Series Solutions Near an Ordinary PointMATH 365 Ordinary Differential Equations

J. Robert Buchanan

Department of Mathematics

Spring 2018

Ordinary Points (1 of 2)

Consider the second order linear homogeneous ODE:

P(t)y ′′ + Q(t)y ′ + R(t)y = 0

where P, Q, and R are polynomials.

DefinitionA point t0 such that P(t0) 6= 0 is called an ordinary point. IfP(t0) = 0 then t0 is called a singular point.

Ordinary Points (2 of 2)

P(t)y ′′ + Q(t)y ′ + R(t)y = 0

If t0 is an ordinary point, then by continuity there exists aninterval (a,b) containing t0 on which P(t) 6= 0 for all t ∈ (a,b).

Thus the functions p(t) =Q(t)P(t)

and q(t) =R(t)P(t)

are defined

and continuous on (a,b) and the ODE can be written as

y ′′ + p(t)y ′ + q(t)y = 0.

If the initial conditions are y(t0) = y0 and y ′(t0) = y ′0 then thereexists a unique solution to the ODE satisfying the initialconditions.

Power Series Solutions (1 of 5)

Consider the ODEy ′′ − 4y = 0

and find a power series solution with positive radius ofconvergence centered at an ordinary point.

Let t0 = 0 be the ordinary point for simplicity and

y(t) =∞∑

n=0

antn

y ′(t) =∞∑

n=1

nantn−1

y ′′(t) =∞∑

n=2

n(n − 1)antn−2.

Substitute into the ODE.

Power Series Solutions (1 of 5)

Consider the ODEy ′′ − 4y = 0

and find a power series solution with positive radius ofconvergence centered at an ordinary point.Let t0 = 0 be the ordinary point for simplicity and

y(t) =∞∑

n=0

antn

y ′(t) =∞∑

n=1

nantn−1

y ′′(t) =∞∑

n=2

n(n − 1)antn−2.

Substitute into the ODE.

Power Series Solutions (1 of 5)

Consider the ODEy ′′ − 4y = 0

and find a power series solution with positive radius ofconvergence centered at an ordinary point.Let t0 = 0 be the ordinary point for simplicity and

y(t) =∞∑

n=0

antn

y ′(t) =∞∑

n=1

nantn−1

y ′′(t) =∞∑

n=2

n(n − 1)antn−2.

Substitute into the ODE.

Power Series Solutions (2 of 5)

0 = y ′′ − 4y

=∞∑

n=2

n(n − 1)antn−2 − 4∞∑

n=0

antn

=∞∑

n=0

(n + 2)(n + 1)an+2tn − 4∞∑

n=0

antn

=∞∑

n=0

[an+2(n + 2)(n + 1)− 4an] tn

0 = an+2(n + 2)(n + 1)− 4an (for n = 0,1,2, . . .)

an+2 =22an

(n + 2)(n + 1)

This last equation is called a recurrence relation.

Power Series Solutions (2 of 5)

0 = y ′′ − 4y

=∞∑

n=2

n(n − 1)antn−2 − 4∞∑

n=0

antn

=∞∑

n=0

(n + 2)(n + 1)an+2tn − 4∞∑

n=0

antn

=∞∑

n=0

[an+2(n + 2)(n + 1)− 4an] tn

0 = an+2(n + 2)(n + 1)− 4an (for n = 0,1,2, . . .)

an+2 =22an

(n + 2)(n + 1)

This last equation is called a recurrence relation.

Power Series Solutions (2 of 5)

0 = y ′′ − 4y

=∞∑

n=2

n(n − 1)antn−2 − 4∞∑

n=0

antn

=∞∑

n=0

(n + 2)(n + 1)an+2tn − 4∞∑

n=0

antn

=∞∑

n=0

[an+2(n + 2)(n + 1)− 4an] tn

0 = an+2(n + 2)(n + 1)− 4an (for n = 0,1,2, . . .)

an+2 =22an

(n + 2)(n + 1)

This last equation is called a recurrence relation.

Power Series Solutions (2 of 5)

0 = y ′′ − 4y

=∞∑

n=2

n(n − 1)antn−2 − 4∞∑

n=0

antn

=∞∑

n=0

(n + 2)(n + 1)an+2tn − 4∞∑

n=0

antn

=∞∑

n=0

[an+2(n + 2)(n + 1)− 4an] tn

0 = an+2(n + 2)(n + 1)− 4an (for n = 0,1,2, . . .)

an+2 =22an

(n + 2)(n + 1)

This last equation is called a recurrence relation.

Power Series Solutions (3 of 5)

Let a0 be arbitrary, then

a2 =22

(2)(1)a0 =

22

2!a0

a4 =22

(4)(3)a2 =

24

4!a0

a6 =22

(6)(5)a4 =

26

6!a0

...

a2n =22n

(2n)!a0.

Power Series Solutions (4 of 5)

Let a1 be arbitrary, then

a3 =22

(3)(2)a1 =

22

3!a1

a5 =22

(5)(4)a3 =

24

5!a1

a7 =22

(7)(6)a5 =

26

7!a1

...

a2n+1 =22n

(2n + 1)!a1.

Power Series Solutions (5 of 5)

Thus the general solution to y ′′ − 4y = 0 can be written as

y(t) =∞∑

n=0

a2nt2n +∞∑

n=0

a2n+1t2n+1

= a0

∞∑n=0

22n

(2n)!t2n + a1

∞∑n=0

22n

(2n + 1)!t2n+1

= a0

∞∑n=0

(2t)2n

(2n)!+

a1

2

∞∑n=0

(2t)2n+1

(2n + 1)!

= a0 cosh(2t) +a1

2sinh(2t)

We can confirm this series converges for all t ∈ R.

Airy’s Equation (1 of 6)

Find a power series solution about the ordinary point t0 = 0 toAiry’s equation:

y ′′ − t y = 0.

y(t) =∞∑

n=0

antn

y ′(t) =∞∑

n=1

nantn−1

y ′′(t) =∞∑

n=2

n(n − 1)antn−2.

Substitute into the ODE.

Airy’s Equation (1 of 6)

Find a power series solution about the ordinary point t0 = 0 toAiry’s equation:

y ′′ − t y = 0.

y(t) =∞∑

n=0

antn

y ′(t) =∞∑

n=1

nantn−1

y ′′(t) =∞∑

n=2

n(n − 1)antn−2.

Substitute into the ODE.

Airy’s Equation (2 of 6)

0 = y ′′ − ty

=∞∑

n=2

n(n − 1)antn−2 − t∞∑

n=0

antn

=∞∑

n=0

(n + 2)(n + 1)an+2tn −∞∑

n=0

antn+1

= 2a2 +∞∑

n=1

(n + 2)(n + 1)an+2tn −∞∑

n=1

an−1tn

= 2a2 +∞∑

n=1

[(n + 2)(n + 1)an+2 − an−1] tn

Airy’s Equation (3 of 6)

Thus

0 = 2a2

0 = (n + 2)(n + 1)an+2 − an−1 (for n = 1,2, . . ..)

From the last equation we obtain the recurrence relation:

an+2 =an−1

(n + 2)(n + 1).

Since a2 = 0 then a5 = a8 = a11 = · · · = 0.

Airy’s Equation (3 of 6)

Thus

0 = 2a2

0 = (n + 2)(n + 1)an+2 − an−1 (for n = 1,2, . . ..)

From the last equation we obtain the recurrence relation:

an+2 =an−1

(n + 2)(n + 1).

Since a2 = 0 then a5 = a8 = a11 = · · · = 0.

Airy’s Equation (4 of 6)

Let a0 be arbitrary then

a3 =a0

2 · 3a6 =

a3

5 · 6=

a0

6 · 5 · 3 · 2a9 =

a6

8 · 9=

a0

9 · 8 · 6 · 5 · 3 · 2...

a3n =a0

(2)(3)(5)(6) · · · (3n − 4)(3n − 3)(3n − 1)(3n).

Airy’s Equation (5 of 6)

Let a1 be arbitrary then

a4 =a1

3 · 4a7 =

a4

6 · 7=

a1

7 · 6 · 4 · 3a10 =

a7

9 · 10=

a1

10 · 9 · 7 · 6 · 4 · 3...

a3n+1 =a1

(3)(4)(6)(7) · · · (3n − 3)(3n − 2)(3n)(3n + 1).

Airy’s Equation (6 of 6)

Thus the solution to Airy’s equation

y ′′ − ty = 0

is

y(t)

= a0

[1 +

∞∑n=1

t3n

(2)(3)(5)(6) · · · (3n − 4)(3n − 3)(3n − 1)(3n)

]

+ a1

[t +

∞∑n=1

t3n+1

(3)(4)(6)(7) · · · (3n − 3)(3n − 2)(3n)(3n + 1)

]= a0y1(t) + a1y2(t).

Illustration

-10 -8 -6 -4 -2 2 4t

-10

-5

5

10

y

y1(t)

y2(t)

Homework

I Read Section 5.2I Exercises: 1, 2, 3, 5, 6, 21

Review

We have determined a method for solving an ODE of the form

P(t)y ′′ + Q(t)y ′ + R(t)y = 0

where P, Q, and R are polynomials.

The solution is a power series of the form y(t) =∞∑

n=0

an(t − t0)n

where t0 is an ordinary point.

Today we extend this work to a broader range of functions thanpolynomial P, Q, and R.

Review

We have determined a method for solving an ODE of the form

P(t)y ′′ + Q(t)y ′ + R(t)y = 0

where P, Q, and R are polynomials.

The solution is a power series of the form y(t) =∞∑

n=0

an(t − t0)n

where t0 is an ordinary point.

Today we extend this work to a broader range of functions thanpolynomial P, Q, and R.

Differentiation of Power SeriesSuppose φ(t) =

∞∑n=0

an(t − t0)n has a positive radius of

convergence, then we may differentiate the seriesterm-by-term.

n φ(n)(t) φ(n)(t0)

0∞∑

n=0

an(t − t0)n a0

1∞∑

n=1

nan(t − t0)n−1 a1

2∞∑

n=2

n(n − 1)an(t − t0)n−2 2a2

......

...

m∞∑

n=m

n(n − 1) · · · (n −m + 1)an(t − t0)n−m m!am

Solution to an ODE

Suppose φ(t) solves the ODE y ′′ + p(t)y ′ + q(t)y = 0 then

0 = φ′′(t) + p(t)φ′(t) + q(t)φ(t)φ′′(t) = −p(t)φ′(t)− q(t)φ(t)

φ′′(t0) = −p(t0)φ′(t0)− q(t0)φ(t0)

2!a2 = −p(t0)a1 − q(t0)a0.

Thus we may find a2 in terms of a0 and a1.

Solution to an ODE

Suppose φ(t) solves the ODE y ′′ + p(t)y ′ + q(t)y = 0 then

0 = φ′′(t) + p(t)φ′(t) + q(t)φ(t)φ′′(t) = −p(t)φ′(t)− q(t)φ(t)φ′′(t0) = −p(t0)φ′(t0)− q(t0)φ(t0)

2!a2 = −p(t0)a1 − q(t0)a0.

Thus we may find a2 in terms of a0 and a1.

Solution to an ODE

Suppose φ(t) solves the ODE y ′′ + p(t)y ′ + q(t)y = 0 then

0 = φ′′(t) + p(t)φ′(t) + q(t)φ(t)φ′′(t) = −p(t)φ′(t)− q(t)φ(t)φ′′(t0) = −p(t0)φ′(t0)− q(t0)φ(t0)

2!a2 = −p(t0)a1 − q(t0)a0.

Thus we may find a2 in terms of a0 and a1.

Solution to an ODE

Suppose φ(t) solves the ODE y ′′ + p(t)y ′ + q(t)y = 0 then

0 = φ′′(t) + p(t)φ′(t) + q(t)φ(t)φ′′(t) = −p(t)φ′(t)− q(t)φ(t)φ′′(t0) = −p(t0)φ′(t0)− q(t0)φ(t0)

2!a2 = −p(t0)a1 − q(t0)a0.

Thus we may find a2 in terms of a0 and a1.

Continued Differentiation (1 of 2)

With patience we can find higher order terms in the seriessolution through continued differentiation.

φ′′(t) = −p(t)φ′(t)− q(t)φ(t)φ′′′(t) = −p′(t)φ′(t)− p(t)φ′′(t)− q′(t)φ(t)− q(t)φ′(t)

φ′′′(t0) = −p′(t0)φ′(t0)− p(t0)φ′′(t0)− q′(t0)φ(t0)− q(t0)φ′(t0)3!a3 = −p′(t0)a1 − 2p(t0)a2 − q′(t0)a0 − q(t0)a1

Substituting the previously determined value of a2 we may finda3 in terms of a0 and a1.

Continued Differentiation (1 of 2)

With patience we can find higher order terms in the seriessolution through continued differentiation.

φ′′(t) = −p(t)φ′(t)− q(t)φ(t)φ′′′(t) = −p′(t)φ′(t)− p(t)φ′′(t)− q′(t)φ(t)− q(t)φ′(t)φ′′′(t0) = −p′(t0)φ′(t0)− p(t0)φ′′(t0)− q′(t0)φ(t0)− q(t0)φ′(t0)

3!a3 = −p′(t0)a1 − 2p(t0)a2 − q′(t0)a0 − q(t0)a1

Substituting the previously determined value of a2 we may finda3 in terms of a0 and a1.

Continued Differentiation (1 of 2)

With patience we can find higher order terms in the seriessolution through continued differentiation.

φ′′(t) = −p(t)φ′(t)− q(t)φ(t)φ′′′(t) = −p′(t)φ′(t)− p(t)φ′′(t)− q′(t)φ(t)− q(t)φ′(t)φ′′′(t0) = −p′(t0)φ′(t0)− p(t0)φ′′(t0)− q′(t0)φ(t0)− q(t0)φ′(t0)

3!a3 = −p′(t0)a1 − 2p(t0)a2 − q′(t0)a0 − q(t0)a1

Substituting the previously determined value of a2 we may finda3 in terms of a0 and a1.

Continued Differentiation (2 of 2)

We can proceed by repeated differentiation to find a4, a5,. . . provided:

I p(t) and q(t) have derivatives of all orders, andI we can show the resulting power series converges.

Continued Differentiation (2 of 2)

We can proceed by repeated differentiation to find a4, a5,. . . provided:I p(t) and q(t) have derivatives of all orders, and

I we can show the resulting power series converges.

Continued Differentiation (2 of 2)

We can proceed by repeated differentiation to find a4, a5,. . . provided:I p(t) and q(t) have derivatives of all orders, andI we can show the resulting power series converges.

ExampleAssuming that y = φ(t) is a solution to the IVP:

y ′′ + t2y ′ + (sin t)y = 0y(0) = 1y ′(0) = −1

find the first four nonzero terms in the power seriesrepresentation of φ(t).

a0 = 1a1 = −1a2 = 0

a3 = − 13!

a4 =13!

ExampleAssuming that y = φ(t) is a solution to the IVP:

y ′′ + t2y ′ + (sin t)y = 0y(0) = 1y ′(0) = −1

find the first four nonzero terms in the power seriesrepresentation of φ(t).

a0 = 1a1 = −1a2 = 0

a3 = − 13!

a4 =13!

Analytic Functions

If p(t) and q(t) are analytic functions at t0, in other words haveTaylor series expansions about t0 which converge to p(t) andq(t) respectively then p and q will have derivatives of all ordersat t0.

p(t) =∞∑

n=0

pn(t − t0)n

q(t) =∞∑

n=0

qn(t − t0)n

Ordinary and Singular Points Revisited

Suppose

P(t)y ′′ + Q(t)y ′ + R(t)y = 0

y ′′ +Q(t)P(t)

y ′ +R(t)P(t)

y = 0

y ′′ + p(t)y ′ + q(t)y = 0

wherep(t) =

Q(t)P(t)

and q(t) =R(t)P(t)

If p(t) and q(t) are analytic at t0, then we say that t0 is anordinary point of the ODE. Otherwise t0 is a singular point.

Ordinary and Singular Points Revisited

Suppose

P(t)y ′′ + Q(t)y ′ + R(t)y = 0

y ′′ +Q(t)P(t)

y ′ +R(t)P(t)

y = 0

y ′′ + p(t)y ′ + q(t)y = 0

wherep(t) =

Q(t)P(t)

and q(t) =R(t)P(t)

If p(t) and q(t) are analytic at t0, then we say that t0 is anordinary point of the ODE. Otherwise t0 is a singular point.

Main Result

TheoremIf t0 is an ordinary point of the ODE

P(t)y ′′ + Q(t)y ′ + R(t)y = 0

then the general solution of the ODE is

y(t) =∞∑

n=0

an(t − t0)n = a0y1(t) + a1y2(t)

where a0 and a1 are arbitrary and y1 and y2 are linearlyindependent series solutions that are analytic at t0. Further theradius of convergence for each of y1 and y2 is at least as largeas the minimum of the radii of convergence for the seriesp(t) = Q(t)/P(t) and q(t) = R(t)/P(t).

Radius of Convergence

If p(t) = Q(t)/P(t) and q(t) = R(t)/P(t) and p(t) and q(t) areanalytic at t0 then from the theory of complex variables we havethe result that

the radius of convergence of p(t) (and similarly q(t))is at least as large as the minimum distance from t0 toany root of P(t) in the complex plane.

Example

The value t0 = 1 is an ordinary point of the ODE

t2y ′′ + (1 + t)y ′ + 3(ln t)y = 0.

Find the radius of convergence of p(t) =1 + t

t2 and

q(t) =3 ln t

t2 .

Example

The value t0 = 0 is an ordinary point of the ODE

(1 + t4)y ′′ + 4ty ′ + y = 0.

Find the radius of convergence of p(t) =4t

1 + t4 and

q(t) =1

1 + t4 .

Using Euler’s Identity:

t4 + 1 = 0t4 = −1 = ei(2n+1)π

t = ei(2n+1)π/4

t = ±√

22± i√

22.

Example

The value t0 = 0 is an ordinary point of the ODE

(1 + t4)y ′′ + 4ty ′ + y = 0.

Find the radius of convergence of p(t) =4t

1 + t4 and

q(t) =1

1 + t4 .

Using Euler’s Identity:

t4 + 1 = 0t4 = −1 = ei(2n+1)π

t = ei(2n+1)π/4

t = ±√

22± i√

22.

Illustration

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

i y

Example

The value t0 = 1 is an ordinary point of the ODE

(1 + t4)y ′′ + 4ty ′ + y = 0.

Find the radius of convergence of p(t) =4t

1 + t4 and

q(t) =1

1 + t4 .

Illustration

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

i y

Homework

I Read Section 5.3I Exercises: 1–7 odd, 10, 11, 22–29