# The Solutions of Sturm-Liouville Boundary-Value Problem for Fourth-Order Impulsive Differential Equation via Variational Methods

Authors: Yu Tian; Dongpo Sun
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

The Sturm-Liouville boundary-value problem for fourth-order impulsive differential equations is studied. The existence results for one solution and multiple solutions are obtained. The main ideas involve variational methods and three critical points theory.


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## Body

## 1. Introduction

The aim of the present paper is to study the following Sturm-Liouville boundary-value problem for the fourth-order impulsive differential equation:(1)u(4)(t)+u(t)=λf(t,u(t)),t≠ti,t∈[0,T],Δu′′′(ti)=λI1i(u(ti)),i=1,2,…,l,-Δu′′(ti)=λI2i(u′(ti)),i=1,2,…,l,au(0)+bu′′′(0)=0,au(T)-bu′′′(T)=0,cu′(0)-du′′(0)=0,cu′(T)+du′′(T)=0,
where a, b, c, and d are real constants, λ is a positive parameter, 0=t0<t1<⋯<tl<tl+1=T, a,b,c,d>0, Δu′′′(ti)=u′′′(ti+)-u′′′(ti-), Δu′′(ti)=u′′(ti+)-u′′(ti-), u′′′(ti+), u′′(ti+)(u′′′(ti-),u′′(ti-)) denote the right (left) limits, respectively, of u′′′(t), u′′(t) at t=ti, and f∈C([0,T]×R,R), I1i,I2i∈C(R;R), i=1,2,…,l.Recently, many authors have studied the existence of solutions for boundary-value problems with impulsive effects [1–16]. Variational methods are powerful tools for them. We refer the readers to [17–19] for related basic information.In [10], the authors studied the following equation with impulsive effects:
(2)-u′′(t)=λu(t)+f(t,u(t)),t≠ti,t∈[0,T],-Δu′(ti)=Ii(u(ti)),i=1,2,…,l,u′(0)=0,u(T)=0.
By applying critical point theory to (2), several existence results are obtained when f is imposed some assumptions and λ lies in suitable interval. In [8], the authors studied the existence of solutions for the following problem:
(3)u(iv)(t)+Au′′(t)+Bu(t)=f(t,u(t)),a.e.t∈[0,T],-Δu′′(tj)=I1j(u′(tj)),j=1,2,…,l,-Δu′′′(tj)=I2j(u(tj)),j=1,2,…,l,u(0)=u(T)=u′′(0+)=u′′(T-)=0.
They essentially proved that when f, I1j, and I2j satisfy some conditions, (3) has at least one solution or infinitely many classical solutions via variational methods.To the best of our knowledge, besides [12, 13] for second-order differential equations, [8] for fourth-order differential equation, limited work has been done in the Sturm-Liouville boundary-value problem, let alone higher order. Motivated by the above facts, we study the existence of solutions for problem (1) by applying variational methods. With the impulse effects and the Sturm-Liouville boundary conditions taken into consideration, the corresponding variational functional J will be more complicated than the ones of any fourth-order boundary-value problems before. In our study, some difficulties such as how to prove that the critical points of J are just the solutions of problem (1) and how to prove the space X and the functional J to satisfy the conditions of the related theorems must be overcome. To verify that the weak solution of problem (1) is just the classical solution of (1), we construct a Fundamental Lemma 5, by which we can easily prove that the critical point of the functional is just the solution of problem (1).This paper is organized as follows. In Section2, we present some preliminaries and establish the variational structure. In Section 3, we discuss the existence results for one solution and multiple solutions. In Section 4, we discuss the existence results for positive solutions. In Section 5, we will give some examples.


## 2. Preliminaries and Variational Structure

First we present some theorems that will be needed in the proof of main results.Theorem 1 (see Theorem 2.2 [19]).
LetE be a real Banach space and I∈C1(E,R) satisfying the Palais-Smale condition (PS). Suppose I(0)=0 and(C1)
there are constantsρ,α>0 such that I|∂Bρ≥α,(C2)
there is ane∈E∖Bρ such that I(e)≤0. Then I possesses a critical value c≥α. Moreover, c can be characterized as
(4)c=inf⁡g∈Γmax⁡u∈g([0,1])I(u),where(5)Γ={g∈(C[0,1],E):g(0)=0,g(1)=e}.Theorem 2 (see Theorem 9.12 [19]).
LetE be an infinite dimensional Banach space and let I∈C1(E,R) be even, satisfy (PS), and I(0)=0. If E=V⨁X, where V is finite dimensional, and I satisfies that(C3)
there exist constantsρ,α>0 such that I|∂Bρ∩X≥α,(C4)
for each finite dimensional subspaceW⊂E, there is an R=R(W) such that I≤0 on W∖BR(W).Then I has unbounded sequence of critical values.LetX be a nonempty set and Φ,Ψ:X→R two functionals. For all r,r1,r2>inf⁡XΦ, r2>r1, r3>0, we define
(6)φ(r):=inf⁡u∈Φ-1(]-∞,r[)(sup⁡u∈Φ-1(]-∞,r[)Ψ(u))-Ψ(u)r-Φ(u),β(r1,r2):=inf⁡u∈Φ-1(]-∞,r1[)sup⁡v∈Φ-1([r1,r2[)Ψ(v)-Ψ(u)Φ(v)-Φ(u),γ(r2,r3):=sup⁡u∈Φ-1(]-∞,r2+r3[)Ψ(u)r3,α(r1,r2,r3):=max⁡{φ(r1),φ(r2),γ(r2,r3)}.Theorem 3 (see Theorem 2.1 [2]).
LetX be a reflexive real Banach space, Φ:X→R a convex, coercive, and continuously Gâteaux differentiable functional whose Gâteaux derivative admits a continuous inverse on X*, and Ψ:X→R a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact, such that(1)
inf⁡XΦ=Φ(0)=Ψ(0)=0;(2)
for everyu1, u2 such that Ψ(u1)≥0 and Ψ(u2)≥0 one has
(7)inf⁡t∈[0,1]Ψ(tu1+(1-t)u2)≥0.Assume that there are three positive constants r1, r2, r3 with r1<r2, such that(i)
φ(r1)<β(r1,r2);(ii)
φ(r2)<β(r1,r2);(iii)
γ(r2,r3)<β(r1,r2).Then, for each λ∈]1/β(r1,r2),1/α(r1,r2,r3)[, the functional Φ-λΨ admits three distinct critical points u1, u2, u3 such that u1∈Φ-1(]-∞,r1[), u2∈Φ-1([r1,r2[), and u3∈Φ-1(]-∞,r2+r3[).Let us define the spaceX=W2,2([0,T],R) equipped with the norm
(8)∥u∥=(∫0T‍|u(t)|2+|u′(t)|2+|u′′(t)|2dt)1/2.
We set the functional J:X→R defined by
(9)J(u)=12∥u∥X2-λ∫0T‍F(t,u(t))dt-λ∑‍i=1l∫0u(ti)‍I1i(s)ds-λ∑‍i=1l∫0u′(ti)‍I2i(s)ds,
where ∥u∥X=(∫0T‍|u′′(t)|2+|u(t)|2dt+(a/b)u2(T)+(a/b)u2(0)+(c/d)|u′(T)|2+(c/d)|u′(0)|2)1/2, F(t,u)=∫0u‍f(t,s)ds. J is differentiable for any u∈X and
(10)J′(u)(v)=∫0T‍(u′′v′′+uv)dt-λ∑‍i=1lI1i(u(ti))v(ti)-λ∑‍i=1lI2i(u′(ti))v′(ti)-λ∫0T‍f(t,u(t))v(t)dt+abu(T)v(T)+abu(0)v(0)+cdu′(T)v′(T)+cdu′(0)v′(0).
Set the usual norm of C1([0,T]), L2(0,T), respectively, as follows:
(11)∥u∥C1=max⁡{max⁡t∈[0,T]|u(t)|,max⁡t∈[0,T]|u′(t)|},∥u∥L2=(∫0T‍u2(t)dt)1/2.Lemma 4.
The norm∥u∥X is equivalent to the usual norm ∥u∥.Proof.
First, we will show that there existsM1>0 such that ∥u∥2≤M1∥u∥X2. Since u′ is absolutely continuous in X, we have u′(t)=u′(0)+∫0t‍u′′(s)ds. So
(12)∫0T‍|u′(t)|2dt=∫0T‍|u′(0)+∫0t‍u′′(s)ds|2dt≤2T|u′(0)|2+2T2∫0T‍|u′′(s)|2ds,
which implies
(13)∥u∥2≤∫0T‍|u(t)|2+|u′′(t)|2dt+2T|u′(0)|2+2T2∫0T‍|u′′(t)|2dt≤(1+2T2)∫0T‍|u′′(t)|2ds+∫0T‍|u(t)|2dt+2T|u′(0)|2≤max⁡{1+2T2,2Tdc}(∫0T‍|u′′(t)|2+|u(t)|2dt+cd|u′(0)|2):=M1∥u∥X2,
where M1=max⁡{1+2T2,2Td/c}.
Next, we prove that there existsM2>0 such that ∥u∥X2≤M2∥u∥2.
Obviously,max⁡t∈[0,T]|u(t)|=u(ξ)=u-+∫ηξ‍u′(s)ds≤u-+∫0T‍|u′(s)|ds, where u-=∫0T‍u(s)ds/T=u(η). Thus, we have
(14)(max⁡t∈[0,T]|u(t)|)2≤2(|u-|2+(∫0T‍|u′(s)|ds)2)≤2T∫0T‍|u(s)|2ds+2T∫0T‍|u′(s)|2ds.
Similar to the above proof, we have
(15)(max⁡t∈[0,T]|u′(t)|)2≤2T∫0T‍|u′(s)|2ds+2T∫0T‍|u′′(s)|2ds.
By (14) and (15), we have
(16)∥u∥X2≤∫0T‍|u(s)|2+|u′′(s)|2ds+2ab(2T∫0T‍|u(s)|2ds+2T∫0T‍|u′(s)|2ds)+2cd(2T∫0T‍|u′(s)|2ds+2T∫0T‍|u′′(s)|2ds)≤(1+4abT)∫0T‍|u(s)|2ds+(1+4cTd)∫0T‍|u′′(s)|2ds+(4aTb+4cdT)∫0T‍|u′(s)|2ds:=M2∥u∥2,
where M2=max⁡{1+4a/bT,1+4cT/d,4aT/b+4c/dT}. By (13) and (16), the proof is complete.Lemma 5 (Fundamental Lemma).
Letu,v∈L1([0,T];R). If for every f∈C1[0,T] with f′′∈L1[0,T], f(0)=f(T)=f′(0)=f′(T)=0 satisfying ∫0T‍u(t)f′′(t)dt=∫0T‍v(t)f(t)dt, then there exist C1,C2∈R such that u(t)=∫0t‍∫0s‍v(θ)dθds+C1t+C2 a.e. on [0,T].Proof.
Definew(t)∈C([0,T];R) by w(t)=∫0t‍∫0s‍v(θ)dθds; we have
(17)∫0T‍w(t)f′′(t)dt=∫0T‍∫0t‍∫0s‍v(θ)dθdsf′′(t)dt.
By the Fubini theorem, we have
(18)∫0T‍w(t)f′′(t)dt=∫0T‍∫sT‍∫0s‍v(θ)dθf′′(t)dtds=∫0T‍∫0s‍v(θ)dθ(f′(T)-f′(s))ds=-∫0T‍∫0s‍v(θ)dθf′(s)ds=-∫0T‍∫θT‍v(θ)f′(s)dsdθ=-∫0T‍v(θ)(f(T)-f(θ))dθ=∫0T‍v(θ)f(θ)dθ.
So
(19)∫0T‍(u(t)-w(t))f′′(t)dt=0.
In particular, we choose f(t)=∫0t‍∫0s‍(u(θ)-w(θ)-C1θ-C2)dθds, where
(20)C1=12T3(∫0T‍t(u(t)-w(t))dt-T2∫0T‍u(s)-w(s)ds),C2=4T∫0T‍u(s)-w(s)ds-6T2∫0T‍t(u-w)dt.
By computation, f∈C01[0,T], f′′∈L1[0,T], f(0)=f(T)=f′(0)=f′(T)=0, and
(21)∫0T‍(C1t+C2)f′′(t)dt=0.
By (19) and (21), ∫0T‍(u(t)-w(t)-C1t-C2)f′′(t)dt=0; that is, ∫0T‍|u(t)-w(t)-C1t-C2|2dt=0, which means u(t)=w(t)+C1t+C2. The proof is complete.Definition 6.
A functionu∈X is said to be a weak solution of (1), if u satisfies J′(u)(v)=0 for all v∈X.Definition 7.
A functionu∈X is said to be a classical solution of problem (1) if u satisfies the equation in (1) for a.e. t∈[0,T]∖{t1,t2,…,tl} and the impulsive condition and boundary condition in (1). Moreover, u is said to be a positive classical solution of problem (1) if u(t)≥0, u(t)≢0, t∈[0,T].Lemma 8.
Ifu∈X is a weak solution of problem (1), then u is a classical solution of problem (1).Proof.
By Definition6, if u∈X is a weak solution of (1), then J′(u)(v)=0 holds for all v∈X and hence for all v∈C0∞(ti,ti+1),  v′∈C0∞(ti,ti+1),  v(t)≡0,  t∈[0,ti]∪[ti+1,T]. So ∫titi+1‍u′′v′′+uv-λf(t,u)vdt=0. By Lemma 5, we have
(22)u′′(t)=∫tit‍(∫0s‍-u(θ)+λf(θ,u(θ))dθ)ds+C1t+C2
for a.e. t∈[ti,ti+1] and some C1,C2∈R. So u(4)+u-λf(t,u)=0, a.e. t∈[ti,ti+1], i=0,1,…,l. Thus u satisfies the equation in problem (1) and u∈C4([0,T]∖{t1,t2,…,tl}). By integration by parts for two times, we have that
(23)∫0T‍(u′′v′′+uv)dt-λ∫0T‍f(t,u(t))v(t)dt-λ∑i=1l‍I1i(u(ti))v(ti)-λ∑i=1l‍I2i(u′(ti))v′(ti)+abu(T)v(T)+abu(0)v(0)+cdu′(T)v′(T)+cdu′(0)v′(0)=∑i=1l‍u′′(t)v′(t)∑i=1l‍u′′(t)v′(t)|t=ti+ti+1-∑i=1l‍u′′′(t)v(t)∑i=1l‍u′′′(t)v(t)|t=ti+ti+1+∫0T‍[u(4)(t)+u(t)-λf(t,u(t))]v(t)dt-λ∑i=1l‍I1i(u(ti))v(ti)-λ∑i=1l‍I2i(u′(ti))v′(ti)+abu(T)v(T)+abu(0)v(0)+cdu′(T)v′(T)+cdu′(0)v′(0)=∑i=1l‍(Δu′′′(ti)-λI1i(u(ti)))v(ti)+∫0T‍[u(4)(t)+u(t)-λf(t,u(t))]v(t)dt-∑i=1l‍(Δu′′(ti)+λI2i(u′(ti)))v′(ti)+u′′(T)v′(T)-u′′(0)v′(0)-u′′′(T)v(T)+u′′′(0)v(0)+abu(T)v(T)+abu(0)v(0)+cdu′(T)v′(T)+cdu′(0)v′(0)
holds for all v∈X. Since u satisfies the equation of problem (1), (23) becomes
(24)∑i=1l‍(Δu′′′(ti)-λI1i(u(ti)))v(ti)-∑i=1l‍(Δu′′(ti)+λI2i(u′(ti)))v′(ti)+[u′′(T)+cdu′(T)]v′(T)-[u′′(0)-cdu′(0)]v′(0)-[u′′′(T)-abu(T)]v(T)+[u′′′(0)+abu(0)]v(0)=0
for all v∈X.
Next we will verify thatu satisfies impulsive condition in (1). If not, without loss of generality, we assume (24) holds for v(t)≡0 for t∈[0,ti]∪[ti+1,T], v′(ti)≠0. So (Δu′′(ti)+λI2i(u′(ti)))v′(ti)=0, and then Δu′′(ti)+λI2i(u′(ti))=0. We assume v(t)≡0 for t∈[0,ti]∪[ti+(ti+1-ti)/2,T], v′(ti)≠0. So (24) becomes (Δu′′(ti)+λI2i(u′(ti)))v′(ti)=0, which means Δu′′(ti)+λI2i(u′(ti)=0. Similarly, by choosing particular v∈X, we can show that u satisfies boundary conditions in problem (1).Lemma 9.
Letu∈X; then ∥u∥C1≤M∥u∥X, where M=max⁡{1/T,T}max⁡{1+2T2,2Td/c}.Proof.
For anyu∈X, it follows from the mean-value theorem that
(25)u(τ)=1T∫0T‍u(s)ds
for some τ∈[0,T]. Hence, for t∈[0,T], using (25) and Hölder’s inequality, we have
(26)|u(t)|=|u(τ)+∫τt‍u′(s)ds|≤1T(∫0T‍|u(s)|2ds)1/2+T(∫0T‍|u′(s)|2ds)1/2≤max⁡{1T,T}∥u∥.
Similarly, we have |u′(t)|≤max⁡{1/T,T}∥u∥. So ∥u∥C1≤max⁡{1/T,T}∥u∥, which together with (13) yields the results.Lemma 10.
Suppose the following conditions hold.(H1)
There exist constantsμ>2 and r≥0 such that, for |ξ|≥r,
(27)0<μF(t,ξ)≤ξf(t,ξ).(H2)
The impulsive functionsI1i satisfy sublinear growth; that is, there exist constants αi>0, βi>0, and γi∈[0,1), i=1,2,…,l, such that
(28)|I1i(u)|≤αi+βi|u|γi.(H3)
The impulsive functionsI2i, i=1,2,…,l, are bounded.(H4)
f(t,u)=o(|u|), I1i(u)=o(|u|), I2i(u)=o(|u|) as |u|→0, i=1,2,…,l.Then the functional J defined by (9) is continuously differentiable. Moreover, it satisfies the Palais-Smale (PS) condition.Proof.
By the continuity off, I1i and I2i, i=1,2,…,l, we know that J is continuously differentiable. Next, we will prove that J satisfies the Palais-Smale condition. Let {J(uk)} be a bounded sequence such that |J′(uk)|→0 as k→∞. Then there exist two constants k,C1>0 such that for k sufficiently large
(29)|J(uk)|≤C1.
By (H3), there exists a constant C2>0 such that
(30)|I2i|≤C2,i=1,2,…,l.
Then for k sufficiently large, by (H1) (H2) and the definitions of J, J′, we have
(31)∥uk∥X2=2J(uk)+2λ∑i=1l‍∫0uk(ti)‍I1i(s)ds+2λ∑i=1l‍∫0uk′(ti)‍I2i(s)ds+2λ∫0T‍F(t,uk(t))dt≤2J(uk)+2λ∑i=1l‍∫0uk(ti)‍I1i(s)ds+2λ∑i=1l‍∫0uk′(ti)‍I2i(s)ds+2λμ∫0T‍f(t,uk(t))uk(t)dt=2J(uk)+2λ∑i=1l‍∫0uk(ti)‍I1i(t)dt+2λ∑i=1l‍∫0uk′(ti)‍I2i(s)ds+2μ∥uk∥X2-2λμ∑i=1l‍I1i(uk(ti))uk(ti)-2λμ∑i=1l‍I2i(uk′(ti))uk′(ti)-2μJ′(uk)(uk).
By Lemma 8, (29), (30), (31), (H2), and (H3), one has
(32)(12-1μ)∥uk∥X2=J(uk)-1μJ′(uk)(uk)+λ∑i=1l‍∫0uk(ti)‍I1i(s)ds+λ∑i=1l‍∫0uk′(ti)‍I2i(s)ds-λμ∑i=1l‍I1i(uk(ti))uk(ti)-λμ∑i=1l‍I2i(uk′(ti))uk′(ti)≤C1+C1μ∥uk∥+λ∥uk∥C1∑i=1l‍(αi+βi∥uk∥C1γi)+lλC2∥uk∥C1+λμ∥uk∥C1∑i=1l‍(αi+βi∥uk∥C1γi)+lλC2μ∥uk∥C1=C1+C1μ∥uk∥+λ(1+1μ)M∥uk∥∑i=1l‍(αi+βiM∥uk∥γi)+lλC2(1+1μ)M∥uk∥.
So {uk} is bounded in X, which implies that the sequence {uk} weakly converges to u.
Next we show that{uk} strongly converges to u in X:
(33)(J′(uk)-J′(u))(uk-u)=∥uk-u∥X2-λ∑i=1l‍[I1i(uk(ti))-I1i(u(ti))][uk(ti)-u(ti)]-λ∑i=1l‍[I2i(uk′(ti))-I2i(u′(ti))][uk′(ti)-u′(ti)]-λ∫0T‍[f(t,uk(t))-f(t,u(t))][uk(ti)-u(ti)]dt.
Similar to the proof of Proposition 1.2 in [18], the weak convergence uk⇀u implies that {uk} uniformly converges to u in C([0,T]). Since uk∈X, uk′ converges to u′ in C[0,T]. Thus
(34)(J′(uk)-J′(u))(uk-u)⟶0,λ∑i=1l‍[I1i(uk(ti))-I1i(u(ti))][uk(ti)-u(ti)]⟶0,λ∑i=1l‍[I2i(uk′(ti))-I2i(u′(ti))][uk′(ti)-u′(ti)]⟶0,λ∫0T‍[f(t,uk(t))-f(t,u(t))][uk(ti)-u(ti)]dt⟶0.
So ∥uk-u∥X→0 as k→∞. In other words, {uk} converges strongly to u in X.Remark 11.
By (H1), there exista1,a2>0 such that
(35)F(t,ξ)≥a1|ξ|μ-a2,
for all t∈[0,T], ξ∈R.


## 3. Existence Results for One Solution and Infinitely Many Solutions

Theorem 12.
Suppose that (H1)–(H3) hold. Furthermore, we assume (H4) holds. Then problem (1) has at least one nontrivial solution.Proof.
We will use Theorem1 to prove the theorem. By Lemma 10, we have known that J satisfies the (PS) condition and it is obvious that J(0)=0. By (H4), for any ε>0, there exists a δ>0 such that |ξ|≤δ, which implies
(36)|F(t,ξ)|≤12ε|ξ|2,|∑i=1l‍∫0ξ‍I1i(s)ds|≤12lε|ξ|2,|∑i=1l‍∫0ξ‍I2i(s)ds|≤12lε|ξ|2
for all t∈[0,T]. Consequently, by Lemma 9, one has, for ∥u∥X≤δ/M,
(37)|∫0T‍F(t,u(t))dt|≤12M2Tε∥u∥X2,|∑i=1l‍∫0u(ti)‍I1i(s)ds|≤12M2lε∥u∥X2,|∑i=1l‍∫0u′(ti)‍I2i(s)ds|≤12M2lε∥u∥X2.
Thus
(38)|∫0T‍F(t,u(t))dt|+|∑i=1l‍∫0u(ti)‍I1i(s)ds|+|∑i=1l‍∫0u′(ti)‍I2i(s)ds|≤12M2(T+2l)ε∥u∥X2=o(∥u∥X2)
as ∥u∥X→0. Therefore,
(39)J(u)=12∥u∥X2-λ∑i=1l‍∫0u(ti)‍I1i(s)ds-λ∑i=1l‍∫0u′(ti)‍I2i(s)ds-λ∫0T‍F(t,u(t))dt≥12∥u∥X2+o(∥u∥X2)
as u→0. Thus (C1) holds.
To verify (C2), we choosee(t)∈X, κ∈R such that K1=∥e∥X>0, K2=∥e∥L2>0, K1, K2 are constants. Then by (H1), (H2), Remark 11, Lemma 8, and (30), one has
(40)J(κe)=κ22∥e∥X2-λ∑i=1l‍∫0κe(ti)‍I1i(s)ds-λ∑i=1l‍∫0κe′(ti)‍I2i(s)ds-λ∫0T‍F(t,κe(t))dt≤κ22∥e∥X2+λ∑i=1l‍(αi∥κe∥C1+βi∥κe∥C1γi+1γi+1)+λlC2|κ|∥e∥C1-λ∫0T‍(a1|κ|μ|e(t)|μ-a2)dt≤κ22K12+λ∑i=1l‍(αi|κ|MK1+βi|κ|γi+1Mγi+1K1γi+1γi+1)+λlC2MK1|κ|-λ∫0T‍(a1|κ|μ|e(t)|μ-a2)dt.
By Hölder’s inequality, we have
(41)∫0T‍|e(t)|μdt≥[∫0T‍e2(t)dtT(2-μ)/μ]μ/2=∥e∥L2μT(2-μ)/2=K2μT(2-μ)/2.
Substituting (41) into (40), we have
(42)J(κe)≤κ22K12+λ[MK1∑i=1l‍(αi+βi|κe|γi+1Mγi+1K1γi+1γi+1)+lC2K1MMK1∑i=1l‍(αi+βi|κe|γi+1Mγi+1K1γi+1γi+1)]|κ|-a1λ|κ|μK2μT(2-μ)/2+a2λT⟶-∞
as |κ|→+∞. Hence (C2) holds. Therefore, applying Theorem 1, we deduce that J admits a critical value c>0 characterized as in the statement of Theorem 1 to J, J possesses critical value c>0 given by(43)c=inf⁡g∈Γmax⁡u∈g([0,1])J(u),
where
(44)Γ={g∈C([0,1],E):g(0)=0,g(1)=e}.
Let u*∈X be a critical point associated with the critical value c of J (i.e., J(u*)=c). Condition c>0 implies that u*≢0. Lemma 8 means thatIBVP (1) has at least one nontrivial solution.Theorem 13.
Suppose that (H1)–(H4) hold. Moreover, assume that the nonlinearityf(t,u) and impulsive functions I1i, I2i are all odd in u. ThenIBVP (1) has infinitely many classical solutions.Proof.
We apply Theorem2 to complete the proof. Clearly J∈C1(X,R) is even and J(0)=0. Lemma 10 shows that J satisfies (PS) condition. The arguments of Theorem 12 show that J satisfies (C3) in Theorem 2. To verify (C4), let W be any finite dimensional space in X. For any u∈W, by (H1), (H2), Remark 11, Lemma 4, and (30), one has
(45)J(u)=12∥u∥X2-λ∑i=1l‍∫0u(ti)‍I1i(s)ds-λ∑i=1l‍∫0u′(ti)‍I2i(s)ds-λ∫0T‍F(t,u(t))dt≤12∥u∥X2+λ∑i=1l‍(αi+βi|u|γi+1γi+1)+lλC2∥u∥C1-λ∫0T‍(a1|u(t)|μ-a2)dt≤12∥u∥X2+λM∥u∥X∑i=1l‍(αi+βiMγi+1∥u∥Xγi+1γi+1)+lλC2M∥u∥X-λ∫0T‍(a1|u(t)|μ-a2)dt.
For finite dimensional space W, the norm ∥·∥X is equivalent to ∥·∥W.
So there existsc0>0 satisfying
(46)∥u∥X≤c0∥u∥Lμ
for u∈W. Thus,
(47)∫0T‍a1|u(t)|μdt≥a1c0-μ∥u∥Xμ.
By (45) (46) (47) we have
(48)J(u)≤12∥u∥X2+λ∑i=1l‍(αi+βiMγi+1∥u∥Xγi+1γi+1)+lλC2M∥u∥X-λa1c0-μ∥u∥Xμ+λa2T⟶-∞
as ∥u∥X→+∞. That is, there exists R>0 such that J(u)<0 for u∈W∖BR(W). The proof is complete.


## 4. Existence Result for Three Nonnegative Solutions

In this part, we need the following conditions.(H5)
ForI1i, I2i, and f,
(C5)
f∈C([0,T]×[0,+∞);[0,+∞)), I1i∈C(R;R), I2i∈C(R;R), i=1,2,…,l;(C6)
f(t,0)=I1i(0)=I2i(0)=0 for almost every t∈[0,T] and I2i(x)x≥0 for all x∈R.Lemma 14 (see Lemma 2.2 [12]).
Foru∈X, let u±=max⁡{±u,0}. Then the following five properties hold:(i)
u∈X⇒u+, u-∈X;(ii)
u=u+-u-;(iii)
∥u+∥X≤∥u∥X;(iv)
if(un) uniformly converges to u in C([0,T]), then (un+) uniformly converges to u+ in C([0,T]);(v)
u+(t)u-(t)=0, (u+)′(t)(u-)′(t)=0 for a.e. t∈[0,T].Lemma 15.
Ifu∈C([0,T]) is a classical solution of problem
(49)u(4)(t)+u(t)=λf(t,u+(t)),t≠ti,t∈[0,T],Δu′′′(ti)=λI1i(u+(ti)),i=1,2,…,l,-Δu′′(ti)=λI2i((u+)′(ti)),i=1,2,…,l,au(0)+bu′′′(0)=0,au(T)-bu′′′(T)=0,cu′(0)-du′′(0)=0,cu′(T)+du′′(T)=0,
then u(t)≥0 for t∈[0,T], and hence it is a nonnegative classical solution of (1).Proof.
Sinceu∈C[0,T] and f∈C([0,T]×[0,+∞);[0,+∞)), we have u(4)∈C[0,T]∖{t1,t2,…,ti}. If u∈C([0,T]) is a classical solution of problem (49), by Lemma 14, (H5) and boundary conditions, we have
(50)0=∫0T‍(u(4)(t)+u(t)-λf(t,u+(t)))u-(t)dt=-λ∑i=1l‍I1i(u+(ti))u-(ti)-λ∑i=1l‍I2i((u+)′(ti))(u-)′(ti)+u′′′(T)u-(T)-u′′′(0)u-(0)-u′′(T)(u-)′(T)+u′′(0)(u-)′(0)+∫0T‍u′′(t)(u-)′′(t)dt+∫0T‍u(t)(u-)(t)dt-λ∫0T‍f(t,u+(t))u-(t)dt=abu(T)u-(T)+abu(0)u-(0)+cdu′(T)(u-)′(T)+cdu′(0)(u-)′(0)+∫0T‍u′′(t)(u-)′′(t)+u(t)(u-)(t)dt≤-∥u-∥X2.
So u-(t)=0 for t∈[0,T]; that is, u(t)≥0. The proof is complete.Remark 16.
By Lemmas14 and 15, in order to obtain the nonnegative solutions of (1), it is sufficient to show the existence of solutions of (49).
For eachu∈X, set
(51)Φ(u)=12∥u∥X2,(52)Ψ(u)=∑i=1l‍∫0u+(ti)‍I1i(s)ds+∑i=1l‍∫0(u+)′(ti)‍I2i(s)ds+∫0T‍F(t,u+(t))dt,J(u)=Φ(u)-λΨ(u).
It is obvious that Φ, Ψ, and J are differentiable for any u∈X. Then we have
(53)Φ′(u)(v)=∫0T‍(u′′v′′+uv)dt+abu(T)v(T)+abu(0)v(0)+cdu′(T)v′(T)+cdu′(0)v′(0),(54)Ψ′(u)(v)=∑i=1l‍I1i(u+(ti))v(ti)+∑i=1l‍I2i((u+)′(ti))v′(ti)+∫0T‍f(t,u+(t))v(t)dt.Definition 17.
A functionu∈X is said to be a weak solution of (49), if u satisfies J′(u)(v)=0 for all v∈X.Lemma 18.
Ifu∈X is a weak solution of (49), then u is a classical solution of (49).Proof.
It is similar to the proof of Lemma8, so we omit it here.Lemma 19.
Φ
′
:
X
→
X
* admits a continuous inverse on X*.Proof.
First, for everyu∈X∖{0}, by (53), we have
(55)lim⁡∥u∥X→+∞Φ′(u)(u)∥u∥X=lim⁡∥u∥X→+∞∥u∥X2∥u∥X=+∞,
which means that Φ′ is coercive. Furthermore, given u,v∈X, one has
(56)(Φ′(u)-Φ′(v))(u-v)≥∥u-v∥X2,
so Φ′ is uniformly monotone. By Theorem 26.A(d) of [20], we have that (Φ′)-1 exists and (Φ′)-1 is continuous on X*. Thus, Φ′:X→X* admits a continuous inverse on X*. The proof is complete.Lemma 20.
Ψ
′
:
X
→
X
* is a continuous and compact operator.Proof.
First we will show thatΨ′ is strongly continuous on X. Let un⇀u as n→∞ on X. By [20], we have (un) converges uniformly to u on [0,T] as n→∞. Since f is continuous, one has f(t,un)→f(t,u) as n→∞. Furthermore, I1i, I2i are all continuous. So Ψ′(un)→Ψ′(u), which implies that Ψ′ is continuous and that Ψ′ is a compact operator by Proposition 26.2 of [20]. The proof is complete.Theorem 21.
Suppose that the condition (H5) holds. Letk:={2M2(1024/T3+(83/240)T)}-1, I(s):=∑i=1l‍|I1i(s)|+∑i=1l‍|I2i(s)|. There exist four positive constants m, n, p, q, with km<n<kp<kq such that(H6)
(C7)
(∫0m‍I(s)ds+∫0T‍F(t,m)dt)/m2<k((-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)/n2),(C8)
(∫0p‍I(s)ds+∫0T‍F(t,p)dt)/p2<k((-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)/n2),(C9)
(∫0q‍I(s)ds+∫0T‍F(t,q)dt)/(q2-p2)<k((-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)/n2).Then, for every
(57)λ∈n22M2k(-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)-1,]n22M2k(-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)-1,min⁡{p22M2(∫0p‍I(s)ds+∫0T‍F(t,p)dt)-1,q2-p22M2(∫0q‍I(s)ds+∫0T‍F(t,q)dt)-1m22M2(∫0m‍I(s)ds+∫0T‍F(t,m)dt)-1,p22M2(∫0p‍I(s)ds+∫0T‍F(t,p)dt)-1,q2-p22M2(∫0q‍I(s)ds+∫0T‍F(t,q)dt)-1}[n22M2k(-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)-1,,
the problem (1) has at least three distinct nonnegative classical solutions ui(i=1,2,3), such that ∥ui∥C1<q, which means that the problem (1) has at least two distinct positive classical solutions.Proof.
The proof is based on Theorem3. First, we will prove that Φ and Ψ satisfy the hypotheses in Theorem 3. On the one hand, Φ is coercive and its Gâteaux derivative admits a continuous inverse by Lemma 19. On the other hand, Φ is obviously convex. Ψ’s Gâteaux derivative is continuous and compact by Lemma 20. In addition, inf⁡XΦ=Φ(0)=Ψ(0)=0. By (H5), we have
(58)∫0s‍I1i(s)ds≥0,∫0s‍I2i(s)ds≥0,F(t,s)=∫0s‍f(t,s)ds≥0,
which deduces that Ψ(u)≥0 for all u∈X.
Next, we will verify the conditions (i), (ii), (iii) in Theorem3. First, we define
(59)v-(t)={32nT2t2,t∈[0,T8],-32nT2(t-T4)2+n,t∈]T8,T4],n,t∈]T4,3T4],-32nT2(t-3T4)2+n,t∈]3T4,7T8],32nT2(t-T)2,t∈]7T8,T].
It is easy to verify that
(60)v-+=v-,v--=0.
By computing,
(61)Φ(v-)=(1024T3+83240T)n2=n22M2k.
Let r1=m2/2M2, r2=p2/2M2, and r3=(q2-p2)/2M2. By km<n<kp<kq, one has r1<Φ(v-)<r2, which means that v-∈Φ-1([r1,r2[) and r3>0. When Φ(u)<r1, by Lemma 9 and (51), we have
(62)max⁡{max⁡t∈[0,T]|u+(t)|,max⁡t∈[0,T]|(u+)′(t)|}≤M∥u+∥X≤M∥u∥X≤2M2Φ(u)<m.
In view of (51) and I(s), we have
(63)sup⁡u∈Φ-1(]-∞,r1[)Ψ(u)≤max⁡|ξ|≤m∑i=1l‍∫0ξ‍|I1i(s)|ds+max⁡|ξ|≤m∑i=1l‍∫0ξ‍|I2i(s)|ds+∫0T‍max⁡|ξ|≤mF(t,ξ)dt=∫0m‍I(s)ds+∫0T‍F(t,m)dt.
Similarly, we have
(64)sup⁡u∈Φ-1(]-∞,r2[)Ψ(u)≤∫0p‍I(s)ds+∫0T‍F(t,p)dt,sup⁡u∈Φ-1(]-∞,r2+r3[)Ψ(u)≤∫0q‍I(s)ds+∫0T‍F(t,q)dt.
Therefore, taking into consideration that 0∈Φ-1(]-∞,r1[) and 0∈Φ-1(]-∞,r2[), by (63) and (64), we have
(65)φ(r1)≤sup⁡u∈Φ-1(]-∞,r1[)Ψ(u)r1≤2M2m2(∫0m‍I(s)ds+∫0T‍F(t,m)dt),φ(r2)≤sup⁡u∈Φ-1(]-∞,r2[)Ψ(u)r2≤2M2p2(∫0p‍I(s)ds+∫0T‍F(t,p)dt),γ(r2,r3)=sup⁡u∈Φ-1(]-∞,r2+r3[)Ψ(u)r3≤2M2q2-p2(∫0q‍I(s)ds+∫0T‍F(t,q)dt).
Furthermore, by (60) and the definition of v-,
(66)Ψ(v-)=∑i=1l‍∫0v-(ti)‍I1i(s)ds+∑i=1l‍∫0v-′(ti)‍I2i(s)ds+∫0T‍F(t,v-(t))dt≥∑i=1l‍∫0min⁡t∈[0,T]v-(t)‍I1i(s)ds+∑i=1l‍∫0min⁡t∈[0,T]|v-′(t)|‍|I2i(s)|ds+∫T/43T/4‍F(t,v-(t))dt=∫T/43T/4‍F(t,n)dt.
Taking v-∈Φ-1([r1,r2[) into consideration, by (61), (63), (66), (C9), and Φ(u)≥0, one has
(67)β(r1,r2)≥inf⁡u∈Φ-1(]-∞,r1[)Ψ(v-)-Ψ(u)Φ(v-)-Φ(u)≥∫T/43T/4‍F(t,n)dt-(∫0m‍I(s)ds+∫0T‍F(t,m)dt)n2/2M2k=2M2kn2(-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt).
By (65), (67), and (C7)–(C9) of (H6), we have
(68)α(r1,r2,r3)<β(r1,r2),
which yields the conditions in Theorem 3. By Theorem 3, it follows that, for each
(69)λ∈n22M2k(-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)-1]n22M2k(-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)-1,min⁡{p22M2(∫0p‍I(s)ds+∫0T‍F(t,p)dt)-1,q2-p22M2(∫0q‍I(s)ds+∫0T‍F(t,q)dt)-1m22M2(∫0m‍I(s)ds+∫0T‍F(t,m)d)-1,p22M2(∫0p‍I(s)ds+∫0T‍F(t,p)dt)-1,q2-p22M2(∫0q‍I(s)ds+∫0T‍F(t,q)dt)-1}[n22M2k(-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)-1,
the functional J=Φ-λΨ has three distinct critical points ui(i=1,2,3) in X with Φ(ui)<r2+r3. By Lemma 9 and (51)
(70)∥ui+∥C1=max⁡{max⁡t∈[0,T]|ui+(t)|,max⁡t∈[0.T]|(ui+)′(t)|}≤M∥ui+∥X≤M∥ui∥X≤2M2Φ(ui)<q.
By Remark 16, ui(i=1,2,3) are three positive solutions of (1). The proof is complete.Remark 22.
If we choose differentv¯, then the constrictions on F, Ii are different.


## 5. Examples

Example 1.
LetT>0, ti∈(0,T), a,b,c,d>0, x1,x2,x3,yi,zi∈C([0,T],R+), i=1,2,…,l. Consider the following Sturm-Liouville boundary-value problem with impulse:
(71)u(4)(t)+u(t)=λf1(t,u),t∈[0,T]∖{t1,t2,…,tl},Δu′′′(ti)=λIi(u(ti)),i=1,2,…,l,-Δu′′(ti)=-λzi(t)sinu′3(ti),i=1,2,…,l,au(0)+bu′′′(0)=0,au(T)-bu′′′(T)=0,cu′(0)-du′′(0)=0,cu′(T)+du′′(T)=0,
where
(72)f1(t,u)=x1(t)u3+x2(t)sinu+x3(t)exp⁡|u|,Ii(u)={yi(t)u3,|u|≤1,yi(t)u1/3,|u|>1.By computing,F(t,u)=(1/4)x1(t)u4-x2(t)cos⁡u+x3(t)sgn⁡uexp⁡|u|. Let μ=6; there exists r≥0 such that, for |ξ|≥r, 0<μF(t,ξ)≤ξf(t,ξ). The conditions (H1)–(H4) are satisfied. Applying Theorem 1, problem (71) has at least one nontrivial solution.Furthermore, the functionsf(t,u), I1i, I2i are all odd in u. Applying Theorem 2, problem (71) has infinitely many classical solutions.Example 2.
LetT=1, ti∈(0,1), i=1,2,…,l, a,b,c,d>0. Consider the following Sturm-Liouville boundary-value problem:
(73)u(4)(t)+u(t)=λf2(t,u(t)),t∈[0,1]∖{t1,t2,…,tl},Δu′′′(ti)=λI1i(u(ti)),i=1,2,…,l,-Δu′′(ti)=λI2i(u′(ti)),i=1,2,…,l,au(0)+bu′′′(0)=0,au(1)-bu′′′(1)=0,cu′(0)-du′′(0)=0,cu′(1)+du′′(1)=0,
where
(74)f2(t,s)=I1i(s)=I2i(s)={0,0<s≤1,t(s-1),1<s≤2,-t(s-3),2<s≤3,0,s>3,11111111111111111foreveryλ∈[7680245843,+∞),
problem (73) has at least two distinct positive classical solutions.In fact, (H5) is fulfilled. By computing,M=max⁡{3,2d/c}, so k=min⁡{40/737529,30c2/245843d2}. Considering km<n<kp<kq, without loss of generality, we choose m=1/2, n=2 and sufficiently large p, q. Then we have
(75)n22M2k(-∫0m‍I(s)ds+∫T/43T/4‍F(t,n)dt-∫0T‍F(t,m)dt)-1=161280245843,m22M2(∫0m‍I(s)ds+∫0T‍F(t,m)dt)-1=∞,p22M2(∫0p‍I(s)ds+∫0T‍F(t,p)dt)-1issufficientlylarge,q2-p22M2(∫0q‍I(s)ds+∫0T‍F(t,q)dt)-1issufficientlylarge,
and that (H6) is satisfied. Applying Theorem 21, problem (73) has at least two distinct positive classical solutions for every λ∈[161280/245843,+∞).



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*Source: 690381-2014-07-21.xml*