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The existence of infinitely many boundary blow-up solutions to the p-k-Hessian equation

  • Meiqiang Feng EMAIL logo and Xuemei Zhang
Published/Copyright: June 2, 2023
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Advanced Nonlinear Studies
From the journal Advanced Nonlinear Studies Volume 23 Issue 1

Abstract

The primary objective of this article is to analyze the existence of infinitely many radial p - k -convex solutions to the boundary blow-up p - k -Hessian problem

σ k ( λ ( D i ( ∣ D u ∣ p − 2 D j u ) ) ) = H ( ∣ x ∣ ) f ( u ) in Ω , u = + ∞ on ∂ Ω .

Here, k ∈ { 1 , 2 , … , N } , σ k ( λ ) is the k -Hessian operator, and Ω is a ball in R N ( N ≥ 2 ) . Our methods are mainly based on the sub- and super-solutions method.

Keywords: p-k-Hessian equation; boundary blow up; sub-supersolution method; p-k-convex solution
MSC 2010: 34B18; 34B15; 34A34

1 Introduction

We study the existence of radial p - k -convex solution to the p - k -Hessian problem

(1.1) σ k ( λ ( D i ( ∣ D u ∣ p − 2 D j u ) ) ) = H ( ∣ x ∣ ) f ( u ) in Ω , u = + ∞ on ∂ Ω ,

where k ∈ { 1 , 2 , … , N } , p ≥ 2 , Ω is a ball in R N ( N ≥ 2 ) , f is a locally Lipschitz continuous, increasing, and positive function, H ∈ C ( Ω ) is positive in Ω and may be singular on ∂ Ω , and ( D i ( ∣ D u ∣ p − 2 D j u ) ) is a matrix with entry

D i ( ∣ D u ∣ p − 2 D j u ) = ∂ ∂ x i ∑ k = 1 N ∂ u ∂ x k 2 p − 2 2 ∂ u ∂ x j

for i , j ∈ { 1 , 2 , … , N } . The boundary condition u = + ∞ on ∂ Ω means

u ( x ) → + ∞ as dist ( x , ∂ Ω ) → 0 .

The equation with such a boundary condition is known as a boundary blow-up problem.

For an arbitrary N × N real symmetric matrix A ,

(1.2) σ k ( λ ( A ) ) = ∑ 1 ≤ i 1 < ⋯ < i k ≤ N λ i 1 ⋯ λ i k

denotes the k th elementary symmetric function, and λ 1 , λ 2 … , λ N are the eigenvalues of A .

The p - k -Hessian operator σ k ( λ ( D i ( ∣ D u ∣ p − 2 D j u ) ) ) was first introduced by Trudinger and Wang [1], which is an important class of perfectly nonlinear operators. It is a k -Hessian operator when p = 2 and a p -Laplacian when k = 1 . In effect, the p - k -Hessian operator may be considered an extension of the Laplacian, p -Laplacian, and k -Hessian operators.

The p - k -Hessian problem was seldom studied in previous articles except [2]. Bao and Feng [2] analyzed the solvability of the p - k -Hessian inequality

σ k ( λ ( D i ( ∣ D u ∣ p − 2 D j u ) ) ) ≥ f k ( u ) in R N

and derived a necessary and sufficient condition

∫ ∞ ∫ 0 s f k ( t ) d t − 1 ( p − 1 ) k + 1 d s = ∞

for the existence of a global positive p - k -convex strong solution

u ∈ C 2 ( R N \ { 0 } ) ∩ Φ p , k ( R N ) ,

where

Φ p , k ( R N ) ≔ { u ∈ W loc 2 , n q ( R N ) : ∣ D u ∣ p − 2 D u ∈ C 1 ( R N ) , λ ( D i ( ∣ D u ∣ p − 2 D j u ) ) ∈ Γ k in R N } ,

here 1 < q < p − 1 p − 2 , and

Γ k ≔ { λ ∈ R N : σ l ( λ ) > 0 , l = 1 , 2 , … , k } .

Bao and Feng [2] generalized the results in [3] and [4] for the case k = 1 , p = 2 , in [5] for the case k = 1 , p > 1 , and in [6] and [7] for the case p = 2 .

Moreover, we note that the study on boundary blow-up problems has recently attracted the attention of many mathematicians and is a topic of current interest, see [8–24] and the references therein. Recently, Zhang and Du [25] studied the boundary blow-up problem for the Monge-Ampère equation

(1.3) M [ u ] = K ( ∣ x ∣ ) f ( u ) , x ∈ B , u = + ∞ , x ∈ ∂ B ,

where M [ u ] = det ( u x i x j ) is the Monge-Ampère operator and B is a ball in R N ( N ≥ 2 ) . The authors proved the multiplicity and nonexistence results of strictly convex solutions of (1.3) by employing the sub- and super-solutions method.

However, to our best knowledge, there is no study investigating the boundary blow-up solution of p - k -Hessian problem on a bounded domain. In this article, we will focus our interest on the existence of a boundary blow-up solution to problem (1.1).

2 Main results

Without losing generality, we assume that Ω is the unit ball. It is not difficult to see that if v = v ( r ) ( r = ∣ x ∣ ) is a radially symmetric solution to problem (1.1), then problem (1.1) is equivalent to

(2.1) C N − 1 k − 1 ( v ′ ) p − 1 r k − 1 ( ( v ′ ) p − 1 ) ′ + C N − 1 k ( v ′ ) p − 1 r k = H ( r ) f ( v ) , r ∈ ( 0 , 1 ) , v ′ ( 0 ) = 0 , v ( 1 ) = + ∞ .

Similar to [2], one can define

Φ p , k ( Ω ) ≔ { u ∈ W loc 2 , n q ( Ω ) : ∣ D u ∣ p − 2 D u ∈ C 1 ( Ω ) , λ ( D i ( ∣ D u ∣ p − 2 D j u ) ) ∈ Γ k in R N } ,

here

1 < q < p − 1 p − 2 .

A function u ∈ Φ p , k ( Ω ) satisfying (1.1) is known as a p - k -convex strong solution.

Suppose that H and f satisfy the following conditions:

  1. H ∈ C [ 0 , 1 ) is increasing and H ( r ) > 0 in [ 0 , 1 ) ;

  2. f ( s ) is locally Lipschitz continuous in ( 0 , ∞ ) , positive and increasing for s > 0 , and satisfies

    (2.2) ∫ ∞ [ F ( s ) ] − 1 ( p − 1 ) k + 1 d s = ∞ ,

    where

    F ( s ) = ∫ 0 s f ( t ) d t .

First, we investigate an initial value problem. For v 0 > 0 , we consider the problem

(2.3) C N − 1 k − 1 ( v ′ ) p − 1 r k − 1 ( ( v ′ ) p − 1 ) ′ + C N − 1 k ( v ′ ) p − 1 r k = H ( r ) f ( v ) , r ∈ ( 0 , 1 ) , v ( 0 ) = v 0 , v ′ ( 0 ) = 0 .

Lemma 2.1

(Corollary 2.2 of [2]) Suppose that v ( r ) ∈ C 0 [ 0 , R ) ∩ C 1 ( 0 , R ) is a positive solution of (2.3) for every v 0 > 0 . Hence, for u ( x ) = v ( r ) ( r < R ) , we can derive that

(2.4) λ ( D i ( ∣ D u ∣ p − 2 D j u ) ) = ( ( v ′ ( r ) ) p − 1 ) ′ , ( v ′ ( r ) ) p − 1 r , … , ( v ′ ( r ) ) p − 1 r ∈ Γ k , r ∈ ( 0 , R ) .

(2.5) σ k ( λ ( D i ( ∣ D u ∣ p − 2 D j u ) ) ) = C N − 1 k − 1 ( v ′ ) p − 1 r k − 1 ( ( v ′ ) p − 1 ) ′ + C N − 1 k ( v ′ ) p − 1 r k , r ∈ ( 0 , R ) ,

and

u ( x ) ∈ C 2 ( B R \ { 0 } ) ∩ W 2 , n q ( B R ) with ∣ D u ∣ p − 2 D u ∈ C 1 ( B R )

is a strong solution to problem (1.1), where

(2.6) 1 < q < p − 1 p − 2 .

Lemma 2.2

Assume that H gratifies ( H ′ ) and f gratifies ( f ′ ) . Then, for every v 0 > 0 , (2.3) admits a unique solution v ( r ) ∈ C 2 ( 0 , a ) ∩ W 2 , q ( 0 , a ) over a maximal interval of existence [ 0 , a ) ⊂ [ 0 , 1 ) , where q and p satisfy (2.6). Moreover, v ′ > 0 in ( 0 , a ) , v ″ > 0 in ( 0 , a ) , and v ( r ) → ∞ as r → a if a < 1 .

Proof

For small δ > 0 , we will demonstrate that (2.3) possesses a unique solution defined on [ 0 , δ ] . It is obvious to see that (2.3) is equivalent to the integral equation

(2.7) v ( r ) = v 0 + ∫ 0 r s k − N ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v ( t ) ) d t 1 / ( p − 1 ) k d s .

Let E = C ( [ 0 , δ ] ) , and define T : E → E by

( T v ) ( r ) = v 0 + ∫ 0 r s k − N ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v ( t ) ) d t 1 / ( p − 1 ) k d s .

We are in a position to verify that for small δ > 0 , T is a contraction mapping on a suitable subset of E and so admits a unique fixed point. It follows that (2.3) admits a unique solution over [ 0 , δ ] .

Set

H ∗ = max r ∈ [ 0 , 1 / 2 ] H ( r ) , h ∗ = min r ∈ [ 0 , 1 / 2 ] H ( r ) ,

and

B δ ( v 0 ) = { v ∈ E : ‖ v − v 0 ‖ E < δ } .

Fix δ 1 ∈ ( 0 , 1 / 2 ) so that v 0 − δ 1 > 0 and

∣ f ( v 1 ) − f ( v 2 ) ∣ ≤ L ∣ v 1 − v 2 ∣ for v 1 , v 2 ∈ [ v 0 − δ 1 , v 0 + δ 1 ] ,

where L denotes the Lipschitz constant of f ( u ) on [ v 0 − δ 1 , v 0 + δ 1 ] . Then,

m ≔ f ( v 0 − δ 1 ) ≤ f ( v ) ≤ M ≔ L δ 1 + f ( v 0 ) for v ∈ [ v 0 − δ 1 , v 0 + δ 1 ] .

It is clear to see that there is δ 2 ∈ ( 0 , δ 1 ) small enough so that

p − 1 p δ 1 p − 1 [ ( C N − 1 k − 1 ) − 1 k H ∗ M N − 1 ] 1 ( p − 1 ) k < 1 for δ ∈ ( 0 , δ 2 ] .

First, we verify that T ( B δ ( v 0 ) ) ⊂ B δ ( v 0 ) for every δ ∈ ( 0 , δ 2 ] .

Indeed, for such δ and any v ∈ B δ ( v 0 ) , we derive

∣ T v − v 0 ∣ = ∫ 0 r s k − N ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v ( t ) ) d t 1 / ( p − 1 ) k d s ≤ ∫ 0 r s k − N ( p − 1 ) k ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ∗ M d t 1 / ( p − 1 ) k d s = p − 1 p r p p − 1 [ ( C N − 1 k − 1 ) − 1 k H ∗ M N − 1 ] 1 ( p − 1 ) k < δ for r ∈ [ 0 , δ ] ,

which shows that

T ( B δ ( v 0 ) ) ⊂ B δ ( v 0 ) , ∀ δ ∈ ( 0 , δ 2 ] .

Next, we demonstrate that T is a contraction mapping on B δ ( v 0 ) for all small δ > 0 .

Using the mean value theorem, for δ ∈ ( 0 , δ 2 ] and v 1 , v 2 ∈ B δ ( v 0 ) , we derive that

J ( s ) ≔ ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v 1 ( t ) ) d t 1 / ( p − 1 ) k − ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v 2 ( t ) ) d t 1 / ( p − 1 ) k = 1 ( p − 1 ) k ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) [ θ f ( v 1 ) + ( 1 − θ ) f ( v 2 ) ] d t 1 ( p − 1 ) k − 1 ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) [ f ( v 1 ) − f ( v 2 ) ] d t

with θ = θ ( s ) ∈ ( 0 , 1 ) . Therefore, for s ∈ [ 0 , δ ] ,

∣ J ( s ) ∣ ≤ 1 p − 1 ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 h ∗ m d t 1 ( p − 1 ) k − 1 ⋅ ∫ 0 s ( C N − 1 k − 1 ) − 1 t N − 1 H ∗ L ‖ v 1 − v 2 ‖ E d t = 1 p − 1 s N ( p − 1 ) k N − 1 ( p − 1 ) k ( C N − 1 k − 1 ) − 1 ( p − 1 ) k ( k h ∗ m ) 1 ( p − 1 ) k − 1 H ∗ L ‖ v 1 − v 2 ‖ E .

Hence, it follows that, for r ∈ [ 0 , δ ] ,

∣ ( T v 1 ) ( r ) − ( T v 2 ) ( r ) ∣ = ∫ 0 r s k − N ( p − 1 ) k J ( s ) d s ≤ 1 p δ p p − 1 ( N C N − 1 k − 1 ) − 1 ( p − 1 ) k ( k h ∗ m ) 1 ( p − 1 ) k − 1 H ∗ L ‖ v 1 − v 2 ‖ E .

Hence, T is a contraction mapping on B δ ( v 0 ) if δ ∈ ( 0 , δ 2 ] is small enough so that

1 p δ p p − 1 ( N C N − 1 k − 1 ) − 1 ( p − 1 ) k ( k h ∗ m ) 1 ( p − 1 ) k − 1 H ∗ L < 1 .

We thus obtain that (2.3) admits a unique solution defined for r ∈ [ 0 , δ ] for small δ > 0 .

Moreover, since

v ′ ( r ) = [ r k − N ∫ 0 r ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v ( t ) ) d t ] 1 / ( p − 1 ) k ≥ 0 for r ∈ ( 0 , δ ] ,

v ( r ) ≥ v ( 0 ) = v 0 > 0 . Since f is increasing on ( 0 , + ∞ ) , we derive

f ( v ( r ) ) ≥ f ( v 0 ) > 0 .

It so follows that v ′ ( r ) > 0 .

By differentiating v ′ ( r ) , we obtain, for r ∈ ( 0 , δ ] ,

v ″ ( r ) = 1 ( p − 1 ) k r k − N ∫ 0 r ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v ( t ) ) d t 1 ( p − 1 ) k − 1 × ( k − N ) r k − N − 1 ∫ 0 r ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v ( t ) ) d t + r k − N ( C N − 1 k − 1 ) − 1 k r N − 1 H ( r ) f ( v ( r ) ) ≥ 1 ( p − 1 ) k r k − N ∫ 0 r ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v ( t ) ) d t 1 ( p − 1 ) k − 1 × ( k − N ) r k − N − 1 ( C N − 1 k − 1 ) − 1 k H ( r ) f ( v ( r ) ) ∫ 0 r t N − 1 d t + r k − 1 ( C N − 1 k − 1 ) − 1 k H ( r ) f ( v ( r ) ) ≥ 1 ( p − 1 ) r k − N ∫ 0 r ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( v ( t ) ) d t 1 ( p − 1 ) k − 1 k N r k − 1 ( C N − 1 k − 1 ) − 1 H ( r ) f ( v ( r ) ) > 0 .

One can also derive that

lim r → 0 v ″ ( r ) r − p − 2 p − 1 = 1 p − 1 ( C N − 1 k − 1 ) − 1 k H ( 0 ) f ( v 0 ) N 1 ( p − 1 ) k .

Since

p − 2 p − 1 q < 1 ,

we derive v ( r ) ∈ W 2 , q ( 0 , δ ] . It indicates that

v ( r ) ∈ C 2 ( 0 , δ ] ∩ W 2 , q ( 0 , δ ] .

To extend the solution v ( r ) to r > δ , we let v ′ = u and

U = u v .

Then, one can discuss the first-order ODE system as follows:

(2.8) U ′ = ( C N − 1 k − 1 ) − 1 r k − 1 H ( r ) f ( v ) − C N − 1 k u p − 1 r k ( p − 1 ) u k ( p − 1 ) − 1 u ≕ F ( r , U ) , U ( δ ) = v ′ ( δ ) v ( δ ) .

It so follows from ( H ′ ) and ( f ′ ) that F ( r , U ) is locally Lipschitz continuous in U in the range u > 0 and v > 0 and continuous in r ∈ [ 0 , 1 ) . It yields that (2.8) admits a unique solution defined for r in a small neighborhood of δ . It is not difficult to see that the v component of U gratifies

C N − 1 k − 1 ( v ′ ) p − 1 r k − 1 ( ( v ′ ) p − 1 ) ′ + C N − 1 k ( v ′ ) p − 1 r k = H ( r ) f ( v ) > 0 , v ( δ ) > 0 , v ′ ( δ ) > 0 .

Then, we have

v ′ ( r ) > v ′ ( δ ) , v ″ ( r ) > 0 for r > δ .

Thus, the solution U ( r ) to problem (2.8) can be extended to r > δ until r reaches 1 or until v ( r ) blows up to ∞ . Then, (2.3) admits 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 . So we complete the proof.□

Lemma 2.3

Suppose that H gratifies ( H ′ ) and f gratifies ( f ′ ) . If u 1 and u 2 are functions in C 1 ( [ 0 , a ) ) ∩ C 2 ( 0 , a ) satisfying

C N − 1 k − 1 ( u 1 ′ ) p − 1 r k − 1 ( ( u 1 ′ ) p − 1 ) ′ + C N − 1 k ( u 1 ′ ) p − 1 r k ≤ H ( r ) f ( u 1 ) for r ∈ ( 0 , a ) ,

C N − 1 k − 1 ( u 2 ′ ) p − 1 r k − 1 ( ( u 2 ′ ) p − 1 ) ′ + C N − 1 k ( u 2 ′ ) p − 1 r k ≥ H ( r ) f ( u 2 ) for r ∈ ( 0 , a ) ,

and

u 1 ′ ( 0 ) = u 2 ′ ( 0 ) = 0 , u 1 ( 0 ) < u 2 ( 0 ) .

Then,

u 1 ( r ) < u 2 ( r ) for r ∈ [ 0 , a ) .

Proof

If u 1 < u 2 in [ 0 , a ) does not hold, then by u 1 ( 0 ) < u 2 ( 0 ) , there is r ¯ ∈ ( 0 , a ) so that

u 1 ( r ¯ ) = u 2 ( r ¯ ) and u 1 ( r ) < u 2 ( r ) for r ∈ [ 0 , r ¯ ) .

Because u 1 and u 2 satisfy (2.7) with the equality sign replaced by inequalities, by the monotonicity of f , we obtain the contradiction:

u 1 ( r ¯ ) ≤ u 1 ( 0 ) + ∫ 0 r ¯ s k − N ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( u 1 ( t ) ) d t 1 / ( p − 1 ) k d s < u 2 ( 0 ) + ∫ 0 r ¯ s k − N ∫ 0 s ( C N − 1 k − 1 ) − 1 k t N − 1 H ( t ) f ( u 2 ( t ) ) d t 1 / ( p − 1 ) k d s ≤ u 2 ( r ¯ ) .

The proof of Lemma 2.3 is finished.□

Now we analyze the existence of radial p - k -convex solution to problem (2.1). For the sake of simplicity, we introduce some notations.

If (2.2) holds, then there is c 0 > 0 so that

(2.9) G ( t ) ≔ ∫ c 0 t [ ( ( p − 1 ) k + 1 ) F ( τ ) ] − 1 ( p − 1 ) k + 1 d τ → ∞ as t → ∞ .

Let g ( t ) denote the inverse of G ( t ) , i.e.,

(2.10) ∫ c 0 g ( t ) [ ( ( p − 1 ) k + 1 ) F ( τ ) ] − 1 ( p − 1 ) k + 1 d τ = t , ∀ t > 0 .

Then,

g ( 0 ) = c 0 , lim t → ∞ g ( t ) = ∞ ,

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

g ″ ( t ) = f ( g ( t ) ) [ ( ( p − 1 ) k + 1 ) F ( g ( t ) ) ] ( p − 1 ) k − 1 ( p − 1 ) k + 1 ,

( g ′ ( t ) ) ( p − 1 ) k − 1 g ″ ( t ) = f ( g ( t ) ) ,

and

(2.11) g ′ ( t ) g ″ ( t ) = [ ( ( p − 1 ) k + 1 ) F ( g ( t ) ) ] ( p − 1 ) k ( p − 1 ) k + 1 f ( g ( t ) ) = − [ G ′ ( t ) ] 2 G ″ ( t ) .

Define

(2.12) R ( s ) = − G ″ ( s ) G ( s ) ( G ′ ( s ) ) 2 .

In order to express the condition on H , let b ∈ C 1 ( 0 , ∞ ) be a positive function and satisfy

b ′ ( t ) < 0 , lim t → 0 + b ( t ) = + ∞ .

Let

B ( τ ) = ∫ τ 1 b ( t ) d t .

If

(2.13) ∫ 0 + [ B ( τ ) ] 1 ( p − 1 ) k d τ = ∞ ,

then we call such a function b is of class B ∞ .

Our main result is the following theorem.

Theorem 2.4

Let H satisfy ( H ′ ) , and assume that there exist constants d 1 , d 2 > 0 and a function p ∈ B ∞ so that

d 1 b ( 1 − r ) ≤ H ( r ) ≤ d 2 b ( 1 − r ) f o r a l l r < 1 c l o s e t o 1 .

Suppose that f satisfies ( f ′ ) and so that (2.2) holds. If lim s → ∞ R ( s ) exists (denoted by R ∞ ), then (1.1) admits infinitely many p-k-convex solutions.

Proof

Since b ∈ B ∞ , (2.13) holds. Let

(2.14) σ ( s ) = ∫ s 1 [ ( p − 1 ) k B ( τ ) ] 1 ( p − 1 ) k d τ .

Then, we obtain

lim s → 0 + σ ( s ) = ∞

and

(2.15) σ ′ ( s ) = − [ ( p − 1 ) k B ( s ) ] 1 ( p − 1 ) k , σ ″ ( s ) = [ ( p − 1 ) k B ( s ) ] 1 ( p − 1 ) k − 1 b ( s ) .

It is obvious to see that

y ( r ) = p − 1 p 1 − r p p − 1

gratifies

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

Define

w ( r ) = g ( c σ ( p − 1 ) k ( p − 1 ) k + 1 ( y ( r ) ) ) , for  r ∈ [ 0 , 1 )  and some constant  c > 0 .

By direct calculation, we derive that

w ′ = c ( p − 1 ) k ( p − 1 ) k + 1 g ′ σ − 1 ( p − 1 ) k + 1 σ ′ y ′ , w ″ = c ( p − 1 ) k ( p − 1 ) k + 1 σ − 1 ( p − 1 ) k + 1 g ′ σ ′ y ″ + g ′ σ ″ ( y ′ ) 2 − 1 ( p − 1 ) k + 1 g ′ σ − 1 ( σ ′ ) 2 ( y ′ ) 2 + c ( p − 1 ) k ( p − 1 ) k + 1 g ″ σ − 1 ( p − 1 ) k + 1 ( σ ′ ) 2 ( y ′ ) 2 .

(2.16) ( w ′ ) ( p − 1 ) k − 1 w ″ = c ( p − 1 ) k + 1 ( p − 1 ) k ( p − 1 ) k + 1 ( p − 1 ) k g ′ ( p − 1 ) k − 1 g ″ ( − σ ′ ) ( p − 1 ) k − 1 σ ″ ( − 1 ) ( p − 1 ) k ( y ′ ) ( p − 1 ) k − 1 y ″ × ( p − 1 ) k ( p − 1 ) k + 1 ( σ ′ ) 2 σ σ ″ − y ′ 2 y ″ − 1 ( p − 1 ) k + 1 g ′ c σ ( p − 1 ) k ( p − 1 ) k + 1 g ″ ( σ ′ ) 2 σ σ ″ − y ′ 2 y ″ + g ′ c σ ( p − 1 ) k ( p − 1 ) k + 1 g ″ − y ′ 2 y ″ + g ′ c σ ( p − 1 ) k ( p − 1 ) k + 1 g ″ ( − σ ′ σ ″ ) .

By the definition of w , we obtain that

(2.17) c σ ( p − 1 ) k ( p − 1 ) k + 1 ( y ( r ) ) = g − 1 ( w ) = G ( w ) .

Combining (2.17) with (2.11), we have

(2.18) g ′ c σ ( p − 1 ) k ( p − 1 ) k + 1 g ″ = 1 R ( w ) .

On the other hand, from the definition of σ and y , we have

(2.19) ( − σ ′ ( t ) ) ( p − 1 ) k − 1 σ ″ ( t ) = b ( t ) , σ ′ ( t ) σ ″ ( t ) = − ( p − 1 ) k B ( t ) b ( t )

and

y ′ = − r 1 p − 1 , y ″ = − 1 p − 1 r 1 p − 1 − 1 ,

(2.20) y ′ y ″ = ( p − 1 ) r , y ′ 2 y ″ = − ( p − 1 ) r p p − 1 .

By (2.16) and (2.18)–(2.20), we derive that

( w ′ ) ( p − 1 ) k − 1 w ″ = c ( p − 1 ) k + 1 ( p − 1 ) k ( p − 1 ) k + 1 ( p − 1 ) k r k − 1 f ( w ) b ( y ) Δ ( r ) ,

with

Δ ( r ) ≔ ( p − 1 ) k ( p − 1 ) k + 1 1 T ( y ) r p p − 1 − 1 ( p − 1 ) k + 1 1 R ( w ) 1 T ( y ) r p p − 1 + 1 R ( w ) r p p − 1 + 1 R ( w ) k B ( y ) b ( y ) ,

where

(2.21) T ( s ) = σ ( s ) σ ″ ( s ) ( σ ′ ( s ) ) 2 .

Thus, we obtain

C N − 1 k − 1 ( w ′ ) p − 1 r k − 1 ( ( w ′ ) p − 1 ) ′ + C N − 1 k ( w ′ ) p − 1 r k = c ( p − 1 ) k + 1 ( p − 1 ) k ( p − 1 ) k + 1 ( p − 1 ) k f ( w ) b ( y ) C N − 1 k − 1 Δ ( r ) + C N − 1 k 1 R ( w ) ( p − 1 ) k B ( y ) b ( y ) .

Because

σ ′ ( t ) 2 σ ( t ) σ ″ ( t ) = [ ( p − 1 ) k B ( t ) ] ( p − 1 ) k + 1 ( p − 1 ) k b ( t ) ∫ t 1 [ ( p − 1 ) k B ( τ ) ] 1 / ( p − 1 ) k d τ = ∫ t 1 ( ( p − 1 ) k + 1 ) [ ( p − 1 ) k B ( s ) ] 1 / ( p − 1 ) k b ( s ) d s ∫ t 1 − b ′ ( s ) ∫ s 1 [ ( p − 1 ) k B ( τ ) ] 1 / ( p − 1 ) k d τ + b ( s ) [ ( p − 1 ) k B ( s ) ] 1 / ( p − 1 ) k d s ≤ ( p − 1 ) k + 1 ,

we obtain that

1 R ( w ) − 1 ( p − 1 ) k + 1 1 R ( w ) 1 T ( y ) ≥ 0 .

We thus have

Δ 1 ( r ) ≔ ( p − 1 ) k ( p − 1 ) k + 1 1 T ( y ) r p p − 1 − 1 ( p − 1 ) k + 1 1 R ( w ) 1 T ( y ) r p p − 1 + 1 R ( w ) r p p − 1 ≥ 0 ,

and

Δ 1 ( r ) > 0 for 0 < r ≤ 1 .

Because

lim t → 0 B ( t ) b ( t ) = 0 and so lim r → 1 B ( y ( r ) ) b ( y ( r ) ) = 0 ,

and R ∞ ≠ ∞ , we see that Δ ( r ) is positive for r ∈ [ 0 , 1 ) . Then, there are positive constants m 1 < m 2 so that

m 1 ≤ C N − 1 k − 1 Δ ( r ) + C N − 1 k 1 R ( w ) ( p − 1 ) k B ( y ) b ( y ) ≤ m 2 for r ∈ [ 0 , 1 ) .

Hence, it follows that

(2.22) C N − 1 k − 1 ( w ′ ) p − 1 r k − 1 ( ( w ′ ) p − 1 ) ′ + C N − 1 k ( w ′ ) p − 1 r k ≤ c ( p − 1 ) k + 1 ( p − 1 ) k ( p − 1 ) k + 1 ( p − 1 ) k f ( w ) b ( y ) m 2 for r ∈ [ 0 , 1 )

(2.23) C N − 1 k − 1 ( w ′ ) p − 1 r k − 1 ( ( w ′ ) p − 1 ) ′ + C N − 1 k ( w ′ ) p − 1 r k ≥ c ( p − 1 ) k + 1 ( p − 1 ) k ( p − 1 ) k + 1 ( p − 1 ) k f ( w ) b ( y ) m 1 for r ∈ [ 0 , 1 ) .

Let b ( t ) be replaced by ε b ( p p − 1 t ) with small ε > 0 . One can suppose that

H ( r ) ≥ b [ p − 1 p ( 1 − r ) ] for r ∈ [ 0 , 1 ) .

Owing to y ( r ) ≥ p − 1 p ( 1 − r ) , we thus derive

b ( y ( r ) ) ≤ b [ p − 1 p ( 1 − r ) ] ≤ H ( r ) for r ∈ [ 0 , 1 ) .

Hence, it follows from (2.22) that

(2.24) C N − 1 k − 1 ( w ′ ) p − 1 r k − 1 ( ( w ′ ) p − 1 ) ′ + C N − 1 k ( w ′ ) p − 1 r k ≤ c ( p − 1 ) k + 1 ( p − 1 ) k ( p − 1 ) k + 1 ( p − 1 ) k f ( w ) H ( r ) m 2 for r ∈ [ 0 , 1 ) .

Define

w 1 ( r ) ≔ g ( c ˜ 1 σ ( p − 1 ) k ( p − 1 ) k + 1 ( y ( r ) ) ) ,

where c ˜ 1 > 0 is a constant. If we take c ˜ 1 small enough, then w 1 satisfies

C N − 1 k − 1 ( w 1 ′ ) p − 1 r k − 1 ( ( w 1 ′ ) p − 1 ) ′ + C N − 1 k ( w 1 ′ ) p − 1 r k ≤ H ( r ) f ( w 1 ) for r ∈ [ 0 , 1 ) .

Next, we will look for a function w 2 ( r ) that gratifies the reversed inequality. Suppose that b ( t ) is replaced by M b ( t ) , where M > 0 is sufficiently large. Then, one can assume that

b ( 1 − r ) ≥ H ( r ) for r ∈ [ 0 , 1 ) .

Owing to y ( r ) ≤ 1 − r , we derive that

b ( y ( r ) ) ≥ b ( 1 − r ) ≥ H ( r ) for r ∈ [ 0 , 1 ) .

Thus, by (2.23) (where σ ( t ) and m 1 are determined by this new function b ( t ) ), we obtain that

(2.25) C N − 1 k − 1 ( w ′ ) p − 1 r k − 1 ( ( w ′ ) p − 1 ) ′ + C N − 1 k ( w ′ ) p − 1 r k ≥ c ( p − 1 ) k + 1 ( p − 1 ) k ( p − 1 ) k + 1 ( p − 1 ) k f ( w ) H ( r ) m 1 for r ∈ [ 0 , 1 ) .

In addition, if we take c = c ˜ 2 large enough, then we can define

w 2 ( r ) ≔ g c ˜ 2 σ ( p − 1 ) k ( p − 1 ) k + 1 ( y ( r ) ) ,

and w 2 gratifies

w 2 ( 0 ) > w 1 ( 0 ) , C N − 1 k − 1 ( w 2 ′ ) p − 1 r k − 1 ( ( w 2 ′ ) p − 1 ) ′ + C N − 1 k ( w 2 ′ ) p − 1 r k ≥ H ( r ) f ( w 2 ) for r ∈ [ 0 , 1 ) .

Let v c denote the unique solution to problem (2.3) with v 0 = c , where c ∈ ( w 1 ( 0 ) , w 2 ( 0 ) ) . Hence, it yields from Lemma 2.3 that

w 1 ( r ) < v c ( r ) < w 2 ( r ) for r ∈ [ 0 , 1 )

and v c ( r ) is defined well. Thus, one can apply Lemma 2.1 to find that v c ( r ) is defined for r ∈ [ 0 , 1 ) and v ′ ( r ) > 0 in ( 0 , 1 ) . Since w 1 ( r ) → ∞ when r → 1 , we obtain that v c ( r ) → ∞ when r → 1 . This shows that v c is a p - k -convex solution to problem (2.1). By altering c , we thus obtain infinitely many solutions to problem (2.1), i.e., (1.1) possesses infinitely many p - k -convex solutions. This finishes the proof of Theorem 2.4.□

Acknowledgements

Both authors would like to express their gratitude to the referee for valuable comments and suggestions.

  1. Funding information: This work is sponsored by the Beijing Natural Science Foundation under grant numbers 1212003 and 1232021.

  2. Conflict of interest: On behalf of all authors, the corresponding author states that there is no conflict of interest.

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Received: 2022-07-26
Revised: 2023-05-14
Accepted: 2023-05-15
Published Online: 2023-06-02

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/ans-2022-0074
Keywords for this article
p-k-Hessian equation; boundary blow up; sub-supersolution method; p-k-convex solution
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