Math 2C03 - Differential EquationsSlides shown in class - Winter 2015

March 25 - April 8, . . . 2015

Series Solutions

Review of Power Series

Definition

(a) A power series in x − x0 (or “centred at x0”) has the form:

∞∑n=0

cn (x − x0)n = c0 + c1 (x − x0) + c2 (x − x0)

2 + . . .

(b) A power series converges at x if a finite limit.

limN→∞

N∑n=0

cn (x − x0)n exists.

(c) A power series is said to converge absolutely at x if

∞∑n=0

|cn| |x − x0|n converges.

Radius of Convergence

Theorem (Radius of convergence, R)

Given a power series∑∞

n=0 cn (x − x0)n centred at x0, there

exists a number R with 0 ≤ R ≤ ∞, called the radius ofconvergence of the series such that

the series converges absolutely if |x − x0| < R.the series diverges if |x − x0| > R.

|x−x0| < R ⇔ x0−R < x < x0+R ⇔ x ∈ (x0−R, x0+R)

A power series centred at x0 always converges at x = x0.Interval of convergence can be:(x0 − R, x0 + R), [x0 − R, x0 + R), (x0 − R, x0 + R],[x0 − R, x0 + R], (−∞,∞), or [x0, x0].Convergence at end points must be tested separately.“Absolute Convergence”⇒ “Convergence”.

Test for Convergence

Theorem (The ratio test)

Consider the power series∑∞

n=0 cn (x − x0)n and assume that

there exists an integer N ≥ 0 such that cn 6= 0 for all n ≥ N.

If limn→∞

∣∣∣∣cn+1

cn

∣∣∣∣ = L, then R =1L.

If L = 0, then R =∞.

If L =∞, then R = 0.

It the limit does not exist other methods, e.g. Root Test can beused.

Examples: Maclaurin Series

11−x =

∑∞n=0 xn = 1 + x + x2 + x3 + . . . if |x | < 1 : R = 1.

(Geometric series.)ex =

∑∞n=0

xn

n! = 1 + x + x2

2! +x3

3! + . . . : R =∞.sin(x) =

∑∞n=0 (−1)n x2n+1

(2n+1)! = x − x3

3! +x5

5! . . . : R =∞cos(x) =

∑∞n=0 (−1)n x2n

(2n)! = 1− x2

2! +x4

4! . . . : R =∞.

Since a power series converges absolutely, within its openinterval of convergence, (x0 − R, x0 + R)

it represents a continuous function,has derivatives of all orders,can be differentiated term-by-term and integratedterm-by-term & the resulting series has radius ofconvergence at least R,its terms can be rearranged.

Theorem

If f (x) =∑∞

n=0 cn (x − x0)n is a power series in x − x0 with a

radius of convergence R > 0, then,(a) f (x) is infinitely differentiable for |x − x0| < R.Within its open interval of convergence:(b) a power series can be differentiated term by term

f ′(x) =∞∑

n=1

n cn (x − x0)n−1, |x − x0| < R;

(c) a power series can be integrated term by term∫ x

x0

f (t)dt =∞∑

n=0

cn(x − x0)

n+1

n + 1, |x − x0| < R;

(d)

cn =f (n)(x0)

n!, n ≥ 0.

More Properties of Power Series:within the Radius of Convergence: can add, subtract, multiply & divide

If f (x) =∑∞

n=0 an (x − x0)n is a power series in x − x0 with a

radius of convergence Rf > 0, &g(x) =

∑∞n=0 bn (x − x0)

n is a power series in x − x0 with aradius of convergence Rg > 0, thenf (x)± g(x) =

∑∞n=0 (an ± bn) (x − x0)

n

with radius of convergence at least min(Rf ,Rg).

f (x)g(x) =∑∞

n=0 an (x − x0)n∑∞

n=0 bn (x − x0)n =

(a0+a1(x−x0)+a2(x−x0)2+. . .)(b0+b1(x−x0)+b2(x−x0)

2+. . .)= a0b0 + (a0b1 + a1b0)(x − x0) + (a0b2 + a1b1 + a2b0)(x − x0)

2

+(. . . )(x − x0)3 + . . .

with radius of convergence at least min(Rf ,Rg).

Can also do f (x)g(x) provided g(x0) 6= 0, BUT here radius of

convergence can be smaller.

Key Observation

Theorem

(a) If∞∑

n=0

an (x − x0)n =

∞∑n=0

bn (x − x0)n

for all x in an open interval that contains x0,then an = bn for all n = 0,1,2, . . ..

(b) If∞∑

n=0

an (x − x0)n = 0

for all x in an open interval that contains x0,then an = 0 for all n = 0,1,2, . . ..

More Manipulations of Power Series

Rearranging into even and odd terms:

∞∑n=0

cn (x − x0)n =

∞∑n=0

c2n (x − x0)2n +

∞∑n=0

c2n+1 (x − x0)2n+1

Rearranging into every 3rd term:∑∞n=0 cn (x − x0)

n

=∞∑

n=0

c3n (x−x0)3n+

∞∑n=0

c3n+1 (x−x0)3n+1+

∞∑n=0

c3n+2 (x−x0)3n+2

Shifting the Index

Shifting the dummy variable:

∞∑n=0

cn (x − x0)n =

∞∑n=k

cn−k (x − x0)n−k

for any integer k .

∞∑n=k

cn (x − x0)n =

∞∑n=0

cn+k (x − x0)n+k

Omitting zero terms:∑∞n=0 n(n − 1)cn (x − x0)

n−2

=∑∞

n=1 n(n − 1)cn (x − x0)n−2

=∑∞

n=2 n(n − 1)cn (x − x0)n−2

6=∑∞

n=3 n(n − 1)cn (x − x0)n−2

Definition (Analytic Functions)

A function f (x) is analytic at x0 if, for some R > 0,

f (x) =∞∑

n=0

cn (x − x0)n, |x − x0| < R,

i.e. f (x) admits a power series expansion in x − x0 near x = x0with a positive radius of convergence.

sin(x), cos(x),& ex are all analytic.A polynomial P(x) = a0 + a1 x + · · ·+ am xm is analytic atany point x0 ∈ R.

A rational function f (x) = P(x)Q(x) (where P(x),Q(x) are

polynomials with no common factor), is analytic at any x0where Q(x0) 6= 0. The corresponding power series hasradius of convergence equal to the distance from x0 to thenearest zero of Q(x) in the complex plane.

Ordinary Points and Singular Points2nd Order Homogeneous linear ODE

Definition

Consider a 2nd -order homogeneous linear DE

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

which we can write in standard form as

y ′′ + P(x) y ′ + Q(x) y = 0.

(a) The number x0 is an ordinary point of (∗) if P(x) and Q(x)are both analytic at x0.

(b) If x0 is not an ordinary point, it is called a singular point of(∗).

Theorem (Existence of power series solutions near ordinarypoints)

If x0 is an ordinary point of the differential equation

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

then (∗) has two linearly independent solutions y1(x) and y2(x)that are both analytic at x0:

y1(x) =∞∑

n=0

bn (x − x0)n; y2(x) =

∞∑n=0

cn (x − x0)n,

& each solution has radius of convergence at least R, whereR is the distance to the nearest singular point to x0 in thecomplex plane.

Useful Notation & Simplifications

[(2)(4)(6) . . . (2n)]︸ ︷︷ ︸n factors

= [(1 · 2)(2 · 2)(3 · 2) · · · (n · 2)] = n!2n

[(3)(6)(9) . . . (3n)]︸ ︷︷ ︸n factors

= [(1 · 3)(2 · 3)(3 · 3) · · · (n · 3)] = n!3n

[f (k)f (k + 1)f (k + 2) · · · f (m)] =m∏

n=k

f (n)

3∏n=1

nn + 1

=

(12

)(23

)(34

)m∏

k=1

1(2k + 1)

=1

[3 · 5 · 7 · · · (2m + 1)]

=[2 · 4 · · · (2m)]

[(2 · 3)(4 · 5) · · · (2m · (2m + 1))]=

2mm!

(2m + 1)!

Regular (RSP) vs Irregular (ISP) Singular Points

Definition

If x0 is a singular point of (∗)

y ′′ + P(x) y ′ + Q(x) y = 0, (∗)

a 2nd -order homogeneous linear ODE in standard form, then(a) x0 is a regular singular point (RSP) of (∗), if the functions

(x−x0)P(x) & (x−x0)2 Q(x) are BOTH analytic at x = x0.

(b) If x0 is not a regular singular point, it is called an irregularsingular point ( ISP).

(a)⇒ limx→x0

(x − x0)P(x) & limx→x0

(x − x0)2Q(x) exist and

are finite.x0 = 0 is a (RSP) of the 2nd order homogeneousCauchy-Euler ODE: ax2y ′′ + bxy ′ + cy = 0.

The Method of FrobeniusPower Series Solutions about RSPs

Theorem

If x0 is a regular singular point of

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

then there exists at least one solution (∗) of the form

y = (x − x0)r∞∑

n=0

cn (x − x0)n =

∞∑n=0

cn (x − x0)n+r , (∗∗)

with c0 6= 0, & cn, n ≥ 1, defined in terms of c0.The series solution in (∗∗) converges at least for0 < |x − x0| < R, where R is the distance from x0 to the nearestsingular point (real or complex) of (∗).

GOAL: To find r & to compute the coefficientscn, n ≥ 1, in terms of c0, in the solution (∗∗).

To find the Indicial Equationif x0 is a RSP of y ′′ + P(x)y ′ + Q(x)y = 0

Since (x − x0)P(x) & (x − x0)2 Q(x) are both analytic at

x = x0 there exists R > 0 such that for |x − x0| < R

(x − x0)P(x) = a0 + a1 (x − x0) + · · · =∞∑

n=0

an (x − x0)n,

(x − x0)2 Q(x) = b0 + b1 (x − x0) + · · · =

∞∑n=0

bn (x − x0)n.

WLOG, assume x0 = 0. If not, substitute v = (x − x0).Therefore, for |x | < R

x P(x) = a0 + a1 x + · · · =∞∑

n=0

an xn,

x2 Q(x) = b0 + b1 x + · · · =∞∑

n=0

bn xn.

Assume a solution of the form:

y =∞∑

n=0

cn xn+r = c0 x r + c1 x r+1 + . . .

y ′ =∞∑

n=0

(n + r) cn xn+r−1 = c0 r x r−1 + c1 (r + 1) x r + . . .

y ′′ =∞∑

n=0

(n + r) (n + r − 1) cn xn+r−2

= c0 r (r − 1)x r−2 + c1 (r + 1) r x r−1 + . . .

Substitute into the ODE.

y ′′ + P(x)y ′ + Q(x)y = 0 ⇔ y ′′ + (xP(x))y ′

x+ (x2Q(x))

yx2 = 0

∞∑n=0

(n + r) (n + r − 1) cn xn+r−2

︸ ︷︷ ︸y ′′

+ (a0 + a1 x + a2 x2 + . . . )︸ ︷︷ ︸x P(x)

( ∞∑n=0

(n + r) cn xn+r−2

)︸ ︷︷ ︸

y′x

+ (b0 + b1 x + b2 x2 + . . . )︸ ︷︷ ︸x2 Q(x)

( ∞∑n=0

cn xn+r−2

)= 0︸ ︷︷ ︸

yx2

,

c0 [r (r − 1) + a0 r + b0] x r−2 + (. . . ) x r−1 + (. . . ) x r + · · · = 0.

Since c0 6= 0, [r (r − 1) + a0 r + b0] = 0, the indicial equation.

The Indicial Equation

[r (r − 1) + a0 r + b0] = 0 OR r2 + (a0 − 1)r + b0 = 0

x P(x) = a0 + a1 x + · · · =∞∑

n=0

an xn,

x2 Q(x) = b0 + b1 x + · · · =∞∑

n=0

bn xn.

a0 = limx→0

xP(x),

b0 = limx→0

x2Q(x).

If ALSO wish to find the solution in the form of a powerseries, y(x) =

∑∞n=0 cnxn+r , can instead find the indicial

equation by substituting into the ODE and setting the factorof c0x r−2 equal to 0.

Theorem (Form of two linearly indep solns)

Assume that x0 is a RSP of y ′′ + p(x)y ′ + q(x)y = 0 and thatr1 & r2 are roots of the indicial equation with Re(r1) ≥ Re(r2).(a) If r1 − r2 is not an integer, then there exist two linearly

independent solutions of the form:

y1(x)=∞∑

n=0

an(x − x0)n+r1 , a0 6= 0,

y2(x) =∞∑

n=0

bn(x − x0)n+r2 , b0 6= 0,

Theorem (Form of two linearly indep solns)

Assume that x0 is a RSP of y ′′ + p(x)y ′ + q(x)y = 0 and thatr1 & r2 are roots of the indicial equation with Re(r1) ≥ Re(r2).(b) If r1 = r2, then there exist two linearly independent

solutions of the form:

y1(x)=∞∑

n=0

an(x − x0)n+r1 , a0 6= 0,

y2(x) = y1(x) ln(x − x0) +∞∑

n=1

bn(x − x0)n+r2 ,

Theorem (Form of two linearly indep solns)

Assume that x0 is a RSP of y ′′ + p(x)y ′ + q(x)y = 0 and thatr1 & r2 are roots of the indicial equation with Re(r1) ≥ Re(r2).(c) If r1 − r2 is a positive integer, then there exist two linearly

independent solutions of the form:

y1(x)=∞∑

n=0

an(x − x0)n+r1 , a0 6= 0,

y2(x) = Cy1(x) ln(x − x0) +∞∑

n=0

bn(x − x0)n+r2 , b0 6= 0,

where C is a constant that could be zero.