Volume 2010, Article ID 459754,19pages doi:10.1155/2010/459754
Research Article
Existence and Numerical Method for Nonlinear
Third-Order Boundary Value Problem in the
Reproducing Kernel Space
Xueqin L ¨u
1, 2and Minggen Cui
11Department of Mathematics, Harbin Institute of Technology, Weihai, Shandong 264209, China 2School of Mathematics and Sciences, Harbin Normal University, Harbin, Heilongjiang 150025, China Correspondence should be addressed to Xueqin L ¨u,[email protected]
Received 14 July 2009; Revised 12 November 2009; Accepted 11 January 2010
Academic Editor: Irena Rach ˚unkova
Copyrightq2010 X. L ¨u and M. Cui. 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 are concerned with general third-order nonlinear boundary value problems. An existence theorem of solution is given under weaker conditions. In the meantime, an iterative algorithm with global convergence is presented. The higher order derivatives of approximate solution is obtained by using this method can approximate the corresponding derivatives of exact solution well.
1. Introduction
The third-order boundary value problems arise in the study of draining and coating flows. In the past few years, third-order boundary value problems with various types of boundary conditions have been studied in many literatures. For details, see1–6and their references therein. In the last decade or so, several papers 7–12have been devoted to the study of third-order differential equations with two point boundary value conditions. In 13, the authors discussed the solvability of a class of particular third-order nonlinear boundary value problems:
u3x−fx, ux, ux, ux0, 0≤x≤1,
u0 0, u0 0, u1 0.
1.1
u3x−fx, ux, u, ux0, 0≤x≤1,
u1 0, u0 0, u1 0. 1.2
In order to derive the existence theorems for solution of1.2, we make the following assumptions H:
H1fx, y, z, w∈0,1×R3is completely continuous function;
H2fx, y, z, w, fxx, y, z, w, fyx, y, z, w, fzx, y, z, w and fwx, y, z, w are
bounded;
H3fx, y, z, w>0 on0,1×R3,
wherefx, y, z, w ∈ W10,1asy yx ∈ W10,1, z zx ∈ W10,1, w wx ∈
W10,1, 0≤x≤1,−∞< y, z, w <∞.
Under weaker conditions, we give an existence theorem. In the meantime, we give an iterative method of obtaining the solution of1.2. The method has following features: firstly, the method is an iterative method in a wide range, that is, the convergence of iterative sequences is independent of the choice of initial values; secondly, the approximate solution obtained by using this method can approximate the higher order derivatives of exact solution well; thirdly, we consider the problem in the reproducing kernel Hilbert space and use sufficiently the good properties of the space. When we choose an appropriate reproducing kernel Hilbert space, we can discuss higher-order boundary value problems in a similar manner.
2. Several Reproducing Kernel Spaces
2.1. The Reproducing Kernel Space
W1
0
,
1
The inner product space W10,1 is defined by W10,1 {ux | ux is absolutely
continuous real value function in0,1, ux ∈ L20,1}. The inner product and norm in
W10,1are given, respectively, by
ux, vxW1u0v0
1
0
uxvxdx, uW1ux, ux1/2. 2.1
In14,15, the author had proved thatW10,1is a complete reproducing kernel space.
That is, there exists a reproducing kernel functionQxy∈W10,1, y∈0,1, for each fixed
x∈0,1and anyuy∈W10,1, such thatuy, QxyW1 ux.The reproducing kernel
Qxycan be denoted by
Qx
y
⎧ ⎨ ⎩
1 y, y≤x,
1 x, y > x.
2.2. The Reproducing Kernel Space
W2
0
,
1
The inner product space W20,1 is defined by W20,1 {ux | u3 are absolutely
continuous real value functions, u4 ∈ L20,1, u1 0, u0 0, u1 0}. The inner
product and norm inW20,1are given, respectively, by
uy, vy W2
3
i0
ui0vi0
1
0
u4v4dy, uW2u, uW2, 2.3
whereu, v∈W20,1.
Theorem 2.1. The spaceW20,1is a reproducing kernel space. That is, for any fixed x ∈ 0,1,
there existsTxy ∈ W20,1, such that for any ux ∈ W20,1,ux uy, TxyW2. The
reproducing kernelTxycan be denoted by
Tx y ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 11975695200
−1 x2504017410 34820x 4350x2−1340x3−835x4
468x5 1260−191520−383040x−30110x2 20200x3−5145x4 38x5y2
−140−892080−1784160x 47340x2−30970x3 7300x4 177x5y3
−35−892080−1784160x 47340x2−30970x3 7300x4 177x5y4
−21−191520−383040x−30110x2 20200x3−5145x4 38x5y5
161−55440−7570x−560x2 410x3−130x4 11x5y6
−17410 34820x 4350x2−1340x3−835x4 468x5y7, y≤x,
− 1
11975695200
−1 y25040
−161x6−55440−7570y−560y2 410y3−130y4 11y5
−1260x2−191520−383040y−30110y2 20200y3−5145y4 38y5 21x5
−191520−383040y−30110y2 20200y3−5145y4 38y5 140x3
−892080−1784160y 47340y2−30970y3 7300y4 177y5 35x4
−892080−1784160y 47340y2−30970y3 7300y4 177y5−5040
17410 34820y 4350y2−1340y3−835y4 468y5 x7
17410 34820y 4350y2−1340y3−835y4 468y5, y > x.
2.4
2.3. The Reproducing Kernel Space
W3
0
,
1
The inner product space W30,1 is defined by W30,1 {ux | u3 are absolutely
continuous real value functions,u4 ∈L20,1, u0 0, u0 0, u1 0}. The definitions
Theorem 2.2. The spaceW30,1is a reproducing kernel space. That is, for any fixed x ∈ 0,1,
there existsRxy ∈ W30,1, such that for any ux ∈ W30,1,ux uy, RxyW3. The
reproducing kernelRxycan be denoted by
Rx
y
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
− y2
4717440
−6552xy4 936y5 420x3360−252y−63y2−6y3 y4
105x4360 60y 15y2−6y3 y4−42x5360 60y 15y2−6y3 y4
7x6360 60y 15y2−6y3 y4
504x2−540 300y 75y2 9y3 5y4, y≤x,
x2
4717440
−936x5 6552x4y−504−540 300x 75x2 9x3 5x4y2
−420360−252x−63x2−6x3 x4y3−105360 60x 15x2−6x3 x4y4
42360 60x 15x2−6x3 x4y5
−7360 60x 15x2−6x3 x4y6, y > x.
2.5
The method of obtaining the reproducing kernel Txy, Rxy and the proofs of Theorem 2.1,Theorem 2.2are given in16.
3. Preliminaries of Constructing Iterative Method
For the convenience of discussion, letL :W2 → W1,Lu u3, then1.2can be converted
into the form as follows:
Lufx, u, u, u, u∈W20,1. 3.1
Now, we fixu ∈W20,1and denotefx,u, u,ubyfu.fu∈W10,1andfu
is a known function. In order to construct the present iterative method, we will consider the following equation:
Lufx,u, u,u, u∈W20,1, f∈W10,1. 3.2
We convert3.1and3.2into the following3.3and3.4, respectively,
Lufu, 3.3
Now, we construct an orthogonal system of functions firstly. Letϕix Qxix, where {xi}
∞
i1 is dense on 0,1. Let ϕ0 fu. The orthogonal
system of functions{ϕix}∞i0can be derived from Gram-Schmidt orthogonalization process of{ϕix}∞i0:
ϕix
i
k0
αikϕkx, 3.5
whereϕ0 ϕ0 fu.In order to emphasize the fact that the first function of orthogonal
system of functions isfu, we denoteϕibyϕifu. Putψix L∗ϕix∈W2, i0,1,2, . . . ,
whereL∗is the conjugate operator ofL. Similarly, we denoteψibyψifu.
Now, we introduce some notations:
Ψnfu Span{ψ1fu, ψ2fu, . . . , ψnfu};
Ψfu Span{ψ1fu, ψ2fu, . . .};
Pn:W2 → Ψnfuis a orthogonal projector;
P :W2 → Ψfuis a orthogonal projector;
Ψ⊥
nfuis the orthogonal complement space ofΨnfu;
Ψ⊥fuis the orthogonal complement space ofΨfu;
gnfudenotes the element inΨn⊥fu, pickgnfu L∗fu−PnL∗fu;
gfudenotes the element inΨ⊥fu, pickgfu L∗fu−P L∗fu.
It is easy to obtain the following lemmas.
Lemma 3.1. Suppose that{xi}∞i1is dense on0,1, then
10{ϕ
ix}∞i0is a complete system ofW1;
20{ϕ
ix}∞i1is not a complete system ofW1;
30{ψix}∞i0is a complete system ofW2;
40{ψ
ix}∞i1is not a complete system ofW2.
Theorem 3.2. uis the solution of 3.4if and only if
u, ψ0
fu W2 fu2W1,
u, ψi
fu W2 0, i1,2, . . . .
Proof .
(Su
ffi
ciency)
Fromu, ψ0fuW2fu 2
W1, it follows that
Lu, ϕ0 W1 u, L∗ϕ0 W2 u, ψ0 W2fu2W1
fu, fu W1fu, ϕ0 W1. 3.7
Fromu, ψifuW20, i1,2, . . .andϕ0, ϕiW10, i /0, we have
Lu, ϕi W1 u, L∗ϕi W2 u, ψi W20
ϕ0, ϕi W1fu, ϕi W1. 3.8
By the completeness of{ϕix}∞i0and3.7,3.8, one obtainsLufu.
(Necessity)
Ifuis the solution of3.4, then
Lu−fu, ϕi W1 0, i0,1,2, . . . . 3.9
Wheni0, we have
u, ψ0 W2
Lu, ϕ0 W1 fu, ϕ0 W1fu2W1. 3.10
Wheni1,2, . . ., we have
u, ψi W2
Lu, ϕi W1ϕ0, ϕi W1 0. 3.11
3.10and3.11yield3.6.
In the same way, we can prove the following corollary.
Corollary 3.3. uis the solution of 3.4if and only if
u g
fufu2W1
ψ0
4. Iterative Formulas
4.1. Iterative Formula I
Pick arbitrarily initial functionuu0in3.4and denote the solution of3.4byun 1. Then,
applyingCorollary 3.3, one sees that
u1
gfu0fu02W1
ψ0
fu0
, gfu0 W2
,
u2
gfu1fu12W1
ψ0
fu1
, gfu1 W2
,
.. .
un 1
gfunfun2W1
ψ0
fun
, gfun W2
,
4.1
where
gfun
L∗fun−P L∗fun∈Ψ⊥
fun
. 4.2
This is the iterative formula I. However,P L∗fucannot be computed through finite steps. Thus, we will revise the formula.
4.2. Iterative Formula II
Replacinggwithgnin iterative formula I, we derive iterative formula II, that is
un 1
gn
funfun2W1
ψ0
fun
, gn
fun W2
, 4.3
where
gn
fun
L∗fun−PnL∗fun∈Ψ⊥n
fun
, n0,1,2, . . . . 4.4
Lemma 4.1. Letfunfun/funW1and
βinf
n
ψ0
fun
−P ψ0
fun W2
ψ0
f∗−P ψ0
f∗W2,
4.5
Proof. Note that
ψ0
fun
−Pnψ0
fun 2
W2
ψ0
fun
−P ψ0
fun
P ψ0
fun
−Pnψ0
fun 2 W2, 4.6 since ψ0 fun
−P ψ0
fun
∈Ψ⊥fu
n
, P ψ0
fun
−Pnψ0
fun
∈Ψfvn
, 4.7
this gives us
ψ0
fun
−P ψ0
fun
⊥P ψ0
fun
−Pnψ0
fun
. 4.8
Thus,4.6leads to
ψ0
fun
−Pnψ0
fun 2
W2
ψ0
fun
−P ψ0
fun 2
W2
P ψ0
fun
−Pnψ0
fun 2
W2
≥ψ0
fun
−P ψ0
fun 2
W2
≥ψ0
f∗−P ψ0
f∗2W2
β2.
4.9
However,
β2ψ0
f∗−P ψ0
f∗2W2
≤ψ0
fun
−P ψ0
fun 2
W2≤
ψ0
fun
−Pnψ0
fun 2
W2
ψ0
fun
−Pnψ0
fvn
, ψ0
fun
−Pnψ0
fun
W2
ψ0
fun
, ψ0
fun
−Pnψ0
fun
W2
−Pnψ0
fun
, ψ0
fun
−Pnψ0
fun
W2.
4.10
BecausePnψ0fun, ψ0fun−Pnψ0funW20, hence
β2≤ψ0
fun
, gn
fun
Now, we proveβ >0. Suppose thatβ0, then4.5gives us
0ψ0
f∗−P ψ0
f∗2W2
ψ0
f∗−P ψ0
f∗, ψ0
f∗−P ψ0
f∗ W2
ψ0
f∗−P ψ0
f∗, gf∗ W2
ψ0
f∗, gf∗ W2−P ψ0
f∗, gf∗ W2.
4.12
Since gf∗ L∗f∗ − P L∗f∗ ∈ Ψ⊥f∗ and P ψ0f∗ P L∗f∗ ∈ Ψf∗, hence
P ψ0f∗, gf∗W2 0, furthermore, ψ0f∗, gf∗W2 0,namely, ψ0f∗ ⊥ gf∗. Note
here thatgf∗∈Ψ⊥f∗, one has
ψ0
f∗∈Ψf∗Spanψ1
f∗, ψ2
f∗, . . .. 4.13
From30ofLemma 3.1, we know that{ψ
ix}∞i0 is a complete system ofW20,1. On the
other hand, byψ0f∗∈Ψf∗, we obtain{ψix}∞i1is a complete system ofW20,1. This is
a contradiction to40ofLemma 3.1, thus,β >0.
FromLemma 4.1we know that iterative formula is significative.
5. Existence of Solution for
3.3
Putuunin3.4, then byCorollary 3.3,Lufuncan be written by
u g
funfun2
ψ0
fun
, gfun W2
, 5.1
Replacinguwithvn, then
Lvnfun, 5.2
vn
gfunfun2
ψ0
fun
, gfun W2
. 5.3
Obviously, iterative formula II can be converted into the form as follows:
Lvnfun,
un 1αnvn,
vn1 vn0 vn1 0, n∈N ,
where
αn
ψ0
fun
, gfun W2
ψ0
fun
, gn
fun W2
·gn
fun
gfun
, 5.5
gn
fun
L∗fun−PnL∗fun∈Ψ⊥n,
gfun
L∗fun−P L∗fun∈Ψ⊥.
5.6
Note that byLemma 4.1we know
ψ0
fun
, gn
fun W2fun2W1
ψ0
fun
, gn
fun
W2≥fun 2 W1β
2>0
5.7
in5.5. Now supposegfunx/0, if there exists anx, such thatgfunx 0,then we
will change definition ofαnx:
αnx
ψ0
fun
, gfun W2
ψ0
fun
, gn
fun W2
·gn
fun
M gfun
M, 5.8
whereMsatisfiesgfun M≥α >0.
Lemma 5.1. A bounded set inW1is a compacted set inC0,1.
Lemma 5.2. Foru∈W2, v∈W1, one has
uix≤μ
iuW2, i0,1,2,3,
|vx| ≤μvW1.
5.9
Lemma 5.3. A bounded set inW2is a compacted set inC30,1.
Lemma 5.4. Letfun fun/funW1, thenαn−1C3 → 0, n → ∞.
Proof. 1Since
ψ0
fun W2
ψ0
fun
fun
W1
W2
1
funW1
ψ0funW2
1
funW1
L∗funW2
≤ 1
funW1
gfun
−gn
fun W2
1
fun
W1
gfun−gnfunW2
1
funW1
P L∗fun
−PnL∗funW2
≤ 1
funW1
P−PnL∗fun
≤ 1
funW1
funW1LP−Pn −→0, n−→ ∞,
5.11
hence, byLemma 4.1, one gets
ψ0 fun
, gfun W2
ψ0
fun
, gn
fun W2
−1 ψ0 fun
, gfun
W2
ψ0
fun
, gn
fun
W2 −1 ψ0 fun
, gfun
−gn
fun
W2
ψ0
fun
, gn
fun
W2 ≤ 1 β2 ψ0 fun
, gfun
−gn
fun
W2
, β >0
≤ 1
β2
ψ0
fun W2
gfun
−gn
fun
W2 −→0, n−→ ∞,
5.12
namely,
ψ0
fun
, gfun W2
ψ0
fun
, gn
fun W2
−→1 n−→ ∞. 5.13
2Note that
gfunk
W2
L∗fu
nk
−P L∗funk
W2
≤L∗fu
nk
W2
P L∗fu
nk
W2≤2L,
5.14
gnk
funk
W2
L∗fu
nk
−PnkL∗
funk
W2
≤L∗funk
W2
PnkL
∗fu
nk
W2 ≤2L,
thus, byLemma 5.3, there exists{nkj} ∞
j1of{nk} ∞
k1, such that
gnkj
funkj
C3
−→g∗x/0. 5.16
By5.11, it follows that
gnkj
funkj
−gfunkj C3 ≤M
gnkj
funkj
−gfunkj W2
Mgn
fun
−gfun
W2−→0, n−→ ∞. 5.17
Therefore,
lim
n→ ∞
gn
fun
gfun
−1
C3 lim
n→ ∞
gnfun
gfun
−1
C3
lim
n→ ∞
gn fun
−gfun
gfun
C3 lim
k→ ∞
gnk
funk
−gfunk
gfunk
C3
limj→ ∞gnkj
funkj
−gfunkj
limj→ ∞g
funkj
C3 1
g∗xC3
lim
j→ ∞
gnkj
funkj
−gfunkj C3.
0.
5.18
In the similar manner, one gets
lim n gn fun
gfun
−1
C3
0. 5.19
Combining the above argument, we obtain
lim n gn fun
gfun
−1
C3
0. 5.20
Lemma 5.5. unW2≤MvnW2.
Lemma 5.6. Iffx, y, z, wsatisfies condition H, thenvnW2,unW2 are bounded. (Technique of
proof is similar to theLemma 7.1in appendix).
Lemma 5.7. Ifθn ∈0,1andvnθn 0, thenθndoes not infinitely go closer to 1, as there exists
θ∈0,1, such thatθn< θ <1.
Proof. Sincevn0 0 andvn1 0, there existsθn∈0,1, such thatvnθn 0.
Suppose θn → 1, asn → ∞. By Lemmas 5.3 and 5.6, there exists vxsuch that
vn−vC3 → 0; thus|vnx−vx| → 0,|vnx−vx| → 0,|vnx−vx| → 0. Since
vn0 0,vn1 0,vnθn 0, it follows thatv0 0,v1 0,v1 0. Hence there exists anη∈0,1, such thatvη 0, thus there exists anξ∈η,1, such thatv3ξ 0. By
5.4,Lvnx fx, unx, unx, unx, taking limit fornon both sides, we have
Lvx fx, ux, ux, ux, 5.21
thusfξ, uξ, uξ, uξ v3ξ 0; this is a contradiction toH3.
Lemma 5.8. If fx, y, z, w satisfies condition H, then un 1−unC2 → 0 as n → ∞. (see
Lemma 7.3in the appendix)
Theorem 5.9Existence. Iffx, y, z, wsatisfies condition H, then the solution of 1.2exists in
W20,1.
Proof. FromLemma 5.6,unW3 ≤ M, so byLemma 5.3,{unx}is compact inC30,1, then
there exists convergent subsequence{unkx}of{unx}, such thatunk−uC3 → 0, in the
similar manner, there existsvx, such that vnk−vC3 → 0, thus |unkx−ux| → 0,
|vnkx−vx| → 0, ask → ∞.
ByLemma 5.4,|αnk−1| → 0, ask → ∞and byLemma 5.8,|unk 1x−unkx| → 0,
then|unk 1x−ux| ≤ |unk 1x−unkx| |unkx−ux| → 0; thus
vx lim
k→ ∞vnkx lim k→ ∞
1
αnk
unk 1 lim k→ ∞
1
αnk
· lim
k→ ∞unk 11· lim
k→ ∞unk ux. 5.22
From5.4and5.22,Lux fx, ux, ux, ux, the solution of1.2exists. Since fx, ux, ux, ux∈W10,1, thusux∈W20,1.
Theorem 5.10. Iffx, y, z, wsatisfies condition H and the solutionuxof 1.2is unique, then the sequence{unx}∞n1is convergent toux, andu
k
n x−ukxC → 0, k0,1,2,3.
Proof. Suppose that{unx}∞n1 is not convergent to ux, since unW2 is bounded, we can
choose two subsequences{unkx} ∞
k1,{unjx} ∞
j1 of {unx} ∞
n1 such thatunkx → u1x,
unjx → u2xand u1x/u2x, FromTheorem 5.9,u1xandu2xare the solutions of
1.2, this contradicts the uniqueness of solution of1.2. consequently,un−uC → 0.
Table 1:Relative errorsExatn10.
Node Ex Node Ex Node Ex
1/10 4.71412E-03 1/5 1.23304E-03 3/10 1.74306E-03
2/5 4.25259E-03 1/2 6.10987E-03 3/5 7.10274E-03
7/10 6.88828E-03 4/5 4.62244E-03 9/10 3.53199E-03
6. Numerical Example
Now we present a numerical example for solving 1.2 in the reproducing kernel space
W20,1. All computations are performed by the Mathematica 5.0.
Example 6.1. Let us consider the following equation:
u3x xux ux xux u2x,
u1 u0 u1 0, 6.1
wherex∈0,1. The true solution isux xx−1. Using our method, 10 points are chosen on0,1and we calculate the relative errorsExdef ux−unx/uxC. The numerical
results are presented inTable 1which shows the method given in the paper is efficient.
7. Appendix
In order to obtain proofs of Lemma 5.6 and Lemma 5.8, firstly, we give an expression of solution of1.1. For any initial value functionu0x∈W30,1,
vnx
x
0
dt t
0
dη
η
0
fξ, un−1ξ, un−1ξ, un−1ξ
dξ
−1
2x
2
1
0
dη
η
0
fξ, un−1ξ, un−1ξ, un−1ξ
dξ,
unx Pnvn, n1,2, . . . ,
7.1
wherePn:W3 → Ψnfuis a orthogonal projector.
Lemma 7.1. Iffx, y, z, wsatisfies condition H, thenvnW3,unW3are bounded.
Proof. ByH2,|fx, un−1x, un−1x, un−1x| ≤M, from7.1,
v
nx
x
0
dη
η
0
fξ, un−1ξ, un−1ξ, un−1ξ
dξ−x
1
0
dη
η
0
fξ, un−1ξ, un−1ξ, un−1ξ
≤
x
0
dη
η
0
fξ, un−1ξ, un−1ξ, un−1ξdξ
|x| 1
0
dη
η
0
fξ, un−1ξ, un−1ξ, un−1ξdξ
≤M1
2
x2 |x|M1
2 ≤M.
7.2
Using this method,|vnx| ≤3/2M,|vn3x| ≤M. ByLemma 5.4,|αni|, i0,1,2,3
are bounded, then|unx||αn−1vn−1x|,|unx||αn−1vn−1 αn−1vn−1|,|unx||αn−1vn−1
2αn−1vn−1 αn−1vn−1|and|u
3
n x||αn3−1vn−1 3αn−1vn−1 3αn−1vn−1 αn−1vn3−1|are bounded,
thus byH2,|vn4x| |fx fyun−1x fzun−1x fwun3−1x|is bounded. Consequently,
we know thatvnW3 is bounded by the definition of · W3, and byLemma 5.5,unW3 is
bounded.
Lemma 7.2. Iffx, y, z, wsatisfies condition H, then|vn 10−vn0| → 0,|vn 10−vn0| → 0,
asn → ∞.
Proof. ByLemma 5.4,αnC3≤M1, and from Lemmas5.2and5.6,vnC3 ≤M2, then
|un 1x−unx||αnvnx−αn−1vn−1x|
|αnvnx−αnvn−1 αnvn−1−αn−1vn−1x|
≤ αnC3|vnx−vn−1x| vn−1C3|αn−αn−1|
≤M1|vnx−vn−1x| M2|αn−αn−1|
≤M1
|vnx−vn−1x| vnx−vn−1x vnx−vn−1x ρn, 7.3
whereρnM2αn−αn−1C3, byLemma 5.4,αn−αn−1C3 → 0, thenρn → 0.
In the similar manner,
un 1x−unx≤M1
|vnx−vn−1x| vnx−vn−1x vnx−vn−1x ρn,
u
n 1x−unx≤2M1
|vnx−vn−1x| vnx−vn−1x vnx−vn−1x 2ρn,
7.4
|un 1x−unx| un 1x−unx un 1x−unx
≤4M1
SinceLvnx vn3x fx, un−1x, un−1x, un−1x, then
vnx
x
θn
fξ, un−1ξ, un−1ξ, un−1ξ
dξ,
vnx
x
0
dη
η
θn
fξ, un−1ξ, un−1ξ, un−1ξ
dξ,
vnx
x
0
dt t
0
dη
η
θn
fξ, un−1ξ, un−1ξ, un−1ξ
dξ,
7.6
whereθn ∈0,1andvnθn 0. Whenx ∈θ,1,θis inLemma 5.7, from H2, letM3
max{fy, fz, fw},
|vn 1x−vnx| ≤
x
0
dt t
0
dη
η
θn1
fξ, unξ, unξ, unξ
dξ
−
x
0
dt t
0
dη
η
θn
fξ, un−1ξ, un−1ξ, un−1ξ
dξ
≤
x
0
dt t
0
dη
η
θn1
f
ξ, unξ, unξ, unξ
−fξ, un−1ξ, un−1ξ, un−1ξdξ
x
0
dt t
0
dη θn
θn1
fξ, un−1ξ, un−1ξ, un−1ξ
dξ
≤
x
0
dt t
0
dη
η
0
fξ, unξ, unξ, unξ
−fξ, un−1ξ, un−1ξ, un−1ξdξ
M|θn 1−θn|
≤M3
1
0
dt 1
0
dη
x
0
|unξ−un−1ξ| unξ−un−1ξ
unξ−un−1ξdξ Mn
≤M3
x
0
|unξ−un−1ξ| unξ−un−1ξ unξ−un−1ξdξ Mn,
7.7
where Mn M|θn 1 − θn|, we suppose {θn} is convergent, otherwise, we can take the
convergent subsequence of{θn}, thusMn → 0n → ∞. By doing this,
v
n 1x−vnx
≤M3
x
0
v
n 1x−vnx
≤M3
x
0
|unξ−un−1ξ| unξ−un−1ξ unξ−un−1ξdξ Mn,
|vn 1x−vnx| vn 1x−vnx vn 1x−vnx
≤3M3
x
0
|unξ−un−1ξ| unξ−un−1ξ unξ−un−1ξdξ 3Mn
≤3M3
x
0
4M1
|vnξ−vn−1ξ| vnξ−vn−1ξ vnξ−vn−1ξ 4ρndξ 3Mn
≤M
x
0
|vnξ−vn−1ξ| vnξ−vn−1ξ vnξ−vn−1ξdξ ρn,
7.8
whereM12M1M3,ρn12M3ρn 3Mn, limn→ ∞ρn0.
Whenn1,|v2x−v1x| |v2x−v1x| |v2x−v1x| ≤Mxv1−v0C2 ρ1.
Whenn2,
|v3x−v2x| v3x−v2x v3x−v2x
≤M
x
0
|v2ξ−v1ξ| v2ξ−v1ξ v2ξ−v1ξdξ ρ2
≤M
x
0
Mξv1−v0C2 ρ1
dξ ρ2≤
xM2
2 v1−v0C2 Mρ1x ρ2.
7.9
Generally,
|vn 1x−vnx| vn 1x−vnx vn 1x−vnx
≤
xMn
n! v1−v0C2
n
k0
xMk
k! ρn−k
≤
Mn
n! v1−v0C2 n/2
k0
Mk
k! ρn−k n
kn/2 1
Mk
k! ρn−k
≤
Mn
n! v1−v0C2 ρn n/2
k0
Mk
k! ρ
n
kn/2 1
Mk
k! −→0, asn−→ ∞,
7.10
where · denotes the integral part of “·”, ρ max{ρ0, ρ1, . . . , ρn−n/2 1}, ρn
max{ρn, ρn−1, . . . , ρn−n/2}, limn→ ∞ρn 0. Thus, vn 1x−vnxC2 → 0, asn → ∞,
Letvnx vn1−x, thenvn 1x−vnxC2 → 0, asvn 11−x−vn1−xC2 →
0,x ∈ θ,1. Thusvn 1x−vnxC2 → 0,x ∈ 0,1−θ. Therefore,|vn 10−vn0| →
0,|vn 10−vn0| → 0, asn → ∞.
After this transformation, the above result holds for 1.2. Next, we will give an equivalent representation of the solution of1.2.
vnx vn0
x
0
dt t
0
dη
η
0
fξ, un−1ξ, un−1ξ, un−1ξ
dξ
−1
2x
2
1
0
dη
η
0
fξ, un−1ξ, un−1ξ, un−1ξ
dξ,
unx Pnvn, n1,2, . . . .
7.11
Lemma 7.3. Iffx, y, z, wsatisfies condition H, thenun 1−unC2 → 0, asn → ∞.
Proof. By7.11,vn0 −10dηη0fξ, un−1ξ, un−1ξ, un−1ξdξ, then
vnx
x
0
dt t
0
dη
η
0
fξ, un−1ξ, un−1ξ, un−1ξ
dξ 1
2x
2v
n0 vn0. 7.12
From Lemma 7.2 and by using the same method, we can prove that
vn 1x−vnxC2 → 0, x ∈ 0,1. By Lemma 5.4 and un αn−1vn−1, it follows that
un 1−unC2 → 0, asn → ∞.
8. Conclusion
In this paper, we have shown the easier verification conditions for the existence of nonlinear third-order two-point boundary value problems. The method presented in this paper is based on the reproducing kernel space. We have presented reproducing kernel theorem for solving the nonlinear third-order two-point boundary value problems. Our results cannot be deduced trivially from any of the earlier published results and it is a pity that most of the people have not paid enough attention to the reproducing kernel theorem.
Acknowledgments
The research is supported by the Doctoral Initial Foundation of Harbin Normal University
KJB200817and the Educational Department Scientific Technology Program of Heilongjiang Province11541097.
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