# Numerical Solutions of a Class of Nonlinear Volterra Integral Equations

Authors: H. S. Malindzisa; M. Khumalo
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We consider numerical solutions of a class of nonlinear (nonstandard) Volterra integral equations. We first prove the existence and
uniqueness of the solution of the Volterra integral equation in the context of the space of continuous functions over a closed interval. We then
use one-point collocation methods with a uniform mesh to construct solutions of the nonlinear (nonstandard) VIE and quadrature rules. We
conclude that the repeated Simpson's rule gives better solutions when a reasonably large value of the stepsize is used.


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## Body

## 1. Introduction

In this paper we study the nonlinear (nonstandard) Volterra integral equation of the second kind of the form(1)u(t)=∑j=1rbj(gj(t)+∫0tkj(t,s)u(s)ds)j,t∈[0,T],
where (r∈N, r≥2), with bj∈R, and gj, kj are continuous functions. Volterra integral equations play an important part in scientific and engineering problems such as population dynamics, spread of epidemics, semiconductor devices, wave propagation, superfluidity, and travelling wave analysis, Saveljeva [1]. In cases where the kernel is of convolution type (K(t,s)=K(t-s)) the solutions to (1) include elliptic functions and natural generalizations of these functions which also have wide applications in the fields of science and engineering [2]. This class of Volterra integral equations was considered by Sloss and Blyth [2] who proved the existence and uniqueness of the solution in the Banach space L2 and applied the Corrington’s Walsh function method to (1).Much work has been done in the study of numerical solutions to Volterra integral equations using collocation methods [1, 3–7]. Benitez and Bolos [8] pointed out that collocation methods have proven to be a very suitable technique for approximating solutions to nonlinear integral equations because of their stability and accuracy. Other authors such as [9–12] used quadrature rules like repeated trapezoidal and repeated Simpson’s rule to solve linear Volterra integral equations. However, collocation methods and quadrature rules have not been used to approximate solutions to (1).


## 2. The Numerical Methods

### 2.1. The Collocation Method

In our work we focus on one-point collocation methods (see [13]).Lettn≔nh(n=0,1,…,N-1) define a uniform partition for I=[0,T] and set ZN≔t0,…,tN,I0≔[t0,t1]In≔(tn,tn+1](1≤n≤N-1). The solution to (1) will be approximated by using collocation in the piecewise constant polynomial space S0-1(ZN).For a given real numberc1, define the set XN≔tn,1 ofcollocation points by
(2)tn,1=tn+c1h(0≤c1≤1,n=0,…,N-1).The collocation solutionun∈S0-1(ZN) is defined by the collocation equation
(3)un(t)=∑j=1rbj(gj(t)+∫0tkj(t,s)u(s)ds)j,t∈XN,
since
(4)un(t)=un(tn+νh)=L1(ν)Un,1,ν∈(0,1],
where L1(ν)=1 and is aLagrange fundamental polynomial.Thus fort=tn,1≔tn+c1h and 0<c1≤1 the collocation equation (3) assumes the form
(5)un(t)=∑j=1rbj(gj(t)+∫0tnkj(t,s)ui(s)hhhhhhhh+h∫0c1kj(t,tn+sh)un(tn+sh)ds)j.Expressing the collocation equation in terms of the stage values we get(6)Un,1=∑j=1rbj(gj(tn,1)+Fjn(tn,1)(∫0c1kj(tn,1,tn+sh)ds)hhhhhhhh+h(∫0c1kj(tn,1,tn+sh)ds)Un,1)j.
Let t∈In and define
(7)Fjn(t)≔∫0tnkj(t,s)ui(s)ds.
Then
(8)Fjn(tn,1)=h∑i=0n-1(∫01kj(tn,1,ti+sh)ds)Ui,1.
The term Fjn(tn,1) is called thelag term corresponding to the collocation solution, [13].Iterated Collocation. The iterated approximation uI corresponding to u is defined by
(9)uI(t)=∑j=1rbj(gj(t)+∫0t‍kj(t,s)u(s)ds)jt∈I
(see [4, 5, 14]).Sett=tn∈ZN- and use (4); we may write (9) in the form
(10)uI(tn)=∑j=1r‍bj(gj(tn)+h∑i-0n-1∫01kj(tn,ti+sh)dsUi,1)j.


### 2.2. Repeated Trapezoidal Rule

Using the trapezoidal rule we construct the solution to the integral equation (1) (see [12]). Let
(11)t0=a,tn=bti=t0+ihi=0,1,2,…,nu(ti)=∑j=1rbj(gj(ti)+∑l=1i∫tl-1tikj(ti,s)u(s)ds)jhhhhhhhhhhhhhhhhhhhhhhhhhhhi=1,2,…,n.The approximation of the integral in (11) by repeated trapezoidal rule will give the following system:
(12)u(t0)=∑j=1rbj(gj)j,u(t1)=∑j=1rbj(gj(t1)h2hhhhhh+h2(kj(t1,t0)u(t0)+kj(t1,t1)u(t1)))j,u(t2)=∑j=1r‍bj(gj(t2)+h2kj(t2,t0)u(t0)hhhhhh+hkj(t2,t1)u(t1)+h2kj(t2,t2)u(t2))j,u3=∑j=1rbj(gj(t3)+h2kj(t3,t0)u(t0)hhhhhh+h(kj(t3,t1)u(t1)hhhhhhihhh+kj(t3,t2)u(t2))hhhhhh+h2kj(t3,t3)u(t3))j,⋮u(tn)=∑j=1rbj(gj(tn)+h2kj(tn,t0)u(t0)hhhhhh+hkj(tn,t1)u(t1)+⋯+hkj(tn,tn-1)hhhhhh×u(tn-1)+h2kj(tn,tn)u(tn))j.


### 2.3. Repeated Simpson’s Rule

We use repeated Simpson’s rule to construct the solution to the integral equation (1) (see [9]).Ifn is even, then Simpson’s rule may be applied to each subinterval [t2i,t2i+1,t2i+2]. For i=0,1,…,(N/2)-1 we have
(13)∫t2it2i+2f(t)ds≃h3[f(t2i)+4f(t2i+1)+f(t2i+2)].Summing up,(14)∫abf(t)dt=h3∑l=0N-1[f(t2i)+4f(t2l+t2l+22)+f(t2l+2)].
We use (14) to solve the nonlinear (nonstandard) VIE. The approximation of (1) in the even nodes t2m is given by
(15)u2m=∑j=1rbj[gj(t2m)+∫at2mkj(t2m,s)u(s)ds]j.Using(16)u(t2l+1)≃u(t2l)+u(t2l+2)2.we obtain(17)u(t2m)=∑j=1rbj[gj(t2m)+h3∑l=0m-1kj(t2m,t2l)u(t2l)hhhhhhh+4kj(t2m,t2l+1)u(t2l)+u(t2l+2)2hhhhhhhgj(t2m)+h3∑l=0m-1kj(t2m,t2l)u(t2l)+kj(t2m,t2l+2)u(t2l+2)]ju(t2m)=∑j=1rbj[∑l=0m-1(kj(t2m,t2l)+2kj(t2m,t2l+1))u(t2l)gj(t2m)HHiHH+h3∑l=0m-1(kj(t2m,t2l)+2kj(t2m,t2l+1))u(t2l)hihhhhh+(kj(t2m,t2l+1)+2kj(t2m,tl+1))HHhiH×u(t2l+2)gj(t2m)∑l=0m-1(kj(t2m,t2l)+2kj(t2m,t2l+1))u(t2l)]ju(t2m)=∑j=1rbj[∑l=0m-1(kj(t2m,t2l-1)+kj(t2m,t2l)gj(t2m)+h3(kj(t2m,t0)+2kj(t2m,t1))HHhHiH+h3(kj(t2m,t2m)+kj(t2m,t2m-1))u(t2m)HHhHiH+2h3∑l=0m-1(kj(t2m,t2l-1)+kj(t2m,t2l)hhhhhhhhhhihhhh+kj(t2m,t2l+1))u(t2l)h3∑l=0m-1(kj(t2m,t2l-1)+kj(t2m,t2l)]j.


## 2.1. The Collocation Method

In our work we focus on one-point collocation methods (see [13]).Lettn≔nh(n=0,1,…,N-1) define a uniform partition for I=[0,T] and set ZN≔t0,…,tN,I0≔[t0,t1]In≔(tn,tn+1](1≤n≤N-1). The solution to (1) will be approximated by using collocation in the piecewise constant polynomial space S0-1(ZN).For a given real numberc1, define the set XN≔tn,1 ofcollocation points by
(2)tn,1=tn+c1h(0≤c1≤1,n=0,…,N-1).The collocation solutionun∈S0-1(ZN) is defined by the collocation equation
(3)un(t)=∑j=1rbj(gj(t)+∫0tkj(t,s)u(s)ds)j,t∈XN,
since
(4)un(t)=un(tn+νh)=L1(ν)Un,1,ν∈(0,1],
where L1(ν)=1 and is aLagrange fundamental polynomial.Thus fort=tn,1≔tn+c1h and 0<c1≤1 the collocation equation (3) assumes the form
(5)un(t)=∑j=1rbj(gj(t)+∫0tnkj(t,s)ui(s)hhhhhhhh+h∫0c1kj(t,tn+sh)un(tn+sh)ds)j.Expressing the collocation equation in terms of the stage values we get(6)Un,1=∑j=1rbj(gj(tn,1)+Fjn(tn,1)(∫0c1kj(tn,1,tn+sh)ds)hhhhhhhh+h(∫0c1kj(tn,1,tn+sh)ds)Un,1)j.
Let t∈In and define
(7)Fjn(t)≔∫0tnkj(t,s)ui(s)ds.
Then
(8)Fjn(tn,1)=h∑i=0n-1(∫01kj(tn,1,ti+sh)ds)Ui,1.
The term Fjn(tn,1) is called thelag term corresponding to the collocation solution, [13].Iterated Collocation. The iterated approximation uI corresponding to u is defined by
(9)uI(t)=∑j=1rbj(gj(t)+∫0t‍kj(t,s)u(s)ds)jt∈I
(see [4, 5, 14]).Sett=tn∈ZN- and use (4); we may write (9) in the form
(10)uI(tn)=∑j=1r‍bj(gj(tn)+h∑i-0n-1∫01kj(tn,ti+sh)dsUi,1)j.


## 2.2. Repeated Trapezoidal Rule

Using the trapezoidal rule we construct the solution to the integral equation (1) (see [12]). Let
(11)t0=a,tn=bti=t0+ihi=0,1,2,…,nu(ti)=∑j=1rbj(gj(ti)+∑l=1i∫tl-1tikj(ti,s)u(s)ds)jhhhhhhhhhhhhhhhhhhhhhhhhhhhi=1,2,…,n.The approximation of the integral in (11) by repeated trapezoidal rule will give the following system:
(12)u(t0)=∑j=1rbj(gj)j,u(t1)=∑j=1rbj(gj(t1)h2hhhhhh+h2(kj(t1,t0)u(t0)+kj(t1,t1)u(t1)))j,u(t2)=∑j=1r‍bj(gj(t2)+h2kj(t2,t0)u(t0)hhhhhh+hkj(t2,t1)u(t1)+h2kj(t2,t2)u(t2))j,u3=∑j=1rbj(gj(t3)+h2kj(t3,t0)u(t0)hhhhhh+h(kj(t3,t1)u(t1)hhhhhhihhh+kj(t3,t2)u(t2))hhhhhh+h2kj(t3,t3)u(t3))j,⋮u(tn)=∑j=1rbj(gj(tn)+h2kj(tn,t0)u(t0)hhhhhh+hkj(tn,t1)u(t1)+⋯+hkj(tn,tn-1)hhhhhh×u(tn-1)+h2kj(tn,tn)u(tn))j.


## 2.3. Repeated Simpson’s Rule

We use repeated Simpson’s rule to construct the solution to the integral equation (1) (see [9]).Ifn is even, then Simpson’s rule may be applied to each subinterval [t2i,t2i+1,t2i+2]. For i=0,1,…,(N/2)-1 we have
(13)∫t2it2i+2f(t)ds≃h3[f(t2i)+4f(t2i+1)+f(t2i+2)].Summing up,(14)∫abf(t)dt=h3∑l=0N-1[f(t2i)+4f(t2l+t2l+22)+f(t2l+2)].
We use (14) to solve the nonlinear (nonstandard) VIE. The approximation of (1) in the even nodes t2m is given by
(15)u2m=∑j=1rbj[gj(t2m)+∫at2mkj(t2m,s)u(s)ds]j.Using(16)u(t2l+1)≃u(t2l)+u(t2l+2)2.we obtain(17)u(t2m)=∑j=1rbj[gj(t2m)+h3∑l=0m-1kj(t2m,t2l)u(t2l)hhhhhhh+4kj(t2m,t2l+1)u(t2l)+u(t2l+2)2hhhhhhhgj(t2m)+h3∑l=0m-1kj(t2m,t2l)u(t2l)+kj(t2m,t2l+2)u(t2l+2)]ju(t2m)=∑j=1rbj[∑l=0m-1(kj(t2m,t2l)+2kj(t2m,t2l+1))u(t2l)gj(t2m)HHiHH+h3∑l=0m-1(kj(t2m,t2l)+2kj(t2m,t2l+1))u(t2l)hihhhhh+(kj(t2m,t2l+1)+2kj(t2m,tl+1))HHhiH×u(t2l+2)gj(t2m)∑l=0m-1(kj(t2m,t2l)+2kj(t2m,t2l+1))u(t2l)]ju(t2m)=∑j=1rbj[∑l=0m-1(kj(t2m,t2l-1)+kj(t2m,t2l)gj(t2m)+h3(kj(t2m,t0)+2kj(t2m,t1))HHhHiH+h3(kj(t2m,t2m)+kj(t2m,t2m-1))u(t2m)HHhHiH+2h3∑l=0m-1(kj(t2m,t2l-1)+kj(t2m,t2l)hhhhhhhhhhihhhh+kj(t2m,t2l+1))u(t2l)h3∑l=0m-1(kj(t2m,t2l-1)+kj(t2m,t2l)]j.


## 3. Existence and Uniqueness of the Solution

The following theorem shows that whenr=2 and b1=0 the integral equation (1) has a unique solution in the space C[0,d]. Theorem 2 gives sufficient conditions for the solution to (1) to exist. We prove the existence and uniqueness of the solution using a procedure analogous to the one used in Sloss and Blyth [2].Theorem 1.
The integral equation(18)z(t)=b(g(t)+∫0tk(t,s)z(s)ds)2,
with g∈C[0,1], b∈R, and k(t,s)∈C([0,1]×[0,1]), has a unique solution u and the solution belongs to Id=[0,d], 0<d≤1, with
(19)0<d<1K[12K|b|-∥g∥∞]-∥u∥∞,
where
(20)K=sup⁡[0,1]×[0,1]|k(t,s)|.Proof.
The existence of the solution is shown in the corollary of Theorem2 (in the next section). Here we prove the uniqueness of the solution. Let u and u+v be solutions of (18).
Then,(21)v(t)=b[(∫0t‍k(t,s)v(s)ds)2g(t)∫0tk(t,s)v(s)dshhhh+2∫0tk(t,s)u(s)ds·∫0tk(t,s)v(s)hhhh+(∫0t‍k(t,s)v(s)ds)2].
DefineTn:C[0,1]×C[0,1] by
(22)Tnz(t)=χnb[(∫0tk(t,s)z(s)ds)22g(t)∫0tk(t,s)z(s)dshhhhhh+2∫0tk(t,s)u(s)ds·∫0tk(t,s)z(s)dshhhhhh+(∫0tk(t,s)z(s)ds)2],
where χn is a sequence of characteristic functions of intervals [0,An]⊂[0,1].
Letv1+v2∈C[0,1]; consider
(23)Tnv1(t)-Tnv2(t)=χn(t)b[2g(t)+2∫0tk(t,s)u(s)dshhhhhhhhhh+∫0tk(t,s)(v1(s)+v2(s))ds]·∫0tk(t,s)(v1(s)-v2(s))ds=defbz(u,v1,v2)(t)∫0tk(t,s)(v1(s)-v2(s))ds;
then,
(24)∥Tnv1(t)-Tnv2(t)∥∞≤∥bz∫01K(t,s)(v1(s)-v2(s))ds∥∞.
Letd=max⁡0≤t≤1(|v1(t)|,|v2(t)|) and take Tn:[0,d]→[0,1].
Then,(25)∥z∥∞≤2∥g∥∞+2K∥u∥∞+2Kd;
therefore
(26)∥Tnv1(t)-Tnv2(t)∥∞≤|b|(2∥g∥∞+2K∥u∥∞+2Kd)K∥v1-v2∥∞.
Thus Tn is contractive if
(27)2|b|K∥g∥∞+2K2∥u∥∞|b|<1.
That is,
(28)0<d<1K[12K|b|-∥g∥∞]-∥u∥∞.
Clearly,Tn maps C[0,d] onto itself if (19) is satisfied. Also, Tn(0)=0.
Supposev(t)≠0 is a solution of (21), such that v may lie outside of [0,d]. Then,
(29)(χnv)(t)=χnb[(∫0t‍k(t,s)v(s)ds)22g(t)∫0t‍k(t,s)v(s)dsHHHH+2∫0t‍k(t,s)u(s)ds·∫0t‍k(t,s)v(s)dsHHiHH+(∫0t‍k(t,s)v(s)ds)2]=b[(χn(t)∫0t‍k(t,s)χn(s)v(s)ds)22g(t)χn(t)∫0t‍k(t,s)χn(s)v(s)dsHHH+2∫0t‍k(t,s)u(s)dsχn(t)HHH×∫0t‍k(t,s)χn(s)v(s)dsHHH+(χn(t)∫0t‍k(t,s)χn(s)v(s)ds)2],=Tn(χnv)(t),
which shows that χnv is a fixed point of Tn for all n. Since χny→0 in C[0,1] as An→0, and for v≠0, we can select χnv∈[0,d]. But this is impossible since 0 is the only solution in [0,d]. Therefore the solution u of (18) is unique in C[0,d] if d>0 exists that satisfies (19).Theorem 2.
There exists a solutionu of (1), where u∈C[0,d] provided that
(30)Nb∑j=1rj|bj|(∥gj∥∞+Kjd)j-1Kj<1,∑j=1r|bj|(∥gj∥∞+Kjd)<d,
where Nb is the number of nonzero bj.Proof.
LetT′v(t)=∑j=1rbj(gj(t)+∫0t‍kj(t,s)v(s)ds)j and v1,v2∈[0,d] for a suitable d, and consider
(31)T′v2(t)-T′v1(t)=∑j=1rbj[(gj(t)+∫0tkj(t,s)v2(s)ds)jhhhhhhhhh-(gj(t)+∫0tkj(t,s)v1(s)ds)j]=∑j=1r‍bj[(gj(t)+∫0tkj(t,s)v1(s)jds)j-1-i(gj(t)+∫0tkj(t,s)v2(s)ds-gj(t)hhhhhhhhhhh-∫0tkj(t,s)v1(s)ds)·∑i=0j-1(gj(t)+∫0tkj(t,s)v2(s)jds)·(gj(t)+∫0tkj(t,s)v1(s)jds)j-1-i]=def∑j=1rbjFj(t,v1,v2)∫0tkj(t,s)(v2(s)-v1(s))ds.
So
(32)∥T′v2-T′v1∥∞≤Nb∑j=1rbjsup⁡Fj(t,v1,v2)·Kj∥v2-v1∥∞;∥Fj∥∞≤j(∥gj∥∞+Kjd)j-1
therefore
(33)∥T′v2-T′v1∥∞≤Nb∑j=1rjbj(∥gj∥∞+Kjd)j-1Kj∥v2-v1∥∞.
ConsequentlyT′ is a contraction mapping if
(34)Nb∑j=1rj|bj|(∥gj∥∞+Kjd)j-1Kj<1.
We need to show thatT′:C[0,d]→C[0,d]. Observe that
(35)∥(gj(t)+∫0t‍kj(t,s)v(s)ds)j∥∞≤(∥gj∥∞+Kj∥v∥∞)j.
Therefore(36)∥T′v∥∞≤∑j=1r|bj|(∥gj∥∞+Kj∥v∥∞)j≤∑j=1r|bj|(∥gj∥∞+Kjd)j;
thusT′:C[0,d]→C[0,d] if
(37)∑j=1r|bj|(∥gj∥∞+Kjd)j<d.
Hence the map T′ is a contraction and maps [0,d] into itself provided (30) is satisfied.Corollary 3.
There exists a solutionu to the integral equation
(38)u(t)=b(g(t)+∫0t‍k(t,s)u(s)ds)2,
where u∈C[0,d] with
(39)0<d<1K[12|b|K-∥g∥∞],
if
(40)∥g∥∞<14K|b|.Proof.
From Theorem2 we get sufficient conditions for the existence of a solution
(41)2K|b|(∥g∥∞+Kd)<1,(42)|b|(∥g∥∞+Kd)2<d.
Inequality (41) is solved by any d<d-, where
(43)d-=1K[12|b|K-∥g∥∞],
and inequality (42) is equivalent to
(44)|b|K2d2+(2K|b|∥g∥∞-1)d+|b|∥g∥∞2<0.
This is satisfied by d∈(d-,d+), where
(45)d±=1-2K|b|∥g∥∞±1-4K|b|∥g∥∞2|b|K2.
If the regularity condition(46)1-4K|b|∥g∥∞>0
is satisfied, d+ and d- are real and positive. Furthermore,
(47)d-=d-+d+2,
so that (46) ensures d∈(d-,d-) satisfies both inequalities (41) and (42) in Theorem 2.


## 4. Numerical Computations

In our work we consider examples of (1) when r=2. We use (6) to approximate the solutions considering two special cases: c1=1/2 (implicit midpoint method) and c1=1 (implicit Euler method). We also use the repeated trapezoidal and repeated Simpson’s rule. Since the methods are implicit we perform an iterative procedure at each step implementing a tolerance of 10-4. For each method we used three different values of h: h=0.01, h=0.005, and h=0.0025.


### 4.1. Example  1

Consider the nonlinear VIE(48)u(t)=2(1+∫0t‍(t-s)u(s)ds)20≤t≤1,
which arises from a nonlinear differential equation in [15] where b1=0 and b2=2.


#### 4.1.1. Using Implicit Euler Method

Whenc1=1 and tn,1=tn+h, the collocation solution of (48) is given by
(49)Un,1=2(1+Fn(tn,1)+Un,1h22)2,
where
(50)Fn(tn,1)=h∑i=0n-1‍(tn-ti+h2)Ui,1.Figure1 shows the solution to (48) at h=0.01, h=0.005, and h=0.0025.Figure 1
The collocation solution of (48) when c1=1.


#### 4.1.2. Using Implicit Midpoint Method

Whenc1=1/2 and tn,1=tn+(h/2), the collocation solution of (48) is given by
(51)Un,1=2(1+Fn(tn,1)+Un,1h28)2,
where
(52)Fn(tn,1)=h∑i=0n-1‍(tn-ti)Ui,1.Figure2 shows the solution to (48) at h=0.01, h=0.005, and h=0.0025.Figure 2
The collocation solution of (48) when c1=1/2.


#### 4.1.3. Using the Iterated Collocation

Forc1=1/2 the iterated collocation solution of (48) is given as
(53)uI(tn)=2(1+h∑i=0n-1‍∫01‍(tn-ti-sh)dsUi1)2.Integrate to obtain(54)uI(tn)=2(1+h∑i=0n-1‍(tn-ti-h2)Ui1)2.The iterated collocation solution of (48) with three different values of h is shown in Figure 3.Figure 3
The iterated collocation solution of (48) when c1=1/2.


#### 4.1.4. Using Repeated Trapezoidal Rule

For the VIE (48) u(0)=2 and
(55)u(tn)=2(1+h2tnu(0)+h∑i=1n-1‍(tn-tn-1)un-1)2.Figure4 shows the solution to the VIE (48) for the three values of h used.Figure 4
The solution of (48) by the repeated trapezoidal rule.


#### 4.1.5. Using Repeated Simpson’s Rule

Whent=0, u(0)=2 for (48) and
(56)u(t2m)=2[1+h3((3t2m-2t1)u(0)∑l=0m-1(3t2m-t2l+1-tl-t2l-1)u(t2l)hhhhhhhhh+(2t2m-2t2m-1)u(t2m))hhhh+2h3∑l=0m-1(3t2m-t2l+1-tl-t2l-1)u(t2l)]2.The solution to (48) using repeated Simpson’s rule is shown in Figure 5.Figure 5
The solution of (48) by the repeated Simpson’s rule.Table1 shows the errors in the solution of the integral equation (48) for the largest value of h used.Table 1
Absolute errors in the solution of (48) when h=0.01.
t
Implicit euler
Implicit midpoint
Repeated trapezoidal
Repeated Simposon
0.1
0.0077
0.0038
0.0074
—
0.2
0.0161
0.0078
0.0154
0.0001
0.3
0.0266
0.0127
0.0252
—
0.4
0.0406
0.0190
0.0377
0.0001
0.5
0.0607
0.0275
0.0548
0.0005
0.6
0.0907
0.0397
0.0180
0.0004
0.7
0.1371
0.0576
0.1146
0.0007
0.8
0.2115
0.0849
0.1689
0.0013
0.9
0.3360
0.1280
0.2549
0.0023
1
0.5530
0.1992
0.3966
0.0044


### 4.2. Example  2

Consider(57)u(t)=(1+∫0t‍(t-s)u(s)ds)+12(1+∫0t‍(t-s)u(s)ds)20≤t≤1,
where b1=1 and b2=1/2. The integral equation (57) arises from nonlinear differential equations that represent conservative systems (see [16]). We used the four methods to approximate the solution to this example and Example 3, and we present tables for the absolute errors in the solution. Table 2 shows the errors in the solution of (57) when h=0.01:Table 2
Absolute errors in the solution of (57) when h=0.01.
t
Implicit euler
Implicit midpoint
Repeated trapezoidal
Repeated Simposon
0.1
0.0028
0.0014
0.0027
—
0.2
0.0057
0.0028
0.0055
—
0.3
0.0089
0.0043
0.0086
—
0.4
0.0127
0.0061
0.0121
0.0001
0.5
0.0171
0.0081
0.0162
0.0001
0.6
0.0226
0.0105
0.0209
0.0001
0.7
0.0295
0.0135
0.0239
—
0.8
0.0385
0.0172
0.0342
0.0001
0.9
0.0501
0.0219
0.0436
0.0001
1.0
0.0659
0.0279
0.0553
0.0002


### 4.3. Example  3

Consider the integral equation(58)u(t)=2(1+∫0t‍(t-s)u(s)ds)+(1+∫0t‍(t-s)u(s)ds)20≤t≤1,
where b1=2 and b2=1. The nonlinear VIE arises from a nonlinear differential equation in [17]. Shown in Table 3 are the errors in the solution of (58) when h=0.01.Table 3
Absolute errors in the solution of (58) when h=0.01.
t
Implicit euler
Implicit midpoint
Repeated trapezoidal
Repeated Simposon
0.1
0.0116
0.0057
0.0110
—
0.2
0.0239
0.0116
0.0229
0.0001
0.3
0.0389
0.0187
0.0369
—
0.4
0.0586
0.0274
0.0543
0.0002
0.5
0.0857
0.0390
0.0774
0.0003
0.6
0.1247
0.0549
0.1089
0.0005
0.7
0.1829
0.0773
0.1537
0.0009
0.8
0.2727
0.1104
0.2196
0.0016
0.9
0.4163
0.1607
0.3196
0.0028
1.0
0.6541
0.2399
0.4771
0.0049


## 4.1. Example  1

Consider the nonlinear VIE(48)u(t)=2(1+∫0t‍(t-s)u(s)ds)20≤t≤1,
which arises from a nonlinear differential equation in [15] where b1=0 and b2=2.


### 4.1.1. Using Implicit Euler Method

Whenc1=1 and tn,1=tn+h, the collocation solution of (48) is given by
(49)Un,1=2(1+Fn(tn,1)+Un,1h22)2,
where
(50)Fn(tn,1)=h∑i=0n-1‍(tn-ti+h2)Ui,1.Figure1 shows the solution to (48) at h=0.01, h=0.005, and h=0.0025.Figure 1
The collocation solution of (48) when c1=1.


### 4.1.2. Using Implicit Midpoint Method

Whenc1=1/2 and tn,1=tn+(h/2), the collocation solution of (48) is given by
(51)Un,1=2(1+Fn(tn,1)+Un,1h28)2,
where
(52)Fn(tn,1)=h∑i=0n-1‍(tn-ti)Ui,1.Figure2 shows the solution to (48) at h=0.01, h=0.005, and h=0.0025.Figure 2
The collocation solution of (48) when c1=1/2.


### 4.1.3. Using the Iterated Collocation

Forc1=1/2 the iterated collocation solution of (48) is given as
(53)uI(tn)=2(1+h∑i=0n-1‍∫01‍(tn-ti-sh)dsUi1)2.Integrate to obtain(54)uI(tn)=2(1+h∑i=0n-1‍(tn-ti-h2)Ui1)2.The iterated collocation solution of (48) with three different values of h is shown in Figure 3.Figure 3
The iterated collocation solution of (48) when c1=1/2.


### 4.1.4. Using Repeated Trapezoidal Rule

For the VIE (48) u(0)=2 and
(55)u(tn)=2(1+h2tnu(0)+h∑i=1n-1‍(tn-tn-1)un-1)2.Figure4 shows the solution to the VIE (48) for the three values of h used.Figure 4
The solution of (48) by the repeated trapezoidal rule.


### 4.1.5. Using Repeated Simpson’s Rule

Whent=0, u(0)=2 for (48) and
(56)u(t2m)=2[1+h3((3t2m-2t1)u(0)∑l=0m-1(3t2m-t2l+1-tl-t2l-1)u(t2l)hhhhhhhhh+(2t2m-2t2m-1)u(t2m))hhhh+2h3∑l=0m-1(3t2m-t2l+1-tl-t2l-1)u(t2l)]2.The solution to (48) using repeated Simpson’s rule is shown in Figure 5.Figure 5
The solution of (48) by the repeated Simpson’s rule.Table1 shows the errors in the solution of the integral equation (48) for the largest value of h used.Table 1
Absolute errors in the solution of (48) when h=0.01.
t
Implicit euler
Implicit midpoint
Repeated trapezoidal
Repeated Simposon
0.1
0.0077
0.0038
0.0074
—
0.2
0.0161
0.0078
0.0154
0.0001
0.3
0.0266
0.0127
0.0252
—
0.4
0.0406
0.0190
0.0377
0.0001
0.5
0.0607
0.0275
0.0548
0.0005
0.6
0.0907
0.0397
0.0180
0.0004
0.7
0.1371
0.0576
0.1146
0.0007
0.8
0.2115
0.0849
0.1689
0.0013
0.9
0.3360
0.1280
0.2549
0.0023
1
0.5530
0.1992
0.3966
0.0044


## 4.1.1. Using Implicit Euler Method

Whenc1=1 and tn,1=tn+h, the collocation solution of (48) is given by
(49)Un,1=2(1+Fn(tn,1)+Un,1h22)2,
where
(50)Fn(tn,1)=h∑i=0n-1‍(tn-ti+h2)Ui,1.Figure1 shows the solution to (48) at h=0.01, h=0.005, and h=0.0025.Figure 1
The collocation solution of (48) when c1=1.


## 4.1.2. Using Implicit Midpoint Method

Whenc1=1/2 and tn,1=tn+(h/2), the collocation solution of (48) is given by
(51)Un,1=2(1+Fn(tn,1)+Un,1h28)2,
where
(52)Fn(tn,1)=h∑i=0n-1‍(tn-ti)Ui,1.Figure2 shows the solution to (48) at h=0.01, h=0.005, and h=0.0025.Figure 2
The collocation solution of (48) when c1=1/2.


## 4.1.3. Using the Iterated Collocation

Forc1=1/2 the iterated collocation solution of (48) is given as
(53)uI(tn)=2(1+h∑i=0n-1‍∫01‍(tn-ti-sh)dsUi1)2.Integrate to obtain(54)uI(tn)=2(1+h∑i=0n-1‍(tn-ti-h2)Ui1)2.The iterated collocation solution of (48) with three different values of h is shown in Figure 3.Figure 3
The iterated collocation solution of (48) when c1=1/2.


## 4.1.4. Using Repeated Trapezoidal Rule

For the VIE (48) u(0)=2 and
(55)u(tn)=2(1+h2tnu(0)+h∑i=1n-1‍(tn-tn-1)un-1)2.Figure4 shows the solution to the VIE (48) for the three values of h used.Figure 4
The solution of (48) by the repeated trapezoidal rule.


## 4.1.5. Using Repeated Simpson’s Rule

Whent=0, u(0)=2 for (48) and
(56)u(t2m)=2[1+h3((3t2m-2t1)u(0)∑l=0m-1(3t2m-t2l+1-tl-t2l-1)u(t2l)hhhhhhhhh+(2t2m-2t2m-1)u(t2m))hhhh+2h3∑l=0m-1(3t2m-t2l+1-tl-t2l-1)u(t2l)]2.The solution to (48) using repeated Simpson’s rule is shown in Figure 5.Figure 5
The solution of (48) by the repeated Simpson’s rule.Table1 shows the errors in the solution of the integral equation (48) for the largest value of h used.Table 1
Absolute errors in the solution of (48) when h=0.01.
t
Implicit euler
Implicit midpoint
Repeated trapezoidal
Repeated Simposon
0.1
0.0077
0.0038
0.0074
—
0.2
0.0161
0.0078
0.0154
0.0001
0.3
0.0266
0.0127
0.0252
—
0.4
0.0406
0.0190
0.0377
0.0001
0.5
0.0607
0.0275
0.0548
0.0005
0.6
0.0907
0.0397
0.0180
0.0004
0.7
0.1371
0.0576
0.1146
0.0007
0.8
0.2115
0.0849
0.1689
0.0013
0.9
0.3360
0.1280
0.2549
0.0023
1
0.5530
0.1992
0.3966
0.0044


## 4.2. Example  2

Consider(57)u(t)=(1+∫0t‍(t-s)u(s)ds)+12(1+∫0t‍(t-s)u(s)ds)20≤t≤1,
where b1=1 and b2=1/2. The integral equation (57) arises from nonlinear differential equations that represent conservative systems (see [16]). We used the four methods to approximate the solution to this example and Example 3, and we present tables for the absolute errors in the solution. Table 2 shows the errors in the solution of (57) when h=0.01:Table 2
Absolute errors in the solution of (57) when h=0.01.
t
Implicit euler
Implicit midpoint
Repeated trapezoidal
Repeated Simposon
0.1
0.0028
0.0014
0.0027
—
0.2
0.0057
0.0028
0.0055
—
0.3
0.0089
0.0043
0.0086
—
0.4
0.0127
0.0061
0.0121
0.0001
0.5
0.0171
0.0081
0.0162
0.0001
0.6
0.0226
0.0105
0.0209
0.0001
0.7
0.0295
0.0135
0.0239
—
0.8
0.0385
0.0172
0.0342
0.0001
0.9
0.0501
0.0219
0.0436
0.0001
1.0
0.0659
0.0279
0.0553
0.0002


## 4.3. Example  3

Consider the integral equation(58)u(t)=2(1+∫0t‍(t-s)u(s)ds)+(1+∫0t‍(t-s)u(s)ds)20≤t≤1,
where b1=2 and b2=1. The nonlinear VIE arises from a nonlinear differential equation in [17]. Shown in Table 3 are the errors in the solution of (58) when h=0.01.Table 3
Absolute errors in the solution of (58) when h=0.01.
t
Implicit euler
Implicit midpoint
Repeated trapezoidal
Repeated Simposon
0.1
0.0116
0.0057
0.0110
—
0.2
0.0239
0.0116
0.0229
0.0001
0.3
0.0389
0.0187
0.0369
—
0.4
0.0586
0.0274
0.0543
0.0002
0.5
0.0857
0.0390
0.0774
0.0003
0.6
0.1247
0.0549
0.1089
0.0005
0.7
0.1829
0.0773
0.1537
0.0009
0.8
0.2727
0.1104
0.2196
0.0016
0.9
0.4163
0.1607
0.3196
0.0028
1.0
0.6541
0.2399
0.4771
0.0049


## 5. Discussion

We approximated the solutions to Examples  1–3 using the implicit Euler method, implicit midpoint method, and repeated trapezoidal and repeated Simpson’s rule using various values of the stepsize. Ath=0.001 and below we obtained a similar solution from all the methods used; hence we take that as our “exact” solution. Therefore, for sufficiently small h we get a good accuracy of the numerical solutions. When the stepsize is greater than 0.001 we obtained different numerical solutions from each of the four methods. We use the “exact” solution and absolute error to study the performance of each method when the stepsize is increased.Tables1–3 show the absolute errors in the solutions when h=0.01. From these tables we observe that the repeated Simpson’s rule performs better followed by the implicit midpoint method then the repeated trapezoidal rule. Among the four methods used, the implicit Euler method gives a larger error as h is increased. We then found an iterated collocation solution for the implicit midpoint method and the accuracy of the method improved as shown in Figure 3. According to our numerical results, we conclude that the repeated Simpson’s rule performs well since it gives better solutions when a reasonably large value of the stepsize is used. These observations are consistent for all three examples used.



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*Source: 652631-2014-07-09.xml*