# Solution of Boundary Value Obstacle Problems Using MQ-RBF and IMQ-RBF

Authors: Feng Gao; Chunmei Chi
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

A kind of numerical method which is based on multiquadric RBF, inverse multiquadric RBF, and Wu-Schaback operators is presented for solving second-order and third-order boundary value problems associated with obstacle, unilateral, and contact problems. The algorithms are proved to be highly accurate and easy to implement. Some numerical tests are also presented to show the efficiency of the algorithms.


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## Body

## 1. Introduction

We consider the numerical solutions of the following second-order boundary value problems:(1)u′′(x)={f(x),a≤x≤c,g(x)u(x)+f(x)+r,c≤x≤d,f(x),d≤x≤b,
with the boundary conditions
(2)u(a)=α,u(b)=β
and the continuity conditions of u and u′ at c and d. We also consider the following third-order boundary value problems:
(3)u′′′(x)={f(x),a≤x≤c,g(x)u(x)+f(x)+r,c≤x≤d,f(x),d≤x≤b,
with the boundary conditions
(4)u(a)=α1,u′(a)=α2,u′(b)=β
and the continuity conditions of u, u′, and u′′ at c and d. In the two problems, f(x) and g(x) are continuous functions and α, β, α1, α2 are real constants. The two problems arise usually in obstacle, contact, unilateral, and equilibrium problems associated with economics, transportation, oceanography, fluid flow through porous media, and many other fields of pure and applied sciences; see [1–12]. In the literature, some techniques were used to solve the previous two systems of second-order and third-order boundary value problems associated with obstacles. For example, for problem (1), Noor and Khalifa [13] used first-order accuracy collocation method with cubic splines as basis functions. Al-Said et al. [14] developed second- and fourth-order finite difference and spline methods. Al-Said [5] developed and analyzed quadratic and cubic splines methods. He proved that both quadratic and cubic spline methods can be used to produce second-order smooth approximations for the solution of (1) and its first derivative over the whole interval [a,b]. Al-Said [7] also used cubic spline polynomial functions for solving problem (1).For the problem (3), Khalifa and Noor [15] applied quintic spline method for contact problems. Noor and Khalifa [16] applied quartic spline method for odd-order obstacle problems. Gao and Chi [17] applied quartic B-spline method for third-order obstacle problems and Siraj-ul-Islam et al. [12] proposed nonpolynomial spline methods. Some of these methods require huge computational work for finding the numerical solutions.In this paper, we use multiquadric RBF (MQs) and inverse multiquadric RBF (IMQs) to solve systems of second-order and third-order boundary value problems associated with obstacle. Recently, radial base functions (RBF) have been widely applied to the numerical solution of many types of PDE; see [18, 19]. The advantage of techniques of using RBF is that mesh grid can be avoided and high accuracy can be achieved. However, to the authors’ knowledge, little work has been done involving the usage of RBF in the solution of boundary value obstacle problems. Among Radial Basis Functions, MQs and IMQs are especially suitable for solving second- and third-order boundary value problems associated with obstacles because we can integrate IMQs to get MQs. Our technique is that we first use IMQs or MQs quasi-interpolation to approximate the derivatives of the solution of problems of (1) and (3); then we get their integrals to approximate the solutions. The following part of this paper is organized as follows. In Section 2, we review some basics of RBFs. In Section 3, we present the numerical schemes for problem (1) and (3). In Section 4, we do some numerical experiments. Section 5 is our conclusion.


## 2. RBF and MQ Quasi-Interpolation

RBF was proposed in 1951 [20] to deal with geological problem.For a given regionΩ∈Rm and a set distinct interpolation points,
(5)X={x1,x2,…xN}⊂Ω.
Given a function f:X→R, construct the interpolation to f(6)Sf,X(x)=∑j=1Nαjφ(∥x-xj∥),forx∈Ω,
where ∥·∥ denote Euclidean norm and φ is a certain RBF. The undetermined coefficients {α1,α2,…,αN}  can be obtained by the linear system
(7)∑j=1Nαjφ(∥xi-xj∥)=f(xi),i=1,2,…,N.
We adopt multiquadric (MQs) RBF
(8)ϕ(r)=s2+r2
and inverse multiquadric (IMQs) RBF
(9)φ(r)=s2(s2+r2)3/2
to solve second-order and third-order boundary value obstacle problems, where s∈R+ is a shape parameter. From [21], we know that the linear system (7) has unique solution. It is easy to see that
(10)(s2+r2)′′=s2(s2+r2)3/2.We also adopt a kind of Quasi-interpolation to numerically solve problem (1) and (3). The Quasi-interpolation of a univariate function f:[a,b]→R with MQs is constructed as follows. On the scattered points
(11)a=x0<x1<⋯<xn=b,h∶=max1≤i≤n(xi-xi-1).
Quasi-interpolant of f usually is as follows:
(12)(Mf)(x)=∑i=0nf(xi)ψi(x),
where ψi(x) is linear combinations of the MQs. Consider
(13)ϕi(|x-xi|)=s2+(x-xi)2,s∈R+,
where s is a shape parameter.Given data{(xi,fi)}i=1n, Wu–Schabacks LD operator is
(14)(LDf)(x)=∑j=0nf(xj)ψj(x),
where
(15)ψ0(x)=12+ϕ1(|x-x1|)-(x-x0)2(x1-x0),ψ1(x)=ϕ2(|x-x2|)-ϕ1(|x-x1|)2(x2-x1)-ϕ1(|x-x1|)-(x-x0)2(x1-x0),ψn-1(x)=(xn-x)-ϕn-1(|x-xn-1|)2(xn-xn-1)-ϕn-1(|x-xn-1|)-ϕn-2(|x-xn-2|)2(xn-1-xn-2),ψn(x)=12+ϕn-1(|x-xn-1|)-(xn-x)2(xn-xn-1),ψj(x)=ϕj-1(|x-xj-1|)-ϕj(|x-xj|)2(xj-1-xj)-ϕj(|x-xj|)-ϕj-1(|x-xj-1|)2(xj-xj-1),hhhhhhhhhhhhhhhhhj=2,…,n-2.
The advantage of LD operator is that high accuracy can be achieved without solving any linear system.For∫ψi(x)dx, we need only to integrate ϕi(|x-xi|),
(16)∫ϕi(|x-xi|)dx=∫s2+(x-xj)2dx=(x-xj)(x-xj)2+s22+s22ln((x-xj)+(x-xj)2+s2).
For  ∫dx∫ψi(x)dx, which is denoted with Φi(x), we have
(17)∫dx∫ϕi(|x-xi|)dx=∫dx∫s2+(x-xj)2dx=((x-xj)2+s2)1.56+s22(x-xj)ln((x-xj)+(x-xj)2+s2)-s22(x-xj)2+s2.


## 3. Numerical Algorithms

To solve problem (1) and (3), we set
(18)a=x1<x2<⋯<xm=c=xm+1<xm+2<⋯<xm+n=d=xm+n+1<xm+n+2<⋯<xm+n+k=b.
Construct (LDf)(x)=∑j=1mf(xj)ψj(x) on [a,c]  to approximate  f(x), x∈[a,c]; and  (LDf)(x)=∑j=1kf(xm+n+j)ψm+n+j(x) on [d,b]  to approximate  f(x), x∈[d,b].On the interval[c,d], let u′′(x)=∑j=1nαjφm+j(|x-xm+j|), where
(19)φm+j(|x-xm+j|)=s2(s2+(x-xm+j)2)3/2,
and s is a shape parameter and αj, j=1,…,n are coefficients to be identified. We can easily verify that
(20)∫φj(|x-xj|)dx=x-xj(s2+(x-xj)2)+λ,(21)∫(x-xj(s2+(x-xj)2)+λ)dx=ϕj(|x-xj|)+λx+μ,
where λ, μ are constants. Therefore,
(22)u(x)=∫dx∫∑j=1nαjφm+j(|x-xm+j|)dx=∑j=1nαjϕm+j(|x-xm+j|)+λx+μ,hhhhhhhhhhhhhhhhhhhhx∈[c,d],
where λ, μ are constants and ϕj(|x-xj|) are defined in (15). Therefore, we have the approximation to the solution to problem (1) as follows:
(23)u={∫dx∫∑j=1mf(xj)ψj(x)dx=∑j=1mf(xj)Φj(x)+c1x+c2,x∈[a,c],∑j=1nαjϕm+j(|x-xm+j|)+λx+μ,x∈[c,d],∫dx∫∑j=1kf(xm+n+j)ψm+n+j(x)dx=∑j=1kf(xm+n+j)Φm+n+j(x)+c3x+c4,x∈[d,b],
where Φj(x)=∫dx∫ψi(x)dx and is computed in (17), and c1, c2, λ, μ, c3, c4 are constants. We need to determine the coefficients c1, c2, α1,…,αn, λ, μ, c3, c4 to identify the numerical solution (23) to problem (1).Let(24)H1(x)=∑j=1mf(xj)Φj(x)+c1x+c2,x∈[a,c],H2(x)=(∑j=1nαjϕm+j(|x-xm+j|)+λx+μ),hhhhhhhhhhhhhhhhhhhhhhhhhhhhx∈[c,d],H3(x)=∑j=1kf(xm+n+j)Φm+n+j(x)+c3x+c4,hhhhhhhhhhhhhhhhhhhhhhhhhhhhx∈[d,b].
Substitute (24) into problem (1) at points (18) to get the following linear system:
(25)H1(a)=α,H1(c)=H2(c),H1′(c)=H2′(c),∑j=1nαjφm+j(|xm+i-xm+j|)=g(xm+i)H2(xm+i)+f(xm+i)+r,hhhhhhhhhhhhhhhhhhhhhi=1,…,n,H2(d)=H3(d),H2′(d)=H3′(d),H3(b)=β,
where H1′(x), H3′(x) can be computed in (16) and H2′(x) can be computed in (20).Solving the linear system (24), we get c1, c2, α1,…,αn, λ, μ, c3, c4 to identify the numerical solution (23) to problem (1).To solve problem (3), we use
(26)∑j=1mαjφj(|x-xj|),∑j=1nαm+jφm+j(|x-xm+j|),∑j=1kαm+n+jφm+n+j(|x-xm+n+j|)
to approximate u′′′ on intervals [a,c], [c,d], and [d,b], respectively, where
(27)φm+j(|x-xm+j|)=s2(s2+(x-xm+j)2)3/2
and s is a shape parameter. To get u′′, u′, and u, we need only to compute
(28)∫φj(|x-xj|)dx=x-xj(s2+(x-xj)2)1/2+λ,(29)∫dx∫φj(|x-xj|)dx=ϕj(|x-xj|)+λx+μ,
where λ,  μ are constants and ϕj is defined in (13). And we also have
(30)∫(ϕj(|x-xj|)+λx+μ)dx=ζj(x)+λ22x2+μ+ν,
where ν is a constant and
(31)ζj(x)=(x-xj)(x-xj)2+s22+s22ln((x-xj)+(x-xj)2+s2).
Therefore, we need to identify the following numerical solution to (3):
(32)u={∑j=1mαjζj(x)+c1x2+c2x+c3,∑j=1nαm+jζm+j(x)+λx2+μx+ν,∑j=1kαm+n+jζm+n+j(x)+c4x2+c5x+c6.
To determine the coefficients c1, c2, c3, α1,…,αm+n+k, λ, μ, ν, c3, c4, c5, denote
(33)H1(x)=∑j=1mαjζj(x)+c1x2+c2x+c3,H2(x)=∑j=1nαm+jζm+j(x)+λx2+μx+ν,H3(x)=∑j=1kαm+n+jζm+n+j(x)+c4x2+c5x+c6,
and H1′(x), H2′(x), H3′(x) can be computed in (29) and H1′′(x), H2′′(x), H3′′(x) can be computed in (28). Substitute (25) into problem (3) at points (18) to get the following linear system:
(34)H1(a)=α1,H1′(a)=α2,H1(c)=H2(c),H1′(c)=H2′(c),H1′′(c)=H2′′(c),∑j=1nαm+jφm+j(|xm+i-xm+j|)=g(xi)H2(xm+i)+f(xm+i)+r,hhhhhhhhhhhhhhhi=2,…,n-1,H2(d)=H3(d),H2′(d)=H3′(d),H2′′(d)=H3′′(d),H3′(b)=β.
Then, we can solve (34) to identify the numerical solution (32) to problem (3).


## 4. Numerical Experiment

We do numerical experiments to test the efficiency of our numerical methods.Example 1.
Consider the following second-order boundary obstacle problem as given in [22]:
(35)u′′={0,0≤x≤π4,u-1,π4≤x≤34π,0,34π≤x≤π,
with boundary condition u(0)=0, u(π)=0. The exact solution of this problem is given as
(36)u={4r1x,0≤x≤π4,1-4r2cosh(π4-x),π4≤x≤34π,4r1(π-x),34π≤x≤π,
where r1=π+4coth(π/4), r2=πsinh(π/4)+4cosh(π/4).
The results are reported in Table1. The errors between numerical and exact solution in terms of L∞ and L2 are listed in Table 1. The accuracy of the proposed method is measured using the L2 and L∞ error norms defined as
(37)L2=1n+1∑i=0n|uiexact-uinum|2,L∞=max0≤i≤n|uiexact-uinum|.
In Table 1, we set
(38)xi=(i-1)h1,i=1,2,…,6,h1=π20,xi=π4+(i-6)h2,i=6,7,…,11,h2=π10,xi=34π+(i-11)h1,i=12,13,…,17,h1=π20.Table 1
Comparison of exact and numerical solutions for Example1.
x
s
=
1.0, m=n=6
Exact solution
s
=
1.2, m=n=9
RBF method
RBF method
x
1
0
0
0
x
3
0.13633712736
0.13598397716525
0.13596345616525
x
6
0.34084281842
0.33995994291313
0.33994454291217
x
7
0.44620383234
0.44535138581350
0.44536178581320
x
8
0.49629810661
0.49554948904960
0.49553678904558
x
9
0.49629810661
0.49554948904960
0.49553578798663
x
10
0.44620383234
0.44535138581350
0.44536480583846
x
11
0.34084281842
0.33995994291313
0.33996457654310
x
14
0.13633712736
0.13598397716525
0.13597686714765
x
16
0
0
0
L
∞ error
8.82875509702
e
-
004
1.54000009599
e
-
005
L
2 error
2.09345530306
e
-
004
3.69805440717
e
-
006Example 2.
Consider the following third-order boundary obstacle problem as given in [17]:
(39)u′′′(x)={0,0≤x≤14,u(x)-1,14≤x≤34,0,34≤x≤1,
with boundary conditions u(0)=0, u′(0)=0, u′(1)=0.
The exact solution of this problem is(40)u(x)={12a1x2,0≤x≤14,1+a2ex+e-x/2(a3cos32x+a4sin32x),14≤x≤34,a5x(12x-1)+a6,34≤x≤1,
where
(41)a1=0.14520742362098,a2=-0.21130240827197,a3=-0.78610085318732,a4=-0.24585768969643,a5=0.05860440434801,a6=0.04768241777632.The results are reported in Table2. The errors between numerical and exact solution in terms of L∞ and L2 are listed in Table 2.Table 2
Comparison of exact and numerical solutions for Example2.
x
s
=
1.0, m=n=6
Exact solution
s
=
1.2, m=n=9
RBF method
RBF method
0.0
0
0
0
0.1
0.00078568857778
0.00072603711810
0.00072504958690
0.2
0.00294888593935
0.00290414847242
0.00290536784209
0.3
0.00653975857589
0.00651360507585
0.00651038335789
0.4
0.01101039579013
0.01105748825987
0.01105193964378
0.5
0.01550387591395
0.01556533144116
0.01556193848997
0.6
0.01909384653746
0.01905044604835
0.01905677849423
0.7
0.02053875436899
0.02053013998469
0.02053284784996
0.8
0.01951387899385
0.01955230368928
0.01955135789349
0.9
0.01862948575889
0.01867323762406
0.01867204968224
1.0
0.01832948588680
0.01838021560231
0.01838334510084
L
∞ error
6.145552721
e
-
005
6.3324458800
e
-
006
L
2 error
1.290788173
e
-
005
9.7350311797
e
-
007From the results of Tables1 and 2, we can see that the shape parameter s plays no significant role in the computation.


## 5. Conclusions

In this paper, based on the multiquadric RBF and inverse multiquadric RBF, numerical schemes are presented for solving system of second-order and third-order boundary value problems associated with obstacle. The schemes are shown to be highly accurate and easy to implement. We point out three notes here. First, to solve second-order problem, we useLD operator to approximate the derivatives on the interval [a,c] and [d,b] because by using LD operator, we can on one hand avoid solving linear systems and on the other hand achieve high accuracy. Second, to solve third-order problem, we do not employ LD operators to approximate the derivatives on the interval [a,c] and [d,b] because it is no easy to integrate LDf for three times and no explicit integral expressions can be found. Third, why we choose MQs and IMQs rather than other RBF? That is because MQs and IMQs can achieve high accuracy and IMQs are the second derivatives of MQs and this property bring much convenience to our calculation. Further study can be carried out for the applications of RBF in the numerical solution of higher-order boundary value obstacle problems.



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