# The Solution of a Class of Singularly Perturbed Two-Point Boundary Value Problems by the Iterative Reproducing Kernel Method

Authors: Zhiyuan Li; YuLan Wang; Fugui Tan; Xiaohui Wan; Tingfang Nie
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

In (Wang et al., 2011), we give an iterative reproducing kernel method (IRKM). The main contribution of this paper is to use an IRKM (Wang et al., 2011), in singular perturbation problems with boundary layers. Two numerical examples are studied to demonstrate the accuracy of the present method. Results obtained by the method indicate that the method is simple and effective.


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## Body

## 1. Introduction

Singularly perturbed problems (SPPs) arise frequently in applications including geophysical fluid dynamics, oceanic and atmospheric circulation, chemical reactions, and optimal control. In this paper, we consider the following singularly perturbed two-point boundary value problem:(1.1)εUxx(x)+a(x)Ux(x)+b(x)U(x)=F(x,U(x)),x∈(0,1),U(0)=α,U(1)=β,
where ε is a positive small parameter, a(x), b(x), and F(x,v) are known functions, and U(x) is a unknown function to be determined. In this paper, we assume that (1.1) has a unique solution that belongs to W23[0,1]. Like in [1–5], we give reproducing kernel spaces W21[a,b] and W23[a,b]. (i) We define the inner product space W21[a,b]={u|uisone-variableabsolutelycontinuousfunction,u′∈L2[a,b]}. The inner product is given by 〈u(x),v(x)〉W21=u(a)v(a)+∫ab‍u′(x)v′(x)dx. The space W21[a,b] is a reproducing kernel space, and its reproducing kernel is Rx{1}(y). (ii) Space W23[a,b]={u|u,u′,u′′isone-variableabsolutelycontinuousfunction,u(a)=u(b)=0,u′′′∈L2[a,b]}.The inner product is given by〈u(x),v(x)〉W23=u′(a)v′(a)+u′′(a)v′′(a)+∫ab‍u′′′(x)v′′′(x)dx. The space W23[a,b] is a reproducing kernel space, and its reproducing kernel is Rx{2}(y).


## 2. Iterative Reproducing Kernel Method (IRKM)

In order to solve (1.1), we first give the analytical and approximate solutions of the following operator equation:
(2.1)(Lu)(x)=F(x,u(x)),
where L:H[a,b]→H1[a,b] is a bounded linear operator and L-1 is existent. H1[a,b] is an RKHS with the reproducing kernel K¯(x,t), H[a,b] is also an RKHS with the reproducing kernel K(x,t).Theorem 2.1.
IfL-1 is existent and {xi}i=1∞ are countable dense points in [a,b], Letting ψ¯i(x)=∑k=1i‍βikψk(x), where the βik are the coefficients resulting from the Gram-Schmidt orthonormalization, ψi(x)=(LyK(x,y))(xi), i=1,2,…, then
(2.2)u(x)=∑i=1∞‍∑k=1i‍βikf(xk,u(xk))ψ¯i(x)
is an analytical solution of (2.1).Proof.
u
(
x
) can be expanded to the Fourier series in terms of normal orthogonal basis {ψ¯i(x)}i=1∞ in H[a,b]:
(2.3)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,u(xk))ψ¯i(x).(i) Linear Problem
Suppose (2.1) is a linear problem, that is, f(x,u)=F(x). We define an approximate solution un(x) by
(2.4)un(x)=∑i=1n‍∑k=1i‍βikF(xk)ψ¯i(x).Theorem 2.2 (convergence analysis).
Letεn2=∥u(x)-un(x)∥2; then the sequence of real numbers εn is monotonously decreasing and εn→0 and the sequence un(x) is convergent uniformly to u(x), k=0,1,2.Proof.
We have(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,
and clearly εn-1≥εn and consequently {εn} is monotone decreasing in the sense of ∥·∥. By Theorem 2.1, we know that ∑i=1∞‍〈u(x),ψ¯i(x)〉ψ¯i(x) is convergent in the norm of ∥·∥, then we have εn2=∥u(x)-un(x)∥2→0.
For anyx∈[a,b], k=0,1,2,
(2.6)|un(k)(x)-u(k)(x)|=|〈un(t)-u(t),∂kK(x,t)∂xk〉|≤‖un(t)-u(t)‖⋅‖∂kK(x,t)∂xk‖,
and by the expression of K(x,t), there exists Ck>0, such that ∥∂kK(x,t)/∂xk∥<Ck; thus
(2.7)|un(k)(x)-u(k)(x)|≤Ck‖un(t)-u(t)‖=Ckεn⟶0.(ii) Nonlinear Problem
Suppose that (1.1) is a nonlinear problem, that is, f(x,u)=N(u)+F(x), where N:H[a,b]→H1[a,b] is a nonlinear operator, and we give an iterative sequence un(x):u0,*(x) is the solution of the linear equation Lu=F(x),un+1,*(x) is the solution of the linear equation Lu=N(un,*)+F(x), n=0,1,2….Lemma 2.3.
Ifun,*(x)→u(x), then u(x), is the solution of (1.1).Theorem 2.4.
Suppose that the nonlinear operatorA≜(L-1N):H1[a,b]→H[a,b] satisfies the contractive mapping principle, that is,
(2.8)‖A(u)-A(v)‖≤λ‖u-v‖,λ<1;
then un,*(x) is convergent.


## 3. Solution of Singularly Perturbed Problems

We notice that a small variation in the parameterε produces a large variation in the solution. In other words, we are treating an ill-posed problem. In this paper, by dividing the domain [0,1] into three subdomains [0,d], [d,1-d], and [1-d,1].(i) Outer Region
We have(3.1)εUxx(x)+a(x)Ux(x)+b(x)U(x)=F(x,U(x)),x∈[d,1-d],u(d)=p,u(1-d)=q.
Letting u¯(x)=U(x)-((x-d)/(1-2d))q+((x+d-1)/(1-2d)) and (L1u¯)(x)=εu¯xx(x)+a(x)u¯x+b(x)u¯, (3.1) can further be converted into
(3.2)(L1u¯)(x)=f¯(x,u¯(x)),
where f¯(x,u¯(x))=((x+d-1)/(1-2d))p). Using IRKM, we can get the solution of the outer region problem.(ii) Left Layer
We have(3.3)εUxx(x)+a(x)Ux(x)+b(x)U(x)=F(x,U(x)),x∈[0,d),U(0)=α,U(d)=pisknown.
Letting u(x)=U(x)-α-(x/d)(p-α), x/d=x1, then x=dx1, u(x)=u(dx1)≐u¯(x1), du/dx=(1/d)(du¯/dx1), and d2u/dx2=(1/d2)(d2u¯/dx12). In space W23[0,1], (3.3) can further be converted into following form:
(3.4)(Lεu¯)(x1)=f¯(x1,u¯(x1)),
where f¯(x1,u¯(x1))=d2(F(x,U(x))-(a(x)/d)(p-α)-b(x)(α+(x/d)(p-α))), (Lεu¯)(x1)=εu¯x1x1(x1)+a(x1)du¯x1(x1)+b(x1)d2u¯(x1). Using IRKM, we can get the solution of the inner region (left layer near) problem.(iii) Right Layer
We have(3.5)εUxx(x)+a(x)Ux(x)+b(x)U(x)=F(x,U(x)),x∈(1-d,1],U(1)=β,U(1-d)=q.
Letting u(x)=U(x)-β+((x-1)/d)(q-β), (x/d)-(1/d)+1=x1, then x=dx1+1-d, u(x)=u(dx1+1-d)≐u¯(x1), du/dx=(1/d)(du¯/dx1), and d2u/dx2=(1/d2)(d2u¯/dx12). In space W23[0,1], (3.5) can further be converted into following form:
(3.6)(Lεu¯)(x1)=f¯(x1,u¯(x1)),
where f(x,u(x))=F(x,U(x))+(a(x)/d)(q-β)-b(x)(β-((x-1)/d)(q-β)), q is known (the outer solution has been given), f¯(x1,u¯(x1))=d2f(x,u(x)), and (Lεu¯)(x1)=εu¯x1x1(x1)+a(x1)du¯x1(x1)+b(x1)d2u¯(x1). Using IRKM, we can get the solution of the inner region (right layer near) problem. After solving the inner and outer region problems, we combine their solutions to obtain an approximate solution to the original problem (1.1) over the interval 0≤x≤1.


## 4. Numerical Examples

Example 4.1.
This example is from [6–8]:
(4.1)-εuxx+u=f(x),0≤x≤1,u(0)=0,u(1)=0.
We determine f(x) to get the true solution, the true solution uT(x)=(e(x-1)/ε+e-x/ε)/(1+e-1/ε)-cos2(πx). The numerical results are given in Tables 1, 2, and 3.Table 1
Comparison of results,x∈[0,d], ɛ=2-10, and d=0.1.
x
u
T
(
x
)
u
100
(
x
)
|
u
T
-
u
100
|
|
u
T
-
u
200
|
0
0
0
0
0
0.01
−0.272864
−0.272866
1.8788×10-6
7.48551×10-7
0.02
−0.468765
−0.46877
5.38727×10-6
1.15911×10-7
0.03
−0.608251
−0.608257
6.10734×10-6
1.17267×10-7
0.04
−0.706254
−0.70626
5.54546×10-6
1.01462×10-7
0.05
−0.773632
−0.773636
4.52354×10-6
8.05672×10-7
0.06
−0.818281
−0.818285
3.45181×10-6
6.04003×10-7
0.07
−0.845955
−0.845958
2.49497×10-6
4.30798×10-7
0.08
−0.860849
−0.86085
1.66887×10-6
2.84958×10-7
0.09
−0.866029
−0.86603
8.9114×10-7
1.50666×10-7
0.1
−0.863746
−0.863746
3.33067×10-16
2.22045×10-7Table 2
Comparison of results,x∈[d,1-d], ɛ=2-30, and d=0.001.
x
u
T
(
x
)
u
100
(
x
)
|
u
T
-
u
10
|
|
u
T
-
u
100
|
0.001
−0.99999
−0.99999
1.51212×10-13
4.71756×10-12
0.1008
−0.903026
−0.903045
1.89196× 10-5
7.69723×10-10
0.2006
−0.652715
−0.652709
5.8384×10-6
1.58393×10-10
0.3004
−0.344297
−0.344299
1.66259×10-6
4.10425×10-10
0.4002
−0.0951225
−0.0951222
3.12214×10-7
2.67338×10-10
0.5998
−0.0951225
−0.0951223
1.7619×10-7
6.92579×10-10
0.6996
−0.344297
−0.344298
1.11205×10-6
2.66992×10-10
0.7994
−0.652715
−0.652711
3.96364×10-6
1.77961×10-10
0.8992
−0.903026
−0.903039
1.27772×10-5
6.22121×10-8
0.999
−0.99999
−0.99999
6.67721×10-12
1.91558×10-12Table 3
Comparison of results,x∈[1-d,1], ɛ=2-10, and  d=0.01.
x
u
T
(
x
)
u
10
(
x
)
|
u
T
-
u
10
|
u
200
(
x
)
|
u
T
-
u
200
|
0.991
−0.249439
−0.249431
8.11552×10-6
−0.249439
2.72812×10-9
0.992
−0.225227
−0.225219
7.75753×10-6
−0.225227
3.72799×10-9
0.993
−0.200201
−0.200194
7.17326×10-6
−0.200201
4.18195×10-9
0.994
−0.174338
−0.174331
6.40087×10-6
−0.174338
4.18298×10-9
0.995
−0.147609
−0.147604
5.47897×10-6
−0.147609
3.82517×10-9
0.996
−0.119989
−0.119984
4.44655×10-6
−0.119989
3.20353×10-9
0.997
−0.0914472
−0.0914438
3.34288×10-6
−0.0914472
2.41376×10-9
0.998
−0.0619555
−0.0619533
2.20746×10-6
−0.0619555
1.55216×10-9
0.999
−0.0314835
−0.0314825
1.07993×10-6
−0.0314835
7.1531×10-10
1.
−2.22045×10-16
−3.62743×10-17
1.8577×10-16
−3.62852×10-17
1.85759×10-16Example 4.2.
Considering the following nonlinear singularly perturbed problem with boundary layers(4.2)εuxx+exxux+eu=f(x),0<x≤1,u(0)=0,u(1)=0,
we determine f(x) to get the true solution, the true solution uT(x)=1+(x-1)e-x/ε-xe(x-1)/ε. The numerical results are given in Tables 3 and 4.Table 4
Comparison of results,x∈[0,d], and  d=0.01
x
u
T
(
x
)
u
3,10
(
x
)
|
u
T
-
u
3,10
|
u
T
(
x
)
u
3,10
(
x
)
|
u
T
-
u
3,10
|
ɛ
=
2
-
10
ɛ
=
2
-
10
ɛ
=
2
-
10
ɛ
=
2
-
5
ɛ
=
2
-
5
ɛ
=
2
-
5
0.001
0.0324619
0.0324564
5.56236×10-6
0.00628391
0.0062835
−4.09563×10-7
0.002
0.063871
0.0638658
5.18059×10-6
0.0125208
0.0125204
−3.72485×10-7
0.003
0.0942614
0.0942567
4.67451×10-6
0.0187109
0.0187106
−3.57056×10-7
0.004
0.123666
0.123662
4.20082×10-6
0.0248545
0.0248541
−3.30969×10-7
0.005
0.152117
0.152113
3.72089×10-6
0.0309517
0.0309514
−3.0024×10-7
0.006
0.179645
0.179642
3.23223×10-6
0.0370029
0.0370026
−2.61703×10-7
0.007
0.20628
0.206277
2.78314×10-6
0.0430082
0.043008
−2.14219×10-7
0.008
0.232051
0.232049
2.18254×10-6
0.048968
0.0489678
−1.56195×10-7
0.009
0.256986
0.256984
2.14645×10-6
0.0548823
0.0548822
−8.66593×10-8


## 5. Conclusions

In this paper, IRKM was employed successfully for solving a class of SPPs with boundary layers. The numerical results show that the present method is an accurate and reliable analytical technique for SPP with boundary layers.



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*Source: 984057-2012-06-21.xml*