Ordinary Differential Equations

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 m^th derivatives of y with respect to x. 

Some definitions:

• the highest derivative in the ode is called the order of the ode.

• the degree of the highest order derivative in the ode is called the degree of the ode.

• the differential equation is linear if no product of the dependent variable y(x) with itself or with any one of its derivatives occur in the equation otherwise is non-linear.

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

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

where c₁, c₂, . . . , c_m are constants and f₁(x), f₂(x), . . . , f_m(x) are m independent 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, . . . , c_m 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 m^th order ode into a system of m first order ordinary differential equations.

Consider the m^th order initial value problem 

y^m(x) = f(x, y, y', . . ., y^(m-1)),       y(x₀) = a₀,   y'(x₀) = a₁,  . . .  ,  y^(m-1)(x₀) = a_m-1

This can be written as 
 

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

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

where U = (u₀, u₁, . . ., u_m-1)^T
           F = (u₁, u₂, . . ., u_m-1, f)^T
          A = (a₀, a₁, . . ., a_m-1)^T

Thus, it is possible to convert an m^th order ivp with a system of m first 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) = y₀ has 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+y⁴ is 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' = y^2/5,  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) = åⁿ_{i = 1} a_i | x - w_i | satisfies the Lipschitz condition with the Lipschitz constant L = åⁿ_{i = 1} |a_i|.
Solution:
                 | f(x₁) - f(x₂) |  = |åⁿ_{i = 1} a_i | x₁ - w_i | - åⁿ_{i = 1} a_i | x₂ - w_i | |
                     = | åⁿ_{i = 1} a_i{ | x₁ - w_i | -  | x₂ - w_i | }|
                     £  åⁿ_{i = 1} |a_i| | | x₁ - w_i | -  | x₂ - w_i | |
                     £  åⁿ_{i = 1} |a_i|  | 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 (x_i, y_i) for i =1, 2, 3, . . . , n, where y_i  is nothing but the value of the dependent variable at the point x_i. Hence in finding the numerical solutions to ivp first one has to identify the points x_i on the abscissa.  This process is called discretisation of the domain and the points x_i are called grid points, mesh points or nodal points.

Discritisation:

The first step in solving any ivp y' = f(x, y), y(x₀) = y₀ is the discritisation of the domain. Since the initial point of the independent variable x is  given as x₀, the interval [x₀, x_n], where x_n  is the point where  the solution of  the ivp is required,  is the domain of interest. Now discritising the domain [x₀, x_n] is nothing but identifying the points between x₀ and x_n at which  the solution is obtained in a marching process. Let the points  be x_i,   i = 0, 1, 2, . . ., n defined as

 x_i = x₀  +  i x h,        i = 0, 1, 2, . . ., n

where h = (x_n - x₀)/n is called the step length. Once the grid points (also called as nodal or mesh points) are identified, different numerical schemes can be used to  find the value of the dependent variable at mesh points x₁, x₂, . . ., x_n.