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The Cardinal Spline Methods for the Numerical Solution of Nonlinear Integral Equations
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.
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:where the kernel and are known functions and is to be determined.
This paper is divided into six sections. In Sections 2 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  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 ). 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 polynomial B-spline (see Figure 1), with the integral iteration formula
We could construct higher-order polynomial spline functions with a higher degree of smoothness. More specifically, has the expression (see Figure 2) as follows:
are called one-dimensional B-splines, which are polynomial splines and have small supports ,i.e., for or and excellent traits (cf ). 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 .
Let . It is proved that
Note that this particular B-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 that exists and is bounded in . Let be an integer; ; let as follows:Then,where denotes the approximation polynomial of .
If and exists and is bounded, let be a real number, and let Then,
3. A Univariate Cardinal Spline
By cardinal conditions (cf ), we mean that let be a function, be interpolation points; then,
Figure 3 shows the graph of .
Then, satisfies the above cardinal condition when . Notice that by the construction . for . is a polynomial of degree in each subinterval of its support.
Furthermore, from direct calculation, we deduct the following two propositions (cf ).
Proposition 2. Let be the cardinal spline constructed above; then,where are any complex number.
Proposition 3. If let be an integer; let
If and is bounded, let be a real number; let
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 . Again, we let and Furthermore, plugging , we get
Let ; we arrive atwhich is a simple system of nonlinear equations of unknowns . Notice that this is a triangular system and it is solvable (the solution may not be unique because it is not linear):
Proposition 4. Given equation (1) and that and exist and are bounded in and exist and are bounded in . Furthermore, satisfies the following condition:where . Let be an integer, ; let satisfies the nonlinear system (17):Then,where is the exact solution of equation (1).
4.2. Method 2-V for Solving the Volterra Integral Equation
For more accurate solution, we apply . Again, we let
Furthermore, let be the cardinal spline given in Section 3, and
Since the support of and is wider than , we need values that define
Let be the unknown coefficients to be determined and and . Plug into the integral equation (1); then, we have
Let ; we arrive atwhich 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 that and exist and are bounded in , . Furthermore, satisfies the following condition:where . Let n be an integer, ; let . satisfies system (35):Then,where is the exact solution of equation (1).
5. Numerical Examples
Example 1. (from ). Solvewhere . For comparison, we found that the exact solution is .
To find the numerical solution in , we let , , and . Plug into integral equation (29), and let ; we obtain the following nonlinear system:Since it is a triangular system, which means, the first equation only contains and the ith equation only contains , it is relatively easy to solve; the solution is not unique because the quadratic nature is unique, so we choose solution that is close to , and so on. We arrive at = , with error: .
Let , , , , , and plug into integral equation (29) and let ; we obtain the following nonlinear system:Since it is still a triangular system, it is relatively easy to solve; the solution is not unique because the quadratic nature is unique, so we choose solution that is close to , and so on. We arrive at = [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 .
Example 2. (from ). Solvewhere and the exact solution is .
To find the numerical solution in , we let , , , and plug into integral equation (32), and let ; we obtain the following nonlinear system:It is a triangular system again, so it is relatively easy to solve. We arrive at = , with the error: (in the paper , the error was for much smaller h).
Let , , , , , and ; plug into integral equation (33) and let ; we obtain the following nonlinear system:We arrive at = with error .
Example 3. (from ). Solve .
By solving 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:To find the numerical solution in , we let , , , , , and ; plug into integral equation (35), and let ; we obtain the following nonlinear system:We arrive at = . The error .
Applying , we still let , , , , ; plug into integral equation (35), and let , we obtain the following nonlinear system:where ; we achieve with error .
We use the same process, and let , the result is = with error: .
Example 4. (from ).
We change it to the linear Volterra integral equation of second kind:where .
Let , , , , , and ; plug into integral equation (38), and let ; we obtain the following nonlinear system: We arrive at = . The error .
Using the same process, let , , , , , and ; plug into integral equation (38), and let ; we obtain the following nonlinear system:We arrive at = . The error:
Applying , we still let , , , , and ; plug into integral equation (38), and let ; we obtain the following nonlinear system: where ; we achieve =