Numerical Solutions of Boundary Value Problems of Non-linear Differential Equations

Sujaul Chowdhury, Syed Badiuzzaman Faruque, Ponkog Kumar Das

• Editorial: Chapman and Hall/CRC
• Año: 2021
• ISBN:
  9781000486148
• Edición: 1
• Categoría: Mathematics
• Categoría: Differential Geometry

The book presents in comprehensive detail numerical solutions to boundary value problems of a number of non-linear differential equations. Replacing derivatives by finite difference approximations in these differential equations leads to a system of non-linear algebraic equations which we have solved using Newton's iterative method. In each case, we have also obtained Euler solutions and ascertained that the iterations converge to Euler solutions. We find that, except for the boundary values, initial values of the 1st iteration need not be anything close to the final convergent values of the numerical solution. Programs in Mathematica 6.0 were written to obtain the numerical solutions.

1

INTRODUCTION

DOI: 10.1201/9781003204916-1

In this chapter, we have narrated the problem and the methodology for the entire book.

1.1 THE NON-LINEAR DIFFERENTIAL EQUATIONS WE SOLVED IN THIS BOOK

We have numerically solved boundary value problems of the following non-linear differential equations:

$\frac{dy}{dx}+y^2=f(x)$ (1.1)

$\frac{dy}{dx}+y^3=f(x)$ (1.2)

$\frac{d^{2}y}{d{x}^2}+y^2=f(x)$ (1.3)

$\frac{d^{2}y}{d{x}^2}=-\frac{y}{{(y^2+r^2(x))}^{3/2}}$ (1.4)

where

(1.5)

where c is a constant parameter and f(x)’s are functions of x. That is, we have obtained tabulated set of values (x_i, y_i) in the interval x = a to b as numerical solutions of these differential equations, provided values of y for x = a and x = b are known or given. Equation (1.4) comes from reference [1].

1.2 APPROXIMATION TO DERIVATIVES

Let us divide the interval x = a to b in n equal parts, each part being h. As such, x₀ = a, x₁ = a + h, x₂ = a + 2h, x₃ = a + 3h, …, x_i = a + ih, …, x_n = b. Let y_i be the value of y for x = x_i.

With e.g. polynomial interpolation in mind, we can regard y as a continuous function of x. As such, Taylor’s series gives us

• y(x + h) = y(x) + h y′(x) (1.6)

if y is a slowly varying function of x. More accurately, we have

• y(x + h) = y(x) + h y′(x) + $\frac{h^2}{2}$ y′′(x) (1.7)

Again, we have

• y(x − h) = y(x) − h y′(x) (1.8)

and

• y(x− h) = y(x) − h y′(x) + $\frac{h^2}{2}$ y′′(x) (1.9)
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