Asymptotic forms of slowly decaying positive solutions of second-order quasilinear ordinary differential equations related to degenerate elliptic equations

J. Math. Anal. Appl. 433 (2016) 1486–1501

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Asymptotic forms of slowly decaying positive solutions of second-order quasilinear ordinary differential equations related to degenerate elliptic equations ✩ Hiroyuki Usami Applied Physics Course, Faculty of Engineering, Gifu University, Gifu, Japan

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a b s t r a c t

Article history: Received 28 March 2015 Available online 12 August 2015 Submitted by Y. Yamada

We determine asymptotic forms of slowly decaying positive solutions of quasilinear ordinary differential equations related to so-called p-Laplace equation. © 2015 Elsevier Inc. All rights reserved.

Keywords: Asymptotic form Quasilinear ordinary differential equation Pseudo-Laplacian

1. Introduction and statement of the main results Let us consider the quasilinear ordinary differential equation (p(t)|u |α−1 u ) + q(t)|u|λ−1 u = 0 near + ∞,

(A)

where we assume that α > 0 and λ > 0 are constants, and p(t) and q(t) are positive continuous functions ∞ defined near +∞ and satisfying p(t)−1/α dt < ∞. Note that equations of the form (A) often arise in the study of pseudo-Laplace equations as seen from Example 1.5 below. It is easily seen from Lemma 1.1 and p. 133 of [8] that every positive solution u of (A) satisfies exactly one of the following three asymptotic properties: u(t) ∼ c1 for some constant c1 > 0 as t → ∞; ∞ u(t) ∼ c2 p(s)−1/α ds as t → ∞ for some constant c2 > 0; t

✩

This work was supported by JSPS KAKENHI Grant Number (C)23540196. E-mail address: [email protected].

http://dx.doi.org/10.1016/j.jmaa.2015.08.018 0022-247X/© 2015 Elsevier Inc. All rights reserved.

(1.1) (1.2)

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and ∞

u(t) → 0 and

t

u(t) →∞ p(s)−1/α ds

as t → ∞.

(1.3)

(Here “f (t) ∼ g(t) as t → ∞” means, as usual, that limt→∞ f (t)/g(t) = 1 for positive functions f and g defined near +∞.) Asymptotic properties of those solutions u which satisfy either (1.1) or (1.2) were widely investigated. For example, necessary and sufficient conditions for the existence of such solutions were established in [4,8]: equation (A) has a positive solution u satisfying (1.1) if and only if ∞

⎛ ⎝ 1 p(t)

t

⎞1/α q(s)ds⎠

dt < ∞;

and equation (A) has a positive solution u satisfying (1.2) if and only if ∞

⎛ ⎝

∞

⎞λ ds ⎠ q(t)dt < ∞. p(s)1/α

t

Nevertheless there seems to be less information about qualitative properties of solutions u satisfying (1.3). Motivated by this fact, in the paper we will discuss asymptotic behavior of solutions u satisfying (1.3). In particular, we will find exact asymptotic forms of such solutions near +∞ when the coefficient functions behave like power functions of t. In what follows we refer solutions u satisfying (1.3) as to slowly decaying solutions, because such solutions decay more slowly than those solutions that satisfy (1.2). ∞ ∞ Remark 1.1. When the condition p(t)−1/α dt < ∞ is violated, that is, when p(t)−1/α dt = ∞, by t putting s = p(ξ)−1/α dξ equation (A) reduces to the simpler form (|us |α−1 us )s + p(t)q(t)|u|λ−1 u = 0

near + ∞.

Studies of such equations were, for example, the main objective of [6]; and asymptotic properties of its solutions have been fully discussed. Asymptotic forms of slowly decaying solutions of (A) may be strongly affected by those of coefficient functions p(t), q(t) and the exponents α and λ. Therefore we will devote ourselves to studying the following equation, which has more restrictive appearance than equation (A): (tβ |u |α−1 u ) + tσ (1 + ε(t))|u|λ−1 u = 0 near + ∞.

(E)

Since the case β < α has been already discussed essentially in [6] as mentioned above, we impose the next assumptions (A1 ) and (A2 ) on (E) in the sequel: (A1 ) α, β, λ and σ are constants satisfying λ > α > 0, β > α, and (β − α) − 1 < σ <

λ (β − α) − 1. α

(A2 ) ε(t) is a continuous or a C 1 -function defined near +∞ satisfying limt→∞ ε(t) = 0.

(1.4)

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Additional conditions will be given later. Notice that a positive solution u(t) of (E) is a slowly decaying solution if and only if lim u(t) = 0 and

t→∞

lim t

β−α α

t→∞

u(t) = ∞.

(1.5)

In order to get an insight into our problems let us consider equation (E) with ε(t) ≡ 0, namely, the equation (tβ |u |α−1 u ) + tσ |u|λ−1 u = 0 near + ∞.

(E0 )

It is easily seen that, under the assumption (A1 ), equation (E0 ) has an exact slowly decaying solution u0 explicitly given by ˆ −k , u0 (t) = Ct

(1.6)

where k=

1 + σ − (β − α) > 0, λ−α

and Cˆ λ−α = kα {β − α(k + 1)} > 0.

(1.7)

Since equation (E) can be regarded as a “perturbed equation” of equation (E0 ), we can conjecture that slowly decaying solutions of equation (E) and those of equation (E0 ) may have the same asymptotic behavior near +∞ in some sense, if ε(t) is sufficiently small. In fact, we can show that our conjecture is true in various cases. To prove this fact is the main purpose of this paper. We note that, if the second inequality in (1.4) is violated, that is, if λ (β − α) − 1, α

σ≥

then both equations (E0 ) and (E) have no positive solutions near +∞; see [8]. So it is natural to assume (1.4) in advance. Our main results are as follows: Theorem 1.2. Let (A1 ) and (A2 ) hold. Suppose further that α ≤ 1, and β − α(k + 1) − k = 0 (⇔ σ = ((β − α)λ + β − 2α − 1)/(1 + α)) hold. If either ∞

ε(t)2 dt < ∞ t

(1.8)

|ε (t)|dt < ∞

(1.9)

or ∞

holds, then every slowly decaying positive solution u of equation (E) satisfies u(t) ∼ u0 (t) as t → ∞, where u0 (t) is given by (1.6). Theorem 1.3. Let (A1 ) and (A2 ) hold. Suppose further that α ≥ 1, and β − α(k + 1) − k = 0 hold. If lim tε (t) = 0

t→∞

(1.10)

and (1.9) hold, then every slowly decaying positive solution u of equation (E) satisfies u(t) ∼ u0 (t) as t → ∞.

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Remark 1.4. Let ε(t) be of class C 1 . If ε(t) tends to 0 as t → ∞ monotonically, then condition (1.9) is satisfied automatically. The following is an application of our results to pseudo-Laplace equations of the Matukuma-type: Example 1.5. Let N > m > 1 and N ≥ 2. Consider the following quasilinear partial differential equation which has radial symmetry in an exterior domain of RN : div(|Du|m−2 Du) +

1 |u|λ−1 u = 0 near 1 + |x|τ

∞,

(1.11)

where let us assume that λ > m − 1, and 0 < τ < m,

and λ >

(m − 1)(N − τ ) ≡ λ∗ . N −m

(1.12)

A radial function u(r), r = |x|, is a solution of (1.11) if and only if u(r) solves the ordinary differential equation (rN −1 |ur |m−2 ur )r + rN −1

1 |u|λ−1 u = 0 near 1 + rτ

+ ∞.

(1.13)

We can rewrite equation (1.13) as (rN −1 |ur |m−2 ur )r + rN −1−τ (1 + ε˜(r))|u|λ−1 u = 0, where ε˜(r) is of class C 1 , and satisfies limr→∞ ε˜(r) = 0, ∞

ε˜(r)2 dr < ∞, r

∞ |˜ εr (r)|dr < ∞,

and

lim r˜ εr (r) = 0.

r→∞

(1.14)

Recall that a positive solution u(r) of (1.13) is a slowly decaying positive solution if and only if N −m

u(r) → 0 and r m−1 u(r) → ∞

as

r → +∞.

Condition (1.12) implies that equation (1.13) satisfies our underlying assumption (A1 ), and condition (1.14) implies that Theorems 1.2 and 1.3 are applicable to equation (1.13). By Case II of Theorem 6.1 of [7], if λ>

(m − 1)N + m − mτ ≡ λ0 , N −m

then equation (1.13) does have slowly decaying positive solutions. Since λ0 > λ∗ , by Theorems 1.2 and 1.3 every slowly decaying positive solution u of equation (1.13) satisfies u(r) ∼ Ar−(m−τ )/(λ−m+1)

as r → +∞,

where A is a positive constant given by  Aλ−m+1 = if λ > λ0 (and 0 < τ < m).

m−τ λ−m+1

m−1 ·

N λ − N m + N + mτ − mλ − τ λ−m+1

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Remark 1.6. For the autonomous equation div(|Du|m−2 Du) + |u|λ−1 u = 0, the assertion of Example 1.5 was established in [1] based on the theory of autonomous dynamical systems. Related results are found in [3,5,7]. The proof of Theorems 1.2 and 1.3, as well as preliminary lemmas, are given in Section 2. 2. Proof of the results We give the proof of Theorems 1.2 and 1.3. We begin with several auxiliary results. Notice that we assume (A1 ) and (A2 ) throughout the paper. Lemma 2.1. Let (A1 ) and (A2 ) hold. Suppose that u(t) is a slowly decaying positive solution of (E). Then (i) limt→∞ tβ (−u (t))α = ∞; (ii) u(t) and u (t) are dominated by u0 (t) and u0 (t), respectively: u(t) = O(u0 (t))

and

u (t) = O(|u0 (t)|)

as

t → ∞,

(2.1)

where u0 (t) is given by (1.6). Proof. (i) By equation (E) we know that u < 0 and tβ (−u )α increases monotonically to a positive constant β−α l ∈ (0, ∞]. If l < ∞, then −u (t) ≤ c1 t−β/α for some constant c1 > 0. It follows that u(t) = O(t− α ) as t → ∞. This gives a contradiction to (1.5). Hence l = ∞, and so (i) holds. (ii) An integration of the both sides of equation (E) on [t0 , t] gives t



t (−u (t)) ≥ β

α

rσ (1 + ε(r))u(r)λ dr, t0

where t0 is a sufficiently large number. Since u is a decreasing function, we have t



t (−u (t)) ≥ u(t) β

α

λ

rσ (1 + ε(r))dr; t0

that is, ⎛ −u (t)u(t)−λ/α ≥ ⎝t−β

t

⎞1/α rσ (1 + ε(r))dr⎠

.

t0

One more integration of the both sides gives the first estimate for u in (2.1). To get the estimates for u , it suffices to notice (i), inequality 

t

t (−u (t)) ≤ C1 β

α

rσ u(r)λ dr, t0

where C1 > 0 is a constant, and the first estimate in (2.1). 2

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Lemma 2.2. Let (A1 ) and (A2 ) hold. Suppose that u(t) is a slowly decaying positive solution of (E), and put t = es , u/u0 = v, and · = d/ds. Then (i) v(s), kv(s) − v(s) ˙ = O(1) as s → ∞, and kv − v˙ > 0 near +∞; (ii) v(s) satisfies the equation {(kv − v) ˙ α } + {β − α(k + 1)}(kv − v) ˙ α − Cˆ λ−α {1 + δ(s)}v λ = 0

near

+ ∞,

(2.2)

where δ(s) = ε(es ) and k and Cˆ are given by (1.7). The proof of this lemma is carried out by direct computations; hence we omit it. Remark 2.3. Equation (2.2) can be rewritten as  v¨ +

 Cˆ λ−α β β − 2k − 1 v˙ − k −k−1 v+ {1 + δ(s)}(kv − v) ˙ 1−α v λ = 0. α α α

(2.3)

For simplicity we further rewrite equation (2.2) (and (2.3)) as v¨ + av˙ − bv + bkα−1 (1 + δ(s))(kv − v) ˙ 1−α v λ = 0,

(2.4)

where a = β/α − 2k − 1 and b = k(β/α − k − 1). Note that b > 0 by (1.6) and (1.4). Lemma 2.4. Let f (s) be a C 1 -function defined near +∞ satisfying f˙(s) = O(1) as s → ∞ and ∞ f (s)2 ds < ∞. Then lims→∞ f (s) = 0. The proof of this lemma will be found in [6]. Proof of Theorem 1.2. By the change of variables (t, u) → (s, v) introduced in Lemma 2.2, we obtain equation (2.2). We note that the integral conditions indicated in (1.8) and (1.9) are equivalent to ∞ δ(s)2 ds < ∞

(2.5)

˙ |δ(s)|ds < ∞,

(2.6)

and ∞

respectively. Since v(s) = u(t)/u0 (t), it suffices to show that v(s) → 1 as s → ∞. The proof is divided into several steps. ∞ v(s) ˙ 2 ds < ∞ and v(s) ˙ → 0 as s → ∞. Step 1. We show that ˙ and integrate the resulting equation on [s0 , s] to obtain We multiply equation (2.2) by v, s

s

α 

{(kv − v) ˙ } vdr ˙ + {β − α(k + 1)} s0

(kv − v) ˙ α vdr ˙ s0

−

Cˆ λ−α λ+1 ˆ λ−α v −C λ+1

s δ(r)v λ vdr ˙ = const. s0

(2.7)

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Since integration by parts implies that s

s

{(kv − v) ˙ α } vdr ˙ =−

s0

{(kv − v) ˙ α } (kv − v)dr ˙ +k

s0

s

{(kv − v) ˙ α }· vdr

s0

α (kv − v) ˙ α+1 + kv(kv − v) ˙ α−k =− α+1

s (kv − v) ˙ α vdr ˙ + const, s0

we obtain from (2.7) α (kv − v) ˙ α+1 + kv(kv − v) − ˙ α + {β − α(k + 1) − k} α+1

s (kv − v) ˙ α vdr ˙ s0

−

Cˆ λ−α λ+1 ˆ λ−α v −C λ+1

s δ(r)v λ vdr ˙ = const. s0

The boundedness of v and v˙ shown in Lemma 2.2 imply that s

s (kv − v) ˙ vdr ˙ − Cˆ λ−α

{β − α(k + 1) − k}

α

s0

s → ∞.

(2.8)

for X, Y ≥ 0 with X + Y > 0

(2.9)

δ(r)v λ vdr ˙ = O(1) as s0

Now, since 0 < α ≤ 1, as stated in [1] for some constant c0 > 0 the inequality (X α − Y α )(X − Y ) ≥ c0 (X − Y )2 (X + Y )α−1 holds. Therefore we obtain {(kv)α − (kv − v) ˙ α }v˙ ≥ c0 {(kv) + (kv − v)} ˙ α−1 v˙ 2 . Employing (i) of Lemma 2.2, we have (kv − v) ˙ α v˙ ≤ −c1 v˙ 2 + kα v α v, ˙

(2.10)

where c1 > 0 is a constant. Let β − α(k + 1) − k > 0. (Recall our assumption β − α(k + 1) − k = 0.) From (2.8) and (2.10) we find that s −c1 {β − α(k + 1) − k}

v˙ 2 dr + {β − α(k + 1) − k}

kα α+1 v α+1

s0

s ≥ Cˆ λ−α

δ(r)v λ vdr ˙ + O(1)

as

s → ∞.

(2.11)

s0

Suppose firstly

∞

[ε(t)2 /t]dt < ∞, that is, (2.5) holds. Schwarz’s inequality and (2.11) imply that

s

s v˙ 2 dr ≤ c3 − c4

c2 s0

s0

⎛ δ(r)v λ vdr ˙ ≤ c3 + c5 ⎝

s s0

⎞1/2 ⎛ δ(r)2 dr⎠

⎝

s s0

⎞1/2 v˙ 2 dr⎠

H. Usami / J. Math. Anal. Appl. 433 (2016) 1486–1501

with some positive constants c2 , c3 , c4 and c5 . We therefore obtain ∞  |ε (t)|dt < ∞, that is, (2.6) holds. We find from (2.11) that s



s v˙ 2 dr ≤ c3 − c4

c2 s0

δ(r)

v λ+1 λ+1

∞

1493

v˙ 2 dr < ∞. Suppose next

 dr

s0

c4 δ(s)v λ+1 − c7 ≤ c6 − λ+1

s

s ˙ |δ(r)|dr,

λ+1 ˙ dr ≤ c8 + c9 δ(r)v s0

s0

∞ 2 where c6 , c7 , c8 and c9 are some positive constants. Hence we obtain v˙ dr < ∞. The case where β − α(k + 1) − k < 0 can be treated similarly. ∞ 2 Since we have just shown v˙ dr < ∞, and α ≤ 1, equation (2.3) shows that v¨ = O(1) as s → ∞. Therefore by Lemma 2.4 we find that lims→∞ v(s) ˙ = 0. Step 2. We show that lim inf s→∞ v(s) > 0. To see this by contradiction, we will derive a contradiction by assuming lim inf s→∞ v(s) = 0. The argument is divided into the two cases: Case (a): v(s) monotonically decreases to 0 (and so, v(s) ˙ ≤ 0). Case (b): v(s) ˙ changes the sign in any neighborhood of +∞, and lim inf s→∞ v(s) = 0. Let Case (a) occur. Put v(s) = x1 (s) and v(s) ˙ = x2 (s), and x = t (x1 , x2 ). Then from (2.4) x satisfies the binary system x˙ = Ax + f (s, x),

(2.12)

where

A=

 0 1 b −a

,

and

f (s, x) =



0 −bkα−1 {1 + δ(s)}(k|x1 | + |x2 |)1−α |x1 |λ

.

Here we have used the fact that v(s) > 0 and v(s) ˙ ≤ 0. Since (k|x1 | + |x2 |)1−α |x1 |λ ≤ (max{1, k})

1−α

(|x1 | + |x2 |)λ−α+1 ,

λ −α+1 > 1, and (v(s), v(s)) ˙ corresponds to a solution x(s) of system (2.12) satisfying lims→∞ x(s) = (0, 0), by Theorem 5 in Chapter 3 of [2], we have log (v(s), v(s)) ˙ = Λ, s→∞ s lim

(2.13)

where · denotes the length of vectors, and Λ is the real part of an eigenvalue of A. The characteristic equation of A is the same as that of the linear part of equation (2.4): X 2 −(2k+1−β/α)X+k(k+1−β/α) = 0. So the eigenvalues of A are k and k+1 −β/α; the former is positive and the latter negative by (1.4) and (1.7). Since (v(s), v(s)) ˙ → 0, we have Λ = −(β/α − k − 1). By assumption (1.4) there is a small η > 0 satisfying σ + λ(−β/α + 1) + λη < −1. We obtain therefore from (2.13)

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v(s) ≤ e{−(β/α−k−1)+η}s

near + ∞.

Returning to the original (t, u)-variables, we find that u(t) ≤ t−β/α+1+η near +∞. (See Lemma 2.2.) Then by (E) finally we have



t

t (−u (t)) ≤ c1 β

α

rσ+λ(−β/α+1)+λη dr = O(1)

as

t → ∞.

t0

This contradicts (i) of Lemma 2.1. Hence Case (a) never occurs. To show that Case (b) never occurs we adapt the argument in the proof of Theorem 1.3 in [6]. Suppose the contrary that Case (b) occurs. Let us introduce the auxiliary function Φ(s) given by Φ(s) = (1+δ(s))−1/(λ−α) which satisfies lims→∞ Φ(s) = 1. If v(s) ˙ = 0 for some s, then from (2.3) we get

v¨(s) = bv(s) 1 − (1 + δ(s))v(s)λ−α . Therefore, if v(s) ˙ = 0 in the region 0 < v < Φ(s), then v¨(s) > 0. Similarly, if v(s) ˙ = 0 in the region v > Φ(s), then v¨(s) < 0. From this simple observation, we find that the solution curve v = v(s) intersects the curve v = Φ(s) in any neighborhood of +∞, and that 0 = lim inf s→∞ v(s) < 1 ≤ lim sups→∞ v(s) < ∞. So there are two sequences {sn } and {sn } satisfying sn < sn < sn+1 ; v(s ˙ n ) = v(s ˙ n ) = 0;

lim s n→∞ n

= lim sn = ∞; n→∞

lim v(sn ) = 0; and v(sn ) ≥ Φ(sn ).

n→∞

Let m ∈ N be an arbitrary number satisfying m > α and we write α = m − ρ, ρ > 0. Let us fix m and ρ for a moment. By the Mean Value Theorem (kv − v) ˙ 1+ρ = (kv)1+ρ − (1 + ρ){kv − θ(s)v} ˙ ρ v˙ holds for some quantity θ(s) ∈ (0, 1). We obtain therefore from (2.4) multiplied by (kv − v) ˙ m v˙ v¨ ˙ v (kv − v) ˙ m + a(kv − v) ˙ m v˙ 2 − b(kv − v) ˙ m v v˙ + kα+ρ bv λ+ρ+1 v˙ − kα−1 b(1 + ρ){kv − θ(s)v} ˙ ρ v λ v˙ 2 + kα−1 bδ(s)(kv − v) ˙ 1+ρ v λ v˙ = 0. Employing the binominal expansion and integrating (2.14) on [sn , sn ], we have    m  m m−i k (−1)i v¨v˙ i+1 v m−i dr + a (kv − v) ˙ m v˙ 2 dr i i=0 sn

sn

sn

−b

m 

(−1)i

i=0

−k

α−1

sn

 sn sn m m−i v˙ i+1 v m−i+1 dr + kα+ρ b v λ+ρ+1 vdr ˙ k i sn

sn sn ρ λ 2 α−1 b(1 + ρ) {kv − θ(r)v} ˙ v v˙ dr + k b δ(r)(kv − v) ˙ 1+ρ v λ vdr ˙ = 0. sn

Since

∞

sn

v˙ 2 dr < ∞ and lims→∞ v(s) ˙ = 0, this implies that

sn

(2.14)

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 sn  sn sn m m m 3 m−1 m m m+1 α+ρ − v˙ v dr − k ˙ dr + k b v λ+ρ+1 vdr ˙ + o(1) = 0; b vv k 2 0 0 sn

sn

sn

that is, −

1 1 v(sn )m+2 + v(sn )λ+ρ+2 + o(1) = 0 as m+2 λ+ρ+2

n → ∞.

From this we have limn→∞ v(sn ) = {(m + 2 + λ − α)/(m + 2)}1/(λ−α) which implies that the limit of the sequence {v(sn )} does vary according to m ∈ N. However, recalling that m ∈ N is chosen arbitrarily, and the choice of m has no effect on the definition of {sn }, we find that this is an obvious contradiction. Hence lim inf s→∞ v(s) > 0. Step 3. We show that lims→∞ v(s) = 1. By the Mean Value Theorem we have (kv − v) ˙ 1−α = (kv)1−α − (1 − α){kv − θ(s)v} ˙ −α v˙ for some constant θ(s) ∈ (0, 1). Hence an integration of (2.4) multiplied by v˙ on [s0 , s] gives v˙ 2 +a 2

s

b bv 2−α+λ +b v˙ dr − v 2 + 2 2−α+λ

s

2

s0

s − (1 − α)bkα−1

δ(r)v 1−α+λ vdr ˙ s0

(1 + δ(r)){kv − θ(r)v} ˙ −α v˙ 2 v λ dr = const.

(2.15)

s0

By the results of Steps 1 and 2, all of the integrals in the left-hand side of (2.15) converge as s → ∞.   Therefore there is a finite limit lims→∞ −v 2 /2 + v 2−α+λ /(2 − α + λ) . Now let us consider the auxiliary function g(z) = −z 2 /2 + z λ−α+2 /(λ − α + 2), z > 0. The function g(z) decreases on the interval (0, 1], increases on the interval [1, ∞), and attains the minimum for z > 0 only at z = 1. As stated above, we have already known that lims→∞ g(v(s)) exists as a finite number. Below we will show that lims→∞ v(s) exists as a finite number. Let 0 < v(s) ≤ 1 near +∞, or v(s) ≥ 1 near +∞. Then by the monotonicity of g(z) on the intervals (0, 1] and [1, ∞), respectively, the existence of lims→∞ g(v(s)) implies the existence of lims→∞ v(s). Let v(s) take values both greater than 1 and less than 1 in any neighborhood of +∞. To see the existence of lims→∞ v(s) let us suppose the contrary that lims→∞ v(s) does not exist. Then, we can find two sequences {cn }, {cn } and a positive number l = 1 satisfying cn < cn < cn+1 ; v(cn ) ≡ 1;

lim cn = lim cn = ∞;

n→∞

n→∞

lim v(cn ) = l.

n→∞

Then, as before, an integration on [cn , cn ] of (2.4) multiplied by v˙ gives bg(v(cn )) − bg(v(cn )) + o(1) = 0 as n → ∞. By letting n → ∞, we have g(l) − g(1) = 0, an obvious contradiction. So lims→∞ v(s) exists. Put L = lims→∞ v(s). Since lims→∞ v˙ = 0, equation (2.4) shows that lims→∞ v¨(s) = b(L − L1+λ−α ). The property lims→∞ v˙ = 0 further implies lims→∞ v¨(s) = 0. So we finally find that L = 1. This completes the proof of Theorem 1.2. 2

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Proof of Theorem 1.3. Essentially as in the proof of Theorem 1.2, we will show that lims→∞ v(s) = 1, where v(s) is the function introduced in Lemma 2.2. Set w = (kv − v) ˙ α.

(2.16)

Recall that k > 0 is defined by (1.7). By equation (2.2) we know that w˙ + {β − α(k + 1)}w − Cˆ λ−α {1 + δ(s)}v λ = 0. Let us rewrite this equation as w˙ + aw − b{1 + δ(s)}v λ = 0,

(2.17)

where we have put β − α(k + 1) = a(> 0) and Cˆ λ−α = b(> 0). Since (2.17) gives v = b−1/λ (1 + δ(s))−1/λ (w˙ + aw)1/λ , (2.16) implies that w(s) satisfies the equation ((1 + δ(s))−1/λ (w˙ + aw)1/λ ) − k(1 + δ(s))−1/λ (w˙ + aw)1/λ + b1/λ w1/α = 0.

(2.18)

Note, by the definition (2.16), equation (2.17), and (i) of Lemma 2.2, that w, w˙ = O(1) as s → ∞. By putting (1 + δ(s))−1/λ = h(s),

1/λ = ρ,

and 1/α = γ,

we can rewrite (2.18) simply as (h(s)(w˙ + aw)ρ ) − kh(s)(w˙ + aw)ρ + bρ wγ = 0.

(2.19)

We notice that ρ < γ < 1 and lims→∞ h(s) = 1 from the standing assumptions. Our assumptions (1.10) and (1.9) are equivalent to ˙ lim h(s) = 0,

s→∞

(2.20)

and ∞ ˙ |h(s)|ds < ∞,

(2.21)

respectively. It should be noted that equation (2.19) is equivalent to  ˙   ˙ a h(s) bρ k h(s) w˙ + −k w+ (w˙ + aw)1−ρ wγ = 0. w ¨+ a− + ρ ρh(s) ρ h(s) ρh(s)

(2.22)

The definition (2.16) is equivalent to (e−ks v)· = −e−ks w1/α ; and so ∞ v(s) = eks s

e−kr w1/α dr.

(2.23)

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(Here we have employed the fact that lims→∞ v(s)/eks = 0.) To see lims→∞ v(s) = 1, it suffices to show that lims→∞ w(s) = k α . In fact, if we can see so, then by (2.23) L’Hôpital’s rule implies that lims→∞ v(s) = 1. The proof is divided into several steps. ∞ Step 1. We show that w(s) ˙ 2 ds < ∞ and w(s) ˙ → 0 as s → ∞. We multiply (2.19) by w, ˙ and integrate the resulting equation on [s0 , s] to obtain s

(h(r)(w˙ + aw)ρ ) wdr ˙ −k

s0

s

s h(r)(w˙ + aw)ρ wdr ˙ + bρ

s0

wγ wdr ˙ = const. s0

We obtain from integration by parts 1 ρ h(s)(w˙ + aw)ρ+1 + ρ+1 ρ+1

s ˙ h(r)( w˙ + aw)ρ+1 dr s0

s h(r)(w˙ + aw)ρ wdr ˙ − ah(s)(w˙ + aw)ρ w +

+ (a − k)

bρ w(s)γ+1 = const. γ+1

s0

By the properties w, ˙ w = O(1) and assumptions (2.20) and (2.21) we find that s (a − k)

h(r)(w˙ + aw)ρ wdr ˙ = O(1) as

s → ∞.

(2.24)

s0

Notice that the assumption β − α(k + 1) − k = 0 means that a − k = 0. Since ρ < 1, inequality (2.9) implies, as before, that {(w˙ + aw)ρ − (aw)ρ )}w˙ ≥ c0 w˙ 2 {(w˙ + aw) + (aw)}ρ−1 ; namely, h(r)(w˙ + aw)ρ w˙ ≥ aρ h(r)wρ w˙ + c1 h(r)w˙ 2 for some constant c1 > 0. Hence by (2.24) and the fact that h(∞) = 1, we find that s

s w˙ 2 dr ≤ c4

ρ

c2

h(r)w wdr ˙ + c3 s0

as

s → +∞

s0

∞ 2 for some positive constants c2 , c3 and c4 . From integration by parts we find that w˙ ds < ∞. Moreover, since ρ < 1, (2.22) implies that w ¨ = O(1). Consequently we see from Lemma 2.4 that lims→∞ w(s) ˙ = 0. Step 2. We show that lim inf s→∞ w(s) > 0. To see this by contradiction, we will derive a contradiction by assuming lim inf s→∞ w(s) = 0. The argument is divided into the two cases as before: Case (a): w(s) monotonically decreases to 0 (and so, w(s) ˙ ≤ 0). Case (b): w(s) ˙ changes the sign in any neighborhood of +∞, and lim inf s→∞ w(s) = 0. Let Case (a) occur. Put w(s) = x1 (s) and w(s) ˙ = x2 (s), and x = t (x1 , x2 ). Then from (2.22) x satisfies the binary system

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x˙ = (A + B(s))x + f (s, x),

(2.25)

where

A=

B(s) =

 0 1 ak/ρ k/ρ − a 0

0

˙

˙

h(s) h(s) − ρh(s) − aρh(s)

,  ,

and

f (s, x) =

0 1−ρ bρ |ax1 + x2 | |x1 |γ − ρh(s)

 .

Since 1−ρ

|ax1 + x2 |

|x1 |γ ≤ (max{1, a})1−ρ (|x1 | + |x2 |)1−ρ+γ ,

∞ ˙ 1 − ρ + γ > 1, |h(s)|ds < ∞, and (w(s), w(s)) ˙ corresponds to a solution x(s) of system (2.25) satisfying lims→∞ x(s) = (0, 0), by Theorem 5 in Chapter 3 of [2], we have lim

s→∞

log (w(s), w(s)) ˙ = Λ. s

Here Λ is the real part of an eigenvalue of A. The characteristic equation of A is X 2 −(k/ρ −a)X −ka/ρ = 0. So the eigenvalues of A are k/ρ and −a(= α(k + 1) − β); the former is positive and the latter negative. Since (w(s), w(s)) ˙ → 0, we have Λ = −a. Therefore for every η > 0 we obtain w(s) ≤ e(−β+α(k+1)+η)s

near

+ ∞.

(2.26)

Combining (2.26) with (2.23), we can get the estimate for solution u(t): tβ |u (t)|α = O(1), as in Step 2 of the proof of Theorem 1.2. Recall that this yields a contradiction to (i) of Lemma 2.1. Hence Case (a) never occurs. To show that Case (b) never occurs suppose the contrary that Case (b) occurs. Let us introduce the auxiliary function H(s) by  H(s) = k

α

λ−α ˙ h(s) h(s) − k λα

which satisfies H(∞) = kα by (2.20). If w(s) ˙ = 0 for some s, then from (2.22) we have   ˙ h(s) ka (α−λ)/λ (λ−α)/(αλ) . w(s) h(s) − −k w(s) ¨ = w(s) ρh(s) k Therefore, if w(s) ˙ = 0 in the region 0 < w < H(s), then w(s) ¨ > 0; if w(s) ˙ = 0 in the region w > H(s), then w(s) ¨ < 0. From this observation, we find that the curve w = w(s) intersects the curve w = H(s) in any neighborhood of +∞, and that 0 = lim inf s→∞ v(s) < kα ≤ lim sups→∞ v(s) < ∞. So we can find two sequences {ξn } and {ηn } satisfying ξn < ηn < ξn+1 < ηn+1 < · · · ;

lim ξn = lim ηn = ∞;

n→∞

n→∞

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and w(ηn ) → 0,

w(ξn ) = H(ξn ) → kα

as n → ∞

and w˙ ≤ 0 on [ξn , ηn ].

Multiplying (2.22) by w˙ and integrating the resulting equation on [ξn , ηn ], we have ηn  ηn ˙ k 1 1 h(r) 2 2 2 2 (w(η ˙ n ) − w(ξ w˙ dr ˙ n) ) + a − w˙ dr + 2 ρ ρ h(r) ξn

a ak (w(ηn )2 − w(ξn )2 ) + − 2ρ ρ

ηn

ξn

ηn ˙ bρ 1 h(r) wwdr ˙ + (w˙ + aw)1−ρ wγ wdr ˙ = 0. h(r) ρ h(r)

ξn

Noting the facts that w(∞) ˙ = 0 and

∞

ξn

w˙ 2 dr < ∞, and that w˙ ≤ 0 on [ξn , ηn ], we have as n → ∞

ηn ˙ ak h(r) 2 w˙ dr − (o(1) − k 2α ) h(r) 2ρ

1 o(1) + ρ

ξn

ηn ˙ ηn a1−ρ bρ a 1 h(r) wwdr ˙ + w1+γ−ρ wdr ˙ ≤ 0. + ρ h(r) ρ h(r) ξn

(2.27)

ξn

Now, let us estimate each term of the above. We have firstly     ηn ∞   h(r) ˙  2  ˙ w˙ dr ≤ C0 sup |h| w˙ 2 dr = o(1)    h(r) [ξn ,∞)  ξn ξn

as

n → ∞;

and        ηn   ηn      h(r)   h(c ˙ n) ˙ n) ˙    h(c   wwdr ˙ = (w(ξn )2 − w(ηn )2 ) = o(1) wwdr ˙  =     h(cn )  h(r)  2h(cn )   ξn  ξn

as

n → ∞.

Here C0 > 0 is a constant, and cn , n ∈ N, are numbers satisfying ξn < cn < ηn . (We have used a variant of the Mean Value Theorem for integrals.) Finally, we obtain ηn ξn

1 w1+γ−ρ wdr ˙ = h(r)

ηn [h(r)−1 − 1]w1+γ−ρ wdr ˙ + ξn −1

= (h(dn )

1 (w(ηn )2+γ−ρ − w(ξn )2+γ−ρ ) 2+γ−ρ

ηn − 1)

w1+γ−ρ wdr ˙ + ξn

= o(1) −

α(2+γ−ρ)

k 2+γ−ρ

as

1 (o(1) − kα(2+γ−ρ) ) 2+γ−ρ

n → ∞.

Here dn , n ∈ N, are numbers satisfying ξn < dn < ηn . Therefore (2.27) can be simplified into a1−ρ bρ kα(2+γ−ρ) ak2α+1 + o(1) ≤ 2ρ ρ(2 + γ − ρ)

as n → ∞,

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which is equivalent to the inequality 1 1 + o(1) ≤ 2 2+γ−ρ

as

n → ∞.

Since 2 + γ − ρ > 2, this is an obvious contradiction. Hence Case (b) does not occur. Step 3. We show that lims→∞ w(s) = kα . By the Mean Value Theorem we have (w˙ + aw)1−ρ = (aw)1−ρ + (1 − ρ)(aw + θ(s)w) ˙ −ρ w˙ for some constant θ(s) ∈ (0, 1). Hence an integration of (2.22) multiplied by w˙ on [s0 , s] gives 1 2 w˙ + 2

 s  s ˙ ˙ k h(r) a h(r) 2 a− + w˙ dr + wwdr ˙ ρ ρh(r) ρ h(r) s0

s0

⎡

bρ a1−ρ ⎣ 1 wγ−ρ+2 ak w2 · + + − ρ 2 ρ h(s) γ − ρ + 2 +

bρ (1 − ρ) ρ

s

⎤ s ˙ h(r) wγ−ρ+2 dr⎦ h(r)2 γ − ρ + 2

s0

1 (aw + θ(r)w) ˙ −ρ w˙ 2 dr = const. h(r)

s0

By the results of Steps 1 and 2, all integrals in the left-hand side of this formula converge as s → ∞. Noting that h(∞) = 1, we find that there is a finite limit   lim −akw2 /(2ρ) + bρ a1−ρ wγ−ρ+2 /{ρ(γ − ρ + 2)} .

s→∞

The existence of this limit is equivalent to the existence of the limit lims→∞ (−w2 /2 + k− λ wγ−ρ+2 / λ−α (γ − ρ + 2)). Let us consider the auxiliary function g(z) = −z 2 /2 + k− λ z γ−ρ+2 /(γ − ρ + 2), z > 0. The function g(z) decreases on the interval (0, kα ], increases on the interval [kα , ∞), and attains the minimum for z > 0 only at z = kα . We have already known that lims→∞ g(w(s)) exists as a finite number. We will show that lims→∞ w(s) exists as a finite number. (The argument is essentially the same as in Step 3 of Proof of Theorem 1.2.) Let 0 < w(s) ≤ kα near +∞, or w(s) ≥ kα near +∞. Then by the monotonicity of g(z) on the intervals (0, kα ] and [kα , ∞), respectively, the existence of lims→∞ g(w(s)) implies the existence of lims→∞ w(s). Let w(s) take values both greater than kα and less than kα in any neighborhood of +∞. To see the existence of lims→∞ w(s) let us suppose the contrary that lims→∞ w(s) does not exist. Then, we can find two sequences {cn }, {cn } and a positive number l = kα satisfying λ−α

cn < cn < cn+1 ;

lim cn = lim cn = ∞;

n→∞ α

w(cn ) ≡ k ;

n→∞

and

lim w(cn ) = l.

n→∞

Then, as before, an integration on [cn , cn ] of (2.22) multiplied by w˙ gives ak [g(w(cn )) − g(w(cn ))] + o(1) = 0 as ρ

n → ∞.

By letting n → ∞, we have g(l) − g(kα ) = 0, an obvious contradiction. So lims→∞ w(s) exists.

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Put L = lims→∞ w(s) ∈ (0, ∞). Since lims→∞ w˙ = 0, equation (2.22) shows that lim w(s) ¨ =

s→∞

ak L(1 − k αρ−1 Lγ−ρ ). ρ

The property lims→∞ w˙ = 0 implies also lims→∞ w ¨ = 0. So we finally find that 1 − kαρ−1 Lγ−ρ = 0; namely, α L = k as desired. 2 Acknowledgment The author would like to thank the referee for his or her careful reading of the manuscript and many helpful comments. References [1] M.-F. Bidaut-Veron, Local and global behavior of solutions of quasilinear equations of Emden–Fowler type, Arch. Ration. Mech. Anal. 107 (1988) 293–324. [2] W.A. Coppel, Stability and Asymptotic Behavior of Differential Equations, D.C. Heath and Company, Boston, 1965. [3] O. Došlý, P. Řehák, Half-Linear Differential Equations, Elsevier, Amsterdam, 2005. [4] Y. Furusho, T. Kusano, A. Ogata, Symmetric positive entire solutions of second-order quasilinear degenerate elliptic equations, Arch. Ration. Mech. Anal. 127 (1994) 231–254. [5] M. Guedda, L. Veron, Local and global properties of solutions of quasilinear elliptic equations, J. Differential Equations 76 (1988) 159–189. [6] K. Kamo, H. Usami, Asymptotic forms of weakly increasing positive solutions for quasilinear ordinary differential equations, Electron. J. Differential Equations 2007 (126) (2007) 1–12. [7] N. Kawano, E. Yanagida, S. Yotsutani, Structure theorems for positive radial solutions to div(|Du|m−2 Du) + K(|x|)uq = 0 in Rn , J. Math. Soc. Japan 45 (1993) 719–742. [8] T. Kusano, A. Ogata, H. Usami, Oscillation theory for a class of second order quasilinear ordinary differential equations with application to partial differential equations, Jpn. J. Math. 19 (1993) 131–147.