MULTIPLE POSITIVE SOLUTIONS FOR KIRCHHOFF PROBLEMS WITH SIGN-CHANGING POTENTIAL

Electronic Journal of Differential Equations, Vol. 015 015), No. 0, pp. 1 10. ISSN: 107-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu MULTIPLE POSITIVE SOLUTIONS FOR KIRCHHOFF PROBLEMS WITH SIGN-CHANGING POTENTIAL GAO-SHENG LIU, CHUN-YU LEI, LIU-TAO GUO, HONG RONG Abstract. In this article, we study the existence and multiplicity of positive solutions for a class of Kirchhoff type equations with sign-changing potential. Using the Nehari manifold, we obtain two positive solutions. 1. Introduction and statement of main result Consider the Kirchhoff type problems with Dirichlet boundary value conditions a + b u + vx)u ) dx) u vx)u) = hx)u p + λfx, u) in, 1.1) u = 0 on, where is a smooth bounded domain in R 3, a > 0, b > 0, λ > 0, 3 < p < 5, h C ), with h + = max{h, 0} 0, v C ) is a bounded function with v > 0, and fx, u) satisfies the following two conditions: F1) fx, u) C 1 R) with fx, 0) 0, and fx, 0) 0. There exists a constant c 1 > 0, such that fx, u) c 1 1 + u q ) for 0 < q < 1 and x, u) R +. F) f u x, u) L R) and for all u H0 1 ), u fx, t u )u has the same sign for every t 0, + ). Remark 1.1. Note that under assumptions F1) and F) hold, we have: F3) there exists a constant c > 0, such that pfx, u) uf u x, u) c 1 + u), for all x, u) R +. F4) F x, u) 1 p+1 fx, u)u c 1 + u ), for all x, u) R +, where F x, u) is defined by F x, u) = u fx, s)ds for x, u R. 0 In recent years, the existence and multiplicity of solutions to the nonlocal problem a + b ) u dx u = gx, u) in, 1.) u = 0, on, have been studied by various researchers and many interesting and important results can be found. For instance, positive solutions could be obtained in [3, 5, 13]. 010 Mathematics Subject Classification. 35D05, 35J60, 58J3. Key words and phrases. Kirchhoff type equation; sign-changing potential; Nehari manifold. c 015 Texas State University - San Marcos. Submitted June, 015. Published August 4, 015. 1

G.-S. LIU, C.-Y. LEI, L.-T. GUO, H. RONG EJDE-015/0 Especially, Chen et al [4] discussed a Kirchhoff type problem when gx, u) = fx)u p u + λgx) u q u, where 1 < q < < p < = N N if N 3, = if N = 1, ), fx) and gx) with some proper conditions are sign-changing weight functions. And they have obtained the existence of two positive solutions if p > 4, 0 < λ < λ 0 a). Researchers, such as Mao and Zhang [], Mao and Luan [1], found sign-changing solutions. As for infinitely many solutions, we refer readers to [11, 1]. He and Zou [14] considered the class of Kirchhoff type problem when gx, u) = λfx, u) with some conditions and proved a sequence of a.e. positive weak solutions tending to zero in L ). In addition, problems on unbounded domains have been studied by researchers, such as Figueiredo and Santos Junior [9], Li et al. [15], Li and Ye [8]. Our main result read as follows. Theorem 1.. Assume that conditions F1) and F) hold. Then there exists λ > 0 such that for any λ 0, λ ), problem 1.1) has at least two positive solutions. The article is organized as following: Section contains notation and preliminaries. Section 3 contains the proof of Theorem 1... Preliminaries Throughout this article, we use the following notation: The space H 1 0 ) is equipped with the norm u = u +vx) u ) dx. Let S r be the best Sobolev constant for the embedding of H 1 0 ) into L r ), where 1 r < 6, then 1 S p+1) p+1 We define a functional I λ u): H0 1 ) R by I λ u) = a u + b 4 u 4 1 Hu) λ where Hu) = u p+1) u p+1 )..1) hx) u p+1 dx. F x, u ) dx for u H 1 0 ),.) The weak solutions of 1.1) is the critical points of the functional I λ. Generally speaking, a function u is called a solution of 1.1) if u H0 1 ) and for all ϕ H0 1 ) it holds a + b u ) u ϕ + vx)uϕ) dx = hx) u u ϕ dx + λ fx, u )ϕ dx. As I λ u) is unbounded below on H 1 0 ), it is useful to consider the functional on the Nehari manifold: N λ ) = {u H 1 0 )\{0} : I λu), u = 0}. It is obvious that the Nehari manifold contains all the nontrivial critical points of I λ, thus, for u N λ ), if and only if a + b u ) u hx) u p+1 dx λ fx, u ) u dx = 0..3) Define ψ λ u) = I λu), u,

EJDE-015/0 MULTIPLE POSITIVE SOLUTIONS 3 then it follows that I λ tu) = a t u + b 4 t4 u 4 tp+1 hx) u p+1 dx λ F x, tu ) dx,.4) ψ λ tu) = at u + bt 4 u 4 t p+1 hx) u p+1 dx λ fx, tu ) tu dx,.5) ψ λtu), tu = at u + 4bt 4 u 4 )t p+1 hx) u p+1 dx.6) λ f u x, tu ) tu dx λ fx, tu ) tu dx. Notice that ψ λ tu) = 0 if and only if tu N λ ). And we divide N λ ) into three parts: Then we have the following results. N λ ) = {u N λ) : ψ λu), u < 0}, N + λ ) = {u N λ) : ψ λu), u > 0}, N 0 λ) = {u N λ ) : ψ λu), u = 0}. Lemma.1. There exists a constant λ 1 > 0, for 0 < λ < λ 1, such that Nλ 0 ) =. Proof. By contradiction, suppose u Nλ 0 ), we obtain ψ λu), u = a u + 4b u 4 ) hx) u p+1 dx λ f u x, u ) u dx λ fx, u ) u dx = 0. On one hand, from.1),.3),.6) and F), one deduces that a u + 3b u 4 = p hx) u p+1 dx + λ f u x, u )u dx L u p+1 + λl u, where L = p h S p+1 p+1, L = f u x, u ) L S, then L u p+1 a λl ) u + 3b u 4 a λl ) u, consequently, a λl u )..7) L On the other hand, by.1),.3),.6) and F3), we obtain ) ap 1) u + bp 3) u 4 λ pfx, u ) f u x, u ) u ) u dx c λ u + u ) dx then thus one has λc 1 S1 u + λc S u, λc 1 S1 u + λc S u ap 1) u, u λc S 1 1/ )..8) ap 1) c λs

4 G.-S. LIU, C.-Y. LEI, L.-T. GUO, H. RONG EJDE-015/0 It follows from.7) and.8) that a λl ) u λc S 1 1/ ), L ap 1) c λs which is a contradiction when λ is small enough. So there exists a constant λ 1 > 0 such that Nλ 0 ) =. The proof is complete. Lemma.. There exists a constant λ > 0, for 0 < λ < λ, such that N ± λ ). Proof. For u H0 1 ), u 0, let A u t) = a t u + b 4 t4 u 4 tp+1 hx) u p+1 dx, K u t) = F x, tu ) dx, then I λ tu) = A u t) λk u t), hence if ψ λ tu) = I λ tu), tu = 0, then A ut) λk ut) = 0, where A ut) = at u + bt 3 u 4 t p hx) u p+1 dx, K ut) = fx, tu ) u dx. By F1), one obtains K ut) = We consider the following two cases: fx, tu ) u dx c 1 + tu q ) u dx..9) Case 1. When Hu) 0 and fx, t u )u dx > 0, we have A ut) > 0, A u 0) = 0 and A u t) increases sharply when t. At the same time, K ut) > 0, K u 0) is a positive constant and K u t) increases relatively slowly when t since.9). When Hu) 0 and fx, t u )u dx 0, we have K ut) 0, K u 0) is a positive constant and K u t) decreases slowly when t since.9). Through the above discussion, we obtain there exists t 1 such that t 1 u N λ ) to every situation. When 0 < t < t 1, one gets ψ λ tu) < 0 and when t > t 1, we have ψ λ tu) > 0, then t 1 u is the local minimizer of I λ u), so t 1 u N + λ ). In conclusion, when Hu) 0, one has N + λ ). Case. When Hu) > 0 and fx, t u )u dx > 0, we have A ut) > 0 as t 0 and A ut) < 0 for t, so A u t) increases as t 0 and then decreases as t. At the same time, K ut) > 0, K u 0) is a positive constant and K u t) increases relatively slowly when t since.9). When Hu) > 0 and fx, t u )u dx < 0, we have A ut) > 0 as t 0 and A ut) < 0 for t, so A u t) increases as t 0 and then decreases as t. At the same time, K ut) < 0, K u 0) is a positive constant and K u t) decreases slowly when t since.9). Through the above discussion, if λ is small enough, there exists t 1 < t, such that ψ λ tu) = 0, for 0 < t < t 1, ψ λ tu) < 0, for t 1 < t < t, ψ λ tu) > 0, and for t > t, ψ λ tu) < 0. Thus t 1 u is the local minimizer of I λ u) and t u is the local maximizer of I λ u). So there exists λ > 0, when λ < λ, one gets t 1 u N + λ ) and t u N λ ). Therefore one concludes that when Hu) > 0 and λ is small enough, N ± λ ). This completes the proof.

EJDE-015/0 MULTIPLE POSITIVE SOLUTIONS 5 Lemma.3. Operator I λ is coercive and bounded below on N λ ). Proof. From.1),.),.3) and F4), one has 1 I λ u) = a 1 ) 1 u + b 4 1 ) u 4 λ F x, u 1 fx, u ) u ) dx 1 a 1 ) 1 u + b 4 1 ) u 4 λc 3 1 + u ) dx 1 a 1 ) 1 u + b 4 1 ) u 4 λc 3 + S u ) ap 1) ) 1 ) λc 3S u + b 4 1 ) u 4 λc 3. By 3 < p < 5, it follows that I λ u) is coercive and bounded below on N λ ). The proof is complete. Remark.4. From Lemmas.1 and., one has N λ ) = N + λ ) N λ ) for all 0 < λ < min{λ 1, λ }. Furthermore, we obtain N + λ ) and N λ ) are non-empty, thus, we may define α + λ = inf I λ u), α u N + λ ) λ = inf I λ u). u N λ ) Lemma.5. If u H 1 0 )\{0}, there exists a constant λ 3 > 0, such that I λ tu) > 0, for λ < λ 3. Proof. For every u H 1 0 ), u 0, if Hu) 0, by.4), we obtain I λ tu) > 0 when t is large enough. Assume Hu) > 0, and let φ 1 t) = a t u tp+1 Hu). Through calculations, one obtains that φ 1 t) takes on a maximum at a u ) 1 t max =. Hu) It follows that φ 1 t max ) = p 1 a u ) p+1 ) 1 ) hx) u p+1 dx) p 1 a p+1 ) 1 := δ ) h + S p+1) 1. p+1 When 1 r < 6, one has a u t max ) r u r dx Sr r ) r u ) r/ Hu) = Sr r a r a u ) p+1 ) r ) Hu)) = S r r a r ) p 1 = c φ 1 t max ) ) r/. ) r/ ) r/ φ 1 t max ).10)

6 G.-S. LIU, C.-Y. LEI, L.-T. GUO, H. RONG EJDE-015/0 Then by F1) and F4), we deduce that F x, t max u ) dx 1 c 4 + t max u ) dx + Since c 1 t max u + t max u q+1 ) B 0 + B 1 φ 1 t max ) + B φ 1 t max )) 1/ + B 3 φ 1 t max ) q+1 I λ t max u) = A u t max ) λk u t max ) φ 1 t max ) λ F x, t max u ) dx,..11) according to.4),.10) and.11), one obtains I λ t max u) φ 1 t max ) λ F x, t max u ) dx ] φ 1 t max ) λ [B 0 + B 1 φ 1 t max ) + B φ 1 t max )) 1/ + B 3 φ 1 t max ) q+1 )] δ 1 [1 λ B 0 δ 1 + B 1 + B δ 1 + B3 δ q 1. So, if λ < λ 3 = B 0 δ 1 +B 1 +B δ 1 +B 3 δ q 1 )) 1, we obtain I λ t max u) > 0. Remark.6. If λ < λ 3 and u N λ ), by F), we conclude that there is a global maximum on u for I λ u), then I λ u) > I λ t max u) > 0. Lemma.7. If u H 1 0 )\{0}, there exists a constant λ 4 > 0 such that ψ λ tu) = I λ tu), tu > 0 when λ < λ 4. Proof. For every u H 1 0 ), u 0, if Hu) 0, by.5), we get ψ λ tu) > 0 when t is large enough. Assume Hu) > 0, and let ψ 1 t) = at u t p+1 Hu). Through calculations, we obtain that ψ 1 t) takes on a maximum at a u t ) 1 max =. )Hu) It follows that a ) p 1 ) u ) p+1 ) 1 ψ 1 t max ) = hx) u p+1 dx) a ) p 1 ) 1 ) 1 := δ h + S p+1). p+1 Similar to the proof of Lemma.5, when 1 r < 6, one obtains t max ) r u r dx c ψ 1 t max ) ) r/..1) According to F1), we deduce that fx, t max u ) t max u dx c 1 t max u + t max u q+1) dx b 0 ψ1 t max ) ) 1/ + b1 ψ1 t max ) ) q+1,.13)

EJDE-015/0 MULTIPLE POSITIVE SOLUTIONS 7 then, by.5),.1) and.13), it follows that ψ λ t max u) ψ 1 t max ) λ fx, t max u ) t max u dx ) ψ 1 t max )) 1+q ψ 1 t max )) 1 q λb 0 ψ 1 t max )) q + b1 ) ) δ 1+q δ 1 q λb 0 δ q + b 1 ), consequently, when λ < λ 4 = δ 1 q /b 0 δ q + b 1 ), we obtain ψ λ t max u) > 0. Remark.8. We claim that: 1) If Hu) 0 for every u H0 1 )\{0}, there exists t 1 such that I λ t 1 u) < 0 for t 1 u N + λ ). Indeed, obviously, in this condition, ψ λ 0) < 0 and lim t ψ λ tu) = +, therefore, there exists t 1 > 0 such that ψ λ tu) = 0. Because of ψ λ tu) < 0 for 0 < t < t 1 and ψ λ tu) > 0 for t > t 1, we obtain that t 1 u N + λ ) and I λt 1 u) < I λ 0) = 0. ) If Hu) > 0 for 0 < λ < λ 1, there exists t 1 < t, such that t 1 u N + λ ), t u N λ ) and I λt 1 u) < 0. Indeed, in this condition, one gets ψ λ 0) < 0 and lim t ψ λ tu) =. By Lemma.7, there exists T > 0 such that ψ λ T u) > 0, therefore, we could obtain there exists 0 < t 1 < T < t, such that ψ λ t 1 u) = ψ λ t u) = 0, t 1 u N + λ ), t u N λ ) and I λt 1 u) < I λ 0) = 0. Lemma.9. Suppose {u n } H 1 0 ) is a P S) c sequence for I λ u), then {u n } is bounded in H 1 0 ). Proof. Let {u n } H 1 0 ) be such that I λ u n ) c, I λu n ) 0 as n. We claim that {u n } is bounded in H 1 0 ). Otherwise, we can suppose that u n as n. It follows from.1),.4),.5) and F4) that 1 + c + o1) u n I λ u n ) 1 I λu n ), u n 1 a 1 ) 1 u n + b 4 1 ) u n 4 λ [F x, u n ) 1 fx, u n ) u n ] dx 1 a 1 ) 1 u n + b 4 1 ) u n 4 λc 3 1 + u n ) dx 1 a 1 ) 1 u n + b 4 1 ) u n 4 λc 3 + S u n ) ap 1) ) 1 ) λc 3S u n + b 4 1 ) u n 4 λc 3. Since 3 < p < 5, it follows that the last inequality is an absurd. Therefore, {u n } is bounded in H0 1 ). So Lemma.9 holds. 3. Proof of Theorem 1. Let λ = min{λ 1, λ, λ 3, λ 4 }, then Lemmas.1.9 hold for every λ 0, λ ). We prove Theorem 1. by three steps.

8 G.-S. LIU, C.-Y. LEI, L.-T. GUO, H. RONG EJDE-015/0 Step 1. We claim that I λ u) has a minimizer on N + λ ). Indeed, from Remark.8, there exists u N + λ ) such that I λu) < 0, so it follows that inf u N + λ ) I λu) < 0. By Lemma.3, let {u n } be a sequence minimizing for I λ u) on N + λ ). Clearly, this minimizing sequence is of course bounded, up to a subsequence still denoted {u n }), there exists u 1 H0 1 ) such that u n u 1, weakly in H 1 0 ), u n u 1, strongly in L p ) 1 p < 6), u n x) u 1, a.e. in. Now we claim that u n u 1 in H 1 0 ). In fact, set lim n u n = l. By the Ekeland s variational principle [7], it follows that o1) = I λu n ), u 1 thus one obtains 0 = a + bl ) u 1 Replacing u 1 with u n, we obtain = a + bl ) n u 1 + vx)u n u 1 ) dx u hx) u n p u 1 dx λ fx, u n ) u 1 dx, o1) = I λu n ), u n = a + bl ) l consequently, one obtains 0 = a + bl )l hx) u 1 p+1 dx λ fx, u 1 ) u 1 dx. 3.1) hx) u n p+1 dx λ fx, u n ) u n dx, hx) u 1 p+1 dx λ fx, u 1 ) u 1 dx. 3.) According to 3.1) and 3.), we obtain u 1 = l = lim n u n, which suggests that u n u 1 in H 1 0 ). Therefore, by Remark.8, one obtains So we proved the claim. I λ u 1 ) = α + λ = lim I λu n ) = inf I λ u) < 0. n u N + λ ) Step. I λ u) has a minimizer on N λ ). As a matter of fact, from Remark.6, we have I λ u) > 0 for u N λ ), so it follows that inf u N λ ) I λu) > 0. Similarly to step 1, we define a sequence {u n } as a minimizing for I λ u) on N λ ), and there exists u H 1 0 ) such that u n u, weakly in H 1 0 ), u n u, strongly in L p ) 1 p < 6), u n x) u, a.e. in. We claim that Hu n ) > 0. By contradiction, assume Hu n ) 0, then phu n ) 0, from u n N λ ), by.1),.4),.5) and F), it follows that a u n < a u n + 3b u n 4 phu n )

EJDE-015/0 MULTIPLE POSITIVE SOLUTIONS 9 < λ f u x, u n ) u n dx λ f u x, u n ) L S u n, which is a contradiction when λ is small enough. We get Hu n ) > 0. Therefore Hu ) > 0 as n. Similar to the proof of step 1, one can get u n u in H 1 0 ). Therefore, I λ u ) = α λ = lim I λu n ) = inf I λ u) > 0. n u N λ ) From above discussion, we obtain that I λ u) has a minimizer on N λ ). By Step 1 and Step, there exist u 1 N + λ ) and u N λ ) such that I λ u 1 ) = α + λ < 0 and I λu ) = α λ > 0. It follows that u 1 and u are nonzero solutions of 1.1). Because of I λ u) = I λ u ), one gets u 1, u 0. Therefore, by the Harnack inequality see [6, Theorem 8.0]), we have u 1, u > 0 a.e. in. Consequently the proof of Theorem 1. is complete. Acknowledgments. This research was supported by the Research Foundation of Guizhou Minzu University No. 15XJS013; No. 15XJS01). References [1] A. M. Mao, S. X. Luan; Sign-changing solutions of a class of nonlocal quasilinear elliptic boundary value problems, J. Math. Anal. Appl. 383 011), 39 43. [] A. M. Mao, Z. T. Zhang; Sign-changing and multiple solutions of Kirchhoff type problems without the P.S. condition, Nonlinear Anal. 70 009), 175 187. [3] C. O. Alves, F. J. S. A. Corréa, T. F. Ma; Positive solutions for a quasilinear elliptic equation of Kirchhoff type, Comput. Math. Appl. 49 005), 85 93. [4] C. Chen, Y. Kuo, T. Wu; The Nehari manifold for a Kirchhoff type problem involving signchanging weight functions, J. Differential Equations. 50 011), 1876 1908. [5] C. T. Cheng, X. Wu; Existence results of positive solutions of Kirchhoff type problems, Nonlinear Anal. 71 009), 4883 489. [6] D. Gilbarg, N. S. Trudinger; Elliptic Partial Differential Equations of Second Order, Springer- Verlag, Berlin, 001. [7] I. Ekeland; On the variational principle, J. Math. Anal. Appl. 47 1974), 34 353. [8] G. B. Li, H. Y. Ye; Existence of positive ground state solutions for the nonlinear Kirchhoff type equations in R 3, J. Differential Equations. 57 014), 566 600. [9] G. M. Figueiredo, J. R. Santos Junior; Multiplicity of solutions for a Kirchhoff equation with subcritical or critical growth, Diff. Integral Equations. 5 01), 853 868. [10] J. J. Nie, X. Wu; Existence and multiplicity of non-trivial solutions for Schrödinger- Kirchhoff-type equations with radial potential, Nonlinear Anal. 75 01), 3470 3479. [11] J. H. Jin, X. Wu; Infinitely many radial solutions for Kirchhoff-type problems in R N, J. Math. Annal. Appl. 369 010), 564 574. [1] L. Wei, X. M. He; Multiplicity of high energy solutions for superlinear Kirchhoff equations, J. Appl. Math. Comput. 39 01), 473 487. [13] T. F. Ma, J. E. M. Rivera; Positive solutions for a nonlinear nonlocal elliptic transmission problem, Appl. Math. Lett. 16 003), 43 48. [14] X. M. He, W. M. Zou; Infinitely many positive solutions for Kirchhoff-type problems, Nonlinear Analysis. 70 009), 1407 1414. [15] Y. H. Li, F. Y. Li, J. P. Shi; Existence of a positive solution to Kirchhoff type problems without compactness conditions, J. Differential Equations. 53 01), 85 94. Gao-Sheng Liu School of Science, Guizhou Minzu University, Guiyang 55005, China E-mail address: 77936104@qq.com

10 G.-S. LIU, C.-Y. LEI, L.-T. GUO, H. RONG EJDE-015/0 Chun-Yu Lei corresponding author) School of Science, Guizhou Minzu University, Guiyang 55005, China E-mail address: leichygzu@sina.cn, Phone +86 15985163534 Liu-Tao Guo School of Science, Guizhou Minzu University, Guiyang 55005, China E-mail address: 35063054@qq.com Hong Rong School of Science, Guizhou Minzu University, Guiyang 55005, China E-mail address: 4045355@qq.com