Let p>1, we take up the existence, the uniqueness and the asymptotic behavior of a positive continuous solution to the following nonlinear problem in (0,+∞), (1/A)(Aϕp(u′))′+q(x)uα=0, limx→0Aϕp(u′)(x)=0, limx→+∞u(x)=0, where α<p−1, ϕp(t)=t|t|p−2 (t∈ℝ), A is a positive differentiable function in (0,+∞) and q is a positive continuous function in (0,+∞) such that there exists c>0 satisfying for each x in (0,+∞), 1/c≤q(x)(1+x)βexp(−∫1x+1(z(s)/s)ds)≤c, β≥p and z∈C([1,+∞)) such that limt→+∞z(t)=0.

1. Introduction

Let p>1 and α<p-1. The following differential equation:
(1)Lpu:=1A(AΦp(u′))′=-q(x)uα,in(0,ω),ω∈(0,+∞]
has been studied with various boundary conditions, where A is a continuous function in [0,+∞), differentiable and positive in (0,+∞), q is a nonnegative continuous function in (0,+∞), and Φp(t)=t|t|p-2, (t∈ℝ) (see [1–15]).

For a function u depending only on t=|x|, x∈ℝn, and A(t)=tn-1, the operator Lp is the radially symmetric p-Laplacian Δpu=div(|∇u|p-2∇u) in ℝn and the equation of type (1) arises also as radial solutions of the Monge-Ampère equation (see [13]).

In this paper, our main purpose is to obtain the existence of a unique positive solution to the following boundary value problem:
(P)1A(AΦp(u′))′+q(x)uα=0,in(0,∞),AΦp(u′)(0):=limx→0+AΦp(u′)(x)=0,limx→+∞u(x)=0,
and to establish estimates on such solution under an appropriate condition on q.

The study of this type of (1) is motivated by [5, 15]. Namely, in the special case A(t)=tn-1g(t), g∈C1(ℝ+) and f(u)=uα, α>-1, the authors in [15] studied (1) and gave some uniqueness results. In this work, we consider a wider class of weights A and we aim to extend the study of (1) in [15], to α<p-1.

For the case α<0, the problem (P) has been studied in [5]. Indeed, the authors of [5] proved the following existence result.

Theorem 1.

The problem (P) has a unique positive solution u∈C([0,+∞)) satisfying for each x∈(0,+∞)(2)1cGpq(x)≤u(x)≤c(Gpq(x))(p-1)/(p-1-α),
where c is a positive constant and Gpq is the function defined on (0,+∞) by
(3)Gpq(x):=∫x+∞(1A(t)∫0tA(s)q(s)ds)1/(p-1)dt.

We shall improve in this paper the above asymptotic behavior of the solution of problem (P) and we extend the study of (P) to 0≤α<p-1.

The pure elliptic problem of type
(Q)-Δu+q(x)uα=0,x∈ℝn(n≥3),α<1,u>0inℝn,lim|x|→+∞u(x)=0
has been investigated by several authors with zero Dirichlet boundary value; we refer the reader to [16–24] and the references therein. More recently, applying Karamata regular variation theory, Chemmam et al. gave in [17] the asymptotic behavior of solutions of problem (Q). In this work, we aim to extend the result established in [17] to the radial case associated to problem (P).

To simplify our statements, we need to fix some notations and make some assumptions. Throughout this paper, we shall use 𝒦, to denote the set of Karamata functions L defined on [1,+∞) by
(4)L(t):=cexp(∫1tz(s)sds),
where c is a positive constant and z∈C([1,∞[) such that limt→+∞z(t)=0.

It is clear that L∈𝒦 if and only if L is a positive function in C1([1,∞[) such that
(5)limt→∞tL′(t)L(t)=0.
For two nonnegative functions f and g on a set S, we write f(x)≈g(x), x∈S, if there exists a constant c>0 such that (1/c)g(x)≤f(x)≤cg(x), for each x∈S.

The letter c will denote a generic positive constant which may vary from line to line.

Furthermore, we point out that if f is a nonnegative continuous function in [0,+∞[, then the function Gpf defined on (0,+∞) by
(6)Gpf(x):=∫x+∞(1A(t)∫0tA(s)f(s)ds)1/(p-1)dt
is the solution of the problem
(7)-1A(AΦp(u′))′=f,in(0,+∞),AΦp(u′)(0)=0,limx→+∞u(x)=0.

As it is mentioned above, our main purpose in this paper is to establish existence and global behavior of a positive solution of problem (P). Let us introduce our hypotheses.

Here, the function A is continuous in [0,+∞), differentiable and positive in (0,+∞) such that
(8)A(x)≈xλ
with λ>p-1.

The function q is required to satisfy the following hypothesis.

(H) q is a positive measurable function on (0,+∞) such that
(9)q(x)≈L(1+x)(1+x)β,
with β≥p and the function L∈𝒦 such that ∫1+∞t(1-β)/(p-1)(L(t))1/(p-1)dt<+∞.

Remark 2.

We need to verify condition
(10)∫1+∞t(1-β)/(p-1)(L(t))1/(p-1)dt<+∞
in hypothesis (H), only if β=p, (see Lemma 6 below).

As a typical example of function q satisfying (H), we quote the following.

Example 3.

Put q(x):=(1+x)-β(log(x+2))-ν,x∈(0,+∞). Then for β>p and ν∈ℝ or β=p and ν>p-1, the function q satisfies (H).

Now, we are ready to state our main result.

Theorem 4.

Assume (H). Then problem (P) has a unique positive continuous solution u satisfying for x∈(0,+∞),
(11)u(x)≈θβ(x),
where θβ is the function defined on [0,+∞) by
(12)θβ(x):={(∫x+1∞(L(s))1/(p-1)sds)(p-1)/(p-1-α),ifβ=p,(L(1+x))1/(p-1-α)(1+x)(β-p)/(p-1-α),ifp<β<(λ+1)(p-1-α)+αpp-1,(1+x)(p-1-λ)/(p-1),ifβ>(λ+1)(p-1-α)+αpp-1,(1+x)(p-1-λ)/(p-1)(∫1x+2L(s)sds)1/(p-1-α),ifβ=(λ+1)(p-1-α)+αpp-1.

The main body of the paper is organized as follows. In Section 2, we establish some estimates and we recall some known results on functions belonging to 𝒦. Theorem 4 is proved in Section 3. The last Section is reserved to some applications.

2. Key Estimates

In what follows, we are going to give estimates on the functions Gpq and Gp(qθβα), where q is a function satisfying (H) and θβ is the function given by (12). First, we recall some fundamental properties of functions belonging to the class 𝒦, taken from [17, 22].

Lemma 5.

Let L1,L2∈𝒦,m∈ℝ and ε>0. Then one has L1L2∈𝒦,L1m∈𝒦 and limt→+∞t-εL1(t)=0.

Lemma 6 (Karamata’s theorem).

Let μ∈ℝ and L be a function in 𝒦. Then one has the following properties:

If μ<-1, then ∫1∞sμL(s)ds converges and
(13)∫t∞sμL(s)ds~t→∞-t1+μL(t)μ+1.

If μ>-1, then ∫1∞sμL(s)ds diverges and
(14)∫1tsμL(s)ds~t→∞t1+μL(t)μ+1.

Lemma 7.

Let L∈𝒦. Then there exists m>0 such that for t≥1 and c>0(15)1(1+c)mL(t)≤L(c+t)≤(1+c)mL(t).

Lemma 8.

Let L∈𝒦, then one has
(16)t→∫1t+1L(s)sds∈𝒦.
If further ∫1∞(L(s)/s)ds converges, one has
(17)t→∫t∞L(s)sds∈𝒦.

Now, we are able to prove the following propositions which play a crucial role in this paper.

Proposition 9.

Let q be a function satisfying (H). Then one has for x∈(0,∞)(18)Gpq(x)≈Ψ(x),
where ψ is the function defined by
(19)Ψ(x):={∫x+1∞(L(s))1/(p-1)sds,ifβ=p,(L(x+1))1/(p-1)(x+1)(β-p)/(p-1),ifp<β<λ+1,(x+1)(p-1-λ)/(p-1)(∫1x+2L(s)sds)1/(p-1),ifβ=λ+1,(x+1)(p-1-λ)/(p-1),ifβ>λ+1.

Proof.

For x∈(0,∞), we have
(20)Gpq(x)≈∫x∞tλ/(1-p)(∫0tsλ(1+s)-βL(1+s)ds)1/(p-1)dt.
Put
(21)h(x)=:∫x∞tλ/(1-p)(∫0tsλ(1+s)-βL(1+s)ds)1/(p-1)dt,x∈(0,∞).
To prove the result, it is sufficient to show that h(x)≈Ψ(x) for x∈(0,∞).

Since the function h is continuous and positive in [0,1], we have
(22)h(x)≈1,x∈[0,1].
Now, assume that x∈[1,∞), then we have
(23)h(x)=∫x∞tλ/(1-p)(∫01sλ(1+s)-βL(1+s)ds+∫1tsλ(1+s)-βL(1+s)ds)1/(p-1)dt.
It follows from Lemma 7 that
(24)h(x)≈∫x∞tλ/(1-p)(1+∫1tsλ-βL(s)ds)1/(p-1)dt.
To reach our estimates, we consider the following cases.

(i) If p<β<λ+1, then it follows from Lemma 6 that
(25)h(x)≈∫x∞t(1-β)/(p-1)L(t)1/(p-1)dt.
Now, using Lemma 5 and again Lemma 6, we deduce that
(26)h(x)≈(L(x))1/(p-1)x(β-p)/(p-1).

(ii) If β>λ+1, then it follows from Lemma 6, that ∫1∞sλ-βL(s)ds<∞. So, since λ>p-1, we have
(27)h(x)≈x(p-λ-1)/(p-1).

(iii) If β=λ+1, then for each t∈[x,+∞), we have
(28)(1+∫1tL(s)sds)1/(p-1)≈(∫1t+1L(s)sds)1/(p-1).
Since the function t↦(∫1t+1(L(s)/s)ds)1/(p-1) is in 𝒦, then using the fact that λ>p-1 and Lemma 6, we obtain that
(29)h(x)≈x(p-1-λ)/(p-1)(∫1x+1L(s)sds)1/(p-1).

(iv) If β=p, we have by Lemma 6 that
(30)∫1tsλ-pL(s)ds≈tλ-p+1L(t),
this yields to
(31)h(x)≈∫x∞L1/(p-1)(t)tdt.
Hence, we reach the result by combining (22) with the estimates stated in each case above. This completes the proof.

Proposition 10.

Let q be a function satisfying (H) and let θβ be the function given by (12). Then for x∈(0,∞), one has
(32)Gp(qθβα)(x)≈θβ(x).

Proof.

Let β≥p and λ>p-1, we obtain by simple calculus that for x∈(0,∞),
(33)q(x)θβα(x)≈q~(x),
where
(34)q~(x)=:{L(1+x)(1+x)p(∫x+1∞(L(s))1/(p-1)sds)α(p-1)/(p-1-α),ifβ=p,(L(1+x))(p-1)/(p-1-α)(1+x)(β-α(p-β)/(p-1-α)),ifp<β<(λ+1)(p-1-α)+αpp-1,L(1+x)(1+x)(β-α(p-1-λ)/(p-1)),ifβ>(λ+1)(p-1-α)+αpp-1,L(1+x)(1+x)(∫1x+2L(s)sds)α/(p-1-α),ifβ=(λ+1)(p-1-α)+αpp-1.
So, one can see that
(35)q~(x)=(1+x)-δL~(1+x),
where δ≥p. Then, using Lemmas 5, 7, and 8, we obtain that L~∈𝒦 and ∫1∞t(1-δ)/(p-1)(L~(t))1/(p-1)dt<+∞. Hence, it follows from Proposition 9 that
(36)Gp(qθβα)(x)≈Gpq~(x)≈ψ~(x),x∈(0,∞),
where ψ~ is the function defined in (19) by replacing L by L~ and β by δ. This ends the proof.

3. Proof of Theorem <xref ref-type="statement" rid="thm2">4</xref>3.1. Existence and Asymptotic Behavior

Let q be a function satisfying (H) and let θβ be the function given by (12). By Proposition 10, there exists a constant m≥1 such that for each x∈(0,∞)(37)1mθβ(x)≤Gp(qθβα)(x)≤mθβ(x).
We look now at the existence of positive solution of problem (P) satisfying (11).

For the case α<0, the existence of a positive continuous solution to problem (P) is due to [5]. Now, we look to the existence result of problem (P) when 0≤α<p-1 and we give precise asymptotic behavior of such solution for α<p-1. For that, we split the proof into two cases.

Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M217"><mml:mi>α</mml:mi><mml:mo><</mml:mo><mml:mn mathvariant="normal">0</mml:mn></mml:math></inline-formula>).

Let u be a positive continuous solution of problem (P). So, in order to obtain estimates (11) on the function u, we need the following comparison result.

Lemma 11.

Let α<0 and u1,u2∈C1((0,∞))∩C([0,∞)) such that
(38)-Lpu1(x)≤q(x)u1α,in(0,∞),AΦp(u1′)(0)=0,limx→+∞u1(x)=0,(39)-Lpu2(x)≥q(x)u2α,in(0,∞),AΦp(u2′)(0)=0,limx→+∞u2(x)=0.
Then u1≤u2.

Proof.

Suppose that u1(x0)>u2(x0) for some x0∈(0,∞). Then there exists x1,x2∈[0,∞], such that 0≤x1<x0<x2≤∞ and for x1<x<x2,u1(x)>u2(x) with u1(x2)=u2(x2),u1(x1)=u2(x1) or x1=0.

By an elementary argument, we have
(40)AΦp(u2′)(x1)≤AΦp(u1′)(x1).
On the other hand, since α<0, then u1α(x)<u2α(x), for each x∈(x1,x2). This yields to
(41)Lpu1-Lpu2≥q(u2α-u1α)≥0,on(x1,x2).

Using further (40), we deduce that the function ω(x):=(AΦp(u1′)-AΦp(u2′))(x) is nondecreasing on (x1,x2) with ω(x1)≥0. Hence, from the monotonicity of Φp, we obtain that the function x↦(u1-u2)(x) is nondecreasing on (x1,x2) with (u1-u2)(x1)≥0 and (u1-u2)(x2)=0. This yields to a contradiction, which completes the proof.

Now, we are ready to prove (11). Put c=m-α/(p-1-α) and v:=Gp(qθβα). It follows from (7) that the function v satisfies
(42)-Lpv=qθβα,in(0,∞).
According to (37), we obtain by simple calculation that (1/c)v and cv satisfy, respectively, (38) and (39). Thus, we deduce by Lemma 11 that
(43)1cv(x)≤u(x)≤cv(x),x∈(0,∞).
This implies (11) by using (37).

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M252"><mml:mn mathvariant="normal">0</mml:mn><mml:mo>≤</mml:mo><mml:mi>α</mml:mi><mml:mo><</mml:mo><mml:mi>p</mml:mi><mml:mo>-</mml:mo><mml:mn mathvariant="normal">1</mml:mn></mml:math></inline-formula>).

Put c0=m(p-1)/(p-1-α) and let
(44)Λ:={u∈C((0,∞));1c0θβ≤u≤c0θβ}.
Obviously, the function θβ belongs to C((0,∞)) and so Λ is not empty. We consider the integral operator T on Λ defined by
(45)Tu(x):=Gp(quα)(x),x∈(0,∞).
We shall prove that T has a fixed point in Λ, to construct a solution of problem (P). For this aim, we look at first that TΛ⊂Λ. Let u∈Λ, then we have for each x∈(0,∞)(46)1c0α(qθβα)(x)≤q(x)uα(x)≤c0α(qθβα)(x).
This together with (37) implies that
(47)1mc0α/(p-1)θβ≤Tu≤mc0α/(p-1)θβ.
Since mc0α/(p-1)=c0 and TΛ⊂C((0,∞)), then T leaves invariant the convex Λ. Moreover, since α≥0, then the operator T is nondecreasing on Λ. Now, let {uk}k be the sequence of functions in C([0,+∞)) defined by
(48)u0=1c0θβ,uk+1=Tuk,fork∈ℕ.
Since TΛ⊂Λ, we deduce from the monotonicity of T that for k∈ℕ, we have
(49)u0≤u1≤⋯≤uk≤uk+1≤c0θβ.
Thanks to the monotone convergence theorem, we deduce that the sequence {uk}k converges to a function u∈Λ which satisfies
(50)u(x)=Gp(quα)(x),x∈(0,∞).
We conclude that u is a positive continuous solution of problem (P) satisfying (11).

3.2. Uniqueness

Assume that q satisfies (H). For α<0, the uniqueness of solution to problem (P) follows from Lemma 11. Thus in the following, we look at the case 0≤α<p-1. Let
(51)Γ={u∈C((0,∞)):u(x)≈θβ(x)}.
Let u and v be two positives solutions of problem (P) in Γ. Then there exists a constant k≥1 such that
(52)1k≤vu≤k.
This implies that the set
(53)J={t∈(1,+∞),1tu≤v≤tu}
is not empty. Now, put c:=infJ, then we aim to show that c=1.

Suppose that c>1, then we have
(54)-Lpv+Lp(c-α/(p-1)u)=q(x)(vα-c-αuα),in(0,∞),limx→0+(AΦp(v′)-AΦp(c-α/(p-1)u′))(x)=0,limx→+∞(v-c-α/(p-1)u)(x)=0.
So, we have -Lpv+Lp(c-α/(p-1)u)≥0 in (0,∞), which implies that the function
(55)θ(x):=(AΦp(c-α/(p-1)u′)-AΦp(v′))(x)
is nondecreasing on (0,+∞) with limx→0+θ(x)=0. Hence from the monotonicity of Φp, we obtain that the function x↦(c-α/(p-1)u-v)(x) is nondecreasing on [0,+∞) with limx→∞(c-α/(p-1)u-v)(x)=0. This implies that c-α/(p-1)u≤v. On the other hand, we deduce by symmetry that v≤cα/(p-1)u. Hence cα/(p-1)∈J. Now, since α<p-1 and c>1, we have cα/(p-1)<c. This yields to a contradiction with the fact that c:=infJ. Hence, c=1 and then u=v.

4. Applications4.1. First Application

Let q be a positive measurable function in (0,∞) satisfying for x∈(0,∞)(56)q(x)≈(log(2+x))-σ(1+x)β,
where the real numbers β and σ satisfy one of the following two conditions:

β>p and σ∈ℝ,

β=p and σ>p-1.

Using Theorem 4, we deduce that problem (P) has a positive continuous solution u in (0,∞) satisfying the following.

If β>((λ+1)(p-1-α)+αp)/(p-1), then for x∈(0,∞)(57)u(x)≈1(1+x)(λ-p+1)/(p-1).

If β=((λ+1)(p-1-α)+αp)/(p-1) and σ=1, then for x∈(0,∞)(58)u(x)≈1(1+x)(λ-p+1)/(p-1)(loglog(3+x))1/(p-1-α).

If β=((λ+1)(p-1-α)+αp)/(p-1) and σ<1, then for x∈(0,∞)(59)u(x)≈1(1+x)(λ-p+1)/(p-1)(log31+x)(1-σ)/(p-1-α).

If β=((μ+1)(p-1-α)+αp)/(p-1) and σ>1, then for x∈(0,∞)(60)u(x)≈1(1+x)(λ-p+1)/(p-1).

If p<β<((λ+1)(p-1-α)+αp)/(p-1), then for x∈(0,∞)(61)u(x)≈1(1+x)(β-p)/(p-1-α)(log(2+x))-σ/(p-1-α).

If β=p and σ>p-1, then for x∈(0,∞)(62)u(x)≈(log(2+x))(p-1-σ)/(p-1-α).

4.2. Second Application

Let q be a function satisfying (H) and let α,γ<p-1. We are interested in the following nonlinear problem:
(63)-1A(AΦp(u′))′+γuΦp(u′)u′=q(x)uα,in(0,∞),AΦp(u′)(0)=0,limx→∞u(x)=0.
Put v=u1-γ/(p-1), then by a simple calculus, we obtain that v satisfies
(64)-1A(AΦp(v′))′=(p-1-γp-1)p-1q(x)v(α-γ)(p-1)/(p-1-γ),in(0,∞),AΦp(v′)(0)=0,limx→∞v(x)=0.
Using Theorem 4, we deduce that problem (64) has a unique solution v such that
(65)v(x)≈θ~β(x):={(∫x+1∞(L(s))1/(p-1)sds)(p-1-γ)/(p-1-α),ifβ=p,1(1+x)(β-p)(p-1-γ)/((p-1)(p-1-α))×(L(1+x))(p-1-γ)/((p-1)(p-1-α)),ifp<β<(λ+1)(p-1-α)+(α-γ)pp-1-γ,1(1+x)(λ-p+1)/(p-1),ifβ>(λ+1)(p-1-α)+(α-γ)pp-1-γ,1(1+x)(λ-p+1)/(p-1)×(∫1x+2L(s)sds)(p-1-γ)/((p-1)(p-1-α)),ifβ=(λ+1)(p-1-α)+(α-γ)pp-1-γ.
Consequently, we deduce that (63) has a unique solution u satisfying
(66)u(x)≈{(∫1+x∞(L(s))1/(p-1)sds)(p-1)/(p-1-α),ifβ=p,(1+x)(p-β)/(p-1-α)(L(1+x))1/(p-1-α),ifp<β<(λ+1)(p-1-α)+(α-γ)pp-1-γ,(1+x)(p-1-λ)/(p-1-γ),ifβ>(λ+1)(p-1-α)+(α-γ)pp-1-γ,(1+x)(p-1-λ)/(p-1-γ)(∫1-xηL(s)sds)1/(p-1-α),ifβ=(λ+1)(p-1-α)+(α-γ)pp-1-γ.

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