# Variational Approach to Impulsive Differential Equations with Singular Nonlinearities

Authors: Naima Daoudi-Merzagui; Abdelkader Boucherif
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We discuss the existence of periodic solutions for nonautonomous second order differential equations with singular nonlinearities. Simple sufficient conditions that enable us to obtain many distinct periodic solutions are provided. Our approach is based on a variational method.


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## Body

## 1. Introduction

Differential equations with impulsive effects appear naturally in the description of many evolution processes whose states experience sudden changes at certain times, called impulse moments. There is an extensive bibliography about the subject. For recent references, see [1].Variational methods have been successfully employed to investigate regular second order differential equations with impulsive effects; See, for instance, [2–8]. In particular, the paper [8] considers the existence of n distinct pairs of nontrivial solutions. However, very few papers have used variational methods to investigate the case of impulsive second order boundary value problems with singular nonlinearities. In fact, it seems that the work [9] is the first paper along this line. Singular boundary value problems without impulses have attracted the attention of many researchers; see [10] for details and references. This paper is devoted to the study of the existence and multiplicity of periodic solutions for impulsive second order differential equations with singular nonlinearities. More specifically, we consider the following impulsive problem:
(1)u′′(t)+λf(t,u(t))=0,t≠tj,t∈I,-Δu′(t)=Ij(u(tj)),j=1,2,…,p,u(0)=u(T),u′(0)=u′(T),
where I denotes the real interval [0,T], with T>0, 0=t0<t1<⋯<tp<tp+1=T,λ is a positive parameter,
(2)Δu′(t)=u′(tj+)-u′(tj-),
and f:I×ℝ→ℝ presents a singularity with respect to its second argument at u=0.Throughout this paper we will use the following notations.Lp(I) is the classical Lebesgue space of measurable functions u:I→ℝ such that |u(·)|p is integrable, and for u∈Lp(I) we define its norm by
(3)∥u∥Lp=(∫0T|u(t)|pdt)1/p.
Let ∥u∥∞=sup{|u(t)|;t∈I} denote the norm of u∈C(I), the space of real-valued continuous functions. W1;2(I) is the classical Sobolev space of functions u∈L2(I) with their distributional derivatives u′∈L2(I). We set HT1={u∈W1,2(I);u(0)=u(T)} and for u∈HT1 we define its norm by
(4)∥u∥HT1=(∥u′∥L2+∥u∥L2)1/2.
Wirtinger inequality
(5)∥u∥L2≤T2π∥u′∥L2
implies that we can consider on HT1 the norm
(6)∥u∥=∥u′∥L2.HT1 endowed with the norm ∥·∥ is a reflexive Banach space.We introduce the following assumptions on the nonlinearity.(H1)
(i)
f:I×(0;+∞)→ℝ is an L1-Carathéodory function;(ii)
limu→0+f(t,u)=-∞ for almost every t∈I;(iii)
there existu0≥1,a,b>0, and γ∈(0,1) such that f(t,u0)=0, and for almost all t∈I and all u∈(u0,+∞)(7)|f(t,u)|≤a+buγ;(iv)
infu>u0+1∫0TF(t,u)dt>0, with F(t;u)=∫u0uf(t,s)ds;(v)
(∂F/∂t)(t,u) exists and is nonpositive for almost all t∈I and all u∈ℝ;(vi)
limu→+∞∫0TF(t,u)dt=+∞.The jump functionsIj, j=1,2,…,p, satisfy the following:(H2)
(i)
Ij:ℝ→ℝ is odd, continuous, and bounded;(ii)
∫0uIj(s)ds≤0.Definition 1.
A solution of (1) is a function u∈C(I) such that for every j=1,…,p,
(8)uj=u|[tj,tj+1]
is absolutely continuous with its derivatives uj′ and uj′′∈L2(tj,tj+1) and satisfies the differential equation in (1) for t∈I∖{t1,t2,…,tp}; the limits u′(tj-) and u′(tj+), j=1,2,…,p, exist; the impulsive conditions and the boundary conditions in (1) hold.


## 2. The Main Result

In this section we state and prove our main result.Theorem 2.
Suppose that conditions (H1) and (H2) hold. Then for anyn∈ℕ, there exists λn such that for λ>λn problem (1) has infinitely many distinct nontrivial solutions.Proof.
To give the proof of the main result, we first modify problem (1) to another one which is not singular.
Forξ∈(0,min(1,u0/4)), define a function fξ on I×[0,+∞) by
(9)fξ(t,u)={f(t,ξ)sin2π(ξ/u0)sin2πuu0,0≤u≤u0,f(t,u),u>u0.
Finally, we requirefξ(t,-u)=-fξ(t,u) for all u∈ℝ and almost t∈I, to make fξ odd. Then fξ satisfies the following:(j)
fξ is an L1-Carathéodory function and is odd in u;(jj)
there exista>0, b>0, b′>0, and γ∈(0,1) such that |fξ(t,u)|≤a+buγ for all u∈(-∞,-u0)∪(u0,+∞) and almost t∈I and |fξ(t,u)|≤b′u for all u∈(-u0,u0) and almost all t∈I;(jjj)
for almost allt∈I and all u∈ℝ, Fξ(t,u):=∫u0ufξ(t,s)ds is such that
(10)Fξ(t,0)=-∫0u0fξ(t,s)ds=-f(t,ξ)sin(2πξ/u0)∫0u0sin2πu0sds22=(f(t,ξ)sin(2πξ/u0)u02π)∫02π-sinζdζ=0,22222222222222222222222foralmostallt∈I,Fξ(t,-u)=∫u0-u‍fξ(t,s)ds2222222=-∫-uu0‍fξ(t,s)ds2222222=-(∫-u0‍fξ(t,s)ds+∫0u0‍fξ(t,s)ds)2222222=-(∫-u0‍fξ(t,s)ds)=∫0-u‍fξ(t,s)ds2222222=-∫0u‍fξ(t,-ζ)dζ=∫0u‍fξ(t,ζ)dζ2222222=∫0u0‍fξ(t,s)ds+∫u0u‍fξ(t,s)ds=Fξ(t,u),222222222222222222foralmostallt∈I,allu∈ℝ,infu>u0+1∫0T‍Fξ(t,u)dt=infu>u0+1∫0T‍F(t,u)dt>0,222222222Fξ(t,u)<0,for0<u<u0,foralmostallt∈I.
We study the modified problem(11)u′′(t)+λfξ(t,u(t))=0,t≠tj,t∈I,-Δu′(t)=Ij(u(tj)),j=1,2,…,p,u(0)=u(T),u′(0)=u′(T).
Consider the functionalφξ:HT1→ℝ, defined by
(12)φξ(u)=12∥u∥2-λ∫0T‍Fξ(t,u(t))dt-∑j=1p∫0u(tj)‍Ij(s)ds.
Clearlyφξ is an even functional, φξ(0)=0, and is Frechet differentiable, whose Frechet derivative at the point u∈HT1 is the functional φξ′(u)∈(HT1)* given by
(13)φξ′(u)·v=∫0T‍u′(t)·v′(t)dt-λ∫0T‍fξ(t,u(t))·v(t)dt-∑j=1pIj(u(tj))v(tj).
Obviously,φξ′ is continuous. If u∈HT1 is a critical point of the functional φξ, then u is a solution of problem (11).
First, we show thatφξ is bounded from below. Define a subset Ω of HT1 as follows:
(14)Ω={u∈HT1,min(u)>1+u0}∪{u∈HT1,max(u)<-(1+u0)},
noticing that
(15)∂Ω={u∈HT1,u(t)≥1+u0∀t∈I,222∃tu∈Iwithu(tu)=1+u0u∈HT1,u(t)≥1+u0∀t∈I}∪{u∈HT1,u(t)≤-(1+u0)∀t∈I,∃tu∈Iwithu(tu)=-(1+u0)u∈HT1,u(t)≥1+u0∀t∈I}.
Foru∈Ω, we have
(16)φξ(u)=12∥u∥2-λ∫0T‍Fξ(t,u(t))dt-∑j=1p∫0u(tj)‍Ij(s)ds≥12∥u∥2-λ∫0T‍(a|u(t)|+b|u(t)|γ+1)dt≥12∥u∥2-λaT3/2∥u∥-bλT(3+γ)/2∥u∥γ+1>-∞.
If u∉Ω, we use the partition of the interval I=I1∪I2∪I3, where
(17)I1={t∈I;u(t)∈[-u0,u0]},I2={t∈I;u(t)∈[u0,1+u0]∪[-(u0+1),-u0]},I3={t∈I;u(t)∈(-∞,-(1+u0]∪[1+u0,-∞)}.
By (H2) and (jjj), we have(18)φξ(u)=12∥u∥2-λ∫0T‍Fξ(t,u(t))dt-∑j=1p∫0u(tj)‍Ij(s)ds≥12∥u∥2-λ∫I3‍(a|u(t)|+b|u(t)|γ+1)dt-λ∫I2‍Fξ(t,u(t))dt-λ∫I1‍Fξ(t,u(t))dt.
From (jjj), Fξ(t,u(t))<0 on I1, which implies
(19)φξ(u)≥12∥u∥2-λ∫I3‍a|u(t)|+b|u(t)|γ+1dt-λ∫I2‍Fξ(t,u(t))dt.
If t∈I2, then u(t)∈[u0,u0+1]∪[-(u0+1),-u0]. This means that u(t) is bounded uniformly in t∈I2. Since fξ is a Caratheodory function, it follows that Fξ is bounded by some positive constant C. Hence,
(20)φξ(u)≥12∥u∥2-λaT3/2∥u∥-bλ·T(3+γ)/2∥u∥γ+1-λC>-∞.
This shows that φξ is bounded from below.
Remark  3. There exists m>0 such that infu∈∂Ωφξ(u)≥-m, whenever ξ∈(0,min(1,u0/4)).
Our next step is to show thatφξ satisfies the Palais-Smale condition. For this purpose, let (uk)k be a sequence in HT1 such that (φ(uk))k is bounded and limk→+∞φ′(uk)=0. Then, there exists M>0 such that |φξ(uk)|≤M. In view of (16), if uk∈Ω, we have
(21)M≥12∥uk∥2-λa′T3/2∥uk∥-bλ·T(3+γ)/2∥uk∥γ+1,
and if uk∉Ω, we can proceed as above to show that
(22)12∥uk∥2-λaT3/2∥uk∥-bλ·T(3+γ)/2∥uk∥γ+1-λC≤φξ(uk)≤|φξ(uk)|≤M.
This shows that ∥uk∥ is bounded. So that, (uk)k∈ℕ is bounded in HT1. From the reflexivity of HT1, we may extract from (uk)k∈ℕ a weakly convergent subsequence, which we label the same; that is, uk⇀u in HT1. Since the injection of HT1 into C(I), with its natural norm, is continuous, it follows that uk⇀u in C(I) and by Banach-Steinhaus theorem, the subsequence (uk)k∈ℕ is bounded in HT1 and hence, in C(I). Moreover, the subsequence (uk)k is uniformly equicontinuous since, for 0≤s≤t≤T, we have
(23)|uk(t)-uk(s)|≤∫st‍|uk′(τ)|dτ≤(t-s)1/2(∫st‍|uk′(τ)|2dτ)1/2≤(t-s)1/2∥uk∥≤C(t-s)1/2.
By Ascoli-Arzela theorem the subsequence (uk)k∈ℕ is relatively compact in C(I). By the uniqueness of the weak limit in C(I), every uniformly convergent subsequence of (uk)k∈ℕ converges to u. Thus, (uk)k∈ℕ converges uniformly on I to u.
Next, we will verify that(uk)k∈ℕ strongly converges to u in HT1. By (13), we have
(24)(φξ′(uk)-φξ′(u))(uk-u)=∥uk-u∥2-λ∫0T[fξ(t,uk(t))-fξ(t,u(t))]‍(uk(t)-u(t))dt-∑j=1p[Ij(uk(tj))-Ij(u(tj))](uk(tj)-u(tj)).
The uniform convergence of (uk)k∈ℕ to u in C(I) implies
(25)λ∫0T‍[fξ(t,uk(t))-fξ(t,u(t))](uk(t)-u(t))dt⟶0,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhask⟶∞,∑j=1p[Ij(uk(tj))-Ij(u(tj))](uk(tj)-u(tj))⟶0,ffffffffffffffffffffffffffffffffffffffffffffffffffffask⟶∞.
Since φ′(uk)→0 and uk⇀u, as k→∞,
(26)limk→∞(φξ′(uk)-φξ′(u))(uk-u)=0.
In view of (24), (25), and (26), we obtain
(27)limk→∞(uk-u)=0.
Thus, φ satisfies the Palais-Smale condition.
Letv1,v2,…,vn denote the eigenfunctions of -u′′=λu, u(0)=u(T), u′(0)=u′(T), corresponding to the eigenvalues μ1,μ2,…,μn. We normalize v1,v2,…,vn, so that
(28)∥vk∥=1=μk∫0Tvk(t)2dt,1≤k≤n.
For r>u0+1, we set
(29)Kn(r)={∑m=1ncmvm;∑m=1ncm2=r2},
and E=Kn(r)∩Ω. Then, for any u∈E, we have for almost every t∈I, u(t)≥u0+1, or u(t)≤-(u0+1), and
(30)Fξ(t,-u(t))=Fξ(t,u(t))=∫u0u(t)fξ(t,s)ds=∫u0u(t)f(t,s)ds=F(t,u(t)).
It follows from (H1)(iv) that, for any u∈E,
(31)∫0TFξ(t,u(t))dt>0.
Let
(32)αn=infu∈E∫0TFξ(t,u(t))dt,βn=infu∈E∑j=1p∫0u(tj)Ij(s)ds.
It is clear that
(33)αn>0,βn≤0.
Letting
(34)λn=(12r2-βn)·αn-1>0,
we see that when λ>λn we have, for any u∈E,
(35)φξ(u)≤12r2-λαn-βn<12r2-λnαn-βn=0.
On the other hand, there is a constant ρ∈(0,u0) such that Bρ={u∈HT1;∥u∥<ρ} is not contained in Ω. So Fξ(t,u)<0 from (jjj). Hence, using (H2)(ii) we obtain, for u∈∂Bρ∩X, where X=[span{v1,v2,…,vn}]⊥,(36)φξ(u)=12∥u∥2-λ∫0TFξ(t,u(t))dt-∑j=1p∫0uk(tj)Ij(s)ds>12ρ2>0.Remark  4. Sobolev inequality implies that E⊂Kn(r)∖Bu0.
Now, we apply Theorem 9.12 in [11] to conclude that φξ possesses infinitely many distinct pairs of nontrivial critical points. That is, problem (11) has infinitely many distinct pairs of distinct nontrivial solutions.
Finally, we must prove that there existsξ0∈(0,min(1,u0/4)) with the property that, for every ξ∈(ξ0,min(1,u0/4)), any positive solution u of (11) satisfying φξ(u)≥-m is such that minu≥u0 and hence u is a solution of (1). We proceed by contradiction. Assume, on the contrary, that there are sequences (ξn)n∈ℕ and (un)n∈ℕ such that
(37)ξn≤1n;unisapositivesolutionof(11)withξ=ξn,φξn(un)≥-m,minun<u0.
First, we have(38)λ∫0Tfξn(t,un(t))dt=-∑j=0p∫tj+tj+1-un′′(s)ds‍=-∑j=0pun′(tj+1-)-un′(tj+)=(un′(0)-un′(t1-))+⋯+(un′(tp)-un′(T))=∑j=1pIj(un(tj)).
Then, by (H1)(i), there existsc1>0 such that
(39)∥fξn(·,un(·))∥L1≤c1,
which implies that
(40)∥un′∥L∞≤c2,
for some constant c2>0. Since φξn(un)≥-m, it follows that there must exist two constants R1 and R2, with 0<R1<R2, such that maxt∈Iun(t)∈[R1,R2]; otherwise, un would tend uniformly to +∞ and, in this case, φξn(un) would go to -∞ (because of (H1)(vi) and (40)) and this contradicts φξn(un)≥-m. Also, minun<u0 implies that there exists an integer kn>1 such that minun≤u0-(u0/kn). Since un is continuous, there exists τn1 such that un(τn1)=u0-(u0/kn). Let τn2∈I be such that, for n large enough,
(41)un(τn1)=u0-u0kn<R1=un(τn2).
Multiplying the differential equation in (11) by un′ and integrating the resulting equation on [τn1,τn2], or on [τn2,τn1], we get
(42)J≔∫τn1τn2un′′(t)un′(t)dt+λ∫τn1τn2fξn(t,un(t))un′(t)dt=0=∑tj∈[τn1,τn2][∫tj+tj+1-un′′(t)·un′(t)dtsssssssss+λ∫tj+tj+1-fξn(t,un(t))un′(t)dt].
It is clear that
(43)J=J1+∑tj∈[τn1,τn2]12(un′(tj+1-)2-un′(tj+)2),
where
(44)J1=λ∑tj∈[τn1,τn2]∫tj+tj+1-fξn(t,un(t))un′(t)dt=λ∑tj∈[τn1,τn2]∫un(tk)un(tk+1)fξn(t,s)ds.
Now, J=0 and ∥un′∥L∞≤c2 imply that J1 is bounded. Since
(45)fξn(t,un(t))un′(t)=ddt[Fξn(t,un(t)]-D1Fξn(t,un(t)),
it follows that
(46)J1=λ∑tk∈(τn1,τn2)∫tk+tk+1-fξn(t,un(t))un′(t)dt=λ∑tk∈(τn1,τn2)∫un(tk)un(tk+1)fξn(t,s)ds=λFξn(τn2,R1)-λFξn(τn1,u0-u0kn)-λ∫τn1τn2D1Fξn(t,un(t))dt≥λFξn(τn2,R1)-λFξn(τn1,u0-u0kn)=λFξn(τn2,R1)-∫u0u0-(u0/kn)fξn(τn1,s)ds.
Now, we have that(47)∫u0u0-(u0/kn)fξn(τn1,s)ds=f(τn1,ξn)sin(2πξn/u0)∫u0u0-(u0/kn)sin2πu0sds=f(τn1,ξn)sin(2πξn/u0)(u02π)[1-cos(1-1kn)].
It follows from (H1)(ii) that(48)Fξn(τn1,u0-u0kn)=∫u0u0-(u0/kn)fξn(τn1,s)ds⟶-∞asn⟶+∞.
This implies that J1 is not bounded. We arrive at a contradiction. This completes the proof of our main result.


## 3. Example

Consider the boundary value problem(49)u′′(t)+λ(t+1+sint)(u(t)3+3)·lnu(t)3=0,hhhhhhhhhhhhhhhhhhhhhhhhhhhht≠tj,t∈I,-Δu′(t)=-sin(u(tj)),j=1,2,3,u(0)=u(T),u′(0)=u′(T).(i)
f:[0,2π]×(0;+∞)→ℝ, given by f(t,u)=(t+1+sint)(u3+3)·lnu3, satisfies (H1) with T=2π, u0=1;a=0;b=8(π+1);γ=2/3.(ii)
Foru>1, lnu3≤u3⇒(t+1+sint)(u3+3)·lnu3≤8(π+1)u2/3.(iii)
Ij(s)=-sin(s) satisfies (H2).We considerr≥e in the definition of Kn(r). We have
(50)∫1u‍(3+s3)·lns3ds=∫1ulnsds+3∫1u1/3x3·lnxdx=ulnu-u+3(u3)416(4lnu3-1)+1916.Then(51)infu>2∫02πF(t,u)·dt>0,limu→+∞∫02πF(t,u)dt=+∞,αn=infu>2∫02πFξ(t,u(t))dt=infu>2∫02π((t+1+sint)·∫1u(t)(3+s3)·lns3ds)dt≥infu>2∫02π(ulnu-u+3(u3)416(4lnu3-1)+1916)×(t)dt≥2πinfu>2[(ulnu-u+1916)+3(u3)416(4lnu3-1)]≥π.Also,(52)αn=infu>2∫02πFξ(t,u(t))·dt=infu>2∫02π((t+1+sint)·∫1u‍(3+s3)·lns3ds)dt≤(2π+1)infu>2∫0T‍[(u3)416(4lnu3-1)(ulnu-u+1916)ddddddddddddddd+3(u3)416(4lnu3-1)]dt≤(2π+1)·∫0T‍34u2(t)dt≤(2π+1)34r2∑k=1n1μk≤32r2∑k=1n1μk,
where ∑k=1n(1/μk)=∑k=1n(2π/2πk)2<∑k=1n(1/k2)<n.It is clear that(53)βn=infu>2-∑j=13∫0u(tj)sinsds=infu>2∑j=13(cos(u(tj))-1)≤0,βn=infu>2-∑j=13∫0u(tj)sinsds≥infu>2-∑j=13∫0u(tj)sds≥infu>2-∑j=13|u(tj)|22ds>-32r2.
Then
(54)λn=(12r2-βn)αn-1<2πr2,λn=(12r2-βn)·αn-1>(12r2)231r2n=13n.
Applying our main result, we see that when λ>λn, for any n∈ℕ*, problem (49) has infinitely many distinct nontrivial solutions.



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*Source: 464393-2013-12-17.xml*