APPROXIMATE SOLUTIONS OF GENERAL SECOND – ORDER INITIAL VALUE PROBLEMS USING DIFFERENTIAL EVOLUTION

In this paper, it is assumed that the solution of the general second order initial value problem u′′ = f (t,u,u′) ; u(t0) = u0, u(t0) = u0, t ∈ [t0, tn] can be approximated by a polynomial u(t). To obtain the values of the coefficients of the terms of u(t), the problem is converted to an optimization problem and the simple stochastic function minimizer called differential evolution is used to obtain the optimal values of the coefficients. Numerical examples show the efficiency and accuracy of the proposed technique compared with some existing classical methods.


Introduction
In recent years, the quest for substantially improved accuracy of numerical methods for the solution of ordinary differential equations is on the increase.One of the approach that is gaining more popularity is the idea of converting the differential equation problem to an optimization problem.The transformed problem is then solved using the techniques of optimization.Interestingly, evolutionary algorithms for solving optimization problems appear to be ideal for this type of transformation.
The author in [4] used the classical genetic algorithm to obtain approximate solutions of secondorder initial value problems.Approximate solutions of firstorder initial value problem was computed via the combination of collocation method (finite elements) and genetic algorithms by the author in [6].In an earlier work, the author in [5] combined the genetic algorithm with the Nelder-Mead method for solving the secondorder initial value problem of the form u = f (x, y).The idea of adapting neural network for the solution of secondorder initial value problems was also proposed by the authors in [2].The use of continuous genetic algorithm for the solution of the twopoint secondorder ordinary differential equation was discussed by the authors in [1].In an early work by the authors in [7], the adaptation of the differential evolution algorithm for the solution of the secondorder initial value problem of the form u + p(t)u + q(t)y = r(t) was proposed.For the second-order two-point boundary value problem u = f (t, u); u(a) = η 1 ; u(b) = η 2 with oscillatory/periodic behaviour, the authors in [3] used the differential algorithm to obtain approximate solutions.In this paper, the second-order initial value problem of the form is considered.It is assumed that the solution can be approximated by a polynomial u(t).The differential evolution algorithm is used to optimize the coefficients of the terms of the solutions u(t).

Basic Notions of Differential Evolution Algorithm
Formally, let f : R n → R be the function which must be optimized.The function takes a candidate solution as argument in the form of a vector of real numbers and produces a real number as output which indicates the fitness of the given candidate solution.The gradient of f is not known.The goal is to find a solution m for which f (m) ≤ f (p) for all p in the search-space, which would mean m is the global minimum.Maximization can be performed by considering the function h := − f instead.Let x ∈ R n designate a candidate solution (agent) in the population.The basic differential evolution algorithm can then be described as follows: • Initialize all agents x with random positions in the search-space.
• Until a termination criterion is met (e.g.number of iterations performed, or adequate fitness reached), repeat the following: -For each agent x in the population do: * Pick three agents a, b, and c from the population at random, they must be distinct from each other as well as from agent x * Pick a random index R ∈ {1, . . ., n} (n being the dimensionality of the problem to be optimized).* Compute the agent's potentially new position y = [y 1 , . . ., y n ] as follows: • For each i, pick a uniformly distributed number r i ≡ U(0, 1) • (In essence, the new position is outcome of binary crossover of agent x with intermediate agent z = a + F(b − c).) * If f (y) < f (x) then replace the agent in the population with the improved candidate solution, that is, replace x with y in the population.
• Pick the agent from the population that has the highest fitness or lowest cost and return it as the best found candidate solution.
Note that F ∈ [0, 2] is called the differential weight and CR ∈ [0, 1] is called the crossover probability, both these parameters are selectable by the practitioner along with the population size NP ≥ 4.

Proposed Method
Consider the second order initial value problem (1), assume a solution of the form where ψ i are parameters to be determined.Substituting (2) and its derivatives into (1) gives Using the initial conditions we have the constraint that At each node point t n , we require that To solve the above problem, we need to find the set {ψ i |i = 0(1)k}, which minimizes the sum of square of the error at each node point given by the expression where N = b−t 0 h and h is the steplength.We now formulate the problem as an optimization problem in the following way: Minimize :

Numerical Experiments
To investigate the performance and efficiency of the technique, two numerical test cases are inplemnted.The propose algorithm is compared with the Runge-Kutta Nystrom scheme for solving (1).The table of "CPU-time" and the maximum error of all computations are also given.
The following parameters are used for all computations: All computations were carried out on a "Core i5 Intel" processor machine.

Problem 1
Consider the initial value problem with the exact solution u(t) = exp(t).

Conclusion
In this paper, we have been able to derive and implement an evolutionary scheme for the solution of the second order initial value problem (1) using the techniques of optimimization (Differential Evolution).With the two test problems, clearly, the technique gave very accurate results compared with the well-known Runge-Kutta method for the direct integration of (1).

TABLE 1 .
Maximum Absolute Error and CPU-time in seconds for Problem with step-size h = 2 −i , i = 3(1)9

TABLE 2 .
Maximum Absolute Error and CPU-time in seconds for Problem with step-size h = 2 −i , i = 3(1)9