# A Note on the Paper “Multipoint BVPs for Second-Order Differential Equations with Impulses” by Xuxin Yang, Zhimin He, and Jianhua Shen

Authors: Yiliang Liu; Jitai Liang
Journal: Mathematical Problems in Engineering
 (2010)

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## Abstract

We give a counter example to the comparison principle for the multipoint BVPs (by
Xuxin Yang, Zhimin He, and Jianhua Shen, in Mathematical Problems in Engineering, Volume 2009,
Article ID 258090, doi:10.1155/2009/258090). Then we suggest and prove a corrected version of
the comparison principle.


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## Body

## 1. Introduction and Preliminaries

Consider the following multipoint BVPs [1]:
(1.1)-u′′(t)=f(t,u(t),u(θ(t))),t≠tk,t∈J=[0,1],Δu′(tk)=Ik(u(tk)),k=1,2,…,m,u(0)-au′(0)=cu(η),u(1)+bu′(1)=du(ξ),
where 0≤θ(t)≤t,θ∈C(J),0=t0<t1<t2<⋯<tk<⋯<tm<tm+1=1, f is continuous everywhere except at {tk}×R2;f(tk+,·,·) and f(tk-,·,·) exist with f(tk-,·,·)=f(tk,·,·);Ik∈C(R,R),  and   Δu′(tk)=u′(tk+)-u′(tk-), a≥0, b≥0, 0≤c≤1, 0≤d≤1, a+c>0,b+d>0,0<η,ξ<1.LetPC(J)={x:J→R;x(t) be continuous everywhere expect for some tk at which x(tk+) and x(tk-) exist and x(tk)=x(tk-),k=1,2,…,m}; PC1(J)={x∈PC(J):x'(t) is continuous everywhere expect for some tk at which x'(tk+) and x′(tk-) exist and x′(tk)=x′(tk-),k=1,2,…,m}. Let J-=J∖{tk,k=1,2,…,m}, and E=PC1(J,R)∩C2(J-,R). a function x∈E is called a solution of BVPS (1.1) if it satisfies (1.1).The purpose of this note is to point out that the results basing on the comparison principle [1, Theorem2.1] are not true. Then we give a new comparison principle.


## 2. Problem and Statement

The authors [1] proved some existence results for multipoint BVPs (1.1) by use of the following comparison principle [1, Theorem 2.1].Assume thatu∈E satisfies
(2.1)-u′′(t)+Mu(t)+Nu(θ(t))≤0,t≠tk,t∈J=[0,1],Δu′(tk)≥Lku(tk),k=1,2,…,m,u(0)-au′(0)≤cu(η),u(1)+bu′(1)≤du(ξ),
where a≥0, b≥0, 0≤c≤1, 0≤d≤1, a+c>0,b+d>0,0<η,ξ<1, Lk≥0, and constants M,N satisfy
(2.2)M>0,N≥0,M+N2+∑k=1mLk≤1.
Then u(t)≤0 for t∈J.However, the comparison principle above is not true.A Counter Example
Let(2.3)u(t)={32t2+20,t∈[0,12],0,∑1252t2+3,t∈(12,1].0,∑12
Then
(2.4)u'(t)={3t,t∈[0,12],0,∑∑125t,t∈(12,1],0,∑12u′′(t)={3,t∈[0,12],0,∑125,t∈(12,1]0,∑12.
And let M=N=1/1000,a=b=c=d=1,m=1,t1=1/2,L1=1/1000,θ(t)=(1/2)t,η=1/3,andξ=1/6.  When t∈[0,1/2), then
(2.5)11000(32t2+20)+11000(32×t24+20)≤3.
When t∈(1/2,1], then
(2.6)11000(52t2+3)+11000(52×t24+3)≤5.
Hence -u′′(t)+Mu(t)+Nu(θ(t))≤0.(2.7)Δu′(12)=u′(12+)-u′(12)=5×12-(3×12)=1,(2.8)11000u(12)=11000(32×14+20)=11000×1638.
Hence          Δu′(t1)≥L1u(t1).(2.9)u(0)-u′(0)=20,u(13)=32×19+20.
Hence        u(0)-au′(0)≤cu(1/3).(2.10)u(1)+u′(1)=52+3+5=212,u(16)=32×136+20.
Hence  u(1)+bu′(1)≤du(1/6).(2.11)M+N2+∑k=1mLk=21000<1.
But we easily show that u(t)>0,forallt∈[0,1], which is a contradiction with (Theorem 2.1) in [1].  In fact,  we can correct Theorem 2.1 in [1] as follows.Theorem 2.1.
Supposeu∈E∩C(J) such that
(2.12)-u′′(t)+Mu(t)+Nu(θ(t))≤0t≠tk,t∈J=[0,1],Δu′(tk)≥Lku(tk),k=1,2,…,m,u(0)-au′(0)≤cu(η),u(1)+bu′(1)≤du(ξ),
where a≥0, b≥0, 0≤c≤1, 0≤d≤1, 0<η,ξ<1,a+c>0,b+d>0,Lk>0, and constants M,N satisfy
(2.13)M>0,N≥0,M+N2+∑k=1mLk≤1.
Then u(t)≤0 for t∈J.Remark 2.2.
In this Theorem, we have to addu∈C(J).Proof.
Suppose to contrary that there exist somet∈J,  such that u(t)>0.
Ifu(1)=maxt∈Ju(t)>0,  we have u′(1)≥0, u(1)≥u(ξ), and
(2.14)du(ξ)≤u(1)≤u(1)+bu′(1)≤du(ξ).
Therefore, d=1 and u(ξ) is maximum value.
Ifu(0)=maxt∈Ju(t)>0, we have u′(0)≤0, u(0)≥u(η), and
(2.15)cu(η)≤u(0)≤u(0)-au′(0)≤cu(η).
Therefore, c=1 and u(η) is maximum value.
So there is aδ∈(0,1) such that
(2.16)u(δ)=maxt∈Ju(t)>0,byΔu=0,thenu′(δ+)≤0,u′(δ-)≥0.
It is obvious to see that  δ∉{tk,k=1,2,…,m} by
(2.17)Δu′(δ)=u′(δ+)-u′(δ)≥Lku(δ)>0
which is a contradiction because of (2.16).
(i)
Suppose thatu(t)≥0 for t∈[0,δ].
Byu(δ)=maxt∈Ju(t)>0,  we get δ∈J-, u′′(δ)≤0. On the other hand, by (2.12), we have
(2.18)0<Mu(δ)+Nu(θ(t))≤u′′(δ)
which is a contradiction.
(ii)
Suppose there existst*∈[0,δ] such that u(t*)=mint∈[0,δ)u(t)<0. By (2.12), we get
(2.19)u′′(t)≥(M+N)u(t*),t∈[0,δ),t≠tk,Δu(tk)=0,Δu′(tk)≥Lku(tk),k=1,2,…,m.Integrating from s(t*≤s≤δ) to δ, we get
(2.20)u′(δ)-u′(s)≥∫sδ(M+N)u(t*)ds+∑s<tk<δLku(tk)=(δ-s)(M+N)u(t*)+∑s<tk<δLku(tk)≥(δ-s)(M+N)u(t*)+∑k=1mLku(t*).
Hence
(2.21)-u′(s)≥(δ-s)(M+N)u(t*)+∑k=1mLku(t*),t*≤s≤δ.
Then integrate from t* to δ to obtain
(2.22)-u(t*)<u(δ)-u(t*)≤∫t*δ(M+N)u(t*)(s-δ)ds-∑k=1mLku(t*)(δ-t*)=(M+N)u(t*)[-(t*-δ)22]-∑k=1mLku(t*)(δ-t*)≤-[M+N2(δ-t*)2+∑k=1mLk]u(t*)≤-(M+N2+∑k=1mLk)u(t*).
By (2.13), we get u(t*)>0 which is a contradiction. We complete the proof.This implies that in order to get the existence results of the multipoint BVPs [1], we have to require an additional continuity hypotheses on the function space.



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*Source: 497029-2010-10-27.xml*