8.1 Section 2 Background .1 Cosmology .11 Math Section 8.3

#### Book 9 Computational Mathematics Chapter 8 Ordinary Differential Equation Initial Value Problems

Section 2 Solution by Taylor Series

Consider the I.V.P. (1.1) with the initial conditions (1.2). Let the required solution y = y(x) be expandable in a Taylor Series in powers of the difference (x - x0):

(2.1)

The initial conditions (1.2) yield directly the values of y(k)(x0), k = 0, 1, 2, ?n-1. The value of y(n)(x0) is found from (1.1) by setting x = x0 and reusing the initial conditions (1.2):

(2.2)

The values of y(n+1)(x0) , y(n+2)(x0), ? can be determined successively by differentiation of Equation (1.1) and setting x=x0. y(k)(x0) = y0k , k = 1, 2, 3, ?. It can be shown that, if the right hand side of Equation (1.1) in the neighbourhood of the point ( x0, y0, y'0, . . .. y0(n-1)) is an analytic function of its arguments, then there exists for values of x sufficiently close to x0 a unique solution of the Cauchy Problem (1.1), which can be expanded in a Taylor Series (2.1), whence the partial sum of this series will be an approximate solution of the problem under consideration. This method can also be applied for the solution of systems of ordinary differential equations.

Example 1: Find the first seven terms of the expansion in a power series of the solution y = y(x) of the equation

for the initial conditions y(0) = 1, y' (0) = 2.

Solution: Look for a solution of the equation in the form of the series

In order to find y"(0), solve the given equation with respect to y":

(2.3)

Thus

Next, successive differentiation of both sides of Equation (2.3) with respect to x yields:

Substitution of the initial conditions and the value of y"(0) yields

Thus, the required approximate solution assumes the form

Example 2: Find the first four terms of the expansions in power series of the solutions y = y(x) and z = z(x) of the system of differential equations

(2.4)

for the initial conditions y0) = 1, z(0) = 0.

Solution: Seek the functions y(x) and z(x) in the form

(2.5, 2.6)

The initial conditions yield directly y(0) =1, z(0) = 0. Setting x = 0 in (2.4) and taking into consideration these conditions, one finds

Differentiation of (2.4) with respect to x yields:

(2.7)

Hence,

Differentiate now System (2.7) with respect to x to obtain:

Hence, y"' (0) = 0, z"' (0) = 3. With these values of the derivatives, the series (2.5 )and (2.6) yield

Note: For a successful application of certain numerical methods of integration of differential equations, one must determine the values of the required functions at several points. These values can be computed with the aid of power series. Thus, expansion in power series can be used in more effective numerical methods of approximate integration of differential equations (cf. the methods of Adams, Milne, etc.). An example of forming a table of values of solutions at certain intervals follows.

If an equation contains a singularity, say, an indeterminate expression of the form 0/0, a numerical solution is impossible. Use of power series may enable circumvention of such irregularities (Exercises 11 to 13).

Example 3: For the function y = y(x) which satisfies the equation

(2.8)

and the initial conditions y(0) = 0, y'(0) = 1, form a table of values in the range [0; 0.2] with intervals h = 0.05 to an accuracy of 0.5?10-4.

Solution: Rewrite the function y = y(x) as the Taylor Series

where y(0) = 0, y'(0) = 1. By (2.8),

Hence y"(0) = -y(0) = 0.

Successive differentiation of Equation (2.8) yields

and, in general,

whence

Moreover,

Hence

This is an alternating series with terms the absolute values of which decrease monotonically on the interval [0; 0.2].

Since the absolute value of the remainder term of such a series is less than that of the first rejected term,
the approximation

yields values of the function on the interval  (0,0.2)  with an error of less than

Hence one finds the required table of values

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Exercises

Applying successive differentiation, find the solutions of the following equations and systems of equations for the given initial conditions in the form of partial sums of series, including up to five terms:

Find the solutions of the following equations using the method of undetermined multipliers:

14. Form a table of the values of the solution on the interval [0; 0.15] with h=0.05, using partial sums:

(a) For Equation 1, (b) for System 7, (c) for System 8.