# An Approximation Method for Solving Volterra Integrodifferential Equations with a Weighted Integral Condition

Authors: A. Guezane-Lakoud; N. Bendjazia; R. Khaldi
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We apply the reproducing kernel Hilbert space (RKHS) method
for getting analytical and approximate solutions for second-order hyperbolic
integrodifferential equations with a weighted integral condition. The
analytical solution is represented in the form of series; thus, then-terms
approximate solutions are obtained. The results of the numerical examples
are compared with the exact solutions to illustrate the accuracy and the
effectivity of this method.


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## Body

## 1. Introduction

In functional analysis, a reproducing kernel Hilbert space is a Hilbert space of functions in which pointwise evaluation is a continuous linear functional. The subject was originally developed by Aronszajn in 1950 (see [1]).Reproducing kernel theory has important applications in numerical analysis, differential equations, integral equations, probability, and statistics [2–4]. Recently, using the RKHS method, the authors in [5–21] have discussed singular linear two-point boundary value problems, singular nonlinear two-point periodic boundary value problems, nonlinear system of boundary value problems, initial value problems, singular integral equations, nonlinear partial differential equations, operator equations, and fourth-order integrodifferential equations (IDEs), differential equations with integral condition.The purpose of this paper is to extend the application of the reproducing kernel Hilbert space method to solve a second-order boundary value problem of IDEs of the following Volterra type:(1)∂2U∂t2-∂2U∂x2=fx,t+∫0tat-sKs,Ux,sds,for all (x,t)∈Q=0,1×0,1, subject to the initial conditions(2)Ux,0=ψ1x,Utx,0=ψ2x,(3)Ux0,t=ψ3tand the weighted integral condition(4)∫01xUx,tdx=0,where f, ψ1, ψ2, ψ3, a, and K are given functions, such that (5)ψ1′0=ψ2′0=0,ψ30=ψ3′0=0.The operator K(t,u) is linear with respect to u and continuous according to both variables t and u. The function a is continuous and denote l=max⁡0≤t≤1a(t).In order to put initial conditions and integral boundary conditions of (1) into the reproducing kernel space HQ constructed in the following sections, we have to homogenize these conditions. For this, let u(x,t)=U(x,t)-ψ1x-tψ2x-x-2/3ψ3t; then the problem can be converted into the following form:(6)∂2u∂t2-∂2u∂x2=Fx,t+∫0tat-sKs,ux,sds,for all (x,t)∈Q, subject to the initial conditions(7)ux,0=0,utx,0=0,ux0,t=0,and the weighted integral condition(8)∫01xux,tdx=0.The theoretical aspects of (1) with conditions (2)–(4) have been studied by Galerkin method in [22]; however, no numerical method was presented. In fact, by this procedure, the authors have found several difficulties to find the discretization of this problem due to the presence of the integral condition. Therefore, to overcome these difficulties, in the present work, we propose the RKHS method which is a simple and effective method for obtaining the numerical solution that converges rapidly to the exact solution with a small error and with a good precision. Moreover, the partial derivatives of unx,t are also convergent to the partial of ux,t. So, from this work, we show the applicability of the RKHS method to solve this kind of problem without discretization such as Galerkin and Rothe’s method. Furthermore, the RKHS method has an advantage that it is the possibility to pick any point in the domain of integration as well as in the approximate solutions and all its partial derivatives up to order two that will be applicable.In recent years, a growing interest has been devoted to the study of IDEs which are a combination of differential and Volterra-Fredholm integral equations. IDEs appear in various fields of science such as physics, biology, and engineering. We should mention also that integrodifferential equations are usually difficult to solve analytically.Recently, various kinds of numerical methods have been used for efficient approximation solution. In [23], authors have presented a practical matrix method for solving nonlinear Volterra-Fredholm integrodifferential equations. In [24], authors have given a numerical solution of nonlinear Volterra-Fredholm integrodifferential equations using the spectral homotopy analysis method. In [25–28], authors have obtained the representation of the exact solution for the nonlinear Volterra-Fredholm integral equations by using the reproducing kernel method.To sum up, in this paper, we define several reproducing kernel spaces in Section2, whereas, in Section 3, we introduce a linear operator, a complete normal orthogonal system, and some essential results. In Section 4, we provide the main results and the exact and approximate solution of (1)–(4) and we also develop an iterative method for this kind of problems. Finally, in Section 5, the results of some numerical examples are presented and are compared with the analytical solution to confirm the good accuracy of the presented method.


## 2. Reproducing Kernel Spaces

In this section, we define some useful reproducing kernel spaces.Definition 1.
Let(9)H30,1=ux∣u,u′,u′′are absolutely continuous real value functions in0,1,u3x∈L20,1,u′0=∫01xuxdx=0.Define, respectively, the inner product and the norm in H3[0,1] by(10)ux,vxH3=∑i=02ui0vi0+∫01u3xv3xdx,uH3=ux,uxH3,u,v∈H30,1.
We have the following result.Theorem 2.
The spaceH3[0,1] is a complete reproducing kernel space; that is, for each fixed x∈[0,1], there exists G^3(y,x)∈H3[0,1](simply G^x(y)) and y∈[0,1] such that u(y),G^x(y)H3=u(x) for any u(y)∈H3[0,1]. The reproducing kernel G^x(y) is given by(11)G^xy=∑i=16cixyi-1-y7Cx5040,y≤x,∑i=16dixyi-1-y7Cx5040,x<y,where (12)c1x=6559727-11349727x2-3789727x3+2529727x4+131233448x5+1848635x7,c2x=-x424,c3x=-11349727+22909727x2+229029181x3+6319454x4-189194540x5+9194540x7,c4x=-3789727-18938908x2-6338908x3+2119454x4-63194540x5+3194540x7,c5x=2520+315x2+105x3-70x4+21x5-x797270,c6x=-3789727-189194540x2-63194540x3+2197270x4-63972700x5+3972700x7,d1x=6559727-11349727x2-3789727x3+2529727x4+131233448x5+1848635x7,d2x=0,d3x=-11349727+22909727x2+229029181x3+6319454x4-189194540x5+9194540x7,d4x=-3789727-18938908x2-6338908x3+2119454x4-63194540x5+3194540x7,d5x=2529727-x24+6319454x2+2119454x3-79727x4+2197270x5-197270x7,d6x=131233448-189194540x2-63194540x3+2197270x4-63972700x5+3972700x7,Cx=-181449727-22689727x2-7569727x3+5049727x4-75648635x5+3648635x7.Proof.
Through several integrations by parts of (10), we obtain (13)uy,G^xyH3=∑i=02ui0∂yiG^x0--12-i∂y5-iG^x0+∑i=02-12-iui1∂y5-iG^x1-∫01uy∂y6G^xydy.
Sinceu(x)∈H3[0,1], we have (14)u′0=∫01xuxdx=0.
Then(15)uy,G^xyH3=u0G^x0-∂y5G^x0+u20∂y2G^x0-∂y3G^x0+∑i=02-12-iui1∂y5-iG^x1-∫01uy∂y6G^xydy-Cx∫01yuydy.
Note that the property of the reproducing kernelu(y),G^x(y)H3=u(x); then G^x(y) is the solution of the following generalized differential equation:(16)∂y6G^xy+Cxy=-δy-x,with the boundary conditions(17)G^x0-∂y5G^x0=0,∂y2G^x0-∂y3G^x0=0,∂y5-iG^x1=0,i=0,1,2.
Asx≠y, it follows that (18)G^xy=∑i=16cixyi-1-y7Cx5040,y≤x,∑i=16dixyi-1-y7Cx5040,x<y.
Let us find the coefficientscix, dix, and Cx.
Since∂y6G^x(y)+C(x)y=-δy-x, we have(19)∂y5G^xx-0-∂y5G^xx+0=1,∂yiG^xx+0=∂yiG^xx-0,i=0,1,2,3,4.
SinceG^x(y)∈H3[0,1], it follows that(20)∂y1G^x0=0,∫01yG^xydy=0.
Through (17)–(20), the unknown coefficients of (11) can be obtained.Definition 3.
Define the space(21)H3′0,1=ut∣u,u′,u′′are absolutely continuous functions in0,1u3t∈L20,1,u′0=u0=0.The inner product and the norm in H3′[0,1] are defined, respectively, by (22)ut,vtH3′=∑i=02ui0vi0+∫01u3xv3xdx,uH3′=ut,utH3′.The space H3′[0,1] is a complete reproducing kernel space, and its reproducing kernel G˘s(t) is given by(23)G˘st=1120s5-124s4t+14s2t2+112s3t2,s≤t,-124s4t+1120t5+112s2t23+t,t<s.Definition 4.
Define(24)HQ=u∣∂4u∂2x∂2tis absolutely continuous real-valued function inQ,∂6u∂3x∂3t∈L2Q,ux,0=∂ux,0∂t=0,∂u0,t∂x=∫01xux,tdx=0.The inner product and the norm in HQ are defined, respectively, by (25)ux,t,vx,tH=∑i=02∫01∂3∂t3∂i∂tiu0,t∂3∂t3∂i∂tiv0,tdt+∑j=02∂j∂tjux,0,∂j∂tjvx,0H3+∫01∫01∂6∂x3∂t3ux,t∂6∂x3∂t3vx,tdtdx,uHQ=ux,t,ux,tH.HQ is a reproducing kernel space and its reproducing kernel function is (26)Gy,sx,t=G^xyG˘stsuch that, for any u(x,t)∈HQ, (27)uy,s=ux,t,Gy,sx,tH,Gx,ty,s=Gy,sx,t.Definition 5.
Define(28)H10,1=ux∣uis absolutely continuous in0,1,u′∈L20,1.The inner product and the norm in H1[0,1] are defined, respectively, by (29)ux,vxH1=u0v0+∫01u′xv′xdx,uH1=ux,uxH1.Definition 6.
Define(30)H^Q=ux,t∣uis absolutely continuous inQ,∂2u∂x∂t∈L2Q.The inner product and the norm in H^Q are defined, respectively, by (31)ux,t,vx,tH^=∫01∂∂tu0,t∂∂tv0,tdt+ux,0,vx,0H1+∫01∫01∂2∂x∂tux,t∂2∂x∂tvx,tdtdx,uH^=ux,t,ux,tH^.H^Q is a complete reproducing kernel space; thus, its reproducing kernel function is (see [11]) (32)Ry,sx,t=R´yxR´st,where(33)R´yx=1+y,y≤x,1+x,x<y.


## 3. A Bounded Linear Operator onHQ

Define differential operatorL: HQ→H^Q by(34)Lux,t=∂2u∂t2-∂2u∂x2;therefore, (1) can be converted into an equivalent form as follows:(35)Lux,t=F~x,t,ux,t,where u(x,t)∈HQ,  F~x,t,u(x,t)∈H^Q, and (36)F~x,t,ux,t=fx,t+∫0tat-sKs,ux,sds.It is easy to show thatL is a bounded linear operator from HQ to H^Q. Now, we choose a countable subset S=x1,t1,x2,t2,… in Q and we define(37)φix,t=Rxi,tix,t,ωix,t=L∗φix,t.Lemma 7.
Assume thatS is dense in Q; then, ωix,ti=1∞ is a complete system in HQ, and (38)ωix,t=Ly,sGy,sx,t,fory,s=xi,ti.Applying Gram-Schmidt process, we obtain an orthogonal basisϖix,ti=1∞ of HQ, such that(39)ϖix,t=∑j=1iβijωix,t,where βij are orthogonal coefficients.


## 4. The Exact and Approximate Solution

Lemma 8.
IfS is dense in Q, then the exact solution of (35) is(40)ux,t=∑i=1∞∑j=1iβijF~xj,tj,uxj,tjϖix,t.Proof.
The exact solutionux,t can be expanded in the Fourier series in terms of normal orthogonal basis ϖix,ti=1∞ in HQ:(41)ux,t=∑i=1∞ux,t,ϖix,tHϖix,t=∑i=1∞∑j=1iβijux,t,ωjx,tHϖix,t=∑i=1∞∑j=1iβijux,t,L∗φix,tHϖix,t=∑i=1∞∑j=1iβijLux,t,φix,tH^Qϖix,t=∑i=1∞∑j=1iβijLux,t,Rxi,tix,tH^Qϖix,t=∑i=1∞∑j=1iβijLuxj,tjϖix,t=∑i=1∞∑j=1iβijF~xj,tj,uxj,tjϖix,t.


### 4.1. Iterative Method

We construct an iterative method to obtain the approximate solutions of (1) in the reproducing kernel space HQ. Let(42)Ai=∑j=1iβijF~xj,tj,uxj,tj.Then(43)ux,t=∑i=1∞Aiϖix,t.Define the initial functionu0x,t=0 and the n-term approximation to ux,t by(44)unx,t=∑i=1nA~iϖix,t,where the coefficients A~i are given as(45)A~1=β11F~x1,t1,u0x1,t1,A~2=∑j=12β2jF~xj,tj,uj-1xj,tj,⋮A~n=∑j=1nβnjF~xj,tj,uj-1xj,tj.Theorem 9.
IfS is dense in Q, then (46)Lunx,t=F~x,t,un-1x,t,where unx,t is given by (44).Theorem 10.
Suppose that∫0t‍K(s,u(x,s))dsH^Q≤CuHQ and 0<lCL-1<1; then (47)unx,t→·Hux,tasn⟶∞.Proof.
From the preceding theorem, we have(48)Lunx,t=F~x,t,un-1x,tso(49)unx,t-ux,tH=L-1F~x,t,un-1x,t-F~x,t,ux,tH≤L-1F~x,t,un-1x,t-F~x,t,ux,tH^Q≤L-1∫0tat-s·Ks,un-1x,s-ux,sdsH^Q≤C·max⁡0≤t≤1⁡atL-1un-1x,t-ux,tH.
FromlCL-1<1 and (49), we get (50)unx,t→·Hux,tasn⟶∞.Proposition 11.
Suppose thatunx,t→·Hux,t as n→∞; then (51)∂i+j∂xi∂tjux,t-∂i+j∂xi∂tjunx,t⟶0,n⟶∞,i,j=0,1,2.Now, the approximate solutionunmx,t can be obtained by taking finitely many terms in the series representation of unx,t and(52)unmx,t=∑i=1m∑j=1iβijF~xj,tj,un-1xj,tjϖix,t.


## 4.1. Iterative Method

We construct an iterative method to obtain the approximate solutions of (1) in the reproducing kernel space HQ. Let(42)Ai=∑j=1iβijF~xj,tj,uxj,tj.Then(43)ux,t=∑i=1∞Aiϖix,t.Define the initial functionu0x,t=0 and the n-term approximation to ux,t by(44)unx,t=∑i=1nA~iϖix,t,where the coefficients A~i are given as(45)A~1=β11F~x1,t1,u0x1,t1,A~2=∑j=12β2jF~xj,tj,uj-1xj,tj,⋮A~n=∑j=1nβnjF~xj,tj,uj-1xj,tj.Theorem 9.
IfS is dense in Q, then (46)Lunx,t=F~x,t,un-1x,t,where unx,t is given by (44).Theorem 10.
Suppose that∫0t‍K(s,u(x,s))dsH^Q≤CuHQ and 0<lCL-1<1; then (47)unx,t→·Hux,tasn⟶∞.Proof.
From the preceding theorem, we have(48)Lunx,t=F~x,t,un-1x,tso(49)unx,t-ux,tH=L-1F~x,t,un-1x,t-F~x,t,ux,tH≤L-1F~x,t,un-1x,t-F~x,t,ux,tH^Q≤L-1∫0tat-s·Ks,un-1x,s-ux,sdsH^Q≤C·max⁡0≤t≤1⁡atL-1un-1x,t-ux,tH.
FromlCL-1<1 and (49), we get (50)unx,t→·Hux,tasn⟶∞.Proposition 11.
Suppose thatunx,t→·Hux,t as n→∞; then (51)∂i+j∂xi∂tjux,t-∂i+j∂xi∂tjunx,t⟶0,n⟶∞,i,j=0,1,2.Now, the approximate solutionunmx,t can be obtained by taking finitely many terms in the series representation of unx,t and(52)unmx,t=∑i=1m∑j=1iβijF~xj,tj,un-1xj,tjϖix,t.


## 5. Numerical Examples

In order to calculate the approximate solutionunmx,t, we take (53)xj,tj=iN,kM,i=0,1,…N,k=0,1,…M,j=0,1,2,…m=N+1M+1,enmx,t=ux,t-unmx,t,enmxxx,t=uxxx,t-unmxxx,t,enmtx,t=utx,t-unmtx,t,enmxt=uxtx,t-unmxtx,t,enmttx,t=uttx,t-unmttx,t,enmxx,t=uxx,t-unmxx,t.Example 1.
Consider the following hyperbolic integrodifferential equation:(54)∂2u∂t2-∂2u∂x2=fx,t+∫0tt-s4ux,sds,for all (x,t)∈Q, subject to the initial conditions (55)ux,0=∂ux,0∂t=0,∂u0,t∂x=t2,and the weighted integral condition(56)∫01xux,tdx=0,where(57)fx,t=-4+2t7105+exp⁡⁡x2-t2-t7105.The exact solution is given by u(x,t)=(exp⁡⁡x-2)t2. In order to homogenize the initial conditions, we put v(x,t)=u(x,t)-x-2/3t2; then the problem can be converted into the following form: (58)∂2v∂t2-∂2v∂x2=Fx,t+∫0tt-s4vx,sds,vx,0=∂vx,0∂t=0,∂v0,t∂x=0,∫01xvx,tdx=0,where (59)Fx,t=-3315t7+105t2-210exp⁡⁡x-33154+3x+-210+t7.The numerical results are presented in Tables1, 2, 3, and 4 and Figures 1 and 2.Table 1
Comparison of results foru (n=2, m=25,121).
(
x
i
,
t
i)
u
u
2
25
e
2
25
u
2
121
e
2
121
(0.00, 0.01)
−0.0000333333
−0.0000333637
3.033825
×
10
-
8
−0.0000333333
1.27745
×
10
-
8
(0.10, 0.05)
−0.000820406
−0.00082375
3.34428
×
10
-
6
−0.000820406
1.2258
×
10
-
6
(0.15, 0.10)
−0.00321499
−0.00323799
0.000022998
−0.00321499
6.31116
×
10
-
6
(0.35, 0.20)
−0.0105706
−0.0106773
0.000106649
−0.0105706
0.0000198708
(0.49, 0.31)
−0.0183567
−0.0185277
0.000170909
−0.0183885
0.0000317253
(0.57, 0.45)
−0.0273509
−0.0275692
0.000218296
−0.0273509
0.0000385202
(0.69, 0.52)
−000800865
−0.00804727
0.0000386212
−0.00800874
8.8295
×
10
-
8
(0.73, 0.71)
0.0059218
0.00598197
0.0000601728
0.00594941
0.0000276115
(0.81, 0.73)
0.0557278
0.0560604
0.000332616
0.0558007
0.0000728933
(0.99, 0.99)
0.36058
0.362213
0.00163333
0.360756
0.000176542Table 2
Comparison of results forux (n=2, m=121).
(
x
i
,
t
i
)
u
x
u
2
121
x
e
2
121
x
u
t
u
2
121
t
e
2
121
t
(0.00, 0.01)
0.00000
0.0000
0.00000
−0.00666667
−0.00667043
3.76
×
10
-
6
(0.10, 0.05)
0.000262927
0.000263362
4.34808
×
10
-
7
−0.0328162
−0.0328162
0.000065
(0.15, 0.10)
0.00161834
0.0016219
3.5546
×
10
-
6
−0.0642998
−0.0644248
0.000124
(0.35, 0.20)
0.0167627
0.0167986
0.0000359083
−0.105706
−0.105901
0.000194
(0.49, 0.31)
0.0607656
0.0608914
0.000125813
−0.118431
−0.118615
0.00018
(0.57, 0.45)
0.155574
0.155866
0.000291519
−0.12156
−0.121689
0.00012
(0.69, 0.52)
0.268701
0.269116
0.000415711
−0.0308025
−0.0307829
0.000019
(0.73, 0.71)
0.541948
0.54252
0.00057215
0.0166811
0.0167237
0.000042
(0.81, 0.73)
0.66501
0.665538
0.000527674
0.152679
0.152725
0.000045
(0.99, 0.99)
1.65758
1.659
0.00142569
0.728444
0.727913
0.00053Table 3
Comparison of results foruxx (n=2, m=121).
(
x
i
,
t
i
)
u
x
x
u
2
121
x
x
e
2
121
x
x
u
t
t
u
2
121
t
t
e
2
121
t
t
(0.00, 0.01)
0.0001000
0.000100171
1.71
×
10
-
7
−0.666667
−0.667391
0.000724
(0.10, 0.05)
0.00276293
0.00277046
7.52
×
10
-
6
−0.656325
−0.658253
0.001928
(0.15, 0.10)
0.0116183
0.0116374
0.0000190
−0.642998
−0.64296
0.0000379
(0.35, 0.20)
0.0567627
0.0568428
0.0000801
−0.528532
−0.528362
0.000169
(0.49, 0.31)
0.156866
0.157201
0.0003349
−0.382034
−0.382073
0.000038
(0.57, 0.45)
0.358074
0.35823
0.0001562
−0.270133
−0.270318
0.0001851
(0.69, 0.52)
0.539101
0.539354
0.0002531
−0.0592356
−0.0059133
0.0001026
(0.73, 0.71)
1.04605
1.04597
0.0000733
0.0234945
0.0232241
0.00027045
(0.81, 0.73)
1.19791016
1.19798
0.00006967
0.209149
0.208289
0.00085993
(0.99, 0.99)
2.63768
2.64076
0.003076
0.735802
0.735802
0.00426859Table 4
Comparison of results foruxt (n=2, m=121).
(
x
i
,
t
i
)
u
x
t
u
2
121
x
t
e
2
121
x
t
(0.00, 0.01)
0.00000
0.0000
0.00000
(0.10, 0.05)
0.0105171
0.0105402
0.0000231397
(0.15, 0.10)
0.0323668
0.0324379
0.0000710837
(0.35, 0.20)
0.167627
0.167991
0.000364067
(0.49, 0.31)
0.392036
0.392836
0.000799951
(0.57, 0.45)
0.69144
0.692595
0.00115446
(0.69, 0.52)
1.03346
1.03456
0.00109998
(0.73, 0.71)
1.52661
1.52694
0.000326263
(0.81, 0.73)
1.82195
1.82199
0.000041932
(0.99, 0.99)
3.34864
3.34924
0.000598899Figure 1
Considere225x,t=ux,t-u225x,t.Figure 2
Considere2121x,t=ux,t-u2121x,t.Example 2.
Consider the following hyperbolic integrodifferential equation:(60)∂2u∂t2-∂2u∂x2=fx,t+∫0t∫0xur,sdrds,ux,0=∂ux,0∂t=0,∂u0,t∂x=0,∫01xux,tdx=0,where (61)fx,t=-2t2+xt36-x3t39+-12+x2.
The exact solution is given byu(x,t)=(x2-1/2)t2.The numerical results are presented in Tables5, 6, 7, and 8 and Figures 3 and 4.Table 5
Comparison of results foru (n=2, m=25).
(
x
i
,
t
i
)
u
u
2
25
e
2
25
e
2
25
/
u
(0.00, 0.01)
−0.000050000
−0.0000500038
3.81237
×
10
-
9
0.0000762474
(0.10, 0.05)
−0.001225000
−0.00122559
5.90641
×
10
-
7
0.000482156
(0.15, 0.10)
−0.004775
−0.0047785
3.49902
×
10
-
6
0.000732779
(0.35, 0.20)
−0.01510000
−0.0151099
9.9186
×
10
-
6
0.000656861
(0.49, 0.31)
−0.0249764
−0.0249918
0.0000153856
0.000616005
(0.57, 0.45)
−0.0354578
−0.0354835
0.0000257761
0.000726954
(0.69, 0.52)
−0.00646256
−0.00645257
9.99319
×
10
-
6
0.00154632
(0.73, 0.71)
0.0165849
0.0166669
0.0000819828
0.00494322
(0.81, 0.73)
0.0831857
0.083387
0.000201347
0.00242045
(0.99, 0.99)
0.470546
0.471707
0.00116079
0.00246689Table 6
Comparison of results forux (n=2, m=25).
(
x
i
,
t
i
)
u
x
u
2
25
x
e
2
25
x
u
t
u
2
25
t
e
2
25
t
(0.00, 0.01)
0.0000
0.0000
0.000
−0.010
−0.0100011
1.13
×
10
-
6
(0.10, 0.05)
0.0005
0.000499228
7.71
×
10
-
7
−0.049
−0.049031
0.00003
(0.15, 0.10)
0.003
0.00300832
8.32
×
10
-
6
−0.0955
−0.0955913
0.00009
(0.35, 0.20)
0.028
0.0280282
0.000028
−0.151
−0.151122
0.00012
(0.49, 0.31)
0.094178
0.0942695
0.00009
−0.161138
−0.161259
0.00012
(0.57, 0.45)
0.23085
0.231107
0.00025
−0.15759
−0.157779
0.00018
(0.69, 0.52)
0.373152
0.37359
0.00043
−0.024856
−0.0247779
0.00007
(0.73, 0.71)
0.735986
0.737218
0.0012
0.046718
0.0471139
0.00039
(0.81, 0.73)
0.863298
0.864699
0.0014
0.227906
0.228805
0.00089
(0.99, 0.99)
1.9406
1.94427
0.0036
0.950598
0.954343
0.0037Table 7
Comparison of results foruxx (n=2, m=25).
(
x
i
,
t
i
)
u
x
x
u
2
25
x
x
e
2
25
x
x
u
t
t
u
2
25
t
t
e
2
25
t
t
(0.00, 0.01)
0.0002
0.000196032
3.96
×
10
-
6
−1.0000
−1.00022
0.0002
(0.10, 0.05)
0.005
0.00504296
0.00004
−0.98
−0.981028
0.00102
(0.15, 0.10)
0.02
0.0201771
0.0001
−0.955
−0.956444
0.0014
(0.35, 0.20)
0.08
0.0799329
0.00006
−0.755
0.0007
0.0001
(0.49, 0.31)
0.1922
0.191932
0.00026
−0.5198
−0.520859
0.0010
(0.57, 0.45)
0.405
0.404939
0.00006
−0.3502
−0.351163
0.0009
(0.69, 0.52)
0.5408
0.541206
0.0004
−0.0478
−0.0473431
0.0004
(0.73, 0.71)
1.0082
1.00936
0.0011
0.0658
0.0670897
0.0012
(0.81, 0.73)
1.0658
1.06519
0.0006
0.3122
0.314928
0.0027
(0.99, 0.99)
1.9406
1.94427
0.0036
0.950598
0.954343
0.0037Table 8
Comparison of results foruxt (n=2, m=25).
(
x
i
,
t
i
)
u
x
t
u
2
25
x
t
e
2
25
x
t
(0.00, 0.01)
0.000
0.000
0.000
(0.10, 0.05)
0.02
0.0199786
0.00002
(0.15, 0.10)
0.06
0.0602084
0.0002
(0.35, 0.20)
0.28
0.280396
0.00039
(0.49, 0.31)
0.6076
0.608402
0.0008
(0.57, 0.45)
1.026
1.02799
0.0019
(0.69, 0.52)
1.4352
1.43788
0.002
(0.73, 0.71)
2.0732
2.07878
0.005
(0.81, 0.73)
2.3652
2.37112
0.005
(0.99, 0.99)
3.9204
3.93184
0.011Figure 3
Considere225x,t=ux,t-u225x,t.Figure 4
Considerux,t-u225x,t/ux,t.


## 6. Conclusion

In this paper, we have applied RKHS method to solve a hyperbolic integrodifferential equation with a weighted integral condition. The numeric results have demonstrated that only a small number of iterations can be used to obtain numeric results with a good precision. From this work, we see that the absolute errors are monotonically decreasing ifm increase. The numerical examples show that the approximate solution and its partial derivatives of a superior order or equal to two converge to the exact solution and its partial derivatives, respectively. Consequently, we confirm that this method is effective for the integrodifferential equation with weighted integral conditions. All the computations are performed using Mathematica 9.0.



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*Source: 758410-2015-05-13.xml*