# The Solution of a Class of Third-Order Boundary Value Problems by the Reproducing Kernel Method

Authors: Zhiyuan Li; Yulan Wang; Fugui Tan
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

This paper expands the application of reproducing kernel method to a class of third-order
boundary value problems with mixed nonlinear boundary conditions. The analytical solution is represented in the form of series in the reproducing kernel space. Then-term approximation is obtained and
is proved to converge to the analytical solution. The numerical examples are given to demonstrate the
computation efficiency of the presented method. Results obtained by the method indicate that the method is
simple and effective.


---

## Body

## 1. Introduction

Third-order differential equations arise in a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross section, a three layer beam, electromagnetic waves, or gravity driven flows. Third-order boundary value problems were discussed in many papers in recent years, for instance, see [1–6] and references therein. In [1–3], the authors used the spline functions to solve boundary value problems. In [4], the authors developed a second-order method for solving third-order three-point boundary value problems based on Padé approximant in a recurrence relation. In [5], the authors introduced Adomian decomposition method for multipoint boundary value problems (BVPs). In this paper, we use reproducing kernel to solve singular third-order boundary value problems with mixed boundary conditions. Recently, the reproducing kernel methods [7–10] emerge one after another. Using the reproducing kernel methods, the authors discussed two-point boundary value problems and periodic boundary value problems. For third-order boundary value problems with mixed nonlinear boundary conditions, however, this method has not yet been applied. In previous work, the reproducing kernel method cannot be used directly to solve third-order boundary value problems with nonlinear boundary conditions. Our work is to present a numerical algorithm for solving a class of singular third-order boundary value problems. By using this method, the analytical solution and approximate solution are given and uniformly converge to the exact solution and its corresponding derivatives. The algorithms are efficiently applied to solve some model problems.Let us consider the following singular problems of third-order ordinary differential equations:(1.1)u′′′(x)+p1(x)u′′(x)+p2(x)u′(x)+p3(x)u(x)=F(x),x∈(0,1),λiu=ri,(i=1,2,3),
where pj(x),f(x)∈L2[0,1],(j=1,2,3) are known functions. λiu,(i=1,2,3) are linear independence conditions of determining the solution. We assume that  has a unique solution which belongs to W24[0,1], where W24[0,1] which is a reproducing kernel space is defined in the second section.In order to solve, let Lu=u′′′(x)+p1(x)u′′(x)+p2(x)u′(x)+p3(x)u(x). It is easy to prove that L:W24[0,1]→L2[0,1] is bounded linear operator.On the other hand, we assume that the conditions of determining the solution can be homogenized; after homogenization of these conditions, we put the conditions into the reproducing kernel space W24[0,1] constructed in the following section. Equation  can be transformed into the following form in W24[0,1]:
(1.2)(Lu)(x)=F(x).
To solve problem (1.2), we give a space W24[0,1]={u∣u∈W¯24[0,1]andλiu=0,i=1,2,3}. The inner product in W24[0,1] is given by 〈u(x),v(x)〉=∑i=03u(i)(0)v(i)(0)+∫01‍u(4)(t)v(4)(t)dt. Like in [8], we can get the following reproducing kernel space.Theorem 1.1.
The spaceW24[0,1] is a reproducing kernel space and its reproducing kernel is K(x,y), and
(1.3)K(x,y)=Ry(x)-h1(x)h1(y)∥h1(x)∥2-h2(x)h2(y)∥h2(x)∥2-h3(x)h3(y)∥h3(x)∥2,
where
(1.4)Ry(x)={5040-y7+35x3y3(4+y)-21x2y2(-60+y3)+7xy(720+y5)5040,x<y5040-x7+5040xy+7x6y+1260x2y2-21x5y2+140x3y3+35x4y35040,y<x,h1(x)=λ1yRy(x),h2(x)=λ2y(R(x,y)-(h1(x)h1(y)/∥h1(x)∥2)),h3(x)=λ3y(R(x,y)-(h1(x)h1(y)/∥h1(x)∥2)-(h2(x)h2(y)/∥h2(x)∥2)),  the symbol λiy indicates that the operator λi applies to the function of y.


## 2. The Reproducing Kernel Method

Letψi(x)=(LyKx(y))(xi),i=1,2,…, Practise Gram-Schmidt orthonomalization for {ψi(x)}i=1∞, we get
(2.1)ψ-i(x)=∑k=1i‍βikψk(x),
where βik are coefficients of Gram-Schmidt orthonormalization.Theorem 2.1.
If{xi}i=1∞ is distinct points dense in [0,1] and L-1 is existent, then
(2.2)u(x)=∑i=1∞‍∑k=1i‍βikF(xk)ψ¯i(x)
is an analytical solution of the problem (1.2).Proof.
Since{ψ-i(x)}i=1∞ is an orthonormal systems, u(x) is expressed as
(2.3)u(x)=∑i=1∞‍〈u(x),ψ-i(x)〉ψ-i(x)=∑i=1∞‍∑k=1i‍βik〈u(x),(LsKx(s))(xk)〉ψ-i(x)=∑i=1∞‍∑k=1i‍βik(Ls〈u(x),Kx(s)〉)(xk)ψ-i(x)=∑i=1∞‍∑k=1i‍βik(Lsu(s))(xk)ψ-i(x)=∑i=1∞‍∑k=1i‍βikF(xk)ψ-i(x).We denote the approximate solution ofun(x) by
(2.4)un(x)=∑i=1n‍∑k=1i‍βikF(xk)ψ¯i(x).Theorem 2.2.
Letεn2=∥u(x)-un(x)∥2 where u(x),  un(x) are given by (2.2) and (2.4), then the sequence of number εn(x) is monotone decreasing and εn(x)→0.Proof.
Because(2.5)εn2=∥u(x)-un(x)∥2=∑i=n+1∞‍〈u(x),ψ-i(x)〉ψ-i(x)=∑i=n+1∞‍(〈u(x),ψ-i(x)〉)2,
clearly εn-1≥εn and consequently {εn} is monotone decreasing in the sense of ∥·∥W24. By Theorem 2.1, we know ∑i=1∞‍〈u(x),ψ-i(x)〉ψ-i(x) is convergent in the norm of ∥·∥W24, then we have
(2.6)εn2=∥u(x)-un(x)∥2→0.
Hence, εn→0.Theorem 2.3 (convergence analysis).
Ifu(x), un(x) are given by (2.2) and (2.4), then un(x) and un(k)(x) uniformly convergent to u(x) and u(k)(x), where k=0,1,2,3.Proof.
For anyx∈[0,1],k=0,1,2,3(2.7)|un(k)(x)-u(k)(x)|=|〈un(t)-u(t),∂kK(x,t)∂xk〉|≤∥un(t)-u(t)∥·∥∂kK(x,t)∂xk∥,
then there exists Ck>0 such that, |un(k)(x)-u(k)(x)|≤Ck∥un(t)-u(t)∥=Ckεn→0.Theorem 2.4.
If{xk}k=1∞ is distinct points dense in [0,1] and u(x),un(x) are given by (2.2) and (2.4), then Lu(xk)=Lun(xk).Proof.
We may set projective operatorPn:W24[0,1]→{∑m=1n‍cmψm(x),cm∈R}. Hence,
(2.8)Lun(xk)=〈un(ξ),LxkKxk(ξ)〉=〈un(ξ),ψk(ξ)〉=〈Pnu(ξ),ψk(ξ)〉=〈u(ξ),Pnψk(ξ)〉=〈u(ξ),ψk(ξ)〉=〈u(ξ),LxkKxk(ξ)〉=Lxk〈u(ξ),Kxk(ξ)〉=Lxku(xk)=Lu(xk).Theorem 2.5 (error estimate).
If{xk}k=1∞ is distinct points dense in [0,1] and u(x), un(x) are given by (2.2) and (2.4), then |u(x)-un(x)|<(M/n), where M=∥∑i=n+1∞‍∑k=1i‍βikF(xk)ψ¯i(x)∥·∥∂Kη(ξ)/∂η∥.Proof.
For every givenx∈[0,1], there is always xi∈{xk}k=1∞ satisfying xi<x and x-xi=1/n. By Theorem 2.5 and xi∈{xk}k=1∞ implying Lu(xi)=Lun(xi), so we obtain
(2.9)|Lu(x)-Lun(x)|=|Lu(x)-Lu(xi)-[Lun(x)-Lun(xi)]|.
For application reproducing kernel property, we have
(2.10)u(x)=〈u(ξ),Kx(ξ)〉,Lu(x)=〈u(ξ),LKx(ξ)〉.
We also have
(2.11)Lu(x)-Lun(x)=Lu(x)-Lu(xi)-[Lun(x)-Lun(xi)]=〈u(ξ),LKx(ξ)-LKxi(ξ)〉-〈un(ξ),LKx(ξ)-LKxi(ξ)〉=〈u(ξ)-un(ξ),LKx(ξ)-LKxi(ξ)〉.
Moreover,
(2.12)|u(x)-un(x)|=|L-1[Lu(x)-Lun(x)]|≤|〈u(ξ)-un(ξ),L-1LKx(ξ)-L-1LKxi(ξ)〉|=|〈u(ξ)-un(ξ),Kx(ξ)-Kxi(ξ)〉|≤∥u-un∥∥Kx(ξ)-Kxi(ξ)∥.
It is noted that we take norm of ∥Kx(ξ)-Kxi(ξ)∥ for variable ξ. The function kx(ξ) is derived on x in the interval of [0,1], so we have Kx(ξ)-Kxi(ξ)=(∂Kη(ξ)/∂η)(x-xi). Hence,
(2.13)|u(x)-un(x)|≤∥u-un∥∥∂Kη(ξ)∂η(x-xi)∥=∥u-un∥∥∂Kη(ξ)∂η∥(x-xi)=∥∑i=n+1∞‍∑k=1i‍βikF(xk)ψ¯i(x)∥∥∂Kη(ξ)∂η∥(x-xi)≤Mn.


## 3. Numerical Experiment

For showing the effectiveness of our method, we consider the following problems.Example 3.1 (see [2, 3]).
Considering the following third-order boundary values problem(3.1)u′′′(x)-xu(x)=(x3-2x2-5x-3)ex,0≤x≤1,u(0)=u(1)=0,u′(0)=1,
where the exact solution is uT(x)=x(1-x)ex. By the homogeneous process of the boundary condition, let v(x)=u(x)-x(1-x), problem  can be transformed into the equivalent form
(3.2)v′′′(x)-xv(x)=(1-x)x2+(x3-2x2-5x-3)ex,0≤x≤1,v(0)=v(1)=0,v′(0)=0.
The numerical results are presented in Tables 1, 2, and 3.Table 1
The absolute error of Example3.1.
Absolute error
x
Present (n=20)
Present (n=40)
Present (n=80)
Present (n=160)
Ref. [3]
Ref. [2]
Present (n=500)
0.0
0
0
0
0
0
0
0
0.1
1.52298
×
1
0
-
5
2.85196
×
1
0
-
6
5.80026
×
1
0
-
7
1.27547
×
1
0
-
7
1
×
1
0
-
7
0
1.18126
×
1
0
-
8
0.3
7.42332
×
1
0
-
5
1.5986
×
1
0
-
5
3.66544
×
1
0
-
6
8.74382
×
1
0
-
7
3
×
1
0
-
7
5
×
1
0
-
3
8.65902
×
1
0
-
8
0.4
1.08810
×
1
0
-
4
2.42167
×
1
0
-
5
5.67330
×
1
0
-
6
1.37023
×
1
0
-
6
5
×
1
0
-
7
7
×
1
0
-
3
1.36944
×
1
0
-
7
0.5
1.40256
×
1
0
-
4
3.19274
×
1
0
-
5
7.58382
×
1
0
-
6
1.84582
×
1
0
-
6
6
×
1
0
-
7
8
×
1
0
-
3
1.85506
×
1
0
-
7
0.7
1.68693
×
1
0
-
4
3.95170
×
1
0
-
5
9.54427
×
1
0
-
6
2.3440
×
1
0
-
6
7
×
1
0
-
7
7
×
1
0
-
3
2.37089
×
1
0
-
7
0.9
9.78854
×
1
0
-
5
2.33293
×
1
0
-
5
5.68883
×
1
0
-
6
1.40422
×
1
0
-
6
4
×
1
0
-
7
4
×
1
0
-
3
1.42538
×
1
0
-
7
1.0
1.50887
×
1
0
-
15
1.74768
×
1
0
-
15
1.57219
×
1
0
-
15
1.90655
×
1
0
-
15
0
1.4
×
1
0
-
2
1.49019
×
1
0
-
15
0 s < CPU runtime ≤ 1 s
CPU runtime ≤ 10 s
CPU runtime = 180 sTable 2
The numerical results of Example3.1.
x
u
T
'
(
x
)
u
500
'
(
x
)
|
u
T
'
(
x
)
-
u
500
'
(
x
)
|
u
T
′′
(
x
)
u
500
′′
(
x
)
|
u
T
′′
(
x
)
-
u
500
′′
(
x
)
|
0.0
1
1
0
0
0.0000123873
0.0000123873
0.1
0.983602
0.983602
2.217025
×
1
0
-
7
-
0.342603
-
0.342601
1.904429
×
1
0
-
6
0.3
0.823414
0.823414
4.842037
×
1
0
-
7
-
1.33636
-
1.33636
6.462399
×
1
0
-
7
0.5
0.41218
0.412181
4.459369
×
1
0
-
7
-
2.88526
-
2.88526
1.127018
×
1
0
-
6
0.7
-
0.382613
-
0.382613
1.172353
×
1
0
-
8
-
5.21562
-
5.21562
3.578872
×
1
0
-
6
0.9
-
1.74632
-
1.74632
1.0452710
×
1
0
-
6
-
8.63321
-
8.63321
6.928269
×
1
0
-
6
1.0
-
2.71828
-
2.71828
1.8404898
×
1
0
-
6
-
10.8731
-
10.8731
9.029439
×
1
0
-
6Table 3
The numerical results of Example3.1.
x
u
T
(
3
)
(
x
)
u
500
(
3
)
(
x
)
|
u
T
(
3
)
(
x
)
-
u
500
(
3
)
(
x
)
|
u
T
(
4
)
(
x
)
u
500
(
4
)
(
x
)
|
u
T
(
4
)
(
x
)
-
u
500
(
4
)
(
x
)
|
0.0
-
3
-
3.01001
0.010006
-
8
-
3.01001
4.98999
0.2
-
4.93447
-
4.93447
8.518423
×
1
0
-
9
-
11.53
-
11.5489
0.0188419
0.4
-
7.69782
-
7.69782
5.477735
×
1
0
-
8
-
16.3504
-
16.3771
0.0266921
0.6
-
11.5887
-
11.5887
1.335948
×
1
0
-
8
-
22.8858
-
22.9229
0.0370827
0.8
-
17.0031
-
17.0031
1.724353
×
1
0
-
8
-
31.6917
-
31.7423
0.050624
1.0
-
24.4645
-
24.4645
2.771117
×
1
0
-
13
-
43.4925
3.021444
×
1
0
-
16
43.4925Example 3.2 (see [11, 14]).
Considering the following third-order obstacle problems:(3.3)u′′′(x)={0,0≤x≤14,14≤x≤14,u(x)-1,3π4≤x≤1,
where u(0)=u'(0)=u'(1)=0, the exact solution is
(3.4)uT(x)={12a1x2,0≤x≤14,1+a2ex+e-x/2[a3cos3x2+a4sin3x2],14≤x≤3412a5x(x-2)+a6,34≤x≤1,
where
(3.5)a1=0.24391096222648,a2=-0.17847234452746,a3=-0.81893573565615,a4=-0.30266818001866,a5=-0.24213890868443,a6=-0.06537630092111.
The numerical results are presented in Table 4.Table 4
The observed maximum errors of Example3.2.
Method
n
=
16
n
=
32
n
=
64
n
=
128
Our method
1.18
×
1
0
-
3
5.47
×
1
0
-
4
2.62
×
1
0
-
4
1.29
×
1
0
-
4
Nonpolynomial spline [11]
7.12
×
1
0
-
4
4.05
×
1
0
-
4
2.22
×
1
0
-
4
1.15
×
1
0
-
4
Quartic spline [12]
1.15
×
1
0
-
3
5.32
×
1
0
-
4
2.56
×
1
0
-
4
1.26
×
1
0
-
4
Finite difference [13]
1.96
×
1
0
-
4
4.89
×
1
0
-
5
1.22
×
1
0
-
5
3.06
×
1
0
-
6
Cubic spline [14]
1.23
×
1
0
-
3
5.53
×
1
0
-
4
2.61
×
1
0
-
4
1.27
×
1
0
-
4
Colloc. quantic spline [15]
1.26
×
1
0
-
3
5.60
×
1
0
-
4
3.10
×
1
0
-
4
1.67
×
1
0
-
4
Finite difference [16]
6.89
×
1
0
-
3
7.11
×
1
0
-
3
7.27
×
1
0
-
3
7.36
×
1
0
-
3
Quartic B spline [17]
1.13
×
1
0
-
3
5.30
×
1
0
-
4
5.52
×
1
0
-
4
1.23
×
1
0
-
4Example 3.3.
Considering the following boundary value problems with nonclassical condition:(3.6)u′′′(x)+1x1-xu′′(x)+1xx(1-x)u′(x)+1x21-xu(x)=f(x),0<x<1,u(0)=0,u′(0)=∫01‍u(x)dx,2u(1)=u′(1).
We determine f(x) to get the true solution, given by uT(x)=xex. The numerical results are presented in Table 5.Table 5
The numerical results of Example3.3.
x
u
T
(
x
)
u
100
(
x
)
|
u
T
(
x
)
-
u
100
(
x
)
|
|
u
T
'
(
x
)
-
u
100
'
(
x
)
|
|
u
T
′′
(
x
)
-
u
100
′′
(
x
)
|
|
u
T
′′′
(
x
)
-
u
100
′′′
(
x
)
|
0.0
0
0
0
2.11892
×
1
0
-
5
5.34582
×
1
0
-
5
2.49157
×
1
0
-
2
0.1
0.110517
0.110518
1.06299
×
1
0
-
6
6.59012
×
1
0
-
6
4.42396
×
1
0
-
6
2.85088
×
1
0
-
4
0.3
0.404958
0.404962
4.25342
×
1
0
-
6
3.27006
×
1
0
-
5
2.16465
×
1
0
-
4
1.15677
×
1
0
-
3
0.5
0.824361
0.824376
1.51582
×
1
0
-
5
7.46045
×
1
0
-
5
1.67531
×
1
0
-
4
8.58014
×
1
0
-
4
0.7
1.40963
1.40966
3.24736
×
1
0
-
5
9.40379
×
1
0
-
5
3.75442
×
1
0
-
5
5.12073
×
1
0
-
4
0.8
1.78043
1.78047
4.20348
×
1
0
-
5
9.71987
×
1
0
-
5
3.89027
×
1
0
-
5
5.59346
×
1
0
-
4
0.9
2.21364
2.21369
5.20699
×
1
0
-
5
1.05201
×
1
0
-
4
1.40742
×
1
0
-
4
1.08743
×
1
0
-
3
1.0
2.71828
2.71835
6.36065
×
1
0
-
5
1.27213
×
1
0
-
4
1.9082
×
1
0
-
4
1.07270
×
1
0
-
2Example 3.4.
Considering the following singular third-order three points boundary value problems with nonlinear condition(3.7)u′′′(x)+pxαu′′(x)-k2xα(1-x)βu′(x)+qsin(x)xα(1-x)βu(x)=f(x),0<x<1,u′(1)=0,2u′(0)=(u′′(0))2,u(12)=0.
We determine f(x) to get the true solution, given by uT(x)=r(k(2x-1)-2sinh(kx)+2(cosh(kx))tanh(k/2))/2k3. The numerical results are presented in Table 6.Table 6
The numerical results of Example3.4, p=q=β=1,α=2,k=2,n=20.
x
u
T
u
20
|
u
T
-
u
20
|
u
T
'
u
20
'
|
u
T
'
-
u
20
'
|
|
u
T
′′
-
u
20
′′
|
0.08
-
0.0286650
-
0.0286650
5.20529
×
1
0
-
9
0.0273871
0.0273873
2.32501
×
1
0
-
7
7.75366
×
1
0
-
6
0.16
-
0.0255710
-
0.0255710
6.94285
×
1
0
-
9
0.0490631
0.0490632
7.15527
×
1
0
-
8
2.10654
×
1
0
-
6
0.24
-
0.0209523
-
0.0209523
9.70246
×
1
0
-
9
0.0655842
0.0655842
1.41703
×
1
0
-
8
5.37958
×
1
0
-
7
0.32
-
0.0152036
-
0.0152035
9.39437
×
1
0
-
9
0.0773741
0.0773741
1.84526
×
1
0
-
8
3.37592
×
1
0
-
7
0.48
-
0.0017588
-
0.0017588
1.71495
×
1
0
-
9
0.0878568
0.0878567
8.12846
×
1
0
-
8
4.48111
×
1
0
-
7
0.56
0.00525584
0.00525583
6.04974
×
1
0
-
9
0.0868185
0.0868184
1.11842
×
1
0
-
7
4.02874
×
1
0
-
7
0.64
0.0120206
0.0120205
1.65310
×
1
0
-
8
0.0815939
0.0815937
1.53109
×
1
0
-
7
6.25311
×
1
0
-
7
0.72
0.0181958
0.0181957
3.10494
×
1
0
-
8
0.0720489
0.0720486
2.13478
×
1
0
-
7
9.11482
×
1
0
-
7
0.88
0.0273346
0.0273346
8.10037
×
1
0
-
8
0.0389010
0.0389005
4.47861
×
1
0
-
7
1.32613
×
1
0
-
6
0.96
0.0295066
0.0295065
1.25247
×
1
0
-
7
0.0144477
0.0144469
8.40832
×
1
0
-
7
9.84025
×
1
0
-
6Example 3.5.
Considering the following boundary value problems with nonlinear condition:(3.8)u′′′(x)-11+xu′(x)+2u(x)=f(x),u(0)=u′(0)=0,(u′(1))2=2∫01‍u(x)dx.
We determine f(x) to get the true solution, given by uT(x)=(1/2)x3. The numerical results are presented in Table 7.Table 7
The numerical results of Example3.5.
x
u
T
u
20
|
u
T
-
u
20
|
u
T
'
u
20
'
|
u
T
'
-
u
20
'
|
|
u
T
′′
-
u
20
′′
|
0.08
0.000256
0.00025618
1.87335
×
1
0
-
7
0.0096
0.009597
3.13184
×
1
0
-
6
1.14007
×
1
0
-
5
0.16
0.002048
0.0020481
1.72967
×
1
0
-
7
0.0384
0.038399
8.39162
×
1
0
-
7
2.89347
×
1
0
-
5
0.24
0.006912
0.0069121
1.38265
×
1
0
-
7
0.0864
0.0864
1.89713
×
1
0
-
7
1.24163
×
1
0
-
5
0.32
0.016384
0.0163841
1.13116
×
1
0
-
7
0.1536
0.1536
2.05885
×
1
0
-
7
5.02804
×
1
0
-
7
0.4
0.032000
0.0320001
9.39829
×
1
0
-
8
0.2400
0.2400
2.21979
×
1
0
-
7
1.40284
×
1
0
-
6
0.48
0.055296
0.0552961
7.77508
×
1
0
-
8
0.3456
0.3456
1.93052
×
1
0
-
7
5.56607
×
1
0
-
7
0.56
0.087808
0.0878081
6.38527
×
1
0
-
8
0.4704
0.4704
1.60088
×
1
0
-
7
3.15123
×
1
0
-
7
0.64
0.131072
0.131072
5.21361
×
1
0
-
8
0.6144
0.6144
1.33773
×
1
0
-
7
3.10240
×
1
0
-
7
0.72
0.186624
0.186624
4.23781
×
1
0
-
8
0.7776
0.7776
1.13440
×
1
0
-
7
1.75032
×
1
0
-
7
0.8
0.256000
0.256000
3.44071
×
1
0
-
8
0.9600
0.9600
9.20785
×
1
0
-
8
3.49872
×
1
0
-
7
0.88
0.340736
0.340736
2.78679
×
1
0
-
8
1.1616
1.1616
4.41977
×
1
0
-
8
3.00596
×
1
0
-
8
0.96
0.442368
0.442368
2.08726
×
1
0
-
8
1.3824
1.3824
2.08167
×
1
0
-
8
9.08609
×
1
0
-
6


## 4. Conclusions and Remarks

In this work, we present an algorithm for solving third-order mixed boundary value problems (BVPs) based on the reproducing kernel method. The method can be generalized to get reproducing kernel of problem with linear conditions. All computations are performed by the Mathematica 7.0 software package.



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*Source: 195310-2012-10-16.xml*