Volume 2011, Article ID 929061,17pages doi:10.1155/2011/929061
Research Article
Approximation of Solutions for Second-Order
m
-Point Nonlocal Boundary Value Problems via
the Method of Generalized Quasilinearization
Ahmed Alsaedi
Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
Correspondence should be addressed to Ahmed Alsaedi,[email protected]
Received 11 May 2010; Revised 29 July 2010; Accepted 2 October 2010
Academic Editor: Gennaro Infante
Copyrightq2011 Ahmed Alsaedi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We discuss the existence and uniqueness of the solutions of a second-orderm-point nonlocal boundary value problem by applying a generalized quasilinearization technique. A monotone sequence of solutions converging uniformly and quadratically to a unique solution of the problem is presented.
1. Introduction
The monotone iterative technique coupled with the method of upper and lower solutions
1–7manifests itself as an effective and flexible mechanism that offers theoretical as well as constructive existence results in a closed set, generated by the lower and upper solutions. In general, the convergence of the sequence of approximate solutions given by the monotone iterative technique is at most linear 8, 9. To obtain a sequence of approximate solutions converging quadratically, we use the method of quasilinearization10. This method has been developed for a variety of problems11–20. In view of its diverse applications, this approach is quite an elegant and easier for application algorithms.
In this paper, we develop the method of generalized quasilinearization to obtain a sequence of approximate solutions converging monotonically and quadratically to a unique solution of the following second-orderm−point nonlocal boundary value problem
−xt ft, xt, xt, t∈0,1, 1.1
px0−qx0 m−2
i1
τix
ηi
, px1 qx1 m−2
i1
σix
ηi
, ηi∈0,1, 1.2
wheref:0,1×R×R → Ris continuous andτi, σi i1,2, . . . , m−2are nonnegative real constants such thatmi−12τi<1,
m−2
i1 σi<1,andp, q >0 withp >1.
Here we remark that26studies1.1with the boundary conditions of the form
δx0−γx0 0, x1 m−2
i1
αix
ηi
, ηi∈0,1. 1.3
A perturbed integral equation equivalent to the problem1.1and1.3considered in26is
xt
1
0
kt, sfs, xs, xsds
m−2
i1
αix
ηi
t2, 1.4
where
kt, s 1 δγ
⎧ ⎨ ⎩
γδt1−s, 0≤t≤s,
δγs1−t, s≤t≤1.
1.5
It can readily be verified that the solution given by1.4does not satisfy1.1. On the other hand, by Green’s function method, a unique solution of the problem1.1and1.3is
xt
1
0
kt, sfs, xs, xsds
m−2
i1
αix
ηi
γδt
δγ, 1.6
2. Preliminaries
Forx∈C10,1,we definex
1xx,wherexmax{|xt|:t∈0,1}.It can easily
be verified that the homogeneous problem associated with1.1-1.2 has only the trivial solution. Therefore, by Green’s function method, the solution of1.1-1.2can be written as
xt
1
0
Gt, sfs, xs, xsds
m−2
i1
τix
ηi
−t
2qp
qp p2qp
m−2
i1
σix
ηi
t
2qp q p2qp
,
2.1
whereGt, sis the Green’s function and is given by
Gt, s 1 pp2q
⎧ ⎨ ⎩
qptqp1−s, 0≤t≤s,
qpsqp1−t, s≤t≤1.
2.2
Note thatGt, s>0 on0,1×0,1.
We say thatα∈C20,1is a lower solution of the boundary value problem1.1and
1.2if
−αt≤ft, αt, αt, t∈0,1,
pα0−qα0≤ m−2
i1
τiα
ηi
, pα1 qα1≤ m−2
i1
σiα
ηi
,
2.3
andβ∈C20,1is an upper solution of1.1and1.2if
−βt≥ft, βt, βt, t∈0,1,
pβ0−qβ0≥ m−2
i1
τiβ
ηi
, pβ1 qβ1≥ m−2
i1
σiβ
ηi
.
2.4
Definition 2.1. A continuous functionh:0,∞ → 0,∞is called a Nagumo function if
∞
λ
sds
hs ∞, 2.5
forλ ≥ 0. We say thatf ∈C0,1 × R × Rsatisfies a Nagumo condition on0,1relative to α, βif for every t ∈ 0,1 andx ∈ mint∈0,1αt,maxt∈0,1βt, there exists a Nagumo functionhsuch that|ft, x, x| ≤h|x|.
Theorem 2.2. Letf : 0,1×R2 → Rbe a continuous function satisfying the Nagumo condition
onE{t, x, y∈0,1×R2 :α≤x≤β}whereα, β:0,1 → Rare continuous functions such
thatαt ≤ βtfor allt ∈ 0,1.Then there exists a constantM >0(depending only onα, β,the Nagumo functionh) such that every solutionxof 1.1-1.2withαt ≤ xt ≤ βt,t ∈ 0,1 satisfies|x| ≤M.
If α, β ∈ C20,1 are assumed to be lower and upper solutions of 1.1-1.2,
respectively, in the statement of Theorem 2.2, then there exists a solution,xtof 1.1and
1.2such thatαt≤xt≤βt,t∈0,1.
Theorem 2.3. Assume thatα, β∈C20,1are, respectively, lower and upper solutions of1.1-1.2.
Ifft, x, y∈C0,1×R×Ris decreasing inxfor eacht, y∈0,1×R,thenα≤βon0,1.
Proof. Let us define ut αt −βt so that u ∈ C20,1 and satisfies the boundary
conditions
pu0−qu0≤ m−2
i1
τiu
ηi
, pu1 qu1≤ m−2
i1
σiu
ηi
. 2.6
For the sake of contradiction, letuhave a positive maximum at somet0∈0,1. Ift0 ∈0,1,
then ut0 0 and ut0 ≤ 0.On the other hand, in view of the decreasing property of
ft, x, yinx,we have
ut0 αt0−βt0≥ −f
t0, αt0, αt0
ft0, βt0, βt0
>0, 2.7
which is a contradiction. If we suppose that uhas a positive maximum at t0 0, then it
follows from the first of boundary conditions2.6that
pu0−qu0≤ m−2
i1
τiu
ηi
≤u0, 2.8
which implies thatp−1u0≤ qu0.Now asp > 1,q > 0,u0 > 0,u0≤ 0,therefore we obtain a contradiction. We have a similar contradiction att0 1.Thus, we conclude that
αt≤βt,t∈0,1.
3. Main Results
Theorem 3.1. Assume that
A1the functions α, β ∈ C20,1are, respectively, lower and upper solutions of 1.1-1.2
such thatα≤βon0,1;
A2the function f ∈ C20,1 × R × R satisfies a Nagumo condition relative to α, β
and fx ≤ 0 on 0,1 × mint∈0,1αt,maxt∈0,1βt × −M, M, where M is a positive constant depending on α, β, and the Nagumo function h. Further, there exists a function φ ∈ C20,1×R2 such that Ψf φ ≥ 0 with Ψφ ≥ 0 on 0,1×
mint∈0,1αt,maxt∈0,1βt×−M, M,where
Ψ x−y2 ∂
2
∂x2 2
x−yx−y ∂
2
∂x∂x
x−y2 ∂
2
Then, there exists a monotone sequence {αn} of approximate solutions converging uniformly to a unique solution of the problems1.1-1.2.
Proof. For y ∈ R, we define ωy max{−M,min{y, M}} and consider the following modifiedm-point BVP
−xt ft, xt, ωxt, t∈0,1,
px0−qx0 m−2
i1
τix
ηi
, px1 qx1 m−2
i1
σix
ηi
.
3.2
We note thatα, βare, respectively, lower and upper solutions of3.2and for everyt, x ∈
0,1×mint∈0,1αt,maxt∈0,1βt,we have
f≤hω
xhx, 3.3
whereh· hω·.As
∞
0
sds
hs
M
0
sds hs
∞
M
sds
hM ∞, 3.4
soh is a Nagumo function. Furthermore, there exists a constantNdepending on α, β, and Nagumo functionhsuch that
M
0
sds
hs ≥
N
0
sds hs >
maxβt:t∈0,1−min{αt:t∈0,1}, 3.5
whereM >max{N,α,β}. Thus, any solutionxof3.2withαt≤xt≤βt,t∈0,1 satisfies|x| ≤Mon0,1and hence it is a solution of1.1-1.2.
Let us define a functionF:0,1×R2 → Rby
Ft, x, xft, x, xφt, x, x−ωx. 3.6
In view of the assumptionA2,it follows thatF ∈C20,1×R2and satisfiesΨF≥0 on
0,1×mint∈0,1αt,maxt∈0,1βt×−M, M.Therefore, by Taylor’s theorem, we obtain
ft, x, ωx≥ft, y, ωyFx
t, y, ωyx−y
Fxt, y, ωyωx−ωy−φt, x,0−φt, y,0
≥ft, y, ωyFx
t, y, ωy−φx
t, β,0x−y
Fxt, y, ωyωx−ωy.
We set
Ht, x, x;y, yft, y, ωyFx
t, y, ωy−φx
t, β,0x−y
Fxt, y, ωyωx−ωy,
3.8
and observe that
ft, x, ωx≥Ht, x, x;y, y,
ft, x, ωxHt, x, x;x, x. 3.9
By the mean value theorem, we can findα≤c1 ≤yandα ≤c2 ≤ yc1, c2depend ony, y,
resp., such that
ft, y, ωy−ft, αt, αtfxt, c1, c2
y−αtfxt, c1, c2ωy−αt. 3.10
Letting
H1
t, x, x;y, yft, αt, αtfxt, c1, c2x−αt fxt, c1, c2ωx−αt, 3.11
we note that
ft, y, ωyH1
t, y, y;y, y,
ft, αt, αtH1
t, αt, αt;y, y.
3.12
Let us defineHas
H
⎧ ⎨ ⎩
Ht, x, x;y, y, forx≥y,
H1
t, x, x;y, y, forx≤y. 3.13
ClearlyHis continuous and bounded on0,1×mint∈0,1αt,maxt∈0,1βt×Rand satisfies a Nagumo condition relative toα, β. For everyαt ≤y ≤ βtandy ∈R, we consider the m-point BVP
−xHt, x, x;y, y, t∈0,1,
px0−qx0 m−2
i1
τix
ηi
, px1 qx1 m−2
i1
σix
ηi
.
Using3.9,3.12and3.13, we have
Ht, αt, αt;y, yH1
t, αt, αt;y, yft, αt, αt≥ −αt,
pα0−qα0≤ m−2
i1
τiα
ηi
, pα1 qα1≤ m−2
i1
σiα
ηi
,
Ht, βt, βt;y, yHt, βt, βt;y, y≤ft, βt, βt≤ −βt,
pβ0−qβ0≥ m−2
i1
τiβ
ηi
, pβ1 qβ1≥ m−2
i1
σiβ
ηi
.
3.15
Thus,α, βare lower and upper solutions of3.14, respectively. SinceHsatisfies a Nagumo condition, there exists a constantM1 > max{α,β}depending onα, βand a Nagumo
functionsuch that any solutionxof3.14withαt≤xt≤βtsatisfies|x|< M1on0,1.
Now, we chooseα0αand consider the problem
−xHt, x, x;α0, α0
, t∈0,1,
px0−qx0 m−2
i1
τix
ηi
, px1 qx1 m−2
i1
σix
ηi
.
3.16
UsingA1,3.9,3.12and3.13, we obtain
Ht, α0, α0;α0, α0
ft, α0, α0
≥ −α0t,
pα00−qα00≤
m−2
i1
τiα0
ηi
, pα01 qα01≤
m−2
i1
σiα0
ηi
,
Ht, βt, βt;α0, α0
Ht, βt, βt;α0, α0
≤ft, βt, βt≤ −βt,
pβ0−qβ0≥ m−2
i1
τiβ
ηi
, pβ1 qβ1≥ m−2
i1
σiβ
ηi
,
3.17
which imply thatα0 andβare lower and upper solutions of3.16. Hence by Theorems2.2
and2.3, there exists a unique solutionα1of3.16such that
Note that the uniqueness of the solution follows by Theorem 2.3. Using 3.9 and 3.13 together with the fact thatα1is solution of3.16, we find thatα1is a lower solution of3.2,
that is,
−α1 Ht, α1, α1;α0, α0
≤ft, α1, ω
α1, t∈0,1,
pα10−qα10
m−2
i1
τiα1
ηi
, pα11 qα11
m−2
i1
σiα1
ηi
.
3.19
In a similar manner, it can be shown by usingA1,3.12,3.13, and3.19thatα1andβare
lower and upper solutions of the followingm-point BVP
−xHt, x, x;α1, α1
, t∈0,1,
px0−qx0 m−2
i1
τix
ηi
, px1 qx1 m−2
i1
σix
ηi
.
3.20
Again, by Theorems2.2and2.3, there exists a unique solutionα2of3.20such that
α1t≤α2t≤βt, α2t≤M1, t∈0,1. 3.21
Continuing this process successively, we obtain a bounded monotone sequence {αn} of solutions satisfying
α1t≤α2t≤α3t≤ · · · ≤αnt≤βt, t∈0,1, 3.22
whereαnis a solution of the problem
−xHt, x, x;αn−1, αn−1
, t∈0,1,
px0−qx0 m−2
i1
τix
ηi
, px1 qx1 m−2
i1
σix
ηi
,
3.23
and is given by
xt
1
0
Gt, sHs, αn, αn;αn−1, αn−1
ds
m−2
i1
τix
ηi
−t
2qp
qp p2qp
m−2
i1
σix
ηi
t
2qp q p2qp
.
3.24
Since H is bounded on 0,1 × mint∈0,1αt, maxt∈0,1βt × R × mint∈0,1αt, maxt∈0,1βt × R, therefore it follows that the sequences {αjn }j 0,1 are uniformly bounded and equicontinuous on 0,1. Hence, by Ascoli-Arzela theorem, there exist the subsequences and a function x ∈ C10,1 such thatαj
n → ∞.Taking the limitn → ∞,we find thatHt, αn, αn;αn−1, αn−1 → ft, x, ωxwhich
consequently yields
xt
1
0
Gt, sfs, xs, ωxsds
m−2
i1
τix
ηi
−t
2qp
qp p2qp
m−2
i1
σix
ηi
t
2qp q p2qp
.
3.25
This proves thatxis a solution of3.2.
Theorem 3.2. Assume thatA1andA2hold. Further, one assumes that
A3the functionF ∈ C20,1×R×Rsatisfiesy∂/∂xFt, x, y my2 ≤ 0for|y| ≥
M,wherem max{|Fxxt, x, y|:t, x, y∈0,1×mint∈0,1αt,maxt∈0,1βt× −M, M},andFfφ.
Then, the convergence of the sequence {αn} of approximate solutions (obtained in Theorem 3.1) is quadratic.
Proof. Let us seten1t xt−αn1t≥0 so thaten1satisfies the boundary conditions
pen10−qen10
m−2
i1
τien1
ηi
, pen11 qen11
m−2
i1
σien1
ηi
. 3.26
In view of the assumption A3,for every t, x ∈ 0,1×mint∈0,1αt,maxt∈0,1βt,it follows that
Fxt, x, M 2mM≤0, Fxt, x,−M−2mM≥0. 3.27
Now, by Taylor’s theorem, we have
−en1t
Ft, x, x−φt, x,0
−ft, αn, ω
αn
Fx
t, α, ωαn
αn1−αn
−φx
t, β,0αn1−αn Fx
t, αn, ω
αn
ωαn1−ωαn
Fx
t, αn, ω
αn
x−αn1 Fxt, αn, ωαnx−ωαn1
1
2
x−αn2Fxxt, z1, z2 2x−αn
x−ωαn
Fxxt, z1, z2
x−ωαn 2
Fxxt, z1, z2
−φt, x,0−φt, αn,0−φx
t, β,0αn1−αn
≤Fx
t, αn, ω
αn
x−ωαn1
M2
2
|x−αn|x−ω
αn
2
where αn ≤ z1 ≤ x, ωαn ≤ z2 ≤ x, αn ≤ ξ ≤ β, M2 max{|Fxx|,|Fxx |,|Fxx|} on 0,1 ×mint∈0,1αt, maxt∈0,1βt ×−M, M and ρ1 ρmax{φxxt, x,0 : t, x,0 ∈ 0,1×mint∈0,1αt, maxt∈0,1βt}withρ >1 satisfyingβ−αn ≤ρx−αnon0,1.Also, in view of3.13, we have
−en1t f
t, x, x−Ht, αn1, αn1;αn, αn
≥ft, x, x−ft, αn1, ω
αn1
fxt, c3, c4en1fxt, c3, c4x−ωαn1
≥ −γen1fxt, c3, c4x−ωαn1,
3.29
where αn1 ≤ c3 ≤ x, ωαn1 ≤ c4 ≤ x and γ max{|fxt, x, y| : t, x, y ∈ 0,1× mint∈0,1αt, maxt∈0,1βt×−M, M}.
Now we show thatωαn1t αn1t.By the mean value theorem, for everyy1 ∈
−M, Mandωαn1t≤c5≤y1,we obtain
Fx
t, αnt, y1
Fx
t, αnt, ω
αn1tFxxt, αnt, c5
y1−ω
αn1t. 3.30
Letαn1> Mfor somet∈0,1.Thenωαn1t Mand3.30becomes
Fx
t, αnt, y1
Fxt, αnt, M Fxxt, αnt, c5
y1−M
≤Fxt, αnt, M−my1−M.
3.31
In particular, takingy1−Mand using3.27, we have
Fxt, αnt,−M≤Fxt, αnt, M 2mM≤0, 3.32
which contradicts thatFxt, αnt,−M ≥ 2mM > 0.Similarly, lettingαn1 < −Mfor some
t∈0,1,we get a contradiction. Thus, it follows that|αn1t| ≤Mfor everyt∈0,1, which implies thatωαn1t αn1tand consequently,3.28and3.29take the form
−en 1t≤Fxt, αn, ωαnten1t M3en21, 3.33
whereM3ρ1 M2/2and
−en 1t≥ −γen1t fxt, c3, c4en1t. 3.34
Now, by a comparison principle, we can obtainen1t≤rton0,1, wherertis a solution
of the problem
−rt Fxt, αn, ωαntrt M3en21,
pr0−qr0 m−2
i1
τien1
ηi
, pr1 qr1 m−2
i1
σien1
ηi
.
SinceFx is continuous and bounded on0,1×mint∈0,1αt,maxt∈0,1βt×R, there exist
ζ2, ζ1>0independent ofnsuch that−ζ1≤Fx ≤ζ2on0,1×mint∈0,1αt,maxt∈0,1βt× −M, M.Sinceζ2−Fxt, αn, ωαn≥0 on0,1,so we can rewrite3.35as
rt ζ2rt
ζ2−Fxt, αn, ωαnrt−M3en21
pr0−qr0 m−2
i1
τien1
ηi
, pr1 qr1 m−2
i1
σien1
ηi
,
3.36
whose solution is given by
rt
1
0
Gζ2t, s
ζ2−Fxt, αn, ωαnrs−M3en21
ds
m−2
i1
τien1
ηi
−t
2qp
qp p2qp
m−2
i1
σien1
ηi
t
2qp q p2qp
3.37
where
Gζ2t, s
−1
ζ2
pqζ2
/p−e−ζ2
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
1−pζ2q
p e
−ζ21−s
pζ2q
p −e
−ζ2t
, 0≤t≤s,
e−ζ2t−s−pζ2q
p e
−ζ21−s
pζ2q
p −e
−ζ2s
, s≤t≤1,
3.38
Introducing the integrating factorμt e0tFxs,αns,ωαnsdssuch thate−ζ1t< μ≤eζ2t,3.34
takes the form
rtμt−M3en21μt. 3.39
Integrating3.39from 0 totand usingr0≥−1/qmi−12τien1ηi,we obtain
rtμt≥ −1 q
m−2
i1
τien1
ηi
−M3en21
t
0
μsds, 3.40
which can alternatively be written as
rt≥ −1 qeζ1t
m−2
i1
τien1
ηi
− M3
ζ2eζ1t
en21
eζ2−1
≥ −1
q
m−2
i1
τien1 −M3
ζ2 en 2 1
eζ2−1
−ρ1en1 −ρ2en21,
whereρ1 1/qmi−12τi,ρ2 M3/ζ2eζ2−1. Using the fact thatGζ2t, s≤0 together with
3.41yields
Gζ2t, sζ2−Fxrt≤Gζ2t, sζ2−Fx
ρ1en1ρ2en21
≤Gζ2t, sζ2ζ1
ρ1en1ρ2en21
,
3.42
which, on substituting in3.37, yields
en1≤rt≤
1
0
Gζ
2t, sζ2ζ1
ρ1en1ρ2en21
M3en21
ds
m−2
i1
τien1
ηi
−t
2qp
qp p2qp
m−2
i1
σien1
ηi
t
2qp q p2qp
≤
1
0
Gζ
2t, sζ2ζ1
ρ1en1ds
1
0
Gζ
2t, sρ2ζ2ζ1 M3
en21
ds
m−2
i1
τi m−2
i1
σi
pq p2qp
en1
ηi ≤ B
m−2
i1
τi m−2
i1
σi
pq p2qp
en1Aen21,
3.43
where
Aρ2ζ2ζ1 M3
max
1
0
Gζ
2t, sds, B ζ2ζ1ρ1max
1
0
Gζ
2t, sds. 3.44
Taking the maximum over0,1and then solving3.43foren1,we obtain
en1 ≤ A
1−B−mi−12τi m−2
i1 σi
pq/p2qpen
2
1. 3.45
Also, it follows from3.33that
en 1μt≥ −M3en21μt≥ −M3eζ2ten21, t∈0,1. 3.46
Integrating3.46from 0 totand usingvn 10≥−1/qmi−12τien1ηi from the boundary conditionpen10−qen10 im−12τien1ηi,we obtain
en 1tμt≥ −1 q
m−2
i1
τien1
ηi
−M3
eζ2t−1
ζ2 en 2
which, in view of the facte−ζ1t< μ≤eζ2tand3.45, yields
en1t≥eζ1t
⎡ ⎢ ⎣ −1 q
m−2
i1
τi ⎛
⎜
⎝ A
1−B−mi−12τi m−2
i1 σi
pq/p2qp
⎞ ⎟ ⎠
− M3
eζ2t−1
ζ2
$
en21≥ −δ1en21,
3.48
where
δ1max
⎧ ⎪ ⎨ ⎪ ⎩e
ζ1t
⎡ ⎢ ⎣ 1 q
m−2
i1
τi ⎛
⎜
⎝ A
1−B−mi−12τi m−2
i1 σi
pq/p2qp
⎞ ⎟ ⎠
M3
eζ2t−1
ζ2
$
, t∈0,1 %
.
3.49
Asen1∈C10,1, there existst∈0,1such that
en 1ten11−en10≤en11
≤ 1
p
m−2
i1
σien1
ηi −q pe n11≤
1
p
m−2
i1
σien1
qδ p en
2 1 ≤ ⎡ ⎢ ⎣ A
p1−B−mi−12σimi−12τipq
/p2qp
m−2
i1 σi qδ p ⎤ ⎥ ⎦en21.
3.50
Integrating3.46fromttott≤tand using3.50, we have
en1t≤eζ1t
⎡ ⎢
⎣ eζ2tA m−2
i1 σi
p1−B−mi−12σi m−2
i1 τipq
/p2qp
qδ
p M3
eζ2t−eζ2t
ζ2
⎤ ⎥ ⎦en21.
3.51
Using3.45in3.34, we obtain
en 1tμ1t
≤ γAμ1t
1−B−mi−12σi m−2
i1 τipq
/p2qpen
2
whereμ1t e
t
0fxs,c3,c4ds. Sincef
x is bounded on0,1×mint∈0,1αt, maxt∈0,1βt×
−M, M,we can chooseζ3, ζ4 > 0 such that −ζ3 ≤ fxt,c3,c4 ≤ ζ4 on0,1×mint∈0,1αt,
maxt∈0,1βt×−M, Mande−ζ3t< μ
1t≤eζ4tso that3.52takes the form
en 1tμ1t
≤ γAeζ4t
1−B−mi−12σimi−12τipq
/p2qpen
2
1. 3.53
Integrating3.53fromttott≥t, and using3.51, we find that
en1t≤ 1 μ1t
⎡ ⎢
⎣en1tμ1
t γA
eζ4t−eζ4t
L2
1−B−mi−12σimi−12τipq
/p2qpen
2 1
⎤ ⎥ ⎦
≤eζ3t
⎡ ⎢
⎣ Aeζ4t m−2
i1 σi
p1−B−mi−12σi m−2
i1 τi
pq/p2qp qδeζ4t
p
γA
eζ4t−eζ4t
ζ4
1−B−mi−12σimi−12τipq
/p2qp
⎤ ⎥ ⎦en21.
3.54
Letting
δ2max
⎧ ⎪ ⎨ ⎪ ⎩max ⎧ ⎪ ⎨ ⎪ ⎩e
ζ1t
⎡ ⎢
⎣ e
ζ2tAm−2
i1 σi
p1−B−mi−12σi m−2
i1 τi
pq/p2qp
qδ
p M3
eζ2t−eζ2t
ζ2
⎤ ⎥
⎦, t∈0, t
⎫ ⎪ ⎬ ⎪ ⎭, max ⎧ ⎪ ⎨ ⎪ ⎩e
ζ3t
⎡ ⎢
⎣ Aeζ4t m−2
i1 σi
p1−B−mi−12σi m−2
i1 τi
pq/p2qp qδeζ4t
p
γA
eζ4t−eζ4t
ζ4
1−B−mi−12σimi−12τipq
/p2qp
⎤ ⎥
⎦, t∈ t,1%%,
it follows from3.51and3.54that
en1t≤δ2en21. 3.56
Hence, from3.48and3.56, it follows that
,,e
n1,,≤δ3en21, 3.57
whereδ3max{δ1, δ2}.From3.45and3.57with
Q A
1−B−mi−12σimi−12τipq
/p2qpδ3, 3.58
we obtain
en11en1,,vn1,,≤Qen21. 3.59
This proves the quadratic convergence inC1norm.
Example 3.3. Consider the boundary value problem
−x− 1
720te x− 1
35x−1−
tx2
161 x2, t∈0,1,
5 4x0−
11 20x
0 1 7x
3 4
1
9x
4 5
, 5
4x1 11 20x
1 1 3x
3 4
.
3.60
Letαt 0 andβt 1tbe, respectively, lower and upper solutions of3.60. Clearlyαt
andβtare not the solutions of 3.60andαt < βt, t ∈ 0,1.Also, the assumptions of Theorem 3.1are satisfied. Thus, the conclusion ofTheorem 3.1applies to the problem3.60.
Acknowledgment
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