# Study of Nonlocal Boundary Value Problem for the Fredholm–Volterra Integro-Differential Equation

Authors: K. R. Raslan; Khalid K. Ali; Reda Gamal Ahmed; Hind K. Al-Jeaid; Amira Abd-Elall Ibrahim
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, the existence and uniqueness of the Fredholm–Volterra integro-differential equation with the nonlocal condition will be studied. Also, we study the continuous dependence of the initial data. The numerical solution of the problem will be studied using the central difference approximations and trapezoidal rule to transform the Volterra–Fredholm integro-differential equation into a system of algebraic equations which can be solved together to get the solution. Finally, we solve some examples numerically to show the accuracy of the proposed method.


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## Body

## 1. Introduction

Recently, some researchers were interested in studying the existence and uniqueness of different types of integro-differential equation with the different conditions. El-Sayed et al. studied the existence of solutions to some integro-differential equations with infinite point and integral conditions, and they have also studied some properties of these solutions [1–4]. There are also many authors interested in studying the numerical solution for integral and integro-differential equations. Mirzaee and Piroozfar used modified Simpson’s quadrature rule for solving linear Fredholm integral equations of the second kind [5]. Rahman et al. solved the system of linear Volterra Integral equations of the second kind using Simpson’s quadrature rule [6]. Garba and Bichi studied the numerical solution for first-order Fredholm integro-differential equation using finite difference-composite Simpson method [7]. Ibrahim et al. studied the existence of a unique solution to nonlinear Fredholm integro-differential equation of the second order, and they introduced the exact solution using the direct computation method, introduced numerical solution using the combination of the finite difference method with the composite Simpson method to transform the Fredholm integro-differential equation into a system of nonlinear algebraic equations, and also computed the error estimation for the scheme to show the accuracy of the presented method [8]. Pandey used the finite difference method and the composite trapezoidal quadrature method to solve the Fredholm integro-differential equation [9]. Saadati et al. solved the linear Volterra and Fredholm integro-differential equation using the combination of the trapezoidal rule and the finite difference method and compared it with the variational iteration method (VIM). The result of comparison shows that VIM is better than the trapezoidal method [10]. Ishak and Norazura Ahmed obtained the numerical solution for the first-order Volterra integro-differential equation using the trapezoidal method and compared the results with the Euler method. The results of comparisons show that the trapezoidal method is better than the Euler method [11]. Raftari used the homotopy perturbation method (HPM) and the finite difference method to solve the Volterra integro-differential equation of the first order. The results of applying these methods demonstrate the validity and applicability of these techniques.In this paper, we study the nonlocal boundary value problem for the Fredholm–Volterra integro-differential equation:(1)u″x=Fx,ux,∫abfx,t,u′tx,t,u′tdt,∫axgx,t,u′tdt,x∈a,b,with the nonlocal condition(2)∑j=0majuτj=μ0,u′a=ρ0,aj≥0,τj∈a,b.We study the existence of solutionux∈Ca,b. We study the continuous dependence of the unique solution on μ0 and on the nonlocal parameter aj.As applications, the nonlocal problem of the Fredholm–Volterra integro-differential equation (1) with the integral condition(3)∫abusdϕs=μ0will be studied.This paper is organized as follows. In Section2, we discuss the integral representation. We discuss the existence of solution and the nonlocal integral condition in Section 3. We discuss the uniqueness of the solution in Section 4. In Section 5, we discuss the continuous dependence on μ0 and aj. In Section 6, we present the methodology of numerical technique and numerical examples. Section 7 gives the conclusion.


## 2. Integral Representation

Consider nonlocal problems (1) and (2) with the following assumptions:(1)
F:a,b×ℝ3⟶ℝ satisfies Caratheodory condition, i.e., F is measurable in x for any ξ,α,γ∈ℝ and continuous for almost all x∈a,b. There exist a function M1x∈L1a,b and a positive constant C1>0, such that(4)Fx,ξ,α,γ≤M1x+C1ξ+C1α+C1γ.(2)
f:a,b×a,b×ℝ⟶ℝ satisfies Caratheodory condition, i.e., f is measurable in x for any vt∈ℝ and continuous for almost all x∈a,b. There exist a function M2x,t∈L1a,b and a positive constant C2>0, such that(5)fx,t,vt≤M2x,t+C2vt.(3)
g:a,b×a,b×ℝ⟶ℝ satisfies Caratheodory condition. There exist a function M3x,t∈L1a,b and a positive constant C3>0, such that(6)gx,t,vt≤M3x,t+C3vt.(4)
(7)supx∈a,b∫axM1θdθ≤N1,supθ∈a,b∫abM2θ,tdt≤N2,supθ∈a,b∫aθM3θ,tdt≤N3.(5)
2C1b2+C1C2b2+C1C3b2<1.Lemma 1.
Letβ=∑j=0maj≠0, and we can represent the solution of nonlocal problems (1) and (2), if it exists by the integral equation(8)ux=β−1μ0−∑j=0maj∫aτjvsds+∫axvsds,where(9)vx=ρ0+∫axFθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ.Proof.
Integrating both sides of (1), we get(10)u′x=u′a+∫axFθ,uθ,∫abfθ,t,u′tdt,∫aθgθ,t,u′tdtdθ,x∈a,b.
Letu′x=vx in (10), and we obtain(11)vx=ρ0+∫axFθ,uθ,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ,x∈a,b,where(12)ux=ua+∫axvsds,x∈a,b,and using nonlocal condition (2), we get(13)∑j=0majuτj=ua∑j=0maj+∑j=0maj∫aτjvsds,and then(14)ua=β−1μ0−∑j=0maj∫aτjvsds.
We obtain (8) and (9) from (11), (12), and (14). This completes the proof.


## 3. Existence of Solution

Definition 1.
By a solution of Fredholm–Volterra integral equation (9), we mean a functionux∈Ca,bthat satisfies (5).Theorem 1.
Let the assumptions (1)–(5) hold. Then, Fredholm–Volterra integral equation (9) has at least one solutionux∈Ca,b.Proof.
Define the operatorE associated with integral equation (9) by(15)Evx=ρ0+∫axFθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ.
LetQr=vx∈ℝ:vC≤r, where r=ρ0+N1+C1bβ−1μ0+C1bN2+C1bN3/1−2C1b2+C1C2b2+C1C3b2.
Then, we have that forvx∈Qr,(16)EvxC≤ρ0+∫axFθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ,≤ρ0+∫axM1θ+C1β−1μ0−∑j=0maj∫aτjvsds+C1∫aθvsds+C1∫abfθ,t,vtdt+C1∫aθgθ,t,vtdtdθ≤ρ0+N1+∫axC1β−1μ0+C1β−1∑j=0maj∫aτjvsds+C1∫aθvsds+C1∫abM2θ,tdt+C1C2∫abvtdt+C1∫aθM3θ,tdt+C1C3∫aθvtdtdθ≤ρ0+N1+∫axC1β−1μ0+C1bv+C1bv+C1N2+C1C2bv+C1N3+C1C3bvdθ≤ρ0+N1+C1bβ−1μ0+2C1b2r+C1bN2+C1C2b2r+C1bN3+C1C3b2r=r.
This proves thatE:Qr⟶Qr and the class of functions Evx is uniformly bounded in Qr.
Now, letx1,x2∈a,b such that x2−x1<δ; then,(17)Evx2−Evx1=ρ0+∫ax2Fθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ−ρ0−∫ax1Fθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ=∫ax1Fθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ+∫x1x2Fθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ−∫ax1Fθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ≤∫x1x2Fθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ≤∫x1x2M1θdθ+C1β−1μ0+2C1br+C1N2+C1C2br+C1N3+C1C3brδ.
This means that the class of functionsEvx is equicontinuous in Qr.Letvnx∈Qr, vnx⟶vxn⟶∞; then, from the continuity of the three functions F,f, and g, we obtain Fx,ξn,αn,γn⟶Fx,ξ,α,γ,fx,t,vnt⟶fx,t,vt and gx,t,vnt⟶gx,t,vt as n⟶∞. Also,(18)limn⟶∞Evnx=limn⟶∞ρ0+∫axFθ,β−1μ0−∑j=0maj∫aτjvnsds+∫aθvnsds,∫abfθ,t,vntdt,∫aθgθ,t,vntdtdθ.Using assumptions (1)–(3) and Lebesgue dominated convergence theorem [13], we obtain(19)limn⟶∞Evnx=ρ0+∫axlimn⟶∞Fθ,β−1μ0−∑j=0maj∫aτjvnsds+∫aθvnsds,∫abfθ,t,vntdt,∫aθgθ,t,vntdtdθ=Evx.Then,Evnx⟶Evx as n⟶∞. This means that the operator E is continuous in Qr. Then, by Schauder fixed point theorem [14], there exists at least one solution vx∈Ca,b of integral equation (9). Thus, based on Lemma 1, nonlocal problems (1) and (2) possess a solution ux∈Ca,b.


### 3.1. Nonlocal Integral Condition

Letvx∈Ca,b be the solution of integral equation (9). Let aj=ϕxj−ϕxj−1, ϕ be increasing function, τj∈xj−1,xj, and a=x0<x1<x2<…<xN=b; then, as m⟶∞, nonlocal condition (2) will be(20)∑j=0mϕxj−ϕxj−1uτj=μ0,(21)limm⟶∞∑j=0mϕxj−ϕxj−1uτj=∫abusds=μ0.Theorem 2.
Let the assumptions (1)–(5) hold; then, nonlocal problems (1) and (3) have at least one solution given by(22)ux=1ϕb−ϕaμ0−∫ab∫aθvsdsdϕθ+∫axvsds,where(23)vx=ρ0+∫axFθ,1ϕb−ϕaμ0−∫ab∫aθvsdsdϕθ+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ.Proof.
Asm⟶∞, the solution of nonlocal problems (1) and (3) will be(24)ux=limm⟶∞β−1μ0−∑j=0maj∫aτjvsds+∫axvsds=1ϕb−ϕaμ0−limm⟶∞∑j=0m∫aτjvsdsϕxj−ϕxj−1+∫axvsds=1ϕb−ϕaμ0−∫ab∫aθvsdsdϕθ+∫axvsds,where(25)vx=ρ0+∫axFθ,1ϕb−ϕaμ0−∫ab∫aθvsdsdϕθ+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ.


## 3.1. Nonlocal Integral Condition

Letvx∈Ca,b be the solution of integral equation (9). Let aj=ϕxj−ϕxj−1, ϕ be increasing function, τj∈xj−1,xj, and a=x0<x1<x2<…<xN=b; then, as m⟶∞, nonlocal condition (2) will be(20)∑j=0mϕxj−ϕxj−1uτj=μ0,(21)limm⟶∞∑j=0mϕxj−ϕxj−1uτj=∫abusds=μ0.Theorem 2.
Let the assumptions (1)–(5) hold; then, nonlocal problems (1) and (3) have at least one solution given by(22)ux=1ϕb−ϕaμ0−∫ab∫aθvsdsdϕθ+∫axvsds,where(23)vx=ρ0+∫axFθ,1ϕb−ϕaμ0−∫ab∫aθvsdsdϕθ+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ.Proof.
Asm⟶∞, the solution of nonlocal problems (1) and (3) will be(24)ux=limm⟶∞β−1μ0−∑j=0maj∫aτjvsds+∫axvsds=1ϕb−ϕaμ0−limm⟶∞∑j=0m∫aτjvsdsϕxj−ϕxj−1+∫axvsds=1ϕb−ϕaμ0−∫ab∫aθvsdsdϕθ+∫axvsds,where(25)vx=ρ0+∫axFθ,1ϕb−ϕaμ0−∫ab∫aθvsdsdϕθ+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdtdθ.


## 4. Uniqueness of the Solution

LetF,f, and g satisfy the following assumptions:(i)
F:a,b×ℝ3⟶ℝ is measurable in x for any ξ,α,γ∈ℝ and satisfies the Lipschitz condition(26)Fx,ξ,α,γ−Fx,ν,α1,γ1≤C1ξ−ν+C1α−α1+C1γ−γ1.(ii)
f:a,b×a,b×ℝ⟶ℝ is measurable in x for any vt∈ℝ and satisfies the Lipschitz condition(27)|fx,t,vt−fx,t,wt|≤C2|vt−wt|.(iii)
g:a,b×a,b×ℝ⟶ℝ is measurable in x for any vt∈ℝ and satisfies the Lipschitz condition(28)gx,t,vt−gx,t,wt≤C3vt−wt.Theorem 3.
Let the assumptionsi−iiihold; then, the solution of Fredholm–Volterra integral equation (9) is unique.Proof.
Letvx,wx be two solutions of Fredholm–Volterra integral equation (9); then,(29)vx−wx≤∫ax|Fθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθfθ,t,vtdtdθ−Fθ,β−1μ0−∑j=0maj∫aτjwsds+∫aθwsds,∫abfθ,t,wtdt,∫aθgθ,t,wtdt|dθ≤∫axC1β−1∑j=0maj∫aτjws−vsds+∫aθvs−wsds+C1∫abfθ,t,vt−fθ,t,wtdt+C1∫aθgθ,t,vt−gθ,t,wtdt≤C1∫axβ−1∑j=0maj∫aτjws−vsds+∫aθws−vsds+∫abfθ,t,vt−fθ,t,wtdt+∫aθ∫abgθ,t,vt−gθ,t,wtdtdθ≤C1w−vb2+C1w−vb2+C1∫ax∫abC2vt−wtdtdθ+C1∫ax∫aθC3vt−wtdtdθ≤2C1w−vb2+C1C2b2w−v+C1C3b2w−v≤2C1b2+C1C2b2+C1C3b2w−v.
Hence,(30)1−2C1b2+C1C2b2+C1C3b2w−v≤0.
Since2C1b2+C1C2b2+C1C3b2<1, then wx=vx and the solution of Fredholm–Volterra integral equation (9) is unique. Thus, based on Lemma 1, nonlocal problems (1) and (2) possess a unique solution ux∈Ca,b.


## 5. Continuous Dependence

### 5.1. Continuous Dependence onμ0

Definition 2.
The solutionux∈Ca,bof nonlocal Fredholm–Volterra problems (1) and (2) depends continuously onμ0, if(31)∀ε>0,∃δεs.tμ0−μ0∗<δ⇒u−u∗<ε,where u∗ is the solution of the nonlocal problem(32)u∗″x=Fx,u∗x,∫abfx,t,u∗′tdt,∫axgx,t,u∗′tdt,x∈a,b,with the nonlocal condition(33)∑j=0maju∗τj=μ0∗,u∗′a=ρ0,aj⩾0,τj∈a,b.Theorem 4.
Let the assumptions (1)–(5) of Theorem 1hold; then, the solution of nonlocal Fredholm–Volterra problems (1) and (2) depends continuously onμ0.Proof.
Letux, u∗x be two solutions of nonlocal Fredholm–Volterra problems (1) and (2) and (23)–(33), respectively. Then,(34)vx−v∗x=|∫axFθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdt−Fθ,β−1μ0∗−∑j=0maj∫aτjv∗sds+∫aθv∗sds,∫abfθ,t,v∗tdt,∫aθgθ,t,v∗tdtdθ|≤∫ax|Fθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdt−Fθ,β−1μ0∗−∑j=0maj∫aτjv∗sds+∫aθv∗sds,∫abfθ,t,v∗tdt,∫aθgθ,t,v∗tdt|dθ≤∫axC1β−1μ0−μ0∗+β−1∑j=0maj∫aτjv∗s−vsds+∫aθvs−v∗sds+C1∫abfθ,t,vt−fθ,t,v∗tdt+C1∫aθgθ,t,vt−gθ,t,v∗tdtdθ≤∫axC1β−1μ0−μ0∗+C1β−1∑j=0maj∫aτjv∗s−vsds+C1∫aθvs−v∗sds+C1∫abfθ,t,vt−fθ,t,v∗tdt+C1∫aθgθ,t,vt−gθ,t,v∗tdtdθ≤C1β−1μ0−μ0∗b+C1v−v∗b2+C1v−v∗b2+C1∫ax∫abC2vt−v∗tdt+C1∫ax∫aθC3vt−v∗tdtdθ≤C1bβ−1δ+2C1v−v∗b2+C1C2b2v−v∗+C1C3b2v−v∗.
Hence,(35)v−v∗≤C1bβ−1δ1−2C1b2+C1C2b2+C1C3b2.
Since(36)ux−u∗x=β−1μ0−∑j=0maj∫aτjvsds+∫axvsds−β−1μ0∗−∑j=0maj∫aτjv∗sds+∫axv∗sdsβ−1μ0−μ0∗+2bv−v∗,then(37)u−u∗≤β−1δ+2C1b2β−1δ1−2C1b2+C1C2b2+C1C3b2=ε.
Therefore, the solution of nonlocal Fredholm–Volterra problems (1) and (2) depends continuously on μ0.


### 5.2. Continuous Dependence onaj

Definition 3.
The solutionux∈Ca,bof nonlocal Fredholm–Volterra problems (1) and (2) depends continuously onaj, if(38)∀ε>0,∃δεs.taj−aj∗<δ⇒u−u∗<ε,where u∗x is the solution of the nonlocal problem(39)u∗″x=Fx,u∗x,∫abfx,t,u∗′tdt,∫axgx,t,u∗′tdt,x∈a,b,with the nonlocal condition(40)∑j=0maj∗u∗τj=μ0,u∗′a=ρ0,aj⩾0,τj∈a,b.Theorem 5.
Let the assumptions (1)–(5) ofTheorem 1hold; then, the solution of nonlocal problems (1) and (2) depends continuously onaj.Proof.
Letβ∗=∑j=0maj∗≠0 and vx,v∗x be two solutions of nonlocal Fredholm–Volterra problems (1) and (2) and (39)–(40), respectively. Then,(41)vx−v∗x≤∫axFθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdt−Fθ,β∗−1μ0−∑j=0maj∗∫aτjv∗sds+∫aθv∗sds,∫abfθ,t,v∗tdt,∫aθgθ,t,v∗tdtdθ≤∫axC1|β−1μ0−β∗−1μ0+β∗−1∑j=0maj∗∫aτjv∗sds−β−1∑j=0maj∫aτjvsds+∫aθvsds−∫aθv∗sds|+C1∫abfθ,t,vt−fθ,t,v∗tdt+C1∫aθgθ,t,vt−gθ,t,v∗tdtdθ≤∫axC1β−1μ0−β∗−1μ0+C1β∗−1∑j=0maj∗∫aτjv∗s−vsds+C1β∗−1∑j=0maj∗−aj∫aτjvsds+C1β−1β∗−1∑j=0maj−aj∗∑j=0maj∫aτjvsds+C1∫aθvs−v∗sds+C1∫abfθ,t,vt−fθ,t,v∗tdt+C1∫aθgθ,t,vt−gθ,t,v∗tdtdθ≤C1β−1β∗−1mδμ0+C1v−v∗b2+C1β∗−1mδvb2+C1β∗−1mδvb2+C1v−v∗b2+C1C2b2v−v∗+C1C3b2v−v∗≤C1β−1β∗−1mδμ0+2C1β∗−1mδvb2+2C1b2+C1C2b2+C1C3b2v−v∗.
Hence,(42)v−v∗≤C1β−1β∗−1mδμ0+2C1β∗−1mδvb21−2C1b2+C1C2b2+C1C3b2.
Since(43)ux−u∗x=|1∑j=0majμ0−∑j=0maj∫aτjvsds+∫axvsds−1∑j=0maj∗μ0−∑j=0maj∗∫aτjv∗sds+∫axv∗sds|≤mδμ0ββ∗+2mδbβ∗−1r+2bv−v∗,then(44)‖u−u∗‖≤mδ|μ0|ββ∗+2mδbβ∗−1r+2bC1β−1β∗−1mδμ0+2C1β∗−1mδ‖v‖b21−2C1b2+C1C2b2+C1C3b2=ϵ.
Thus, the solution of nonlocal Fredholm–Volterra problems (1) and (2) depends continuously on aj.


## 5.1. Continuous Dependence onμ0

Definition 2.
The solutionux∈Ca,bof nonlocal Fredholm–Volterra problems (1) and (2) depends continuously onμ0, if(31)∀ε>0,∃δεs.tμ0−μ0∗<δ⇒u−u∗<ε,where u∗ is the solution of the nonlocal problem(32)u∗″x=Fx,u∗x,∫abfx,t,u∗′tdt,∫axgx,t,u∗′tdt,x∈a,b,with the nonlocal condition(33)∑j=0maju∗τj=μ0∗,u∗′a=ρ0,aj⩾0,τj∈a,b.Theorem 4.
Let the assumptions (1)–(5) of Theorem 1hold; then, the solution of nonlocal Fredholm–Volterra problems (1) and (2) depends continuously onμ0.Proof.
Letux, u∗x be two solutions of nonlocal Fredholm–Volterra problems (1) and (2) and (23)–(33), respectively. Then,(34)vx−v∗x=|∫axFθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdt−Fθ,β−1μ0∗−∑j=0maj∫aτjv∗sds+∫aθv∗sds,∫abfθ,t,v∗tdt,∫aθgθ,t,v∗tdtdθ|≤∫ax|Fθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdt−Fθ,β−1μ0∗−∑j=0maj∫aτjv∗sds+∫aθv∗sds,∫abfθ,t,v∗tdt,∫aθgθ,t,v∗tdt|dθ≤∫axC1β−1μ0−μ0∗+β−1∑j=0maj∫aτjv∗s−vsds+∫aθvs−v∗sds+C1∫abfθ,t,vt−fθ,t,v∗tdt+C1∫aθgθ,t,vt−gθ,t,v∗tdtdθ≤∫axC1β−1μ0−μ0∗+C1β−1∑j=0maj∫aτjv∗s−vsds+C1∫aθvs−v∗sds+C1∫abfθ,t,vt−fθ,t,v∗tdt+C1∫aθgθ,t,vt−gθ,t,v∗tdtdθ≤C1β−1μ0−μ0∗b+C1v−v∗b2+C1v−v∗b2+C1∫ax∫abC2vt−v∗tdt+C1∫ax∫aθC3vt−v∗tdtdθ≤C1bβ−1δ+2C1v−v∗b2+C1C2b2v−v∗+C1C3b2v−v∗.
Hence,(35)v−v∗≤C1bβ−1δ1−2C1b2+C1C2b2+C1C3b2.
Since(36)ux−u∗x=β−1μ0−∑j=0maj∫aτjvsds+∫axvsds−β−1μ0∗−∑j=0maj∫aτjv∗sds+∫axv∗sdsβ−1μ0−μ0∗+2bv−v∗,then(37)u−u∗≤β−1δ+2C1b2β−1δ1−2C1b2+C1C2b2+C1C3b2=ε.
Therefore, the solution of nonlocal Fredholm–Volterra problems (1) and (2) depends continuously on μ0.


## 5.2. Continuous Dependence onaj

Definition 3.
The solutionux∈Ca,bof nonlocal Fredholm–Volterra problems (1) and (2) depends continuously onaj, if(38)∀ε>0,∃δεs.taj−aj∗<δ⇒u−u∗<ε,where u∗x is the solution of the nonlocal problem(39)u∗″x=Fx,u∗x,∫abfx,t,u∗′tdt,∫axgx,t,u∗′tdt,x∈a,b,with the nonlocal condition(40)∑j=0maj∗u∗τj=μ0,u∗′a=ρ0,aj⩾0,τj∈a,b.Theorem 5.
Let the assumptions (1)–(5) ofTheorem 1hold; then, the solution of nonlocal problems (1) and (2) depends continuously onaj.Proof.
Letβ∗=∑j=0maj∗≠0 and vx,v∗x be two solutions of nonlocal Fredholm–Volterra problems (1) and (2) and (39)–(40), respectively. Then,(41)vx−v∗x≤∫axFθ,β−1μ0−∑j=0maj∫aτjvsds+∫aθvsds,∫abfθ,t,vtdt,∫aθgθ,t,vtdt−Fθ,β∗−1μ0−∑j=0maj∗∫aτjv∗sds+∫aθv∗sds,∫abfθ,t,v∗tdt,∫aθgθ,t,v∗tdtdθ≤∫axC1|β−1μ0−β∗−1μ0+β∗−1∑j=0maj∗∫aτjv∗sds−β−1∑j=0maj∫aτjvsds+∫aθvsds−∫aθv∗sds|+C1∫abfθ,t,vt−fθ,t,v∗tdt+C1∫aθgθ,t,vt−gθ,t,v∗tdtdθ≤∫axC1β−1μ0−β∗−1μ0+C1β∗−1∑j=0maj∗∫aτjv∗s−vsds+C1β∗−1∑j=0maj∗−aj∫aτjvsds+C1β−1β∗−1∑j=0maj−aj∗∑j=0maj∫aτjvsds+C1∫aθvs−v∗sds+C1∫abfθ,t,vt−fθ,t,v∗tdt+C1∫aθgθ,t,vt−gθ,t,v∗tdtdθ≤C1β−1β∗−1mδμ0+C1v−v∗b2+C1β∗−1mδvb2+C1β∗−1mδvb2+C1v−v∗b2+C1C2b2v−v∗+C1C3b2v−v∗≤C1β−1β∗−1mδμ0+2C1β∗−1mδvb2+2C1b2+C1C2b2+C1C3b2v−v∗.
Hence,(42)v−v∗≤C1β−1β∗−1mδμ0+2C1β∗−1mδvb21−2C1b2+C1C2b2+C1C3b2.
Since(43)ux−u∗x=|1∑j=0majμ0−∑j=0maj∫aτjvsds+∫axvsds−1∑j=0maj∗μ0−∑j=0maj∗∫aτjv∗sds+∫axv∗sds|≤mδμ0ββ∗+2mδbβ∗−1r+2bv−v∗,then(44)‖u−u∗‖≤mδ|μ0|ββ∗+2mδbβ∗−1r+2bC1β−1β∗−1mδμ0+2C1β∗−1mδ‖v‖b21−2C1b2+C1C2b2+C1C3b2=ϵ.
Thus, the solution of nonlocal Fredholm–Volterra problems (1) and (2) depends continuously on aj.


## 6. Methodology of Numerical Technique

In this section, we wish to determine the numerical solution of equation (1). We divide the domain a,x and a,b of equation (1) into N finite points as a=x0<x1<…<xN−1<xN=x=b. We use uniform step length h=b−a/N =  xi−a/i,i≥1, as xj=a+jh=tj, j=0,1,2,…,N. Then, we use the trapezoidal rule to approximate the integral parts of (1) as follows [10]:(45)∫abkxi,tju′tjdt≃h2kxi,t0u′t0+2∑j=1N−1kxi,tju′tj+kxi,tNu′tN,(46)∫axiKxi,tju′tjdt≃hi2Kxi,t0u′t0+2∑j=1N−1Kxi,tju′tj+Kxi,tNu′tN,where Kxi,tj=0 for tj≤xi,j≥1.Then, we use central difference approximations to approximate the derivative parts of (1) as(47)ui″≃ui+1−2ui+ui−1h2,ui′≃ui+1−ui−12h,where ui″=u′′xi, ui′=u′xi.


### 6.1. Numerical Examples

Now, we apply Theorem1 on some examples of the nonlocal Fredholm–Volterra integro-differential equation and we solve it numerically by using the finite difference-trapezoidal method. The results obtained are tabulated in Tables 1–4, and all results for these examples are obtained by using Wolfram Mathematica.Table 1
The exact and numerical solutions of example 1.
xiNumerical solutionExact solutionAbsolute error0.01.000071.000007.1195E-50.10.995070.995006.7119E-50.20.980120.980075.5076E-50.30.955370.955343.6189E-50.40.921080.921061.3545E-50.50.877580.877586.6020E-60.60.825320.825341.3545E-50.70.764850.764849.2025E-60.80.696790.696718.5149E-50.90.621860.621612.4584E-41.00.540830.540305.3145E-4Table 2
The exact and numerical solutions of example 2.
xiApproximate solutionExact solutionAbsolute error−1.0−0.845787−0.8414714.31629E-3−0.8−0.720890−0.7173563.53409E-3−0.6−0.567342−0.5646422.69934E-3−0.4−0.391242−0.3894181.82331E-3−0.2−0.199588−0.1986699.18883E-40.00.0000000.0000000.000000.20.1995880.1986699.18883E-40.40.3912420.3894181.82331E-30.60.5673420.5646422.69934E-30.80.7208900.7173563.53409E-31.00.8457870.8414714.31629E-3Table 3
The exact and numerical solutions of example 3.
xiApproximate solutionExact solutionAbsolute error0.00.999671.000003.20617E-40.11.004681.005003.24861E-40.21.019721.020073.49559E-40.31.044931.045344.06892E-40.41.080561.081075.09384E-40.51.126961.127636.70039E-40.61.184561.185479.02486E-40.71.253951.255171.22112E-30.81.335791.337431.64126E-30.91.430911.433092.17931E-31.01.540231.543082.85299E-3Table 4
The exact and numerical solutions of example 4.
xiApproximate solutionExact solutionAbsolute error0.01.000941.9.4143E-40.11.105941.105177.7156E-40.21.2221.22145.9439E-40.31.350271.349864.0876E-40.41.492041.491822.1332E-40.51.648731.648726.5459E-60.61.821911.822122.1332E-40.72.01332.013754.4832E-40.82.224842.225547.0085E-40.92.458632.45969.7372E-41.02.717012.718281.2703E-3Example 1.
Consider the equation(48)u″x−18u2x=1240−2x5−6x2sin2x+9xsin2x+3sin3xcosx−148x32−e−sin1−18cos2x−cosx+148∫01tx+xcosteu′tdt+140∫0xtx+sinxu′2tdt,u0.4+u0.6=1.746,u′0=0.
The exact solution of this problem isux=cosx.
Firstly, we apply the assumptions of Theorem1 to prove that this example has a continuous solution:(49)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt=1240−2x5−6x2sin2x+9xsin2x+3sin3xcosx−148x32−e−sin1−18cos2x−cosx+18u2x+148∫01tx+xcosteu′tdt+140∫0xtx+sinxu′2tdt.
Then,(50)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt≤1240−2x5−6x2sin2x+9xsin2x+3sin3xcosx−148x32−e−sin1−18cos2x−cosx+18u2x+18∫0116tx+xcosteu′tdt+18∫0x15tx+sinxu′2tdt,and also(51)fx,t,u′tdt≤16xt+16eu′t,gx,t,u′tdt≤15xt+15u′2t,where M1x=1/240−2x5−6x2sin2x+9xsin2x+3sin3xcosx−1/48x3/2−e−sin1−1/8cos2x−cosx∈L1a,b, M2x,t=1/6xt∈L1a,b,M3x,t=1/5xt∈L1a,b,C1=1/8,C2=1/6,C3=1/5,b=1; then, 2C1b2+C1C2b2+C1C3b2=2/8+1/48+1/40=71/240<1. It is clear that the assumptions (1)–(5) of Theorem 1 hold; therefore, the given nonlocal problem has a continuous solution.
Now, we use the finite difference-trapezoidal method withN=10 to find the numerical solution of this problem. Table 1 and Figure 1 give the comparison between the numerical and exact solutions of this problem.
Through our observation of Table1, the interval 0,1 was divided into 10 subintervals of equal length. We obtain solutions at the endpoints of subintervals and show that the method used is effective, and this is evident from the absolute error that was calculated for the difference between the numerical and real solutions. Also, by looking at Figure 1, we find that the numerical solution and the real solution are very close, which means that the numerical solutions are good.Figure 1
Comparison between the numerical and exact solutions of example 1.Example 2.
Consider the equation(52)u″x−17ux=156−12x2+2sinx−1cosx−sinxxsinx−sin1−cos1−8sinx7+170∫−11tx+sinxtu′tdt+156∫−1xtcosx+tsinxu′tdt,u−1+u1=0,u′−1=cos−1.
The exact solution of this problem isux=sinx.
Firstly, we apply the assumptions of Theorem1 to prove that this example has a continuous solution:(53)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt=156−12x2+2sinx−1cosx−sinxxsinx−sin1−cos1−8sinx7+17ux+170∫−11tx+sinxtu′tdt+156∫−1xtcosx+tsinxu′tdt.
Then,(54)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt≤156−12x2+2sinx−1cosx−sinxxsinx−sin1−cos1−8sinx7+17ux+17∫−11110tx+sinxtu′tdt+17∫−1x18tcosx+tsinxu′tdt,and also(55)fx,t,u′tdt≤110tx+110u′t,gx,t,u′tdt≤18tcosx+18u′t,where(56)M1x=156−12x2+2sinx−1cosx−sinxxsinx−sin1−cos1−8sinx7∈L1a,b,M2x,t=110tx∈L1a,b,M3x,t=18tcosx∈L1a,b,C1=17,C2=110,C3=18,b=1.
Then,2C1b2+C1C2b2+C1C3b2=2/7+1/70+1/56=89/280<1. It is clear that the assumptions (1)–(5) of Theorem 1 hold; therefore, the given nonlocal problem has a continuous solution.
Now, we use the finite difference-trapezoidal method withN=10 to find the numerical solution of this problem. Table 2 and Figure 2 give the comparison between the numerical and exact solutions of this problem.
Through our observation of Table2, the interval −1,1 was divided into 10 subintervals of equal length. We obtain solutions at the endpoints of subintervals and show that the method used is effective, and this is evident from the absolute error that was calculated for the difference between the numerical and real solutions. Also, by looking at Figure 2, we find that the numerical solution and the real solution are very close, which means that the numerical solutions are good.Figure 2
Comparison between the numerical and exact solutions of example 2.Example 3.
Consider the equation(57)u″x−112ux=160−x32−sinhxxcoshx−sinhx+11coshx12+184−sinhx2−coshxe+184∫01tsinhx+tcoshxu′tdt+160∫0xtx+tsinhxu′tdt,∫01uxdx=sinh1,u′0=0.
The exact solution of this problem isux=coshx.
Firstly, we apply the assumptions of Theorem1 to prove that this example has a continuous solution:(58)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt=160−x32−sinhxxcoshx−sinhx+11coshx12+184−sinhx2−coshxe+112ux+184∫01tsinhx+tcoshxu′tdt+160∫0xtx+tsinhxu′tdt.
Then,(59)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt≤160−x32−sinhxxcoshx−sinhx+11coshx12+184−sinhx2−coshxe+112ux+112∫0117tsinhx+tcoshxu′tdt+112∫0x15tx+tsinhxu′tdt,and also(60)fx,t,u′tdt≤17tsinhx+17u′t,gx,t,u′tdt≤15tx+15u′t,where(61)M1x=160−x32−sinhxxcoshx−sinhx+11coshx12+184−sinhx2−coshxe∈L1a,b,M2x,t=17tsinhx∈L1a,b,M3x,t=15tx∈L1a,b,C1=112,C2=17,C3=15,b=1.
Then,2C1b2+C1C2b2+C1C3b2=2/12+1/84+1/60=41/210<1. It is clear that the assumptions (1)–(5) of Theorem 1 hold; therefore, the given nonlocal problem has a continuous solution.
Now, we use the finite difference-trapezoidal method withN=10 to find the numerical solution of this problem. Table 3 and Figure 3 give the comparison between the numerical and exact solutions of this problem.
Through our observation of Table3, the interval 0,1 was divided into 10 subintervals of equal length. We obtain solutions at the endpoints of subintervals and show that the method used is effective, and this is evident from the absolute error that was calculated for the difference between the numerical and real solutions. Also, by looking at Figure 3, we find that the numerical solution and the real solution are very close, which means that the numerical solutions are good.Figure 3
Comparison between the numerical and exact solutions of example 3.Example 4.
Consider the equation(62)u″x−19u2x=−1144x+2ex−2x2−1324e2x+x+2x+ex−e2x9+181∫01xt+x2tu′2tdt+172∫0xtx+x2u′tdt,u0.4+u0.6=3.31394,u′0=1.
The exact solution of this problem isux=ex.
Firstly, we apply the assumptions of Theorem1 to prove that this example has a continuous solution:(63)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt=−1144x+2ex−2x2−1324e2x+x+2x+ex−e2x9+19u2x+181∫01xt+x2tu′2tdt+172∫0xtx+x2u′tdt.
Then,(64)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt≤−1144x+2ex−2x2−1324e2x+x+2x+ex−e2x9+19u2x+19∫0119xt+x2tu′2tdt+19∫0x18tx+x2u′tdt,and also(65)fx,t,u′tdt≤19tx+19u′t,gx,t,u′tdt≤18tx+18u′t,where(66)M1x=−1144x+2ex−2x2−1324e2x+x+2x+ex−e2x9∈L1a,b,M2x,t=19tx∈L1a,b,M3x,t=18tx∈L1a,b,C1=19,C2=19,C3=18,b=1.
Then,2C1b2+C1C2b2+C1C3b2=2/9+1/81+1/72=161/684<1. It is clear that the assumptions (1)–(5) of Theorem 1 hold; therefore, the given nonlocal problem has a continuous solution.
Now, we use the finite difference-trapezoidal method withN=10 to find the numerical solution of this problem. Table 4 and Figure 4 give the comparison between the numerical and exact solutions of this problem.
Through our observation of Table4, the interval 0,1 was divided into 10 subintervals of equal length. We obtain solutions at the endpoints of subintervals and show that the method used is effective, and this is evident from the absolute error that was calculated for the difference between the numerical and real solutions. Also, by looking at Figure 4, we find that the numerical solution and the real solution are very close, which means that the numerical solutions are good.Figure 4
Comparison between the numerical and exact solutions of example 4.


## 6.1. Numerical Examples

Now, we apply Theorem1 on some examples of the nonlocal Fredholm–Volterra integro-differential equation and we solve it numerically by using the finite difference-trapezoidal method. The results obtained are tabulated in Tables 1–4, and all results for these examples are obtained by using Wolfram Mathematica.Table 1
The exact and numerical solutions of example 1.
xiNumerical solutionExact solutionAbsolute error0.01.000071.000007.1195E-50.10.995070.995006.7119E-50.20.980120.980075.5076E-50.30.955370.955343.6189E-50.40.921080.921061.3545E-50.50.877580.877586.6020E-60.60.825320.825341.3545E-50.70.764850.764849.2025E-60.80.696790.696718.5149E-50.90.621860.621612.4584E-41.00.540830.540305.3145E-4Table 2
The exact and numerical solutions of example 2.
xiApproximate solutionExact solutionAbsolute error−1.0−0.845787−0.8414714.31629E-3−0.8−0.720890−0.7173563.53409E-3−0.6−0.567342−0.5646422.69934E-3−0.4−0.391242−0.3894181.82331E-3−0.2−0.199588−0.1986699.18883E-40.00.0000000.0000000.000000.20.1995880.1986699.18883E-40.40.3912420.3894181.82331E-30.60.5673420.5646422.69934E-30.80.7208900.7173563.53409E-31.00.8457870.8414714.31629E-3Table 3
The exact and numerical solutions of example 3.
xiApproximate solutionExact solutionAbsolute error0.00.999671.000003.20617E-40.11.004681.005003.24861E-40.21.019721.020073.49559E-40.31.044931.045344.06892E-40.41.080561.081075.09384E-40.51.126961.127636.70039E-40.61.184561.185479.02486E-40.71.253951.255171.22112E-30.81.335791.337431.64126E-30.91.430911.433092.17931E-31.01.540231.543082.85299E-3Table 4
The exact and numerical solutions of example 4.
xiApproximate solutionExact solutionAbsolute error0.01.000941.9.4143E-40.11.105941.105177.7156E-40.21.2221.22145.9439E-40.31.350271.349864.0876E-40.41.492041.491822.1332E-40.51.648731.648726.5459E-60.61.821911.822122.1332E-40.72.01332.013754.4832E-40.82.224842.225547.0085E-40.92.458632.45969.7372E-41.02.717012.718281.2703E-3Example 1.
Consider the equation(48)u″x−18u2x=1240−2x5−6x2sin2x+9xsin2x+3sin3xcosx−148x32−e−sin1−18cos2x−cosx+148∫01tx+xcosteu′tdt+140∫0xtx+sinxu′2tdt,u0.4+u0.6=1.746,u′0=0.
The exact solution of this problem isux=cosx.
Firstly, we apply the assumptions of Theorem1 to prove that this example has a continuous solution:(49)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt=1240−2x5−6x2sin2x+9xsin2x+3sin3xcosx−148x32−e−sin1−18cos2x−cosx+18u2x+148∫01tx+xcosteu′tdt+140∫0xtx+sinxu′2tdt.
Then,(50)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt≤1240−2x5−6x2sin2x+9xsin2x+3sin3xcosx−148x32−e−sin1−18cos2x−cosx+18u2x+18∫0116tx+xcosteu′tdt+18∫0x15tx+sinxu′2tdt,and also(51)fx,t,u′tdt≤16xt+16eu′t,gx,t,u′tdt≤15xt+15u′2t,where M1x=1/240−2x5−6x2sin2x+9xsin2x+3sin3xcosx−1/48x3/2−e−sin1−1/8cos2x−cosx∈L1a,b, M2x,t=1/6xt∈L1a,b,M3x,t=1/5xt∈L1a,b,C1=1/8,C2=1/6,C3=1/5,b=1; then, 2C1b2+C1C2b2+C1C3b2=2/8+1/48+1/40=71/240<1. It is clear that the assumptions (1)–(5) of Theorem 1 hold; therefore, the given nonlocal problem has a continuous solution.
Now, we use the finite difference-trapezoidal method withN=10 to find the numerical solution of this problem. Table 1 and Figure 1 give the comparison between the numerical and exact solutions of this problem.
Through our observation of Table1, the interval 0,1 was divided into 10 subintervals of equal length. We obtain solutions at the endpoints of subintervals and show that the method used is effective, and this is evident from the absolute error that was calculated for the difference between the numerical and real solutions. Also, by looking at Figure 1, we find that the numerical solution and the real solution are very close, which means that the numerical solutions are good.Figure 1
Comparison between the numerical and exact solutions of example 1.Example 2.
Consider the equation(52)u″x−17ux=156−12x2+2sinx−1cosx−sinxxsinx−sin1−cos1−8sinx7+170∫−11tx+sinxtu′tdt+156∫−1xtcosx+tsinxu′tdt,u−1+u1=0,u′−1=cos−1.
The exact solution of this problem isux=sinx.
Firstly, we apply the assumptions of Theorem1 to prove that this example has a continuous solution:(53)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt=156−12x2+2sinx−1cosx−sinxxsinx−sin1−cos1−8sinx7+17ux+170∫−11tx+sinxtu′tdt+156∫−1xtcosx+tsinxu′tdt.
Then,(54)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt≤156−12x2+2sinx−1cosx−sinxxsinx−sin1−cos1−8sinx7+17ux+17∫−11110tx+sinxtu′tdt+17∫−1x18tcosx+tsinxu′tdt,and also(55)fx,t,u′tdt≤110tx+110u′t,gx,t,u′tdt≤18tcosx+18u′t,where(56)M1x=156−12x2+2sinx−1cosx−sinxxsinx−sin1−cos1−8sinx7∈L1a,b,M2x,t=110tx∈L1a,b,M3x,t=18tcosx∈L1a,b,C1=17,C2=110,C3=18,b=1.
Then,2C1b2+C1C2b2+C1C3b2=2/7+1/70+1/56=89/280<1. It is clear that the assumptions (1)–(5) of Theorem 1 hold; therefore, the given nonlocal problem has a continuous solution.
Now, we use the finite difference-trapezoidal method withN=10 to find the numerical solution of this problem. Table 2 and Figure 2 give the comparison between the numerical and exact solutions of this problem.
Through our observation of Table2, the interval −1,1 was divided into 10 subintervals of equal length. We obtain solutions at the endpoints of subintervals and show that the method used is effective, and this is evident from the absolute error that was calculated for the difference between the numerical and real solutions. Also, by looking at Figure 2, we find that the numerical solution and the real solution are very close, which means that the numerical solutions are good.Figure 2
Comparison between the numerical and exact solutions of example 2.Example 3.
Consider the equation(57)u″x−112ux=160−x32−sinhxxcoshx−sinhx+11coshx12+184−sinhx2−coshxe+184∫01tsinhx+tcoshxu′tdt+160∫0xtx+tsinhxu′tdt,∫01uxdx=sinh1,u′0=0.
The exact solution of this problem isux=coshx.
Firstly, we apply the assumptions of Theorem1 to prove that this example has a continuous solution:(58)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt=160−x32−sinhxxcoshx−sinhx+11coshx12+184−sinhx2−coshxe+112ux+184∫01tsinhx+tcoshxu′tdt+160∫0xtx+tsinhxu′tdt.
Then,(59)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt≤160−x32−sinhxxcoshx−sinhx+11coshx12+184−sinhx2−coshxe+112ux+112∫0117tsinhx+tcoshxu′tdt+112∫0x15tx+tsinhxu′tdt,and also(60)fx,t,u′tdt≤17tsinhx+17u′t,gx,t,u′tdt≤15tx+15u′t,where(61)M1x=160−x32−sinhxxcoshx−sinhx+11coshx12+184−sinhx2−coshxe∈L1a,b,M2x,t=17tsinhx∈L1a,b,M3x,t=15tx∈L1a,b,C1=112,C2=17,C3=15,b=1.
Then,2C1b2+C1C2b2+C1C3b2=2/12+1/84+1/60=41/210<1. It is clear that the assumptions (1)–(5) of Theorem 1 hold; therefore, the given nonlocal problem has a continuous solution.
Now, we use the finite difference-trapezoidal method withN=10 to find the numerical solution of this problem. Table 3 and Figure 3 give the comparison between the numerical and exact solutions of this problem.
Through our observation of Table3, the interval 0,1 was divided into 10 subintervals of equal length. We obtain solutions at the endpoints of subintervals and show that the method used is effective, and this is evident from the absolute error that was calculated for the difference between the numerical and real solutions. Also, by looking at Figure 3, we find that the numerical solution and the real solution are very close, which means that the numerical solutions are good.Figure 3
Comparison between the numerical and exact solutions of example 3.Example 4.
Consider the equation(62)u″x−19u2x=−1144x+2ex−2x2−1324e2x+x+2x+ex−e2x9+181∫01xt+x2tu′2tdt+172∫0xtx+x2u′tdt,u0.4+u0.6=3.31394,u′0=1.
The exact solution of this problem isux=ex.
Firstly, we apply the assumptions of Theorem1 to prove that this example has a continuous solution:(63)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt=−1144x+2ex−2x2−1324e2x+x+2x+ex−e2x9+19u2x+181∫01xt+x2tu′2tdt+172∫0xtx+x2u′tdt.
Then,(64)Fx,ux,∫abfx,t,u′tdt,∫axgx,t,u′tdt≤−1144x+2ex−2x2−1324e2x+x+2x+ex−e2x9+19u2x+19∫0119xt+x2tu′2tdt+19∫0x18tx+x2u′tdt,and also(65)fx,t,u′tdt≤19tx+19u′t,gx,t,u′tdt≤18tx+18u′t,where(66)M1x=−1144x+2ex−2x2−1324e2x+x+2x+ex−e2x9∈L1a,b,M2x,t=19tx∈L1a,b,M3x,t=18tx∈L1a,b,C1=19,C2=19,C3=18,b=1.
Then,2C1b2+C1C2b2+C1C3b2=2/9+1/81+1/72=161/684<1. It is clear that the assumptions (1)–(5) of Theorem 1 hold; therefore, the given nonlocal problem has a continuous solution.
Now, we use the finite difference-trapezoidal method withN=10 to find the numerical solution of this problem. Table 4 and Figure 4 give the comparison between the numerical and exact solutions of this problem.
Through our observation of Table4, the interval 0,1 was divided into 10 subintervals of equal length. We obtain solutions at the endpoints of subintervals and show that the method used is effective, and this is evident from the absolute error that was calculated for the difference between the numerical and real solutions. Also, by looking at Figure 4, we find that the numerical solution and the real solution are very close, which means that the numerical solutions are good.Figure 4
Comparison between the numerical and exact solutions of example 4.


## 7. Conclusion

The existence and uniqueness of the nonlocal boundary value problem for the Fredholm–Volterra integro-differential equation with the nonlocal condition and the integral condition have been studied. The continuous dependence of the solution onμ0 and aj has been introduced. Also, we used the central difference approximations and trapezoidal rule to obtain a numerical solution for problems. The error estimation has been derived in this paper. Finally, we solve some numerical examples to illustrate the accuracy of the proposed method [12].



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