On boundary blow-up solutions to equations involving the ∞ Laplacian

Nonlinear Analysis 74 (2011) 5238–5252

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Nonlinear Analysis

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On boundary blow-up solutions to equations involving the ∞-LaplacianAhmed Mohammed a,∗, Seid Mohammed b

a Department of Mathematical Sciences, Ball State University, Muncie, IN 47306, USAb Department of Mathematics, Addis Ababa University, Addis Ababa, Ethiopia

a r t i c l e i n f o

Article history:Received 3 February 2011Accepted 3 April 2011Communicated by S. Ahmad

Keywords:Infinity LaplacianBoundary blow-upBoundary asymptotic estimateUniquenessComparison principle

a b s t r a c t

Given a non-negative, continuous function h onΩ × R such that h(x, 0) = 0 for all x ∈ Ω ,h(x, t) > 0 in Ω × (0,∞), and h(x, t) non-decreasing in t for each x ∈ Ω , we study theboundary value problem

∆∞u = h(x, u) inΩu = ∞ on ∂Ω

where Ω ⊆ RN , N ≥ 2 is a bounded domain and ∆∞ is the ∞-Laplacian, a degenerateelliptic operator. We provide sufficient conditions on h under which the above problemadmits a solution, or fails to admit a solution. A necessary and sufficient condition on f isgiven for a solution to exist in the special case when h(x, t) = b(x)f (t). In the latter casean asymptotic boundary behavior of solutions will be studied. As an application a sufficientcondition on f will be given to ensure the uniqueness of solutions in case b is a constant.

© 2011 Elsevier Ltd. All rights reserved.

1. Introduction

LetΩ ⊆ RN ,N ≥ 2 be a bounded domain. In this paper we study the problem∆∞u = h(x, u) inΩu = ∞ on ∂Ω. (1.1)

The boundary condition u = ∞ on ∂Ω is to be understood in the following sense.

limx→z

u(x) = ∞ for each z ∈ ∂Ω.

Here∆∞ is the degenerate elliptic operator given by

∆∞u =

N−i,j=1

uxiuxixjuxj .

This operator, called the∞-Laplacian, first appeared in thework of Aronsson [1] in connectionwith the geometric problemoffinding the so-called absolutelyminimizing functions inΩ .We refer the reader to the papers [2–4] for an in-depth discussionof the problem and other many interesting related properties.

The ∞-Laplacian has been the subject of extensive investigation since the fundamental work of Jensen [5] in which heestablished that solutions of∆∞u = 0 inΩ characterize absolutelyminimizing functions inΩ . Since absolutelyminimizingfunctions are not necessarily smooth, the equation∆∞u = 0 inΩ is understood in aweak sense to be described in Section 2.

∗ Corresponding author. Tel.: +1 765 285 8813; fax: +1 765 285 1721.E-mail addresses: [email protected] (A. Mohammed), [email protected] (S. Mohammed).

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A. Mohammed, S. Mohammed / Nonlinear Analysis 74 (2011) 5238–5252 5239

The non-linearity h in (1.1) is assumed to be continuous on Ω × R and h(x, t) is non-decreasing in t for each x ∈ Ω .Furthermore h(x, 0) = 0 on Ω × 0 and h(x, t) > 0 on Ω × (0,∞). One good example of such non-linearities ish(x, t) = tq(x) where q ∈ C(Ω, (0,∞)). We will study conditions on the non-linearity h that would guarantee the existenceof a solution u ∈ C(Ω) to problem (1.1).

A function u ∈ C(Ω) that satisfies (1.1) in the viscosity sense (see definition below) is called a boundary blow-up solution.problem (1.1) with non-linearities of the form h(x, t) = b(x)f (t) where b ∈ C(Ω, (0,∞)) and f is a continuous, non-

decreasing function defined on R such that f (0) = 0 and f (t) > 0 for t > 0 has been studied extensively in the contextof the Laplace operator and other elliptic operators, including some degenerate ones. Therefore as a special case of ourconsideration of problem (1.1) we will establish some existence as well as non-existence results to the following problem.

∆∞u = b(x)f (u) inΩu = ∞ on ∂Ω. (1.2)

Under suitable conditions we will also investigate some boundary asymptotic behavior of solutions to (1.2). The result onboundary asymptotic behavior will naturally lead to uniqueness of boundary blow-up solutions when b is a constant.

Our investigation in this paper is motivated by the recent work [6] of Juutinen and Rossi in which they consideredproblem (1.1) with the nonlinearity h(x, t) = tp for p > 1 and with the operator∆∞ replaced by the so-called normalized∞-Laplacian∆N

∞u := |Du|−2∆∞u.

The investigation of boundary blow-up problems for elliptic equations has a long history. As mentioned above, earlystudies focused on problems involving the Laplace operator ∆. The main areas of investigations over the years haverevolved around the question of existence, asymptotic boundary behavior and uniqueness. These questions were addressedby Bieberbach for the equation 1u = eu in smooth bounded domains in R2. Given a bounded domain Ω ⊆ R2 withC2 boundary, Bieberbach showed in [7] that there is a unique u ∈ C2(Ω) which satisfies the above equation and suchthat u(x) − log(dΩ(x)) tends to zero as dΩ(x) := dist(x, ∂Ω) approaches zero. Later Rademacher, in [8], extended theseresults to smooth bounded domains in R3. Bieberbach’s investigation was motivated by a problem in differential geometrywhile Rademacher considered this in connection with a problem in mathematical physics, see [9]. In 1957, Keller [10]and Osserman [11] gave, independently, a necessary and sufficient condition on f for the equation 1u = f (u) to admita boundary blow-up solution in a bounded domain in RN . Boundary blow-up problems for elliptic problems continue toattract a lot of attention and research output in recent years has increased dramatically. This being the case, here we wishto mention the works of Bandle, Marcus in [12,13], Lazer and McKenna [14], Cîrstea and Radulescu [15,16], Garcia-Milianet al. [17–19], Lair [20,21], Marcus and Veron [22,23], Veron [24,25], Zhang [26] and Zhang et al. [27]. We refer the reader tothemonograph [28] for an extensive list of references.We should alsomention that similar investigations have been carriedout where the Laplace operator has been replaced by other elliptic operators. See [29–36].

The paper is organized as follows. In Section 2we recall some results from the literature thatwill be useful in ourwork andmake some useful observations. In Section 3 we explore some necessary and sufficient conditions that would guarantee theexistence of viscosity solutions to problem (1.1). An asymptotic boundary estimatewill be established for viscosity solutionsof problem (1.2). This is done in Section 4. The estimate will be used to establish uniqueness of solutions of such problemswhen f satisfies a strong monotonicity condition and b is a constant. Finally, for the readers’ convenience we have includedan Appendix that contains proofs of two results stated and used in the paper. These results, while known in the literature,are not stated in the form we need in the present work.

2. Preliminaries

Let us introduce some notations that we shall use throughout the remainder of the paper. We will use Ω to denote abounded open set in RN and diam(Ω) its diameter. Given r > 0, and x0 ∈ RN we will use B(x0, r) to denote the ball in RN

of radius r and centered at x0. The set C(Ω) will stand for the space of continuous functions inΩ . Given a subset I of R, wewrite C(Ω, I) to denote the class of continuous functions inΩ that take values in I . We use the notation O ⊂⊂ Ω to denotean open subset O ofΩ whose closure O is contained inΩ . Finally for δ > 0 we set

Ωδ := x ∈ Ω : dΩ(x) < δ

where dΩ(x) stands for the distance function dist(x, ∂Ω).Let us consider the following PDE inΩ ⊆ RN .

∆∞u = h(x, u), (x ∈ Ω). (2.1)

Definition 2.1. A function u ∈ C(Ω) is a viscosity sub-solution of the PDE (2.1) if for every ψ ∈ C2(Ω) with the propertythat u − ψ has a local maximum at some x0 ∈ Ω , then

∆∞ψ(x0) ≥ h(x0, u(x0)).

A function u ∈ C(Ω) is a viscosity super-solution of the PDE (2.1) if for everyψ ∈ C2(Ω), with the property that u −ψ hasa local minimum at some x0 ∈ Ω , then

∆∞ψ(x0) ≤ h(x0, u(x0)).

5240 A. Mohammed, S. Mohammed / Nonlinear Analysis 74 (2011) 5238–5252

A function u ∈ C(Ω) is a viscosity solution of (2.1) if it is both a sub-solution and a super-solution of (2.1). If u ∈ C(Ω) is asub-solution (or a super-solution) of (2.1) inΩ we indicate this by writing

∆∞u ≥ h(x, u) inΩ (or∆∞u ≤ h(x, u) inΩ).

Finally, by a solution (sub-solution or super-solution) u of (1.1) we mean a solution (sub-solution or super-solution) u of(2.1) such that u = ∞ on ∂Ω .

Remark 2.2. If u ∈ C(Ω) is a sub-solution (or a super-solution) of (2.1) inΩ we indicate this by writing

∆∞u ≥ h(x, u) inΩ (or∆∞u ≤ h(x, u) inΩ).

Throughout the paper, solutions, sub-solutions and super-solutions of (2.1) are all to be understood in the viscosity sense,as given in Definition 2.1 above.

In studying boundary value problems associatedwith the PDE (2.1) comparison principles are often used as indispensabletools. In what follows we recall a few comparison principles. The first is due to a pioneering work of Robert Jensen [5]. Werefer to the papers [37,3] for alternative proofs. The recent paper [38] gives a short and elementary proof.

Lemma 2.3 (Comparison Principle). Let u, v ∈ C(Ω) satisfy∆∞u ≥ 0 and∆∞v ≤ 0 inΩ in the viscosity sense. Then

supΩ

(u − v) = sup∂Ω

(u − v).

We will also need the following comparison lemma (see [39]).

Lemma 2.4 (Comparison Principle). Suppose k ∈ C(Ω) ∩ L∞(Ω) is positive in Ω or negative in Ω . Let u, v ∈ C(Ω) satisfy∆∞u ≥ k(x) inΩ , and∆∞v ≤ k(x) inΩ . Then

supΩ

(u − v) = sup∂Ω

(u − v).

We now state a simple comparison lemma which suits our purpose. In the lemma we assume that h : Ω × R → Ris continuous, h(x, t) is non-decreasing in t for each x ∈ Ω and h(x, 0) = 0 for all x ∈ Ω while h(x, t) > 0 for all(x, t) ∈ Ω × (0,∞).

Lemma 2.5 (Comparison Principle). Suppose u, v ∈ C(Ω) such that

∆∞u ≥ h(x, u) inΩ,∆∞v ≤ h(x, v) inΩ. (2.2)

If u ≤ v on ∂Ω and 0 ≤ v on ∂Ω , then u ≤ v inΩ .

Proof. First we observe that any viscosity solution w ∈ C(Ω) of∆∞w ≤ h(x, w) inΩ with w ≥ 0 on ∂Ω is non-negativeinΩ . For ifw(x) < 0 for some x ∈ Ω , then consider the non-empty open set O := x ∈ Ω : w(x) < 0. Then we have

∆∞w ≤ h(x, w) ≤ 0 in O andw = 0 on ∂O.

But then Lemma 2.3 implies that w ≥ 0 in O, which is an obvious contradiction. Therefore w ≥ 0 in Ω as claimed. Let usnow proceed to prove the lemma. Since v ≥ 0 on ∂Ω , by hypothesis, the above observation allows us to conclude that v ≥ 0in Ω . Suppose O := x ∈ Ω : u(x) > v(x) is non-empty. Since v ≥ 0 in Ω we see that u > 0 in O. Thus ∆∞u ≥ k(x) inO, where k(x) := h(x, u(x)) is positive in O. Moreover in O we have∆∞v ≤ h(x, v(x)) ≤ h(x, u(x)) = k(x). Since u = v on∂O, by Lemma 2.4, we conclude that u ≤ v in O which is a contradiction. This establishes the lemma.

We will also need the following result on existence of solutions to the following Dirichlet problem (see [40]) for a givenk ∈ C(∂Ω).

∆∞u = h(x, u) inΩu = k on ∂Ω. (2.3)

Lemma 2.6. Suppose h ∈ C(Ω × R, [0,∞)) and h(x, t) is non-decreasing in t for each x ∈ Ω . Given k ∈ C(∂Ω), thenproblem (2.3) admits a solution u ∈ C(Ω).

Given g ∈ C(R, [0,∞)), not identically zero, we will also need the following growth condition on g at infinity.∫∞

1

14√G(t)

dt < ∞, where G(t) =

∫ t

0g(s)ds. (2.4)


This is a Keller–Osserman type condition on g , see [12,10,20,11,24,25] for instance. In this paper we will refer to condition(2.4) as the Keller–Osserman condition.

The next lemma concerns the solvability of the following boundary blow-up problem on a ball B := B(x0, R).∆∞u = g(u) in Bu = ∞ on ∂B. (2.5)

In (2.5) we allow the radius of the ball B to be infinity, that is B = RN , in which case the boundary condition is interpretedto mean u(x) → ∞ as |x| → ∞.

The following lemma will serve as the crucial step in proving existence of solutions to the PDE (1.1).

Lemma 2.7. Let g be a non-negative, non-decreasing, continuous function defined on [0,∞) such that g(0) = 0 and g(t) > 0for t > 0.(1) Let g satisfy the Keller–Osserman condition (2.4). Then there is 0 < R∗

≤ ∞ such that problem (2.5) admits a solutionv ∈ C(B) for each ball B = B(x0, R) with 0 < R < R∗.

(2) Suppose g fails to satisfy the Keller–Osserman condition (2.4). Then given (a, x0) ∈ (0,∞) × RN , problem (2.5) admits asolution v ∈ C(RN) such that v(x0) = a.

Proof. Let us consider the initial value problem

(φ′)2φ′′= g(φ), φ(0) = a, φ′(0) = 0, (2.6)

where a > 0. Given x0 ∈ RN , first we show that v(x) := φ(|x − x0|) is a viscosity solution of ∆∞v = g(v) in B = B(x0, R)assuming that φ ∈ C2((0, R)) ∩ C1([0, R)) is a solution of (2.6) in the interval (0, R) for some 0 < R ≤ ∞. Here B = RN incase R = ∞.

To this end, first we note that integrating the equation in (2.6) in (0, r) shows that φ′ > 0 in (0, R). Next we multiplyboth sides of the differential equation in (2.6) by φ′ and integrate on (0, r). We find

φ′(r)4

= 4∫ φ(r)

ag(s)ds ≥ g(a)(φ(r)− a).

Recalling that φ′ > 0 in (0, R) this leads to

φ′(r)(φ(r)− a)1/4

≥ g(a)1/4.

Integrating this again on (0, r)we find

φ(r)− a ≥ Cr4/3, r ∈ (0, R), (2.7)

where C is a positive constant. We will come back to this inequality shortly.Now we proceed to show that if v(x) := φ(|x − x0|), x ∈ B(x0, R), then v is a viscosity solution of problem (2.5).Let us note that v ∈ C2(B \ x0) ∩ C1(B). Therefore by virtue of (2.6) it is clear that v is a classical solution in B \ x0

and therefore v is a viscosity solution there. First we remark that, since φ′(0) = 0 we have Dv(x0) = 0. It is easy to verifythat v is actually a viscosity super-solution in B. Indeed let ϕ ∈ C2(Ω) such that v − ϕ has a minimum at x0 ∈ Ω . Note thatDϕ(x0) = Dv(x0) = 0. Therefore∆∞ϕ(x0) = 0 ≤ g(v(x0)) as claimed. Now we show that v is a sub-solution in B. For thiswe will show that there is no ϕ ∈ C2(Ω) such that v − ϕ has a local maximum at x0. Assume on the contrary that there isϕ ∈ C2(B) such that v − ϕ has a maximum at x0. Then, recalling that Dϕ(x0) = 0, we have

v(x)− v(x0) ≤ ϕ(x)− ϕ(x0) =12

D2ϕ(x0)(x − x0), x − x0

+ o(|x − x0|2).

Taking x = x0 + te, where e is a unit vector and t > 0, we see from (2.7) that

Ct4/3 ≤ φ(t)− a = v(x0 + te)− v(x0) ≤12⟨D2ϕ(x0)e, e⟩t2 + o(t2).

However, this contradicts the fact that D2ϕ(x0) exists. Thus there is no ϕ ∈ C2(B) such that v − ϕ has a maximum at x0.Therefore v is a sub-solution, and hence a solution, as claimed.

To complete the proof of the lemma it remains to show that problem (2.6) admits an appropriate solutionφ ∈ C2((0, R))∩C([0, R))such that φ(r) → ∞ as r → R− (or r → ∞), depending on whether g satisfies the Keller–Osserman condition ornot, so that the function v(x) = φ(|x − x0|) has the stated properties.

Let us first suppose that g satisfies the Keller–Osserman condition (2.4). Then there is 0 < R∗≤ ∞ such that for any

given 0 < R < R∗, one can find a > 0 for which problem (2.6) has a solution φ ∈ C2((0, R)) ∩ C1([0, R))which satisfies

φ(r) → ∞ as r → R.

On the other hand, suppose now that g fails to satisfy the Keller–Osserman condition. Then for each a > 0, problem (2.6) hasa solution φ ∈ C2((0,∞))∩C1([0,∞)) such that φ(r) → ∞ as r → ∞. A proof of the above results runs similarly to thosethat can be found in [6,41]. For completeness, as well as for the readers’ convenience, we give a proof in the Appendix.


3. Existence of boundary blow-up solutions

In this section we investigate conditions on the non-linearity h in problem (1.1) that would lead to existence andconditions that imply non-existence of viscosity solutions.

We start with the following lemma. A proof, which we supply below for completeness, is an adaptation of the proof ofLemma 5.1 in [40].

Lemma 3.1. Let h ∈ C(Ω × R, [0,∞)) and uk be a sequence of functions in C(Ω) that satisfy the Eq. (2.1) inΩ . Suppose thatfor any O ⊂⊂ Ω there are constants mO and MO such that mO ≤ uk ≤ MO in O for all k. Then there is a subsequence of uk

that converges locally uniformly onΩ to a solution u ∈ C(Ω) of (2.1).Proof. We start out by showing that uk is equicontinuous on any E ⊂⊂ Ω . Let δ =: dist (E, ∂Ω), and let O := x ∈ Ω :

dist(x, ∂Ω) > δ/8 so that E ⊆ O ⊆ O ⊆ Ω . By hypothesis, there are constants mO and MO such that mO ≤ uk ≤ MO onO for all k ≥ 1. Thus

supO

uk − infO

uk ≤ M := MO − mO for all k.

We take R > 0 such that 4R < δ. Since∆∞uk ≥ 0 onΩ , by a well known result (see [2,3] for instance), we see that for anyz ∈ E

|uk(x)− uk(y)| ≤ M|x − y|

R, x, y ∈ BR/2(z).

Since E is compactwe see that uk is equicontinuous on E. On taking an exhaustion ofΩ by subdomains compactly containedinΩ , we apply the standard method of Cantor diagonalization to extract a subsequence of uk that converges uniformly oncompact subsets ofΩ . For simplicity we will continue to denote such subsequence by uk. We set

u(x) := limk→∞

uk(x), (x ∈ Ω).

Then u ∈ C(Ω).It remains to show that ∆∞u = h(x, u), in the viscosity sense. We first show that ∆∞u ≥ h(x, u). For this suppose that

ϕ ∈ C2(Ω) and u − ϕ has a local maximum at some x0 ∈ Ω , that is,

u(x)− ϕ(x) ≤ u(x0)− ϕ(x0) x ∈ Br(x0) ⊂⊂ Ω

for some r > 0. Let ε > 0 be small. Suppose that xk is a point of maximum of

uk(x)−

ϕ(x)+

ε

2|x − x0|2

on Br(x0). In particular,

uk(xk)−

ϕ(xk)+

ε

2|xk − x0|2

≥ uk(x0)− ϕ(x0). (3.1)

Since xk ∈ Br(x0), by passing to a subsequence if necessary, xk → x for somex ∈ Br(x0). On taking the limit in (3.1) ask → ∞ we obtain

u(x)−

ϕ(x)+

ε

2|x − x0|2

≥ u(x0)− ϕ(x0).

That isε

2|x − x0|2 ≤ u(x)− ϕ(x)− (u(x0)− ϕ(x0)) ≤ 0,

showing thatx = x0. Thus xk ∈ Br/2(x0) for sufficiently large k. Since uk is a sub-solution of (2.1) and xk is a point of localmaximum of uk − (ϕ + ε|x − x0|2/2) in Br(x0), we have

∆∞ϕ(xk)+ O(ε) ≥ h(xk, uk(xk)). (3.2)

Taking the limit in (3.2) and recalling that uk → u locally uniformly onΩ we find that

∆∞ϕ(x0)+ O(ε) ≥ h(x0, u(x0)).

We now take the limit ε → 0 to conclude that ∆∞u ≥ h(x, u) in the viscosity sense. A similar argument shows that u is asuper-solution. For details, suppose that ϕ ∈ C2(Ω) and u − ϕ has a local minimum at a point x0 ∈ Ω , that is

u(x)− ϕ(x) ≥ u(x0)− ϕ(x0) x ∈ Br(x0) ⊂⊂ Ω

for some r > 0. Let ε > 0 be small. Suppose that xk is a point of minimum of

uk(x)−

ϕ(x)−

ε

2|x − x0|2


in Br(x0). In particular,

uk(xk)−

ϕ(xk)−

ε

2|xk − x0|2

≤ uk(x0)− ϕ(x0). (3.3)

Since xk ∈ Br(x0), by passing to a subsequence if necessary, xk → x for somex ∈ Br(x0). On taking the limit in (3.3) ask → ∞ we obtain

u(x)−

ϕ(x)−

ε

2|x − x0|2

≤ u(x0)− ϕ(x0).

That isε

2|x − x0|2 ≤ u(x0)− ϕ(x0)− (u(x)− ϕ(x)) ≤ 0,

showing thatx = x0. Thus xk ∈ Br/2(x0) for sufficiently large k. Since uk is a super-solution of (2.1) and xk is a point of localminimum of uk − (ϕ + ε|x − x0|2/2) in Br(x0), we have

∆∞ϕ(xk)+ O(ε) ≤ h(xk, uk(xk)). (3.4)

Taking the limit in (3.4) and recalling uk → u locally uniformly onΩ we find that

∆∞ϕ(x0)+ O(ε) ≤ h(x0, u(x0)).

We now take the limit ε → 0 to conclude that∆∞u(x) ≤ h(x, u) in the viscosity sense.

Throughout the remainder of this section we assume that h : Ω × [0,∞) → [0,∞) is continuous, h(x, 0) = 0 for allx ∈ Ω and that h(x, t) > 0 for all (x, t) ∈ Ω × (0,∞). We also assume that h(x, t) is non-decreasing in t for each x ∈ Ω .

For each (x, r) ∈ Ω × (0,∞) defineh∗(t; x, r) := minh(z, t) : z ∈ B(x, r) ∩Ω.

Given (x, r) ∈ Ω × (0,∞) it is clear that h∗(t) := h∗(t; x, r) is non-decreasing with h∗(0) = 0 and h∗(t) > t for t > 0. Italso follows easily that h∗ is continuous on [0,∞). Let us now set

H∗(t; x, r) :=

∫ t

0h∗(s; x, r)ds, t > 0.

We say that h satisfies the Keller–Osserman condition at x ∈ Ω if∫∞

1

14√H∗(t; x, r)

dt < ∞ for some r > 0. (3.5)

We should remark that if condition (3.5) holds for some (x, r) then the condition continues to hold at (x, s) for all 0 < s < r .We now state and prove the following existence result.

Theorem 3.2. Suppose that h satisfies the Keller–Osserman condition (3.5) at each x ∈ Ωδ for some δ > 0. Thenproblem (1.1) admits a non-negative solution.Proof. For each positive integer k let uk be a viscosity solution of

∆∞u = h(x, u) inΩu = k on ∂Ω.

The existence of a solution uk follows from Lemma 2.6 (one may extend h(x, t) to be zero for t ≤ 0). By the Comparisonprinciple, Lemma 2.5, we see that 0 ≤ uk ≤ uk+1 for each k ≥ 1.

Let 0 < ε < δ/2 and consider the domain Dε = Ω \Ωε . Since ∂Dε ⊆ Ωδ we note that h satisfies the Keller–Ossermancondition (3.5) at every point of ∂Dε . Thus given z ∈ ∂Dε consider 0 < rz ≤ ε such that condition (3.5) holds. Therefore (1)of Lemma 2.7 shows that the following problem admits a solution vz .

∆∞w = h∗(w; z, rz) in B(z, rz)w = ∞ on ∂B(z, rz).

Since∆∞uk = h(x, uk) ≥ h∗(uk; z, rz) in B(z, rz), by comparison we have uk ≤ vz on B(z, rz) for all k. Therefore uk ≤ Mz inB(z, rz/2) for all k ≥ 1, where

Mz := maxvz(x) : x ∈ B(z, rz/2).

From the open cover U := B(z, rz/2) : z ∈ ∂Dε of ∂Dε we pick a finite subcover B(zj, rzj/2) : j = 1, 2, . . . ,m. Therefore

uk ≤ M on ∂Dε,

where M := maxMzj : j = 1, . . . ,m. Since M is a super-solution of ∆∞u = h(x, u) we see that u1 ≤ uk ≤ M in Dε . As ε> 0 is arbitrary by Lemma 3.1, we conclude that the sequence uk converges to a solution u ∈ C(Ω). On observing thatuk ≤ u inΩ for all k, it follows that u(x) → ∞ as x → ∂Ω . Thus u is a solution of (1.1) as asserted.


We state the following special casewhen h(x, t) = b(x)f (t) and f : R → [0,∞) is a continuous, non-decreasing functionwith f (0) = 0 and f (t) > 0 for t > 0.

Corollary 3.3. Let b ∈ CΩ, (0,∞)

. If f satisfies the Keller–Osserman condition (2.4), then problem (1.2) admits a solution

u ∈ C(Ω).

Proof. Note that if h(x, t) = b(x)f (t), then h satisfies all the hypothesis of Theorem 3.2, and hence the conclusion holds.

Next we prove a non-existence result. For this we introduce the following definition.For each (x, r) ∈ Ω × (0,∞), define

h∗(t; x, r) := maxh(z, t) : z ∈ B(x, r) ∩Ω.

When no possible ambiguity ariseswewill suppress the dependence on the parameters x and r and simply denote h∗(t; x, r)by h∗(t). Again we note that h∗ is continuous and non-decreasing on [0,∞)with h∗(0) = 0. Given x ∈ Ω , we will need thefollowing growth condition at infinity on h∗(t; x, r) for some r > 0∫

∞

1

13√h∗(t; x, r)

dt = ∞. (3.6)

Once again, it is easy to see that if condition (3.6) holds at x ∈ Ω for some r > 0, then it holds at (x, s) for all 0 < s < r .We introduce the following function:

β(t) =1t

∫ 2t

th∗(s; x, r)ds, t > 0.

One can verify that β is a non-decreasing C1 function on (0,∞) with β(0) = 0, and β(t) > 0 for t > 0. As a consequenceof the monotonicity of h∗ we note that

h∗(t) ≤ β(t) ≤ h∗(2t), t > 0. (3.7)

By virtue of the second inequality in (3.7), condition (3.6) holds for β whenever it holds for h∗. This remark will be useful inproving the following non-existence result.

Theorem 3.4. Suppose that h satisfies condition (3.6) at some x0 ∈ ∂Ω . Then problem (1.1) has no super-solution u ∈ C(Ω).

Proof. Assume to the contrary that problem (1.1) has a super-solution u ∈ C(Ω). Let (x0, r) ∈ ∂Ω × (0,∞) be as in thehypothesis of the theorem. We assume r is sufficiently small that u > ε > 0 inΩB := B(x0, r)∩Ω for some ε. We considerthe following smooth increasing function η : [ε,∞) → [0,∞).

η(t) :=

∫ t

ε

13√β(s)

ds, t ≥ ε. (3.8)

We set

v(x) := η(u(x)), x ∈ ΩB.

We now claim that∆∞v ≤ 1 inΩB.To prove the assertionwe proceed as follows. Let ϕ ∈ C2(ΩB) and suppose v−ϕ has a local minimum at z ∈ ΩB. Without

loss of generality we suppose that v(z) = ϕ(z) so that, for some small δ > 0,

v(x) ≥ ϕ(x), ∀ x ∈ N(z) := Bδ(z) ⊆ ΩB.

Since v(z) > 0, we may assume that δ is small enough that ϕ > 0 on N(z).Since η is increasing, we have

u(z) = η−1(ϕ(z)) and u(x) ≥ η−1(ϕ(x)) x ∈ N(z).

Note that η−1(ϕ) is in C2(N(z)) and we extend it to ψ ∈ C2(ΩB) so that u − ψ has a local minimum at z. Since u is asuper-solution of (1.1) we see that

∆∞ψ(z) ≤ h(z, u(z)). (3.9)

Computing in N(z)we find

∆∞ψ = (η−1)′′(ϕ)((η−1)′(ϕ))2|Dϕ|4+ ((η−1)′(ϕ))3∆∞ϕ

=13(β(η−1(ϕ)))1/3β ′(η−1(ϕ))|Dϕ|

4+ β(η−1(ϕ))∆∞ϕ. (3.10)

Since β is non-decreasing it follows from (3.10) that

∆∞ψ ≥ β(η−1(ϕ))∆∞ϕ x ∈ N(z).


Recalling (3.9) and evaluating the above inequality at z, we find that

h(z, u(z)) ≥ ∆∞ψ(z) ≥ β(ψ(z))∆∞ϕ(z).

Due to the first inequality in (3.7) we obtain

∆∞ϕ(z) ≤h(z, u(z))β(u(z))

≤ 1.

Thus we have, in the viscosity sense,

∆∞v ≤ 1 inΩB (3.11)

as claimed. Recalling that β satisfies condition (3.6) and u = ∞ on ∂Ω , we see that v = ∞ on B(x0, r) ∩ ∂Ω .We now proceed to show that the conclusion in (3.11) leads to a contradiction.Let χ be a continuous function supported on ∂ΩB such that 0 ≤ χ ≤ 1. Further assume that χ ≡ 1 in B(x0, r/3) ∩ ∂Ω

and χ ≡ 0 on ∂ΩB \ (B(x0, r/2) ∩ ∂Ω). Now letw be the solution of∆∞w = 1 inΩBw = χ on ∂ΩB.

By the Comparison principle, Lemma 2.4 and recalling (3.11) we note that

kw ≤ v inΩB, k ≥ 1.

By choosing x1 ∈ ΩB such thatw(x1) > 0, we see in particular that

kw(x1) ≤ v(x1)

for all k. This is an obvious contradiction.

We illustrate the above theorems with a simple example. Let h(x, t) = tq(x) where q ∈ C(Ω, (0,∞)). That is we considerthe problem

∆∞u = uq(x) inΩu = ∞ on ∂Ω. (3.12)

This problemwas studied in the paper [18] in the case when the infinity Laplacian∆∞ is replaced by the Laplacian,∆. Morequestions about problem (3.12) are explored in [18] than considered here.

For t ≥ 1, we observe that

h∗(t; x, r) = tminq(z):z∈B(x,r) and h∗(t; x, r) = tmaxq(z):z∈B(x,r).

Suppose q(x0) < 3 at some point x0 ∈ ∂Ω . Since q ∈ C(Ω, (0,∞)), there is r0 > 0 such that q(x) ≤ γ < 3 on B(x0, r0)∩Ω .Then h∗(t; x0, r0) ≤ tγ for all t ≥ 1, and hence condition (3.6) holds. Therefore Theorem 3.4 shows that problem (3.12) hasno super-solution. That is for problem (3.12) to admit a super-solution, it is necessary that q(x) ≥ 3 on ∂Ω .

Assume now that q(x) > 3 for all x ∈ ∂Ω . There is δ > 0 such that q(x) ≥ α > 3 in Ωδ . Thus, for each x ∈ Ωδ thereis rx > 0 such that h∗(t; x, rx) ≥ tα for all t ≥ 1. As a result condition (3.5) holds, and therefore Theorem 3.2 implies thatproblem (3.12) has a solution inΩ .

Going back to the separable case h(x, t) = b(x)f (t) we get the following non-existence result, an improvement ofTheorem 3.4. For instance the following theorem does not follow from Theorem 3.4 when h(x, t) = t3 log4(t + 1). Togetherwith Corollary 3.3, the following theorem provides a necessary and sufficient condition for problem (1.2) to have a solution.This is reminiscent of similar results in [20], in the context of the Laplace operator, and more generally the p-Laplacian.

Theorem 3.5. Suppose that f fails to satisfy the Keller–Osserman condition (2.4), and that b ∈ C(Ω,R). Then (1.2) has nosuper-solution in C(Ω).

Proof. By way of contradiction, let us assume that there is such u. Then ∆∞u ≤ g(u) where M > maxΩ b is positive andg(t) := Mf (t). Let x0 ∈ Ω . Since f fails to satisfy the Keller–Osserman condition (2.4), given a > u(x0)we apply Lemma 2.7to find a solution v ∈ C(RN) of ∆∞v = g(v) in RN with v(x0) = a. Since Ω is bounded, we see that v ≤ u in Ω by theComparison principle, Lemma 2.5. In particular a = v(x0) ≤ u(x0). This is a contradiction.

4. Boundary asymptotic estimates and uniqueness

In this section we wish to establish boundary asymptotic estimates to solutions of problem (1.2) in bounded smoothdomains Ω ⊆ RN . Throughout this section we will always assume that f is a non-negative, non-decreasing continuousfunction defined on [0,∞) such that f (0) = 0 and f (t) > 0 for t > 0. Furthermore we assume that b ∈ C(Ω) is positiveinΩ .


The asymptotic boundary estimates developed will then be used to study the uniqueness question associated withproblem (1.2) when b is a positive constant. More specifically the boundary estimate will lead to the proof of uniqueness ofsolutions of such problem provided that f satisfies the following strong monotonicity condition on f .

f (t)t3

is non-decreasing on (0,∞). (4.1)

We start with the following comparison principle for solutions of problem (1.2).

Lemma 4.1. Let f satisfy condition (4.1). Let u ∈ C(Ω) be a sub-solution and v ∈ C(Ω) a super-solution of (1.2) such that

lim supx→∂Ω

u(x)v(x)

≤ 1. (4.2)

Then u ≤ v inΩ .

Proof. Let ε > 0 be given. Then by the assumption in (4.2), there is δ > 0 such that u ≤ (1 + ε)v inΩδ , where we recall

Ωδ := x ∈ Ω : dist (x, ∂Ω) < δ.

By virtue of (4.1) we observe that

∆∞((1 + ε)v) = (1 + ε)3∆∞v = (1 + ε)3b(x)f (v) ≤ b(x)f ((1 + ε)v).

Let O := Ω \Ωδ . Since u ≤ (1 + ε)v on ∂O, by the comparison principle we see that u ≤ (1 + ε)v on O. Therefore

u ≤ (1 + ε)v inΩ.

Since ε > 0 is arbitrary, the claim follows.

Throughout this section we will suppose that Ω is a bounded C1 domain in RN . Let us fix some notations and state a fewfacts that we shall need in this section. Let us recall that

dΩ(x) := dist (x, ∂Ω) (x ∈ Ω).

It is known, [42], that there is µ > 0, depending on Ω , such that dΩ ∈ C1(Ωµ) where Ωµ := x ∈ Ω : dΩ(x) < µ.Moreover |DdΩ(x)| = 1 on Ωµ, and consequently ∆∞d = 0 on Ωµ. The fact that the distance function d is ∞-harmonic(i.e. ∆∞d = 0) in Ωµ allows us to obtain the following estimate. In the proof of the estimate we will use the followingnotations

b∗ = 4

infx∈Ω

b(x) and b∗= 4

supx∈Ω

b(x).

Suppose f satisfies the Keller–Osserman condition (2.4). Let ϑ be the inverse of the decreasing function ψ : (0,∞) →

(0, ψ(0))

ψ(t) :=

∫∞

t

14√4F(s)

ds (t > 0). (4.3)

We observe that ϑ is a decreasing function with ϑ(t) → ∞ as t → 0+. Furthermore, a direct computation shows that

(ϑ ′(t))2ϑ ′′(t) = f (ϑ(t)), 0 < t < ψ(0).

We shall also need the following condition in our subsequent discussion.

b∗(diam(Ω)) < ψ(0) :=

∫∞

0

14√4F(t)

dt. (4.4)

Remark 4.2. We remark that if∫ 1

0

14√4F(t)

dt = ∞, (4.5)

then (4.4) does not put any restriction on the domainΩ or the function b. In particular, we note that if f satisfies the strongmonotonicity condition (4.1), then (4.5) holds and hence condition (4.4) is true for any bounded domainΩ , and b ∈ C(Ω).

We now state and prove a boundary estimate on any solution u ∈ C(Ω) of (1.2).

Theorem 4.3. Let Ω be a bounded C1 domain, b ∈ C(Ω, (0,∞)) and suppose f satisfies the Keller–Osserman condition (2.4).Further assume that condition (4.4) holds. There are constants γ > 0 and 0 < ρ ≤ µ such that the following estimate holds forany positive solution u ∈ C(Ω) of (1.2).


ϑ

4supΩ

bdΩ(x)

≤ u(x) ≤ ϑ

4infΩ

bdΩ(x)

+ γ (x ∈ Ωρ). (4.6)

Proof. Let us take a fixed but arbitrary z ∈ ∂Ω and consider

w(x) := ϑ(b∗|x − z| + ε) (x ∈ Ω),

for any ε > 0 such that b∗diam(Ω)+ε < ψ(0). Note thatw depends on z and ε, but we choose to suppress this dependencefor notational simplicity as this would not cause any confusion. By virtue of (4.4) we observe thatw is defined onΩ , and issmooth there. Computation shows that

∆∞w(x) = (b∗)4ϑ ′(b∗

|x − z| + ε)2ϑ ′′(b∗

|x − z| + ε)|D(|x − z|)|4 + (ϑ ′(b∗|x − z| + ε))3∆∞(b∗

|x − z| + ε)

= (b∗)4f (w) ≥ b(x)f (w)

holds inΩ . Note also thatw(x) ≤ ϑ(ε) inΩ . Therefore if u ∈ C(Ω) is a solution of (1.2) then by the comparison principlewesee thatw(x) ≤ u(x) inΩ . For each x ∈ Ω wepick zx ∈ ∂Ω such that dΩ(x) = |x−zx|. Sincew(y) = ϑ(b∗

|y−zx|+ε) ≤ u(y)for all y ∈ Ω , in particular we have ϑ(b∗ dΩ(x)+ ε) = w(x) ≤ u(x) inΩ . As ε is arbitrary we get the lower bound in (4.6).

To get the upper bound we proceed as follows. First we take 0 < ρ ≤ µ such that b∗ dΩ(x) ≤ ψ(0) for all x ∈ Ωρ .Let us now estimate u on x ∈ Ω : dΩ(x) = ρ as follows. Fix x0 ∈ Ω with d(x0) = ρ. For 0 < r < ρ, sufficientlysmall, consider the ball B(x0, r) ⊆ Ω . Let v(x) = φ(|x − x0|) be the radial function (see Lemma 2.7 and its proof) such that∆∞v = b4

∗f (v) in B(x0, r) and satisfies v(x) → ∞ as x → ∂B. By the comparison principle we see that u ≤ v in B. In

particular u(x0) ≤ v(x0) = φ(0). Therefore we conclude that

u(x) ≤ γ for all x ∈ Ω with dΩ(x) = ρ,where γ := φ(0). (4.7)

Note that γ := φ(0) depends only on r, infΩ b and f .For 0 < ε < ρ let Dε := x ∈ Ω : ε < dΩ(x) < ρ and dΩ, ε(x) := dΩ(x) − ε. We set w(x) := ϑ( b∗ dΩ, ε(x)) + γ for

x ∈ Dε where γ is the constant in (4.7). Obviously we see that

u(x) < ϑ(b∗µ)+ γ for x ∈ Ω with dΩ(x) = ρ.

We observe that w(x) → ∞ as dΩ(x) → ε. Therefore, recalling (4.7), we note that u < w on x ∈ Ω : dΩ(x) = ε foreach 0 < ε < ρ. We now proceed to show that ∆∞w ≤ b(x)f (w) in Dε . Let x0 ∈ Dε and ϕ ∈ C2(Dε) such thatw ≥ ϕ in a neighborhood N of x0, and w(x0) = ϕ(x0). Then ψ := ζ (ϕ) ∈ C2(Dε) where ζ is the inverse of the functionη(s) := ϑ(b∗s) + γ . Furthermore dΩ,ε(x) ≤ ψ(x) in N , and dΩ,ε(x0) = ψ(x0). Since dΩ(x) is infinity-harmonic in Dε , wenote that∆∞ψ(x0) ≥ 0. Direct computation shows that

∆∞ψ = (ζ ′(ϕ))2ζ ′′(ϕ)|Dϕ|4+ (ζ ′(ϕ))3∆∞ϕ.

Since∆∞ψ(x0) ≥ 0 and ζ ′ < 0 we see that

∆∞ϕ(x0) ≤ −(ζ ′(ϕ(x0)))−1ζ ′′(ϕ(x0))|Dϕ(x0)|4.

Note that since |DdΩ(x)| = 1 for x ∈ Dε , and dΩ,ε − ψ has a local maximum at x0, we have

1 = |DdΩ,ε(x0)| = |ζ ′(ϕ(x0))Dϕ(x0)|.

Therefore

∆∞ϕ(x0) ≤ −(ζ ′(ϕ(x0)))−5ζ ′′(ϕ(x0)). (4.8)

We now compute ζ ′ and ζ ′′.

ζ ′(s) =1

η′(ζ (s))=

b∗ϑ

′(b∗ζ (s))−1

ζ ′′(s) = −b2∗

b∗ϑ

′(b∗ζ (s))−2

ϑ ′′(b∗ζ (s))ζ ′(s)

= −b2∗

b∗ϑ

′(b∗ζ (s))−3

ϑ ′′(b∗ζ (s)).

On recalling that ϑ ′′(s) = f (ϑ(s))(−ϑ ′(s))−2, the inequality in (4.8) becomes

∆∞ϕ(x0) ≤ b4∗

ϑ ′(b∗ζ (ϕ(x0)))

2ϑ ′′(b∗ζ (ϕ(x0)))

= b4∗f (ϑ(b∗ζ (ϕ(x0))))

≤ b4∗f (ϑ(b∗ζ (ϕ(x0)))+ γ )

= b4∗f (η(ζ (ϕ(x0))))

= b4∗f (ϕ(x0)) ≤ b(x)f (w(x0)).


Therefore ∆∞w ≤ b(x)f (w) in Dε in the viscosity sense. Since u ≤ w on ∂Dε , we conclude from the comparison principlethat u ≤ w in Dε . Letting ε → 0 we conclude that

u(x) ≤ ϑ(b∗dΩ(x))+ γ for x ∈ Ωρ . (4.9)

This provides the upper estimate in (4.6).

Theorem 4.4. Let b(x) ≡ c and suppose f satisfies (2.4) and (4.1). Then problem (1.2) has a unique solution u ∈ C(Ω).

Proof. We only need to show uniqueness. Let u and v be solutions of (1.2). Then, by takingµ sufficiently small if necessary,we can assume that u > 0 and v > 0 inΩµ. By Theorem 4.3, and keeping Remark 4.2 in mind, we see that

u(x)v(x)

≤ϑ( 4

√cdΩ(x))+ γ

ϑ( 4√cdΩ(x))

.

This implies that

lim supx→∂Ω

u(x)v(x)

≤ 1.

By Lemma 4.1 we conclude that u ≤ v inΩ . By reversing the roles of u and v in the above argument, we also conclude thatv ≤ u inΩ . This establishes uniqueness.

Under an additional assumption on f , we can also compare two solutions of (1.2). The following condition, introducedin [13], will be very useful in investigating some boundary behavior of solutions to (1.2).

lim inft→∞

ψ(γ t)ψ(t)

> 1, for any 0 < γ < 1,where ψ(t) =

∫∞

t

ds4√4F(s)

. (4.10)

The following lemma, whose proof is given in the Appendix, will be useful. See also [13].

Lemma 4.5. Suppose (4.10) holds. Then given 0 < α ≤ β , there is a positive number δ = δ(α, β) and a positive integerk = k(α, β) such that

ϑ(αt) ≤

β

α

k

ϑ(βt) for all 0 < t < δ.

We use the above lemma to prove the following.

Theorem 4.6. Suppose b ∈ C(Ω, (0,∞)), and f satisfies conditions (2.4), (4.1) and (4.10). There is a positive number σ , thatdepends on f and b only such that for any positive solutions u, v ∈ C(Ω) of (1.2) we have inf

Ωb

supΩ

b

σ

v ≤ u ≤

supΩ

b

infΩ

b

σ

v inΩ. (4.11)

Proof. Let u and v be solutions of (1.2). Then, by taking µ sufficiently small if necessary, we can assume that u > 0 andv > 0 inΩµ. Using Theorem 4.3 (noting Remark 4.2 as well), followed by Lemma 4.5, we see that

u(x)v(x)

≤

ϑ

4infΩ

bdΩ(x)

+ γ

ϑ

4supΩ

bdΩ(x)

≤

4supΩ

b

4infΩ

b

k

+γ

ϑ

4supΩ

bdΩ(x) inΩδ for sufficiently small δ > 0.

Therefore, recalling that ϑ(t) → ∞ as t → 0, we have

limx→∂Ω

u(x)v(x)

≤

4supΩ

b

4infΩ

b

k

.


That is

limx→∂Ω

u(x)ρv(x)

≤ 1 where 1 ≤ ρ :=

4supΩ

b

4infΩ

b

k

.

Since f satisfies (4.1) we note that ρv is a super-solution of (1.2) and hence by Lemma 4.1 we have

u ≤

supΩ

b

infΩ

b

k/4

v inΩ.

Reversing the roles of u and v leads to the desired inequality (4.11).

Acknowledgments

We would like to thank the anonymous referee for reading the manuscript carefully and for the useful suggestions thatimproved the clarity of the paper. The first author would also like to express his thanks to Tilak Bhattacharya for manyhelpful discussions about the Infinity Laplacian and its properties.

Appendix

Consider the initial value problem (2.6). By a solution φ to (2.6) we mean φ ∈ C2((0, λ)) ∩ C1([0, λ)) for some λ > 0that satisfies the integral equation

φ(r) = a +

∫ r

0

3

∫ t

0g(φ(s))ds

1/3

dt, 0 < r < λ. (A.1)

It follows from the existence theory of ordinary differential equations that problem (A.1) has a solution in the intervalI = [0, r) for some 0 < r ≤ ∞.

In this Appendix we prove the following.

Lemma A.1. Suppose g is a non-decreasing, continuous function defined on [0,∞) such that g(0) = 0, and g(t) > 0 for t > 0.

(1) If g satisfies the Keller–Osserman condition (2.4), then there is 0 < R∗≤ ∞ such that for any given 0 < R < R∗, one can

find a > 0 for which problem (2.6) has a solution φ ∈ C2((0, R)) ∩ C1([0, R)) which satisfies

φ(r) → ∞ as r → R.

(2) If g does not satisfy the Keller–Osserman condition (2.4), then for each a > 0 problem (2.6) has a solution φ ∈ C2((0,∞))∩C([0,∞)) such that

φ(r) → ∞ as r → ∞.

Proof. Let a > 0. The initial value problem (2.6) is known to have a solution in [0, λ) for some λ > 0. Let 0 < Ra ≤ ∞

be such that [0, Ra) is the maximal interval of existence. Note that φ is increasing in [0, Ra) and φ(r) → ∞ as r → Ra.Multiplying both sides of the ODE in (2.6) by φ′ and integrating on (0, r) for 0 < r < Ra we find that

14φ′(r)4 =

∫ φ(r)

ag(s)ds = G(φ(r))− G(a), where G(t) =

∫ t

0g(s)ds.

Since φ′ > 0 we see that

φ′(r)(4(G(φ(r))− G(a)))1/4

= 1 0 < r < Ra.

We now integrate this once more in (0, r) and we get∫ φ(r)

a

1(4(G(t)− G(a)))1/4

dt = r, 0 < r < Ra. (A.2)

We observe that G(t)− G(a) ≥ g(a)(t − a) holds for t ≥ a. Therefore for t ≥ 2a, we see that G(t)− G(a) ≥ ag(a) ≥ G(a)so that

12G(t)− G(a) ≥ 0, or G(t)− G(a) ≥

12G(t), (t ≥ 2a).


Therefore we see that

14√4

∫ φ(r)

2a

1G(t)1/4

dt ≤

∫ φ(r)

2a

1(4(G(t)− G(a)))1/4

dt ≤14√2

∫ φ(r)

2a

1G(t)1/4

dt. (A.3)

Thus if the Keller–Osserman condition (2.4) fails to hold, it follows from the first inequality in (A.3) and the relation (A.2)that Ra = ∞ for each a > 0. We also note that any φ that satisfies the relation (A.2) also satisfies the initial value problem(2.6).

Now let us suppose that f satisfies the Keller–Osserman condition (2.4). Then again, from the second inequality in (A.3)we see that lima→∞ H(a) = 0, where

H(a) :=

∫∞

a

1(4(G(t)− G(a)))1/4

dt, a > 0.

Wemake the change of variable s = G(t)− G(a) to rewrite the above integral as

H(a) =

∫∞

0

1g(G−1(s + G(a)))

1(4s)1/4

ds. (A.4)

We note that themonotonicity of g(G−1) implies thatH is decreasing. Let us now show thatH is continuous on (0,∞). Notethat

1g(G−1(s + G(a)))

1(4s)1/4

≤1

g(G−1(s))1

(4s)1/4,

and ∫∞

ε

1g(G−1(s))

1(4s)1/4

ds =

∫∞

G−1(ε)

1(4G(s))1/4

ds < ∞,

for any ε > 0. Therefore the Lebesgue Dominated Convergence theorem shows that H is continuous at any a > 0.Furthermore, from (A.4) and the Monotone Convergence Theorem, we see that

lima→0

H(a) =

∫∞

0

1g(G−1(s))

1(4s)1/4

ds =

∫∞

0

1(4G(s))1/4

ds.

Note that eitherH(0) < ∞ orH(0) = ∞ is possible. For instance,H(0) = ∞ for g(t) = t4, andH(0) < ∞ for g(t) = et −1.Now we go back to proving the lemma. Let R∗

:= H(0). Given 0 < R < R∗, we choose a > 0 such that H(a) = R, that is∫∞

a

1(4(G(t)− G(a)))1/4

dt = R.

Set

ψ(t) =

∫ t

a

1(4(G(t)− G(a)))1/4

dt, a ≤ t < ∞.

Note that ψ : [a,∞) → [0, R) is an increasing, continuous function that belongs to C2((0,∞)). Let φ be the inverse of ψ .Clearly φ ∈ C2((0, R)) ∩ C1([0, R)) and note that∫ φ(r)

a

1(4(G(t)− G(a)))1/4

dt = r, 0 ≤ r < R. (A.5)

Furthermore we observe that φ(R) = ∞ and φ(0) = a. On differentiating (A.5) we find that

φ′(r) = (4(G(φ(r))− G(a)))1/4,φ′′(r) = (4(G(φ(r))− G(a)))−3/4g(φ(r))φ′(r).

Therefore

(φ′(r))2φ′′(r) = g(φ(r)), 0 < r < R.

Let ψ : (0,∞) → (0,∞) be a decreasing function such that


lim inft→∞

ψ(γ t)ψ(t)

> 1 for each 0 < γ < 1. (A.6)

The following lemma was used in Section 4 to prove Theorem 4.6. As it may be of independent interest, we state it in amore general form without any reference to the function f used in problem (1.2).

Lemma A.2. Suppose ψ : (0,∞) → (0,∞) is a decreasing function such that (A.6) holds. Then given 0 < α ≤ β , there is apositive number δ = δ(α, β) and a positive integer k = k(α, β) such that

ϑ(αt) ≤

β

α

k

ϑ(βt) for all 0 < t < δ,

where ϑ is the inverse of ψ .

Proof. Obviously the inequality is truewhenα = β . Thereforewe assumeα < β , so that γ := αβ−1 < 1. Then by condition(A.6), we see that

lim inft→∞

ψ(γ t)ψ(t)

= ν for some ν = ν(γ ) > 1.

We fix 1 < ω < ν. Then there is tγ > 0 such that

ψ(γ t) ≥ ωψ(t) for all t ≥ tγ . (A.7)

Let τγ := ψ(tγ )α−1. Then for 0 < t < τγ we see that ϑ(αt) > ϑ(ψ(tγ )) = tγ and hence by (A.7) we get

ψ(γϑ(αt)) ≥ ωψ(ϑ(αt)) = ωαt.

That is

ϑ(αt) ≤ γ−1ϑ(ωαt) 0 < t < τγ . (A.8)

Now let ℓ be the smallest positive integer such that ωℓα ≥ β and let

δ :=τγ

ωℓ.

On noting that 0 < ωkt < τγ for k = 1, 2, . . . , ℓ, we now iterate the inequality (A.8) for 0 < t < δ. We get

ϑ(αt) ≤ γ−1ϑ(ωαt) = γ−1ϑ(α(ωt))≤ γ−2ϑ(ω(α(ωt))) = γ−2ϑ(α(ω2t))...

≤ γ−ℓϑ(ωℓ(αt)) = γ−ℓϑ((ωℓα)t).

Since ϑ is decreasing and ωℓα ≥ β we get the desired inequality.

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On boundary blow-up solutions to equations involving the ∞ Laplacian

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