# Quasilinearization for the Boundary Value Problem of Second-Order Singular Differential System

Authors: Peiguang Wang; Tiantian Kong
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

We study the boundary value problems of second-order singular differential equations. At first, we reduce the BVPs to initial value problems of first-order singular integrodifferential equations and then we employ the quasilinearization method in studying the IVPs and obtain two monotone iterative sequences, which converge uniformly and quadratically to the unique solution of the IVPs. Finally, we get the similar result for the given BVPs.


---

## Body

## 1. Introduction

It is well known that quasilinearization method is a powerful tool for proving the existence of approximate solutions of nonlinear systems and the converge quadratically to the unique solution of the given problems (see [1]). Recently, Abd-Ellateef Kamar et al. investigated the first-order singular systems of differential equations with initial value problem [2]. In [3], Wang and Liu developed monotone iterative technique combined with the method of upper and lower solutions for studying the second-order singular systems with boundary value problems (BVPs).In this paper, we extend quasilinearization method to study the second-order singular systems with the boundary conditions:(1)AX¨=f(t,X,X˙),X˙(0)=X0,X(b)=X1,
where A∈Rn×n is a singular matrix, f∈C[I×Rn×Rn,Rn], I=[0,b], and X0 and X1 are two constant vectors. By using the existence result [4] for linear singular systems and the comparison result [5], we investigate two monotone iterative sequences which converge uniformly and quadratically to the solution of the problem.


## 2. Preliminaries

Consider the following initial value problem:(2)-AU˙=f(t,SU,-U),U(0)=-X0,
where A is a singular n×n matrix, f∈C[I×Rn×Rn,Rn], I=[0,b], SU=X1+∫tb‍U(t)ds is an increasing operator, and X0 is a constant vector.In order to obtain two monotone sequences, we introduce an existence result for the corresponding linear singular systems and a comparison result.The existence of the solution of the linear initial value problem of the form(3)-AU˙+M(t)U=g(t),U(0)=-X0
is well known and is given by the following lemma.Lemma 1 (see [4]).
Assume that the nonhomogeneous linear system (3) exists and if(i)
there exists aλ∈R such that (-λA+M(t))-1 exists,(ii)
X0 is the solution of  (I-A^A^D)(-X0-W^(0))=0, where
(4)A^=(-A+M(t))-1(-A),M^=(-A+M(t))-1M(t),g^(t)=(-A+M(t))-1g(t),W^(t)=(I-A^DA^)M^Dg^(t),and A^D,M^D are the Drazin inverses of A^,M^(t), respectively, then the solution is given by
(5)U(t)=e-A^DM^tA^DA^(-X0)+e-A^DM^t∫0t‍eA^DM^sA^Dg^(s)ds+(I-A^DA^)M^Dg^(t).
Now one gives the following assumptions for convenience.(H2.1)
LetA and M(t) be matrices such that (-λA+M(t))-1 exists and is nonnegative for some λ∈R1. Suppose further that T,T-1 exist and are nonnegative, such that
(6)T-1A^T=(C000),T-1M^T=(I1-λC00I2),
where A^=(-λA+M(t))-1A, M^=(-λA+M(t))-1M(t), and C is a diagonal square matrix with C-1>0 and C-1(I1-λC)>0.(H2.2)
There existα0(t),β0(t)∈C1[I,Rn] with α0(t)≤β0(t) on I, such that
(7)-Aα˙0≤f(t,Sα0,-α0),α0(0)≤-X0,-Aβ˙0≥f(t,Sβ0,-β0),β0(0)≥-X0.(H2.3)
All second-order derivatives off(t,X,U) exist and are bounded, f(t,X,U) is convex in X for each (t,U), f(t,X,U) is convex in U for each (t,X), fx(t,X,U) is nonincreasing in U for each (t,X), and fy(t,X,U) is nonincreasing in X for each (t,U).(H2.4)
MoreoverM(t)=-[fx(t,Sη,-η)b-fy(t,Sη,-η)].
To obtain the results, one needs the following comparison theorem.Lemma 2.
Let-AP˙+M(t)P≤0 such that A and M(t) satisfy assumptions (H2.1) and (H2.4). Then P(0)≤0 implies P(t)≤0 on I.
The proof is similar to [5] and one omits it.


## 3. Main Results

Firstly, we develop the following result which is important for the final result.Theorem 3.
Suppose that assumptions(H2.1)–(H2.4) hold and, for X0∈Rn,(i)
(I-A^A^D)(-X0-W^(0))=0, where W^(t)=(I-A^DA^)M^Dg^(t).
Then there exist two monotone sequences{αn} and {βn}, which converge uniformly on I to the unique solution of problem (2) and the convergence rate is quadratic.Proof.
From(H2.3) we find that fxx(t,X,U)≥0, and
(8)f(t,X1,U)≥f(t,X2,U)+fx(t,X2,U)(X1-X2),
for all X1,X2,U∈Rn and t∈I. The convexity of f(t,X,U) in U implies that
(9)f(t,X,U1)≥f(t,X,U2)+fy(t,X,U2)(U1-U2),
forU1,U2∈Rn, t∈I, and X∈Rn.
Now consider the following linear problems. Fork=0,1,2,3…,
(10)-Aα˙k+1=f(t,Sαk,-αk)+fx(t,Sαk,-αk)(Sαk+1-Sαk)+fy(t,Sαk,-αk)(-αk+1+αk),αk+1(0)=-X0,(11)-Aβ˙k+1=f(t,Sβk,-βk)+fx(t,Sαk,-αk)(Sβk+1-Sβk)+fy(t,Sαk,-αk)(-βk+1+βk),βk+1(0)=-X0.
For (10), set
(12)g(t,Sx,-x)≡f(t,Sα0,-α0)+fx(t,Sα0,-α0)(Sx-Sα0)+fy(t,Sα0,-α0)(-x+α0).
Then, we obtain that -Aα˙0≤f(t,Sα0,-α0)=g(t,Sα0,-α0). Furthermore, we can get
(13)-Aβ˙0≥f(t,Sβ0,-β0)≥f(t,Sα0,-α0)+fx(t,Sα0,-α0)(Sβ0-Sα0)+fy(t,Sα0,-α0)(-β0+α0)=g(t,Sβ0,-β0).
Hence α0 and β0 are lower and upper solutions of (12), respectively. Thus (12) has a solution α1 on I, and we have α0≤α1≤β0. Similarly, set
(14)g*(t,Sx,-x)≡f(t,Sβ0,-β0)+fx(t,Sα0,-α0)(Sx-Sβ0)+fy(t,Sα0,-α0)(-x+β0).
Then we obtain that (14) has a solution β1 on I, and we have α0≤β1≤β0.
Now we claim that(15)α0≤α1≤β1≤β0onI.
Let p=α1-β1. We get that
(16)-Ap˙=-Aα˙1-(-Aβ˙1)=f(t,Sα0,-α0)+fx(t,Sα0,-α0)(Sα1-Sα0)+fy(t,Sα0,-α0)(-α1+α0)-f(t,Sβ0,-β0)-fx(t,Sα0,-α0)(Sβ1-Sβ0)-fy(t,Sα0,-α0)(-β1+β0)=f(t,Sα0,-α0)-f(t,Sβ0,-α0)+f(t,Sβ0,-α0)-f(t,Sβ0,-β0)+fx(t,Sα0,-α0)(Sα1-Sα0)+fy(t,Sα0,-α0)(-α1+α0)-fx(t,Sα0,-α0)(Sβ1-Sβ0)-fy(t,Sα0,-α0)(-β1+β0)≤fx(t,Sα0,-α0)(Sα0-Sβ0)+fy(t,Sα0,-α0)(-α0+β0)+fx(t,Sα0,-α0)(Sα1-Sα0)+fy(t,Sα0,-α0)(-α1+α0)-fx(t,Sβ0,-β0)(Sβ1-Sβ0)-fy(t,Sβ0,-β0)(-β1+β0)≤fx(t,Sα0,-α0)×(Sα0-Sβ0+Sα1-Sα0-Sβ1+Sβ0)+fy(t,Sα0,-α0)×(-α0+β0-α1+α0+β1-β0)=fx(t,Sα0,-α0)(Sα1-Sβ1)-fy(t,Sα0,-α0)(α1-β1)≤-M(t)(α1-β1).
Noticing that p(0)=α1(0)-β1(0)≤0, we get α1(t)≤β1(t), on I.
Now, assume that, forn=0,1,2,…,k, (10) and (11) admit solutions αn and βn, respectively, such that
(17)α0≤α1≤⋯≤αk≤βk≤⋯≤β1≤β0,onI.
Then setting n=k in (10) and (11), we observe that the assumptions in Theorem 3 are satisfied. Thus, there exist solutions αk+1(t) and βk+1(t) for (10) and (11), respectively, and we now will show that the following relation
(18)αk(t)≤αk+1(t)≤βk+1(t)≤βk(t),t∈I,
holds. Firstly, we can easily know that αk(t)≤αk+1(t)≤βk(t) and αk(t)≤βk+1(t)≤βk(t).
To prove thatαk+1≤βk+1, consider p=αk+1-βk+1. Then
(19)-Ap˙=-Aα˙k+1-(-Aβ˙k+1)=f(t,Sαk,-αk)+fx(t,Sαk,-αk)(Sαk+1-Sαk)+fy(t,Sαk,-αk)(-αk+1+αk)-f(t,Sβk,-βk)-fx(t,Sβk,-βk)(Sβk+1-Sβk)-fy(t,Sβk,-βk)(-βk+1+βk)=f(t,Sαk,-αk)-f(t,Sβk,-αk)+f(t,Sβk,-αk)-f(t,Sβk,-βk)+fx(t,Sαk,-αk)(Sαk+1-Sαk)+fy(t,Sαk,-αk)(-αk+1+αk)-fx(t,Sβk,-βk)(Sβk+1-Sβk)-fy(t,Sβk,-βk)(-βk+1+βk)≤fx(t,Sαk,-αk)(Sαk-Sβk)+fy(t,Sβk,-αk)(-αk+βk)+fx(t,Sαk,-αk)(Sαk+1-Sαk)+fy(t,Sαk,-αk)(-αk+1+αk)-fx(t,Sβk,-βk)(Sβk+1-Sβk)-fy(t,Sβk,-βk)(-βk+1+βk)≤fx(t,Sαk,-αk)×(Sαk-Sβk+Sαk+1-Sαk-Sβk+1+Sβk)+fy(t,Sαk,-αk)×(-αk+βk-αk+1+αk-βk+1+βk)=fx(t,Sαk,-αk)(Sαk+1-Sβk+1)+fy(t,Sαk,-αk)(-αk+1+βk+1)=-M(t)p.
We know that p(0)=αk+1(0)-βk+1(0)≤0. Hence, from Lemma 2, we deduce that αk+1(t)≤βk+1(t), on I. Thus, we have monotone sequences {αn},{βn} such that
(20)α0≤α1≤⋯≤αn≤βn≤⋯≤β1≤β0,onI.
Now, employing Ascoli-Arzela's theorem we conclude that the sequences converge uniformly and monotonically to the unique solution U(t) of (2) on I.
To show that the convergence rate is quadratic, we begin withpn+1=x-αn+1. Then
(21)-Ap˙n+1=-Ax˙-(-Aα˙n+1)=f(t,Sx,-x)-f(t,Sαn,-αn)-fx(t,Sαn,-αn)(Sαn+1-Sαn)-fy(t,Sαn,-αn)(-αn+1+αn)=f(t,Sx,-x)-f(t,Sαn,-αn)-fx(t,Sαn,-αn)(Sαn+1-Sx+Sx-Sαn)-fy(t,Sαn,-αn)(-αn+1+x-x+αn)=f(t,Sx,-x)-f(t,Sαn,-αn)+fx(t,Sαn,-αn)Spn+1-fx(t,Sαn,-αn)Spn-fy(t,Sαn,-αn)pn+1+fy(t,Sαn,-αn)pn=fx(t,Sαn,-αn)Spn+1-fy(t,Sαn,-αn)pn+1+f(t,Sx,-x)-f(t,Sαn,-αn)-fx(t,Sαn,-αn)Spn+fy(t,Sαn,-αn)pn≤-M(t)p+f(t,Sx,-x)-f(t,Sαn,-αn)-fx(t,Sαn,-αn)Spn+fy(t,Sαn,-αn)pn=-M(t)p+f(t,Sx,-x)-f(t,Sαn,-x)+f(t,Sαn,-x)-f(t,Sαn,-αn)-fx(t,Sαn,-αn)Spn+fy(t,Sαn,-αn)pn=-M(t)p+∫01‍fx(t,σSx+(1-σ)Sαn,-x)(Sx-Sαn)dσ+∫01‍fy(t,Sαn,σ(-x)+(1-σ)(-αn))×(-x+αn)dσ-fx(t,Sαn,-αn)Spn+fy(t,Sαn,-αn)pn=-M(t)p+∫01‍[fx(t,σSx+(1-σ)Sαn,-x)-M(t)1p[+∫01‍-fx(t,Sαn,-αn)]Spndσ-∫01‍[fy(t,Sαn,σ(-x)+(1-σ)(-αn))-fy(t,Sαn,-αn)]pndσ=-M(t)p+A1+B1,
where
(22)A1=∫01‍[fx(t,σSx+(1-σ)Sαn,-x)-fx(t,Sαn,-αn)]Spndσ,B1=-∫01‍[fy(t,Sαn,σ(-x)+(1-σ)(-αn))-fy(t,Sαn,-αn)]pndσ.
Set ξ1=σSx+(1-σ)Sαn. Then
(23)A1=∫01‍[fx(t,ξ1,-x)-fx(t,Sαn,-αn)]Spndσ=∫01‍[fx(t,ξ1,-x)-fx(t,Sαn,-x)+fx(t,Sαn,-x)-fx(t,Sαn,-αn)]Spndσ=∬01[fxx(t,τξ1+(1-τ)Sαn,-x)(ξ1-Sαn)+fxy(t,Sαn,τ(-x)+(1-τ)(-αn))×(-x+αn)]Spndτdσ=∬01[-fxy(t,Sαn,ξ3)(x-αn)fxx(t,ξ2,-x)σ(Sx-Sαn)-fxy(t,Sαn,ξ3)(x-αn)]Spndτdσ=12N1Spn2+N2pnSpn≤(12N1b2+N2b)pn2,
where ξ2=τξ1+(1-τ)Sαn, ξ3=τ(-x)+(1-τ)(-αn), fxx(t,U,X)≤N1, and -fxy(t,U,X)≤N2.
Next, setη1=σ(-x)+(1-σ)(-αn). Then
(24)B1=-∫01‍[fy(t,Sαn,η1)-fy(t,Sαn,-αn)]pndσ=-∬01[fyy(t,Sαn,τη1+(1-τ)(-αn))×(η1+αn)fyy(t,Sαn,τη1+(1-τ)(-αn))]pndσdτ=∬01fyy(t,Sαn,η2)σpn2dσdτ≤12N3pn2,
where fyy(t,U,X)≤N3 and η2=τη1+(1-τ)(-αn).
Furthermore, we have that(25)-Ap˙n+1≤-M(t)pn+1+N4pn2,
where N4=(1/2)N1b2+N2b+(1/2)N3.
Using Lemma2, we show that pn+1(t)≤U(t) on I, where U(t) is the solution of
(26)-AU˙+M(t)U=N4pn2,U(0)=0.
Thus, using Lemma 1, the solution of the previously mentioned equation is given as
(27)U(t)=e-A^DM^t∫0t‍eA^DM^sA^D(-λA+M(s))-1N4pn2(s)ds+(I-A^DA^)M^D(-λA+M(t))-1N4pn2(t).
After taking suitable estimates, we obtain
(28)pn+1≤K1pn2,
where K1=e-A^DM^t∫0t‍eA^DM^sA^D(-λA+M(s))-1N4ds+(I-A^DA^)M^D(-λA+M(t))-1N4.
Setqn+1=βn+1-x. Then we can get
(29)-Aq˙n+1=-Aβ˙n+1-(-Ax˙)=f(t,Sβn,-βn)+fx(t,Sαn,-αn)(Sβn+1-Sβn)+fy(t,Sαn,-αn)(-βn+1+βn)-f(t,Sx,-x)=fx(t,Sαn,-αn)(Sβn+1-Sx+Sx-Sβn)+fy(t,Sαn,-αn)(-βn+1+x-x+βn)+f(t,Sβn,-βn)-f(t,Sx,-x)=fx(t,Sαn,-αn)(Sβn+1-Sx)-fy(t,Sαn,-αn)(βn+1-x)+f(t,Sβn,-βn)-f(t,Sx,-x)-fx(t,Sαn,-αn)(Sβn-Sx)+fy(t,Sαn,-αn)(βn-x)≤-M(t)qn+1+f(t,Sβn,-βn)-f(t,Sx,-βn)+f(t,Sx,-βn)-f(t,Sx,-x)-fx(t,Sαn,-αn)(Sβn-Sx)+fy(t,Sαn,-αn)(βn-x)=-M(t)qn+1+∫01‍fx(t,σSβn+(1-σ)Sx,-βn)(Sβn-Sx)dσ+∫01‍fy(t,Sx,σ(-βn)+(1-σ)(-x))×(-βn+x)dσ-fx(t,Sαn,-αn)(Sβn-Sx)+fy(t,Sαn,-αn)(βn-x)=-M(t)qn+1+∫01‍[fx(t,σSβn+(1-σ)Sx,-βn)-M(t)qn+11[+∫01‍-fx(t,Sαn,-αn)]Sqndσ-∫01‍[fy(t,Sx,σ(-βn)+(1-σ)(-x))-fy(t,Sαn,-αn)]qndσ=-M(t)qn+1+A2+B2,
in which
(30)A2=∫01‍[fx(t,σSβn+(1-σ)Sx,-βn)-fx(t,Sαn,-αn)]Sqndσ,B2=-∫01‍[fy(t,Sx,σ(-βn)+(1-σ)(-x))-fy(t,Sαn,-αn)]qndσ.
Set θ1=σSβn+(1-σ)Sx. Then we get that
(31)A2=∫01‍[fx(t,θ1,-βn)-fx(t,Sαn,-αn)]Sqndσ=∫01‍[fx(t,θ1,-βn)-fx(t,Sαn,-βn)+fx(t,Sαn,-β)-fx(t,Sαn,-αn)]Sqndσ=∬01[fxx(t,τθ1+(1-τ)Sαn,-βn)(θ1-Sαn)+fxy(t,Sαn,τ(-βn)+(1-τ)(-αn))×(-βn+αn)]Sqndσdτ=∬01[-fxy(t,Sαn,θ3)(qn+pn)fxx(t,θ2,-βn)(σqn+pn)-fxy(t,Sαn,θ3)(qn+pn)]Sqndσdτ≤N1σqnSqn+N1pnSqn+N2qnSqn+N2pnSqn≤12N1qn2b+bN112(pn2+qn2)+N2bqn2+N2b12(pn2+qn2)=(N1+32N2)bqn2+12b(N1+N2)pn2,
where θ2=τθ1+(1-τ)Sαn and θ3=τ(-βn)+(1-τ)(-αn).
Similarly, setϑ1=σ(-βn)+(1-σ)(-x). Then
(32)B2=-∫01‍[fy(t,Sx,ϑ1)-fy(t,Sαn,-αn)]qndσ=-∫01‍[fy(t,Sx,ϑ1)-fy(t,Sαn,ϑ1)+fy(t,Sαn,ϑ1)-fy(t,Sαn,-αn)]qndσ=-∬01[fyx(t,τSx+(1-τ)Sαn,ϑ1)(Sx-Sαn)+fyy(t,Sαn,τϑ1+(1-τ)(-αn))×(ϑ1+αn)fyx]qndσdτ=-∬01[fyx(t,ϑ2,ϑ1)Spn+fyy(t,Sαn,ϑ3)×(σ(-βn+x)-(x-αn))fyx]qndσdτ=∬01[-fyx(t,ϑ2,ϑ1)Spn+fyy(t,Sαn,ϑ3)(σqn+pn)fyx]qndσdτ≤N5bqnpn+12N3qn2+N3pnqn≤(12bN5+N3)qn2+12(bN5+N3)pn2,
where ϑ2=τSx+(1-τ)Sαn,ϑ3=τϑ1+(1-τ)(-αn), and -fyx(t,X,U)≤N5.
Then we get that(33)-Aq˙n+1≤-M(t)qn+1+N6qn2+N7pn2,
where N6=(N1+(3/2)N2)b+((1/2)bN5+N3) and N7=(1/2)b(N1+N2)+(1/2)(bN5+N3). Thus we have that
(34)qn+1(t)≤U(t)=e-A^DM^t∫0t‍eA^DM^sA^D(-λA+M(s))-1≤U(t)=e-A^DM^t∫0t‍×(N6qn(s)2+N7pn(s)2)ds+(I-A^DA^)M^D(-λA+M(t))-1N6qn(t)2+N7pn(t)2.
Furthermore, we obtain after taking suitable estimates
(35)pn+1≤K2qn2+K3pn2,
where K2=e-A^DM^t∫0t‍eA^DM^sA^D(-λA+M(s))-1N6ds+(I-A^DA^)M^D(-λA+M(t))-1N6 and K3=e-A^DM^t∫0t‍eA^DM^sA^D(-λA+M(s))-1N7ds+(I-A^DA^)M^D(-λA+M(t))-1N7.
Hence we proved that the convergence rate is quadratic.Next, we consider singular differential systems BVPs and prove the following main result.Theorem 4.
Let assumptions(H2.1)–(H2.4) hold. Suppose further the following.(H3.1)
There existV0(t),W0(t)∈C2[I,Rn] with V0(t)≤W0(t) and V˙0(t)≥W˙0(t) on I and
(36)AV¨0≤f(t,V0,V˙0),V˙0(0)≥X0,V0(b)≤X1,AW¨0≥f(t,W0,W˙0),W˙0(0)≤X0,W0(b)≥X1.(H3.2)
ForX0∈Rn, (I-A^A^D)(-X0-W^(0))=0, where W^(t)=(I-A^DA^)M^Dg^(t).Then there exist monotone sequences {Vn} and {Wn} which converge uniformly on I to the unique solution of (1) and the convergence rate is quadratic.Proof.
Using the transformationU=-X˙, we have that
(37)SU(t)=X(t)=X1+∫tb‍U(s)ds,t∈I.
BVPs (1) can be transformed into IVPs of singular system:
(38)-AU˙=f(t,SU,-U),U(0)=-X0.
Let α0(0)=-V˙0(0) and β0(0)=-W˙0(0). By using assumption (H3.1), we get that
(39)α0(t)≤-V˙0(t)≤-W˙0(t)=β0(t)
andα0(0)=-V˙0(0)≤-X0≤-W˙0(0)=β0(0).
Noticing thatα0(0)=-V˙0(0), we have that
(40)V0(t)=V0(b)+∫tb‍α0(s)ds,t∈I,[Sα0](t)=X1+∫tb‍α0(s)ds,t∈I.
Then V0(t)≤[Sα0](t) from V0(b)≤X1, and
(41)-Aα˙0=AV¨0≤f(t,V0,V˙0)≤f(t,Sα0,V˙0)=f(t,Sα0,-α0).
A similar argument shows that(42)-Aβ˙0=AW¨0≥f(t,W0,W˙0)≥f(t,Sβ0,W˙0)=f(t,Sβ0,-β0).
By Theorem3, there exist monotone sequences {αn},{βn} such that
(43)αn⟶U⟵βn
and the convergence rate is quadratic. Again set Vn=Sαn and Wn=Sβn. Then
(44)V0≤V1≤V2≤⋯≤Vn≤Wn≤⋯≤W2≤W1≤W0,Vn⟶U⟵Wn,
and also the convergence rate is quadratic. Noticing that
(45)x(t)-Vn+1(t)=X1+∫tb‍U(s)ds-(X1+∫0b‍αn+1(s)ds)=∫0b‍(U(s)-αn+1(s))ds≤∫0b‍[K1(U(s)-αn(s))2]ds=K1(x(t)-Vn(t))2,
we can obtain the similar result that Wn converges quadratically to the solution of (1). The proof is complete.



---

*Source: 308413-2013-12-17.xml*