On non negative solutions of some quasilinear elliptic inequalities

Transcription

1 On non negative solutions of some quasilinear elliptic inequalities Lorenzo D Ambrosio and Enzo Mitidieri September Abstract Let f : R R be a continuous function. We prove that under some additional assumptions on f and A : R R +, weak C solutions of the differential inequality diva u ) u) fu) on R N are non negative. Some extensions of the result in the framework of subelliptic operators on Carnot Groups are considered. Introduction In this paper we shall study the following problem. Let L be a second order differential operator and let f : R R be a continuous function. Find additional assumptions on L, f) that imply the positivity of the possible solutions of the differential inequality Lu) fu) on R N..) Some partial answers to this problem have been obtained in [5, 7]. In those papers, the authors deal with elliptic inequalities of the form.) in the case when L is the Laplacian operator or the polyharmonic operator ) k in the Euclidean setting or, more generally L is a sub elliptic Laplacian on a Carnot group and f is non negative. The main strategy used in [5, 7] for proving positivity results was via integral representation formulae. One essential Dipartimento di Matematica, via E. Orabona, 4 - Università degli Studi di Bari, I-7025 Bari, Italy dambros@dm.uniba.it Dipartimento di Matematica e Informatica, via A. Valerio, 2/ - Università degli Studi di Trieste, I-3427 Trieste, Italy mitidier@units.it

2 difficulty using this approach is that no assumptions on the behavior of the solutions at infinity are known. A typical example in this direction is given by, where N 3 and q >. The following result holds see [5]). u u q on R N,.2) Theorem. Let N 3 and q >. Let u L q loc RN ) be a distributional solution of.2) and let Lebu) be the set of its Lebesgue points. If x Lebu), then q uy) ux) C N dy, x y N 2 where C N is an explicit positive constant. R N From this result it follows that, if u is a solution of.2) then, either ux) = 0 or ux) > 0 a.e. on R N. Obviously, the approach via representation formulae cannot be applied to quasilinear problems. In this paper we shall consider a class of quasilinear model problems for which the positivity property mentioned at the beginning of this introduction holds. More precisely, we shall deal with the case when L is the p-laplacian operator, namely p = div p 2 ), or the mean curvature operator div ). In this cases + 2 some results on positivity of solutions of.) are proved by using a suitable comparison Lemma see Lemma 2.8 below). This paper is organized as follows. In section 2 we state and prove our main results in the Euclidean setting for the p-laplacian and the mean curvature operator. In addition we point out some consequences. In section 3 we briefly indicate some possible generalizations to other quasilinear operators and to differential inequalities on Carnot groups. 2 Main results Throughout this paper we shall assume that N 2 and that we deal with weak C solutions of the problems under consideration. See Definition 2.7 below for further details. Our main results are the following. 2

3 Theorem 2. Let p > and N > Let f : R R be a continuous function such that and If u is a solution of ft) > 0 if t < 0, f is non increasing on ], 0[ 2.3) t ) p fs) ds dt < ) div u p 2 u) fu) on R N, 2.5) then u 0 on R N. Moreover if ft) 0 for t 0 then, either u 0 or u > 0 on R N. Corollary 2.2 Let p > and q > p. Let f : R R be a continuous function such that ft) C t q for t < 0. Let u be a solution of div u p 2 u) fu) on R N. 2.6) If q > p then u 0 on R N. Moreover if ft) 0 for t 0 then, either u 0 or u > 0 on R N. In the case of the mean curvature operator the above results can be improved. Indeed, the claim follows without the assumption 2.4) on f. Theorem 2.3 Let f : R R be a continuous function satisfying 2.3). Let u be a solution of u div ) fu) on R N. + u 2 Then u 0 on R N. The following Liouville theorems are an easy consequence of the fact that the only non negative functions u such that p u 0 on R N with N p are the constants, and the u only non negative functions u such that div ) 0 on R 2 are the constants, + u 2 see [6]. Corollary 2.4 Let p N > and f : R [0, + [ be a continuous function satisfying 2.3) and 2.4). If u is a solution of 2.5) then u is constant on R N. More precisely u α 0 and fα) = 0. Moreover if ft) > 0, then 2.5) has no solutions. 3

4 Corollary 2.5 Let f : R [0, + [ be a continuous function satisfying 2.3). Let u be a solution of u div ) fu) on R ) + u 2 Then u is constant on R 2. More precisely u α 0 and fα) = 0. Moreover if ft) > 0, then 2.7) has no solutions. Remark 2.6 The above assumptions on f are sharp in the following sense. If p = 2 and q = = p the result is false. Indeed the equation admits the explicit negative solution u = u on R N, ux) := Expx ), x R N, or solutions that changes sign see [5]). In the general case q = p the equation p u = u p on R N, admits a positive solution see for instance [8]). Therefore, the equation has a negative solution. p u = u p on R N, Let us briefly describe the idea of the proof of our main result. Let u be a solution of 2.5). Without loss of generality we will show that u0) 0. The function U := u satisfies the inequality Let v be a positive solution of div U p 2 U) f U) on R N. div v p 2 v) = f v) on B R, such that v0) = a > 0 and vx) + as x R. The assumptions on f imply the existence of v. Since Ux) vx) for x close to R, by a comparison Lemma see Lemma below) it follows that Ux) vx) for any x < R. In particular U0) v0) = a. Letting a 0 we have U0) 0. Hence u0) 0. Finally, if ft) 0 for t 0, by the weak Harnack inequality we get that, either u 0 or u > 0 on R N. 4

5 2. A comparison Lemma In this section, we shall prove a comparison Lemma. This Lemma will be useful when dealing with more general operators then the p-laplacian and the mean curvature operator, thus we shall present it in a general form. To this end let us introduce some notations. Let µ = µx) i,j ) be a matrix with N columns and l N) rows with entries belonging, for simplicity, to C R N ). We denote by L := µ and by div L = L = divµt ). The isotropic Euclidean case corresponds to the choice µ = I N the unitary matrix of dimension N. We shall assume that if L u = 0 on a connected region Ω then u const in such a region. Definition 2.7 Let Ω R N be an open set, let A : R R and h : Ω R be a continuous functions. We say that u is a solution of div L A L u ) L u) h on Ω, if u C Ω) and for any non negative φ C 0 Ω), we have A L u ) L u L φ hφ. and Ω In a similar manner we can define solutions of the inequalities div L A L u ) u) h and div L A L u ) u) h. The following Lemma is useful when considering solutions of inequalities of the form, Ω div L A L u ) L u) g x, u) on Ω, 2.8) div L A L v ) L v) g 2 x, v) on Ω. 2.9) Here, A is a continuous function such that At) > 0 for t > 0 and for i =, 2, g i : Ω R R is continuous. Lemma 2.8 Let Ω be a bounded open set and let u and v be solutions of 2.8) and 2.9) respectively. Assume that. a) For any x Ω, t s 0 there holds g x, t) g 2 x, s), g x, ) is not decreasing on ]0, + [ and v 0; or 5

6 b) For any x Ω, t s there holds g x, t) g 2 x, s) and g x, ) is not decreasing; 2. The function tat) is increasing and positive for t > 0; 3. u v on Ω. Then u v on Ω. Proof. Let u and v be solutions of 2.8) and 2.9) respectively. Let ɛ > 0 be fixed and set v ɛ := v + ɛ. It is a simple matter to check that the function v ɛ satisfies the inequality div L A L w ) L w) g x, w) on Ω. Therefore, for any non negative φ C 0 Ω) we have A L u ) L u A L v ɛ ) L v ɛ ) L φ g x, u) g x, v ɛ ))φ. 2.0) Ω Ω Next we choose φ as follows: φ := u v ɛ ) + ) 2. It is clear that φ is non negative and φ C Ω). Moreover, since v ɛ u ɛ > 0 on Ω, it follows that φ has compact support. Substituting φ in 2.0), we obtain A L u ) L u A L v ) L v) L u L v)2u v ɛ ) + g x, u) g x, v ɛ ))φ. Ω We claim that Indeed, A L u ) L u A L v ) L v) L u L v) 0. 2.) A L u ) L u A L v ) L v) L u L v) = A L u ) L u 2 + A L v ) L v 2 A L u ) + A L v )) L u L v) ) ) = A L u ) L u A L v ) L v L u L v + ) ) + A L u ) + A L v ) L u L v L u L v =: I + I ) Since A 0, we have I 2 0. From the monotonicity of tat), it follows that ) ) I = A L u ) L u A L v ) L v L u L v 0. Ω 6

7 Assume first that g x,.) is strictly increasing. Therefore, the inequality g x, u) g x, v ɛ )u v ɛ ) + ) 2 0 holds for every x Ω. As a consequence, g x, u) g x, v ɛ )u v ɛ ) + ) 2 = 0 on Ω and hence u v ɛ. This completes the proof in case g x,.) is strictly increasing. For the general case we need of an extra argument. Indeed, from 2.) and 2.2) we have that I Ω + I 2 )u v ɛ ) + = 0. Let x Ω be such that ux) v ɛ x). Since I 0 and I 2 0 we have I x) = 0 = I 2 x). We claim that L ux) = L vx). Indeed, if L ux) L vx), from I 2 x) = 0, we deduce that L ux) L vx). Thus from I x) = 0, the fact that tat) is injective, it follows that 0 = A L u ) L u A L v ) L v 0. This implies L u v ɛ ) + ) 2 = 0 on Ω, that is u v ɛ ) + ) 2 = φ = 0.main Theorems Therefore, letting ɛ 0 in u v + ɛ the claim follows. Remark 2.9 It is possible to extend the above result to situations where the function A depends on x. Namely, when A : Ω R R is continuous and for any x Ω the function t ]0, + [ tax, t) is increasing and strictly positive. 2.2 Proofs of the Main Results The proof of Theorem 2. relies on the following. Theorem 2.0 Let g : R R be a continuous function such that, gt) > 0 if t < 0, g is non decreasing on ]0, + [, and + t ) p gs) ds dt < ) For any p >, a > 0, D >, there exists a function ϕ and R > 0 such that, ϕ is a solution of r D ϕ r) p 2 ϕ r)) = r D gϕr)), ϕ0) = a, ϕ 0) = 0, 2.4) φ is increasing on ]0, R[ and ϕr) + as r R. See [] for a proof in the case p = 2 and [8] for the quasilinear case p 2. Proof of Theorem 2.. Let u be a solution of 2.5). Since the inequality is invariant under translations, it is sufficient to prove that u0) 0. if t, s are two different vectors in a Hilbert space such that s t) = t s, then 0 < t s = t 2 + s 2 2s t) = t 2 + s 2 2 s t = s t ) 2. 7

8 Let gt) := f t). The function g satisfies the assumptions of Theorem 2.0. Let D = N >, a > 0 and let ϕ be a solution of 2.4) such that ϕr) + as r R. We set vx) := ϕ x ). Therefore, the function v satisfies the differential equation div L L v p 2 L v) = gv) on Ω R, where Ω R := {x x < R}. On the other hand the function U := u satisfies the inequality div L L U p 2 L U) g U) := gu) on R N. Since Ux) vx) for x close to R we are in the position to apply the comparison Lemma 2.8. As a consequence, Ux) vx) for any x Ω R. In particular U0) v0) = a. Letting a 0 it follows that U0) 0. Hence u0) 0. Next, if f 0, then u is a non negative solution of the inequality, div L L u p 2 L u) 0 on R N. Hence, by the weak Harnack inequality see [2]) it follows that, either u 0 or u > 0 on R N. The argument for proving Theorem 2.3 is the same of the one used in the proof of Theorem 2., so we shall be brief. Proof. Let gt) := f t). Under the assumptions of Theorem 2.3, there exists a radial v solution v of div ) gv) such that v0) = a > 0 and vr) + as r R, see 2 + v [8]. By the comparison Lemma 2.8 we get u0) a. Letting a 0, the claim follows. 3 Some extensions of the main results We extend our main results to more general quasilinear operator in Section 3.. In Section 3.2 we extend Theorem 2. in Carnot group setting. The final Section 3.3 deals with a quasilinear inequality related to the porous medium equation. 3. A class of differential inequalities In this section we shall consider inequalities of the type diva u ) u) fu) on R N, 3.5) where we shall assume that { A C]0, + [), At) > 0 for t > 0, ta t ) CR) C ]0, [) and tat)) > 0 for t > ) 8

9 We shall distinguish two cases accordingly to the asymptotic behavior of the function tat). Namely, lim t + tat) < + or lim t + tat) = +. Theorem 3. Let A as in 3.6) and such that lim t + tat) < +. Let f : R R be a continuous function satisfying 2.3). Let u be a solution of 3.5), then u 0 on R N. We observe that Theorem 2.3) is a particular case of the above theorem. In the case lim t + tat) = +, then accordingly with [9, 0], we can construct the following function G and H. Let G be defined as Gt) := t 2 At) t 0 sas)ds, t 0. The function G is continuous, strictly increasing, G0) = 0 and G+ ) = +, see [9, 0, 8]. Let H be its inverse: H is increasing and H+ ) = +. Theorem 3.2 Let A as in 3.6) and such that lim t + tat) = +. Let f : R R be a continuous function satisfying 2.3) and Let u be a solution of 3.5), then u 0 on R N. ) dt < ) H fs) ds t If At) = t p 2, then the function H is Ht) = p P ) /p t /p and the above theorem is indeed Theorem 2.. We leave the proof of the above results to the interested reader since it is based on the same idea already discussed above by taking into account the non existence Theorem and Theorem 2 in [8]. Condition 3.7) is sharp in the following sense. If the integral in 3.7) diverges, then 3.5) admits a negative solution. Indeed from Theorem 3 in [8] it follows that equation 2.4) with gt) = f t) has a positive solution. This implies the claim. Example 3.3 Let At) := ln+t) and fs) c s q for s < 0. We claim that if q > 0 the t solutions of div ln + u ) u ) fu) on R N 3.8) u are non negative. We are in the position to apply Theorem 3.2. In this case Gt) = t ln + t). In order to prove the claim it is enough to show that the function HT ) behaves at infinity as T. Since HT ) is the solution of t ln + t) = T by the change 9

10 of variable z = + T ) and x = + t) x the equation becomes z = and we +x ln x have to study the zero of the function xz) defined implicitly. It is easy to recognize that xz) = z + oz), and this implies the claim. Using the same argument as above one can prove the following. We omit the details. Theorem 3.4 Let f : R R be a continuous function such that fs) c ln + s ) q for s < 0. If q >, then the solutions of 3.8) are non negative. 3.2 Inequalities on Carnot Groups Let R N G be a Carnot group and let L be the horizontal gradient on G and Q > the homogeneous dimension see [] for details on these structures). Let Γ p be the fundamental solution of the quasilinear operator L,p u = div L L u p 2 L u) at the origin. Set { p p Q Γp N p := exp Γ p ) p >, p Q p = Q It is known that N p is a homogeneous norm on G. From [2, 3], it is known that N p is Hölder continuous. In what follows we shall assume that N p is smooth. This assumption is satisfied for example for Heisenberg type groups. See [3]. With the above notation, we have that if ζ : R R is a smooth function, then the radial function v := ζ N p : G R satisfies div L L v p 2 L v) = p )ψ p ζ p 2 ζ r) + Q ) ζ r), r r=n P where ψ := L N p is a bounded function, see [4]. Hence we can apply the same arguments used in the preceding section obtaining an analog of Theorem 2. in this more general setting. Theorem 3.5 Let p >. Let f : R R be a continuous function satisfying 2.3) and 2.4). Let u be a solution of div L L u p 2 L u) ψ p fu) on R N, 3.9) then u 0 on R N. Moreover if ft) 0 for t 0 then, either u 0 or u > 0 on R N. Proof. Let u be a solution of 2.5). Since the inequality is invariant under translations, it is sufficient to prove that u0) 0. 0

11 Let gt) := f t). The function g satisfies the assumptions of Theorem 2.0. Let D = Q > be the homogeneous dimension. Let a > 0 and let ϕ be a solution of 2.4) such that ϕr) + as r R. We set vx) := ϕn p x)). By computation we have, div L L v p 2 L v) = p )ψ p v p 2 v r) + Q ) v r) r = ψ p Np Q r Q v r) p 2 v r) Therefore, the function v satisfies the differential equation div L L v p 2 L v) = g 2 x, v) := ψ p gv) r=n p ). r=n p on Ω R := {x N p x) < R}. On the other hand the function U := u satisfies the inequality div L L U p 2 L U) g x, U) := ψ p gu) on R N. Since g = g 2 and Ux) vx) for N P x) close to R we are in the position to apply the comparison Lemma 2.8. As a consequence, Ux) vx) for any x Ω R. In particular U0) v0) = a. Letting a 0 it follows that U0) 0. Hence u0) 0. Next, if f 0, then u is non negative and solves the inequality div L L u p 2 L u) 0 on R N. Hence, by the weak Harnack inequality see [2]) it follows that, either u 0 or u > 0 on R N. 3.3 A porous medium type inequality We end this paper by pointing out the following slight modification of Theorem 2.. Theorem 3.6 Let γ. Let f : R R be a continuous function satisfying 2.3) and ) t γ fs) s γ 2 ds dt < ) Let u be a solution of t u γ u) fu) on R N, then u 0 on R N. In particular if u is a solution and ft) C t q for t < 0 with q > γ, then either u 0 or u > 0 on R N. Acknowledgment. We warmly thank Patrizia Pucci for pointing out an error in a earlier version of this paper entitled, Positivity property of solutions of some quasilinear elliptic inequalities, Functional Analysis and Evolution Equations, The Günter Lumer Volume, Birkhäuser 2008),

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