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Haar wavelet method for some nonlinear Volterra integral equations of the first kind
Accepted Manuscript Haar wavelet method for some nonlinear Volterra integral equations of the first kind Inderdeep Singh, Sheo Kumar PII: DOI: Reference:
S0377-0427(15)00384-2 http://dx.doi.org/10.1016/j.cam.2015.07.022 CAM 10240
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Journal of Computational and Applied Mathematics
Received date: 7 May 2014 Revised date: 22 June 2015 Please cite this article as: I. Singh, S. Kumar, Haar wavelet method for some nonlinear Volterra integral equations of the first kind, Journal of Computational and Applied Mathematics (2015), http://dx.doi.org/10.1016/j.cam.2015.07.022 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
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Haar wavelet method for some nonlinear Volterra integral equations of the first kind Inderdeep Singh and Sheo Kumar Department of Mathematics Dr. B. R. Ambedkar National Institute of Technology Jalandhar, Punjab-144011, India Email: [email protected], [email protected] Abstract: We present here a simple efficient Haar wavelet method for numerical solution of a class of nonlinear Volterra integral equations of the first kind. The present method is based on converting nonlinear Volterra integral equations of the first kind into linear Volterra integral equations of the second kind. Numerical examples are given to illustrate efficiency and accuracy of the present method. Comparison with numerical methods in the past has also been done. M SC: 65R20; 45G10; 45D05; 41A30 Keywords: Nonlinear Volterra integral equation; Haar wavelet; Operational matrix. 1. Introduction In literature, many numerical methods using wavelets have been presented for solving nonlinear Volterra integral equations of the first kind. In [1, 2, 3], numerical methods based on collocation and implicit Runge-Kutta were described for the solution of first and second kind Volterra integral equations. Methods based on product integrations were described in [4, 5, 6, 7, 8] for solution of second kind Volterra integral equations with singular, nonsingular and periodic kernels. Method described in [7] can be used directly for first kind Volterra integral equations. An operational Haar wavelet method for solving fractional Volterra integral equations is also presented in [9]. A survey of numerical methods for solving nonlinear integral equations have been presented in [10]. In mathematical physics and engineering, problems are often reduced to Volterra integral equations of the first kind. Therefore, we present here Haar wavelet method for 1
solution of these equations. Volterra integral equations of the first kind are inherently ill-posed. Due to this, small variations in the problem can cause very large variations in the solution obtained, see, for example [11, 12, 13]. From past literatures, we know that it is more difficult to find the numerical solutions of ill-posed problems. Unbounded errors are obtained due to small errors in observations of data of ill-posed problems arising in science, engineering, medicines and ecology. In [11, 14] different regularization methods have been proposed to overcome with the difficulty of ill-posed problems. Consider the nonlinear Volterra integral equation of the form Z x K(x, t)S(y(t))dt = g(x), x∈ [a, b], (1) a
where K, S and g are the given smooth functions and S is invertible and nonlinear in y. The solution y is determined under the assumption that g(a) = 0. As a illustration, some forms of S(y(t)) are given below: • y (n) (t), it is assumed that y = y (1) = y (2) = ........ = y (n−1) = 0 at t = a, where y (n) represents the nth derivative of y with respect to t, • y n (t), where y n represents the nth power of y(t), p • ln(y(t)) and ln( y(t)), qp p y(t), • y(t) and p p • sin(y(t)) and cos(y(t)).
Equations of the form (1) have been studied in [15, 16, 17, 18]. Babolian et al. [15, 16] suggested direct method for solving Volterra integral equation of the first kind using operational matrix with block pulse functions and operational matrices of piecewise constant orthogonal functions respectively. Biazar et al. [17, 18] described a method for solving nonlinear Volterra integral equation of the first kind using Adomian decomposition method and homotopy perturbation method. Lepik [19, 20] have described methods for solution of integral and differential equations using Haar wavelets. Chen and Hsiao [21] have described numerical methods for solving lumped and distributed parameter systems using Haar wavelets. In Section 2, we briefly describe Haar wavelet method. In Section 3, we have given function approximation. In Section 4, error analysis is presented. In Section 5, we propose the method for solving nonlinear Volterra integral equations of the first kind using Haar wavelet and in Section 6, numerical examples have been solved using the present method to illustrate the efficiency and accuracy of present method. In this section, we also compare the present method with numerical methods presented in [15, 16].
2
2. Haar wavelet method The Haar functions are an orthogonal family of switched rectangular waveforms where amplitudes can differ from one function to another. They are defined in the interval [0, 1] as below: α ≤ x < β, 1, hi (x) = −1, β ≤ x < γ, (2) 0, elsewhere,
k , β = k+0.5 and γ = k+1 . Integer m = 2j , (j = 0, 1, 2, 3, 4, .......J) indicates where α = m m m the level of the wavelet, and k = 0, 1, 2, 3, ....., m − 1 is the translation parameter. Maximal level of resolution is J. The index i = m + k + 1. In case of minimal values, m = 1, k = 0 we have i = 2. The maximal value of i is i = 2M , where M = 2J . It is assumed that for value i = 1, the corresponding scaling function in [0, 1] is as: ( 1, 0 ≤ x ≤ 1, h1 (x) = (3) 0, elsewhere.
Let us define the collocation points xl = (l−0.5) , where l = 1, 2, 3, ..., 2M and discretise 2M the Haar function hi (x). The Haar coefficient matrix is defined as HAAR(i, l) = (hi (xl )), which is a square matrix of order 2M ×2M . The Haar coefficient matrix of order 8 is given below: 1 1 1 1 1 1 1 1 1 1 1 1 −1 −1 −1 −1 1 1 −1 −1 0 0 0 0 0 0 0 0 1 1 −1 −1 . HAAR8 (x) = (4) 1 −1 0 0 0 0 0 0 0 0 1 −1 0 0 0 0 0 0 0 0 1 −1 0 0 0 0 0 0 0 0 1 −1
The operational matrix of integration, which is a 2M × 2M square matrix, is defined by the relations: Z x Pi,1 (x) = hi (t)dt, (5) 0
Pi,n+1 (x) =
Z
x
Pi,n (t)dt,
(6)
0
where n = 1, 2, 3, 4, .... These integrals can be four of them are given below: x − α, Pi,1 (x) = γ − x, 0, 3
evaluated using Equation (2) and first x∈[α, β), x∈[β, γ), elsewhere;
(7)
1 (x − α)2 , x∈[α, β), 2 1 − 1 (γ − x)2 , x ∈ [β, γ), 2 2 Pi,2 (x) = 4m1 , x∈[γ, 1), 2 4m 0, elsewhere; 1 (x − α)3 , x∈[α, β), 6 1 (x − β) − 1 (γ − x)3 , x∈[β, γ), 2 6 Pi,3 (x) = 4m1 (x − β), x∈[γ, 1), 2 4m 0, elsewhere; 1 (x − α)4 , x∈[α, β), 24 1 (x − β)2 − 1 (γ − x)4 + 1 , x∈[β, γ), 2 24 192m4 Pi,4 (x) = 8m1 1 2 x∈[γ, 1), 2 (x − β) + 192m4 , 8m 0, elsewhere.
The Heaviside step function F (x) is defined as: ( 1, x ≥ 0, F (x) = 0, x < 0.
(8)
(9)
(10)
(11)
This function has the property that F (x − α)F (x − β) = F (x − max{α, β}),
α, β ∈ R.
By using Heaviside step function, we can also write h0 = F (x) − F (x − 1), hn (x) = F (x −
(12)
k + 0.5 k+1 k ) − 2F (x − ) + F (x − ), j j 2 2 2j n = 2j + k, j, k ∈ N ∪ {0}, 0 ≤ k < 2j . (13)
3. Function approximation We know that all the Haar wavelets are orthogonal to each other: ( Z 1 2−j , i = l = 2j + k, hi (x)hl (x)dx = 0, i 6= l. 0
4
(14)
Therefore, they construct a very good orthogonal transform basis. Any square integrable function y(x) in the interval [0, 1] can be expanded by a Haar series of infinite terms: j −1 ∞ 2X X y(x) = c0 h0 (x) + c2j +k h2j +k (x), x ∈ [0, 1], (15) j=0 k=0
where the Haar coefficients ci are determined as: Z 1 c0 = y(x)h0 (x)dx,
(16)
0
j
ci = 2
Z
1
y(x)hi (x)dx,
(17)
0
where i = 2j + k, j≥0 and 0≤k < 2j , x∈[0, 1] such that the following integral : ε=
Z
0
1
[y(x) −
m X
ci hi (x)]2 dx,
(18)
i=1
of square error ε is minimized, where m = 2j and j = 0, 1, 2, 3, .... By using (13), the Haar wavelet coefficients can be written as: Z k+1 Z k+0.5 2j 2j j y(x)dx − y(x)dx], (19) ci = 2 [ k+0.5 2j
k 2j
where i = 2j + k, j, k ∈ N ∪ {0} and 0 ≤ k < 2j . Usually the series expansion of (15) 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 m finite terms. This means j
y(x) ∼ = c0 h0 (x) +
−1 J 2X X
c2j +k h2j +k (x),
(20)
j=0 k=0
= cm T hm (x) = ym (x), x ∈ [0, 1], where the coefficients cm T and the Haar function vectors hm (x) are defined as: cm T = [c1 , c2 , c3 , ..........., cm ] and hm (x) = [h1 (x), h2 (x), h3 (x), ..........., hm (x)]T . 4. Error analysis
5
Consider the nonlinear Volterra integral equation (1) and convert this nonlinear Volterra integral equation of first kind into linear Volterra integral equation of second kind. Let y(x) is the exact solution and ym (x) is the approximate solution of linear Volterra integral equation of second kind. ERRORm (x) be the corresponding error function and is defined as: j
ERRORm (x) = |y(x) − ym (x)| '
∞ 2X −1 X
j=J+1 k=0
c2j +k h2j +k (x), x ∈ [0, 1].
(21)
Theorem: Suppose that y(x) satisfies a Lipschitz’s condition on [0, 1], then there exists constant K > 0 (dependent on both y(x) and interval), such that |y(x1 ) − y(x2 )| ≤ K|x1 − x2 |,
∀x1 , x2 ∈ [0, 1].
(22)
The Haar wavelet method will be convergent in the sense that ERRORm (x) goes to zero as m goes to infinity. The order of convergence is: kERRORm (x)k2 ' O(
1 ). m
(23)
Proof: Squaring the integrand and breaking the summation 2
kERRORm (x)k2 =
Z
j
1
(
0
∞ 2X −1 X
c2j +k h2j +k (x))2 dx,
(24)
j=J+1 k=0
we obtain j
2
kERRORm (x)k2 =
∞ 2X −1 X
c22j +k
j=J+1 k=0
Z
1
h22j +k (x)dx
0
j
+
p −1 2X
∞ 2X −1 X ∞ X
c2j +k c2p +q
Z
1
h2j +k (x)h2p +q (x)dx. (25)
0
j=J+1 k=0 p=J+1 q=0,q6=k
Using orthogonality described in (14), from (25), we obtain j
=
∞ 2X −1 X
j=J+1 k=0
c22j +k (
1 ). 2j
(26)
Writing (17) for i = 2j + k, we obtain j
c2j +k = 2
Z
1
y(x)h2j +k (x)dx.
0
6
(27)
From mean value theorem and Equation (19), there exist two points described below xjk 11 ∈ (
k k + 0.5 , ), 2j 2j
xjk 22 ∈ (
k + 0.5 k + 1 , j ), 2j 2
(28)
such that from (19) and (27), we obtain c2j +k = 2j [(
k k + 1 k + 0.5 k + 0.5 − j )y(xjk − )y(xjk 11 ) − ( 22 )]. j 2 2 2j 2j
(29)
Simplifying (29) and using Lipschitz’s condition, we obtain 1 1 1 1 1 jk jk jk c2j +k = [y(xjk 11 ) − y(x22 )] ≤ K(x11 − x22 ) ≤ ( )K( j ) ' K( j+1 ). 2 2 2 2 2
(30)
From (26) and (30),we obtain j
∞ 2X −1 X
2
kERRORm (x)k2 =
j=J+1 k=0
∞ 1 K2 X j 1 K 2j+2 ( j ) = 2 ( 3j ). 2 2 4 j=J+1 2 2
1
(31)
Applying series summation on (31), we obtain kERRORm (x)k2 2 =
K2 1 2 ( ). 3 2J+1
(32)
But, m = 2J+1 , therefore, we obtain, kERRORm (x)k2 ' O(
1 ). m
(33)
Thus, when m goes to infinity, then ERRORm (x) tends to zero. 5. Method for solving nonlinear Volterra integral equation Consider the nonlinear Volterra integral equation (1). First, we convert the nonlinear Volterra integral equation of first kind into linear Volterra integral equation of second kind by substituting S(y(t)) = w(t), (34) in (1). We get, the linear Volterra integral equation of the form: Z x K(x, t)w(t)dt = g(x).
(35)
a
After differentiating the above equation with respect to x, we obtain Z x ∂ (K(x, t))w(t)dt = g 0 (x). K(x, x)w(x) + ∂x 0 7
(36)
With assumption K(x, x) 6= 0, Equation (36) is converted to Z x ∂K(x, t) [ /K(x, x)]w(t)dt = g 0 (x)/K(x, x). w(x) + ∂x 0
(37)
By setting k(x, t) = ∂K(x,t) /K(x, x) and g10 (x) = g 0 (x)/K(x, x), Equation (37) can be ∂x written in the following form Z x k(x, t)w(t)dt = g10 (x). (38) w(x) + 0
Taking the discrete form of (38), we obtain Z xl w(xl ) + k(xl , t)w(t)dt = g10 (xl ).
(39)
0
Now, we apply Haar wavelet method taking the solution w(x) of (39) in the form: w(x) =
2M X
ai hi (x).
(40)
i=1
Substituting the value of w(x) in Equation (39), we obtain 2M X
ai [hi (xl ) + Iil ] = g10 (xl ),
(41)
i=1
where Iil =
Z
xl
(k(xl , t))hi (t)dt.
(42)
a
The integral Iil is determined as : R xl k(xl , t)dt, α R β k(x , t)dt − R xl k(x , t)dt, l l Rβγ Iil = Rαβ k(xl , t)dt − β k(xl , t)dt, α 0,
α ≤ xl < β, β ≤ xl < γ, γ ≤ xl < 1, xl < α.
(43)
From (41), wavelet coefficients are calculated and solution w(x) of equation (38) is obtained. The solution y(t) of the given problem is obtained from (34). The accuracy of the results are estimated by the maximum absolute error (MAE) M AE = max
1≤l≤2M
{|yapproximate (xl ) − yexact (xl )|}.
8
(44)
6. Numerical examples Here, we take few examples of nonlinear Volterra integral equations of the first kind to show the efficiency and accuracy of the Haar wavelet method. We use MATLAB package to perform all computational work. Example 1: Consider the following nonlinear Volterra integral equation Z x cos(x − t)(y 00 (t))dt = 2sinx, x∈[0, 1],
(45)
0
with initial conditions y(0) = 0 and y 0 (0) = 0. Exact solution is y(x) = x2 . First we convert the nonlinear Volterra integral equation of first kind into second kind by substituting y 00 (t) = w(t) and differentiating it partially with respect to x, we get linear Volterra integral equation of second kind in unknown w(t). Discrete form of the above is: Z xl w(xl ) − sin(xl − t)w(t)dt = 2cosxl . (46) 0
Consider the solution of above equation as:w(x) =
2M X
ai hi (x).
(47)
i=1
By substituting w(x) in (46), we get Z xl 2M X sin(xl − t)hi (t)dt] = 2cosxl . ai [hi (xl ) −
(48)
0
i=1
We calculate wavelet coefficients from here. Using these, we get the solution w(t) of (46) and the solution of (45) is obtained from the relation y 00 (t) = w(t) by taking integration twice with respect to x from 0 to x; that is solution is obtained from y(x) =
2M X
ai Pi,2 (x).
(49)
i=1
The maximum absolute errors are shown in Table 1, for different values of J. By the direct method in [15], the computational error for N = 8 is: N
1X 2 e N (xi ))1/2 ≈ 1.7E − 14. ( N i=1
(50)
But, in the proposed method , the computational error for 2M = 8 is: 2M
1 X 2 ( e 2M (xi ))1/2 ≈ 2.1E − 16. 2M i=1 9
(51)
0.9
Haar wavelet solutions 0.8
Exact solution
Numerical solutions (y(x))
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 1: Comparison of numerical and exact solutions of Example 1 for J=2.
In Equations (50) and (51), we have used the notation of [15]. Therefore, proposed method is better in comparison of numerical results obtained by direct method in [15]. The numerical errors of Table 1 and in [15] are very good i.e of order 10−16 and 10−14 respectively, because the second derivative of y(x) = x2 is 2, and Haar wavelet method and the method presented in [15] are exact for the constants. Figure 1 shows the comparison of numerical and exact solutions of Example 1 for J = 2. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 0 16 2.2E-016 32 4.4E-016 64 5.5E-016 128 5.5E-016 256 8.8E-016 512 3.4E-016 1024 5.5E-016
Table 1: Maximum absolute error for Example 1. Example 2: Consider the following nonlinear Volterra integral equation Z x e(x−t) (y 000 (t))dt = 6ex , x∈[0, 1], 0
10
(52)
0.9
0.8
Haar wavelet solution Exact solution
0.7
Numerical solutions (y(x))
0.6
0.5
0.4
0.3
0.2
0.1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 2: Comparison of numerical and exact solutions of Example 2 for J=2.
with initial conditions y(0) = 0, y 0 (0) = 0 and y 00 (0) = 0. Exact solution is y(x) = x3 . We solved Example 2, by the method of Example 1. The maximum absolute errors are given in Table 2. The numerical results in Table 2, establishes the facts given in Example 1. Figure 2 shows the comparison of numerical and exact solutions of Example 2 for J = 2. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 8.6E-019 16 1.1E-016 32 1.1E-016 64 1.1E-016 128 3.3E-016 256 3.3E-016 512 3.3E-016 1024 6.6E-016
Table 2: Maximum absolute error for Example 2. Example 3: Consider the following nonlinear Volterra integral equation Z x cos(x − t)(y 00 (t))dt = 6(1 − cos(x)), x∈[0, 1], 0
11
(53)
0.9
Haar wavelet solution Exact solution
0.8
0.7
Numerical solutions (y(x))
0.6
0.5
0.4
0.3
0.2
0.1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 3: Comparison of numerical and exact solutions of Example 3 for J=2.
with initial conditions y(0) = 0 and y 0 (0) = 0. Exact solution is y(x) = x3 . Solving Example 3, by the same method of Example 1, we see that maximum absolute errors are much higher as compared to Example 1 and Example 2. The result here is poor, because the second derivative of x3 is 6x, which is not a constant. The maximum absolute errors are given in Table 3. Figure 3 shows the comparison of numerical and exact solutions of Example 3 for J = 2. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 8.4E-003 16 2.1E-003 32 5.5E-004 64 1.4E-004 128 3.5E-005 256 8.8E-006 512 2.2E-006 1024 5.5E-007
Table 3: Maximum absolute error for Example 3. Example 4: Consider the following linear Volterra integral equation Z x e(x−t) (y 2 (t))dt = e2x − ex , x∈[0, 1]. 0
12
(54)
whose exact solution is y(x) = ex . First we convert the nonlinear Volterra integral equation of first kind into second kind bypsubstituting y 2 (t) = w(t). The solution of (54) is obtained from the relation y(x) = (w(x)). The maximum absolute errors for different values of J are shown in the Table 4. J
2M
Maximum absolute error(MAE) 8 2.8E-003 16 7.3E-004 32 1.8E-004 64 4.6E-005 128 1.1E-005 256 2.9E-006 512 7.3E-007 1024 1.8E-007
2 3 4 5 6 7 8 9
Table 4: Maximum absolute error for Example 4. Table 5, shows the comparison of absolute errors of proposed method and the method presented in [16] of Example 4. The absolute errors of proposed method is low as compared to the absolute errors in [16]. This shows that proposed method is better than method given in [16]. Figure 4 shows the comparison of numerical and exact solutions of Example 4 for J = 2. Collocation Absolute errors in Absolute errors points proposed method in [16] 1/16 0.00195 0.0328 3/16 0.00198 0.0374 5/16 0.00204 0.0425 7/16 0.00214 0.0485 9/16 0.00226 0.0552 11/16 0.00243 0.0628 13/16 0.00263 0.0716 15/16 0.00287 0.0815 Table 5: Comparison of absolute errors for Example 4, at different collocation points. Example 5: Consider the following Volterra integral equation Z x ex−t y(t)dt = x, x∈[0, 1]. 0
13
(55)
2.6
Haar wavelet solution 2.4
Exact solution
Numerical solutions (y(x))
2.2
2
1.8
1.6
1.4
1.2
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 4: Comparison of numerical and exact solutions of Example 4 for J=2.
The exact solution of this problem is y(x) = 1 − x. Table 6, shows the maximum absolute errors of Example 5 for different values of J. Again, in Table 7, comparison of absolute errors of proposed method and the method presented in [16] of Example 5 at different collocation points is given. This again establishes the fact that the absolute errors of the proposed method is very low as compared to the absolute errors in [16]. Figure 5 shows the comparison of numerical and exact solutions of Example 5 for J = 2. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 1.6E-003 16 4.5E-004 32 1.1E-004 64 2.9E-005 128 7.5E-006 256 1.8E-006 512 4.7E-007 1024 1.1E-007
Table 6: Maximum absolute error for Example 5. Example 6: Consider the following nonlinear Volterra integral equation Z x e(x−t) ln(y(t))dt = ex − x − 1, x∈[0, 1], 0
14
(56)
Collocation Absolute errors in Absolute errors points proposed method in [16] 1/16 0.0016 0.0313 3/16 0.0012 0.0313 5/16 0.0009 0.0313 7/16 0.0006 0.0313 9/16 0.0004 0.0313 11/16 0.0003 0.0313 13/16 0.0002 0.0313 15/16 0.0001 0.0313 Table 7: Comparison of absolute errors for Example 5, at different collocation points.
1
Haar wavelet solution Exact solution
0.9
Numerical solutions (y(x))
0.8
0.7
0.6
0.5
0.4
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 5: Comparison of numerical and exact solutions of Example 5 for J=2.
15
2.6
Haar wavelet solution Exact solution
2.4
Numerical solutions (y(x))
2.2
2
1.8
1.6
1.4
1.2
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 6: Comparison of numerical and exact solutions of Example 6 for J=2.
whose exact solution is y(x) = ex . First, we convert the nonlinear Volterra integral equation of first kind into second kind by substituting ln(u(t)) = w(t). The solution of (56) is obtained from the relation y(x) = ew(x) . The maximum absolute errors are shown in Table 8, for different values of J. Comparison of numerical and exact solutions of Example 6 for J = 2 are shown in Figure 6.
J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 3.4E-003 16 8.7E-004 32 2.1E-004 64 5.5E-005 128 1.3E-005 256 3.4E-006 512 8.6E-007 1024 2.1E-007
Table 8: Maximum absolute error for Example 6. Example 7: Consider the following nonlinear Volterra integral equation Z x xsinx (sin(x − t) + 1)cos(y(t))dt = + sinx, x∈[0, 1], 2 0 16
(57)
1
0.9
Haar wavelet solution Exact solution
0.8
Numerical solutions (y(x))
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 7: Comparison of numerical and exact solutions of Example 7 for J=2.
whose exact solution is y(x) = x. First we convert the nonlinear Volterra integral equation of first kind into second kind by substituting cos(y(t)) = w(t). The solution of (57) is obtained from the relation y(x) = cos−1 (w(x)). The maximum absolute errors for different values of J are shown in Table 9. Comparison of numerical and exact solutions of Example 7 for J = 2 are shown in Figure 7. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 1.2E-003 16 3.1E-004 32 8.0E-005 64 2.0E-005 128 5.0E-006 256 1.2E-006 512 3.1E-007 1024 7.9E-008
Table 9: Maximum absolute error for Example 7.
Conclusion We conclude here from the above, that the Haar wavelet method is more accurate, simple and fast than other known methods to solve nonlinear Volterra integral equations. 17
The above examples demonstrate the simplicity of the Haar wavelet solution. The comparison of the present method with [15, 16] shows that the present method gives better results. For getting the necessary accuracy the number of calculation points may be increased. References
[1] P. Linz, Numerical methods for Volterra integral equations with singular kernels, SIAM J. Numer. Anal. 6 (1969) 365-374. [2] F. de Hoog, R. Weiss, Impilict Runge- Kutta methods for second kind Volterra integral equations, Numer. Math. 23 (1975) 199-213. [3] K.E. Atkinson, J. Flores, The collocation method for nonlinear integral equations, IMA Journal of Numerical Analysis 13 (1993) 195-213. [4] K.E. Atkinson, The numerical solution of an Abel integral equation by a product trapezoidal method, SIAM J. Numer. Anal. 11 (1974) 97-101. [5] Sheo Kumar, On a method of Noble for second kind Volterra integral equations, BIT 19 (1979) 482-488. [6] Sheo Kumar, On modified increment methods of Garey for nonlinear second kind Volterra integral equations, BIT 20 (1980) 346-350. [7] Sheo Kumar, A recurrence relation for solution of singular Volterra integral equations using Chebyshev polynomials, BIT 21 (1981) 123-125. [8] Sheo Kumar, Modifications of Linz methods for nonlinear second kind Volterra integral equations with singular or periodic kernels, Journal of Mathematical and Physical Sciences 26 (1992) 591-597. [9] H. Saeedi, N. Mollahasani, M. M. Moghadam, G. N. Chuev, An operational Haar wavelet method for solving fractional Volterra integral equations, Int. J. Appl. Math. Comput. Sci. 21(3) (2011) 535-547. [10] K.E. Atkinson, A survey of numerical methods for solving nonlinear integral equations, J. Integral Equation Appl. 4 (1992) 15-40. [11] E. Babolian, L.M. Delves, An augmented Galerkin method for first kind Fredholm equations, J. Inst. Math. Appl. 24 (1979) 157-174.
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[12] L.M. Delves, J.L. Mohamed, Computational methods for integral equations, Cambridge University Press, Cambridge, 1985. [13] P.K. Kythe, P. Puri, Computational methods for integral equations, Birkh¨auser, Boston, 2002. [14] A.N. Tikhonov, V.Y. Arsenin, Solutions of ill-posed problems, Winston, Washington DC, 1977. [15] E. Babolian, Z. Masouri, Direct method to solve Volterra integral equation of first kind using operational matrix with block-pulse functions, Journal of Computational and Applied Mathematics 220 (2008) 51-57. [16] E. Babolian, A. Salimi Shamloo, Numerical solution of Volterra integral and integro-differential equation of convolution type by using operational matrices of piecewise constant orthogonal functions, Journal of Computational and Applied Mathematics 214 (2008) 495-508. [17] J. Biazar, E. Babolian, R. Islam, Solution of system of Volterra integral equations of the first kind by Adomian method, Applied Mathematics and Computation 139 (2003) 249-248. [18] J. Biazar, M. Islami, H. Aminkhah, Application of homotopy perturbation method for system of Volterra integral equations of the first kind, Chaos, Solutions and Fractals 42 (2009) 3020-3026. ¨ Lepik, Solving differential and integral equations by Haar wavelet method, [19] U. revisted, International Journal of Mathematics and Computation 1(8) (2008) 4352. ¨ Lepik, Application of Haar wavelet transform for solving integral and differen[20] U. tial equations, Proc. Estonian Acad. Sci. Phys. Math. 56(1) (2007) 28-46. [21] C.F. Chen, C.H. Hsiao, Haar wavelet methods for solving lumped and distributedparameter systems, J. IEE. Proc. Control Theory Appl. 144 (1997) 87-94.
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HIGHLIGHTS
Solving some nonlinear first kind Volterra integral equations using Haar wavelet. Haar wavelet method is simple and fast. Solved examples demonstrate the accuracy and efficiency of the present method. Present method gives better results than the numerical methods described in past. Necessary accuracy may be obtained by increasing the number of calculation points.
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Haar wavelet method for some nonlinear Volterra integral equations of the first kind Inderdeep Singh and Sheo Kumar Department of Mathematics Dr. B. R. Ambedkar National Institute of Technology Jalandhar, Punjab-144011, India Email: [email protected], [email protected] Abstract: We present here a simple efficient Haar wavelet method for numerical solution of a class of nonlinear Volterra integral equations of the first kind. The present method is based on converting nonlinear Volterra integral equations of the first kind into linear Volterra integral equations of the second kind. Numerical examples are given to illustrate efficiency and accuracy of the present method. Comparison with numerical methods in the past has also been done. M SC: 65R20; 45G10; 45D05; 41A30 Keywords: Nonlinear Volterra integral equation; Haar wavelet; Operational matrix. 1. Introduction In literature, many numerical methods using wavelets have been presented for solving nonlinear Volterra integral equations of the first kind. In [1, 2, 3], numerical methods based on collocation and implicit Runge-Kutta were described for the solution of first and second kind Volterra integral equations. Methods based on product integrations were described in [4, 5, 6, 7, 8] for solution of second kind Volterra integral equations with singular, nonsingular and periodic kernels. Method described in [7] can be used directly for first kind Volterra integral equations. An operational Haar wavelet method for solving fractional Volterra integral equations is also presented in [9]. A survey of numerical methods for solving nonlinear integral equations have been presented in [10]. In mathematical physics and engineering, problems are often reduced to Volterra integral equations of the first kind. Therefore, we present here Haar wavelet method for 1
solution of these equations. Volterra integral equations of the first kind are inherently ill-posed. Due to this, small variations in the problem can cause very large variations in the solution obtained, see, for example [11, 12, 13]. From past literatures, we know that it is more difficult to find the numerical solutions of ill-posed problems. Unbounded errors are obtained due to small errors in observations of data of ill-posed problems arising in science, engineering, medicines and ecology. In [11, 14] different regularization methods have been proposed to overcome with the difficulty of ill-posed problems. Consider the nonlinear Volterra integral equation of the form Z x K(x, t)S(y(t))dt = g(x), x∈ [a, b], (1) a
where K, S and g are the given smooth functions and S is invertible and nonlinear in y. The solution y is determined under the assumption that g(a) = 0. As a illustration, some forms of S(y(t)) are given below: • y (n) (t), it is assumed that y = y (1) = y (2) = ........ = y (n−1) = 0 at t = a, where y (n) represents the nth derivative of y with respect to t, • y n (t), where y n represents the nth power of y(t), p • ln(y(t)) and ln( y(t)), qp p y(t), • y(t) and p p • sin(y(t)) and cos(y(t)).
Equations of the form (1) have been studied in [15, 16, 17, 18]. Babolian et al. [15, 16] suggested direct method for solving Volterra integral equation of the first kind using operational matrix with block pulse functions and operational matrices of piecewise constant orthogonal functions respectively. Biazar et al. [17, 18] described a method for solving nonlinear Volterra integral equation of the first kind using Adomian decomposition method and homotopy perturbation method. Lepik [19, 20] have described methods for solution of integral and differential equations using Haar wavelets. Chen and Hsiao [21] have described numerical methods for solving lumped and distributed parameter systems using Haar wavelets. In Section 2, we briefly describe Haar wavelet method. In Section 3, we have given function approximation. In Section 4, error analysis is presented. In Section 5, we propose the method for solving nonlinear Volterra integral equations of the first kind using Haar wavelet and in Section 6, numerical examples have been solved using the present method to illustrate the efficiency and accuracy of present method. In this section, we also compare the present method with numerical methods presented in [15, 16].
2
2. Haar wavelet method The Haar functions are an orthogonal family of switched rectangular waveforms where amplitudes can differ from one function to another. They are defined in the interval [0, 1] as below: α ≤ x < β, 1, hi (x) = −1, β ≤ x < γ, (2) 0, elsewhere,
k , β = k+0.5 and γ = k+1 . Integer m = 2j , (j = 0, 1, 2, 3, 4, .......J) indicates where α = m m m the level of the wavelet, and k = 0, 1, 2, 3, ....., m − 1 is the translation parameter. Maximal level of resolution is J. The index i = m + k + 1. In case of minimal values, m = 1, k = 0 we have i = 2. The maximal value of i is i = 2M , where M = 2J . It is assumed that for value i = 1, the corresponding scaling function in [0, 1] is as: ( 1, 0 ≤ x ≤ 1, h1 (x) = (3) 0, elsewhere.
Let us define the collocation points xl = (l−0.5) , where l = 1, 2, 3, ..., 2M and discretise 2M the Haar function hi (x). The Haar coefficient matrix is defined as HAAR(i, l) = (hi (xl )), which is a square matrix of order 2M ×2M . The Haar coefficient matrix of order 8 is given below: 1 1 1 1 1 1 1 1 1 1 1 1 −1 −1 −1 −1 1 1 −1 −1 0 0 0 0 0 0 0 0 1 1 −1 −1 . HAAR8 (x) = (4) 1 −1 0 0 0 0 0 0 0 0 1 −1 0 0 0 0 0 0 0 0 1 −1 0 0 0 0 0 0 0 0 1 −1
The operational matrix of integration, which is a 2M × 2M square matrix, is defined by the relations: Z x Pi,1 (x) = hi (t)dt, (5) 0
Pi,n+1 (x) =
Z
x
Pi,n (t)dt,
(6)
0
where n = 1, 2, 3, 4, .... These integrals can be four of them are given below: x − α, Pi,1 (x) = γ − x, 0, 3
evaluated using Equation (2) and first x∈[α, β), x∈[β, γ), elsewhere;
(7)
1 (x − α)2 , x∈[α, β), 2 1 − 1 (γ − x)2 , x ∈ [β, γ), 2 2 Pi,2 (x) = 4m1 , x∈[γ, 1), 2 4m 0, elsewhere; 1 (x − α)3 , x∈[α, β), 6 1 (x − β) − 1 (γ − x)3 , x∈[β, γ), 2 6 Pi,3 (x) = 4m1 (x − β), x∈[γ, 1), 2 4m 0, elsewhere; 1 (x − α)4 , x∈[α, β), 24 1 (x − β)2 − 1 (γ − x)4 + 1 , x∈[β, γ), 2 24 192m4 Pi,4 (x) = 8m1 1 2 x∈[γ, 1), 2 (x − β) + 192m4 , 8m 0, elsewhere.
The Heaviside step function F (x) is defined as: ( 1, x ≥ 0, F (x) = 0, x < 0.
(8)
(9)
(10)
(11)
This function has the property that F (x − α)F (x − β) = F (x − max{α, β}),
α, β ∈ R.
By using Heaviside step function, we can also write h0 = F (x) − F (x − 1), hn (x) = F (x −
(12)
k + 0.5 k+1 k ) − 2F (x − ) + F (x − ), j j 2 2 2j n = 2j + k, j, k ∈ N ∪ {0}, 0 ≤ k < 2j . (13)
3. Function approximation We know that all the Haar wavelets are orthogonal to each other: ( Z 1 2−j , i = l = 2j + k, hi (x)hl (x)dx = 0, i 6= l. 0
4
(14)
Therefore, they construct a very good orthogonal transform basis. Any square integrable function y(x) in the interval [0, 1] can be expanded by a Haar series of infinite terms: j −1 ∞ 2X X y(x) = c0 h0 (x) + c2j +k h2j +k (x), x ∈ [0, 1], (15) j=0 k=0
where the Haar coefficients ci are determined as: Z 1 c0 = y(x)h0 (x)dx,
(16)
0
j
ci = 2
Z
1
y(x)hi (x)dx,
(17)
0
where i = 2j + k, j≥0 and 0≤k < 2j , x∈[0, 1] such that the following integral : ε=
Z
0
1
[y(x) −
m X
ci hi (x)]2 dx,
(18)
i=1
of square error ε is minimized, where m = 2j and j = 0, 1, 2, 3, .... By using (13), the Haar wavelet coefficients can be written as: Z k+1 Z k+0.5 2j 2j j y(x)dx − y(x)dx], (19) ci = 2 [ k+0.5 2j
k 2j
where i = 2j + k, j, k ∈ N ∪ {0} and 0 ≤ k < 2j . Usually the series expansion of (15) 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 m finite terms. This means j
y(x) ∼ = c0 h0 (x) +
−1 J 2X X
c2j +k h2j +k (x),
(20)
j=0 k=0
= cm T hm (x) = ym (x), x ∈ [0, 1], where the coefficients cm T and the Haar function vectors hm (x) are defined as: cm T = [c1 , c2 , c3 , ..........., cm ] and hm (x) = [h1 (x), h2 (x), h3 (x), ..........., hm (x)]T . 4. Error analysis
5
Consider the nonlinear Volterra integral equation (1) and convert this nonlinear Volterra integral equation of first kind into linear Volterra integral equation of second kind. Let y(x) is the exact solution and ym (x) is the approximate solution of linear Volterra integral equation of second kind. ERRORm (x) be the corresponding error function and is defined as: j
ERRORm (x) = |y(x) − ym (x)| '
∞ 2X −1 X
j=J+1 k=0
c2j +k h2j +k (x), x ∈ [0, 1].
(21)
Theorem: Suppose that y(x) satisfies a Lipschitz’s condition on [0, 1], then there exists constant K > 0 (dependent on both y(x) and interval), such that |y(x1 ) − y(x2 )| ≤ K|x1 − x2 |,
∀x1 , x2 ∈ [0, 1].
(22)
The Haar wavelet method will be convergent in the sense that ERRORm (x) goes to zero as m goes to infinity. The order of convergence is: kERRORm (x)k2 ' O(
1 ). m
(23)
Proof: Squaring the integrand and breaking the summation 2
kERRORm (x)k2 =
Z
j
1
(
0
∞ 2X −1 X
c2j +k h2j +k (x))2 dx,
(24)
j=J+1 k=0
we obtain j
2
kERRORm (x)k2 =
∞ 2X −1 X
c22j +k
j=J+1 k=0
Z
1
h22j +k (x)dx
0
j
+
p −1 2X
∞ 2X −1 X ∞ X
c2j +k c2p +q
Z
1
h2j +k (x)h2p +q (x)dx. (25)
0
j=J+1 k=0 p=J+1 q=0,q6=k
Using orthogonality described in (14), from (25), we obtain j
=
∞ 2X −1 X
j=J+1 k=0
c22j +k (
1 ). 2j
(26)
Writing (17) for i = 2j + k, we obtain j
c2j +k = 2
Z
1
y(x)h2j +k (x)dx.
0
6
(27)
From mean value theorem and Equation (19), there exist two points described below xjk 11 ∈ (
k k + 0.5 , ), 2j 2j
xjk 22 ∈ (
k + 0.5 k + 1 , j ), 2j 2
(28)
such that from (19) and (27), we obtain c2j +k = 2j [(
k k + 1 k + 0.5 k + 0.5 − j )y(xjk − )y(xjk 11 ) − ( 22 )]. j 2 2 2j 2j
(29)
Simplifying (29) and using Lipschitz’s condition, we obtain 1 1 1 1 1 jk jk jk c2j +k = [y(xjk 11 ) − y(x22 )] ≤ K(x11 − x22 ) ≤ ( )K( j ) ' K( j+1 ). 2 2 2 2 2
(30)
From (26) and (30),we obtain j
∞ 2X −1 X
2
kERRORm (x)k2 =
j=J+1 k=0
∞ 1 K2 X j 1 K 2j+2 ( j ) = 2 ( 3j ). 2 2 4 j=J+1 2 2
1
(31)
Applying series summation on (31), we obtain kERRORm (x)k2 2 =
K2 1 2 ( ). 3 2J+1
(32)
But, m = 2J+1 , therefore, we obtain, kERRORm (x)k2 ' O(
1 ). m
(33)
Thus, when m goes to infinity, then ERRORm (x) tends to zero. 5. Method for solving nonlinear Volterra integral equation Consider the nonlinear Volterra integral equation (1). First, we convert the nonlinear Volterra integral equation of first kind into linear Volterra integral equation of second kind by substituting S(y(t)) = w(t), (34) in (1). We get, the linear Volterra integral equation of the form: Z x K(x, t)w(t)dt = g(x).
(35)
a
After differentiating the above equation with respect to x, we obtain Z x ∂ (K(x, t))w(t)dt = g 0 (x). K(x, x)w(x) + ∂x 0 7
(36)
With assumption K(x, x) 6= 0, Equation (36) is converted to Z x ∂K(x, t) [ /K(x, x)]w(t)dt = g 0 (x)/K(x, x). w(x) + ∂x 0
(37)
By setting k(x, t) = ∂K(x,t) /K(x, x) and g10 (x) = g 0 (x)/K(x, x), Equation (37) can be ∂x written in the following form Z x k(x, t)w(t)dt = g10 (x). (38) w(x) + 0
Taking the discrete form of (38), we obtain Z xl w(xl ) + k(xl , t)w(t)dt = g10 (xl ).
(39)
0
Now, we apply Haar wavelet method taking the solution w(x) of (39) in the form: w(x) =
2M X
ai hi (x).
(40)
i=1
Substituting the value of w(x) in Equation (39), we obtain 2M X
ai [hi (xl ) + Iil ] = g10 (xl ),
(41)
i=1
where Iil =
Z
xl
(k(xl , t))hi (t)dt.
(42)
a
The integral Iil is determined as : R xl k(xl , t)dt, α R β k(x , t)dt − R xl k(x , t)dt, l l Rβγ Iil = Rαβ k(xl , t)dt − β k(xl , t)dt, α 0,
α ≤ xl < β, β ≤ xl < γ, γ ≤ xl < 1, xl < α.
(43)
From (41), wavelet coefficients are calculated and solution w(x) of equation (38) is obtained. The solution y(t) of the given problem is obtained from (34). The accuracy of the results are estimated by the maximum absolute error (MAE) M AE = max
1≤l≤2M
{|yapproximate (xl ) − yexact (xl )|}.
8
(44)
6. Numerical examples Here, we take few examples of nonlinear Volterra integral equations of the first kind to show the efficiency and accuracy of the Haar wavelet method. We use MATLAB package to perform all computational work. Example 1: Consider the following nonlinear Volterra integral equation Z x cos(x − t)(y 00 (t))dt = 2sinx, x∈[0, 1],
(45)
0
with initial conditions y(0) = 0 and y 0 (0) = 0. Exact solution is y(x) = x2 . First we convert the nonlinear Volterra integral equation of first kind into second kind by substituting y 00 (t) = w(t) and differentiating it partially with respect to x, we get linear Volterra integral equation of second kind in unknown w(t). Discrete form of the above is: Z xl w(xl ) − sin(xl − t)w(t)dt = 2cosxl . (46) 0
Consider the solution of above equation as:w(x) =
2M X
ai hi (x).
(47)
i=1
By substituting w(x) in (46), we get Z xl 2M X sin(xl − t)hi (t)dt] = 2cosxl . ai [hi (xl ) −
(48)
0
i=1
We calculate wavelet coefficients from here. Using these, we get the solution w(t) of (46) and the solution of (45) is obtained from the relation y 00 (t) = w(t) by taking integration twice with respect to x from 0 to x; that is solution is obtained from y(x) =
2M X
ai Pi,2 (x).
(49)
i=1
The maximum absolute errors are shown in Table 1, for different values of J. By the direct method in [15], the computational error for N = 8 is: N
1X 2 e N (xi ))1/2 ≈ 1.7E − 14. ( N i=1
(50)
But, in the proposed method , the computational error for 2M = 8 is: 2M
1 X 2 ( e 2M (xi ))1/2 ≈ 2.1E − 16. 2M i=1 9
(51)
0.9
Haar wavelet solutions 0.8
Exact solution
Numerical solutions (y(x))
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 1: Comparison of numerical and exact solutions of Example 1 for J=2.
In Equations (50) and (51), we have used the notation of [15]. Therefore, proposed method is better in comparison of numerical results obtained by direct method in [15]. The numerical errors of Table 1 and in [15] are very good i.e of order 10−16 and 10−14 respectively, because the second derivative of y(x) = x2 is 2, and Haar wavelet method and the method presented in [15] are exact for the constants. Figure 1 shows the comparison of numerical and exact solutions of Example 1 for J = 2. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 0 16 2.2E-016 32 4.4E-016 64 5.5E-016 128 5.5E-016 256 8.8E-016 512 3.4E-016 1024 5.5E-016
Table 1: Maximum absolute error for Example 1. Example 2: Consider the following nonlinear Volterra integral equation Z x e(x−t) (y 000 (t))dt = 6ex , x∈[0, 1], 0
10
(52)
0.9
0.8
Haar wavelet solution Exact solution
0.7
Numerical solutions (y(x))
0.6
0.5
0.4
0.3
0.2
0.1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 2: Comparison of numerical and exact solutions of Example 2 for J=2.
with initial conditions y(0) = 0, y 0 (0) = 0 and y 00 (0) = 0. Exact solution is y(x) = x3 . We solved Example 2, by the method of Example 1. The maximum absolute errors are given in Table 2. The numerical results in Table 2, establishes the facts given in Example 1. Figure 2 shows the comparison of numerical and exact solutions of Example 2 for J = 2. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 8.6E-019 16 1.1E-016 32 1.1E-016 64 1.1E-016 128 3.3E-016 256 3.3E-016 512 3.3E-016 1024 6.6E-016
Table 2: Maximum absolute error for Example 2. Example 3: Consider the following nonlinear Volterra integral equation Z x cos(x − t)(y 00 (t))dt = 6(1 − cos(x)), x∈[0, 1], 0
11
(53)
0.9
Haar wavelet solution Exact solution
0.8
0.7
Numerical solutions (y(x))
0.6
0.5
0.4
0.3
0.2
0.1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 3: Comparison of numerical and exact solutions of Example 3 for J=2.
with initial conditions y(0) = 0 and y 0 (0) = 0. Exact solution is y(x) = x3 . Solving Example 3, by the same method of Example 1, we see that maximum absolute errors are much higher as compared to Example 1 and Example 2. The result here is poor, because the second derivative of x3 is 6x, which is not a constant. The maximum absolute errors are given in Table 3. Figure 3 shows the comparison of numerical and exact solutions of Example 3 for J = 2. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 8.4E-003 16 2.1E-003 32 5.5E-004 64 1.4E-004 128 3.5E-005 256 8.8E-006 512 2.2E-006 1024 5.5E-007
Table 3: Maximum absolute error for Example 3. Example 4: Consider the following linear Volterra integral equation Z x e(x−t) (y 2 (t))dt = e2x − ex , x∈[0, 1]. 0
12
(54)
whose exact solution is y(x) = ex . First we convert the nonlinear Volterra integral equation of first kind into second kind bypsubstituting y 2 (t) = w(t). The solution of (54) is obtained from the relation y(x) = (w(x)). The maximum absolute errors for different values of J are shown in the Table 4. J
2M
Maximum absolute error(MAE) 8 2.8E-003 16 7.3E-004 32 1.8E-004 64 4.6E-005 128 1.1E-005 256 2.9E-006 512 7.3E-007 1024 1.8E-007
2 3 4 5 6 7 8 9
Table 4: Maximum absolute error for Example 4. Table 5, shows the comparison of absolute errors of proposed method and the method presented in [16] of Example 4. The absolute errors of proposed method is low as compared to the absolute errors in [16]. This shows that proposed method is better than method given in [16]. Figure 4 shows the comparison of numerical and exact solutions of Example 4 for J = 2. Collocation Absolute errors in Absolute errors points proposed method in [16] 1/16 0.00195 0.0328 3/16 0.00198 0.0374 5/16 0.00204 0.0425 7/16 0.00214 0.0485 9/16 0.00226 0.0552 11/16 0.00243 0.0628 13/16 0.00263 0.0716 15/16 0.00287 0.0815 Table 5: Comparison of absolute errors for Example 4, at different collocation points. Example 5: Consider the following Volterra integral equation Z x ex−t y(t)dt = x, x∈[0, 1]. 0
13
(55)
2.6
Haar wavelet solution 2.4
Exact solution
Numerical solutions (y(x))
2.2
2
1.8
1.6
1.4
1.2
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 4: Comparison of numerical and exact solutions of Example 4 for J=2.
The exact solution of this problem is y(x) = 1 − x. Table 6, shows the maximum absolute errors of Example 5 for different values of J. Again, in Table 7, comparison of absolute errors of proposed method and the method presented in [16] of Example 5 at different collocation points is given. This again establishes the fact that the absolute errors of the proposed method is very low as compared to the absolute errors in [16]. Figure 5 shows the comparison of numerical and exact solutions of Example 5 for J = 2. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 1.6E-003 16 4.5E-004 32 1.1E-004 64 2.9E-005 128 7.5E-006 256 1.8E-006 512 4.7E-007 1024 1.1E-007
Table 6: Maximum absolute error for Example 5. Example 6: Consider the following nonlinear Volterra integral equation Z x e(x−t) ln(y(t))dt = ex − x − 1, x∈[0, 1], 0
14
(56)
Collocation Absolute errors in Absolute errors points proposed method in [16] 1/16 0.0016 0.0313 3/16 0.0012 0.0313 5/16 0.0009 0.0313 7/16 0.0006 0.0313 9/16 0.0004 0.0313 11/16 0.0003 0.0313 13/16 0.0002 0.0313 15/16 0.0001 0.0313 Table 7: Comparison of absolute errors for Example 5, at different collocation points.
1
Haar wavelet solution Exact solution
0.9
Numerical solutions (y(x))
0.8
0.7
0.6
0.5
0.4
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 5: Comparison of numerical and exact solutions of Example 5 for J=2.
15
2.6
Haar wavelet solution Exact solution
2.4
Numerical solutions (y(x))
2.2
2
1.8
1.6
1.4
1.2
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 6: Comparison of numerical and exact solutions of Example 6 for J=2.
whose exact solution is y(x) = ex . First, we convert the nonlinear Volterra integral equation of first kind into second kind by substituting ln(u(t)) = w(t). The solution of (56) is obtained from the relation y(x) = ew(x) . The maximum absolute errors are shown in Table 8, for different values of J. Comparison of numerical and exact solutions of Example 6 for J = 2 are shown in Figure 6.
J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 3.4E-003 16 8.7E-004 32 2.1E-004 64 5.5E-005 128 1.3E-005 256 3.4E-006 512 8.6E-007 1024 2.1E-007
Table 8: Maximum absolute error for Example 6. Example 7: Consider the following nonlinear Volterra integral equation Z x xsinx (sin(x − t) + 1)cos(y(t))dt = + sinx, x∈[0, 1], 2 0 16
(57)
1
0.9
Haar wavelet solution Exact solution
0.8
Numerical solutions (y(x))
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Collocation points (x)
Figure 7: Comparison of numerical and exact solutions of Example 7 for J=2.
whose exact solution is y(x) = x. First we convert the nonlinear Volterra integral equation of first kind into second kind by substituting cos(y(t)) = w(t). The solution of (57) is obtained from the relation y(x) = cos−1 (w(x)). The maximum absolute errors for different values of J are shown in Table 9. Comparison of numerical and exact solutions of Example 7 for J = 2 are shown in Figure 7. J 2 3 4 5 6 7 8 9
2M
Maximum absolute error(MAE) 8 1.2E-003 16 3.1E-004 32 8.0E-005 64 2.0E-005 128 5.0E-006 256 1.2E-006 512 3.1E-007 1024 7.9E-008
Table 9: Maximum absolute error for Example 7.
Conclusion We conclude here from the above, that the Haar wavelet method is more accurate, simple and fast than other known methods to solve nonlinear Volterra integral equations. 17
The above examples demonstrate the simplicity of the Haar wavelet solution. The comparison of the present method with [15, 16] shows that the present method gives better results. For getting the necessary accuracy the number of calculation points may be increased. References
[1] P. Linz, Numerical methods for Volterra integral equations with singular kernels, SIAM J. Numer. Anal. 6 (1969) 365-374. [2] F. de Hoog, R. Weiss, Impilict Runge- Kutta methods for second kind Volterra integral equations, Numer. Math. 23 (1975) 199-213. [3] K.E. Atkinson, J. Flores, The collocation method for nonlinear integral equations, IMA Journal of Numerical Analysis 13 (1993) 195-213. [4] K.E. Atkinson, The numerical solution of an Abel integral equation by a product trapezoidal method, SIAM J. Numer. Anal. 11 (1974) 97-101. [5] Sheo Kumar, On a method of Noble for second kind Volterra integral equations, BIT 19 (1979) 482-488. [6] Sheo Kumar, On modified increment methods of Garey for nonlinear second kind Volterra integral equations, BIT 20 (1980) 346-350. [7] Sheo Kumar, A recurrence relation for solution of singular Volterra integral equations using Chebyshev polynomials, BIT 21 (1981) 123-125. [8] Sheo Kumar, Modifications of Linz methods for nonlinear second kind Volterra integral equations with singular or periodic kernels, Journal of Mathematical and Physical Sciences 26 (1992) 591-597. [9] H. Saeedi, N. Mollahasani, M. M. Moghadam, G. N. Chuev, An operational Haar wavelet method for solving fractional Volterra integral equations, Int. J. Appl. Math. Comput. Sci. 21(3) (2011) 535-547. [10] K.E. Atkinson, A survey of numerical methods for solving nonlinear integral equations, J. Integral Equation Appl. 4 (1992) 15-40. [11] E. Babolian, L.M. Delves, An augmented Galerkin method for first kind Fredholm equations, J. Inst. Math. Appl. 24 (1979) 157-174.
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[12] L.M. Delves, J.L. Mohamed, Computational methods for integral equations, Cambridge University Press, Cambridge, 1985. [13] P.K. Kythe, P. Puri, Computational methods for integral equations, Birkh¨auser, Boston, 2002. [14] A.N. Tikhonov, V.Y. Arsenin, Solutions of ill-posed problems, Winston, Washington DC, 1977. [15] E. Babolian, Z. Masouri, Direct method to solve Volterra integral equation of first kind using operational matrix with block-pulse functions, Journal of Computational and Applied Mathematics 220 (2008) 51-57. [16] E. Babolian, A. Salimi Shamloo, Numerical solution of Volterra integral and integro-differential equation of convolution type by using operational matrices of piecewise constant orthogonal functions, Journal of Computational and Applied Mathematics 214 (2008) 495-508. [17] J. Biazar, E. Babolian, R. Islam, Solution of system of Volterra integral equations of the first kind by Adomian method, Applied Mathematics and Computation 139 (2003) 249-248. [18] J. Biazar, M. Islami, H. Aminkhah, Application of homotopy perturbation method for system of Volterra integral equations of the first kind, Chaos, Solutions and Fractals 42 (2009) 3020-3026. ¨ Lepik, Solving differential and integral equations by Haar wavelet method, [19] U. revisted, International Journal of Mathematics and Computation 1(8) (2008) 4352. ¨ Lepik, Application of Haar wavelet transform for solving integral and differen[20] U. tial equations, Proc. Estonian Acad. Sci. Phys. Math. 56(1) (2007) 28-46. [21] C.F. Chen, C.H. Hsiao, Haar wavelet methods for solving lumped and distributedparameter systems, J. IEE. Proc. Control Theory Appl. 144 (1997) 87-94.
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HIGHLIGHTS
Solving some nonlinear first kind Volterra integral equations using Haar wavelet. Haar wavelet method is simple and fast. Solved examples demonstrate the accuracy and efficiency of the present method. Present method gives better results than the numerical methods described in past. Necessary accuracy may be obtained by increasing the number of calculation points.