Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 27, pp. 1–9. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu NON-TRIVIAL SOLUTIONS FOR TWO-POINT BOUNDARY-VALUE PROBLEMS OF FOURTH-ORDER STURM-LIOUVILLE TYPE EQUATIONS SHAPOUR HEIDARKHANI Abstract. Using critical point theory due to Bonanno [3], we prove the existence of at least one non-trivial solution for a class of two-point boundary-value problems for fourth-order Sturm-Liouville type equations. 1. Introduction In this note, we prove the existence of at least one non-trivial solution for the two-point boundary-value problem of fourth-order Sturm-Liouville type: (pi (x)u00i (x))00 − (qi (x)u0i (x))0 + ri (x)ui (x) = λFui (x, u1 , . . . , un ) x ∈ (0, 1), ui (0) = ui (1) = u00i (0) = u00i (1) = 0 (1.1) for 1 ≤ i ≤ n, where n ≥ 1 is an integer, pi , qi , ri ∈ L∞ ([0, 1]) with p− i := ess inf x∈[0,1] pi (x) > 0 for 1 ≤ i ≤ n, λ is a positive parameter, F : [0, 1] × Rn → R is a function such that F (., t1 , . . . , tn ) is measurable in [0, 1] for all (t1 , . . . , tn ) ∈ Rn , F (x, ., . . . , .) is C 1 in Rn for every x ∈ [0, 1] and for every % > 0, sup n X |Fti (x, t1 , . . . , tn )| ∈ L1 ([0, 1]), |(t1 ,...,tn )|≤% i=1 and Fui denotes the partial derivative of F with respect to ui for 1 ≤ i ≤ n. Due to importance of fourth-order two-point boundary-value problems in describing a large class of elastic deflection, many authors have studied the existence and multiplicity of solutions for such a problem; we refer the reader to [1, 2, 4, 5, 6, 7, 11, 14, 17] and references therein. In [4], the authors, employing a three critical point theorem due to Bonanno and Marano [8, Theorem 2.6], determined an exact open interval of the parameter λ for which system (1) in the case n = 1, admits at least three distinct weak solutions. The aim of this article is to prove the existence of at least one non-trivial weak solution for (1) for appropriate values of the parameter λ belonging to a precise real interval, which extend the results in [7]. Our motivation comes from the recent 2000 Mathematics Subject Classification. 35J35, 47J10, 58E05. Key words and phrases. Fourth-order Sturm-Liouville type problem; multiplicity results; critical point theory. c 2012 Texas State University - San Marcos. Submitted September 12, 2011. Published February 15, 2012. 1 2 S. HEIDARKHANI EJDE-2012/27 paper [5]. For basic notation and definitions, and also for a thorough account on the subject, we refer the reader to [9, 12, 13]. 2. Preliminaries and basic notation First we recall for the reader’s convenience [16, Theorem 2.5] as given in [3, Theorem 5.1] (see also [3, Proposition 2.1] for related results) which is our main tool to transfer the question of existence of at least one weak solution of (1) to the existence of a critical point of the Euler functional: For a given non-empty set X, and two functionals Φ, Ψ : X → R, we define the following two functions: supu∈Φ−1 (]r1 ,r2 [) Ψ(u) − Ψ(v) , r2 − Φ(v) v∈Φ−1 (]r1 ,r2 [) Ψ(v) − supu∈Φ−1 (]−∞,r1 [) Ψ(u) ρ(r1 , r2 ) = sup Φ(v) − r1 −1 v∈Φ (]r1 ,r2 [) β(r1 , r2 ) = inf for all r1 , r2 ∈ R, r1 < r2 . Theorem 2.1 ([3, Theorem 5.1]). Let X be a reflexive real Banach space, Φ : X → R be a sequentially weakly lower semicontinuous, coercive and continuously Gâteaux differentiable functional whose Gâteaux derivative admits a continuous inverse on X ∗ and Ψ : X → R be a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. Put Iλ = Φ − λΨ and assume that there are r1 , r2 ∈ R, r1 < r2 , such that β(r1 , r2 ) < ρ(r1 , r2 ). Then, for each λ ∈] ρ(r11,r2 ) , β(r11,r2 ) [ there is u0,λ ∈ Φ−1 (]r1 , r2 [) such that Iλ (u0,λ ) ≤ Iλ (u) ∀u ∈ Φ−1 (]r1 , r2 [) and Iλ0 (u0,λ ) = 0. Let us introduce notation that will be used later. Assume that q− r− q− r− min i2 , i4 , i2 + i4 > −p− i , π π π π where p− i := ess inf x∈[0,1] pi (x) > 0, qi− := ess inf x∈[0,1] qi (x), (2.1) ri− := ess inf x∈[0,1] ri (x), for 1 ≤ i ≤ n. Moreover, set q qi− ri− qi− ri− p− , , + , 0 , δ := i i + σi , π2 π4 π2 π4 for 1 ≤ i ≤ n. Let Y := H 2 ([0, 1]) ∩ H01 ([0, 1]) be the Sobolev space endowed with the usual norm. We recall the following Poincaré type inequalities (see, for instance, [15, Lemma 2.3]): σi := min 1 00 2 ku k 2 , (2.2) π 2 i L ([0,1]) 1 kui k2L2 ([0,1]) ≤ 4 ku00i k2L2 ([0,1]) (2.3) π for all ui ∈ Y for 1 ≤ i ≤ n. Therefore, taking into account (2.1)-(2.3), the norm Z 1 1/2 kui k = (pi (x)|u00i (x)|2 + qi (x)|u0i (x)|2 + ri (x)|ui (x)|2 )dx ku0i k2L2 ([0,1]) ≤ 0 EJDE-2012/27 NON-TRIVIAL SOLUTIONS 3 for 1 ≤ i ≤ n is equivalent to the usual norm and, in particular, one has 1 ku00i kL2 ([0,1]) ≤ kui k (2.4) δi for 1 ≤ i ≤ n. We need the following proposition in the proof of Theorem 3.1. Proposition 2.2 ([4, Proposition 2.1]). Let ui ∈ Y for 1 ≤ i ≤ n. Then kui k∞ ≤ 1 kui k 2πδi for 1 ≤ i ≤ n. Put 1/2 1 1 ki := kpi k∞ + 2 kqi k∞ + 4 kri k∞ π π for 1 ≤ i ≤ n. It is easy to see that ki > 0 and δi < ki for 1 ≤ i ≤ n. Set δ := min{δi ; 1 ≤ i ≤ n} and k := max{ki ; 1 ≤ i ≤ n}. Here and in the sequel, X := Y × · · · × Y . We say that u = (u1 , . . . , un ) is a weak solution to the (1) if u = (u1 , . . . , un ) ∈ X and n Z 1 X (pi (x)u00i (x)vi00 (x) + qi (x)u0i (x)vi0 (x) + ri (x)ui (x)vi (x)) dx i=1 −λ 0 n Z X i=1 1 Fui (x, u1 , . . . , un )vi (x)dx = 0 0 for every v = (v1 , . . . , vn ) ∈ X. For γ > 0 we denote the set n X K(γ) = (t1 , . . . , tn ) ∈ Rn : |ti | ≤ γ . (2.5) i=1 3. Results 2 For a given non-negative constant ν and a positive constant τ with 2( δπν n ) 6= put R1 R5 sup(t1 ,...,tn )∈K(ν) F (x, t1 , . . . , tn )dx − 38 F (x, τ, . . . , τ )dx 0 8 aτ (ν) := 2 − 4096 n(kτ )2 2( δπν ) n 54 Pn n where K(ν) = (t1 , . . . , tn ) ∈ R : i=1 |ti | ≤ ν (see (2.5)). We formulate our main result as follows: 4096 2 54 n(kτ ) , Theorem 3.1. Assume that there constant ν1 and two positive q qexist a non-negative constants ν2 and τ with πν1 < n 4096 108 nτ and n 4096 108 nkτ < δπν2 such that (A1) F (x, t1 , . . . , tn ) ≥ 0 for each (x, t1 , . . . , tn ) ∈ ([0, 3/8] ∪ [5/8, 1]) × [0, τ ]n ; (A2) aτ (ν2 ) < aτ (ν1 ). 1 1 , aτ (ν [, system (1) admits at least one non-trivial weak Then, for each λ ∈] aτ (ν 1) 2) solution u0 = (u01 , . . . , u0n ) ∈ X such that n 4( δπν1 2 X δπν2 2 ) < ku0i k2 < 4( ) . n n i=1 4 S. HEIDARKHANI EJDE-2012/27 Proof. To apply Theorem 2.1 to our problem, arguing as in [7, 10], we introduce the functionals Φ, Ψ : X → R for each u = (u1 , . . . , un ) ∈ X, as follows Φ(u) = n X kui k2 i=1 Z Ψ(u) = 2 , 1 F (x, u1 (x), . . . , un (x))dx. 0 It is well known that Φ and Ψ are well defined and continuously differentiable functionals whose derivatives at the point u = (u1 , . . . , un ) ∈ X are the functionals Φ0 (u), Ψ0 (u) ∈ X ∗ , given by n Z 1 X 0 Φ (u)(v) = (pi (x)u00i (x)vi00 (x) + qi (x)u0i (x)vi0 (x) + ri (x)ui (x)vi (x)) dx, i=1 0 Ψ0 (u)(v) = Z 0 n 1X Fui (x, u1 (x), . . . , un (x))vi (x)dx i=1 for every v = (v1 , . . . , vn ) ∈ X, respectively. Moreover, Φ is sequentially weakly lower semicontinuous, Φ0 admits a continuous inverse on X ∗ as well as Ψ is sequentially weakly upper semicontinuous. Furthermore, Ψ0 : X → X ∗ is a compact operator. Indeed, it is enough to show that Ψ0 is strongly continuous on X. For this, for fixed (u1 , . . . , un ) ∈ X let (u1m , . . . , unm ) → (u1 , . . . , un ) weakly in X as m → +∞, then we have (u1m , . . . , unm ) converges uniformly to (u1 , . . . , un ) on [0, 1] as m → +∞(see [18]). Since F (x, ., . . . , .) is C 1 in Rn for every x ∈ [0, 1], the derivatives of F are continuous in Rn for every x ∈ [0, 1], so for 1 ≤ i ≤ n, Fui (x, u1m , . . . , unm ) → Fui (x, u1 , . . . , un ) strongly as m → +∞ which follows Ψ0 (u1m , . . . , unm ) → Ψ0 (u1 , . . . , un ) strongly as m → +∞. Thus we proved that Ψ0 is strongly continuous on X, which implies that Ψ0 is a compact operator by Proposition 26.2 of [18]. Set w(x) = (w1 (x), . . . , wn (x)) such that for 1 ≤ i ≤ n, 3 64τ 2 x ∈ [0, 38 [, − 9 (x − 4 x) wi (x) = τ x ∈ [ 38 , 85 ], 64τ 2 5 1 − 9 (x − 4 x + 4 ) x ∈] 85 , 1], δπν2 2 1 2 r1 = 2( δπν n ) and r2 = 2( n ) . It is easy to verify that w = (w1 , . . . , wn ) ∈ X, and in particular, one has 4096 2 2 4096 2 2 δ τ ≤ kwi k2 ≤ k τ 27 i 27 i for 1 ≤ i ≤ n. So, from the definition of Φ, we have n n X X 4096 4096 2 2 4096 2 2 4096 n(δτ )2 ≤ δi τ ≤ Φ(w) ≤ ki τ ≤ n(kτ )2 . 54 27 27 54 i=1 i=1 q q 4096 From the conditions πν1 < n 4096 108 nτ and n 108 nkτ < δπν2 , we obtain r1 < Φ(w) < r2 . Moreover, from Proposition 2.2 one has n X sup |ui (x)|2 ≤ x∈[0,1] i=1 n 1 X kui k2 (2πδ)2 i=1 EJDE-2012/27 NON-TRIVIAL SOLUTIONS 5 for each u = (u1 , . . . , un ) ∈ X, so from the definition of Φ, we observe that Φ−1 (] − ∞, r2 [) = {(u1 , . . . , un ) ∈ X; Φ(u1 , . . . , un ) < r2 } n X = (u1 , . . . , un ) ∈ X; kui k2 < 2r2 i=1 ⊆ (u1 , . . . , un ) ∈ X; n X |ui (x)|2 < i=1 ⊆ (u1 , . . . , un ) ∈ X; n X ν22 for all x ∈ [0, 1] 2 n |ui (x)| < ν2 for all x ∈ [0, 1] , i=1 from which it follows Z sup Ψ(u) = (u1 ,...,un )∈Φ−1 (]−∞,r2 [) (u1 ,...,un )∈Φ−1 (]−∞,r2 [) Z ≤ 1 F (x, u1 (x), . . . , un (x))dx sup 0 1 sup F (x, t1 , . . . , tn )dx. 0 (t1 ,...,tn )∈K(ν2 ) Since for 1 ≤ i ≤ n, 0 ≤ wi (x) ≤ τ for each x ∈ [0, 1], the condition (A1) ensures that Z 38 Z 1 F (x, w1 (x), . . . , wn (x))dx ≥ 0. F (x, w1 (x), . . . , wn (x))dx + 5 8 0 So, one has supu∈Φ−1 (]−∞,r2 [) Ψ(u) − Ψ(w) r2 − Φ(w) R1 sup(t1 ,...,tn )∈K(ν2 ) F (x, t1 , . . . , tn )dx − Ψ(w) ≤ 0 r2 − Φ(w) ≤ aτ (ν2 ). β(r1 , r2 ) ≤ On the other hand, by similar reasoning as before, one has Ψ(w) − supu∈Φ−1 (]−∞,r1 [) Ψ(u) Φ(w) − r1 R1 Ψ(w) − 0 sup(t1 ,...,tn )∈K(ν1 ) F (x, t1 , . . . , tn )dx ≥ Φ(w) − r1 ≥ aτ (ν1 ). ρ(r1 , r2 ) ≥ Hence, from Assumption (A2), one has β(r1 , r2 ) < ρ(r1 , r2 ). Therefore, from Theorem 2.1, taking into account that the weak solutions of the system (1) are exactly the solutions of the equation Φ0 (u) − λΨ0 (u) = 0, we have the conclusion. Now we point out the following consequence of Theorem 3.1. Theorem 3.2. Assume that Assumption (A1) inqTheorem 3.1 holds. Suppose that there exist two positive constants ν and τ with n 4096 108 nkτ < δπν such that 6 S. HEIDARKHANI EJDE-2012/27 (A3) R1 R 58 sup(t1 ,...,tn )∈K(ν) F (x, t1 , . . . , tn )dx 108 δπ 2 38 F (x, τ, . . . , τ )dx < ( ) ; ν2 4096n3 k τ2 (A4) F (x, 0, . . . , 0) = 0 for every x ∈ [0, 1] Then, for each 4096 h i 2 n(kτ )2 2( δπν n ) , R1 , λ ∈ R 5 54 8 sup F (x, t , . . . , t )dx 1 n (t ,...,t )∈K(ν) 3 F (x, τ, . . . , τ )dx 1 n 0 0 8 system (1)Padmits at least one non-trivial weak solution u0 = (u01 , . . . , u0n ) ∈ X n such that i=1 kui k∞ < ν. Proof. The conclusion follows from Theorem 3.1, by taking ν1 = 0 and ν2 = ν. Indeed, owing to our assumptions, one has 4096n3 τ k 2 R ( ) 1 1 − 108 ν 2 δπ sup(t1 ,...,tn )∈K(ν) F (x, t1 , . . . , tn )dx 0 aτ (ν2 ) < 4096 2 2 2( δπν n ) − 54 n(kτ ) 2 4096 R1 n(τ k) 1 − 54 δπν 2 sup(t1 ,...,tn )∈K(ν) F (x, t1 , . . . , tn )dx 0 2( n ) = δπν 2 2 2( n ) − 4096 54 n(kτ ) R1 sup(t1 ,...,tn )∈K(ν) F (x, t1 , . . . , tn )dx = 0 . 2 2( δπν n ) On the other hand, taking Assumption (A4) into account, one has R 58 3 F (x, τ, . . . , τ )dx 8 = aτ (ν1 ). 4096 2 54 n(kτ ) Moreover, since n X n 1 X sup |ui (x)| ≤ kui k2 (2πδ)2 i=1 x∈[0,1] i=1 2 Pn for each u = (u1 , . . . , un ) ∈ X, an easy computation ensures that i=1 kui k∞ < ν whenever Φ(u) < r2 . Now, owing to Assumption (A3), it is sufficient to invoke Theorem 3.1 for concluding the proof. Now, we point out a simple version of Theorem 3.1 in the case n = 1. Let p1 = p, q1 = q, r1 = r, δ1 = δ and k1 = k. Let f : [0, 1] × R → R be an L2 Rt Carathéodory function. Let F be the function defined by F (x, t) = 0 f (x, s)ds for each (x, t) ∈ [0, 1] × R. For a given non-negative constant ν and a positive constant 2 τ with 2(δπν)2 6= 4096 54 (kτ ) , put R1 R5 sup|t|≤ν F (x, t)dx − 38 F (x, τ )dx 0 8 bτ (ν) := . 2(δπν)2 − 4096 (kτ )2 54 Then, we have the following result. Theorem 3.3. Assume that there constant ν1 and two positive q exist a non-negative q constants ν2 and τ with πν1 < 4096 108 τ and 4096 108 kτ < δπν2 such that EJDE-2012/27 NON-TRIVIAL SOLUTIONS 7 (B1) F (x, t) ≥ 0 for each (x, t) ∈ ([0, 3/8] ∪ [5/8, 1]) × [0, τ ];; (B2) bτ (ν2 ) < bτ (ν1 ). 1 1 Then, for each λ ∈] bτ (ν , bτ (ν [, the problem 1) 2) (p(x)u00 (x))00 − (q(x)u0 (x))0 + r(x)u(x) = λf (x, u) 00 x ∈ (0, 1), 00 u(0) = u(1) = u (0) = u (1) = 0 (3.1) admits at least one non-trivial weak solution u0 ∈ Y such that 2δπν1 < ku0 k < 2δπν2 . We remark that if p(x) = 1, q(x) = −A and r(x) = B for every x ∈ [0, 1], then Theorem 3.3 gives [7, Theorem 3.1]. The following result gives the existence of at least one non-trivial weak solution in Y to the problem (3.1) in the autonomous case. Let f : R → R be a continuous Rt function. Put F (t) = 0 f (ξ)dξ for all t ∈ R. We have the following result as a direct consequence of Theorem 3.3. Theorem 3.4. Assume that there constant ν1 and two positive q q exist a non-negative constants ν2 and τ with πν1 < 4096 108 τ and 4096 108 kτ < δπν2 such that (C1) f (t) ≥ 0 for each t ∈ [−ν2 , max{ν2 , τ }]; F (ν )− 41 F (τ ) F (ν )− 14 F (τ ) (C2) 2(δπν )22 − 4096 < 2(δπν )12 − 4096 . (kτ )2 (kτ )2 2 Then, for each λ 54 1 54 2 2(δπν2 )2 − 4096 2(δπν )2 − 4096 (kτ )2 54 (kτ ) [, ∈] F (ν1 )− 154F (τ ) , 1 F (τ ) F (ν )− 1 2 4 4 the problem (p(x)u00 (x))00 − (q(x)u0 (x))0 + r(x)u(x) = λf (u) 00 x ∈ (0, 1), 00 u(0) = u(1) = u (0) = u (1) = 0 (3.2) 0k admits at least one non-trivial weak solution u0 ∈ Y such that ν1 < ku 2δπ < ν2 . q q 4096 4096 Proof. Since δ < k, from the conditions πν1 < 108 τ and 108 kτ < δπν2 , 4096 2 2 2 2 we obtain 2(δπν1 )2 < 4096 54 (δτ ) and 54 (kτ ) < 2(δπν2 ) , and so 2(δπν1 ) < 2 2(δπν2 ) , and so ν1 < ν2 . Therefore, Assumptions (C1) means f (t) ≥ 0 for each t ∈ [−ν1 , ν1 ] and f (t) ≥ 0 for each t ∈ [−ν2 , ν2 ], which follow maxt ∈ [−ν1 , ν1 ]F (t) = F (ν1 ) and maxt ∈ [−ν2 , ν2 ]F (t) = F (ν2 ). So, from Assumptions (C1) and (C2) we arrive at assumptions (B1) and (B2), respectively. Hence, we achieve the stated assertion by applying Theorem 3.3 observing that the problem (1) reduces to the problem (3.2). As an example, we point out the following special case of our main result. Theorem 3.5. Let g : R → R be a nonnegative continuous function such that 2 2 R ν limt→0+ g(t) t = +∞. Then, for each λ ∈]0, 2(δπ) supν>0 ν g(ξ)dξ [, the problem 0 (p(x)u00 (x))00 − (q(x)u0 (x))0 + r(x)u(x) = λg(u) 00 x ∈ (0, 1), 00 u(0) = u(1) = u (0) = u (1) = 0 admits at least one non-trivial weak solution in Y . Proof. For fixed λ as in the conclusion, there exists positive constant ν such that λ < 2(δπ)2 R ν 0 ν2 . g(ξ)dξ 8 S. HEIDARKHANI EJDE-2012/27 Rt g(t) t g(ξ)dξ = +∞ implies limt→0+ 0 t2 = +∞. Therefore, a posiq tive constant τ satisfying 4096 108 kτ < δπν can be chosen such that Moreover, limt→0+ 1 4 × 4096 2 ( k )< λ 54 Rτ 0 g(ξ)dξ . τ2 Hence, arguing as in the proof of Theorem 3.2, the conclusion follows from Theorem 3.4 with ν1 = 0, ν2 = ν and f (t) = g(t) for every t ∈ R. 2 ν Remark 3.6. For fixed ρ put λρ := 2(δπ)2 supν∈]0,ρ[ R ν g(ξ)dξ . The result of Theo0 rem 3.5 for every λ ∈]0, λρ [ holds with |u0 (x)| < ρ for all x ∈ [0, 1] where u0 is the ensured non-trivial weak solution in Y (see [7, Remark 4.3]). We close this article by presenting the following examples to illustrate our results. example 3.7. Consider the problem + (3ex u00 )00 − ((x2 − π 2 )u0 )0 + (x2 − π 4 )u = λ(1 + e−u (u+ )2 (3 − u+ )) x ∈ (0, 1), u(0) = u(1) = 0, u00 (0) = u00 (1) = 0 (3.3) where u+ = max{u, 0}. Let + g(t) = 1 + e−t (t+ )2 (3 − t+ ) for all t ∈ R where t+ = max{t, 0}. It is clear that limt→0+ g(t) t = +∞. Note that p− = 3, q − = −π 2 and r− = −π 4 , we have σ = −2, and so δ = 1. Hence, taking 2 e ν2 , by Remark 3.6 into account, since supν∈]0,1[ R ν g(ξ)dξ = supν∈]0,1[ ν+eν−ν ν 3 = 1+e 0 2 e applying Theorem 3.5, for every λ ∈]0, 2π 1+e [ the problem (3.3) has at least one non-trivial classical solution u0 ∈ Y such that ku0 k∞ < 1. example 3.8. Put p(x) = 1, q(x) = π 2 , r(x) = x − π for all x ∈ [0, 1] and 1 g(t) = (1 + t)et for every t ∈ R. Clearly, one has σ = (1 − π13 ) 2 . Hence, since sup R ν ν∈]0,1[ 0 ν2 ν2 1 = sup = , e g(ξ)dξ ν∈]0,1[ νeν from Theorem 3.5, taking Remark 3.6 into account, for every λ ∈]0, problem uiv − π 2 u00 + (x − π)u = λ(1 + u)eu u(0) = u(1) = 0, 00 2π 2 (1− π13 ) [ e x ∈ (0, 1), 00 u (0) = u (1) = 0 the (3.4) has at least one non-trivial classical solution u0 ∈ Y such that ku0 k∞ < 1. Acknowledgements. 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Comput. Appl. Math. 113 (2000) 401-410. [17] Y.-H. Wang, S. Heidarkhani, L. Li; Multiple solutions for a fourth-order Sturm–Liouville type boundary value problem, preprint. [18] E. Zeidler; Nonlinear functional analysis and its applications, Vol. II, III. Berlin-HeidelbergNew York 1985. Shapour Heidarkhani Department of Mathematics, Faculty of Sciences, Razi University, 67149 Kermanshah, Iran School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box: 19395-5746, Tehran, Iran E-mail address: sh.heidarkhani@yahoo.com, s.heidarkhani@razi.ac.ir