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An Initial-Value Technique for Singularly Perturbed Boundary Value Problems via Cubic Spline
Manoj Kumar a; P. Singh a; Hradyesh Kumar Mishra a a Department of Mathematics, Motilal Nehru National Institute of Technology, Allahabad, India Online Publication Date: 01 November 2007 To cite this Article: Kumar, Manoj, Singh, P. and Mishra, Hradyesh Kumar (2007) 'An Initial-Value Technique for Singularly Perturbed Boundary Value Problems via Cubic Spline', International Journal for Computational Methods in Engineering Science and Mechanics, 8:6, 419 - 427 To link to this article: DOI: 10.1080/15502280701587999 URL: http://dx.doi.org/10.1080/15502280701587999
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International Journal for Computational Methods in Engineering Science and Mechanics, 8:419–427, 2007 c Taylor & Francis Group, LLC Copyright ISSN: 1550–2287 print / 1550–2295 online DOI: 10.1080/15502280701587999
An Initial-Value Technique for Singularly Perturbed Boundary Value Problems via Cubic Spline Manoj Kumar, P. Singh, and Hradyesh Kumar Mishra Department of Mathematics, Motilal Nehru National Institute of Technology, Allahabad, India
ear singular perturbation problems. However, there can be difficulties in applying these methods, such as the matching of the coefficients of the inner and outer expansions. Success may depend on finding the proper scaling or the proper transformation to express the dependent and independent variables. Recently, a non-asymptotic method, also called boundaryvalue technique, has been introduced by Roberts [14–16] to solve certain classes of singular perturbation problems. He also discussed the analytical and approximate solutions of several numerical examples. The concept of replacing singularly perturbed two-point boundary value problem by initial-value technique is presented by many authors; see for example [4, 8–11, 20 and 21]. In view of the wealth of literature on singular perturbation problems, we raise the question of whether there are other ways Keywords Singular Perturbation, Ordinary Differential Equation, Two- to attack singular perturbation problems, namely ways that are Point Boundary Value Problems, Boundary Layers, Initial- very easy to use and ready for computer implementation. In reValue Technique, Cubic Spline sponse to this need for a fresh approach to singular perturbation problems, we propose and illustrate in this paper an initial-value technique for singularly perturbed two-point boundary value 1. INTRODUCTION Singular perturbation problems are of common occurrence problems via cubic spline with a boundary layer on the left (or in fluid mechanics and other branches of applied mathematics. right) end of the underlying interval. It is distinguished by the There are a wide variety of techniques for the solution of sin- following fact: The original second order problem is replaced gular perturbation problems (cf.[1–3, 5–7, 12, 22–24]). Many by an asymptotically equivalent three first order initial–value of these techniques consist of: (a) dividing the problem into an problems and these initial-value problems are solved by cubic inner region (boundary layer) problem and an outer region prob- spline. Numerical experience with several linear and non-linear lem; (b) expressing the inner and outer solutions as asymptotic examples is described. expansions; (c) equating various terms in the inner and outer expressions to determine the constants in these expressions; and 2. INITIAL-VALUE TECHNIQUE (d) combining the inner and outer solutions in some fashion to For convenience, we call our method the initial-value techobtain a uniformly valid solution. Typically, the inner region nique. To set the stage for the initial-value technique, we consider problems are obtained from the original problem by rescaling a linear singularly perturbed two-point boundary value problem the independent variable. These techniques and their variations of the form: have been used successfully on a variety of linear and nonlinεu (x) + p(x)u (x) + q(x)u(x) = r (x), x ∈ [a, b] (1) A new way to solve singular perturbation problems is introduced. It is designed for the practicing engineer or applied mathematician who needs a practical tool for these problems (easy to use, modest problem preparation and ready computer implementation). In this paper, we consider singularly perturbed two-point boundary value problems with the boundary layer at one end (left or right) point. An initial-value technique is used for its solution. However, this technique is based on the boundary layer behavior of the solution. It is distinguished by the following fact: The original second order problem is replaced by an asymptotically equivalent first-order problem and is solved as an initial-value problem via cubic spline. Numerical experience with several linear and non-linear examples is described. It is observed that the present method approximates the exact solution very well.
Received 22 July 2007; accepted 23 July 2007. Address correspondence to Manoj Kumar, Department of Mathematics, Motilal Nehru National Institute of Technology, Allahabad 211004, India. E-mail:
[email protected]
with
419
u(a) = α u(b) = β
(2) (3)
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where ε is a small positive parameter (0 < ε 1) and α, β are known constants. We assume that p(x), q(x) and r (x) are sufficiently continuously differentiable functions in [a, b]. Furthermore, we assume that p(x) ≥ M > 0 throughout the interval [a, b], where M is some positive constant. This assumption merely implies that the boundary layer will be in the neighborhood of x = a. Since singular perturbation problems exhibits boundary layer behavior of the solution. Therefore, the solution of (1), (2) and (3) is given by u(x) = v(x) + w(x)e−t(x)/ε
+ −
∞
wn (x)εn (e−t(x)/ε )( p (x)) + p(x)
+ p(x)
∞
wn (x)ε n
− ( p(x))
∞
∞
wn (x)ε
n−1
(e
−t(x)/ε
) + q(x)
wn (x)ε
∞
vn (x)ε
n
n=0
n
vn (x)ε n
(e−t(x)/ε )
n=0
n=0
n=0 2
∞
(e−t(x)/ε ) = r (x)
(9)
x
By restricting these series to their first terms, we get
p(x)d x
∞
u(x) =
(e−t(x)/ε )( p(x))2
n=0
(4)
a n n=0 vn (x)ε
∞
wn (x)ε
n−1
n=0
t(x) = where v(x) = (cf.[18,p.292]) , i.e.,
n=0 ∞
+ q(x)
with
∞
and w(x) =
vn (x)ε + n
n=0
∞
n n=0 wn (x)ε
wn (x)ε
×w0 (x)e−t(x)/ε + q(x)v0 (x) + q(x)w0 (x)e−t(x)/ε = r (x),
i.e.,
n
−2 p(x)e−t(x)/ε w0 (x) − w0 (x)e−t(x)/ε p (x) + p(x)v0 (x) + p(x)
e
−t(x)/ε
p(x)v0 (x) + q(x)v0 (x) + [−2 p(x)w0 (x) − p (x)w0 (x) + p(x)w0 (x) + q(x)w0 (x)]e−t(x)/ε = r (x).
(5)
n=0
Therefore, we have
with
t(x) =
x
p(x)d x
(6)
Differentiating (5) with respect to x, we get ∞ ∞ n n u (x) = vn (x)ε + wn (x)ε e−t(x)/ε n=0
−
n=0
∞
wn (x)ε
n=0
u (x) =
∞ n=0
n
vn (x)ε n
+
e
∞
−t(x)/ε
p(x) ε
wn (x)ε n (e−t(x)/ε )
Substituting (5), (7) and (8) in (1), we get ∞ ∞ vn (x)εn+1 + wn (x)εn+1 (e−t(x)/ε )
−2
n=0 ∞ n=0
(7)
d [ p(x)w0 (x)] = q(x)w0 (x) dx
(11)
wn (x)ε n (e−t(x)/ε )( p(x))
v0 (a) + w0 (a) = α
(12)
v0 (b) = β,
(13)
and
p(x) ε n=0
∞ p(x) 2 + wn (x)ε n (e−t(x)/ε ) ε n=0
∞ p (x) n −t(x)/ε − wn (x)ε )(e ε n=0 −2
(10)
The representation (5) and (6) can be inserted to the boundary conditions (2) and (3). Now the boundary conditions become
wn (x)εn e−t(x)/ε
n=0 ∞
p(x)v0 (x) + q(x)v0 (x) = r (x) and
a
n=0
(8)
where we have neglected the exponentially small term e−t(b)/ε (which is asymptotically zero) in obtaining the boundary condition (13) at x = b. First the differential equation (10) can be solved along with the boundary condition (13) to determine v0 (x). Now w0 (x) is determined by solving Eq. (11) subject to the condition w0 (a) = α − v0 (a) where x v0 (a) is determined already. Now from (6) we have t(x) = a p(x)d x, i.e., t (x) = p(x) with t (a) = 0. Therefore, the three initial-value problems corresponding to (1), (2) and (3) are given by p(x)v0 (x) + q(x)v0 (x) = r (x) d (IVP. II) [ p(x)w0 (x)] = q(x)w0 (x) dx with w0 (a) = α − v0 (a)
(IVP. I)
with v0 (b) = β (14)
(15)
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BOUNDARY VALUE PROBLEMS VIA CUBIC SPLINE
and
and (IVP. III)t (x) = p(x)
with t(a) = 0
(16)
Remark 1: Here it is to be noted that these initial-value problems are independent of perturbation parameterε and we solve these initial-value problems by cubic spline method.
y(x0 ) = y0 then from (22), we obtain
S(x) =
m i−1 (xi − x)2 (x − xi−1 ) m i (x − xi−1 )2 − h2 h2 2 yi−1 (xi − x) [2(x − xi−1 ) + h] + h3 2 yi (x − xi−1 ) [2(xi − x) + h] + h3
(17)
(18)
Differentiating (17) with respect to x and simplifying, we obtain S (x) =
m i−1 (xi − x)(2xi−1 + xi − 3x) h2 m i (x − xi−1 )(xi−1 + 2xi − 3x) − h2 6(yi − yi−1 )(x − xi−1 )(xi − x) + h3
(19)
Again, differentiating (19) with respect to x, we have S (x) =
−2m i−1 (xi−1 + 2xi − 3x) h2 2m i (2xi−1 + xi − 3x) − h2 6(yi − yi−1 )(xi−1 + xi − 2x) + h3
(20)
which gives S (xi ) =
2m i−1 4m i 6(yi − yi−1 ) + − h h h2
S (xi ) =
2m i−1 4m i 6(Si − Si−1 ) + − h h h2
(21)
If we consider the initial-value problem dy = f (x, y) dx
∂f ∂x
+
∂ f dy ∂y dx
or
∂f (xi , yi ) f (xi , yi ) ∂y ∂f (xi , Si ) f (xi , Si ) ∂y
(24)
Equating (21) and (24), we have 2m i−1 4m i 6(Si − Si−1 ) ∂f ∂f + − (xi , Si )+ (xi , Si ) f (xi , Si ) = 2 h h h ∂x ∂y (25) from which Si can be computed. Substituting these values of Si in Eq. (17), we get S(x).
where h = xi − xi−1 = mesh size
=
∂f (xi , yi ) + ∂x ∂f = (xi , Si ) + ∂x
y (xi ) = 3. CUBIC SPLINE METHOD The cubic spline function S(x) in xi−1 ≤ x ≤ xi in terms of its first derivatives S (xi ) = m i is given by [17]
d2 y dx2
(23)
(22)
Remark 2: First we solve Eq.(14) at x = b and then we find the solution of Eq. (14) at x = a. We use this value in the initial condition of Eq. (15) and solve Eq. (15) at x = a. Hence we simply divide the interval a to x and x to b and putting ε = 0 we obtain the solution of Eq. (14), which gives the outer solution and Eq. (15), (16) gives inner solution. Combining these solutions we get the complete solution. So integration of (14)–(16) is just for finding the solution of first-order differential equation. It clearly indicates that the integration of equation (14) and (15) goes in opposite direction and equation (15) can be solved only if the solution of equation (14) is known at x = a. Since Eq. (16) is independent of the first two initial-value problems (14) and (15). So this can be solved independently. We used cubic spline method for solving these initial-value problems. After finding v0 (x), w0 (x) and t(x), we obtain the solution of (1), (2) and (3) from (4) as follows: u(x) = v0 (x) + w0 (x)e−t(x)/ε . Remark 3: The solution of Eq. (14) will be generally nonstiff, but we may expect that the solution of Eq. (15) changes character in the interval of integration. In fact, even if the initial-value problem (15) is nonstiff within the boundary layer, it may be stiff outside the layer where the exact solution is regular. So, while the problem (14) can be solved using sufficiently accurate methods for nonstiff problems, the numerical integration of (15) requires a scheme automatically determining whether the problem can be solved more efficiently using a class of methods designed for nonstiff problems or a class of methods suitable for stiff problems.
4. NUMERICAL EXAMPLES To illustrate how the initial-value method works, we will discuss three linear and three non-linear examples solved by the method. These examples have been chosen because they have been discussed widely in the literature and approximate solutions are available for comparison.
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5. LINEAR PROBLEMS To demonstrate the applicability of the method, we have applied it to three linear singular perturbation problems with leftend boundary layer. Example 5.1: Consider the following homogeneous singular perturbation problem from Bender and Orszag [1, p.480, Problem 9.17 with α = 0] εy (x) + y (x) − y(x) = 0 x ∈ [0, 1]
with
y(0) = 1 and y(1) = 1.
The exact solution is given by [(em 2 − 1)em 1 x + (1 − em 1 )em 2 x ] , [em 2 − em 1 ] √ √ where m 1 = (−1 + 1 + 4ε)/(2ε) and m 2 = (−1 − 1 + 4ε) /(2ε). y(x) =
The numerical results are given in Tables 1 and 2 for ε = 10−3 and10−4 , respectively. Example 5.2: Now consider the following non-homogeneous singular perturbation problem from fluid dynamics for fluid of small viscosity, [13, Example 2] εy (x) + y (x) = 1 + 2x x ∈ [0, 1] with y(0) = 0 and
x 0.0000 0.0001 0.0010 0.0020 0.0030 0.0040 0.0050 0.1000 0.3000 0.5000 0.7000 0.9000 1.0000
0.000 0.001 0.010 0.020 0.030 0.040 0.050 0.100 0.300 0.500 0.700 0.900 1.000
Exact solution
1.0000000 0.6004374 0.3682764 0.3686162 0.3689850 0.3693542 0.3697237 0.4065699 0.4965858 0.6065312 0.7408186 0.9048377 1.0000000
1.0000000 0.6004604 0.3683130 0.3686527 0.3690215 0.3693907 0.3697602 0.4066062 0.4966200 0.6065609 0.7408404 0.9048464 1.0000000
Example 5.3: Finally, we consider the following variable coefficient singular perturbation problem from Kevorkian and Cole [5, p. 33, Eqs. (2.3.26) and (2.3.27) with α = −1/2]
x 1 εy (x) + 1 − y (x) − y(x) = 0, 2 2
x ∈ [0, 1]
with y(0) = 0 and y(1) = 1. (2ε − 1)(1 − e−x/ε ) . (1 − e−1/ε )
We have chosen to use uniformly valid approximation (which is obtained by the method given by Nayfeh [12, p.148,
TABLE 1 Numerical results of Example 5.1 with ε = 10−3 , h = 10−3 x
y(x)
The numerical results are given in Tables 3 and 4 for ε = 10−3 and 10−4 , respectively.
y(1) = 1.
The exact solution is given by y(x) = x(x + 1 − 2ε) +
TABLE 2 Numerical results of Example 5.1 with ε = 10−4 , h = 10−4
y(x)
Exact solution
1.0000000 0.6005594 0.3716051 0.3753111 0.3790831 0.3828929 0.3867410 0.4065696 0.4965852 0.6065305 0.7408181 0.9048373 1.0000000
1.0000000 0.6007918 0.3719724 0.3756784 0.3794502 0.3832599 0.3871079 0.4069350 0.4969324 0.6068334 0.7410401 0.9049277 1.0000000
TABLE 3 Numerical results of Example 5.2 with ε = 10−3 , h = 10−3 x 0.000 0.001 0.010 0.020 0.030 0.040 0.050 0.100 0.300 0.500 0.700 0.900 1.000
y(x)
Exact solution
0.0000000 −0.6311195 −0.9898545 −0.9795999 −0.9690999 −0.9583998 −0.9474999 −0.8899999 −0.6099999 −0.2499999 0.1900001 0.7099999 1.0000000
0.0000000 −0.6298573 −0.9878747 −0.9776400 −0.9671600 −0.9564800 −0.9456000 −0.8882000 −0.6086000 −0.2490000 0.1906001 0.7102001 1.0000000
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TABLE 4 Numerical results of Example 5.2 with ε = 10−4 , h = 10−4 x 0.0000 0.0001 0.0010 0.0020 0.0030 0.0040 0.0050 0.1000 0.3000 0.5000 0.7000 0.9000 1.0000
TABLE 6 Numerical results of Example 5.3 with ε = 10−4 , h = 10−4
y(x)
Exact solution
x
0.0000000 −0.6320207 −0.9989538 −0.9979961 −0.9969911 −0.9959841 −0.9949751 −0.8900002 −0.6100004 −0.2500001 0.1899999 0.7099998 1.0000000
0.0000000 −0.6318942 −0.9987538 −0.9977964 −0.9967916 −0.9957848 −0.9947760 −0.8898200 −0.6098600 −0.2499000 0.1900600 0.7100199 1.0000000
0.0000 0.0001 0.0010 0.0020 0.0030 0.0040 0.0050 0.1000 0.3000 0.5000 0.7000 0.9000 1.0000
0.0000000 0.3160796 0.5002186 0.5004917 0.5007424 0.5009933 0.5012443 0.5263079 0.5882292 0.6666627 0.7692288 0.9090901 1.0000000
0.0000000 0.3160807 0.5002274 0.5005005 0.5007511 0.5010020 0.5012531 0.5263158 0.5882353 0.6666667 0.7692308 0.9090909 1.0000000
εy (x) + 2y (x) + e y(x) = 0 x ∈ [0, 1] with y(0) = 0 and y(1) = 0.
1 1 2 y(x) = − e−(x−x /4)/ε . (2 − x) 2 The numerical results are given in Tables 5 and 6 for ε = 10−3 and 10−4 , respectively. 6. NON-LINEAR PROBLEMS Here, we consider three non-linear singular perturbation problems with left-end boundary layer. First we convert non-linear singular perturbation problems as a sequence of linear singular perturbation problems by using quasilinearization method. Then the outer solution (the solution of the given problem by putting ε = 0) is taken to be the initial approximation. TABLE 5 Numerical results of Example 5.3 with ε = 10−3 , h = 10−3
0.000 0.001 0.010 0.020 0.030 0.040 0.050 0.100 0.300 0.500 0.700 0.900 1.000
Nayfeh solution
Example 6.1. Consider the following singular perturbation problem from Bender and Orszag [1, p.463, Eq. (9.7.1)]
Eq. (4.2.32)] as our “exact” solution;
x
y(x)
y(x)
Nayfeh solution
0.0000000 0.3162644 0.5024892 0.5050504 0.5076141 0.5102040 0.5128204 0.5263155 0.5882348 0.6666664 0.7692305 0.9090911 1.0000000
0.0000000 0.3162644 0.5024893 0.5050505 0.5076142 0.5102041 0.5128205 0.5263158 0.5882353 0.6666667 0.7692308 0.9090909 1.0000000
We have chosen to use Bender and Orszag’s uniformly valid approximation [1, p. 463, Eq. (9.7.6)] for comparison, y(x) = loge (2/(1 + x)) − (loge 2)e−2x/ε . For this example, we have boundary layer of thickness O(ε) at x = 0 [1]. The numerical results are given in Tables 7 and 8 for ε = 10−3 and 10−4 , respectively. Example 6.2. Now consider the following singular perturbation problem from Kevorkian and Cole [5, p. 56, Eq. (2.5.1)] εy (x) + y(x)y (x) − y(x) = 0,
x ∈ [0, 1] with y(0) = −1 and y(1) = 3.9995.
TABLE 7 Numerical results of Example 6.1 with ε = 10−3 , h = 10−3 x 0.000 0.001 0.010 0.020 0.030 0.040 0.050 0.100 0.300 0.500 0.700 0.900 1.000
y(x)
Bender solution
0.0000000 0.5985410 0.6832713 0.6734177 0.6636603 0.6539972 0.6444266 0.5979007 0.4308263 0.2877095 0.1625336 0.0512977 0.0000000
0.0000000 0.5983404 0.6831968 0.6733446 0.6635884 0.6539265 0.6443570 0.5978370 0.4307829 0.2876821 0.1625189 0.0512933 0.0000000
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TABLE 8 Numerical results of Example 6.1 with ε = 10−4 , h = 10−4 x 0.0000 0.0001 0.0010 0.0020 0.0030 0.0040 0.0050 0.1000 0.3000 0.5000 0.7000 0.9000 1.0000
y(x)
Bender solution
0.0000000 0.5992603 0.6921556 0.6911569 0.6901595 0.6891628 0.6881672 0.5978435 0.4307855 0.2876833 0.1625209 0.0512937 0.0000000
0.0000000 0.5992399 0.6921477 0.6911492 0.6901517 0.6891552 0.6881596 0.5978370 0.4307829 0.2876821 0.1625189 0.0512933 0.0000000
We have chosen to use the Kevorkian and Cole’s uniformly valid approximation [5, pp. 57–58, Eqs. (2.5.5), (2.5.11) and (2.5.14)] for comparison, y(x) = x + c1 tanh(c1 (x/ε + c2 )/2),
TABLE 10 Numerical results of Example 6.2 with ε = 10−4 , h = 10−4 x 0.0000 0.0001 0.0010 0.0020 0.0030 0.0040 0.0050 0.1000 0.3000 0.5000 0.7000 0.9000 1.0000
y(x)
Kevorkian solution
−1.0000000 3.0006261 3.0015251 3.0025241 3.0035231 3.0045220 3.0055210 3.1004235 3.3002183 3.5000131 3.6998078 3.8996026 3.9995000
−1.0000000 2.4560400 3.0005000 3.0015000 3.0025000 3.0035000 3.0045000 3.0995000 3.2995000 3.4995000 3.6995000 3.8995000 3.9995000
We have chosen to use O’Malley’s approximate solution [6, pp. 9–10, Eqs. (1.13) and (1.14)] for comparison, y(x) = −(1 − e−(x+1)/ε )/(1 + e−(x+1)/ε ). For this example, we have a boundary layer of width O (ε) at x = −1 (cf. [6, pp. 9-10, Eqs. (1.10), (1.13), (1.14), case 2] and [15]). The numerical results are given in Tables 11 and 12 for ε = 10−3 and 10−4 , respectively.
where c1 = 2.9995 andc2 = (1/c1 ) loge [(c1 − 1)/(c1 + 1)]. For this example also we have a boundary layer of width O (ε) at x = 0 [5, pp. 56–66]. The numerical results are given in Tables 9 and 10 for ε = 10−3 and 10−4 , respectively. Example 6.3. Finally, we consider the following singular perturbation problem from O’Malley [6, p. 9, Eq. (1.10) case 2]; εy (x) − y(x)y (x) = 0, x ∈ [−1, 1] with y(−1) = 0 and y(1) = −1.
7. RIGHT-END BOUNDARY LAYER PROBLEMS Now, we discuss our method for singularly perturbed twopoint boundary value problems with right-end boundary layer
TABLE 9 Numerical results of Example 6.2 with ε = 10−3 , h = 10−3
TABLE 11 Numerical results of Example 6.3 with ε = 10−3 , h = 10−3
x 0.000 0.001 0.010 0.020 0.030 0.040 0.050 0.100 0.300 0.500 0.700 0.900 1.000
y(x)
Kevorkian solution
−1.0000000 3.0005724 3.0095718 3.0195710 3.0295703 3.0395696 3.0495688 3.0995652 3.2995507 3.4995362 3.6995217 3.8995072 3.9995000
−1.0000000 2.4569400 3.0095000 3.0195000 3.0295000 3.0395000 3.0495000 3.0995000 3.2995000 3.4995000 3.6995000 3.8995000 3.9995000
x −1.000 −0.999 −0.980 −0.960 −0.940 −0.920 −0.900 −0.800 −0.400 0.000 0.400 0.800 1.000
y(x)
O’Malley solution
0.0000000 −0.6321205 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000
0.0000000 −0.4621121 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000
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TABLE 12 Numerical results of Example 6.3 with ε = 10−4 , h = 10−4 x −1.0000 −0.9999 −0.9980 −0.9960 −0.9940 −0.9920 −0.9900 −0.8000 −0.4000 0.0000 0.4000 0.8000 1.0000
y(x)
O’Malley solution
0.0000000 −0.6321205 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000
0.0000000 −0.4621824 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000 −1.0000000
u(b) = β
with
(IVP. I)
t(x) =
v0 (a) = α d (IVP. II) [ p(x)w0 (x)] = q(x)w0 (x) dx with w0 (b) = β − v0 (b)
x ∈ [a, b] (26) (27) (28)
(IVP. III) t (x) = p(x)
(30)
∞ n n where v(x) = ∞ n=0 vn (x)ε and w(x) = n=0 wn (x)ε . Restricting these series to their first terms and substituting (29) in (26), we get p(x)v0 (x) + q(x)v0 (x) = r (x)
(31)
d [ p(x)w0 (x)] = q(x)w0 (x). dx
(32)
and
(36)
with
t(b) = 0.
(37)
We use cubic spline method to solve these initial-value problems. After findingv0 (x), w0 (x) and t(x), we obtain the solution of (26), (27) and (28) from (29) as u(x) = v0 (x) + w0 (x)e−t(x)/ε . 8. EXAMPLES WITH RIGHT-END BOUNDARY LAYER To demonstrate the applicability of the method, we have applied it to two linear singular perturbation problems with rightend boundary layer. Example 8.1: Consider the following singular perturbation problem εy (x) − y (x) = 0, x ∈ [0, 1]
with
y(0) = 1 and y(1) = 0.
TABLE 13 Numerical results of Example 8.1 with ε = 10−3 , h = 10−3 x
b
(35)
and
(29)
x
p(x)d x
p(x)v0 (x) + q(x)v0 (x) = r (x)
with
with
(33)
v0 (b) + w0 (b) = β (34) x Now from (30), we have t(x) = b p(x)d x, i.e. t (x) = p(x) with t(b) = 0. Therefore the three initial-value problems corresponding to (26), (27) and (28) are given by
where ε is a small positive parameter (0 < ε 1) and α, β are known constants. We assume that p(x), q(x) and r (x) are sufficiently continuously differentiable functions in [a, b]. Furthermore, we assume that p(x) ≤ M < 0 throughout the interval [a, b], where M is some negative constant. This assumption merely implies that the boundary layer will be in the neighborhood of x = b. Because of the boundary layer behavior of the solution of singular perturbation problems, it is known that the solution of (26), (27) and (28) is given by u(x) = v(x) + w(x)e−t(x)/ε
v0 (a) = α and
of the underlying interval. To be specific, we consider a class of singular perturbation problem of the form εu (x) + p(x)u (x) + q(x)u(x) = r (x), with u(a) = α
The boundary conditions become
0.000 0.200 0.400 0.600 0.800 0.900 0.920 0.940 0.960 0.980 0.999 1.000
y(x)
Kevorkian solution
1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 0.6321205 0.0000000
1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 0.6320939 0.0000000
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M. KUMAR ET AL.
TABLE 14 Numerical results of Example 8.1 with ε = 10−4 , h = 10−4 x 0.0000 0.2000 0.4000 0.6000 0.8000 0.9000 0.9200 0.9400 0.9600 0.9800 0.9999 1.0000
y(x)
Kevorkian solution
1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 0.6321205 0.0000000
1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 1.0000000 0.6321816 0.0000000
Clearly, this problem has a boundary layer at x = 1, i.e., at the right-end of the underlying interval. The exact solution is given by y(x) =
(e − 1) . (e−1/ε − 1) (x−1)/ε
TABLE 16 Numerical results of Example 8.2 with ε = 10−4 , h = 10−4 x 0.0000 0.2000 0.4000 0.6000 0.8000 0.9000 0.9200 0.9400 0.9600 0.9800 0.9999 1.0000
y(x)
Exact solution
1.0000000 0.8187149 0.6702932 0.5487787 0.4492931 0.4065330 0.3984824 0.3905911 0.3828562 0.3752743 0.7357356 1.3678790
1.0000000 0.8187308 0.6703200 0.5488117 0.4493290 0.4065697 0.3985191 0.3906278 0.3828929 0.3753111 0.7356979 1.3678790
Clearly this problem has a boundary layer at x = 1. The exact solution is given by y(x) = e(1+ε)(x−1)/ε + e−x
.
The numerical results are given in Tables 15 and 16 for ε = 10−3 and 10−4 , respectively.
The numerical results are given in Tables 13 and 14 for ε = 10−3 and 10−4 , respectively.
9. DISCUSSION AND CONCLUSIONS We have described and illustrated an initial-value technique Example 8.2: Now we consider the following singular perturfor solving singularly perturbed two-point boundary value probbation problem lems. The solution of the given singularly perturbed boundary value problem is computed numerically by solving three initialεy (x) − y (x) − (1 + ε)y(x) = 0, x ∈ [0, 1] value problems, which are deduced from the original problem. This technique is very easy to implement on any computer with y(0) = 1 + exp(−(1 + ε)/ε); and y(1) = 1 + 1/e. with minimum problem preparation. We have implemented the present method on three linear examples, three non-linear examples with left-end boundary layer, and two examples with rightTABLE 15 end boundary layer by taking different values of ε .To solve the Numerical results of Example 8.2 with ε = 10−3 , h = 10−3 initial-value problems, we used the cubic spline method. It can be seen from the numerical results that the given initial-value x y(x) Exact solution technique for singularly perturbed boundary value problem is accurate and suitable to solve linear, non-linear problems with 0.000 1.0000000 1.0000000 extremely thin layers. Further, in this way, we can take full ad0.200 0.8185673 0.8187308 vantage of well-known results about the numerical solution of 0.400 0.6700525 0.6703200 initial-value problems for ordinary differential equations. 0.600 0.5484830 0.5488116 0.800 0.4489703 0.4493290 0.900 0.4062046 0.4065697 REFERENCES 0.920 0.3981533 0.3985190 1. C.M. Bender and S.A. Orszag, Advanced Mathematical Methods for Sci0.940 0.3902616 0.3906278 entists and Engineers, McGraw-Hill, New York, 1978. 0.960 0.3825263 0.3828929 2. D.J. Fyfe, The Use of Cubic Splines in the Solution of Two-point Boundary 0.980 0.3749442 0.3753111 Value Problems, The Computer Journal, vol. 12, pp. 188–192, 1969. 3. J. Jayakumar, Improvement of Numerical Solution by Boundary Value Tech0.999 0.7355268 0.7357859 nique for Singularly Perturbed One Dimensional Reaction Diffusion Prob1.000 1.3678790 1.3678790 lem, Applied Mathematics and Computation, vol. 142, pp. 417–447, 2003.
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