# Multiple Solutions of Second-Order Damped Impulsive Differential Equations with Mixed Boundary Conditions

Authors: Jian Liu; Lizhao Yan
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We use variational methods to investigate the solutions of damped impulsive differential equations with mixed boundary conditions. The conditions for the multiplicity of solutions are established. The main results are also demonstrated with examples.


---

## Body

## 1. Introduction

Impulsive effect exists widely in many evolution processes in which their states are changed abruptly at certain moments of time. The theory of impulsive differential systems has been developed by numerous mathematicians [1–6]. Applications of impulsive differential equations with or without delays occur in biology, medicine, mechanics, engineering, chaos theory, and so on [7–11].In this paper, we consider the following second-order damped impulsive differential equations with mixed boundary conditions:(1)-u′′(t)+g(t)u′(t)-λu(t)=f(t,u(t)),hhhhhhhhhhhhhht≠tj,a.e.t∈[0,T],-Δu′(tj)=Ij(u(tj)),j=1,2,…,n,u′(0)=0,u(T)=0,
where 0=t0<t1<t2<⋯<tn<tn+1=T, g∈C[0,T], f:[0,T]×R→R is continuous, Ij:R→R, j=1,2,…,n are continuous, and Δu′(tj)=u′(tj+)-u′(tj-) for u′(tj±)=lim⁡t→tj±u′(t),j=1,2,…,n.The characteristic of (1) is the presence of the damped term g(t)u′. Most of the results concerning the existence of solutions of these equations are obtained using upper and lower solutions methods, coincidence degree theory, and fixed point theorems [12–15]. On the other hand, when there is no presence of the damped term, some researchers have used variational methods to study the existence of solutions for these problems [16–21]. However, to the best of our knowledge, there are few papers concerned with the existence of solutions for impulsive boundary value problems like problem (1) by using variational methods.For this nonlinear damped mixed boundary problem (1), the variational structure due to the presence of the damped term g(t)u′ is not apparent. However, inspired by the work [22, 23], we will be able to transform it into a variational formulation. In this paper, our aim is to study the existence of n distinct pairs of nontrivial solutions of problem (1). Our main results extend the study made in [22, 23], in the sense that we deal with a class of problems that is not considered in those papers.


## 2. Preliminaries and Statements

Letm=min⁡t∈[0,T]eG(t), M=max⁡t∈[0,T]eG(t), G(t)=-∫0t‍g(s)ds, t∈[0,T]. We transform (1) into the following equivalent form:
(2)-(eG(t)u′(t))′-λeG(t)u(t)=eG(t)f(t,u(t)),hhhhhhhhhhhhhhhhhht≠tj,a.e.t∈[0,T],-Δu′(tj)=Ij(u(tj)),j=1,2,…,n,u′(0)=0,u(T)=0.Obviously, the solutions of (2) are solutions of (1).Define the spaceX={u(t)∣u(t)isabsolutelycontinuouson⁡[0,T],u′(·)∈L2[0,T],u(T)=0}. It is easy to see that H01(0,T)⊂X⊂H1(0,T) and X is a closed subset of H1(0,T). So X is a Hilbert space with the usual inner product in H1(0,T).Consider the Hilbert spacesX with the inner product
(3)(u,v)=∫0TeG(t)u′(t)v′(t)dt,
inducing the norm
(4)∥u∥=(∫0TeG(t)|u′(t)|2dt)1/2.
We also consider the inner product
(5)(u,v)=∫0T‍u′(t)v′(t)dt,
inducing the norm
(6)∥u∥X=(∫0T|u′(t)|2dt‍)1/2.Consider the problem(7)-u′′(t)=λu(t),t∈[0,T],u′(0)=0,u(T)=0.
As is well known, (7) possesses a sequence of eigenvalues (λi)(λi=[(2i-1)π/2T]2) with
(8)0<λ1<λ2<⋯<λj<⋯.
The corresponding eigenfunctions are normalized so that ∥φj∥X=1=λj∫0T‍|φj(t)|2dt; here
(9)φj(t)=2Tλjcos⁡⁡(λjt),j=1,2,….Now multiply (2) by v∈X and integrate on the interval [0,T]:
(10)∫0TeG(t)u′(t)v′(t)dt-λ∫0TeG(t)u(t)v(t)dt=∑j=1neG(tj)Ij(u(tj))v(tj)+∫0TeG(t)f(t,u(t))v(t)dt.
Then, a weak solution of (2) is a critical point of the following functional:
(11)E(u)=12∫0TeG(t)|u′(t)|2dt-λ2∫0TeG(t)|u(t)|2dt-∑j=1neG(tj)∫0u(tj)Ij(t)dt-∫0TeG(t)F(t,u(t))dt,
where F(t,u)=∫0u‍f(t,ξ)dξ.We say thatu∈C[0,T] is a classical solution of IBVP (1) if it satisfies the following conditions: u satisfies the first equation of (1) a.e. on [0,T]; the limits u′(tj+),u′(tj-), j=1,2,…,n, exist and impulsive condition of (1) holds; u satisfies the boundary condition of (1).Lemma 1.
Ifu∈X is a weak solution of (1), then u is a classical solution of (1).Proof.
Ifu∈X is a weak solution of (1), then u is a weak solution of (2), so (E′(u),v)=0 holds for all v∈X; that is,
(12)∫0T‍[eG(t)u′(t)v′(t)+λeG(t)u(t)v(t)]dt-∑j=1neG(tj)Ij(u(tj))v(tj)‍-∫0TeG(t)f(t,u(t))v(t)dt=0.‍
By integrating by part, we have
(13)∫0T‍[eG(t)u′(t)v′(t)+λeG(t)u(t)v(t)]dt-∑j=1neG(tj)Ij(u(tj))v(tj)-∫0TeG(t)f(t,u(t))v(t)dt=∫0T[-(eG(t)u′(t))′+λeG(t)u(t)-eG(t)f(t,u(t))]v(t)dt-∑j=1neG(tj)[Δu′(tj)+Ij(u(tj))]v(tj)-eG(0)u′(0)v(0)+eG(T)u′(T)v(T)=∫0T[-(eG(t)u′(t))′+λeG(t)u(t)-eG(t)f(t,u(t))]v(t)dt-∑j=1neG(tj)[Δu′(tj)+Ij(u(tj))]v(tj)-u′(0)v(0).
Thus
(14)∫0T[-(eG(t)u′(t))′+λeG(t)u(t)-eG(t)f(t,u(t))]v(t)dt-∑j=1neG(tj)[Δu′(tj)+Ij(u(tj))]v(tj)-u′(0)v(0)=0
holds for all v∈X. Without loss of generality, for any j={1,2,…,n} and v∈X with v(t)≡0, for every t∈[0,tj]∪[tj+1,T], then substituting v into (14), we get
(15)-(eG(t)u′(t))′+λeG(t)u(t)-eG(t)f(t,u(t))=0,hhhhhhhhhhhhhhhhhhhhhhhhhhhhht∈(tj,tj+1).
Hence u satisfies the first equation of (2). Therefore, by (14) we have
(16)-∑j=1neG(tj)[Δu′(tj)+Ij(u(tj))]v(tj)-u′(0)v(0)=0.
Next we will show thatu satisfies the impulsive and the boundary condition in (2). If the impulsive condition in (2) does not hold, without loss of generality, we assume that there exists j∈{1,2,…,n} such that
(17)Δu′(tj)+Ij(u(tj))≠0.
Let v(t)=∏i=0,i≠jn+1‍(t-ti); then
(18)-∑j=1neG(tj)[Δu′(tj)+Ij(u(tj))]v(tj)-u′(0)v(0)=-eG(tj)[Δu′(tj)+Ij(u(tj))]v(tj)≠0,
which contradicts (16). So u satisfies the impulsive condition in (2) and (16) implies
(19)u′(0)v(0)=0.
If u′(0)≠0, pick v(t)=∏i=1n+1‍(t-ti); one has
(20)u′(0)∏i=1n+1‍(t0-ti)≠0,
which contradicts (19), so u satisfies the boundary condition. Therefore, u is a solution of (1).Lemma 2.
Letu∈X. Then there exists a constant σ>0, such that
(21)∥u∥∞≤σ∥u∥,
where ∥u∥∞=max⁡t∈[0,T]|u(t)|.Proof.
By Hölder inequality, foru∈X,
(22)|u(t)|=|u(T)-∫tTu′(s)ds‍|≤(∫0T1eG(s)ds)1/2(∫0TeG(s)|u′(s)|2ds)1/2≤Tm∥u∥=σ∥u∥.Lemma 3 (see [24, Theorem 9.1]).
LetE be a real Banach space, I∈C1(E,R) with I even, bounded from below, and satisfying P.S. condition. Suppose I(0)=0; there is a set K⊂E such that K is homeomorphic to Sj-1 by an odd map and sup⁡KI<0. Then I possesses at least j distinct pairs of critical points.


## 3. Main Results

Theorem 4.
Suppose that the following conditions hold.(H1)
There existu1>0, r>Mλk/m,λk which is the kth eigenvalue of (7) such that
(23)rMu1+eG(t)f(t,u1)=0,rMu1+eG(t)f(t,u)>0hhhhhhhhhhhhhhhhhhhhhhhhforeveryu∈(0,u1).(H2)
There existaj,bj>0 and rj∈[0,1)(j=1,2,…,n) such that
(24)|Ij(u)|≤aj+bj|u|rjforanyu∈R.(H3)
f(t,u) and Ij(u)(j=1,2,…,n) are odd about u.(H4)
f(t,u)=o(|u|), Ij(u)=o(|u|), as |u|→0, j=1,2,…,n.Then, for λ∈(Mλk/m,r], problem (1) has at least k distinct pairs of solutions.Proof.
Set(25)h1(λ,t,u)={λeG(t)u+eG(t)f(t,u)u∈[-u1,u1],λeG(t)u1+eG(t)f(t,u1),u∈[u1,+∞),-λeG(t)u1-eG(t)f(t,-u1),u∈(-∞,-u1].
Consider
(26)-(eG(t)u′(t))′=h1(λ,t,u(t)),hhhhhht≠tj,a.e.t∈[0,T],-Δu′(tj)=Ij(u(tj)),j=1,2,…,n,u′(0)=0,u(T)=0.
Next, we will verify that the solutions of problem (26) are solutions of problem (1).
In fact, letu0(t) be the solution of problem (26). If max⁡0≤t≤Tu0(t)>u1, then there exists an interval [a,b]⊂[0,T] such that
(27)u0(a)=u0(b)=u1,u0(t)>u1foranyt∈(a,b).
When t∈[a,b], by (H1), we have
(28)-(eG(t)u0′(t))′=h1(λ,t,u)=λeG(t)u1+eG(t)f(t,u1)≤rMu1+eG(t)f(t,u1)=0.
That is, eG(t)u0′(t) is nondecreasing in [a,b]. By u0′(a)≥0 and u0′(b)≤0, we have
(29)0≤eG(t)u0′(a)≤eG(t)u0′(t)≤eG(t)u0′(b)≤0hhhhhhhhhhhhhhhhhhhforeveryt∈[a,b].
That is, eG(t)u0′(t)≡0 for any t∈[a,b]. Since eG(t)≠0, then u0′(t)≡0. So, there exists a constant ϵ such that u0(t)≡ϵ, which contradicts (27). Then max⁡0≤t≤Tu0(t)≤u1. Similarly, we can prove that min⁡0≤t≤Tu0(t)>-u1.
Therefore, any solution of (26) is a solution of (1). Hence to prove Theorem 4, it suffices to produce at least k distinct pairs of critical points of
(30)E1(u)=12∫0TeG(t)|u′(t)|2dt‍-∫0TH1(λ,t,u(t))dt-∑j=1neG(tj)∫0u(tj)Ij(t)dt,
where H1(λ,t,u(t))=∫0u‍h1(λ,t,s)ds.
We will apply Lemma3 to finish the proof.
By (30) and (H3), E1∈C′(X,R) is even and E1(0)=0.
Next, we will show thatE1 is bounded from below.
LetC1=max⁡{a1,a2,…,an}, C2=max⁡{b1,b2,…,bn}. By (H1) and (H3), we have uh1(λ,t,u(t))≤0 for |u|≥u1; thus
(31)∫0TH1(λ,t,u(t))dt=∫0T∫0u(t)h1(λ,t,s)dsdt≤∫0T∫0u1h1(λ,t,s)dsdt≤∫0T∫0u1[rMs+f(t,s)]dsdt=ρ>0.
So, we have
(32)E1(u)=12∥u∥2-∫0TH1(λ,t,u(t))dt-∑j=1neG(tj)∫0u(tj)Ij(t)dt≥12∥u∥2-ρ-nσC1M∥u∥-C2M∑j=1nσrj+1∥u∥rj+1>-∞,
for any u∈X. Therefore, E1 is bounded from below.
In the following we will show thatE1 satisfies the P.S. condition. Let {uk}⊂X such that {E1(uk)} is a bounded sequence and lim⁡k→∞E1′(uk)=0; then there exists C3>0 such that
(33)|E1(uk)|≤C3.
By (32), we have
(34)12∥uk∥2≤C3+ρ+nσC1M∥uk∥+C2M∑j=1nσrj+1∥u∥rj+1.
So {uk} is bounded in X. From the reflexivity of X, we may extract a weakly convergent subsequence that, for simplicity, we call {uk},uk⇀u in X. In the following we will verify that {uk} strongly converges to u:
(35)(E1′(uk)-E1′(u))(uk-u)=∥uk-u∥2-∫0T‍[h1(λ,t,uk(t))-h1(λ,t,u(t))]hhhhhhhhhhhhhhhh×(uk(t)-u(t))dt+∑j=1neG(tj)[Ij(uk(tj))-Ij(u(tj))]×(uk(tj)-u(tj)).
By uk⇀u in X, we see that {uk} uniformly converges to u in C[0,T]. So
(36)∫0T[h1(λ,t,uk(t))-h1(λ,t,u(t))]×(uk(t)-u(t))dt⟶0,∑j=1neG(tj)[Ij(uk(tj))-Ij(u(tj))]×(uk(tj)-u(tj))⟶0,(E1′(uk)-E1′(u))(uk-u)⟶0,ask⟶+∞.
So we obtain ∥uk-u∥→0, as k→+∞. That is, {uk} strongly converges to u in X, which means that E1 satisfies the P.S. condition.
Now setK={∑i=1k‍ciφi:∑i=1k‍ci2=c2}, where φi is defined in (9). It is clear that K is homeomorphic to Sk-1 by an odd map for any c>0. In the following we verify that E1|K<0 if c is sufficiently small.
For anyu∈K,u=∑i=1k‍ciφi. By (H4) and (30), we have
(37)E1(u)=12∫0TeG(t)[(∑i=1kciφi(t))′]2dt-∫0TH1(λ,t,u(t))dt-∑j=1neG(tj)∫0u(tj)Ij(t)dt=12∑i=1kci2∫0TeG(t)[φi′(t)]2dt-λ2∑i=1kci2∫0TeG(t)[φi(t)]2dt-∫0TeG(t)F(t,u(t))dt-∑j=1neG(tj)‍∫0u(tj)Ij(t)dt‍≤M2∑i=1kci2∫0T‍[φi′(t)]2dt-mλ2∑i=1kci2∫0T[φi(t)]2dt‍-∫0TeG(t)F(t,u(t))dt-∑j=1neG(tj)∫0u(tj)Ij(t)dt=12∑i=1kci2(M-mλλi)-∫0TeG(t)F(t,u(t))dt-∑j=1neG(tj)∫0u(tj)Ij(t)dt≤12(M-mλλk)c2+o(c2)+o(c2),
for small c>0. Since λ∈(Mλk/m,r], E1(u)<0 and the proof is complete.Theorem 5.
Suppose that the following conditions hold.(H1)
There existu1>0, r>Mλk/m,λk which is the kth eigenvalue of (7) such that
(38)rMu1+eG(t)f(t,u1)=0,rMu1+eG(t)f(t,u)>0hhhhhhhhhhhhhhhhhhhhhhhhforeveryu∈(0,u1).(H2)
∫0u‍Ij(s)ds≤0 for any u∈R(j=1,2,…,n).(H3)
f(t,u) and Ij(u)(j=1,2,…,n) are odd about u.(H4)
f(t,u)=o(|u|), Ij(u)=o(|u|), as |u|→0, j=1,2,…,n.Then, for λ∈(Mλk/m,r], problem (1) has at least k distinct pairs of solutions.Proof.
The proof is similar to the proof of Theorem4, and therefore we omit it.Theorem 6.
Suppose that the following conditions hold.(H1)
There existu2>0, r>Mλk/m,λk which is the kth eigenvalue of (7) such that
(39)rMu2+eG(t)f(t,u2)≤0,Ij(u2)≤0,hhhhhhhhhhhhhhhhhhhhj=1,2,…,n.(H2)
f(t,u) and Ij(u)(j=1,2,…,n) are odd about u.(H3)
f(t,u)=o(|u|), Ij(u)=o(|u|), as |u|→0, j=1,2,…,n.Then, for λ∈(Mλk/m,r], problem (1) has at least k distinct pairs of solutions.Proof.
Set(40)h2(λ,t,u)={λeG(t)u+eG(t)f(t,u),u∈[-u2,u2],λeG(t)u2+eG(t)f(t,u2),u∈[u2,+∞),-λeG(t)u2-eG(t)f(t,-u2),u∈(-∞,-u2],Tj(u)={Ij(u),u∈[-u2,u2],Ij(u2),u∈[u2,+∞),Ij(-u2),u∈(-∞,-u2].
Consider
(41)-(eG(t)u′(t))′=h2(λ,t,u(t)),hhhhhht≠tj,a.e.t∈[0,T],-Δu′(tj)=Tj(u(tj)),j=1,2,…,n,u′(0)=0,u(T)=0.
Next, we will verify that the solutions of problem (41) are solutions of problem (1).
In fact, letω1={t∈(a1,b1)⊆[0,T]:u(t)>u2}. By the definitions of h2(λ,t,u) and Tj(u), (41) is reduced to
(42)-(eG(t)u′(t))′=h2(λ,t,u2)=λeG(t)u2+eG(t)f(t,u2)≤rMu2+eG(t)f(t,u2)≤0,t≠tj,a.e.t∈(a1,b1),-Δu′(tj)=Tj(u(tj))=Ij(u2)≤0,j=1,2,…,n,u(a1)=u(b1)=u2.
The solution u(t) of (42) satisfies u(t)≤u2,t∈(a1,b1). So ω1=∅ and u(t)≤u2.
Letω2={t∈(a2,b2)⊆[0,T]:u(t)<-u2}. By the definitions of h2(λ,t,u) and Tj(u), (41) is reduced to
(43)-(eG(t)u′(t))′=h2(λ,t,-u2)=-λeG(t)u2+eG(t)f(t,-u2)≥-rMu2-eG(t)f(t,u2)≥0,hhhhhhhhhhhhhhhht≠tj,a.e.t∈(a2,b2),-Δu′(tj)=Tj(u(tj))=-Ij(u2)≥0,j=1,2,…,n,u(a2)=u(b2)=-u2.
The solution u(t) of (43) satisfies u(t)≥-u2, t∈(a2,b2). So ω2=∅ and u(t)≥-u2.
Therefore, the solutions of (41) are solutions of (1). Hence to prove Theorem 6, it suffices to produce at least k distinct pairs of critical points of
(44)E2(u)=12∫0TeG(t)|u′(t)|2dt-∫0TH2(λ,t,u(t))dt-∑j=1neG(tj)∫0u(tj)Tj(t)dt,
where H2(λ,t,u(t))=∫0u‍h2(λ,t,s)ds.
We will apply Lemma3 to finish the proof.
By (44) and (H2), E2∈C′(X,R) is even and E2(0)=0.
Next, we will show thatE2 is bounded from below.
By (H1) and (H2), we haveuh2(λ,t,u(t))≤0 and uTj(u)≤0 for |u|≥u2; thus
(45)∫0TH2(λ,t,u(t))dt=∫0T∫0u(t)h2(λ,t,s)dsdt≤∫0T∫0u2h2(λ,t,s)dsdt‍≤∫0T∫0u2[rMs+f(t,s)]dsdt=ρ>0,∫0u(tj)‍Tj(t)dt≤∫0u2‍Tj(t)dt=δ>0.
So, we have
(46)E2(u)=12∥u∥2-∫0TH2(λ,t,u(t))dt-∑j=1neG(tj)∫0u(tj)Tj(t)dt≥12∥u∥2-ρ-nMδ>-∞,
for any u∈X. Therefore, E2 is bounded from below.
In the following we will show thatE2 satisfies the P.S. condition. Let {uk}⊂X such that {E2(uk)} is a bounded sequence and lim⁡k→∞E2′(uk)=0; then there exists C4>0 such that
(47)|E2(uk)|≤C4.
By (46), we have
(48)12∥uk∥2≤C4+ρ+nMδ.
So {uk} is bounded in X. From the reflexivity of X, we may extract a weakly convergent subsequence that, for simplicity, we call {uk},uk⇀u in X. In the following we will verify that {uk} strongly converges to u:
(49)(E2′(uk)-E2′(u))(uk-u)=∥uk-u∥2-∫0T‍[h2(λ,t,uk(t))-h2(λ,t,u(t))]×(uk(t)-u(t))dt+∑j=1neG(tj)[Tj(uk(tj))-Tj(u(tj))]×(uk(tj)-u(tj)).
By uk⇀u in X, we see that {uk} uniformly converges to u in C[0,T]. So
(50)∫0T[h2(λ,t,uk(t))-h2(λ,t,u(t))](uk(t)-u(t))dt⟶0,∑j=1neG(tj)[Tj(uk(tj))-Tj(u(tj))](uk(tj)-u(tj))⟶0,(E2′(uk)-E2′(u))(uk-u)→0,ask⟶+∞.
So we obtain ∥uk-u∥→0, as k→+∞. That is, {uk} strongly converges to u in X, which means E2 satisfies the P.S. condition.
Now setK={∑i=1k‍ciφi:∑i=1k‍ci2=c2}, where φi is defined in (9). It is clear that K is homeomorphic to Sk-1 by an odd map for any c>0. In the following we verify that E2|K<0 if c is sufficiently small.
For anyu∈K, u=∑i=1k‍ciφi. By (H3) and (44), we have
(51)E2(u)=12∫0TeG(t)[(∑i=1kciφi(t))′]2dt-∫0TH2(λ,t,u(t))dt-∑j=1neG(tj)∫0u(tj)Tj(t)dt=12∑i=1kci2∫0TeG(t)[φi′(t)]2dt-λ2∑i=1kci2∫0TeG(t)‍[φi(t)]2dt-∫0TeG(t)F(t,u(t))dt-∑j=1neG(tj)∫0u(tj)Tj(t)dt≤M2∑i=1kci2∫0T[φi′(t)]2dt-mλ2∑i=1kci2∫0T‍[φi(t)]2dt-∫0TeG(t)F(t,u(t))dt-∑j=1neG(tj)∫0u(tj)Tj(t)dt=12∑i=1kci2(M-mλλi)-∫0TeG(t)F(t,u(t))dt-∑j=1neG(tj)∫0u(tj)Tj(t)dt≤12(M-mλλk)c2+o(c2)+o(c2),
for small c>0. Since λ∈(Mλk/m,r], E2(u)<0 and the proof is complete.


## 4. Example

To illustrate how our main results can be used in practice we present the following example.Example 1.
LetT=π/4,g(t)=-2t, and consider the following problem:
(52)-u′′(t)-2tu′(t)-λu(t)=(1+t)(u-u2)-1000e(π2/16)-t2,t∈[0,π4],t≠tj,-Δu′(tj)=2-u(tj)3,j=1,2,…,n,u′(0)=0,u(π4)=0.
Compared with (1), f(t,u)=(1+t)(u-u2)-1000e(π2/16)-t2, Ij(u)=2-u(t)3. Obviously (H2), (H3), and (H4) are satisfied. Let u1=1, r=1000; then (H1) is satisfied. By Theorem 4, for (Mλk/m,1000]=(4eπ2/16(2k-1)2π2,1000], k=1,2, problem (52) has at least k distinct pairs of solutions.Example 2.
LetT=π/2,g(t)=-t/4, and consider the following problem:
(53)-u′′(t)-t4u′(t)-λu(t)=(1+t2)(2u-u2)-1000e(π2/32)-t2,hhhhhhhhhhhhhhht∈[0,π2],t≠tj,-Δu′(tj)=-u(tj),j=1,2,…,n,u′(0)=0,u(π2)=0.
Compared with (1), f(t,u)=(1+t2)(2u-u2)-1000e(π2/32)-t2, Ij(u)=-u(tj). Obviously (H2), (H3), and (H4) are satisfied. Let u1=2, r=500; then (H1) is satisfied. By Theorem 5, for (Mλk/m,500]=(eπ2/32(2k-1)2π2,500], k=1,2,3,4, problem (53) has at least k distinct pairs of solutions.Example 3.
LetT=π/2,g(t)=-t/2, and consider the following problem:
(54)-u′′(t)-t2u′(t)-λu(t)=-eπ2/16(1+t2)u3(t),t∈[0,π2],t≠tj,-Δu′(tj)=-3u3(tj),j=1,2,…,n,u′(0)=0,u(π2)=0.
Compared with (1), f(t,u)=-eπ2/16(1+t2)u3(t), Ij(u)=-3u3(tj). Obviously (H2) and (H3) are satisfied. Let u2=25,r=625; then (H1) is satisfied. By Theorem 6, for (Mλk/m,625]=(eπ2/16(2k-1)2π2,625], k=1,2,3, problem (54) has at least k distinct pairs of solutions.



---

*Source: 356745-2014-04-23.xml*