using reproducing kernel for solving a class of singular weakly nonlinear boundary value problems

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Using reproducing kernel for solvinga class of singular weakly nonlinearboundary value problemsYulan Wang a , Temuer Chaolu b a & Zhong Chen ca Department of Mathematics, Inner Mongolia University ofTechnology, Hohhot, People's Republic of Chinab Institute of Applied Mathematics and Physics of ShanghaiMaritime University, Shanghai, People's Republic of Chinac Department of Mathematics, Harbin Institute of Technology,WeiHai, ShanDong, People's Republic of ChinaVersion of record first published: 27 Sep 2008.

To cite this article: Yulan Wang , Temuer Chaolu & Zhong Chen (2010): Using reproducing kernelfor solving a class of singular weakly nonlinear boundary value problems, International Journal ofComputer Mathematics, 87:2, 367-380

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International Journal of Computer MathematicsVol. 87, No. 2, February 2010, 367–380

Using reproducing kernel for solving a class of singular weaklynonlinear boundary value problems

Yulan Wanga*, Temuer Chaolub,a and Zhong Chenc

aDepartment of Mathematics, Inner Mongolia University of Technology, Hohhot, People’s Republic ofChina; bInstitute of Applied Mathematics and Physics of Shanghai Maritime University, Shanghai,People’s Republic of China; cDepartment of Mathematics, Harbin Institute of Technology, WeiHai,

ShanDong, People’s Republic of China

(Received 21 February 2007; revised version received 28 July 2007; accepted 29 February 2008)

Cui et al. [M. Cui and F. Geng, Solving singular two point boundary value problems in reproducingkernel space, J. Comput. Appl. Math. 205 (2007), pp. 6–15; H. Yao and M. Cui, A new algorithm for aclass of singular boundary value problems, Appl. Math. Comput. 186 (2007), pp. 1183–1191] presents analgorithm to solve a class of singular linear boundary value problems in the reproducing kernel space. Inthis paper, we will present three new algorithms to solve a class of singular weakly nonlinear boundaryvalue problems in reproducing kernel space. The algorithms are efficiently applied to solving some modelproblems. It is demonstrated by the numerical examples that those algorithms are highly accurate.

Keywords: exact solution; approximate solution; singular boundary value problem; weakly nonlinearproblems; reproducing kernel space

2000 AMS Subject Classification: 34B16; 65L10

1. Introduction

In applied mathematics, many problems lead to the following singular boundary value problem

p(x)u′′(x) + 1

q(x)u′(x) + 1

r(x)u(x) = N(u(x)) + f (x), 0 < x < 1

u(0) = 0, u(1) = 0 (1)

where the functions 1/q(x), 1/r(x) ∈ C(0, 1), q(0)q(1) = 0, r(0)r(1) = 0, N(u(x)), p(x),f (x) ∈ W 1

2 [0, 1], N(u) is a nonlinear function of u, u ∈ W 32 [0, 1] is the unknown function to

be determined. In recent years, there has been a growing interest in the singular boundary valueproblems. Many authors present methods for solving singular boundary value problems by usingTAGE methods, Wavelet–Galerkin, fixed point theorem, spline functions, radial basis functions(RBFs) [1,3,5–17,19]. In [4,18], Cui et al. present an algorithm to solve singular linear boundary

*Corresponding author. Email: [email protected]

ISSN 0020-7160 print/ISSN 1029-0265 online© 2010 Taylor & FrancisDOI: 10.1080/00207160802047640http://www.informaworld.com

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368 Y. Wang et al.

value problems in the reproducing kernel space. In this paper, we will present three new algorithmsto solve weakly nonlinear problems in reproducing kernel space. The algorithms are efficientlyapplied to solving some model problems. It is demonstrated by the numerical examples that therealgorithms are highly accurate.

2. Several reproducing kernel spaces

Definition 2.1 Let H be a real Hilbert spaces of functions f : � → R. A functionK : � × � → R is called reproducing kernel for H is

(i) K(x, ·) ∈ H for all x ∈ �,(ii) f (x) = 〈f, K(·, x)〉H for all f ∈ H and all x ∈ �.

Definition 2.2 A real Hilbert spaces H of functions on a set � is called a reproducing kernelHilbert space if there exists a reproducing kernel K of H .

It is known that the reproducing kernel of a Hilbert space is unique and that existence of areproducing kernel is due to the Riesz representation theorem. The reproducing kernel K of aHilbert space H completely determines the space H . For the solution of the above boundary valueproblem, we give two reproducing kernel spaces.

(i) Space W 12 [0, 1] is defined by

W 12 [0, 1] = {u|u is one-variable absolutely continuous function, u′ ∈ L2[0, 1]}.

The inner product and the norm in W 12 [0, 1] are defined, respectively, by

〈u(x), v(x)〉W 12

=∫ 1

0u(x)v(x) + u′(x)v′(s)dx, u, v ∈ W 1

2 [0, 1].

and ‖u‖W 12

= 〈u(x), u(x)〉1/2W 1

2, u ∈ W 1

2 [0, 1]. In [18], the authors had proved that W 12 [0, 1] is

a reproducing kernel space and its reproducing kernel is given by

R{1}x (y) = 1

2 sinh(1)[cosh(x + y − 1) + cosh(|x − y| − 1)]. (2)

As in [18], we give reproducing kernel space W 32 [0, 1].

(ii) Space W 32 [0, 1] is defined by

W3[0, 1] = {u|u, u′, u′′ is absolutely continuous function,

u(0) = u(1) = 0, u′′′ ∈ L2[0, 1]}.The inner product and the norm in W 3

2 [0, 1] are defined respectively, by

〈u(x), v(x)〉W 32

=∫ b

a

u(x)v(x) + 3u′(x)v′(x) + 3u′′(x)v′′(x) + u′′′(x)v′′′(x)dx,

‖u‖W 32

= 〈u(x), u(x)〉1/2W 3

2, u, v ∈ W 3

2 [0, 1].W 3

2 [0, 1] is a complete reproducing kernel space and its reproducing kernel is given by

Rx(y) ={

R1(x, y), y < x

R2(x, y), x < y,(3)

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International Journal of Computer Mathematics 369

where R1(x, y) = e−x−y(225e6+2y(3 + x2 + x(3 − 2y) − 3y + y2) + 225e2x(3 + x2 + 3y +y2 − x(3 + 2y)) + 15e2(2+x+y)(−187 + 101y − 19y2 + x(101 − 28y + 2y2) + x2(−19 + 2y +2y2)) + 15e2(17 − 25y − 7y2 − 5x(5 − 4y + 2y2) + x2(−7 − 10y + 2y2)) + 15e6(−15(3 +3y + y2) + x2(−15 − 6y + 2y2) − 3x(15 + 4y + 2y2)) + 15e2(x+y) (− 15(3 − 3y + y2) + 3x

(15 − 4y + 2y2) + x2(−15 + 6y + 2y2)) − 2e4+2y (−186 − 459y + 263y2 + x2(261 − 14y +16y2) + x(465 − 514y − 18y2)) + 2e4(−186 + 465y + 261y2 + x2(261 + 140y − 26y2) + 5x

(93 + 8y + 28y2)) + 10e2+2x(−312 − 249y − 85y2 + x2(−75 − 14y + 4y2) + x(279 + 134y +6y2)) − 2e2(1+x+y) (− 15(104 − 93y + 25y2) + x2(− 375 + 88y + 26y2) + x(1395 − 496y +88y2)) + 5e2+2y (− 51 − 15y + 41y2 + x2(21 + 12y + 8y2) + x(75 − 114y − 28y2)) + e4+2x

(2805 + 1503y + 289y2 + x2(285 + 36y − 32y2) + x(−1515 − 594y + 28y2)))/(16(−225 +955e2 − 811e4 + 225e6)).

R2(x, y) = e−x−y(225e2y(3 + x2 + x(3 − 2y) − 3y + y2) + 225e6+2x (3 + x2 + 3y + y2 −x(3 + 2y)) + 15e2(2+x+y)(−187 + 101y − 19y2 + x(101 − 28y + 2y2) + x2(−19 + 2y + 2y2))

+ 15e2(17 − 25y − 7y2 − 5x(5 − 4y + 2y2) + x2(− 7 − 10y + 2y2)) + 15e6(− 15(3 + 3y +y2) + x2(− 15 − 6y + 2y2) − 3x(15 + 4y + 2y2)) + 15e2(x+y)(−15(3 − 3y + y2) + 3x(15 −4y + 2y2) + x2(−15 + 6y + 2y2)) + e4+2y(15(187 − 101 + 19y2) + x2(289 + 28y − 32y2) +9x(167 − 66y + 4y2)) + 2e4(− 186 + 465y + 261y2 + x2(261 + 140y − 26y2) + 5x(93 + 8y

+ 28y2)) + 10e2+2y(−312 + 279y − 75y2 + x(−249 + 134y + 14y2) + x2(−85 + 6y + 4y2))

− 2e2(1+x+y)(−15(104 − 93y + 25y2) + x2(−375 + 88y + 26y2) + x(1395 − 496y + 88y2))

+ 5e2+2x(3x(−5 − 38y + 4y2) + x2(41 − 28y + 8y2) + 3(−17 + 25y + 7y2)) − 2e4+2x(−186+ 465y + 261y2) + x2 (263 − 18y + 16y2) − x(459 + 514y + 14y2)))/(16(− 225 + 955e2

− 811e4 + 225e6)).

3. The exact solution of Equation (1)

In order to solve Equation (1), we define the linear operator L : W 32 [0, 1] → W 1

2 [0, 1]

(Lu)(x)def= p(x)u′′(x) + 1

q(x)u′(x) + 1

r(x)u(x).

Equation (1) can be equivalently turned into the following form:

(Lu)(x) = N(u(x)) + f (x). (4)

Let ϕi(x) = R{1}xi

(x), ri(x) = Rxi(x), where {xi}∞i=1 is dense in the interval [0, 1], and

ψi(x) = (L∗ϕi)(x) = 〈(L∗yR

{1}xi

(y))(s), Rx(s)〉W 32

= 〈R{1}xi

(y), (LsRx(s))(y)〉W 32

= (LsRx(s))(xi), i = 1, 2, . . . ,

where L∗ is a conjugate operator L and the symbol Ls indicates that the operator L applies to thefunction of s. Practice Gram–Schmidt orthonomalization for {ψi(x)}∞i=1

ψi(x) =i∑

k=1

βikψk(x), (5)

where βik are the coefficients of Gram–Schmidt orthonomalization and {ψi(x)}∞i=1 is anorthonormal system.

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Theorem 3.1 Let {xi}∞i=1 is dense in [0, 1]. {ψi(x)}∞i=1 is a complete function system in W 32 [0, 1]

if L−1 is existent.

Proof For each fixed u(x) ∈ W 32 [0, 1], if 〈u(x), ψi(x)〉W 3

2= 0, then

〈u(x), ψi(x)〉W 32

= 〈Lu(x), ϕi(x)〉W 12

= Lu(xi) = 0. (6)

Taking into account the density of {xi}∞i=1, it results in Lu(x) = 0. It follows that u(x) ≡ 0 fromthe existence of L−1. �

Lemma 3.2 Let {xi}∞i=1 be dense on [0, 1] and u(x) is the solution of Equation (1). Then u(x)

satisfies the form

u(x) =∞∑i=1

i∑k=1

βik[N(u(xk)) + f (xk)]ψi(x). (7)

Proof Let u(x) is the solution of Equation (1). Since {ψi(x)}∞i=1 is an orthonormal system, u(x)

is expressed as

u(x) =∞∑i=1

< u(x), ψi(x) >W 32

ψi(x)

=∞∑i=1

i∑k=1

βik < u(x), ψk(x) >W 32

ψi(x)

=∞∑i=1

i∑k=1

βik < Lu(x), ϕk(x) >W 12

ψi(x)

=∞∑i=1

i∑k=1

βik < N(u(x)) + f (x), ϕk(x) >W 12

ψi(x)

=∞∑i=1

i∑k=1

βik[N(u(xk)) + f (xk)]ψi(x). �

Remark Case (i): Equation (1) is linear, that is, N(u(x)) = 0. Then the exact solution toEquation (1) can be obtained directly from Equation (7).

Case (ii): Equation (1) is nonlinear. In this case, the approximate solution to Equation (1) canbe obtained using Method 1.

4. The approximate solution of Equation (1)

In the section, we present three methods for solving Equation (1).

4.1. Method 1

We denote Equation (7) by

u(x) =∞∑i=1

Aiψi(x),

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where

Ai =i∑

k=1

βik(f (xk) + Nu(xk)).

In fact, Ai is unknown. We will approximate Ai using known Bi .For a numerical computation, we define initial function u1(x) and the n-term approximation to

u(x) by

un+1(x) =n∑

i=1

Biψi(x), (8)

where

B1 = β11(Nu1(x1) + f (x1)), u2(x) = B1ψ1(x),

B2 =2∑

k=1

β2k(Nu2(xk) + f (xk)), u3(x) = B1ψ1(x) + B2ψ2(x),

... (9)

Bn =n∑

k=1

βnk(Nun(xk) + f (xk)), un+1(x) =n∑

k=1

Bkψk(x)

...

Lemma 4.1

‖u‖W 32 [0,1] ≤ M‖u‖C[0,1].

Proof From

u(x) = u(y) +∫ x

y

u′(t)dt, x, y ∈ [0, 1], u(x) ∈ W 32 [a, b], (10)

we have

|u(x)| = |u(y)| +∫ x

y

|u′(t)|dt. (11)

Integrating both sides of Equation (11) for y from 0 to 1 and applying Cauchy inequality, one gets

‖u‖W 32 [0,1] ≥ 2‖u‖C[0,1], �

Lemma 4.2 If un(x)‖·‖

W32−→ u(x)(n → ∞) and xn → y, then

un(xn) → u(y), (n → ∞), (12)

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Proof Since

|un(xn) − u(y)| = |un(xn) − un(y) + un(y) − u(y)|≤ |un(xn) − un(y)| + |un(y) − u(y)|.

From the definition of the reproducing kernel, we have

un(xn) = 〈un(x), Rxn(x)〉, un(y) = 〈un(x), Ry(x)〉. (13)

It follows that

|un(xn) − un(y)| = |〈un(x), Rxn(x) − Ry(x)〉|

≤ ‖un(x)‖W 32‖Rxn

(x) − Ry(x)‖W 32.

From the convergence of un(x), there exists a constant K , such that ‖un(x)‖W 32

≤ 2‖u(x)‖W 32

assoon as n ≥ N . At the same time, we can prove ‖Rxn

(x) − Ry(x)‖W 32

→ 0 as soon as n → ∞using Equation (2). Hence |un(xn) − un(y)| → 0 as soon as xn → y from Lemma 4.1.

On the other hand, for any y ∈ [0, 1], it holds that |un(y) − u(y)|C[0,1] → 0(n → ∞) assoon as ‖un(y) − u(y)‖W 3

2 [0,1] → 0(n → ∞) from Lemma 4.1. Hence, it follows that un(xn) →u(y)(n → ∞) as soon as xn → y. �

By means of the continuation of N(·), it is obtained that N(un(xn)) → N(u(y))(n → ∞), Itshows that

Nun(xn) −→ Nu(y), (n → ∞). (14)

The convergence theorem is established for the method mentioned above.

Theorem 4.3 Assume ‖un(x)‖W 32

is bounded in Equation (8), if {xi}∞i=1 is dense in [0, 1], thenthe n-term approximate solution un(x) converges to the exact solution u(x) of Equation (1) andthe exact solution is expressed as

u(x) =∞∑i=1

Biψi(x), (15)

where Bi is given by Equation (9).

Proof (i) We prove the convergence of Equation (8). From Equation (8), one gets

un+1(x) = un(x) + Bnψn(x). (16)

Using the orthogonality of {ψi}∞i=1, it follows that

‖un+1(x)‖2W 3

2= ‖un(x)‖2

W 32+ (Bn)

2.

The sequence ‖un(x)‖W 32

is monotone decreasing. Due to ‖un(x)‖W 32

being bounded,{‖un(x)‖W 3

2} is convergent as soon as n → ∞. Then there exists a constant c such that

∞∑i=1

(Bi)2 = c. (17)

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It implies that

Bi =i∑

k=1

βik(f (xk) + Nui(xk)) ∈ l2, i = 1, 2, . . . . (18)

If m > n, using the orthogonality of un+1(x) − un(x), n = 2, 3, . . . , then one gets

‖um(x) − un(x)‖2W 3

2= ‖um(x) − um−1(x) + um−2(x) + · · · + un+1(x) − un(x)‖2

W 32

= ‖um(x) − um−1(x)‖2W 3

2+ · · · + ‖un+1(x) − un(x)‖2

W 32

=m∑

i=n+1

(Bi)2 → 0 (n → ∞).

Considering the completeness of W 32 [0, 1], we get

un(x)‖·‖

W32−−−→ u(x) (n → ∞).

Hence

u(x) =∞∑i=1

Biψi(x). (19)

(ii) Define the projection operator

Pnu(x) =n∑

i=1

Biψi(x).

Then

un+1(x) = Pnu(x). (20)

We can prove

un+1(xk) = u(xk), k ≤ n. (21)

In fact, it follows that

un+1(xk) = 〈un+1(x), ψk(x)〉= 〈Pnu(x), ψk(x)〉= 〈u(x), Pnψk(x)〉= 〈u(x), ψk(x)〉= u(xk).

Hence

Nun+1(xk) = Nu(xk), k ≤ n. (22)

(ii) We will prove that u(x) is the solution of Equation (1).

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From Equation (19), one gets

u(xj ) =∞∑i=1

Bi〈ψi(x), ψj (x)〉. (23)

Multiplying by βnj on both sides of Equation (23) and summing for j from 1 to n, we get

n∑j=1

βnj u(xj ) =∞∑i=1

Bi

⟨ψi(x),

n∑j=1

βnjψj

⟩=

∞∑i=1

Bi〈ψi(x), ψn〉 = Bn. (24)

From Equations (8) and (24), if n = 1, then

u(x1) = B1 = f (x1) + Nu1(x1).

If n = 2, then

β21u(x1) + β22u(x2) = B2

= β21(f (x1) + Nu2(x1)) + β22(f (x2) + Nu2(x2))

= β21(f (x1) + Nu(x1)) + β22(f (x2) + Nu2(x2))

= β21u(x1) + β22(f (x2) + Nu2(x2)).

We get

u(x2) = f (x2) + Nu2(x2). (25)

In the same way, we have

u(xn) = f (xn) + Nun(xn). (26)

For any y ∈ [0, 1], there exists a subsequence {xnk}∞k=1 converging to y since {xi}∞i=1 is dense

in [0, 1]. From Lemma 4.2 and the above form, it holds that

u(y) = f (y) + Nu(y). (27)

That is, u(x) is the solution of Equation (1) and

u(x) =∞∑i=1

Biψi(x). �

Theorem 4.4 Assume u(x) is the solution of Equation (1) and rn(x) is the error in the approx-imate solution un+1(x), where un+1(x) is given by Equation (8). Then the error rn is monotonedecreasing in the sense of ‖ · ‖W 3

2.

Proof Suppose that u(x) and un+1(x) are given by Equations (15) and (8) respectively. We have

‖rn(x)‖2W 3

2= ‖u(x) − un+1(x)‖2

W 32

=∥∥∥∥∥

∞∑i=n+1

Biψi(x)

∥∥∥∥∥2

W 32

(28)

=∞∑

i=n+1

(Bi)2

It shows that the error rn is monotone decreasing in the sense of ‖ · ‖W 32. �

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4.2. Method 2

Theorem 4.5 If {xi}∞i=1 is dense on [0, 1], then {ri(x)}∞i=1 is a complete function system inW 3

2 [0, 1].

Proof For each fixed u(x) ∈ W 32 [0, 1], if 〈u(x), ri(x)〉 = 0, then

〈u(x), ri(x)〉 = u(xi) = 0. (29)

Taking into account the density of {xi}∞i=1, it results in u(x) = 0. �

Using reproducing kernel for u(x) and Lu(x) as follows:

u(x) =∞∑i=1

ciri(x), (30)

Lu(x) =∞∑i=1

ciLri(x), (31)

where ci, i = 1, 2, . . . , is the unknown number to be determined. Substituting form (31) intoEquation (1), we get

∞∑i=1

ciLri(x) = N(u(x)) + f (x). (32)

If Equation (1) is linear, that is N(u(x)) = 0. Let

um(x) =m∑

i=1

ciri(x). (33)

where the coefficients ci, i = 1, . . . , m, are determined by equations

m∑i=1

ciLri(x)|x=xj= f (xj ), j = 1, 2, . . . , m. (34)

If Equation (1) is nonlinear, we give u1(x), m, n and let

un,m(x) =∞∑i=1

ci,n,mri(x), n = 2, 3, . . . . (35)

where the coefficients ci,n,m, i = 1, . . . , m, n = 2, 3, . . . are determined by equations

m∑i=1

ci,n,mLri(x)|x=xj= N(un−1,m(x))|x=xj

+ f (xj ),

j = 1, 2, . . . , m, n = 2, 3, . . . . (36)

Using strictly positive definite property of the reproducing kernel [2], we get Theorem 4.6.

Theorem 4.6 If L−1 is existent, then the solution of Equations (36) and (37) is existence anduniqueness.

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Using Equations (33) and (34), we have (Lum)(xj ) = f (xj ), j = 1, 2, . . . , m, So um(x) is theapproximation solution of linear Equation (1).

Next, we will prove the convergence of un,m(x).

Lemma 4.7 A = {un,m| ‖un,m‖W 32 [0,1] ≤ λ} is a compact set in space C[0, 1], where λ is a

constant.

Proof From the definition of the reproducing kernel, one gets

‖Rx(y)‖ = 〈Rx(y), Rx(y)〉 = Rx(x) < λ1.

For any un,m ∈ A, we have

|un,m(x)| = |〈un,m(y), Rx(y)〉| ≤ ‖un,m(y)‖(Rx(x))1/2 ≤ λλ1/21 , (37)

where λ1 is a constant. Hence, A is a bounded set in space C[0, 1]. On the other hand, for anyun,m ∈ A and ε > 0, it holds that

|un,m(x + h) − un,m(x)| = |〈un,m(y), R(x+h)(y) − Rx(y)〉|≤ ‖un,m(y)‖ × ‖R(x+h)(y) − Rx(y)‖

≤ λ ·(∥∥∥∥∥∂Rx(y)

∂x

∣∣∣∣x=μ

∥∥∥∥∥ · |h|)

, μ ∈ [x, x + h].

From the definition of Rx(y), we have ∥∥∥∥∂Rx(y)

∂x

∥∥∥∥ ≤ λ2

where λ2 is a constant. It exists δ > 0, take δ = ε/λλ2. When |h| < δ, we get

|un,m(x + h) − un,m(x)| < ε

It shows that the function un,m(x) is equicontinuous in set A. Consequently, A is a compact setin space C[0, 1]. �

Theorem 4.8 Assume ‖un,m(x)‖ is bounded in Equations (35) and (36), if {xi}∞i=1 is dense in[0, 1] and L(u) is a invertible continuous function of u, N(u) is a continuous function of u, thenthe exact solution u(x) and the approximate solution for Equation (1) are existent.

Proof Using Equations (35) and (36), we have

(Lun,m)(xj ) = N(un−1,m(x))|x=xj+ f (xj ), j = 1, 2, . . . , m, n = 2, 3, . . . . (38)

From Lemma 4.7, it is known that A = {un,m(x)| ‖un,m(x)‖ ≤ λ} is a compact set. Sothere exists a convergent subsequence {unl,m(x)}∞l=1 of {un,m(x)}∞n=1, such that unl,m(x) → u(x)

(l → ∞, m → ∞) in space C[0, 1], and

(Lunl,m)(xj ) = N(unl−1,m(x))|x=xj+ f (xj ), j = 1, 2, . . . , m, n = 2, 3, . . . . (39)

It’s easy to see that if L(u) is a invertible continuous function of u, N(u) is a continuous functionof u, then after taking limits for both sides of Equation (39), so u(x) is the exact solution ofEquation (1). So un1,m(x) is the approximate solution of Equation (1). �

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Theorem 4.9 Assume u(x) is the solution of Equation (1) and enl,m(x) is the error in theapproximate solution unl,m(x), where unl,m(x) is given by Theorem 4.8. Then the absolute error|enl,m(x)| ≤ (

∑∞i=m+1(ci,nl ,m)2)(1/2)(

∑∞i=m+1(ri(x))2)(1/2).

Proof Suppose that u(x) and unl,m(x) are given by Theorem 4.8, respectively. We have

|en1,m(x)| = |u(x) − unl,m(x)|

=∣∣∣∣∣

∞∑i=m+1

ci,nl ,mri(x)

∣∣∣∣∣≤

( ∞∑i=m+1

(ci,nl ,m)2

)(1/2) ( ∞∑i=m+1

(ri(x))2

)(1/2)

. (40)�

Similarly, We get Method 3.

4.3. Method 3

Using theme (4) for u(x) and Lu(x) as follows, we have

u(x) =∞∑i=1

diψi(x), (41)

Lu(x) =∞∑i=1

diLψi(x), (42)

where di, i = 1, 2, . . . , is the unknown number to be determined. Substituting form (42) inEquation (1), we get

∞∑i=1

diLψi(x) = (Nu)(x) + f (x). (43)

Similarly, if Equation (1) is linear, that is, (Nu)(x) = 0. Then the approximation solution toEquation (1) can be obtained using the following method

um(x) =m∑

i=1

diψi(x), (44)

where the coefficients di, i = 1, . . . , m, are determined by equations

m∑i=1

diLψi(x)|x=xj= f (xj ), j = 1, 2, . . . , m. (45)

If Equation (1) is nonlinear, in this case, the approximation solution to Equation (1) can beobtained using a finite section and iterative method. We give u1(x), k, l and define the k-termapproximation, l-times iteration to u(x) by

ul,k(x) =k∑

i=1

di,l,kψi(x), l = 2, 3, . . . . (46)

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378 Y. Wang et al.

where the coefficients di,l,k, i = 1, . . . , k, l = 2, 3, . . . , are determined by equations:

k∑i=1

di,l,kLψi |x=xj= N(ul−1,k(x))|x=xj

= f (xj ), j = 1, 2, . . . , k, l = 2, 3, . . . . (47)

Similarly, we can get Theorems 4.10 and 4.11.

Theorem 4.10 If L−1 is existent, then the solution of Equations (45) and (47) is existence anduniqueness.

Theorem 4.11 Assume ‖ul,k(x)‖ is bounded in Equations (46) and (47), if {xi}∞i=1 is dense in[0, 1], L(u) is an invertible continuous function of u and N(u) is a continuous function of u, thenthe exact solution u(x) and the approximate solution ul,k(x) for Equation (1) are existent.

5. Numerical experiment

Example 1 Consider the following singular boundary values problem [4]:

u′′(x) + 1

xu′(x) + u(x) = f (x), x ∈ (0, 1),

wheref (x) = 4 − 9x + x2 − x3, with the boundary conditionsu(0) = u(1) = 0.u(x) = x2 − x3

is the true solution.

Example 2 Consider the following singular nonlinear boundary values problem:

u′′(x) + 1√x

u′(x) + 1

xu(x) = f (x) + sin(u2(x)) + eu9(x) − u11(x), x ∈ (0, 1),

where f (x) = 1/√

x((2√

x + x − 4x√

x − x2) cos x + (1 − √x − 2x − 2x

√x + x2√x sin x))

− sin(((x − x2) sin x)2) − e((x−x2) sin x)9 + ((x − x2) sin x)11, with the boundary conditionsu(0) = u(1) = 0. The true solution is u(x) = (x − x2) sin x. Using our method, let u1(x) = 0.

Example 3 Consider the following singular boundary values problem [18]:

εu′′(x) + 1

xu′(x) + (1 + x2)u(x) = f (x),

where f (x) = (2ex2x(150001 + 50002x2) − ex(100000 + 100001x + 100000x2))/100000x

with the boundary conditions u(0) = u(1) = 0, the true solution is u(x) = ex2 − ex .

Example 4 Consider the following singular boundary values problem:

εu′′(x) + 1

xu′(x) + (1 + x2)u(x) = (u(x))9 + f (x), x ∈ (0, 1),

where f (x) = −2ε + (1 − x) + (1 + x2)(x − x2) − (x − x2)9, with the boundary conditionsu(0) = u(1) = 0. The true solution is u(x) = (x − x2). Using our method, let u1(x) = 0.In calculation, let xj = j (1/N), j = 0, 1, 2, . . . , N, N is positive integer.

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Table 1. Comparison of the absolute error (Example 1).

N = 26 N = 51

Node x [4] Method 2 [4] Method 2 Method 3

0.08 2.2E−6 4.28232E−5 4.8E−7 6.22399E−6 1.97618E−40.16 1.1E−5 5.23266E−5 2.7E−6 7.28666E−6 2.31897E−40.32 5.5E−5 6.08702E−5 1.3E−5 8.26185E−6 2.44972E−40.48 1.1E−4 6.51065E−5 2.9E−5 8.73036E−6 2.17772E−40.64 1.6E−4 6.75371E−5 4.1E−5 8.98548E−6 1.6299E−40.80 1.5E−4 6.9121E−5 3.9E−5 9.13935E−6 9.17922E−40.96 4.7E−5 7.81774E−5 1.1E−5 9.19078E−6 1.73727E−5

Table 2. Absolute error of Example 2 (10 times iteration).

N = 40 N = 80 N = 120

Node x Method 1 Method 2 Method 3 Method 2 Method 2


Table 3. Comparison of the absolute error (Example 3).

N = 200 N = 260

Node x [15] [18] Method 2 Method 3 Method 3

1/8 3.18851E−3 9.17222E−7 4.83975E−8 1.68984E−81/16 8.381E−2 3.7987E−4 8.26781E−7 3.77595E−8 1.17483E−81/32 8.014E−2 4.10187E−5 7.52859E−7 3.22602E−8 9.20198E−91/64 7.772E−2 4.02599E−5 9.16553E−7 2.98198E−8 7.91241E−91/128 7.221E−2 3.47918E−7 8.13045E−7 2.87733E−8 7.04046E−91/256 6.655E−2 2.99306E−8 9.31758E−7 2.99769E−8 7.94164E−91/512 5.876E−2 3.9648E−9 6.38449E−7 1.96443E−8 5.8722E−9

Table 4. Absolute error of Example 4 (N = 80, 10 times iteration).

ε = 10−5 ε = 10−10

Node x Method 1 Method 2 Method 3 Method 2 Method 3


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380 Y. Wang et al.

6. Conclusions and remarks

In this paper, the exact solution is given for the above boundary value problem in reproducingkernel space. The exact solution is represented in the form of series. For a computation, wepresent three new methods for solving the above boundary value problem in reproducing kernelspace. Tables 1–4 show that our methods are effective. It is worthy to note that the methodcan be employed to solve the equation L(u)(x) = N(u)(x) + f (x) in reproducing kernel space,where L(u) is a invertible linear continuous function of u and N(u) is a nonlinear continuousfunction of u.

Acknowledgements

The author thanks the reviewers for their valuable suggestions, which greatly improved the paper. This paper is supportedby the key project of Inner Mongolia University of Technology (No. ZD200709), the key project of Inner Mongolia(No. NJZY07066) and the NSF of China (No. 10461005).

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using reproducing kernel for solving a class of singular weakly nonlinear boundary value problems

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