# Solving a Class of Singular Fifth-Order Boundary Value Problems Using Reproducing Kernel Hilbert Space Method

Authors: Yulan Wang; Shuai Lu; Fugui Tan; Mingjing Du; Hao Yu
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

We use the reproducing kernel Hilbert space method to solve the fifth-order boundary value problems. The exact solution to the fifth-order boundary value problems is obtained in reproducing kernel space. The approximate solution is given by using an iterative method and the finite section method. The present method reveals to be more effective and convenient compared with the other methods.


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## Body

## 1. Introduction

The reproducing kernel Hilbert space method has been shown [1–7] to solve effectively, easily, and accurately a large class of linear and nonlinear, ordinary, partial differential equations. However, in [1–7], it cannot be used directly  boundary value problems with mixed boundary conditions, since it is very difficult to obtain a reproducing kernel function satisfying mixed nonlinear boundary conditions. The aim of this work is to fill this gap. In [8], we give a new reproducing kernel Hilbert space for solving singular linear fourth-order boundary value problems with mixed boundary conditions. In this paper, we use the new reproducing kernel Hilbert function space method to solve the nonlinear fifth-order boundary value problems.Singular fifth-order boundary value problems arise in the fields of gas dynamics, Newtonian fluid mechanics, fluid mechanics, fluid dynamics, elasticity, reaction-diffusion processes, chemical kinetics, and other branches of applied mathematics.Let us consider the following class of singular fifth-order mixed boundary value problems:(1)u(5)(x)+p1(x)xα1(1-x)β1u(4)(x)+⋯+p4(x)xα4(1-x)β4u′(x)+p5(x)xα5(1-x)β5u(x)=f(x),x∈(0,1),Biu=ri,(i=1,2,…,5),
where pj(x),f(x)∈L2[0,1] (j=1,…,5) are known functions. Biu (i=1,2,…,5) are linear conditions. We assume that (1) has a unique solution which belongs to W26[0,1], where W26[0,1] is a reproducing kernel space.Remark 1.
IfBiu=u(i)(0) (i=1,…,m), then (1) is an initial value problem. If Biu=u(xi) (i=1,2,…,m), then (1) is a multipoint problem and so on. That is, problem (1) has a rather general form.
Letα=max1≤i≤5{αi} and β=max1≤i≤5{βi}, F(x)=xα(1-x)βf(x).
Consider(2)(Lu)(x)=xα(1-x)βu(5)(x)+xα-α1(1-x)β-β1p1(x)u(4)(x)+⋯+xα-α4(1-x)β-β4p4(x)u′(x)+xα-α5(1-x)β-β5p5(x)u(x).
It is easy to prove that L:W26[0,1]→L2[0,1] is a bounded linear operator. On the other hand, we suppose that the linear conditions can also always be homogenized; after homogenization of these conditions, we put these conditions into the reproducing kernel space W26[0,1] constructed in the following section. Equation (1) can be transformed into the following form in W26[0,1]:
(3)(Lu)(x)=F(x).


## 2. Reproducing Kernel Hilbert Space

Definition 2.
LetH be a real Hilbert space of functions f:Ω→R. A function K:Ω×Ω→R is called reproducing kernel for H if(i)
K(x,·)∈H for all x∈Ω,(ii)
f(x)=〈f,K(·,x)〉H for all f∈H and all x∈Ω.Definition 3.
A real Hilbert spaceH of functions on a set Ω is called a reproducing kernel Hilbert space if there exists a reproducing kernel K of H.One defines that the inner product spaceW¯2m+1[0,1]={u∣u,u′,…,u(m) are  absolutely  continuous  function, u(m+1)∈L2[0,1]}.The inner product inW¯2m+1[0,1] is given by
(4)〈u(x),v(x)〉=∑i=0m‍u(i)(0)v(i)(0)+∫01‍u(m+1)(t)v(m+1)(t)dt.Theorem 4 (see [8]).
The spaceW¯2m+1[0,1] is a reproducing kernel space, and its reproducing kernel is
(5)R{m+1}(x,y)={∑i=0m‍1(i!)2xiyi+1(m!)2∫0x(x-t)m(y-t)mdt,x<y,∑i=0m‍1(i!)2xiyi+1(m!)2∫0y(x-t)m(y-t)mdt,y<x.For studying the solution of (1) in the homogenized form, we give space (6) as follows:
(6)W26[0,1]={u∣u∈W¯26[0,1],Biu=0,i=1,2,…,5}.W26[0,1] is equipped with the same inner product W¯26[0,1]. In the following, we construct a reproducing kernel for the space W26[0,1], and we give Lemmas 5 and 6.Lemma 5.
LetA:H[a,b]→L2[a,b] be a bounded linear operator; function Rx(y) is the reproducing kernel of space H[a,b]. Let gz(x)=(AsRx(s))(z); then ∥gz(x)∥2=(As(AtRs(t))(z))(z), where H[a,b] denotes any reproducing kernel space of functions over [a,b], the symbol As indicates that the operator A applies to functions of the variable s, and the symbol (AsRx(s))(z) indicates that the operator A applies to function Rx(s) of the variable s and s=z.Proof.
Consider(7)∥gz(x)∥2=〈(AsRx(s))(z),(AtRx(t))(z)〉=(As(At〈Rx(s),Rx(t)〉)(z))(z)=(As(AtRs(t))(z))(z).Lemma 6.
IfA, gz(x), and Rx(y) are defined as in Lemma 5, let Kx(y)=Rx(y)-gz(x)gz(y)/∥gz(x)∥2; consider the  space  H1={u(y)∣u(y)∈H[a,b],and(Ayu(y))(z)=0}, then, Kx(y) is the reproducing kernel of space H1.Proof.
For anyu(y)∈H1, next, we will prove 〈u(y),Kx(y)〉=u(x).
Consider(8)〈u(y),Kx(y)〉=〈u(y),Rx(y)-gz(x)gz(y)∥gz(x)∥2〉=〈u(y),Rx(y)〉-〈u(y),gz(x)gz(y)∥gz(x)∥2〉=u(x)-gz(x)〈u(y),(AsRy(s))(z)〉∥gz(x)∥2=u(x)-gz(x)(As〈u(y),Ry(s)〉)(z)∥gz(x)∥2=u(x)-gz(x)(Asu(s))(z)∥gz(x)∥2=u(x).Leth1(x)=B1yR{6}(x,y), h2(x)=B2y(R{6}(x,y)-h1(x)h1(y)/∥h1(x)∥2),  h3(x)=B3y(R{6}(x,y)-h1(x)h1(y)/∥h1(x)∥2-h2(x)h2(y)/∥h2(x)∥2), h4(x)=B4y(R{6}(x,y)-h1(x)h1(y)/∥h1(x)∥2-h2(x)h2(y)/∥h2(x)∥2-h3(x)h3(y)/∥h3(x)∥2),  and h5(x)=B5y(R{6}(x,y)-h1(x)h1(y)/∥h1(x)∥2-h2(x)h2(y)/∥h2(x)∥2-h3(x)h3(y)/∥h3(x)∥2-h4(x)h4(y)/∥h4(x)∥2),  where the symbol Biy (i=1,2,3,4,5) indicates that the operator Bi (i=1,2,3,4,5) applies to functions of the variable y. Using Lemma 6, we get Theorem 7.Theorem 7.
The spaceW26[0,1] is a reproducing kernel space, and its reproducing kernel is
(9)K(x,y)=R{6}(x,y)-h1(x)h1(y)∥h1(x)∥2-h2(x)h2(y)∥h2(x)∥2-h3(x)h3(y)∥h3(x)∥2-h4(x)h4(y)∥h4(x)∥2-h5(x)h5(y)∥h5(x)∥2.


## 3. Analytical Solution

Letψi(x)=(LyK(x,y))(xi),   i=1,2,…. Via Gram-Schmidt orthonormalization for {ψi(x)}i=1∞, we get
(10)ψ-i(x)=∑k=1i‍βikψk(x),
where the βik are the coefficients resulting from Gram-Schmid orthonormalization.Lemma 8.
If{xi}i=1∞ are  distinct points dense in [0,1] and L-1 is existent, then {ψi(x)}i=1∞ is a complete function system in W2m+1[0,1].Proof.
For each fixedu(x)∈W2m+1[0,1], if 〈u(x),ψi(x)〉=0, then
(11)〈u(x),ψi(x)〉=(Ly〈u(x),K(x,y)〉)(xi)=(Lyu(y))(xi)=0.
Taking into account the density of {xi}i=1∞, it results in (Lyu(y))(x)=0. It follows that u(x)≡0 from the existence of L-1.Theorem 9.
If{xi}i=1∞  are  distinct points dense in [0,1] and L-1 is existent, then
(12)u(x)=∑i=1∞‍∑k=1i‍βikF(xk)ψ¯i(x)
is an analytical solution of (3).Proof.
u
(
x
) can be expanded to Fourier series in terms of normal orthogonal basis {ψ-i(x)}i=1∞ in W2m+1[0,1] as follows:
(13)u(x)=∑i=1∞‍〈u(x),ψ-i(x)〉ψ-i(x)=∑i=1∞‍∑k=1i‍βik〈u(x),ψk(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).


## 4. Numerical Solution

We define an approximate solutionun(x) by
(14)un(x)=∑i=1n‍∑k=1i‍βikF(xk)ψ¯i(x).Theorem 10.
Letεn2=∥u(x)-un(x)∥2, where u(x) and un(x) are given by (12) and (14); then the sequence of real numbers εn(x) is monotonously decreasing and εn(x)→0.Proof.
We have(15)εn2=∥u(x)-un(x)∥2=∥∑i=n+1∞‍〈u(x),ψ-i(x)〉ψ-i(x)∥2=∑i=n+1∞‍(〈u(x),ψ-i(x)〉)2.
Clearly, εn-1≥εn and consequently {εn} is monotonously decreasing in the sense of ∥·∥. By Theorem 9, we know that ∑i=1∞‍〈u(x),ψ-i(x)〉ψ-i(x) is convergent in the norm of ∥·∥; then we have
(16)εn2=∥u(x)-un(x)∥2⟶0.
Hence, εn→0.Theorem 11 (convergence analysis).
u
n
(
x
) and un(k)(x) are uniformly convergent to u(x) and u(k)(x), k=0,1,2,…,m, where u(x) and un(x) are given by (12) and (14).Proof.
For anyx∈[0,1], k=0,1,2,…,5,
(17)|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
(18)|un(k)(x)-u(k)(x)|≤Ck∥un(t)-u(t)∥=Ckεn⟶0.The numerical solution to (3) can be obtained using the following method:
(19)un(x)=∑i=1n‍diψi(x),
where the coefficients di, i=1,…,m, are determined by the equation
(20)∑i=1n‍diLψi(x)|x=xj=F(xj),j=1,2,…,n.
Using (19) and (20), we have (Lum)(xj)=F(xj), j=1,2,…,n. So, un(x) is the approximation solution of (3).


## 5. Numerical Experiment

In this section, two numerical examples are studied to demonstrate the accuracy of the present method.Example 12.
Consider the following fifth-order boundary value problem with nonclassical side condition (the right-hand side of this problem has a singularity atx=0, x=1):
(21)u(5)(x)-exx1-xu′′(x)+exx1-xu(x)=f(x),0<x<1,u(0)=u′(0)=u(14)=0,5u′′(0)+42∫01‍e-xu(x)dx=0,4u′′′(1)+u′(1)=10u′′(1),
where f(x)=ex(-45+195x-750x2+320x5/2-600x3-32(5+ex)x7/2+40(20+3ex)x4-16(9+4ex)x9/2+16(23+10ex)x5-16x11/2+32x6)/32(-1+x)x5/2. The exact solution is uT(x)=x2(x-1/2)ex. The numerical results are presented in Table 1.Table 1
The numerical results of Example12.
x
u
T
(
x
)
u
100
(
x
)
|
u
20
(
x
)
-
u
T
(
x
)
|
|
u
40
(
x
)
-
u
T
(
x
)
|
|
u
100
(
x
)
-
u
T
(
x
)
|
|
u
100
′
(
x
)
-
u
T
′
(
x
)
|
0
0
0
0
0
0
0
0.08
−0.00150556
−0.00149854
3.11533
×
1
0
-
4
2.12134
×
1
0
-
4
7.01613
×
1
0
-
6
4.26411
×
1
0
-
5
0.16
−0.00300419
−0.00299716
5.86206
×
1
0
-
4
4.03105
×
1
0
-
4
7.02601
×
1
0
-
6
3.96519
×
1
0
-
5
0.24
−0.000739712
−0.000738437
1.36893
×
1
0
-
4
1.08655
×
1
0
-
4
1.27471
×
1
0
-
6
1.06360
×
1
0
-
4
0.32
0.00926282
0.00925363
1.38316
×
1
0
-
3
9.02756
×
1
0
-
4
9.18729
×
1
0
-
6
1.59425
×
1
0
-
4
0.4
0.0316161
0.0315926
4.00907
×
1
0
-
3
2.65016
×
1
0
-
3
2.34247
×
1
0
-
5
1.93257
×
1
0
-
4
0.48
0.0717954
0.0717548
7.49348
×
1
0
-
3
4.96714
×
1
0
-
3
4.06485
×
1
0
-
5
2.28313
×
1
0
-
4
0.56
0.136337
0.136276
1.13395
×
1
0
-
3
7.52215
×
1
0
-
3
6.03524
×
1
0
-
5
2.58664
×
1
0
-
4
0.64
0.23304
0.232959
1.48395
×
1
0
-
3
9.84568
×
1
0
-
3
8.01808
×
1
0
-
5
2.77257
×
1
0
-
4
0.72
0.371189
0.371084
1.71134
×
1
0
-
3
1.13541
×
1
0
-
3
1.04886
×
1
0
-
4
2.93384
×
1
0
-
4
0.8
0.561801
0.561673
1.71481
×
1
0
-
3
1.13757
×
1
0
-
3
1.27883
×
1
0
-
4
3.40704
×
1
0
-
4
0.88
0.817902
0.817739
1.38345
×
1
0
-
3
9.1769
×
1
0
-
3
1.63033
×
1
0
-
4
6.89140
×
1
0
-
4
0.96
1.15484
1.15458
5.99608
×
1
0
-
3
3.97735
×
1
0
-
3
2.60877
×
1
0
-
4
1.51714
×
1
0
-
3Example 13 (see [9]).
Consider the following fifth-order boundary value problem (the right-hand side of this problem has a singularity atx=0):
(22)u(5)(x)-e-xu(x)=-e-xx92(1-x)+945(1-11x)32x,0<x<1,u(0)=u′(0)=u′′(0)=u(1)=0,u′(1)=-1,
where the exact solution is uT(x)=x9/2(1-x). By the homogeneous process of the boundary condition, letting v(x)=u(x)-x3(1-x), the problem can be transformed into the equivalent form
(23)v(5)(x)-e-xv(x)=-e-x(x92-x3)(1-x)+945(1-11x)32x,0≤x≤1,v(0)=v′(0)=v′′(0)=v(1)=v′(1)=0.
The numerical results are presented in Tables 2, 3, and 4.Table 2
Comparison of the absolute error of Example13.
x
Solution
Absolute error
u
T
(
t
,
x
)
Reference [9]
u
100
(
x
)
|
u
50
(
x
)
-
u
T
(
x
)
|
Reference [9]
|
u
100
(
x
)
-
u
T
(
x
)
|
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1249
0.0000752
0.0000754
0.0000754
8.11621
×
1
0
-
7
2
×
1
0
-
7
2.74983
×
1
0
-
7
0.2431
0.0013039
0.0013043
0.0013037
3.70493
×
1
0
-
6
4
×
1
0
-
7
1.13843
×
1
0
-
7
0.3806
0.0080242
0.0080249
0.0080244
7.8819
×
1
0
-
6
7
×
1
0
-
7
2.32339
×
1
0
-
7
0.4195
0.0116531
0.0116538
0.0116533
8.85202
×
1
0
-
6
7
×
1
0
-
7
2.59035
×
1
0
-
7
0.5
0.0220970
0.0220978
0.0220972
1.01895
×
1
0
-
5
8
×
1
0
-
7
2.94629
×
1
0
-
7
0.6923
0.0588207
0.0588201
0.0588209
8.5994
×
1
0
-
6
6
×
1
0
-
7
2.44068
×
1
0
-
7
0.7854
0.0723723
0.0723726
0.0723724
5.68292
×
1
0
-
6
4
×
1
0
-
7
1.60302
×
1
0
-
7
0.8917
0.0646361
0.0646363
0.0646366
1.96562
×
1
0
-
6
2
×
1
0
-
7
5.51306
×
1
0
-
7
1.0
0.0
0.0
0.0
0.0
0.0
0.0Table 3
The numerical results of Example13.
x
u
T
′
(
x
)
u
100
′
(
x
)
|
u
100
′
(
x
)
-
u
T
′
(
x
)
|
u
T
′
′
(
x
)
u
100
′
′
(
x
)
|
u
100
′
′
(
x
)
-
u
T
′
′
(
x
)
|
0.0
0
0
0
0
0
0
0.1
0.0012491
0.00125306
3.95623
×
1
0
-
6
0.0419792
0.0420291
4.98366
×
1
0
-
5
0.2
0.0121642
0.0121721
7.93764
×
1
0
-
6
0.193196
0.193223
2.63540
×
1
0
-
5
0.3
0.0421473
0.0421562
8.95432
×
1
0
-
6
0.410381
0.410375
6.05312
×
1
0
-
6
0.4
0.0930975
0.0931043
6.87917
×
1
0
-
6
0.591978
0.591944
3.40825
×
1
0
-
5
0.5
0.15468
0.154682
2.49761
×
1
0
-
6
0.596621
0.59657
5.13938
×
1
0
-
5
0.6
0.200775
0.200773
2.91343
×
1
0
-
6
0.250969
0.250915
5.41587
×
1
0
-
5
0.7
0.186533
0.186526
7.76184
×
1
0
-
6
−0.645692
−0.645732
3.97775
×
1
0
-
5
0.8
0.0457947
0.0457844
1.023370
×
1
0
-
5
−2.31836
−2.31836
6.35516
×
1
0
-
6
0.9
−0.311216
−0.311224
8.34947
×
1
0
-
6
−5.01403
−5.01398
4.75597
×
1
0
-
5
1.0
−1
−1
6.12843
×
1
0
-
14
−9
−8.99988
1.23122
×
1
0
-
5Table 4
The numerical results of Example13.
x
u
T
(
3
)
(
x
)
u
100
(
3
)
(
x
)
|
u
100
(
3
)
(
x
)
-
u
T
(
3
)
(
x
)
|
u
T
(
4
)
(
x
)
u
100
(
4
)
(
x
)
|
u
100
(
4
)
(
x
)
-
u
T
(
4
)
(
x
)
|
0.1
0.971215
0.971117
9.72528
×
1
0
-
5
11.8289
11.8244
4.50235
×
1
0
-
3
0.2
1.97221
1.9719
3.12435
×
1
0
-
4
7.04361
7.0429
7.15499
×
1
0
-
4
0.3
2.19979
2.19947
3.16007
×
1
0
-
4
−3.23499
−3.2345
4.83768
×
1
0
-
4
0.4
1.19534
1.19511
2.34524
×
1
0
-
4
−17.4321
−17.431
1.09059
×
1
0
-
3
0.5
−1.39212
−1.39222
1.05489
×
1
0
-
4
−34.8029
−34.8014
1.46428
×
1
0
-
3
0.6
−5.85595
−5.8559
5.44543
×
1
0
-
5
−54.8995
−54.8978
1.72026
×
1
0
-
3
0.7
−12.4526
−12.4524
2.36294
×
1
0
-
4
−77.4172
−77.4153
1.90772
×
1
0
-
3
0.8
−21.4126
−21.4122
4.34553
×
1
0
-
4
−102.132
−102.13
2.05168
×
1
0
-
3
0.9
−32.9466
−32.9459
6.45658
×
1
0
-
4
−128.873
−128.871
2.16643
×
1
0
-
3
1.0
−47.25
−47.2491
9.67158
×
1
0
-
4
−157.5
−157.498
2.26074
×
1
0
-
3


## 6. Conclusions and Remarks

In this paper, a new reproducing kernel space satisfying mixed boundary value conditions is constructed skillfully. This makes it easy to solve such kind of problems. Furthermore, the exact solution of the problem can be expressed in series form. The numerical results demonstrate that the new method is quite accurate and efficient for singular problems of fifth-order ordinary differential equations. All computations have been performed using the Mathematica 7.0 software package.



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*Source: 925192-2013-07-11.xml*