Introduction to Ordinary Differential Equations pter is about the most basic con

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Introduction to Ordinary Differential Equations pter is about the most basic concepts of the theory of differential equations. We will answer some fundamental questions: What is a differen- ial equation? Do differential equations always have solutions? Are solutions of differential equations unique? But, the most important goal of this chap- ter is to introduce a geometric interpretation for the space of solutions of a differential equation. Using this geometry, we will introduce some of the el- ements of the subject: rest points, periodic orbits, and invariant manifolds. Finally, we will to prove the classic theorems on the existence, uniqueness, and extension of solutions. References for this chapter include [1o,13), 157), 150), 101), 100, [123), [141), 183), [209), and [226. 1.1 Existence and Uniqueness Let J-R, U-R", and Rk be open subsets, and suppose that f : J × U × Rn is a smooth function. Here the tern "smooth" mieans that the function f is continuously differentiable. An ordinary differential equation (ODE) is an equation of the form = f(t,x,A) where the dot denotes differentiation with respect to the independent vari- able t (usually a measure of time), the dependent variable r is a vector of state variables, and is a vector of parameters. As convenient terminology

Explanation / Answer

Introduction

A most general form of an ordinary differential equation (ode) is given by

f( x, y, y', . . ., y(m) ) = 0

where x is the independent variable and y is a function of x.  y', y'' . . . y(m) are respectively, first, second and mth derivatives of y with respect to x.

Some definitions:

The general solution of any mth order equation is a linear combination of m linearly independent functions of the form

f(x) = c1f1(x) + c2f2(x) + . . . + cmfm(x)

where c1, c2, . . . , cm are constants and f1(x), f2(x), . . . , fm(x) are mindependent functions also satisfy the differential equation

There are m conditions on the dependent variable y or on its derivatives are required to find the m constants c1, c2, . . . , cm in the general solution. If these conditions are given at the initial point say at x = x0 then the differential equation along with the conditions is called an initial value problem (ivp). On the other hand if the conditions are given at more than one point the problem is called a boundary value problem (bvp).

Another important concept in ode is the reduction of an mth order ode into a system of m first order ordinary differential equations.

Consider the mth order initial value problem

ym(x) = f(x, y, y', . . ., y(m-1)),       y(x0) = a0,   y'(x0) = a1, . . . , y(m-1)(x0) = am-1

This can be written as  

or in the vector form    U' = F(x, U),   U(x0) = A

where U = (u0, u1, . . ., um-1)T
           F = (u1, u2, . . ., um-1, f)T
          A = (a0, a1, . . ., am-1)T

Thus, it is possible to convert an mth order ivp with a system of mfirst order ivps and solve them together. So to solve ordinary differential equations numerically, it is important to understand the methods of solutions to the first order equations and first order system of equations. Hence the remaining part of this section is devoted to explain the schemes to solve the first order initial value problems and extension of the same to solve first order initial value systems.

Consider the first order initial value problem

y' = dy/dx = f(x, y) ,   y(x0) = y0                                                        (1)

Existence of the solution:

It is not necessary that any ivp of the form y' = f(x,y), y(x0) = y0has a solution.

For example, consider y' = 1+y4,   y(0) = 0, Since the slope is positive at x = 0, y(x) is increasing near x = 0. Therefore, 1+y4is also increasing. Hence, y' is increasing. Since y and y' are both increasing and are related by y' = 1+y4, one can expect that at some finite value of x there will be no solution. That is, the solution may not exists or even if it exists it may exists only in a particular interval. For the existence of the solution the right hand side function f(x,y) of the ivp has to satisfy certain conditions and those conditions are given in the following theorems.

Theorem 1:If f is continuous in a rectangle R centered at (x0, y0), say

R = { (x, y) : |x - x0| £a, |y - y0| £ b}

then the ivp (1) has a solution y(x) for |x - x0| £ min(a, b/M), where M is the maximum of |f(x,y)| in the rectangle R.

Example: Prove that the initial value problem y' = (x + sin y)2, y(0) = 3 has a solution on the interval -1 £ x £ 1.

Solution: Here  f(x, y) = (x + sin y)2 , (x0, y0) = (0, 3) and the rectangle

R = { (x, y) : | x | £a, |y - 3| £ b}

f(x, y) is bounded by |f(x, y)| £ (a +1)2 = M

if  a to be equal to 1 then   min(a, b/M)has to be greater or equal to 1.     Þ        M = 4.

Then by theorem 1 there exists a solution on the interval -1 £ x £ 1 for the given ivp by letting
b ³ 4.

Example: Prove that the initial value problem y' = tan x, y(0) = 0 has a solution on the interval -p/4 £ x £ p/4.

Uniqueness:Theorem 1 guarantees the existence but not the uniqueness of the solution to the given ivp. for example consider the ivp y' = y2/5, y(0) = 0. Clearly, y = 0 is a solution for this ivp. On the other hand if we solve analytically the above ivp we get y = (3x/5)5/3. That is, there exists atleast two solutions for the given ivp. Hence for the existence of an unique solution to the given ivp it is necessary to impose another condition on f.

Theorem 2: If f and ¶f/¶y are continuous in the rectangle

R = { (x, y) : | x | £a, |y - 3| £ b}

then the initial value problem

y' = f(x,y), y(x0) = y0

has an unique solution in the interval |x - x0| £ min(a, b/M).

Note:  In theorems 1 and 2 the interval on x - axis for existence and uniqueness of the solution may be smaller than the base of the rectangle in which f(x, y) has been defined. This can be avoided if f(x, y) satisfies the following theorem.

Theorem 3: If f is continuous in the strip a £ x £ b, -¥ < y <¥ , and also satisfy the inequality

| f(x, y1) - f(x, y2) | £ L| y1 - y2 |

then the initial value problem y' = f(x, y), y(x0) = y0 has a unique solution in the interval [a, b].

Note: The condition on f in the theorem 3 is called Lipschitz condition with Lipschitz constant L.

Example: Show that the function f(x) = åni = 1 ai | x - wi | satisfies the Lipschitz condition with the Lipschitz constant L = åni = 1 |ai|.
Solution:
                 | f(x1) - f(x2) | = |åni = 1 ai | x1 - wi | - åni = 1 ai | x2 - wi| |
                     = | åni = 1 ai{ | x1 - wi | - | x2 - wi | }|
                     £  åni = 1 |ai| | | x1 - wi | - | x2 - wi | |
                     £  åni = 1 |ai| | x1 - x2 |
                                  £  L | x1 - x2 |

Thus if the function f in the given initial value problem satisfies theorems 1 and 2 or theorem 3 then there exists an unique solution to the given ivp and the aim of the present chapter is to explain various numerical schemes to solve this kind of first order ivp for which there exists an unique solution.

Numerical solutions of ivps can't give analytical expressions of the form y(x) as solutions, instead generate data namely (xi, yi) for i =1, 2, 3, . . . , n, where yi is nothing but the value of the dependent variable at the point xi. Hence in finding the numerical solutions toivp first one has to identify the points xi on the abscissa. This process is called discretisation of the domain and the points xi are called grid points, mesh points or nodal points.

u0 = y u0' = y' = u1 u1' = y'' = u2 .
.
. um-2' = um-1 u'm-1 = f(x, u0, u1, . . ., um-1) and initial conditions
u0(x0) = a0, u1(x0) = a1, . . . , um-1(x0) = am-1

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