EXISTENCE OF SOLUTIONS FOR KIRCHHOFF TYPE EQUATIONS WITH UNBOUNDED POTENTIAL. 1. Introduction In this article, we consider the Kirchhoff type problem
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1 Electronic Journal of Differential Equations, Vol. 207 (207), No. 84, pp. 2. ISSN: URL: or EXISTENCE OF SOLUTIONS FOR KIRCHHOFF TYPE EQUATIONS WITH UNBOUNDED POTENTIAL YUELIANG DUAN, YINGGAO ZHOU Communicated by Paul H. Rabinowitz Abstract. In this article, we study the Kirchhoff type equation Z Z a + u 2 + b u 2 [ u + bu] = K(x) u p u, in, where a, b > 0, p (2, 5), 0 is a parameter, and K may be an unbounded potential function. By using variational methods, we prove the existence of nontrivial solutions for the above equation. A cut-off functional and some estimates are used to obtain the bounded Palais-Smale sequences.. Introduction In this article, we consider the Kirchhoff type problem ( ) a + u 2 + b u 2 [ u + bu] = K(x) u p u, in, u H r ( ), (.) where a > 0, b > 0, p (2, 5), 0 is a parameter and K(x) is a given potential satisfying the following conditions: (H) K : R is a nonnegative continuous function, K is radial (that is k(x) = k( x )) and K 0; (H2) there exists C 0 > 0 and 0 l < p 2 such that K(x) C 0 ( + x l ) for all x. When Ω is a bounded domain in R N, the Kirchhoff type problem ( a + b u 2) u = f(x, u), in Ω, Ω (.2) u = 0, on Ω, has attracted a lot of attention in recent years. Equation (.2) is a nonlocal problem because of the appearance of the term Ω u 2. This causes some mathematical difficulties but makes the study of (.2) very interesting. One can refer to [, 4, 3, 9, 5, 25, 27, 23, 28, 7, 8, 3] and references therein for the existence and multiplicity of positive solutions to (.2). Mao and Luan [2], Mao and Luan [20], 200 Mathematics Subject Classification. 35A5, 35A20, 35J20. Key words and phrases. Cut-off functional; Kirchhoff type problem; unbounded potential; variational method. c 207 Texas State University. Submitted March 7, 207. Published July 9, 207.
2 2 Y. DUAN, Y. ZHOU EJDE-207/84 Zhang and Perera [29] have investigated the existence of sign-changing solutions of (.2). When Ω = R N, the existence and multiplicity of solutions to (.2) have been treated in [6, 2, 4, 26, 6, 5]. In particular, Nie and Wu [22] have studied the Schrödinger-Kirchhoff type problem ( ) a + b u 2 u + V ( x )u = Q( x )f(u), R N (.3) u(x) 0, as x, where N 2, a, b > 0, f C(R, R), V and Q are both radial functions in R N. Under some conditions on V and Q, the existence of nontrivial solutions and a sequence of high energy solutions for problem (.3) has been proved by the Mountain Pass theorem and symmetric Mountain Pass theorem. From [22, Remarks and 2], we see that the potential Q is bounded in R N. In [7], Li, Li and Shi showed the existence of nontrivial solutions of the following problem with zero mass ( ) a + b u 2 u = K(x)f(u), in R N, R N u D,2 (R N ), where the potential function K(x) is a nonnegative continuous function, K [L s (R N ) L (R N )]\{0} for some s 2N/(N + 2) and x K(x) αk(x) for a.e. x R N and some α (0, 2). To be best of our knowledge, there is no result about the Kirchhoff type problems with unbounded potential at infinity. Motivated by the above work, in this article, we are lead to study problem (.) with unbounded potential K. Our main results are as follows: Theorem.. Assume (H) and (H2) hold. Then there exists 0 > 0 such that for any [0, 0 ), problem (.) has at least one nontrivial solution in H r ( ). Corollary.2. Assume (H) and (H2) hold. Then the local problem has at least one nontrivial solution in H r ( ). u + bu = K(x) u p u, in (.4) We remark that in [7], a cut-off functional and Pohozaev type identity are utilized to obtain the bounded Palais-Smale sequences. However, we find that when f(u) = u p u can be expressed explicitly, it is not necessary to establish the Pohozaev type identity. In the case that K is unbounded, we need the Radial Lemma (see Lemma 2.) to overcome the loss of compactness, which is different from [7]. The remaining of this paper is organized as follows. In Section 2, we give some notations and elementary lemmas which will be used in the paper. In Section 3, we are devoted to the proof of our main results. 2. Preliminaries In this paper, we shall use the following notation: C stands for different positive constants.
3 EJDE-207/84 EXISTENCE OF SOLUTIONS 3 For r > 0, B r (x) is an open ball in with radius r centered at x. H ( ) denotes the usual Sobolev space equipped with the inner product (u, v) = ( u v + buv)dx, and the corresponding norm u = (u, u) /2. Hr ( ) is the set of all radial functions in H ( ). For Ω and q <, L q (Ω) denotes the Lebesgue space with the norm u L q (Ω) = ( u q dx) q. When Ω =, we write u q = u L q ( ) for simplicity of notations. Furthermore, when q =, we write u = ess sup x u(x). H ( ) L q ( ) continuously for q [2, 6], while Hr ( ) L q ( ) compactly for q (2, 6). The dual space of Hr ( ) is denoted by Hr ( ). Let, be the duality pairing between Hr ( ) and Hr ( ). Before establishing the variational setting for (.), we need the following lemma. Lemma 2. ([2]). Let N 2. Then for any radial function u H r (R N ), u(r) C u r N 2, for r, where C is a positive constant and only depends on N. Remark 2.2. Since H ( ) L q ( ) continuously for q [2, 6], for any u Hr ( ), by (H) and (H2), we have 0 K(x) u p+ dx C 0 ( + x R l ) u p+ dx 3 = C 0 ( + x l ) u p+ dx + C 0 C( B (0) B (0) C u p+ + C u p+ dx + Ω \B (0) r l+2 u(r) p+ dr) r l+2 u(r) p+ dr. ( + x l ) u p+ dx Note that 0 l < p 2. Then by Lemma 2., we deduce r l+2 u(r) p+ dr C u p+ r l p+ C dr p 2 l u p+. Thus, 0 K(x) u p+ dx < C u p+. In view of Remark 2.2, we consider the Sobolev space Hr ( ). As usual, the energy functional J : Hr ( ) R associated with (.) is well defined by J (u) = 2 a u u 4 K(x) u p+ dx. p +
4 4 Y. DUAN, Y. ZHOU EJDE-207/84 It is easy to check that J is C -functional, whose Gateaux derivative is given by J (u), v = a(u, v) + u 2 (u, v) K(x) u p uvdx u, v Hr ( ). In the weak sense, solutions to (.) correspond to the critical points of the functional J. 3. Proof of our main results In this section, we prove the existence of nontrivial solutions to (.). We first have the following lemma. Lemma 3.. If u is a nontrivial weak solution of (.), then u r for some r > 0. Proof. Since u is a nontrivial weak solution of (.), by Remark 2.2, we have a u 2 (a + u 2 ) u 2 = K(x) u p+ dx C u p+. Because p + > 2, there exists r > 0 such that u > r. To find a bounded Palais-Smale sequence for the energy functional J, by following [9, ] (also see [7, 8]), we introduce a cut-off function φ C (R +, R) such that φ(t) =, t [0, ], 0 φ(t), t (, 2), φ(t) = 0, φ 2. t [2, ), So we can define the functional J T : H r ( ) R by J T (u) = 2 a u h T (u) u 4 K(x) u p+ dx, u Hr ( ), p + where for every T > 0, h T (u) = φ( u 2 T 2 ). Moreover, for every u, v H r ( ), we have (J T ) (u), v = a(u, v) + h T (u) u 2 (u, v) + 2T 2 φ ( u 2 T 2 ) u 4 (u, v) K(x) u p uvdx. (3.) It is easy to see that if u is a critical point of J T such that u T, then (J T ) (u), v = J (u), v. So the arbitrary of v yields J (u) = 0 and thus u is also a critical point of J. Next we recall a theorem, for which a corollary was proved by Struwe [24]. Theorem 3.2 ([0]). Let (X, ) be a Banach space and I R + an interval. Consider the family of C functional on X J µ (u) = A(u) µb(u), µ I,
5 EJDE-207/84 EXISTENCE OF SOLUTIONS 5 with B nonnegative and either A(u) or B(u) as u and such that J µ (0) = 0. For every µ I, we set Γ µ = {γ C([0, ], X) : γ(0) = 0, J µ (γ()) < 0}. If for every µ I, the set Γ µ is nonempty and c µ = inf γ Γ µ max J µ(γ(t)) > 0, t [0,] then for almost every µ I, there is a sequence {u n } X such that (i) {u n } is bounded; (ii) J µ (u n ) c µ ; (iii) J µ(u n ) 0 in the dual space X of X. In our case, X = H r ( ), A(u) = 2 a u h T (u) u 4, B(u) = K(x) u p+ dx, p + and the associated perturbed functional we study is J,µ(u) T = 2 a u h T (u) u 4 µ K(x) u p+ dx, u Hr ( ). p + It is clear that this functional is C -functional and for every u, v H r ( ), (J,µ) T (u), v = a(u, v) + h T (u) u 2 (u, v) + 2T 2 φ ( u 2 T 2 ) u 4 (u, v) µ K(x) u p uvdx. (3.2) Notice that J T,µ (0) = 0, B is nonnegative in H r ( ) and A(u) + as u. Next we shall prove the following two lemmas which show that the functional J T,µ satisfies the other conditions of Theorem 3.2. Lemma 3.3. We have Γ µ for all 0 and µ I := [ 2, ]. Proof. By (H), there exist R, C 2 > 0 and ϕ C 0 ( ) such that ϕ 0, ϕ =, supp(ϕ), K(x)ϕ p+ dx C 2. Then for t 2 2T 2, J,µ(tϕ) T = 2 at2 ϕ 2 + ϕ 2 4 φ(t2 T 2 )t 4 ϕ 4 µ p + t p+ K(x)ϕ p+ dx = 2 at2 µ p + t p+ K(x)ϕ p+ dx 2 at2 2(p + ) t p+ K(x)ϕ p+ dx (3.3)
6 6 Y. DUAN, Y. ZHOU EJDE-207/84 2 at2 C 2 2(p + ) t p+. Since p + > 2, we can choose t 0 > 0 large such that J T,µ (t 0ϕ) < 0. We set γ(t) = tt 0 ϕ, t [0, ]. So γ Γ µ and Γ µ. The proof is complete. Lemma 3.4. There exists a positive constant α such that c µ α for all µ I. Proof. For any µ I and u Hr ( ), by Remark 2.2, we have that J,µ(u) T = 2 a u h T (u) u 4 µ K(x) u p+ dx p + 2 a u 2 µ p + C u p+ 2 a u 2 C p + u p+. Take ρ := ( ) ap+a p 4C and α := 2 aρ2 C p+ ρp+ > 0. Then J,µ T (u) > 0 for any µ I and 0 < u ρ. Moreover, J,µ T (u) α if u = ρ. By the definition of Γ µ, for each γ Γ µ, there is t (0, ) such that γ(t ) = ρ. Therefore, the arbitrary of µ I yields that c µ inf γ Γ µ J T,µ(γ(t )) α. The proof is complete. J T,µ Next we need some compactness on Palais-Smale sequences of the functional in order to prove our main results. Lemma 3.5. Assume 4T 2 < a. For any µ I, any bounded Palais-Smale sequence of J,µ T contains a convergent subsequence. Proof. Fix µ I. We assume {u n } be a bounded Palais-Smale sequence of J T,µ, namely {u n } and {J T,µ(u n )} are bounded, (J T,µ) (u n ) 0 in H r ( ). Up to a subsequence, there exists u H r ( ) such that u n u, in H r ( ), u n u, in L s ( ), s (2, 6), u n u, in L s loc( ), s [, 6), u n (x) u(x), a.e. x.
7 EJDE-207/84 EXISTENCE OF SOLUTIONS 7 For R >, from (H2) we obtain K(x) u n p u n (u n u)dx K(x) u n p u n u dx C 0 ( + x R l ) u n p u n u dx 3 = C 0 ( + x l ) u n p u n u dx + C 0 =: I + I 2. \ By Hölder s inequality, we have I = C 0 ( + x l ) u n p u n u dx C(R)( ( u n p+ dx) p p+ ( + x l ) u n p u n u dx ) u n u p+ p+ dx (3.4) (3.5) = C(R) u n p L p+ () u n u L p+ () C(R) u n p u n u L p+ (). Moreover, from Lemma 2. we deduce that I 2 = C 0 ( + x l ) u n p u n u dx 2C 0 \ \ C( u n p+ + u n p u ) x l ( u n p+ + u n p u )dx R r l p+ dr = C Rl p+2 p 2 l ( u n p+ + u n p u ). (3.6) Since 3 < p + < 6, u n u in L p+ (). Therefore, by (3.4)-(3.6), we can find R > large enough such that K(x) u n p u n (u n u)dx 0 as n. (3.7) Note that (J T,µ ) (u n ) 0. Then by (3.7), we deduce that So 0 (J T,µ) (u n ), u n u = a(u n, u n u) + h T (u n ) u n 2 (u n, u n u) + T 2 ) u n 4 (u n, u n u) µ K(x) u n p u n (u n u)dx = [a + h T (u n ) u n 2 + T 2 ) u n 4 ](u n, u n u) + o(). (a + h T (u n ) u n 2 + T 2 ) u n 4 )(u n, u n u) 0. (3.8)
8 8 Y. DUAN, Y. ZHOU EJDE-207/84 Next, we claim that (u n, u n u) 0 as n. (3.9) Note that if u n 2 > 2T 2, then h T (u n ) = 0 and φ ( un 2 T 2 ) = 0. If u n 2 2T 2, then φ ( u n 2 T 2 ) u n 4 8T 4. Since 4T 2 < a, we have and a + h T (u n ) u n 2 + a + h T (u n ) u n 2 + T 2 ) u n 4 a + T 2 ) u n 4 a + 2T 2 ( 8T 4 ) = a 4T 2 > 0 T 2 ) u n 4 a + u n 2 + T 2 ) u n 4 a + 2T 2 + = a + 6T 2 5a 2. 2T 2 (8T 4 ) This combined with (3.8) yields that (3.9) holds. Since u n u in H r ( ), we infer from the above claim that up to a subsequence, The proof is complete. u n u in H r ( ). Lemma 3.6. Let 4T 2 < a. For almost every µ I, there exists u µ H r ( )\{0} such that (J T,µ) (u µ ) = 0 and J T,µ(u µ ) = c µ. (3.0) Proof. By Lemma 3.3, Lemma 3.4 and Theorem 3.2, for almost every µ I, we can find a bounded sequence {u µ n} H r ( ) satisfying J T,µ(u µ n) c µ, (J T,µ) (u µ n) 0. Furthermore, by using Lemma 3.5, we conclude that there is u µ Hr ( ) such that u µ n u µ in Hr ( ). So the continuity of J,µ T and (J,µ T ) imply that (3.0) holds. This completes the proof. According to Lemma 3.6, we obtain that there exist sequences {µ n } I with µ n and {u n } H r ( ) such that J T,µ n (u n ) = c µn, (J T,µ n ) (u n ) = 0. (3.) In what follows, we shall show u n T, which is a critical key in the proof of existence of solutions to (.). Lemma 3.7. Assume {µ n } I with µ n and {u n } satisfies (3.). Then for T > 0 sufficiently large, there exists 0 = 0 (T ) with 4 0 T 2 < a such that for any [0, 0 ), up to a subsequence, u n T n.
9 EJDE-207/84 EXISTENCE OF SOLUTIONS 9 Proof. The proof consists of three steps. Step. By the definition of c µ and Lemma 3.4, we conclude that α c µn sup J,µ T n (tϕ) t [0, ) [ = sup t [0, ) 2 at2 + t2 φ( 4 T 2 )t4 µ n p + t p+ K(x)ϕ p+ dx ] R 3 [ max t [0, ) 2 at2 2(p + ) t p+ K(x)ϕ p+ dx ] t2 + sup φ( t [0, ) 4 T 2 )t4 a(p ) = 2(p + ) ( 2a K(x)ϕp+ dx ) 2 p + sup t2 =: A + sup φ( t [0, ) 4 T 2 )t4, where ϕ is defined in Lemma 3.3. t [0, ) t2 φ( 4 T 2 )t4 Step 2. It is easy to see that if t 2 2T 2, then φ( t2 T 2 ) = 0 and so while if t 2 < 2T 2, then sup t [ 2T, ) sup t [0, 2T ) This, combined with (3.2), yields t2 φ( 4 T 2 )t4 = 0, t2 φ( 4 T 2 )t4 < T 4. 0 < α c µn A + T 4. (3.2) Step 3. Note that and c µn = J T,µ n (u n ) = 2 a u n h T (u n ) u n 4 µ n p + 0 = (J T,µ n ) (u n ), u n K(x) u n p+ dx = a(u n, u n ) + h T (u n ) u n 2 (u n, u n ) + T 2 ) u n 4 (u n, u n ) µ n K(x) u n p+ dx. Thus, we have that c µn = J T,µ n (u n ) p + (J T,µ n ) (u n ), u n = a( 2 p + ) u n 2 + ( 4 p + )h T (u n ) u n 4 2(p + )T 2 φ ( u n 2 T 2 ) u n 6. (3.3)
10 0 Y. DUAN, Y. ZHOU EJDE-207/84 We claim: there exists a subsequence of {u n } which is uniformly bounded by T. By way of contradiction, we distinguish two cases to prove the claim. Case. Up to a subsequence, u n 2 > 2T 2 for all n. It is easy to check that h T (u n ) = 0 and φ ( un 2 T ) = 0. Then we deduce from (3.3) and Step 2 that ( 2 2 ) a u n 2 = c µn A + T 4. p + Since 4T 4 < a, [( 2T 2 < u n 2 2 ) (A [( a] + T 4 ) p + 2 ) ] (A a a + p + 4 ), which contradicts with the assumption that T is sufficiently large. Case 2. Up to a subsequence, T 2 < u n 2 2T 2 for all n. By (3.3) and Step 2, we have c µn A + T 4. Furthermore, since p (2, 5), we obtain ( 2 p + )at 2 < ( 2 p + )a u n 2 = c µn + A + T 4 + 2(p + )T 2 φ ( u n 2 T 2 ) u n 6 ( 4 p + )h T (u n ) u n 4 φ 2(p + )T 2 ( u n 2 T 2 ) un 6 + ( 4 + p + )h T (u n ) u n 4 A + T p + T 4 + ( 4 + p + ) 4T 4 A + a(p + 4) 4(p + ), which contradicts the assumption that T is sufficiently large. Then the claim holds and we complete the proof. Proof of Theorem.. We assume T, 0 are defined as in Lemma 3.7 and for each µ n I, u n is a critical point for J,µ T n at level c µn. By using Lemma 3.7, up to a subsequence, we can also assume u n T. Hence, h T (u n ) = and J,µ T n (u n ) = 2 a u n u n 4 µ n K(x) u n p+ dx. p + Next, we claim that {u n } is a Palais-Smale sequence of J. Indeed, since u n is bounded for all n, we conclude that J (u n ) is bounded for all n. Moreover, we have J (u n ), v = (J,µ T n ) (u n ), v + (µ n ) K(x) u n p u n vdx, for any v Hr ( ). Note that (J,µ T n ) (u n ) = 0 and µ n. Then we obtain J (u n) 0 and the claim holds. By Lemma 3.5, {u n } has a convergent subsequence. We may assume that u n ū in Hr ( ). Therefore, J (ū) = 0 and by Lemma 3.4, J (ū) = lim n J (u n ) = lim n J T,µ n (u n ) = c µn α > 0. Consequently, ū is nontrivial solution of (.). The proof is complete.
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