numerical solution of differential equations by using haar wavelets
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007
NUMERICAL SOLUTION OF DIFFERENTIAL EQUATIONS BY USING HAAR WAVELETS ZHI SHI, LI-YUAN DENG, QING-JIANG CHEN School of Science, Xi’an Univ. of Arch. & Tech., Xi’an 710055,China E-MIAL:[email protected], [email protected], [email protected]
Abstract This paper establishes a clear procedure for finite-length beam problem and convection-diffusion equation solution via Haar wavelet technique. An operational matrix of integration based on the Haar wavelet is established,and the procedure for applying the matrix to solve the differential equations is formulated. The fundamental idea of Haar wavelet method is to convert the differential equations into a group of algebra equations which involves a finite number of variables. Illustrative examples are given to demonstrate the fast and flexible of the method ,in the mean time,it is found that the trouble of Daubechies wavelets for solving the differential equations which need to calculate the correlation coefficients is avoided. The method can be used to deal with all the other differential and integral equations. Keywords: Haar wavelets; the differential equation; finiteLength beam; convection-diffusion equation 1.
Introduction
Haar wavelets have been applied extensively for signal processing in communications and physics research, and more mathematically focused on differential equations and even nonlinear problems. After discretizing the differential equation in a convential way like the finite difference approximation,wavelets can be used for algebraic manipulations in the system of equations obtained which may lead to better condition number of the resulting system. The previous work in system analysis via Haar wavelets was led by Chen and Hsiao[1], who first derived a Haar operational matrix for the integrals of the Haar function vector and put the application for the Haar analysis into the dynamic systems. Then, the pioneer work in state analysis of linear time delayed systems via Haar wavelets was laid down by Hsiao[2], who first proposed a Haar product matrix and a coefficient matrix. In order to take the advantages of the local property,many authors researched the Haar wavelet to solve the differential and integral equations[3,4,5,6].
This paper is focused on the solution of finite-length beam and convection-diffusion via Haar wavelets. We established a clear procedure for solving the differential equations via Haar wavelet. We organized our paper as follows. In sections 2, the Haar wavelet is introduced and an operational matrix is established. Section 3, we use Haar wavelets to solve two differential equations. Because of the local property of the powerful Haar wavelet, the new method is simpler in reasoning as well as in calculation. 2.
Some properties of Haar wavelets
2.1.
Haar wavelets
In 1910, Haar showed that certain square wave function could be translated and scaled to create a basis set that span L .Years later, it was seen that the system of Haar is a particular wavelet system. If we choose our scaling function to have compact support over 0 ≤ x < 1 , that is ⎧1, 0 ≤ x < 1 h0 ( x ) = ⎨ ⎩0, otherwise then the mother wavelet function to be 2
⎧ 1, ⎪ h ( x ) = ⎨ −1, ⎪ 0, ⎩ 1
0 ≤ x < 1/ 2 1/ 2 ≤ x < 1 otherwise
All the other subsequent functions are generated from h1 ( x ) with two operations: translation and dilation. That is j
hn ( x ) = h1 (2 x − k ) j
j
where n = 2 + k , j ≥ 0 , 0 ≤ k < 2 . We have noticed that all the Haar wavelets are orthogonal to each other, that is
1-4244-1066-5/07/$25.00 ©2007 IEEE 1039
⎧2 − j , m = n = 2 j + k ∫0 hm ( x)hn ( x)dx = 2 δ mn = ⎨ 0, m≠n ⎩ 1
−j
(1)
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007 2.2.
Function approximation
The integration of the vector h ( m ) ( x ) is given by
∫
Any function y ( x ) ∈ L [ 0,1) can be decomposed as 2
0
+∞
y ( x ) = ∑ cn hn ( x )
(2)
P( m ) =
where the coefficients cn are determined by
∫
j
1
0
(3)
y ( x ) hn ( x )dx
j
j
where n = 2 + k , j ≥ 0 , 0 ≤ k < 2 .Speciall, c0 =
∫
1
0
y ( x )dx .
y ( x ) = ∑ cn hn ( x ) = c ( m ) h ( m ) ( x ) T
(4)
T
where the coefficient vector c ( m ) and the Haar function vector h ( m ) ( x ) are defined as c ( m ) = [ c0 , c1 , " , cm −1 ] T
and h ( m ) ( x ) = [ h0 ( x ), h1 ( x ), " , hm −1 ( x ) ]
T
j
where “T” means transpose and m = 2 . The product operational matrix of the Haar wavelet
3.
The first four Haar function vectors which x = n / 8 , n = 1, 3, 5, 7 can be expressed the following h ( 4 ) (1 / 8) = [1,1,1, 0 ]
T
h ( 4 ) (5 / 8) = [1, −1, 0,1]
T
,
h ( 4 ) (3 / 8) = [1,1, −1, 0 ]
,
h ( 4 ) (7 / 8) = [1, −1, 0, −1]
T
[
=
]
⎡1 1 1 1 ⎤ ⎢1 1 −1 −1⎥ ⎢ ⎥ ⎢1 −1 0 0 ⎥ ⎢ ⎥ ⎣ 0 0 1 −1⎦
In general, we have H( m ) = [h ( m ) (1/ 2 m),
h ( m ) (3 / 2 m),
h ( m ) ((2m − 1) / 2m)]
⎡1
⎢ -1 2m ⎣ H ( m / 2 )
−H ( m / 2 ) ⎤ 0
⎥ ⎦
(7)
3.1.
Analysis of the differential equation via Haar wavelets Finite-length beam via Haar wavelets
By the medium of Winkler, the basic differential equation for the flexibility of the beam is
T
which can be written in matrix form as H ( 4 ) = h (4) (1/ 8), h (4) (3 / 8), h (4) (5 / 8), h (4) (7 / 8)
1 ⎡ 2 mP( m / 2 )
⎡ 8 −4 −2 −2 ⎤ ⎢ ⎥ 1 ⎡ 2 −1⎤ 1 4 0 −2 2 ⎢ ⎥, P( 2 ) = ⎢ , P = (4) 0 0⎥ 4 ⎣ 1 0 ⎥⎦ 16 ⎢ 1 1 ⎢ ⎥ ⎣ 1 −1 0 0 ⎦ ⎡32 −16 −8 −8 −4 −4 −4 −4 ⎤ ⎢16 0 −8 8 −4 −4 4 4 ⎥ ⎢ ⎥ 4 0 0 −4 4 0 0⎥ ⎢4 ⎢ ⎥ 4 0 0 −4 4 0 0 1 ⎢4 ⎥. P(8) = 1 2 0 0 0 0 0⎥ 64 ⎢ 1 ⎢ ⎥ −2 0 1 0 0 0 0⎥ ⎢1 ⎢ 1 −1 0 2 0 0 0 0 ⎥ ⎢ ⎥ ⎣ 1 −1 0 −2 0 0 0 0 ⎦
n=0
2.3.
(6)
(Proof of formula (7) can be found in [1]) where P(1) = [1/ 2 ] , so
The series expansion of y ( x ) contains infinite terms. If y ( x ) is piecewise constant by itself,or may be approximated as piecewise constant during each subinterval, then y ( x ) will be terminated at finite terms, that is m −1
h ( m ) (t )dt = P( m )h ( m ) ( x ), x ∈ [ 0,1)
where P( m ) is the m × m operational matrix and is given by
n=0
cn = 2
x
dω 4
+ k 0 bω = bp ( x ) 4 dx where Eb I is the elasticity modulus and section moment of
Eb I
inertia for the material of the beam; b is the width of the beam; k0 is the centralized bedding coefficient for the unit
",
(5)
length of the beam; p ( x ) is the external load; ω is the flexibility of the beam. The boundary conditions of finite-length beam can not only be divided into the free end, simple-supported and fixed end, but also be divided into any two terms among the three kinds conditions. All kinds of the conditions could be expressed as
1⎤
where H (1) = [1] , H ( 2 ) = ⎢ ⎥. ⎣1 −1⎦
1040
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007 x=0
fixed end
or
x=0
simple - supported
or
x=0
free end
ω = ω ′ = 0;
1 :
ω =ω
1 :
or
1 :
ω
(2)
( 2)
=ω
ω ( x ) + 4ω ( x ) = cos(2 x ) into a matrix equation, that is T 4 T T ( 3) 3 c ( m ) ⎣⎡I( m ) + 4P( m ) ⎦⎤ = d ( m ) − 4f ω (0)P( m ) ( 4)
= 0;
(3)
= 0.
T
For clarity in presentation, in the illustrative example, we assume that Eb I = 1 , b = 1 , k0 = 4 , p ( x ) = cos(2 x ) .The basic differential equation for the flexibility of the beam is
ω ( x ) + 4ω ( x ) = cos(2 x ) (8) For solving this problem by the Haar wavelet method, (4) we assume that ω ( x ) can be expanded in terms of Haar wavelets as formula (4).That is ( 4)
m −1
ω ( x ) = ∑ cn hn ( x ) = c ( m ) h ( m ) ( x ) ( 4)
T
(9)
n=0
The function cos(2 x ) can be expanded into Haar series over the interval [0,1) , m −1
cos(2 x ) = ∑ d n hn ( x ) = d ( m )h ( m ) ( x ) T
(10)
T
where the vector d ( m ) can be obtained using formula (3). Integrating formula (9) from 0 to x and using formula ( 3) (2) (6), the variable ω ( x ) , ω ( x ) , ω ′( x ) and ω ( x ) can be expressed as
ω ( x) ( 3)
∫
=
0
ω ( x)
∫
=
ω ( x)
T
( 3)
x
0
∫
x
0
T
( 3)
2
T
ω (0) =
T
2
T
T
T
2
(12)
3
T
f P( m ) f - f P( m ) f
ω (0) = −c ( m ) P( m ) f ( 2)
(11)
T
c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ − c ( m ) P( m ) f T
T
T
2
3
T
f P( m ) f - f P( m ) f
Case 3 fixed end at one and free end at the other one Using the boundary conditions ( 2) ( 3) ω (0) = ω ′(0) = 0 , ω (1) = ω (1) = 0
we can gain
ω (0) = −c ( m ) f ( 3)
T
ω (0) = −c ( m ) P( m ) f − c ( m ) f (2)
(13)
T
c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ − c ( m ) P( m ) f T
ω (0) =
(14)
2
T
T
ω (0) = −c ( m ) P( m ) f ( 2)
where the vector f is defined as
T
T
f P( m ) f - ⎡⎣f P( m ) f ⎤⎦ 2
2
T
2
2
2
T
T
3
f P( m ) f - ⎡⎣f P( m ) f ⎤⎦ T
T
2
T
T
2
Substituting ω (0) , ω (0) , ω ′(0) and ω (0) into ( 3)
(m-1) elements
Substituting formula (9), (10) and (14) into formula (8), we transfer the equation
3
c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ − c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ T
T
T
Case 4 fixed end at both ends Using the boundary conditions ω (0) = ω ′(0) = 0 , ω (1) = ω ′(1) = 0 we can gain ( 3)
f ω (0) ⎤⎦ h ( m ) ( x ) f = [1, 0, ", 0 ] N
T
T
3
f ω (0)P( m ) + f ω ′(0)P( m ) (2)
3
c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ − c ( m ) P( m ) f
( 3)
( 2)
4
( 2)
ω ′(0) = − c ( m ) P( m ) f + f ⎡⎣ c ( m ) P( m ) f ⎤⎦ P( m ) f Case 2 fixed end at one and simple-supported at the other one Using the boundary conditions ( 2) ω (0) = ω ′(0) = 0 , ω (1) = ω (1) = 0 we can gain
ω (t )dt + ω ′(0)
⎡⎣ c ( m ) P( m ) + f ω (0)P( m ) T
(2)
T
ω ′(t )dt + ω (0)
T
T
ω (0) = ω (0) = 0 , ω (1) = ω (1) = 0 we can gain ( 3) T ω (0) = −c ( m ) P( m ) f
( 2)
⎡⎣ c T( m ) P(3m ) + f Tω ( 3) (0)P(2m ) T ( 2) T f ω (0)P( m ) + f ω ′(0) ⎤⎦ h ( m ) ( x )
T
Then, we choose four cases which involve to the ordinary differential equation (8) to preform the Haar wavelet method. Case 1 simple-supported at both ends Using the boundary conditions
ω (t )dt + ω (0) ( 3)
2
whereI( m ) is identity matrix.
(3)
⎡⎣ c T( m ) P(2m ) + f Tω ( 3) (0)P( m ) T ( 2) f ω (0) ⎤⎦ h ( m ) ( x )
∫
= =
x
0
= =
(4)
⎡⎣ c ( m ) P( m ) + f ω (0) ⎤⎦ h ( m ) ( x )
=
ω ′( x )
ω (t )dt + ω (0)
T
= ( 2)
x
(2)
T
n=0
(15)
−4f ω (0)P( m ) − 4f ω ′(0)P( m ) − 4f ω (0)
(2)
T
formula (15),we can gain c ( m ) via solving the matrix equation, then we have ω ( x ) by formula (14) respectively.
1041
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007
The comparison of the analytic solution and the Haar solution is shown in Table.1.It is seen that even when m=32 and m = 64 the Haar direct method is quite satisfactory. x 0 0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375 1
Analytic solution
Haar solution (m=32) 0 0.0014 0.0038 0.0055 0.0059 0.0060 0.0047 0.0033 0.0011 0
Haar solution (m=64) 0 0.0013 0.0037 0.0054 0.0061 0.0059 0.0048 0.0032 0.0010 0
Haar solution (m=32) 0 0.00014 0.00072 0.00140 0.00200 0.00210 0.00190 0.00130 0.00052 0
Haar solution (m=64) 0 0.00013 0.00078 0.00150 0.00210 0.00230 0.00200 0.00140 0.00056 0
Haar solution (m=32) 0 0.00012 0.00130 0.00250 0.00380 0.00530 0.00610 0.00690 0.00760 0.00780 (c)
Haar solution (m=64) 0 0.00016 0.00120 0.00260 0.00390 0.00510 0.00620 0.00690 0.00770 0.00780
0 0.0013 0.0037 0.0054 0.0061 0.0059 0.0048 0.0031 0.0010 0 (a)
x
Analytic solution
0 0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375 1
0 0.00012 0.00080 0.00160 0.00210 0.00230 0.00200 0.00140 0.00057 0 (b)
x
Analytic solution
0 0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375 1
0 0.00017 0.00120 0.00260 0.00400 0.00520 0.00620 0.00700 0.00770 0.00790
Analytic solution
x 0 0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375 1
Haar solution (m=32) 0 0.0002 0.0005 0.0012 0.0016 0.0012 0.0008 0.0005 0.0002 0
0 0.0001 0.0006 0.0011 0.0014 0.0013 0.0009 0.0004 0.0001 0
Haar solution (m=64) 0 0.0001 0.0006 0.0011 0.0014 0.0013 0.0009 0.0004 0.0001 0
(d) Table 1. comparison of the analytic solution and the Haar solution:(a)simple-supported at both ends,(b)fixed end at one and simple-supported at the other one,(c)fixed end at one and free end at the other end,(d)f ixed end at both ends 3.2.
Convection-diffusion equation via Haar wavelet
We will consider the one-dimensional convectiondiffusion equation with constant coefficients ∂u
∂u
2
=α
∂u
,0 < x < 1,0 < t ≤ T 2 ∂t ∂x ∂x with initial condition u ( x , 0) = f ( x ) , 0 ≤ x ≤ 1 and boundary conditions u (0, t ) = g 0 (t ) , u (1, t ) = g1 (t ) , 0 < t ≤ T +a
(16)
Let us divide the interval (0, T ] into N equal parts of ∆t = ( 0, T ] / N
length
and
denote
t s = ( s − 1) ∆t , s = 1, 2, " , N .We assume that u ′′( x, t ) can
be expanded in terms of Haar wavelets as formula (4) m −1
T u ′′( x, t ) = ∑ cs ( n ) hn ( x ) = c ( m )h ( m ) ( x )
(17)
n=0
where · and ' means differentiation with respect to t and x T
respectively, the row vector c ( m ) is constant in the subinterval
(
t ∈ t , t s +1
].
Integrating formula (17) with respect to t from t s to t and twice with respect to x from 0 to x, and using formula (6), the variable u ′′( x, t ) , u ′( x, t ) , u ( x, t ) and u ( x, t ) can be expressed as T u ′′( x, t ) = (t − t s )c ( m ) h ( m ) ( x ) + u ′′( x, t s ) (18) u ′( x , t )
=
(t − t s )c ( m ) P( m )h ( m ) ( x ) + u ′( x, t s ) T
−u ′(0, t s ) + u ′(0, t )
1042
(19)
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007 T
=
u ( x, t )
u ( xl , t s +1 ) = −0.8u ′( xl , t s +1 ) + 0.1u ′′( xl , t s +1 )
2
(t − t s )c ( m ) P( m )h ( m ) ( x ) + u ( x, t s ) − u (0, t s ) + x [ u ′(0, t ) − u ′(0, t s ) ] + u (0, t )
T 2 u ( x, t ) = c ( m ) P( m )h ( m ) ( x ) + xu ′(0, t ) + u (0, t )
(20)
which leads us from the time layer t s to t s +1 is used. Substituting formula (24)-(27) into formula (28),we gain
(21)
T
Putting x = 1 in formula (20) and (21), we have T u ′(0, t ) − u ′(0, t s ) = −(t − t s )c ( m ) P( m ) h ( m ) ( x ) g1 (t ) − g 0 ( t ) − g1 (t s ) + g 0 (t s ) T 2 u ′(0, t ) = -c ( m ) P( m ) f + g1′ (t ) − g 0′ (t )
=
(22)
⎡ ( xl − 2) 2 ⎤ ⎥ 8 ⎣ ⎦ x −2 ⎡ ( x − 2) 2 ⎤ u ′( xl , t1 ) = − l exp ⎢ − l ⎥ 4 8 ⎣ ⎦ 2 x − 4 xl ⎡ ( x − 2) 2 ⎤ u ′′( xl , t1 ) = − l exp ⎢ − l ⎥ 16 8 ⎣ ⎦ u ( xl , t1 ) = exp ⎢ −
(24)
T
=
(t s +1 − t s )c ( m ) P( m ) h ( m ) ( xl ) +u ′( xl , t s ) − (t s +1 − t s )c ( m ) P( m ) f T
(25)
+ g1 (t s +1 ) − g 0 (t s +1 ) − g1 (t s ) + g 0 (t s ) u ( xl , t s +1 )
T
=
2
(t s +1 − t s )c ( m ) P( m )h ( m ) ( xl ) + u ( xl , t s ) − g 0 (t s ) + g 0 (t s +1 ) T
+ xl [ − (t s +1 − t s )c ( m ) P( m ) f + g1 (t s +1 )
(26)
Computer simulation was carried out in the cases m=32 and m=64,the computed results were compared with the exact solution, more accurate results can be obtained by using a larger m. For the time step was taken ∆t = 0.001 , further diminution of ¢t did not give any essential effect. Some results of computation are presented in Table.2. Comparison with these algorithms shows that the Haar wavelet method is competitive and efficient. The advantages of our method are its simplicity and speed of convergence.
− g 0 (t s +1 ) − g1 (t s ) + g 0 (t s )] u ( xl , t s +1 )
T 2 c ( m ) P( m )h ( m ) ( x ) + g 0′ (t s +1 )
=
+ xl [ − c ( m ) P( m ) f + g1′ (t s +1 ) − g 0′ (t s +1 )] T
(27)
where the vector f is defined as f = [1, 0, ", 0 ] N
T
(m-1) elements
For clarity in the illustrative example, we assume that a = 0.8 , α = 0.1 ,
⎡ ( x − 2) 2 ⎤ f ( x ) = exp ⎢ − 8 ⎥⎦ ⎣
20
u ( x, t ) =
20
⎡ 2(5 + 2t ) 2 ⎤ ⎥ ⎣ 5(t + 20) ⎦
exp ⎢ −
20 + t for which the exact solution is
⎡ ( x − 2 − 0.8t ) 2 ⎤ ⎥ ⎣ 0.4(t + 20) ⎦
exp ⎢ −
20 + t In the following the scheme
x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Exact solution 0.5765716 0.6065073 0.6364239 0.6661694 0.6955854 0.7245093 0.7527748 0.7802143 0.9066598
Haar solution (m=32) 0.5762574 0.6062752 0.6361158 0.6658236 0.6953754 0.7242356 0.7525858 0.7800047 0.8064472 (a)
⎡ (5 + 4t ) 2 ⎤ g 0 (t ) = exp ⎢ − ⎥ 20 + t ⎣ 10(t + 20) ⎦ 20
g1 ( t ) =
−0.8u ′( xl , t s ) + 0.1u ′′( xl , t s )
successively calculated. This process is started with
t = t s +1 ,we obtain
u ′( xl , t s +1 )
(29)
T
Substituting formula (22) and (23) into formula (18)-(21),and discretizising the results by assuming x = xl ， T
2
From formula (29) the wavelet coefficients c ( m ) can be
(23)
u ′′( xl , t s +1 ) = (t s +1 − t s )c ( m )h ( m ) ( xl ) + u ′′( xl , t s )
T
+ g1′ (t s +1 ) − g 0′ (t s +1 )] + g 0′ (t s +1 )
u (1, t ) = g1′ (t )
,
2
c ( m ) P( m )h ( m ) ( xl ) + xl [ −c ( m ) P( m ) f
By the boundary conditions, we obtain u (0, t s ) = g 0 (t s ) , u (1, t s ) = g1 (t s ) u (0, t ) = g 0′ (t )
(28)
1043
Haar solution (m=64) 0.5765247 0.6064857 0.6363998 0.6660952 0.6955841 0.7245062 0.7527289 0.7801957 0.8066249
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007
x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Exact solution 0.5227473 0.5520503 0.5815743 0.6111833 0.6407336 0.6700747 0.6990506 0.7275012 0.7552636
Haar solution (m=32) 0.5224653 0.5517783 0.5813547 0.6109966 0.6405742 0.6704588 0.6988549 0.7272821 0.7550917
sparse matrices of representation, fast transformation and possibility of implementation of fast algorithms. The main advantages of this method is its simplicity and small computation costs, it is due to the sparcity of the transform matrices and to the small number of significant wavelet coefficients.
Haar solution (m=64) 0.5227218 0.5519745 0.5815227 0.6111046 0.6407871 0.6701562 0.6990986 0.7275663 0.7552478
References
[1] C.F. Chen, C.H. Hsiao, “Haar wavelet method for solving lumped and distributed-parameter systems”, IEE Proc. Control Theory Appl, 144, pp. 87-94, 1997. [2] C.H. Hsiao, “States analysis of linear time delayed systems via Haar wavelets”, Math. Comput. Simulat, 44, pp.457-470, 1997. [3] Chun-Hui Hsiao, “Haar wavelet direct method for solving variational problems”, Mathematics and Computers in Simulation, 64, pp. 569-585,2004. [4] Ulo Lepik, “Numerical solution of evolution equations by the Haar wavelet method”, Applied Mathematics and Computation,185, pp. 695-704, 2007. [5] C.F. Chen, C.H. Hsiao, “Wavelet approach to optimizing dynamic systems”, IEE Proc. Control Theory Appl,146, pp. 213-219,1997. [6] C.H. Hsiao, W.J. Wang, “Haar wavelet approach to nonlinear stiff systems”, Math. Comput. Simulat, 57, pp. 347-353, 2001.
(b) x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Exact solution 0.5135925 0.5427477 0.5721621 0.6017027 0.6312285 0.6605916 0.6896381 0.7182096 0.7461446
Haar solution (m=32) 0.5133896 0.5424384 0.5718349 0.6014876 0.6310182 0.6602749 0.6893567 0.7180179 0.7458774
Haar solution (m=64) 0.5135481 0.5427598 0.5721547 0.6017536 0.6312178 0.6605749 0.6896124 0.7182073 0.7461267
(c) x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Exact solution 0.4408986 0.4684767 0.4965875 0.5251243 0.5539709 0.5830024 0.6120858 0.6410807 0.6698409
Haar solution (m=32) 0.4406249 0.4681895 0.4962283 0.5249834 0.5537685 0.5828859 0.6118746 0.6411492 0.6694583
Haar solution (m=64) 0.4408764 0.4684373 0.4965538 0.5251097 0.5538918 0.5830183 0.6120752 0.6410286 0.6697782
(d) Table 2. comparison of the exact solution and the Haar solution:(a)solution for t=0.25,(b)solution for t=0.48,(c)solution for t=0.52,(d)solution for t=0.85 4.
Conclusions
It has been well demonstrated that in applying the nice properties of Haar wavelets, the differential equations can be solved conveniently and accurately by using Haar wavelet method systematically. It has been also shown that the key idea is to transform the differential equation into a group of algebra equations which involves a finite number of variables. The benefits of the Haar wavelet approach are
1044
NUMERICAL SOLUTION OF DIFFERENTIAL EQUATIONS BY USING HAAR WAVELETS ZHI SHI, LI-YUAN DENG, QING-JIANG CHEN School of Science, Xi’an Univ. of Arch. & Tech., Xi’an 710055,China E-MIAL:[email protected], [email protected], [email protected]
Abstract This paper establishes a clear procedure for finite-length beam problem and convection-diffusion equation solution via Haar wavelet technique. An operational matrix of integration based on the Haar wavelet is established,and the procedure for applying the matrix to solve the differential equations is formulated. The fundamental idea of Haar wavelet method is to convert the differential equations into a group of algebra equations which involves a finite number of variables. Illustrative examples are given to demonstrate the fast and flexible of the method ,in the mean time,it is found that the trouble of Daubechies wavelets for solving the differential equations which need to calculate the correlation coefficients is avoided. The method can be used to deal with all the other differential and integral equations. Keywords: Haar wavelets; the differential equation; finiteLength beam; convection-diffusion equation 1.
Introduction
Haar wavelets have been applied extensively for signal processing in communications and physics research, and more mathematically focused on differential equations and even nonlinear problems. After discretizing the differential equation in a convential way like the finite difference approximation,wavelets can be used for algebraic manipulations in the system of equations obtained which may lead to better condition number of the resulting system. The previous work in system analysis via Haar wavelets was led by Chen and Hsiao[1], who first derived a Haar operational matrix for the integrals of the Haar function vector and put the application for the Haar analysis into the dynamic systems. Then, the pioneer work in state analysis of linear time delayed systems via Haar wavelets was laid down by Hsiao[2], who first proposed a Haar product matrix and a coefficient matrix. In order to take the advantages of the local property,many authors researched the Haar wavelet to solve the differential and integral equations[3,4,5,6].
This paper is focused on the solution of finite-length beam and convection-diffusion via Haar wavelets. We established a clear procedure for solving the differential equations via Haar wavelet. We organized our paper as follows. In sections 2, the Haar wavelet is introduced and an operational matrix is established. Section 3, we use Haar wavelets to solve two differential equations. Because of the local property of the powerful Haar wavelet, the new method is simpler in reasoning as well as in calculation. 2.
Some properties of Haar wavelets
2.1.
Haar wavelets
In 1910, Haar showed that certain square wave function could be translated and scaled to create a basis set that span L .Years later, it was seen that the system of Haar is a particular wavelet system. If we choose our scaling function to have compact support over 0 ≤ x < 1 , that is ⎧1, 0 ≤ x < 1 h0 ( x ) = ⎨ ⎩0, otherwise then the mother wavelet function to be 2
⎧ 1, ⎪ h ( x ) = ⎨ −1, ⎪ 0, ⎩ 1
0 ≤ x < 1/ 2 1/ 2 ≤ x < 1 otherwise
All the other subsequent functions are generated from h1 ( x ) with two operations: translation and dilation. That is j
hn ( x ) = h1 (2 x − k ) j
j
where n = 2 + k , j ≥ 0 , 0 ≤ k < 2 . We have noticed that all the Haar wavelets are orthogonal to each other, that is
1-4244-1066-5/07/$25.00 ©2007 IEEE 1039
⎧2 − j , m = n = 2 j + k ∫0 hm ( x)hn ( x)dx = 2 δ mn = ⎨ 0, m≠n ⎩ 1
−j
(1)
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007 2.2.
Function approximation
The integration of the vector h ( m ) ( x ) is given by
∫
Any function y ( x ) ∈ L [ 0,1) can be decomposed as 2
0
+∞
y ( x ) = ∑ cn hn ( x )
(2)
P( m ) =
where the coefficients cn are determined by
∫
j
1
0
(3)
y ( x ) hn ( x )dx
j
j
where n = 2 + k , j ≥ 0 , 0 ≤ k < 2 .Speciall, c0 =
∫
1
0
y ( x )dx .
y ( x ) = ∑ cn hn ( x ) = c ( m ) h ( m ) ( x ) T
(4)
T
where the coefficient vector c ( m ) and the Haar function vector h ( m ) ( x ) are defined as c ( m ) = [ c0 , c1 , " , cm −1 ] T
and h ( m ) ( x ) = [ h0 ( x ), h1 ( x ), " , hm −1 ( x ) ]
T
j
where “T” means transpose and m = 2 . The product operational matrix of the Haar wavelet
3.
The first four Haar function vectors which x = n / 8 , n = 1, 3, 5, 7 can be expressed the following h ( 4 ) (1 / 8) = [1,1,1, 0 ]
T
h ( 4 ) (5 / 8) = [1, −1, 0,1]
T
,
h ( 4 ) (3 / 8) = [1,1, −1, 0 ]
,
h ( 4 ) (7 / 8) = [1, −1, 0, −1]
T
[
=
]
⎡1 1 1 1 ⎤ ⎢1 1 −1 −1⎥ ⎢ ⎥ ⎢1 −1 0 0 ⎥ ⎢ ⎥ ⎣ 0 0 1 −1⎦
In general, we have H( m ) = [h ( m ) (1/ 2 m),
h ( m ) (3 / 2 m),
h ( m ) ((2m − 1) / 2m)]
⎡1
⎢ -1 2m ⎣ H ( m / 2 )
−H ( m / 2 ) ⎤ 0
⎥ ⎦
(7)
3.1.
Analysis of the differential equation via Haar wavelets Finite-length beam via Haar wavelets
By the medium of Winkler, the basic differential equation for the flexibility of the beam is
T
which can be written in matrix form as H ( 4 ) = h (4) (1/ 8), h (4) (3 / 8), h (4) (5 / 8), h (4) (7 / 8)
1 ⎡ 2 mP( m / 2 )
⎡ 8 −4 −2 −2 ⎤ ⎢ ⎥ 1 ⎡ 2 −1⎤ 1 4 0 −2 2 ⎢ ⎥, P( 2 ) = ⎢ , P = (4) 0 0⎥ 4 ⎣ 1 0 ⎥⎦ 16 ⎢ 1 1 ⎢ ⎥ ⎣ 1 −1 0 0 ⎦ ⎡32 −16 −8 −8 −4 −4 −4 −4 ⎤ ⎢16 0 −8 8 −4 −4 4 4 ⎥ ⎢ ⎥ 4 0 0 −4 4 0 0⎥ ⎢4 ⎢ ⎥ 4 0 0 −4 4 0 0 1 ⎢4 ⎥. P(8) = 1 2 0 0 0 0 0⎥ 64 ⎢ 1 ⎢ ⎥ −2 0 1 0 0 0 0⎥ ⎢1 ⎢ 1 −1 0 2 0 0 0 0 ⎥ ⎢ ⎥ ⎣ 1 −1 0 −2 0 0 0 0 ⎦
n=0
2.3.
(6)
(Proof of formula (7) can be found in [1]) where P(1) = [1/ 2 ] , so
The series expansion of y ( x ) contains infinite terms. If y ( x ) is piecewise constant by itself,or may be approximated as piecewise constant during each subinterval, then y ( x ) will be terminated at finite terms, that is m −1
h ( m ) (t )dt = P( m )h ( m ) ( x ), x ∈ [ 0,1)
where P( m ) is the m × m operational matrix and is given by
n=0
cn = 2
x
dω 4
+ k 0 bω = bp ( x ) 4 dx where Eb I is the elasticity modulus and section moment of
Eb I
inertia for the material of the beam; b is the width of the beam; k0 is the centralized bedding coefficient for the unit
",
(5)
length of the beam; p ( x ) is the external load; ω is the flexibility of the beam. The boundary conditions of finite-length beam can not only be divided into the free end, simple-supported and fixed end, but also be divided into any two terms among the three kinds conditions. All kinds of the conditions could be expressed as
1⎤
where H (1) = [1] , H ( 2 ) = ⎢ ⎥. ⎣1 −1⎦
1040
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007 x=0
fixed end
or
x=0
simple - supported
or
x=0
free end
ω = ω ′ = 0;
1 :
ω =ω
1 :
or
1 :
ω
(2)
( 2)
=ω
ω ( x ) + 4ω ( x ) = cos(2 x ) into a matrix equation, that is T 4 T T ( 3) 3 c ( m ) ⎣⎡I( m ) + 4P( m ) ⎦⎤ = d ( m ) − 4f ω (0)P( m ) ( 4)
= 0;
(3)
= 0.
T
For clarity in presentation, in the illustrative example, we assume that Eb I = 1 , b = 1 , k0 = 4 , p ( x ) = cos(2 x ) .The basic differential equation for the flexibility of the beam is
ω ( x ) + 4ω ( x ) = cos(2 x ) (8) For solving this problem by the Haar wavelet method, (4) we assume that ω ( x ) can be expanded in terms of Haar wavelets as formula (4).That is ( 4)
m −1
ω ( x ) = ∑ cn hn ( x ) = c ( m ) h ( m ) ( x ) ( 4)
T
(9)
n=0
The function cos(2 x ) can be expanded into Haar series over the interval [0,1) , m −1
cos(2 x ) = ∑ d n hn ( x ) = d ( m )h ( m ) ( x ) T
(10)
T
where the vector d ( m ) can be obtained using formula (3). Integrating formula (9) from 0 to x and using formula ( 3) (2) (6), the variable ω ( x ) , ω ( x ) , ω ′( x ) and ω ( x ) can be expressed as
ω ( x) ( 3)
∫
=
0
ω ( x)
∫
=
ω ( x)
T
( 3)
x
0
∫
x
0
T
( 3)
2
T
ω (0) =
T
2
T
T
T
2
(12)
3
T
f P( m ) f - f P( m ) f
ω (0) = −c ( m ) P( m ) f ( 2)
(11)
T
c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ − c ( m ) P( m ) f T
T
T
2
3
T
f P( m ) f - f P( m ) f
Case 3 fixed end at one and free end at the other one Using the boundary conditions ( 2) ( 3) ω (0) = ω ′(0) = 0 , ω (1) = ω (1) = 0
we can gain
ω (0) = −c ( m ) f ( 3)
T
ω (0) = −c ( m ) P( m ) f − c ( m ) f (2)
(13)
T
c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ − c ( m ) P( m ) f T
ω (0) =
(14)
2
T
T
ω (0) = −c ( m ) P( m ) f ( 2)
where the vector f is defined as
T
T
f P( m ) f - ⎡⎣f P( m ) f ⎤⎦ 2
2
T
2
2
2
T
T
3
f P( m ) f - ⎡⎣f P( m ) f ⎤⎦ T
T
2
T
T
2
Substituting ω (0) , ω (0) , ω ′(0) and ω (0) into ( 3)
(m-1) elements
Substituting formula (9), (10) and (14) into formula (8), we transfer the equation
3
c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ − c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ T
T
T
Case 4 fixed end at both ends Using the boundary conditions ω (0) = ω ′(0) = 0 , ω (1) = ω ′(1) = 0 we can gain ( 3)
f ω (0) ⎤⎦ h ( m ) ( x ) f = [1, 0, ", 0 ] N
T
T
3
f ω (0)P( m ) + f ω ′(0)P( m ) (2)
3
c ( m ) P( m ) f ⎡⎣f P( m ) f ⎤⎦ − c ( m ) P( m ) f
( 3)
( 2)
4
( 2)
ω ′(0) = − c ( m ) P( m ) f + f ⎡⎣ c ( m ) P( m ) f ⎤⎦ P( m ) f Case 2 fixed end at one and simple-supported at the other one Using the boundary conditions ( 2) ω (0) = ω ′(0) = 0 , ω (1) = ω (1) = 0 we can gain
ω (t )dt + ω ′(0)
⎡⎣ c ( m ) P( m ) + f ω (0)P( m ) T
(2)
T
ω ′(t )dt + ω (0)
T
T
ω (0) = ω (0) = 0 , ω (1) = ω (1) = 0 we can gain ( 3) T ω (0) = −c ( m ) P( m ) f
( 2)
⎡⎣ c T( m ) P(3m ) + f Tω ( 3) (0)P(2m ) T ( 2) T f ω (0)P( m ) + f ω ′(0) ⎤⎦ h ( m ) ( x )
T
Then, we choose four cases which involve to the ordinary differential equation (8) to preform the Haar wavelet method. Case 1 simple-supported at both ends Using the boundary conditions
ω (t )dt + ω (0) ( 3)
2
whereI( m ) is identity matrix.
(3)
⎡⎣ c T( m ) P(2m ) + f Tω ( 3) (0)P( m ) T ( 2) f ω (0) ⎤⎦ h ( m ) ( x )
∫
= =
x
0
= =
(4)
⎡⎣ c ( m ) P( m ) + f ω (0) ⎤⎦ h ( m ) ( x )
=
ω ′( x )
ω (t )dt + ω (0)
T
= ( 2)
x
(2)
T
n=0
(15)
−4f ω (0)P( m ) − 4f ω ′(0)P( m ) − 4f ω (0)
(2)
T
formula (15),we can gain c ( m ) via solving the matrix equation, then we have ω ( x ) by formula (14) respectively.
1041
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007
The comparison of the analytic solution and the Haar solution is shown in Table.1.It is seen that even when m=32 and m = 64 the Haar direct method is quite satisfactory. x 0 0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375 1
Analytic solution
Haar solution (m=32) 0 0.0014 0.0038 0.0055 0.0059 0.0060 0.0047 0.0033 0.0011 0
Haar solution (m=64) 0 0.0013 0.0037 0.0054 0.0061 0.0059 0.0048 0.0032 0.0010 0
Haar solution (m=32) 0 0.00014 0.00072 0.00140 0.00200 0.00210 0.00190 0.00130 0.00052 0
Haar solution (m=64) 0 0.00013 0.00078 0.00150 0.00210 0.00230 0.00200 0.00140 0.00056 0
Haar solution (m=32) 0 0.00012 0.00130 0.00250 0.00380 0.00530 0.00610 0.00690 0.00760 0.00780 (c)
Haar solution (m=64) 0 0.00016 0.00120 0.00260 0.00390 0.00510 0.00620 0.00690 0.00770 0.00780
0 0.0013 0.0037 0.0054 0.0061 0.0059 0.0048 0.0031 0.0010 0 (a)
x
Analytic solution
0 0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375 1
0 0.00012 0.00080 0.00160 0.00210 0.00230 0.00200 0.00140 0.00057 0 (b)
x
Analytic solution
0 0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375 1
0 0.00017 0.00120 0.00260 0.00400 0.00520 0.00620 0.00700 0.00770 0.00790
Analytic solution
x 0 0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375 1
Haar solution (m=32) 0 0.0002 0.0005 0.0012 0.0016 0.0012 0.0008 0.0005 0.0002 0
0 0.0001 0.0006 0.0011 0.0014 0.0013 0.0009 0.0004 0.0001 0
Haar solution (m=64) 0 0.0001 0.0006 0.0011 0.0014 0.0013 0.0009 0.0004 0.0001 0
(d) Table 1. comparison of the analytic solution and the Haar solution:(a)simple-supported at both ends,(b)fixed end at one and simple-supported at the other one,(c)fixed end at one and free end at the other end,(d)f ixed end at both ends 3.2.
Convection-diffusion equation via Haar wavelet
We will consider the one-dimensional convectiondiffusion equation with constant coefficients ∂u
∂u
2
=α
∂u
,0 < x < 1,0 < t ≤ T 2 ∂t ∂x ∂x with initial condition u ( x , 0) = f ( x ) , 0 ≤ x ≤ 1 and boundary conditions u (0, t ) = g 0 (t ) , u (1, t ) = g1 (t ) , 0 < t ≤ T +a
(16)
Let us divide the interval (0, T ] into N equal parts of ∆t = ( 0, T ] / N
length
and
denote
t s = ( s − 1) ∆t , s = 1, 2, " , N .We assume that u ′′( x, t ) can
be expanded in terms of Haar wavelets as formula (4) m −1
T u ′′( x, t ) = ∑ cs ( n ) hn ( x ) = c ( m )h ( m ) ( x )
(17)
n=0
where · and ' means differentiation with respect to t and x T
respectively, the row vector c ( m ) is constant in the subinterval
(
t ∈ t , t s +1
].
Integrating formula (17) with respect to t from t s to t and twice with respect to x from 0 to x, and using formula (6), the variable u ′′( x, t ) , u ′( x, t ) , u ( x, t ) and u ( x, t ) can be expressed as T u ′′( x, t ) = (t − t s )c ( m ) h ( m ) ( x ) + u ′′( x, t s ) (18) u ′( x , t )
=
(t − t s )c ( m ) P( m )h ( m ) ( x ) + u ′( x, t s ) T
−u ′(0, t s ) + u ′(0, t )
1042
(19)
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007 T
=
u ( x, t )
u ( xl , t s +1 ) = −0.8u ′( xl , t s +1 ) + 0.1u ′′( xl , t s +1 )
2
(t − t s )c ( m ) P( m )h ( m ) ( x ) + u ( x, t s ) − u (0, t s ) + x [ u ′(0, t ) − u ′(0, t s ) ] + u (0, t )
T 2 u ( x, t ) = c ( m ) P( m )h ( m ) ( x ) + xu ′(0, t ) + u (0, t )
(20)
which leads us from the time layer t s to t s +1 is used. Substituting formula (24)-(27) into formula (28),we gain
(21)
T
Putting x = 1 in formula (20) and (21), we have T u ′(0, t ) − u ′(0, t s ) = −(t − t s )c ( m ) P( m ) h ( m ) ( x ) g1 (t ) − g 0 ( t ) − g1 (t s ) + g 0 (t s ) T 2 u ′(0, t ) = -c ( m ) P( m ) f + g1′ (t ) − g 0′ (t )
=
(22)
⎡ ( xl − 2) 2 ⎤ ⎥ 8 ⎣ ⎦ x −2 ⎡ ( x − 2) 2 ⎤ u ′( xl , t1 ) = − l exp ⎢ − l ⎥ 4 8 ⎣ ⎦ 2 x − 4 xl ⎡ ( x − 2) 2 ⎤ u ′′( xl , t1 ) = − l exp ⎢ − l ⎥ 16 8 ⎣ ⎦ u ( xl , t1 ) = exp ⎢ −
(24)
T
=
(t s +1 − t s )c ( m ) P( m ) h ( m ) ( xl ) +u ′( xl , t s ) − (t s +1 − t s )c ( m ) P( m ) f T
(25)
+ g1 (t s +1 ) − g 0 (t s +1 ) − g1 (t s ) + g 0 (t s ) u ( xl , t s +1 )
T
=
2
(t s +1 − t s )c ( m ) P( m )h ( m ) ( xl ) + u ( xl , t s ) − g 0 (t s ) + g 0 (t s +1 ) T
+ xl [ − (t s +1 − t s )c ( m ) P( m ) f + g1 (t s +1 )
(26)
Computer simulation was carried out in the cases m=32 and m=64,the computed results were compared with the exact solution, more accurate results can be obtained by using a larger m. For the time step was taken ∆t = 0.001 , further diminution of ¢t did not give any essential effect. Some results of computation are presented in Table.2. Comparison with these algorithms shows that the Haar wavelet method is competitive and efficient. The advantages of our method are its simplicity and speed of convergence.
− g 0 (t s +1 ) − g1 (t s ) + g 0 (t s )] u ( xl , t s +1 )
T 2 c ( m ) P( m )h ( m ) ( x ) + g 0′ (t s +1 )
=
+ xl [ − c ( m ) P( m ) f + g1′ (t s +1 ) − g 0′ (t s +1 )] T
(27)
where the vector f is defined as f = [1, 0, ", 0 ] N
T
(m-1) elements
For clarity in the illustrative example, we assume that a = 0.8 , α = 0.1 ,
⎡ ( x − 2) 2 ⎤ f ( x ) = exp ⎢ − 8 ⎥⎦ ⎣
20
u ( x, t ) =
20
⎡ 2(5 + 2t ) 2 ⎤ ⎥ ⎣ 5(t + 20) ⎦
exp ⎢ −
20 + t for which the exact solution is
⎡ ( x − 2 − 0.8t ) 2 ⎤ ⎥ ⎣ 0.4(t + 20) ⎦
exp ⎢ −
20 + t In the following the scheme
x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Exact solution 0.5765716 0.6065073 0.6364239 0.6661694 0.6955854 0.7245093 0.7527748 0.7802143 0.9066598
Haar solution (m=32) 0.5762574 0.6062752 0.6361158 0.6658236 0.6953754 0.7242356 0.7525858 0.7800047 0.8064472 (a)
⎡ (5 + 4t ) 2 ⎤ g 0 (t ) = exp ⎢ − ⎥ 20 + t ⎣ 10(t + 20) ⎦ 20
g1 ( t ) =
−0.8u ′( xl , t s ) + 0.1u ′′( xl , t s )
successively calculated. This process is started with
t = t s +1 ,we obtain
u ′( xl , t s +1 )
(29)
T
Substituting formula (22) and (23) into formula (18)-(21),and discretizising the results by assuming x = xl ， T
2
From formula (29) the wavelet coefficients c ( m ) can be
(23)
u ′′( xl , t s +1 ) = (t s +1 − t s )c ( m )h ( m ) ( xl ) + u ′′( xl , t s )
T
+ g1′ (t s +1 ) − g 0′ (t s +1 )] + g 0′ (t s +1 )
u (1, t ) = g1′ (t )
,
2
c ( m ) P( m )h ( m ) ( xl ) + xl [ −c ( m ) P( m ) f
By the boundary conditions, we obtain u (0, t s ) = g 0 (t s ) , u (1, t s ) = g1 (t s ) u (0, t ) = g 0′ (t )
(28)
1043
Haar solution (m=64) 0.5765247 0.6064857 0.6363998 0.6660952 0.6955841 0.7245062 0.7527289 0.7801957 0.8066249
Proceedings of the 2007 International Conference on Wavelet Analysis and Pattern Recognition, Beijing, China, 2-4 Nov. 2007
x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Exact solution 0.5227473 0.5520503 0.5815743 0.6111833 0.6407336 0.6700747 0.6990506 0.7275012 0.7552636
Haar solution (m=32) 0.5224653 0.5517783 0.5813547 0.6109966 0.6405742 0.6704588 0.6988549 0.7272821 0.7550917
sparse matrices of representation, fast transformation and possibility of implementation of fast algorithms. The main advantages of this method is its simplicity and small computation costs, it is due to the sparcity of the transform matrices and to the small number of significant wavelet coefficients.
Haar solution (m=64) 0.5227218 0.5519745 0.5815227 0.6111046 0.6407871 0.6701562 0.6990986 0.7275663 0.7552478
References
[1] C.F. Chen, C.H. Hsiao, “Haar wavelet method for solving lumped and distributed-parameter systems”, IEE Proc. Control Theory Appl, 144, pp. 87-94, 1997. [2] C.H. Hsiao, “States analysis of linear time delayed systems via Haar wavelets”, Math. Comput. Simulat, 44, pp.457-470, 1997. [3] Chun-Hui Hsiao, “Haar wavelet direct method for solving variational problems”, Mathematics and Computers in Simulation, 64, pp. 569-585,2004. [4] Ulo Lepik, “Numerical solution of evolution equations by the Haar wavelet method”, Applied Mathematics and Computation,185, pp. 695-704, 2007. [5] C.F. Chen, C.H. Hsiao, “Wavelet approach to optimizing dynamic systems”, IEE Proc. Control Theory Appl,146, pp. 213-219,1997. [6] C.H. Hsiao, W.J. Wang, “Haar wavelet approach to nonlinear stiff systems”, Math. Comput. Simulat, 57, pp. 347-353, 2001.
(b) x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Exact solution 0.5135925 0.5427477 0.5721621 0.6017027 0.6312285 0.6605916 0.6896381 0.7182096 0.7461446
Haar solution (m=32) 0.5133896 0.5424384 0.5718349 0.6014876 0.6310182 0.6602749 0.6893567 0.7180179 0.7458774
Haar solution (m=64) 0.5135481 0.5427598 0.5721547 0.6017536 0.6312178 0.6605749 0.6896124 0.7182073 0.7461267
(c) x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Exact solution 0.4408986 0.4684767 0.4965875 0.5251243 0.5539709 0.5830024 0.6120858 0.6410807 0.6698409
Haar solution (m=32) 0.4406249 0.4681895 0.4962283 0.5249834 0.5537685 0.5828859 0.6118746 0.6411492 0.6694583
Haar solution (m=64) 0.4408764 0.4684373 0.4965538 0.5251097 0.5538918 0.5830183 0.6120752 0.6410286 0.6697782
(d) Table 2. comparison of the exact solution and the Haar solution:(a)solution for t=0.25,(b)solution for t=0.48,(c)solution for t=0.52,(d)solution for t=0.85 4.
Conclusions
It has been well demonstrated that in applying the nice properties of Haar wavelets, the differential equations can be solved conveniently and accurately by using Haar wavelet method systematically. It has been also shown that the key idea is to transform the differential equation into a group of algebra equations which involves a finite number of variables. The benefits of the Haar wavelet approach are
1044
