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Modified Wavelet Full-Approximation Scheme for the Numerical Solution of Nonlinear Volterra integral and integro-differential Equations

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Introduction

Integral equations arise in various fields of science and engineering. Nowadays integral equations and their applications is an important subject in applied mathematics. In some cases, it is difficult to solve them, especially analytically. There are many analytical approaches were introduced, such as A domian decomposition method, successive substitutions, Laplace transformation method, Picard’s method, etc. [1]]. The analytical approaches have limited applicability either because the analytical solutions of some of the problems those arise in application do not exists or because they are more time consuming. Therefore, many integral equations arising in mathematical modelling of physical world problem demands efficient numerical techniques. A number of numerical techniques such as Galerkin method [2]], collocation method [2]], Nystrom interpolation [3]], etc. have been proposed by various authors. In numerical analysis, solving integral equations is reduced to solving a system of algebraic equations. Iterative techniques are available to solve a system of algebraic equations, such as Newton, method, Jacobi iterative method, Gauss-Seidel method, etc. In the past several years, considerable attention has been made for obtaining solutions of nonlinear integral equations. Nonlinearity is one of the interesting topics among the physicists, mathematicians, engineers, etc. Since most physical systems are inherently nonlinear in nature.

The full-approximation scheme (FAS) is largely applicable in increasing the efficiency of the iterative methods used to solve nonlinear system of algebraic equations. In the historical three decades the development of effective iterative solvers for nonlinear systems of algebraic equations has been a significant research topic in numerical analysis. Nowadays it is recognized that FAS iterative solvers are highly efficient for nonlinear differential equations introduced by Brandt [4]]. For a detailed treatment of FAS is given in Briggs et al. [5]]. An introduction of FAS is found in Hackbusch and Trottenberg [6]], Wesseling [7]] and Trottenberg et al. [8]]. Many authors applied the FAS to some class of differential equations; Lubrecht [9]], Venner and Lubrecht [10]], Zargari [11]] and others have significant contributions in EHL problems. Lee [12]] has introduced a multigrid method for solving the nonlinear Urysohn integral equations. In this paper, we introduce the full-approximation scheme (FAS) for the numerical solution of nonlinear Volterra integral and integro-differential equations.

Wavelet based numerical methods are used for solving the system of equations with faster convergence and lower computational cost. Since 1991, the various types of wavelet based methods have been applied for the numerical solution of different kinds of integral equations, a detailed survey on these papers can be found in [13]–15]. In recent years, wavelet analysis is gaining considerable attention in the numerical solution of differential equations. Recently, many authors (Leon [16]], Bujurke et al. [17]–19], Avudainayagam and Vani [32]]) have worked on wavelet multigrid method for the numerical solution of differential equations. Wang et al. [21]] have applied a fast wavelet multigrid algorithm for the solution of electromagnetic integral equations. In this paper, we introduce the modified wavelet full-approximation scheme (MWFAS) for the numerical solution of nonlinear Volterra integral and integro-differential equations. Thus, the proposed scheme can be applied to a wide range of science and engineering problems.

The organization of this paper is as follows. In Section 2, Daubechies wavelets is given. In Section 3 intergrid operators are discussed. In Section 4, method of solution is discussed. In Section 5, presents the numerical experiments and results are given. Finally, conclusion of the proposed work is given in Section 6.

Daubechies Wavelets

The refinement relation of scaling function ϕ(t) is given by,

ϕ(t)=2k=0L1hkϕ(2tk)$$\begin{array}{} \displaystyle \phi (t) = \sqrt 2 \mathop \Sigma \limits_{k = 0}^{L - 1} {h_k}\phi (2t - k) \end{array}$$

where ϕ(t) is normalized: ϕ(t)dt=1$\begin{array}{} \displaystyle \int_{-\infty}^{\infty} \phi(t) dt =1 \end{array}$.

Based on the scaling function ϕ(t) , the wavelet function can be written as, The refinement relation of wavelet function ψ(t) is given by,

ψ(t)=2k=0L1gkϕ(2tk)$$\begin{array}{} \displaystyle \psi(t)=\sqrt{2}\sum\limits_{k=0}^{L-1} g_{k} \phi(2t-k) \end{array}$$

where {hk}kεZ and {gk}kεZ are known respectively as the low and high pass filter coefficients, and they are related by gk = (−1)khL−k for k = 0, 1,...,L − 1. where, L is an even integer, is the order of the wavelet. The compactly supported wavelet bases were introduced by Daubechies [22]]. They are an orthonormal bases for functions in L2(R). A family of orthogonal Daubechies wavelets with compactly supported property has been constructed by Daubechies in [23]]. Due to excellent properties of orthogonality and minimum compact support, Daubechies wavelets can be useful and convenient in a wide variety of situations.

In this paper, we use Daubechies filter coefficients for L = 4 which are, h0 = 0.4830, h(1) = 0.8365, h(2) = 0.2241, h(3) = −0.1294 are low pass filter coefficients and g0 = 0.1294, g(1) = −0.2241, g(2) = 0.8365, g(3) = −0.4830 are the high pass filter coefficients.

Intergrid operators
Full-Approximation Scheme (FAS) operators

Brandt [4]] was one of the first to introduce nonlinear multigrid, which seeks to use concepts from the linear multigrid iteration and apply them directly in the nonlinear setting. Since the early application to elliptic partial differential equations, multigrid methods have been applied successfully to a large and growing class of problems. Classical multigrid begins with a two-grid process. First, iterative relaxation is applied, whose effect is to smooth the error. In this paper, we describe how to apply multigrid to nonlinear problems. Applying multigrid method directly to the nonlinear problems by employing the method so-called Full Approximation Scheme (FAS). Full approximation scheme suitable for nonlinear problems is the FAS [5],8] which treats directly the nonlinear equations on finer and coarser grids. In FAS, a nonlinear iteration, such as the nonlinear Gauss-Seidel method is applied to smooth the error. In FAS, the residual is passed from the fine grids to the coarser grids. Vectors from fine grids are transferred to coarser grids with Restriction operator (R), while vectors are transferred from coarse grids to the finer grids with a Prolongation operator (P) respectively given in Section 4.1. The detail survey on FAS is given in [6]].

Wavelet Full-Approximation Scheme (WFAS) operators

Discrete wavelet transform (DWT) matrix: The DWT matrix, which play an important part in the wavelet method. This is highly expedient and informative, particularly for the numerical computations. As we already know about the DWT matrix and its applications in the wavelet method and is given in [17]] as,

D1=(h0h1h2h30000g0g1g2g3000000h0h1h2h30000g0g1g2g30000h2h3000h0h1g2g3000g0g1)N×N$$\begin{array}{} \displaystyle D_{1} = \left(\begin{array}{ccccccccc} h_{0} & h_{1} & h_{2} & h_{3} & 0 & 0 & \ldots & 0 & 0\\ g_{0} & g_{1} & g_{2} & g_{3} & 0 & 0 & \ldots & 0 & 0\\ 0 & 0 & h_{0} & h_{1} & h_{2} & h_{3} & \ldots & 0 & 0\\ 0 & 0 & g_{0} & g_{1} & g_{2} & g_{3} & \ldots & 0 & 0\\ \vdots & & \ddots & & \cdots & & \cdots & 0 & 0\\ h_{2} & h_{3} & 0 & 0 & \cdots & \cdots & 0 & h_{0} & h_{1}\\ g_{2} & g_{3} & 0 & 0 & \cdots & \cdots & 0 & g_{0} & g_{1}\\ \end{array} \right)_{N \times N} \end{array}$$

Using this matrix author’s [24]] used the restriction operator and prolongation operators respectively given in Section 4.2.

Modified Wavelet Full-Approximation Scheme (MWFAS) operators

Modified Discrete wavelet transform (MDWT) matrix: Here, we defined modified DWT matrix from DWT matrix in which we have added rows and columns consecutively with diagonal element as 1, which is built as,

D2=(h00h10h20h3000001000000g00g10g20g30000000100000000g20g300g00g100001)N×N$$\begin{array}{} \displaystyle D_{2} = \left(\begin{array}{cccccccccccc} h_{0} & 0 & h_{1} & 0 & h_{2} & 0 & h_{3} & 0 & 0 & \ldots & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & \cdots & & & \cdots & 0 & 0\\ g_{0} & 0 & g_{1} & 0 & g_{2} & 0 & g_{3} & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & & \cdots & 0 & 0\\ \vdots & & \ddots & & \cdots & & \cdots & & & \cdots & 0 & 0\\ g_{2} & 0 & g_{3} & 0 & \cdots & & \cdots & 0 & g_{0} & 0 & g_{1} & 0\\ 0 & 0 & \cdots & & \cdots & & \cdots & & \cdots & & 0 & 1\\ \end{array} \right)_{N \times N} \end{array}$$

Using this matrix we defined the new restriction operator and new prolongation operators respectively given in Section 4.3.

Method of solution
Full-Approximation Scheme (FAS)

Consider the Nonlinear Volterra integral equation of the second kind,

u(t)=f(t)+0tK(t,s,u(s))ds,0t,s1,$$\begin{array}{} \displaystyle u(t)=f(t)+\int_{0}^{t} K(t,s,u(s)) ds, 0 \leq t, s \leq 1, \end{array}$$

where k(t, s, u(s)) is a nonlinear function defined on [0], 1] × [0], 1]. The known function k(t, s, u(s)) is called the kernel of the integral equation, while the unknown function u(t) represents the solution of the integral equation. After discretizing the integral equation through the trapezoidal discretization method (TDM) [25]], we get the system of nonlinear equations of the form,

A(u)=b,$$\begin{array}{} \displaystyle A\left( u \right) = b, \end{array}$$

It has N equations with N unknowns. Solving (4) through the iterative method that is Gauss Seidel (GS), we get approximate solution ν.

Now, we are deliberating about the Full-Approximation Scheme (FAS) of solutions given by Briggs et. al [5]] is as follows the procedure,

Step 1: From the system (4), we get the approximate solution ν for u. Now we find the residual as,

rN×1=bN×1A(ν)N×1$$\begin{array}{} \displaystyle r_{N \times 1} = b_{N \times 1} - A(\nu)_{N \times 1} \end{array}$$

Reduce the matrices in the finer level to coarsest level using Restriction operator, i.e.,

R=14(12100000012100000012)(N/2)×N$$\begin{array}{} \displaystyle R = \dfrac{1}{4} \left(\begin{array}{cccccccc} 1 & 2 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & 2 & 1 & \cdots & 0 & 0 \\ \vdots & & \ddots & & & \cdots & 0 & 0 \\ 0 & 0 & & \cdots & & \cdots & 1 & 2 \\ \end{array} \right)_{(N/2) \times N} \end{array}$$

and then construct the matrices back to finer level from the coarsest level using Prolongation operator, i.e.,

P=12(10002000110020100010002)(N)×(N/2)$$\begin{array}{} \displaystyle P = \dfrac{1}{2} \left(\begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ 2 & 0 & 0 & \cdots & 0 \\ 1 & 1 & 0 & \cdots & \vdots \\ 0 & 2 & \vdots & \cdots & \vdots \\ 0 & 1 & & & \\ \vdots & \vdots & \ddots & & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 2 \\ \end{array} \right)_{(N) \times (N/2)} \end{array}$$

Step 2:

r(N/2)×1=R(N/2)×NrN×1$$\begin{array}{} \displaystyle r_{(N/2) \times 1} = R_{(N/2) \times N} r_{N \times 1} \end{array}$$

Similarly,

ν(N/2)×1=R(N/2)×NνN×1$$\begin{array}{} \displaystyle \nu_{(N/2) \times 1}= R_{(N/2) \times N} \nu_{N \times 1} \end{array}$$

and

A(ν(N/2)×1+e(N/2)×1)A(ν(N/2)×1)=r(N/2)×1$$\begin{array}{} \displaystyle A(\nu_{(N/2) \times 1}+ e_{(N/2) \times 1})-A(\nu_{(N/2) \times 1})=r_{(N/2) \times 1} \end{array}$$

Solving (6) with initial guess 0, we get e(N/2)×1.

Step 3:

r(N/4)×1=R(N/4)×(N/2)r(N/2)×1$$\begin{array}{} \displaystyle r_{(N/4) \times 1} = R_{(N/4) \times (N/2)} r_{(N/2) \times 1} \end{array}$$

Similarly,

ν(N/4)×1=R(N/4)×(N/2)ν(N/2)×1$$\begin{array}{} \displaystyle \nu_{(N/4) \times 1}= R_{(N/4) \times (N/2)} \nu_{(N/2) \times 1} \end{array}$$

and

A(ν(N/4)×1+e(N/4)×1)A(ν(N/4)×1)=r(N/4)×1$$\begin{array}{} \displaystyle A(\nu_{(N/4) \times 1}+ e_{(N/4) \times 1})-A(\nu_{(N/4) \times 1})=r_{(N/4) \times 1} \end{array}$$

Solving (7) with initial guess 0, we get e(N/4)×1.

Step 4: The procedure is continue up to the coarsest level, we have,

r1×1=R1×2r2×1$$\begin{array}{} \displaystyle r_{1 \times 1} = R_{1 \times 2} r_{2 \times 1} \end{array}$$

Similarly,

ν1×1=R1×2ν2×1$$\begin{array}{} \displaystyle \nu_{1 \times 1}= R_{1 \times 2} \nu_{2 \times 1} \end{array}$$

and

A(ν1×1+e1×1)A(ν1×1)=r1×1$$\begin{array}{} \displaystyle A(\nu_{1 \times 1}+ e_{1 \times 1})-A(\nu_{1 \times 1})=r_{1 \times 1} \end{array}$$

Solving (8), we get e1×1.

Step 5: Interpolate error upto the finer level, i.e.,

e2×1=P2×1e1×1,e4×1=P4×2e2×1,$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{e_{2 \times 1}}} \hfill & = \hfill & {{P_{2 \times 1}}{e_{1 \times 1}},} \hfill \\ {{e_{4 \times 1}}} \hfill & = \hfill & {{P_{4 \times 2}}{e_{2 \times 1}},} \hfill \\ \end{array} \end{array}$$

and so on we have,

eN×1=PN×(N/2)e(N/2)×1.$$\begin{array}{} \displaystyle e_{N \times 1} = P_{N \times (N/2)} e_{(N/2) \times 1}. \end{array}$$

Step 6: Correct the solution with error,

uN×1=νN×1+eN×1.$$\begin{array}{} \displaystyle u_{N \times 1} = \nu_{N \times 1}+e_{N \times 1}. \end{array}$$

This is the required solution of the given integral equation.

Wavelet Full-Approximation Scheme (WFAS)

In this paper, we applied WFAS for the numerical solution of nonlinear Volterra integral equations. The same procedure is applied as explained in the FAS in the above Section 4.1. But replacing W and WT as the restriction and prolongation operators in place of R and P, i.e.,

W=(h0h1h2h30000g0g1g2g3000000h0h1h2h30000g0g1g2g3000000g0g1g2g30)(N/2)×N$$\begin{array}{} \displaystyle W = \left(\begin{array}{ccccccccc} h_{0} & h_{1} & h_{2} & h_{3} & 0 & 0 & \cdots & 0 & 0\\ g_{0} & g_{1} & g_{2} & g_{3} & 0 & 0 & \ldots & 0 & 0\\ 0 & 0 & h_{0} & h_{1} & h_{2} & h_{3} & \cdots & 0 & 0\\ 0 & 0 & g_{0} & g_{1} & g_{2} & g_{3} & \cdots & 0 & 0\\ \vdots & & \ddots & & \cdots & & \cdots & 0 & 0\\ 0 & 0 &\cdots & g_{0} & g_{1} & g_{2} & g_{3} & \cdots & 0\\ \end{array} \right)_{(N/2) \times N} \end{array}$$

and WT respectively.

Modified Wavelet Full-Approximation Scheme (MWFAS)

In this paper, we introduced a modified wavelet full-approximation scheme (MWFAS) for the numerical solution of nonlinear Volterra integral equations. The same procedure is applied as explained in the FAS in Section 4.1, using MW and MWT as the new restriction and new prolongation operators in place of R and P, i.e.,

MW=(h00h10h20h3000001000000g00g10g20g300000001000000000g00g10g20g300000001000)(N/2)×N$$\begin{array}{} \displaystyle MW = \left(\begin{array}{cccccccccccc} h_{0} & 0 & h_{1} & 0 & h_{2} & 0 & h_{3} & 0 & 0 & \ldots & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & \cdots & & & \cdots & 0 & 0\\ g_{0} & 0 & g_{1} & 0 & g_{2} & 0 & g_{3} & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & & \cdots & 0 & 0\\ \vdots & & \ddots & & \cdots & & \cdots & & & \cdots & 0 & 0\\ 0 & \cdots & g_{0} & 0 & g_{1} & 0 & g_{2} & 0 & g_{3} & 0 & \cdots & 0\\ 0 & 0 & \cdots & 0 & 0 & 0 & 1 & 0 & \cdots & \cdots & 0 & 0\\ \end{array} \right)_{(N/2) \times N} \end{array}$$

and MWT respectively.

Illustrative problems

In this section, we implemented FAS, WFAS and MWFAS for the numerical solution of nonlinear Volterra integral and integro-differential equations and subsequently presented the efficiency of the MWFAS in the form of tables and figures, here error analysis is considered as,

Emax=max|ueua|,$$\begin{array}{} \displaystyle E_{max}=max \vert u_{e} - u_{a} \vert, \end{array}$$

where ue and ua are the exact and approximate solution respectively.

Test problem 5.1.

Let us consider the nonlinear Volterra integral equation [28]],

u(t)=tt2t54+2t65+t76+0ttsu2(s)ds,0t1$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {u(t) = t - {t^2} - \frac{{{t^5}}}{4} + \frac{{2{t^6}}}{5} + \frac{{{t^7}}}{6} + \int_0^t t s{u^2}(s)ds,} & {0 \le t \le 1} \\ \end{array} \end{array}$$

which has the exact solution u(t) = tt2. After discretizing the (9) through the trapezoidal discretization method (TDM), we get system of nonlinear algebraic equations of the form (for N = 8),

[A]8×8[u]8×1=[b]8×1$$\begin{array}{} \displaystyle [A]_{8 \times 8}[u]_{8 \times 1} = [b]_{8 \times 1} \end{array}$$

Solving (10) through the iterative method, we get the approximate solution ν of u. i.e., u = e + νν = ue, where e is (8 × 1 matrix) error to be determined. The implementation of MWFAS is discussed in Section 4 is as follows,

From (10), we find the residual as

r8×1=[b]8×1[A]8×8[ν]8×1$$\begin{array}{} \displaystyle r_{8 \times 1}=[b]_{8 \times 1}-[A]_{8 \times 8}[\nu]_{8 \times 1} \end{array}$$

we get, = r8×1[0] 0 2.30e-06 2.01e-05 6.32e-05 8.89e-05 3.98e-05 -2.03e-07] we reduce the matrices in the finer level to coarsest level using Restriction operator MW and then construct the matrices back to finer level from the coarsest level using Prolongation operator MWT .

From (11),

r4×1=[MW]4×8[r]8×1$$\begin{array}{} \displaystyle r_{4 \times 1}=[MW]_{4 \times 8}[r]_{8 \times 1} \end{array}$$

Similarly,

ν4×1=[MW]4×8[ν]8×1$$\begin{array}{} \displaystyle \nu_{4 \times 1}=[MW]_{4 \times 8}[\nu]_{8 \times 1} \end{array}$$

and

A(ν4×1+e4×1)A(ν4×1)=r4×1.$$\begin{array}{} \displaystyle A(\nu_{4 \times 1}+ e_{4 \times 1})-A(\nu_{4 \times 1})=r_{4 \times 1}. \end{array}$$

Solving (13) with initial guess 0, we get e4×1.

From (12),

r2×1=[MW]2×4[r]4×1$$\begin{array}{} \displaystyle r_{2 \times 1}=[MW]_{2 \times 4}[r]_{4 \times 1} \end{array}$$

Similarly,

ν2×1=[MW]2×4[ν]4×1$$\begin{array}{} \displaystyle \nu_{2 \times 1}=[MW]_{2 \times 4}[\nu]_{4 \times 1} \end{array}$$

and

A(ν2×1+e2×1)A(ν2×1)=r2×1.$$\begin{array}{} \displaystyle A(\nu_{2 \times 1}+ e_{2 \times 1})-A(\nu_{2 \times 1})=r_{2 \times 1}. \end{array}$$

Solving (15) with initial guess 0, we get e2×1.

From (14),

r1×1=[MW]1×2[r]2×1$$\begin{array}{} \displaystyle r_{1 \times 1}=[MW]_{1 \times 2}[r]_{2 \times 1} \end{array}$$

Similarly,

ν1×1=[MW]1×2[ν]2×1$$\begin{array}{} \displaystyle \nu_{1 \times 1}=[MW]_{1 \times 2}[\nu]_{2 \times 1} \end{array}$$

and

A(ν1×1+e1×1)A(ν1×1)=r1×1.$$\begin{array}{} \displaystyle A(\nu_{1 \times 1}+ e_{1 \times 1})-A(\nu_{1 \times 1})=r_{1 \times 1}. \end{array}$$

Solving (17), we get e1×1.

From e1×1 , Interpolate error up to the finer level, i.e.,

e2×1=[MWT]2×1[e]1×1,e4×1=[MWT]4×2[e]2×1,$$\begin{array}{} \displaystyle \begin{array}{*{20}{l}} {{e_{2 \times 1}}}& = &{{{[M{W^T}]}_{2 \times 1}}{{[e]}_{1 \times 1}},}\\ {{e_{4 \times 1}}}& = &{{{[M{W^T}]}_{4 \times 2}}{{[e]}_{2 \times 1}},} \end{array} \end{array}$$

and lastly we have,

e8×1=[MWT]8×4[e]4×1,$$\begin{array}{} \displaystyle e_{8 \times 1} = [MW^T]_{8 \times 4}[e]_{4 \times 1}, \end{array}$$

we get e8×1 = [1].31e − 06 0 2.28e − 06 2.06e − 05 3.04e − 05 0 − 1.75e − 05 0]

From (18), correct the solution with error u8×1 = ν8×1 + e8×1.

Lastly, we get u8×1 is the required solution of (9). The numerical solutions of the given equation is obtained through the method as explained in Section 4 and are presented in comparison with the exact solution are shown in Table 1 and in Figure 1 for N = 64. Maximum error analysis and CPU time is shown in Table 2.

Numerical results of test problem 5.1, for N = 8.

tExactFASWFASMWFAS
00000
0.14280.12240.12240.12240.1224
0.28570.20400.20430.20430.2043
0.42850.24480.24570.24570.2457
0.57140.24480.24630.24630.2463
0.71420.20400.20550.20550.2055
0.85710.12240.12310.12310.1231
100.00010.00010.0001

Fig. 1

Comparison of numerical solutions with exact solution of test problem 5.1, for N=64.

Maximum error and CPU time (in seconds) of the methods of test problem 5.1.

NMethodsEmaxSetup timeRunning timeTotal time
16 FASWFASMWFAS7:52e − 047:35e − 047:35e − 040.00540.00310.00190.03210.02640.02540.03750.02950.0273
32FASWFASMWFAS3:66e − 043:62e − 043:62e − 040.00990.00310.00190.14590.14790.14700.15580.15100.1489
64FASWFASMWFAS1:80e − 041:79e − 041:79e − 040.12610.00310.00190.15810.03690.03571.28420.04000.0376
128FASWFASMWFAS8:99e − 058:96e − 058:96e − 050.13070.00850.00190.21850.14190.11620.34920.15040.1182

Test problem 5.2

Next, consider [29]]

u(t)=sin(t)+18sin(2t)14t+120tu2(s)ds,0t1$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {u(t) = sin(t) + \frac{1}{8}sin(2t) - \frac{1}{4}t + \frac{1}{2}\int_0^t {{u^2}} (s)ds,}&{0 \le t \le 1} \end{array} \end{array}$$

which has the exact solution u(t) = sin(t). The numerical solutions of (19) are presented in Table 3 for N = 16 and in Figure 2 for N = 64. Maximum error analysis and CPU time is shown in Table 4.

Numerical results of test problem 5.2, for N = 16.

tExactFASWFASMWFAS
0 0 0 0 0
0.0666 0.0666 0.0667 0.0667 0.0667
0.1333 0.1329 0.1332 0.1332 0.1332
0.2000 0.1986 0.1994 0.1994 0.1994
0.2666 0.2635 0.2648 0.2648 0.2648
0.3333 0.3271 0.3291 0.3291 0.3291
0.4000 0.3894 0.3922 0.3922 0.3922
0.4666 0.4499 0.4537 0.4537 0.4537
0.5333 0.5084 0.5133 0.5130 0.5130
0.6000 0.5646 0.5708 0.5703 0.5703
0.6666 0.6183 0.6259 0.6253 0.6253
0.7333 0.6693 0.6784 0.6776 0.6776
0.8000 0.7173 0.7280 0.7270 0.7270
0.8666 0.7621 0.7746 0.7734 0.7734
0.9333 0.8036 0.8174 0.8164 0.8164
1 0.8414 0.8449 0.8441 0.8441

Fig. 2

Comparison of numerical solutions with exact solution of test problem 5.2, for N=64.

Maximum error and CPU time (in seconds) of the methods of test problem 5.2.

NMethodsEmaxSetup timeRunning timeTotal time
16FASWFASMWFAS1:38e − 021:28e − 021:28e − 020.06240.00480.00190.06360.04430.02560.12600.04910.0275
32FASWFASMWFAS7:08e − 036:80e − 036:80e − 030.00990.00310.00190.14570.14690.14750.15560.14990.1494
64FASWFASMWFAS3:56e − 033:49e − 033:49e − 030.24630.00310.00191.17611.13851.14561.42251.14161.1475
128FASWFASMWFAS1:79e − 031:77e − 031:77e − 030.18190.01250.001910.25919.53019.322010.44109.54269.3239

Test problem 5.3

Next, consider the Nonlinear Volterra-Hammerstein integral equations [30]],

u(t)=1+sin2(t)0t3sin(ts)u2(s)ds,0t1$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {u(t) = 1 + si{n^2}(t) - \int_0^t 3 sin(t - s){u^2}(s)ds,}&{0 \le t \le 1} \end{array} \end{array}$$

which has the exact solution u(t) = cos(t). The numerical solutions of (20) are presented in Table 5 for N = 16 and in Figure 3 for N = 64. Maximum error analysis and CPU time is shown in Table 6.

Numerical results of test problem 5.3, for N = 16.

tExactFASWFASMWFAS
0 1 1.0357 1.0357 1.0357
0.0666 0.9977 0.9972 0.9972 0.9972
0.1333 0.9911 0.9901 0.9901 0.9901
0.2000 0.9800 0.9786 0.9786 0.9786
0.2666 0.9646 0.9628 0.9628 0.9628
0.3333 0.9449 0.9427 0.9427 0.9427
0.4000 0.9210 0.9185 0.9185 0.9185
0.4666 0.8930 0.8903 0.8903 0.8903
0.5333 0.8611 0.8582 0.8582 0.8582
0.6000 0.8253 0.8223 0.8223 0.8223
0.6666 0.7858 0.7829 0.7829 0.7829
0.7333 0.7429 0.7400 0.7400 0.7400
0.8000 0.6967 0.6939 0.6939 0.6939
0.8666 0.6473 0.6448 0.6448 0.6448
0.9333 0.5951 0.5929 0.5929 0.5929
1 0.5403 0.5383 0.5383 0.5383

Fig. 3

Comparison of numerical solutions with exact solution of test problem 5.3, for N=64.

Maximum error and CPU time (in seconds) of the methods of test problem 5.3.

NMethodsEmaxSetup timeRunning timeTotal time
16FASWFASMWFAS3:57e − 023:57e − 023:57e − 020.00540.00310.00190.01210.01210.01080.01750.01520.0127
32FASWFASMWFAS1:66e − 021:66e − 021:66e − 020.00990.00310.00190.01720.01640.01560.02710.01950.0175
64FASWFASMWFAS8:06e − 038:06e − 038:06e − 030.02810.00310.00190.03700.03600.03570.06510.03900.0375
128FASWFASMWFAS3:96e − 033:96e − 033:96e − 030.10090.00440.00200.11680.13290.13000.21770.13730.1320

Test problem 5.4

Next, consider [31]]

u(t)=1556t8+1314t71110t6+920t5+t2t+0t(t+s)u3(s)ds,0t1$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {u(t) = - \frac{{15}}{{56}}{t^8} + \frac{{13}}{{14}}{t^7} - \frac{{11}}{{10}}{t^6} + \frac{9}{{20}}{t^5} + {t^2} - t + \int_0^t {(t + s)} {u^3}(s)ds,}&{0 \le t \le 1} \end{array} \end{array}$$

which has the exact solution u(t) = t2t. The numerical solutions of (21) are presented in Table 7 for N = 16 and in Figure 4 for N = 64. Maximum error analysis and CPU time is shown in Table 8.

Numerical results of test problem 5.4, for N = 16.

tExactFASWFASMWFAS
0 0 0 0 0
0.0666 -0.0622 -0.0622 -0.0622 -0.0622
0.1333 -0.1155 -0.1155 -0.1155 -0.1155
0.2000 -0.1600 -0.1600 -0.1600 -0.1600
0.2666 -0.1955 -0.1957 -0.1957 -0.1957
0.3333 -0.2222 -0.2224 -0.2224 -0.2224
0.4000 -0.2400 -0.2403 -0.2403 -0.2403
0.4666 -0.2488 -0.2493 -0.2493 -0.2493
0.5333 -0.2488 -0.2494 -0.2494 -0.2494
0.6000 -0.2400 -0.2405 -0.2405 -0.2405
0.6666 -0.2222 -0.2227 -0.2227 -0.2227
0.7333 -0.1955 -0.1959 -0.1959 -0.1959
0.8000 -0.1600 -0.1602 -0.1602 -0.1602
0.8666 -0.1155 -0.1156 -0.1156 -0.1156
0.9333 -0.0622 -0.0622 -0.0622 -0.0622
1 0 0 0 0

Fig. 4

Comparison of numerical solutions with exact solution of test problem 5.4, for N=64.

Maximum error and CPU time (in seconds) of the methods of test problem 5.4.

NMethodsEmaxSetup timeRunning timeTotal time
16FASWFASMWFAS5:72e − 045:60e − 045:60e − 040.00540.00300.00180.04040.03340.03150.04580.03650.0333
32FASWFASMWFAS2:81e − 042:77e − 042:76e − 040.01050.00300.00180.20840.20560.20570.21890.20860.2075
64FASWFASMWFAS1:38e − 041:37e − 041:37e − 040.03060.00300.00180.04840.04710.04560.07900.05020.0474
128FASWFASMWFAS6:87e − 056:85e − 056:85e − 050.11140.00440.00180.18100.17800.18040.29240.18230.1822

Test problem 5.5

Lastly, consider the Nonlinear Volterra integro-differential equation [32]],

u(t)=1+0tu2(s)ds,u(0)=0,0t1$$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {u'(t) = - 1 + \int_0^t {{u^2}} (s)ds,u(0) = 0,}&{0 \le t \le 1} \end{array} \end{array}$$

which has the exact solution u(t)=t+t412t7252+t106048t13157248$\begin{array}{} \displaystyle u(t)=-t+\dfrac{t^4}{12}-\dfrac{t^7}{252}+\dfrac{t^10}{6048}-\dfrac{t^13}{157248} \end{array}$.

We convert the Volterra integro-differential equation to equivalent Volterra integral equation by using the well-known formula, which converts multiple integrals into a single integral. i.e.,

0t0t...0tu(t)dtnntimes=1(n1)!0t(ts)n1u(s)ds$$\begin{array}{} \displaystyle \underbrace{\int_{0}^{t} \int_{0}^{t} . . . \int_{0}^{t} u(t) dt^n}_{n-times} = \dfrac{1}{(n-1)!} \int_{0}^{t} (t-s)^{n-1} u(s) ds \end{array}$$

Integrating (22) on both sides from 0 to t and using the initial condition and also converting the double integral to the single integral, we obtain

u(t)=f(t)+0tk(t,s)u2(s)ds,$$\begin{array}{} \displaystyle u(t)=f(t)+\int_{0}^{t} k(t,s) u^2(s) ds, \end{array}$$

where k(t, s) = (ts) and f(t) = −t. The numerical solutions of (22) are presented in Table 9 for N = 32 and in Figure 5 for N = 64. Maximum error analysis and CPU time is shown in Table 10.

Numerical results of test problem 5.5, for N = 32.

tExactFASWFASMWFAS
0 0 0 0 0
0.0322 -0.0322 -0.0322 -0.0322 -0.0322
0.0645 -0.0645 -0.0645 -0.0645 -0.0645
0.0967 -0.0967 -0.0967 -0.0967 -0.0967
0.1290 -0.1290 -0.1290 -0.1290 -0.1290
0.1612 -0.1612 -0.1612 -0.1612 -0.1612
0.1935 -0.1934 -0.1934 -0.1934 -0.1934
0.2258 -0.2255 -0.2255 -0.2255 -0.2255
0.2580 -0.2577 -0.2577 -0.2577 -0.2577
0.2903 -0.2897 -0.2897 -0.2897 -0.2897
0.3225 -0.3216 -0.3216 -0.3216 -0.3216
0.3548 -0.3535 -0.3535 -0.3535 -0.3535
0.3870 -0.3852 -0.3852 -0.3852 -0.3852
0.4193 -0.4167 -0.4168 -0.4168 -0.4168
0.4516 -0.4481 -0.4481 -0.4481 -0.4481
0.4838 -0.4793 -0.4793 -0.4793 -0.4793
0.5161 -0.5102 -0.5102 -0.5102 -0.5102
0.5483 -0.5409 -0.5409 -0.5409 -0.5409
0.5806 -0.5712 -0.5712 -0.5712 -0.5712
0.6129 -0.6012 -0.6013 -0.6013 -0.6013
0.6451 -0.6309 -0.6309 -0.6309 -0.6309
0.6774 -0.6601 -0.6601 -0.6601 -0.6601
0.7096 -0.6888 -0.6889 -0.6889 -0.6889
0.7419 -0.7171 -0.7172 -0.7172 -0.7172
0.7741 -0.7449 -0.7449 -0.7449 -0.7449
0.8064 -0.7720 -0.7721 -0.7721 -0.7721
0.8387 -0.7986 -0.7986 -0.7986 -0.7986
0.8709 -0.8244 -0.8245 -0.8245 -0.8245
0.9032 -0.8496 -0.8497 -0.8497 -0.8497
0.9354 -0.8740 -0.8741 -0.8741 -0.8741
0.9677 -0.8976 -0.8977 -0.8977 -0.8977
1 -0.9204 -0.9205 -0.9205 -0.9205

Fig. 5

Comparison of numerical solutions with exact solution of test problem 5.5, for N=64.

Maximum error and CPU time (in seconds) of the methods of test problem 5.5.

NMethodsEmaxSetup timeRunning timeTotal time
16FASWFASMWFAS2:81e − 042:81e − 042:81e − 040.00540.00390.00250.01250.01130.01080.01800.01520.0133
32FASWFASMWFAS6:56e − 056:56e − 056:56e − 050.01000.00300.00190.01690.01610.01580.02680.01900.0177
64FASWFASMWFAS1:57e − 051:57e − 051:57e − 050.02860.00300.00190.03680.03580.03530.06540.03880.0372
128FASWFASMWFAS3:69e − 063:69e − 063:69e − 060.10420.00330.00190.12400.11560.11670.22810.11890.1186

Conclusion

In this paper, we proposed a modified wavelet full-approximation scheme for the numerical solution of nonlinear Volterra integral and integro-differential equations. The modified wavelet intergrid operators of prolongation and restrictions are used in this paper, a modified wavelet based FAS, has been shown to be effective and versatile. The numerical solutions obtained agree well with the exact ones. Convergence is also observed in the numerical solutions when the calculation is refined by increasing the number N used. The standard FAS and WFAS converge slowly with larger computational cost, whereas MWFAS does ensure such slower convergence with lesser computational cost. Test problems are justified through the error analysis, as the level of resolution N increases, larger the accuracy increases. Hence, the new scheme is very convenient and efficient than the existing standard methods.

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