Multiplicity of weak solutions for a class of nonuniformly elliptic equations of p-laplacian type

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1 Nonlinear Analysis Multilicity of weak solutions for a class of nonuniformly ellitic equations of -Lalacian tye Hoang Quoc Toan, Quô c-anh Ngô Deartment of Mathematics, College of Science, Viêt Nam National University, Hà Nôi, Viêt Nam Received 6 October 27; acceted 4 February 28 Abstract This aer deals with the multilicity of weak solutions in W to a class of nonuniformly ellitic equations of the form divax, u = hx u r u + gx u s u in a bounded domain of R N. Here a satisfies ax, ξ c h x + h x ξ for all ξ R N, a.e. x, h L, h L loc, h x for a.e. x in, < r < < s < N N + /N. c 28 lsevier Ltd. All rights reserved. Keywords: -Lalacian; Nonuniform ellitic equations; Multilicity. Introduction Let be a bounded domain in R N. In the resent aer we study the multilicity of nontrivial solutions of the following Dirichlet ellitic roblem: divax, u = hx u r u + gx u s u where ax, ξ c h x + h x ξ for any ξ in R N and a.e. x, h x and h x for any x in. For h and h belonging to L, the roblem has been studied. Here we study the case in which h and h belong to L and L loc resectively. The equation now may be nonuniformly ellitic. To our knowledge, such equations were first studied by Duc et al. and Vu [3,6]. In both aers, the authors studied the following roblem: divax, u = f x, u where the nonlinear term f verifies the Ambrosetti Rabinowitz tye condition, see []. They then obtained the existence of a weak solution by using a variation of the Mountain-Pass Theorem introduced in [2]. We also oint out the fact that for the case when h, our roblem was studied in [4]. Our goal is to extend the results of [4] for the nonuniform case and of [3,6] the existence of at least two weak solutions. 2 Corresonding author. Tel.: mail address: bookworm vn@yahoo.com Q.-A. Ngô X/$ - see front matter c 28 lsevier Ltd. All rights reserved. doi:.6/j.na

2 H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis In order to state our main theorem, let us introduce our hyotheses on the structure of roblem. Assume that N 3 and 2 < N. Let be a bounded domain in R N having C 2 boundary. Consider a : R N R N R N, a = ax, ξ, as the continuous derivative with resect to ξ of the continuous function A : R N R N R, A = Ax, ξ, that is, ax, ξ = Ax,ξ ξ. Assume that there are a ositive real number c and two nonnegative measurable functions h, h on such that h L loc, h L, h x for a.e. x in. Suose that a and A satisfy the hyotheses below: A ax, ξ c h x + h x ξ for all ξ R N, a.e. x. A 2 There exists a constant k > such that A x, ξ + ψ 2 2 Ax, ξ + 2 Ax, ψ k h x ξ ψ for all x, ξ, ψ, that is, A is -uniformly convex. A 3 A is -subhomogeneous, that is, ax, ξξ Ax, ξ for all ξ R N, a.e. x. A 4 There exists a constant k > such that Ax, ξ k h x ξ for all ξ R N, a.e. x. A 5 Ax, = for all x. A 6 Ax, ξ = Ax, ξ for all ξ R N, a.e. x. xamle. i Ax, ξ = ξ, ax, ξ = ξ 2 ξ with 2. We get the -Lalacian oerator. ii Ax, ξ = ξ +θx + ξ 2, ax, ξ = ξ 2 ξ +θx We get the oerator div u 2 u + div u θx + u 2 ξ + ξ 2 with 2 and θ a suitable function. which can be regarded as the sum of the -Lalacian oerator and a degenerate form of the mean curvature oerator. iii Ax, ξ = θ 2 x + ξ 2 2 θ x, ax, ξ = θ 2 x + ξ ξ with 2 and θ a suitable function. Now we get the oerator div θ 2 2 x + u 2 2 u which is a variant of the generalized mean curvature oerator div + u u. Regarding the functions h and g, we assume that H hx for all x and h L r L, where r + + r =, N N r+n. that is r = G gx > a.e. x and g L s L, where that is s = s + + s =, N N s+n.

3 538 H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis Let W, be the usual Sobolev sace. Next, we define X := W, as the closure of C u = u dx. We now consider the following subsace of W, : { } = u W, : h x u dx < +. The sace can be endowed with the norm u = h x u dx. under the norm As in [3], it is known that is an infinite dimensional Banach sace. We say that u is a weak solution for roblem if ax, u ϕdx hx u r uϕdx gx u s uϕdx = for all ϕ. Let Ju = hx u r+ dx + gx u s+ dx, r + s + Λu = A x, u dx, and I u = Λu Ju for all u. The following remark lays an imortant role in our arguments. Remark 2. i u u for all u since h x. Thus the continuous embeddings X L i, i hold true. ii By A 4 and i in Lemma 5, it is easy to see that = {u W, : Λu < + } = {u W, : I u < + }. iii C since u is in C c for any u C and h L loc. Our main results are included in a coule of theorems below. Theorem 3. Assume < r < < s < N N + /N and conditions A A 5, H, and G are fulfilled. Then roblem has at least two nontrivial weak solutions in rovided that the roduct h s+ r L r g L s is small enough. Theorem 4. Assume < r < < s < N N + /N and conditions A A 6, H, and G are fulfilled. Then roblem has infinitely many nontrivial generalized solutions in. To see the ower of the theorems, we comare our assumtions to those considered in [5,3,4,6]. Our roblem covers the following cases which have been considered in the literature: 3

4 i Ax, ξ = ξ with 2. ii Ax, ξ = + ξ 2 2 with 2. H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis Moreover, our assumtion includes the following situations which could not be handled in [5,4]. i Ax, ξ = hx ξ with 2 and h L loc. ii Ax, ξ = hx + ξ 2 2 with 2 and h L. 2. Auxiliary results In this section we recall certain roerties of functionals Λ and J. But firstly, we list here some roerties of A. Lemma 5 See [3]. i A verifies the growth condition Ax, ξ c h x ξ + h x ξ for all ξ R N, a.e. x. ii Ax, zξ Ax, ξz for all z, x, ξ R N. Due to the resence of h, the functional Λ does not belong to C, R. This means that we cannot aly directly the Mountain-Pass Lemma of Ambrosetti and Rabinowitz. In this situation, we recall the following concet of weakly continuous differentiability. Our aroach is based on a weak version of the Mountain-Pass Lemma introduced by Duc [2]. Definition 6. Let F be a ma from a Banach sace Y to R. We say that F is weakly continuous differentiable on Y if and only if the following two conditions are satisfied: i For any u Y there exists a linear ma DFu from Y to R such that Fu + tv Ju lim = DFuv t t for every v Y. ii For any v Y, the ma u DFuv is continuous on Y. Denote by C w Y the set of weakly continuously differentiable functionals on Y. It is clear that C Y C w Y where we denote by C Y the set of all continuously Frechet differentiable functionals on Y. For simlicity of notation, we shall denote DFu by F u. The following lemma concerns the smoothness of the functional Λ. Lemma 7 See [3]. i If {u n } is a sequence weakly converging to u in X, denoted by u n u, then Λu lim inf n Λu n. ii For all u, z, u + z Λ 2 2 Λu + 2 Λz k u z. iii Λ is continuous on. iv Λ is weakly continuously differentiable on and Λ u, v = ax, u vdx for all u, v. v Λu Λv Λ v, u v for all u, v. The following lemma concerns the smoothness of the functional J. The roof is standard and simle, so we omit it.

5 54 H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis Lemma 8. i If u n u in X, then lim n J u n = Ju. ii J is continuous on. iii J is weakly continuously differentiable on and J u, v = hx u r uvdx + gx u s uvdx for all u, v. Our main tool is a variation of the Mountain-Pass Theorem introduced in [2] and the Z 2 version of it introduced in [6]. Lemma 9 Mountain-Pass Lemma. Let F be a continuous function from a Banach sace into R. Let F be weakly continuously differentiable on and satisfy the Palais Smale condition. Assume that F = and there exist a ositive real number ρ and z such that i z > ρ, Fz F. ii α = inf{fu : u, u = ρ} >. Put G = {φ C[, ], : φ =, φ = z }. Assume that G. Set β = inf{max Fφ[, ] : φ G}. Then β α and β is a critical value of F. Lemma Symmetric Mountain-Pass Lemma. Let be an infinite dimensional Banach sace. Let F be weakly continuously differentiable on and satisfy the Palais Smale condition. Assume that F = and: i There exist a ositive real number α and ρ such that inf Fu α > u B ρ where B ρ is an oen ball in of radius ρ centered at the origin and B ρ is its boundary. ii For each finite dimensional linear subsace Y in, the set { u Y : Fu } is bounded. Then F ossesses an unbounded sequence of critical values. 3. Proofs We remark that the critical oints of the functional I corresond to the weak solutions of. In order to aly Lemma 9 we need to verify the following facts. Lemma. i I is a continuous function from to R. ii I be weakly continuously differentiable on and I u, v = a x, u vdx hx u r uvdx gx u s uvdx for all u, v. iii I =. iv There exist two ositive real numbers ρ and α such that inf{i u : u, u = ρ} > α. v There exists ψ such that lim t I tψ =. vi I satisfies the Palais Smale condition on. vii There exists z such that z > ρ, I z.

6 viii The set H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis G = {ϕ C [, ], : ϕ =, ϕ = z } is not emty. Proof. i This comes from iii in Lemma 7 and ii in Lemma 8. ii This comes from iv in Lemma 7 and iii in Lemma 8. iii This comes from the definition of I. iv First, let S be the best Sobolev constant of the embedding W, L, that is, S = Thus, we obtain inf u W, \{} S v L v u dx u dx. for all v. Since N r = N r + N then + = r r+ N r + N N + r + N N By Hölder s inequality and the above relation we deduce hx u r+ dx h L r u r+ L h L r S r+ h L r With similar arguments, we have Thus, we obtain S r+ gx u s+ dx s + ν u s+. I u min =: { k, } u µ u r+ ν u s+ λ µ u r+ ν u s+ We show that there exists t > such that λ µt r+ νt s+ >. To do that, we define the function Q t = µt r+ + νt s+, t >. =. S h L r+ 5 u r+ =: r + µ u r u. 8 Since lim t Q t = lim t Q t =, it follows that Q ossesses a ositive minimum, say t >. In order to find t, we have to solve equation Q t =, where Q t = r + µt r + s + νt s. A simle comutation yields r µ t =. s + ν 4 9

7 542 H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis Thus, relation 9 holds rovided that r+ r µ µ + ν s + ν or equivalently or µ s+ ν r+ r s + s+ r h L r g L s r+ r s + + µ s+ ν r+ s+ µ ν < λ s+ r s + is small enough. v Let ψ C, ψ, ψ. Then using Lemma 5, we have I tψ = A x, t ψ dx tr+ hx ψ r+ dx r + ts+ gx ψ s+ dx s + t A x, ψ dx ts+ gx ψ s+ dx. 2 s + Thus, lim t I tψ =. vi Let {u n } be a sequence in X and β be a real number such that I u n β and I u n in. We rove that {u n } is bounded in. We assume by contradiction that u n as n. It follows from conditions A 3 and A 4 that for n large enough β + + u n I u n I u n, u n 3 s + = A x, u n s + a x, u n u n dx 4 s r hx u n r+ dx 5 r + s + or s r β + + u n + hx u n r+ dx 6 r + s + k u n dx. 7 s + Hence s r β + + u n + r + s + h L r u n r+ 8 k u n s +. S r+ Since < r < and u n as n, dividing the above inequality by u n and assing to the limit as n we obtain a contradiction. Hence {u n } is bounded in. By Remark 2, we deduce that {u n } is bounded in X. Since X is reflexive, then by assing to a subsequence, still denoted by {u n }, we can assume that the sequence {u n } converges weakly to some u in X. We shall rove that the sequence {u n } converges strongly to u in. We observe by Remark 2 that u. Hence { u n u } is bounded. Since { I u n u } converges to, then I u n u, u n u converges to. < λ 9

8 H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis Since is continuously embedded in L, then we deduce that u n converges weakly to u in L. Then it is clear that u n r u n converges weakly to u r u in L r. Now define the oerator U : L r R by U, w = hxuwdx. We remark that U is linear. Moreover, it follows from + r + = r that U is continuous since h L r. All the above ieces of information imly U, u n r u n U, u r u, that is, lim hx u n r u n udx = hx u r+ dx. n With the same arguments we can show that lim gx u n s u n udx = gx u s+ dx, 2 n lim hx u n r+ dx = hx u r+ dx, 2 n lim gx u n s+ dx = gx u s+ dx. 22 n Therefore, lim hx u n r u n u n u dx = lim hx u n r+ dx hx u r+ dx n n lim hx u n r u n udx hx u r+ dx n which yields lim hx u n r u n u n u dx =. n Similarly, we obtain lim gx u n s u n u n u dx =. n On the other hand, J u n, u n u = hx u n r u n u n u dx 23 + gx u n s u n u n u dx. 24 Thus lim J u n, u n u =. n This and the fact that Λ u n, u n u = I u n, u n u + J u n, u n u gives lim Λ u n, u n u =. n

9 544 H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis By using v in Lemma 7, we get Λu lim Λ u n = lim Λu Λ u n lim Λ u n, u u n =. n n n This and i in Lemma 7 give lim Λ u n = Λu. n Now if we assume by contradiction that u n u does not converge to, then there exists ε > and a subsequence { } u nm of {un } such that unm u ε. By using relation ii in Lemma 7, we get 2 Λu + 2 Λ unm + u u nm Λ k unm u 2 k ε. Letting m we find that unm + u lim su Λ Λu k ε. m 2 We also have that u nm +u 2 converges weakly to u in. Using i in Lemma 7 again, we get Λu lim inf Λ unm + u. m 2 That is a contradiction. Therefore {u n } converges strongly to u in. vii The existence of z such that z > ρ, I z is followed from the fact that lim t I tψ =. viii We consider a function ϕ C [, ], defined by ϕ t = tz, for every t [, ]. It is clear that ϕ G. Proof of Theorem 3. Using Lemmas 9 and we deduce the existence of u as a nontrivial generalized solution of. We rove now that there exists a second weak solution u 2 such that u 2 u. By Lemma, it follows that there exists a ball centered at the origin B, such that inf I >. B On the other hand, by the lemma there exists φ such that I tφ <, for all t > small enough. Recalling that relation 7 holds for all u, that is, I u λ u µ u r+ ν u s+ we get that We let now < c := inf B I <. < ε < inf B I inf B I. Alying keland s Variational Princile for the functional I : B R, there exists u ε B such that Since I u ε < inf I + ε 25 B I u ε < I u + ε u u ε, u u ε. 26 I u ε < inf I + ε < inf I + ε < inf I B B B it follows that u ε B. Now we define M : B R by Mu = I u + ε u u ε. It is clear that u ε is a minimum oint of M and thus M u ε + tν M u ε t

10 H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis for a small t > and ν in the unit shere of. The above relation yields I u ε + tν I u ε t Letting t it follows that I u ε, ν + ε ν > + ε ν. and we infer that I u ε ε. We deduce that there exists {u n } B such that I u n c and I u n. Using the fact that J satisfies the Palais Smale condition on we deduce that {u n } converges strongly to u 2 in. Thus, u 2 is a weak solution for and since > c = I u 2 it follows that u 2 is nontrivial. Finally, we oint out the fact that u u 2 since I u = c > > c = I u 2. The roof is comlete. Proof of Theorem 4. In view of A 6, I is even. In order to aly Lemma, it is enough to verify condition ii in Lemma. It is known that A x, u n dx c h x dx u n + u n 27 This gives =: c u n + c u n. 28 I u c u n + c u n s + gx u s+ dx. Suose that Ẽ is a finite dimensional subsace of. Setting u Ẽ = gx u s+ s+ dx for all u Ẽ, we see that. Ẽ is a norm in Ẽ. We also note that in Ẽ the norms are equivalent. Thus, there exists a ositive constant K such that u K u Ẽ. This imlies that I u c u n + c u n K s + u s+. Since < s + then { u Ẽ : I u } is bounded. Hence, I ossesses an unbounded sequence of critical values. Therefore, I ossesses infinitely many critical oints in. This comletes the roof. References [] A. Ambrosetti, P. Rabinowitz, Dual variational methods in critical oint theory and alications, J. Funct. Anal [2] D.M. Duc, Nonlinear singular ellitic equations, J. London Math. Soc [3] D.M. Duc, N.T. Vu, Nonuniformly ellitic equations of -Lalacian tye, Nonlinear Anal [4] M. Mihailescu, xistence and multilicity of weak solutions for a class of degenerate nonlinear ellitic equations, Bound. Value Probl Article ID [5] P. de Náoli, M.C. Mariani, Mountain ass solutions to equations of -Lalacian tye, Nonlinear Anal [6] N.T. Vu, Mountain ass theorem and nonuniformly ellitic equations, Vietnam J. Math

11 546 H.Q. Toan, Q.-A. Ngô / Nonlinear Analysis Further reading [] R.A. Adams, Sobolev Saces, Academic Press, London, 975. [2] G. Dinca, P. Jebelean, J. Mawhin, Variational and toological methods for Dirichlet roblems with -Lalacian, Protugaliae Math [3] M. Struwe, Variational Methods, Sringer, New York, 996.