MULTIPLE SOLUTIONS FOR BIHARMONIC ELLIPTIC PROBLEMS WITH THE SECOND HESSIAN

Electronic Journal of Differential Equations, Vol 2016 (2016), No 289, pp 1 16 ISSN: 1072-6691 URL: http://ejdemathtxstateedu or http://ejdemathuntedu MULTIPLE SOLUTIONS FOR BIHARMONIC ELLIPTIC PROBLEMS WITH THE SECOND HESSIAN FEI FANG, CHAO JI, BINLIN ZHANG Abstract In this article, we study the biharmonic elliptic problem with the secondnd Hessian 2 u = S 2 (D 2 u) + λf(x) u p 1 u, in R 3, u = u n = 0, on, where f(x) C( ) is a sign-changing weight function By using variational methods and some properties of the Nehari manifold, we prove that the biharmonic elliptic problem has at least two nontrivial solutions 1 Introduction Let be a bounded domain in R 3, 0 < p < 1 In this work, we consider the problem 2 u = S 2 (D 2 u) + λf(x) u p 1 u, in, u = u n = 0, on, (11) where f(x) C( ) is a sign-changing weight function, S 2 (D 2 u)(x) = λ i (x)λ j (x), λ i, (i = 1,, N) are the solutions of the equation det(λi D 2 u(x)) = 0, and 2 the bi-laplacian operator The case N = 2 appears as the stationary part of a model of epitaxial growth of crystals (see [6, 15]) initially studied in [7] In dimension N = 3 the model can be seen as the stationary part of a 3-dimensional growth problem driven by the scalar curvature For the case n = 2, the equation is expressed by the formula 2 u = det(d 2 u) + λf(x)u, in R 2 (12) In this case, (12) was studied by Escudero and Peral [7] For a Dirichlet boundary condition, they used variational methods to prove that (12) has at least two 2010 Mathematics Subject Classification 35J50, 35J60, 35J62, 35J96 Key words and phrases Biharmonic elliptic problem; second Hessian; variational methods; Nehari manifold c 2016 Texas State University Submitted August 18, 2016 Published October 26, 2016 1

2 F FANG, C JI, B ZHANG EJDE-2016/289 solutions However under the Navier boundary condition, (12) does not have a variational characteristic, so the authors used fixed point arguments to obtain existence of solutions For the evolution formula of problem (12), Escudero, Gazzola, and Peral [9] proved existence of local solutions for arbitrary data and existence of global solutions for small data Moreover, by exploiting the boundary conditions and the variational structure of the equation, according to the size of the data the authors proved finite time blow-up of the solution and (or) convergence to a stationary solution for global solutions For problem (11), Ferrari, Medina and Peral [12] obtained the following results for f(x) 1: (1) If p < 1 there exists a λ 0 > 0 such that if 0 < λ < λ 0, problem (11) has at least two nontrivial solutions (2) If p > 1 problem (11) has at least one nontrivial solution for every λ 0 (3) If p = 1 problem (11) has at least one nontrivial solution whenever 0 < λ < λ 1, where λ 1 denotes the first eigenvalue of 2 in with Dirichlet boundary conditions In the high dimensional case, Escudero and Torres [11] proved the existence of radial solutions for the problem 2 u = ( 1) k S k [u] + λf(x), in B 1 (0) R N, provided either with Dirichlet boundary conditions or Navier boundary conditions, where the k-hessian S k [u] is the k-th elementary symmetric polynomial of eigenvalues of the Hessian matrix We can state now the following result Theorem 11 Let 0 < p < 1 There exists λ 0 > 0 such that for each λ (0, λ 0 ), problem (11) has at least two nontrivial solutions As in [12], we will use variational methods and some properties of the Nehari manifold to obtain two nontrivial solutions For a study on variational methods and their applications, we refer the reader to [4, 17, 18, 20, 21] The Nehari manifold was introduced by Nehari in [19] and has been widely used; see [1, 2, 3, 13, 14, 16, 22, 23, 24, 25] The main idea for the proof or theorem 11 is dividing the Nehari manifold into two parts and then considering the minimum of the functional on each part This article is organized as follows In Section 2, we give some preliminary lemmas In Section 3, we present the proof of Theorem 11 2 Preliminaries To use variational methods and some properties of the Nehari manifold, we firstly define the corresponding functional and Nehari manifold with respect to problem (11) The energy functional for problem (11) is I(u) = 1 u 2 dx ij u i u j u dx λ f(x) u p+1 dx, (21) 2 p + 1

EJDE-2016/289 MULTIPLE SOLUTIONS 3 u W 2,2 0 () From [12], we know that (I (u), v) = u v dx ( i u j u ij v + j u ij u i v + i u ji u j v) dx λf(x) u p v J(u) = (I (u), u) = u 2 dx 3 (J (u), u) = 2 u 2 dx 9 i u j u ij u dx λf(x) u p+1 dx, i u j u ij u dx (p + 1) λf(x) u p+1 dx As the energy functional I is not bounded on W 2,2 0 (), it is useful to consider the functional on the Nehari manifold N = {u : (I (u), u) = 0} Furthermore, we consider the minimization problem: for λ > 0 α = inf{i(u) : u N } The Nehari manifold N can be split three parts: N + = {u : (J (u), u) > 0}, N 0 = {u : (J (u), u) = 0}N = {u : (J (u), u) < 0} Lemma 21 There exists λ 1 > 0 such that for each λ (0, λ 1 ), N 0 = Proof We consider the following two cases Case 1 Assume that u N and λf(x) u p+1 dx = 0 This implies (I (u), u) = u 2 dx 3 i u j u ij u dx = 0 Hence, and so u N 0 (J (u), u) = u 2 dx < 0 Case 2 u N and λf(x) u p+1 dx 0 Assume that N 0 for all λ > 0 If u N 0, then 0 = (J (u), u) = 2 u 2 dx 9 i u j u ij u dx Therefore, (p + 1) λf(x) u p+1 dx = (1 p) u 2 dx (6 3p) u 2 dx = (6 3p) (1 p) i u j u ij u dx (22) i u j u ij u dx, (23)

4 F FANG, C JI, B ZHANG EJDE-2016/289 λf(x) u p+1 dx = = 3 1 p u 2 dx 3 i u j u ij u dx i u j u ij u dx Moreover, using Hölder s inequality, one has 1 u 2 dx = u 2 dx 3 i u j u ij u dx (2 p) = λ f(x) u p+1 dx λ f L m u p+1 1+q λ f L ms p+1( u 2 dx where m = 1+q q p (so the conjugate index m = 1+q p+1 have ( or ( ) p+1 2, (24) (25) 2N ), q + 1 < N 4 By (25), we ) 1 p u 2 2 dx λ(2 p) f L ms p+1 (26) ) u 2 dx ( λ(2 p) f L ms p+1) 2 p 1 Define the following functional on W 2,2 0 (), [ ( A(u) = K(p, q) u 2 dx) q iju i u j u dx where K(p, q) = 3 ( 1 p ) q q 1 1 p 6 3p Then by (23) and (24), we have A(u) = 0 On the other hand, for u W 2,2 0 (), we have ( ) 1/2 ( ij u i u j u dx C u 2 dx ( C ] 1 q 1 λf(x) u p+1 dx, ) 1/2 u 4 dx (27) ) 3/2 u 2 dx Then using (25), (26), the Holder inequality and Sobolev inequality, for u N 0, we deduce [ ( A(u) K(p, q) u 2 dx ) q ] 1 q 1 λ f L m u p+1 1+q iju i u j u dx [ ( K(p, q) u 2 dx ) q ] 1 ( ) p+1 C ( u 2 dx ) q 1 Cλ f 3/2 L m u 2 2 dx ( ) p+1 [ ( ) q 2 pq+p ] u 2 2 dx K(p, q) u 2 2(q 1) dx Cλ f L m ( ) p+1 [ u 2 2 dx K(p, q) ((λ(2 p) f L ms p+1 ) 2 p 1 ) q(1 p)+p 2 2(q 1)

EJDE-2016/289 MULTIPLE SOLUTIONS 5 Cλ f L m ] Since q(1 p) + p 2 2 2(q 1) p 1 < 0, for λ sufficiently small, we have A(u) > 0 This contradicts A(u) = 0 Hence we can conclude that there exits λ 1 > 0 such that for λ (0, λ 1 ), N 0 = Lemma 22 If u N +, then λf(x) u p+1 dx > 0 Proof For u N +, we have u 2 dx 3 2 u 2 dx 9 i u j u ij u dx λf(x) u p+1 dx = 0, i u j u ij u dx (p + 1) Combining the above two formulas, we have (2 p) λf(x) u p+1 dx > This completes the proof u 2 dx > 0 λf(x) u p+1 dx > 0 According to Lemma 22, for λ (0, λ 1 ), we can write N = N + N and define α + = inf I(u), α = inf I(u) u N + u N Next we show that the minimizers on N are the critical points for I We denote the dual space of W 2,2 0 () by ( W 2,2 0 () ) Lemma 23 For λ (0, λ 1 ), if u 0 is a local minimizer for I(u) on N, then I (u 0 ) = 0 in ( (W 2,2 0 () ) Proof If u 0 is a local minimizer for I(u) on N, then u 0 is a solution of the optimization problem minimize I(u) subject to J(u) = 0 Hence, by the theory of Lagrange multipliers, there exists θ R such that Thus, I (u 0 ) = θj (u 0 ) in ( W 2,2 0 () ) (I (u 0 ), u 0 ) = θ(j (u 0 ), u 0 ) (28) Since u 0 N, (I (u 0 ), u 0 ) = 0 Moreover, since N =, (J (u 0 ), u 0 ) 0 and by (28), θ = 0 This completes the proof For u W 2,2 0 (), we write Lemma 24 (1 p) t max = u 2 dx (6 3p) iju i u j u dx iju i u j u dx < 0 ( 0), then there exits a (1) If unique t > 0 (t + > 0) such that t u N + (t + u N ) and I(t u) = min t>0 I(tu) (I(t u) = max t>0 I(tu)); (2) t (u) is a continuous function for nonzero u;

6 F FANG, C JI, B ZHANG EJDE-2016/289 (3) Proof (1) We firstly define N + = { u W 2,2 0 () \ {0} : t ( u ) 1 u u = 1} i(t) := I(tu) = t2 2 We easily compute i (t) := I (tu) = t and (I (tu), tu) = t 2 u 2 dx 3t 3 = ti (t) t p+1 t p u 2 dx t 3 λf(x) p + 1 u p+1 dx u 2 dx 3t 2 λf(x) u p+1 dx i u j u ij u dx t p+1 ij u i u j u dx ij u i u j u dx λf(x) u p+1 dx (29) (210) (211) We distinguish the following two cases Case i iju i u j u dx < 0 In this case, i(t) is convex and achieves its minimum at t and t 0 Thus, using (29) and (211), we obtain t u N + and I (t) > 0 for t = t Case ii iju i u j u dx > 0 Let s(t) = t 1 p u 2 dx 3t 2 p ij u i u j u dx It is easy to show that s(0) = 0, s(t) as t + is convex and achieves its maximum at (1 p) t max = u 2 dx (6 3p) iju i u j u dx Then, using (27) we obtain s(t max ) = s(t) ( (1 p) = u 2 dx ) (6 3p) 1 p u 2 dx iju i u j u dx ( (1 p) 3 u 2 dx (6 3p) ) 2 p iju i u j u dx ij u i u j u dx

EJDE-2016/289 MULTIPLE SOLUTIONS 7 = (3p 2) u 2 dx ( C 1 u 2 dx ) 1+p 2 ( (1 p) u 2 dx (6 3p) iju i u j u dx ) 1 p From the above inequality, there exists a λ 0 such that for λ (0, λ 0 ) small, s(0) = 0 < λ f(x) u p+1 dx = λ f(x) u p+1 dx λ f L m u p+1 1+q λ f L ms p+1( u 2 dx ) p+1 2 s(t max ) (212) where m = 1+q q p (so the conjugate index m = 1+q 2N p+1 ), q + 1 < N 4 Using (212), we easily deduce that there are unique values t + and t such that 0 < t + < t max < t, s(t + ) = λ f(x) u p+1 dx = s(t ), s (t + ) > 0 > s (t ) We have t + u N +, t u N, and I(t u) I(tu) I(t + u) for each t [t +, t ] and I(t + u) I(tu) for each t [0, t + ] Thus I(t u) = max t t max I(tu), I(t + u) = min 0 t t I(tu) In this case, i(t) is concave and achieves its maximum at t + and t + 0 Thus, using (29) and (211), we obtain t + u N and I (t) < 0 for t = t + (2) By the uniqueness of t (u) and the external property of t (u), we have that t (u) is a continuous function of u 0 (3) For u N +, let v = u u Using the discussion (1), there exists an t > ( ) 0 such that t v N +, that is t u u u u N + Since u N +, we obtain t ( u u ) 1 u = 1 This shows that N + { u W 2,2 0 () \ {0} : t ( u ) 1 u u = 1} ( Conversely, let u W 2,2 0 () \ {0} such that t Hence, u u t ( u ) u u u N + ) 1 u = 1, then N + = { u W 2,2 0 () \ {0} : t ( u ) 1 u u = 1}

8 F FANG, C JI, B ZHANG EJDE-2016/289 Now we consider the degenerate equation 2 u = S 2 (D 2 u), in, u = u n = 0, on (213) The functional corresponding to (213) is H(u) = 1 u 2 dx ij u i u j u dx 2 We consider the minimization problem β = inf{h(u) : u N}, where N = {u : u W 2,2 0 () \ {0} : (H (u), u) = 0} Next we show that problem (213) has a nontrivial solution ω 0 such that H(ω 0 ) = β > 0 Lemma 25 For any u W 2,2 0 () \ {0}, there exits an unique t(u) > 0 such t(u)u N The maximum of H(tu) for t 0 is achieved at t = t(u) The function W 2,2 0 ()\{0} (0, + ) : u t(u) is continuous and defines a homeomorphism of the unit sphere of W 2,2 0 () with N Proof Let u W 2,2 0 ()\{0} be fixed and define the function g(t) := H(tu) on [0, ) Obviously, we obtain g (t) = 0 tu N (214) u 2 dx = 3t i u j u ij u dx = 0 (215) If for all u W 2,2 0 (), it holds iju i u j u dx 0, then 0 is an unique critical point of H(u) And if iju i u j u dx > 0, using the mountain pass theorem, we can show that H(u) has a nontrivial critical point So for each u W 2,2 0 ()\{0}, it is easy to verify that g(0) = 0 and g(t) > 0 for t > 0 small and g(t) < 0 for t > 0 large Therefore max [0, ) g(t) is achieved at an unique t = t(u) such that g (t(u)) = 0 and t(u)u N To prove the continuity of t(u), assume that u n u in W 2,2 0 ()\{0} It is easy to verify that {t(u n )} is bounded If a subsequence of {t(u n )} converges to t 0, it follows from (214) that t 0 = t(u), but then t(u n ) t(u) Finally the continuous map from the unit sphere of W 2,2 0 ()\{0} N, u t(u)u, is inverse of the retraction u u u a Define where c 1 := inf max H(tu), u W 2,2 0 ()\{0} t 0 c := inf max H(γ(tu)), r Γ t [0,1] Γ := { γ C[0, 1], W 2,2 0 () : γ(0) = 0, H(γ(1)) < 0 } Lemma 26 c 1 = c = β > 0 and c is a critical value of H(u) Proof From Lemma 25, we easily know that β = c 1 Since H(tu) < 0 for u W 2,2 0 ()\{0} and t large, we have c c 1 The manifold N separates W 2,2 0 () into two components The component containing the origin also contains a small ball around the origin Moreover H(u) 0 for all u in this component, because

EJDE-2016/289 MULTIPLE SOLUTIONS 9 (H(tu), u) 0 for all 0 t t(u) Thus every γ Γ has to cross N and β c Since the embedding W 2,2 0 () L m () (m < 2 ) is compact, it is easy to prove that c > 0 is a critical value of H(u) and ω 0 a nontrivial solution corresponding to c Lemma 27 (1) There exist ˆt > 0 such that α α + < p 1 6p + 6 ˆt 2 β < 0 (2) I(u) is coercive and bounded below on N for λ sufficiently small Proof (1) Let ω 0 be a nontrivial solution of problem (213)) such that H(ω 0 ) = β > 0 Then ω 0 2 dx 3 i ω 0 j ω 0 ij ω 0 dx = 0 Set ˆt = t + () as defined by Lemma 24 Hence ˆtω 0 N + and I(ˆtω 0 ) = ˆt 2 ω 0 2 dx ˆt 3 ij ω 0 i ω 0 j ω 0 dx 2 ˆt p+1 λf(x) p + 1 ω 0 p+1 dx = ( 1 2 1 )ˆt 2 ω 0 2 dx p + 1 + ( 3 p + 1 1)ˆt 3 ij ω 0 i ω 0 j ω 0 dx This yields < p 1 6p + 6 ˆt 2 β (2) For u N, we have J(u) = (I (u), u) = u 2 dx 3 α α + < p 1 6p + 6 ˆt 2 β < 0 Then by Hölder and Young inequalities, I(u) = 1 u 2 dx ij u i u j u dx 2 = 1 6 1 6 ( 1 6 u 2 dx ( λ p + 1 λ ) f(x) u p+1 dx 3 u 2 dx ( λ p + 1 λ ) f L ms p+1( 3 2λ(2 p) ) 3(p + 1) 2 u 2 λ(2 p) dx 3(p + 1) i u j u ij u dx λ f(x) u p+1 dx = 0 λ f(x) u p+1 dx p + 1 ) p+1 u 2 2 dx ( f L ms p+1) 2 1 p (216) (217)

10 F FANG, C JI, B ZHANG EJDE-2016/289 In (217), since p < 1, for λ small, we have I(u) > 0 on N So we easily know that I(u) is coercive and bounded below on N for λ sufficiently small We need the following lemmas 3 Proof of Theorem 11 Lemma 31 For each u N, there exist ε > 0 and a differentiable function ξ : B(0, ε) W 2,2 0 () R + such that ξ(0) = 1, the function ξ(v)(u v) N and (ξ (0), v) = 2 u 2 dx 9 iu j u ij u dx λ(p + 1) f(x) u p+1 dx (1 p) u 2 dx (6 3p), iu j u ij u dx for all v W 2,2 0 () Proof For u N, define a function by F : R W 2,2 0 () R by F u (ξ, ω) = (I(ξ(u ω), ξ(u ω))) = ξ 2 (u ω) 2 dx 3ξ 3 λξ p+1 f(x) (u ω) p+1 dx i (u ω) j (u ω) ij (u ω)dx Then F u (1, 0) = (I (u), u) = 0 and d dt F u(1, 0) = 2 u 2 dx 9 i u j u ij u dx λ(p + 1) f(x) u p+1 dx = (1 p) u 2 dx (6 3p) i u j u ij u dx 0 (31) (32) According to the implicit function theorem, there exist ε > 0 and a differentiable function ξ : B(0, ε) W 2,2 0 () R + such that ξ(0) = 1 and and (ξ (0), v) = 2 u 2 dx 9 iu j u ij u dx λ(p + 1) f(x) u p+1 dx (1 p) u 2 dx (6 3p) iu j u ij u dx that is, ξ(v)(u v) N F u (ξ(v), v) = 0 Similarity, we have the following result for all v B(0, ε); Lemma 32 For each u N, there exist ε > 0 and a differentiable function ξ : B(0, ε) W 2,2 0 () R + such that ξ (0) = 1, the function ξ (v)(u v) N and (ξ (0), v)

EJDE-2016/289 MULTIPLE SOLUTIONS 11 = 2 u 2 dx 9 iu j u ij u dx λ(p + 1) f(x) u p+1 dx (1 p) u 2 dx (6 3p), iu j u ij u dx for all v W 2,2 0 () Proof As in the proof in Lemma 31, there exist ε > 0 and a differentiable function ξ : B(0, ε) W 2,2 0 () R + such that ξ 0 = 1 and ξ (v)(u v) N for all v B(0, ε) Since (J (u), u) = (1 p) u 2 dx (6 3p) i u j u ij u dx < 0 Thus, by the continuity of the function J (u) and ξ, we have (J (ξ (v)(u v)), ξ (v)(u v)) = (1 p) ( ξ (v)(u v) ) 2 dx ( (6 3p) i ξ (v)(u v) ) ( j ξ (v)(u v) ) ij ( ξ (v)(u v) ) dx < 0 For ε sufficiently small, this implies ξ (v)(u v) N Lemma 33 Let λ 0 = inf{λ 1, λ 2 } (1) There exists a minimizing sequences {u n } N such that (33) I(u n ) = α + o(1), I (u n ) = o(1) for ( W 2,2 0 () ) (34) (2) There exists a minimizing sequences {u n } N such that I(u n ) = α + o(1), I (u n ) = o(1) for ( W 2,2 0 () ) (35) Proof Using Lemma 27 and Ekeland variational principle [5], there exists a minimizing sequence {u n } N such that I(u n ) < α + 1 n, (36) I(u n ) < I(ω) + 1 n ω u n for each ω N (37) By taking n enough large, from Lemma 27 (1), we have I(u n ) = 1 u n 2 (2 p)λ dx f(x) u n p+1 dx 2 p + 1 < α + 1 n < p 1 6p + 6 ˆt 2 β < 0 This implies ) p+1 2 (38) 1 p 6λ(2 p) ˆt 2 β (39) f L ms p+1( u n 2 dx f(x) u n p+1 dx > Consequently u n 0 and combining the above two estimates and the Holder inequality, we obtain u n 2 dx > [ 1 p 6λ(2 p) ˆt 2 β f 1 L ms p 1] 2 p+1, (310)

12 F FANG, C JI, B ZHANG EJDE-2016/289 Next we show that u n 2 dx < [ (4 2p)λ 2 (p + 1) f L msp+1] 1 p (311) I (u n ) (W 2,2 0 ()) 0 as n + Applying Lemma 31 with u n to obtain the function ξ n : B(0, ε n ) W 2,2 0 () R + for some ε n > 0, such that ξ n (ω)(u n ω) N Choose 0 < ρ < ε n Let u W 2,2 0 () with u 0 and let ω ρ = ρu u We set η ρ = ξ n (ξ ρ )(u ω ρ ) Since η ρ N, we deduce that from (37) that and by the mean value theorem, we have Thus I(η ρ ) I(u n ) 1 n η ρ u n, (I (u n ), η ρ u n ) + o( η ρ u n 1 n η ρ u n ) (312) (I (u n ), ω ρ ) + (ξ n (ω ρ ) 1)(I (u n ), (u n ω ρ )) 1 n η ρ u n + o( η ρ u n ) (313) It follows from (ξ n (ω ρ ))(u n ω ρ ) N and (313) that ρ ( I u ) (u n ), + (ξn (ω ρ ) 1)(I (u n ) I (η ρ ), (u n ω ρ )) u 1 n η ρ u n + o( η ρ u n ) (314) Thus ( I u ) (ξ n (ω ρ ) 1) (u n ), (I (u n ) I (η ρ ), (u n ω ρ )) u ρ + 1 nρ η ρ u n + o( η ρ u n ) ρ Since η ρ u n ξ n (ω ρ 1) u n + ρ ξ n (ω ρ ) and ξ n (ω ρ 1) lim ξ ρ 0 ρ n(0) (315) If we let ρ 0 in (315) for a fixed n, then by (311) we can find a constant C > 0, independent of ρ, such that ( I (u n ), u ) C u n (1 + ξ n(0) ) (316) We are done once we show that ξ n(0) is uniformly bounded in n By (311) and Lemma 31 and Hölder inequality, we have (ξ b v (0), v) = (1 p) u 2 dx (6 3p) iu j u ij u dx, for some b > 0 We only need to show that (1 p) u 2 dx (6 3p) i u j u ij u dx > 0, (317)

EJDE-2016/289 MULTIPLE SOLUTIONS 13 for some c > 0 and n large enough We argue by contradiction Assume that there exists a subsequence {u n } such that (1 p) u 2 dx (6 3p) i u j u ij u dx = o(1) (318) Using (27), (318) and (310), we can find a constant d > 0 such that i u n j u n ij u n dx d (319) for n sufficiently large In addition (318), and the fact {u n } N also give λ f(x) u n p+1 dx = u n 2 dx 3 i u n j u n ij u n dx (320) = 3 1 p i u n j u n ij u n dx + o(1) (321) and [ (4 2p)λ u n 2 dx < (p + 1) f L msp+1] 2 1 p + o(1) (322) This implies [ ( A(u) = K(p, q) u 2 dx ) q ] 1 q 1 λf(x) u p+1 dx, iju i u j u dx 3 ( 1 p ) q [ ( 6 3p q 1 1 p iu j u ij u dx ) q ] 1 q 1 1 p 6 3p (323) iju i u j u dx 3 i u n j u n ij u n dx + o(1) = o(1) 1 p However, from (319) and (322), for λ small, we have [ ( A(u) K(p, q) u 2 dx ) q ] 1 q 1 λ f L m u p+1 1+q iju i u j u dx [ ( K(p, q) u 2 dx ) q ] 1 ( ) p+1 C ( u 2 dx ) q 1 Cλ f 3/2 L m u 2 2 dx ( ) p+1 [ ( ) q 2 pq+p ] u 2 2 dx K(p, q) u 2 2(q 1) dx Cλ f L m ( ) p+1 [ ( u 2 2 (λ(2 dx K(p, q) p) f L ms p+1) 2 p 1 Cλ f L m ] This contradicts (323) We deduce that ( I u ) (u n ), u ) q(1 p)+p 2 2(q 1) (324) C n (325) The proof is complete (2) Similar to the proof of (1), we may prove (2) Now we establish the existence of a local minimum for I on N +

14 F FANG, C JI, B ZHANG EJDE-2016/289 Lemma 34 For λ small, the functional I has a minimizer u + 0 N + and it satisfies (1) I(u + 0 ) = α = α+ ; (2) u + 0 is a nontrivial nonnegative solution of problem (11); (3) I(u + 0 ) 0 as λ 0 Proof Let {u n } N be a minimizing sequence for I on N such that I(u n ) = α + o(1), I (u n ) = o(1), for ( W 2,2 0 () ) (326) Then by Lemma 27 and the compact embedding theorem, there exists a subsequence {u n } and u + 0 W 2,2 0 () such that u n u + 0 in W 2,2 0 (), u n u + 0 in L h (), where 1 < h < 2 We now show that f(x) u 0 p+1 dx 0 If not, by (326), we can conclude that f(x) u n p+1 dx = 0, f(x) u n p+1 dx 0 as n Thus, u n 2 dx = 3 i u n j u n ij u n dx + o(1), and I(u n ) = 1 u n 2 dx ij u n i u n j u n dx 2 λ f(x) u n p+1 dx p + 1 = 1 (327) ij u n i u n j u n dx + o(1) 2 1 ij u 0 i u 0 j u 0 dx as n +, 2 Similar to Lemma 25, we can see that iju 0 i u 0 j u 0 dx > 0 So (327) contradicts I(u n ) α < 0 as n + In particular, u + 0 N + is a nontrivial solution of problem (11) and I(u + 0 ) α Similar to the proof of [12, Lemma 31], we can prove that u n u + 2,2 0 strongly in W0 ) In fact, if u + 0 N, by Lemma 24, there are unique t + 0 and t 0 such that t+ 0 u+ 0 N + and t 0 u+ 0 N, we have t + 0 < t 0 = 1 Since d dt I(t+ 0 u+ 0 ) = 0 and d 2 dt 2 I(t+ 0 u+ 0 ) > 0, there exists t + 0 < t t 0 such that I(t+ 0 u+ 0 ) < I( tu + 0 ) By I(t + 0 u+ 0 ) < I( tu + 0 ) I(t 0 u+ 0 ) = I(u+ 0 ),

EJDE-2016/289 MULTIPLE SOLUTIONS 15 which is a contradiction By Lemma 23, we know that u + 0 Moreover, from (217), we know that 0 > I(u + 0 It is clear that I(u + 0 ) 0 as λ 0 p) ( ) λ(2 f L ms p+1) 2 1 p 3(p + 1) is a nontrivial solution As in the proof of Lemma 34, we establish the existence of a local minimum for I on N Lemma 35 For λ small, the functional I has a minimizer u 0 N and it satisfies (1) I(u 0 ) = α ; (2) u 0 is a nontrivial nonnegative solution of problem (11) Combining Lemma 34 and 35, for problem (11), there exist two nontrivial solutions u + 0 and u0 such that u + 0 N +, u 0 N Since N + N =, this shows that u + 0 and u0 are different Acknowledgements The first author is supported by Young Teachers Foundation of BTBU (No QNJJ2016-15) The second author is supported by NSFC (No 11301181) and China Postdoctoral Science Foundation The third author is supported by Research Foundation of Heilongjiang Educational Committee (No 12541667) References [1] K J Brown; The Nehari manifold for a semilinear elliptic equation involving a sublinear term, Calc Var Partial Differential Equations 22 (2005) 483-494 [2] K J Brown, Y P Zhang; The Nehari manifold for a semilinear elliptic equation with a sign-changing weight function, J Differential Equations 193 (2003) 481-499 [3] J Chabrowski, D G Costa; On a class of Schrodinger-type equations with indefinite weight functions, Comm Partial Differential Equations 33 (2008) 1368-1394 [4] L M Del Pezzo, J D Rossi, N Saintier, A Salort; An optimal mass transport approach for limits of eigenvalue problems for the fractional p-laplacian, Adv Nonlinear Anal 4 (3) (2015) 235-249 [5] I Ekeland; On the variational principle, J Math Anal Appl 17 (1974) 324-353 [6] C Escudero; Geometric principles of surface growth Phys Rev Lett 101, 196102 (2008) [7] C Escudero, I Peral; Some fourth order nonlinear elliptic problems related to epitaxial growth, J Differential Equations 254 (2013) 2515-2531 [8] C Escudero, R Hakl, I Peral, P J Torres; On radial stationary solutions to a model of nonequilibrium growth, Eur J Appl Math, 24 (2013), 437-453 [9] C Escudero, F Gazzola, I Peral; Global existence versus blow-up results for a fourth order parabolic PDE involving the Hessian, J Math Pures Appl 103 (2015), 924-957 [10] C Escudero, R Hakl, I Peral, P J Torres; Existence and nonexistence results for a singular boundary value problem arising in the theory of epitaxial growth, Math Methods Appl Sci 37 (2014), 793-807 [11] C Escudero, P J Torres; Existence of radial solutions to biharmonic k-hessian equations, J Differential Equations 259 (7) (2015), 2732-2761 [12] F Ferrari, M Medina, I Peral; Biharmonic elliptic problems involving the 2-nd Hessian operator, Calc Var Partial Differential Equations 51 (2014), 867-886 [13] C Ji; The Nehari manifold for a degenerate elliptic equation involving a sign-changing weight function, Nonlinear Anal 75 (2012) 806-818 [14] C Ji, W H Wang; On the p-biharmonic equation involving concave-convex nonlinearities and sign-changing weight function, Electron J Qual Theorty Differ Equ No2, (2012), 1-17

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