Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 147, pp. 1–11. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu EXISTENCE OF SOLUTIONS TO DIRICHLET IMPULSIVE DIFFERENTIAL EQUATIONS THROUGH A LOCAL MINIMIZATION PRINCIPLE GHASEM A. AFROUZI, SAEID SHOKOOH, ARMIN HADJIAN Abstract. A critical point theorem (local minimum result) for differentiable functionals is used for proving that a Dirichlet impulsive differential equation admits at least one non-trivial solution. Some particular cases and a concrete example are also presented. 1. Introduction In this article, we study the existence of at least one non-trivial classical solution to the nonlinear Dirichlet boundary-value problem −(p(t)u0 (t))0 + q(t)u(t) = λf (t, u(t)), t ∈ [0, T ], t 6= tj , u(0) = u(T ) = 0, ∆u0 (tj ) = λIj (u(tj )), (1.1) j = 1, 2, . . . , n, where T > 0, p ∈ C 1 ([0, T ], ]0, +∞[), q ∈ L∞ ([0, T ]), λ ∈]0, +∞[, f : [0, T ] × R → R is an L1 -Carathéodory function, 0 = t0 < t1 < t2 < · · · < tn < tn+1 = T , 0 − ∆u0 (tj ) = u0 (t+ u0 (t) − limt→t− u0 (t) and Ij : R → R are j ) − u (tj ) = limt→t+ j j continuous for every j = 1, 2, . . . , n. The study of impulsive boundary-value problems is important due to its various applications in which abrupt changes at certain times in the evolution process appear. The dynamics of evolving processes is often subjected to abrupt changes such as shocks, harvesting, and natural disasters. Often these short-term perturbations are treated as having acted instantaneously or in the form of impulses. Such problems arise in physics, population dynamics, biotechnology, pharmacokinetics, industrial robotics. Recently, many researchers pay their attention to impulsive differential equations by variational method and critical point theory, and we refer the reader to [12, 15, 17, 18, 19, 20] and references cited therein. Our analysis is mainly based on the critical point theorem by Bonanno [2], contained in Theorem 2.1 below. This theorem has been used in several works 2000 Mathematics Subject Classification. 34B37, 34B15, 58E05. Key words and phrases. Impulsive differential equations; Dirichlet condition; classical solution; variational methods. c 2014 Texas State University - San Marcos. Submitted January 20, 2014. Published June 24, 2014. 1 2 G. A. AFROUZI, S. SHOKOOH, A. HADJIAN EJDE-2014/147 to obtain existence results for different kinds of problems (see, for instance, [1, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14]). 2. Preliminaries Our main tool is Ricceri’s variational principle [16, Theorem 2.5] as given in [2, Theorem 5.1] which is below recalled (see also [2, Proposition 2.1]). For a given non-empty set X, and two functionals Φ, Ψ : X → R, we define the functions supu∈Φ−1 (]r1 ,r2 [) Ψ(u) − Ψ(v) β(r1 , r2 ) := inf −1 r2 − Φ(v) v∈Φ (]r1 ,r2 [) and Ψ(v) − supu∈Φ−1 (]−∞,r1 ]) Ψ(u) ρ(r1 , r2 ) := sup Φ(v) − r1 v∈Φ−1 (]r1 ,r2 [) for all r1 , r2 ∈ R, with r1 < r2 . Theorem 2.1 ([2, 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 function whose Gâteaux derivative admits a continuous inverse on X ∗ ; Ψ : X → R be a continuously Gâteaux differentiable function whose Gâteaux derivative is compact. Assume that there are r1 , r2 ∈ R, with r1 < r2 , such that β(r1 , r2 ) < ρ(r1 , r2 ). Then, setting Iλ := Φ−λΨ, for each such that Iλ (u0,λ ) ≤ Iλ (u) for all u λ ∈] ρ(r11,r2 ) , β(r11,r2 ) [ there is u0,λ ∈ ∈ Φ−1 (]r1 , r2 [) and Iλ0 (u0,λ ) = 0. (2.1) Φ −1 (]r1 , r2 [) In the Sobolev space X := H01 (0, T ), consider the inner product Z T Z T 0 0 (u, v) := p(t)u (t)v (t) dt + q(t)u(t)v(t) dt , 0 which induces the norm Z kuk := 0 T 0 2 Z p(t)(u (t)) dt + 0 T 1/2 . q(t)(u(t))2 dt 0 Then the following Poincaré-type inequality holds: Z i1/2 hZ T i1/2 Th T 0 2 (u ) (t) dt . u2 (t) dt ≤ π 0 0 (2.2) 2 − Let us introduce some notation that will be used later. Assume that T πq2 > −p− , where p− := ess inf t∈[0,T ] p(t) > 0, q − := ess inf t∈[0,T ] q(t). p Moreover, put σ0 := min{T 2 q − /π 2 , 0} and δ := p− + σ0 . Then, we have the following useful proposition. Proposition 2.2. Let u ∈ X. Then 1 ku0 kL2 ([0,T ]) ≤ kuk, √ δ T kuk∞ ≤ kuk. 2δ (2.3) (2.4) EJDE-2014/147 DIRICHLET IMPULSIVE DIFFERENTIAL EQUATIONS Proof. First we prove (2.3). To this end, let q − ≥ 0. Then, σ0 = 0 and δ = Therefore, Z T 1 ku0 k2L2 ([0,T ]) ≤ − p(t)|u0 (t)|2 dt p 0 Z T 1 1 ≤ − p(t)|u0 (t)|2 + q(t)|u(t)|2 dt = 2 kuk2 . p δ 0 3 p p− . Thus, the desired inequality (2.3) follows. On the other hand, if q − < 0, we have q 2 − 2 − σ0 = T πq2 and δ = p− + T πq2 . Obviously, Z T − 2 q(t)|u(t)|2 dt. q kukL2 ([0,T ]) ≤ 0 Now, applying inequality (2.2) and bearing in mind that q − < 0, one has T 2 q− 0 2 ku kL2 ([0,T ]) ≤ q − kuk2L2 ([0,T ]) . π2 By the above inequalities we have Z T T 2 q− 0 2 ku k ≤ q(t)|u(t)|2 dt. L2 ([0,T ]) π2 0 This inequality together with − p ku0 k2L2 ([0,T ]) Z ≤ T p(t)|u0 (t)|2 dt, 0 imply (2.3). In view of Hölder’s inequality and (2.3), one has √ √ T 0 T kuk∞ ≤ ku kL2 ([0,T ]) ≤ kuk, 2 2δ which completes and the proof. 1/2 2 Put k := kpk∞ + Tπ2 kqk∞ . Then, from (2.2) we have kuk ≤ kku0 kL2 ([0,T ]) . (2.5) Here and in the sequel f : [0, T ] × R → R is an L1 -Carathéodory function, namely: (a) the mapping t 7→ f (t, x) is measurable for every x ∈ R; (b) the mapping x 7→ f (t, x) is continuous for almost every t ∈ [0, T ]; (c) for every ρ > 0 there exists a function lρ ∈ L1 ([0, T ]) such that sup |f (t, x)| ≤ lρ (t) |x|≤ρ for almost every t ∈ [0, T ]. Corresponding to f we introduce the function F : [0, T ] × R → R as follows Z x F (t, x) := f (t, ξ) dξ, 0 for all (t, x) ∈ [0, T ] × R. By a classical solution of problem (1.1), we mean a function u ∈ w ∈ C([0, T ]) : w|[tj ,tj+1 ] ∈ H 2 ([tj , tj+1 ]) 4 G. A. AFROUZI, S. SHOKOOH, A. HADJIAN EJDE-2014/147 that satisfies the equation in (1.1) a.e. on [0, T ] \ {t1 , . . . , tn }, the limits u0 (t+ j ), 0 u0 (t− ), j = 1, . . . , n, exist, satisfy the impulsive conditions ∆u (t ) = λI (u(t j j j )) j and the boundary condition u(0) = u(T ) = 0. We say that a function u ∈ X is a weak solution of problem (1.1), if u satisfies Z T Z T 0 0 p(t)u (t)v (t) dt + q(t)u(t)v(t) dt 0 0 T Z −λ f (t, u(t))v(t) dt − 0 n X p(tj )Ij (u(tj ))v(tj ) = 0, j=1 for any v ∈ X. Lemma 2.3 ([3, Lemma 2.3]). The function u ∈ X is a weak solution of problem (1.1) if and only if u is a classical solution of (1.1). Lemma 2.4 ([3, Lemma 3.1]). Assume that (A1) there exist constants η, θ > 0 and σ ∈ [0, 1[ such that |Ij (x)| ≤ 2η|x| + θ|x|σ+1 for all x ∈ R, j = 1, 2, . . . , n. Then, for any u ∈ X, we have Z u(tj ) n n X X p(tj ) p(tj ) ηkuk2∞ + Ij (x) dx ≤ 0 j=1 j=1 Also put p̃ := n X j=1 √ p(tj ), µ(τ ) := 2kτ , δ Γc := θ kukσ+2 . ∞ σ+2 (2.6) η θ σ−1 + c , c σ+2 where η, θ, σ are given by (A1) and c, τ are positive constants. We assume throughout, and without further mention, that the assumption (A1) holds. 3. Main results For a given non-negative constant ν and a positive constant τ with δ 2 ν 2 6= 2k 2 τ 2 , put RT R 3T /4 max|x|≤ν F (t, x) dt + p̃ν 3 Γν + p̃(µ(τ ))3 Γµ(τ ) − T /4 F (t, τ ) dt 0 aτ (ν) := . δ 2 ν 2 − 2k 2 τ 2 Theorem 3.1. Assume that there √ exist a non-negative constant ν1 and two positive constants ν2 and τ , with ν1 < 2τ < δν2 /k, such that (A2) F (t, ξ) ≥ 0 for all (t, ξ) ∈ ([0, T4 ] ∪ [ 3T 4 , T ]) × [0, τ ]; (A3) aτ (ν2 ) < aτ (ν1 ). Then, for each λ ∈] T aτ2(ν1 ) , T aτ2(ν2 ) [, problem (1.1) admits at least one non-trivial classical solution u0 ∈ X such that 2δν1 2δν √ < ku0 k < √ 2 . T T Proof. The aim is to apply Theorem 2.1 to our problem. To this end, we introduce the functionals Φ, Ψ : X → R by setting 1 Φ(u) := kuk2 , 2 EJDE-2014/147 DIRICHLET IMPULSIVE DIFFERENTIAL EQUATIONS T Z F (t, u(t)) dt − Ψ(u) := 0 n X Z 5 u(tj ) p(tj ) Ij (x) dx, 0 j=1 for every u ∈ X, and put Iλ (u) := Φ(u) − λΨ(u) ∀ u ∈ X. Clearly, Φ and Ψ are well defined and continuously Gâteaux differentiable functionals whose Gâteaux derivatives at the point u ∈ X are the functionals Φ0 (u), Ψ0 (u) ∈ X ∗ , given by Z T Z T q(t)u(t)v(t) dt, p(t)u0 (t)v 0 (t) dt + Φ0 (u)(v) = Ψ0 (u)(v) = 0 0 T Z f (t, u(t))v(t) dt − 0 n X p(tj )Ij (u(tj ))v(tj ), j=1 for every v ∈ X. Moreover, Φ is coercive and sequentially weakly lower semicontinuous and Ψ is sequentially weakly upper semicontinuous. Also, Φ0 admits a continuous inverse on X ∗ and Ψ0 is compact. Note that the critical points of the functional Iλ in X are exactly the weak solutions of problem (1.1). We verify condition (2.1) of Theorem 2.1. To this end, we put 2δ 2 2 2δ 2 2 r1 := ν1 , r2 := ν , T 2 T 4τ if t ∈ [0, T /4[, T t, w(t) := τ, if t ∈ [T /4, 3T /4], 4τ T (T − t), if t ∈]3T /4, T ]. It is easy to verify that w ∈ X and, in particular, taking (2.3) and (2.5) into account, one has 8k 2 τ 2 8δ 2 τ 2 ≤ kwk2 ≤ . T T So, we have 4δ 2 τ 2 4k 2 τ 2 ≤ Φ(w) ≤ . T T √ From the condition ν1 < 2τ < δν2 /k, we obtain r1 < Φ(w) < r2 . Moreover, for all u ∈ X such that u ∈ Φ−1 (] − ∞, r2 [), from (2.4), one has |u(t)| < ν2 for all t ∈ [0, T ], from which it follows Z u(tj ) n Z T X sup Ψ(u) = sup F (t, u(t)) dt − p(tj ) Ij (x) dx u∈Φ−1 (]−∞,r2 [) u∈Φ−1 (]−∞,r2 [) Z ≤ 0 0 j=1 0 T max F (t, x) dt + p̃ν23 Γν2 . |x|≤ν2 Arguing as before, we obtain Z sup u∈Φ−1 (]−∞,r1 ]) Ψ(u) ≤ 0 T max F (t, x) dt + p̃ν13 Γν1 . |x|≤ν1 Since 0 ≤ w(t) ≤ τ for each t ∈ [0, T ], assumption (A2) ensures that Z T /4 Z T F (t, w(t)) dt + F (t, w(t)) dt ≥ 0, 0 3T /4 6 G. A. AFROUZI, S. SHOKOOH, A. HADJIAN EJDE-2014/147 and so Z 3T /4 Ψ(w) ≥ F (t, τ ) dt − T /4 Z n X Z w(tj ) p(tj ) Ij (x) dx 0 j=1 3T /4 ≥ F (t, τ ) dt − p̃ (µ(τ ))3 Γµ(τ ) . T /4 Therefore, we have β(r1 , r2 ) ≤ ≤ supu∈Φ−1 (]−∞,r2 [) Ψ(u) − Ψ(w) r2 − Φ(w) RT R 3T /4 max|x|≤ν2 F (t, x) dt + p̃ν23 Γν2 + p̃(µ(τ ))3 Γµ(τ ) − T /4 F (t, τ ) dt 0 2δ 2 ν22 T − 4k2 τ 2 T T aτ (ν2 ). 2 On the other hand, we have = ρ(r1 , r2 ) ≥ ≥ Ψ(w) − supu∈Φ−1 (]−∞,r1 ]) Ψ(u) Φ(w) − r1 R 3T /4 RT F (t, τ ) dt − p̃(µ(τ ))3 Γµ(τ ) − p̃ν13 Γν1 − 0 max|x|≤ν1 F (t, x) dt T /4 4k2 τ 2 T − 2δ 2 ν12 T T aτ (ν1 ). 2 So, from assumption (A3), it follows that β(r1 , r2 ) < ρ(r1 , r2 ). Therefore, from Theorem 2.1, for each λ ∈] T aτ2(ν1 ) , T aτ2(ν2 ) [, the functional Iλ admits at least one critical point u0 such that r1 < Φ(u0 ) < r2 ; that is, = 2δν 2δν1 √ < ku0 k < √ 2 , T T and the proof is complete. Now, we point out the following consequence of Theorem 3.1. √ Theorem 3.2. Assume that there are two positive constants ν and τ , with 2kτ < δν, such that assumption (A2) in Theorem 3.1 holds. Furthermore, suppose that RT max F (t,x) dt+p̃ν 3 Γ 2 R 3T /4 F (t,τ ) dt−p̃ (µ(τ ))3 Γµ(τ ) ν |x|≤ν T /4 δ < 2k . (A4) 0 2 ν2 τ2 Then, for each 4k2 τ 2 2δ 2 ν 2 i h T T λ ∈ R 3T /4 , RT , F (t, τ ) dt − p̃ (µ(τ ))3 Γµ(τ ) 0 max|x|≤ν F (t, x) dt + p̃ν 3 Γν T /4 problem (1.1) admits at least one non-trivial classical solution u0 ∈ X such that |u0 (t)| < ν for all t ∈ [0, T ]. Proof. The conclusion follows from Theorem 3.1, by taking ν1 = 0 and ν2 = ν. Indeed, owing to assumption (A4), one has 2 2 RT τ 1 − 2k max|x|≤ν F (t, x) dt + p̃ν 3 Γν ν 2 δ2 0 aτ (ν) < δ 2 ν 2 − 2k 2 τ 2 EJDE-2014/147 DIRICHLET IMPULSIVE DIFFERENTIAL EQUATIONS RT = 0 7 max|x|≤ν F (t, x) dt + p̃ν 3 Γν . δ2 ν 2 On the other hand, R 3T /4 T /4 F (t, τ ) dt − p̃(µ(τ ))3 Γµ(τ ) . 2k 2 τ 2 Now, owing to assumption (A4) and (2.4), it is sufficient to invoke Theorem 3.1 for concluding the proof. aτ (0) = The following result gives the existence of at least one non-trivial classical solution in X to problem (1.1) R x in the autonomous case. Let f : R → R be a continuous function. Put F (x) := 0 f (ξ) dξ for all x ∈ R. We have the following result as a direct consequence of Theorem 3.1. Theorem 3.3. Assume that there √ exist a non-negative constant ν1 and two positive constants ν2 and τ , with ν1 < 2τ < δν2 /k, such that (A5) f (x) ≥ 0 for all x ∈ [−ν2 , max{ν2 , τ }]; (A6) T F (ν2 ) + p̃ν2 3 Γν2 + p̃ (µ(τ ))3 Γµ(τ ) − δ 2 ν22 − 2k 2 τ 2 < T 2 F (τ ) T F (ν1 ) + p̃ν1 3 Γν1 + p̃(µ(τ ))3 Γµ(τ ) − δ 2 ν12 − 2k 2 τ 2 T 2 F (τ ) . Then, for each λ∈ i T T F (ν2 ) + 2δ 2 ν22 − 4k 2 τ 2 + p̃ (µ(τ ))3 Γµ(τ ) − p̃ ν2 3 Γν2 2δ 2 ν22 − 4k 2 τ 2 T T F (ν1 ) + p̃ ν1 3 Γν1 + p̃ (µ(τ ))3 Γµ(τ ) − T 2 T 2 F (τ ) , F (τ ) h , the problem −(p(t)u0 (t))0 + q(t)u(t) = λf (u(t)), t ∈ [0, T ], t 6= tj , u(0) = u(T ) = 0, 0 ∆u (tj ) = λIj (u(tj )), (3.1) j = 1, 2, . . . , n, admits at least one non-trivial classical solution u0 ∈ X such that 2δν1 2δν √ < ku0 k < √ 2 . T T √ Proof. Since δ ≤ k, from the condition ν1 < 2τ < δνk2 we obtain ν1 < ν2 . Therefore, assumption (A5) means f (x) ≥ 0 for each x ∈ [−ν1 , ν1 ] and f (x) ≥ 0 for each x ∈ [−ν2 , ν2 ], which implies max x∈[−ν1 ,ν1 ] F (x) = F (ν1 ) and max x∈[−ν2 ,ν2 ] F (x) = F (ν2 ). So, from assumptions (A5) and (A6), we arrive at assumptions (A2) and (A3), respectively. Hence, Theorem 3.1 yields the conclusion. Theorem 3.4. Let f : R → R be a non-negative continuous function such that (A7) limξ→0+ f (ξ) ξ = +∞. 8 G. A. AFROUZI, S. SHOKOOH, A. HADJIAN 2 EJDE-2014/147 2 ν Then, for each λ ∈]0, 2δT supν>0 T F (ν)+ p̃ν 3 Γν [, problem (3.1) admits at least one non-trivial classical solution u0 ∈ X. Proof. For fixed λ as in the conclusion, there exists a positive constant ν such that λ(T F (ν) + p̃ν 3 Γν ) < 2δ 2 ν 2 . T (3.2) = +∞. On the other hand, Moreover, assumption (A7) implies that limt→0+ Fξ(ξ) 2 ( √ 2 (µ(ξ))3 Γµ(ξ) η δ2 k , if 0 < σ < 1, √ 2 = lim+ 2 ξ ξ→0 Γ1 δ2 k , if σ = 0. Therefore, F (ξ) − (µ(ξ))3 Γµ(ξ) = +∞. ξ2 ξ→0 √ So, a positive constant τ satisfying 2kτ < δν can be chosen such that lim+ λ T F (τ ) − p̃ (µ(τ ))3 Γ 2 µ(τ ) τ2 > 4k 2 . T (3.3) Hence, taking (3.2) and (3.3) into account, Theorem 3.1 ensures the conclusion. Remark 3.5. Taking (A7) into account, fix ρ > 0 such that f (ξ) > 0 for all ξ ∈]0, ρ[. Then, put 2δ 2 ν2 λρ := sup . T ν∈]0,ρ[ T F (ν) + p̃ν 3 Γν The result of Theorem 3.4 for every λ ∈]0, λρ [ holds with |u0 (t)| < ρ for all t ∈ [0, T ], where u0 is the ensured non-trivial classical solution in X. Example 3.6. Let I(u(t1 )) = u(t1 ) for some t1 ∈ (0, 1). Then I : R → R is a continuous function satisfying the sublinear growth condition (A1) with η = θ = 31 2 ξ and σ = 0. Now, put p(t) = 1, q(t) = −π 2 for all t ∈ [0, 1] and f (ξ) = (1 + ξ)e for 1 every ξ ∈ R. Clearly, one has δ = √2 . Hence, since ν2 ν2 2 = sup = , ν 3Γ 3Γ νe + ν 2e +1 f (ξ) dξ + ν ν ν∈]0,1[ ν∈]0,1[ 0 ν 2 from Remark 3.5, for every λ ∈ 0, 2e+1 the problem sup R ν −u00 (t) − π2 u(t) = λ(1 + u(t))eu(t) , 2 u(0) = u(1) = 0, a.e. in [0, 1], ∆u0 (t1 ) = λu(t1 ), has at least one non-trivial classical solution u0 ∈ H01 (0, 1) such that |u0 (t)| < 1 for all t ∈ [0, 1]. Here, we point out a special situation of our main result when the nonlinear term has separable variables. To be precise, let α ∈ L1 ([0, T ]) such that α(t) ≥ 0 EJDE-2014/147 DIRICHLET IMPULSIVE DIFFERENTIAL EQUATIONS 9 a.e. t ∈ [0, T ], α 6≡ 0, and let g : R → R be a nonnegative continuous function. Consider the following Dirichlet boundary-value problem −(p(t)u0 (t))0 + q(t)u(t) = λα(t)g(u(t)), t ∈ [0, T ], t 6= tj , u(0) = u(T ) = 0, (3.4) ∆u0 (tj ) = λIj (u(tj )), j = 1, 2, . . . , n. Rx RT Put G(x) := 0 g(ξ) dξ for all x ∈ R, and set kαk1 := 0 α(t) dt and α0 := R 3T /4 α(t) dt. T /4 Corollary 3.7. Let Ij (x) ≤ 0 for all x ∈ R, j = 1, . . . , n. Assume that there √ exist a non-negative constant ν1 and two positive constants ν2 and τ , with ν1 < 2τ < δν2 /k, such that (A8) G(ν2 )kαk1 + p̃ν23 Γν2 + p̃(µ(τ ))3 Γµ(τ ) − G(τ )α0 δ 2 ν22 − 2k 2 τ 2 G(ν1 )kαk1 + p̃ν13 Γν1 + p̃(µ(τ ))3 Γµ(τ ) − G(τ )α0 . < δ 2 ν12 − 2k 2 τ 2 Then, for each 2δ 2 ν12 − 4k 2 τ 2 , T G(ν1 )kαk1 + p̃ν13 Γν1 + p̃(µ(τ ))3 Γµ(τ ) − G(τ )α0 h 2δ 2 ν22 − 4k 2 τ 2 , 3 T G(ν2 )kαk1 + p̃ν2 Γν2 + p̃(µ(τ ))3 Γµ(τ ) − G(τ )α0 λ∈ i problem (3.4) admits at least one positive classical solution u0 ∈ X, such that 2δν1 2δν √ < ku0 k < √ 2 . T T Proof. Put f (t, ξ) := α(t)g(ξ) for all (t, ξ) ∈ [0, T ] × R. Clearly, F (t, x) = α(t)G(x) for all (t, x) ∈ [0, T ] × R. Therefore, taking into account that G is a non-decreasing function, Theorem 3.1 and [3, Lemma 3.6] ensure the conclusion. An immediate consequence of Corollary 3.7 is the following. Corollary 3.8. Let Ij (x) ≤ 0 for all√x ∈ R, j = 1, . . . , n. Assume that there exist two positive constants ν and τ , with 2kτ < δν, such that (A9) G(ν)kαk1 +p̃ν 3 Γν ν2 < 3 δ 2 G(τ )α0 −p̃(µ(τ )) Γµ(τ ) . 2k2 τ2 Then, for each λ∈ i h 4k 2 τ 2 2δ 2 ν 2 , , 3 3 T G(τ )α0 − p̃(µ(τ )) Γµ(τ ) T G(ν)kαk1 + p̃ν Γν problem (3.4) admits at least one positive classical solution u0 ∈ X, such that |u0 (t)| < ν for all t ∈ [0, T ]. The above corollary follows directly from Theorem 3.2 and [3, Lemma 3.6]. 10 G. A. AFROUZI, S. SHOKOOH, A. HADJIAN EJDE-2014/147 Now, consider the nonlinear Dirichlet boundary-value problem −u00 (t) + a(t)u0 (t) + b(t)u(t) = λh(t, u(t)), t ∈ [0, T ], t 6= tj , u(0) = u(T ) = 0, 0 ∆u (tj ) = λIj (u(tj )), (3.5) j = 1, 2, . . . , n, where h : [0, T ] × R → R is an L -Carathéodory function and a, b ∈ L∞ ([0, T ]) satisfy the following conditions π2 ess inf t∈[0,T ] a(t) ≥ 0, ess inf t∈[0,T ] b(t)e−A(t) > − 2 e−A(T ) , T where A(t) be a primitive of a(t). It is easy to see that the solutions of (1.1) are solutions of (3.5) if 1 p(t) = e− Let H(t, x) := Rt 0 a(ξ) dξ Rx 0 , q(t) = b(t)e− Rt 0 a(ξ) dξ , f (t, u) = h(t, u)e− Rt 0 a(ξ) dξ . h(t, ξ) dξ. Then, by a simple computation, we obtain F (t, x) = e−A(t) H(t, x), ∀(t, x) ∈ [0, T ] × R. Set 1/2 T2 T2 k̃ := 1 + 2 kbe−A k∞ ess inf t∈[0,T ] b(t)e−A(t) , 0 , , σ̃0 := min 2 π π p −A(T ) δ̃ := e + σ̃0 . For a given non-negative constant ν and a positive constant τ with δ̃ 2 ν 2 6= 2k̃ 2 τ 2 , √ put µ̃(τ ) := 2k̃τ /δ̃ and Z T ãτ (v) := e−A(t) max H(t, x) dt + p̃ν 3 Γν + p̃ (µ̃(τ ))3 Γµ̃(τ ) Z 0 3T /4 − |x|≤ν δ̃ 2 ν 2 − 2k̃ 2 τ 2 . e−A(t) H(t, τ ) dt T /4 With the above notation and by Theorem 3.1, we obtain the following existence property for problem (3.5). Theorem 3.9. Assume that there √ exist a non-negative constant ν1 and two positive constants ν2 and τ , with ν1 < 2τ < δ̃ν2 /k̃, such that (A10) H(t, ξ) ≥ 0 for all (t, ξ) ∈ ([0, T4 ] ∪ [ 3T 4 , T ]) × [0, τ ]; (A11) ãτ (ν2 ) < ãτ (ν1 ). 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Afrouzi Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran E-mail address: afrouzi@umz.ac.ir Saeid Shokooh Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran E-mail address: saeid.shokooh@stu.umz.ac.ir Armin Hadjian Department of Mathematics, Faculty of Basic Sciences, University of Bojnord, P.O. Box 1339, Bojnord 94531, Iran E-mail address: hadjian83@gmail.com