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Advances in Nonlinear Analysis

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Boundary blow-up solutions to the Monge-Ampère equation: Sharp conditions and asymptotic behavior

Xuemei Zhang
  • Corresponding author
  • School of Mathematics and Physics, North China Electric Power University, Beijing, 102206, PR China
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  • De Gruyter OnlineGoogle Scholar
/ Meiqiang Feng
Published Online: 2019-07-24 | DOI: https://doi.org/10.1515/anona-2020-0023

Abstract

Consider the boundary blow-up Monge-Ampère problem

M[u]=K(x)f(u) for xΩ,u(x)+ as dist(x,Ω)0.

Here M[u] = det (uxixj) is the Monge-Ampère operator, and Ω is a smooth, bounded, strictly convex domain in ℝN (N ≥ 2). Under K(x) satisfying appropriate conditions, we first prove that the boundary blow-up Monge-Ampère problem has a strictly convex solution if and only if f satisfies Keller-Osserman type condition. Then the asymptotic behavior of strictly convex solutions to the boundary blow-up Monge-Ampère problem is considered under weaker condition with respect to previous references. Finally, if f does not satisfy Keller-Osserman type condition, we show the existence of strictly convex solutions under different conditions on K(x). The proof combines standard techniques based upon the sub-supersolution method with non-standard arguments, such as the Karamata regular variation theory.

Keywords: Monge-Ampère equation; boundary blow up; sub-supersolution method; sharp conditions; strictly convex solution; asymptotic behavior

MSC 2010: 35J60; 35J96

1 Introduction

Monge-Ampère problems are fully nonlinear problems, which can describe Weingarten curvature, or reflector shape design (see [1]). In the past years, increasing attention has been paid to the study of Monge-Ampère problems by various approaches. We list here, for example, papers [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]. The other recent results concerning fully nonlinear uniformly elliptic equations can be found in [15, 16, 17, 18, 19, 20]

In this article, we consider the boundary blow-up problem for the Monge-Ampère equation

M[u]=K(x)f(u) in Ω,u=+ on Ω,(1.1)

where M[u] = det (uxixj) is the Monge-Ampère operator, Ω is a smooth, bounded, strictly convex domain in ℝN (N ≥ 2), and K(x), f(u) are smooth positive functions. The boundary blow-up condition u = +∞ on ∂Ω means

u(x)+asdist(x,Ω)0.

We aim to study the existence and asymptotic behavior of strictly convex solution to (1.1). Suppose K(x) and f(u) satisfy

  • (K)

    KC(Ω) and K(x) > 0 in Ω;

  • (f1)

    there exists η ∈ ℝ1 ∪ {−∞} such that

    1. fC(η, ∞) is positive and strictly increasing in (η, ∞),

    2. if η ∈ℝ1 then additionally f(η) := limsηf(s) = 0.

The boundary blow-up problems were first studied by Cheng and Yau [21, 22] with f(u) an exponential function of u, due to their applications in geometry. The case f(u) = up (p > 0) and K(x) is a smooth positive function over Ω was considered by Lazer and McKenna [23]. Further results can be found in [24, 25, 26, 27, 28, 29, 30, 31, 32], and especially papers [33, 34, 35, 36] which have mainly motivated us.

Mohammed [33] proved that if K(x) satisfies (K) and is such that the Dirichlet problem

M[u]=K(x) in Ω,u=0 on Ω(1.2)

has a strictly convex solution, then (1.1) has a strictly convex solution if f satisfies (f1) and the Keller-Osserman ([37] and [38]) type condition

Ψ(r)=r[(N+1)F(s)]1/(N+1)ds<,r>η.(1.3)

Here

F(s)=ηsf(t)dt ifηR1,F(s)=0sf(t)dt if η=.

In [34], the authors showed that, in the case that η > −∞, (1.3) alone does not guarantee the existence of a strictly convex solution to (1.1). One needs additionally

η+[(N+1)F(s)]1/(N+1)ds=.(1.4)

Here ∫η+ Φ(s)ds = ∞ means that

ηη+ϵΦ(s)ds= for all small positive ϵ.

Obviously, (1.4) is equivalent to limrη+ Ψ(r) = ∞.

From Theorem 1.1 and Theorem 1.2 of [34] we know, if K(x) satisfies (K) and KL(Ω), the Keller-Osserman type condition is necessary and sufficient (combing with (1.4) if η ∈ ℝ1) for the existence of strictly convex solution, but if K(x) satisfies (K) and is such that (1.2) has a strictly convex solution, it is only sufficient. We would like to prove the necessity in this paper.

So the first main result of this paper is the following.

Theorem 1.1

Suppose that K(x) satisfies (K) and is such that (1.2) has a strictly convex solution. Suppose that f(u) satisfies (f1), and when η ∈ℝ1, it satisfies additionally (1.4). Then (1.1) has a strictly convex solution if and only if (1.3) holds.

At the same time, in [34], the authors did not consider the boundary asymptotic behavior of the strictly convex solution. The study of boundary asymptotic behavior of blow-up solutions is also a hot topic, see [37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51], and the references therein. Recently, in [35], Zhang studied the boundary behavior of the strictly convex solution to (1.1) with K(x) ∈ C(Ω̄). Very recently, in [36], Zhang studied the boundary behavior of the strictly convex solution to (1.1) with f(u) including gradient terms and K(x) is in general case or borderline case.

In [35] and [36], there is an important condition on f, i.e.

(f2) there exists Cf ∈ (0, ∞] such that

limsH(s)sdτH(τ)=Cf,H(s):=[(N+1)F(s)]1/(N+1),s>0.

However, we find that it is not necessary. It is implied by other conditions. By Lemma AP.2 in Appendix we can see (f1) and (1.3) imply (f2).

Let

I(s)=Ψ(s)Ψ(s)(Ψ(s))2,I=limsI(s).(1.5)

We can see Cf has the same meaning of I.

By Theorem 1.1 the existence of strictly convex solution to problem (1.2) is the key point for the existence of strictly convex solution to problem (1.1). If K(x) is bounded on Ω, according to Theorem 1 of [52], (1.2) always has a strictly convex solution. If K(x) is unbounded near ∂Ω, problem (1.2) is not always having solution. The existence of solution depends on the increasing speed of K(x) when x approaches ∂ Ω. In [34], the authors gave a sufficient condition for the existence of strictly convex solutions. For ease of composition, we first introduce some notations.

For a positive function p(t) in C1(0, ∞) satisfying p′(t) < 0 and limt→0+ p(t) = +∞, to distinguish its behavior near t = 0 we set P(τ)=τ1p(t)dt. We say such a function p(t) is of class 𝓟finite if

0+[P(τ)]1Ndτ<,

and is of class 𝓟 if

0+[P(τ)]1Ndτ=.

In Theorem 1.5 of [34], the author proved that, if K(x) satisfies (K), then (1.2) has no strictly convex solution if there exists a function p(t) of class 𝓟 such that K(x) ≥ p(d(x)) near ∂ Ω, and has a strictly convex solution if there exists a function p(t) of class 𝓟finite such that K(x) ≤ p(d(x)) near ∂ Ω.

If p(t) is of class 𝓟finite, we may modify p(t) for large t and assume that p(t) = c0et for some positive constant c0 and all large t, say tM0. With p(t) modified as above, if we define

P~(τ)=τp(t)dt,

then we still have

0+[P~(τ)]1Ndτ<.(1.6)

Moreover,

P~(t)=c0et,P~(t)/p(t)=1 for tM0,P~(t)/p(t)0 as t0.(1.7)

Set

ω(t):=0t(NP~(τ))1Ndτ for t>0.(1.8)

Let

J(s)=ω(s)ω(s)(ω(s))2,J0=lims0+J(s).(1.9)

The second main result of this paper is the following.

Theorem 1.2

Suppose that K(x) satisfies (K) and there exists a function p(t) of class 𝓟finite such that

k2p(d(x))K(x)k1p(d(x))nearΩ,(1.10)

where k1, k2 are positive constants. Suppose that f(u) satisfies (f1), (1.3) and when η ∈ℝ1, it satisfies additionally (1.4). f, K are such that I = ∞ and J0 = ∞ can not hold at same time, then for any strictly convex solution u(x) of (1.1), it holds

1lim infxΩ,d(x)0u(x)ψ[ξ_(ω(d(x)))NN+1],lim supxΩ,d(x)0u(x)ψ[ξ¯(ω(d(x)))NN+1]1,

where ψ is the inverse of Ψ, i.e. ψ satisfies

ψ(t)[(N+1)F(s)]1/(N+1)ds=t,t>η(1.11)

and

ξ¯={(N+1)Nk2NNM0[NN+11J0+1(N+1)J0I+1I]}1N+1,ξ_={(N+1)Nk1NNm0[NN+11J0+1(N+1)J0I+1I]}1N+1.

Here

M0=maxx¯Ωκ1(x¯)κ2(x¯)κN1(x¯),m0=minx¯Ωκ1(x¯)κ2(x¯)κN1(x¯),

κ1(x̄), κ2(x̄), … κN−1(x̄) are the principal curvatures of ∂Ω at x̄.

Corollary 1.3

In Theorem 1.2, if Ω is a ball of radius R, I ∈ [1, ∞) and k1 = k2 = k̄, then

limxΩ,d(x)0u(x)ψ[(ω(d(x)))NN+1]=ξ01I,

where

ξ0={k¯RN1(NN+1)N[NN+11J0+1N+11J0I+1I]}1N+1.

For more articles about boundary blow-up solutions in a ball, please see [53, 54, 55, 56, 57].

Remark 1.4

We can determine that the condition imposed on b(x) in [36] is equivalent to

0+[(N+1)p(τ)]1N+1dτ<.(1.12)

We can see that (1.6) is a weaker condition than (1.12).

For example, letting p(s) = sN−1(− ln s)β, 0 < s < s0 < 1, then P̃(s) ≅ sN(−ln s)β as s → 0+. Then we have

0+[(N+1)p(s)]1N+1ds=0+(N+1)1N+1s1(lns)βN+1ds=(N+1)1N+1N+1N+1β(lns)N+1βN+1|0+0,β>N+1,,β<N+1,ass0+.

0+[NP~(s)]1Nds0+N1Ns1(lns)βNds=N1NNNβ(lns)NβN|0+0,β>N,,β<N,ass0+.

Meanwhile, by Theorem 1.5 of [34] we know (1.6) is sharper than (1.12) for the existence of strictly convex solutions of (1.2) and the existence of strictly convex solutions of (1.1).

If K(x) is such that (1.2) has no strictly convex solution, then (1.1) may have or have no strictly convex solution, depending on the behavior of f. In [34], the authors only examined some such cases for the radially symmetric situation. In this paper, we'll consider the general case. But we have to impose some sufficient condition such that (1.2) has no strictly convex solution. It is

(K1) there exists a function p(t) of class 𝓟 such that k4p(d(x)) ≤ K(x) ≤ k3p(d(x)) near ∂ Ω, where k3, k4 are positive constants.

Suppose that K(x) satisfies (K) and (K1), f(u) satisfies (f1). We’ll prove that (1.1) has strictly convex solution if (1.3) does not hold.

If (1.3) does not hold, there exists c0 > 0 such that

G(t):=c0t[(N+1)F(τ)]1N+1dτ as t.(1.13)

Let

R(s)=G(s)G(s)(G(s))2,R=limsR(s).(1.14)

We have

Theorem 1.5

Suppose that K(x) satisfies (K) and (K1). Suppose that f(u) satisfies (f1) and R ≠ ∞. Then (1.1) has strictly convex solution if (1.3) does not hold.

The rest of the paper is organized in the following way. In Section 2 we will collect some known results to be used in the subsequent sections. Section 3 is devoted to the proofs of the Theorem 1.1 and Theorem 1.2. In Section 4 we prove that Theorem 1.5 holds. In Appendix we will introduce the theory of regular variation for the proof of Corollary 1.3.

2 Some preliminary results

In this section, we collect some results for the convenience of later use and reference.

Lemma 2.1

(Lemma 2.1 of [23]) Let Ω be a bounded domain inN, N ≥ 2, and let ukC2(Ω) ∩ C(Ω) for k = 1, 2. Let f(x, u) be defined for xΩ and u in some interval containing the ranges of u1 and u2 and assume that f(x, u) is strictly increasing in u for all xΩ. Suppose

  1. the matrix (uxixj1) is positive definite in Ω,

  2. M[u1](x) ≥ f(x, u1(x)), ∀ xΩ,

  3. M[u2](x) ≤ f(x, u2(x)), ∀ xΩ,

  4. u1(x) ≤ u2(x), ∀ x∂ Ω.

Then u1(x) ≤ u2(x) in Ω.

Remark 2.2

From the proof in [23], it is easily seen that the condition “f(x, u) is strictly increasing in u for all xΩ” in Lemma 2.1 can be relaxed to “f(x, u) is nondecreasing in u for all xΩ” provided that one of the inequalities in (ii) and (iii) is replaced by a strict inequality. This observation will be used later in the paper.

Lemma 2.3

(Proposition 2.1 of [24]) Let uC2(Ω) be such that the matrix (uxixj) is invertible for xΩ, and let g be a C2 function defined on an interval containing the range of u. Then

M[g(u)]=M[u]{[g(u)]N+[g(u)]N1g(u)(u)TB(u)u},(2.1)

where AT denotes the transpose of the matrix A, B(u) denotes the inverse of the matrix (uxixj), and

u=(ux1,ux2,,uxN)T.

If u = d(x), then

M[g(d(x))]=[g(d(x))]N1g(d(x))i=1N1κi(x¯)1d(x)κi(x¯),xΩδ1,(2.2)

where

Ωδ1={xΩ:0<d(x)<δ1},κ1(x¯),κ2(x¯),κN1(x¯)

are the principal curvatures of ∂Ω at x̄.

The following interior estimate for derivatives of smooth solutions of Monge-Ampère equations is a simple variant of Lemma 2.2 in [23], which follows from [58, 59].

Lemma 2.4

Let Ω be a bounded domain inN, N ≥ 2, with ∂ ΩC. Let η ∈ [−∞, +∞) and fC(Ω×(η, ∞)) with f(x, u) > 0 for (x, u) ∈ Ω×(η, ∞). Let uC(Ω) be a solution of the Dirichlet problem

M[u](x)=f(x,u),xΩ,u(x)=c=constant,xΩ,(2.3)

with η < u(x) < c in Ω. Let Ωbe a subdomain of Ω with Ω′ ⊂ Ω and assume that η < au(x) ≤ b for xΩand let k ≥ 1 be an integer. Then there exists a constant C which depends only on k, a, b, bounds for the derivatives of f(x, u) for (x, u) ∈ Ω′ × [a, b], and dist(Ω′, ∂Ω) such that

||u||Ck(Ω¯)C.

The existence result below is a variant of Lemma 2.3 in [23], which is a special case of Theorem 7.1 in [52].

Lemma 2.5

Let Ω be a strictly convex, bounded domain inN, N ≥ 2, with ∂ ΩC. Let f(x, u) be a positive C function on Ω × (η, c], where c > η ≥ −∞. If there exists a function u*C2(Ω), which is convex on Ω, such that u* > η and

M[u](x)f(x,u(x)),xΩ,u(x)=c,xΩ,

then there exists a solution u of (2.3) with uC(Ω) and u strictly convex. Moreover, u(x) ≥ u*(x) on Ω.

Let Ω be a smooth, bounded, strictly convex domain in ℝN, by Theorem 1.1 of [52], there exists u0C(Ω) which is the unique strictly convex solution to

M[u0]=1 in Ω,u0=1 on Ω.

Set z(x) := 1 − u0(x). Then z(x) > 0 in Ω and it is the unique strictly concave solution to

(1)NM[z]=1 in Ω,z=0 on Ω.(2.4)

Since (zxixj) is negative definite on Ω, its trace is negative, that is Δ z < 0, and hence one can apply the Hopf boundary lemma to conclude that |∇ z| > 0 for x∂Ω. It follows that there exist positive constants b1 and b2 such that

b1d(x)z(x)b2d(x) for xΩ.(2.5)

3 Proof of Theorem 1.1 and Theorem 1.2

Proof of Theorem 1.1

Sufficiency. It was proved in [34].

Necessity. Assume to the contrary that (1.1) has a strictly convex solution u. We aim to derive a contradiction.

Denote by g(t) the inverse of G(t), i.e.,

c0g(t)[(N+1)F(τ)]1N+1dτ=t,t>0,(3.1)

where G(t) is defined by (1.13).

Then

g(0)=c0,limtg(t)=

and

g(t)=[(N+1)F(g(t))]1N+1,g(t)=f(g(t))[(N+1)F(g(t))]N1N+1,(g(t))N1g(t)=f(g(t)),g(t)g(t)=[(N+1)F(g(t))]NN+1f(g(t)).(3.2)

Since Ω is bounded in RN, there exists R0 such that ΩB(0, R0). Then define

y(x):=i=1N12(xi+R0+K1)2 for xΩ,

where K1 is a positive constant to be determined.

Clearly

[y(x)]T=(x1+R0+K1,,xN+R0+K1)>0,

(yxixj) is the identity matrix, and M[y] = 1.

Let w1(x) be a strictly convex solution of (1.2) and K1=maxxΩ¯|w1|+1. Let

w(x)=w1(x)+y(x)+H,

where H=maxxΩ¯(w1(x))+1. Then

|w|=|w1+y|>1,M[w]>M[w1].

For c > 0 define

v(x):=gcw(x),xΩ.

Then we obtain, for xΩ,

M[v]=M[cw]{[g(cw)]N+(g(cw))N1g(cw)((cw))TB(cw)(cw)}=cNM[w](g(cw))N1g(cw)g(cw)g(cw)+c(w)TB(w)w>cNM[w1]f(v)g(cw)g(cw)+cλ1|w|2>cN+1λ1K(x)f(v),

where B(w) is the inverse matrix of (wxixj), λ1 is the minimal eigenvalue of B(w). Since w is strictly convex, all the eigenvalue of B(w) is positive. We thus obtain

M[v]>K(x)f(v) in Ω

provided that c is chosen large enough.

Fix x1Ω and by further enlarging c if necessary we may assume that

v(x1)>u(x1)andM[v]>K(x)f(v)inΩ.

Since u(x) → ∞ as d(x) → 0, while v(x) is continuous on Ω, there exists an open connected set D such that

x1D,D¯Ω,u(x)<v(x)inDandu(x)=v(x)onD.

On the other hand, since

M[u]=K(x)f(u) inDandv=uonD,

and the matrix (vxixj) is positive definite on D (since w1(x), y(x) are strictly convex in Ω and g′, g″ > 0), we can apply Lemma 3.1 to conclude that v(x) ≤ u(x) in D. This contradiction completes our proof. □

Proof of Theorem 1.2

For small δ1 > 0, let

Ωδ1={xΩ|0<d(x)<δ1}.

For an arbitrary ε ∈ (0, min{1/4, k2}), let

ξ¯ε={k2(1ε)2(NN+1)NM0[NN+11J0+1N+11J0I+1I]}1N+1,ξ_ε={k1(1+ε)2(NN+1)Nm0[NN+11J0+1N+11J0I+1I]}1N+1.

where m0, M0, k1, k2 are given in Theorem 1.2, I, J0 are given (1.5) and (1.9).

From the definition of J0, I, ξε, ξε we see that

limd(x)0[NP~(d(x))]N+1Np(d(x))0d(x)(NP~(τ))1Ndτ=1J0,lims[(N+1)F(s)]NN+1f(s)Ψ(s)=1I,limd(x)0i=1N1(1d(x)κi(x¯))=1,

ξ¯εN+1(NN+1)N[NN+11J0+1N+11J0I+1I]M0k2(1ε)1=ε,ξ_εN+1(NN+1)N[NN+11J0+1N+11J0I+1I]m0k1(1+ε)1=ε.

For xΩδ1, define

u¯ε(x)=ψ[ξ¯ε[ω¯(d(x))]],ω¯(d(x))=[ω(d(x))]NN+1[ω(σ)]NN+1,d(x)>σ>0,u_ε(x)=ψ[ξ_ε[ω_(d(x))]],ω_(d(x))=[ω(d(x))]NN+1+[ω(σ)]NN+1,σ(0,δε),

where δε ∈ (0, min{1, δ1/2}) is sufficiently small such that for xΩ2δε

1ε<i=1N1(1d(x)κi(x¯))<1+ε,

ξ¯εN+1(NN+1)N[NN+11J(d(x))+1N+11I(u¯ε)1J(d(x))+1I(u¯ε)]M0k2(1ε)1<0,ξ_εN+1(NN+1)N[NN+11J(d(x))+1N+11I(u_ε)1J(d(x))+1I(u_ε)]m0k1(1+ε)1>0.

Let

Dσ=Ω2δε/Ω¯σ,Dσ+=Ω2δεσ.

By (2.2) we have for xDσ

M[u¯ε]K(x)f(u¯ε)=(NN+1)Nξ¯εN+1(1)N1ψN1ψωN1ω[NN+1ω2ωω1N+1ψψξ¯εωNN+1ω2ωw+ψψξ¯εωNN+1]i=1N1κi(x¯)1d(x)κi(x¯)K(x)f(u¯ε)=(NN+1)Nξ¯εN+1f(ψ)p(d(x))[NN+1ω2ωω+1N+1ψψξ¯εωNN+1ω2ωwψψξ¯εωNN+1]i=1N1κi(x¯)1d(x)κi(x¯)K(x)f(u¯ε)<(NN+1)Nξ¯εN+1f(ψ)p(d(x))[NN+1[NP~(d(x))]N+1Np(d(x))0d(x)(NP~(τ))1Ndτ+1N+1[(N+1)F(ψ)]NN+1f(ψ)Ψ(ψ)[NP~(d(x))]N+1Np(d(x))0d(x)(NP~(τ))1Ndτ+[(N+1)F(ψ)]NN+1f(ψ)Ψ(ψ)]i=1N1κi(x¯)1d(x)κi(x¯)K(x)f(u¯ε)=[ξ¯εN+1(NN+1)N[NN+11J(d(x))+1N+11I(u¯)1J(d(x))+1I(u¯)]M0k2(1ε)1]K(x)f(u¯ε)<0,

i.e. ūε is a supersolution to (1.1) in Dσ.

Similarly, we can prove uε is a subsolution to (1.1) in Dσ+.

By (1.10) and Theorem F, (1.2) has a strictly convex solution. It follows from Theorem 1.1 that (1.1) has a strictly convex solution u. Let A be large enough such that

uu¯ε+Aond(x)=2δε

and

u_εu+Aond(x)=2δεσ.

By the definition of ūε and uε, we know that ūε(x) → ∞ as d(x) → σ and uε|∂Ω < u|∂Ω. By Lemma 2.1 we have

uu¯ε+AinDσ

and

u_εu+AinDσ+.

Then

u(x)ψ[ξ¯ε[ω¯(d(x))]]1+Aψ[ξ¯ε[ω¯(d(x))]],xDσ,

and

1Aψ[ξ_ε[ω_(d(x))]]u(x)ψ[ξ_ε[ω_(d(x))]],xDσ+.

Then, for xDσDσ+, the two formulas above hold. Letting σ → 0, then we obtain

u(x)ψ[ξ¯ε[ω(d(x))]NN+1]1+Aψ[ξ¯ε[ω(d(x))]NN+1],

and

1Aψ[ξ_ε[ω(d(x))]NN+1]u(x)ψ[ξ_ε[ω(d(x))]NN+1].

Letting d(x) → 0, ε → 0, we get

lim supxΩ,d(x)0u(x)ψ[ξ¯(ω(d(x)))NN+1]1,

and

1lim infxΩ,d(x)0u(x)ψ[ξ_(ω(d(x)))NN+1].

Proof of Corollary 1.3

By (1.11) we have

ψ(t)=[(N+1)F(ψ(t))]1N+1.

Then

limt0tψ(t)ψ(t)=limt0t[(N+1)F(ψ(t))]1N+1ψ(t)=limsΨ(s)Ψ(s)s=1limsΨ(s)Ψ(s)Ψ2(s)=1I.

It follows from Proposition AP.3 of Appendix that ψNRV1–I.

Combing this with Theorem 1.2 we obtain

limxΩ,d(x)0u(x)ψ[(ω(d(x)))NN+1]=limxΩ,d(x)0u(x)ψ[ξ0(ω(d(x)))NN+1]ψ[ξ0(ω(d(x)))NN+1]ψ[(ω(d(x)))NN+1]=ξ01I.

4 Proof of Theorem 1.5

For the proof of Theorem 1.5, we first introduce a lemma which is about radial solutions. Let K(x) = K1(r), u(x) = v(r), r = |x|, B is a ball with radius R in ℝN (N ≥ 2), then

M[u]=K(x)f(u),xB,u=,xB,(4.1)

is equivalent to

(v)N1v=rN1K1(r)f(v),r(0,R),v(0)=0,v(R)=.(4.2)

In the radially symmetric setting, the smoothness requirements for K and f can be greatly relaxed. But for convenience, we still use (K), (K1) and (f1). In the case (K1) can be state as:

there exist constants d3, d4 > 0 and a function p(t) of class 𝓟 such that

d4p(Rr)K1(r)d3p(Rr) for allr<Rclose to R.

We modify p(t) as in Section 1 and define σ(t) by

σ(s)=s(NP~(τ)]1Ndτ,(4.3)

we have

σ(s)=[NP~(s)]1N,σ(s)=[NP~(s)]1N1p(s).(4.4)

It follows that

1σ(s)=[NP~(s)]1N.

Let

T(s)=σ(s)σ(s)σ2(s).(4.5)

Lemma 4.1

Suppose that K satisfies (K) and (K1). Suppose that f satisfies (f1) and R ≠ ∞. If (1.3) does not hold, then (4.2) has infinitely many strictly convex solutions.

Proof

Let w(r)=g(cσNN+1(y(r))) for r ∈ [0, R), where y(r) satisfies

(1)NyN1y=rN1,r(0,R),y(0)=0,y(R)=0.

and g, σ is defined by (3.1), (4.3).

Similar to the proof of Theorem 5.3 of [34], we can prove the Lemma 4.1. So we omit it here.□

Proof of Theorem 1.5

  • Step1

    Let

    w(x)=g(cσNN+1(1b2z(x))),xΩ,

    where g, σ, z is defined by (3.1), (4.3), (2.4), respectively, and b2 is defined in (2.5). Then by (1.14), (4.4), (4.5), (3.2) we have

    M[w]=cN+1b2N(NN+1)NgN1gσN1σM[z]{gcσNN+1gσσb21(NN+1σ2σσ1N+1gcσNN+1gσ2σσ+gcσNN+1g)(z)TB(z)z}=cN+1b2N(NN+1)Nf(w)p(1b2z){1R(w)(σσ)b21(NN+11T(z)1N+11R(w)1T(z)+1R(w))(z)TB(z)z},(4.6)

    where R, T is defined by (1.14) and (4.5) respectively.

    Since

    σ(t)2σ(t)σ(t)=[NP~(t)]N+1Np(t)t[NP~(τ)]1/Ndτ=t(N+1)[NP~(s)]1/Np(s)dst{p(s)s[NP~(τ)]1/Ndτ+p(s)[NP~(s)]1/N}ds(N+1).

    We have

    1R(w)1N+11R(w)1T(z)0.

    Then

    NN+11T(z)1N+11R(w)1T(z)+1R(w)>0.

    Let

    1=1R(w)(σσ)b21(NN+11T(z)1N+11R(w)1T(z)+1R(w))(z)TB(z)z.

    By the definition of z we have (zxixj) is negative definite. It follows that there exist e1, e2 > 0 such that

    e1||z||2(z)TB(z)ze2||z||2,

    and trace(zxixj) = Δz < 0. Therefore, since Δ(–z) > 0 on Ω and –z attains its maximum on Ω̄ at each point of ∂Ω, it follows from the maximum principle that there exists an open set U containing ∂Ω such that

    ||z||e>0.

    On the other hand, it is easy to see that z is bounded below by a positive constant on ΩU. Then

    wasconΩU.

    Combining this with the fact that R ≠ ∞, we can conclude that Δ1 is positive on Ω. By (4.6) we have

    M[w]cN+1b2N(NN+1)Nf(w)p(d(x))Δ1cN+1b2N(NN+1)Nf(w)1k3K(x)Δ1K(x)f(w)(4.7)

    for large c1, i.e. w1(x)=g(c1σNN+1(1b2z(x))) is a subsolution of (1.1).

  • Step2

    The existence of a solution u(x) ∈ C(Ω).

    Let {σn}1 be a strictly increasing sequence of positive numbers such that σn → ∞ as n → ∞, and let Ωn = {xΩ|w1(x) < σn}. Since any level surface of w1 is a level surface of z, for each n ≥ 1, ∂Ωn is a strictly convex C-submanifold of RN of dimension N – 1.

By Lemma 2.5 there exists unC (Ω̄n) for n ≥ 1 such that

M[un]=K(x)f(un),xΩn,unΩn=σn=w1Ωn.(4.8)

By Lemma 2.1 and (4.7)

un(x)w1(x)xΩ¯n.

Clearly, for n > 1

Ω¯nΩn+1

and

Ω=n=1Ωn.

We claim that

un(x)un+1(x),xΩn.(4.9)

Indeed, since un and un+1 are both positive solutions of M[u] = K(x)f(u) on Ω̄n, un is strictly convex in Ωn and for x∂ΩnΩn+1,

un+1(x)w1(x)=un(x),

the inequality (4.9) is a consequence of Lemma 2.1.

Fix m. For each x0Ω̄m, let R be small to have (x0; R) ⊂ Ωm+1. By Lemma 4.1, there exists a solution v of (4.2). Let d(x, ∂B) = d(x, ∂B(x0; R)). Define

w2(x)=v(Rd(x,B)),xB(x0;R).

Then by (2.2) and (4.2) we have

M[w2]=(v)N1vi=1N1κi(x¯)1d(x)κi(x¯)=rN1K1(r)f(w2)i=1N1κi(x¯)1d(x)κi(x¯)K(x)f(w2)

for small R.

It follows from Lemma 2.1 un(x) ≤ w2(x), xB(x0; R) for all nm + 1. Then un have an uniform bound from above in B(x0; R/2) for nm + 1. Covering Ωm with finite ball of this kind, one gets the uniform bound Cm, i.e. there exists Cm > 0 such that

un(x)Cm for xΩ¯m,nm+1.

This implies that, for every xΩ,

u(x):=limnun(x) exists 

and

um(x)u(x)Cmfor xΩ¯m.

As we also have un(x) ≥ c0 > 0 in Ωm for nm + 1, and for such n, ΩmΩn,

0<dist(Ω¯m,Ωm+1)dist(Ω¯m,Ωn)<dist(Ωm,Ω),

we are in a position to apply Lemma 2.4 to conclude that, for any fixed integer k ≥ 1, there exists a constant C = Ck,m independent of n such that for all n > m,

unCk(Ω¯m)C.

It follows that the convergence un(x) → u(x) holds in Clock(Ω) for every k ≥ 1, and uC(Ω). Moreover, for xΩ,

M[u](x)=limnM[un](x)=K(x)limnf(un(x))=K(x)f(u(x)).

Since each un is strictly convex, u(x) is strictly convex in Ω. Thus u is a strictly convex solution of (1.1).□

Appendix

We present some basic facts of Karamata regular variation theory (refer to [60], [61]) here.

Definition AP.1

A positive measurable function f defined on [A, ∞), for some A > 0, is called regularly varying at infinity with index ρR, written fRVρ, if for all ξ > 0,

limsf(ξs)f(s)=ξρ.(AP1)

In particular, when ρ = 0, f is called slowly varying at infinity.

Clearly, if fRVρ, then L(s)=f(s)sρ is slowly varying at infinity.

Definition AP.2

A positive measurable function f defined on [A, ∞), for some A > 0, is called rapidly varying at infinity if for each ρ > 1

limsf(s)sρ=.(AP2)

or

A positive measurable function f defined on [A, ∞), for some A > 0, is called rapidly varying at infinity if for each ξ > 1

limsf(ξs)f(s)=.

Proposition AP.1

(Uniform convergence theorem). If fRVρ, then (AP1) holds uniformly for ξ ∈ [c1, c2] with 0 < c1 < c2. Moreover, if ρ < 0, then uniform convergence holds on intervals (c1, ∞) with c1 > 0; if ρ > 0, then uniform convergence holds on intervals (0, c2] provided f is bounded on (0, c2] with c2 > 0.

Proposition AP.2

(Representation theorem). A function L is slowly varying at infinity if and only if it may be written in the form

L(s)=ψ(s)expA1sy(τ)τdτ,sA1,

for some A1A, where the function ψ and y are continuous and for s → ∞, y(s) → 0 and ψ(s) → c0, with c0 > 0.

We say that

L^(s)=c0expA1sy(τ)τdτ

is normalized slowly varying at infinity and

f(s)=sρL^(s),sA1,

is normalized regularly varying at infinity with index ρ and write fNRVρ.

Proposition AP.3

A function fRVρ belongs to NRVρ if and only if

fC1[A1,),forsomeA1>0andlimssf(s)f(s)=ρ.

Proposition AP.4

If function f, g, L are slowly varying at infinity, then

  1. fp for every pR, c1f + c2g(c1, c2 ≥ 0), fg(if g(s) → 0 as s → 0+) are also slowly varying at infinity.

  2. For every ρ > 0 and s → ∞,

    sρL(s)0,sρL(s).

  3. For ρR and s,ln(L(s))lns0andln(sρL(s))lnsρ.

Proposition AP.5

If f1RVρ1, f2RVρ2, then f1f2RVρ1+ρ2 and f1f2RVρ1ρ2.

Proposition AP.6

(Asymptotic behavior) If a function L is slowly varying at infinity, then for a ≥ 0 and t → ∞,

  1. at sρ L(s)ds ≅ (1 + ρ)–1t1+ρL(t), for ρ > –1;

  2. t sρ L(s)ds ≅ (–1 – ρ)–1 t1+ρ L(t), for ρ < –1.

Remark AP.1

The result of Proposition AP.6 remains true for ρ = –1 in the sense that

ats1L(s)dsL(t)ast.

The way to remember roposition AP.6 is that L(s) can be taken out of the integral as if it were L(t), thus

atsρL(s)dsL(t)atsρds(t).

When ρ = –1, let z(s) = s–1L(s), we have

Proposition AP.7

(Asymptotic behavior)(See Karamata’s Theorem 1.5.9b in [60].) If a function zRV–1 and s z(τ) < ∞, s > 0, then sz(τ) is slowly varying at infinity and

limssz(s)sz(τ)dτ=0.

By (1.3) we have

Ψ(s)=[(N+1)F(s)]1N+1,Ψ(s)=[(N+1)F(s)]1N+11f(s).

It follows that

1Ψ(s)=[(N+1)F(s)]1N+1.(AP3)

By (1.5) we have

I(s)=f(s)Ψ(s)[(N+1)F(s)]NN+1.

Then we have

Lemma AP.1

Let f satisfy (f1) and (1.3). Then I ≥ 1.

Proof

It is easy to prove by integrating I(s) from a(a > 0) to v. So we omit it.□

Lemma AP.2

Let f satisfy (f1), (1.3). We have

  1. I ∈ (1, ∞) if and only if FNRVq+1 with q > N. In this case, fRVq;

  2. I = 1 if and only if F is rapidly varying at infinity;

  3. I = ∞ if and only if FNRVN+1. In this case, fRVN.

Proof

It can be proved by the definition of regularly varying and rapidly varying at infinity. So we omit it.□

Acknowledgements

This work is sponsored by the National Natural Science Foundation of China (11301178), the Beijing Natural Science Foundation (1163007), the key research and cultivation project of the improvement of scientific research level of BISTU (2018ZDPY18/521823903), and the teaching reform project of BISTU (2018JGYB32). The authors are grateful to anonymous referees for their constructive comments and suggestions, which has greatly improved this paper.

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About the article

Received: 2018-12-12

Accepted: 2019-03-04

Published Online: 2019-07-24

Published in Print: 2019-03-01


Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 729–744, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2020-0023.

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