# The Cardinal Spline Methods for the Numerical Solution of Nonlinear Integral Equations

Authors: Xiaoyan Liu; Jin Xie; Zhi Liu; Jiahuan Huang
Journal: Journal of Chemistry
 (2020)

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## Abstract

In this study, an effective technique is presented for solving nonlinear Volterra integral equations. The method is based on application of cardinal spline functions on small compact supports. The integral equation is reduced to a system of algebra equations. Since the matrix for the system is triangular, it is relatively straightforward to solve for the unknowns and an approximation of the original solution with high accuracy is accomplished. Several cardinal splines are employed in the paper to enhance the accuracy. The sufficient condition for the existence of the inverse matrix is examined, and the convergence rate is analyzed. We compare our method with other methods proposed in recent papers and demonstrated the advantage of our method with several examples.


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## Body

## 1. Introduction

Integral equations appear in many fields, including dynamic systems, mathematical applications in economics, communication theory, optimization and optimal control systems, biology and population growth, continuum and quantum mechanics, kinetic theory of gases, electricity and magnetism, potential theory, geophysics, and theoretical-model chemistry. Many differential equations with boundary values can be reformulated as integral equations. There are also some problems that can be expressed only in terms of integral equations. Many differential equations with boundary values can be reformulated as integral equations; for example, for chemical integral equations with boundaries, Polyanin summarized different solutions of integral equations in [1–3], published in 2013 and 2016. In [4–6], we discussed numerical methods using cardinal splines in solving systems of linear integral equations. In this paper, we are going to explore the applications of cardinal splines in solving nonlinear integral equations.We are interested in the Volterra integral equations of the second kind:(1)yx=gx+∫axKx,t,ytdt,x∈a,b,where the kernel Kx,t,y and gx are known functions and yx is to be determined.This paper is divided into six sections. In Sections2 and 3, two univariate cardinal continuous splines on small compact supports are constructed and properties are given. In Section 4, the applications of cardinal splines on solving integral equations are explored. The unknown function is expressed as a linear combination of horizontal translations of a cardinal spline function. Then, a system of equations on the coefficients are deducted. We can solve the system, and a good approximation of the original solution is obtained. The sufficient condition for the existence of the inverse matrix is discussed, and the convergence is investigated. In Section 5, the numerical examples are given. The nonlinear system on unknowns is solved, and an accurate approximation of the original solution is obtained in each case. Section 6 contains the conclusion remarks.


## 2. Cardinal Splines with Small Compact Supports

Since the paper [7] by Schoenberg published in 1946, spline functions have been studied by many scholars. Spline functions have excellent properties, and applications are endless (for examples, cf [2]). The spline functions on uniform partitions are simple to construct and easy to apply and are sufficient for a variety of applications.The starting point is frequently the zero degree polynomialB-spline (see Figure 1), with the integral iteration formula(2)B0x=1,−12<x<12,0,elsewhere,Bnx=∫−1/21/2Bn−1x+tdt,n=1,2,3,….Figure 1
The graph ofB0x.We could construct higher-order polynomial spline functions with a higher degree of smoothness. More specifically,B1x has the expression (see Figure 2) as follows:(3)B1x=12x−1+12x+1−x.Figure 2
The graph ofB1x.B
n
x are called one-dimensional B-splines, which are polynomial splines and have small supports −n+1/2,n+1/2,i.e., Bnx=0 for x>n+1/2 or <−n+1/2 and excellent traits (cf [8]). In my previous papers [4, 5], low-degree orthonormal spline and cardinal spline functions with small compact supports were applied in solving the second kind of Volterra integral equations. In this paper, we use the notation uA=maxx∈Aux.LetB1,hx=B1x/h. It is proved that(4)∑m=−∞∞a+bmhB1,hx−mh=a+bx.Note that this particularB-spline is also a cardinal spline; therefore, it is straightforward to apply it in interpolations. As far as the convergence rate of interpolation is concerned, we have the following proposition (cf [9–12]).Proposition 1.
Given thatfx∈Ca,b,f′x exists and is bounded in a,b. Let n be an integer; h=b−a/n; let xi=a+ih;fi=fxi,i=0,1,2,…,n, as follows:(5)Ωfx=∑k=0nfkB1,hx−kh.Then,(6)Ωfx−fxa,b≤3f′xa,bh2,where Ω denotes the approximation polynomial of fx.
Iffx∈C−∞,∞ and f′x exists and is bounded, let h be a real number, and let fi=fih,i=0,1,2,…:(7)Ωfx=∑k=−∞∞fkB1,hx−kh.Then,(8)Ωfx−fx−∞,∞≤3f′x−∞,∞h2.


## 3. A UnivariateC2 Cardinal Spline

By cardinal conditions (cf [13]), we mean that let Lx be a function, xi,i=0,±1,±2,±3,… be interpolation points; then,(9)Lxi=1,i=0,0,i≠0,i=0,±1,±2,±3,±4,….The cardinal spline that was originally given in [10] is based on B3x using the similar process as in Section 2. Let(10)L3x=6B3x−60∫−1/21/2t2B2x+tdt=∫−1/21/26−60t2B2x+tdt.Figure3 shows the graph of L3x.Figure 3
The graph ofL3x.Then,L3x/h satisfies the above cardinal condition when xi=ih,i=0,±1,±2,±3,…. Notice that by the construction L3x∈C2−∞,∞. L3x=0 for x∈−∞,−2∪2,∞. L3x is a polynomial of degree ≤5 in each subinterval −2,−1,−1,0,0,1,and1,2 of its support.Furthermore, from direct calculation, we deduct the following two propositions (cf [10]).Proposition 2.
LetL3x be the cardinal spline constructed above; then,(11)∑m=−∞∞L3xh−mαih2+βih+γ=αx2+βx+γ,∑m=−∞∞L3xh−mih3−x3≤16h3,where α,β,andγ are any complex number.Proposition 3.
Iffx∈C3a,b, let h=b−a/n,n be an integer; let fi=fa+ih,i=0,1,2,…:(12)Ωfx=3f0−3f1+f2L3xh+1+∑k=0nfL3xh−ki+3fn−3fn−1+fn−2L3xh−n−1.Then,(13)Ωfx−fxa,b≤7f‴xa,bh3.Iffx∈C3−∞,∞ and is bounded, let h be a real number; let fi=fih,i=0,1,2,…:(14)Ωfx=∑k=−∞∞fkL3xh−k.Then,(15)Ωfx−fx−∞,∞≤7f‴x−∞,∞h3.


## 4. Numerical Methods Solving Integral Equations

### 4.1. Method 1-V for Solving the Volterra Integral Equation

As for the Volterra integral equations (1), we solve it in an interval a,b. Again, we let h=b−a/n and xi=a+ih,i=0,1,…,n. Furthermore, plugging fx=∑k=0nckB1,hx−xk,Kx,t,yt=∑i=0n∑j=0nKxi,xj,cjB1,hx−xiB1,ht−xj,andgx=∑k=0ngxkB1,hx−xk, we get(16)∑k=0nckB1,hx−xk−∑i=0n∑j=0nB1,hx−xi∫axKxi,xj,cjB1,ht−xjdt=∑k=0ngxkB1,hx−xk.Letx=xs; we arrive at(17)cs−∑j=0nKxs,xj,cj∫axsB1,ht−xjdt=gxs,s=0,1,2,3,4,…,n,which is a simple system of n+1 nonlinear equations of unknowns c0,c1,…,cn. Notice that this is a triangular system and it is solvable (the solution may not be unique because it is not linear):(18)c0=gx0,c1−h2∑j=01Kx1,xj,cj=gx1,cs−h2Kxs,x0,c0+2∑j=1s−1Kxs,xj,cj+Kxs,xs,cs=gxs,s=2,3,4,…,n.Proposition 4.
Given equation (1) and that fx,gx∈Ca,b,f′x and g′x exist and are bounded in a,b,Kx,y,u∈Ca,b×a,b×a,band∂/∂xKx,y,u and ∂/∂yKx,y,u exist and are bounded in a,b×a,b×a,b. Furthermore, Kx,y,u satisfies the following condition:(19)∫abKx,t,ft−Kx,t,utdt<LMmaxx∈a,bfx−ux,where LM<1. Let n be an integer, h=b−a/n; let xi=a+ih,fi=fxi,i=0,1,2,…,n.c0,c1,…,cn satisfies the nonlinear system (17):(20)f∗x=∑k=0nciB1,hx−kh.Then,(21)f∗x−fxa,b=Oh2,where fx is the exact solution of equation (1).


### 4.2. Method 2-V for Solving the Volterra Integral Equation

For more accurate solution, we applyL3x/h. Again, we let h=b−a/nandxi=a+ih,i=−1,0,…,n+1Furthermore, letL3,hx=L3x/h be the cardinal spline given in Section 3, and(22)fx=∑k=−1n+1ckL3,hx−xk,Kx,t,yt,Kx,t,yt=∑i=−1n+1∑j=−1n+1Kxi,xj,cjL3,hx−xiL3,ht−xj,gx=∑k=−1n+1gxkL3,hx−xk.Since the support ofL3,hx−x−1 and L3,hx−xn+1 is wider than a,b, we need values that define(23)Kx−1,xj=3Kx0,xj−3Kx1,xj+Kx2,xj,j=0,1,…,n,Kxi,x−1=3Kxi,x0−3Kxi,x1+Kxi,x2,i=0,1,…,n,Kx−1,x−1=3Kx0,x0−3Kx1,x0+Kx2,x0+3Kx0,x1−3Kx1,x1+Kx2,x1−3Kx0,x2−3Kx1,x2+Kx2,x2,gx−1=3gx0−3gx1+gx2,Kxn+1,xj=3Kxn,xj−3Kxn−1,xj+Kxn−2,xj,j=0,1,…,n,Kxi,xn+1=3Kxi,xn−3Kxi,xn−1+Kxi,x2,i=0,1,…,n,Kxn+1,xn+1=3Kxn,xn−3Kxn−1,xn+Kxn−2,xn+3Kxn,xn−1−3Kxn−1,xn−1+Kxn−2,xn−1−3Kxn,xn−2−3Kxn−1,xn−2+Kxn−2,xn−2,gxn+1=3gxn−3gxn−1+gxn−2.Letc0,c1,…,cn be the unknown coefficients to be determined and c−1=3c0−3c1+c2 and cn+1=3cn−3cn−1+cn−2. Plug into the integral equation (1); then, we have(24)∑k=−1n+1ckL3,hx−xk−λ∫ax∑i=−1n+1∑j=−1n+1Kxi,xj,cjL3,hx−xiL3,ht−xjdt=∑k=−1n+1gxkL3,hx−xk,x∈a,b.Letx=xi; we arrive at(25)cs−∑j=−1n+1Kxs,xj,cj∫axsL3,ht−xjdt=gxs,s=0,1,2,3,4,…,n,which is still a relatively simple system of equations. For the convergence rate of solution of Volterra integral equation (1), we have a similar result.Proposition 5.
Given thatfx,gx∈C3a,b,f4x and g4x exist and are bounded in a,b, Kx,t,y∈C3a,b×a,b×c,d. Furthermore, Kx,t,y satisfies the following condition:(26)∫abKx,t,ft−Kx,t,utdt<LMmaxx∈a,bfx−ux.where LM<1. Let n be an integer, h=b−a/n; let xi=a+ih,fi=fxi,i=0,1,2,…,n. c0,c1,…,cn satisfies system (35):(27)f∗x=∑k=−1n+1ckL3,hx−xk.Then,(28)f∗x−fxa,b%=Oh3,f∗x−fxa,b=Oh3,where fx is the exact solution of equation (1).


## 4.1. Method 1-V for Solving the Volterra Integral Equation

As for the Volterra integral equations (1), we solve it in an interval a,b. Again, we let h=b−a/n and xi=a+ih,i=0,1,…,n. Furthermore, plugging fx=∑k=0nckB1,hx−xk,Kx,t,yt=∑i=0n∑j=0nKxi,xj,cjB1,hx−xiB1,ht−xj,andgx=∑k=0ngxkB1,hx−xk, we get(16)∑k=0nckB1,hx−xk−∑i=0n∑j=0nB1,hx−xi∫axKxi,xj,cjB1,ht−xjdt=∑k=0ngxkB1,hx−xk.Letx=xs; we arrive at(17)cs−∑j=0nKxs,xj,cj∫axsB1,ht−xjdt=gxs,s=0,1,2,3,4,…,n,which is a simple system of n+1 nonlinear equations of unknowns c0,c1,…,cn. Notice that this is a triangular system and it is solvable (the solution may not be unique because it is not linear):(18)c0=gx0,c1−h2∑j=01Kx1,xj,cj=gx1,cs−h2Kxs,x0,c0+2∑j=1s−1Kxs,xj,cj+Kxs,xs,cs=gxs,s=2,3,4,…,n.Proposition 4.
Given equation (1) and that fx,gx∈Ca,b,f′x and g′x exist and are bounded in a,b,Kx,y,u∈Ca,b×a,b×a,band∂/∂xKx,y,u and ∂/∂yKx,y,u exist and are bounded in a,b×a,b×a,b. Furthermore, Kx,y,u satisfies the following condition:(19)∫abKx,t,ft−Kx,t,utdt<LMmaxx∈a,bfx−ux,where LM<1. Let n be an integer, h=b−a/n; let xi=a+ih,fi=fxi,i=0,1,2,…,n.c0,c1,…,cn satisfies the nonlinear system (17):(20)f∗x=∑k=0nciB1,hx−kh.Then,(21)f∗x−fxa,b=Oh2,where fx is the exact solution of equation (1).


## 4.2. Method 2-V for Solving the Volterra Integral Equation

For more accurate solution, we applyL3x/h. Again, we let h=b−a/nandxi=a+ih,i=−1,0,…,n+1Furthermore, letL3,hx=L3x/h be the cardinal spline given in Section 3, and(22)fx=∑k=−1n+1ckL3,hx−xk,Kx,t,yt,Kx,t,yt=∑i=−1n+1∑j=−1n+1Kxi,xj,cjL3,hx−xiL3,ht−xj,gx=∑k=−1n+1gxkL3,hx−xk.Since the support ofL3,hx−x−1 and L3,hx−xn+1 is wider than a,b, we need values that define(23)Kx−1,xj=3Kx0,xj−3Kx1,xj+Kx2,xj,j=0,1,…,n,Kxi,x−1=3Kxi,x0−3Kxi,x1+Kxi,x2,i=0,1,…,n,Kx−1,x−1=3Kx0,x0−3Kx1,x0+Kx2,x0+3Kx0,x1−3Kx1,x1+Kx2,x1−3Kx0,x2−3Kx1,x2+Kx2,x2,gx−1=3gx0−3gx1+gx2,Kxn+1,xj=3Kxn,xj−3Kxn−1,xj+Kxn−2,xj,j=0,1,…,n,Kxi,xn+1=3Kxi,xn−3Kxi,xn−1+Kxi,x2,i=0,1,…,n,Kxn+1,xn+1=3Kxn,xn−3Kxn−1,xn+Kxn−2,xn+3Kxn,xn−1−3Kxn−1,xn−1+Kxn−2,xn−1−3Kxn,xn−2−3Kxn−1,xn−2+Kxn−2,xn−2,gxn+1=3gxn−3gxn−1+gxn−2.Letc0,c1,…,cn be the unknown coefficients to be determined and c−1=3c0−3c1+c2 and cn+1=3cn−3cn−1+cn−2. Plug into the integral equation (1); then, we have(24)∑k=−1n+1ckL3,hx−xk−λ∫ax∑i=−1n+1∑j=−1n+1Kxi,xj,cjL3,hx−xiL3,ht−xjdt=∑k=−1n+1gxkL3,hx−xk,x∈a,b.Letx=xi; we arrive at(25)cs−∑j=−1n+1Kxs,xj,cj∫axsL3,ht−xjdt=gxs,s=0,1,2,3,4,…,n,which is still a relatively simple system of equations. For the convergence rate of solution of Volterra integral equation (1), we have a similar result.Proposition 5.
Given thatfx,gx∈C3a,b,f4x and g4x exist and are bounded in a,b, Kx,t,y∈C3a,b×a,b×c,d. Furthermore, Kx,t,y satisfies the following condition:(26)∫abKx,t,ft−Kx,t,utdt<LMmaxx∈a,bfx−ux.where LM<1. Let n be an integer, h=b−a/n; let xi=a+ih,fi=fxi,i=0,1,2,…,n. c0,c1,…,cn satisfies system (35):(27)f∗x=∑k=−1n+1ckL3,hx−xk.Then,(28)f∗x−fxa,b%=Oh3,f∗x−fxa,b=Oh3,where fx is the exact solution of equation (1).


## 5. Numerical Examples

Example 1.
(from [1]). Solve(29)ux=fx+∫0xexu2t1+ex+tdt,where fx=ex/2−ln1+e2x/1+ex. For comparison, we found that the exact solution is ux=ex/2.
To find the numerical solution in0,1, we let h=0.1,n=10,xi=ih,i=0,1,2,…,10, rx,t=ex/1+ex+t,Kx,t,ut=exu2t/1+ex+t=∑k=010∑j=010rxk,tjcj2B1,h%x−xk⋅B1,ht−xj, and fx=∑k=010fxkB1,hx−xk,ux=∑k=010ckB1,hx−xk. Plug into integral equation (29), and let x=xi; we obtain the following nonlinear system:(30)ci=fxi+∫0xi∑j=010rxi,tjcj2⋅B1,ht−xjdt,i=0,1,2,…,10.
Since it is a triangular system, which means, the first equation only containsc0 and the ith equation only contains c0,c1,…,cn−1,i=0,1,2,…,10, it is relatively easy to solve; the solution is not unique because the quadratic nature c0 is unique, so we choose solution c1 that is close to c0, and so on. We arrive at c0,c1,c2,c3,c4,c5,c6,c7,c8,c9,c10 = 1,1.05127,1.10516,1.16182,1.221378,1.28399,1.34980,1.41899,1.49173,1.568120,1.64859, with error: ci−exi/2<1.5×10−4.
Leth=1/20,n=20,xi=ih, i=0,1,2,…,20, rx,t=ex/1+ex+t, Kx,t,ut=exu2t/1+ex+t=∑k=020∑j=020rxk,tjcj2B1,hx−xk⋅B1,ht−xj, fx=∑k=020fxkB1,hx−xk, and ux=∑k=020ckB1,hx−xk; plug into integral equation (29) and let x=xi; we obtain the following nonlinear system:(31)ci=fxi+∫0xi∑j=020rxi,tjcj2⋅B1,ht−xjdt,i=0,1,2,…,20.
Since it is still a triangular system, it is relatively easy to solve; the solution is not unique because the quadratic naturec0 is unique, so we choose solution c1 that is close to c0, and so on. We arrive at c0,c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12,c13,c14,c15,c16,c17,c18,c19,c20 = [1, 1.025315, 1.051271, 1.077883, 1.105169, 1.133146, 1.161831, 1.191241, 1.221396, 1.252315, 1.284015, 1.316519, 1.349845, 1.384014, 1.419049, 1.454971, 1.491801, 1.529565, 1.568284, 1.607983, 1.648688] with error <3.6×10−5.Example 2.
(from [1]). Solve(32)ux=fx−∫0xu2t+utdt,where fx=3/2−1/2e−2x and the exact solution is ux=e−x.
To find the numerical solution in0,1, we let h=0.1,n=10,xi=ih,i=0,1,2,…,10, rx,t=1, Kx,t,ut=∑k=010∑j=010rxk,tjcj2+cjB1,hx−xk⋅B1,ht−xj, fx=∑k=010fxkB1,hx−xk,and ux=∑k=0n+1ckB1,hx−xk; plug into integral equation (32), and let x=xi; we obtain the following nonlinear system:(33)ci=fxi−∫0xi∑j=010rxi,tjcj2+cj⋅B1,ht−xjdt,i=0,1,2,…,10.
It is a triangular system again, so it is relatively easy to solve. We arrive atc0,c1,c2,c3,c4,c5,c6,c7,c8,c9,c10 = 1,0.904602,0.818183,0.740144,0.669608,0.605795,0.548075,0.495864,0.448631,0.405902,0.367245, with the error: ci−uxi<8×10−4 (in the paper [1], the error was 10−3 for much smaller h).
Leth=0.05,n=20,xi=ih,i=0,1,2,…,20, rx,t=sinx−t+1, fx=cosx+1/2sinx+1/2xcosx, Kx,t,ut=∑k=020∑j=010rxk,tjcjB1,hx−xk⋅B1,ht−xj, fx=∑k=020fxkB1,hx−xk, and ux=∑k=020ckB1,hx−xk; plug into integral equation (33) and let x=xi; we obtain the following nonlinear system:(34)ci=fxi−∫0xi∑j=020rxi,tjcj⋅B1,ht−xjdt,i=0,1,2,…,10.
We arrive atc0,c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12,c13,c14,c15,c16,c17,c18,c19,c20 = 1,0.951144,0.904745,0.860597,0.818598,0.778651,0.740656,0.704517,0.670143,0.637447,0.606348,0.576766,0.548628,0.521864,0.496405,0.472189,0.449155,0.427245,0.406403,0.386579,0.367721 with error <2×10−4.Example 3.
(from [3]). Solve ∫0xsinx−t+1cosytdt=1/2xsinx+sinx,0≤x≤1.
By solvingux=cosyx first, it becomes a linear equation. To change it into Volterra integral equation of the second kind, we differentiate the equation and obtain the following:(35)ux+∫0xcosx−tutdt=cosx+12sinx+12xcosx.
To find the numerical solution in0,1, we let h=0.05,n=20,xi=ih,i=0,1,2,…,20, rx,t=cosx−t, fx=cosx+1/2sinx+1/2xcosx, Kx,t,ut=∑k=020∑j=020rxk,tjcjB1,hx−xk⋅B1,ht−xj, fx=∑k=020fxkB1,hx−xk, and ux=∑k=020ckB1,hx−xk; plug into integral equation (35), and let x=xi; we obtain the following nonlinear system:(36)ci=fxi−∫0xi∑j=020rxi,tjcj⋅B1,ht−xjdt,i=0,1,2,…,10.
We arrive atc0,c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12,c13,c14,c15,c16,c17,c18,c19,c20 = 1,0.999166,0.994962,0.989190,0.979979,0.969336,0.955202,0.939805,0.920877,0.900891,0.877347,0.852984,0.825046,0.796563,0.764495,0.732193,0.696299,0.660518,0.621139,0.582251,0.539763. The error ci−uxi<5.4×10−4.
ApplyingL3, we still let h=0.1,n=10,xi=ih,i=0,1,2,…,10, rx,t=sinx−t+1, fx=cosx+1/2sinx+1/2xcosx, Kx,t,ut=∑k=−111∑j=−111rxk,tjcjL3,hx−xk⋅L3,ht−xj, fx=∑k=−111fxkL3,hx−xk,andux=∑k=−111ckL3,hx−xk; plug into integral equation (35), and let x=xi, we obtain the following nonlinear system:(37)∑k=−111ckL3,hx−xk−∫ax∑k=−111∑j=−111rxk,tjcj⋅L3,hx−xkL3,ht−xjdt=∑k=−1n+1gxkL3,hx−xk,x∈a,b,where c−1=3c0−3c1+c2andcn+1=3cn−3cn−1+cn−2; we achieve c0,c1,c2,c3,c4,c5,c6,c7,c8,c9,c10=1,0.99501,0.98007,0.95534,0.92107,0.87759,0.82534,0.76485,0.69671,0.62162,0.54031 with error ci−uxi<1.2×10−5.
We use the same process, and leth=0.05andxi=ih,i=0,1,2,…,20, the result is c0,c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12,c13,c14,c15,c16,c17,c18,c19,c20 = 1.0,0.99875,0.995,0.98877,0.98007,0.96891,0.95534,0.93937,0.92106,0.90045,0.87758,0.85252,0.82534,0.79608,0.76484,0.73169,0.69671,0.65998,0.62161,0.58168,0.5403 with error: <4.6×10−4.Example 4.
(from [3]). ∫0xex−tlnytdt=ex−x−1,0≤x≤1.yx=ex
We change it to the linear Volterra integral equation of second kind:(38)wx+∫0xex−twtdt=ex−1,0≤x≤1,where wt=lnyt.
Leth=0.1,n=10,xi=ih,i=0,1,2,…,10, rx,t=ex−t, fx=ex−1, Kx,t,ut=∑k=010∑j=010rxk,tjcjB1,hx−xk⋅B1,ht−xj, fx=∑k=010fxkB1,hx−xk, and ux=∑k=0n+1ckB1,hx−xk; plug into integral equation (38), and let x=xi; we obtain the following nonlinear system: (39)ci=fxi−∫0xi∑j=010rxi,tjcj⋅B1,ht−xjdt,i=0,1,2,…,10.
We arrive atc0,c1,c2,c3,c4,c5,c6,c7,c8,c9,c10 = 0,0.100163,0.200317,0.300463,0.400601,0.500730,0.600851,0.700964,0.801068,0.901164,1.00125. The error ci−uxi<4×10−3.
Using the same process, leth=0.05,n=20,xi=ih,i=0,1,2,…,20, rx,t=ex−t, fx=ex−x−1, Kx,t,ut=∑k=020∑j=020rxk,tjcjB1,hx−xk⋅B1,ht−xj, fx=∑k=020fxkB1,hx−xk, and ux=∑k=0n+1ckB1,hx−xk; plug into integral equation (38), and let x=xi; we obtain the following nonlinear system:(40)ci=fxi−∫0xi∑j=020rxi,tjcj⋅B1,ht−xjdt,i=0,1,2,...,20.
We arrive atc01,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12,c13,c14,c15,c16,c17,c18,c19,c20 = 0,0.0500206,0.100041,0.150060,0.200079,0.250098,0.300116,0.350133,0.400150,0.450166,0.50018,0.550198,0.600213,0.650227,0.700241,0.750254,0.800267,0.850279,0.900291,0.950302,1.00031. The error: ci−uxi<8.5×10−4.
ApplyingL3, we still let h=0.05,n=20,xi=ih,i=0,1,2,…,20, rx,t=sinx−t+1fx=cosx+1/2sinx+1/2xcosx, Kx,t,ut=∑k=−121∑j=−121rxk,tjcjL3,hx−xk⋅L3,ht−xj, fx=∑k=−121fxkL3,hx−xk, and ux=∑k=−121ckL3,hx−xk; plug into integral equation (38), and let x=xi; we obtain the following nonlinear system: (41)∑k=−121ckL3,hx−xk−∫ax∑k=−121∑j=−121rxk,tjcj⋅L3,hx−xkL3,ht−xjdt=∑k=−121gxkL3,hx−xk,x∈a,b,where c−1=3c0−3c1+c2andcn+1=3cn−3cn−1+cn−2; we achieve c0,c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12,c13,c14,c15,c16,c17,c18,c19,c20 = 0.0,0.04,0.999998,0.09999996,0.1499999,0.1999999,0.2499999,0.2999999,0.3499999,0.39999986,0.44999984,0.49999982,0.54999980,0.59999979,0.64999977,0.69999976,0.74999974,0.79999972,0.84999971,0.89999969,0.94999968,0.99999967. The error ci−uxi<3.5×10−7 (in paper [1], the error was 10−4 for a much smaller h=1/32, they achieve the error 4.7×10−7 with h=1/512).


## 6. Conclusion

The proposed method is a simple and effective procedure for solving nonlinear Volterra integral equations of the second kind. The methods can be adapted easily to the Volterra integral equations of the first kind. The methods can also be extended to the Fredholm and Volterra integral equations of the first kind or the second kind, where the integral is on an infinite set. When the higher degree cardinal splines are applied to nonlinear integral equations, the resulting system of coefficients will be a little more complicated nonlinear system, which takes more time and effort to solve. Compared with the recent papers [1] and [3], our method is more effective and efficient. Furthermore, we could use higher dimensional spline functions to solve higher dimensional integral equations.



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*Source: 3236813-2020-02-12.xml*