Exhaustive existence and non-existence results for some ... - arXiv

EXHAUSTIVE EXISTENCE AND NON-EXISTENCE RESULTS FOR SOME PROTOTYPE POLYHARMONIC EQUATIONS

arXiv:1802.05956v1 [math.AP] 16 Feb 2018

´ˆ ANH NGO, ˆ VAN HOANG NGUYEN, QUOC HUNG PHAN, AND DONG YE QUOC

A BSTRACT. We consider the polyharmonic equations ∆m u = ±uα in Rn with n > 1, m > 1 and α ∈ R. We study the existence of entire non-trivial non-negative solutions and/or entire positive solutions. In each case, we provide necessary and sufficient conditions on the exponent α to guarantee the existence of such classical solutions in Rn .

1. I NTRODUCTION In this paper, we are interested in the existence and non-existence results for the following equations ∆m u = ±uα in the whole Euclidean space Rn , (1.1) where n, m > 1, and α ∈ R is a parameter. It is easy to see that equations (1.1) can be rewritten in the form (−∆)m u = ±uα . However, we intend to keep rather the notation ∆m instead of (−∆)m for the convenience of presentation. Among others, one basic reason that we are interested in such equations is that (1.1) are the simplest models of polyharmonic equations with power-type nonlinearity. In the literature, equations of the form (1.1) have attracted much attention in various mathematical directions, including existence and non-existence results, the multiplicity, the regularity, the stability of solutions, the asymptotic behaviors at infinity of entire solutions, as well as Liouville-type results etc. For interested readers, we refer to the monograph [GGS10] for further motivations and results. The exponent α here can take any value in R. Regarding the nonlinearity uα , it is usually called superlinear, sublinear, or singular respectively if α > 1, α ∈ (0, 1), or α < 0. As we can imagine, the existence of solutions to (1.1) strongly depends on the range of the exponent α. It is well known that the semilinear polyharmonic equations arise in many physics phenomena. For example, several particular cases of (1.1) have their origins such as the elasticity, the equilibrium states in thin films, the modeling of electrostatic actuations etc. The equations (1.1) have also their root in conformal geometry. The equation n+2

−∆u = k(x)u n−2 with n > 3 is closely related to the famous Yamabe problem and the prescribing scalar curvature problem. The geometric aspect of higher order cases m > 2 are related the problem of prescribing Q-curvature on Riemannian manifolds. Loosely speaking, given a Riemanng ian manifold (Mn , g) of dimension n, we denote by Pm the GJMS operator of order m, Date: 19th Feb, 2018 at 01:24. 2010 Mathematics Subject Classification. Primary 35B53, 35J91, 35B33; Secondary 35B08, 35B51, 35A01. Key words and phrases. Polyharmonic equation; Existence and non-existence; Liouville type results. 1

2

ˆ V.H. NGUYEN, Q.H. PHAN, AND D. YE Q.A. NGO,

constructed by Graham–Jenne–Mason–Sparling in the celebrated work [GJMS92]. The prescribing Q-curvature problem asks us to look for a positive solution u to the following partial differential equation on Mn n+2m

g P2m (u) = Q(x)u n−2m .

Under conformal projection or as the limit equation of blow-up analysis, we are often led to understand the problems like n+2m

(−∆)m u = ±u n−2m in Rn , so a special case of (1.1). If the second order case m = 1 is well understood, the situation of polyharmonic problems m > 2 is much less clear. For example, as far as we know, we cannot find exhaustive results on the existence or non-existence of positive solutions to (1.1) for all exponents α ∈ R. To be clear, by solutions in this paper, we mean the classical solutions. Our main purpose here is to give a complete answer to this question, that is, to establish the existence or the non-existence of positive solutions to (1.1) for any m, n > 1 and α ∈ R. We will handle also the case of non-trivial non-negative solutions when α > 0, with the natural convention 00 = 1. In other words, we will find the necessary and sufficient conditions on the exponent α to confirm the non-existence, i.e. the Liouville-type results for positive solutions with real exponents α; and the Liouville-type results for non-negative solutions in Rn provided α > 0. Such results are sometimes called optimal Liouville-type theorems. The reason to consider separately the two classes of solutions is due to the lack of the strong maximum principle for high order elliptic operators, when m > 2. In recent years, the Liouville property has emerged as an important subject in the analysis of nonlinear partial differential equations. In particular, Polacik, Quittner and Souplet [PQS07a, PQS07b] developed a general method to derive universal pointwise estimates of local solutions from Liouville-type results. Their approach is based on rescaling arguments combined with a key doubling property, which is different from the classical rescaling method of Gidas and Spruck [GS81b]. It turns out that one can obtain from Liouville-type theorems a variety of results on qualitative properties of solutions, such as a priori estimates, universal bounds, universal singularity and decay estimates... For this reason, we expect to see many applications of Liouville-type theorems obtained here. Before closing this section, we would like to mention the outline of the paper. The next section is devoted to the statement of our main results, which consist of two theorems. Theorem 1 concerns the solvability of (1.1) with a negative sign, that is ∆m u = −uα , while Theorem 2 concerns solutions to ∆m u = uα . The proofs of Theorems 1 and 2 are presented in Section 4, where we used several important approaches, including a priori integral estimates derived for local solutions, interpolation inequalities, the comparison principle for radial solutions, the derivation of sub/super polyharmonic properties and the Moser’s iteration. 2. S TATEMENT OF

MAIN RESULTS

2.1. Some known results. Let us start by reviewing some well-known results concerning the existence and non-existence of solutions to the problems (1.1). We recall the Sobolev exponent   n + 2m if n > 2m + 1, pS (m) = n − 2m (2.1) ∞ if n 6 2m.

EXHAUSTIVE EXISTENCE AND NON-EXISTENCE RESULTS FOR ∆m u = ±uα IN Rn

3

The first known result is for positive solutions to (1.1) in the singular case α < 0. Proposition A. Let m > 2 be an integer and n > 3. Assume α < −1/(m − 1). Then both problem in (1.1) possesses positive solutions. Proposition A was proved by Kusano, Naito and Swanson via the Schauder–Tychonoff fixed-point theorem. More precisely, as a special case of Theorem 1 in [KNS88], if n > 3 and Z ∞ t(1 + t2m−2 )α dt < ∞, (2.2) 0

then (1.1) possesses infinitely many positive radial solutions satisfying the growth condition C1 (1 + |x|2m−2 ) 6 u(x) 6 C2 (1 + |x|2m−2 ) in Rn , where C1 and C2 are positive constants. It is obvious that the integral in (2.2) is finite when α < −1/(m − 1). Remark 2.1. We stress that the restriction on dimension n > 3 in Proposition A is necessary for problem ∆m u = −uα ; see the Proposition 4.1 below for the nonexistence result when n 6 2. Next we collect some known results for the equation (−∆)m u = uα

in Rn ,

(2.3)

in the superlinear case α > 1. These results can be summarized as follows. Proposition B. Let m be a positive integer. We have the following claims: (i) If 1 < α < pS (m) then the problem (−∆)m u = uα has no non-trivial nonnegative solution in Rn . (ii) If n > 2m and α > pS (m), then the problem (−∆)m u = uα possesses positive radial solutions in Rn . Let us comment on Proposition B. First for the Part (i) on the subcritical exponent range, the second order case, i.e. m = 1 was first established by Gidas and Spruck in [GS81a] via the technique of nonlinear integral estimates and the Bochner formula. Chen and Li gave a different proof in [CL91], using the moving plane method combined with the Kelvin transform. Later on, Lin resolved in [Lin98] the case m = 2, it was generalized finally by Wei and Xu in [WX99] for any m > 2 via the argument of moving planes. For n 6 2m and arbitrary m, the non-existence result (i) can be deduced from the method of rescaled test-function [MP01]; and also from the method of representation formula as presented in [CDM08]. Now we turn to the Part (ii). The case m = 1 can be proved easily by applying the Pohozaev identity [Poh65] with radial solutions on balls, or it can be deduced from the shooting argument [JL73]. In addition, if α = pS (1) = (n + 2)/(n − 2), the critical exponent, it was showed by Caffarelli, Gidas and Spruck in [CGS89] that any positive n+2 solution of −∆u = u n−2 in Rn (n > 3) is radially symmetric, up to a translation. This result was extended to the case of biharmonic equation by Lin [Lin98] and to the case m > 2 by Wei and Xu [WX99]. More precisely, it was shown in [WX99] that any positive solution u to n+2m (−∆)m u = u n−2m in Rn with n > 2m > 2 is of the following form n−2m 2 2λ u(x) = with x0 ∈ Rn , λ > 0. 1 + λ2 |x − x0 |2


4

For supercritical exponents α > pS (m) with m > 2, the existence of positive solutions to (2.3) was completely resolved by Liu, Guo and Zhang in [LGZ06, Theorem 1.1]. They used a combination of the shooting method together with degree theory and the Pohozaev identity. However, as already mentioned, it seems that there did not exist exhaustive study on the existence or non-existence of solution to (1.1) for all m > 2, n > 1 and α ∈ R. Our aim in this paper is to consider all the situations, and determine whether positive or non-negative solutions of (1.1) exist. For the sake of transparent presentation, we shall present our results in two different subcases according to the sign of the right-hand side. We will see that the range of α insuring the existence of solutions to equations (1.1) strongly depends on the parity of m. 2.2. Problem ∆m u = −uα in Rn . As mentioned above, the results depend on the parity of m, it’s more convenient to split the study for two equations: ∆2k u = −uα

in Rn

(P− 2k )

and ∆2k−1 u = −uα

in Rn ,

(P− 2k−1 )

where k is a positive integer. As far as we know, for (P− 2k ), there are many results only if k = 1, i.e. for the biharmonic equation ∆2 u = −uα

in Rn .

(2.4)

Here, the non-existence of positive solutions to (2.4) with α ∈ [−1, 0] was first proved by Choi and Xu in [CX09, Theorem 1.1] for dimension n = 3. This result was extended to all dimensions by Lai and Ye in [LY16, Theorem 1.3]. On the other hand, Proposition A (see also [MR03, Theorem 3.1]) ensures the existence of positive solutions for any α < −1. For the problem (P− 2k−1 ), when k = 1, it is worth noticing that the class of positive solutions coincides with the one of non-trivial non-negative solutions, due to the strong maximum principle. For k = 1, the non-existence of positive solution when α 6 1 is a consequence of result of Armstrong and Sirakov in [AS11], where the author made use of a new argument of the maximum principle; while the situations α > 1 are well known. Recently, it was proved in [DN17] that ∆3 u = −uα has no positive solution in R3 if α ∈ [−1/2, 0). We give here a complete answer to the question of existence for ∆m u = −uα . Theorem 1. Let m, k ∈ N∗ . We have then: (i) The problem (P− 2k ) possesses a positive solution if and only if n > 3 and α < −1/(2k − 1). (ii) The problem (P− 2k−1 ) possesses a positive solution if and only if n > 3 and α satisfies α < −1/(2k − 2) or α > pS (2k − 1). (iii) The problem (P− 2k ) has no non-trivial non-negative solution for any α > 0. (iv) The problem (P− 2k−1 ) possesses a non-trivial non-negative solution if and only if α > pS (2k − 1). The existence of positive or non-trivial non-negative solution to ∆m u = −uα can be summarized in the following table.


α< u > 0, n 6 2

u > 0, n > 3

−1 m−1

−1 m−1

6α pS (m)

Prop. 4.3

Props. B, 4.4

u 0

TABLE 1. Existence results for problem ∆m u = −uα in Rn Clearly, our contributions in Theorem 1 are multifold. • By determining the sign of ∆m−1 u with Lemma 3.3, we show quickly the nonexistence of positive solution to ∆m u = −uα in Rn with n = 1, 2, for any α ∈ R; see Proposition 4.1. • For any m > 1 and n > 3, we obtain the non-existence of positive solution in the range α ∈ [−1/(m − 1), 1]; see Propositions 4.2 and 4.3. • We obtain the non-existence of non-trivial non-negative solution in the range α ∈ [0, ∞) for the equation (P− 2k ); see Propositions 4.3 and 4.4. The proof of the non-existence in the singular case α ∈ [−1/(m − 1), 0) with n > 3 relies on the convexity of uα and comparison principle; In the superlinear case α > 1 for (P− 2k ), we made use of the integral estimate and a Liouville type result; see Lemma 3.4. However, the case α ∈ [0, 1] is significantly more delicate, it seems that many wellknown approaches, such as the standard rescaled test-function method [MP01], the moving plane technique [CL91], the argument of maximum principle [AS11], the representation formula method [CDM08], or the derivation technique of super/sub polyharmonic property [Lin98, WX99, CL13, Ngo17], cannot be applicable. Our proof in the sublinear case is inspired by the idea of Serrin and Zou [SZ96] and Souplet [Sou09], it is based on the integral estimates and Moser’s iteration method. 2.3. Problem ∆m u = uα in Rn . Again, we will split our study following the parity of m, that is, ∆2k u = uα

in Rn

(P+ 2k )

and ∆2k−1 u = uα

in Rn ,

(P+ 2k−1 )

where k is a positive integer. Here are our main results. Theorem 2. Let k be a positive integer. Then we have the following claims. (i) (ii) (iii) (iv)

The problem (P+ 2k ) possesses positive solutions if and only if α 6 1 or α > pS (2k). The problem problem (P+ 2k−1 ) possesses positive solutions if and only if α 6 1. ) has no non-trivial non-negative solutions if 1 < α < pS (2k). The problem (P+ 2k The problem (P+ ) has no non-trivial non-negative solution if α > 1. 2k−1

The results of Theorem 2 are summarized in the following table.


6

u>0

α pS (m)

Prop. 4.5

Props. B, 4.4

u 0

TABLE 2. Existence results for the problem ∆m u = uα in Rn Our contribution are twofold here. • We give a unified proof of the existence of positive solutions for all α 6 1; see Proposition 4.5. • We prove the non-existence of non-trivial non-negative solutions for (P+ 2k−1 ) with any α > 1; see Proposition 4.4. In the second order case, it is known that the problem ∆u = uα in Rn has no positive solution if α > 1, but it possesses a positive one if α 6 1. More precisely, the nonexistence for the superlinear case α > 1 is a consequence of the so-called Keller–Osserman criteria developed by Keller [Kel57] and Osserman [Oss57]. This theory was employed in [Bre84, Lemma 2] to show that the equation ∆u = uα admits no non-trivial non-negative entire solution whenever α > 1. When α 6 1, the existence of radial solutions can be easily obtained by the monotonicity of u(r). In what follows, the notation ∆i u stands for u when i = 0. Also, we denote always by u(r) the spherical average of u centered at the origin on the sphere ∂Br , that is Z 1 udσ. u(r) = |∂Br | ∂Br

When u is a radial function, we use also the notation u(r). The symbol C denotes a generic positive constant whose value could be different from one line to another. 3. P RELIMINARIES Here are some basic results, which will be useful for our analysis. Lemma 3.1. Let m > 1 and v1 , v2 : BR → I ⊂ R be two C 2m radial functions verifying ∆m v1 > f (v1 ),

∆m v2 6 f (v2 ) in BR

and ∆i v1 (0) > ∆i v2 (0), ∀ 0 6 i 6 m − 1. If f is non-decreasing in I, then v1 > v2 in BR . In other words, v1 (r) > v2 (r) for all r ∈ [0, R). The above comparison principle is a special form of more general well-known results; see for instance [FF16, Proposition A.2] or [LS09, Remark 2.3]. An easy consequence is Lemma 3.2. Let u be in C 2m (Rn ) satisfying ∆m u 6 0 in Rn , then we have u(r) 6 u(0) +

m−1 X i=1

Q

∆i u(0)r2i , 16k6i [2k(n + 2k − 2)]

∀ r > 0.

(3.1)


7

Proof. Let Φ be the radial function defined by Φ(r) := u(0) +

m−1 X i=1

There hold then

∆i u(0)r2i . 16k6i [2k(n + 2k − 2)]

Q

∆m Φ ≡ 0 > ∆m u = ∆m u

in Rn

and ∆i Φ(0) = ∆i u(0) = ∆i u(0) for any 0 6 i 6 m − 1. Applying Lemma 3.1 with f ≡ 0, there holds u 6 Φ in Rn . The following result is a simple but important fact of our approach. Lemma 3.3. Let m > 1. Then we have the following claims: (i) If u be a positive function satisfying ∆m u < 0 in Rn , then ∆m−1 u > 0 in Rn . (ii) If u be a non-negative function satisfying ∆m u 6 0 in Rn , then ∆m−1 u > 0 in Rn . Proof. Consider first the claim (ii). Set w = ∆m−1 u, suppose that there was a point x0 ∈ Rn such that w(x0 ) < 0. By translation, we may assume that x0 = 0. Moreover, it follows from Lemma 3.2 that u satisfies the estimate (3.1). As ∆m−1 u(0) < 0, there holds u(r) < 0 for r large enough. This is impossible because u is non-negative in Rn . The point (ii) holds true. Now we consider (i). Set again w = ∆m−1 u, we have w > 0 in Rn by (ii). If w vanishes at some x1 ∈ Rn , then w attains its minimum at x1 . However, this contradicts the fact that ∆w(x1 ) < 0, we are done. The following Liouville type result is a crucial step in the proof of Proposition 4.4. Lemma 3.4. Assume that u is a C 2m non-negative function in Rn , verifying (−∆)m u 6 0 and Z udx = o(Rn ), as R → ∞. (3.2) BR

n

Then u ≡ 0 in R . Proof. Let vk = (−∆)k u, for 0 6 k 6 m. We shall prove by backward induction that vk 6 0 in Rn

(3.3)

for k = m, m − 1, . . . , 0. It is obvious that (3.3) is true for k = m. Suppose now (3.3) is true for j + 1 6 k 6 m with some j > 0, we shall show that vj 6 0 in Rn . We have two possible cases. Case 1: j is odd. As ∆j+1 u = vj+1 6 0, Lemma 3.3 (ii) gives vj 6 0. Case 2: j is even. We will prove vj 6 0 by contradiction. Assume that there exists x0 ∈ Rn such that vj (x0 ) > 0. Up to a translation, let x0 = 0. Then ∆v j = ∆vj = −v j+1 > 0

in Rn .

We have then v ′j (r) > 0 for any r > 0, hence v j (r) > v j (0) = vj (0) > 0. Let ψ be a smooth, radial, cut-off function satisfying 0 6 ψ 6 1 and ( 0 if |x| > 2, ψ(x) = (3.4) 1 if |x| 6 1.


8

For any R > 0, Z Z Z x x v j (x)ψ v j (x)dx > v j (0)Rn ωn , vj (x)ψ dx = dx > R R Rn Rn BR

(3.5)

where ωn denotes the volume of the unit ball B1 ⊂ Rn . On the other hand, there holds Z Z x x vj (x)ψ (−∆)j u(x)ψ dx = dx R R Rn Rn Z x = R−2j u(x)(−∆)j ψ dx (3.6) R Rn Z 6 CR−2j u(x)dx. B2R

Putting (3.5) and (3.6) together gives 0 < v j (0) 6 CR−2j−n

Z

u(x)dx.

B2R

Letting R → ∞ and using (3.2) we meet a contradiction. We get then vj 6 0 in Rn . Therefore, by induction principle, (3.3) is true as claimed. Taking k = 0 in (3.3), we have u 6 0 in Rn , hence u ≡ 0. The last result in this subsection is a classical interpolation-type estimate. Lemma 3.5. Let m be a positive integer. Let z be a function in W 2m,ℓ (B2R ) for some ℓ > 1. Then for any exponent q > 1 such that 1 2m 1 > − , q ℓ n there holds Z Z 1/q n n z q dx 6 CR q +2m− ℓ

B2R

BR

where C = C(m, n, ℓ, q).

Z 1/ℓ n |∆m z|ℓ dx + CR q −n

B1

and

Z

zdx,

(3.8)

B2R

Proof. By the dilation w(x) = z(Rx), we obtain Z Z Z wq dx, z q dx = Rn BR

(3.7)

zdx = Rn

wdx,

B2

B2R

|∆m z|ℓ dx = R−2mℓ+n

Z

Z

|∆m w|ℓ dx.

B2

B2R

From these identities, it’s equivalent to prove that for w ∈ W 2m,ℓ (B2 ), kwkLq (B1 ) 6 Ck∆m wkLℓ (B2 ) + CkwkL1 (B2 ) .

(3.9)

This follows from (3.7) and standard elliptic estimate; see for instance [GGS10, Theorem 2.20]. The Lemma is proved. 4. P ROOF

OF MAIN RESULTS

This section is devoted to the proof of our main results. We prove some Liouville type results in subsections 4.1 and 4.2, while some existence results are proved in subsection 4.3. It is worth noticing that for each case in Tables 1 and 2 above, we have already included the name of the main proposition yielding the result in the case. Therefore, there is no need to write a proof for Theorems 1 and 2.


9

4.1. Non existence results for ∆m u = −uα . This subsection is devoted to the nonexistence results in Theorem 1, and we do not consider specially the situations under applications of Propositions A and B. 4.1.1. For dimension 1 and 2. Let us start with the case n 6 2. We will prove that in dimension one and two, the equation in Rn

∆m u = −uα

(4.1)

has no positive solution for any α ∈ R; and has no non-trivial non-negative solution for any α > 0. In fact, these claims are trivial consequence of the following result. Proposition 4.1. Let n 6 2. If u is a non-negative C 2m function verifying ∆m u 6 0 in Rn , then ∆m u ≡ 0 in Rn . Proof. As ∆m u 6 0 in Rn , Lemma 3.3 (ii) shows that ∆m−1 u =: w > 0. This means that w is a non-negative super-harmonic function in Rn . It is well-known that such w must be constant in Rn if n 6 2, hence ∆m u = ∆w ≡ 0. 4.1.2. For negative values of α. Here we prove the non-existence of positive solution to (4.1) for n > 3 and suitable negative values of α. Proposition 4.2. Let n > 3. Then the equation (4.1) has no positive solution for any α ∈ [−1/(m − 1), 0) if m > 1; or for any α < 0 if m = 1. Proof. Assume that n > 3, m > 1 and α ∈ [−1/(m − 1), 0] ∩ R. By way of contradiction, suppose that u is a positive solution to (4.1). Using Lemma 3.2, we have u(r) 6 u(0) +

m−1 X i=1

Q

∆i u(0)r2l , 16k6i [2k(n + 2k − 2)]

∀ r > 0.

Hence, there exists a constant C > 0 such that u(r) 6 Cr2(m−1)

for any r > 1.

Set w = ∆m−1 u. By Lemma 3.3 (i), there holds w > 0. Moreover, as uα is convex in (0, ∞) with α 6 0, Jensen’s inequality implies −∆w = uα > uα

in Rn ,

so that ′ − rn−1 w ′ (r) > rn−1 uα (r) > Crn−1+2(m−1)α ,

∀ r > 1.

Integrating over (1, r), taking into account n > 3 and (m − 1)α > −1, there holds w ′ (1) − rn−1 w ′ (r) > Cr2(m−1)α+n − C. Therefore, w ′ (r) 6 −Cr2(m−1)α+1 + Cr−n+1 ,

∀ r > 1.

(4.2)

If α ∈ (−1/(m − 1), 0), then integrating (4.2) over [1, r] gives w(r) − w(1) 6 −Cr2(m−1)α+2 + C,

∀ r > 1.

We have then w(r) → −∞ as r → ∞, which is a contradiction with w > 0. If α = −1/(m − 1), then integrating of (4.2) over [1, r] gives Z r Z r −1 w(r) − w(1) 6 −C r dr + Cr−n+1 dr = −C log r + C, 1

1

which implies also w(r) → −∞ as r → ∞. We reach again a contradiction.


10

4.1.3. For 0 6 α 6 1. Proposition 4.3. For any m, n > 1 and α ∈ [0, 1], the equation (4.1) has no non-trivial non-negative solution. Proof. When α = 0, as ∆m u ≡ −1, the non-existence of non-negative solutions in Rn is a direct consequence of (3.1). From now on, α ∈ (0, 1]. For convenience, we divide the proof into three steps. Step 1. Suppose that u is a non-trivial non-negative solution to ∆m u = −uα in Rn . By Lemma 3.2, we have u(R) 6 CR2(m−1)

for R > 1.

Hence Z

udx =: F (R) 6 CRn+2(m−1)

for any R > 1.

(4.3)

B2R

Here C is a constant independent of R. We use the rescaled test-function argument to estimate F (R). Let ψ be a smooth cut-off function satisfying 0 6 ψ 6 1 and (3.4). For any R > 0, let x φR (x) = ψ 2m+1 . R There holds then x |∆m φR (x)| = R−2m ∆m (ψ 2m+1 ) R x 1/(2m+1) 6 CR−2m ψ (x). = CR−2m φR R Hence Z Z α −∆m u φR dx u φR dx = Rn Rn Z Z 1/(2m+1) m −2m = −u ∆ φR dx 6 CR dx. u φR Rn

Rn

This yields

Moreover, we claim that

Z

uα dx 6 CR−2m F (R),

∀ R > 0.

(4.4)

BR

∃ M > 0 and Ri → ∞ such that F (2Ri ) 6 M F (Ri )

∀ i.

(4.5)

Indeed, assume that (4.5) is false. Let us fix M0 > 2n+2m−2 , then there exists R0 > 0 such that F (2R) > M0 F (R),

∀ R > R0 .

(4.6)

Let R1 > 1 be sufficiently large verifying F (R1 ) > 0, such a R1 exists since u is nontrivial. Denote R∗ := max{R1 , R0 }. Iterating (4.6) and thanks to (4.3), we arrive at, for any i, M0i F (R∗ ) 6 F (2i R∗ ) 6 C(2i R∗ )n+2m−2 , that is

M0 2n+2m−2

i

6

CR∗n+2m−2 , F (R∗ )

∀ i.

But this is a just impossible when i → ∞ by the choice of M0 . So the claim (4.5) holds true.


11

Step 2. Consider first the case α = 1. It follows from (4.4) with R = 2Ri and (4.5) that F (Ri ) 6 CRi−2m F (2Ri ) 6 CM Ri−2m F (Ri ),

∀ i.

Therefore F (Ri ) = 0 for i large enough because Ri → ∞, which is a contradiction since u is non-trivial. Step 3. Here we handle the case α ∈ (0, 1). Let (qh ) be the sequence defined as follows q0 = 1,

1 α 2m = − , h = 1, 2, ... qh qh−1 n

By induction, we can compute qh explicitly as 1 2m(1 − αh ) = αh − , qh n(1 − α)

whenever qh is well defined.

(4.7)

Obviously, the sequence (qn−1 ) is decreasing since α ∈ (0, 1), and there exists a unique integer j∗ > 0 such that 1 6 0 < qj∗ . qj∗ +1 We will estimate kukLqh (BR ) successively. We claim that for all h 6 j∗ and R > 1, Z

BR

1/qh h uqh dx 6 CR(n+2m−2)α .

(4.8)

Firstly, (4.8) for h = 0 follows from (4.3). Assume that (4.8) is true for h − 1 with h 6 j∗ . Using the equation ∆m u = −uα , Lemma 3.5 and (4.3), we get that Z 1/qh qh u dx BR Z Z α/qh−1 nα n m qh−1 /α n/qh −n qh +2m− qh−1 |∆ u| dx udx + CR 6 CR B2R B2R (4.9) Z n

6 CR qh Z =C

+2m−n( q1 + 2m n )

uqh−1 dx

h

α/qh−1

+ CRn/qh +2(m−1)

B2R

B2R

α/qh−1 + CRn/qh +2(m−1) . uqh−1 dx

Thanks to the induction hypothesis on kukLqh−1 (BR ) , (4.9) implies that for any R > 1, q α Z 1/qh Z uqh−1 dx h−1 + CRn/qh +2(m−1) 6C uqh dx B2R

BR

= CR

(n+2m−2)αh

+ CRn/qh +2(m−1)

h

6 CR(n+2m−2)α . For the last line, we have used n 2m (n + 2m − 2)αh − + 2m − 2 = (1 − αh ) − 2(m − 1) > 0. qh 1−α Hence the claim (4.8) holds true for all h 6 j∗ and R > 1. Furthermore, by the definition of j∗ , there holds α 2m − 6 0. qj∗ n

12


Therefore, for any q > 1, the condition (3.7) is always fulfilled with ℓ = α/qj∗ . Applying Lemma 3.5, (4.8) with h = j∗ and (4.3), we get Z 1/q Z α/qj∗ n +2m− qnα j∗ uq dx 6 CR q |∆m u|qj∗ /α dx BR B2R Z udx + CRn/q−n B2R Z α/qj∗ n n +2m− qnα j∗ uqj∗ dx + CR q +2(m−1) 6 CR q (4.10) B2R

6 CR

n nα q +2m− qj ∗

R

(n+2m−2)αj∗ +1

n

j∗ +2 j∗ +1 +2m 1−α 1−α

n

j∗ +2 j∗ +1 +2m 1−α 1−α

= CR q −2α 6 CR q −2α

n

+ CR q +2(m−1) n

+ CR q +2(m−1) .

On the other hand, using now Hölder’s inequality with a=α

q−α q−1 ∈ [0, 1) and p = > 1. q−α q−1

We obtain, with also (4.4), Z Z udx 6 F (R) = B2R

B2R

=

Z

B2R

p1 Z uap dx

B2R

q−1 Z q−α α u dx

p−1 p p u(1−a) p−1 dx

B2R

Z q−1 6 C R−2m F (2R) q−α

1−α q−α uq dx

B2R

(4.11)

1−α q−α . uq dx

It follows that for the sequence (Ri ) given by (4.5), there holds 1−α Z q−α q−1 − 2m(q−1) . F (Ri ) 6 CRi q−α F (Ri ) q−α uq dx BRi

Combining the above estimate with (4.10), we get Z − 2m(q−1) 1−α F (Ri ) 6 CRi uq dx BRi

− 2m(q−1) 1−α

6 CRi

2m

= CRi1−α

+n−

2mα 1−α +2

We fix q > 1 large enough such that 2m +n− 1−α

j∗ +2

n+ −2αj∗ +1 +2m 1−α 1−α

Ri

αj∗ +1 q

q

(4.12)

.

2mα + 2 αj∗ +1 q < 0. 1−α

The estimate (4.12) means then F (Ri ) → 0 as i → ∞. Immediately, this is a contradiction with the fact that u is non-trivial. 4.2. Non existence results for (−∆)m u = −uα with α > 1. In this subsection, we consider non-negative classical solutions of (−∆)m u = −uα

in Rn .

(4.13)

Proposition 4.4. For any m > 1 and α > 1, the equation (4.13) has no non-trivial nonnegative solution.


13

Proof. Assume that u is a non-negative solution in Rn of (4.13) with α > 1. We first derive an integral estimate of u over BR . Let ψ be a smooth, radial, cut-off function satisfying 0 6 ψ 6 1 and (3.4). For any R > 0, let φR (x) = ψ p (R−1 x) with p = 2mα α−1 > 2m. We have then x 1/α = CR−2m φR (x). |∆m φR (x)| 6 CR−2m ψ p−2m R Therefore, Z Z uα φR dx = − (−∆)m uφR dx Rn Rn Z Z (4.14) 1/α m −2m =− u(−∆) φR dx 6 CR uφR dx. Rn

Rn

Using Hölder’s inequality and noting the support of φR , Z Z 1/α n(α−1) 1/α α uφR dx 6 CR uα φR dx . Rn

(4.15)

Rn

Putting together (4.14) and (4.15), we arrive at Z Z 2m 1/α udx 6 uφR dx 6 CRn− α−1 . Rn

BR

Applying Lemma 3.4, we deduce that u ≡ 0 in Rn .

4.3. Existence results for ∆m u = uα . This subsection is devoted to the existence results for the equation ∆m u = uα

in Rn .

(4.16)

We do not consider specially the situations under applications of Propositions A and B. Proposition 4.5. For any m, n ∈ N∗ and α 6 1, the equation (4.16) has infinitely many positive radial solutions. Proof. We look for radial solutions of (4.16). To this purpose, consider the following initial value problem ( m ∆ u(r) = uα (r) in [0, R), (4.17) u(0) = 1, ∆i u(0) = ai > 0, i = 1, ..., m − 1. Clearly, using ODE theory, (4.17) has a unique positive solution in a maximal interval [0, Rmax ). We need only to prove that Rmax = ∞. To do this, we construct suitable sub and super-solutions to (4.17), and apply then the comparison principle. Indeed, let u∗ (r) ≡ 1. Trivially, u∗ is a sub-solution to (4.17) and ∆i u∗ (0) 6 ∆i u(0) for any 0 6 i 6 m − 1. Hence u(r) > 1 = u∗ (r) in [0, Rmax ) by Lemma 3.1 with f ≡ 0. On the other hand, let v(r) = er

2

/2

and denote

∆k v(r) = Pk (r)er

2

/2

.

A direct computation yields that Pk+1 (r) = (r2 + n)Pk + rPk′ (r) + ∆Pk (r)

∀ k > 0, r > 0.

We can prove readily that Pk ∈ N[X], degPk = 2k and Pk (0) > 1. Let now u∗ (r) = λv(r), with λ = max(1, a1 , . . . am−1 ) > 1, it follows that ∆m u∗ (r) = λPm (r)er i ∗

i

2

/2

> λPm (0)er

2

/2

> λer

and ∆ u (0) = λPi (0) > λ > ∆ u(0) for 0 6 i 6 m − 1.

2

/2

,

∀r>0

(4.18)

14

ˆ V.H. NGUYEN, Q.H. PHAN, AND D. YE Q.A. NGO, 2

• If α ∈ [0, 1], (4.18) yields ∆m u∗ (r) > λα eαr /2 = u∗ (r)α for r > 0. We get then u(r) 6 u∗ (r) in [0, Rmax ) by Lemma 3.1 with f (t) = tα in R+ . • If α < 0, (4.18) yields ∆m u∗ > 1 > uα = ∆m u in BRmax as u > 1. Applying this time Lemma 3.1 with f ≡ 1, there holds again u(r) 6 u∗ (r) in [0, Rmax ). To conclude, we have always u∗ 6 u 6 u∗ in BRmax , where u∗ and u∗ are smooth positive functions in Rn . The local boundedness of u implies readily that Rmax = ∞. The infinity of solutions can be obtained by choosing different values of ai if m > 2; or at least by the natural scaling of the equation (4.16). Remark 4.1. If n > 2, we can also put solutions of lower dimensions in Rn , to get infinitely many non radial solutions of (4.16) with α 6 1. Similar remark goes to the equation (4.1) when n > 4. ACKNOWLEDGMENTS This work was initiated while QAN was visiting the Vietnam Institute for Advanced Study in Mathematics (VIASM) in 2017. He gratefully acknowledges the institute for its hospitality and support. The researches of QAN and QHP are funded by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant numbers 101.02-2016.02 and 101.02-2017.307 respectively. R EFERENCES [AS11]

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OF

M ATHEMATICS , C OLLEGE

OF

S CIENCE , V I Eˆ T N AM N ATIONAL U NIVER -

SITY,

E-mail address: [email protected], bookworm [email protected] (V.H. Nguyen) I NSTITUTE OF R ESEARCH AND D EVELOPMENT, D UY TAN U NIVERSITY, D A N ANG , V IETNAM .

E-mail address: [email protected] (Q.H. Phan) I NSTITUTE OF R ESEARCH AND D EVELOPMENT, D UY TAN U NIVERSITY, D A N ANG , V IETNAM .

E-mail address: [email protected] (D. Ye) IECL, UMR 7502, D E´ PARTEMENT AUGUSTIN F RESNEL , 57073 M ETZ , F RANCE .

DE

M ATH E´ MATIQUES , U NIVERSIT E´ DE L ORRAINE , 3

E-mail address: [email protected]

RUE