The existence and asymptotic behavior of boundary blow-up solutions to the k-Hessian equation

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ScienceDirect J. Differential Equations 267 (2019) 4626–4672 www.elsevier.com/locate/jde

The existence and asymptotic behavior of boundary blow-up solutions to the k-Hessian equation Xuemei Zhang a,∗ , Meiqiang Feng b a Department of Mathematics and Physics, North China Electric Power University, Beijing, 102206, PR China b School of Applied Science, Beijing Information Science & Technology University, Beijing, 100192, PR China

Received 10 April 2018; revised 4 November 2018 Available online 10 May 2019

Abstract In this paper we consider the existence and asymptotic behavior of k-convex solution to the boundary blow-up k-Hessian problem Sk (D 2 u) = H (x)f (u) for x ∈ , u(x) → +∞ as dist(x, ∂) → 0, where k ∈ {1, 2, · · · , N }, Sk (D 2 u) is the k-Hessian operator,  is a smooth, bounded, strictly convex domain in RN (N ≥ 2), H ∈ C ∞ () is positive in , but is not necessarily bounded on ∂, and f is a smooth positive function that satisfies the so-called Keller-Osserman condition. Further results are obtained for the special case that  is a ball. Our approach to show the existence and asymptotic behavior, exploits the method of sub- and super-solutions and Karamata regular variation theory. © 2019 Elsevier Inc. All rights reserved. MSC: 34B18; 34B15; 34A34 Keywords: k-Hessian equation; Boundary blow up; Sub-supersolution method; k-convex solution; Existence and asymptotic behavior

* Corresponding author.

E-mail addresses: [email protected] (X. Zhang), [email protected] (M. Feng). https://doi.org/10.1016/j.jde.2019.05.004 0022-0396/© 2019 Elsevier Inc. All rights reserved.

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1. Introduction Let  ⊂ RN (N ≥ 2) be a smooth, bounded, strictly convex domain. We consider the boundary blow-up problem for the k-Hessian equation Sk (D 2 u) = H (x)f (u) in , u = +∞ on ∂,

(1.1)

where H and f are smooth positive functions. The boundary blow-up condition u = +∞ on ∂ means u(x) → +∞ as dist(x, ∂) → 0. Sk (D 2 u)(k ∈ {1, 2, . . . , N }) denotes the kth elementary symmetric function of the eigenvalues of D 2 u, the Hessian of u, i.e. Sk (D 2 u) = Sk (λ1 , λ2 . . . , λN ) =



λi 1 . . . λ i k ,

1≤i1 <···
where λ1 , λ2 . . . , λN are the eigenvalues of D 2 u. It is not hard to observe that {Sk : k ∈ {1, 2, . . . , N }} contains a family of operators: for k = 1, the Laplace operator; for k = N , the Monge-Ampère operator; and many other well known operators. Many authors have demonstrated increasing interest in the subject of Laplace problems and Monge-Ampère problems, and many excellent results for Laplace problems and Monge-Ampère problems have been obtained, for instance, see ([1], [8], [9], [10], [11], [18], [21], [22]) and the references cited therein. The k-Hessian problem arises in geometric problems, fluid mechanics and other applied subjects. For example, under the case k = N , the k-Hessian problem can describe Weingarten curvature, or reflector shape design (see [2]). In contrast to numerous results on the case k = 1 or k = N less is known about the situation k ∈ {2, . . . , N − 1}, since the case of k ∈ {2, . . . , N − 1} becomes even more complicated. Only in recent years, increasing attention has been paid to the study of the k-Hessian problems by different methods (see by instance [3], [4], [5], [12], [13], [14], [15], [16], [17], [19], [20], [23], [24], [25]). For k ∈ {1, 2, . . . , N }, letting k be the component of {λ ∈ RN : Sk (λ) > 0} ⊂ RN containing the positive cone + = {λ ∈ RN : λi > 0, i = 1, 2, . . . , N }, then it follows from [4] that + = N ⊂ · · · ⊂ k+1 ⊂ k ⊂ · · · ⊂ 1 . Definition 1.1. (See [16]) Let k ∈ {1, 2, . . . , N}, and let  be an open bounded subset of RN ; a function u ∈ C 2 () is k-convex if (λ1 , λ2 , . . . , λN ) ∈ k for every x ∈ , where λ1 , λ2 , . . . , λN are the eigenvalues of D 2 u. Equivalently, we can say that u is k-convex if Si (D 2 u) ≥ 0 in  for i = 1, . . . , k.

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Definition 1.2. (See [16]) Let  ⊂ RN be an open set with boundary of class C 2 and let k ∈ {1, . . . , N −1}; we say that  is strictly convex if Si (κ1 (x), . . . , κN−1 (x)) > 0, for i = 1, . . . , N − 1 and for every x ∈ ∂, where κi (x), i = 1, . . . , N − 1, are the principal curvatures of ∂ at x. Notice that a C 2 function u is N -convex if and only if it is convex, and it is 1-convex if and only if it is subharmonic; analogously a C 2 domain is (N − 1)-convex if and only if it is convex. It is generally known that the concept of k-convexity comes naturally from the involved op¯ Sk (D 2 u) changes into erator Sk . In fact, when  is strictly (k − 1)-convex and H ∈ C ∞ (), elliptic in the class of k-convex functions and the related Dirichlet problem  Sk (D 2 u) = H (x) > 0 in , (1.2) u|∂ = φ ∈ C(∂) admits a unique k-convex solution. Moreover the (k − 1)-convexity of the domain is necessary if φ is constant (see [4]). Meanwhile, we observe that the subject of blow-up solutions has received much attention starting with the pioneering work of Bieberbach [6]. It was about the following model with the classical Laplace operator  u = H (x)f (u) in , (1.3) u = +∞ on ∂ with H (x) ≡ 1 in , f (u) = eu and N = 2. If f (u) = up , H (x) is growing like a negative power of d(x) near ∂, the radial case was completely discussed in [7], the general case was discussed when p > 1 in [26], the authors obtained existence, nonexistence, uniqueness, multiplicity and estimates for all positive solutions. When f may exhibit non monotone behavior, in [27], Dumont, Dupaigne, Goubet, and Radulescu studied the existence, asymptotic behavior, uniqueness and numerical approximation of solutions of (1.3). An overview of the asymptotic behavior of solutions of elliptic problems can be found in Ghergu and Radulescu [28]. Recently, in [29], Yang and Chang considered the boundary blow up solutions of MongeAmpère equations M[u] = K(x)up in , u = +∞ on ∂

(1.4)

with K(x) growing like a negative power of d(x) near ∂. They showed some results on the uniqueness, nonexistence, and the exact boundary blow up rate of the strictly convex solutions of (1.4). The latest results of blow-up solutions of partial differential equations can be found in ([30], [31]). Moreover, for k ≥ 2 we know that the k-Hessian operator is a fully nonlinear partial differential operator, and we notice that some fully nonlinear degenerate elliptic operators have attracted the attention of Harvey and Lawson ([32–35]), Caffarelli, Li and Nirenberg ([36], [37]), Amendola, Galise and Vitolo [38], Galise and Vitolo [39], Capuzzo-Dolcetta, Leoni and Vitolo [40], and Vitolo [41]. However, in literature there aren’t articles on boundary blow-up solutions to the k-Hessian equation with singular weights. More precisely, the study of H may be unbounded near ∂ is still open for the k-Hessian problem. In this paper, we aim to analyze the existence and the asymptotic behavior of k-convex solution to (1.1). Suppose H (x) and f (u) satisfy

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(H): H ∈ C ∞ () and H (x) > 0 in ; (f1): For η ∈ R, f ∈ C ∞ (η, ∞) is positive, strictly increasing and lim f (s) = 0; s→η

(f2): f satisfies Keller-Osserman type condition ∞

(r) = [(k + 1)F (s)]−1/(k+1) ds < ∞, ∀r > η, r

where s F (s) =

f (t)dt. η

By the arguments of Keller [18] and Osserman [42], it is easily checked (see, for example, section 6.1 in [1]) that if H ∈ C ∞ () is positive, and f satisfies (f1), then (1.3) has a strictly convex solution if and only if ∞ [F (s)]−1/2 ds < ∞.

(1.5)

When k = N , Matero [43] and Mohammed [44] showed that (i) (1.1) has a strictly convex solution if f satisfies (f1) and (f2)1 ; (ii) (1.1) has no strictly convex solution if ∞

f (s)−1/N ds = ∞.

(1.6)

Mohammed [44] proved that if H (x) satisfies (H) and is such that the Dirichlet problem M[u] = H (x) in , u = 0 on ∂

(1.7)

has a strictly convex solution, then (1.1) has a strictly convex solution if f satisfies (f1) and (f2),2 where M[u] = SN (D 2 u) is the Monge-Ampère operator. In [45], Zhang and Du showed that, in the case that η > −∞, (f2) alone does not guarantee the existence of a strictly convex solution to (1.1). One needs to require additionally 

[(k + 1)F (s)]−1/(N+1) ds = ∞.

η+

1 As explained below, the condition (f2) alone is actually not sufficient. 2 See footnote 1.

(1.8)

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Here

X. Zhang, M. Feng / J. Differential Equations 267 (2019) 4626–4672

 η+

(s)ds = ∞ means that η+ (s)ds = ∞ for all small positive . η

¯ satisfies (H), and f satisfies (f1), then In [45], Zhang and Du also proved that if H ∈ C ∞ () (f2) is also a necessary condition for (1.1) to have a strictly convex solution. ¯ is positive, it follows from Salani [16] that If H ∈ C ∞ () (i) (1.1) has a k-convex solution if f satisfies (f2); (ii) (1.1) has no k-convex solution if lim sup t→∞

f (t) = c0 < ∞. tk

(1.9)

In [12] and [23], the authors studied (1.1) with H (x) ≡ 1, η = 0 (in [23], η = −∞) and got a Keller-Osserman type condition, i.e., they showed that (1.1) has a solution if and only if (f2) holds. However, we find that, similar to Monge-Ampère case, when η > −∞, (f2) alone cannot not guarantee the existence of a k-convex solution to (1.1). We also need to require additionally 

[(k + 1)F (s)]−1/(k+1) ds = ∞.

(1.10)

η+

Obviously, (1.10) is equivalent to lim (r) = ∞. r→η+

Let us observe that if f (u) = up with p > k, or f (u) = uk [ln(1 + u)]β with β > k + 1, then both (f2) and (1.10) are satisfied, for detail to see ([13], [46]). However, we find that the existence of solution and the boundary behavior was not considered in references when H may be unbounded near ∂. We would like to do some work on this topic. For getting better global estimate and boundary asymptotic behavior of solution we impose a stronger condition than (1.10). By replacing f (t) with f (t + η) and u with u − η, we may assume that η = 0. Let ψ satisfy ∞

[(k + 1)F (s)]−1/(k+1) ds = t, ∀t > 0.

ψ(t)

Then ψ is the inverse function of . Throughout the paper, the following notations are used: I (s) =

 (s) (s) (  (s))2

(1.11)

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and I0 = lim I (s), I∞ = lim I (s). s→0+

s→∞

The first main result of this paper is the following Theorem 1.3. ¯ is positive and that f satisfies (f1). Theorem 1.3. Suppose that H satisfies (H) and H ∈ C ∞ () (i) If f satisfies (f2) and I0 = ∞, then Problem (1.1) has a k-convex solution u such that ψ(c1 d(x)) ≤ u(x) ≤ ψ(c2 d(x)), x ∈ 

(1.12)

for some constants 0 < c2 < c1 . (ii) If Problem (1.1) has a k-convex solution then f satisfies (f2). Remark 1.4. We will prove in Section 3 that I0 = ∞ implies (1.10). If we replace I0 = ∞ with (1.10) in Theorem 1.3, then we can also obtain the existence of solution. But we can not obtain the global estimate (1.12). If H (x) is unbounded near ∂, problem (1.1) is not always having solution. The existence of solution depends on the increasing speed of H (x) and the increasing speed of f (u) when x approaches to ∂. For ease of composition, we first introduce some notations. For a positive function p(t) in C 1 (0, ∞) satisfying p  (t) < 0 and limt→0+ p(t) = +∞, to 1 distinguish its behavior near t = 0 we set P (τ ) = τ p(t)dt. We say such a function p is of class Pf inite if 

1

[P (τ )] k dτ < ∞, 0+

and is of class P∞ if 

1

[P (τ )] k dτ = ∞. 0+

If p is of class Pf inite , we may modify p(t) for large t and assume that p(t) = c0 e−t for some positive constant c0 and all large t , say t ≥ M0 . With p modified as above, if we define P˜ (τ ) =

∞ p(t)dt, τ

then we still have

 0+

Moreover,

[P˜ (τ )] k dτ < ∞. 1

(1.13)

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P˜ (t) = c0 e−t , P˜ (t)/p(t) = 1 for t ≥ M0 , P˜ (t)/p(t) → 0 as t → 0.

(1.14)

Set t

(k P˜ (τ )) k dτ for t > 0,

(1.15)

ω(s)ω (s) , J0 = lim J (s). (ω (s))2 s→0+

(1.16)

ω(t) :=

1

0

and J (s) = −

The following theorem deals with the case H (x) is unbounded near ∂. Theorem 1.5. Assume f satisfies (f1), (f2) and such that I0 = ∞ and that H satisfies (H) and there exists a function p of class Pf inite such that k2 p(d(x)) ≤ H (x) ≤ k1 p(d(x)) near ∂, here k1 > k2 > 0. If I∞ = ∞ and J0 = ∞ can not hold at same time and (f2) holds, then Problem (1.1) has a k-convex solution u, which satisfies k

k

ψ(c1 [ω(d(x))] k+1 ) ≤ u(x) ≤ ψ(c2 [ω(d(x))] k+1 ), x ∈ 

(1.17)

for some constants 0 < c2 < c1 . Remark 1.6. If we replace I0 = ∞ with (1.10) in Theorem 1.5, then we can also obtain the existence of solution. But we can not obtain the global estimate (1.17). The asymptotic behavior of boundary blow-up solutions is also a hot topic. When H ∈ ¯ [48] studied the asymptotic behavior of boundary blow-up solutions of (1.1). HowC ∞ (), ever, there is no such results when H ∈ C ∞ () is unbounded near ∂. Our next result will do this work. Theorem 1.7. Suppose that H, f satisfy the conditions of Theorem 1.5, then for the k-convex solution u obtained in Theorem 1.5, it holds u(x)

1 ≤ lim inf

x∈, d(x)→0

ψ[ξ (ω(d(x)))

k k+1

]

, lim sup x∈, d(x)→0

u(x) k

ψ[ξ (ω(d(x))) k+1 ]

where  ξ=



k2 k k k 1 ( k+1 ) Mk−1 [ k+1 J0 +

1 1 k+1 J0 I∞

+

1 I∞ ]

1 k+1

,

≤ 1,

X. Zhang, M. Feng / J. Differential Equations 267 (2019) 4626–4672

 ξ=



k1 k k k 1 ( k+1 ) mk−1 [ k+1 J0 +

1 1 k+1 J0 I∞

+

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1 k+1

,

1 I∞ ]

and Mk−1 = max Sk−1 (κ1 (y), κ2 (y), . . . , κN−1 (y)), y∈∂

mk−1 = min Sk−1 (κ1 (y), κ2 (y), . . . , κN−1 (y)), y∈∂

κ1 (y), κ2 (y), . . . κN−1 (y) are the principal curvatures of ∂ at y. ¯ then Corollary 1.8. In Theorem 1.7, if  is a ball of radius R, I∞ ∈ [1, ∞) and k1 = k2 = k, lim

x∈, d(x)→0

u(x) k

ψ[(ω(d(x))) k+1 ]

= ξ01−I∞ ,

where  ξ0 =



¯ k−1 kR k k k 1 ( k+1 ) [ k+1 J0 +

1 1 k+1 J0 I∞

+

1 I∞ ]

1 k+1

.

Remark 1.9. It is interesting to know what happens to Problem (1.1) if H (x) ∼ p(d(x)) near ∂ with p ∈ P∞ . We will examine some cases for the radially symmetric situation, and show that (1.1) may have infinitely many k-convex solutions or no such solution, depending on the behavior of f ; see Theorems 5.3 and 5.4 for details. The remainder of the article is arranged as follows. In Section 2 we collect several well-known results to be used in the proof of our main results. To study the asymptotic behavior of solution, we will introduce the theory of regular variation in Section 3. Section 4 provides the proof of Theorems 1.3 and 1.5, while in Section 5 we consider radial solutions and discuss the cases mentioned in Remark 1.9. Section 6 gives the proof of Theorem 1.7 and Corollary 1.8. 2. Some preliminary results In this part, we give some well-known results for the convenience of later utilization and reference. Lemma 2.1. (See [25], Lemma 2.1) Suppose that  ⊂ RN is a bounded domain, and u, v ∈ C 2 () are k-convex. If (1) (2) (3) (4)

ψ(x, z, p) ≥ φ(x, z, p), ∀ (x, z, p) ∈  × R × RN ; Sk (D 2 u) ≥ ψ(x, u, Du) and Sk (D 2 v) ≤ φ(x, v, Dv) in ; u ≤ v on ∂; ψz (x, z, p) > 0 or φz (x, z, p) > 0,

then u ≤ v in .

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Remark 2.2. It is not difficult to find that “ψz (x, z, p) > 0 or φz (x, z, p) > 0” in Lemma 2.1 can be relaxed to “ψz (x, z, p) ≥ 0 or φz (x, z, p) ≥ 0” provided that one of the inequalities in (2) is replaced by a strict inequality. This observation will be used later in the article. Lemma 2.3. Let u ∈ C 2 () be such that all of the principal submatrix of (uxi xj ) are invertible for x ∈ , and let g be a C 2 function defined on an interval containing the range of u. Then k

Sk (D 2 g(u)) = Sk (D 2 u)[g  (u)]k + [g  (u)]k−1 g  (u)

CN 

det (uxis xij )(∇ui )T B(ui )∇ui ,

(2.1)

i=1

where AT denotes the transpose of the matrix A, B(ui ) denotes the inverse of the i-th principal submatrix (uxis xij ), det (uxis xij ) is the determinant of (uxis xij ) and k ∇ui = (uxi1 , uxi2 , · · · , uxik )T , i = 1, 2, . . . , CN , k = here CN

N! (N−k)!k! .

k distinct principal minors, and S (D 2 g(u)) Proof. As is well known, the matrix (D 2 g(u)) has CN k equals to the sum of all the principal minor of size k of matrix (D 2 g(u)). For convenience, we only compute the i-th principal minor of size k. Denote i-th principal minor of size k by Di , then

   g (u)uxi1 uxi1 + g  (u)uxi1 xi1 · · · g  (u)uxi1 uxik + g  (u)uxi1 xik   g  (u)u u + g  (u)u · · · g  (u)uxi2 uxik + g  (u)uxi2 xik xi2 xi1 xi2 xi1  Di =  . .  .. .. ... ... ...   g  (u)u u + g  (u)u · · · g  (u)uxik uxik + g  (u)uxik xik xik xi1 xik xi1

     .   

The j th column of Di is the sum of two columns. The first of these two has entries g  (u)uxis xij (s = 1, 2, . . . , k) and the second has entries g  (u)uxis uxij (s = 1, 2, . . . , k). Since the determinant of a matrix is linear in each of its columns, Di can be expressed as a sum of 2k determinants where each summand has as its j th column one of the two types given above. Since for j = t the two columns col(uxi1 uxij , uxi2 uxij , . . . , uxik uxij ) and col(uxi1 uxit , uxi2 uxit , . . . , uxik uxit )

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are proportional, any of the 2k summands which have two different columns of the second type are zero. Therefore Di = [g  (u)]k D¯ i + [g  (u)]k−1 g  (u)

k 

D¯ ij ,

j =1

where D¯ i is the determinant whose (s, j )-th entry is uxis xij and D¯ ij (j = 1, 2 . . . , k) is the determinant obtained from D¯ i by replacement of the j th column of D¯ i by the column with entries uxis uxij (s = 1, 2, . . . , k). Therefore, if we denote the cofactor of the (s, j )-th entry of D¯ ij by Csj (ui ), then Di = [g  (u)]k det (uxis xij ) + [g  (u)]k−1 g  (u)

k  k 

Csj (ui )uxis uxij ,

j =1 s=1

where det (uxis xij ) is the i-th principal minor of size k of matrix (D 2 u). Since the matrix (uxis xij ) is symmetric, if we write ∇ui = col(uxi1 , uxi2 , · · · , uxik ), then, by the formula for the inverse of a matrix, k k  

Csj (ui )uxis uxij = det (uxis xij )(∇ui )T B(ui )∇ui ,

j =1 s=1

where B(ui ) is the inverse of the i-th principal submatrix (uxis xij ). Thus k

Sk (D g(u)) = 2

CN 

Di

i=1 CN     k  k−1  T [g (u)] det (uxis xij ) + [g (u)] g (u)det (uxis xij )(∇ui ) B(ui )∇ui = k

i=1 k





k−1 

= Sk (D u)[g (u)] + [g (u)] 2

k

g (u)

CN  i=1

det (uxis xij )(∇ui )T B(ui )∇ui .

2

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¯ : d(x, ∂) < μ} for μ > 0. In the following lemma, we denote μ by μ = {x ∈  Lemma 2.4. (Corollary 2.3 of [15]) Let  be bounded with ∂ ∈ C l for l ≥ 2. Assume that μ > 0 is small such that d ∈ C 2 (μ ) and g is a C 2 -function on (0, μ). Let x0 ∈ μ \ ∂ and y0 ∈ ∂ be such that d(x0 ) = |x0 − y0 |. Then we have Sk (D 2 g(d(x0 ))) = [−g  (d(x0 )]k Sk (ε1 , . . . , εN−1 ) + [−g  (d(x0 )]k−1 g  (d(x0 ))Sk−1 (ε1 , . . . , εN−1 ),

(2.2)

where εi =

κi (y0 ) , i = 1, 2, . . . , N − 1, 1 − κi (y0 )d(x0 )

and κ1 (y0 ), . . . , κN−1 (y0 ) are the principal curvatures of ∂ at y0 . The following interior estimate for derivatives of smooth solutions is a simple variant of Lemma 2.2 in [21], which can be proved by following the idea of Theorem 3.1 and Remark 3.1 of [3]. Lemma 2.5. Let  be a bounded strictly (k − 1)-convex domain in RN , N ≥ 2, with ∂ ∈ C ∞ . Let η ∈ [−∞, +∞) and f ∈ C ∞ ( × (η, ∞)) with f (x, u) > 0 for (x, u) ∈  × (η, ∞). Let u ∈ C ∞ () be a solution of the Dirichlet problem 

Sk (D 2 u) = f (x, u),

x ∈ , u(x) = c = constant, x ∈ ∂

(2.3)

with η < u(x) < c in . Let  be a subdomain of  with  ⊂  and assume that η < a ≤ u(x) ≤ b for x ∈  and let τ ≥ 1 be an integer. Then there exists a constant C which depends only on τ, a, b, bounds for the derivatives of f (x, u) for (x, u) ∈  × [a, b], and dist( , ∂) such that ||u||C τ ( ) ≤ C. The existence result below is a special case of Theorem 1.1 in [16,17]. Lemma 2.6. Let  be an open domain in RN with boundary of class C ∞ and let g(x, t) be a C ∞ function such that g > 0 and gt ≥ 0 in  × R. The problem 

Sk (D 2 u) = g(x, u), u|∂ = ϕ ∈ C(∂)

x ∈ ,

has a unique k-convex solution if there exists a k-convex strict subsolution v, i.e. a k-convex function v such that v|∂ = ϕ and Sk (v) ≥ g(x, v) + δ in , for some δ > 0.

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For the convenience of proofs in Sections 3-5, we introduce a function z(x). Let  be a smooth, bounded, strictly convex domain in RN , by Theorem 1.1 of [5], there exists u0 ∈ C ∞ (), which is the unique strictly convex solution to M[u0 ] = 1 in , u0 = 1 on ∂, where M[u0 ] = det (u0ij ) is the Monge-Ampère operator. Set z(x) := 1 − u0 (x). Then z(x) > 0 in  and it is the unique strictly concave solution to (−1)N M[z] = 1 in , z = 0 on ∂.

(2.4)

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

(2.5)

Remark 2.7. Since (zxi xj ) is negative definite, the eigenvalues are negative, then, for all principal minor k of size k to det (D 2 z), we have (−1)k k > A for some positive constant A, and then k. (−1)k Sk (D 2 z) > ACN 3. Regular variation theory To consider the asymptotic behavior of k-convex solution to problem (1.1), we will use Karamata regular variation theory which was introduced and established by Karamata in 1930, and it is a fundamental tool in stochastic processes (see [49], [50], [51]). In this part, we present some basic facts of Karamata regular variation theory. We also give the properties of I0 , I∞ , J0 , which are defined in Section 1. Definition 3.1. A positive measurable function f defined on [A, ∞), for some A > 0, is called regularly varying at infinity with index ρ ∈ R, written f ∈ RVρ , if for all ξ > 0, lim

s→∞

f (ξ s) = ξρ. f (s)

(3.1)

In particular, when ρ = 0, f is called slowly varying at infinity. Clearly, if f ∈ RVρ , then L(s) = fs(s) ρ is slowly varying at infinity. Definition 3.2. A positive measurable function f defined on [A, ∞), for some A > 0, is called rapidly varying at infinity if for each ξ > 1 lim

s→∞

f (ξ s) = ∞. f (s)

(3.2)

Proposition 3.3 (Uniform convergence theorem). If f ∈ RVρ , then (3.1) 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.

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Proposition 3.4 (Representation theorem). A function L is slowly varying at infinity if and only if it may be written in the form ⎛ ⎞ s ⎜ y(τ ) ⎟ L(s) = ψ(s)exp ⎝ dτ ⎠ , s ≥ A1 , τ A1

for some A1 ≥ A, where the function ψ and y are continuous and for s → ∞, y(s) → 0 and ψ(s) → c0 , with c0 > 0. We say that ⎞ ⎛ s ⎜ y(τ ) ⎟ ˆ dτ ⎠ L(s) = c0 exp ⎝ τ A1

is normalized slowly varying at infinity and ˆ f (s) = s ρ L(s), s ≥ A1 , is normalized regularly varying at infinity with index ρ and write f ∈ N RVρ . Proposition 3.5. A function f ∈ RVρ belongs to N RVρ if and only if sf  (s) = ρ. s→∞ f (s)

f ∈ C 1 [A1 , ∞) for some A1 > 0 and lim

Proposition 3.6. If function f, g, L are slowly varying at infinity, then 1) f p for every p ∈ R, c1 f + c2 g(c1 , c2 ≥ 0), f ◦ g(if g(s) → ∞ as s → ∞) are also slowly varying at infinity. 2) For every ρ > 0 and s → ∞, s ρ L(s) → ∞, s −ρ L(s) → 0. 3) For ρ ∈ R and s → ∞,

ln(L(s)) ln s

→ 0 and

ln(s ρ L(s)) ln s

→ ρ.

Proposition 3.7. If f1 ∈ RVρ1 , f2 ∈ RVρ2 , then f1 f2 ∈ RVρ1 +ρ2 and f1 ◦ f2 ∈ RVρ1 ρ2 . Proposition 3.8 (Asymptotic behavior). If a function L is slowly varying at infinity, then for a ≥ 0 and t → ∞, t 1) a s ρ L(s)ds ∼ = (1 + ρ)−1 t 1+ρ L(t), for ρ > −1; ∞ ρ 2) t s L(s)ds ∼ = (−1 − ρ)−1 t 1+ρ L(t), for ρ < −1.

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Remark 3.9. The result of Proposition 3.8 remains true for ρ = −1 in the sense that t a

s −1 L(s)ds → ∞ as t → ∞. L(t)

The way to remember Proposition 3.8 is that L(s) can be taken out of the integral as if it were L(t). Thus t

t s L(s)ds ∼ L(t)

s ρ ds(t → ∞).

ρ

a

a

When ρ = −1, let z(s) = s −1 L(s), we have Proposition 3.10 Karamata’s Theorem 1.5.9b in [49].) If a function  ∞(Asymptotic behavior). (See ∞ z ∈ RV−1 and s z(τ )dτ < ∞, s > 0, then s z(τ )dτ is slowly varying at infinity and sz(s) = 0. lim  ∞ s z(τ )dτ

s→∞

Similar to Definition 3.1 and Definition 3.2, we also say that a positive continuous function f defined on (0, a1 ) for some a1 > 0, is regularly varying, slowly varying, rapidly varying at 0 if we replace s → ∞ by s → 0+ . If f is regularly varying at 0 with index ρ, we denote it by f ∈ RV Zρ . It also has properties of Proposition 3.3-3.10. For convenience, we give the following Proposition 3.11. If a function z ∈ RV Z−1 and slowly varying at zero and

 a1 s

z(τ )dτ < ∞, s > 0, then

 a1 s

sz(s) = 0. lim  a1 s z(τ )dτ

s→0+

Proof. It is similar to the proof of Theorem 1.5.9b in [49], so we omit it here. 2 We list some examples of regularly varying, slowly varying, rapidly varying at 0. 1. [ln(1 + x)]β is regularly varying at 0 with index β. 2. 11 is slowly varying at 0. ln

x 1

3. e− x is rapidly varying at 0. Now we apply regularly varying theory to investigate the property of I0, I∞ , J0 . By (f2) we have 1

1

 (s) = −[(k + 1)F (s)]− k+1 ,  (s) = [(k + 1)F (s)]− k+1 −1 f (s). It follows that

z(τ )dτ is

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1

−

1

 (s)

= [(k + 1)F (s)] k+1 ,

 (s) (s) f (s) (s) = , k (  (s))2 [(k + 1)F (s)] k+1

I (s) =

(3.3)

and then, by the definitions of I0 and I∞ , we get the following lemma. Lemma 3.12. Let f satisfy (f1), (f2). Then I0 ≥ 1, I∞ ≥ 1. Proof. Integrating from 0 to v we have v

v

 (s) (s) ds = (  (s))2

I (s)ds = 0

=

0

(s) v | −

 (s) 0

v

(s) d  (s) (  (s))2

0

v

(  (s))3 − 2 (s)  (s)  (s) ds (  (s))3

0

(v) =  −v+2

(v)

v I (s)ds. 0

Since  (v) < 0, we have

(v) 0 ≥ lim = 1 − lim v→0+ v  (v) v→+ i.e. I0 ≥ 1. Similarly we can prove I∞ ≥ 1.

v 0

I (s)ds = 1 − lim I (v) = 1 − I0 , v v→0+

2

The following Lemma 3.13, Lemma 3.14 are important for our proof. Some ideas of the proof are from [47]. For convenience of reader, we give the proof here even we can find it in [47]. Lemma 3.13. Let f satisfy (f1), (f2). We have (1) I0 ∈ (1, ∞) if and only if F ∈ N RV Zq+1 with q > k. In this case, f ∈ RV Zq ; (2) I0 = 1 if and only if F is rapidly varying at 0; (3) I0 = ∞ if and only if F ∈ N RV Zk+1 . In this case, f ∈ RV Zk . Proof. (1) Necessity. By Lemma 3.12, we see that

lim

s→0+

− 1(s) s(− 1(s) )

= − lim

s→0+

1

 (s) (s)  (s) s (

 (s)) 2 (s)

=−

1 lim I0 s→0+

(  (s))2 − (s)  (s) I0 − 1 1 lim = . =− I0 s→0+ I0 (  (s))2

(s)

 (s)

s

(3.4)

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By Proposition 3.5 we have − 1(s) ∈ N RVI0 /(I0 −1) . Then F ∈ N RV Z(k+1)I0 /(I0 −1) . Denote (k + 1)I0 /(I0 − 1) by q + 1, then q + 1 > k + 1, i.e. q > k. Sufficiency. If F ∈ N RV Zq+1 with q > k, then − 1 ∈ N RV Z(q+1)/(k+1) , and s(− 1(s) )

lim

− 1(s)

s→0+

=

q+1 q +1 1 ˆ , −  = s k+1 L(s), ∀s ∈ (0, a1 ), k+1

(s)

ˆ where L(s) is normalized slowly varying at 0. By Proposition 3.8, lim (−

s→0+

= lim

s→0+

1

 (s)

) (s)

s(− 1(s) ) − 1(s)

lim

− 1(s)

s→0+

q−k q +1 ˆ = lim s k+1 L(s) + k + 1 s→0

s

(s)

+∞ q+1 τ − k+1 Lˆ −1 (τ )dτ s

q−k q+1 q +1 q +1 ˆ lim s k+1 L(s)( − 1)−1 s 1− k+1 Lˆ −1 (s). = + k + 1 s→0 k+1

It follows that I0 = In this case,

q+1 q−k

> 1. ˆ F (s) = s q+1 L(s), ∀s ∈ (0, a1 ).

Then by Proposition 3.4 we obtain ˆ f (s) = s q [(q + 1) + y(s)]L(s) ∈ RV Zq , ∀s ∈ (0, a1 ), where lim y(s) = 0. s→0+

(2) If I0 = 1, by (3.4) we see lim

s(− 1(s) )

s→0+

− 1(s)

= ∞.

Let s(− 1(s) ) − 1(s)

:= y(s), ∀s > 0,

i.e. (− 1(s) ) − 1(s)

=

y(s) , ∀s > 0. s

(3.5)

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Integrating (3.5) from s to a1 , we have 1 −  = c0 exp (−

(s)

a1

y(τ ) dτ ), s ∈ (0, a1 ), τ

s 1 where c0 = −  (a . 1) By lim y(s) = ∞, we have that for each ξ > 1, s→0+

ξ s ξ y(τ ) y(sv) 1 1 −  /−  = exp ( dτ ) = exp ( dv) → ∞ as s → 0+ .

(ξ s)

(s) τ v s

(3.6)

1

Then − 1(s) is rapidly varying at 0. It follows that F is rapidly varying at 0. Conversely, if F is rapidly varying at 0. We suppose that I0 = 1. By (3.4) we have lim y(s) < s→0+

∞. It follows from (3.6) that −

1

 (ξ s)

/−

1

 (s)

→ C < ∞ as s → 0+ .

Then − 1(s) is not rapidly varying at 0, i.e. F is not rapidly varying at 0. (3) If I0 = ∞, by (3.4) we have lim

s→0+

s(− 1(s) ) − 1(s)

= 1.

It follows from Proposition 3.5 that − 1(s) ∈ N RV1 . Then F ∈ N RVk+1 .

If F ∈ NRVk+1 , by Proposition 3.5 we see that − 1(s) ∈ N RV1 ,  (s) ∈ N RV−1 . Then lim

s→0+

s(− 1(s) ) − 1(s)

= 1.

By Proposition 3.11 we have s  (s) = 0. lim  a1  s→0+ s (τ )dτ Thus s(− 1(s) )

 (s) (s) I0 = lim = lim s→0+ (  (s))2 s→0+ − 1

(s)

 a1 s

(−  (τ ))dτ +

∞ a1

−s  (s)

Similarly to Lemma 3.13 we can prove the following lemma.

(−  (τ ))dτ

= ∞.

2

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Lemma 3.14. Let f satisfy (f1), (f2). We have (1) I∞ ∈ (1, ∞) if and only if F ∈ N RVq+1 with q > k. In this case, f ∈ RVq ; (2) I∞ = 1 if and only if F is rapidly varying at infinity; (3) I∞ = ∞ if and only if F ∈ N RVk+1 . In this case, f ∈ RVk . Lemma 3.15. Let f be a positive measurable function defined on (0, a1 ) for some a1 > 0 and lim f (s) = 0. s→0+

(1) If f is rapidly varying at 0 or f ∈ RV Zq with q > k, then (1.10) holds; (2) If f is slow varying at 0 or f ∈ RV Zq with q < k, then (1.10) does not hold. Proof. (1) If f is rapidly varying at 0 and lim f (s) = 0, then by Definition 3.2 we have s→0+

lim

s→0+

f (s) sk

= 0. Then there exist δ > 0 such that f (s) < s k , ∀ 0 < s < δ.

It follows that F (s) <

1 k+1 s , ∀ 0 < s < δ. k+1

We can see δ

1

F (s)− k+1 ds = ∞.

0

Then (1.10) holds. If f ∈ RV Zq with q > k, then F ∈ RV Zq+1 with q + 1 > k + 1. It follows that F (s) = s q+1 L(s), where L(s) is slow varying at 0. By Proposition 3.8 we have ∞ ∞ q+1 1 1 1 − k+1 [(k + 1)F (s)] ds = (k + 1)− k+1 s − k+1 (L(s))− k+1 ds t

t 1

= (k + 1)− k+1 (

1 q − k −1 k−q ) t k+1 (L(t))− k+1 → ∞ as t → 0. k+1

Then (1.10) holds. (2) Similar to the proof of (1), one can also show that (2) holds, so we omit it here. 2 Remark 3.16. If f ∈ RV Zk , then we can not conclude if or not (1.10) hold. For example, let F (s) = s k+1 (ln 1s )β , then F ∈ RV Zk+1 , f ∈ RV Zk . It follows that

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X. Zhang, M. Feng / J. Differential Equations 267 (2019) 4626–4672

a1 1 [(k + 1)F (s)]− k+1 ds t

a1 β 1 1 = (k + 1)− k+1 s −1 (ln )− k+1 ds s t 1

= −(k + 1)− k+1  →

k+1 1 k+1−β (ln ) k+1 |at 1 k+1−β s

0, β > k + 1, as t → 0+ . ∞, β < k + 1,

Similar to Lemma 3.15, we have Lemma 3.17. Let f be a positive measurable function defined on (A1 , ∞) for some A1 > 0 and lim f (s) = ∞. s→∞

(1) If f is rapidly varying at infinity or f ∈ RVq with q > k, then (f2) holds; (2) If f is slow varying at infinity or f ∈ RVq with q < k, then (f2) does not hold. Remark 3.18. If f ∈ RVq with q = k, then we can not conclude if or not (f2) hold. For example, F (s) = s k+1 (ln s)β , then F ∈ RVk+1 , f ∈ RVk . Then we have t

1

[(k + 1)F (s)]− k+1 ds

a1

t =

β

1

(k + 1)− k+1 s −1 (ln s)− k+1 ds

a1 1

= (k + 1)− k+1  →

k+1−β k+1 (ln s) k+1 |ta1 k+1−β

0, β > k + 1, ∞, β < k + 1,

as t → ∞.

Now we consider the property of J (s). By (1.15), (1.16) we have J (s) =

p(s)

s

˜

0 [k P (τ )]

[k P˜ (s)]

k+1 k

1 k

dτ

.

Lemma 3.19. Let p ∈ Pf inite be defined as in Section 1. Then J0 ≥ 0. Lemma 3.20. Let p ∈ Pf inite be defined as in Section 1. We have (1) J0 ∈ (0, ∞) if and only if P˜ ∈ N RV Zq+1 with −1 > q > −k − 1. In this case, p ∈ RV Zq ;

X. Zhang, M. Feng / J. Differential Equations 267 (2019) 4626–4672

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(2) J0 = 0 if and only if P˜ is slowly varying at zero; (3) J0 = ∞ if and only if P˜ ∈ N RV Z−k . In this case, p ∈ RV Z−k−1 . Proof. (1) Necessity. By the definition of J0 , we see that

lim

s→0+

1 ω (s) s( ω1(s) )

= lim

s→0+

1 lim J0 s→0+

=

1 ω (s) ω(s) −ω (s) s (ω  (s))2 ω(s)

(ω (s))2

=

1 lim J0 s→0+

− ω(s)ω (s)

(ω (s))2

=

ω(s) ω (s)

s

(3.7)

J0 + 1 . J0

By Proposition 3.5 we have ω1(s) ∈ N RV ZJ0 /(J0 +1) . Then P˜ ∈ N RV Z−kJ0 /(J0 +1) . Denote −kJ0 /(J0 + 1) by q + 1, then 0 > q + 1 > −k, i.e. −1 > q > −k − 1. Sufficiency. If P˜ ∈ N RV Zq+1 with −1 > q > −k − 1, then ω1 ∈ N RV Z−(q+1)/k , and lim

s( ω1(s) )

s→0+

1 ω (s)

=−

q+1 q +1 1 ˆ ∀s ∈ (0, a1 ), ,  = s − k L(s), k ω (s)

ˆ where L(s) is normalized slowly varying at 0. By Proposition 3.8, lim (

s→0+

1  ) ω(s) ω (s)

= lim

s→0+

s( ω1(s) ) 1 ω (s)

lim

s→0+

1 ω (s)

s

ω(s)

q+1 q +1 ˆ lim s − k −1 L(s) =− + k s→0

s τ

q+1 k

Lˆ −1 (τ )dτ

0 q+1 q+1 q +1 q +1 ˆ =− lim s − k −1 L(s)( + 1)−1 s 1+ k Lˆ −1 (s) + k s→0 k q +1 . =− k+q +1

q+1 It follows that J0 = − k+q+1 > 0. In this case,

ˆ P˜ (s) = s q+1 L(s), ∀s ∈ (0, a1 ). Then by Proposition 3.4 we obtain ˆ p(s) = s q [(q + 1) + y(s)]L(s) ∈ RV Zq , ∀ s ∈ (0, a1 ), where lim y(s) = 0. s→0+

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X. Zhang, M. Feng / J. Differential Equations 267 (2019) 4626–4672

(2) If J0 = 0, by (3.7) we see lim

s( ω1(s) ) 1 ω (s)

s→0+

= 0.

Let s( ω1(s) )

:= y(s), ∀s > 0,

1 ω (s)

i.e. ( ω1(s) ) 1 ω (s)

=

y(s) , ∀s > 0. s

(3.8)

Integrating (3.8) from s to a1 , we have 1 = c0 exp (−  ω (s)

a1

y(τ ) dτ ), s ∈ (0, a1 ), τ

s 1 where c0 = ω (a . 1) By lim y(s) = 0, we have that for each ξ > 1, s→0+

ξ s ξ 1 y(τ ) y(sv) 1 /  = exp ( dτ ) = exp ( dv) → 1 as s → 0+ .  ω (ξ s) ω (s) τ v s

(3.9)

1

is slowly varying at 0. It follows that P˜ is slowly varying at 0. Conversely, if P˜ is slowly varying at 0. We suppose that J0 = 0. By (3.7) we have lim y(s) =

Then

1 ω (s)

s→0+

m = 0. It follows from (3.9) that 1 1 / → ξ m as s → 0+ . ω (ξ s) ω (s) Then ω1(s) ∈ N RV Zm , i.e. P˜ ∈ N RV Z−mk . It is a paradox. (3) If J0 = ∞, by (3.7) we have lim

s→0+

It follows from Proposition 3.5 that

1 ω (s)

s( ω1(s) ) 1 ω (s)

= 1.

∈ N RV Z1 . Then P˜ ∈ N RV Z−k .

X. Zhang, M. Feng / J. Differential Equations 267 (2019) 4626–4672

If P˜ ∈ N RV Z−k , by Proposition 3.5 we see that lim

s( ω1(s) ) 1 ω (s)

s→0+

1 ω (s)

4647

∈ N RV Z1 , ω (s) ∈ N RV Z−1 . Then

= 1.

It follows from Remark 3.9 that sω (s) = 0.  0 ω (τ )dτ

lim  s

s→0+

Thus s( ω1(s) ) ω (s)ω(s) J0 = − lim = lim 1 s→0+ (ω (s))2 s→0+ ω (s)

s 0

ω (τ )dτ = ∞. sω (s)

2

Lemma 3.21. Let p be defined as in Section 1. (1) If p is slow varying at 0 or p ∈ RV Zq with q > −k − 1, then (1.13) holds; (2) If p is rapidly varying at 0 or p ∈ RV Zq with q < −k − 1, then (1.13) does not hold. Proof. (1) If p is slow varying at 0, then by Proposition 3.6 lim s 2 p(s) = 0.

s→0+

Therefore, there exists δ > 0 such that p(s) < s −2 , ∀ 0 < s < δ. It follows that P˜ (s) < s −1 , ∀ 0 < s < δ. We can see s

[k P˜ (τ )] k dτ < ∞. 1

0

Then (1.13) holds. If p ∈ RV Zq with q > −k − 1, then P˜ ∈ RV Zq+1 with q + 1 > −k. It follows that P˜ (s) = q+1 L(s), where L(s) is slow varying at 0. By Proposition 3.8 we have s t

[k P˜ (s)] k ds = 1

0

1

kks

q+1 k

1

(L(s)) k ds

0 1

=kk( Then (1.13) holds.

t

q+1 1 q +1 + 1)−1 t k +1 (L(t)) k → 0 as t → 0. k

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X. Zhang, M. Feng / J. Differential Equations 267 (2019) 4626–4672

(2) If p is rapidly varying at 0, then by Definition 3.2 we have lim p(s)s k+1 = ∞. Then s→0+

there exist small δ > 0 and large M > 0 such that p(s) > Ms −k−1 , ∀ 0 < s < δ. It follows that M P˜ (s) > s −k , ∀ 0 < s < δ. k We can see s

[k P˜ (τ )] k dτ = ∞. 1

0

Then (1.13) does not hold. If p ∈ RV Zq with q < −k − 1, then P˜ ∈ RV Zq+1 with q + 1 < −k. It follows that P˜ (s) = q+1 L(s), where L(s) is slow varying at 0. By Proposition 3.8 we have s t

[k P˜ (s)] k ds = 1

0

t

1

kks

q+1 k

1

(L(s)) k ds

0 1

= k k (− Then (1.13) does not hold.

q+1 1 q +1 − 1)−1 t k +1 (L(t)) k → ∞ as t → 0. k

2

Remark 3.22. If p ∈ RV Zq with q = −k − 1, then we can not conclude if or not (1.13) hold. For example, P˜ (s) = s −k (ln s)−β , then P˜ ∈ RV Z−k , p ∈ RV Z−k−1 . Then we have t

[k P˜ (s)] k ds 1

0

t =

1

β

k k s −1 (ln s)− k ds

0 1

=kk

 →

k−β k (ln s) k |t0 k−β

0, β > k, as t → 0+ . ∞, β < k,

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4. Proof of Theorem 1.3 and Theorem 1.5 4.1. Proof of Theorem 1.3 We will need the following lemma whose proof uses results in Section 5.1. Lemma 4.1. Suppose that D is a bounded domain in RN and H ∈ C ∞ (D) is positive on D. Suppose that f satisfies (f1), (f2) and (1.10). Then for any δ > 0 there exists a k-convex function u ∈ C ∞ (D) such that Sk (D 2 u) ≥ H (x)f (u), δ > u(x) > 0 in D. Proof. Let H∗ := maxx∈D H (x), and for > 0 define ∞ T :=



−1/(k+1) (k + 1)H∗ [F (t) − F ( )] dt.



Since  −1/(k+1) −1/(k+1) 1 F (t) − F ( ) ≤ F (t) for all large t, 2



by (f2) we see that ∞

 −1/(k+1) (k + 1)H∗ [F (t) − F ( )] dt < ∞.

We also have F (t) − F ( ) ≥ f ( )(t − ) for t > . It follows that



 −1/(k+1) (k + 1)H∗ [F (t) − F ( )] dt < ∞.

+

Hence T is a finite positive number for any > 0. On the other hand, due to (1.10) and 

−1/(k+1)  −1/(k+1) F (t) − F ( ) > F (t) for t > ,

we have ∞ T >



−1/(k+1) (k + 1)H∗ F (t) dt → ∞ as → 0.

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Therefore we can choose 0 > 0 sufficiently small such that 2k

T > R k+1 for ∈ (0, 0 ], where R > 0 is chosen such that D ⊂ BR := {x ∈ RN : |x| < R}. For ∈ (0, 0 ], we define v(r) = v (r) by v(r)  −1/(k+1) 1 k−1 k−1 − k+1 (k + 1)H∗ [F (t) − F ( )] dt = (CN−1 ) R k+1 r, r ∈ (0, R ),

with 1

k−1

k−1 k+1 − k+1 R := T (CN−1 ) R > R.

It is easily checked that v is smooth in (0, R ), v(0) = , v  (0) = 0, v  (r) > 0 for r ∈ (0, R ), v(r) → ∞ as r → R and k−1 CN−1 (v  )k−1 v  = R k−1 H∗ f (v) ≥ r k−1 H∗ f (v) for r ∈ (0, R].

Then k−1 CN−1 (

v  k−1  v k ( )k ≥ H∗ f (v) for r ∈ (0, R]. ) v + CN−1 r r

Moreover, since 1 k−1 k−1 − k+1 (CN−1 ) R k+1 r

>

v(r) 

−1/(k+1) (k + 1)H∗ F (t) dt,



by (1.10) we deduce v(r) → 0 uniformly for r ∈ [0, R] as → 0.

(4.1)

Since v  (0) = ∞, to obtain a smooth function u with the required properties we consider the initial value problem k−1 CN−1 (

u k−1  u k ( )k = H∗ f (u) for r > 0, u(0) = /2, u (0) = 0. ) u + CN−1 r r

By Lemmas 5.1 and 5.2 we see that u(r) is defined for r ∈ [0, R] and /2 < u(r) < v(r) for r ∈ (0, R), u (r) > 0 for r ∈ [0, R]. Thus Sk (D 2 u(|x|)) = H∗ f (u(|x|)) in BR .

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In particular, u(|x|) is a strictly convex function in C ∞ (B R ), u(|x|) ≥ /2 in B R and Sk (D 2 u(|x|)) = H∗ f (u(|x|)) ≥ H (x)f (u(|x|)) in D. By (4.1), for any δ > 0 by shrinking > 0 further we have 0 < u(|x|) ≤ v(|x|) < δ for x ∈ D ⊂ BR . This completes the proof of Lemma 4.1. 2 We are now ready to prove Theorem 1.3. We will make use of Lemma 4.1 above. Part I. Proof of (i). Step 1. Sub-supersolution. ¯ is positive, we suppose h1 ≤ H (x) ≤ h2 for h1 , h2 > 0. Since H ∈ C ∞ () By (1.11) we have 1

f (ψ(t))

ψ  (t) = −[(k + 1)F (ψ(t))] k+1 ,

ψ  (t) =

.

(4.2)

(−ψ  (t))k−1 ψ  (t) = f (ψ(t)), −

ψ  (t) [(k + 1)F (ψ(t))] k+1 = .  ψ (t) f (ψ(t))

(4.3)

k−1

[(k + 1)F (ψ(t))] k+1

Then k

Let v(x) = ψ(az(x)), x ∈ , where a is a positive constant to be determined and ψ is defined by (1.11), z is defined by (2.4). By (2.1), (3.3) and (4.3) we have k

Sk (D 2 v) = a k ψ  k Sk (D 2 z) + a k+1 ψ  k−1 ψ 

CN 

det (zxis xij )(∇zi )T B(zi )∇zi

i=1 Ck

=a

k+1

 k−1

(−ψ )

N  ψ ψ [(−1) Sk (D z)(−  )z − (−1)k det (zxis xij )(∇zi )T B(zi )∇zi ] ψ az



2

k

i=1

=a

Ck

N  [(k + 1)F (v)] k+1 f (v(x))[(−1) Sk (D z) (−1)k det (zxis xij )(∇zi )T B(zi )∇zi ] z− f (v) (v) k

k+1

k

2

i=1

k CN

= a k+1 f (v(x))[(−1)k Sk (D 2 z)

 1 (−1)k det (zxis xij )(∇zi )T B(zi )∇zi ]. z− I (v)

(4.4)

i=1

Let Ck

N  1 1 = (−1) Sk (D z) (−1)k det (zxis xij )(∇zi )T B(zi )∇zi ]. z−a I (v)

k

2

i=1

By the definition of z and Remark 2.7 we have (zxi xj ) is negative definite, then all of its principal submatrix of size k are negative definite. It follows that there exist e1 , e2 > 0 such that

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−e1 ||∇zi ||2 ≤ (∇zi )T B(zi )∇zi ≤ −e2 ||∇zi ||2 , and trace (zxis xij ) = zxi1 xi1 + · · · + zxik xik = zi < 0. Therefore, since (−zi ) > 0 on  and −zi ¯ at each point of ∂, it follows from the maximum principle that there attains its maximum on  exists an open set U containing ∂ such that ||∇zi || ≥ e > 0 for all i. On the other hand, it is easy to see that z is bounded below by a positive constant on  − U . Then v → 0 as a → ∞ on  − U. By Lemma 3.12, I0 ≥ 1. Combining this with the fact that k (−1)k det (zxis xij ) > A, (−1)k Sk (D 2 z) > ACN ,

and I0 = ∞, we can conclude that 1 is positive on . By (4.4) we can see there exists a1 > 0 large enough such that Sk (D 2 v)(x) ≥ H (x)f (v(x)), x ∈ 

(4.5)

and there exists a2 > 0 small enough such that Sk (D 2 v)(x) ≤ H (x)f (v(x)), x ∈ .

(4.6)

Let v1 (x) = ψ(a1 z(x)), v2 (x) = ψ(a2 z(x)). Then v1 (x) and v2 (x) are subsolution and supersolution of (1.1) respectively. Step 2. Existence and global estimation. Let {σn }∞ 1 be a strictly increasing sequence of positive numbers such that σn → ∞ as n → ∞, and let n = {x ∈ |v1 (x) < σn }. Since any level surface of v1 is a level surface of z, for each n ≥ 1, ∂n is a strictly convex C ∞ -submanifold of RN of dimension N − 1. Since I0 = ∞, we note that according to Lemma 3.13 and Lemma 3.15 the nonlinearity f satisfies (1.11). Let wn be the k-convex function obtained in Lemma 4.1 with D = n satisfying Sk (D 2 wn ) ≥ H (x)f (wn ), σn > wn (x) > 0 in n . Set n = min wn (x). x∈n

We now let f˜n (t) be a function satisfying (f1) with η = −∞ and f˜n (t) = f (t) for t ≥ n . Then we consider the problem Sk (D 2 u) = H (x)f˜n (u) in n , u = σn on ∂n .

(4.7)

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By Theorem 1 of [4], (4.7) has a unique k-convex solution zn when H (x)f˜(u) is replaced by H (x)f˜(σn ). It follows that Sk (D 2 zn ) = H (x)f˜n (σn ) > H (x)f˜n (zn ) in n , zn = σn on ∂n . Therefore we can apply Lemma 2.6 to conclude that (4.7) has a k-convex solution un ∈ C ∞ (n ). Since wn is k-convex and Sk (D 2 wn ) ≥ H (x)f (wn ) = H (x)f˜n (wn ) in n , wn < σn = un on ∂n , by Lemma 2.1 we deduce un ≥ wn in n and in particular, un ≥ n in n . Hence f˜n (un ) = f (un ) in n and Sk (D 2 un ) = H (x)f (un ) in n , un = σn = v1 on ∂n . By Lemma 2.1 and (4.5) ¯ n. un (x) ≥ v1 (x) x ∈  From the fact that un (x) = v1 (x) ≤ v2 (x), ∀x ∈ ∂n , (4.6) and Lemma 2.1, we see that ¯ n. v2 (x) ≥ un (x), x ∈  Clearly, for n > 1 ¯ n ⊂ n+1  and =

∞ 

n .

n=1

We claim that un (x) ≤ un+1 (x), ∀x ∈ n .

(4.8)

¯ un is k-convex in n and for Indeed, since un and un+1 are both positive solutions of (4.7) on , x ∈ ∂n ⊂ n+1 , un+1 (x) ≥ v1 (x) = un (x), the inequality (4.8) is a consequence of Lemma 2.1.

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Let x0 ∈  be fixed. If m is so large that x0 ∈ m , then for all n ≥ m, we have v1 (x0 ) ≤ un (x0 ) ≤ un+1 (x0 ) ≤ v2 (x0 ). Therefore, lim un (x0 ) exists. Since x0 is arbitrary we see that for x ∈ , n→∞

lim un (x) = u(x)

n→∞

exists and v1 (x) ≤ u(x) ≤ v2 (x), x ∈ . Next we prove u ∈ C ∞ () satisfies (1.1). Fix an integer m. For n > m ¯ m ⊂ n .  ¯ m , q1 ≤ u(x) ≤ q2 , where q1 is the minimum of Suppose un is a solution of (4.7), then for x ∈  ¯ ¯ m . Moreover, for n > m, v1 (x) on m and q2 is the maximum of v2 (x) on  ¯ m , ∂m+1 ) ≤ dist ( ¯ m , ∂n ) < dist (m , ∂). 0 < dist ( ¯ n , it follows from Lemma 2.5 that there exists a Let j ≥ 3 be an integer. Since un is convex on  constant C ∗ such that if n > m, then ¯ m, |D γ un (x)| ≤ C ∗ , ∀x ∈  where D γ un is any partial derivative of un of order ≤ j . It follows from Ascoli’s lemma that there ∞ η exists a subsequence {unj (x)}∞ 1 of {un (x)}1 such that if D is any partial derivative operator of ∞ η ¯ m . Hence u ∈ C j −1 () ¯ order ≤ j − 1, then the sequence {D unj (x)}1 converges uniformly on  ¯ and for x ∈ m , Sk (D 2 (x)) = H (x) lim Sk (D 2 unj )(x) j →∞

= H (x) lim f (unj (x)) j →∞

= H (x)f (u(x)). Since j ≥ 3 was arbitrary and m ≥ 1 is arbitrary, this argument proves that u ∈ C ∞ () satisfies (1.1). Combining (2.5) with the definition of v1 (x), v2 (x) we have ψ(c1 d(x)) ≤ u(x) ≤ ψ(c2 d(x)), x ∈ , where c2 = a2 b1 < a1 b2 = c1 .

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Part II. Proof of (ii). Assume, contrary to the assertion of the theorem, that there exists c0 > 0 such that t G(t) :=

1

[(k + 1)F (τ )]− k+1 dτ → ∞ as t → ∞,

c0

and (1.1) has a k-convex solution u. We aim to derive a contradiction. Denote by g(t) the inverse of G(t), i.e., g(t) 1 [(k + 1)F (τ )]− k+1 dτ = t, ∀t > 0.

(4.9)

c0

Then g(0) = c0 , lim g(t) = ∞ t→∞

and 1

g  (t) = [(k + 1)F (g(t))] k+1 ,

g  (t) =

f (g(t)) k−1

[(k + 1)F (g(t))] k+1

,

k



k−1 

(g (t))

g (t) = f (g(t)),

g  (t) [(k + 1)F (g(t))] k+1 = .  g (t) f (g(t))

Take x0 ∈ RN \  so there exists d0 > 0 such that |x − x0 | ≥ d0 for x ∈ . Then define 1 y(x) := |x − x0 |2 for x ∈ . 2 Clearly k [∇y(x)]T = x − x0 , (yxi xj ) is the identity matrix, and Sk (D 2 y) = CN .

For c > 0 define w(x) := g (cy(x)) , x ∈ . By (2.1) we obtain, for x ∈ , Sk (D w) =c 2

k

k

CN   k  k−1  det (yxis xij )(∇(yi ))T B(yi )∇(yi ) Sk (D y)g (cy) + c(g (cy)) g (cy)



2



⎧ ⎨

i=1

⎫ k ⎬  k 2 (x − x ) + c =ck (g  (cy))k−1 g  (cy) CN ij 0ij ⎩ ⎭ g  (cy) g  (cy)

k CN

i=1 j =1

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   k g (cy) 2 ≥ck (g  (cy))k−1 g  (cy) CN | + c|x − x 0 g  (cy) >f (w)ck+1 d02 , where we have used g  (cy(x)) > 0, |x − x0 | ≥ d0 for x ∈ . g  (cy(x)) ¯ is positive, We thus obtain, in view of H ∈ C ∞ () Sk (D 2 w) > H (x)f (w) in  provided that c is chosen large enough. Fix x1 ∈  and by further enlarging c if necessary we may assume that w(x1 ) > u(x1 ) and Sk (D 2 w) > H (x)f (w) in . Since u(x) → ∞ as d(x) → 0, while w(x) is continuous on , there exists an open connected set D such that x1 ∈ D, D ⊂ , u(x) < w(x) in D and u(x) = w(x) on ∂D. On the other hand, since Sk (D 2 u) = H (x)f (u) in D and w = u on ∂D, and the matrix (wxi xj ) is positive definite on D (since y(x) is strictly convex in  and g  , g  > 0), we can apply Lemma 2.1 to conclude that w(x) ≤ u(x) in D. This contradiction completes the proof of necessity of Theorem 1.3. 2 4.2. Proof of Theorem 1.5 By (1.15) we have (ω (t))k−1 ω (t) = −p(t),

ω (t) k P˜ (t) = − . ω (t) p(t)

k

(4.10)

Let v(x) = ψ(c[ω(lz(x))] k+1 ), where c, l are positive constants to be determined and ψ is defined by (1.11), z is defined by (2.4). By (1.16), (2.1), (3.3), (4.3) and (4.10) we have Sk (D 2 v) k k k k − k k k  k−1  k−1 − k−1 ω ω k+1 ) ψ ω ω k+1 Sk (D 2 z) + ck l k ( ) ψ k+1 k+1 2 1 1 k l × [cl ω− k+1 ω 2 ψ  − ψ  ω 2 ω− k+1 −1 + lψ  ω ω− k+1 ] k+1 k+1 = ck l k (

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k

×

CN  i=1

=c

det (zxis xij )(∇zi )T B(zi )∇zi  ω k k ψ kl ω 2  k−1   k−1  l ( ω ) (−1)k Sk (D 2 z) − [ ) ) (−ψ ) ψ (−ω (− k  k+1 k + 1 ωω ψ  cω k+1 ω

k+1 k

k CN   l ω 2 lψ  ψ k T + + (− )] (−1) det (z )(∇z ) B(z )∇z ] x x i i i is ij k k k + 1 ψ  cω k+1 ωω ψ  cω k+1 i=1  k k 1 k P˜ (lz) kl 1 = ck+1 l k ( ) f (v)p(lz) (−1)k Sk (D 2 z) −[ k+1 I (v) p(lz) k + 1 J (lz) k CN  l  1 1 l k T (−1) det (zxis xij )(∇zi ) B(zi )∇zi ] . + ] + k + 1 I (v) J (lz) I (v)

(4.11)

i=1

Let 2 = (−1)k Sk (D 2 z)

1 k P˜ (lz) kl 1 −[ I (v) p(lz) k + 1 J (lz) Ck

N l  1 1 l (−1)k det (zxis xij )(∇zi )T B(zi )∇zi ]. + ] + k + 1 I (v) J (lz) I (v)

i=1

Because I0 = ∞, I∞ = ∞ and J0 = ∞ can not hold at same time, similar to the proof of Theorem 1.3, we can show that 2 is positive. We may now choose l = 1/b2 , where b2 is the positive constant in (2.5), and use z(x) ≤ b2 d(x) to deduce p(lz(x)) ≥ p(d(x)) ≥

1 H (x) in . k1

Therefore Sk (D 2 v) ≥ H (x)f (v) for large c = c1 . Similarly, we choose l = 1/b1 and use z(x) ≥ b1 d(x) to deduce p(lz(x)) ≤ p(d(x)) ≤

1 H (x) in . k2

Then Sk (D 2 v) ≤ H (x)f (v) for small c = c2 . Other process is similar to that of Theorem 1.3. So we omit it here. 2

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4.3. Supplementary specification 1. As we mentioned in Remark 1.4, I0 = ∞ is a little bit stronger than (1.10). By Lemma 3.13 and Lemma 3.15 we can see if I0 = ∞, then (1.10) holds. But if (1.10) holds, we can not conclude I0 = ∞. For example, in Remark 3.18, F (s) = s k+1 (ln s)β , β > k + 1, then (1.10) holds. However, by Lemma 3.13, we have I0 = ∞. Then, if we replace I0 = ∞ with (1.10) in Theorem 1.3, we can not obtain the subsolution by (4.4). 2. By Lemma 3.14 (3) and Lemma 3.17 we can see I∞ = ∞ is a critical case of f such that (f2) holds. By Lemma 3.20 (3) and Lemma 3.21 we can see J0 = ∞ is a critical case of p such that (1.13) holds. Therefore “I∞ = ∞ and J0 = ∞ can not hold at same time” in Theorem 1.5 means that (1.1) has k-convex solution as long as f and p do not take the critical situation at the same time. 5. Some results for radial solutions In this section, we consider such a situation in the special case that  is a ball and H = H (|x|) is radially symmetric. Our approach in this section is motivated by ideas in [7]. So we consider the problem 

Sk (D 2 u) = H (|x|)f (u), u = +∞,

x ∈ B, x ∈ ∂B,

(5.1)

where B is a ball in RN (N ≥ 2). For simplicity, and without loss of generality, we assume that B is the unit ball. By a direct calculation, it is seen (and well-known) that if v = v(r) (r = |x|) is a radially symmetric solution of (5.1), then (1.1) becomes 





k−1 v k−1  k−1 1−N N−k  k  k ( r ) v + CN−1 ( vr )k = k −1 CN−1 r [r (v ) ] = H (r)f (v), r ∈ (0, 1), CN−1

v  (0) = 0, v(1) = +∞.

(5.2)

In the radially symmetric setting, the smoothness requirements for H and f can be greatly relaxed. We assume that H and f satisfy, respectively (H ): H ∈ C([0, 1)) and H (r) > 0 in [0, 1); (f ): for some η ∈ (−∞, ∞), f (s) is locally Lipschitz continuous in (η, ∞), positive and increasing for s > η. 5.1. The initial value problem and a comparison result For v0 > η, consider the following initial value problem, 





k−1 v k−1  k ( r ) v + CN−1 ( vr )k = H (r)f (v), r ∈ (0, 1), CN−1

v(0) = v0 , v  (0) = 0.

(5.3)

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Lemma 5.1. Assume that H satisfies (H ) and f satisfies (f ). Then for every v0 > η, (5.3) has a unique solution v(r) over a maximal interval of existence [0, a) ⊂ [0, 1). Moreover, v  > 0 in (0, a), v  > 0 in [0, a) and v(r) → ∞ as r → a if a < 1. Proof. We first show that (5.3) has a unique solution defined over [0, δ] for δ > 0 sufficiently small. It is easy to see that (5.3) is equivalent to the following integral equation r v(r) = v0 +

⎡ ⎣s k−N

0

s

⎤1/k k−1 −1 N−1 (CN−1 ) kt H (t)f (v(t))dt ⎦

ds.

(5.4)

0

Let E = C([0, δ]) with δ > 0 small to be specified, and define T : E → E by r (T v)(r) = v0 +

⎡ ⎣s k−N

0

s

⎤1/k k−1 −1 N−1 (CN−1 ) kt H (t)f (v(t))dt ⎦

ds.

0

We are going to show that if δ > 0 is sufficiently small, then T is a contraction mapping on a suitable subset of E and hence has a unique fixed point, which gives a unique solution to (5.3) over [0, δ]. Let H∗ = max H (r), h∗ = min H (r) and r∈[0,1/2]

r∈[0,1/2]

Bδ (v0 ) = {v ∈ E : v − v0 E < δ}. Fix δ1 ∈ (0, 1/2) such that v0 − δ1 > η, and let L be the Lipschitz constant of the function f (u) over [v0 − δ1 , v0 + δ1 ]: |f (v1 ) − f (v2 )| ≤ L|v1 − v2 | for v1 , v2 ∈ [v0 − δ1 , v0 + δ1 ]. Then m := f (v0 − δ1 ) ≤ f (v) ≤ M := Lδ1 + f (v0 ) for v ∈ [v0 − δ1 , v0 + δ1 ]. Clearly there exists δ2 ∈ (0, δ1 ) sufficiently small such that 1 1 1 2 k−1 − 1 δ (CN−1 ) k (kH∗ M) k N − k < δ for δ ∈ (0, δ2 ]. 2

We prove that T (Bδ (v0 )) ⊂ Bδ (v0 ) for every δ ∈ (0, δ2 ]. Indeed, for such δ and any v ∈ Bδ (v0 ), we have r |T v − v0 | = 0

⎡ ⎣s k−N

s 0

⎤1/k k−1 −1 N−1 (CN−1 ) kt H (t)f (v(t))dt ⎦

ds

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⎡

r ≤

s

k−N k

0

⎤1/k

s

⎣ (C k−1 )−1 kt N−1 H∗ Mdt ⎦ N−1

ds

0

1 1 1 k−1 − 1k ) (kH∗ M) k N − k < δ for r ∈ [0, δ]. = δ 2 (CN−1 2

Hence T (Bδ (v0 )) ⊂ Bδ (v0 ) for every δ ∈ (0, δ2 ]. Next we show that T is a contraction mapping on Bδ (v0 ) for all small δ > 0. We first observe that, by the mean value theorem, for δ ∈ (0, δ2 ] and v1 , v2 ∈ Bδ (v0 ), ⎡

⎤1/k

s

k−1 −1 N−1 J (s) := ⎣ (CN H (t)f (v1 (t))dt ⎦ −1 ) kt

⎡ s ⎤1/k  − ⎣ (C k−1 )−1 kt N−1 H (t)f (v2 (t))dt ⎦ N−1

0

0

⎤ 1 −1 ⎡ s k   1⎣ k−1 −1 N−1 ⎦ = (CN−1 ) kt H (t) θf (v1 ) + (1 − θ )f (v2 )]dt k 0

s

  k−1 −1 N−1 (CN H (t) f (v1 ) − f (v2 ) dt −1 ) kt

0

with θ = θ (s) ∈ (0, 1). Therefore, for s ∈ [0, δ], ⎤ 1 −1 s ⎡ s k   1 k−1 −1 N−1 k−1 −1 N−1 |J (s)| ≤ ⎣ (CN−1 ) kt h∗ mdt ⎦ · (CN−1 ) kt H∗ Lv1 − v2 E dt k 0 N k

=s N

0 − 1k

− 1k

k−1 (CN−1 )

(kh∗ m)

1 k −1

H∗ Lv1 − v2 E .

It follows that, for r ∈ [0, δ],  |(T v1 )(r) − (T v2 )(r)| = 

r s

k−N k

 1 1 1 k−1 − 1k J (s)ds  ≤ δ 2 N − k (CN−1 ) (kh∗ m) k −1 H∗ Lv1 − v2 E . 2

0

Hence T is a contraction mapping on Bδ (v0 ) if δ ∈ (0, δ2 ] is small enough such that 1 1 2 − 1 k−1 − 1 δ N k (CN−1 ) k (kh∗ m) k −1 H∗ L < 1. 2

We fix such a small δ > 0 and have thus proved that (5.3) has a unique solution defined for r ∈ [0, δ]. Moreover, since k−1 CN−1 (

v  k−1  v k ( )k = H (r)f (v) > 0 for r ∈ (0, δ], and v  (0) = 0, ) v + CN−1 r r

we further have v  (r) > 0, v  (r) > 0 for r ∈ (0, δ], and v  (0) := limr→0 v  (r) > 0.

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To extend the solution v(r) to r > δ we let v  = u and " # u U= . v Then we consider the first order ODE system $ 

U =

k−1 −1 k−1 k (CN−1 ) r [H (r)f (v)−CN−1 ( ur )k ] uk−1

%

$ =: F (r, U ), U (δ) =

v  (δ)

%

v(δ)

u

.

(5.5)

By (H ) and (f ), F (r, U ) is locally Lipschitz continuous in U in the range u > 0 and v > η, and continuous in r ∈ [0, 1). Hence (5.5) has a unique solution defined for r in a small neighbourhood of δ. Clearly the v component of U satisfies k−1 CN −1 (

v  k−1  v k ( )k = H (r)f (v) > 0, v(δ) > 0, v  (δ) > 0. ) v + CN−1 r r

It follows that v  (r) > v  (δ), v  (r) > 0 for r > δ. Hence the solution U (r) of (5.5) can be extended to r > δ until r reaches 1 or until v(r) blows up to ∞. It follows that (5.3) has a unique solution v(r) on some maximal interval of existence [0, a) with a ≤ 1, and v(r) → ∞ as r → a if a < 1. The proof of the lemma is now complete. 2 Lemma 5.2. Assume that H satisfies (H ) and f satisfies (f ). If u1 and u2 are functions in C 1 ([0, a)) ∩ C 2 ((0, a)) satisfying u1 k−1  u k ( 1 )k ≤ H (r)f (u1 ) for r ∈ (0, a), ) u1 + CN−1 r r  u k−1 u2 k−1  k CN−1 ( ) u2 + CN−1 ( 2 )k ≥ H (r)f (u2 ) for r ∈ (0, a), r r

k−1 CN−1 (

and u1 (0) = u2 (0) = 0, u1 (0) < u2 (0). Then u1 (r) < u2 (r) for r ∈ [0, a). Proof. If u1 < u2 in [0, a) does not hold, then due to u1 (0) < u2 (0), there exists r ∈ (0, a) such that u1 (r) = u2 (r) and u1 (r) < u2 (r) for r ∈ [0, r). Since u1 and u2 satisfy (5.4) with the equality sign replaced by inequalities, by the monotonicity of f , we have the following contradiction: r u1 (r) ≤ u1 (0) + 0

r < u2 (0) + 0

≤ u2 (r). The proof is complete.

2

⎡ ⎣s k−N ⎡ ⎣s k−N

s 0

s 0

⎤1/k k−1 −1 N−1 (CN−1 ) kt H (t)f (u1 (t))dt ⎦

ds

⎤1/k k−1 −1 N−1 (CN−1 ) kt H (t)f (u2 (t))dt ⎦

ds

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5.2. Multiplicity and non-existence results for (5.2) The theorems below deal with a case that f does not satisfy (f2) and (5.2) has infinitely many solutions. Theorem 5.3. Suppose that H satisfies (H ), there exist constants d1 , d2 > 0 and a function p(t) of class P∞ such that d1 p(1 − r) ≤ H (r) ≤ d2 p(1 − r) for all r < 1 close to 1. Suppose that f satisfies (f ) and there exist constants α ≤ k and c1 , c2 > 0 such that c1 uα ≤ f (u) ≤ c2 uα for u > 0. Then (5.2) has infinitely many k-convex solutions. Proof. It is obvious that y(r) = 12 (1 − r 2 ) satisfies 

(−1)k y  k−1 y  = r k−1 , r ∈ (0, 1), y  (0) = 0, y(1) = 0.

We modify p(t) as in Section 1 and define σ (t) by ∞ 1 σ (t) = (k P˜ (τ )) k dτ for t > 0.

(5.6)

t

If α < k, then we set  k/(k−α) w(r) := c σ (y(r)) for r ∈ [0, 1) and some constant c > 0. We calculate ck α/(k−α)   w = σy, σ k−α   ck α/(k−α)   α (σ  )2  2    2 w = σ (y ) , σ y + σ (y ) + k−α k−α σ " #   ck k kα  k−1   k−1  σ  (y  )2 α (σ  )2 (y  )2  k−1  k−α σ (σ ) σ (y ) y +  + , (w ) w = k−α σ  y k − α σ σ  y  w  k−1  w k ( )k ) w + CN−1 r r #k "   k−1 y   σ  kα (y  )2 α (σ  )2 (y  )2 ck k−1  k−1  (y ) k−α = CN−1 σ (σ ) σ +  + k−α σ  y k − α σ σ  y  r k−1 #k " kα y k ck k σ k−α (σ  )k k . + CN−1 k−α r

k−1 CN−1 (

X. Zhang, M. Feng / J. Differential Equations 267 (2019) 4626–4672

Using (σ  (t))k−1 σ  (t) = (−1)k−1 p(t),

k P˜ (t) σ  (t) = − σ  (t) p(t)

and y  = −r, y  = −1, we can simplify the above expression to obtain k−1 CN −1 (

w  k−1  w k k ) w + CN ) = ck−α −1 ( r r

"

k k−α

#k w α p(y)(r),

with & k−1 (r) := CN −1

' k P˜ (y) α (σ  )2 2 k P˜ (y) 2 k +r + . r + C N −1 p(y) k − α σ σ  p(y)

We have k+1 [σ  (t)]2 [k P˜ (t)] k =  σ (t)σ  (t) p(t) t∞ [k P˜ (τ )]1/k dτ

∞ (k + 1)[k P˜ (s)]1/k p(s)ds =

t

∞

∞  − p (s) [k P˜ (τ )]1/k dτ + p(s)[k P˜ (s)]1/k ds 

t

s

≤ (k + 1). It follows that 0≤

[σ  (y)]2 2 α α r ≤ M0 := (k + 1) for r ∈ [0, 1). k − α σ (y)σ  (y) k−α

Since P˜ (t) P˜ (y(r)) = 0 and so lim = 0, t→0 p(t) r→1 p(y(r)) lim

the function k−1 ¯ (r) := CN −1

k P˜ (y(r)) k P˜ (y) k + r 2 + CN −1 p(y(r)) p(y)

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k−1 ¯ is positive and continuous for r ∈ [0, 1) with (r) → CN−1 as r → 1. Therefore we can find positive constants m1 < m2 , depending on the function p, such that

¯ m1 ≤ (r) ≤ m2 for r ∈ [0, 1). It follows that k−1 m1 ≤ (r) ≤ m2 + M0 CN−1 for r ∈ [0, 1).

Therefore k−1 CN −1 (

w  k−1  w k ( )k ≤ ck−α ) w + CN−1 r r

"

k k−α

#k k−1 w α p(y)(m2 + M0 CN−1 ) for r ∈ [0, 1),

(5.7) k−1 ( CN−1

w  k−1  w k ( )k ≥ ck−α ) w + CN−1 r r

"

k k−α

#k w α p(y)m1 for r ∈ [0, 1).

(5.8)

Replacing p(t) by p(2t) with > 0 sufficiently small, we may assume that H (r) ≥ p((1 − r)/2) for r ∈ [0, 1). Therefore, due to y(r) ≥ (1 − r)/2, we have p(y(r)) ≤ p((1 − r)/2) ≤ H (r) for r ∈ [0, 1). It then follows from (5.7) that k−1 CN−1 (

w  k−1  w k ( )k ≤ ck−α ) w + CN−1 r r

"

k k−α

#k k−1 w α H (r)(m2 + M0 CN−1 ) for r ∈ [0, 1).

If we take c small enough such that " c

k−α

k k−α

#k k−1 (m2 + M0 CN−1 ) ≤ c1 ,

k

then w1 (r) = c[σ (y(r))] k−α satisfies k−1 CN−1 (

w1 k−1  w k ( 1 )k ≤ H (r)f (w1 ) for r ∈ [0, 1). ) w + CN−1 r r

Next we construct a function w2 (r) that satisfies the reversed inequality. By replacing p(t) with Mp(t), with M > 0 sufficiently large, we may assume that p(1 − r) ≥ H (r) for r ∈ [0, 1). Then, due to y(r) ≤ 1 − r, we get

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p(y(r)) ≥ p(1 − r) ≥ H (r) for r ∈ [0, 1). Thus by (5.8) (with σ (t) and m1 determined by this new function p(t)), we have k−1 CN −1 (

w  k−1  w k ( )k ≥ ck−α ) w + CN−1 r r

"

k k−α

#k w α H (r)m1 for r ∈ [0, 1).

If we take c large enough such that " ck−α

k k−α

#k m1 ≥ c2 ,

k

then w2 (r) = c[σ (y(r))] k−α satisfies k−1 w2 (0) > w1 (0), CN−1 (

w2 k−1  w k ( 2 )k ≥ H (r)f (w2 ) for r ∈ [0, 1). ) w + CN−1 r r

For any c ∈ (w1 (0), w2 (0)), let vc denote the unique solution of (5.3) with v0 = c. By Lemma 5.2 we have w1 (r) < vc (r) < w2 (r) for r ∈ [0, 1) and such that vc (r) is defined. Hence we can use Lemma 5.1 to see that vc (r) is defined for r ∈ [0, 1) and vc (r) > 0, vc (r) > 0 in (0, 1). Since w1 (r) → ∞ as r → 1, we have vc (r) → ∞ as r → 1. Hence vc is a strictly convex solution to (5.2). By varying c we thus obtain infinitely many solutions to (5.2). If α = k, letting w(x) = ecσ (y(r)) , then we can obtain the same result as above. 2 Theorem 5.4. Suppose that f satisfies (f ) and there exist α > k and b > 0 such that f (u) ≥ buα for all large u > 0. Suppose that H satisfies (H ) and for some β ≥ k + 1, c > 0, H (r) ≥ c(1 − r)−β for all r < 1 close to 1. Then (5.2) has no solution. Proof. Suppose (5.2) has a solution v(r). Then v  (r) > 0 and v  (r) > 0 in (0, 1). Choose r0 ∈ ( 12 , 1) close to 1 such that f (v(r)) ≥ bv α (r), H (r) ≥ c(1 − r)−β for r ∈ [r0 , 1). Then for r ∈ [r0 , 1), we have k−1 CN −1 (

v  k−1  v k ( )k = H (r)f (v) ≥ bc(1 − r)−β v α ≥ bc(1 − r0 )−β v α . ) v + CN−1 r r

Set 1

c0 := [bc(1 − r0 )k+1−β ] α−k

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and w(r) := c0 v(r0 + (1 − r0 )r), r ∈ [0, 1). Then clearly w(0) = c0 v(r0 ) > 0, w  (0) = c0 (1 − r0 )v  (r0 ) > 0 and with s = r0 + (1 − r0 )r, w  k−1  w k ( )k ) w + CN−1 r r  v v k−1 k k = CN−1 c0 (1 − r0 )k+1 ( )k−1 v  + CN−1 c0k (1 − r0 )k ( )k r r   v k−1 v k−1  k ≥ c0k (1 − r0 )k+1 [CN−1 ( ) v + CN−1 ( )k ] r r

k−1 CN−1 (

≥ c0k bc(1 − r0 )k+1−β v α (s) = wα . Since α > k, by Theorem 3.1 of [16], the problem 





k−1 W k−1  k CN−1 ( r ) W + CN−1 ( Wr )k = W α , r ∈ (0, 12 ),

W  (0) = 0, W ( 12 ) = +∞ has a positive, strictly convex solution W . We show next that w ≤ W in (0, 1/2). Indeed, the function z(r) := w(r) − W (r) satisfies z (0) > 0, z( 12 ) = −∞. Hence the maximum of z(r) over (0, 1/2) is achieved at some r ∗ ∈ (0, 1/2). It follows that z (r ∗ ) = 0, z (r ∗ ) ≤ 0, and so 0 < w  (r ∗ ) = W  (r ∗ ), 0 < w  (r ∗ ) ≤ W  (r ∗ ). We thus obtain w  (r ∗ ) k−1  ∗ w  (r ∗ ) k k ) w (r ) + C ( ) N−1 r∗ r∗  ∗ W  (r ∗ ) k k−1 W (r ) k−1  ∗ k ≤ CN−1 ( ) W (r ) + C ( ) = W α (r ∗ ), N−1 r∗ r∗ k−1 w α (r ∗ ) ≤ CN−1 (

which leads to w(r ∗ ) ≤ W (r ∗ ), and hence w(r) ≤ W (r) in [0, 1/2), as we wanted. From w(0) ≤ W (0) and the definition of w we obtain ( ) 1 k−α v(r0 ) ≤ bc(1 − r0 )k+1−β W (0). Since α > k, β ≥ k + 1, it follows that v(r0 ) ≤ (bc)1/(k−α) 2(k+1−β)/(α−k) W (0) for all r0 ∈ (1/2, 1) close to 1.

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But as a solution to (5.2), we have limr→1 v(r) = ∞. This contradiction completes the proof. 2 6. Asymptotic behavior of the blow up solution 6.1. Proof of Theorem 1.7 For δ > 0, we set δ = {x ∈  : 0 < d(x) < δ}, δ = {x ∈  : d(x) = δ}. Since |Dd(x)| = 1 in δ , we can choose a coordinate system such that Dd(x) = (0, 0, . . . , 0, 1), D 2 d(x) = diag(−ε1 (x), . . . , −εN−1 (x), 0), where εi (x) = κi (y)/(1 − κi (y)d(x)), and y ∈ ∂ is such that |x − y| = d(x) as in Lemma 2.4. By the definition of k-Hessian, we have 

Sk (ε1 , ε2 , . . . , εN−1 ) =

k *

1≤i1 <···


Sk−1 (ε1 , ε2 , . . . , εN−1 ) =

κij (y) 1 − κij (y)d(x)

k−1 *

1≤i1 <···
κij (y) 1 − κij (y)d(x)

Because k *

lim

d(x)→0

lim

d(x)→0

(1 − d(x)κij (x)) ¯ = 1,

j =1 k−1 *

(1 − d(x)κij (x)) ¯ = 1,

j =1

we chose ε small such that for x ∈ δε (where δε is corresponding to ε), 1−ε<

k *

(1 − d(x)κij (x)) ¯ < 1 + ε,

j =1

1−ε<

k−1 * j =1

Let

,

(1 − d(x)κij (x)) ¯ < 1 + ε.

.

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Mk = max Sk (κ1 (y), κ2 (y), . . . , κN−1 (y)), mk = min Sk (κ1 (y), κ2 (y), . . . , κN−1 (y)), y∈∂

y∈∂

Mk−1 = max Sk−1 (κ1 (y), κ2 (y), . . . , κN−1 (y)), mk−1 = min Sk−1 (κ1 (y), κ2 (y), . . . , κN−1 (y)). y∈∂

y∈∂

Then mk Mk ≤ Sk (ε1 , ε2 , . . . , εN−1 ) ≤ , x ∈ δ ε , 1+ε 1−ε

(6.1)

mk−1 Mk−1 ≤ Sk−1 (ε1 , ε2 , . . . , εN−1 ) ≤ , x ∈  δε . 1+ε 1−ε

(6.2)

For small ε > 0 (we can choose smaller ε if needed), μ > 0, let  ξε =

k k 1 (( k+1 + μ)k Mk−1 [( k+1 + μ) J10 + ( k+1 − μ) J01I∞ +

 ξε =



k2 (1 − ε)2

,

1 I∞ ]



k1 (1 + ε)2 k k 1 (( k+1 − μ)k mk−1 [( k+1 − μ) J10 + ( k+1 + μ) J01I∞ +

1 k+1

1 I∞ ]

1 k+1

,

where k1 , k2 , J0 , I∞ are given in Theorem 1.5. From Lemma 3.13 and Lemma 3.20 and the definition of ξ ε , ξ ε we see that [k P˜ (t)] k 1 lim = , t 1 + J0 t→0 p(t) (k P˜ (τ )) k dτ 0 k+1

k

[(k + 1)F (s)] k+1 1 , = lim s→∞ f (s) (s) I∞

  1 k 1 1 k 1 1 1 k + ]Mk−1 − 1 = −ε, + μ) [( + μ) + ( − μ) k+1 k2 (1 − ε) k+1 J0 k+1 I∞ J0 I∞   1 k 1 1 k 1 1 1 k+1 k + ]mk−1 − 1 = ε. − μ) [( − μ) + ( + μ) ξε ( k+1 k1 (1 + ε) k+1 J0 k+1 I∞ J0 I∞

k+1 ξε (

Define k

k

u(x) ¯ = ψ[ξ ε [ω(d(x))] k+1 +μ ], u(x) = ψ[ξ ε [ω(d(x))] k+1 −μ ], x ∈ δε , where δε is sufficiently small such that (6.1), (6.2) and  1 k Mk k P˜ (d(x)) k 1 + μ)k + [( + μ) k+1 k2 (1 − ε) I (u) p(d(x)) k+1 J (d(x))  1 1 1 1 +( − μ) + ]Mk−1 − 1 < 0, k+1 I (u) J (d(x)) I (u) k+1

ξε

(

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 1 k mk k P˜ (d(x)) k 1 k − μ) + [( − μ) +1 k1 (1 + ε) I (u) p(d(x)) k+1 J (d(x))  1 1 1 1 +( + μ) + ]mk−1 − 1 > 0 k+1 I (u) J (d(x)) I (u) ξ k+1 ( ε k

hold. By (2.1) we have for x ∈ δε Sk (D 2 u) − H (x)f (u) k k k + μ)k ψ  k ω k ω− k+1 +kμ Sk (ε1 , ε2 , . . . , εN−1 ) = (−1)k ξ ε ( k+1 k−1 k k + (−1)k−1 ξ ε ( + μ)k ψ  k−1 ω k−1 ω− k+1 +(k−1)μ k+1 2 1 1 k 1 × [ξ ε ( + μ)ω− k+1 +μ ω 2 + ( − μ)ψ  ω 2 ω− k+1 −1+μ ψ  ω ω− k+1 +μ ] k+1 k+1 × Sk−1 (ε1 , ε2 , . . . , εN−1 ) − H (x)f (u) k k+1 = ξε ( + μ)k (−ψ  )k−1 ψ  (−ω k−1 ω ) k+1  ω ψ k ω 2 + [( ) + μ)(− × Sk (ε1 , ε2 , . . . , εN−1 ) k  k+1 ωω ψ  ξ ε ω k+1 +μ ω  1 ω 2 ψ ψ +( + (− )]S (ε , ε , . . . , ε )] − H (x)f (u) − μ) k−1 1 2 N−1 k k  k+1 ψ  ξ ε ω k+1 +μ ωω ψ  ξ ε ω k+1 +μ  k 1 N P˜ (d(x)) k+1 = ξε ( + μ)k f (u)p(d(x)) Sk (ε1 , ε2 , . . . , εN−1 ) k+1 I (u) p(d(x))  k 1 1 1 1 1 + [( + μ) +( − μ) + ]Sk−1 (ε1 , ε2 , . . . , εN−1 )] k+1 J (d(x)) k+1 I (u) J (d(x)) I (u) − H (x)f (u)   1 k Mk N P˜ (d(x)) k 1 k+1 + μ)k + [( + μ) ≤ ξε ( k+1 k2 (1 − ε) I (u) p(d(x)) k+1 J (d(x))   1 1 1 1 +( − μ) + ]Mk−1 − 1 H (x)f (u) k+1 I (u) J (d(x)) I (u) < 0,

(6.3)

i.e. u is a supersolution to (1.1) in δε . By (1.17), for any μ > 0, lim (u − u(x)) = +∞.

d(x)→0

Thus u(x) ≤ u(x), x ∈ δε .

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Letting μ → 0, ε → 0, we get lim sup x∈, d(x)→0

u(x) k

ψ[ξ (ω(d(x))) k+1 ]

≤ 1.

Similarly, we can prove that Sk (D 2 u) − H (x)f (u) ≥ 0 in δε , and 1 ≤ lim inf

x∈, d(x)→0

u(x) k

ψ[ξ (ω(d(x))) k+1 ]

.

6.2. Proof of Corollary 1.8 It follows from (1.11) that 1

ψ  (t) = −[(k + 1)F (ψ(t))] k+1 . Then 1

tψ  (t) t[(k + 1)F (ψ(t))] k+1 lim = − lim t→0 ψ(t) t→0 ψ(t) = lim

(s)

 (s)

s→∞

s

(s)  (s) = 1 − I∞ . s→∞  2 (s)

= 1 − lim

This shows that ψ ∈ N RV Z1−I∞ . Combining this with Theorem 1.7 and Definition 3.1 we obtain lim

x∈, d(x)→0

u(x) k

ψ[(ω(d(x))) k+1 ]

= lim

x∈, d(x)→0

k

ψ[ξ0 (ω(d(x))) k+1 ]

u(x) k

k

ψ[ξ0 (ω(d(x))) k+1 ] ψ[(ω(d(x))) k+1 ]

= ξ01−I∞ . This gives the proof of Corollary 1.8.

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Acknowledgments This work is sponsored by the National Natural Science Foundation of China (11301178), the Beijing Natural Science Foundation (1163007), the Scientific Research Project of Construction for Scientific and Technological Innovation Service Capacity (KM201611232017), 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. References [1] Y. Du, Order Structure and Topological Methods in Nonlinear Partial Differential Equations, vol. 1, Maximum Principles and Applications, World Scientific, 2006. [2] N. Trudinger, X. Wang, The Monge-Ampère equation and its geometric applications, in: Handbook of Geometric Analysis, vol. 1, 2008, pp. 467–524. [3] X. Wang, A class of fully nonlinear elliptic equations and related functionals, Indiana Univ. Math. J. 43 (1994) 25–54. [4] L. Caffarelli, L. Nirenberg, J. Spruck, The Dirichlet problem for nonlinear second order elliptic equations, III: functions of the eigenvalues of the Hessian, Acta Math. 155 (1985) 261–301. [5] L. Caffarelli, L. Nirenberg, J. Spruck, The Dirichlet problem for nonlinear second-order elliptic equations I. MongeAmpère equations, Commun. Pure Appl. Math. 37 (1984) 369–402. [6] L. Bieberbach, u = eu und die authomorphen Funktionen, Math. Ann. 77 (1916) 173–212. [7] M. Chuaqui, C. Cortázar, M. Elgueta, C. Flores, J. García-Melián, R. Letelier, On an elliptic problem with boundary blow-up and a singular weight: the radial case, Proc. R. Soc. Edinb. 133 (2003) 1283–1297. [8] F. Gladiali, G. Porru, Estimates for explosive solutions to p-Laplace equations, in: Progress in Partial Differential Equations, vol. 1, Pont-á-Mousson, 1997, in: Pitman Res. Notes Math. Series, vol. 383, Longman, 1998, pp. 117–127. [9] Z. Zhang, Large solutions to the Monge-Ampère equations with nonlinear gradient terms: existence and boundary behavior, J. Differ. Equ. 264 (2018) 263–296. [10] Z. Zhang, K. Wang, Existence and non-existence of solutions for a class of Monge-Ampère equations, J. Differ. Equ. 246 (2009) 2849–2875. [11] K. Tso, On a real Monge-Ampère functional, Invent. Math. 101 (1990) 425–448. [12] D. Covei, The Keller-Osserman problem for the k-Hessian operator, arXiv preprint arXiv:1508.04653. [13] X. Zhang, M. Feng, Boundary blow-up solutions to the k-Hessian equation with singular weights, Nonlinear Anal. 167 (2018) 51–66. [14] H. Sun, M. Feng, Boundary behavior of k-convex solutions for singular k-Hessian equations, Nonlinear Anal. 176 (2018) 141–156. [15] Y. Huang, Boundary asymptotical behavior of large solutions to Hessian equations, Pac. J. Math. 244 (2010) 85–98. [16] P. Salani, Boundary blow-up problems for Hessian equations, Manuscr. Math. 96 (1998) 281–294. [17] B. Guan, The Dirichlet problem for a class of fully nonlinear elliptic equations, Commun. Partial Differ. Equ. 19 (1994) 399–416. [18] J.B. Keller, On solutions of u = f (u), Commun. Pure Appl. Math. 10 (1957) 503–510. [19] A. Colesanti, P. Salani, E. Francini, Convexity and asymptotic estimates for large solutions of Hessian equations, Differ. Integral Equ. 13 (2000) 1459–1472. [20] K. Takimoto, Solution to the boundary blowup problem for k-curvature equation, Calc. Var. Partial Differ. Equ. 26 (2006) 357–377. [21] A.C. Lazer, P.J. McKenna, On singular boundary value problems for the Monge-Ampère operator, J. Math. Anal. Appl. 197 (1996) 341–362. [22] A. Mohammed, Existence and estimates of solutions to a singular Dirichlet problem for the Monge-Ampère equation, J. Math. Anal. Appl. 340 (2008) 1226–1234. [23] X. Ji, J. Bao, Necessary and sufficient conditions on solvability for Hessian inequalities, Proc. Am. Math. Soc. 138 (2010) 175–188.

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[24] Q. Jin, Y. Li, H. Xu, Nonexistence of positive solutions for some fully nonlinear elliptic equations, Methods Appl. Anal. 12 (2005) 441–450. [25] H. Jian, Hessian equations with infinite Dirichlet boundary, Indiana Univ. Math. J. 55 (2006) 1045–1062. [26] M. Chuaqui, C. Cortazar, M. Elgueta, et al., Uniqueness and boundary behavior of large solutions to elliptic problems with singular weight, Commun. Pure Appl. Anal. 3 (2004) 653–662. [27] S. Dumont, L. Dupaigne, O. Goubet, V. Radulescu, Back to the Keller-Osserman condition for boundary blow-up solutions, Adv. Nonlinear Stud. 7 (2007) 271–298. [28] M. Ghergu, V. Radulescu, Singular Elliptic Problems: Bifurcation and Asymptotic Analysis, Oxford Lecture Series in Mathematics and Its Applications, vol. 37, The Clarendon Press, Oxford University Press, Oxford, 2008. [29] H. Yang, Y. Chang, On the blow-up boundary solutions of the Monge-Ampère equation with singular weights, Commun. Pure Appl. Anal. 11 (2012) 697–708. [30] N. Abatangelo, Very large solutions for the fractional Laplacian: towards a fractional Keller-Osserman condition, Adv. Nonlinear Anal. 6 (2017) 383–405. [31] A. Mohammed, G. Porru, Large solutions to non-divergence structure semilinear elliptic equations with inhomogeneous term, Adv. Nonlinear Anal. 8 (1) (2019) 517–532, https://doi.org/10.1515/anona-2017-0065. [32] F.R. Harvey, H.B. Lawson Jr., Dirichlet duality and the nonlinear Dirichlet problem, Commun. Pure Appl. Math. 62 (2009) 396–443. [33] F.R. Harvey, H.B. Lawson Jr., Plurisubharmonicity in a general geometric context, Geom. Anal. 1 (2010) 363–401. [34] F.R. Harvey, H.B. Lawson Jr., Geometric plurisubharmonicity and convexity: an introduction, Adv. Math. 230 (2012) 2428–2456. [35] F.R. Harvey, H.B. Lawson Jr., Removable singularities for nonlinear subequations, Indiana Univ. Math. J. 63 (2014) 1525–1552. [36] L. Caffarelli, Y. Li, L. Nirenberg, Some remarks on singular solutions of nonlinear elliptic equations, I , J. Fixed Point Theory Appl. 5 (2009) 353–395. [37] L. Caffarelli, Y. Li, L. Nirenberg, Some remarks on singular solutions of nonlinear elliptic equations, I I I : viscosity solutions including parabolic operators, Commun. Pure Appl. Math. 66 (2013) 109–143. [38] M.E. Amendola, G. Galise, A. Vitolo, Riesz capacity, maximum principle and removable sets of fully nonlinear second-order elliptic operators, Differ. Integral Equ. 26 (2013) 845–866. [39] G. Galise, A. Vitolo, Removable singularities for degenerate elliptic Pucci operators, Adv. Differ. Equ. 22 (2017) 77–100. [40] I. Capuzzo-Dolcetta, F. Leoni, A. Vitolo, On the inequality F (x, D 2 u) ≥ f (u) + q(u)|Du|q , Math. Ann. 365 (2016) 423–448. [41] A. Vitolo, Removable singularities for degenerate elliptic equations without conditions on the growth of the solution, Trans. Am. Math. Soc. 370 (2018) 2679–2705. [42] R. Osserman, On the inequality u ≥ f (u), Pac. J. Math. 7 (1957) 1641–1647. [43] J. Matero, The Bieberbach-Rademacher problem for the Monge-Ampère operator, Manuscr. Math. 91 (1996) 379–391. [44] A. Mohammed, On the existence of solutions to the Monge-Ampère equation with infinite boundary values, Proc. Am. Math. Soc. 135 (2007) 141–149. [45] X. Zhang, Y. Du, Sharp conditions for the existence of boundary blow-up solutions to the Monge-Ampère equation, Calc. Var. Partial Differ. Equ. 57 (2018) 1–24. [46] X. Zhang, M. Feng, Boundary blow-up solutions to the k-Hessian equation with singular weights, J. Math. Anal. Appl. 464 (2018) 456–472. [47] Z. Zhang, Boundary behavior of large solutions to the Monge-Ampère equations with weights, J. Differ. Equ. 259 (2015) 2080–2100. [48] Z. Zhang, A unified boundary behavior of large solutions to Hessian equations, Chin. Ann. Math., Ser. B (2017), acception. [49] N.H. Bingham, C.M. Goldie, J.L. Teugels, Regular Variation, Encyclopedia Math. Appl., vol. 27, Cambridge University Press, 1987. [50] J. Karamata, Sur un mode de croissance régulière. Théorèmes fondamentaux, Bull. Soc. Math. Fr. 61 (1933) 55–62. [51] R. Seneta, Regular Varying Functions, Lecture Notes in Math., vol. 508, Springer-Verlag, 1976.