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Theoretical Error Analysis of Solution for Two-Dimensional Stochastic Volterra Integral Equations by Haar Wavelet

  • M. Fallahpour
  • M. KhodabinEmail author
  • K. Maleknejad
Open Access
Original Paper
  • 156 Downloads

Abstract

The finding an efficient way to the approximate solutions of the stochastic integral equations is an essential requirement. In this paper we discuss the convergence analysis of the two-dimensional Haar wavelet functions (2D-HWFs) method for solve 2D linear stochastic Volterra integral equation. The illustrative examples are included to demonstrate the validity and applicability of this numerical method.

Keywords

Haar wavelet function Two-dimensional integral equation Stochastic integral equation Volterra integral equation Brownian motion process Ito integral 

Mathematics Subject Classification

65C20 60H20 45Axx 45Dxx 65T60 

Introduction

In very of the engineering problems, we see the important role of 2D integral equations [1, 2], where are produced from a differential equation. Also if we import the statistical noise into a general hyperbolic differential equation we can obtain 2D linear stochastic Volterra integral equation of the second kind, i.e.
$$\begin{aligned} g(x,y)= & {} f(x,y)+\int _{0}^{ y}\int _{0}^{x}K_{1}(x,y,s,t) g(s,t)dsdt \nonumber \\&+\int _{0}^{y}\int _{0}^{x}K_{2}(x,y,s,t) g(s,t) dB(s)dB(t)\nonumber \\&\quad (x,y)\in [0,1]\times [0,1], \ \ \ {s}\leqslant {x<t}\leqslant {y.} \end{aligned}$$
(1)
where B(t) is Brownian motion process. In the last years, the numerous numerical methods have been introduced to estimate the solution of 2D ordinary integral equations [3, 4, 5, 6, 7, 8] and many other papers, whereas it is worked few for 2D stochastic integral equations. Recently, some authors have proposed the methods to solve 2D stochastic integral equations [9, 10, 11, 12, 13, 14]. Especially, Fallahpour et al. [9], included HWFs method to solve such equations without the error analysis. In this paper, we use HWFs method with the error and convergence analysis to derive approximate solution of (1).

This paper is organized as follows:

In next section we present HWFs. In section “Haar Wavelets Numerical Method”, the method is applied to solve Eq. (1). The error analysis of this method is discussed in section “Error Analysis”. Section “Numerical Example”, uses some numerical examples to show the convergence of the proposed method and compares it with the block-pulse functions (BPFs) method as proposed in [10, 11]. Finally, conclusion is given, in section “Conclusion”.

Haar Wavelets

As we know, for HWFs we can write
$$\begin{aligned} \psi _j,_i\left( z\right) =2^{j/2}\psi \left( 2^{j}z-i\right) , \end{aligned}$$
so on the interval [0, 1) we have
$$\begin{aligned} h_{1}(z)=\left\{ \begin{array}{cc} 1, &{} for \ z\in [0,1)\\ 0, &{} otherwise,\end{array}\right. \end{aligned}$$
and
$$\begin{aligned} h_{i}(z)=\left\{ \begin{array}{cc} 1, &{} for \ z \in [\alpha ,\beta ) \\ -1, &{}for \ z \in [\beta ,\gamma )\\ 0, &{}\ otherwise, \end{array}\right. \end{aligned}$$
for \(i=2,3, \ldots ,\) where
$$\begin{aligned} \alpha =\frac{n}{m},\ \ \ \ \beta =\frac{(n+0.5)}{m},\ \ \ \ \gamma =\frac{(n+1)}{m}, \end{aligned}$$
so
$$\begin{aligned} m=2^{\ell },\ \ell =0, 1, \ldots ,\ n=0, 1, \ldots , m-1. \end{aligned}$$
The integer \( \ell \) and n, are the wavelet level and the translation parameter, respectively. Also Haar wavelets \(h_{i}(z)\) are pairwise orthonormal in the interval [0, 1) as
$$\begin{aligned} \int _{0}^{1}h_{i}(z)h_{j}(z)dy=\delta _{ij}, \end{aligned}$$
(2)
where \(\delta _{ij}\) is Kronecker delta. We display the maximum value of \( \ell \) by J where \( M=2^{J}\). Therefore for any square integrable function f(z) we have
$$\begin{aligned} f(z)\thickapprox \sum _{i=1}^{2M }a_{i} h_{i} (z). \end{aligned}$$
We show the ordinary Volterra ihtegral of Haar wavelet as
$$\begin{aligned} p_{i}(z)=\int _{0}^{z}h_{i}(u)du, \end{aligned}$$
where by using HWFs we get
$$\begin{aligned} p_{i}(z)=\left\{ \begin{array}{cl} z-\alpha , &{}\quad for\quad z \in [\alpha ,\beta ) \\ \gamma -z, &{}\quad for\quad z \in [\beta ,\gamma ) \\ 0, &{}\quad elsewhere. \end{array}\right. \end{aligned}$$
(3)
Also for the stochastic Volterra integral we can introduce
$$\begin{aligned} q_{i}(z)=\int _{0}^{z}h_{i}(u)dB(u), \end{aligned}$$
where similarly by Haar wavelet definition we have
$$\begin{aligned} q_{i}(z)=\left\{ \begin{array}{ll} B(z)-B(\alpha ), &{} for \, z \, \in [\alpha ,\beta )\ \ \\ 2B(\beta )-B(\alpha )-B(z), &{} for \,z\, \in [\beta ,\gamma )\ \ \\ 0,&{} elsewhere.\ \ \end{array}\right. \end{aligned}$$
(4)

Haar Wavelets Numerical Method

In this section, we solve Eq. (1) by using 2D-HWFs. For Haar wavelet approximation of a function f(xy) , we use the collocation points
$$\begin{aligned} x_{m}= & {} \frac{m-0.5}{2M}, m=1,2, \ldots ,2M, \end{aligned}$$
(5)
$$\begin{aligned} y_{n}= & {} \frac{n-0.5}{2N}, n=1,2, \ldots ,2N. \end{aligned}$$
(6)

2D-HWFs System

A real-valued function E(xy) can be approximated using 2D-HWFs as
$$\begin{aligned} E(x,y)\thickapprox \sum _{d=1}^{2M}\sum _{e=1}^{2N}b_{d,e} h_{d} (x)h_{e} (y), \end{aligned}$$
where the unknown coefficients \( b_{d,e} \)’s, have calculated in [9]. Also we consider a function E(xyst) where is approximated using 2D-HWFs as
$$\begin{aligned} E(x,y,s,t)\thickapprox \sum _{d=1}^{2M}\sum _{e=1}^{2N}b_{d,e}(x,y) h_{d}(s)h_{e}(t), \end{aligned}$$
so by substituting the collocation points
$$\begin{aligned} s_{i}=\frac{i-0.5}{2M},\ i=1,2, \ldots ,2M, \end{aligned}$$
and
$$\begin{aligned} t_{j}=\frac{j-0.5}{2N},\ j=1,2, \ldots ,2N, \end{aligned}$$
we get
$$\begin{aligned} E(x,y,s_{i},t_{j})\thickapprox \sum _{d=1}^{2M}\sum _{e=1}^{2N}b_{d,e}(x,y)h_{d} (s_{i})h_{e} (t_{j}), \end{aligned}$$
(7)
where the solution of System (7) is calculated from Corollary (1) in [9].

2D-Linear Stochastic Volterra Integral Equation

We approximate the functions \( K_{1}g \) and \(K_{2}g\) in (1) by using 2D-HWFs as
$$\begin{aligned} K_{1}(x,y,s,t)g(s,t)\thickapprox \sum _{i=1}^{2M}\sum _{j=1}^{2N}b_{i,j}(x,y) h_{i}(s)h_{j}(t), \end{aligned}$$
and
$$\begin{aligned} K_{2}(x,y,s,t)g(s,t)\thickapprox \sum _{i=1}^{2M}\sum _{j=1}^{2N}c_{i,j}(x,y) h_{i}(s)h_{j}(t), \end{aligned}$$
respectively. With these approximations, Eq. (1) can be writen as
$$\begin{aligned} g(x,y)= & {} f(x,y)+\int _{0}^{ y}\int _{0}^{x}\sum _{i=1}^{2M}\sum _{j=1}^{2N}b_{i,j}(x,y) h_{i}(s)h_{j}(t)dsdt \nonumber \\&+\int _{0}^{y}\int _{0}^{x}\sum _{i=1}^{2M}\sum _{j=1}^{2N}c_{i,j}(x,y) h_{i}(s)h_{j}(t) dB(s)dB(t). \end{aligned}$$
(8)
By applying Eqs. (3) and (4) in (8), we have
$$\begin{aligned} g(x,y)=f(x,y)+\sum _{i=1}^{2M}\sum _{j=1}^{2N}b_{i,j}(x,y)p_{i}(x)p_{j}(y)+\sum _{i=1}^{2M}\sum _{j=1}^{2N}c_{i,j}(x,y)q_{i}(x)q_{j}(y). \end{aligned}$$
The collocation points (5) and (6), give
$$\begin{aligned} g(x_{m},y_{n})= & {} f(x_{m},y_{n})+\sum _{i=1}^{2M}\sum _{j=1}^{2N}b_{i,j}(x_{m},y_{n})p_{i}(x_{m})p_{j}(y_{n})\\&+\sum _{i=1}^{2M}\sum _{j=1}^{2N}c_{i,j}(x_{m},y_{n})q_{i}(x_{m})q_{j}(y_{n}). \end{aligned}$$
Now if we replace \( b_{i,j} \)’s and \( c_{i,j} \)’s by the given expressions of Corollary 1 in [9], we get
$$\begin{aligned} g(x_{m},y_{n})= & {} f(x_{m},y_{n})+\frac{p_{1}(x_{m})p_{1}(y_{n})}{2M\times 2N}\sum _{d=1}^{2M}\sum _{e=1}^{2N}K_{1}(x_{m},y_{n},s_{d},t_{e}) g(s_{d},t_{e})\nonumber \\&+\sum _{i=2}^{2M}\frac{p_{i}(x_{m})p_{1}(y_{n})}{\rho _{1}\times 2N}\left( \sum _{d=\alpha _{1}}^{\beta _{1}}\sum _{e=1}^{2N}K_{1}(x_{m},y_{n},s_{d},t_{e})g(s_{d},t_{e}) \right. \nonumber \\&\left. \ -\sum _{d=\beta _{1}+1}^{\gamma _{1}}\sum _{e=1}^{2N}K_{1}(x_{m},y_{n},s_{d},t_{e}) g(s_{d},t_{e})\right) +\sum _{j=2}^{2N}\frac{p_{1}(x_{m})p_{j}(y_{n})}{2M\times \rho _{2}}\nonumber \\&\times \left( \sum _{d=1}^{2M}\sum _{e=\alpha _{2}}^{\beta _{2}}K_{1}(x_{m},y_{n},s_{d},t_{e}) g(s_{d},t_{e})\right. \nonumber \\&\left. -\sum _{d=1}^{2M}\sum _{e=\beta _{2}+1}^{\gamma _{2}}K_{1}(x_{m}, y_{n},s_{d},t_{e})g(s_{d},t_{e})\right) \nonumber \\&+\sum _{i=2}^{2M}\sum _{j=2}^{2N}\frac{p_{i}(x_{m})p_{j}(y_{n})}{\rho _{1} \times \rho _{2}}\left( \sum _{d=\alpha _{1}}^{\beta _{1}}\sum _{e=\alpha _{2}}^{\beta _{2}}K_{1}(x_{m},y_{n},s_{d},t_{e})g(s_{d},t_{e})\right. \nonumber \\&\left. -\sum _{d=\alpha _{1}}^{\beta _{1}}\sum _{e=\beta _{2}+1}^{\gamma _{2}}K_{1} (x_{m},y_{n},s_{d},t_{e}) g(s_{d},t_{e})-\sum _{d=\beta _{1}+1}^{\gamma _{1}}\sum _{e=\alpha _{2}}^{\beta _{2}}K_{1}(x_{m},y_{n},s_{d},t_{e})g(s_{d},t_{e})\right. \nonumber \\&\left. +\sum _{d=\beta _{1}+1}^{\gamma _{1}}\sum _{e=\beta _{2}+1}^{\gamma _{2}}K_{1} (x_{m},y_{n},s_{d},t_{e})g(s_{d},t_{e})\right) +\frac{q_{1}(x_{m})q_{1}(y_{n})}{2M\times 2N} \nonumber \\&\times \sum _{i=1}^{2M}\sum _{j=1}^{2N}K_{2}(x_{m},y_{n},s_{d},t_{e}) g(s_{d},t_{e})+\sum _{i=2}^{2M}\frac{q_{i}(x_{m})q_{1}(y_{n})}{\rho _{1}\times 2N}\nonumber \\&\times \left( \sum _{d=\alpha _{1}}^{\beta _{1}}\sum _{e=1}^{2N}K_{2}(x_{m},y_{n},s_{d},t_{e})g(s_{d},t_{e})- \sum _{d=\beta _{1}+1}^{\gamma _{1}}\sum _{e=1}^{2N}K_{2}(x_{m},y_{n}, s_{d},t_{e})g(s_{d},t_{e})\right) \nonumber \\&+\sum _{j=2}^{2N}\frac{q_{1}(x_{m})q_{j}(y_{n})}{2M\times \rho _{2}}\left( \sum _{d=1}^{2M}\sum _{e=\alpha _{2}}^{\beta _{2}}K_{2}(x_{m},y_{n},s_{d},t_{e})g(s_{d},t_{e}) \right. \nonumber \\&\left. - \sum _{d=1}^{2M}\sum _{e=\beta _{2}+1}^{\gamma _{2}}K_{2}(x_{m},y_{n},s_{d},t_{e})g(s_{d},t_{e})\right) +\sum _{i=2}^{2M}\sum _{j=2}^{2N}\frac{q_{i}(x_{m})q_{j}(y_{n})}{\rho _{1}\times \rho _{2}}\nonumber \\&\times \left( \sum _{d=\alpha _{1}}^{\beta _{1}}\sum _{e=\alpha _{2}}^{\beta _{2}}K_{2}(x_{m},y_{n},s_{d},t_{e})g(s_{d},t_{e})-\sum _{d=\alpha _{1}}^{\beta _{1}}\sum _{e=\beta _{2}+1}^{\gamma _{2}}K_{2}(x_{m},y_{n},s_{d},t_{e}) \right. \nonumber \\&\left. \times g(s_{d},t_{e})-\sum _{d=\beta _{1}+1}^{\gamma _{1}}\sum _{e=\alpha _{2}}^{\beta _{2}}K_{2}(x_{m},y_{n},s_{d},t_{d})g(s_{d},t_{e})\right. \nonumber \\&\left. +\sum _{d=\beta _{1}+1}^{\gamma _{1}} \sum _{e=\beta _{2}+1}^{\gamma _{2}}K_{2}(x_{m},y_{n},s_{d},t_{e})g(s_{d},t_{e})\right) , \end{aligned}$$
(9)
where \( \alpha _{1},\beta _{1},\gamma _{1} \), \( \rho _{1} \) and \( \alpha _{2},\beta _{2},\gamma _{2} \), \( \rho _{2} \) are defined in [9] and Eq. (9) shows a system \( 2M\times 2N \).

Error Analysis

In this section, we investigate the convergence and error analysis of HWFs method for solution (1). Here norm 2 is defined as
$$\begin{aligned} \Vert f \Vert _{2}=\left[ \underbrace{\int _{a}^{ b}\int _{a}^{ b}\ldots \int _{a}^{ b}}_{m\ times}\vert f(x_{1},x_{2},\ldots ,x_{m}) \vert ^{2}dx_{1}dx_{2}\ldots dx_{m}\right] ^{1/2}. \end{aligned}$$
To obtaining the typical convergence rate of HWFs method, we have:

Theorem 1

[15] For the differentiable function \(p:D\subset R^2\longrightarrow R\), with
$$\begin{aligned} \Vert p^{\prime } \Vert \le M, \end{aligned}$$
where D and M are a convex open set and a real number, respectively, we have
$$\begin{aligned} \vert p(b)-p(a) \vert \le M \vert b-a \vert \end{aligned}$$
for all \( a, b \in D \).

Theorem 2

[16] Suppose that \( f(s,t) \in {\textit{L}}^{2}([0,1)\times [0,1))\) with bounded partial derivatives, \( \vert \dfrac{\partial ^{2}f}{\partial s \partial t}\vert \le M, \) and
$$\begin{aligned} e_{m}(s,t)= f(s,t)-\sum _{i=0}^{m-1}\sum _{j=0}^{m-1}f_{i,j}h_{i}(s)h_{j}(t). \end{aligned}$$
Then for \( (s,t)\in D^{2} \), we have
$$\begin{aligned} \Vert e_{m} \Vert \le \dfrac{M}{3m^{2}}. \end{aligned}$$

Theorem 3

Consider \( K(x,y,s,t) \in {\textit{L}}^{2}([0,1)\times [0,1)\times [0,1)\times [0,1))\) with bounded partial derivatives,
$$\begin{aligned} \vert \dfrac{\partial ^{4}f}{\partial x \partial y\partial s \partial t}\vert \le U, \end{aligned}$$
and
$$\begin{aligned} {\hat{K}}_{m}(x,y,s,t)=\sum _{a=0}^{m-1}\sum _{b=0}^{m-1}\sum _{c=0}^{m-1} \sum _{d=0}^{m-1}K_{abcd}h_{a}(x)h_{b}(y)h_{c}(s)h_{d}(t), \end{aligned}$$
is 4D-HWFs expansion of K(xyst) where
$$\begin{aligned} e_{a,b,c,d}(x,y,s,t)=K(x,y,s,t)-{\hat{K}}_{m}(x,y,s,t), \end{aligned}$$
then
$$\begin{aligned} \Vert e_{a,b,c,d} \Vert \le \dfrac{U}{9m^{4}}, \end{aligned}$$
for every \( (x,y,s,t)\in D^{4} \).

Proof

We can write
$$\begin{aligned} \Vert e_{a,b,c,d} \Vert ^{2}= & {} \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\int _{0}^{1} \nonumber \\&\left( K(x,y,s,t) -\sum _{a=0}^{m-1}\sum _{b=0}^{m-1}\sum _{c=0}^{m-1}\sum _{d=0}^{m-1} K_{abcd}h_{a}(x)h_{b}(y)h_{c}(s)h_{d}(t)\right) ^{2} dt ds dy dx,\nonumber \\= & {} \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\int _{0}^{1} \left( \sum _{a=m}^{\infty }\sum _{b=m}^{\infty }\sum _{c=m}^{\infty }\sum _{d=m}^{\infty } K_{abcd}h_{a}(x)h_{b}(y)h_{c}(s)h_{d}(t)\right) ^{2} dt ds dy dx,\nonumber \\= & {} \sum _{a=m}^{\infty }\sum _{b=m}^{\infty }\sum _{c=m}^{\infty }\sum _{d=m}^{\infty } \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}K^{2}_{abcd}h^{2}_{a}(x)h^{2}_{b}(y)h^{2}_{c}(s)h^{2}_{d}(t) dt ds dy dx,\nonumber \\ \end{aligned}$$
(10)
where \( a=2^{j_{1}}+k \), \( b=2^{j_{2}}+k \), \( c=2^{j_{3}}+k \), \( d=2^{j_{4}}+k \), \( m=2^{J} \), \( J>0 \) and
$$\begin{aligned} K_{abcd}=\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}h_{a}(x)h_{b}(y)h_{c}(s)h_{d}(t)K(x,y,s,t)dt ds dy dx. \end{aligned}$$
Based on Haar wavelet definition, mean value theorem, Theorems 1 and 2 there are
$$\begin{aligned} \eta _{j_{4}}, \alpha , \alpha ^{\prime } \ \ , \ \ \eta _{j_{3}}, \beta , \beta ^{\prime } \ \ , \ \ \eta _{j_{2}}, \gamma , \gamma ^{\prime } \ \ , \ \ \eta _{j_{1}}, \theta , \theta ^{\prime }, \end{aligned}$$
where
$$\begin{aligned}&\alpha , \alpha ^{\prime }\in \left[ k2^{-j_{4}},\left( k+\dfrac{1}{2}\right) 2^{-j_{4}}\right] \ \ , \ \ \beta , \beta ^{\prime }\in \left[ k2^{-j_{3}},\left( k+\dfrac{1}{2}\right) 2^{-j_{3}}\right] ,\\&\gamma , \gamma ^{\prime }\in \left[ k2^{-j_{2}},\left( k+\dfrac{1}{2}\right) 2^{-j_{2}}\right] \ \ , \ \ \theta , \theta ^{\prime }\in \left[ k2^{-j_{1}},\left( k+\dfrac{1}{2}\right) 2^{-j_{1}}\right] , \end{aligned}$$
such that
$$\begin{aligned}&K_{abcd}\\&\quad =\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}h_{a}(x)h_{b}(y)h_{c}(s)\left( \int _{0}^{1}h_{d}(t)K(x,y,s,t)dt\right) ds dy dx\\&\quad =\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}h_{a}(x)h_{b}(y)h_{c}(s)\left( 2^{\dfrac{-j_{4}}{2}-1}(\alpha - \alpha ^{\prime })\dfrac{\partial K(x,y,s,\eta _{j_{4}})}{\partial t}\right) ds dy dx \\&\quad =\int _{0}^{1}\int _{0}^{1}h_{a}(x)h_{b}(y)\times 2^{\dfrac{-j_{4}}{2}-1}(\alpha - \alpha ^{\prime })\left( \int _{0}^{1}h_{c}(s) \dfrac{\partial K(x,y,s,\eta _{j_{4}})}{\partial t} ds\right) dy dx \\&\quad =\int _{0}^{1}\int _{0}^{1}2^{\dfrac{-j_{4}}{2}-\dfrac{-j_{3}}{2}-2}\times (\alpha - \alpha ^{\prime })(\beta - \beta ^{\prime })(\gamma - \gamma ^{\prime })\dfrac{\partial ^{2} K(x,y,\eta _{j_{3}},\eta _{j_{4}})}{\partial s\partial t}h_{a}(x)h_{b}(y) dy dx. \end{aligned}$$
Finally we conclude
$$\begin{aligned} K_{abcd}= & {} 2^{\dfrac{-j_{4}}{2}-\dfrac{-j_{3}}{2}-\dfrac{-j_{2}}{2}-\dfrac{-j_{1}}{2}-4} \nonumber \\&\times (\alpha - \alpha ^{\prime })(\beta - \beta ^{\prime })(\gamma - \gamma ^{\prime })(\theta - \theta ^{\prime })\dfrac{\partial ^{4} K(\eta _{j_{1}},\eta _{j_{2}},\eta _{j_{3}},\eta _{j_{4}})}{\partial x\partial y\partial s\partial t}. \end{aligned}$$
(11)
By using Eq. (2) and substituting Eq. (11) in (10) we get
$$\begin{aligned} \Vert e_{a,b,c,d}\Vert ^{2}= & {} \sum _{a=m}^{\infty }\sum _{b=m}^{\infty }\sum _{c=m}^{\infty }\sum _{d=m}^{\infty }2^{-j_{4}-j_{3}-j_{2}-j_{1}-8}\\&\times (\alpha -\alpha ^{\prime })^{2}(\beta - \beta ^{\prime })^{2}(\gamma - \gamma ^{\prime })^{2}(\theta - \theta ^{\prime })^{2}\left| \dfrac{\partial ^{4} K(\eta _{j_{1}},\eta _{j_{2}},\eta _{j_{3}},\eta _{j_{4}})}{\partial x\partial y\partial s\partial t}\right| ^{2}\\\le & {} \sum _{a=m}^{\infty }\sum _{b=m}^{\infty }\sum _{c=m}^{\infty } \sum _{d=m}^{\infty }2^{-j_{4}-j_{3}-j_{2}-j_{1}-8}\times 2^{-2j_{1}}\times 2^{-2j_{2}}\times 2^{-2j_{3}}\times 2^{-2j_{4}}\times U^{2}\\= & {} \sum _{a=m}^{\infty }\sum _{b=m}^{\infty }\sum _{c=m}^{\infty } \sum _{d=m}^{\infty }2^{-3j_{4}-3j_{3}-3j_{2}-3j_{1}-8}\times U^{2} \\= & {} U^{2} \sum _{a=m}^{\infty }\sum _{b=m}^{\infty }2^{-3j_{2}-3j_{1}-4} \sum _{c=m}^{\infty }\sum _{d=m}^{\infty }2^{-3j_{4}-3j_{3}-4},\ \end{aligned}$$
therefore we can derive
$$\begin{aligned} \Vert e_{a,b,c,d} \Vert ^{2}\le U^{2}\times \dfrac{1}{9m^{4}}\times \dfrac{1}{9m^{4}}=\dfrac{U^{2}}{81m^{8}}. \end{aligned}$$
In other words
$$\begin{aligned} \Vert e_{a,b,c,d} \Vert \le \dfrac{U}{9m^{4}}. \end{aligned}$$
\(\square \)

Theorem 4

If g(st) be the exact solution of Eq. (1) and \( {\hat{g}}_{m}(s,t) \) be HWFs approximate solution of (1) that is obtained by (9) and \(U_{i}\)’s be defined in Theorem 3, then by using hypothesises
  1. 1.

    \(\Vert g \Vert \le A, \ \ \ (s,t) \in [0,1) \times [0,1)\),

     
  2. 2.

    \(\Vert K_{i} \Vert \le Q_{i}, \ i=1,2, \ (x,y,s,t) \in [0,1) \times [0,1)\times [0,1) \times [0,1)\),

     
  3. 3.

    \(O(x)=\sup _{x\in [0,1)} \vert B(x) \vert \) ,

     
  4. 4.

    \(\left( Q_{1}+\dfrac{U_{1}}{9m^{4}}+\left( Q_{2}+\dfrac{U_{2}}{9m^{4}}\right) \times O(x) \times O(y) \right) <1\),

     
we have
$$\begin{aligned} \Vert g-{\hat{g}}_{m} \Vert \le \dfrac{\dfrac{M}{3m^{2}}+\dfrac{U_{1}A}{9m^{4}}+\dfrac{U_{2}A}{9m^{4}} \times O(x) \times O(y) }{1-\left[ Q_{1}+\dfrac{U_{1}}{9m^{4}}+\left( Q_{2}+\dfrac{U_{2}}{9m^{4}}\right) \times O(x) \times O(y) \right] }. \end{aligned}$$

Proof

We can obtain
$$\begin{aligned}&g(x,y)-{\hat{g}}_{m}(x,y)\\&\quad =f(x,y)-{\hat{f}}_{m}(x,y)\\&\qquad +\int _{0}^{y}\int _{0}^{x}\left( K_{1}(x,y,s,t)g(s,t)-{\hat{K}}_{1,m}(x,y,s,t) {\hat{g}}_{m}(s,t)\right) dsdt\\&\qquad +\int _{0}^{y}\int _{0}^{x}\left( K_{2}(x,y,s,t)g(s,t)-{\hat{K}}_{2,m}(x,y,s,t){\hat{g}}_{m}(s,t)\right) dB(s)dB(t). \end{aligned}$$
The mean value theorem conclude
$$\begin{aligned} \Vert g-{\hat{g}}_{m}\Vert\le & {} \Vert f-{\hat{f}}_{m}\Vert +xy \Vert K_{1}g-{\hat{K}}_{1,m} {\hat{g}}_{m}\Vert \nonumber \\&+B(x) B(y)\Vert K_{2}g-{\hat{K}}_{2,m}{\hat{g}}_{m}\Vert . \end{aligned}$$
(12)
By using two first hypothesis and Theorem 3 we obtain
$$\begin{aligned} \Vert K_{1}g-{\hat{K}}_{1,m} {\hat{g}}_{m}\Vert\le & {} \Vert K_{1}\Vert \Vert g-{\hat{g}}_{m}\Vert +\Vert K_{1}-{\hat{K}}_{1,m}\Vert \left( \Vert g-{\hat{g}}_{m}\Vert +\Vert g\Vert \right) \nonumber \\\le & {} Q_{1}\Vert g-{\hat{g}}_{m}\Vert +\dfrac{U_{1}}{9m^{4}}\left( \Vert g-{\hat{g}}_{m}\Vert +A \right) \nonumber \\= & {} \left( Q_{1}+\dfrac{U_{1}}{9m^{4}}\right) \Vert g-{\hat{g}}_{m}\Vert +\dfrac{U_{1}}{9m^{4}}A. \end{aligned}$$
(13)
Similarly for the stochastic case we get
$$\begin{aligned} \Vert K_{2}g-{\hat{K}}_{2,m} {\hat{g}}_{m}\Vert\le & {} \Vert K_{2}\Vert \Vert g-{\hat{g}}_{m}\Vert +\Vert K_{2}-{\hat{K}}_{2,m}\Vert \left( \Vert g-{\hat{g}}_{m}\Vert +\Vert g\Vert \right) \nonumber \\= & {} \left( Q_{2}+\dfrac{U_{2}}{9m^{4}}\right) \Vert g-{\hat{g}}_{m}\Vert +\dfrac{U_{2}}{9m^{4}}A. \end{aligned}$$
(14)
By Theorem 2 and substituting Eqs. (12) and (13) in (14), we have
$$\begin{aligned} \Vert g-{\hat{g}}_{m} \Vert\le & {} \dfrac{M}{3m^{2}}+xy\left[ \left( Q_{1}+\dfrac{U_{1}}{9m^{4}}\right) \Vert g-{\hat{g}}_{m}\Vert +\dfrac{U_{1}}{9m^{4}}A\right] \nonumber \\&+B(x) B(y)\left[ \left( Q_{2}+\dfrac{U_{2}}{9m^{4}}\right) \Vert g-{\hat{g}}_{m}\Vert +\dfrac{U_{2}}{9m^{4}}A\right] . \end{aligned}$$
(15)
By taking sup we have
$$\begin{aligned}&\Vert g-{\hat{g}}_{m}\Vert \\&\quad \le \dfrac{M}{3m^{2}}+\sup _{x\in [0,1)} x \times \sup _{y\in [0,1)} y \left[ \left( Q_{1}+\dfrac{U_{1}}{9m^{4}}\right) \sup _{s\le x \ , \ t\le y}\Vert g-{\hat{g}}_{m}\Vert +\dfrac{U_{1}}{9m^{4}}A \right] \\&\qquad +\sup _{x\in [0,1)} \vert B(x) \vert \times \sup _{y\in [0,1)} \vert B(y) \vert \left[ \left( Q_{2}+\dfrac{U_{2}}{9m^{4}}\right) \sup _{s\le x \ , \ t\le y}\Vert g-{\hat{g}}_{m}\Vert +\dfrac{U_{2}}{9m^{4}}A\right] , \end{aligned}$$
so
$$\begin{aligned} \Vert g-{\hat{g}}_{m}\Vert \le \dfrac{\dfrac{M}{3m^{2}}+\dfrac{U_{1}A}{9m^{4}}+\dfrac{U_{2}A}{9m^{4}} \times O(x)\times O(y) }{1-\left[ Q_{1}+\dfrac{U_{1}}{9m^{4}}+\left( Q_{2}+\dfrac{U_{2}}{9m^{4}}\right) \times O(x)\times O(y) \right] }, \end{aligned}$$
finally Theorem 3 of [11] complete the proof. \(\square \)

Numerical Example

To illustrate the effectiveness of the proposed method, are carried out three numerical examples in this section as the solution mean, the error mean, %95 confidence interval and the length of the confidence interval are shown by \({\bar{g}}(x,y)\), \({\bar{e}}(x,y)\), CI and LCI, respectively in Tables 1, 2 and 3. According to these tables by applying HWFs method, as J increases, \({\bar{e}}(x,y)\) and LCI decrease, nearly. Also the results of this method are compared with the BPFs method [10, 11] and are displayed in Table 4. According to Table 4 we can say that HWFs mehod is better than the BPFs method at some points and also there are some other points that the BPFs method is an able method in those. In addition, 3D graphs of the these examples are shown in Figs. 12 and 3. Consider the following 2D-linear stochastic Volterra integral equations of the second kind:
Table 1

The numerical results to Example 1

J

2M

(xy)

\({\bar{g}}(x,y)\)

\({\bar{e}}(x,y)\)

(LU)

LCI

1

2

(0.12, 0.37)

0.9136

0.52231

(0.82015, 2.12111)

1.30096

(0.62, 0.87)

1.0256

0.61874

(0.90361, 2.52031)

1.61670

2

8

(0.19, 0.31)

0.8564

0.58270

(0.36914, 1.50694)

1.13780

(0.69, 0.94)

0.6846

0.49231

(0.53103, 1.14062)

0.60959

3

16

(0.16, 0.34)

1.6913

0.21044

(1.54109, 1.90764)

0.36655

(0.53, 0.91)

0.7318

0.17455

(0.52810, 0.85127)

0.32317

4

32

(0.30, 0.76)

0.8456

0.22661

(0.81154, 1.20124)

0.38970

(0.58, 0.92)

0.5962

0.09197

(0.54120, 0.71061)

0.16941

Table 2

The numerical results to Example 2

J

2M

(xy)

\({\bar{g}}(x,y)\)

\({\bar{e}}(x,y)\)

(LU)

LCI

1

2

(0.12, 0.37)

0.5201

0.62163

(0.48156, 1.85401)

1.37245

(0.62, 0.87)

0.4545

0.34438

(0.41086, 1.71244)

1.30158

2

8

(0.19, 0.31)

0.4891

0.44962

(0.25116, 1.34201)

1.09085

(0.69, 0.94)

0.5709

0.45031

(0.22164, 1.31132)

1.08963

3

16

(0.16, 0.34)

0.5228

0.16420

(0.48511, 0.92120)

0.43609

(0.53, 0.91)

0.6814

0.18010

(0.43160, 0.87492)

0.44332

4

32

(0.30, 0.76)

0.7001

0.08460

(0.65320, 0.89418)

0.24098

(0.58, 0.92)

0.4667

0.11020

(0.42293, 0.69409)

0.27116

Table 3

The numerical results to Example 3

J

2M

(xy)

\({\bar{g}}(x,y)\)

\({\bar{e}}(x,y)\)

(LU)

LCI

1

2

(0.12, 0.37)

0.6143

0.21438

(0.51617, 0.91564)

0.39947

(0.62, 0.87)

0.2182

0.25460

(0.19851, 0.75694)

0.55843

2

8

(0.19, 0.31)

0.5149

0.19460

(0.36180, 0.50934)

0.14754

(0.69, 0.94)

0.3939

0.18281

(0.34108, 0.58447)

0.24339

3

16

(0.16, 0.34)

0.5196

0.19400

(0.48156, 0.71601)

0.23445

(0.53, 0.91)

0.5994

0.13327

(0.48485, 0.75462)

0.26977

4

32

(0.30, 0.76)

0.6996

0.14025

(0.66451, 0.73124)

0.06673

(0.58, 0.92)

0.4984

0.05612

(0.44640, 0.52681)

0.08041

Table 4

Comparison of the numerical results for HWFs method with BPFs method for Examples 1, 2 and 3

Example

J

(xy)

Method

\({\bar{e}}(x,y)\)

LCI

1

1

(0.37, 0.87)

BPFs

0.12101

1.23060

HWFs

0.11569

0.85423

3

(0.16, 0.91)

BBFs

0.01892

0.85446

HWFs

0.21520

0.23148

4

(0.64, 0.89)

BBFs

0.20236

0.75129

HWFs

0.13740

0.81453

2

1

(0.37, 0.87)

BPFs

0.23001

0.46501

HWFs

0.01475

0.15047

3

(0.16, 0.91)

BBFs

0.11254

0.42718

HWFs

0.23160

0.08297

4

(0.64, 0.89)

BBFs

0.04156

0.13451

HWFs

0.01266

0.02828

3

1

(0.37, 0.87)

BPFs

0.19007

0.25425

HWFs

0.10105

0.36061

3

(0.16, 0.91)

BBFs

0.00892

0.56490

HWFs

0.05141

0.54001

4

(0.64, 0.89)

BBFs

0.15230

0.31640

HWFs

0.05857

0.03114

Fig. 1

Solutions of Example 1

Fig. 2

Solutions of Example 2

Fig. 3

Solutions of Example 3

Example 1

$$\begin{aligned} g(x,y)=f(x,y)+\int _{0}^{ y}\int _{0}^{x}xyst g(s,t)dsdt +\int _{0}^{ y}\int _{0}^{x}(x+y+s+t)g(s,t)dB(s)dB(t), \end{aligned}$$
where
$$\begin{aligned} f(x,y)= & {} (x+y)\left[ 1-\dfrac{1}{6}x^{3}y^{3}-\left( x B(x)-\int _{0}^{x}B(s)ds\right) B(y)\right. \\&\left. -\left( y B(y)-\int _{0}^{y}B(t)dt\right) B(x)\right] -\left( x^{2} B(x)-2\int _{0}^{x}sB(s)ds\right) B(y) \\&-2\left( x B(x)-\int _{0}^{x}B(s)ds\right) \left( y B(y)-\int _{0}^{y}B(t)dt\right) \\&-\left( y^{2} B(y)-2\int _{0}^{y}tB(t)dt\right) B(x), \end{aligned}$$
with the exact solution \( g(x,y)=x+y \).

Example 2

$$\begin{aligned} g(x,y)=f(x,y)+\int _{0}^{ y}\int _{0}^{x}(x+y+s+t) g(s,t)dsdt +\int _{0}^{ y}\int _{0}^{x}xystg(s,t)dB(s)dB(t), \end{aligned}$$
where
$$\begin{aligned} f(x,y)= & {} B(x)B(y)-(x+y)\int _{0}^{y}B(s)ds\int _{0}^{x}B(s)ds-y\int _{0}^{x}sB(s)ds \\&-x\int _{0}^{y}sB(s)ds-xy\left( \dfrac{y}{2}B^{2}(y)-\dfrac{y^{2}}{4}-\dfrac{1}{2}\int _{0}^{y}B^{2}(s)ds\right) \\&\times \left( \dfrac{x}{2}B^{2}(x)-\dfrac{x^{2}}{4}-\dfrac{1}{2}\int _{0}^{x}B^{2}(s)ds\right) , \end{aligned}$$
by the exact solution \( g(x,y)=B(x)B(y) \).

Example 3

$$\begin{aligned} g(x,y)= & {} f(x,y)+\int _{0}^{ y}\int _{0}^{x}t\sqrt{x+y} Sin(s) g(s,t)dsdt\\&+\int _{0}^{y}\int _{0}^{x}st Cos(xy) g(s,t) dB(s)dB(t), \end{aligned}$$
where
$$\begin{aligned} f(x,y)= & {} (xy) e^{B(x)+B(y)}-\int _{0}^{ y}\int _{0}^{x}t^{2}s\sqrt{x+y} Sin(s) e^{B(s)+B(t)}dsdt\\&-Cos(xy)\left( xe^{B(x)}-\int _{0}^{x}e^{B(s)}\left( 1+\dfrac{s}{2}\right) ds)\right) \\&\quad \left( ye^{B(y)}-\int _{0}^{y}e^{B(s)}\left( 1+\dfrac{s}{2}\right) ds)\right) , \end{aligned}$$
by the exact solution \(g(x,y)=(xy) e^{B(x)+B(y)}\).

Conclusion

In this paper, we developed HWFs numerical method for approximate a solution of Eq. (1). The error analysis and the numerical examples show accuracy of this method. Furthermore, the current method can be run with increasing J until the results settle down to an appropriate accuracy that it leads to solving \( 2^{2J+2} \) linear systems, that have its difficulties. Finally, this method can be extended and applied to 2D linear or non-linear multi-noise stochastic Volterra- Fredholm integral equations of the first or second kind.

Notes

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Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  1. 1.Department of Mathematics, Karaj BranchIslamic Azad UniversityKarajIran

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