# Multiple Solutions for a Second-Order Impulsive Sturm-Liouville Equation

Authors: Jingli Xie; Zhiguo Luo
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the existence of solutions to a boundary value problem of a second-order impulsive Sturm-Liouville equation with a control parameterλ. By employing some
existing critical point theorems, we find the range of the control parameter in which the
boundary value problem admits at least three solutions. It is also shown that, under certain
conditions, there exists an interval of the control parameter in which the boundary value
problem possesses infinitely many solutions. Some examples are given to demonstrate the
main results in this paper.


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## Body

## 1. Introduction

The aim of this paper is to investigate the existence of multiple solutions of the following Neumann boundary value problem with impulsive Sturm-Liouville type equation:(1)-(p(t)u(t)′)′+q(t)u(t)=λf(t,u(t)),t≠tj,t∈[0,1],Δ(p(tj)u′(tj))=Ij(u(tj)),j=1,2,…,m,u′(0)=u′(1)=0,wherep∈C1([0,1], q∈C([0,1]), f∈C([0,1]×R,R), Ij∈C(R,R), 0=t0<t1<t2<⋯<tm<tm+1=1, and λ is a positive real parameter. Δu′(tj)=u′(tj+)-u′(tj-), where u′(tj+) and u′(tj-) denote the right and left limits, respectively, of u′(t) at t=tj, j=1,2,…,m.In the last few years, the existence of multiple solutions to Neumann problems has been widely investigated [1–5]. But little research has focused on the existence of multiple solutions for impulsive Sturm-Liouville equations whose right-hand side nonlinear term is depending on a parameter λ. Processes subject to sudden changes in their states are modeled by the impulsive differential equations and have been investigated in various fields of science and technology. In the motion of spacecraft, one has to consider instantaneous impulses at a position with jump discontinuities in velocity, but no change in the position [6, 7]. This motivates us to consider problem (1).In the literature, tools employed to establish the existence of solutions of impulsive differential equations include fixed point theorems, the upper and lower solutions method, the degree theory, critical point theory, and variational methods. See, for example, [8–20]. In this paper, our focus is on the existence of solutions of problem (1) with λ being the parameter. The problem is first transformed into the existence of critical points of some variational structure. Then with the help of critical point theory, results on the existence of at least three solutions and infinitely many solutions are established.The rest of this paper is organized as follows. In Section2 we present some preliminary results. Our main results and their proofs are given in Section 3.


## 2. Preliminaries

Throughout we assume thatp(t) and q(t) satisfy
(2)p0=mint∈[0,1]p(t)>0,q0=mint∈[0,1]q(t)>0.TakeX=W1,2([0,1]) and define
(3)∥u∥X=(∫01‍(p(t)|u′(t)|2+q(t)|u(t)|2)dt)1/2,u∈X.For the norm inC([0,1]), we put
(4)∥u∥∞=maxt∈[0,1]|u(t)|.We have the following relation.Lemma 1.
LetM1=min{p0,q0}. Then
(5)∥u∥∞≤2M1∥u∥X,∀u∈X.Proof.
For anyu∈X, it follows from the mean value theorem that
(6)∫01‍u(s)ds=u(τ)
for someτ∈[0,1]. Hence, for t∈[0,1], using Hölder inequality and (2), we have
(7)|u(t)|=|u(τ)+∫τt‍u′(s)ds|≤|u(τ)|+∫01‍|u′(s)|ds|u(t)|≤∫01‍|u(s)|ds+∫01‍|u′(s)|ds|u(t)|≤(1q0)1/2(∫01‍q(t)|u(t)|2dt)1/2|u(t)|≤+(1p0)1/2(∫01‍p(t)|u′(t)|2dt)1/2|u(t)|≤2M1∥u∥X.Define a functionalφλ as
(8)φλ(u)=Φ(u)-λΨ(u),u∈X,where(9)Φ(u)=12∥u∥X2+∑j=1m‍∫0u(tj)‍Ij(s)ds,(10)Ψ(u)=∫01‍F(t,u)dt,with(11)F(t,u)=∫0u(t)‍f(t,s)ds.Note thatφλ is Fréchet differentiable at any u∈X, and for any v∈X, we have
(12)φ′λ(u)(v)=limh→0φλ(u+hv)-φλ(u)hφλ′(u)(v)=∫01‍(p(t)u′(t)v′(t)+q(t)u(t)v(t))dtφλ′(u)(v)=+∑j=1m‍Ij(u(tj))v(tj)φλ′(u)(v)=-λ∫01‍f(t,u(t))v(t)dt.Lemma 2.
Ifu∈X is a weak solution of problem (1), then u is a classical solution of problem (1).Proof.
For any functionv∈X, we get
(13)0=∫01‍(p(t)u′(t)v′(t)+q(t)u(t)v(t))dt0=+∑j=1m‍Ij(u(tj))v(tj).
By the regularity theory, the weak solutionu is a classical solution of problem (1).Next we show that a critical point of the functionalφλ is a solution of problem (1).Lemma 3.
Ifu∈X is a critical point of φλ, then u is a classical solution of problem (1).Proof.
Ifu∈X is a critical point of φλ, by (12) and Lemma 2, we have u is a classical solution of problem (1)Forr∈ℝ, we define
(14)ρ(r)=infv∈Φ-1((-∞,r))supu∈Φ-1((-∞,r))Ψ(u)-Ψ(v)r-Φ(v),(15)γ1:=liminfr→+∞ρ(r),δ:=liminfr→(infXΦ)+ρ(r).


## 3. Main Results

### 3.1. Existence of At Least Three Solutions

In this section we derive conditions under which problem (1) admits at least three solutions. For this purpose, we introduce the following assumptions.  (H1)
Assume that there exists a positive constantk1 such that for each u∈X(16)0≤∑j=1m‍∫0u(tj)‍Ij(s)ds≤k1maxj∈{1,2,…,m}|u(tj)|2.(H2)
Assume that there exist positive constantsa, b, and γ∈[0,1), such that
(17)0<f(t,u)≤a+b|u|γ,for(t,u)∈[0,1]×ℝ.Letk0=∫01‍q(t)dt and let k2=(1/2)k0+k1 with k1 given in (16). Clearly, M1≤k0. For constants c1, c, we define
(18)λ1=4M1∫01‍max|u|≤c1F(t,u)dtc12,λ2=23∫01‍F(t,c)dt.k2c2.Theorem 4.
Assume that (H1), (H2) are satisfied. If there exist two positive constantsc1, c satisfying c1<c, and
(19)λ1<λ2,
then, for eachλ∈(1/λ2,1/λ1), problem (1) admits at least three solutions.Proof.
By Lemma3, it suffices to show that the functional φλ defined in (8) has at least three critical points. We prove this by verifying the conditions given in [21, Theorem  3.2]. Note that Φ defined in (9) is a nonnegative Gâteaux differentiable, coercive, and sequentially weakly lower semicontinuous functional and its Gâteaux derivative admits a continuous inverse on X*. Moreover, Ψ defined in (10) is a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. Set
(20)r=M1c124.
Note thatu1(t)=c∈X. It then follows from (H1) that
(21)Φ(u1)=12∥u1∥X2+∑j=1m‍∫0u1(tj)‍Ij(s)dsΦ(u1)=12c2∫01‍q(s)ds+∑j=1m‍∫0c‍Ij(s)dsΦ(u1)≤12k0c2+k1c2Φ(u1)≤k2c2,Φ(u1)≥12∥u1∥X2=12k0c2.
Then, we have(22)2r=M1c122<12k0c2≤Φ(u1).
Foru∈X satisfying Φ(u)<r, by Lemma 1, one has
(23)|u|2≤∥u∥∞2≤2M1∥u∥X2≤4M1Φ(u)<4M1r=c12,t∈[0,1],
which implies that(24)Ψ(u)=∫01‍F(t,u(t))dt≤∫01‍max|u|≤c1F(t,u)dt.
Hence(25)supu∈Φ-1((-∞,r))Ψ(u)≤∫01‍max|u|≤c1F(t,u)dt.
So, one has(26)0≤supu∈Φ-1((-∞,r))Ψ(u)r≤4M1∫01‍max|u|≤c1F(t,u)dtc12,23Ψ(u1)Φ(u1)≥23∫01‍F(t,c)dt.k2c2.
Making use of (19), we obtain
(27)supu∈Φ-1((-∞,r))Ψ(u)r<23Ψ(u1)Φ(u1).
By (H1) and (H2), whenλ∈(1/λ2,1/λ1), we easily obtain that the functional φλ is coercive.
Thus all conditions in [21, Theorem  3.2] are verified, and hence for each λ∈(1/λ2,1/λ1), the functional φλ=Φ-λΨ admits at least three critical points. Consequently, problem (1) admits at least three solutions.Example 5.
Consider the boundary value problem(28)-(etu′(t))′+u(t)=λf(t,u),t≠t1,t∈[0,1],Δ(et1u′(t1))=u(t1),t1=12,u′(0)=u′(1)=0,
where(29)f(t,u)={eu,u≤16,t∈[0,1],u1/2+e16-4,u>16,t∈[0,1].
Here,I1(s)=s, p(t)=et, q(t)=1, and m=1. Note that (H1), (H2) are satisfied. Moreover, we have k0=1, k1=1/2, k2=1, M1=1, and
(30)F(t,u)={eu-1,u≤16,t∈[0,1],23u3/2+(e16-4)u+613-15e16,u>16,t∈[0,1].
Choosec1=1, c=16. Direct calculations give
(31)λ1=4M1∫01‍max|u|≤c1F(t,u)dtc12=4(e-1)<8,λ2=23∫01‍F(t,c)dt.k2c2=1384(e16-1)>100.Therefore it follows from Theorem4 that (28) admits at least three solutions in X provided that λ∈(0.01,0.125).


### 3.2.  Existence of Infinitely Many Solutions

In this section, we derive some conditions under which problem (1) admits infinitely many distinct solutions. To this end, we need the following assumption.  (H3)
Assume that(32)F(t,u)≥0,(t,u)∈[0,1]×ℝ.Set(33)μ1=4M1liminfξ→+∞∫01‍max|u|≤ξF(t,u)dtξ2,μ2=1k2limsupξ→+∞∫01‍F(t,ξ)dtξ2.Theorem 6.
Assume that (H3) is satisfied. If(34)μ1<μ2
holds, then for eachλ∈(1/μ2,1/μ1), problem (1) has an unbounded sequence of solutions in X.Proof.
We apply [22, Theorem  2.1] to show that the functional φλ defined in (8) has an unbounded sequence of critical points.
We first show thatγ1<+∞. Let {ξn} be a sequence of positive numbers such that ξn→+∞ as n→∞ and
(35)limn→+∞∫01‍max|u|≤ξnF(t,u)dtξn2=liminfξ→+∞∫01‍max|u|≤ξF(t,ξ)dtξ2.
For any positive integern, we let rn=M1ξn2/4. For u∈X satisfying Φ(u)<rn, similar to the proof of Theorem 4, one can show that
(36)∥u∥∞2≤4M1Φ(u)<ξn2,t∈[0,1],
which implies that(37)Ψ(u)=∫01‍F(t,u)dt≤∫01‍max|u|≤ξnF(t,u)dt.
Note thatΨ(0)=Φ(0)=0; thus we have
(38)ρ(rn)=infv∈Φ-1((-∞,rn))supu∈Φ-1((-∞,rn))Ψ(u)-Ψ(v)rn-Φ(v)≤supu∈Φ-1((-∞,rn))Ψ(u)-Ψ(0)rn-Φ(0)=supu∈Φ-1((-∞,rn))Ψ(u)rn≤4∫01‍max|u|≤ξnF(t,u)dtM1ξn2,
which, together with (15), gives us
(39)γ1≤4M1liminfξ→+∞∫01‍max|u|≤ξF(t,u)dtξ2=μ1<+∞.
This shows that(1/μ2,1/μ1)⊆(0,1/γ1). For any fixed λ∈(1/μ2,1/μ1), it follows from [22, Theorem  2.1] that either φλ=Φ-λΨ has a global minimum or there is a sequence {un} of critical points (local minima) of φλ such that limn→+∞∥un∥X=+∞.
Next we show that the functionalφλ has no global minimum for λ∈(1/μ2,1/μ1). Since λ>1/μ2=k2/limsupξ→+∞(∫01‍F(t,ξ)dt/ξ2), we can choose a constant M2 such that, for each n∈N={1,2,…},
(40)supξ≥n∫01‍F(t,ξ)dtξ2>M2>k2λ.
Thus, there existsξn≥n such that
(41)∫01‍F(t,ξn)dtξn2>M2.
Defineun(t) as follows:
(42)un(t)=ξn,t∈[0,1].
This yields(43)Φ(un)=12k0ξn2+∑j=1m‍∫0ξn‍Ij(s)dsΦ(un)≤12k0ξn2+k1ξn2Φ(un)=k2ξn2.
Then(44)φλ(un)=Φ(un)-λΨ(un)φλ(un)≤k2ξn2-λ∫01‍F(t,ξn)dtφλ(un)≤ξn2(k2-λM2).
Note thatk2-λM2<0. Thus the functional φλ has no lower bound and hence it has no global minimum and the proof is complete.Let(45)μ3=4M1liminfω→0+∫01‍max|u|≤ωF(t,u)dtω2,μ4=1k2limsupω→0+∫01‍F(t,ω)dtω2.Theorem 7.
Assume that (H3) is satisfied. If(46)μ3<μ4
holds, then, for eachλ∈(1/μ4,1/μ3), problem (1) has a sequence of nonzero solutions in X, which weakly converges to 0.Proof.
The proof is similar to that of Theorem6 by showing that δ<+∞ and 0 is not a local minimum of the functional φλ=Φ-λΨ.Example 8.
Consider the boundary value problem(47)-(etu′(t))′+u(t)=λf(t,u),t≠t1,t∈[0,1],Δ(et1u′(t1))=u(t1),t1=12,u′(0)=u′(1)=0,
wheref(t,u)=8tu(1-sinu)-4tu2cosu.
Here,I1(s)=s, p(t)=et, q(t)=1, and m=1. Hence we have k0=1, k1=1/2, k2=1, M1=1, and
(48)F(t,u)=4t(1-sinu)u2,
so (H3) is satisfied. Moreover, we have(49)limξ→+∞inf∫01‍max|u|≤ξF(t,u)dtξ2=0,limξ→+∞sup∫01‍F(t,ξ)dtξ2=2.
Therefore, condition (34) holds and Theorem 6 applies: for λ∈(0.5,+∞), problem (47) admits an unbounded sequence of solutions in X.Example 9.
Consider the boundary value problem(50)-(etu′(t))′+u(t)=λf(t,u),t≠t1,t∈[0,1],Δ(et1u′(t1))=u(t1),t1=12,u′(0)=u′(1)=0,
where(51)f(t,u(t))={tu(0.35+14sin(ln|u|)+18cos(ln|u|))+18cos(ln|u|)u∈R∖{0},t∈[0,1]0,u=0.
In this example,I1(s)=s, p(t)=et, q(t)=1, and m=1. Hence we have k0=1, k1=1/2, k2=1, M1=1, and
(52)F(t,u(t))={12tu2(0.35+14sin(ln|u|)),+14sin(ln|u|)u∈R∖{0},t∈[0,1],0,u=0.
Hence, one has(53)4M1limw→0+inf∫01‍max|u|≤wF(t,u)dtw2=0.1<1k2limw→0+sup∫01‍F(t,w)dtw2=0.15.
Therefore (46) holds. Owing to Theorem 7, when λ∈(6.67,10) problem (50) admits a sequence of pairwise distinct classical solutions strongly converging at 0 in X.


## 3.1. Existence of At Least Three Solutions

In this section we derive conditions under which problem (1) admits at least three solutions. For this purpose, we introduce the following assumptions.  (H1)
Assume that there exists a positive constantk1 such that for each u∈X(16)0≤∑j=1m‍∫0u(tj)‍Ij(s)ds≤k1maxj∈{1,2,…,m}|u(tj)|2.(H2)
Assume that there exist positive constantsa, b, and γ∈[0,1), such that
(17)0<f(t,u)≤a+b|u|γ,for(t,u)∈[0,1]×ℝ.Letk0=∫01‍q(t)dt and let k2=(1/2)k0+k1 with k1 given in (16). Clearly, M1≤k0. For constants c1, c, we define
(18)λ1=4M1∫01‍max|u|≤c1F(t,u)dtc12,λ2=23∫01‍F(t,c)dt.k2c2.Theorem 4.
Assume that (H1), (H2) are satisfied. If there exist two positive constantsc1, c satisfying c1<c, and
(19)λ1<λ2,
then, for eachλ∈(1/λ2,1/λ1), problem (1) admits at least three solutions.Proof.
By Lemma3, it suffices to show that the functional φλ defined in (8) has at least three critical points. We prove this by verifying the conditions given in [21, Theorem  3.2]. Note that Φ defined in (9) is a nonnegative Gâteaux differentiable, coercive, and sequentially weakly lower semicontinuous functional and its Gâteaux derivative admits a continuous inverse on X*. Moreover, Ψ defined in (10) is a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. Set
(20)r=M1c124.
Note thatu1(t)=c∈X. It then follows from (H1) that
(21)Φ(u1)=12∥u1∥X2+∑j=1m‍∫0u1(tj)‍Ij(s)dsΦ(u1)=12c2∫01‍q(s)ds+∑j=1m‍∫0c‍Ij(s)dsΦ(u1)≤12k0c2+k1c2Φ(u1)≤k2c2,Φ(u1)≥12∥u1∥X2=12k0c2.
Then, we have(22)2r=M1c122<12k0c2≤Φ(u1).
Foru∈X satisfying Φ(u)<r, by Lemma 1, one has
(23)|u|2≤∥u∥∞2≤2M1∥u∥X2≤4M1Φ(u)<4M1r=c12,t∈[0,1],
which implies that(24)Ψ(u)=∫01‍F(t,u(t))dt≤∫01‍max|u|≤c1F(t,u)dt.
Hence(25)supu∈Φ-1((-∞,r))Ψ(u)≤∫01‍max|u|≤c1F(t,u)dt.
So, one has(26)0≤supu∈Φ-1((-∞,r))Ψ(u)r≤4M1∫01‍max|u|≤c1F(t,u)dtc12,23Ψ(u1)Φ(u1)≥23∫01‍F(t,c)dt.k2c2.
Making use of (19), we obtain
(27)supu∈Φ-1((-∞,r))Ψ(u)r<23Ψ(u1)Φ(u1).
By (H1) and (H2), whenλ∈(1/λ2,1/λ1), we easily obtain that the functional φλ is coercive.
Thus all conditions in [21, Theorem  3.2] are verified, and hence for each λ∈(1/λ2,1/λ1), the functional φλ=Φ-λΨ admits at least three critical points. Consequently, problem (1) admits at least three solutions.Example 5.
Consider the boundary value problem(28)-(etu′(t))′+u(t)=λf(t,u),t≠t1,t∈[0,1],Δ(et1u′(t1))=u(t1),t1=12,u′(0)=u′(1)=0,
where(29)f(t,u)={eu,u≤16,t∈[0,1],u1/2+e16-4,u>16,t∈[0,1].
Here,I1(s)=s, p(t)=et, q(t)=1, and m=1. Note that (H1), (H2) are satisfied. Moreover, we have k0=1, k1=1/2, k2=1, M1=1, and
(30)F(t,u)={eu-1,u≤16,t∈[0,1],23u3/2+(e16-4)u+613-15e16,u>16,t∈[0,1].
Choosec1=1, c=16. Direct calculations give
(31)λ1=4M1∫01‍max|u|≤c1F(t,u)dtc12=4(e-1)<8,λ2=23∫01‍F(t,c)dt.k2c2=1384(e16-1)>100.Therefore it follows from Theorem4 that (28) admits at least three solutions in X provided that λ∈(0.01,0.125).


## 3.2.  Existence of Infinitely Many Solutions

In this section, we derive some conditions under which problem (1) admits infinitely many distinct solutions. To this end, we need the following assumption.  (H3)
Assume that(32)F(t,u)≥0,(t,u)∈[0,1]×ℝ.Set(33)μ1=4M1liminfξ→+∞∫01‍max|u|≤ξF(t,u)dtξ2,μ2=1k2limsupξ→+∞∫01‍F(t,ξ)dtξ2.Theorem 6.
Assume that (H3) is satisfied. If(34)μ1<μ2
holds, then for eachλ∈(1/μ2,1/μ1), problem (1) has an unbounded sequence of solutions in X.Proof.
We apply [22, Theorem  2.1] to show that the functional φλ defined in (8) has an unbounded sequence of critical points.
We first show thatγ1<+∞. Let {ξn} be a sequence of positive numbers such that ξn→+∞ as n→∞ and
(35)limn→+∞∫01‍max|u|≤ξnF(t,u)dtξn2=liminfξ→+∞∫01‍max|u|≤ξF(t,ξ)dtξ2.
For any positive integern, we let rn=M1ξn2/4. For u∈X satisfying Φ(u)<rn, similar to the proof of Theorem 4, one can show that
(36)∥u∥∞2≤4M1Φ(u)<ξn2,t∈[0,1],
which implies that(37)Ψ(u)=∫01‍F(t,u)dt≤∫01‍max|u|≤ξnF(t,u)dt.
Note thatΨ(0)=Φ(0)=0; thus we have
(38)ρ(rn)=infv∈Φ-1((-∞,rn))supu∈Φ-1((-∞,rn))Ψ(u)-Ψ(v)rn-Φ(v)≤supu∈Φ-1((-∞,rn))Ψ(u)-Ψ(0)rn-Φ(0)=supu∈Φ-1((-∞,rn))Ψ(u)rn≤4∫01‍max|u|≤ξnF(t,u)dtM1ξn2,
which, together with (15), gives us
(39)γ1≤4M1liminfξ→+∞∫01‍max|u|≤ξF(t,u)dtξ2=μ1<+∞.
This shows that(1/μ2,1/μ1)⊆(0,1/γ1). For any fixed λ∈(1/μ2,1/μ1), it follows from [22, Theorem  2.1] that either φλ=Φ-λΨ has a global minimum or there is a sequence {un} of critical points (local minima) of φλ such that limn→+∞∥un∥X=+∞.
Next we show that the functionalφλ has no global minimum for λ∈(1/μ2,1/μ1). Since λ>1/μ2=k2/limsupξ→+∞(∫01‍F(t,ξ)dt/ξ2), we can choose a constant M2 such that, for each n∈N={1,2,…},
(40)supξ≥n∫01‍F(t,ξ)dtξ2>M2>k2λ.
Thus, there existsξn≥n such that
(41)∫01‍F(t,ξn)dtξn2>M2.
Defineun(t) as follows:
(42)un(t)=ξn,t∈[0,1].
This yields(43)Φ(un)=12k0ξn2+∑j=1m‍∫0ξn‍Ij(s)dsΦ(un)≤12k0ξn2+k1ξn2Φ(un)=k2ξn2.
Then(44)φλ(un)=Φ(un)-λΨ(un)φλ(un)≤k2ξn2-λ∫01‍F(t,ξn)dtφλ(un)≤ξn2(k2-λM2).
Note thatk2-λM2<0. Thus the functional φλ has no lower bound and hence it has no global minimum and the proof is complete.Let(45)μ3=4M1liminfω→0+∫01‍max|u|≤ωF(t,u)dtω2,μ4=1k2limsupω→0+∫01‍F(t,ω)dtω2.Theorem 7.
Assume that (H3) is satisfied. If(46)μ3<μ4
holds, then, for eachλ∈(1/μ4,1/μ3), problem (1) has a sequence of nonzero solutions in X, which weakly converges to 0.Proof.
The proof is similar to that of Theorem6 by showing that δ<+∞ and 0 is not a local minimum of the functional φλ=Φ-λΨ.Example 8.
Consider the boundary value problem(47)-(etu′(t))′+u(t)=λf(t,u),t≠t1,t∈[0,1],Δ(et1u′(t1))=u(t1),t1=12,u′(0)=u′(1)=0,
wheref(t,u)=8tu(1-sinu)-4tu2cosu.
Here,I1(s)=s, p(t)=et, q(t)=1, and m=1. Hence we have k0=1, k1=1/2, k2=1, M1=1, and
(48)F(t,u)=4t(1-sinu)u2,
so (H3) is satisfied. Moreover, we have(49)limξ→+∞inf∫01‍max|u|≤ξF(t,u)dtξ2=0,limξ→+∞sup∫01‍F(t,ξ)dtξ2=2.
Therefore, condition (34) holds and Theorem 6 applies: for λ∈(0.5,+∞), problem (47) admits an unbounded sequence of solutions in X.Example 9.
Consider the boundary value problem(50)-(etu′(t))′+u(t)=λf(t,u),t≠t1,t∈[0,1],Δ(et1u′(t1))=u(t1),t1=12,u′(0)=u′(1)=0,
where(51)f(t,u(t))={tu(0.35+14sin(ln|u|)+18cos(ln|u|))+18cos(ln|u|)u∈R∖{0},t∈[0,1]0,u=0.
In this example,I1(s)=s, p(t)=et, q(t)=1, and m=1. Hence we have k0=1, k1=1/2, k2=1, M1=1, and
(52)F(t,u(t))={12tu2(0.35+14sin(ln|u|)),+14sin(ln|u|)u∈R∖{0},t∈[0,1],0,u=0.
Hence, one has(53)4M1limw→0+inf∫01‍max|u|≤wF(t,u)dtw2=0.1<1k2limw→0+sup∫01‍F(t,w)dtw2=0.15.
Therefore (46) holds. Owing to Theorem 7, when λ∈(6.67,10) problem (50) admits a sequence of pairwise distinct classical solutions strongly converging at 0 in X.



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*Source: 527082-2013-07-10.xml*