Firstly, we develop the following result which is important for the final result.
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),
for U1,U2∈Rn, t∈I, and X∈Rn.
Now consider the following linear problems. For k=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≤β0 on I.
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, for n=0,1,2,…,k, (10) and (11) admit solutions αn and βn, respectively, such that
(17)α0≤α1≤⋯≤αk≤βk≤⋯≤β1≤β0, on I.
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 that αk+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, on I.
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 with pn+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)]Spndτ dσ=∬01[-fxy(t,Sαn,ξ3)(x-αn)fxx(t,ξ2,-x)σ(Sx-Sαn) -fxy(t,Sαn,ξ3)(x-αn)]Spndτ dσ=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, set η1=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))]pndσ dτ=∬01fyy(t,Sαn,η2)σpn2dσ dτ≤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 Lemma 2, 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.
Set qn+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)]Sqndσ dτ=∬01[-fxy(t,Sαn,θ3)(qn+pn)fxx(t,θ2,-βn)(σqn+pn) -fxy(t,Sαn,θ3)(qn+pn)]Sqndσ dτ≤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, set ϑ1=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]qndσ dτ=-∬01[fyx(t,ϑ2,ϑ1)Spn+fyy(t,Sαn,ϑ3) ×(σ(-βn+x)-(x-αn))fyx]qndσ dτ=∬01[-fyx(t,ϑ2,ϑ1)Spn +fyy(t,Sαn,ϑ3)(σqn+pn)fyx]qndσ dτ≤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.