Power Series Solution of Ordinary Differential Equations

Power series solution of ordinary differential equations

Objective

To study Series solution of ordinary differential equations

Modules

Module I- Introduction

Module II- Power series solution

Module III- Frobenius method

Module I- Introduction

The solution of ordinary differential equations which can be termed as closed form solutions. However it is often not possible to obtain closed form solutions for differential equations with variable coefficients. In such cases the solution is obtained as an infinite series in terms of independent variable. In this session we shall consider two methods namelypower series method and Frobenius method, which is an extension of the power series method, to solve differential equations with variable coefficients.

Classification of singularities

Consider a homogeneous linear second order differential equation with variable coefficients:

(1)

Assuming, the above equation is written in the standard form as

(2)

where and .

Analytic function

A function f(x) is said to be analytic at x0 if f(x) has Taylor’s series expansion about x0 given by

exists and converges to f(x) for all x in some open interval including x0.

If a function f(x) is not analytic at x0, then it is said to be singular at x0.

Regular or ordinary point

A point x0 is said to be a regular or ordinary point of the differential equation (2), equivalently (1), if both P(x) and Q(x) are analytic at x0.

Singular point

A point x0 is said to be a singular point of the differential equation (2) if either of P(x) or Q(x) or both are not analytic at x0.

Singular points are classified as regular singular point and irregular singular point as follows:

A singular point x0 of differential equation (2) is called a regular singular point if both (x – x0)P(x) and (x – x0)2Q(x) are analytic at x0, otherwise it is called an irregular singular point.

Example1.Consider the equation.

Here and are polynomials, both are analytic everywhere.

Therefore any point x is an ordinary or regular point of the given differential equation.

Example 2. Consider the equation .

This can be written in the standard form as .

Here and , both are not analytic at .

Therefore the given differential equation has two singular points at x = +1 and x = -1.

Further since both and are analytic at x = 1, x = 1 is a regular singular point.

Similarly since both and are analytic at x = -1, x = -1 is also a regular singular point.

Example 3. Consider the equation.

Here and

both are not analytic at x = 0 and x = 1. Since is not analytic at x = 0, it is an irregular singular point. Further since both and are analytic at x = 1, it is a regular singular point.

Module II- Power series solution

An expression of the form c0 + c1(x – x0) + c2(x – x0)2 + … + cn(x – x0)n + …

or in the summation form

(3)

is known as a power series of the variable x in powers of (x – x0) or about the point x0. The constants c0, c1, …, cn, … are known as the coefficients and x0 is known as the centre of the power series. Since n takes only positive integral values, the power series does not contain negative or fractional powers. So the power series contains only positive powers.

Theorem. If x0 is a regular (or ordinary) point of the differential equation (2), then a general solution of (2) is obtained as a linear combination of two linearly independent power series solutions of the form

and these power series both converge in some interval (with R > 0).

Steps in power series method

Step I: Assume that (4)

be the solution of (2).

Step II: Substitute obtained by differentiating (4) term wise, in (2). Collect the coefficients of like powers of (x – x0). This converts the differential equation (2) into the form

(5)

where ki (i = 0, 1, 2, 3, …) are functions of certain coefficients cn.

Step III: If (4) is a solution of (2), all ki’s must be zero. Solve k0 = 0, k1 = 0, k2 = 0, … for the unknown coefficients cn’s.

Generally this leads to a recurrence relation between cn’s, which helps to determine unknown coefficients in terms of the other known coefficients.

Thus cn’s are determined by equating to zero each power of (x – x0) in (5).

Step IV: Substitution of these cn’s in (4) gives the required power series solution of (2).

Example 4.Find apower series solution in powers of x of the differential equation

Solution.

Let(1)

be the solution.

Therefore

and

From the given equation we obtain

(2)

Equating each of the coefficients to zero, we obtain the identities

, , , …

From which we find that

, , , ,…

Substituting these values into equation (1), we obtain the power series solution to the given differential equation.

Module III- Frobenius method

In power series method the solution was obtained only when x0 is an ordinary point of the given differential equation. Solution near regular singular point x0 can be obtained by an extension of the power series method known as Frobenius method (or generalized power series method). In this method we consider a series of the form

(1)

known as Frobenius series. Here r is an unknown constant to be determined.

Theorem. Let x0 be a regular singular point of the differential equation

(2)

Then (2) has at least one nontrivial solution of the form

(3)

which is convergent in some deleted interval about x0, with R > 0.

Note. In the following analysis, for simplicity, we consider the interval as . Solutions valid in the interval can be obtained by replacing by .

Steps in Frobenius method

Step I: Assume a solution of the form

(4)

for the differential equation (2) with x0 as a regular singular point. The series (4) is valid in . Here the exponent r is chosen so that, which simply means that the highest possible power of x is factored out.

Step II: Substitute obtained by differentiating (4) term wise, in (2). Collecting the coefficients of like powers of (x – x0), equation (2) takes the form

(5)

where the coefficients ki (i = 0, 1, 2, 3, …) are functions of the exponent r and the constant coefficients cn. Also k is an integer.

Step III: Since (4) is assumed to be the solution of (2), we must have

k0 = k1 = k2 = … = 0

Here k0 is the coefficient of the lowest power r + k of (x – x0). The equation k0 = 0 is a quadratic equation in r, known as the indicial equation of the differential equation (2). The roots r1 and r2 of the indicial equation are known as the exponents of (2) and are the only possible values for the constant r in the assumed solution (4). Generally r1 is taken as the larger root so that r1r2.

Step IV: Now that r is known, solving the equations k1 = 0, k2 = 0, k3 = 0, …, we determine completely the unknown constant coefficients cn’s.

Step V: Let y1(x) and y2(x) be the two non trivial linearly independent solutions of the differential equation (2). Then the general solution of (2) is y(x) = Ay1(x) + By2(x), where A and B are arbitrary constants. Then with the known exponent r and the known coefficients cn’s, y1(x), one of the two solutions of (2) is of the form (4). The form of the second solution y2(x) may be similar to (4) or may contain a logarithmic term. The form of y2(x) will be indicated by the indicial equation. There are three cases:

Case 1. Distinct roots not differing by an integer 1, 2, 3, ..

Suppose , where N is a non-negative integer (i.e., ). Then

and

Here .

Case 2. Roots differing by an integer 1, 2, 3, ..

Suppose , where N is a positive integer. Then

and

Here and the constant A* may or may not be zero.

Case 3. Double root r1 = r2

Suppose . Then

with and

Note: If , where N = 1, 2, 3, …sometimes it is possible to obtain the general solution using the smaller root alone, without bothering to find explicitly the solution corresponding to the larger root.

In case 2, the second solution may or may not contain the logarithmic term . In some cases A* is zero, so y2 is same as y2 in case 1.

In case 3, y2 always contain the logarithmic term and is never of the simple form in case 1.

In case (2), take the first solution y1(x, r), then the second solution is . i.e., derivative of y1(x, r) partially with respect to the exponent r. Then taking r as the smaller root r2, we get two linearly independent solutions y1(x, r2) and . Here the second solution y2 contains the logarithmic term.

Example 5. Find the general solution of .

Solution.

The given equation can be written as

So x = 0 is a regular singular point, since and are both analytic at the point x = 0. Assume the solution of the given equation as

Differentiating y with respect x twice, we get

and.

Substituting the values of in the given equation, we obtain

Equating to zero the coefficient of lowest power of x, we get

[8(0 + r)(0 + r -1)+10(0 + r) - 1]c0 = 0

i.e.,8r2 + 2r – 1 = 0, since ,

which is the indicial equation, whose roots are and .

Here the two roots are real, distinct and not differ by an integer.

To obtain the recurrence relation the last summation in the above equation is rewritten.

Then

Thus the recurrence relation is

To obtain y1(x), put . Then the recurrence relation reduces to

For n = 1,

For n = 2,

For n = 3,

For n = n - 1,

For n = n, .

Then multiplying these, we can express cn in terms of c0 as

Hence the first solution is

i.e.,.

Generally c0 is assumed to be equal to 1.

In a similar way, to obtain the second solution y2(x), put in the recurrence relation which then reduces to

Then multiplying c1, c2, c3,…, we can express cn in terms of c0 as

Choose c0 = 1, we have

Thus the required general solution is

Example 6. Obtain the series solution of the equation .

Solution.

Here x = 0 is a regular singular point.

Let

be a solution.

Then

and.

Substituting the values of in the given equation, we obtain

The lowest power of x is xm. Therefore the indicial equation is

m(m - 1) + m – 1 = 0 m2 = 1 .

Hence the roots are differing by an integer. Let us start with the smallest root. i.e., m = -1.

On re writing the summation, we have

Therefore we obtain the following recurrence relation

and at m = -1

For r = 2,

But for m = -1, c2 is undefined. Also the consequent terms depending on c2 will be undefined.

Letc0 = d0(r + 1)

Therefore

For r = 3,

Similarly, c5 = c7 = … = 0

For r = 4, .

Hence one independent solution is

Form = -1,

Now

At m = -1,

Hence the complete solution is

Summary

In this session we havedefined regular point, regular singular point and irregular singular point of a differential equation Also we discussedpower series method and Frobenius method to solve differential equations with variable coefficients. In Frobenius method we have seen that the solution depends on the nature of the roots of indicial equation.

Assignments

1. Locate and classify the ordinary points, regular singular point and irregular singular point of the following differential equations:

(i)

(ii)

(iii)

2. Obtain power series solution in powers of x for the differential equation

3. Using Frobenius method, obtain two linearly independent solutions about x0 = 0 for the equation

Reference

Advanced Engineering Mathematics by E. Kreyszig, John Wylie & Sons, New York (1999).
Differential Equations by F. Simmons, Tata McGraw-Hill, New Delhi (1996).
Advanced Engineering Mathematics by Michael D. Greenberg, Pearson Education pvt. ltd. Delhi (2002).

Quiz

1. In if both and are analytic at x0, then

(a) The point x0 is said to be a regular point of the differential equation

(b) The point x0 is said to be a singular point of the differential equation

2. For the equation , x = 1 is

(a) an irregular singular point

(b) a regular singular point

3. A point x0 is said to be a singular point of the differential equation where and if

(a) both P(x) and Q(x) are analytic at x0.

(b) either of P(x) or Q(x) or both are analytic at x0.

4. A singular point x = 0 of differential equation where and is called a regular singular point

(a) if both x P(x) and x2 Q(x) are analytic at x0

(b) if both x2P(x) and xQ(x) are analytic at x0

5. The power series solution of is

(a)

(b)

(c)

6. The indicial equation of is

(a) a linear equation

(b) a cubic equation

Answers

1.a2.b3.c4.a5.b6.c

Glossary

Function: A relation from a set X to another set Y which associates every element of X with a unique element of Y is called a function.

Differential equation: An equation containing ordinary differential coefficients is called an ordinary differential equation.

Solution: The relation between independent variable and the dependent variable from which a differential equation is formed is called the solution of the differential equation.

Quadratic equation: An equation whose degree is two is called a quadratic equation.

Roots of equation: A solution of equation f(x) = 0 is called a root of the equation.

FAQs

1. Define Regular point, Singular point, regular singular point and irregular singular point of where and .

Answer

Regular point: A point x0 is said to be a regularpoint of the given differential equation, if both P(x) and Q(x) are analytic at x0.

Singular point: A point x0 is said to be a singular point of the given differential equation if either of P(x) or Q(x) or both are not analytic at x0.

A singular point x0 of the given differential equation is called a regular singular point if both (x – x0)P(x) and (x – x0)2Q(x) are analytic at x0, otherwise it is called an irregular singular point.

2. Solve in series the equation .

Answer

Here x = 0 is a regular point. Let

be the solution.

Therefore

and

Substitute these values in the given equation and equate to zero the coefficients of various powers of x, we get

Equate to zero the coefficient of xn, we have

i.e.,.

Putting n = 2, 3, 4, 5, … successively

;;;

; and so on.

Substituting these values, we get the solution as

3. Solve in series the equation .

Answer

Here x = 0 is a regular singular point. Assume the solution of the given equation as

Differentiating y with respect x twice, we get

and.

Substituting the values of in the given equation, we obtain

Equating to zero the coefficient of lowest power of x, (i.e., coeff. of xr-1) we get

[ r( r -1) + r]c0 = 0

i.e.,r2 = 0, since ,

which is the indicial equation, whose roots are r = 0, 0 which are equal.

Equating to zero the coefficient of xr, we get

Equating to zero the coefficient of xr+1, we get

Equating to zero the coefficient of xr+2, we get

and so on.

Therefore the solution is given by

Putting r = 0, the first solution is

To get the second independent solution, differentiate y partially with respect to r

The second independent solution is

Hence the complete solution is

y = c1y1 + c2y2.