Non-Polynomial Spline Solution of Fourth-Order Obstacle ... - CiteSeerX
World Academy of Science, Engineering and Technology 52 2009
Non-Polynomial Spline Solution of Fourth-Order Obstacle Boundary-Value Problems Jalil Rashidinia, Reza Jalilian
of (1) for arbitrary choices of f (x), g(x). A special form of problem (1) have been considered by the numbers of authors [1-4,13,14] they used finite difference, collocation and spline methods. In the present paper, we apply non-polynomial spline functions [16,17,20] that have a polynomial and trigonometric part to develop numerical methods for obtaining smooth approximation to the solutions of such system. These methods are based on a nonKeywords—Quintic non-polynomial spline, Boundary for- polynomial spline space. The spline functions we mula, Convergence, Obstacle problems. propose in this paper have the form Abstract— In this paper we use quintic non-polynomial spline functions to develop numerical methods for approximation to the solution of a system of fourth-order boundaryvalue problems associated with obstacle, unilateral and contact problems. The convergence analysis of the methods has been discussed and shown that the given approximations are better than collocation and finite difference methods. Numerical examples are presented to illustrate the applications of these methods, and to compare the computed results with other known methods.
a sin(kx) + b cos(kx) + cx3 + dx2 + ex + f. I. I NTRODUCTION In this paper, we apply non-polynomial spline functions to develop numerical methods for obtaining smooth approximations to the solution of a system of fourth-order boundary-value problem of the form: f (x),
a ≤ x ≤ c, u(4) = g(x)u(x) + f (x) + r, c ≤ x ≤ d, f (x), d ≤ x ≤ b,
(1)
subjected to the boundary and continuity conditions u(a) = u(b) = α1 , u′′ (a) = u′′ (b) = α2 , u(c) = u(d) = β1 , u′′ (c) = u′′ (d) = β2 ,
We develop the class of various methods. Our method perform better than the other collocation, finite difference and spline methods of same order. This approach has the advantage over finite difference methods that it provides continuous approximations to not only for u(x) but also for u(i) (x), i = 1, 2, 3, at every point of the range of integration. Also, the c∞ -differentiability of the trigonometric part of non-polynomial spline compensates for the loss smoothness inherited by polynomial spline. The spline function we propose in this paper has the form Span{1, x, x2 , x3 , sin(|k|x), cos(|k|x)},
(2)
where f (x) and g(x) are continuous functions on [a, b] and [c, d], respectively. The parameters r, αi and βi , (i = 1, 2) are real constants. Such type of system arise in the study of obstacle, unilateral, moving and free boundary-value problems and has important applications in other branches of pure and applied science [1-5,10,12-14]. In general it is not possible to obtain the analytical solution
where k is the frequency of trigonometric part of the spline function,when k → 0 our spline reduce to the form: Span{1, x, x,2 , x3 , x4 , x5 },
(when k=0).
The above fact is evident when correlation between polynomial and non-polynomial splines basis is investigated in the following manner, T5 = span{1, x, x2 , x3 , sin(kx), cos(kx)} 24 (kx)2 ), = span{1, x, x2 , x3 , 4 (cos(kx) − 1 + k 2 120 (kx)3 (sin(kx) − (kx) + )}. k5 6
J. Rashidinia is with the School of Mathematics, Iran University of Science & Technology Narmak, Tehran 16844, Iran e-mail: [email protected]. R. Jalilian is with the Department of Mathematics, Ilam University, PO Box 69315516, Ilam, Iran, e-mail: [email protected]
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From the above equation it follows that limk→0 T5 = {1, x, x2 , x3 , x4 , x5 }, so that the Usmani’s method [18], based on quintic splines is a special case (k = 0) of our approach.
ej =
II. C LASS OF M ETHODS
uj+1 − uj h 3 h [(6 − 2θ2 )Sj − (6 + θ2 )Sj+1 ] + 6θ4 h(mj+1 + 2mj ) , (6) − 6
and θ = kh. Using the continuity of the first and third derivatives at (xj , uj ), we get the following relation for j = 1, 2, . . . , n − 1:
For simplicity we first develop the quintic nonpolynomial spline for solving the fourth-order boundary value problem d4 u = g(x)u + f (x), for x ∈ [c, d], dx4 u′′ (c) = u′′ (d) = β2 . (3) u(c) = u(d) = β1 ,
6(uj+1 − 2uj + uj−1 ) h2 6(Sj+1 − 2Sj cos θ + Sj ) 3(Sj + Sj−1 ) − + hk 3 sin θ k2 2 2 (6 + θ )Sj+1 − (12 − θ )Sj − k2 θ2 2 (6 − 2θ )Sj−1 + , (7) k2 θ2
mj+1 + 4mj + mj−1 =
For this purpose, we divide the interval [c,d] into n equal subintervals using the grid points. Let u(x) be the exact solution of the boundary-value problem (3) and uj be an approximation to u(xj ), in order to develop the numerical method for approximating solution of differential equations (3), we introduce and the set {xj } so that xj = c + jh, h = d−c ,j = n 0, 1 . . . , n, the non-polynomial quintic spline pj (x) in subinterval xj ≤ x ≤ xj+1 , has the form
2Sj − Sj−1 − Sj+1 k2 h(Sj+1 − 2Sj cos θ + Sj−1 ) . (8) + k sin θ
mj+1 − 2mj + mj−1 =
pj (x) = aj sin k(x − xj ) + bj cos k(x − xj ) Using equations (7) and (8), we get the follow+cj (x − xj )3 + dj (x − xj )2 ing scheme: +ej (x − xj ) + lj , j = 0, 1, . . . , n, (4) uj+2 − 4uj+1 + 6uj − 4uj−1 + uj−2 = h4 [α(Sj+2 + Sj−2 ) +β(Sj+1 + Sj−1 ) + γSj ],
where aj , bj , cj , dj , ej and lj are constants and k is free parameter. If k → 0 then pj (x) reduces to quintic spline in [c,d]. By using continuity conditions at the common nodes (xj , uj ), and to derive expression for the coefficients of (4) in terms of uj , uj+1 , mj , mj+1 , Sj and Sj+1 we have:
where j = 2, 3, . . . , n − 2 and α=
pj (xj ) = uj , pj (xj+1 ) = uj+1 , (2) (2) pj (xj ) = mj , pj (xj+1 ) = mj+1 , (4) (4) pj (xj ) = Sj , pj (xj+1 ) = Sj+1 .
β=
(5)
Using the (5) we get the following expressions: h4 Sj h4 Sj , l = u − , j j θ4 θ4 Sj+1 − Sj cos θ k 2 m j + Sj , d = , aj = j k 4 sin θ 2k 2 h(Sj+1 − Sj ) + θk(mj+1 − mj ) cj = , 6θ2
(9)
γ=
bj =
θ3 − 6(θ − sin θ) , 6θ4 sin θ
12θ(1 + cos θ) − 2θ3 (cos θ − 2) − 24 sin θ , 6θ4 sin θ
36 sin θ − 12θ(1 + 2 cos θ) − 2θ3 (4 cos θ − 1) . 6θ4 sin θ
1 26 66 , 120 , 120 ). If θ → 0 then (α, β, γ) → ( 120 (4) Using uj = gj uj + fj + r, fj ≡ f (xj ), uj ≡ u(xj ), gj ≡ g(xj ), at nodal points xj and by Taylor expansion, the local truncation errors tj , j =
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2, 3, ...n − 2, associated with our scheme is:
III. D EVELOPMENT OF THE B OUNDARY F ORMULAS For discretization of boundary conditions we define:
(4)
tj = (1 − 2(α + β) − γ)h4 uj 1 (6) +( − (4α + β))h6 uj 6 1 16 1 (8) +( − ( α + β))h8 uj 80 12 12 17 1 (10) +( − (64α + β))h10 uj 30240 360 31 4α 2β (12) +( −( + ))h12 uj (ζj ) 1814400 315 20160 +O(h13 ).
(i) (ii)
1 6 h M6 +O(h7 ), j = 2, 3, ..., n−2. 12 (10) (ii)Second-Order Method −6 72 For α = 4319 , β = 400 and γ = 1 − 2α − 2β gives: (4)
2519 6 h M6 + O(h7 ), (11) 323925 j = 2, 3, ..., n − 2. (4)
δ 4 uj = h4 uj −
(iii) Fourth-Order Method For α = 0, β = 16 − 4α and γ = 1 − 2α − 2β gives: (4)
(4)
+O(h9 ), j = 2, 3, ..., n − 2.
1 8 h M8 720 (12)
(iv) Sixth-Order Method −1 3 For α = 720 , β = 20 − 16α and γ = 1 − 2α − 2β gives: h4 (4) (4) (4) (4) [−(uj+2 + uj−2 ) + 124(uj+1 + uj−1 ) 720 1 10 (4) +474(uj )] + h M10 3024 +O(h11 ), j = 2, 3, ..., n − 2, (13) δ 4 uj =
where M6 = maxc≤x≤d | u(6) (x) |, M8 = maxc≤x≤d | u(8) (x) |, M10 = maxc≤x≤d | u(10) (x) |. Each of the above recurrence relations gives n − 2 linear equations in n unknowns, we need two more equations at each end of the rang of integration.
k=0 3
′′
3
(4)
d′k uk + t1 = 0,
k=0
b′k un−k
k=0
δ 4 uj = h4 uj −
(4)
b′k uk + c′ h2 u0 + h4
+tn = 0,
For different choices of parameters α, β and γ we get the class of methods such as: (i) Second-Order Method 1 26 , β = 120 and γ = 1 − 2α − 2β gives: For α = 120
δ 4 uj = h4 [uj+1 + 4uj + uj−1 ] −
3
′ 2
′′
4
+ c h un + h
3
(4)
d′k un−k
k=0
(14)
where b′k , c′ and d′k are arbitrary parameters to be determined. In order to obtain the second-order method we find that: (b′0 , b′1 , b′2 , b′3 , c′ ) = (−2, 5, −4, 1, 1), 1 , −1, 0, 0). (d′0 , d′1 , d′2 , d′3 ) = ( 12 We obtain the second order boundary formulas as follows: 1 (5 − h4 g1 )u1 − 4u2 + u3 = (2 − h4 g0 )β1 12 2 4 1 −h β2 − h ( (f0 + r) − (f1 + r)) 12 59 6 (6) + h u (x1 ) + O(h7 ), 360 un−3 − 4un−2 + (5 − h4 gn−1 )un−1 1 = (2 − h4 gn )β1 − h2 β2 − h4 (−(fn−1 + r) 12 1 59 6 (6) + (fn + r)) + h u (xn ) + O(h7 ). (15) 12 360 For four-order method we find that: (b′0 , b′1 , b′2 , b′3 , c′ ) = (−2, 5, −4, 1, 1), −1 (d′0 , d′1 , d′2 , d′3 ) = 360 (28, 245, 56, 1), and 56 4 245 4 h g1 )u1 + (−4 − h g2 )u2 (5 − 360 360 h4 28 4 +(1 − g3 )u3 = (2 + h g0 )β1 − h2 β2 360 360 h4 + (28(f0 + r) + 245(f1 + r) + 56(f2 + r) 360 241 8 (8) +(f3 + r) − h u (ζ1 ) + O(h9 ), 60480 1 4 56 4 (1 − h gn−3 )un−3 + (−4 − h gn−2 )un−2 360 360 245 4 28 4 h gn−1 )un−1 = (2 + h gn )β1 +(5 − 360 360 h4 (28(fn + r) + 245(fn−1 + r) −h2 β2 + 360 +56(gn−1 + r) + (gn−3 + r))
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241 8 (8) h u (ζn ) + O(h9 ). (16) actual solution u(xj ) at x = xj and A is nine-band 60480 matrix which can be described as For discretization of boundary conditions for A = M +BG, G = h4 diag(gj ), j = 1, 2, . . . , n−1, sixth-order method we define: (20) 2 here M = P , where P = (pij ), is a tridigonal and 5 3 ′ ′ 2 ′′ 4 ′ (4) monotone matrix defined by: b u =ch u +h d u +t , (i) −
0
k k
(ii)
k=0 3
k k
1
2,
k=0
i = j = 1, 2, ..., N − 1, pij = −1, |i − j| = 1, 0, otherwise,
b′k un−k = c′∗ h2 un ′′
k=0
+h4
5
(4)
d′5−k un−5+k + tn .
and
(17)
k=0
B=
In order to obtain the truncation errors of t1 and t2 we find that: (b′0 , b′1 , b′2 , b′3 , c′ , c′∗ ) = (−2, 5, −4, 1, 1, −1), (d′0 , d′1 , d′2 , d′3 , d′4 , d′5 ) −1 = (4233, 43274, 5662, 3432, −1391, 230), 60480 and (5 − d′1 h4 g1 )u1 + (−4 − d′2 h4 g2 )u2 +(1 − d′3 h4 g3 )u3 − (d′4 h4 g4 )u4 − (d′5 h4 g5 )u5 = (2 − d′0 h4 g0 )β1 − h2 β2 + h4 (d′0 (f0 + r) +d′1 (f1 + r) + d′2 (f2 + r) + d′3 (f3 + r) +d′4 (f4 + r) + d′5 (f5 + r)) 167 10 (10) − h u (ζ1 ) + O(h11 ), 50400 (5 − d′1 h4 gn−1 )un−1 + (−4 − d′2 h4 gn−2 )un−2 +(1 − d′3 h4 gn−3 )un−3 − (d′4 h4 gn−4 )un−4 −(d′5 h4 gn−5 )un−5 = (2 − d′0 h4 gn )β1 + h2 β2 +h4 (d′0 (fn + r) + d′1 (fn−1 + r) + d′2 (fn−2 + r) +d′3 (fn−3 + r) + d′4 (fn−4 + r) + d′5 (fn−5 + r)) 167 10 (10) h u (ζn ) + O(h11 ). − (18) 50400 IV. C ONVERGENCE A NALYSIS Here we prove the convergence of the methods. Let us write the error equation of the methods as follows: AE = T,
d1 d2 d3 d4 d5 β γ β α α β γ β α α β γ β α ... ... ... ... ... .. .. .. .. .. . . . . . α β γ β α α β γ β d5 d4 d3 d2 d1
(21)
,
(22)
E = A−1 T = [M + BG]−1 T, |E ≤ [I + M −1 BG]−1 M −1 T . (23) By using (I + A)−1 ≤ (1 − A)−1 and Usmani et. al. [19] we obtain M −1 ≤
5(d − c)4 + 4(d − c)2 h2 , 384h4
(24)
M −1 T . 1 − M −1 BG
(25)
E ≤
Provided that M −1 BG < 1. Also we can obtain 724 B ≤ , G ≤ h4 Mg , Mg = maxc≤x≤d |g(x)|. 720 (26) For second order we obtain 59h6 M6 , M6 = maxc≤ζ≤d |u(6) (ζ)|, 360 using (24)-(26) we obtain T ≤
(19)
E ≤
118ωh6 M6 ≡ O(h2 ), 276480h4 − 724ωG
where E = (ej ), is the (n-1)-dimensional column vector with ej , the error of discretization defined by for fourth order we get ej = u(xj ) − uj . In other words ej is the amount 241h8 M8 , M8 = maxc≤ζ≤d |u(8) (ζ)|, T ≤ by which computed solution uj deviates from the 60480 203
(27)
World Academy of Science, Engineering and Technology 52 2009
using (24)-(26 we obtain 17352ωh8 M8 ≡ O(h4 ), 1672151040h4 − 4378752ωG (28) and for sixth order we get E ≤
167h10 M10 , M10 = maxc≤ζ≤d |u(10) (ζ)|. 50400 By using (24)-(26,we get T ≤
[-1,1] by using the Quintic non-polynomial spline methods with step lengths h = 2−m , m = 3, 4, 5. The maximum absolute errors in solution for our various methods are listed in tables 1 and also the maximum absolute errors in the solution at middle points of interval are tabulated in table 2. To compare our computed results obtained by second and fourth order methods with the results obtained by other known methods in [1-4,13,14], the maximum absolute errors in the solution of example 1 are listed in tables 3,4 and 5.
12024ωh10 M10 ≡ O(h6 ), 1393459200h4 − 3648960ωG (29) Spline approach has the advantage over finite where ω = 5(d − c)4 + 4(d − c)2 h2 , G = difference method that it provides continuous max|g(x)|, c ≤ x ≤ d provided approximations to u(i) (x), i = 1, 2, 3, at every point 4 69120h of the range of integration beside approximation G < . to u(x). Following [20] to obtain the necessary 181ω It follows E → 0 as h → 0. Therefore the formula for computing values of first, second and convergence of the methods have been established. third derivatives of solution of example 1, by using equation (7), (8) and solving the resulting identity for mj , j = 1, ..., n we have V. N UMERICAL R ESULTS E ≤
We consider the system of differential equations [1-4,13-15]
u
(4)
=
, 1, −1 ≤ x ≤ −1 2 −1 1 2 − 4u, 2 ≤ x ≤ 2 ,
1 2
≤ x ≤ 1,
and the conditions of continuity of u and u′′ atx = −1 and 12 . 2
Γ1 (x),
u(x) = Γ2 (x), Γ (x), 3
(uj+1 −2uj +uj−1 ) h2
h2 (Sj −Sj−1 ) 2θ2
−
−
+
h2 (Sj+1 −2Sj cos θ+Sj−1 ) 3 sin θ
h2 [(6+θ2 )Sj+1 −(12−θ2 )Sj +(6−2θ2 )Sj−1 ] 6θ4
(2Sj −Sj−1 −Sj+1 ) 6θ2
(30) −
with the boundary conditions: 1 −1 u(−1) = u( ) = u( ) = u(1) = 0, 2 2 −1 1 u′′ (−1) = u′′ ( ) = u′′ ( ) 2 2 = u′′ (1) = 0, (31)
The analytical solution problem is
mj =
−
h2 (Sj+1 −2Sj cos θ+Sj−1 ) , 6θ2 sin θ
with m0 = mn+1 = β2 being known from the boundary conditions. Having computed uj , mj , Sj , j = 0, ..., n + 1, it is possible to evaluate the coefficient of the spline function (4) as given by (6). Since yj′ = p′j (xj ), j = 0, ..., n and ′ yn+1 = p′n (xn+1 ), it follows that
aj k + ej , j = 0, ..., n, Ψn + 3cn h2 + 2dn h + en , j = n + 1, for this boundary value (33) where Ψn = an k cos θ − bn k sin θ. Similarly, from ′′′ ′′′ yj′′′ = p′′′ j (xj ), j = 0, ..., n and yn+1 = pn (xn+1 ), we −1 ≤ x ≤ −1 , 2 −1 ≤ x ≤ 12 , (32) can obtain 2 1 ≤ x ≤ 1, 2
where 1 4 3 1 Γ1 (x) = 24 x + 18 x3 + 18 x2 + 64 x + 192 , 1 Γ2 (x) = 0.5− ϕ1 [ϕ2 sin x sinh x+ϕ3 cos x cosh x], 1 4 3 1 x − 18 x3 + 18 x2 − 64 x + 192 , Γ3 (x) = 24 1 ϕ1 = cos(1) + cosh(1), ϕ2 = sin( 2 ) sinh( 12 ), ϕ3 = cos( 12 ) cosh( 12 ). We solved this example over the whole interval
yj′
≈
−aj k 3 + 6cj , j = 0, ..., n, −an k 3 cos θ + bn k 3 sin θ + 6cn , j = n + 1. (34) (i) The values of u (x), i = 1, 2, 3 have been computed by our second order method (i). To compare with the method in [1] the maximum absolute errors are listed in tables 6.
yj′′′
204
≈
World Academy of Science, Engineering and Technology 52 2009
TABLE III M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1
Example 2: We consider the system of differential equation solved by Al-Said and Noor [1].
u
(4)
=
, 0, −1 ≤ x ≤ −1 2 1 1 − 4u, −1 ≤ x ≤ , 2 2
1 2
m
Our fourth-order
3 4 5
1.7×10−8 4.1×10−10 1.3×10−11
Fourth-order[4]
Finite difference[14]
≤ x ≤ 1, (35)
with the boundary conditions: 1 −1 ) = u( ) = u(1) = 0, 2 2 −1 1 u′′ (−1) = −u′′ ( ) = u′′ ( ) 2 2 = −u′′ (1) = ǫ, (36)
1.3×10−6 8.7×10−8 6.8×10−9
TABLE IV M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1
u(−1) = u(
where ǫ → 0. The analytical solution for this boundary value problem is Λ1 (x),
for − 1 ≤ x ≤ −1 , 2 1 for −1 ≤ x ≤ , u(x) = Λ2 (x), 2 2 Λ (x) =, for 1 ≤ x ≤ 1, 3 2
2.4×10−8 1.5×10−9 9.5×10−11
h
Second-order(i)
Second-order(ii)
1/12 1/24 1/48
4.5×10−6 1.2×10−6 3.3×10−7
1.2×10−7 2.3×10−7 3.2×10−8
Second-order[1]
1.2×10−5 2.8×10−6 6.9×10−7
VI. C ONCLUSION (37)
where x3 − 32 x2 − 13 x − 14 )ǫ, Λ1 (x) = ( −2 3 12 1 Λ2 (x) = 0.25− 2ϕ1 [ϕ2 sin x sinh x+ϕ3 cos x cosh x], x3 + 32 x2 − 13 x + 14 )ǫ, Λ3 (x) = ( −2 3 12 ϕ1 = cos(1) + cosh(1), ϕ2 = sin( 12 ) sinh( 12 ), ϕ3 = cos( 12 ) cosh( 12 ). We solved this example over the whole interval [-1,1] by using our second order methods (i),(ii), with step lengths h = 2−m , m = 3, 4, 5. The maximum absolute error in solution are listed in table 7 and 8, our results compared with the results obtained in [1,2,11]. The results shows superiority of our second orders methods. TABLE I M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1 m
O(h2 )(i)
O(h2 )(ii)
O(h4 )
O(h6 )
3 4 5
8.19×10−6 2.73×10−6 7.44×10−7
9.04×10−7 2.26×10−7 7.19×10−8
1.69×10−8 4.08×10−10 1.30×10−11
7.65×10−11 3.31×10−13 6.41×10−15
TABLE II M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1 IN MIDDLE POINTS
m
O(h2 )(i)
O(h2 )(ii)
O(h4 )
O(h6 )
3 4 5
1.25×10−6 1.56×10−7 1.95×10−8
2.26×10−7 7.19×10−8 1.84×10−8
4.40×10−10 1.30×10−11 4.06×10−13
5.95×10−13 5.49×10−15 –
We have developed a new non-polynomial quintic spline for solving a system of fourth-order boundary-value problems. This approach has the advantages over finite difference methods that it provides continuous approximations to not only for u(x) but also for u(i) (x), i = 1, 2, 3, at every point of the range of integration. Our numerical results are better than those produced by collocation and finite difference methods for solution of equation(1). R EFERENCES [1] AL-SAID, E.A. and NOOR, M.A., Quartic Spline Method for Solving Fourth-Order Obstacle Boundary Value Problems, Journal of Computational and Applied Mathematics Vol.143,pp.107116,2002. [2] AL-SAID,E.A. NOOR,M.A. Computational Methods for FourthOrder Obstacle Boundary Value Problems, Comm. Appl. Nonlinear. Anal. Vol.2,pp.73-83,1995. [3] AL-SAID, E.A. NOOR,M.A. and RASSIAS,T.M., Cubic Splines Method for Solving Fourth-Order Obstacle Problems, Appl. Math. Comput.,Vol.174, pp.180-187,2006. [4] AL-SAID,E.A., NOOR,M.A., KAYA,D., Al-KHALED,K., Finite difference method for solving fourth-order obstacle problems, Int. J. Comput. Math. Vol.81,pp.741-748,2004. [5] BAIOCCHI,C.C. and CALEO,A., Variational and quasivariational Inequalities, John Wiley and Sons, New York, 1984. [6] CHAWLA,M.M. and SUBRAMANIAN, High accuracy quintic spline solution of fourth-order two-Point boundary value problems, Int. J. Computer. Math. Vol.31,pp.87-94,1989. [7] HENRICI,P., Discrete variable method in ordinary differential equations, John Wiley, New York, 1961. [8] JAIN,M.K., Numerical solution of differential equations, Second Editions,Wiley Eastern Limited, 1984. [9] JAIN,M.K., IYANGER,S.R.K. and SOLDHANHA,J.S.V., Numerical solution of a fourth-order ordinary differential equation, J. Engg. Math. Vol.11,pp.373-380,1977. [10] KIKUCHI,N., and ODEN,J.T., Contact problem in elasticity, SIAM, Publishing Co. Philadelphia,1988.
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TABLE VII M AXIMUM ABSOLUTE ERRORS FOR EXAMPLE 2
TABLE V M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1 h
Second-order In [2]
1/12 1/24 1/48
m 3 4 5
Second-order[3]
6.2×10−5 1.6×10−5 3.9×10−6
7.8×10−6 1.9×10−6 4.9×10−7
Second-order(i)
Second-order(ii)
4.66×10−6 1.39×10−6 3.74×10−7
4.45×10−7 1.32×10−7 4.25×10−8
AND
||u′′ (xi ) − u′′ i ||∞
u′′′
FOR EXAMPLE
m
||u′ (xi ) − u′i ||∞
3 4 5
6.28×10−5 1.77×10−5 4.71×10−6
Our O(h2 )(i) 1.94×10−4 5.51×10−5 1.46×10−5
2.83×10−4 8.35×10−5 2.25×10−5
3 4 5
8.81×10−5 3.17×10−5 1.06×10−5
O(h2 ) in[1] 1.30×10−3 3.66×10−4 9.66×10−5
2.08×10−2 5.76×10−3 1.72×10−3
m 3 4 5
1
||u′′′ (xi ) − u′′′ i ||∞
ǫ = 10−6
Second-order In [1]
1.3×10−5 3.2×10−6 8.1×10−7
TABLE VIII M AXIMUM ABSOLUTE ERRORS FOR EXAMPLE 2 TABLE VI M AXIMUM ABSOLUTE ERRORS OF u′ , u′′
WHEN
WHEN
Second-order In [11]
Second-order In [2]
3.0×10−4 7.0×10−5 1.4×10−5
1.4×10−4 3.6×10−5 8.9×10−6
ǫ = 10−6
Jalil Rashidinia, Associate professor, School of Mathematics, Iran University of Science & Technology Narmak, Tehran 16844, Iran e-mail: [email protected] Nationality: Iranian Research Interests: Applied mathematics and computational sciences Numerical solution of ODE, PDE and IE Spline approximations Numerical linear algebra
[11] KHALIFA,A.K. and NOOR,M.A., Quintic spline solutions of a class of contact problems, Math. Comput. Modlelling Vol.13,pp.51-58,1990. [12] LEWY,H. and STAMPACCHIA,G., On the regularity of the solution of the variational inequalities, Comm. pure and Appl. Math. Vol.22,pp.153-188,1969. [13] NOOR,M.A. and Al-SAID,E.A., Fourth order obstacle problems.In: Th.M.Rassias and H.M.Srivastava(Eds), Analytic and geometric inequalities and applications, kluwer Academic Publishers, Dordrecht, Netherlands,pp.277-300,1999. [14] NOOR,M.A. and AL-SAID,E.A., Numerical solution of fourth order variational inequalities, Inter. J. Comput. Math. Vol.75,pp.107-116,2000. [15] PAPAMICHEL,N. and WORSEY,E.A., A cubic spline method for the solution of linear fourth-order two-point boundary value problem, J. Comput. Appl. Math. Vol.7,pp.187-189,1981. [16] RASHIDINIA, J., Applications of splines to the numerical solution of differential equations, Ph.D. thesis, Aligarh Muslim University, Aligarh, India, 1994. [17] SIRAJ-UL-ISLAM, TIRMIZI,S.I.A. and SAADAT ASHRAF, A class of method based on non-polynomial spline function for the solution of a special fourth-order boundary-value problems with engineering applications, Appl. Math. Comput. Vol.174,pp.11691180,2006. [18] USMANI,R.A. and WARSI,S.A., Smooth spline solutions for boundary value problems in plate deflection thoery, Comput. Maths. with Appls. Vol.6, pp.205-211,1980. [19] USMANI,R.A., Discrete variable method for a boundary value problem with engineering applications, Math. Comput. Vol.32,1087-1096,1978. [20] Van DAELE,M., VANDEN BERGHE,G. and De MEYER,H., A smooth approximation for the solution of a fourth-order boundary value problem based on non-polynomial splines, J. Comput. Appl. Math. Vol.pp.51,383-394,1994.
Reza Jalilian Assistant professor, Department of Mathematics, Ilam University, PO Box 69315516, Ilam, Iran e-mail: [email protected] Research Interests: Spline approximations
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Non-Polynomial Spline Solution of Fourth-Order Obstacle Boundary-Value Problems Jalil Rashidinia, Reza Jalilian
of (1) for arbitrary choices of f (x), g(x). A special form of problem (1) have been considered by the numbers of authors [1-4,13,14] they used finite difference, collocation and spline methods. In the present paper, we apply non-polynomial spline functions [16,17,20] that have a polynomial and trigonometric part to develop numerical methods for obtaining smooth approximation to the solutions of such system. These methods are based on a nonKeywords—Quintic non-polynomial spline, Boundary for- polynomial spline space. The spline functions we mula, Convergence, Obstacle problems. propose in this paper have the form Abstract— In this paper we use quintic non-polynomial spline functions to develop numerical methods for approximation to the solution of a system of fourth-order boundaryvalue problems associated with obstacle, unilateral and contact problems. The convergence analysis of the methods has been discussed and shown that the given approximations are better than collocation and finite difference methods. Numerical examples are presented to illustrate the applications of these methods, and to compare the computed results with other known methods.
a sin(kx) + b cos(kx) + cx3 + dx2 + ex + f. I. I NTRODUCTION In this paper, we apply non-polynomial spline functions to develop numerical methods for obtaining smooth approximations to the solution of a system of fourth-order boundary-value problem of the form: f (x),
a ≤ x ≤ c, u(4) = g(x)u(x) + f (x) + r, c ≤ x ≤ d, f (x), d ≤ x ≤ b,
(1)
subjected to the boundary and continuity conditions u(a) = u(b) = α1 , u′′ (a) = u′′ (b) = α2 , u(c) = u(d) = β1 , u′′ (c) = u′′ (d) = β2 ,
We develop the class of various methods. Our method perform better than the other collocation, finite difference and spline methods of same order. This approach has the advantage over finite difference methods that it provides continuous approximations to not only for u(x) but also for u(i) (x), i = 1, 2, 3, at every point of the range of integration. Also, the c∞ -differentiability of the trigonometric part of non-polynomial spline compensates for the loss smoothness inherited by polynomial spline. The spline function we propose in this paper has the form Span{1, x, x2 , x3 , sin(|k|x), cos(|k|x)},
(2)
where f (x) and g(x) are continuous functions on [a, b] and [c, d], respectively. The parameters r, αi and βi , (i = 1, 2) are real constants. Such type of system arise in the study of obstacle, unilateral, moving and free boundary-value problems and has important applications in other branches of pure and applied science [1-5,10,12-14]. In general it is not possible to obtain the analytical solution
where k is the frequency of trigonometric part of the spline function,when k → 0 our spline reduce to the form: Span{1, x, x,2 , x3 , x4 , x5 },
(when k=0).
The above fact is evident when correlation between polynomial and non-polynomial splines basis is investigated in the following manner, T5 = span{1, x, x2 , x3 , sin(kx), cos(kx)} 24 (kx)2 ), = span{1, x, x2 , x3 , 4 (cos(kx) − 1 + k 2 120 (kx)3 (sin(kx) − (kx) + )}. k5 6
J. Rashidinia is with the School of Mathematics, Iran University of Science & Technology Narmak, Tehran 16844, Iran e-mail: [email protected]. R. Jalilian is with the Department of Mathematics, Ilam University, PO Box 69315516, Ilam, Iran, e-mail: [email protected]
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From the above equation it follows that limk→0 T5 = {1, x, x2 , x3 , x4 , x5 }, so that the Usmani’s method [18], based on quintic splines is a special case (k = 0) of our approach.
ej =
II. C LASS OF M ETHODS
uj+1 − uj h 3 h [(6 − 2θ2 )Sj − (6 + θ2 )Sj+1 ] + 6θ4 h(mj+1 + 2mj ) , (6) − 6
and θ = kh. Using the continuity of the first and third derivatives at (xj , uj ), we get the following relation for j = 1, 2, . . . , n − 1:
For simplicity we first develop the quintic nonpolynomial spline for solving the fourth-order boundary value problem d4 u = g(x)u + f (x), for x ∈ [c, d], dx4 u′′ (c) = u′′ (d) = β2 . (3) u(c) = u(d) = β1 ,
6(uj+1 − 2uj + uj−1 ) h2 6(Sj+1 − 2Sj cos θ + Sj ) 3(Sj + Sj−1 ) − + hk 3 sin θ k2 2 2 (6 + θ )Sj+1 − (12 − θ )Sj − k2 θ2 2 (6 − 2θ )Sj−1 + , (7) k2 θ2
mj+1 + 4mj + mj−1 =
For this purpose, we divide the interval [c,d] into n equal subintervals using the grid points. Let u(x) be the exact solution of the boundary-value problem (3) and uj be an approximation to u(xj ), in order to develop the numerical method for approximating solution of differential equations (3), we introduce and the set {xj } so that xj = c + jh, h = d−c ,j = n 0, 1 . . . , n, the non-polynomial quintic spline pj (x) in subinterval xj ≤ x ≤ xj+1 , has the form
2Sj − Sj−1 − Sj+1 k2 h(Sj+1 − 2Sj cos θ + Sj−1 ) . (8) + k sin θ
mj+1 − 2mj + mj−1 =
pj (x) = aj sin k(x − xj ) + bj cos k(x − xj ) Using equations (7) and (8), we get the follow+cj (x − xj )3 + dj (x − xj )2 ing scheme: +ej (x − xj ) + lj , j = 0, 1, . . . , n, (4) uj+2 − 4uj+1 + 6uj − 4uj−1 + uj−2 = h4 [α(Sj+2 + Sj−2 ) +β(Sj+1 + Sj−1 ) + γSj ],
where aj , bj , cj , dj , ej and lj are constants and k is free parameter. If k → 0 then pj (x) reduces to quintic spline in [c,d]. By using continuity conditions at the common nodes (xj , uj ), and to derive expression for the coefficients of (4) in terms of uj , uj+1 , mj , mj+1 , Sj and Sj+1 we have:
where j = 2, 3, . . . , n − 2 and α=
pj (xj ) = uj , pj (xj+1 ) = uj+1 , (2) (2) pj (xj ) = mj , pj (xj+1 ) = mj+1 , (4) (4) pj (xj ) = Sj , pj (xj+1 ) = Sj+1 .
β=
(5)
Using the (5) we get the following expressions: h4 Sj h4 Sj , l = u − , j j θ4 θ4 Sj+1 − Sj cos θ k 2 m j + Sj , d = , aj = j k 4 sin θ 2k 2 h(Sj+1 − Sj ) + θk(mj+1 − mj ) cj = , 6θ2
(9)
γ=
bj =
θ3 − 6(θ − sin θ) , 6θ4 sin θ
12θ(1 + cos θ) − 2θ3 (cos θ − 2) − 24 sin θ , 6θ4 sin θ
36 sin θ − 12θ(1 + 2 cos θ) − 2θ3 (4 cos θ − 1) . 6θ4 sin θ
1 26 66 , 120 , 120 ). If θ → 0 then (α, β, γ) → ( 120 (4) Using uj = gj uj + fj + r, fj ≡ f (xj ), uj ≡ u(xj ), gj ≡ g(xj ), at nodal points xj and by Taylor expansion, the local truncation errors tj , j =
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2, 3, ...n − 2, associated with our scheme is:
III. D EVELOPMENT OF THE B OUNDARY F ORMULAS For discretization of boundary conditions we define:
(4)
tj = (1 − 2(α + β) − γ)h4 uj 1 (6) +( − (4α + β))h6 uj 6 1 16 1 (8) +( − ( α + β))h8 uj 80 12 12 17 1 (10) +( − (64α + β))h10 uj 30240 360 31 4α 2β (12) +( −( + ))h12 uj (ζj ) 1814400 315 20160 +O(h13 ).
(i) (ii)
1 6 h M6 +O(h7 ), j = 2, 3, ..., n−2. 12 (10) (ii)Second-Order Method −6 72 For α = 4319 , β = 400 and γ = 1 − 2α − 2β gives: (4)
2519 6 h M6 + O(h7 ), (11) 323925 j = 2, 3, ..., n − 2. (4)
δ 4 uj = h4 uj −
(iii) Fourth-Order Method For α = 0, β = 16 − 4α and γ = 1 − 2α − 2β gives: (4)
(4)
+O(h9 ), j = 2, 3, ..., n − 2.
1 8 h M8 720 (12)
(iv) Sixth-Order Method −1 3 For α = 720 , β = 20 − 16α and γ = 1 − 2α − 2β gives: h4 (4) (4) (4) (4) [−(uj+2 + uj−2 ) + 124(uj+1 + uj−1 ) 720 1 10 (4) +474(uj )] + h M10 3024 +O(h11 ), j = 2, 3, ..., n − 2, (13) δ 4 uj =
where M6 = maxc≤x≤d | u(6) (x) |, M8 = maxc≤x≤d | u(8) (x) |, M10 = maxc≤x≤d | u(10) (x) |. Each of the above recurrence relations gives n − 2 linear equations in n unknowns, we need two more equations at each end of the rang of integration.
k=0 3
′′
3
(4)
d′k uk + t1 = 0,
k=0
b′k un−k
k=0
δ 4 uj = h4 uj −
(4)
b′k uk + c′ h2 u0 + h4
+tn = 0,
For different choices of parameters α, β and γ we get the class of methods such as: (i) Second-Order Method 1 26 , β = 120 and γ = 1 − 2α − 2β gives: For α = 120
δ 4 uj = h4 [uj+1 + 4uj + uj−1 ] −
3
′ 2
′′
4
+ c h un + h
3
(4)
d′k un−k
k=0
(14)
where b′k , c′ and d′k are arbitrary parameters to be determined. In order to obtain the second-order method we find that: (b′0 , b′1 , b′2 , b′3 , c′ ) = (−2, 5, −4, 1, 1), 1 , −1, 0, 0). (d′0 , d′1 , d′2 , d′3 ) = ( 12 We obtain the second order boundary formulas as follows: 1 (5 − h4 g1 )u1 − 4u2 + u3 = (2 − h4 g0 )β1 12 2 4 1 −h β2 − h ( (f0 + r) − (f1 + r)) 12 59 6 (6) + h u (x1 ) + O(h7 ), 360 un−3 − 4un−2 + (5 − h4 gn−1 )un−1 1 = (2 − h4 gn )β1 − h2 β2 − h4 (−(fn−1 + r) 12 1 59 6 (6) + (fn + r)) + h u (xn ) + O(h7 ). (15) 12 360 For four-order method we find that: (b′0 , b′1 , b′2 , b′3 , c′ ) = (−2, 5, −4, 1, 1), −1 (d′0 , d′1 , d′2 , d′3 ) = 360 (28, 245, 56, 1), and 56 4 245 4 h g1 )u1 + (−4 − h g2 )u2 (5 − 360 360 h4 28 4 +(1 − g3 )u3 = (2 + h g0 )β1 − h2 β2 360 360 h4 + (28(f0 + r) + 245(f1 + r) + 56(f2 + r) 360 241 8 (8) +(f3 + r) − h u (ζ1 ) + O(h9 ), 60480 1 4 56 4 (1 − h gn−3 )un−3 + (−4 − h gn−2 )un−2 360 360 245 4 28 4 h gn−1 )un−1 = (2 + h gn )β1 +(5 − 360 360 h4 (28(fn + r) + 245(fn−1 + r) −h2 β2 + 360 +56(gn−1 + r) + (gn−3 + r))
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241 8 (8) h u (ζn ) + O(h9 ). (16) actual solution u(xj ) at x = xj and A is nine-band 60480 matrix which can be described as For discretization of boundary conditions for A = M +BG, G = h4 diag(gj ), j = 1, 2, . . . , n−1, sixth-order method we define: (20) 2 here M = P , where P = (pij ), is a tridigonal and 5 3 ′ ′ 2 ′′ 4 ′ (4) monotone matrix defined by: b u =ch u +h d u +t , (i) −
0
k k
(ii)
k=0 3
k k
1
2,
k=0
i = j = 1, 2, ..., N − 1, pij = −1, |i − j| = 1, 0, otherwise,
b′k un−k = c′∗ h2 un ′′
k=0
+h4
5
(4)
d′5−k un−5+k + tn .
and
(17)
k=0
B=
In order to obtain the truncation errors of t1 and t2 we find that: (b′0 , b′1 , b′2 , b′3 , c′ , c′∗ ) = (−2, 5, −4, 1, 1, −1), (d′0 , d′1 , d′2 , d′3 , d′4 , d′5 ) −1 = (4233, 43274, 5662, 3432, −1391, 230), 60480 and (5 − d′1 h4 g1 )u1 + (−4 − d′2 h4 g2 )u2 +(1 − d′3 h4 g3 )u3 − (d′4 h4 g4 )u4 − (d′5 h4 g5 )u5 = (2 − d′0 h4 g0 )β1 − h2 β2 + h4 (d′0 (f0 + r) +d′1 (f1 + r) + d′2 (f2 + r) + d′3 (f3 + r) +d′4 (f4 + r) + d′5 (f5 + r)) 167 10 (10) − h u (ζ1 ) + O(h11 ), 50400 (5 − d′1 h4 gn−1 )un−1 + (−4 − d′2 h4 gn−2 )un−2 +(1 − d′3 h4 gn−3 )un−3 − (d′4 h4 gn−4 )un−4 −(d′5 h4 gn−5 )un−5 = (2 − d′0 h4 gn )β1 + h2 β2 +h4 (d′0 (fn + r) + d′1 (fn−1 + r) + d′2 (fn−2 + r) +d′3 (fn−3 + r) + d′4 (fn−4 + r) + d′5 (fn−5 + r)) 167 10 (10) h u (ζn ) + O(h11 ). − (18) 50400 IV. C ONVERGENCE A NALYSIS Here we prove the convergence of the methods. Let us write the error equation of the methods as follows: AE = T,
d1 d2 d3 d4 d5 β γ β α α β γ β α α β γ β α ... ... ... ... ... .. .. .. .. .. . . . . . α β γ β α α β γ β d5 d4 d3 d2 d1
(21)
,
(22)
E = A−1 T = [M + BG]−1 T, |E ≤ [I + M −1 BG]−1 M −1 T . (23) By using (I + A)−1 ≤ (1 − A)−1 and Usmani et. al. [19] we obtain M −1 ≤
5(d − c)4 + 4(d − c)2 h2 , 384h4
(24)
M −1 T . 1 − M −1 BG
(25)
E ≤
Provided that M −1 BG < 1. Also we can obtain 724 B ≤ , G ≤ h4 Mg , Mg = maxc≤x≤d |g(x)|. 720 (26) For second order we obtain 59h6 M6 , M6 = maxc≤ζ≤d |u(6) (ζ)|, 360 using (24)-(26) we obtain T ≤
(19)
E ≤
118ωh6 M6 ≡ O(h2 ), 276480h4 − 724ωG
where E = (ej ), is the (n-1)-dimensional column vector with ej , the error of discretization defined by for fourth order we get ej = u(xj ) − uj . In other words ej is the amount 241h8 M8 , M8 = maxc≤ζ≤d |u(8) (ζ)|, T ≤ by which computed solution uj deviates from the 60480 203
(27)
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using (24)-(26 we obtain 17352ωh8 M8 ≡ O(h4 ), 1672151040h4 − 4378752ωG (28) and for sixth order we get E ≤
167h10 M10 , M10 = maxc≤ζ≤d |u(10) (ζ)|. 50400 By using (24)-(26,we get T ≤
[-1,1] by using the Quintic non-polynomial spline methods with step lengths h = 2−m , m = 3, 4, 5. The maximum absolute errors in solution for our various methods are listed in tables 1 and also the maximum absolute errors in the solution at middle points of interval are tabulated in table 2. To compare our computed results obtained by second and fourth order methods with the results obtained by other known methods in [1-4,13,14], the maximum absolute errors in the solution of example 1 are listed in tables 3,4 and 5.
12024ωh10 M10 ≡ O(h6 ), 1393459200h4 − 3648960ωG (29) Spline approach has the advantage over finite where ω = 5(d − c)4 + 4(d − c)2 h2 , G = difference method that it provides continuous max|g(x)|, c ≤ x ≤ d provided approximations to u(i) (x), i = 1, 2, 3, at every point 4 69120h of the range of integration beside approximation G < . to u(x). Following [20] to obtain the necessary 181ω It follows E → 0 as h → 0. Therefore the formula for computing values of first, second and convergence of the methods have been established. third derivatives of solution of example 1, by using equation (7), (8) and solving the resulting identity for mj , j = 1, ..., n we have V. N UMERICAL R ESULTS E ≤
We consider the system of differential equations [1-4,13-15]
u
(4)
=
, 1, −1 ≤ x ≤ −1 2 −1 1 2 − 4u, 2 ≤ x ≤ 2 ,
1 2
≤ x ≤ 1,
and the conditions of continuity of u and u′′ atx = −1 and 12 . 2
Γ1 (x),
u(x) = Γ2 (x), Γ (x), 3
(uj+1 −2uj +uj−1 ) h2
h2 (Sj −Sj−1 ) 2θ2
−
−
+
h2 (Sj+1 −2Sj cos θ+Sj−1 ) 3 sin θ
h2 [(6+θ2 )Sj+1 −(12−θ2 )Sj +(6−2θ2 )Sj−1 ] 6θ4
(2Sj −Sj−1 −Sj+1 ) 6θ2
(30) −
with the boundary conditions: 1 −1 u(−1) = u( ) = u( ) = u(1) = 0, 2 2 −1 1 u′′ (−1) = u′′ ( ) = u′′ ( ) 2 2 = u′′ (1) = 0, (31)
The analytical solution problem is
mj =
−
h2 (Sj+1 −2Sj cos θ+Sj−1 ) , 6θ2 sin θ
with m0 = mn+1 = β2 being known from the boundary conditions. Having computed uj , mj , Sj , j = 0, ..., n + 1, it is possible to evaluate the coefficient of the spline function (4) as given by (6). Since yj′ = p′j (xj ), j = 0, ..., n and ′ yn+1 = p′n (xn+1 ), it follows that
aj k + ej , j = 0, ..., n, Ψn + 3cn h2 + 2dn h + en , j = n + 1, for this boundary value (33) where Ψn = an k cos θ − bn k sin θ. Similarly, from ′′′ ′′′ yj′′′ = p′′′ j (xj ), j = 0, ..., n and yn+1 = pn (xn+1 ), we −1 ≤ x ≤ −1 , 2 −1 ≤ x ≤ 12 , (32) can obtain 2 1 ≤ x ≤ 1, 2
where 1 4 3 1 Γ1 (x) = 24 x + 18 x3 + 18 x2 + 64 x + 192 , 1 Γ2 (x) = 0.5− ϕ1 [ϕ2 sin x sinh x+ϕ3 cos x cosh x], 1 4 3 1 x − 18 x3 + 18 x2 − 64 x + 192 , Γ3 (x) = 24 1 ϕ1 = cos(1) + cosh(1), ϕ2 = sin( 2 ) sinh( 12 ), ϕ3 = cos( 12 ) cosh( 12 ). We solved this example over the whole interval
yj′
≈
−aj k 3 + 6cj , j = 0, ..., n, −an k 3 cos θ + bn k 3 sin θ + 6cn , j = n + 1. (34) (i) The values of u (x), i = 1, 2, 3 have been computed by our second order method (i). To compare with the method in [1] the maximum absolute errors are listed in tables 6.
yj′′′
204
≈
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TABLE III M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1
Example 2: We consider the system of differential equation solved by Al-Said and Noor [1].
u
(4)
=
, 0, −1 ≤ x ≤ −1 2 1 1 − 4u, −1 ≤ x ≤ , 2 2
1 2
m
Our fourth-order
3 4 5
1.7×10−8 4.1×10−10 1.3×10−11
Fourth-order[4]
Finite difference[14]
≤ x ≤ 1, (35)
with the boundary conditions: 1 −1 ) = u( ) = u(1) = 0, 2 2 −1 1 u′′ (−1) = −u′′ ( ) = u′′ ( ) 2 2 = −u′′ (1) = ǫ, (36)
1.3×10−6 8.7×10−8 6.8×10−9
TABLE IV M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1
u(−1) = u(
where ǫ → 0. The analytical solution for this boundary value problem is Λ1 (x),
for − 1 ≤ x ≤ −1 , 2 1 for −1 ≤ x ≤ , u(x) = Λ2 (x), 2 2 Λ (x) =, for 1 ≤ x ≤ 1, 3 2
2.4×10−8 1.5×10−9 9.5×10−11
h
Second-order(i)
Second-order(ii)
1/12 1/24 1/48
4.5×10−6 1.2×10−6 3.3×10−7
1.2×10−7 2.3×10−7 3.2×10−8
Second-order[1]
1.2×10−5 2.8×10−6 6.9×10−7
VI. C ONCLUSION (37)
where x3 − 32 x2 − 13 x − 14 )ǫ, Λ1 (x) = ( −2 3 12 1 Λ2 (x) = 0.25− 2ϕ1 [ϕ2 sin x sinh x+ϕ3 cos x cosh x], x3 + 32 x2 − 13 x + 14 )ǫ, Λ3 (x) = ( −2 3 12 ϕ1 = cos(1) + cosh(1), ϕ2 = sin( 12 ) sinh( 12 ), ϕ3 = cos( 12 ) cosh( 12 ). We solved this example over the whole interval [-1,1] by using our second order methods (i),(ii), with step lengths h = 2−m , m = 3, 4, 5. The maximum absolute error in solution are listed in table 7 and 8, our results compared with the results obtained in [1,2,11]. The results shows superiority of our second orders methods. TABLE I M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1 m
O(h2 )(i)
O(h2 )(ii)
O(h4 )
O(h6 )
3 4 5
8.19×10−6 2.73×10−6 7.44×10−7
9.04×10−7 2.26×10−7 7.19×10−8
1.69×10−8 4.08×10−10 1.30×10−11
7.65×10−11 3.31×10−13 6.41×10−15
TABLE II M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1 IN MIDDLE POINTS
m
O(h2 )(i)
O(h2 )(ii)
O(h4 )
O(h6 )
3 4 5
1.25×10−6 1.56×10−7 1.95×10−8
2.26×10−7 7.19×10−8 1.84×10−8
4.40×10−10 1.30×10−11 4.06×10−13
5.95×10−13 5.49×10−15 –
We have developed a new non-polynomial quintic spline for solving a system of fourth-order boundary-value problems. This approach has the advantages over finite difference methods that it provides continuous approximations to not only for u(x) but also for u(i) (x), i = 1, 2, 3, at every point of the range of integration. Our numerical results are better than those produced by collocation and finite difference methods for solution of equation(1). R EFERENCES [1] AL-SAID, E.A. and NOOR, M.A., Quartic Spline Method for Solving Fourth-Order Obstacle Boundary Value Problems, Journal of Computational and Applied Mathematics Vol.143,pp.107116,2002. [2] AL-SAID,E.A. NOOR,M.A. Computational Methods for FourthOrder Obstacle Boundary Value Problems, Comm. Appl. Nonlinear. Anal. Vol.2,pp.73-83,1995. [3] AL-SAID, E.A. NOOR,M.A. and RASSIAS,T.M., Cubic Splines Method for Solving Fourth-Order Obstacle Problems, Appl. Math. Comput.,Vol.174, pp.180-187,2006. [4] AL-SAID,E.A., NOOR,M.A., KAYA,D., Al-KHALED,K., Finite difference method for solving fourth-order obstacle problems, Int. J. Comput. Math. Vol.81,pp.741-748,2004. [5] BAIOCCHI,C.C. and CALEO,A., Variational and quasivariational Inequalities, John Wiley and Sons, New York, 1984. [6] CHAWLA,M.M. and SUBRAMANIAN, High accuracy quintic spline solution of fourth-order two-Point boundary value problems, Int. J. Computer. Math. Vol.31,pp.87-94,1989. [7] HENRICI,P., Discrete variable method in ordinary differential equations, John Wiley, New York, 1961. [8] JAIN,M.K., Numerical solution of differential equations, Second Editions,Wiley Eastern Limited, 1984. [9] JAIN,M.K., IYANGER,S.R.K. and SOLDHANHA,J.S.V., Numerical solution of a fourth-order ordinary differential equation, J. Engg. Math. Vol.11,pp.373-380,1977. [10] KIKUCHI,N., and ODEN,J.T., Contact problem in elasticity, SIAM, Publishing Co. Philadelphia,1988.
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TABLE VII M AXIMUM ABSOLUTE ERRORS FOR EXAMPLE 2
TABLE V M AXIMUM ABSOLUTE ERRORS IN SOLUTION OF EXAMPLE 1 h
Second-order In [2]
1/12 1/24 1/48
m 3 4 5
Second-order[3]
6.2×10−5 1.6×10−5 3.9×10−6
7.8×10−6 1.9×10−6 4.9×10−7
Second-order(i)
Second-order(ii)
4.66×10−6 1.39×10−6 3.74×10−7
4.45×10−7 1.32×10−7 4.25×10−8
AND
||u′′ (xi ) − u′′ i ||∞
u′′′
FOR EXAMPLE
m
||u′ (xi ) − u′i ||∞
3 4 5
6.28×10−5 1.77×10−5 4.71×10−6
Our O(h2 )(i) 1.94×10−4 5.51×10−5 1.46×10−5
2.83×10−4 8.35×10−5 2.25×10−5
3 4 5
8.81×10−5 3.17×10−5 1.06×10−5
O(h2 ) in[1] 1.30×10−3 3.66×10−4 9.66×10−5
2.08×10−2 5.76×10−3 1.72×10−3
m 3 4 5
1
||u′′′ (xi ) − u′′′ i ||∞
ǫ = 10−6
Second-order In [1]
1.3×10−5 3.2×10−6 8.1×10−7
TABLE VIII M AXIMUM ABSOLUTE ERRORS FOR EXAMPLE 2 TABLE VI M AXIMUM ABSOLUTE ERRORS OF u′ , u′′
WHEN
WHEN
Second-order In [11]
Second-order In [2]
3.0×10−4 7.0×10−5 1.4×10−5
1.4×10−4 3.6×10−5 8.9×10−6
ǫ = 10−6
Jalil Rashidinia, Associate professor, School of Mathematics, Iran University of Science & Technology Narmak, Tehran 16844, Iran e-mail: [email protected] Nationality: Iranian Research Interests: Applied mathematics and computational sciences Numerical solution of ODE, PDE and IE Spline approximations Numerical linear algebra
[11] KHALIFA,A.K. and NOOR,M.A., Quintic spline solutions of a class of contact problems, Math. Comput. Modlelling Vol.13,pp.51-58,1990. [12] LEWY,H. and STAMPACCHIA,G., On the regularity of the solution of the variational inequalities, Comm. pure and Appl. Math. Vol.22,pp.153-188,1969. [13] NOOR,M.A. and Al-SAID,E.A., Fourth order obstacle problems.In: Th.M.Rassias and H.M.Srivastava(Eds), Analytic and geometric inequalities and applications, kluwer Academic Publishers, Dordrecht, Netherlands,pp.277-300,1999. [14] NOOR,M.A. and AL-SAID,E.A., Numerical solution of fourth order variational inequalities, Inter. J. Comput. Math. Vol.75,pp.107-116,2000. [15] PAPAMICHEL,N. and WORSEY,E.A., A cubic spline method for the solution of linear fourth-order two-point boundary value problem, J. Comput. Appl. Math. Vol.7,pp.187-189,1981. [16] RASHIDINIA, J., Applications of splines to the numerical solution of differential equations, Ph.D. thesis, Aligarh Muslim University, Aligarh, India, 1994. [17] SIRAJ-UL-ISLAM, TIRMIZI,S.I.A. and SAADAT ASHRAF, A class of method based on non-polynomial spline function for the solution of a special fourth-order boundary-value problems with engineering applications, Appl. Math. Comput. Vol.174,pp.11691180,2006. [18] USMANI,R.A. and WARSI,S.A., Smooth spline solutions for boundary value problems in plate deflection thoery, Comput. Maths. with Appls. Vol.6, pp.205-211,1980. [19] USMANI,R.A., Discrete variable method for a boundary value problem with engineering applications, Math. Comput. Vol.32,1087-1096,1978. [20] Van DAELE,M., VANDEN BERGHE,G. and De MEYER,H., A smooth approximation for the solution of a fourth-order boundary value problem based on non-polynomial splines, J. Comput. Appl. Math. Vol.pp.51,383-394,1994.
Reza Jalilian Assistant professor, Department of Mathematics, Ilam University, PO Box 69315516, Ilam, Iran e-mail: [email protected] Research Interests: Spline approximations
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