,+∞)such that

v1(0) > v2(0).

For1<0<2, integrating (2.1) from1to2, we get v_i⁻¹²

1

= −1

₂

1

(q(v_i)) d= −1 q(v_i)

2

1

+1

₂

1

q(v_i)d (i=1,2).

(3.4) Sincev_i(i=1,2)are solutions of problems (2.2) and(2.3),

1q(v_i(1))=1q(v_i(−∞))+1

₁

−∞q(v_i)v_id

=1q(v₋)+1

₁

−∞q(v_i)v_id (i=1,2) (3.5) and

2q(vi(2))=2q(vi(+∞))−2

_+∞

2

q(vi)v_id

=2q(v+)−2

_+∞

2

q(vi)v_id (i=1,2). (3.6) We distinguish four cases to complete the proof.

(i) If

v1() > v2(), ∀∈(−∞,+∞).

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From (3.4)–(3.6), we can get v1−1²

1

−v2−1²

1

=2

_+∞

2

q (v1)v1d+1

₁

−∞q(v1)v1d

−2

_+∞

2

q(v2)v2d−1

₁

−∞q(v2)v2d +1

₂

1

(q(v1)−q(v2))d.

Since v1() > v2()in (−∞,+∞)andq (s) >0 for s ∈ R+(p), for 10−1 and 20+1, we have

v1−1²

1

−v2−1 ²

1

−2

_+∞

2

q (v1)v1d−1

₁

−∞q(v1)v1d +2

_+∞

2

q(v2)v2d+1

₁

−∞q(v2)v2d

=1

₂

1

(q(v1)−q(v2))d

¹

₀₊₁

0−1

(q(v1)−q(v2))d.

Letting1→ −∞and2→ +∞, andnoticing that each term in the left converges to 0 from Lemma 2.1, we see that

0¹

₀₊₁

0−1

(q(v1)−q(v2))d,

which contradicts thatv1() > v2()in(−∞,+∞)andq(s) >0 fors∈R+(p). (ii) If

v1() > v2(), ∀∈(−∞,0), but

v1() > v2(), ∀∈(0,+∞) is invalid. Let

^∗=inf{>0:v1()v2()}.

Then^∗>0,v1(^∗)=v2(^∗)andv1() > v2()in(−∞,^∗). We ﬁrst prove thatv₁(^∗)= v₂(^∗). On the one hand, ifv₁(^∗) > v₂(^∗), then Lemma 2.2 implies that

v₁() > v₂(), ∀^∗,

which contradicts thatv1(^∗)=v2(^∗)andv1(+∞)=v2(+∞)=v₊. On the other hand, ifv₁(^∗) < v₂(^∗), then Lemma 2.2 implies that

v1() < v2(), ∀^∗,

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which contradicts thatv1(^∗)=v2(^∗)andv1(+∞)=v2(+∞)=v₊too. Hencev₁(^∗)= v₂(^∗).

Choosing2=^∗in (3.4), by (3.5) we get

−v1−1

(1)+v2−1

(1)=1

₁

−∞q (v1)v1d−1

₁

−∞q(v2)v2d +1

^∗

1

(q(v1)−q(v2))d.

Sincev1() > v2()in(−∞,^∗)andq (s) >0 fors∈R₊(p), we have

−v1−1

(1)+v2−1

(1)−1

₁

−∞q(v1)v1d+1

₁

−∞q(v2)v2d

=1

^∗

1

(q(v1)−q(v2))d

¹

^∗

0

(q(v1)−q(v2))d. Letting1→ −∞, we see that

0¹

^∗

0

(q(v1)−q(v2))d,

which contradicts thatv1() > v2()in(−∞,^∗)andq(s) >0 fors∈R₊(p). (iii) If

v1() > v2(), ∀∈(0,+∞), but

v1() > v2(), ∀∈(−∞,0) is invalid. Let

_∗=sup{<0:v1()v2()}.

Then_∗<0,v1(_∗)=v2(_∗)andv1() > v2()in(_∗,+∞). Similar to (ii), we can prove thatv₁(_∗)=v₂(_∗)by Lemma 2.2.

Choosing1=_∗in (3.4), by (3.6) we get v1−1

(2)−v2−1

(2)=2

_+∞

2

q(v1)v1d−2

_+∞

2

q (v2)v2d +1

₂

∗

(q(v1)−q(v2))d.

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Sincev1() > v2()in(_∗,+∞)andq (s) >0 fors∈R₊(p), we have v₁⁻¹(2)−v₂⁻¹(2)−2

_+∞

2

q(v1)v₁d+2

_+∞

2

q (v2)v₂d

=1

₂

∗

(q(v1)−q(v2))d

¹

₀

∗

(q(v1)−q(v2))d. Letting2→ +∞, we see that

0¹ ₀

∗

(q(v1)−q(v2))d,

which contradicts thatv1() > v2()in(_∗,+∞)andq(s) >0 fors∈R+(p). (iv) If

v1() > v2(), ∀∈(−∞,0) and

v1() > v2(), ∀∈(0,+∞) are both invalid. Let

_∗=sup{<0:v1()v2()}, ^∗=inf{>0:v1()v2()}.

Then_∗<0<^∗^,v1(_∗)=v2(_∗),v1(^∗)=v2(^∗)andv1() > v2()in(_∗,^∗). Similar to (ii), we can prove thatv₁(_∗)=v₂(_∗)andv₁(^∗)=v₂(^∗)by Lemma 2.2.

Choosing1=_∗^and2=^∗in (3.4), we get 0=1

^∗

_∗ (q(v1)−q(v2))d,

which contradicts thatv1() > v2()in(_∗,^∗)andq(s) >0 fors∈R+(p). The proof is complete.

Proposition 3.3. Assumev₋< v₊andvis the solution of problems (2.2) and (2.3). Then vis strictly increasing and satisﬁes

→−∞lim ||^/(²⁻⁾(v()−v₋)=2−

2(−1) (2−)q(v−)

1/(2−)

, (3.7)

→+∞lim ^/(²⁻⁾(v+−v())=2−

2(−1) (2−)q(v₊)

1/(2−)

. (3.8)

Proof. The monotone property ofv can be obtaineddirectly from Proposition 2.1. The proof of (3.7) and(3.8) is similar. Here we only prove (3.8). Sinceq ∈ C¹(R+(p))and

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q is monotone decreasing, for any >0, there exists a constanta ∈ (v₋, v₊)such that q(v₊)q(a)q (v₊)+ . Choose0>0 such thatv(0)a. Sincev >0, Eq. (2.2) can be transformedto

(v⁻²) = 2−

(−1)q(v).

Integrating from0to^{, we get} v⁻²()=v⁻²(0)+ 2−

(−1)

0

sq(v(s))ds, ∀>0.

Sincev >0,

av(0) < v() < v₊, ∀>0. Thus

0< q(v+)q (v())q(a)q (v+)+ , ∀>0. Therefore,

(C1+1²)¹^/(−²⁾v()(C2+2²)¹^/(−²⁾, ∀>0, where

1=(2−)(q(v₊)+ )

2(−1) , C1=v⁻²(0)−1²0, 2=(2−)q(v₊)

2(−1) , C2=v⁻²(0)−2²0. Hence for any>0,

_∞

(C1+1s²)^1/(−2)dsv₊−v() _∞

(C2+2s²)^1/(−2)ds.

By L’Hospital’s Rule, we see that

→+∞lim ^/(²⁻⁾ _∞

(Ci +is²)¹^/(−²⁾ds

=2−

→+∞^lim ²^/(²⁻⁾(C_i+i²)¹^/(−²⁾

=2−

¹_i^/(−²⁾ (i=1,2).

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Therefore,

→+∞lim ^/(²⁻⁾ _∞

(C1+1s²)¹^/(−²⁾ds

^lim

→+∞^/(²⁻⁾(v₊−v())

^lim

→+∞^/(²⁻⁾(v₊−v())

^lim

→+∞^/(²⁻⁾ _∞

(C2+2s²)¹^/(−²⁾ds, i.e.

2−

(2−)(q(v+)+ ) 2(−1)

1/(−2)

^lim

→+∞^/(²⁻⁾(v₊−v())

^lim

→+∞^/(²⁻⁾(v₊−v())

²⁻

(2−)q(v₊) 2(−1)

1/(−2)

.

Owing to the arbitrariness of >0, we get (3.8). The proof is complete.

Since problems (1.2) and(1.3) are equivalent to problems (2.2) and(2.3), we get Theorems 1 and2 with 0< w−< w+from Proposition 3.1–3.3 directly.

4. The case with possible degeneracy

In this section, we consider the case with degeneracy, i.e.v_± ∈ R(p)andv₋=p(0). By Proposition 2.1, problems (2.2) and(2.3) have a unique solution forv₋=v₊=p(0). Therefore, we needonly investigate the casev₋=p(0)andv₊∈R₊(p). Without generality, we assumev₋=p(0)=0 andv₊=p(w₊)=1, andinvestigate the inﬁnite two-point boundary value problem of (2.2) with

v(−∞)=0, v(+∞)=1. (4.1)

Proposition 4.1. There exists at least one solution of problems (2.2) and (4.1).

Proof. We denotevn()the solution of Eq. (2.2) with the boundary value v(−∞)=1

n, v(+∞)=1,

wherenis a positive integer. By Proposition 3.1, Lemma 2.2 andthe proof of Proposition 3.2,v_nare existent and

0< vn+1() < vn() <1, ∀∈(−∞,+∞) (n=1,2, . . .).

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Let

v()= lim

n→∞v_n(), ∀∈(−∞,+∞).

Since 0< v_n<1 andv_nare strictly increasing in(−∞,+∞), 0v <1 andvis monotone increasing in(−∞,+∞).

Now we show thatvis a solution of problems (2.2) and(4.1). Sincev_n>0 in(−∞,+∞) andv_n>0 in(−∞,0),

0< v_n(−1) <1 (n=1,2, . . .).

For any1<2, integrating Eq. (2.1) withv=vnfrom1to2, we get v_n⁻¹²

1

= −1

₂

1

(q(v_n)) d= 1

₂

1

q(v_n)d−1 q(v_n)

2

1

. (4.2)

Choosing1= −1 and2=0, we get v_n⁻¹(0)=v_n⁻¹(−1)+1

₀

−1

q(vn)d−1

^q(vⁿ⁽⁻¹⁾⁾¹+q(1) ^. By Lemma 2.1,

0< v_n()C(1+²)¹^/(−²⁾, ∀∈(−∞,+∞), (4.3) whereC >0 is a constant independent ofn. This implies thatv_n are uniformly bounded andequi-continuous in(−∞,+∞). Therefore,v∈C(−∞,+∞)and

v(−∞)=0, v(+∞)=1.

(i) Assumev(0) >0 for some0∈(−∞,+∞). Then

vn() > v()v(0) >0, ∀0 (n=1,2, . . .). (4.4) For any2>10, from (4.2) we get

|v_n⁻¹(2)−v_n⁻¹(1)| =1

₂

1

(q(vn)) d ^max^{|1|,|2|}

₂

1

|(q(v_n))|d

=max{|1|,|2|}

₂

1

(q(v_n)) d ^max^{|1|,|2|}

^(q(vⁿ⁽2))−q(v_n(1))) ^max^{|1|,|2|}

^q^(v(0))v(0)(2−1).

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This estimate and(4.3) imply thatv_nare uniformly boundedandequi-continuous on any compact subset of[0,+∞). Therefore,v∈C¹([0,+∞)). Noticing (4.4), we see thatv∈ C²([0,+∞))andsatisﬁes Eq. (2.2) on[0,+∞)sincev_nsatisfy Eq. (2.2) in(−∞,+∞). (ii) Assumev(0)=0 for some0 ∈ (−∞,+∞). Thenv() ≡ 0 on(−∞,0] by Proposition 2.1. Thusv∈C²((−∞,0])andsatisﬁes Eq. (2.2) on(−∞,0].

(iii) Assume there is0∈ (−∞,+∞)such thatv(0)=0 andv() >0 for all>0. Choosing1=0in (4.2), we see that for any2>0,

v_n⁻¹(2)−v_n⁻¹(0)=1

₂

0

q(vn)d−1

2q(vn(2))+1

0q(vn(0)).

Lettingn→ ∞, we get v⁻¹(2)=1

₂

0

q(v)d−1

2q(v(2)), ∀2>0. Letting2→⁺₀, we see that lim

→⁺0

v()=0.

From the above discussion, we see thatv ∈C¹(−∞,+∞)andvis a solution of problems (2.2) and(4.1). The proof is complete.

Lemma 4.1. Letvbe a solution of problems (2.2) and (4.1) and ˆ=inf{∈(−∞,+∞):v() >0}.

Then v is monotone increasing in (−∞,+∞) and ˆ<0. In addition, for ˆ ^and

1

0 q⁻¹^/(−¹⁾(s)dswe have the following two cases:

(i) Ifˆ>− ∞, then₁

0 q⁻¹^/(−¹⁾(s)ds <+ ∞. (ii) Ifˆ= −∞, then₁

0 q⁻¹^/(−¹⁾(s)ds= +∞.

Proof. From (2.2) andProposition 2.1,vis monotone increasing andv()v(0)for all ∈(−∞,+∞). Henceˆ<0.

(i) The caseˆ>− ∞. For anyˆ<<0, integrating (2.1) fromˆ^to^{, we get} v⁻¹_ˆ

= −1

ˆ s(q(v(s))) ds.

Thus

v⁻¹()= −1

ˆ s(q(v(s))) ds ^|ˆ|

ˆ |(q(v(s)))|ds=|ˆ|

ˆ (q(v(s))) ds=|ˆ|

^q(v()).

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Therefore,

q⁻¹^/(−¹⁾(v())v() |ˆ|

1/(−1)

, ∀∈(ˆ,0).

Integrating fromˆto 0, we get _v(0)

0

q⁻¹^/(−¹⁾(s)ds= 0

ˆ q⁻¹^/(−¹⁾(v())v()d |ˆ|

1/(−1)

|ˆ|.

Therefore,₁

0q⁻¹^/(−¹⁾(s)ds <+ ∞.

(ii) The caseˆ= −∞. For any1<<0, integrating (2.1) from1to^{, we get} v⁻¹

1

= −1

1

s(q(v(s))) ds.

Letting1→ −∞, we get v⁻¹()= −1

−∞s(q(v(s))) ds

= −1

−∞s|(q(v(s)))|ds ^||

−∞(q(v(s))) ds=||

^q(v()). (4.5)

Therefore,

q⁻¹^/(−¹⁾(v())v() ||

₁_/(−₁₎

, ∀<0. Integrating from−∞to 0, we get

_v(₀₎

0

q⁻¹^/(−¹⁾(s)ds= ₀

−∞q⁻¹^/(−¹⁾(v())v()d

0

−∞

||

1/(−1)

d= +∞.

Therefore,₁

0q⁻¹^/(−¹⁾(s)ds= +∞. The proof is complete.

Proposition 4.2. Problems (2.2) and (4.1) admit at most one solution.

Proof. Assume v1 and v2 are two different solutions of problems (2.2) and (4.1). If ₁

0 q⁻¹^/(−¹⁾(s)ds= +∞, thenv1(), v2()∈R₊(p)for all∈(−∞,+∞)by Lemma 4.1. Letting→ −∞in (4.5), by Lemma 2.1, we get

→−∞lim ||q(v())=0,