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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco


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On Cauchy–Liouville-type theorems

Ataklti Araya / Ahmed Mohammed
Published Online: 2017-08-24 | DOI: https://doi.org/10.1515/anona-2017-0158

Abstract

In this paper we explore Liouville-type theorems to solutions of PDEs involving the ϕ-Laplace operator in the setting of Orlicz–Sobolev spaces. Our results extend Liouville-type theorems that have been obtained recently.

Keywords: Liouville-type theorems; Orlicz–Sobolev spaces

MSC 2010: 35J60; 35J62; 35J70; 35J75

1 Introduction

Any real-valued harmonic function on N (N2) which is bounded either from above or below is a constant. This is perhaps the earliest formulation of what is commonly known as the Cauchy–Liouville theorem (or simply Liouville’s theorem). Today there is an extensive amount of work that extends this basic result in many different directions. A beautiful account of the history, techniques and some recent results on Liouville-type problems can be found in [9].

Our goal in this paper is to investigate classes of functions f: for which the quasilinear PDE

Δϕu=f(u)(1.1)

admits constants as the only possible non-negative solutions in the entire Euclidean space N (N2). In this case we will say that the PDE has a Liouville-type property. Here Δϕu:=div(ϕ(|u|)u) is the so-called ϕ-Laplacian which reduces to the well-known p-Laplacian when ϕ(t)=ptp-2 for p>1.

In our attempt to understand the Liouville-type property of (1.1), we will focus on two classes of functions f. In the first part of the paper we investigate problem (1.1) when f belongs to a class of non-decreasing and non-negative functions whose growth at infinity is dictated by a Keller–Osserman-type condition. To elaborate on this, suppose that f is a non-decreasing continuous real-valued function defined on such that f(0)=0 and f(t)>0 for t>0. It is well known (see [9]) that the only non-negative solution of Δu=f(u) in N is the trivial solution u0 if and only if

1(0tf(s)ds)-1/2dt<.(1.2)

This result is essentially due to Keller [14] and Osserman [22], obtained independently. We will use an adaptation of condition (1.2) to the ϕ-Laplacian to obtain a sufficient condition on f in order for (1.1) to have the Liouville-type property.

In the second part of the paper we study another class of functions f that will lead to the PDE (1.1) having a Liouville-type property. To motivate this aspect of our investigation, we mention the paper [29], where the Liouville-type property of the p-Laplacian has been discussed extensively. Among many important results developed therein, we would like to mention the following result which is relevant to the work at hand. Suppose f:0+0+ is subcritical in the sense that f satisfies

f(t)(α-1)f(t)t,t>0

for some 1<α<p*, where p* is the critical Sobolev exponent of p>1. If there exists q>p such that f(t)tq-1 at infinity, then any non-negative solution of Δpu=-f(u) in N is the trivial solution, see [29]. In the recent paper [7], an adaptation of a method in [21] was used to obtain a result which complements the aforementioned Liouville-type result. More specifically, the following result was proved. Suppose f: is a differentiable function that has a root and satisfies

f(t)(p-1)N+1N-1f(t)t,t>0.(1.3)

Then any positive solution of Δpu=-f(u) in N must be a constant. As pointed out above, the method used in [7] is based on the work of McCoy [21, 20], who investigated the Liouville-type property for Δu=-f(u). This result in [21] is just one among many other Liouville-type results studied for elliptic equations, including some fully nonlinear higher-order equations. We also call attention to the nice paper [3] which discusses, among other things, Liouville-type results for general anisotropic quasilinear equations. The paper [15] contains Liouville comparison principles for quasilinear singular parabolic second-order partial differential inequalities.

The paper is organized as follows. In Section 2 we begin by stating some general assumptions on ϕ and by recalling some results from the literature used in our work. In Section 3 we show that (1.1) has the Liouville-type property when ϕ meets a suitable condition and f is continuous and non-decreasing, and satisfies a generalized Keller–Osserman condition. Some examples will be presented to illustrate the applicability of the main result of the section. Finally, in Section 4, we discuss a Liouville-type property of (1.1) when f belongs to a class of differentiable functions f:, which are not necessarily monotonic, may change sign and satisfy a condition that generalizes (1.3). An appropriate condition on ϕ will also be required. We conclude the section with some illustrative examples.

2 Preliminaries

Given ϕ:(0,)(0,), let us set

Ψ(t):=tϕ(t)andΦ(t):=0tΨ(s)𝑑s,t>0.

We make the following assumptions:

  • (ϕ1)

    Ψ is a strictly increasing C1 function in +:=(0,).

  • (ϕ2)

    lims0+Ψ(s)=0 and limsΨ(s)=.

  • (ϕ3)

    There exist constants 0<σρ such that

    σΦ′′(t)tΦ(t)ρfor all t>0.

Equations involving the ϕ-Laplacian have been used to model different physical phenomena. For instance, they appear in quantum physics [4] and in the modeling of nonlinear elasticity problems or in plasticity [12]. We refer to the papers [26, 2, 6, 10, 11, 24, 25, 28, 27] for more details and other applications.

In the first part of this section we discuss some immediate but useful consequences of conditions (ϕ1)(ϕ3) listed above.

We begin with inequalities which follow directly from (ϕ3), namely,

λ(s)Ψ(t)Ψ(st)Λ(s)Ψ(t)for all s,t0+:=[0,),(2.1)

for some increasing functions λΛ. In fact,

λ(s):=min{sσ,sρ}andΛ(s):=max{sσ,sρ}for s0+.

Inequalities (2.1), in turn, imply

Λ-1(ϱ)Ψ-1(τ)Ψ-1(ϱτ)λ-1(ϱ)Ψ-1(τ)for all ϱ,τ0+.(2.2)

Another immediate consequence of (2.1) is

λ~(s)Φ(t)Φ(st)Λ~(s)Φ(t),s,t0,(2.3)

where

λ~(s)=λ(s)sandΛ~(s)=Λ(s)s.

The following inequality, a direct consequence of (2.3), will be useful in Section 3:

Λ~-1(s)Φ-1(t)Φ-1(st)λ~-1(s)Φ-1(t).(2.4)

As a result of assumptions (ϕ1)(ϕ2) one can show that Φ is an N-function. From condition (ϕ3) we deduce that Φ satisfies the global Δ2-condition, namely, there exists a constant c>0 such that

Φ(2t)cΦ(t)for all t>0.

In fact, this is a direct consequence of (2.3).

Given an open set ΩN, the Orlicz space

LΦ(Ω):={u:Ω|u is measurable and ΩΦ(|u(x)|)dx<}

is a Banach space under the so-called Luxemburg norm

uΦ:=inf{k>0:ΩΦ(|u(x)|k)𝑑x1}.

The Orlicz–Sobolev space W1,Φ(Ω) is defined as the set of all weakly differentiable uLΦ(Ω) such that DαuLΦ(Ω) for all multi-indices α=(α1,,αN) with |α|1. This is a Banach space under the norm

uW1,Φ(Ω)=uΦ+uΦ.

As with the usual Sobolev spaces, W01,Φ(Ω) is defined as the closure of Cc(Ω) in W1,Φ(Ω).

We define the local spaces LlocΦ(Ω) and Wloc1,Φ(Ω) by

LlocΦ(Ω):={u:uLΦ(𝒪) for all 𝒪Ω}andWloc1,Φ(Ω):={u:uW1,Φ(𝒪) for all 𝒪Ω}.

The dual space (LΦ(Ω))* is equal to LΦ~(Ω), where Φ~ is the N-function given by

Φ~(t)=0tΨ-1(s)𝑑s,

and is called the complement of Φ. The assumption (ϕ3) shows that Φ~ satisfies a global Δ2-condition. This can be seen by integrating the right inequality in (2.2).

For more discussion of Orlicz–Sobolev spaces we refer the reader to [1, 5, 24] and the references therein.

Let ΩN be an open set and g:Ω× be a continuous function. We say a weakly differentiable function v:Ω is a sub-solution of the PDE

Δϕu=g(x,u),xΩ,(2.5)

if and only if for every open and bounded subset 𝒪Ω, we have vW1,Φ(𝒪), with g(x,v(x))LΦ~(𝒪), such that

𝒪ϕ(|v|)vφ-𝒪g(x,v)φfor all 0φW01,Φ(𝒪).(2.6)

A weakly differentiable function w:Ω is said to be a super-solution of (2.5) in Ω if and only if for every open and bounded subset 𝒪Ω, we have wW1,Φ(𝒪), with g(x,w(x))LΦ~(𝒪), such that the reverse inequality holds in (2.6) for all non-negative φW01,Φ(𝒪). A weakly differentiable function u:Ω is said to be a solution of (2.5) in Ω if u is both a sub-solution and a super-solution of (2.5) in Ω.

We follow common practice and write

Δϕvg(x,v)in Ω  and  Δϕwg(x,w)in Ω

to indicate that v is a sub-solution and w is a super-solution of (2.5), respectively, in Ω.

As noted in the introduction, in this paper we are interested in the investigation of Liouville-type property for the PDE

Δϕu=±f(u),

where the absorption term will be required to satisfy appropriate conditions.

We recall two results that will be useful in this paper. We begin with the following comparison principle, which is taken from [23, Theorem 2.4.1 and Proposition 2.4.2].

Theorem A (Comparison principle).

Let ΩRN be a bounded domain. Suppose that ϕ satisfies (ϕ1) and (ϕ2), and gLloc(Ω×R) is such that tg(x,t) is a non-decreasing function in R for each xΩ. Suppose u,vW1,Φ(Ω)C(Ω) satisfy

Δϕug(x,u)xΩ  𝑎𝑛𝑑  Δϕvg(x,v)in Ω.

If uv on Ω, then uv in Ω.

The following regularity result of Lieberman [17, Theorem 1.7], reminiscent of the regularity results for p-Laplacian-type quasilinear equations of DiBenedetto [8], Lieberman [16] and Tolksdorff [30], will also be useful.

Theorem B (Regularity theorem).

Let ΩRN be a bounded open set. Suppose that conditions (ϕ1) and (ϕ3) hold, and gL(Ω×R). If uW1,Φ(Ω)L(Ω) is a solution of (2.5), then uC1,α(Ω) for some constant 0<α<1.

3 Monotonic non-negative absorption term

Let f: be a non-decreasing continuous function such that f(0)=0 and f(t)>0 for t>0. In this section we extend a classical result of Keller [14] and Osserman [22] who showed independently that Δu=f(u) can not have a non-trivial solution in N if f satisfies a growth condition at infinity that has come to be known as the Keller–Osserman condition. We refer to the paper [9] for an excellent account on this and other related Liouville-type results.

Since σρ, where σ and ρ are the parameters in condition (ϕ3), we note that for s>1, we have

Λ-1(s)=min{s1/σ,s1/ρ}=s1/ρandλ-1(s)=max{s1/σ,s1/ρ}=s1/σ.

Therefore, for s>ϱ>1, we have

ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1)=ϱ(N-1)(ρ+1)/ρs(N-1)(σ+1)/σ.(3.1)

Since (ρ+1)/ρ(σ+1)/σ, we see that the right-hand side of (3.1) is less than one whenever s>ϱ>1. Now let us note that

Λ~-1(s)=min{s1/(σ+1),s1/(ρ+1)},

so that for 0<s<1, we have

Λ~-1(s)=s1/(σ+1).

Applying this to (3.1), we find that

Λ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))=ϱ(N-1)(ρ+1)/ρ(σ+1)s(N-1)/σ.

If σN-1, then it is clear that

limrϱrΛ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))𝑑s=ϱ(N-1)(ρ+1)/ρ(σ+1)limrϱrs-(N-1)/σ𝑑s=.

On the other hand suppose σ<N-1. If, in addition, ρ-σρσ(σ+1)<1N-1, then we have

ϱΛ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))𝑑s=ϱ(N-1)(ρ+1)/ρ(σ+1)ϱs-(N-1)/σ𝑑s=σN-1-σϱ1-(N-1)(ρ-σ)/ρσ(σ+1).

Consequently, in this case we find

limϱϱΛ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))𝑑s=.

The above discussion leads us to consider the following condition on the parameters σ and ρ in condition (ϕ3):

  • (ϕ4)

    0ρ-σρ(σ+1)<σN-1.

For easy reference, let us summarize the above discussion with regard to condition (ϕ4) in the following remark.

Remark 3.1.

Let us suppose that condition (ϕ4) holds. Then direct computations lead to the following conclusions:

  • (i)

    If σN-11, then for each ϱ>1,

    limrϱrΛ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))𝑑s=.

  • (ii)

    If σN-1<1, then

    limϱϱΛ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))𝑑s=.

Let us now consider a non-decreasing function f:(0,)(0,) such that

1dsΦ-1(F(s))<,where F(t):=0tf(s)𝑑s.(3.2)

Condition (3.2) is easily recognized as a generalization of the well-known Keller–Osserman condition when ϕ1.

We point out that condition (3.2) is equivalent to

tdsΦ-1(F(s)-F(t))<for all t>0.(3.3)

That (3.3) implies (3.2) is obvious. Therefore, we only need to show that (3.3) is implied by (3.2). For this, we first observe that

F(s)-F(t)f(t)(s-t)andF(s)-F(t)F(s-t)for all st.

Let t>0 and fix 0<θ<1. Using (2.4), together with the above inequalities, we have

tdsΦ-1(F(s)-F(t))=tθ+tdsΦ-1(F(s)-F(t))+θ+tdsΦ-1(F(s)-F(t))1Φ-1(f(t))tθ+tdsΛ~-1(s-t)+θ+tdsΦ-1(F(s-t))1Φ-1(f(t))tθ+tds(s-t)1/(σ+1)+θdsΦ-1(F(s))=θσ/(σ+1)Φ-1(f(t))+θdsΦ-1(F(s)).(3.4)

Thus, condition (3.2) on f implies that the right-hand side of (3.4) is finite for any t>0. Therefore, (3.3) holds.

In this section we study Liouville-type properties of solutions to the following equation:

ΔΦu=f(u).(3.5)

We assume that f:0+0+ is a continuous function such that f(0)=0 and f(t)>0 for t>0. Our first Liouville-type result in this section is provided by the following theorem.

Theorem 3.2.

Suppose that (ϕ1)(ϕ4) hold and that f is a non-decreasing function that satisfies condition (3.2). If uWloc1,Φ(RN)C(RN) is a non-negative sub-solution of (3.5) in RN, then u0 in RN.

Proof.

Let zN and let ε>0 be arbitrary. Let v be a solution of

{(rN-1ϕ(|v|)v)=rN-1f(v(r)),r>0,v(0)=ε,v(0)=0.

Let (0,R) be the maximal interval of existence of v. Note that v cannot be a constant. We shall show that R<. Note that

rN-1ϕ(|v(r)|)v(r)=0rsN-1f(v(s))𝑑s,0<r<R.(3.6)

Therefore, v>0 in (0,R), and rN-1Ψ(v) is a non-decreasing function in (0,R), and hence Ψ-1(rN-1Ψ(v)) is non-decreasing on (0,R) as well. On multiplying both sides of (3.6) by Ψ-1(rN-1Ψ(v)), for any ϱ<r<R, we estimate

Ψ-1(rN-1Ψ(v))rN-1Ψ(v)=Ψ-1(rN-1Ψ(v(r)))0rsN-1f(v(s))𝑑s0rsN-1Ψ-1(sN-1Ψ(v(s)))f(v(s))𝑑sϱrsN-1Λ-1(sN-1)f(v(s))v(s)𝑑s(3.7)ϱN-1Λ-1(ϱN-1)ρrf(v(s))v(s)𝑑s=ϱN-1Λ-1(ϱN-1)(F(v(r))-F(v(ϱ))).(3.8)

We have used (2.2) in obtaining (3.7). Inequality (3.8), together with

Ψ-1(rN-1Ψ(v))rN-1Ψ(v)λ-1(rN-1)rN-1Ψ(v(r))v(r),ϱ<r<R,

which again is a consequence of (2.2), shows that

Ψ(v(r))v(r)ϱN-1Λ-1(ϱN-1)rN-1λ-1(rN-1)(F(v(r))-F(v(ϱ))),ϱ<r<R.

On noting that Φ(2v(r))Ψ(v(r))v(r), we use (2.4) to get

v(r)12Φ-1(ϱN-1Λ-1(ϱN-1)rN-1λ-1(rN-1)(F(v(r))-F(v(ϱ))))12Λ~-1(ϱN-1Λ-1(ϱN-1)rN-1λ-1(rN-1))Φ-1(F(v(r))-F(v(ϱ))),0<ϱ<r.

Integrating this last inequality on (ϱ,r) yields

v(ϱ)v(r)dsΦ-1(F(s)-F(v(ϱ)))12ϱrΛ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))𝑑s.

Consequently, we see that

12ϱrΛ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))𝑑sv(ϱ)dsΦ-1(F(s)-F(v(ϱ)))for all r>ϱ.(3.9)

Note that

𝒬(t)=tdsΦ-1(F(s)-F(t))=0dxΦ-1(x)F-1(x+F(t))

is a non-increasing function. Therefore, since ε=v(0)v(ϱ), we have 𝒬(ε)𝒬(v(ϱ)). Hence, from inequality (3.9), we obtain

ϱrΛ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))𝑑s2εdsΦ-1(F(s)-F(ε))for all ϱ<r<R.(3.10)

Thus, condition (3.2) on f implies that the left-hand side of (3.10) is finite.

Now assume that R=. Then we can take ϱ>1 in (3.10). If σN-1, then taking the limit as r in (3.10) leads to a contradiction, by Remark 3.1.

Now let us suppose that σ<N-1. Taking the limit in (3.10) as r, we find that

ϱΛ~-1(ϱN-1Λ-1(ϱN-1)sN-1λ-1(sN-1))𝑑s2εdsΦ-1(F(s)-F(ε)).(3.11)

Then we use condition (ϕ4) and Remark 3.1 to see that inequality (3.11) leads to a contradiction upon taking the limit as ϱ.

Therefore, we conclude that indeed R<, and as a consequence v(r) as rR-. Let us set w(x):=v(|x-z|) for xN, and suppose that u is any non-negative sub-solution of (3.5) in N. Then, for 0<δ<1 such that 1-δ is sufficiently small, we have u<w on B(z,δR). Since w is a solution of (3.5) in the ball B(z,δR) we invoke the comparison principle, Theorem A, to conclude

u(x)w(x),xB(z,δR).

In particular, we have 0u(z)ε. Since ε>0 is arbitrary, we find that u(z)=0, and since zN is arbitrary, we conclude u0 in N. ∎

Theorem 3.3.

Suppose that (ϕ1)(ϕ4) hold. Let f:RR be such that f(t)>0 for t>0 and f(0)=0. Assume also that f satisfies (3.2).

  • (a)

    If u is a sub-solution of ( 3.5 ), then u0 in N.

  • (b)

    If f is an odd function and u satisfies ( 3.5 ), then u0 in N.

Proof.

(a)  Let u be any sub-solution of (3.5) in N. To show that u0 in N, it suffices to demonstrate that u+ is a sub-solution of (3.5) in N. Once this is proven, then we can invoke Theorem 3.2 to conclude that u+0, and hence u=-u-0 in N. First, since uWloc1,Φ(N), let us notice that u+Wloc1,Φ(N). To see that u+ is a sub-solution of (3.5), let Ω:={xN:u(x)>0}. We suppose that Ω is non-empty for otherwise there is nothing to prove. For each positive integer j, set ϑj(t):=ϑ(jt), t, where ϑC1(), with

ϑ(t)>0on (0,1),ϑ1on [1,)  and  ϑ0on (-,0].

Given 𝒪N, let φW01,Φ(𝒪) with φ0 in 𝒪. We have

𝒪ϕ(|u+|)u+,φ=𝒪Ωϕ(|u|)u,φ=limj𝒪ϑj(u)ϕ(|u|)u,φ.(3.12)

Let us note that ϑj(u)φW01,Φ(𝒪) and

ϑj(u)ϕ(|u|)u,φ=ϕ(|u|)u,ϑj(u)φ=ϕ(|u|)u,(ϑj(u)φ)-ϑj(u)φϕ(|u|)u,uϕ(|u|)u,(ϑj(u)φ).(3.13)

Using (3.12), (3.13) and recalling that u is a sub-solution of (3.5) in N, we find

𝒪ϕ(|u+|)u+,φ-limj𝒪ϑj(u)f(u)φ=-𝒪Ωf(u)φ=-𝒪f(u+)φ,

since f(0)=0. Therefore, u+ is a sub-solution of (3.5), as claimed.

(b)  Suppose that f is an odd function, and that u is a solution of (3.5) in N. By what was proved in (a), we observe that u0 in N. On the other hand, it is easily seen that -u is a solution of (3.5) in N. Therefore, -u0 in N again, completing the proof that u0 in N. ∎

3.1 Some examples

Here we consider examples to illustrate Theorems 3.2 and 3.3. It would be convenient to use the following equivalent formulation of condition (ϕ4).

Remark 3.4.

  • (i)

    Suppose 0<σ<12(4N-3-1). Then (ϕ4) holds if and only if

    0<σρ<σ(N-1)N-1-σ(σ+1).

  • (ii)

    Suppose σ12(4N-3-1). Then (ϕ4) holds if and only if ρσ.

In the examples below, conditions (ϕ1)(ϕ3) are easily verifiable.

(1)  Let ϕ(t)=ptp-2, with p>1. Then Δϕ is the standard p-Laplacian. In this case, σ=ρ=p-1, and therefore condition (ϕ4) holds. Condition (3.2) reduces to the requirement that

tds(F(s))1/p<for all t>0.(3.14)

Therefore, if (3.14) holds then Theorems 3.2 and 3.3 hold for any p>1.

(2)  Let us now consider ϕ(t)=ptp-2+qtq-2 for 1<p<q. Computation shows that σ=p-1, ρ=q-1 and

Φ(t)=tp+tq2tq,t>1.

As a consequence, we see that

tdsΦ-1(F(s))21/qtds(F(s))1/qfor all t>0.

Therefore, if the right-hand side in the above inequality is finite for some t>0, then condition (3.2) holds, and therefore Theorems 3.2 and 3.3 hold provided that condition (ϕ4) is satisfied. According to Remark 3.4, if

1<p<12(4N-1-1),(3.15)

then (ϕ4) holds if and only if

1<pq<(p-1)(N-1)N-1-p(p-1).

On the other hand, if

p12(4N-1-1),(3.16)

then (ϕ4) holds if and only if qp.

(3)  Let ϕ(t)=ptp-1logq(1+t)+qtp-1(1+t)-1logq-1(1+t) for p>1 and q>0. Then σ=p-1 and ρ=p+q-1. Moreover, we see that Φ(t)=tplogq(1+t). Note that given ε>0, there exists a constant tε>0, sufficiently large, such that

Φ(t)tp+εfor all t>tε.

Therefore, if there exists r>p such that

tds(F(s))1/r<

for some t>0, then condition (3.2) holds. Thus, if (ϕ4) is satisfied, then Theorems 3.2 and 3.3 apply. Here, again, we invoke Remark 3.4. Suppose that (3.15) holds. Then (ϕ4) holds if and only if

0<q<(p-1)(N-1)N-1-p(p-1).

If on the other hand (3.16) holds, then (ϕ4) holds if and only if q>0.

4 Sign-changing absorption term

This section is devoted to the study of the Liouville-type property of the following equation for a given f::

Δϕu=-f(u).(4.1)

We do not make any monotonicity assumption on f, but we require f to be a C1 function that satisfies a sub-critical-type condition (see condition (Cf) below).

We remark that the solutions of (4.1) are invariant under rotations, in the sense that if u is a solution of (4.1) in N, then v(x):=u(Ax) is also a solution of (4.1) on N for any orthogonal matrix A.

To study the Liouville-type property of (4.1), we start by making some suitable assumptions on ϕ. Specifically, we require that ϕC2(0,) and satisfies the following condition:

  • (ϕ5)

    We have

    -<inft>0ϖ(t)supt>0ϖ(t)<(NN-1)σ2-3ρ+2,where ϖ(t):=ϕ′′(t)t2ϕ(t),

    and σ,ρ are the constants in condition (ϕ3).

The main result of this section is the following Liouville-type theorem for the solutions of (4.1). This theorem extends the result in [7], where the special case ϕ(t)=tp-2, p>1, was considered. In the proof of the theorem we will observe the Einstein summation convention over repeated indices.

Theorem 4.1.

Suppose that conditions (ϕ1)(ϕ3) and (ϕ5) hold, fCloc1,γ(R) for some 0<γ<1, and

f(t)σ(N+1N-1)f(t)tfor all t>0.(Cf)

If uWloc1,Φ(RN)C(RN) is a positive solution of (4.1) in RN, then u is a constant in RN.

Remark 4.2.

Suppose that ϕ satisfies the assumptions of Theorem 4.1. If f(t)=tθ-tϑ for some 0θσ(N+1)/(N-1)ϑ, then f satisfies (Cf), and therefore the only positive solution of (4.1) in N is u1. We should also note that if f is a non-negative and non-increasing function on , then f satisfies (Cf), and therefore in this case only constants are the possible positive solutions of (4.1).

Proof of Theorem 4.1.

Let us first note that, as a consequence of the continuity of u and Theorem B (see also [13, Lemma 3.3]), we conclude that uCloc1,α(N). Therefore, it follows that equation (4.1) is uniformly elliptic on open sets that are compactly contained in 𝒪:={xN:|u(x)|>0}. Hence, we observe that uC3 in any open set that is compactly contained in 𝒪 (see [18, Corollary 2.2], with p=q=2). See also [19].

Let x0N be an arbitrary but fixed point. We wish to show u(x0)=0. Given a>0, let us set

J(x):=(a2-r2)2Θ,where Θ=|u|2u2 and r=|x-x0|.

We note that J0 in B:=B(x0,a) and J||x-x0|=a=0. Therefore, J attains its maximum value on B¯ at some interior point x*B. Suppose u(x*)=0. Then Θ (and hence J) will be zero at x*. But then J (and therefore Θ) will be zero on B¯. This would imply |u|=0 in B¯, and in particular u(x0)=0. Therefore, we assume that |u|>0 at x*. We recall that u is 𝒞loc3(𝒪), where 𝒪:={xΩ:|u|>0}.

Now, at x*, we have

0=Jj=-2(a2-r2)(r2)jΘ+(a2-r2)2Θj,j=1,,N,(4.2)

and D2J0.

Let us first show that the N×N matrix

H:=IN+ϕ(|u|)|u|ϕ(|u|)|u|-2uuT

is positive definite. Here IN is the N×N identity matrix, and AT stands for the transpose of the matrix A. To see that H is positive definite, we first note that for any ξN{0},

ξTHξ=|ξ|2+ϕ(|u|)|u|ϕ(|u|)(ξTu)2|u|2.

If ϕ(|u|)|u|ϕ(|u|)0, then ξTHξ|ξ|2>0. If, on the other hand, -1<ϕ(|u|)|u|ϕ(|u|)<0, then on noting that |ξTu||ξ||u| and therefore

(ξTu)2|u|2|ξ|2,

we have (recalling condition ϕ4)

ξTHξ|ξ|2+ϕ(|u|)|u|ϕ(|u|)|ξ|2=|ξ|2(1+ϕ(|u|)|u|ϕ(|u|))>0.

Thus, in any case, we have shown that H>0.

Now recalling that D2J0 at x*, we have HD2J0 at x*. Therefore, at x*, we have

0j=1NejT(HD2J)ej=ΔJ+ϕ(|u|)|u|ϕ(|u|)Jijuiuj|u|-2,

where ej is the unit vector in N with 1 in the jth position. In the last equation we have used the simple fact that ekTBek=bkk, where B=[bij] is an N×N matrix. Thus, at x*, we have

ΔJ+ϕ(|u|)|u|ϕ(|u|)Jijuiuj|u|-20.(4.3)

Recalling that (4.1) is invariant under rotation, we choose coordinates so that at the point x*, we have

|u|=u1,uj=0,j=2,,N.(4.4)

Through direct computation, we note that

ΔJ+ϕ(u1)u1ϕ(u1)J11=2|(r2)|2Θ-2(a2-r2)Δ(r2)Θ-4(a2-r2)(r2)Θ+(a2-r2)2ΔΘ+ϕ(u1)u1ϕ(u1)[2((r2)1)2Θ-4(a2-r2)Θ-4(a2-r2)(r2)1Θ1+(a2-r2)2Θ11].

This, together with inequality (4.3), implies

ΔΘ+ϕ(u1)u1ϕ(u1)Θ11-8r2(a2-r2)2Θ+4N(a2-r2)Θ+4(r2)Θa2-r2+ϕ(u1)u1ϕ(u1)[-2((r2)1)2(a2-r2)2Θ+4a2-r2Θ+4(r2)1a2-r2Θ1].(4.5)

To obtain (4.5), we have used the easily verifiable identities

|(r2)|2=4r2andΔ(r2)=2N.(4.6)

Moreover, from (4.2), we obtain

Θ1=2(r2)1a2-r2ΘandΘ=2Θa2-r2(r2).

Using these and (4.6) in (4.5), we find

ΔΘ+ϕ(u1)u1ϕ(u1)Θ11Θ[24r2(a2-r2)2+4Na2-r2+ϕ(u1)u1ϕ(u1)(6((r2)1)2(a2-r2)2+4a2-r2)].

Therefore, at x*, we have

1Θ(ΔΘ+ϕ(u1)u1ϕ(u1)Θ11)24r2(a2-r2)2+4Na2-r2+|ϕ(u1)u1ϕ(u1)|(24r2(a2-r2)2+4a2-r2).

Since r2<a2 in B(x0,a), we have

1Θ(ΔΘ+ϕ(u1)u1ϕ(u1)Θ11)24a2(a2-r2)2+4Na2(a2-r2)2+|ϕ(u1)u1ϕ(u1)|(24a2(a2-r2)2+4a2(a2-r2)2).(4.7)

Let us now notice that for i,j=1,,N, we have

Θi=2uijuju-2-2|u|2uiu-3,(4.8)Θii=2uiijuju-2+2uijuiju-2-8uijuiuju-3-2|u|2u-3uii+6|u|2u-4ui2.(4.9)

Upon using (4.4), from (4.9) we obtain

Θ11=2u111u1u-2+2u1ju1ju-2-10u11u12u-3+6u14u-4,ΔΘ=2(Δu)1u1u-2+2uijuiju-2-8u11u12u-3-2u12u-3Δu+6u14u-4.

From the above computation we see that at x*,

ΔΘ+ϕ(u1)u1ϕ(u1)Θ11=2u1u-2[(Δu)1+ϕ(u1)u1ϕ(u1)u111]+2u-2(uijuij+ϕ(u1)u1ϕ(u1)u1ju1j)-2u12u-3[Δu+(4+5ϕ(u1)u1ϕ(u1))u11]+6(1+ϕ(u1)u1ϕ(u1))u14u-4.(4.10)

Let us now proceed to find alternative forms for the expressions in the parentheses in (4.10). We start by rewriting equation (4.1) in any open set 𝒪 that contains x*, that is,

ϕ(|u|)Δu+ϕ(|u|)|u|uijuiuj=-f(u).(4.11)

where u is C3-smooth. On recalling (4.4), at x*, we obtain the following from (4.11):

ϕ(u1)Δu+ϕ(u1)u11u1=-f(u),

that is,

Δu+ϕ(u1)u1ϕ(u1)u11=-f(u)ϕ(u1).(4.12)

We now differentiate (4.11) with respect to the variable x1. On evaluating the resulting expression at x*, we find

ϕ(u1)(Δu)1+ϕ(u1)u11Δu+ϕ′′(u1)u1u112-ϕ(u1)u112+ϕ(u1)u111u1+2ϕ(u1)j=1Nu1j2=-u1f(u),

that is

ϕ(u1)(Δu)1+ϕ(u1)u1u111=-2ϕ(u1)j=2Nu1j2-ϕ(u1)u11Δu-ϕ′′(u1)u1u112-ϕ(u1)u112-u1f(u).(4.13)

On dividing both sides of (4.13) by ϕ(u1), using (4.12) to replace Δu in the resulting expression, and rearranging, we find

(Δu)1+ϕ(u1)u1ϕ(u1)u111=-2ϕ(u1)ϕ(u1)j=2Nu1j2+[f(u)ϕ(u1)-(1-ϕ(u1)u1ϕ(u1))u11]ϕ(u1)ϕ(u1)u11-ϕ′′(u1)u1ϕ(u1)u112-u1f(u)ϕ(u1).(4.14)

Let us observe the following.

uijuij+ϕ(u1)u1ϕ(u1)u1ju1j=2j=2Nu1j2+u112+j=2Nujj2+i,j2,ijNuij2+ϕ(u1)u1ϕ(u1)u112+ϕ(u1)u1ϕ(u1)j=2Nu1j2(2+ϕ(u1)u1ϕ(u1))j=2Nu1j2+(1+ϕ(u1)u1ϕ(u1))u112+j=2Nujj2.(4.15)

To proceed further, we need the following inequality.

Using Lemma 4.3 in inequality (4.15), we obtain

uijuij+ϕ(u1)u1ϕ(u1)u1ju1j(2+ϕ(u1)u1ϕ(u1))j=2Nu1j2+(NN-1+ϕ(u1)u1ϕ(u1))u112-2N-1u11Δu+1N-1(Δu)2.(4.16)

We use (4.14) and (4.16) in (4.10) to get

ΔΘ+ϕ(u1)u1ϕ(u1)Θ112u1u-2{-2ϕ(u1)ϕ(u1)j=2Nu1j2+[f(u)ϕ(u1)-(1-ϕ(u1)u1ϕ(u1))u11]ϕ(u1)ϕ(u1)u11-ϕ′′(u1)u1ϕ(u1)u112-u1f(u)ϕ(u1)}   +2u-2[(2+ϕ(u1)u1ϕ(u1))j=2Nu1j2+(NN-1+ϕ(u1)u1ϕ(u1))u112-2N-1u11Δu+1N-1(Δu)2]   -2u12u-3[Δu+(4+5ϕ(u1)u1ϕ(u1))u11]+6(1+ϕ(u1)u1ϕ(u1))u14u-4.

Rearranging the terms in the last inequality, we find

ΔΘ+ϕ(u1)u1ϕ(u1)Θ112u-2{(2-ϕ(u1)u1ϕ(u1))j=2Nu1j2+ϕ(u1)u1ϕ(u1)f(u)ϕ(u1)u11-fu12ϕ(u1)-2N-1(Δu)u11+1N-1(Δu)2+[(ϕ(u1)u1ϕ(u1))2-ϕ′′(u1)u12ϕ(u1)+NN-1]u112-[Δu+(4+5ϕ(u1)u1ϕ(u1))u11]u12u-1+3(1+ϕ(u1)u1ϕ(u1))u14u-2}.(4.17)

We recall from (4.8) that for j=2,,N, we have Θj=2u1ju1u-2, and therefore

2u-2j=2Nu1j2=j=2NΘj22(u1u)2=j=2NΘj22Θ|Θ|22Θ.

We use the above inequality and (4.12) to estimate inequality (4.17) as follows:

ΔΘ+ϕ(u1)u1ϕ(u1)Θ11-|2-ϕ(u1)u1ϕ(u1)||Θ|22Θ+2u-2{ϕ(u1)u1ϕ(u1)f(u)ϕ(u1)u11-fu12ϕ(u1)+[(ϕ(u1)u1ϕ(u1))2-ϕ′′(u1)u12ϕ(u1)+NN-1]u112+2N-1(ϕ(u1)u1ϕ(u1)u11+f(u)ϕ(u1))u11+1N-1(ϕ(u1)u1ϕ(u1)u11+f(u)ϕ(u1))2-[-ϕ(u1)u1ϕ(u1)u11-f(u)ϕ(u1)+(4+5ϕ(u1)u1ϕ(u1))u11]u12u-1+3(1+ϕ(u1)u1ϕ(u1))u14u-4}.

Rearranging the terms in the above inequality yields

ΔΘ+ϕ(u1)u1ϕ(u1)Θ11-|2-ϕ(u1)u1ϕ(u1)||Θ|22Θ+2u-2{-fu12ϕ(u1)+[NN-1(ϕ(u1)u1ϕ(u1))2+2N-1ϕ(u1)u1ϕ(u1)-ϕ′′(u1)u12ϕ(u1)+NN-1]u112+1N-1(f(u)ϕ(u1))2+[2N-1(1+ϕ(u1)u1ϕ(u1))+ϕ(u1)u1ϕ(u1)]f(u)ϕ(u1)u11-[4(1+ϕ(u1)u1ϕ(u1))u11-f(u)ϕ(u1)]u12u-1+3(1+ϕ(u1)u1ϕ(u1))u14u-2}.

We recall from (4.8) that Θ1=2u1u11u-2-2u13u-3, that is

u11=Θ1u22u1+u12u.

Inserting this in the last inequality gives

ΔΘ+ϕ(u1)u1ϕ(u1)Θ11-|2-ϕ(u1)u1ϕ(u1)||Θ|22Θ+2u-2{-fu12ϕ(u1)+[NN-1(ϕ(u1)u1ϕ(u1))2+2N-1ϕ(u1)u1ϕ(u1)-ϕ′′(u1)u12ϕ(u1)+NN-1](Θ1u22u1+u12u)2+1N-1(f(u)ϕ(u1))2+[2N-1(1+ϕ(u1)u1ϕ(u1))+ϕ(u1)u1ϕ(u1)](Θ1u22u1+u12u)f(u)ϕ(u1)-[4(1+ϕ(u1)u1ϕ(u1))(Θ1u22u1+u12u)-f(u)ϕ(u1)]u12u-1+3(1+ϕ(u1)u1ϕ(u1))u14u-2}.

Further rearrangement leads to

ΔΘ+ϕ(u1)u1ϕ(u1)Θ11-|2-ϕ(u1)u1ϕ(u1)||Θ|22Θ+[NN-1(ϕ(u1)u1ϕ(u1))2+2N-1ϕ(u1)u1ϕ(u1)-ϕ′′(u1)u12ϕ(u1)+NN-1](2Θ1u1u+2u14u4)+2N-1(f(u)uϕ(u1))2+[2N-1(1+ϕ(u1)u1ϕ(u1))+ϕ(u1)u1ϕ(u1)](Θ1f(u)u1ϕ(u1))-4(1+ϕ(u1)u1ϕ(u1))Θ1u1u-2(1+ϕ(u1)u1ϕ(u1))u14u-4+2[NN-1(ϕ(u1)u1ϕ(u1))2+2N-1ϕ(u1)u1ϕ(u1)-ϕ′′(u1)u12ϕ(u1)+NN-1](Θ1u22uu1)2+2u-2[-f(u)+N+1N-1(1+ϕ(u1)u1ϕ(u1))f(u)u]u12ϕ(u1).(4.18)

Condition (Cf), together with condition (ϕ5), implies that

-f(u)+N+1N-1(1+ϕ(u1)u1ϕ(u1))f(u)u-f(u)+σ(N+1N-1)f(u)u0.(4.19)

To continue it will be convenient to introduce the following notations:

A(t):=-|2-ϕ(t)tϕ(t)|,B(t):=2NN-1(ϕ(t)tϕ(t))2-2(N-3)N-1ϕ(t)tϕ(t)-2ϕ′′(t)t2ϕ(t)+2N-1,C:=2N-1,D(t):=2N-1(1+ϕ(t)tϕ(t))+ϕ(t)tϕ(t),E(t):=2NN-1(ϕ(t)tϕ(t))2-4(N-2)N-1ϕ(t)tϕ(t)-2ϕ′′(t)t2ϕ(t)-2(N-2)N-1,F(t):=NN-1(ϕ(t)tϕ(t))2+2N-1ϕ(t)tϕ(t)-ϕ′′(t)u12ϕ(t)+NN-1.

Using inequality (4.19) in the last inequality (4.18), dividing both sides of (4.18) by Θ, and rearranging we find

1Θ(ΔΘ+ϕ(u1)u1ϕ(u1)Θ11)-|2-ϕ(u1)u1ϕ(u1)||Θ|22Θ2+[2NN-1(ϕ(u1)u1ϕ(u1))2-2(N-3)N-1ϕ(u1)u1ϕ(u1)-2ϕ′′(u1)u12ϕ(u1)+2N-1]u12u2   +2N-1(f(u)u1ϕ(u1))2+[2N-1(1+ϕ(u1)u1ϕ(u1))+ϕ(u1)u1ϕ(u1)](Θ1f(u)Θu1ϕ(u1))   +[2NN-1(ϕ(u1)u1ϕ(u1))2-4(N-2)N-1ϕ(u1)u1ϕ(u1)-2ϕ′′(u1)u12ϕ(u1)-2(N-2)N-1]Θ1u1Θu   +2[NN-1(ϕ(u1)u1ϕ(u1))2+2N-1ϕ(u1)u1ϕ(u1)-ϕ′′(u1)u12ϕ(u1)+NN-1]Θ124Θ2=A(u1)|Θ|22Θ2+B(u1)u12u2+C(f(u)u1ϕ(u1))2+D(u1)(Θ1f(u)Θu1ϕ(u1))+E(u1)Θ1u1Θu+F(u1)Θ124Θ2A(u1)|Θ|22Θ2+B(u1)u12u2+C(f(u)u1ϕ(u1))2+D(u1)(Θ1f(u)Θu1ϕ(u1))+E(u1)Θ1u1Θu.(4.20)

Note that we have used condition (ϕ5) and FB>0 in the penultimate relation.1 Using the Cauchy–Schwarz inequality we estimate (4.20) as follows (we suppress the dependence of A,B,D,E and F on u1):

1Θ(ΔΘ+ϕ(u1)u1ϕ(u1)Θ11)A|Θ|22Θ2+Bu12u2+C(f(u)u1ϕ(u1))2-D22C(Θ122Θ2)-C(f(u)u1ϕ(u1))2-E2B(Θ122Θ2)-B2u12u2=A|Θ|22Θ2+B2u12u2-12(D2C+2E2B)Θ122Θ2.(4.21)

We now combine inequalities (4.7) and (4.21) to get

A|Θ|22Θ2+B2u12u2-12(D2C+2E2B)Θ122Θ224a2(a2-r2)2+4Na2(a2-r2)2+|ϕ(u1)u1ϕ(u1)|(24a2(a2-r2)2+4a2(a2-r2)2).

In other words, we have

0u12u22B[4a2(6+N)(a2-r2)2+|ϕ(u1)u1ϕ(u1)|28a2(a2-r2)2+|12(D2C+2E2B)-A||Θ|22Θ2].

From (4.2), we recall that at x*, we have

ΘΘ=r2(a2-r2)2,so that |Θ|22Θ2=8r2(a2-r2)2.

Therefore, at x*, we have the estimate

0|u|2u2C0(u1)a2(a2-r2)2,

where

C0(u1):=8B[6+N+7(ρ+1)+2|12(D2(u1)C+2E2(u1)B(u1))-A(u1)|].

Note that by conditions (ϕ3) and (ϕ5), C0(u1) is bounded by a positive . Moreover, we observe that

J(x*)=|u|2u2(a2-r2)2a2.

Since J(x0)J(x*), we conclude that, at x0, we have

|u|2u2a4=J(x0)J(x*)a2.

Thus, at x0, we have the estimate

|u|2u2a2.

Letting a, we find that |u|=0 at x0. Since x0 was arbitrary, we conclude that u0 on N, as desired. The proof is complete. ∎

4.1 An example

Let us illustrate the above theorem with a couple of examples. The simplest case occurs when ϕ(t)=ptp-2 for some p>1. In this case, σ=ρ=p-1, ϖ(t)=(p-2)(p-3) and

NN-1σ3-3ρ+2=(p-1)2N-1+(p-2)(p-3).

Therefore, we immediately see that condition (ϕ5) holds. This has been investigated in [7].

Now let us consider ϕ(t)=ptp-2+qtq-2 for 1<p<q. Recall that in this case σ=p-1 and ρ=q-1. Let us note that2

ϖ(t):=ϕ′′(t)t2ϕ(t)=p(p-2)(p-3)tp-1+q(q-2)(q-3)tq-1ptp-1+qtq-1max{(p-2)(p-3),(q-2)(q-3)}.

We now proceed to find conditions under which (ϕ5) holds. We have

NN-1σ2-3ρ+2-ϖNN-1(p-1)2-3(q-1)+2-max{(p-2)(p-3),(q-2)(q-3)}.(4.22)

Let us observe that

(q-2)(q-3)=(q-p+p-2)(q-p+p-3)=(q-p)2+(q-p)(2p-5)+(p-2)(p-3)=(q-p)(q+p-5)+(p-2)(p-3)(q-p)(2p-5)+(p-2)(p-3).

Therefore, we see that

max{(p-2)(p-3),(q-2)(q-3)}={(p-2)(p-3)if p<5/2,(q-2)(q-3)if p5/2.

So let us first suppose that 1<p<5/2. Then inequality (4.22) reduces to

NN-1σ2-3ρ+2-ϖNN-1(p-1)2-3(q-1)+2-(p-2)(p-3).

Thus, if

p<q<13[NN-1(p-1)2-(p-2)(p-3)+5]=13[NN-1(p-1)2-(p-1)2+3p]=13[(p-1)2N-1+3p],

then condition (ϕ5) holds. Now suppose p5/2. Then inequality (4.22) becomes

NN-1σ2-3ρ+2-ϖNN-1(p-1)2-3(q-1)+2-(q-2)(q-3)=NN-1(p-1)2-q2+2q-1=NN-1(p-1)2-(q-1)2.

Therefore, if

p<q<(p-1)NN-1+1,

then condition (ϕ5) holds.

Now, given p>1, let us set

p*:={13[(p-1)2N-1+3p]if p<52,(p-1)NN-1+1if p52.

We remark that p<p* for p>1. We now summarize the above discussion in the following corollary.

Corollary 4.4.

Given p>1, suppose that f satisfies condition (Cf) with σ=p-1. If 1<p<q<p*, then any non-negative entire solution of

Δpu+Δqu=-f(u)

is a constant on RN.

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Footnotes

  • 1

    Let ω:=ϕ(t)tϕ(t) and ϖ:=ϕ′′(t)t2ϕ(t). Then we see that B=2NN-1ω2-2(N-3)N-1ω-2ϖ+2N-1=2[NN-1(1+ω)2-3(1+ω)+2-ϖ] and F=2NN-1ω2+4N-1ω-2ϖ+2NN-1=2[NN-1(1+ω)2-2(1+ω)+2-ϖ]. 

  • 2

    Let a,b. Since min{a,b}a,bmax{a,b}, we have min{a,b}θa+(1-θ)bmax{a,b} for all 0θ1. 

About the article

Received: 2017-07-04

Accepted: 2017-07-12

Published Online: 2017-08-24


This work was supported by ISP (International Science Program) of Uppsala University, Sweden.


Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 725–742, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2017-0158.

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© 2019 Walter de Gruyter GmbH, Berlin/Boston. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0

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