# Solving Volterra Integrodifferential Equations via Diagonally Implicit Multistep Block Method

Authors: Nur Auni Baharum; Zanariah Abdul Majid; Norazak Senu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

The performance of the numerical computation based on the diagonally implicit multistep block method for solving Volterra integrodifferential equations (VIDE) of the second kind has been analyzed. The numerical solutions of VIDE will be computed at two points concurrently using the proposed numerical method and executed in the predictor-corrector (PECE) mode. The strategy to obtain the numerical solution of an integral part is discussed and the stability analysis of the diagonally implicit multistep block method was investigated. Numerical results showed the competence of diagonally implicit multistep block method when solving Volterra integrodifferential equations compared to the existing methods.


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## Body

## 1. Introduction

Consider the Volterra integrodifferential equation of the second kind(1)y′x=Fx,yx,zx,yx0=y0,where(2)zx=∫0xKx,sysds,0≤s≤x.The numerical methods are generated to solve (1) which is a standard algorithm for ordinary differential equations and Newton-Cote integration formulae are required for solving the integral part since it cannot be solved explicitly. These equations usually appeared in physics, biology, and engineering applications such as biological models, neutron diffusion, wind ripple in the desert, heat transfer, and many more.For many years, several methods had been applied to solve first-order problem of VIDE. Day [1] proposed Newton-Cotes integration formula of the trapezoidal rule for the solutions of outer and inner integral to obtain approximate solutions of integrodifferential equations. A comparison between the variational iteration method and trapezoidal rule revealing that the variational iteration method is more efficient and convenient to solve linear VIDE was discussed by Saadati et al. [2].In [3], finite difference method is used for solving linear VIDE by Raftari. He transforms the Volterra integrodifferential equation in a matrix form and solved it by using finite difference method based on Simpson’s rule and trapezoidal rule. A fourth-order robust numerical method was presented by Filiz [4] with a combination of the trapezoidal rule and Simpson’s 1/3 rule to evaluate the solution of VIDE for kernel equal to one. Then, he extended his work with a Runge-Kutta-Verner method in [5] and used higher rules of numerical integration method for solving the integral part.The extended trapezoidal method [6] was proposed for the numerical solution of VIDE of the second kind and implemented the method in PECE scheme. Mohamed and Majid [7] had solved the second kind of VIDE using one-step block method and the Newton-Cotes quadrature formula was employed for finding the solution of the integral part. The multistep block method in [8] had implemented two approaches for solving VIDE for K(x,s)=1 and K(x,s)≠1.


## 2. Numerical Method

The proposed numerical method is in the form of block method and it will generate two or more solutions at the same time. The proposed method is a two-point block method; hence it will generate two solutions in one block.In Figure1, the two approximate values of yn+1 and yn+2 will be computed simultaneously in a block. The approximate values of yn+1 can be developed by integrating (1) over the interval [xn,xn+1] while the interval for values yn+2 is [xn,xn+2]. Hence, the formulae of yn+1 and yn+2 can be obtained as(3)∫xnxn+ry′xdx=∫xnxn+rFx,y,zdx.Therefore,(4)yn+r-yn=∫xnxn+rFx,y,zdx,where r=1,2. Then, function F(x,y,z) in (4) will be approximated using Lagrange interpolating polynomial, and the interpolation points involved in obtaining the corrector formula of yn+1 are {xn-1,xn,xn+1}. Taking x=xn+1+sh, dx=hds, and replacing into (4), the limit of integration in (4) will be -1 to 0.Figure 1
Two-point multistep block method.The formulation ofyn+2 can be obtained when three points are involved in the interpolation polynomial, that is, {xn-1,xn+1,xn+2}. Considering x=xn+2+sh, dx=hds in (4) and the limit of integration will be changed from -2 to 0. The corrector formulae of yn+1 and yn+2 will be obtained usingMAPLE software. The predictor formulae are one order less than the corrector formulae and the same process of derivation is applied.Diagonally Implicit Multistep Block MethodPredictor
(5)
y
n
+
1
p
=
y
n
+
h
3
2
F
n
-
1
2
F
n
-
1
,
y
n
+
2
p
=
y
n
+
h
4
F
n
-
2
F
n
-
1
.Corrector
(6)
y
n
+
1
c
=
y
n
+
h
5
12
F
n
+
1
+
2
3
F
n
-
1
12
F
n
-
1
,
y
n
+
2
c
=
y
n
+
h
2
9
F
n
+
2
+
5
3
F
n
+
1
+
1
9
F
n
-
1
.The matrix form of the corrector formulae is(7)1001yn+1yn+2=0-10-1yn-1yn+h-112231953Fn-1Fn+h51205329Fn+1Fn+2which is equivalent to the difference equations(8)A0Ym=A1Ym-1+hB0Fm-1+B1Fm,where A0, A1, B0, and B1 are the coefficients with m-vectors Ym, Ym-1, Fm-1, and Fm defined as(9)Ym=yn+1yn+2,Ym-1=yn-1yn,Fm-1=Fn-1Fn,Fm=Fn+1Fn+2.


## 3. Analysis of Diagonally Implicit Multistep Block Method

### 3.1. Order and Convergence of the Method

The order of the method can be obtained by referring [9](10)∑j=0kαjyx+jh-hβjy′x+jh=Cpyp+Ohp+1,where p is the order of the linear multistep method, Ohp+1 is the local truncation error, and Cp is defined as(11)Cp=∑j=0kjpαjp!-jp-1βjp-1!.Definition 1.
The numerical method is said to be in orderp if the linear operator of numerical method is(12)C0=C1=C2=⋯=Cp=0,Cp+1≠0,where Cp+1 is called as an error constant of the method.The order of diagonally implicit multistep block method in (7) can be determined by applying the formula in (11); hence the values of α and β are obtained as follows:(13)α0=00,α1=-1-1,α2=10,α3=01,β0=-11219,β1=230,β2=51253,β3=029.Substitute the values of α and β into (11) and obtain(14)C0=∑j=0310!j0αj=00,C1=∑j=0311!j1αj-∑j=0310!j0βj=00,C2=∑j=0312!j2αj-∑j=0311!j1βj=00,C3=∑j=0313!j3αj-∑j=0312!j2βj=00,C4=∑j=0314!j4αj-∑j=0313!j3βj=-12419.Therefore, the diagonally implicit multistep block method is third-order where the coefficient of error constant is(15)Cp+1=C4=-12419T≠00T.Definition 2.
The local truncation error atxn+k of the method is defined to be expression Lyxn;h, when y(x) is the theoretical solution of the initial value problem(16)Lyxn;h=∑j=0kαjyx+jh-hβjy′x+jh.For the formulayn+1c,(17)yn+1c=yn+h512Fn+1+23Fn-112Fn-1.Since Fn+1=y′xn+1, Taylor expansion will be applied to the derivatives y′xn+1 and y′xn-1 where Fn-1=y′xn-1,(18)y′xn+1=y′xn+hy′′xn+h22!y′′′xn+Oh3,y′xn-1=y′xn-hy′′xn+h22!y′′′xn+Oh3.Then, since Fn=y′xn we have(19)yn+1=yn+hyn′+h22yn′′+h36yn′′′+Oh4.So, the local truncation error for yn+1 is Oh4. For the formula yn+2c,(20)yn+2c=yn+h29Fn+2+53Fn+1+19Fn-1.The Taylor expansion for y′xn+2=Fn+2 is given as(21)y′xn+2=y′xn+2hy′′xn+2h22!y′′′xn+Oh3.Then, we will have(22)yn+2=yn+2hyn′+2h2yn′′+43h3yn′′′+Oh4.This shows that the local truncation error for yn+2 is Oh4.Definition 3.
The numerical method is said to be consistent if the order of method isp≥1 and the method is consistent if and only if(23)∑j=0kαj=0,∑j=0kjαj=∑j=0kβj.Proof.
(i)∑j=0kαj=0,(24)∑j=03αj=α0+α1+α2+α3=00+-1-1+10+01=00.
(ii)∑j=0kj·αj=∑j=0kβj,(25)∑j=03j·αj=0·α0+1·α1+2·α2+3·α3=000+1-1-1+210+301=12,∑j=03βj=β0+β1+β2+β3=-11219+230+51253+029=12.
Therefore,(26)∑j=03j·αj=∑j=03βj=12.By Definitions 1 and 3, the diagonally implicit multistep block method is consistent.Definition 4.
A block method is said to be zero-stable if and only if providing the roots ofRj, j=1(1)k of the first characteristic polynomial, ρ(R), specified as(27)ρR=det⁡∑j=0kAiRk-i=0satisfies |Rj|≤1 and those roots with |Rj|=1.Proof.
The values ofA can be obtained in (7):(28)ρr=det⁡RA0-A1=det⁡R1001-0-10-1=det⁡R10R+1=RR+1.The diagonally implicit multistep block method is zero stable since |R|≤1.Theorem 5.
The method is said to be convergent if and only if the method is consistent and zero-stable.Proof.
By Definitions1, 3, and 4, the diagonally implicit multistep block method is convergent.


### 3.2. Stability Region of the Method

In this section, the stability region of the diagonally implicit multistep block method of order three is discussed for the numerical solution of VIDE. The test equation for first-order VIDE of the second kind [10] is(29)y′x=ξyx+η∫0xytdt,where ξ and η are real constants, ξ=λ+μ and η=-λμ. Therefore,(30)y′x=λ+μyx-λμ∫0xytdt.Definition 6.
The method is said to beA-stable if and only if the region of absolute stability contains at the quarter plane hξ<0, h2η<0.From the proposed method for the numerical solution, the characteristics polynomialsρ(r), σ(r), ρ~(r), and σ~(r) can be developed as follows.Corrector formula foryn+1 is(31)ρr=r2-r,σr=512r2+23r-112.Corrector formula foryn+2 is(32)ρr=r3-r,σr=29r3+53r2+19.Simpson’s rule is(33)ρ~r=r2-1,σ~r=13r2+43r+13.The stability polynomial of the diagonally implicit multistep block method can be determined by substituting (31), (32), and (33) into this particular formula,(34)πr,hξ,h2η=ρ~rρr-H1σr-H2σ~rσr,where H1=hξ and H2=h2μ. Thus, the stability polynomial of the proposed method is obtained:(35)-736H1+7108H2-227H12-2243H22+554H12r4-7154H12r3+4118H12r2-5354H12r+32H1r2-79H1r-2336H1r4+19H1r3+481H1H2+5486H22r4-131486H22r3+9554H22r2-5486H22r-23108H2r4-20554H2r3+11627H2r2-1954H2r-6181H1H2r3+6181H1H2r-19H1H2r2+581H1H2r4+r4-3r3+3r2-r=0.The region of stability polynomial can be illustrated in Figure 2.Figure 2
Stability region inH1, H2 plane.Regarding Definition6, the third-order of diagonally implicit multistep block method in Figure 2 is A-stable within the shaded region.


## 3.1. Order and Convergence of the Method

The order of the method can be obtained by referring [9](10)∑j=0kαjyx+jh-hβjy′x+jh=Cpyp+Ohp+1,where p is the order of the linear multistep method, Ohp+1 is the local truncation error, and Cp is defined as(11)Cp=∑j=0kjpαjp!-jp-1βjp-1!.Definition 1.
The numerical method is said to be in orderp if the linear operator of numerical method is(12)C0=C1=C2=⋯=Cp=0,Cp+1≠0,where Cp+1 is called as an error constant of the method.The order of diagonally implicit multistep block method in (7) can be determined by applying the formula in (11); hence the values of α and β are obtained as follows:(13)α0=00,α1=-1-1,α2=10,α3=01,β0=-11219,β1=230,β2=51253,β3=029.Substitute the values of α and β into (11) and obtain(14)C0=∑j=0310!j0αj=00,C1=∑j=0311!j1αj-∑j=0310!j0βj=00,C2=∑j=0312!j2αj-∑j=0311!j1βj=00,C3=∑j=0313!j3αj-∑j=0312!j2βj=00,C4=∑j=0314!j4αj-∑j=0313!j3βj=-12419.Therefore, the diagonally implicit multistep block method is third-order where the coefficient of error constant is(15)Cp+1=C4=-12419T≠00T.Definition 2.
The local truncation error atxn+k of the method is defined to be expression Lyxn;h, when y(x) is the theoretical solution of the initial value problem(16)Lyxn;h=∑j=0kαjyx+jh-hβjy′x+jh.For the formulayn+1c,(17)yn+1c=yn+h512Fn+1+23Fn-112Fn-1.Since Fn+1=y′xn+1, Taylor expansion will be applied to the derivatives y′xn+1 and y′xn-1 where Fn-1=y′xn-1,(18)y′xn+1=y′xn+hy′′xn+h22!y′′′xn+Oh3,y′xn-1=y′xn-hy′′xn+h22!y′′′xn+Oh3.Then, since Fn=y′xn we have(19)yn+1=yn+hyn′+h22yn′′+h36yn′′′+Oh4.So, the local truncation error for yn+1 is Oh4. For the formula yn+2c,(20)yn+2c=yn+h29Fn+2+53Fn+1+19Fn-1.The Taylor expansion for y′xn+2=Fn+2 is given as(21)y′xn+2=y′xn+2hy′′xn+2h22!y′′′xn+Oh3.Then, we will have(22)yn+2=yn+2hyn′+2h2yn′′+43h3yn′′′+Oh4.This shows that the local truncation error for yn+2 is Oh4.Definition 3.
The numerical method is said to be consistent if the order of method isp≥1 and the method is consistent if and only if(23)∑j=0kαj=0,∑j=0kjαj=∑j=0kβj.Proof.
(i)∑j=0kαj=0,(24)∑j=03αj=α0+α1+α2+α3=00+-1-1+10+01=00.
(ii)∑j=0kj·αj=∑j=0kβj,(25)∑j=03j·αj=0·α0+1·α1+2·α2+3·α3=000+1-1-1+210+301=12,∑j=03βj=β0+β1+β2+β3=-11219+230+51253+029=12.
Therefore,(26)∑j=03j·αj=∑j=03βj=12.By Definitions 1 and 3, the diagonally implicit multistep block method is consistent.Definition 4.
A block method is said to be zero-stable if and only if providing the roots ofRj, j=1(1)k of the first characteristic polynomial, ρ(R), specified as(27)ρR=det⁡∑j=0kAiRk-i=0satisfies |Rj|≤1 and those roots with |Rj|=1.Proof.
The values ofA can be obtained in (7):(28)ρr=det⁡RA0-A1=det⁡R1001-0-10-1=det⁡R10R+1=RR+1.The diagonally implicit multistep block method is zero stable since |R|≤1.Theorem 5.
The method is said to be convergent if and only if the method is consistent and zero-stable.Proof.
By Definitions1, 3, and 4, the diagonally implicit multistep block method is convergent.


## 3.2. Stability Region of the Method

In this section, the stability region of the diagonally implicit multistep block method of order three is discussed for the numerical solution of VIDE. The test equation for first-order VIDE of the second kind [10] is(29)y′x=ξyx+η∫0xytdt,where ξ and η are real constants, ξ=λ+μ and η=-λμ. Therefore,(30)y′x=λ+μyx-λμ∫0xytdt.Definition 6.
The method is said to beA-stable if and only if the region of absolute stability contains at the quarter plane hξ<0, h2η<0.From the proposed method for the numerical solution, the characteristics polynomialsρ(r), σ(r), ρ~(r), and σ~(r) can be developed as follows.Corrector formula foryn+1 is(31)ρr=r2-r,σr=512r2+23r-112.Corrector formula foryn+2 is(32)ρr=r3-r,σr=29r3+53r2+19.Simpson’s rule is(33)ρ~r=r2-1,σ~r=13r2+43r+13.The stability polynomial of the diagonally implicit multistep block method can be determined by substituting (31), (32), and (33) into this particular formula,(34)πr,hξ,h2η=ρ~rρr-H1σr-H2σ~rσr,where H1=hξ and H2=h2μ. Thus, the stability polynomial of the proposed method is obtained:(35)-736H1+7108H2-227H12-2243H22+554H12r4-7154H12r3+4118H12r2-5354H12r+32H1r2-79H1r-2336H1r4+19H1r3+481H1H2+5486H22r4-131486H22r3+9554H22r2-5486H22r-23108H2r4-20554H2r3+11627H2r2-1954H2r-6181H1H2r3+6181H1H2r-19H1H2r2+581H1H2r4+r4-3r3+3r2-r=0.The region of stability polynomial can be illustrated in Figure 2.Figure 2
Stability region inH1, H2 plane.Regarding Definition6, the third-order of diagonally implicit multistep block method in Figure 2 is A-stable within the shaded region.


## 4. Implementation

The one-step methods are required for finding the first starting point atxn+1 since Volterra integrodifferential equations of the second kind have two types of kernels. For Kx,s=1, Runge-Kutta method is involved in solving differential part of VIDE, while, midpoint method is needed to solve VIDE when Kx,s≠1. Hence, the predictor and the corrector formula can be implemented until the end of the interval(36)yn+1p=yn+h32Fn-12Fn-1,yn+2p=yn+h4Fn-2Fn-1,yn+1c=yn+h512Fn+1+23Fn-112Fn-1,yn+2c=yn+h29Fn+2+53Fn+1+19Fn-1.Since z(x) in Fx,yx,zx is the integral term in VIDE and cannot be solved explicitly, therefore, Simpson’s rule is adapted for solving the integral part. (i)
ForK(x,s)=1, Simpson’s 1/3 rule is applied to solve the integral term in VIDE:(37)zn+2=zn+h13yn+43yn+1+13yn+2.(ii)
ForK(x,s)≠1, composite Simpson’s rule is employed for solving the integral part:(38)zn+2=h3∑i=0n+2ωisKxn+2,xi,yi,zn+3=h3∑i=0n+2ωisKxn+3,xi,yi+h6Kxn+3,xn+2,yn+2+4Kxn+2,xn+5/2,yn+5/2+Kxn+3,xn+3,yn+3,where ωis are Simpson’s rule weights 1,4,2,4,…,2,4,1. The unknown value of yn+5/2 in (38) can be estimated by using Lagrange interpolation at the point xn,xn+1,xn+2,xn+3,(39)yn+5/2=116yn-516yn+1+1516yn+2+516yn+3.


### 4.1. Algorithm of the Method

The input of the programming is the endpoints ofa and b and the integer, N. The developed algorithm for the method is given as follows.Step 1.
Setx0=a;y0=α;z0=0;h=b-a/N;OUTPUT:(x0,y0,z0).Step 2.
Fori=1,WhenK(x,s)=1, using RK3 to evaluate the value of y.WhenK(x,s)≠1, applying Midpoint Method.Step 3.
Fori=2,…,N/2, do Steps 4–6.Step 4.
Setx=a+ih.Step 5.
Calculate foryn+1p and zn+1p, yn+2p and zn+2p.Step 6.
Compute the solution foryn+1c and zn+1c, yn+2c and zn+2c.Step 7.
Calculate the error.Step 8.
OUTPUT:(x,y,z) and the absolute error.Step 9.
STOP.


## 4.1. Algorithm of the Method

The input of the programming is the endpoints ofa and b and the integer, N. The developed algorithm for the method is given as follows.Step 1.
Setx0=a;y0=α;z0=0;h=b-a/N;OUTPUT:(x0,y0,z0).Step 2.
Fori=1,WhenK(x,s)=1, using RK3 to evaluate the value of y.WhenK(x,s)≠1, applying Midpoint Method.Step 3.
Fori=2,…,N/2, do Steps 4–6.Step 4.
Setx=a+ih.Step 5.
Calculate foryn+1p and zn+1p, yn+2p and zn+2p.Step 6.
Compute the solution foryn+1c and zn+1c, yn+2c and zn+2c.Step 7.
Calculate the error.Step 8.
OUTPUT:(x,y,z) and the absolute error.Step 9.
STOP.


## 5. Numerical Results

Four tested problems of first-order Volterra integrodifferential equations were considered in order to study the performance of the diagonally implicit multistep block method.Example 1 ((K(x,s)=1) linear VIDE).
(40)
y
′
x
=
1
-
∫
0
x
y
s
d
s
,
y
0
=
0
,
0
≤
x
≤
1
.
Exact solution isy(x)=sin(x).
Source: [4].Example 2 ((K(x,s)≠1) linear VIDE).
(41)
y
′
x
=
-
sin
⁡
x
-
cos
⁡
x
+
∫
0
x
2
cos
⁡
x
-
s
y
s
d
s
,
y
0
=
1
,
0
≤
x
≤
5
.
Exact solution isy(x)=exp(-x).
Source: [11].Example 3 ((K(x,s)≠1) nonlinear VIDE).
(42)
y
′
x
=
x
exp
⁡
1
-
y
x
-
1
1
+
x
2
-
x
-
∫
0
x
x
1
+
s
2
exp
⁡
1
-
y
s
d
s
y
0
=
1
,
0
≤
x
≤
4
.
Exact solution isy(x)=1/1+x.
Source: [12].Example 4 ((K(x,s)≠1) nonlinear VIDE).
(43)
y
′
x
=
2
x
-
1
2
sin
⁡
x
4
+
∫
0
x
x
2
s
cos
⁡
x
2
y
s
d
s
,
y
0
=
0
,
0
≤
x
≤
2
.
Exact solution isy(x)=x2.
Source: [13].Notations used in Tables1–4 are as follows:h: step sizeMAXE: maximum errorTS: total stepsTFC: total functions callTime: the execution time takenRK3: Runge-Kutta method of order 3 with Simpson’s1/3 rule by Filiz [4]ABM3: Adam Bashforth Moulton Order 3 with Simpson’s rule.BVMs: Combination of BVMs and third-order Generalized Adams Method by Chen and Zhang [11]2PMBM: Two points Multistep Block Method of Order 3 with Simpson’s rule by Mohamed and Majid [8]DIMBM: Diagonally implicit multistep block method with Simpson’s rule proposed in this paperTable 1
Numerical results for Example1.
h
0.025
0.0125
0.00625
0.003125
MAXE
RK3
8.3229(−08)
1.0910(−08)
1.3953(−09)
1.7637(−10)
ABM3
1.9468(−06)
2.4548(−07)
3.0813(−08)
3.8593(−09)
2PMBM
4.2079(−07)
4.9165(−08)
5.9272(−09)
7.2715(−10)
DIMBM
4.1354(−07)
4.7815(−08)
5.8390(−09)
7.2152(−10)
TFC
RK3
120
240
480
960
ABM3
82
162
322
642
2PMBM
43
83
163
323
DIMBM
42
82
162
322
TS
RK3
40
80
160
320
ABM3
40
80
160
320
2PMBM
21
41
81
161
DIMBM
21
41
81
161
Time
RK3
0.1035
0.1840
0.2970
0.4527
ABM3
0.0860
0.1410
0.2264
0.3454
2PMBM
0.0670
0.1200
0.2024
0.3204
DIMBM
0.0450
0.1120
0.1600
0.2660Table 2
Numerical results for Example2.
h
0.25
0.125
0.0625
0.03125
MAXE
BVMs
2.1607(−01)
2.8411(−02)
3.6378(−03)
4.6011(−04)
ABM3
1.5401(−02)
3.1321(−03)
5.3778(−04)
7.9127(−05)
2PMBM
5.7923(−02)
1.2209(−03)
3.7500(−04)
8.4172(−05)
DIMBM
1.8211(−02)
3.4489(−03)
4.6803(−04)
6.0807(−05)
TFC
BVMs
-
-
-
-
ABM3
84
164
324
644
2PMBM
50
90
170
330
DIMBM
22
42
82
162
TS
BVMs
-
-
-
-
ABM3
20
40
80
160
2PMBM
11
21
41
81
DIMBM
11
21
41
81
Time
BVMs
-
-
-
-
ABM3
0.0715
0.1546
0.2433
0.3546
2PMBM
0.0460
0.0780
0.1710
0.2738
DIMBM
0.0140
0.0410
0.1170
0.1480Table 3
Numerical results for Example3.
h
0.025
0.0125
0.00625
0.003125
MAXE
ABM3
2.3797(−06)
3.2252(−07)
4.2061(−08)
5.3727(−09)
2PMBM
8.0020(−06)
9.7319(−07)
1.2000(−07)
1.4862(−08)
DIMBM
3.6545(−06)
4.6274(−07)
5.8216(−08)
7.2999(−09)
TFC
ABM3
644
1284
2564
5124
2PMBM
330
650
1290
2570
DIMBM
322
642
1282
2562
Time
ABM3
0.3543
0.5677
1.1650
1.9855
2PMBM
0.2810
0.4210
0.7496
1.3570
DIMBM
0.2610
0.3010
0.6740
1.1050Table 4
Numerical results for Example4.
h
2/9
2/17
2/33
2/65
MAXE
ABM3
5.0218(−02)
1.8761(−03)
1.1046(−04)
5.8996(−06)
2PMBM
2.7425(−02)
6.5258(−04)
1.7781(−04)
6.6888(−06)
DIMBM
8.8008(−03)
8.6068(−04)
1.7703(−04)
6.6994(−06)
TFC
ABM3
40
72
136
246
2PMBM
22
38
70
134
DIMBM
10
18
34
66
Time
ABM3
0.0670
0.0723
0.1291
0.2030
2PMBM
0.0140
0.0352
0.0662
0.1336
DIMBM
0.0050
0.0280
0.0350
0.0420Tables1–4 display the numerical results for the four tested problems when solved using the proposed block method and the code was written in C language.The numerical results for Examples1–4 displayed in Tables 1–4 are solved numerically using the proposed numerical method with Simpson’s rule. In Table 1, the numerical results will be obtained when the step size h=0.025, 0.0125, 0.00625, and 0.003125 for the case when K(x,s)=1. The maximum error of DIMBM is comparable compared to 2PMBM at all tested h but the order of accuracy is the same or one order less compared to RK3 and ABM3. The performance of DIMBM is better in terms of total functions call and total number of steps compared to RK3 and ABM3.In Tables2, 3, and 4, the numerical results are solved using the proposed numerical method via composite Simpson’s for the integral part when K(x,s)≠1. In Table 2, the maximum error of DIMBM is comparable compared to 2PMBM and ABM3. The DIMBM manage to obtain less total functions call compared to ABM3 and 2PMBM. For the nonlinear Examples 3 and 4, we could observe that ABM3 and 2PMBM are expensive in terms of total functions call, respectively. Figures 3–6 display the numerical results of maximum error versus total functions call when solving the tested problems. This has shown the advantage of DIMBM in the form of a standard multistep method because the cost per step is cheaper and the numerical results are more accurate when the step size is reduced. In terms of timing, DIMBM gave faster results compared to ABM3 and 2PMBM.Figure 3
Graph of the numerical results when solving Example1.
(a)
Graph of TFC versus MAXE
(b)
Graph of Time versus MAXEFigure 4
Graph of the numerical results when solving Example2.
(a)
Graph of TFC versus MAXE
(b)
Graph of time versus MAXEFigure 5
Graph of the numerical results when solving Example3.
(a)
Graph of TFC versus MAXE
(b)
Graph of Time versus MAXEFigure 6
Graph of the numerical results when solving Example4.
(a)
Graph of TFC versus MAXE
(b)
Graph of time versus MAXE


## 6. Conclusion

In this research, we proposed the diagonally implicit multistep block method for solving linear and nonlinear Volterra integrodifferential equations and comparisons were made with the existing method. Comparisons with existing method reveal that the diagonally implicit multistep block method is more efficient and cheaper.



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*Source: 7392452-2018-01-01.xml*