Entire solutions of certain fourth order elliptic problems and related inequalities

Lorenzo D’Ambrosio; Enzo Mitidieri

doi:10.1515/anona-2021-0217

Open Access Published by De Gruyter February 14, 2022

Entire solutions of certain fourth order elliptic problems and related inequalities

Lorenzo D’Ambrosio and Enzo Mitidieri

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2021-0217

Abstract

We study distributional solutions of semilinear biharmonic equations of the type

Δ2u+f(u)=0 onℝN,

where f is a continuous function satisfying f (t)t ≥ c |t|^q+1 for all t ∈ ℝ with c > 0 and q > 1. By using a new approach mainly based on careful choice of suitable weighted test functions and a new version of Hardy- Rellich inequalities, we prove several Liouville theorems independently of the dimension N and on the sign of the solutions.

Keywords: Liouville theorems; biharmonic operator; Hardy–Rellich inequalities

MSC 2010: 35G20; 31B30; 35B53; 26D10

Dedicated to the memory of our dear friend Stanislav Pohozaev

1 Introduction

In a well known work Brezis [5] proved the following result.

Theorem 1.1

Let q > 1. If u∈Llocq(ℝN) is a distributional solution of

(1.1) Δu≥|u|q−1u onℝN,

then u ≤ 0 a.e. in ℝ^N.

In particular this implies.

Theorem 1.2

Let q > 1. If u∈Llocq(ℝN) is a distributional solution of

(1.2) Δu=|u|q−1u onℝN,

then u ≡ 0 a.e. in ℝ^N.

The interesting point here, besides the quite general functional framework, is that no assumptions on the behavior nor on the sign of the possible solutions of (1.2) are made. Brezis's technique is based on the following form of Kato inequality (see [5, Lemma A.1]),

u,f∈Lloc1(Ω) such that Δu≥f, then Δu+≥sign+(u)f,

and on a construction of a suitable barrier function. These tools are typically second order in nature, so, in general it is hopeless to use them when dealing with problems of order higher than two. Comprehensive results in the Brezis’ spirit for quasilinear elliptic inequalities of second order on ℝ^N have been obtained in a series of papers by Farina and Serrin [15, 16] and the Authors [10,11,12]. In addition these results are also studied in the subelliptic framework [12, 13] and in the Riemannian setting [4].

These results suggest a general natural problem for higher order elliptic equations and inequalities.

General problem: What are the necessary conditions that guarantee the existence of non trivial solutions for higher order nonlinear elliptic coercive^[1] problems on ℝ^N?

The results for the second order case cited above, altogether are proved in the spirit of [27]. For higher order problem the beneficial of a systematic approach for studying coercive elliptic problems is still missing. The aim of this paper is to give a contribution to develop a possible unitary method for higher order elliptic equation and inequalities of coercive type and it represents a first step in this direction.

In concrete situations for fourth order semilinear elliptic equations with simple power nonlinearities, the problem is connected to find a so called critical exponent. Here, by critical exponent we mean the existence of q^*(N) > 1, depending on the dimension N such that there are no non trivial solutions for q < q^*(N) and there exist non trivial solutions for q > q^*(N). Then from Theorem 1.2 we can say that for equation (1.2) the critical exponent is q^*(N) = ∞.

Let us consider the fourth order analogue of (1.2), that is

(1.3) −Δ2u=|u|q−1u onℝN, q>1.

It is well known that these kind of problems have strong connections with differential geometry [7, 8], higher order Schrödinger equations [20, 24, 31] and models for suspension bridges [17]. In this regard see [19] for further results on related applications of polyharmonic elliptic equations.

Looking at solutions of (1.3) in the natural global space H²(ℝ^N), it is clear that for q≤N+4N−4 the only solution u is given by u ≡ 0 a.e. in ℝ^N . Of course, the interesting problem is when u does not belong to the global space H²(ℝ^N), so there is no a priori knowledge of the behavior at infinity of the solutions.

Notice that if in (1.3) we have q = 1, it is easy to see that the equation admits nontrivial solutions in dimension N = 1, and hence in any dimension.

It is well known that the literature on nonlinear higher order coercive equations is far from complete. However there exist notable results due to Bernis [2] that for some particular biharmonic problem reads as follows.

Theorem 1.3

Let u∈Hloc2(ℝN)∩Llocq+1(ℝN) be a solution of (1.3). If q(N − 4) ≤ N + 4, then u ≡ 0 a.e. in ℝ^N.

A first immediate observation is that from [2] it appears that for (1.3) the Sobolev exponent N+4N−4 is critical (in our sense) when N > 4. However, and this is the main motivation to write this paper, this is not true.

Even if our main interest is in possible changing sign solution of (1.3), we present here a simple result concerning solutions of (1.3) which do not change sign. Its proof relies on our integral representation results obtained in [6] and it serves as a motivation to focus our attention on the possible sign changing solutions.

Theorem 1.4

Let u∈Llocq(ℝN) be a distributional solution of (1.3). If u does not change sign, then u ≡ 0 a.e. in ℝ^N.

More generally we have the following unexpected result for the equation (1.3).

Theorem 1.5

Let u∈Llocq+1(ℝN) be a distributional solution of (1.3). If

(1.4) N=1,…,7, and q>1,orN≥8 and 1<q≤qN,

where

(1.5) qN:=N2+2N−28+44−2 N+2 N2N2−10N+20,

then u ≡ 0 a.e. in ℝ^N.

The above result shows that within the class of distributional solutions, if N ≤ 7 then there is no critical exponent (i.e. q^*(N) = ∞), while if N > 7 we qN>N+4N−4 . We believe that the value of q_N is not sharp and it can be improved. Indeed we can state the following conjecture.

Conjecture

For any N ≥ 1 and q > 1 the only solution of (1.3) is given by u ≡ 0 a.e. in ℝ^N.

The methods used in this paper apply to more general problems than (1.3). More precisely, some of our results are still valid for distributional solutions of the double inequality

g(u)≥−Δ2u≥f(u), onℝN, (gPf).

where f , g ∈ 𝒞(ℝ). Throughout this paper we shall denote by H the following function

(1.6) H(t):={f(t)t,for t≥0,g(t)t,for t<0.

In what follows we shall deal with the autonomous case. The non-autonomous one, that is f = f (x, u) and g = g(x, u) can be studied in similar way. However, for sake of simplicity we limit ourselves to the autonomous case.

We have the following result.

Theorem 1.6

Let f, g ∈ 𝒞(ℝ) and let H be defined by (1.6). Assume that

(1.7) H(t)≥cHmin{|t|q+1,|t|p+1}, ∀ t∈ℝ , for some q≥p>1

with c_H > 0. Let u be a distributional solution of (_gP_f ) such that u∈Llocs(ℝN) , 2 ≤ s ≤ +∞ and f(u),g(u)∈Llocs′(ℝN) .

If N = 1, . . . , 7 or N ≥ 8 and 1 < q ≤ q_N with q_N defined in (1.5), then u ≡ 0 a.e. in ℝ^N.

Notice that (1.7) is an assumption on the behavior of f and g for nonnegative and nonpositive values of the independent variable, respectively. No assumptions are required on f and g for negative and nonpositive values of the independent variable.

Hypothesis (1.7) allows us to handle nonlinearities that behave differently for positive and negative values of the independent variable or behave differently at the origin and at infinity.

Theorem 1.6 contains several Liouville results for the equation

−Δ2u=f(u) onℝN.

For instance if f (t) = |t|^s−1t + |t|^p−1t for any p > 1 ≥ s > 0, or f (t) = te^|t|, or f (t) = sinh(t), then the above problem has only the trivial solution. Notice that as byproduct, we can deduce that the defocusing Schrödinger equation

ivt+Δ2v+|v|q−1u=0,

has no nontrivial standing wave solutions of the form v(t, x) = e^−iω²tu(x) with ω ∈ ℝ. See [31] and reference therein for further results on this equation, and its connection with several models from physics.

A common feature of the above results is that we do not require any assumption on the behavior of the solutions at infinity. We also point out that for higher order coercive problems Kato's inequality does not hold and in general is not possible to use comparison principles. Thus the main idea to study problems like (1.3) is first to obtain suitable a priori bounds on the various quantities involved in the analysis. These estimates yield a Liouville result for q running in the range of values of Bernis’ result i.e. 1<q≤N+4N−4. .

To improve this result, i.e. going above the Sobolev exponent for equation (1.3), we develop some machinery by demonstrating functional inequalities related to some quadratic forms. By using the positivity of a particular quadratic form along the solutions of our problems, and taking into account of suitable a priori estimates, we are able to achieve our goal. The involved argument is quite intricate and this is the reason why we begin illustrating the method for the second order prototype equation (1.2) in Section 2.

In Section 3 we study inequalities related to (1.3), obtaining some information on the sign of the possible solution of (1.3). For analogous results see also Section 7.2.

In Section 4 we discuss the different notions of solutions and justify why the study of solutions of (_gP_f) is reduced to the study of

−u Δ2u≥h, onℝN, (Ph)

where h∈Lloc1(ℝN) .

In Section 5 we develop a number of functional inequalities and positive quadratic forms.

Section 6 is devoted to prove some a priori estimates on the solutions of our problems, which combined with the results of Section 5 yields the Liouville theorems.

Section 7 contains some applications of the results obtained in the preceding sections. In Section 7.1 we prove a special representation formula of u², being u a possible solutions of (P_h). Section 7.2 deals with some results on the sign of possible solutions of the problem under consideration and their Laplacian. In Section 7.3 we apply our Liouville theorems to the uniqueness problem.

Appendix A recalls a known result on the integral representation of the solutions of some higher order elliptic equation.

Notation

B_R will denote the Euclidean ball of radius R centered at the origin B_R := {|x| ≤ R}. By ω_N we denote the measure of the unit Euclidean ball, ω_N = |B₁|.

Throughout this paper ϕ₁ : ℝ → [0, +∞[ stands for a standard cut off function, that is ϕ₁ is a smooth function on ℝ such that ϕ₁(t) = 1 for |t| ≤ 1, ϕ₁(t) = 0 fot |t| ≥ 2 and 0 ≤ ϕ₁(t) ≤ 1. We set

(1.8) ϕR(x):=ϕ1(|x|/R),

the support of ϕ_R is contained in B_2R = {|x| ≤ 2R}, while the support of any derivative of ϕ_R is contained in

AR:={R≤|x|≤2R}.

Furthermore without loss of generality we shall assume that ϕ₁ is an admissible test function, that is for a fixed p > 1 there exists c₁ > 0 such that

|||∇ϕ1|pϕ1p−1||∞<c1, |||Δϕ1|pϕ1p−1||∞<c1, |||∇Δϕ1|pϕ1p−1||∞<c1.

Indeed if ϕ₁ is not admissible, then it follows that for large γ, ϕ1γ is admissible.

Finally, in what follows c stands for a positive constant which can vary from line to line and it is independent from the solution u and R. Writing c₁ we always mean a positive constant depending only on the test function ϕ₁, that is c₁ = c₁(ϕ₁).

For N > 4, by C_N we denote the normalization positive constant in the relation

(1.9) Δ2|x|4−N=CNδ0,

that is CN:=14Γ(N−42)π−N/2 .

Finally, in what follows an integral without the indication of the domain of integration, is understood on the whole space ℝ^N.

2 A detour on the second order case: the quadratic form approach

The purpose of this section is to illustrate a specific method for handling nonexistence theorems for a class of second order coercive equations on ℝ^N . For simplicity we restrict our attention to smooth solutions and to the simple prototype equation

Δu=|u|q−1u, onℝN.

The general scheme of our method develops in several steps.

Step 1: Functional identity

Lemma 2.1

Let u ∈ 𝒞¹(ℝ^N), φ∈𝒞 0 2(ℝN) and v ∈ 𝒞²(ℝ^N). We have

(2.1) ∫(∇u⋅∇(uvφ))=∫|∇u|2vφ−12∫u2Δv φ+E(u2,v,φ),

where

E(u2,v,φ):=−∫u2(∇v⋅∇φ)−12∫u2vΔφ.

In particular if φ = ϕ_R, with ϕ_R defined in (1.8), and for k = 0, 1, |v^(k)| ≤ c_vR^α−N−k holds on A_R, then

(2.2) |E(u2,v,φ)|≤cvc1(ϕ1)Rα−2⨍AR|u|2.

Proof

By computation, we get

∫∇u⋅∇(uvφ)=∫|∇u|2vφ+∫(∇u⋅∇v) uφ+∫(∇u⋅∇φ) uv.

Next an integrating by parts gives,

∫(∇u⋅∇v) uφ=−∫(∇u⋅∇v) uφ−∫u2Δv φ−∫u2(∇v⋅∇φ),

that is

∫(∇u⋅∇v) uφ=−12∫u2Δv φ−12∫u2(∇v⋅∇φ).

Analogously we obtain

∫(∇u⋅∇φ) uv=−12∫u2Δφ v−12∫u2(∇v⋅∇φ).

Gluing together these identities, we deduce (2.1).

In order to prove (2.2), we first observe that the domain of integration of the functional E is given by supp(∇ϕ_R) = A_R. Next by using the hypotheses on v, we get

| E ( u 2 , v , φ ) | ≤ ∫ A R u 2 | ∇ v | | φ | + 1 2 ∫ A R u 2 | v | | Δ φ | ≤ c v R α − N − 1 c 1 R − 1 ∫ A R u 2 + 1 2 c v R α − N c 1 R − 2 ∫ A R u 2 .

We notice that as direct consequence of Lemma 2.1 we deduce that if v ∈ 𝒞²(ℝ^N) is nonnegative and superharmonic, then for any nonnegative φ∈𝒞 0 2(ℝN) , the quadratic form,

u∈𝒞 1(ℝN)↦∫(∇u⋅∇(uvφ))−E(u2,v,φ),

is positive.

Step 2: A functional inequality

In what follows we set r := |x| and for ɛ > 0,

rϵ:=(ε2+|x|2)1/2=(ε2+r2)1/2,

and for x ∈ ℝ^N, ɛ > 0 and α ∈ ℝ, we define

vε(x)=vε(|x|)=vε(r):=1(ε2+r2)N−α2=rεα−N.

We have that −Δv_ɛ ≥ 0 for N > α ≥ 2. Hence, by choosing v = v_ɛ in Lemma 2.1, we obtain.

Lemma 2.2

Let N > α ≥ 2. Let u ∈ 𝒞¹(ℝ^N), and let φ∈𝒞 0 2(ℝN) be nonnegative. We have

(2.3) ∫(∇u⋅∇(uφrεN−α))≥∫|∇u|2φrεN−α+E(u2,rεα−N,φ),

and for φ = ϕ_R, the estimate (2.2) holds.

Remark 2.3

If u∈𝒞 0 1(ℝN) , then taking φ = ϕ_R with R large enough, it follows that E(u², v, φ) = 0. In addition for any N ≥ α ≥ 2 the quadratic form

u∈𝒞 0 1(ℝN)↦∫(∇u⋅∇urεN−α),

is positive.

Step 3: A priori estimate on the solutions

Lemma 2.4

Let q > 1 and let u ∈ 𝒞²(ℝ^N) be a solution of

Δu=|u|q−1u, onℝN.

Then

(2.4) (⨍BRu2)1/2≤(⨍BR|u|q+1)1/q+1≤cR−2q−1.

Proof

Multiplying the equation by uϕ_R and using Young inequality we obtain,

∫|u|q+1ϕR=−∫(∇u⋅∇(uϕR))=−∫|∇u|2ϕR−∫u(∇u⋅∇ϕR)≤−∫|∇u|2ϕR+12∫|∇u|2ϕR+12∫u2|∇ϕR|2ϕR.

By Hölder inequality with exponent x := (q + 1)/2, it follows that

∫|u|q+1ϕR≤12∫u2|∇ϕR|2ϕR≤12(∫|u|q+1ϕR)1/x(∫|∇ϕR|2x′ϕRx′−1)1/x′,

which in turn implies

∫ B R | u | q + 1 ≤ ∫ | u | q + 1 ϕ R ≤ 2 − x ′ ∫ | ∇ ϕ R | 2 x ′ ϕ R x ′ − 1 ≤ c 1 R N − 2 x ′ .

A simple application of Jensen inequality gives (2.4).

Step 4: A Liouville theorem

Theorem 2.5

Let q > 1 and let u ∈ 𝒞²(ℝ^N) be a solution of

Δu=|u|q−1u, onℝN.

Then u ≡ 0 a.e. in ℝ^N.

Proof

Let N ≥ α ≥ 2 and φ = ϕ_R. Multiplying the equation by uφrεN−α , from (2.3) we obtain

(2.5) 0≥−∫|u|q+1φrεN−α=∫(∇u⋅∇(uφrεN−α))≥∫|∇u|2φrεN−α+E(u2,rεα−N,φ).

Choosing α = 2, from (2.4) and (2.2) it follows that |E(u2,rε2−N,φ)|→0 as R → ∞. Therefore by letting R → +∞ in (2.5), we obtain

0≥−∫|u|q+11rεN−2=∫(∇u⋅∇(u1rεN−2))≥∫|∇u|21rεN−2≥0.

This last inequality implies the claim.

3 Some simple results on biharmonic problems

Theorem 3.1

Let q > 1 and let u∈Llocq(ℝN) be a distributional solution of

(3.1) −Δ2u≥|u|q onℝN.

1. If q(N − 4) ≤ N, then u ≡ 0 a.e. in ℝ^N.
2. If u is nontrivial, then u < 0 a.e. in ℝ^N and Δu ≥ 0 in distributional sense.

Theorem 3.2

Let q > 1 and let u∈Llocq(ℝN) be a distributional solution of

(3.2) −Δ2u=|u|q onℝN.

1. If q(N − 4) ≤ N, then u ≡ 0 a.e. in ℝ^N.
2. If u u∈LlocN(q−1)/4(ℝN) and 1<q<N+4N−4 , then u ≡ 0 a.e. in ℝ^N.
3. If u is nontrivial, then u < 0 a.e. and Δu ≥ 0 in distributional sense.
4. If q>NN−4 , then (3.2) has nontrivial solutions (which is negative and subharmonic) in ℝ^N.
5. If q≥N+4N−4 , then (3.2) has nontrivial smooth solutions (which is negative and subharmonic).

Notice that the above theorems imply that problems (3.1) and (3.2) do not admit nontrivial nonnegative solutions.

From Theorem 3.1 and Theorem 1.1 we can conjecture that if u is a solution of

(3.3) −Δ2u≥|u|q−1u onℝN

then u ≤ 0 a.e. in ℝ^N . However this conjecture is false as the following simple example shows. Let u:=1−x14/24 . The function u changes sign, it is superharmonic and −Δ²u = 1. Let q > 1 and considering the function f (t) = |t|^q−1 t. Since 1 ≥ u and f is increasing, we get that −Δ²u = 1 ≥ |u|^q−1 u. This example shows that the conjecture is false even if we assume a sign on the Laplacian of the solution. Moreover, the above example also shows that a Kato inequality of the type

(3.4) u,f∈Lloc1(Ω) such that −Δ2u≥f, then −Δ2u+≥sign+(u)f,

in distributional sense, cannot hold. Indeed if (3.4) holds, then u⁺ solves (3.1), and by Theorem 3.1 we obtain u⁺ ≡ 0, and this contradicts our counterexample.

Further remarks on the sign of the solutions of biharmonic inequalities and on the sign of their Laplacian will be considered in Section 7.2.

Proof of Theorem 3.1

Let u be a distributional solution of (3.1). Multiplying by a test function ϕ_R as in (1.8), we have

(3.5) ∫|u|qϕR≤−∫uΔ2ϕR≤∫|u||Δ2ϕR|≤∫AR|u|ϕR1/q|Δ2ϕR|ϕR1/q ≤(∫AR|u|qϕR)1q(∫AR|Δ2ϕR|q′ϕRq′−1)1q′≤c1(∫AR|u|qϕR)1qR−4+Nq′.

Therefore, we have

(3.6) ∫BR|u|q≤c1R−4q′+N,

which in turn implies that

(3.7) ⨍BR|u|≤(⨍BR|u|q)1q≤cR−4q−1.

Proof of 1. If N ≤ 4, from (3.6), by letting R → +∞ it follows that u ≡ 0 a.e. in ℝ^N . Let N > 4 and q ≤ N/(N − 4). From (3.6) by letting R → +∞, we deduce ∫_ℝ^N |u|^q ≤ c₁ < ∞. This implies that

limR∫AR|u|q=0,

which plugged into (3.5) yields |u|^q ≡ 0 a.e. in ℝ^N.

Proof of 2. Let N > 4 and let u be a nontrivial solution of (3.1). By translation invariance from (3.7) we have,

(3.8) lim infR→+∞1RN∫R≤|x−y|≤2R|u(y)|dy=0 for any x∈ℝN.

Hence, by Theorem A.1 applied to v := −u, it follows that

(3.9) −u(x)≥1CN∫|u(y)|q|x−y|N−4dy, for any x∈ℝN,

where C_N is defined by (1.9). Clearly this implies that −u is superharmonic in distributional sense in ℝ^N.

Proof of Theorem 3.2

Since (3.2) is a particular case of (3.1), statements 1. and 3. are a direct consequence of Theorem 3.1.

Arguing as in the proof of Theorem 3.1, it follows that the solutions of (3.2) can be represented by (3.9) with the equality sign. Therefore the function v := −u is a nonnegative superharmonic solution of

(3.10) Δ2v=vq onℝN,

and of the integral equation

(3.11) v(x)=1CN∫v(y)q|x−y|N−4dy, for any x∈ℝN.

Proof of 2. Since v∈LlocN(q−1)/4(ℝN) and 1<q<N+4N−4 , from Theorem 1.4 in [21], it follows that v ≡ 0 a.e. in ℝ^N.

Proof of 4. Equation (3.10) admits singular solutions of the form v(x) = c |x|^−4/(q−1) for a suitable c > 0.

Proof of 5. From [32] it follows that there exist infinitely many nontrivial radial positive smooth solutions of (3.10) (see also [18]), which yields our claim.

Similar results of those of Theorems 3.1 and Theorem 3.2 (1., · · · , 4.) with the same proofs, can be proved for higher order problems of the type

(3.12) −(−Δ)mu≥|u|q onℝN,

and

(3.13) −(−Δ)mu=|u|q onℝN.

Let us to emphasize that for problems (3.12) and (3.13), the corresponding point 2. of Theorem 3.1 (and a fortiori, point 3. of Theorem 3.2) can be written as

(−Δ)iu≤0 in distributional sense, for i=0,…,m.

The existence result for (3.13) for m ≥ 3, like in 5. of Theorem 3.2, that is for q>N+2mN−2m is an open problem.

4 On the notion of solution: other related problems

In this paper we are mainly interested to the study possible solutions of the prototype equation

−Δ2u=|u|q−1u, onℝN

which is, clearly a special case of the double inequality

g(u)≥−Δ2u≥f(u), onℝN (gPf)

where f , g : ℝ → ℝ are given functions satisfying suitable assumptions. We emphasize that the methods that we are going to develop can be fruitfully used to study the solution of the one side inequality

−u Δ2u≥h onℝN. (Ph)

We begin noticing that if u ∈ 𝒞⁴(ℝ^N) is a solution of (_gP_f ), then u solves

(4.1) g(u)u+−f(u)u−≥−u Δ2u≥f(u)u+−g(u)u− onℝN,

where u⁺ and u⁻ are the positive and negative part of u respectively. Indeed multiplying (_gP_f ) by u⁺ and −u⁻, we have

g(u)u+≥ −Δ2u u+ ≥f(u)u+,−f(u)u−≥−Δ2u (−u−)≥−g(u)u−.

Summing these last two inequalities we obtain (4.1). Therefore, in what follow we shall study also possible solutions of the inequality (P_h).

Having in mind that in our main Liouville Theorems we are going to assume that

f,g∈𝒞 (ℝ ), f(t)t≥0, g(t)t≥0, for any t∈ℝ , (f0)

we see that h = H(u) = f (u)u⁺ − g(u)u⁻ ≥ 0. However, it will be useful to study (P_h) without any assumption on the sign of h∈Lloc1(ℝN) . This extra generality, beside the fact that is interesting in itself, it will be essential when studying the distributional solutions of (_gP_f ).

Definition 4.1

A function u∈Lloc1(ℝN) is a distributional solution of (_gP_f ), if f(u),g(u)∈Lloc1(ℝN) and

∫g(u)φ≥−∫uΔ2φ≥∫f(u)φ,

for any nonnegative φ∈𝒞 0 4(ℝN) .

A function u∈Hloc2(ℝN) is a weak solution of (P_h), if h∈Lloc1(ℝN) and

(4.2) ∫−ΔuΔ(uφ)≥∫hφ,

for any nonnegative φ∈𝒞 0 2(ℝN) .

Analogously, u∈Hloc2(ℝN) is a weak solution of (4.1), if f(u)u,g(u)u∈Lloc1(ℝN) and

∫(g(u)u+−f(u)u−)φ≥−∫ΔuΔ(uφ)≥∫(f(u)u+−g(u)u−)φ,

for any nonnegative φ∈𝒞 0 2(ℝN) .

Theorem 4.2

Let f,g∈ C ( R N) and let u∈Hloc2(ℝN) be a distributional solution of (_gP_f ) such that u∈Llocs(ℝN) , 2 ≤ s ≤ ∞ and f(u),g(u)∈Llocs′(ℝN) . Then u is a weak solution of (4.1).

Proof

Let (m_η)_η>0 be a family of standard mollifier cutoff functions. Let u⁺ and u⁻ be the positive and negative part of u respectively. Then setting

uη:=u⋆mη, uη+:=(u+)η:=u+⋆mη, uη−:=(u−)η:=u−⋆mη,

we have that u_η → u in Hloc2(ℝN) , u_η → u, uη+→u+ , uη−→u− in Llocs(ℝN) and a.e. in ℝ^N.

Now using φuη+ and φuη− as test functions in (_gP_f ) we get

∫g(u)uη+φ≥−∫ΔuΔ(uη+φ)≥∫f(u)uη+φ,∫g(u)uη−φ≥−∫ΔuΔ(uη−φ)≥∫f(u)uη−φ,

hence

∫(g(u)uη+−f(u)uη−)φ≥−∫ΔuΔ((uη+−uη−)φ)≥∫(f(u)uη+−g(u)uη−)φ.

Since uη=uη+−uη−→u in Hloc2(ℝN) , we have ∫ΔuΔ((uη+−uη−)ϕ)→∫ΔuΔ(uϕ) . On the other hand, we know that ∫f(u)uη+φ→∫f(u)u+φ , ∫f(u)uη−φ→∫f(u)u−φ , ∫g(u)uη+φ→∫g(u)u+φ and ∫g(u)uη−φ→∫g(u)u−φ , yielding

∫(g(u)u+−f(u)u−)φ≥−∫ΔuΔ(uφ)≥∫(f(u)u+−g(u)u−)φ,

which is the claim.

Remark 4.3

From the above theorem we easily deduce that if u∈Hloc2(ℝN) is a weak solution of (_gP_f) with f, g ∈ 𝒞(ℝ^N), then u is a weak solution of (4.1) provided one of the following conditions is satisfied

1. u∈Hloc2(ℝN)∩Llocq+1(ℝN) and C₁t^q ≥ g(t) ≥ f (t) ≥ 0 for t ≥ 0, 0 ≥ g(t) ≥ f (t) ≥ −C₂|t|^q for t < 0, with suitable C₁, C₂ > 0;
2. g(u) = f (u) = |u|^q−1 u, u∈Hloc2(ℝN)∩Llocq+1(ℝN) ;
3. f(u)∈Llocp1(ℝN) , g(u)∈Llocp2(ℝN) for some p1,p2≥2NN+4 ;
4. u∈Hloc2(ℝN)∩Lloc∞(ℝN) .

Remark 4.4

When studying distributional solutions u∈Lloc1(ℝN) of (_gP_f ), we encounter several difficulties that can be overcome by analyzing the general problem (P_h) without the extra assumption on the sign of h.

5 Asymptotic Hardy-Rellich type inequalities

In order to develop the scheme described in Section 2 to our fourth order problem, we need to prove the counterpart of inequality (2.3). To this end an important step is to obtain some inequalities that we name Asymptotic Hardy-Rellich type inequalities. Let us point out why we call these inequalities asymptotic. It well known that for u∈𝒞 0 ∞(ℝN) , N > 4, the following inequalities

∫|Δu|2≥N2(N−4)216∫|u|21|x|4, and ∫|Δu|2≥N24∫|∇u|21|x|2,

holds (see for instance [30], [33]). Usually, in the literature the above inequalities hold for compactly supported functions u and are known as Rellich type inequalities. If u has not compact support and does not belong to some appropriate function space, say D, the above inequalities are not necessarily valid. However a version of these inequalities are satisfied by localization and by adding an error term, say E₁(u). The latter may vanishes under suitable conditions.

The precise relation between the vanishing property of our error E₁(u) and the fact that the function u belongs to a suitable space D is an interesting problem however we will not investigate this question in this paper.

Now we present some general inequalities that can be useful for further investigation. To the best of our knowledge the results in this section are new.

From the definition of weak solution (4.2) it is clear that we need to develop some estimates for integrals of the type

∫ΔuΔ(uφ).

To this end we observe that from Corollary 2.1 in [25] we deduce the following.

Lemma 5.1

Let Ω ⊂ ℝ^N be an open set, u ∈ 𝒞²(Ω), H = (H¹, . . . , H^N) ∈ 𝒞¹(Ω; R^N) with u or H having compact support in . Then

2∫ΩΔu(H⋅∇u)=∫Ωdiv(H)|∇u|2−2∑i,j=1N∫Ω∂Hj∂xi∂u∂xi∂u∂xj.

Lemma 5.2

Let Ω ⊂ ℝ^N be an open set, u, v ∈ 𝒞²(Ω), φ ∈ 𝒞¹(Ω) with v and φ radial functions with at least one of them having compact support contained in Ω. Then

(5.1) 2∫ΩΔu(∇u⋅∇v)φ=∫Ω|∇u|2Δvφ+∫Ω|∇u|2v′φ′ −2∫Ω{(∇u⋅x|x|)2v′φ′+(∇u⋅x|x|)2v″φ−(∇u⋅x|x|)2v′|x|φ+|∇u|2v′|x|φ}.

Proof

Choose H := φ∇v in Lemma 5.1.

Our first main result is the following.

Theorem 5.3

Let Ω ⊂ ℝ^N be an open set. Let v ∈ 𝒞²(Ω) be a radial function such that v′2|v|∈L1(Ω) . For any u ∈ 𝒞²(Ω), for any nonnegative radial φ∈𝒞 0 1(Ω) , and δ > 0, we have

(5.2) ∫Ω|Δu|2|v|φ≥∫Ω(∇u⋅x|x|)22δ2[v″−v′|x|−δ22v′2|v|]φ +2δ2∫Ω|∇u|2[−Δv2+v′|x|]φ+2δ2E1(∇u,v′,φ′),

where

(5.3) E1(∇u,v′,φ′):=∫Ω(∇u⋅x|x|)2v′φ′−12∫Ω|∇u|2v′φ′.

In particular if φ=ϕR2 , with ϕ_R defined in (1.8), and the estimate |v′| ≤ c_vR^α−N−1 holds on A_R, then

(5.4) |E1(∇u,v′,φ′)|≤cvc1(ϕ1)Rα−2⨍AR|∇u|2ϕR.

Moreover, if in addition −Δv + 2v^′/|x| ≥ 0, then

(5.5) ∫Ω|Δu|2|v|φ≥∫Ω(∇u⋅x|x|)2δ2[v″−(N−1)v′|x|−δ2v′2|v|]φ +2δ2E1(∇u,v′,φ′).

Proof

By Cauchy-Schwarz inequality we have

∫ΩΔu(∇u⋅∇v)φ≤(∫Ω|Δu|2vφ)12(∫Ω(∇u⋅x|x|)2v′2vφ)12≤12δ2∫Ω|Δu|2vφ+δ22∫Ω(∇u⋅x|x|)2v′2vφ,

which plugged into (5.1) yields the inequality (5.2).

From the estimate

|E1(∇u,v′,φ′)|≤32∫Ω|∇u|2|v′||φ′|=3∫AR|∇u|2ϕR|v′||ϕR′|

we deduce (5.4).

Finally, since −Δv+2v′|x|≥0 , the inequality

|∇u|2≥(∇u⋅x|x|)2,

combined with (5.2) yields inequality (5.5).

For compactly supported function we have the following

Corollary 5.4

Let v ∈ 𝒞²(ℝ^N) be a radial function such that v′2|v|∈Lloc1(ℝN) . For any u∈𝒞 0 2(ℝN) , and δ > 0, we have

(5.6) ∫ℝN|Δu|2|v|≥∫ℝN(∇u⋅x|x|)22δ2[v″−v′|x|−δ22v′2|v|] +2δ2∫ℝN|∇u|2[−Δv2+v′|x|].

Moreover, if in addition −Δv + 2v′/|x| ≥ 0, then

(5.7) ∫ℝN|Δu|2|v|≥∫ℝN(∇u⋅x|x|)2δ2[v″−(N−1)v′|x|−δ2v′2|v|].

Proof

Let u∈𝒞 0 2(ℝN) . We choose φ = ϕ_R with R large enough such that the support of u is contained in the ball of radius R. With this choice of ϕ, the term E₁ which appears in (5.2) and (5.5) vanishes. Taking into account that φ = 1 on the support of u, the claim follows.

Remark 5.5

Since in Theorem 5.3 and Corollary 5.4 there is no assumption on the sign of v, we notice that if

−Δv+2v′|x|≤0,

then (5.5) and (5.7) still hold replacing v by −v (as well as its derivatives).

In what follows we deal with a particular weight v_ɛ that we are going to define below. From now on we set

(5.8) r:=|x|, and for ε>0, rε:=(ε2+|x|2)1/2=(ε2+r2)1/2,

and define

(5.9) vε(x)=vε(|x|)=vε(r):=1(ε2+r2)N−α2=rεα−N,

where α ∈ ℝ. A simple computation gives

(5.10) −Δvε=−Δrεα−N=(N−α)(α−2)r2+Nε2(ε2+r2)N−α+42,Δ2vε=Δ2rεα−N=(N−α)(N+2−α)× ×(α−2)(α−4)r4+2(α−4)(N+2)ε2r2+N(N+2)ε4(ε2+r2)N−α+82,|∇vε|=|N−α|r(ε2+r2)N−α+22.

Therefore, by choosing N > α ≥ 4, it follows that v_ɛ is a positive super-biharmonic function namely, −Δv_ɛ > 0, Δ²v_ɛ > 0. Furthermore on A_R = B_2R \ B_R the following estimates hold

vε(x)≤|x|α−N, |∇vε(x)|≤|N−α||x|α−1−N,|Δvε(x)|≤c|x|α−2−N, |∇Δvε(x)|≤c|x|α−3−N,

where c = c(N, α) > 0 is a suitable positive constant independent of ɛ.

As particular case of Theorem 5.3 we have the following.

Theorem 5.6

Let N ≥ 1 and α ∈ ℝ. For any u∈Hloc2(ℝN) , t ∈ ℝ, ɛ > 0, and a radial nonnegative function φ∈𝒞 0 1(ℝN) , we have

(5.11) ∫ℝN|Δu|2φrεN−α≥t(2N−2α+4−t)∫ℝN(∇u⋅x|x|)2φr2rεN−α+4 + t(α−4)∫ℝN|∇u|2φr2rεN−α+4+ t(N−2)ε2∫ℝN|∇u|2φ1rεN−α+4− 2tE1(∇u,rrεN−α+2,φ′).

In particular if 2N > α ≥ 4 or N = 1 and 4 > α > 2, we have

(5.12) ∫ℝN|Δu|2φrεN−α≥(2N−α2)2∫ℝN(∇u⋅x|x|)2φr2rεN−α+4 −(2N−α)E1(∇u,rrεN−α+2,φ′),

(5.13) ∫ℝN|Δu|2φrεN−α≥Hα∫ℝN|∇u|2φr2rεN−α+4−2tαE1(∇u,rrεN−α+2,φ′),

where

(5.14) if 2N>α≥2N+83, thentα:=2N−α2,Hα:=tα(2N−α−tα)=(2N−α2)2,

(5.15) if 2N+83>α>4, thentα:=2(N+2−α),Hα:=tα(2N−α−tα)=2(N+2−α)(α−4).

Moreover, if φ=ϕR2 then

(5.16) |E1(∇u,rrεN−α+2,φ′)|≤|E1(∇u,1|x|N−α+1,φ′)|≤c1(ϕ1)Rα−2−∫AR|∇u|2ϕR.

A version of the above inequalities for singular weight is contained in the following.

Theorem 5.7

Let N ≥ 1 and α ∈ ℝ. Let u ∈ 𝒞²(ℝ^N), t ∈ ℝ, and let φ∈𝒞 0 1(ℝN) be a radial nonnegative function. If one of the following cases holds

1. u ≡ 0 or φ ≡ 0 in a neighborhood of 0, N ≥ 1, α ∈ ℝ, t ∈ ℝ,
2. N ≥ 2, α > 2, t ≥ 0,

then we have

(5.17) ∫ℝN|Δu|2φ|x|N−α≥t(2N−2α+4−t)∫ℝN(∇u⋅x|x|)2φ|x|N−α+2 + t(α−4)∫ℝN|∇u|2φ|x|N−α+2−2tE1(∇u,1|x|N−α+1,φ′).

In particular if 2N > α ≥ 4, then

(5.18) ∫ℝN|Δu|2φ|x|N−α≥(2N−α2)2∫ℝN(∇u⋅x|x|)2φ|x|N−α+2 −(2N−α)E1(∇u,1|x|N−α+1,φ′),

(5.19) ∫ℝN|Δu|2φ|x|N−α≥Hα∫ℝN|∇u|2φ1|x|N−α+2−2tαE1(∇u,1|x|N−α+1,φ′),

where H_α and t_α are defined in (5.14) and (5.15).

Moreover, if φ=ϕR2 then (5.16) holds.

For radial functions we have the following.

Corollary 5.8

Let 2N > α > 2. For any radial u ∈ 𝒞²(ℝ^N), the inequality (5.18) holds.

Proof

Since (∇u⋅x|x|)2=|∇u|2 , and N > 2 from (5.17), for t > 0, we have

∫|Δu|2φ|x|N−α≥t(2N−α−t)∫|∇u|2φ|x|N−α+2−2tE1(∇u,1|x|N−α+1,φ′).

The conclusion follows by choosing t = (2N − α)/2.

Dealing with compactly supported functions we have the following.

Corollary 5.9

Let N ≥ 2, α > 2, t ≥ 0. For any u∈𝒞 0 2(ℝN) , we have

(5.20) ∫ℝN|Δu|21|x|N−α≥t(2N−2α+4−t)∫ℝN(∇u⋅x|x|)21|x|N−α+2+t(α−4)∫ℝN|∇u|21|x|N−α+2

In particular if 2N > α ≥ 4, then

(5.21) ∫ℝN|Δu|21|x|N−α≥(2N−α2)2∫ℝN(∇u⋅x|x|)21|x|N−α+2,

(5.22) ∫ℝN|Δu|21|x|N−α≥Hα∫ℝN|∇u|21|x|N−α+2,

where H_α is defined in (5.14) and (5.15).

If 2N > α > 2 and u∈𝒞 0 2(ℝN) is radial, then (5.21) holds.

The proof is similar to the proof of Corollary 5.4.

Remark 5.10

Observe that if 2N > α ≥ (2N +8)/3, then H_α = (2N − α)²/4 and (5.13) ⇒ (5.12), (5.19) ⇒ (5.18), and (5.22) ⇒ (5.21).

Remark 5.11

In [33, Theorems 1.7 and 6.4] the authors prove that the inequality

(5.23) ∫ℝN|Δu|21|x|N−α≥Bα∫ℝN|∇u|21|x|N−α+2, ∀ u∈𝒞 0 ∞(ℝN),

holds for N ≥ α > 4, where the best constant B_α is given by Bα=(2N−α2)2 whenever 4(N+1)−2N2−N+13≤α≤N , while 0<Bα<(2N−α2)2 for 4<α<4(N+1)−2N2−N+13 .

From (5.21), and the fact that |∇u|2≥(∇u⋅x|x|) , we deduce that for 4<α<4(N+1)−2N2−N+13 , the minimizing sequence related to (5.23) is not radial.

We observe that the range 2N > α > N has not been considered in [33]. However, inequalities (5.21) and (5.22) are still valid in that range.

It seems an interesting problem to study the sharpness of the constant appearing in (5.21).

Proof of Theorems 5.6 and 5.7

We begin proving the results for u ∈ 𝒞²(ℝ^N) and for α ≠ N. The results follows from Theorem 5.3 by choosing v = γv_ɛ where v_ɛ is defined in (5.9) and γ can assume only two values, γ = 1 or γ = −1. With this choice, we have

v″−v′r−δ22v′2|v|=(N−α)γ(N−α+2−δ22γ(N−α))r2rεN−α+4,−Δv2+v′r=(N−α)γ(α−4)2r2rεN−α+4+ε2(N−α)γ(N−2)21rεN−α+4.

Plugging these quantity in (5.2), setting t := δ²(N − α)γ we deduce (5.11) for any t ≠ 0 (since δ² runs on all positive numbers and γ can be choose in {−1, 1}). The case t = 0 is trivial. The case α = N follows by letting α → N in (5.11).

With the choice t=2N−α2 in (5.11), since t(α − 4) ≥ 0, t(N − 2) ≥ 0 and taking into account that

(5.24) |∇u|2≥(∇u⋅x|x|)2,

we deduce (5.12).

In order to prove (5.13), we use (5.11) by choosing t = t_α with t_α as defined in (5.14) or in (5.15). With this choice, the coefficient of the integral involving (∇u⋅x|x|)2 is nonpositive, t_α(2N − 2α + 4 − t_α) ≤ 0, and by (5.24) we obtain the claim.

The inequalities in Theorem 5.7 follow by letting ɛ → 0 in (5.11), (5.12) and (5.13). To this end, we notice that in the case α = 4 the term containing the weight rε−N+α−4 is too singular when ɛ → 0, however taking into account that it is nonnegative, it can be ignored.

Finally, the estimate (5.16), easily follows from the definition of E₁.

The case u∈Hloc2(ℝN) in Theorem 5.6 follows by a regularization argument.

5.1 An integral identity

Lemma 5.12

Let u∈Hloc2(ℝN) , φ∈𝒞 0 4(ℝN) and v ∈ 𝒞⁴(ℝ^N). We have

(5.25) ∫ΔuΔ(uvφ)=∫(Δu)2vφ+12∫u2Δ2v φ−∫|∇u|2Δvφ+2∫Δu φ(∇v,∇u)+12∫u2ΔvΔφ+∫u2(∇Δv,∇φ)+∫Δu u v Δφ+2∫Δu u(∇v,∇φ)+2∫Δu v(∇u,∇φ).

Furthermore if φ and v are assumed to be radial, then

(5.26) ∫ΔuΔ(uvφ)=∫(Δu)2vφ+12∫u2Δ2v φ−2∫(∇u⋅x|x|)2v″φ −2∫|∇u|2v′|x|φ+2∫(∇u⋅x|x|)2v′|x|φ−2E1(∇u,v′,φ′)+E2(u,v,φ)

where E₁ is defined in (5.3) and E₂ is defined as

(5.27) E2(u,v,φ):=∫u2ΔvΔφ+∫u2(∇Δv,∇φ)+∫ΔuuvΔφ +2∫Δuu(∇v⋅∇φ)+2∫Δuv(∇u⋅∇φ).

In particular if φ=ϕR4 , with ϕ_R defined in (1.8), and

(5.28) for k=0,1,2,3, |v(k)|≤cvRα−N−k

holds on A_R, then

(5.29) |E2(u,v,φ)|≤cvc1(ϕ1)Rα××[R−4⨍AR|u|2ϕR+(⨍AR|Δu|2ϕR)12(R−4⨍AR|u|2ϕR)12+(⨍AR|Δu|2ϕR)12(R−2⨍AR(∇u⋅x|x|)2ϕR)12].

Proof

We prove the claim for u ∈ 𝒞²(ℝ^N). The general case follows by a regularization argument. Let φ∈𝒞 0 4(ℝN) and v ∈ 𝒞⁴(ℝ^N). We have

(5.30) ∫ΔuΔ(uvφ) =∫(Δu)2vφ+∫Δu uΔv φ+∫Δu u v Δφ+ +2∫Δu u(∇v,∇φ)+2∫Δu v(∇u,∇φ)+2∫Δu φ(∇v,∇u).

By using the identity

Δu2=2uΔu+2|∇u|2,

the second term in right hand side of (5.30) can be rewritten as

∫Δu u Δv φ=12∫Δ(u2)Δv φ−∫|∇u|2Δv φ =12∫u2Δ(Δv φ)−∫|∇u|2Δv φ =12∫u2Δ2v φ+12∫u2Δv Δφ+∫u2(∇Δv,∇φ)−∫|∇u|2Δv φ.

Hence, (5.30) can be rewritten as in the claim.

Taking into account the identity in Lemma 5.1, from (5.25) we get (5.26).

The estimate (5.29) can be proved by Cauchy-Schwarz inequality and by using the estimates (5.28) and the estimates on the derivatives of ϕ_R. For instance, since

|ΔϕR4|=|4ϕR3ΔϕR+12ϕR2|∇ϕR|2|≤c1(ϕ1)ϕRR−2

(we have used the fact that ϕRk≤ϕR for any k ≥ 1), we have

|∫ΔuuvΔϕR4|≤c1∫|Δu|ϕRR−2|u|cvRα−N≤c1cv(Rα⨍AR|Δu|2ϕR)12(Rα−4⨍AR|u|2ϕR)12.

The other terms can be estimated in a similar way.

Choosing v = v_ɛ in the above Lemma and arguing as in the proof of Proposition 5.6, it follows that

Proposition 5.13

Let N > 4, α ≥ 4, and u ∈ 𝒞⁴(ℝ^N). For any radial φ∈𝒞 0 4(ℝN) , we have

(5.31) ∫ΔuΔ(uφ|x|N−α)=∫(Δu)2φ|x|N−α+2(N−α)∫|∇u|2φ|x|N−α+2−2(N−α)(N+2−α)∫(∇u⋅x|x|)2φ|x|N−α+2+(N−α)(N+2−α)P(u2φ,α)+2(N−α)E1(∇u,1|x|N−α+1,φ′)+E2(u,|x|α−N,φ)

where E₁ and E₂ are defined respectively in (5.3) and (5.32) and P is defined as

(5.32) P(u2φ,α):={(α−2)(α−4)2∫u2φ|x|N−α+4,if α>4,CN2(N−4)(N−2)u2(0)φ(0),if α=4,

where C_N > 0 is the positive constant defined in (1.9).

5.2 Some functional weighted quadratic inequalities

Gluing the identity (5.26) and the inequality (5.2) we deduce an inequality for the bilinear form

u∈𝒞 2(ℝN)↦∫ΔuΔ(uvφ)

for a general radial nonnegative weight v ∈ 𝒞⁴(ℝ^N).

Theorem 5.14

Let u ∈ 𝒞²(ℝ^N). Let φ∈𝒞 0 4(ℝN) , v ∈ 𝒞⁴(ℝ^N) be radial nonnegative functions and δ > 0 be such that

(5.33) δ2(−Δv+2v′|x|)−2v′|x|≥0, v′2v∈Lloc1(ℝN).

Then

(5.34) ∫ΔuΔ(uvφ)≥12∫u2Δ2v φ+∫(∇u⋅x|x|)2[2δ2(v″−12Δv−δ22v′2v)−2v″]φ+2(δ2−1)E1(∇u,v′,φ′)+E2(u,v,φ).

Proof

Gluing the identity (5.26) and the inequality (5.2) we obtain

∫ΔuΔ(uvφ)≥ 12∫u2Δ2v φ+∫(∇u⋅x|x|)2[2δ2(v″−v′|x|−δ22v′2v)−2v″+2v′|x|]φ+∫|∇u|2[2δ2(−12Δv+v′|x|)−2v′|x|]φ+ 2(δ2−1)E1(∇u,v′,φ′)+E2(u,v,φ).

By using the hypothesis (5.33) together with (5.24), we get the claim.

Theorem 5.15

Let N > 4 and N+N2−4N≥α≥N−N2−4N . Let u∈Hloc2(ℝN) and let r, r_ɛ be as in (5.8). For any φ∈𝒞 0 4(ℝN) , radial and nonnegative functions, we have

(5.35) ∫ΔuΔ(uφrεN−α)≥12∫u2Δ2(rεα−N) φ+C(N,α)4∫ℝN(∇u⋅x|x|)2r2rεN−α+4φ+ε22((2N−α)(N−2)+4(N−α))∫ℝN|∇u|21rεN−α+4φ−αE1(∇u,rrεN−α+2,φ′)+E2(u,rεα−N,φ)

where

(5.36) C(N,α):=−7α2+4(2+3N)α−4N(N+2),

and E₁, and E₂ are defined in (5.3) and (5.27) respectively.

Remark 5.16

Let us analyze the positivity of some terms appearing in (5.35).

i) Let
(5.37) α1:=27(3N+2−2(N2−N+2)),α2:=27(3N+2+2(N2−N+2)),
be the two roots of C(N, α). Clearly, the constant C(N, α) is non negative for α₁ ≤ α ≤ α₂.
ii) The coefficient ε22((2N−α)(N−2)+4(N−α)) is nonnegative for α≤2N2N+2 .
iii) From (5.10), we have Δ2rεα−N is nonnegative provided α ≥ 4 and (N − α)(N + 2 − α) ≥ 0.
iv) For N > 4, we have
N−N2−4N<α1<N<α2<2N2N+2<N+N2−4N,
with 4 < α₁ for N ≥ 8 and α₁ < 4 for N = 5, 6, 7.

Therefore for N ≥ α ≥ max{4, α₁}, it follows C(N, α) is nonnegative as well as the coefficient ε22((2N−α)(N−2)+4(N−α)) and the integral involving Δ2rεα−N . This remark allow us to establish the positivity of the quadratic form u↦∫ΔuΔ(uφrεN−α) under suitable conditions. See Section 5.2.1 below.

Specializing Theorem 5.14 with the weight v = v_ɛ and δ2=2N−α2(N−α) , we can deduce (5.35) of Theorem 5.15 under a more restrictive hypothesis on the parameter α. This is the reason why we do not deduce Theorem 5.15 as a consequence of Theorem 5.14.

Proof of Theorem 5.15

We prove the claim for u ∈ 𝒞²(ℝ^N). The general case follows by a regularization argument. By choosing v = v_ɛ in (5.26), and plugging in (5.11), we obtain

∫ΔuΔ(uφrεN−α)≥ ∫(|∇u|2p1(t)+(∇u⋅x|x|)2p2(t))φr2rεN−α+4+ε2(t(N−2)+2(N−α))∫|∇u|2φrεN−α+4+12∫u2Δ2(rεα−N) φ+2(N−α−t)E1(∇u,rrεN−α+2,φ′)+E2(u,rεα−N,φ),

for any t ∈ ℝ, where

p1(t):=2(N−α)+t(α−4), p2(t):=t(2N−2α+4−t)−2(N−α)(N−α+2).

Choosing t=tα:=2N−α2 , we get p₁(t_α) ≥ 0 (since the hypothesis on α). By using (5.24) we complete the proof by taking into account that p₁(t_α) + p₂(t_α) = C(N, α)/4.

The analog of Theorem 5.15 for singular weight is the following.

Theorem 5.17

Let N > 4 and 2N2N+2≥α≥4 then for any u ∈ 𝒞²(ℝ^N), and any nonnegative radial φ∈𝒞 0 4(ℝN) , we have

(5.38) ∫ΔuΔ(uφ|x|N−α)≥C(N,α)4∫(∇u⋅x|x|)2φ|x|N−α+2+(N−α)(N+2−α)P(u2φ,α)−αE1(∇u,1|x|N−α+1,φ′)+E2(u,|x|α−N,φ)

where C(N, α), E₁, E₂ and P are defined in (5.36), (5.3), (5.27), and (5.32) respectively.

Proof

Since α≤2N2N+2 , we have ɛ²((2N − α)(N − 2) + 4(N − α)) ≥ 0, and from (5.35) we obtain

(5.39) ∫ΔuΔ(uφrεN−α)≥12∫u2Δ2(rεα−N) φ+C(N,α)4∫(∇u⋅x|x|)2r2rεN−α+4φ−αE1(∇u,rrεN−α+2,φ′)+E2(u,rεα−N,φ).

Letting ɛ → 0, we conclude the proof.

5.2.1 Some results on functional positive quadratic forms

The topic of this section is a brief detour from our main scope: namely the positivity of certain integral quadratic forms. From Theorem 5.17 we can deduce that

∀ u∈𝒞 0 2(ℝN): ∫ΔuΔ(u|x|N−α)≥0,

for max{4, α₁} ≤ α ≤ N or N + 2 ≤ α ≤ α₂. It remains to consider the case N < α < N + 2. In order to fill this gap, we need an extra argument based on the following asymptotic Hardy inequality.

In what follow, for brevity, we give only a sketch of the proofs of the results and we consider only smooth functions.

Theorem 5.18

Let γ ∈ ℝ, u ∈ 𝒞¹(ℝ^N). For any ɛ > 0, t ∈ ℝ and a nonnegative radial function φ∈𝒞 0 1(ℝN) , we have

(5.40) ∫(∇u⋅x|x|)2r2rεγ+2φ≥t(N−γ−2−t)∫u2r2rεγ+4φ +ε2t(N−t)∫u21rεγ+4φ+tE3(u2,rrεγ+2,φ′),

and

(5.41) ∫(∇u⋅x|x|)21rεγφ≥t(N−γ−2−t)∫u2r2rεγ+4φ +ε2tN∫u21rεγ+4φ+tE3(u2,rrεγ+2,φ′),

where

(5.42) E3(u2,rrεγ+2,φ′):=∫u2rrεγ+2(∇φ⋅x|x|).

In particular, if N − γ − 2 ≥ 0, we have

(5.43) ∫(∇u⋅x|x|)21rγφ≥(N−γ−2)24∫u21rγ+2φ+N−γ−22E3(u2,1rγ+1,φ′).

Finally, if φ=ϕR2 , we have that

|E3(u2,rrϵγ+2,φ′)|≤c1(ϕ1)RN−γ−2−∫ARu2ϕR.

The proof of the above theorem is based on an application of the vector field method (see [26]).

Proof

Let t ≠ 60 and set s := sign(t). Let H be the vector field defined by H(x):=srε−γ−2x . Since

0=∫div(H u2 φ)=∫div(H)u2φ+∫u2(H⋅∇φ)+2∫uφ(H⋅∇u),

by computation we have

(5.44) s(N−γ−2)∫u2r2rεγ+4φ+ε2Ns∫u21rεγ+4φ+s∫u2rrεγ+2(∇φ⋅x|x|)

(5.45) =−2s∫uφrrεγ+2(∇u⋅x|x|)≤δ2∫u21rεγ+2φ+δ−2∫(∇u⋅x|x|)2r2rεγ+2φ.

Since

1rεγ+2=r2rεγ+4+ε2rεγ+4,

by choosing δ² = ts = |t|, from (5.44), (5.45) and (5.42) we deduce (5.40).

The proof of (5.41) is similar to the proof of (5.40). Indeed, from the left hand side of (5.45) it follows that

−2s∫uφrrεγ+2(∇u⋅x|x|)≤δ2∫u2r2rεγ+4φ+δ−2∫(∇u⋅x|x|)21rεγφ.

So we can proceed as above.

For t ≥ 0, since the coefficient ɛ²tN is nonnegative, in (5.41) we can ignore the term containing it, and hence by choosing t=N−γ−22 , and letting ɛ → 0 in (5.41), we obtain (5.43).

Theorem 5.19

Let u ∈ 𝒞²(ℝ^N), N > 4 and max{4, α₁} ≤ α ≤ α₂. For any nonnegative radial function φ∈𝒞 0 2(ℝN) , we have

(5.46) ∫ΔuΔ(uφ|x|N−α)≥D116∫u2φ|x|N−α+4−αE1(∇u,1|x|N−α+1,φ′)+E2(u,|x|α−N,φ)+C(N,α)(α−4)8E3(u2,1|x|N−α+3,φ′),

where

D1:=D1(N,α):=α(α−4)(2N−α)(2N−α−4),

and for any ɛ > 0 we have

(5.47) ∫ΔuΔ(uφrεN−α)≥116∫u2φD1r4+D2ε2r2+D3ε4rεN−α+8−αE1(∇u,rrεN−α+2,φ′)+E2(u,rεα−N,φ)+C(N,α)(α−4)8E3(u2,rrεN−α+4,φ′)+ε22((2N−α)(N−2)+4(N−α))tE3(u2,rrϵN−α+6,φ′),

where t ∈ ℝ and

D2:=2(α−4)[NC(N,α)+8(N−α)(N+2−α)(N+2)]+8((2N−α)−2α)(α−6−t)t,D3:=C(N,α)(α−4)(2N−α+4)+8(N−α)(N+2−α)N(N+2)+8((2N−α)−2α)Nt.

Proof

Since C(N, α) ≥ 0, by choosing γ = N − α + 2 and plugging (5.43) in (5.38), we easily obtain (5.46).

Next we prove (5.47). Again, since C(N, α) ≥ 0, and ɛ²((2N − α)(N − 2) + 4(N − α)) ≥ 0, by inserting (5.40) in (5.35) with γ = N − α + 2 and t = (α − 4)/2, and (5.41) in (5.35) with the choice γ = N − α + 4 and t ∈ ℝ, we obtain (5.47).

For compactly supported functions we have the following.

Corollary 5.20

Let N = 5 and 4 ≤ α < 6, or N = 6 and max{4, α₁} ≤ α ≤ α². Then we have

(5.48) ∀ u∈𝒞 0 2(ℝN): ∫ΔuΔ(u1|x|N−α)≥0.

Moreover if N = 5 and 4 ≤ α < 6, N = 6 and 4 ≤ α < 8, or N ≥ 7 and max{4, α₁} ≤ α ≤ α₂,

(5.49) ∀ u∈𝒞 0 2(ℝN) u≠0: ∫ΔuΔ(u1|x|N−α)>0.

Proof

Let u∈𝒞 0 2(ℝN) and let R > 0 be large enough such that the support of u is contained in B_R. Let φ = ϕ_R. With this choice, it follows that E₁, E₂ and E₃ in (5.46) vanish. This implies that

∫ΔuΔ(u1|x|N−α)≥D116∫u2|x|N−α+4.

Now, if N ≥ 6, we have max{4, α₁} ≤ α ≤ α₂ ≤ 2N − 4 < 2N and this implies that D₁ = α(α − 4)(2N − α)(2N − α − 4) is nonnegative. Analogously, if N = 5 and 4 ≤ α ≤ 6 < α₂, we conclude again that D₁ ≥ 0 and (5.48) holds.

In order to show the strict inequality in (5.49), we argue as follows. If D₁ > 0 the conclusion is obvious. It remains to analyze only the case when α = 4, and hence N = 5, 6, 7. In this situation it follows that C(N, 4) > 0 and then, the claim follows from Theorem 5.17.

For a non singular weight we have the following.

Corollary 5.21

Let N = 5 and 4 ≤ α ≤ 6, or N ≥ 6 and max{4, α₁} ≤ α ≤ α₂. Then we have

(5.50) ∀ u∈𝒞 0 2(ℝN) u≠0: ∫ΔuΔ(u1rεN−α)>0.

Proof

Let u∈𝒞 0 2(ℝN) , by using the same argument used in the proof of Corollary 5.20, from (5.47) we are reduced to study

∫ΔuΔ(uφrεN−α)≥116∫u2φD1r4+2D2ε2r2+D3ε4rεN−α+8,

and hence to establish that D₁r⁴ + 2D₂ɛ²r² + D₃ɛ⁴ is positive for some t ∈ ℝ. We note that the case α = N is trivial. Furthermore, from Remark 5.16 it follows that it is enough to perform the analysis only for N < α ≤ N + 2 when N ≥ 6 and for 5 < α ≤ 6 when N = 5.

We begin analyzing the case N ≥ 7. In this case the choice t = 0 assures that D₁, D₂, D₃ > 0 and our claim is proved.

Let N = 6 and 6 < α < 8. Even in this case the choice t = 0 implies that D₁, D₂, D₃ > 0 and our claim holds. For the case α = 8, we have D₁ = 0 and D₂, D₃ > 0 for t > 0 small enough (0 < t < α − 6) which implies (5.50).

Let N = 5. By the change of variable α = N + x, we have to check the positivity of D₁ = (x + 5)(x + 1)(1 − x)(5 − x), D₂ = 42 x³ − 202 x² + 6 x + 250 + 8 t²(7 x − 15) − 8(7 x² − 22 x + 15)t, and D₃ = 7 x⁴ − 54 x³ + 176 x² − 378 x + 225 − 40 t(7 x − 15), for 0 < x ≤ 1 and for some t ∈ ℝ. The choice t = 1/8 accomplishes the claim. We leave the detailed computations to the interested reader.

Remark 5.22

From the proof of (5.48)–(5.50), we can deduce the positivity of those quadratic forms, and that they can be controlled from below by quadratic integrals depending on u. For instance, if u∈𝒞 0 2(ℝN) and u ≢ 0, if D₁ > 0 and max{4, α₁} < α ≤ α₂. we have

∫ΔuΔ(u1|x|N−α)≥D116∫u2|x|N−α+4>0,

while for N > α ≥ 4 and α > α₁, we have

∫ΔuΔ(u1|x|N−α)≥C(N,α)4∫(∇u⋅x|x|)21|x|N−α+2>0.

We leave the analysis of the remaining cases to the interested reader.

The results on the positivity of the quadratic form in Corollary 5.20 and 5.21 can be extended relaxing the request that the functions have compact support. The idea is to assume that for the functions u the quantities E1(∇u,|x|α−N−1,ϕR′) , E²(u, |x|^α−N , ϕ_R) and E3(u2,|x|α−N−3,ϕR′) vanish as R → +∞. A sample of these kind of results is the following.

Theorem 5.23

Let N = 5 and 4 ≤ α < 6, or N = 6 and 4 ≤ α < 8, or N ≥ 7 and max{4, α₁} ≤ α ≤ α₂. For any u ∈ 𝒞⁴(ℝ^N), u ≢ 0 such that

(5.51) Δ2u u|x|N−α∈L1(ℝN), lim supR→+∞Rα−∫AR|Δu|2<∞limR→+∞Rα−2−∫AR|∇u|2=0, limR→+∞Rα−4−∫ARu2=0,

then

(5.52) ∫ΔuΔu|x|N−αdx=∫Δ2u u|x|N−αdx>0.

In particular, if (5.51) holds with α = 4, and N = 5, 6, 7, then

(5.53) ∫ΔuΔu|x|N−4dx=∫Δ2u u|x|N−4dx>0.

Theorem 5.23 extends a result of [22] where the inequality (5.53) is proved for compactly supported functions. Indeed in [22] the author proves that (5.53) holds for u∈𝒞 0 ∞(ℝN) and N = 5, 6, 7, while in dimension N ≥ 8, the inequality (5.53) is not satisfied. See also [23] for further extension to the higher order case.

6 A priori estimates and Liouville theorems

6.1 A priori estimates

In this section we deduce some a priori estimates on the solution of the inequality (P_h), that we remind for reader convenience

−uΔ2u≥h, onℝN, (Ph)

whose definition of weak solution is given in (4.2). The following results refines some estimates obtaind earlier in [2].

Theorem 6.1

For any d < 1 there exists c_d = c(d) > 0 such that for any h∈Lloc1(ℝN) (without any assumption on its sign), for any weak solution u∈Hloc2(ℝN) of (P_h) and any nonnegative φ∈𝒞 0 2(ℝN) , we have

(6.1) ∫ℝNhφ+d∫ℝN(Δu)2φ≤cd[∫ℝNu2|Δφ|2φ+∫ℝNu2|Δ|∇φ|2φ|+∫ℝNu2|∇φ|4φ3].

In particular if φ = ϕ_R, an admissible test function as in (1.8), there exists c₁(d) = c₁(ϕ₁, d) > 0 constant, and we have

(6.2) ∫ℝNhϕR+d∫ℝN(Δu)2ϕR≤c1(d)R−4∫ARu2,

(6.3) ∫ℝN|∇u|2ϕR≤c1R−2∫ARu2+(∫B2Ru2ϕR)1/21d(c1(d)R−4∫ARu2⨍B2RhϕR)1/2,

(6.4) ∫AR|∇u|2≤c1R−2∫AR/2∪A2Ru2+(∫AR/2∪AR∪A2Ru2)1/21d(c1(d)R−4∫A2Ru2−∫B4Rhϕ2R)1/2.

If h ≥ 0 on B_4R for some R > 0, then

(6.5) d∫BR(Δu)2≤∫BRhϕR+d∫ℝN(Δu)2ϕR≤c1R−4∫ARu2,

(6.6) ∫BR|∇u|2≤c1R−2∫B2Ru2,

(6.7) ∫AR|∇u|2≤c1R−2∫AR/2∪AR∪A2Ru2.

In particular if h ≥ c_h |u|^q+1 for some c_h > 0 and q > 1, then there exists c₁ = c₁(c_h , q) > 0 such that for any u∈Hloc2(ℝN) weak solution of (P_h) and any R > 0 there holds

(6.8) ch(⨍BR|u|2)q+12≤ch⨍BR|u|q+1≤⨍BRh+⨍BR(Δu)2≤c1R−4q+1q−1,

(6.9) ⨍BR|∇u|2≤c1R−2q+3q−1.

Remark 6.2

Notice that the constant c_d in the above theorem, does not depend on u nor on R, and the quantity c_dR⁻⁴ ∫_{A_R} u² − ∫_{B_2R} hϕ_R is nonnegative for any R and any u.

The following Lemma contains an Hloc2 version of the identity

(6.10) ΔΦ2=2ΦΔΦ+2|∇Φ|2.

Lemma 6.3

Let Φ∈Hloc2(Ω) . For any φ∈𝒞 0 2(Ω) , we have

(6.11) ∫ΩΦ2 Δφ=2∫ΩΦΔΦ φ+2∫Ω|∇Φ|2φ,

and

(6.12) ∫Ω|∇Φ|2φ≤12∫ΩΦ2 Δφ+(∫ΩΦ2|φ|)1/2(∫Ω|ΔΦ|2|φ|)1/2.

Proof

Let (m_η)_η>0 be a family of standard mollifier cutoff functions. Setting

uη:=Φ⋆mη,

we have that u_η → Φ, ∇u_η → ∇Φ and Δu_η → ΔΦ in Lloc2(ℝN) . Hence for any φ∈𝒞 0 2(ℝN) we have

2∫ΩuηΔuηφ+2∫Ω|∇uη|2φ=∫ΩΔuη2φ=∫Ωuη2Δφ.

Letting η → 0 identity (6.11) follows.

Inequality (6.12) can be deduced by applying Cauchy-Schwarz inequality to the identity (6.11).

Proof of Theorem 6.1

Proof of estimate (6.1). By choosing v ≡ 1 in the identity (5.25), we obtain

∫hφ+∫(Δu)2φ≤−∫ΔuΔ(uφ)+∫(Δu)2φ=−∫Δu u Δφ−2∫Δu (∇u,∇φ)=:I1+I2.

Next, by Young inequality with ɛ > 0, we have

|I1|≤ε22∫(Δu)2φ+12ϵ2∫u2|Δφ|2φ.

Analogously for any δ > 0 we have

|I2|≤δ2∫(Δu)2φ+1δ2∫|∇u|2|∇φ|2φ.

For simplicity we denote by Θ the quantity Θ:=|∇φ|2φ . Using the identity (6.11) with Φ = u, integrating by parts and by Young's inequality, we obtain

|I2|≤δ2∫(Δu)2φ+1δ2∫|∇u|2Θ=δ2∫(Δu)2φ+12δ2∫Δ(u2)Θ−1δ2∫uΔuΘ=δ2∫(Δu)2φ+12δ2∫u2ΔΘ−1δ2∫uΔuΘ≤δ2∫(Δu)2φ+12δ2∫u2|ΔΘ|+γ22δ2∫(Δu)2φ+12γ2δ2∫u2Θ2φ.

A suitable choice of the parameter ɛ, δ and γ gives the estimate (6.1).

Proof of (6.2). We begin noticing that each term in the right hand side of (6.1) has the form

(6.13) ∫u2|ψ|,

where ψ stands for one of the functions |Δφ|2φ , Δ|∇φ|2φ or |∇φ|4φ3 .

Taking φ := ϕ_R = ϕ₁(|x| /R) an admissible test function the function, |ψ| has support in A_R and can be estimate as |ψ| ≤ c₁R⁻⁴. This concludes the proof of (6.2).

Proof of (6.3). From (6.12) with Φ = u, φ = ϕ_R, and from the estimate of |Δu|², that can be deduced from (6.2), we have

∫|∇u|2ϕR≤12∫u2 ΔϕR+(∫u2ϕR)1/2(∫|Δu|2ϕR)1/2 ≤c1R−2∫ARu2+(∫B2Ru2ϕR)1/21d(c1(d)R−4∫ARu2−∫B2RhϕR)1/2,

that is the claim.

In order to prove estimate (6.4) we use (6.12) with Φ = u and as test function φ(x) = ρ_R(|x|) := ϕ_2R(t)−ϕ_R/2(t). The support of ρ_R is G_R := A_R/2 ∪ A_R ∪ A_2R. We have

∫AR|∇u|2≤∫ℝN|∇u|2ρR≤∫AR/2∪A2Ru2|ΔρR|+(∫GRu2ρR)1/2(∫GR|Δu|2ρR)1/2≤c1R−2∫AR/2∪A2Ru2+(∫GRu2)1/2(∫B4R|Δu|2ϕ2R)1/2≤c1R−2∫AR/2∪A2Ru2+(∫GRu2)1/21d(c1(d)R−4∫A2Ru2−∫B4Rhϕ2R)1/2,

where in the last inequality we have used the estimate (6.2), obtaining the claim.

Estimates (6.5), (6.6) and (6.7) are an immediate consequence of (6.2), (6.3) (6.4) and the hypothesis h ≥ 0.

Finally, to prove the missing inequalities (6.8) and (6.9) arguing again as in the proof of (6.2), and with the same notation, we notice that each term in the right hand side of (6.1) has the form (6.13). By using Hölder and Young inequalities, with exponent x=q+12 , we obtain

(6.14) ∫u2|ψ|≤(∫|u|q+1φ)1x(∫|ψ|x′φx′−1)1x′≤εxxch∫hφ+1x′εx′∫|ψ|x′φx′−1.

Taking φ := ϕ_R = ϕ₁(|x| /R) the term ∫|ψ|x′φx′−1 behaves as RN−4x′=RN−4q+1q−1 . Using (6.14) in (6.1), with a suitable choice of the parameter d and ɛ we get

(6.15) ∫BRh+∫BR(Δu)2≤cRN−4q+1q−1 and ⨍BR|∇u|2≤cR−2q+3q−1,

which in turn, by our assumption h ≥ c_h |u|^q+1 and Hölder inequality, yields (6.8). The inequality (6.9) is a consequence of the estimates (6.8) and (6.6).

6.2 Some glimpses on Liouville theorems: weak solutions

In this section we continue to study some Liouville theorems for weak solutions of the inequality,

−uΔ2u≥h, onℝN, (Ph)

where h ≡ 0 or h ≥ |u|^q+1.

In our first result we consider the homogeneous case.

Theorem 6.4

Let u∈Hloc2(ℝN) be a solution of

(6.16) −uΔ2u≥0, onℝN.

Assume that either,

(6.17) N=1,…,7, and limR→∞⨍AR|u|2=0,

(6.18) N≥8, and Rα1−4⨍AR|u|2≤C,for any R large,

where α1=27(3N+2−2(N2−N+2)) is defined in (5.37), then u ≡ 0 a.e. in ℝ^N.

A first consequence of the above theorem is the following corollary which is reminiscent of a result proved by Ambrosio and Cabré (see [1] for details and applications).

Corollary 6.5

Let u∈Hloc2(ℝN) be a solution of (6.16) with N ≥ 5. If for any R large

(6.19) ∫AR|u|2≤cR4,

with c independent of R, then u ≡ 0 a.e. in ℝ^N.

Proof of Corollary 6.5

From the hypothesis (6.19), we have that

−∫AR|u|2≤cR4−N→0 as R→∞.

It is enough to check that the assumptions of Theorem 6.4 hold. Indeed, if N = 5, 6, 7, then the hypothesis (6.17) is verified. While, for N ≥ 8, since α₁ < N it follows that

Rα1−4−∫AR|u|2≤cRα1−4+4−N→0 as R→∞,

and (6.18) holds.

A further consequence of Theorem 6.4, under the stronger assumption of global integrability of the solutions of (6.16), is the following Liouville theorem that can be obtained directly by using the Hölder inequality.

Corollary 6.6

Let u∈Hloc2(ℝN)∩Ls(ℝN\B1) be a solution of (6.16). If N = 1, . . . , 7 and 2 ≤ s < ∞ or N ≥ 8 and 2≤s≤2Nα1−4 , then u ≡ 0 a.e. in ℝ^N.

Notice that the expression

2Nα1−4=3N2−12N+N2N2−2N+4N2−10N+20,

as function of N is decreasing and converges to 3+2 as N → +∞.

An essential ingredient of the proof of Theorem 6.4 is the following.

Lemma 6.7

Let u be an harmonic function in ℝ^N. Then the following inequality holds

(6.20) ⨍BR|u−u(0)|2≤c⨍ARu2,

where c > 0 does not depend on u nor R.

In particular if

(6.21) lim infR→∞⨍ARu2=0,

then u ≡ 0 in ℝ^N.

Proof

Let ϕ_R be defined in (1.8). Since u is harmonic, from (6.10), we have

(6.22) ∫BR|∇u|2≤∫|∇u|2ϕR=12∫u2ΔϕR≤c1R−2∫ARu2.

Next by Poincaré inequality, setting uBR=⨍BRu . we have

∫BR|u−uBR|2≤cR2∫BR|∇u|2≤c∫ARu2.

Now by using the fact that u has the mean value property, that is u(0)=⨍BRu , we obtain (6.20).

Next, we observe that for an harmonic function u, for any x ∈ ℝ^N, we have

u(x)=⨍B2R(x)\BR(x)u.

Indeed, by the mean value property, we have

⨍B2R(x)\BR(x)u=1ωNRN(2N−1)(∫B2R(x)u−∫BR(x)u)=2N2N−1⨍B2R(x)u−12N−1⨍BR(x)u=2N2N−1u(x)−12N−1u(x)=u(x).

Next, from the inequalities

|u(0)|=|⨍ARu|≤⨍AR|u|≤(⨍AR|u|2)1/2,

and (6.21), we deduce that u(0) = 0. Finally, from (6.20) and (6.21), we have

⨍BRu2≤c⨍ARu2→0 as R→∞.

Now since u² is subharmonic, we know that

⨍BRu2↗supℝNu2 as R→∞,

and we deduce u² ≡ 0.

Proof of of Theorem 6.4

Let u be a solution of (6.16). Set α := 4 if N = 1, . . . , 7 and α := α₁ in the remaining cases N ≥ 8.

We begin proving the claim under the hypothesis

(6.23) limR→∞Rα−4⨍AR|u|2=0.

First we examine the cases N = 1, 2, 3, 4. From (6.2) it follows that

∫(Δu)2ϕR≤cR−4∫ARu2=cRN−4⨍ARu2.

Since α = 4, by our assumption (6.23) it follows that by letting R → ∞, we deduce that u is an harmonic function and Lemma 6.7 applies.

Next we consider the cases N ≥ 5. An application of (5.38), taking into account that C(N, α) ≥ 0, yields

(6.24) 0≥∫ΔuΔ(uφ|x|N−α)≥C(N,α)4∫(∇u⋅x|x|)2φ|x|N−α+2+(N−α)(N+2−α)P(u2φ,α)−αE1(∇u,1|x|N−α+1,φ′)+E2(u,|x|α−N,φ).

By using (6.23) in (6.7) and (6.5), we deduce

limR→∞Rα−2⨍AR|∇u|2=0 and limR→∞Rα⨍AR|Δu|2=0.

Next by choosing φ=ϕR4 as in (1.8), from estimate (5.4) and (5.29) we deduce that E₁ → 0 and E₂ → 0 as R → ∞.

Since P(u2ϕR4,α) is nonnegative and non decreasing with respect to R, by the monotone convergence theorem we obtain P(u2ϕR4,α)→P(u2,α)≥0 as R → ∞. Finally, from (6.24) we have

0≥∫ΔuΔ(u1|x|N−α)≥C(N,α)4∫(∇u⋅x|x|)21|x|N−α+2+(N−α)(N+2−α)P(u2,α)≥0,

hence

∫ΔuΔ(u1|x|N−α)=C(N,α)4∫(∇u⋅x|x|)21|x|N−α+2=P(u2,α)=0.

In the case α = α₁ > 4, from the definition of P in (5.32), it follows that u² ≡ 0. This complete the proof in the case α₁ > 4.

Next we consider the case α = 4. Since C(N, 4) > 0 we get

∫(∇u⋅x|x|)21|x|N−α+2=0,

that is (∇u⋅x|x|)=0 on ℝ^N. Therefore u is a constant functionwhose mean vanishes at infinity, that is u ≡ 0.

Finally we consider the cases when N ≥ 8 and (6.18) holds. Arguing as above, using (6.18) in (6.7) and (6.5), we deduce that

Rα1−2⨍AR|∇u|2≤C and Rα1⨍AR|Δu|2≤C.

By choosing φ=ϕR4 as in (1.8), from estimate (5.4) and (5.29) it follows that |E₁| ≤ c and |E₂| ≤ c for some constant c > 0. Since C(N, α₁) = 0, from (5.46) we have

c≥∫ΔuΔ(uϕR4|x|N−α)≥(N−α)(N+2−α)P(u2ϕR2,α1),

and by letting R → ∞, we obtain

c≥P(u2,α1)=(α1−2)(α1−4)2∫ℝNu21|x|N−α1+4

that is u2|x|N−α1+4∈L1(ℝN) .

Therefore,

Rα1−4⨍AR|u|2≤C∫ARu21|x|N−α1+4→0, as R→∞.

That is u satisfies the stronger assumption (6.23), and the claim follows.

One of the main result of this paper within the class of weak solutions is the following.

Theorem 6.8

Let u∈Hloc2(ℝN)∩Llocq+1(ℝN) be a solution of

−uΔ2u≥c|u|q+1 onℝN.

(6.25) N=1,…,7, and q>1orN≥8 and 1<q≤qN

where

(6.26) qN:=α1+4α1−4=3N+16−4−2 N+2 N23N−12−4−2 N+2 N2=N2+2N−28+44−2 N+2 N2N2−10N+20,

then u ≡ 0 a.e. in ℝ^N.

Proof

We distinguish various cases. Let N = 1, . . . , 4. Under this assumption the claim follows directly form (6.8). Indeed, we have

∫BR|u|q+1+∫BR(Δu)2≤CRN−4q+1q−1→0 as R→+∞.

Next we consider the case N ≥ 5. From (6.8), we deduce that

Rα−4⨍AR|u|2≤Rα−4q+1q−1.

Now if N = 5, 6, 7, the choice α = 4 is admissible in Theorem 6.4 and since 4−4q+1q−1<0 , the claim follows.

Let N ≥ 8. In this case, α = α₁ > 4 is an admissible choice in Theorem 6.4 and since α1−4q+1q−1≤0 , we conclude the proof.

6.3 Weighted a priori estimates

In this section we shall prove some a priori estimates for solutions of

−uΔ2u≥h, onℝN. (Ph)

These estimates will be useful in the study of distributional solutions of the fourth order problem (_gP_f ).Notice that there is no hypothesis on the sign of the function h.

Theorem 6.9

Let u ∈ 𝒞⁴(ℝ^N) and h∈Lloc∞(ℝN) satisfy (P_h). Let α ≥ 4, N > 4.

For any nonnegative and radial function φ∈𝒞 0 4(ℝN) , we have

(6.27) 2(N−α)(N+2−α)∫(∇u⋅x|x|)2φ|x|N−α+2≥∫hφ|x|N−α +∫(Δu)2φ|x|N−α+2(N−α)∫|∇u|2φ|x|N−α+2 +(N−α)(N+2−α)P(u2φ,α) +2(N−α)E1(∇u,1|x|N−α+1,φ′)+E2(u,|x|α−N,φ),

and for N+N2−4N≥α≥4 , for any ɛ > 0 by setting r := |x|, r_ɛ := (ɛ² + |x|²)^1/2, we have,

(6.28) 0≥∫hφrεN−α+C(N,α)4∫(∇u⋅x|x|)2r2rεN−α+4φ+12∫u2Δ2(rεα−N) φ −αE1(∇u,rrεN−α+2,φ′)+E2(u,rεα−N,φ).

Furthermore, for any δ > 0, let ϕ₁ be an admissible test function, then there exists c₁ = c(ϕ₁, δ) such that for any R large, by setting φ=ϕR4 we have

(6.29) |E1(∇u,rrεN−α+2,φ′)|≤c1(δ)Rα−4⨍B2Ru2ϕR+c1(δ)Rα−4⨍ARu2−δ2Rα⨍B2RhϕR,

(6.30) |E1(∇u,rrεN−α+2,φ′)|≤c1(δ)Rα−4⨍AR/2∪AR∪A2Ru2−δ2Rα⨍B4Rhϕ2R,

(6.31) |E2(u,rεα−N,φ)|≤c1(δ)Rα−4⨍B2Ru2ϕR+c1(δ)Rα−4⨍ARu2−δ2Rα⨍B2RhϕR,

(6.32) |E2(u,rεα−N,φ)|≤c1(δ)Rα−4⨍AR/2∪AR∪A2Ru2−δ2Rα⨍B4Rhϕ2R,

and the involved constants do not depend on ɛ nor on u.

Proof

We begin by observing that if u solves (P_h), then for any nonnegative test function φ, we have

∫−ΔuΔ(uvεφ)≥∫hvεφ.

where v_ɛ is defined in (5.9). Now letting ɛ → 0, an application of the identity (5.31) yields (6.27), while from the inequality (5.35) we deduce (6.28).

In order to estimate E₁, from (5.4), it suffices to estimate Rα−2⨍AR|∇u|2ϕR . To this end from the estimate (6.3) with d = 1/2, and by Young's inequality, we deduce

(6.33) R−2⨍ℝN|∇u|2ϕR≤c1R−4⨍ARu2 +(R−4⨍B2Ru2ϕR)1/22(c1R−4⨍ARu2−⨍B2RhϕR)1/2≤c1R−4⨍ARu2 +12δ2R−4⨍B2Ru2ϕR+c1δ2R−4⨍ARu2−δ2⨍B2RhϕR,

that is the estimate (6.29). The estimate (6.30) follows similarly from (6.4).

To estimate E₂, it suffices to estimate the terms in (5.29) containing Δu and ∇u. Arguing as before, by Young's inequality and the estimate (6.2) with d = 1/2, we have

(6.34) (⨍AR|Δu|2ϕR)12(R−4⨍AR|u|2)12≤δ22⨍AR|Δu|2ϕR+12δ2R−4⨍AR|u|2 ≤δ2c1R−4⨍AR|u|2−δ2⨍hϕR+12δ2R−4⨍AR|u|2.

The last term in the estimate (5.29) of E₂, can be controlled similarly by using (6.33) where δ is replaced by η, obtaining

(6.35) (⨍AR|Δu|2ϕR)12(R−2⨍AR|∇u|2ϕR)12≤≤δ22⨍AR|Δu|2ϕR+12δ2R−2⨍AR|∇u|2ϕR≤δ2c1R−4⨍AR|u|2−δ2⨍hϕR+14δ2η2R−4⨍B2Ru2ϕR +c11+η22δ2R−4⨍ARu2−η22δ2⨍B2RhϕR.

Now, choosing η² = 2δ⁴, gluing together the estimates (5.29), (6.34) and (6.35) and rescaling δ we get the estimate (6.31). Similarly we deduce (6.32), concluding the proof.

Corollary 6.10

Let N > 4 and α ≥ 4 such that C(N, α) ≥ 0. Then for any δ > 0 and any admissible function ϕ₁, there exists c₁(δ) > 0 such that for any u ∈ 𝒞⁴(ℝ^N) and h∈Lloc∞(ℝN) satisfying (P_h), and any R large, we have

c1(δ)Rα−4⨍B2Ru2ϕR+c1(δ)Rα−4⨍ARu2 ≥δ2Rα⨍B2RhϕR+∫hϕR|x|N−α+(N−α)(N+2−α)P(u2ϕR4,α),

where ϕ_R is defined as in (1.8).

Notice that the last inequality is an estimate on the possible solutions of (P_h) which does not involve the derivatives of u. This will be useful when dealing with distributional solutions.

6.4 Distributional solutions: a priori estimates

The a priori estimates contained in the following result will play a crucial role in what follows.

Theorem 6.11

Assume that f , g : ℝ → ℝ are continuous functions satisfying

f(t)t≥0, g(t)t≥0, for all t∈ℝ ,(f0)

and set H as in (1.6), that is

H(t):={f(t)t,for t≥0,g(t)t,for t<0.

Let u∈Llocs(ℝN) be such that f(u),g(u)∈Llocs′(ℝN) with 2 ≤ s ≤ +∞, and let u be a distributional solution of

g(u)≥−Δ2u≥f(u) onℝN. (gPf)

Then,

1. For any R we have
(6.36) ∫BRH(u)≤cR−4∫ARu2.
2. Let N > α ≥ 4 be such that C(N, α) ≥ 0, we have that |x|α−NH(u)∈Lloc1(ℝN) , and for any R, we have
(6.37) ∫BRH(u)1rϵN−α+12∫BRu2Δ2(rϵα−N)≤cRα−4⨍AR/2∪AR∪A2Ru2,

(6.38) ∫BRH(u)1|x|N−α≤cRα−4⨍AR/2∪AR∪A2Ru2,

(6.39) ∫BRH(u)1|x|N−α≤cRα−4⨍B2Ru2.
3. If H(t) ≥ c_q |t|^q+1 for some q > 1 and c_q > 0, we have
(6.40) ⨍BRH(u)≤cR−4q+1q−1, ⨍BRu2≤cR−8q−1.
Hence, if N > α ≥ 4 and C(N, α) ≥ 0, it follows that
(6.41) ∫BRH(u)1rϵN−α+12∫BRu2Δ2(rϵα−N)≤cRα−4q+1q−1
(where c > 0 is independent of u and R).

Proof

Let (m_n)_n>0 be a family of standard radial mollifier. By using m_n(x − ·) as test function for (_gP_f ), and setting

un:=u⋆mn, fn:=f(u)⋆mn, gn:=g(u)⋆mn,

we shall deal with a sequence of functions u_n, f_n and g_n satisfying the following properties

un→u in Llocs(ℝN)∩Lloc2(ℝN) and a.e. as n→+∞ fn→f(u), gn→g(u) in Llocs′(ℝN) and a.e. as n→+∞.

Since the functions t ↦ t⁺ and t ↦ t⁻ are Lipschitz functions we have (up to a subsequence) that

un±→u± in Llocs(ℝN)∩Lloc2(ℝN) and a.e. as n→+∞.

Therefore u_n is a smooth functions satisfying

gn≥−Δ2un≥fn, onℝN.

Multiplying by un+ , then by −un− and then adding the inequalities we have

gnun+−fnun−≥−unΔ2un≥fnun+−gnun−.

That is, setting hn:=fnun+−gnun− , it follows that u_n is a smooth solution of

(6.42) −unΔ2un≥hn,

and hence all the a priori estimates of the previous sections apply. Before to going on, let us to notice that

hn→f(u)u+−g(u)u−=H(u) in Lloc1(ℝN) and a.e.,

and furthermore, since g(u)(−u⁻) ≥ 0 we get that

(6.43) hn→H(u)=f(u)u+−g(u)u−≥0.

Applying Theorem (6.1) to inequality (6.42), from (6.2) with d = 1/2„ we deduce that,

(6.44) ∫ℝNhnϕR≤c1R−4∫ARun2.

Letting n → +∞, and taking into account (6.43) we have (6.36).

Next, since C(N, α) ≥ 0, from (6.28) we have

(6.45) ∫hnϕRrεN−α+12∫un2Δ2(rεα−N) ϕR ≤α|E1(∇un,rrεN−α+2,ϕR′)|+|E2(un,rεα−N,ϕR)|.

By using the estimates (6.30) and (6.32) with δ = 1 we have

∫hnϕRrεN−α+12∫un2Δ2(rεα−N) ϕR≤ ≤(N+1)(c1Rα−4⨍AR/2∪AR∪A2Run2−2Rα⨍B4Rhnϕ2R).

Letting n → ∞, and taking into account (6.43), we deduce

∫BRH(u)1rεN−α+12∫u2Δ2(rεα−N) ϕR ≤(N+1)(c1Rα−4⨍AR/2∪AR∪A2Ru2−Rα⨍B4RH(u)ϕ2R) ≤cRα−4⨍AR/2∪AR∪A2Ru2,

that is (6.37). Estimate (6.38) follows taking into account that Δ2(rεα−N)≥0 , and by the monotone convergence theorem by letting ɛ → 0.

The proof of (6.39) follows similarly to (6.37), from (6.45), by using estimates (6.29) and (6.31) with δ = 1, obtaining

∫hnϕRrεN−α+12∫un2Δ2(rεα−N) ϕR ≤(N+1)(c1Rα−4⨍B2Run2−Rα⨍B4Rhnϕ2R,),

and concluding again by letting, first n → ∞, and then ɛ → 0.

Proof of (6.40). Applying Theorem (6.1) to (6.42), from (6.1) we deduce that

(6.46) ∫hnφ≤c[∫un2|Δφ|2φ+∫un2|Δ|∇φ|2φ|+∫un2|∇φ|4φ3].

By letting n → ∞, we have

(6.47) ∫H(u)φ≤c[∫u2|Δφ|2φ+∫u2|Δ|∇φ|2φ|+∫u2|∇φ|4φ3].

Now, since

cq∫|u|q+1φ≤∫H(u)φ,

arguing as in the proof of Theorem 6.1, and using the same chain of inequalities (6.14), we get the estimates (6.40).

Plugging the second estimate of (6.40) in (6.37), we obtain (6.41), concluding the proof.

6.5 Liouville theorems: distributional solutions

In this section we shall prove some Liouville theorems within the class of distributional solutions for the problem (_gP_f ).

Theorem 6.12

Assume that f , g : ℝ → ℝ are continuous functions satisfying

f(t)t≥0, g(t)t≥0, for all t∈ℝ . (f0)

Let u be a distributional solution of

g(u)≥−Δ2u≥f(u) onℝN, (gPf)

such that u∈Llocs(ℝN) , 2 ≤ s ≤ +∞ and g(u),f(u)∈Llocs′(ℝN) .

Assume that

N=5,6,7 and limR→∞⨍AR|u|2=0,

(6.48) N≥8 and Rα1−4⨍AR|u|2≤C,for any R large,

where α₁ is defined in (5.37), then u ≡ 0 a.e. in ℝ^N.

Proof

We shall argue as in the proof of Theorem 6.4. Set α := 4 if N = 5, 6, 7 and α := α₁ in the remaining cases N ≥ 8.

Step 1. We begin proving the claim under the hypothesis

(6.49) limR→∞Rα−4⨍AR|u|2=0.

From (6.37), it follows that

(6.50) 12∫BRu2Δ2(rεα−N)≤cRα−4⨍AR/2∪AR∪A2Ru2.

Hence, by letting R → +∞ we get ∫ℝNu2Δ2(rεα−N)=0 , that is u ≡ 0 a.e. in ℝ^N.

Step 2. If (6.48) holds, from (6.50) by letting R → +∞ we have,

∫ℝNu2Δ2(rεα−N)<∞,

which implies

limR→+∞∫ARu2Δ2(rεα−N)=0.

Now, since in this case α = α₁ > 4, and Δ2(rεα−N)≥crα−4−N≥cRα−4−N on A_R, the hypotheses (6.49) holds and the claim follows from Step 1.

Remark 6.13

If N = 1, 2, 3, 4, Theorem 6.12 still holds provided

f(t)=0 if and only if g(t)=0,

and

limR→∞⨍AR|u|2=0.

Indeed from (6.36), we have

∫BRH(u)≤cRN−4⨍ARu2→0, as R→∞.

Hence f (u)u = 0 a.e. whenever u ≥ 0 and g(u)u = 0 a.e. for u < 0. Therefore g(u) = f (u) = 0 whenever u ≠ 0, and since f and g are continuous we deduce that g(u) = f (u) ≡ 0 a.e. Since u solves (_gP_f ) it follows that Δ²u = 0 in distributional sense. Hence by a standard argument u is smooth. Now by using Theorem 6.4 we achieve the claim.

Theorem 6.14

Assume that f , g : ℝ → ℝ are continuous functions satisfying

f(t)t≥0, g(t)t≥0, for all t∈ℝ . (f0)

Let H be defined as (1.6) and assume that

(6.51) H(t)≥cq|t|q+1, for all t∈ℝ , and for some q>1,cq>0.

Let u be a distributional solution of

g(u)≥−Δ2u≥f(u) onℝN, (gPf)

such that u∈Llocs(ℝN) , 2 ≤ s ≤ +∞, and f(u),g(u)∈Llocs′(ℝN) .

If N = 1, . . . , 7 and q > 1 or

N≥8 and 1<q≤qN,

where q_N is defined in (6.26), then u ≡ 0 a.e. in ℝ^N.

Proof

The proof is similar to the proof of Theorem 6.8. Let N = 1, . . . , 4. The claim in this case follows directly form (6.40). Indeed we have,

∫BR|u|q+1≤CRN−4q+1q−1→0 as R→+∞.

Let N ≥ 5. From the a priori estimate (6.40) we deduce that

Rα−4⨍AR|u|2≤Rα−4q+1q−1.

If N = 5, 6, 7, the choice α = 4 is admissible for an the application of Theorem 6.12 and since 4−4q+1q−1<0 the claim follows.

Let N ≥ 8. In this case, α = α₁ > 4 is admissible in Theorem 6.12 and since α1−4q+1q−1≤0 we conclude the proof.

Remark 6.15

Notice that if instead of the double inequality (_gP_f ) we deal with the one side inequality

(6.52) −Δ2u≥|u|q−1u onℝN,

it is easy to see that (6.52) admits the non trivial solution u(x)=1−x14/24 for any q > 0.

This example can be extended to the more general case of the one side inequality

(6.53) −Δ2u≥f(u) onℝN,

with f ∈ 𝒞(ℝ) satisfying the following assumption: there exists k > 0 such that

(6.54) f(t)≤f(k), ∀ t≤k.

Indeed (6.53) admits the solution u(x) = k − c |x|⁴ with c > 0 such that 8N(N + 2)c ≥ f (k). Examples of functions satisfying (6.54) are the nondecreasing functions, f (t) = |t|^q−1t + γ sin t for q > 0 and γ ∈ ℝ or f (t) = |t|^q−1t sin t for q > 0.

Corollary 6.16

Let f , g : ℝ → ℝ are continuous functions. Assume that H satisfies

H(t)>0 for t≠0,lim inft→0H(t)t2>0 (possibly +∞),lim inft→±∞H(t)|t|p+1>0 (possibly +∞) for some p>1.

Let u is a distributional solution of (_gP_f ) such that u∈Llocs(ℝN) , 2 ≤ s ≤ +∞ and f(u),g(u)∈Llocs′(ℝN) .

Then u ≡ 0 a.e. in ℝ^N.

Proof

Let q > 1 be such that q ≤ min{p, q_N} if N ≥ 8, while q := p for N = 1, . . . , 7.

From the hypotheses on H, it follows that H(t) ≤ c_q |t|^q+1 for any t ∈ ℝ and a suitable c_q > 0. An application of Theorem 6.14 completes the proof.

Example 6.17

When dealing with the equation

−Δ2u=f(u) onℝN,

we have that g = f and H(t) = f (t)t. Examples of functions f such that the corresponding function H satisfies (6.51) are the following.

f (u) = λ₁u + λ₂ |u|^p−1 u with λ₁, λ₂ > 0. In this case f satisfies (6.51) with any 1 < q ≤ p, and Corollary 6.16 applies.
Let p > 1 and λ > 0, the function f (u) := u + sin(u) + λ₂ |u|^p−1 u, satisfies (6.51) with any 1 < q ≤ p.
Let 0 < p₁ < p₂ < p₃, λ₁, λ₂ > 0, and μ ∈ ℝ. The functions f₁(u) := λ₁ |u|^p₁−1 u + μ |u|^p₂−1 u + λ₂ |u|^p₃−1 u, and f₂(u) := λ₁ |u|^p₁−1 u + μ |u|^p₂ + λ₂ |u|^p₃−1 u, satisfy (6.51) with any p₁ ≤ q ≤ p³, provided μ is small enough. In particular if 0 < p₁ ≤ 1 < p₃, then Corollary 6.16 applies.

6.6 A generalization

Results similar to Theorem 6.14 can be formulated for more general nonlinearities f and g. For instance when the nonlinearity f and g behave differently for positive and negative values of the independent variable, that is when

H(t)≥{c1|u|q+1, if t≥0,c2|u|p+1, if t<0,

with c₁, c₂ > 0. More generally we have.

Theorem 6.18

Let f, g : ℝ → ℝ be continuous functions. Assume that H satisfies

(6.55) H(t)≥cHmin{|t|q+1,|t|p+1}, ∀ t∈ℝ , for some q≥p>1,

with c_H > 0. Let u is a distributional solution of (_gP_f ) such that u∈Llocs(ℝN) , 2 ≤ s ≤ +∞ and f(u),g(u)∈Llocs′(ℝN) .

Then for large R > 0, (6.40) and (6.41) hold.

Moreover, if N = 1, . . . , 7 or N ≥ 8 and 1 < q ≤ q_N, then u ≡ 0 a.e. in ℝ^N.

Proof

Let

h(u):=min{|u|q+1,|u|p+1}={|u|q+1if |u|≤1,|u|p+1if |u|>1.

Arguing as in the proof of Theorem 6.11 we obtain inequality (6.47), which in turn yields

(6.56) cH∫h(u)φ≤∫ℝNH(u)φ≤c[∫u2|Δφ|2φ+∫u2|Δ|∇φ|2φ|+∫u2|∇φ|4φ3].

Next denoting with χ₁ and χ₂ the characteristic functions of Ω₁ := {x : |u(x)| ≤ 1} and Ω₂ := {x : |u(x)| > 1} respectively, we have h(u) = |u|^q+1 χ₁ + |u|^p+1 χ₂. Arguing as in the proof of (6.2), and with the same notation, we observe that each term in the right hand side of (6.56) has the form (6.13). By using Hölder and Young inequalities, with exponents x=q+12 and y=p+12 and parameter ɛ and δ, we have

∫ℝNH(u)φ≤c∫ARu2χ1|ψ|+∫ARu2χ2|ψ| ≤(∫|u|q+1χ1φ)1x(∫|ψ|x′φx′−1)1x′+(∫|u|p+1χ2φ)1y(∫|ψ|y′φy′−1)1y′ ≤ϵxxc1∫|u|q+1χ1φ+1x′ϵx′∫|ψ|x′φx′−1+δyyc1∫|u|p+1χ2φ+1y′δy′∫|ψ|y′φy′−1 ≤(ϵxxc1+δyyc1)∫h(u)φ+1x′ϵx′∫|ψ|x′φx′−1+1y′δy′∫|ψ|y′φy′−1.

Taking φ := ϕ_R = ϕ₁(|x| /R), and with a suitable choice of ɛ and δ, since q ≥ p, we have

(6.57) ∫ℝNH(u)φ≤cRN−4q+1q−1+cRN−4p+1p−1≤cRN−4q+1q−1,

which is the first estimate in (6.40). To obtain the second estimate in (6.40), by Hölder inequality and estimate (6.57), we argue as follows

(⨍BRu2)1/2≤(⨍BRu2χ1)1/2+(⨍BRu2χ2)1/2 ≤(⨍BR|u|q+1χ1)1q+1+(⨍BR|u|p+1χ2)1p+1 ≤(1cf⨍BRH(u))1q+1+(1cf⨍BRH(u))1p+1 ≤cR−41q−1+cR−4p+1q+1q−1≤cR−41q−1.

Finally, the claim follows arguing as in the proof of Theorem 6.14.

7 Further remarks and results on the solutions

The main purpose of this Section is to show further qualitative properties on the possible solutions of our prototype equation (1.3) and for more general problems. In order to simplify the presentation we consider only smooth solutions.

7.1 Representation formula for u²

Theorem 7.1

Let h ∈ 𝒞(ℝ^N) and u ∈ 𝒞⁴(ℝ^N) be such that −u Δ²u = h. Let x ∈ ℝ^N.

The representation formula

(7.1) −CN2u2(x)=∫ℝNh(y)|x−y|N−4dy+∫ℝN|Δu|2(y)|x−y|N−4dy +2(N−4)∫ℝN|∇u|2(y)|x−y|N−2dy −2(N−4)(N−2)∫ℝN(x−y⋅∇u(y))2|x−y|Ndy,

holds ^[2] provided one of the following assumptions is satisfied

1. h ≥ 0,
(7.2) limR⨍R<|x−y|<2Ru2(y)dy=0,
and
(7.3) ∫ℝN|∇u|2|x|N−2<∞.
2. (7.2) holds and
(7.4) ∫ℝN|h||x|N−4<+∞, ∫ℝN(Δu)2|x|N−4<+∞, and ∫ℝN|∇u|2|x|N−2<∞.
3. −Δ²u = f (u) with f (u)u ≥ c_q |u|^q+1 for some q > 1 and c_q > 0.

Moreover, if one of claims 1., 2. or 3. holds, then all the integrals in the representation formula (7.1) are finite.

Proof

By translation, it suffices to prove the claim for x = 0. From (5.31) of Proposition 5.13 with α = 4 and φ=φR=ϕR4 , we have

(7.5) ∫h|x|N−4φR+∫(Δu)2φR|x|N−4+ +2(N−4)∫|∇u|2φR|x|N−2+CN2u2(0)φR(0)=2(N−4)(N−2)∫(∇u⋅x|x|)2φR|x|N−2 −2(N−4)E1(∇u,|x|3−N,φR′)−E2(u,|x|4−N,φR).

The representation (7.1) will follows letting R → ∞ and showing that E1(∇u,|x|3−N,φR′)→0 and E₂(u, |x|^4−N , φ_R) → 0.

Estimates (6.30) and (6.32), by hypothesis (7.2) and h ≥ 0, assure that E1(∇u,|x|3−N,φR′)→0 and E₂(u, |x|^4−N , φ_R) → 0 as R → ∞. Furthermore, all the integrals in (7.5) have a limit as R → ∞ by monotone convergence theorem. All the integrals are finite since all of them are nonnegative and the integral in right hand side of (7.5) is convergent because of (7.3).
From the hypotheses, we deduce
limRR2⨍AR|∇u|2=0, and limRR4⨍AR|Δu|2=0.
Plugging this information in (5.4) and (5.29), we deduce respectively that E₁ → 0 and E₂ → 0, concluding the proof.
Arguing as in the case 1. from the estimates in Theorem 6.1 we deduce that E₁ → 0 and E₂ → 0, and the representation (7.1) holds. It remains to prove that the integrals are finite.

Plugging the estimate (6.40) in (6.7), we obtain
R2⨍AR|∇u|2≤cR−8q−1,
which in turn yields
∫B2R\BR|∇u|2|x|N−2≤ωN(2N−1)R2⨍AR|∇u|2≤cωN(2N−1)R−8q−1.

Let k ≥1, we have
∫B2k+1|∇u|2|x|N−2=∫B1|∇u|2|x|N−2+∑j=0k∫B2j+1\B2j|∇u|2|x|N−2 ≤∫B1|∇u|2|x|N−2+c∑j=0k(2j)−8q−1 <∫B1|∇u|2|x|N−2+c1−2−8q−1<∞.
By letting k → ∞, it follows that (7.3) holds.

The following Lemma, which we believe is interesting in itself, provides a sufficient condition for the validity of (7.4).

Lemma 7.2

Let h∈Lloc1(ℝN) be a nonnegative function and let u∈Hloc2(ℝN) be a solution of (P_h). Assume that there exist θ :]0, +∞[→]0, +∞[ be such that for R > 0 large there holds

(7.6) ⨍ARu2≤θ(R) for R large.

If ∑j=0∞θ(2j)<∞ , then

(7.7) h|x|N−4, (Δu)2|x|N−4, |∇u|2|x|N−2∈L1(ℝN\B1).

Moreover

1. If h∈Lloc∞ , we have h|x|N−4,∈L1(ℝN) .
2. If u ∈ 𝒞⁴(ℝ^N), then the functions in (7.7) belong to L¹(ℝ^N).

Remark 7.3

Special cases of solutions of (P_h) that fulfill the hypotheses of the above Lemma are the following.

1. By simple computation it is easy to see that if there exist c, σ > 0 such that
(7.8) ⨍ARu2≤cR−σ,
then Lemma 7.2 applies and (7.7) holds.
2. From the a priori estimates in Theorem (6.8) and (6.9), it is immediate to verify that if h ≥ c |u|^q+1 with q > 1 and c > 0, then Lemma 7.2 applies and (7.7) holds and |u|q+11+|x|N−4∈L1(ℝN) .

Proof

Let us prove that h|x|N−4∈L1(ℝN\B1) . The proofs of the other claims follow similarly.

Since h|x|N−4 is nonnegative it is enough to show that R↦∫BR\B1h1+|x|N−4 is bounded for a subsequence R_k → ∞.

To this end we choose R_k := 2^k+1. We have,

∫BRk\B1h|x|N−4=∑j=0k∫B2j+1\B2jh|x|N−4.

Now, since by using (6.5) and the assumption (7.6) each addendum of the right hand side of the above identity can be estimate as

∫B2R\BRh|x|N−4≤ωNR4|B2R|∫B2R\BRh≤ωNc1−∫A2Ru2≤cθ(2R),

it follows that

∫BRk\B1h|x|N−4≤∑j=0kc θ(2j+1)≤c∑j=0∞θ(2j)<∞.

This completes the proof of the first claim in (7.7).

The proof of other cases follows by using similar argument and the estimates (6.5) and (6.7).

7.2 Remarks on the sign of the solutions

Theorem 7.4

Let u ∈ 𝒞⁴(ℝ^N) be a solution of (_gP_f ) with f , g continuous functions and H defined as in (1.6) and satisfying (6.55). If one of the following condition holds

1. u has a sign;
2. Δu has a sign;
3. uΔu has a sign;

then u ≡ 0 in ℝ^N.

Proof

Clearly the interesting cases occour for N ≥ 8. From Theorem 6.18,we know that (6.40) holds. Plugging (6.40) in (6.2) and since the problem is invariant by translation, we deduce

(7.9) limR⨍BR(x)|u|2=0, and limR⨍BR(x)(Δu)2=0, for any x∈ℝ .

Which in turns implies

(7.10) lim infR⨍BR(x)|u|=0.

The above limits will play a crucial role to get information on the sign of the solutions.

1. Without loss of generality we may assume that u ≥ 0. In this case u solves

−Δ2u≥f(u)≥cHmin{|u|q,|u|p}≥0.

Arguing as in the proof of Theorem 3.1, since −u satisfies the ring condition (7.10), from the representation formula of Theorem A.1, it follows that u ≤ 0 ≤ u. Hence u ≡ 0 in ℝ^N.

In the case u ≤ 0 we deduce that −u solves

−Δ2(−u)≥−g(u)≥cHmin{|u|q,|u|p}≥0,

which by the same argument above yields −u ≤ 0.

2. Without loss of generality we assume that −Δu ≥ 0. Since u is superharmonic and satisfies the ring condition (7.10), arguing as in the proof of Theorem 3.1, by the representation formula of Theorem A.1, it follows that u is nonnegative and by point 1. we get the claim (the case Δu ≥ 0 can be handled similarly).

3. First we consider the case uΔu ≥ 0. Indeed, by (6.10) with Φ = u we have that Δu² ≥ 0. Since u² satisfies (7.9), by the representation formula of Theorem A.1, we obtain that u² ≤ 0 and hence the claim.

Assume that −uΔu ≥ 0. Observing that −u, −f (u), −g(u) and Δ²u have the same sign, we deduce that Δ²u Δu ≥ 0. Therefore from (6.10) with Φ = Δu we have that Δ(Δu)² ≥ 0. Since (Δu)² satisfies the ring condition (7.9), again by the the representation formula of Theorem A.1, we deduce Δu = 0 that is u is harmonic, which implies that u ≡ 0 because u satisfies the ring condition (7.10).

The statements 1. and 2. of Theorem 7.4 and their proofs still hold for the higher order problem

(7.11) (−Δ)mu+|u|q−1u=0 onℝN,

as well as for its generalizations in the same spirit of (_gP_f ).

7.3 Uniqueness

Theorem 7.5

Let f ∈ 𝒞(ℝ) be a continuous function such that

(7.12) (f(t)−f(s))(t−s)≥cf|t−s|q+1, ∀ t,s∈ℝ ,

for some q > 1 and c_f > 0. Let k ∈ 𝒞(ℝ^N) and let u ∈ 𝒞⁴(ℝ^N) be a solution of

(7.13) −Δ2u=f(u)+k onℝN.

If N = 1, . . . , 7 or N ≥ 8 and 1 < q ≤ q_N, where q_N is defined in (6.26), then u is unique.

Proof

Let u and v be solutions of (7.13) and set w := u − v. We have that w solves the problem

−Δ2w w=((−Δ2u)−(−Δ2v))(u−v)=(f(u)−f(v))(u−v)≥cf|u−v|q+1=cf|w|q+1.

From Theorem 6.8 we get the claim.

Remark 7.6

Condition (7.12) implies that f is increasing. Without the increasing property of f , the uniqueness results is in general false. Indeed, let f be defined as

f(t):={119t,if 0<t<9,10−|t−10|N+4N−4−1(t−10),if 9≤t≤11,9+(t−11)q,if t>11,−|t|q,if t≤0,

where q is any number q > 1. See Figure 1. Clearly this f is not increasing and hence (7.12) does not hold, while f satisfies

f(t)t≥cq|t|q+1, ∀ t∈ℝ , and cq>0 sufficiently small.

The problem

−Δ2u=f(u)−10, onℝN, N>4,

admits the constants solution u(x) = 10 as well as the function

v(x)=10+c(ε2+|x|2)N−42

for suitable ɛ, c > 0 (namely, c^8/(N−4) = (N − 4)(N² − 4)Nɛ⁴ and ɛ⁴ ≥ (N − 4)(N² − 4)N).

Fig. 1

f (t) in continuous line, c|t|^q−1t in dotted line.

Remark 7.7

(On the symmetry preserving property). As usual the uniqueness result implies several symmetry properties on the solutions of (7.13). For instance, assuming that f satisfies the hypotheses in the uniqueness result Theorem 7.5, we have

if k ∈ 𝒞(ℝ^N) is a radial function, then the solution of (7.13) is radial;
if k ∈ 𝒞(ℝ^N) is even is some direction, then the solution of (7.13) shares the same symmetry;
if k ∈ 𝒞(ℝ^N) depends only on j < N variables, say x₁, . . . , x_j, then also the solution of (7.13) depends only on x₁, . . . , x_j.

Remark 7.8

(On the sign preserving property). The prototype case related to (7.13) for the second order case is

Δu=|u|q−1u−k.

In this case the problem present a sign preserving property, namely, if k is nonnegative then also the solution is nonnegative. See for instance [9] where a discussion of the quasilinear case is presented.

For the higher order case this property cannot be expected (in general the maximum principle fails and a Kato's inequality does not hold). Indeed, for instance, consider the problem

(7.14) −Δ2u=|u|q−1u−k,

and for simplicity consider the 1-dimensional case N = 1 (by a lifting argument our examples are still valid in higher dimension). Choosing k(x) = 1, the only solution of (7.14) is the constant function u(x) = 1, which has the same sign of k. While by choosing k(x) = |x⁴ − 1|^q−1(x⁴ − 1) + 24, which is positive, it follows that (7.14) is solved by the changing sign function u(x) = x⁴ − 1.

A Representation formula

Here we state some results from [6] for the reader convenience. The main equation is

(A.1) (−Δ)mu=μ onℝN,

where μ is a positive Radon measure.

Theorem A.1

Let m ≥ 1 be an integer and N > 2m. Let μ be a positive Radon measure on ℝ^N and l ∈ ℝ. The following statements are equivalent:

a) u is a distributional solution of (A.1) and for a.e. x ∈ ℝ^N ,
lim infR→+∞1RN∫R≤|x−y|≤2R|u(y)−l|dy=0.
b) u is a distributional solution of (A.1), essinf u = l and u is weakly superharmonic.
c) u∈Lloc1(ℝN) and we have
u(x)=l+c(2m)∫ℝNdμ(y)|x−y|N−2m a.e. x∈ℝN,
where, for general α > 0 with 0 < α < N, c(α):=Γ(N−α2)2απN/2Γ(α2) .

Moreover, if a), b) or c) holds, then for i = 1, . . . , m the distribution (−Δ)ⁱu is a positive Radon measure and can be represented by

〈(−Δ)iu,φ〉=∫ℝNu(−Δ)iφ=c(2(m−i))∫ℝNφ(x)∫ℝNdμ(y)|x−y|N−2(m−i)dx.

Acknowledgments

The authors thank the referees for the careful reading of the manuscript and for suggesting the possible connections with the papers [3], [28], [29], as well as with [14]. The authors acknowledge the support of FRA 2018 Università di Trieste.

Conflict of interest:
Authors state no conflict of interest.

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Received: 2021-11-16

Accepted: 2021-11-21

Published Online: 2022-02-14

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