Series Solution of Linear Ordinary DifferentialEquations

Department of MathematicsIIT Guwahati

SU/KSK MA-102 (2018)

Aim: To study methods for determining series expansions forsolutions to linear ODE with variable coefficients.

In particular, we shall obtain

• the form of the series expansion,

• a recurrence relation for determining the coefficients, and

• the interval of convergence of the expansion.

SU/KSK MA-102 (2018)

Review of power series

A series of the form

∞∑n=0

an(x− x0)n = a0 + a1(x− x0) + a2(x− x0)

2 + · · · , (1)

is called a power series about the point x0. Here, x is avariable and an’s are constants.

The series (1) converges at x = c if∑∞

n=0 an(c− x0)n

converges. That is, the limit of partial sums

limN→∞

N∑n=0

an(c− x0)n <∞.

If this limit does not exist, the power series is said to divergeat x = c.

SU/KSK MA-102 (2018)

Note that∑∞

n=0 an(x− x0)n converges at x = x0 as

∞∑n=0

an(x0 − x0)n = a0.

Q. What about convergence for other values of x?

Theorem: (Radius of convergence)For each power series of the form (1), there is a number R(0 ≤ R ≤ ∞), called the radius of convergence of the powerseries, such that the series converges absolutely for|x− x0| < R and diverge for |x− x0| > R.

If the series (1) converges for all values of x, then R =∞.When the series (1) converges only at x0, then R = 0.

SU/KSK MA-102 (2018)

Theorem: (Ratio test) If

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = L,

where 0 ≤ L ≤ ∞, then the radius of convergence (R) of thepower series

∑∞n=0 an(x− x0)

n is

R =

1L

if 0 < L <∞,∞ if L = 0,0 if L =∞.

Remark. If the ratio∣∣∣an+1

an

∣∣∣ does not have a limit, then

methods other than the ratio test (e.g. root test) must beused to determine R.

SU/KSK MA-102 (2018)

Example: Find R for the series∑∞

n=0(−2)nn+1

(x− 3)n.

Note that an = (−2)nn+1

. We have

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣(−2)n+1(n + 1)

(−2)n(n + 2)

∣∣∣∣ = limn→∞

2(n + 1)

(n + 2)= 2 = L.

Thus, R = 1/2. The series converges absolutely for|x− 3| < 1

2and diverge for |x− 3| > 1

2.

Next, what happens when |x− 3| = 1/2?

At x = 5/2, the series becomes the harmonic series∑∞

n=01

n+1,

and hence diverges. When x = 7/2, the series becomes analternating harmonic series, which converges.

Thus, the power series converges for each x ∈ (5/2, 7/2].

SU/KSK MA-102 (2018)

Given two power series

f(x) =∞∑n=0

an(x− x0)n, g(x) =

∞∑n=0

bn(x− x0)n,

with nonzero radii of convergence. Then

f(x) + g(x) =∞∑n=0

(an + bn)(x− x0)n

has common interval of convergence.The formula for the product is

f(x)g(x) =∞∑n=0

cn(x− x0)n, where cn :=

n∑k=0

akbn−k. (2)

This power series in (2) is called the Cauchy product and willconverge for all x in the common interval of convergence forthe power series of f and g.

SU/KSK MA-102 (2018)

Differentiation and integration of power series

Theorem: If f(x) =∑∞

n=0 an(x− x0)n has a positive radius of

convergence R, then f is differentiable in the interval|x− x0| < R and termwise differentiation gives the powerseries for the derivative:

f ′(x) =∞∑n=1

nan(x− x0)n−1 for |x− x0| < R.

Furthermore, termwise integration gives the power series forthe integral of f :∫

f(x)dx =∞∑n=0

ann + 1

(x− x0)n+1 + C for |x− x0| < R.

SU/KSK MA-102 (2018)

Example: A power series for

1

1− x= 1 + x + x2 + x3 + · · ·+ · · · .

Since ddx{1/(1− x)} = 1

(1−x)2 , we obtain a power series for

1

(1− x)2= 1 + 2x + 3x2 + 4x3 + · · ·+ nxn−1 + · · · .

A power series for

1

1 + x2= 1− x2 + x4 − x6 + · · ·+ (−1)nx2n + · · · .

Since tan−1 x =∫ x

01

1+t2dt, integrate the series for 1

1+x2

termwise to obtain

tan−1 x = x− 1

3x3 +

1

5x5 − 1

7x7 + · · ·+ (−1)nx2n+1

2n + 1+ · · · .

SU/KSK MA-102 (2018)

Shifting the summation index

The index of a summation in a power series is a dummy indexand hence

∞∑n=0

an(x− x0)n =

∞∑k=0

ak(x− x0)k =

∞∑i=0

ai(x− x0)i.

Shifting the index of summation is particularly important whenone has to combine two different power series.

Example:

∞∑n=2

n(n− 1)anxn−2 =

∞∑k=0

(k + 2)(k + 1)ak+2xk.

x3

∞∑n=0

n2(n− 2)anxn =

∞∑n=3

(n− 3)2(n− 5)an−3xn.

SU/KSK MA-102 (2018)

Adding two power series

Problem: Write Σ∞n=12ncnxn−1 + Σ∞n=06cnx

n+1 as one series.In order to add the series, we require that both summationindices start with the same number and that the powers of xin each series be “in phase”;that is, if one series starts with amultiple of x1 say, then we want the other series also to startwith the same power of x. By writing

Σ∞n=12ncnxn−1 + Σ∞n=06cnx

n+1

= 2.1.c1x0 + Σ∞n=22ncnx

n−1 + Σ∞n=06cnxn+1

we have both the series Σ∞n=22ncnxn−1 and Σ∞n=06cnx

n+1 ofthe right hand side of the above equation start with x1.

SU/KSK MA-102 (2018)

To get the same summation index, we are inspired by the exponents of x.We let k = n− 1 in the first series and let k = n+ 1 in the second series.Thus the right hand side of the above equation becomes

2c1 + Σ∞k=12(k + 1)ck+1xk + Σ∞k=16ck−1x

k

= 2c1 + Σ∞k=1[2(k + 1)ck+1 + 6ck−1]xk

which is the required form (as a single series) of the sum of the two givenseries.

SU/KSK MA-102 (2018)

Definition: (Analytic function)A function f is said to be analytic at x0 if it has a power seriesrepresentation

∑∞n=0 an(x− x0)

n in an neighborhood aboutx0, and has a positive radius of convergence.

Example: Some analytic functions and their representations:

ex =∞∑n=0

xn

n!.

sinx =∞∑n=0

(−1)n

(2n + 1)!x2n+1.

lnx =∞∑n=1

(−1)n−1

n(x− 1)n, x > 0.

SU/KSK MA-102 (2018)

Power series solutions to linear ODEs

Consider linear ODE of the form:

a2(x)y′′(x) + a1(x)y′(x) + a0(x)y(x) = 0, a2(x) 6= 0. (∗)

Writing in the standard form

y′′(x) + p(x)y′(x) + q(x)y(x) = 0,

where p(x) := a1(x)/a2(x) and q(x) := a0(x)/a2(x).

Definition: A point x0 is called an ordinary point of (∗) if bothp(x) = a1(x)/a2(x) and q(x) = a0(x)/a2(x) are analytic atx0. If x0 is not an ordinary point, it is called a singular pointof (∗).

SU/KSK MA-102 (2018)

Fact from Analysis: If p(x) and q(x) are polynomilas in x having no

common factors, then p(x)q(x) is analytic at all points where q(x) 6= 0. And

at the points where q(x) = 0, the function p(x)q(x) is not analytic.

SU/KSK MA-102 (2018)

Example: The differential equation xy′′ + (Sin x)y = 0 has anordinary point at x = 0 because

Sin x

x= 1− x2

3!+

x4

5!− · · · .

The singular points of the equation (x2− 1)y′′+ 2xy′+ 6y = 0are 1 and −1. All other points are ordinary points.The equation (x2 + 1)y′′ + xy′ − y = 0 has two singular pointsgiven by i,−i. All other finite values of x (real or complex),are ordinary points.

SU/KSK MA-102 (2018)

Power series method about an ordinary point

Consider the equation

2y′′ + xy′ + y = 0. (∗∗)

Let’s find a power series solution about x = 0. Seek a powerseries solution of the form

y(x) =∞∑n=0

anxn,

and then attempt to determine the coefficients an’s.Differentiate termwise to obtain

y′(x) =∞∑n=1

nanxn−1, y′′(x) =

∞∑n=2

n(n− 1)anxn−2.

SU/KSK MA-102 (2018)

Substituting these power series in (∗∗), we find that

∞∑n=2

2n(n− 1)anxn−2 +

∞∑n=1

nanxn +

∞∑n=0

anxn = 0.

By shifting the indices, we rewrite the above equation as

∞∑k=0

2(k + 2)(k + 1)ak+2xk +

∞∑k=1

kakxk +

∞∑k=0

akxk = 0.

Combining the like powers of x in the three summation toobtain

4a2 + a0 +∞∑k=1

[2(k + 2)(k + 1)ak+2 + kak + ak]xk = 0.

SU/KSK MA-102 (2018)

Equating the coefficients of this power series equal to zeroyields

4a2 + a0 = 0

2(k + 2)(k + 1)ak+2 + (k + 1)ak = 0, k ≥ 1.

This leads to the recurrence relation

ak+2 =−1

2(k + 2)ak, k ≥ 1.

Thus,

a2 =−1

22a0, a3 =

−1

2 · 3a1

a4 =−1

2 · 4a2 =

1

22 · 2 · 4a0, a5 =

−1

2 · 5a3 =

1

22 · 3 · 5a1

· · · · · ·

SU/KSK MA-102 (2018)

With a0 and a1 as arbitrary constants, we find that

a2n =(−1)n

22nn!a0, n ≥ 1,

and

a2n+1 =(−1)n

2n[1 · 3 · 5 · · · (2n + 1)]a1, n ≥ 1.

From this, we have two linearly independent solutions as

y1(x) =∞∑n=0

(−1)n

22nn!x2n,

y2(x) =∞∑n=0

(−1)n

2n[1 · 3 · 5 · · · (2n + 1)]x2n+1.

SU/KSK MA-102 (2018)

Hence the general solution is

y(x) = a0y1(x) + a1y2(x).

Remark. Suppose we are given the value of y(0) and y′(0),then a0 = y(0) and a1 = y′(0). These two coefficients leads toa unique power series solution for the IVP.

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SU/KSK MA-102 (2018)