7.5 Power series solutions, Legendre and Bessel

When the coefficients of a linear equation change with x, assume a power series (or a power series times a power of x) and the equation turns into a recurrence for the coefficients. At an ordinary point the series converges as far as the coefficients do. At a regular singular point the indicial equation says which power to start with. Legendre's equation gives orthogonal polynomials, Bessel's gives the function J0, and Airy's equation is the model of a turning point.

The recipe of 7.2 needs constant coefficients: guess $e^{rt}$ and read the exponent off a polynomial. Change the coefficients and the recipe fails. The equation $y'' = xy$ looks harmless, but none of $e^{rx}$ , $\sin$ , $\cos$ or any other function you know solves it. The substitute is a more flexible guess, an infinite polynomial $\sum a_n x^n$ , which turns the equation into a rule for computing the numbers $a_n$ one after another. This chapter asks when that works, how far the series reaches, and what to do at a point where the coefficients themselves blow up. The answers produce three families of functions that physics uses constantly: Legendre polynomials, Bessel functions and the Airy function.

Before you start

You need power series and their radius of convergence from 1.6, the inner product and orthogonality from 2.9 and 3.1, and the Wronskian from 7.3. Three quick checks:

Quick checkA power series

\sum a_nx^n

has

a_{n+1}/a_n \to \frac12

. What is its radius of convergence?

Quick checkAfter putting

y = \sum_{n \ge 0} a_nx^n

into an equation, the term

y'' = \sum_{n \ge 2} n(n-1)a_nx^{n-2}

is rewritten as a sum of multiples of

x^k

. What is the coefficient of

x^k

Quick checkWith the inner product

\langle f, g \rangle = \int_{-1}^{1} f(x)g(x)\,dx

, are

f(x) = x

and

g(x) = x^2 - \frac13

orthogonal?

The picture

Below, each equation is solved by a power series. The black dashed curve is the true solution, computed numerically (or from a closed form). The blue curve is the partial sum of the series up to degree $N$ . Raise $N$ and watch the blue curve lock onto the black one from the middle outwards. For the arctangent equation and for the Legendre equation with $\nu = 1.5$ the red lines mark the radius of convergence: the blue curve never crosses them successfully, however large $N$ is. For Legendre with $\nu = 2$ the series stops by itself.

Three things to notice. For the Airy and Bessel equations the series converges at every $x$ , but the number of terms needed grows with $\lvert x \rvert$ . For the arctangent equation the radius is exactly $1$ , although nothing is wrong with the equation at $x = 1$ . And for Legendre with $\nu = 2$ , a polynomial appears. Each of these is explained below.

Where this came from· Spheroids, planets and a rainbow

Series solutions come from astronomy. Legendre was computing the gravitational pull of a spheroid such as the flattened Earth, and the polynomials that carry his name appear in a memoir on that problem which he read to the Paris Académie in the early 1780s (it was printed in 1785). The closed formula for them was found by Olinde Rodrigues around 1816. Bessel studied planetary perturbations in a memoir of 1824 and worked out the general properties of the functions that now carry his name. Special cases were older: Daniel Bernoulli met them in a hanging chain in the 1730s and Euler in a vibrating membrane in 1764. Frobenius gave the general method for equations with a singular point in 1873, in Crelle’s Journal. Airy introduced his function in 1838, in a paper on the light near a caustic, to explain the faint bands inside a rainbow.

The idea, in words

Why· Replace the guess e^(rx) by an infinite polynomial

For constant coefficients the derivative of $e^{rx}$ is again a multiple of $e^{rx}$ , so substituting turns the equation into a polynomial equation for $r$ . When a coefficient depends on $x$ , the factor $x$ spoils this. A power series $y = \sum a_nx^n$ is flexible enough to absorb it: differentiating lowers the power of $x$ by one, and multiplying by $x$ raises it by one, so after substitution every power $x^k$ gets a coefficient made of a few of the $a_n$ . Setting each coefficient to zero gives a recurrence, a rule for $a_{k+2}$ in terms of earlier ones. Two numbers ( $a_0$ and $a_1$ ) stay free, and they are the starting values $y(0)$ and $y'(0)$ of 7.1.

Two questions remain. First, does the series converge, and where? The answer (below) is that it converges at least as far as the coefficients of the equation do. Second, what if the coefficients are infinite at the point we expand about, as in Bessel’s equation at $x = 0$ ? Then close to that point the equation looks like an Euler equation $x^2y'' + p_0xy' + q_0y = 0$ , whose solutions are the powers $x^s$ (7.2, problem on $t^2y'' + ty' - y = 0$ ). The Frobenius method multiplies the series by one such power $x^s$ , and the number $s$ comes from a quadratic equation, the indicial equation.

Definitions and theorems

Matching coefficients

Every power series method ends with a step like “so each coefficient must be zero”. That step needs a reason.

Theorem T MATH 376 lecture of April 22 (Example 8.12); standard

Matching coefficients

Suppose $f(x) = \sum_{n \ge 0} c_n(x - x_0)^n$ converges for $\lvert x - x_0 \rvert < r$ and $f(x) = 0$ on an interval around $x_0$ . Then $c_n = 0$ for all $n$ . Equivalently: two power series that agree on an interval have the same coefficients.

In one lineIf a power series $\sum c_n(x - x_0)^n$ equals $0$ at every point of an interval around $x_0$ , then every coefficient $c_n$ is $0$ .

Why it is trueA power series can be differentiated term by term, and the $k$ -th derivative at the centre is $k!\,c_k$ . A function that is zero on an interval has all derivatives zero.

ProofMATH 376 lecture of April 22; 1.6

The idea: Differentiate k times at the centre to read off c_k, then use that f is the zero function.

f^{(k)}(x) = \sum_{n \ge k} n(n-1)\cdots(n-k+1)\,c_n(x - x_0)^{n-k}

A power series may be differentiated term by term inside its radius of convergence (1.6); each differentiation lowers every power by one.

f^{(k)}(x_0) = k!\,c_k

x = x_0

every term with

n > k

contains the factor

(x - x_0)^{n-k} = 0

, and only

n = k

survives.

Hence

k!\,c_k = 0

, and

c_k = 0

for every

k

f

is the zero function on an interval, so all its derivatives there are

0

\blacksquare

The April 22 notes justify this step loosely (they cite the fundamental theorem of algebra, which is about roots of polynomials). The reason is the theorem above, which is about derivatives of series.

Analytic functions and ordinary points

Definition A

Analytic function

A function $f$ is analytic on the interval $(x_0 - r, x_0 + r)$ if there are numbers $a_0, a_1, \dots$ with $f(x) = \sum_{n \ge 0} a_n(x - x_0)^n$ for every $x$ in the interval. Apostol (6.17) uses this definition. Example: $\frac{1}{1 - x^2} = \sum_{n \ge 0} x^{2n}$ is analytic on $(-1, 1)$ (geometric series, 1.6) and nowhere else around $0$ , because the function is infinite at $\pm 1$ .

In one lineA function $f$ is analytic on $(x_0 - r, x_0 + r)$ if it equals a power series $\sum a_n(x - x_0)^n$ at every point of that interval.

Why it existsIt is the class of functions that a power series method can handle, and it contains every polynomial, $e^x$ , $\sin$ , $\cos$ , and every quotient of those where the denominator is not zero.

Consider a second-order equation in the form with leading coefficient $1$ :

y'' + p(x)\,y' + q(x)\,y = 0. \tag{$\ast$}

(If the equation is given as $A(x)y'' + B(x)y' + C(x)y = 0$ , divide by $A$ where $A(x) \ne 0$ : $p = B/A$ and $q = C/A$ .)

Definition A

Ordinary point

The point $x_0$ is an ordinary point of $(\ast)$ if $p$ and $q$ are both analytic on some interval $(x_0 - r, x_0 + r)$ .

In one lineA point $x_0$ is an ordinary point of $y'' + py' + qy = 0$ if both $p$ and $q$ are analytic on an interval around $x_0$ .

Why it existsAt an ordinary point the equation is as well behaved as a constant-coefficient one, and a power series centred there solves it.

Definition A

Singular point

A point $x_0$ that is not an ordinary point of $(\ast)$ is a singular point. Example: for $(1 - x^2)y'' - 2xy' + \nu(\nu + 1)y = 0$ we get $p = \frac{-2x}{1 - x^2}$ and $q = \frac{\nu(\nu + 1)}{1 - x^2}$ , which are infinite at $x = \pm 1$ , so $\pm 1$ are singular points, and every other point is ordinary.

In one lineA point that is not an ordinary point of $y'' + py' + qy = 0$ : $p$ or $q$ fails to be analytic there, usually because it is infinite.

Why it existsSingular points are where the equation can change character, and where the interesting special functions live.

The main theorem at an ordinary point

Theorem A Apostol 6.17 (Theorem 6.13)

Series solutions at an ordinary point

Let $p(x) = \sum_{k \ge 0} b_k(x - x_0)^k$ and $q(x) = \sum_{k \ge 0} c_k(x - x_0)^k$ on $(x_0 - r, x_0 + r)$ . Then the equation $y'' + py' + qy = 0$ has two linearly independent solutions $u_1, u_2$ , each a power series $\sum a_n(x - x_0)^n$ that converges on the whole interval. They are given by $a_0 = 1, a_1 = 0$ for $u_1$ and $a_0 = 0, a_1 = 1$ for $u_2$ , and for all $n \ge 0$

(n+2)(n+1)\,a_{n+2} = -\sum_{k=0}^{n}\Big[(k+1)\,a_{k+1}\,b_{n-k} + a_k\,c_{n-k}\Big]. \tag{R}

In one lineIf $p$ and $q$ are analytic on $(x_0 - r, x_0 + r)$ , then $y'' + py' + qy = 0$ has two independent solutions that are power series in $x - x_0$ converging on the whole interval. The coefficients come from a recurrence, with $a_0$ and $a_1$ free.

Why it is trueSubstituting a series gives a recurrence for the coefficients. The coefficients of $p$ and $q$ are bounded by a geometric sequence, and that makes the solution's coefficients bounded by a geometric sequence too, so the series converges as far as $p$ and $q$ do.

ProofApostol 6.17; the induction is arranged differently from Apostol's, to avoid the ratio test

The idea: Define the coefficients by the recurrence (R). Show that |a_n| is at most K times rho^(-n) for every rho below r, by induction, using that b_k and c_k are bounded by M times s^(-k). Then the series converges on the interval, and term-by-term differentiation shows it solves the equation.

The coefficient of

(x - x_0)^n

y'' + py' + qy

(n+2)(n+1)a_{n+2} + \sum_{k=0}^n[(k+1)a_{k+1}b_{n-k} + a_kc_{n-k}]

. By matching coefficients, the equation holds exactly when this is

0

for every

n

, which is (R).

Put

y = \sum a_n(x - x_0)^n

in the equation and use term-by-term differentiation:

y' = \sum (n+1)a_{n+1}(x - x_0)^n

and

y'' = \sum (n+2)(n+1)a_{n+2}(x - x_0)^n

. The products

py'

and

qy

are products of absolutely convergent series (Cauchy product, 1.6), so their coefficient of

(x - x_0)^n

\sum_{k=0}^n (k+1)a_{k+1}b_{n-k}

, respectively

\sum_{k=0}^n a_kc_{n-k}

Choose

a_0, a_1

and define

a_2, a_3, \dots

by (R).

(R) gives

a_{n+2}

from

a_0, \dots, a_{n+1}

, so after the free choice of

a_0

and

a_1

every coefficient is determined. This is a definition of the numbers

a_n

, and it remains to show the series converges.

There is a number

M

with

\lvert b_k \rvert \le M s^{-k}

and

\lvert c_k \rvert \le Ms^{-k}

for all

k

Pick

0 < \rho < s < r

. The series

\sum b_ks^k

and

\sum c_ks^k

converge (inside the radius), so their terms tend to

0

and are bounded.

Choose

N_0

so that

\frac{M\rho(n+1+\rho)}{(1-\theta)(n+2)(n+1)} \le 1

for all

n \ge N_0

. Put

K = \max_{j \le N_0 + 1}\lvert a_j \rvert\rho^j

Let

\theta = \rho/s < 1

. The expression

\frac{M\rho(n+1+\rho)}{(1-\theta)(n+2)(n+1)}

tends to

0

n \to \infty

(the numerator grows like

n

, the denominator like

n^2

), so it is at most

1

from some

n

on.

(n+2)(n+1)\lvert a_{n+2} \rvert \le MK\Big[\rho^{-n-1}\sum_{k=0}^n(k+1)\theta^{n-k} + \rho^{-n}\sum_{k=0}^n\theta^{n-k}\Big]

Strong induction on

n

. Suppose

\lvert a_j \rvert \le K\rho^{-j}

for all

j \le n + 1

, where

n \ge N_0

(true for

j \le N_0 + 1

by the choice of

K

). Take absolute values in (R), use

\lvert b_{n-k} \rvert, \lvert c_{n-k} \rvert \le Ms^{k-n}

and the hypothesis. Since

s^{k-n} = \rho^{k-n}\theta^{n-k}

, the term

(k+1)\rho^{-k-1}s^{k-n}

equals

(k+1)\rho^{-n-1}\theta^{n-k}

and the term

\rho^{-k}s^{k-n}

equals

\rho^{-n}\theta^{n-k}

.This is the step where the geometric decay of the coefficients of

p

and

q

is turned into geometric decay of the

a_n

\lvert a_{n+2} \rvert\rho^{n+2} \le K\,\frac{M\rho(n+1+\rho)}{(1-\theta)(n+2)(n+1)} \le K

, which completes the induction:

\lvert a_n \rvert \le K\rho^{-n}

for all

n

Geometric series:

\sum_{k=0}^n\theta^{n-k} \le \frac{1}{1-\theta}

and

\sum_{k=0}^n(k+1)\theta^{n-k} \le \frac{n+1}{1-\theta}

. Multiply by

\rho^{n+2}

and divide by

(n+2)(n+1)

, then use the choice of

N_0

\sum a_n(x - x_0)^n

converges absolutely for

\lvert x - x_0 \rvert < r

: its radius of convergence is at least

r

Comparison with the geometric series

\sum K(\lvert x - x_0 \rvert/\rho)^n

, which converges for

\lvert x - x_0 \rvert < \rho

. Since

\rho < r

was arbitrary, this covers the whole interval.

The sum

y

satisfies the equation on

(x_0 - r, x_0 + r)

Inside the radius the series can be differentiated term by term, and the Cauchy products are legitimate. The computation of the first step then shows that every coefficient of

y'' + py' + qy

is zero by (R), so the sum is zero.

W(u_1, u_2)(x_0) = 1\cdot1 - 0\cdot0 = 1 \ne 0

, so

u_1, u_2

are independent.

The Wronskian of 7.3 at

x_0

u_1(x_0) = 1

u_1'(x_0) = a_1 = 0

u_2(x_0) = 0

u_2'(x_0) = 1

. A non-zero Wronskian at one point proves independence.

\blacksquare

Proof workshopRebuild the convergence proof for series at an ordinary pointApostol 6.17

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Beyond the course: for understanding· B: the radius is set by the complex plane

The theorem says “at least $r$ ”. It can be larger, and it cannot always be larger than the nearest singularity in the complex plane. The equation $(1 + x^2)y'' + 2xy' = 0$ has no singular point on the real line, yet its solution $\arctan x$ has radius $1$ around $0$ : the coefficient $\frac{2x}{1+x^2}$ is infinite at $x = \pm i$ , which are at distance $1$ from the origin. This is the first picture’s second equation. Over the complex numbers (MATH 521) the guarantee is sharp: the radius of convergence is the distance to the nearest singular point of the coefficients.

Quick checkThe equation

(1 + x^2)y'' + 2xy' = 0

is expanded about

x_0 = 0

. What does the theorem guarantee?

Example: the Airy equation

Worked example MATH 376 worksheet 26 #5 (the method); Apostol 6.17

Find two independent power series solutions of y'' = xy (Airy's equation) and say where they converge.

Solution. Use the stepper to watch the recurrence appear.

Answer. $y = a_0u_1 + a_1u_2$ with $u_1, u_2$ as above, converging for all real $x$ . The first picture shows $u_1$ . For $x > 0$ the solution curves away from the axis (because $y'' = xy$ has the sign of $y$ ), and for $x < 0$ it oscillates (because $y'' = xy$ has the opposite sign to $y$ ). This is the whole character of the Airy function.

Your answerFor

y'' = xy

with

y(0) = 1

y'(0) = 0

the recurrence gives

a_3 = \frac16

. What is

a_6

? (A fraction.)

Quick checkWithout solving

y'' = xy

: for large positive

x

, what do the solutions do?

The Legendre equation

Definition A

Legendre polynomial

Legendre’s equation is

(1 - x^2)\,y'' - 2x\,y' + \nu(\nu + 1)\,y = 0,

where $\nu$ (Greek nu) is a constant. The Legendre polynomial of degree $n$ ( $n = 0, 1, 2, \dots$ ) is defined by Rodrigues’ formula

P_n(x) = \frac{1}{2^n\,n!}\,\frac{d^n}{dx^n}\big(x^2 - 1\big)^n .

The first ones are $P_0 = 1$ , $P_1 = x$ , $P_2 = \frac{3x^2 - 1}{2}$ , $P_3 = \frac{5x^3 - 3x}{2}$ , $P_4 = \frac{35x^4 - 30x^2 + 3}{8}$ .

In one lineThe polynomial $P_n(x) = \dfrac{1}{2^n n!}\dfrac{d^n}{dx^n}(x^2 - 1)^n$ of degree $n$ (Rodrigues' formula). It solves Legendre's equation with $\nu = n$ and has $P_n(1) = 1$ .

Why it existsLegendre's equation $(1-x^2)y'' - 2xy' + \nu(\nu+1)y = 0$ describes gravity and electric fields around a sphere. Its polynomial solutions are the angular patterns, and they are mutually orthogonal.

We find the series first, then show the polynomials are among its solutions.

Theorem A Apostol 6.18 (formula 6.36)

Recurrence for Legendre's equation

For Legendre’s equation, with $y = \sum a_nx^n$ , the coefficients satisfy $a_{n+2} = \frac{(n - \nu)(n + \nu + 1)}{(n+1)(n+2)}a_n$ for $n \ge 0$ , and $a_0, a_1$ are free. The even chain ( $a_0, a_2, a_4, \dots$ ) and the odd chain ( $a_1, a_3, \dots$ ) never mix. Both series converge for $\lvert x \rvert < 1$ . If $\nu = m$ is a non-negative integer, the chain that contains $a_m$ stops: $a_{m+2} = a_{m+4} = \cdots = 0$ , and it is a polynomial of degree $m$ .

In one lineSeries solutions $y = \sum a_nx^n$ of Legendre's equation satisfy $a_{n+2} = \dfrac{(n-\nu)(n+\nu+1)}{(n+1)(n+2)}\,a_n$ . They converge for $\lvert x \rvert < 1$ , and they stop (are polynomials) when $\nu$ is an integer of the right parity.

Why it is trueThe two singular points $\pm1$ limit the radius to $1$ . The factor $(n - \nu)$ is zero when $n = \nu$ , and then the chain of coefficients ends.

ProofApostol 6.18

The idea: Compute each of the three terms as a sum of multiples of x^n, collect, and read off the recurrence. The radius comes from the theorem of this chapter, since 1/(1-x^2) is a geometric series.

-2xy' = \sum -2na_nx^n

\ y'' = \sum (n+2)(n+1)a_{n+2}x^n

and

-x^2y'' = \sum -n(n-1)a_nx^n

Term-by-term differentiation:

y' = \sum na_nx^{n-1}

y'' = \sum n(n-1)a_nx^{n-2}

. Multiplying by

x

x^2

shifts the power, and the shift for

y''

is the one from the quiz.

(n+2)(n+1)a_{n+2} - n(n-1)a_n - 2na_n + \nu(\nu+1)a_n = 0

Matching coefficients: the coefficient of each

x^n

must be

0

(n+2)(n+1)a_{n+2} = \big[n(n+1) - \nu(\nu+1)\big]a_n = (n - \nu)(n + \nu + 1)a_n

Algebra:

n(n-1) + 2n = n(n+1)

, and

n(n+1) - \nu(\nu+1) = (n - \nu)(n + \nu + 1)

(expand the right side).

a_{n+2} = \dfrac{(n-\nu)(n+\nu+1)}{(n+1)(n+2)}\,a_n

Divide by

(n+1)(n+2)

The radius is at least

1

, and the chain stops when

n = \nu = m

The equation in the form

(\ast)

has

p = \frac{-2x}{1-x^2}

and

q = \frac{\nu(\nu+1)}{1-x^2}

, and

\frac{1}{1-x^2} = \sum x^{2n}

is analytic on

(-1, 1)

. Series solutions converge there by the main theorem. If

\nu = m

then the factor

n - \nu

0

n = m

, so

a_{m+2} = 0

and every later coefficient of the same chain is

0

\blacksquare

Worked example Apostol 6.18 (formulas 6.38 to 6.40)

Legendre's equation for ν = 2 and for ν = 1/2: find the first coefficients of the even solution.

Solution.

Write the recurrence for the even chain. $a_{n+2} = \frac{(n-\nu)(n+\nu+1)}{(n+1)(n+2)}a_n$ for $n = 0, 2, 4, \dots$ with $a_0 = 1$ . The odd coefficients belong to a different chain and are not needed for the even solution.
Take $\nu = 2$ . $a_2 = \frac{(0-2)(0+3)}{1\cdot2}a_0 = -3$ , and $a_4 = \frac{(2-2)(2+3)}{3\cdot4}a_2 = 0$ . The factor $n - \nu$ is $0$ at $n = 2$ , so the chain ends.
Read off the solution. $u_1 = 1 - 3x^2$ . Check: $y'' = -6$ , $y' = -6x$ , so $(1-x^2)(-6) - 2x(-6x) + 6(1 - 3x^2) = -6 + 6x^2 + 12x^2 + 6 - 18x^2 = 0$ . Substituting the answer into the equation is the check.
Normalise. $P_2(1) = 1$ forces the multiple $-\frac12(1 - 3x^2) = \frac{3x^2 - 1}{2}$ . $P_n$ is defined to equal $1$ at $x = 1$ , and every multiple of a solution is a solution.
Take $\nu = \frac12$ . $a_2 = \frac{(0 - \frac12)(0 + \frac32)}{2}a_0 = -\frac38$ , $a_4 = \frac{(2 - \frac12)(2 + \frac32)}{12}a_2 = \frac{5.25}{12}\cdot\left(-\frac38\right) = -\frac{21}{128}$ . Now $n - \nu$ is never $0$ for $n$ even, so no coefficient is zero and the series does not stop.

Answer. For $\nu = 2$ : $1 - 3x^2 = -2P_2$ . For $\nu = \frac12$ : $1 - \frac38x^2 - \frac{21}{128}x^4 - \cdots$ , an infinite series with radius $1$ .

Quick checkFor which

\nu

is the EVEN series solution of Legendre's equation (the one starting

a_0 = 1

a_1 = 0

) a polynomial?

Theorem A Apostol 6.19 and 6.20; the proof by Leibniz's rule is the standard one

Legendre polynomials solve Legendre's equation

For every integer $n \ge 0$ the Rodrigues polynomial $P_n = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$ has degree $n$ , satisfies Legendre’s equation with $\nu = n$ , and has $P_n(1) = 1$ . It is the only polynomial solution with these properties: the other series solution for $\nu = n$ does not stop.

In one line $P_n$ has degree $n$ and satisfies $(1 - x^2)P_n'' - 2xP_n' + n(n+1)P_n = 0$ and $P_n(1) = 1$ .

Why it is trueDifferentiating the identity $(x^2-1)w' = 2nxw$ for $w = (x^2-1)^n$ exactly $n + 1$ times gives Legendre's equation for $w^{(n)}$ .

Proofstandard

The idea: Write w for (x^2 - 1)^n. It satisfies a first-order identity. Differentiate that identity n + 1 times with Leibniz's product rule and recognise Legendre's equation for the n-th derivative of w.

Let

w = (x^2-1)^n

. Then

(x^2 - 1)\,w' = 2nx\,w

Product and chain rule:

w' = n(x^2-1)^{n-1}\cdot2x

. Multiply by

x^2 - 1

Differentiate

n + 1

times:

(x^2-1)w^{(n+2)} + (n+1)\,2x\,w^{(n+1)} + \tfrac{(n+1)n}{2}\,2\,w^{(n)} = 2n\big[x\,w^{(n+1)} + (n+1)\,w^{(n)}\big]

Leibniz's rule:

(fg)^{(m)} = \sum_j\binom mj f^{(j)}g^{(m-j)}

. For

f = x^2 - 1

only

f, f' = 2x, f'' = 2

are non-zero, so only three terms occur. For

f = x

only two terms occur.

(x^2-1)w^{(n+2)} + 2x\,w^{(n+1)} - n(n+1)\,w^{(n)} = 0

Collect the terms:

2(n+1)x - 2nx = 2x

, and

n(n+1) - 2n(n+1) = -n(n+1)

y = w^{(n)}

satisfies

(1 - x^2)y'' - 2xy' + n(n+1)y = 0

P_n

is a constant multiple of

y = w^{(n)}

. Multiply the equation by

-1

The degree of

w^{(n)}

n

w

has degree

2n

, and each derivative lowers the degree by one.

x = 1

w^{(n)}(1) = n!\,(1+1)^n = 2^nn!

, so

P_n(1) = 1

Write

w = (x-1)^n(x+1)^n

and apply Leibniz's rule to the

n

-th derivative. Every term that differentiates

(x-1)^n

fewer than

n

times still contains a factor

(x-1)

and vanishes at

x = 1

. The one remaining term has all

n

derivatives on

(x-1)^n

.This is why the factor

\frac{1}{2^nn!}

is in the definition: it makes the value at

1

equal to

1

P_n

is the unique polynomial solution with

P_n(1) = 1

The solution space is spanned by the even and the odd series. For

\nu = n

one of them is a polynomial and the other is not, because for the other parity no factor

(m - \nu)

ever vanishes. So up to a multiple there is one polynomial solution, and

P_n(1) = 1

fixes the multiple.

\blacksquare

Worked example Apostol 6.19

Compute P_3 by Rodrigues' formula and check that it solves Legendre's equation with ν = 3.

Solution.

Expand $(x^2 - 1)^3$ . $x^6 - 3x^4 + 3x^2 - 1$ . The binomial theorem with $a = x^2$ , $b = -1$ .
Differentiate three times. First $6x^5 - 12x^3 + 6x$ , then $30x^4 - 36x^2 + 6$ , then $120x^3 - 72x$ . Each step lowers every power by one and multiplies by it.
Divide by $2^3\,3! = 48$ . $P_3 = \frac{120x^3 - 72x}{48} = \frac{5x^3 - 3x}{2}$ . This is Rodrigues’ formula.
Check the value at $1$ . $P_3(1) = \frac{5 - 3}{2} = 1$ . The normalisation.
Check the equation. $P_3' = \frac{15x^2 - 3}{2}$ , $P_3'' = 15x$ . Then $(1-x^2)(15x) - 2x\cdot\frac{15x^2-3}{2} + 12\cdot\frac{5x^3-3x}{2} = 15x - 15x^3 - 15x^3 + 3x + 30x^3 - 18x = 0$ . $\nu(\nu+1) = 12$ for $\nu = 3$ .

Answer. $P_3(x) = \frac{5x^3 - 3x}{2}$ .

Your answerUsing

P_4(x) = \frac{35x^4 - 30x^2 + 3}{8}

, what is

P_4\!\left(\frac12\right)

? (A fraction.)

Orthogonality

Now the property that makes Legendre polynomials useful. Use the inner product $\langle f, g \rangle = \int_{-1}^{1}f(x)g(x)\,dx$ of 2.9. Legendre’s equation can be written $\big((1 - x^2)y'\big)' = -\nu(\nu+1)y$ , which says that $y$ is an eigenvector of the operator $T(y) = \big((1 - x^2)y'\big)'$ with eigenvalue $-\nu(\nu+1)$ . That operator is symmetric in the sense of 3.1, and eigenvectors for different eigenvalues of a symmetric operator are orthogonal (3.1). The proof below is that argument written for this operator.

Theorem A Apostol 6.19 and 6.20

Orthogonality of the Legendre polynomials

For non-negative integers $m \ne n$ : $\int_{-1}^{1}P_mP_n\,dx = 0$ . For every $n$ : $\int_{-1}^{1}P_n^2\,dx = \frac{2}{2n+1}$ .

In one line $\int_{-1}^{1}P_m(x)P_n(x)\,dx = 0$ for $m \ne n$ , and $\int_{-1}^{1}P_n(x)^2\,dx = \dfrac{2}{2n+1}$ .

Why it is trueEach $P_n$ is an eigenvector of the symmetric operator $y \mapsto ((1-x^2)y')'$ with eigenvalue $-n(n+1)$ . The eigenvalues for different $n$ are different, and the boundary terms of integration by parts vanish because $1 - x^2 = 0$ at $\pm1$ .

ProofApostol 6.19 and 6.20

The idea: Write Legendre's equation for P_m and P_n in the form (p y')' = -lambda y, multiply by the other polynomial, subtract and integrate. The left side integrates to boundary terms that vanish. For the norm, integrate Rodrigues' formula by parts n times.

\big((1-x^2)P_n'\big)' = -n(n+1)P_n

and

\big((1-x^2)P_m'\big)' = -m(m+1)P_m

Legendre's equation is

\big((1 - x^2)y'\big)' = -\nu(\nu+1)y

, because

\big((1-x^2)y'\big)' = (1-x^2)y'' - 2xy'

. Use it for

P_m

with

\nu = m

and for

P_n

with

\nu = n

P_m\big((1-x^2)P_n'\big)' - P_n\big((1-x^2)P_m'\big)' = \big[m(m+1) - n(n+1)\big]P_mP_n

Multiply the first equation by

P_m

, the second by

P_n

, and subtract. The right sides give

\big[m(m+1) - n(n+1)\big]P_mP_n

The left side equals

\dfrac{d}{dx}\Big[(1 - x^2)\big(P_mP_n' - P_nP_m'\big)\Big]

The left side is the derivative of

(1-x^2)(P_mP_n' - P_nP_m')

: expand the derivative, the terms

P_m'(1-x^2)P_n'

cancel in pairs.This is the picture of a symmetric operator: moving the derivative from one function to the other costs only a boundary term.

\big[m(m+1) - n(n+1)\big]\displaystyle\int_{-1}^{1}P_mP_n\,dx = \Big[(1-x^2)(\cdots)\Big]_{-1}^{1} = 0

Integrate from

-1

1

. The bracket contains the factor

1 - x^2

, which is

0

at both ends, and the polynomials are finite there.

For

m \ne n

\int_{-1}^{1}P_mP_n\,dx = 0

m(m+1) = n(n+1)

means

(m - n)(m + n + 1) = 0

, and

m + n + 1 > 0

. So for

m \ne n

the bracket is non-zero and can be divided out.

\displaystyle\int_{-1}^{1}P_n^2\,dx = \frac{1}{(2^nn!)^2}\int_{-1}^{1}w^{(n)}w^{(n)}\,dx = \frac{(-1)^n}{(2^nn!)^2}\int_{-1}^{1}w\,w^{(2n)}\,dx

For the norm write

P_n = \frac{1}{2^nn!}w^{(n)}

with

w = (x^2-1)^n

and integrate by parts

n

times, moving one derivative at a time from one factor to the other. Each boundary term contains some

w^{(k)}

with

k < n

\pm1

, and these vanish because

w

has a zero of order

n

\pm 1

= \dfrac{(2n)!}{(2^nn!)^2}\displaystyle\int_{-1}^{1}(1-x^2)^n\,dx

w

is a polynomial of degree

2n

with leading coefficient

1

, so

w^{(2n)} = (2n)!

. Also

(-1)^nw = (1-x^2)^n

(2n+1)I_n = 2nI_{n-1}

, so

I_n = \dfrac{2^{2n+1}(n!)^2}{(2n+1)!}

Let

I_n = \int_{-1}^{1}(1-x^2)^n dx

. Integrate by parts with

u = (1-x^2)^n

dv = dx

: the boundary term vanishes and

I_n = 2n\int x^2(1-x^2)^{n-1}dx = 2n\,(I_{n-1} - I_n)

, because

x^2 = 1 - (1 - x^2)

. Solve for

I_n

and start from

I_0 = 2

\displaystyle\int_{-1}^{1}P_n^2\,dx = \dfrac{2}{2n+1}

Multiply the two constants:

\frac{(2n)!}{(2^nn!)^2}\cdot\frac{2^{2n+1}(n!)^2}{(2n+1)!} = \frac{2\,(2n)!}{(2n+1)!} = \frac{2}{2n+1}

\blacksquare

In the picture above: for $m = 1$ , $n = 2$ the product of an odd and an even function is odd, and it cancels by symmetry. For $m = 1$ , $n = 3$ or $m = 2$ , $n = 4$ the product is even, so nothing cancels by symmetry: the areas above and below the axis are equal for a reason that comes from the argument above.

Proof workshopRebuild: Legendre polynomials of different degree are orthogonalApostol 6.19

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$\big((1-x^2)P_n'\big)' = -n(n+1)P_n$ and $\big((1-x^2)P_m'\big)' = -m(m+1)P_m$ .
$P_m\big((1-x^2)P_n'\big)' - P_n\big((1-x^2)P_m'\big)' = \big[m(m+1) - n(n+1)\big]P_mP_n$ .
The left side equals $\frac{d}{dx}\big[(1-x^2)(P_mP_n' - P_nP_m')\big]$ .
Integrating from $-1$ to $1$ , the left side gives $0$ .
So $\big[m(m+1) - n(n+1)\big]\int_{-1}^{1}P_mP_n\,dx = 0$ , and the bracket is non-zero when $m \ne n$ .

Quick checkWhich of these integrals is NOT zero?

Your answerWhat is

\int_{-1}^{1}P_3(x)^2\,dx

? (A fraction.)

Corollary A Apostol 6.20 (properties list)

Expansion in Legendre polynomials

(a) If $q$ is a polynomial of degree less than $n$ then $\int_{-1}^{1}P_nq\,dx = 0$ . (b) If $f$ is a polynomial of degree at most $N$ , then $f = \sum_{k=0}^{N}c_kP_k$ with $c_k = \frac{2k+1}{2}\int_{-1}^{1}f(x)P_k(x)\,dx$ .

In one line $P_n$ is orthogonal to every polynomial of degree less than $n$ . A polynomial $f$ of degree at most $N$ equals $\sum_{k=0}^{N}c_kP_k$ with $c_k = \frac{2k+1}{2}\int_{-1}^{1}fP_k\,dx$ .

Why it is true $P_0, \dots, P_{N}$ have degrees $0, \dots, N$ , so they are a basis of the polynomials of degree at most $N$ . Taking the inner product of $f = \sum c_jP_j$ with $P_k$ kills every term but one.

ProofApostol 6.20

The idea: Change basis from powers of x to Legendre polynomials, then use orthogonality.

A polynomial

q

of degree below

n

is a combination

q = \sum_{k<n}d_kP_k

P_k

has degree exactly

k

, so

P_0, \dots, P_{n-1}

are

n

independent polynomials in the

n

-dimensional space of polynomials of degree below

n

, hence a basis (2.2).

\int P_nq = \sum_{k<n}d_k\int P_nP_k = 0

. This is (a).

Linearity of the integral and orthogonality of

P_n

P_k

for

k < n

\int fP_k = c_k\cdot\frac{2}{2k+1}

, so

c_k = \frac{2k+1}{2}\int fP_k

\blacksquare

The same basis argument for degree at most

N

gives

f = \sum_{j \le N}c_jP_j

for some numbers

c_j

. Multiply by

P_k

and integrate: orthogonality leaves only

j = k

, and the norm formula gives

\int P_k^2 = \frac{2}{2k+1}

This also says what the Legendre polynomials are in the language of 2.9. Run the Gram–Schmidt process on $1, x, x^2, \dots$ with the inner product $\int_{-1}^{1}fg\,dx$ . The $n$ -th output is a polynomial of degree $n$ orthogonal to all lower degrees, and by part (a) so is $P_n$ . Such a polynomial is unique up to a constant multiple, so Gram–Schmidt produces $P_n$ up to a scale. (If you never rescale, Gram–Schmidt on $1, x, x^2, \dots$ gives the version with leading coefficient $1$ , while Rodrigues’ formula gives the version with value $1$ at $x = 1$ .)

Worked example Apostol 6.20; new numbers

Expand x² in Legendre polynomials, and check the answer at x = 1.

Solution.

Choose the range of $k$ . $x^2$ has degree $2$ , so $c_k = 0$ for $k > 2$ . By part (a): $\int x^2P_k = 0$ when $\deg x^2 < k$ .
Compute $c_0$ . $c_0 = \frac12\int_{-1}^{1}x^2\,dx = \frac12\cdot\frac23 = \frac13$ . $P_0 = 1$ and the formula with $k = 0$ .
Compute $c_1$ . $x^2P_1 = x^3$ is odd, so $c_1 = 0$ . An odd function integrates to $0$ over $[-1, 1]$ .
Compute $c_2$ . $c_2 = \frac52\int_{-1}^{1}x^2\cdot\frac{3x^2 - 1}{2}\,dx = \frac54\left(3\cdot\frac25 - \frac23\right) = \frac54\cdot\frac{8}{15} = \frac23$ . $\int x^4 = \frac25$ and $\int x^2 = \frac23$ over $[-1, 1]$ .
Assemble and check. $x^2 = \frac13P_0 + \frac23P_2 = \frac13 + \frac{3x^2 - 1}{3}$ . At $x = 1$ the right side is $\frac13 + \frac23 = 1$ . $P_n(1) = 1$ for every $n$ , so the coefficients must add up to $f(1)$ .

Answer. $x^2 = \frac13P_0(x) + \frac23P_2(x)$ .

Beyond the course: for understanding· B: why physics only uses integer ν

For $\nu$ not an integer, neither the even nor the odd series stops. The ratio of consecutive coefficients in a chain is $\frac{a_{n+2}}{a_n} = 1 - \frac2n + O(n^{-2})$ , so the coefficients behave like $\frac1n$ , as in the harmonic series, and $\sum a_n$ diverges at $x = \pm1$ . In fact the solutions are unbounded there (like $\ln(1 - x)$ ). When Legendre’s equation comes from a problem on a sphere, $x = \cos\theta$ and the points $x = \pm1$ are the north and south poles, where the answer must be finite. That forces $\nu = 0, 1, 2, \dots$ , and the allowed angular patterns are exactly the Legendre polynomials. A sketch of the divergence is problem S1 below.

Regular singular points and the Frobenius method

Bessel’s equation $x^2y'' + xy' + (x^2 - \nu^2)y = 0$ has $p(x) = \frac1x$ and $q(x) = 1 - \frac{\nu^2}{x^2}$ , which are infinite at $x = 0$ . So $0$ is a singular point, and the theorem of this chapter says nothing there. Yet the infinity is mild: $x\,p(x)$ and $x^2q(x)$ are fine. That is the pattern to isolate.

Definition A

Regular singular point

A singular point $x_0$ of $(\ast)$ is regular if $(x - x_0)\,p(x)$ and $(x - x_0)^2\,q(x)$ are both analytic around $x_0$ . Equivalently, the equation can be written

(x - x_0)^2\,y'' + (x - x_0)\,P(x)\,y' + Q(x)\,y = 0, \qquad P, Q \text{ analytic around } x_0,

with $P = (x - x_0)p$ and $Q = (x - x_0)^2q$ . A singular point that is not regular is irregular. From now on $x_0 = 0$ (shift the variable otherwise), and $P = \sum p_nx^n$ , $Q = \sum q_nx^n$ .

In one lineA singular point $x_0$ of $y'' + py' + qy = 0$ where $(x - x_0)p(x)$ and $(x - x_0)^2q(x)$ are analytic. Any other singular point is irregular.

Why it existsIt marks the singular points where the equation is close to an Euler equation, and so a power $x^s$ times a series can still solve it.

The simplest equation of this type has $P \equiv p_0$ and $Q \equiv q_0$ constant. It is the Euler equation $x^2y'' + p_0xy' + q_0y = 0$ , and the guess $y = x^s$ gives $x^s\big[s(s-1) + p_0s + q_0\big] = 0$ . This is the origin of the next definition.

Definition A

Indicial equation

The indicial equation of the regular singular point $0$ is $F(s) = s(s-1) + p_0s + q_0 = 0$ , that is $s^2 + (p_0 - 1)s + q_0 = 0$ . Its roots $\alpha_1, \alpha_2$ (with $\alpha_1 \ge \alpha_2$ if real) are the indicial roots. Note that $\alpha_1 + \alpha_2 = 1 - p_0$ .

In one lineFor $x^2y'' + xP(x)y' + Q(x)y = 0$ with $P(0) = p_0$ and $Q(0) = q_0$ : the quadratic $F(s) = s(s-1) + p_0s + q_0 = 0$ . Its roots are the possible starting exponents $s$ of a Frobenius series.

Why it existsIt is the characteristic equation of the Euler equation that the real equation resembles near $0$ . It tells you which power of $x$ the solution begins with.

Definition A

Frobenius series

A Frobenius series is a function of the form $y = x^s\sum_{n=0}^{\infty}a_nx^n$ with $a_0 \ne 0$ . For $s = 0, 1, 2, \dots$ it is an ordinary power series.

In one lineA function $x^s\sum_{n \ge 0}a_nx^n$ with $a_0 \ne 0$ , defined for $x > 0$ (use $\lvert x \rvert^s$ for $x < 0$ ). The exponent $s$ need not be a whole number.

Why it existsThe factor $x^s$ carries the singular behaviour at $0$ , and the power series carries the rest.

Theorem A Apostol 6.22 and 6.23 (the computation is done there for Bessel's equation)

The Frobenius recurrence

Let $P = \sum p_nx^n$ and $Q = \sum q_nx^n$ near $0$ . A Frobenius series $y = x^s\sum a_nx^n$ with $a_0 = 1$ solves $x^2y'' + xPy' + Qy = 0$ on an interval $(0, r)$ if and only if $F(s) = 0$ and for all $n \ge 1$

F(s + n)\,a_n = -\sum_{k=0}^{n-1}\Big[(s + k)\,p_{n-k} + q_{n-k}\Big]a_k . \tag{F}

In one lineThe Frobenius series $x^s\sum a_nx^n$ with $a_0 = 1$ solves $x^2y'' + xPy' + Qy = 0$ exactly when $F(s) = 0$ and, for every $n \ge 1$ , $F(s+n)\,a_n = -\sum_{k=0}^{n-1}\big[(s+k)p_{n-k} + q_{n-k}\big]a_k$ .

Why it is trueEach power $x^{s+k}$ is sent to a multiple of itself by the part $x^2y'' + p_0xy' + q_0y$ , namely $F(s+k)x^{s+k}$ , and the remaining terms of $P$ and $Q$ push contributions to higher powers. So the coefficients can be solved one at a time.

ProofApostol 6.23

The idea: Apply the operator to one term x^(s+k), expand P and Q as series, collect the coefficient of x^(s+n), and set each to zero.

For one term,

x^2(x^{s+k})'' + xP\,(x^{s+k})' + Q\,x^{s+k} = x^{s+k}\Big[(s+k)(s+k-1) + (s+k)\sum_jp_jx^j + \sum_jq_jx^j\Big]

For a power

x^m

x^2(x^m)'' = m(m-1)x^m

and

x(x^m)' = mx^m

. With

P = \sum p_jx^j

and

Q = \sum q_jx^j

this gives the expression on the right.

The coefficient of

x^{s+n}

in the result is

a_n\big[(s+n)(s+n-1) + (s+n)p_0 + q_0\big] + \sum_{k<n}\big[(s+k)p_{n-k} + q_{n-k}\big]a_k

Sum over

k

with coefficients

a_k

, and collect the power

x^{s+n}

: it comes from

k = n

with

j = 0

and from

k < n

with

j = n - k

. Term-by-term differentiation is allowed inside the radius of convergence.

F(s+n)a_n + \sum_{k<n}[\cdots]a_k = 0

for every

n \ge 0

The bracket in front of

a_n

F(s+n)

by the definition of

F

. Matching coefficients: each must be

0

. For

n = 0

there is no sum and

a_0 \ne 0

, so

F(s) = 0

. For

n \ge 1

solve for

F(s+n)a_n

\blacksquare

(F) can be solved for $a_n$ whenever $F(s + n) \ne 0$ . Now see what that means for each root. Write $F(s) = (s - \alpha_1)(s - \alpha_2)$ with $\alpha_1 \ge \alpha_2$ . Then

F(\alpha_1 + n) = n\,(n + \alpha_1 - \alpha_2) \ne 0 \quad (n \ge 1), \qquad F(\alpha_2 + n) = n\,(n - (\alpha_1 - \alpha_2)).

The larger root $\alpha_1$ always works: (F) can be solved for every $n$ . The smaller root $\alpha_2$ fails at $n = N$ exactly when $\alpha_1 - \alpha_2 = N$ is a positive whole number. That is the origin of the three cases.

Theorem A Apostol 6.22 (Theorems 6.14 and 6.15); Frobenius 1873

The Frobenius theorem

Let $0$ be a regular singular point, with $P, Q$ analytic on $(-r, r)$ , and let $\alpha_1 \ge \alpha_2$ be the indicial roots (if they are complex, see below). For $0 < x < r$ :

$\alpha_1 - \alpha_2$ not a whole number. There are two independent solutions $u_1 = x^{\alpha_1}\sum a_nx^n$ and $u_2 = x^{\alpha_2}\sum b_nx^n$ , with $a_0 = b_0 = 1$ .
$\alpha_1 = \alpha_2$ . There is $u_1 = x^{\alpha_1}\sum a_nx^n$ and a second solution $u_2 = u_1\ln x + x^{\alpha_1}\sum b_nx^n$ .
$\alpha_1 - \alpha_2 = N$ , a positive whole number. There is $u_1 = x^{\alpha_1}\sum a_nx^n$ and a second solution $u_2 = C\,u_1\ln x + x^{\alpha_2}\sum b_nx^n$ with $b_0 = 1$ , where the constant $C$ may or may not be $0$ .

The series converge for $\lvert x \rvert < r$ . (Complex roots $\alpha \pm i\beta$ differ by $2i\beta$ , which is not a whole number, so case 1 applies; real solutions are the real and imaginary parts of $x^{\alpha+i\beta}\sum a_nx^n$ .)

In one lineAt a regular singular point with indicial roots $\alpha_1 \ge \alpha_2$ , the larger root always gives a Frobenius series solution $u_1$ . If $\alpha_1 - \alpha_2$ is not a whole number, the smaller root gives a second one. If the roots are equal, or differ by a whole number, a second solution contains $u_1\ln x$ .

Why it is trueThe recurrence (F) can be solved for the smaller root unless $F(\alpha_2 + N) = 0$ for some $N \ge 1$ . When it fails, the second solution needs a logarithm, as the reduction-of-order computation below shows.

The theorem is stated without a full proof here, as in Apostol, who refers to Coddington and Hille for the convergence argument (a majorant argument like the proof of the ordinary-point theorem). The recurrence (F) shows how $u_1$ is built and why the three cases arise. What (F) does not show is that the series converge on the whole interval, which the references prove, and why the second solution carries a logarithm, which is explained now. For Bessel’s equation the convergence is checked directly below.

Why· Why a logarithm appears exactly when the roots differ by a whole number

Use Abel’s formula of 7.3. Any two solutions of $y'' + py' + qy = 0$ have Wronskian $W = c\,e^{-\int p}$ . Here $p = \frac{p_0}{x} + (\text{series})$ , so $W = c\,x^{-p_0}(1 + e_1x + \cdots)$ . If $u_1 = x^{\alpha_1}(1 + \cdots)$ is known, reduction of order gives a second solution $u_2 = u_1\int\frac{W}{u_1^2}\,dx$ . Insert $W$ and $u_1$ , and use $-p_0 = \alpha_1 + \alpha_2 - 1$ :

\frac{W}{u_1^2} = c\,x^{\alpha_2 - \alpha_1 - 1}\big(1 + d_1x + d_2x^2 + \cdots\big).

Integrate term by term. The term $d_jx^{\alpha_2-\alpha_1-1+j}$ integrates to a power $x^{\alpha_2-\alpha_1+j}/(\alpha_2 - \alpha_1 + j)$ , unless the exponent $\alpha_2 - \alpha_1 - 1 + j$ equals $-1$ , in which case the integral is $d_j\ln x$ . That happens for some $j \ge 0$ exactly when $\alpha_1 - \alpha_2 = j$ is a whole number. If the roots are equal, $j = 0$ and $d_0 = 1 \ne 0$ , so the logarithm is unavoidable. If they differ by $N > 0$ , the logarithm has the coefficient $d_N$ , which may happen to be $0$ . That is the constant $C$ of the theorem. The powers that remain, times $u_1$ , form $x^{\alpha_2}\sum b_nx^n$ .

Worked example Apostol 6.22; new

Find and classify the singular points at x = 0 of: x²y'' + 3xy' + y = 0; x²y'' + xy' + (x² − 4)y = 0; x²y'' + y' + y = 0; and at x = 1 of Legendre's equation.

Solution.

Equation	$P$ , $Q$ (standard form)	$p_0$ , $q_0$	Indicial polynomial $F(s)$	Type
$x^2y'' + 3xy' + y = 0$	$P = 3$ , $Q = 1$	$3$ , $1$	$s^2 + 2s + 1 = (s+1)^2$	regular, equal roots $-1, -1$
$x^2y'' + xy' + (x^2 - 4)y = 0$	$P = 1$ , $Q = x^2 - 4$	$1$ , $-4$	$s^2 - 4$	regular, roots $2, -2$ (differ by $4$ )
$x^2y'' + y' + y = 0$	$xP = 1$ gives $P = \frac1x$	none	none	irregular
Legendre at $x = 1$	$P = \frac{2x}{1+x}$ , $Q = -\frac{(x-1)\nu(\nu+1)}{1+x}$	$1$ , $0$	$s(s-1) + s = s^2$	regular, equal roots $0, 0$

Put the equation in standard form. Divide so that $y''$ has coefficient $x^2$ , then read off $xP$ as the coefficient of $y'$ and $Q$ as the coefficient of $y$ . The indicial equation is stated for $x^2y'' + xPy' + Qy = 0$ . A common mistake is to read $p_0$ off the original coefficient of $y'$ .
Test analyticity. In the third row the coefficient of $y'$ is $1 = xP$ , so $P = \frac1x$ , which is not analytic at $0$ : irregular. An irregular singular point has no indicial equation, and power series methods may fail (see the traps).
Legendre at $x = 1$ . Use the variable $x - 1$ . With $p = \frac{-2x}{1-x^2}$ we get $(x - 1)p = \frac{2x}{1+x}$ , and with $q = \frac{\nu(\nu+1)}{1-x^2}$ we get $(x-1)^2q = -\frac{(x-1)\nu(\nu+1)}{1+x}$ . At $x = 1$ : $p_0 = 1$ , $q_0 = 0$ . Both are analytic at $1$ (the denominator $1 + x$ is $2$ there), so $1$ is a regular singular point.
Read the consequence. Equal indicial roots mean case 2: Legendre’s equation has one solution analytic at $x = 1$ (the series in powers of $x - 1$ ) and a second with $\ln(x - 1)$ . That second solution is the one that is unbounded at the pole. This matches the box on non-integer $\nu$ .

Answer. As in the table.

Quick checkWhat is the indicial equation of the Euler equation

x^2y'' + 2xy' - 6y = 0

Worked example Apostol 6.22 style; new numbers

Case 1: solve 2xy'' + y' − y = 0 for x > 0 by the Frobenius method.

Solution.

Standard form. Multiply by $\frac x2$ : $x^2y'' + \frac x2\,y' - \frac x2\,y = 0$ . So $P = \frac12$ and $Q = -\frac x2$ , giving $p_0 = \frac12$ , $q_0 = 0$ , $q_1 = -\frac12$ , and all other $p_n, q_n$ equal $0$ . Read $P$ and $Q$ off the standard form.
Indicial equation. $F(s) = s(s-1) + \frac s2 = s\left(s - \frac12\right)$ , with roots $\alpha_1 = \frac12$ and $\alpha_2 = 0$ . They differ by $\frac12$ , not a whole number: case 1, two Frobenius series.
Recurrence. Only $q_1$ is non-zero, so (F) reads $F(s+n)a_n = -q_1a_{n-1} = \frac12a_{n-1}$ . The sum in (F) has only the term $k = n - 1$ , with $q_{n-k} = q_1$ .
Root $s = 0$ . $F(n) = n\left(n - \frac12\right)$ , so $a_n = \frac{a_{n-1}}{n(2n-1)}$ : $a_1 = 1$ , $a_2 = \frac16$ , $a_3 = \frac1{90}$ . $n\left(n - \frac12\right)a_n = \frac12a_{n-1}$ ; multiply by $2$ .
Root $s = \frac12$ . $F\left(\frac12 + n\right) = \left(n + \frac12\right)n$ , so $a_n = \frac{a_{n-1}}{n(2n+1)}$ : $a_1 = \frac13$ , $a_2 = \frac1{30}$ . Same computation with the other exponent.
Recognise the sums. $u_2 = 1 + x + \frac{x^2}{6} + \frac{x^3}{90} + \cdots = \cosh\sqrt{2x}$ and $x^{1/2}\left(1 + \frac x3 + \frac{x^2}{30} + \cdots\right) = \frac{\sinh\sqrt{2x}}{\sqrt2}$ . Expand $\cosh t = \sum\frac{t^{2n}}{(2n)!}$ and $\sinh t = \sum\frac{t^{2n+1}}{(2n+1)!}$ with $t = \sqrt{2x}$ : the coefficients of $x^n$ are $\frac{2^n}{(2n)!}$ and $\frac{2^n}{(2n+1)!}$ , the values above.
Check one by hand. For $y = \cosh\sqrt{2x}$ : $y' = \frac{\sinh\sqrt{2x}}{\sqrt{2x}}$ and $y'' = \frac{\cosh\sqrt{2x}}{2x} - \frac{\sinh\sqrt{2x}}{(2x)^{3/2}}$ . Then $2xy'' + y' - y = \cosh\sqrt{2x} - \frac{\sinh\sqrt{2x}}{\sqrt{2x}} + \frac{\sinh\sqrt{2x}}{\sqrt{2x}} - \cosh\sqrt{2x} = 0$ . Substitution is the final check.

Answer. $y = A\cosh\sqrt{2x} + B\sinh\sqrt{2x}$ for $x > 0$ .

Your answerThe Frobenius series for

2xy'' + y' - y = 0

at the root

s = 0

has

a_n = \frac{a_{n-1}}{n(2n-1)}

with

a_0 = 1

. What is

a_3

? (A fraction.)

Bessel’s equation

Definition A

Bessel function

For a whole number $n \ge 0$ the Bessel function of the first kind and order $n$ is the series above. The case $n = 0$ is

J_0(x) = \sum_{k=0}^{\infty}\frac{(-1)^k}{(k!)^2}\left(\frac x2\right)^{2k} = 1 - \frac{x^2}{4} + \frac{x^4}{64} - \frac{x^6}{2304} + \cdots .

(For orders that are not whole numbers, $(k+n)!$ is replaced by the gamma function; see the box below.)

In one line $J_n(x) = \sum_{k \ge 0}\dfrac{(-1)^k}{k!\,(k+n)!}\left(\dfrac x2\right)^{2k+n}$ , the solution of Bessel's equation $x^2y'' + xy' + (x^2 - n^2)y = 0$ that is finite at $x = 0$ (for $n = 0, 1, 2, \dots$ ).

Why it existsBessel's equation is what the wave equation becomes in polar coordinates, so $J_n$ describes the shape of a vibrating drumhead and of waves in a cylinder.

Theorem A Apostol 6.23

J0 solves Bessel's equation of order zero

For $x^2y'' + xy' + x^2y = 0$ : the point $0$ is regular singular with $F(s) = s^2$ . The Frobenius series for $s = 0$ is $J_0(x)$ , which converges for all $x$ , satisfies $J_0(0) = 1$ , and is an even function. Every solution that is not a multiple of $J_0$ is unbounded as $x \to 0^+$ .

In one lineThe series $J_0$ converges for every real $x$ and solves $x^2y'' + xy' + x^2y = 0$ . The indicial equation is $s^2 = 0$ (a double root), so a second solution contains $\ln x$ .

Why it is trueBessel's equation has a regular singular point at $0$ with $p_0 = 1$ and $q_0 = 0$ (order zero). The recurrence $n^2a_n = -a_{n-2}$ links every second coefficient, and the factorials make the series converge everywhere.

ProofApostol 6.23

The idea: Read P and Q off the equation, apply the Frobenius recurrence, solve it, and apply the ratio test. For the unboundedness use Abel's formula.

F(s) = s(s-1) + s = s^2

. Double root

s = 0

Standard form: the equation already has the form

x^2y'' + xPy' + Qy = 0

with

P = 1

and

Q = x^2

. So

p_0 = 1

q_2 = 1

, and all other

p_n, q_n

are

0

F(1)a_1 = 0

, so

a_1 = 0

; and

n^2a_n = -a_{n-2}

for

n \ge 2

(F) with

s = 0

. For

n = 1

the sum has only

k = 0

and the coefficients

p_1, q_1

are

0

. For

n \ge 2

the only non-zero term in the sum is

k = n - 2

, with

q_2 = 1

With

a_0 = 1

y = \sum_{k \ge 0}\frac{(-1)^k}{4^k(k!)^2}x^{2k} = J_0(x)

Odd coefficients all vanish because

a_1 = 0

. For even indices,

a_{2k} = -\frac{a_{2k-2}}{(2k)^2} = -\frac{a_{2k-2}}{4k^2}

, so by induction

a_{2k} = \frac{(-1)^k}{4^k(k!)^2}

The series converges for all

x

Ratio test:

\left\lvert\frac{a_{2k+2}x^{2k+2}}{a_{2k}x^{2k}}\right\rvert = \frac{x^2}{4(k+1)^2} \to 0

for every fixed

x

, so the radius of convergence is infinite.

u = J_0\int\frac{c\,dx}{xJ_0^2} = cJ_0\ln x + (\text{bounded})

near

0

, and

\lvert u \rvert \to \infty

unless

c = 0

\blacksquare

Abel's formula (7.3): in standard form

y'' + \frac1xy' + y = 0

has

p = \frac1x

, so any Wronskian is

\frac cx

. For a second solution

u

this gives

(u/J_0)' = \frac{W}{J_0^2} = \frac{c}{xJ_0^2}

near

0

, where

J_0 \ne 0

. As

J_0(x) \to 1

, the right side is about

\frac cx

The unbounded second solution is a combination of $J_0\ln x$ and a power series. Its name is $Y_0$ (the Bessel function of the second kind), which differs from the one computed below by a constant multiple and a multiple of $J_0$ .

Worked example Apostol 6.23; check against tables

Evaluate J0(2) from the series and find the zero of J0 near 2.4; then find the second solution of Bessel's equation of order 0 in two ways.

Solution.

Write the terms at $x = 2$ . $\left(\frac x2\right)^{2k} = 1$ , so $J_0(2) = \sum_k\frac{(-1)^k}{(k!)^2} = 1 - 1 + \frac14 - \frac1{36} + \frac1{576} - \frac1{14400} + \cdots$ . The factor $\frac x2$ is $1$ .
Add. The partial sums are $1,\ 0,\ 0.25,\ 0.22222,\ 0.22396,\ 0.22389$ , converging to $J_0(2) = 0.22389$ . This is an alternating series, so the error is less than the first omitted term ( $\frac{1}{518400}$ after the last one shown).
Locate the zero. At $x = 2.4$ the terms $\frac{(1.2)^{2k}}{(k!)^2}$ are $1,\ 1.44,\ 0.5184,\ 0.0829,\ 0.0075,\ 0.0004$ , with alternating signs, and $J_0(2.4) = 0.0025 > 0$ . The slope is $J_0'(2.4) = -J_1(2.4) = -0.52$ , so the graph reaches zero about $\frac{0.0025}{0.52} \approx 0.005$ further on, at $x \approx 2.405$ . One step of Newton’s method on the series. The exact zero is $2.4048$ .
Second solution, method 1: the logarithm ansatz. By case 2, $u = J_0\ln x + v$ with $v = \sum_{n \ge 2}b_nx^n$ . Let $L[y] = x^2y'' + xy' + x^2y$ . Then $L[J_0\ln x] = \ln x\cdot L[J_0] + 2xJ_0' = 2xJ_0'$ , so $L[v] = -2xJ_0' = x^2 - \frac{x^4}{8} + \frac{x^6}{192} - \cdots$ . Differentiating $J_0\ln x$ twice gives $J_0''\ln x + \frac{2J_0'}{x} - \frac{J_0}{x^2}$ ; the terms with $\ln x$ vanish because $L[J_0] = 0$ , and $-J_0 + J_0$ cancel.
Match coefficients. $L[x^n] = n^2x^n + x^{n+2}$ , so the coefficient of $x^2$ : $4b_2 = 1$ ; of $x^4$ : $16b_4 + b_2 = -\frac18$ ; of $x^6$ : $36b_6 + b_4 = \frac1{192}$ . This gives $b_2 = \frac14$ , $b_4 = -\frac3{128}$ , $b_6 = \frac{11}{13824}$ . The solution is unique up to adding multiples of $J_0$ ; choosing no $x^0$ term fixes it.
Method 2: Abel and reduction of order. $u = J_0\int\frac{dx}{xJ_0^2}$ . Since $J_0^2 = 1 - \frac{x^2}{2} + \frac{3x^4}{32} - \cdots$ , we get $\frac{1}{J_0^2} = 1 + \frac{x^2}{2} + \frac{5x^4}{32} + \cdots$ , so $\int\frac{dx}{xJ_0^2} = \ln x + \frac{x^2}{4} + \frac{5x^4}{128} + \cdots$ . Multiply by $J_0 = 1 - \frac{x^2}{4} + \cdots$ : $u = J_0\ln x + \frac{x^2}{4} + \left(\frac5{128} - \frac1{16}\right)x^4 + \cdots = J_0\ln x + \frac{x^2}{4} - \frac{3x^4}{128} + \cdots$ . Both methods give the same series, which is the check.

Answer. $J_0(2) = 0.2239$ ; the first zero of $J_0$ is $2.4048$ ; the second solution is $J_0(x)\ln x + \frac{x^2}{4} - \frac{3x^4}{128} + \frac{11x^6}{13824} - \cdots$ .

Your answerSum the series for

J_0(2)

until the terms are below

10^{-5}

. What is

J_0(2)

to four decimals?

Worked example Apostol 6.22 (Theorem 6.15); new numbers

Case 3: solve x y'' + y = 0 near x = 0 (the roots differ by 1 and a logarithm is needed), and show that Bessel's equation of order 1/2 has no logarithm.

Solution.

Standard form. Multiply by $x$ : $x^2y'' + xy = 0$ , so $P = 0$ and $Q = x$ ( $q_1 = 1$ ). The indicial polynomial is $F(s) = s(s-1)$ , with roots $\alpha_1 = 1$ , $\alpha_2 = 0$ , differing by $N = 1$ . A whole-number difference is case 3.
Larger root. (F) gives $F(1+n)a_n = -a_{n-1}$ with $F(1+n) = n(n+1)$ , so $a_n = -\frac{a_{n-1}}{n(n+1)}$ and $u_1 = x - \frac{x^2}{2} + \frac{x^3}{12} - \frac{x^4}{144} + \cdots$ . The larger root always works.
Smaller root fails. For $s = 0$ and $n = 1$ , (F) says $F(1)a_1 = -q_1a_0$ , that is $0\cdot a_1 = -1$ . This is impossible. $F(\alpha_2 + N) = 0$ at $N = 1$ , and the right side is not $0$ .
Allow a logarithm. Try $u_2 = Cu_1\ln x + v$ with $v = 1 + b_1x + b_2x^2 + \cdots$ . Since $xu_1'' + u_1 = 0$ , we get $xu_2'' + u_2 = C\left(2u_1' - \frac{u_1}{x}\right) + xv'' + v$ . Differentiating $u_1\ln x$ twice gives $u_1''\ln x + \frac{2u_1'}{x} - \frac{u_1}{x^2}$ ; multiply by $x$ .
Match the constant term. $2u_1' - \frac{u_1}{x} = 1 - \frac32x + \cdots$ , and $xv'' + v = 1 + (b_1 + 2b_2)x + \cdots$ . The constant term gives $C + 1 = 0$ , so $C = -1$ . The logarithm is forced: $C \ne 0$ .
Match the next term. The $x$ coefficient: $\frac32 + b_1 + 2b_2 = 0$ . With $b_1 = 0$ (a free choice, adding a multiple of $u_1$ ), $b_2 = -\frac34$ . The next terms give $b_3 = \frac7{36}$ .
Compare with Bessel of order $\frac12$ . Here $F(s) = s^2 - \frac14$ , roots $\pm\frac12$ , difference $1$ . For $s = -\frac12$ and $n = 1$ : (F) says $F(\frac12)a_1 = -(\ldots)a_0 = 0$ , and $F(\frac12) = 0$ too. So $0\cdot a_1 = 0$ holds for any $a_1$ : nothing goes wrong and $C = 0$ . The right side happens to vanish because $P$ and $Q$ have no $x^1$ terms. The two free numbers $a_0, a_1$ give $x^{-1/2}\cos x$ and $x^{-1/2}\sin x$ .

Answer. $u_1 = x - \frac{x^2}{2} + \frac{x^3}{12} - \cdots$ and $u_2 = 1 - u_1\ln x - \frac34x^2 + \frac7{36}x^3 - \cdots$ for $xy'' + y = 0$ . For Bessel’s equation of order $\frac12$ the solutions are $\frac{\sin x}{\sqrt x}$ and $\frac{\cos x}{\sqrt x}$ , with no logarithm.

Quick checkThe equation

x^2y'' + xy' + \left(x^2 - \frac14\right)y = 0

has indicial roots

\pm\frac12

, which differ by the whole number

1

. What does the Frobenius theorem predict?

Beyond the course: for understanding· B: other orders, the gamma function, and what J0 looks like for large x

For an order $\nu$ that is not a whole number, $(k + \nu)!$ is replaced by $\Gamma(k + \nu + 1)$ , where the gamma function $\Gamma(s) = \int_0^\infty t^{s-1}e^{-t}dt$ satisfies $\Gamma(s+1) = s\Gamma(s)$ and $\Gamma(n+1) = n!$ . Then $J_\nu(x) = \sum_k\frac{(-1)^k}{k!\,\Gamma(k+\nu+1)}\left(\frac x2\right)^{2k+\nu}$ , and $J_{\pm\nu}$ are independent when $2\nu$ is not a whole number. Half-integer orders are elementary: $J_{1/2}(x) = \sqrt{\frac{2}{\pi x}}\sin x$ . For large $x$ one has $J_0(x) \approx \sqrt{\frac{2}{\pi x}}\cos\left(x - \frac\pi4\right)$ : an oscillation with a slowly decaying amplitude. At $x = 10$ the exact value is $-0.2459$ and the formula gives $-0.2468$ . The zeros $2.405, 5.520, 8.654, 11.792$ are almost equally spaced by $\pi$ , but not quite; that explains the sound of a drum (see below).

Beyond the course: for understanding· B: the Airy functions Ai and Bi

The series $u_1, u_2$ of the Airy equation $y'' = xy$ both grow as $x \to +\infty$ . One special combination does not: $\operatorname{Ai}(x) = \operatorname{Ai}(0)\,u_1 + \operatorname{Ai}'(0)\,u_2$ with $\operatorname{Ai}(0) = 0.35503$ and $\operatorname{Ai}'(0) = -0.25882$ . It decays like $\frac{e^{-\frac23x^{3/2}}}{2\sqrt\pi\,x^{1/4}}$ for large positive $x$ . For large negative $x$ it oscillates, like $\frac{\sin\left(\frac23\lvert x\rvert^{3/2} + \frac\pi4\right)}{\sqrt\pi\,\lvert x\rvert^{1/4}}$ , with zeros at $-2.338, -4.088, -5.521, \dots$ . A second solution $\operatorname{Bi}$ grows. The Airy function is the model of a turning point: on one side of $x = 0$ the equation has oscillating solutions and on the other exponentials.

Python labThe Airy series against a numerical solution

import numpy as np

# Power series for y'' = x*y with y(0) = 1, y'(0) = 0, from the recurrence a[n+3] = a[n] / ((n+3)(n+2))
def coefficients(N):
    a = np.zeros(N + 1)
    a[0] = 1.0
    for n in range(N - 2):
        a[n + 3] = a[n] / ((n + 3) * (n + 2))
    return a

def partial_sum(a, x):
    return sum(c * x**k for k, c in enumerate(a))

# The same equation solved numerically (RK4), with no series anywhere
def rk4(x_end, h=5e-4):
    n = int(round(abs(x_end) / h)); h = x_end / n
    x, y, v = 0.0, 1.0, 0.0
    for _ in range(n):
        k1y, k1v = v, x * y
        k2y, k2v = v + h/2 * k1v, (x + h/2) * (y + h/2 * k1y)
        k3y, k3v = v + h/2 * k2v, (x + h/2) * (y + h/2 * k2y)
        k4y, k4v = v + h * k3v, (x + h) * (y + h * k3y)
        y += h/6 * (k1y + 2*k2y + 2*k3y + k4y)
        v += h/6 * (k1v + 2*k2v + 2*k3v + k4v)
        x += h
    return y

a = coefficients(120)
print("    x        RK4 value   rel. error of the series with degree 10, 30, 60, 120")
for x in [1.0, 5.0, -5.0, 12.0, -12.0]:
    ref = rk4(x)
    errs = [abs(partial_sum(a[:N + 1], x) - ref) / abs(ref) for N in (10, 30, 60, 120)]
    print(f"{x:6.1f}  {ref:14.6g}   " + "  ".join(f"{e:9.2e}" for e in errs))

Try this. Run it. At $x = 1$ ten terms give 7 digits ( $5 \times 10^{-7}$ ) and 30 terms reach the floor set by the numerical solution itself ( $4 \times 10^{-15}$ ). At $x = 5$ ten terms are 52% off and 60 terms are exact to 14 digits. The farther out you go, the more terms you need: at $x = 12$ even 60 terms are 0.8% off, and 120 terms are good to $10^{-12}$ . Now look at $x = -12$ , where the solution oscillates and has size $0.33$ . Even with 120 terms the error is $6 \times 10^{-5}$ , because the individual terms $a_nx^n$ reach about $4 \times 10^{10}$ and cancel down to $0.33$ , and double precision holds only 16 digits. Add -13.0 to the list of $x$ values to see the loss get worse. A convergent series is not automatically a good algorithm; for the Airy function the library routines switch to other formulas for large $\lvert x \rvert$ .

Where it is used

Used inside maths· Inside maths: Gauss–Legendre quadrature, with orthogonality doing the work

Chapter 15.1 built quadrature rules from interpolation: with $n$ nodes, the rule is exact for polynomials of degree below $n$ . Put the nodes at the roots of $P_n$ and the rule becomes exact up to degree $2n - 1$ , twice as good. The reason is the corollary above.

Claim. Let $x_1, \dots, x_n$ be the roots of $P_n$ (all in $(-1, 1)$ , problem P2 below) and let $w_i = \int_{-1}^{1}\ell_i(x)\,dx$ , where $\ell_i$ is the Lagrange basis polynomial of the nodes (15.1). Then $\sum_iw_if(x_i) = \int_{-1}^{1}f\,dx$ for every polynomial $f$ of degree at most $2n - 1$ .

Proof. Divide: $f = qP_n + r$ with $\deg q \le n - 1$ and $\deg r \le n - 1$ . Then $\int f = \int qP_n + \int r = 0 + \int r$ by orthogonality. At each node $P_n(x_i) = 0$ , so $f(x_i) = r(x_i)$ and $\sum w_if(x_i) = \sum w_ir(x_i) = \int r$ , because the rule is exact for polynomials of degree below $n$ . So both sides equal $\int r$ . $\blacksquare$

For $n = 2$ the roots of $P_2 = \frac{3x^2-1}{2}$ are $\pm\frac1{\sqrt3}$ with weights $1, 1$ . For $\int_{-1}^{1}e^x\,dx = 2.3504$ : two Gauss–Legendre nodes give $e^{1/\sqrt3} + e^{-1/\sqrt3} = 2.3427$ (error $0.008$ ). Simpson’s rule with three nodes gives $\frac{e^{-1} + 4 + e}{3} = 2.3621$ (error $0.012$ ). Three Gauss–Legendre nodes ( $0, \pm\sqrt{0.6}$ , weights $\frac89, \frac59, \frac59$ ) give $2.35034$ (error $0.00007$ ).

Your answerThe two-point Gauss–Legendre rule on

[-1, 1]

has nodes

\pm\frac{1}{\sqrt3}

and weights

1, 1

. What does it give for

\int_{-1}^{1}e^x\,dx

? (The true value is

2.3504

Used inside maths· Inside maths: other orthogonal polynomials solve similar equations

The Chebyshev polynomials of 15.1 solve $(1 - x^2)y'' - xy' + n^2y = 0$ (check with $T_2 = 2x^2 - 1$ : $4(1-x^2) - 4x^2 + 4(2x^2 - 1) = 0$ ), and they are orthogonal for the inner product $\int_{-1}^{1}fg\,\frac{dx}{\sqrt{1-x^2}}$ . The Hermite polynomials solve $y'' - 2xy' + 2\alpha y = 0$ (problem U1), orthogonal with weight $e^{-x^2}$ on the whole line, and they describe the quantum harmonic oscillator. Each family comes from a second-order equation of the form $\big(p(x)y'\big)' + \lambda w(x)y = 0$ , whose operator is symmetric, so the proof of orthogonality above works for all of them. This is the Sturm–Liouville theory that MATH 521 and PDE courses develop.

Used in the real world· Quantum mechanics and optics: the Airy function in tunnelling, in a bouncing neutron and in the rainbow

A particle in a constant force. With potential $V = Fx$ the Schrödinger equation $-\frac{\hbar^2}{2m}\psi'' + Fx\,\psi = E\psi$ rearranges to $\psi'' = \frac{2mF}{\hbar^2}\left(x - \frac EF\right)\psi$ . Put $\ell = \left(\frac{\hbar^2}{2mF}\right)^{1/3}$ and $z = \frac{x - E/F}{\ell}$ : then $\frac{d^2\psi}{dz^2} = z\psi$ , which is Airy’s equation. The solution that stays finite on the forbidden side ( $z > 0$ ) is $\operatorname{Ai}(z)$ . For an electron in a field of $10^9$ V/m (a force of $1$ eV per nanometre, a field of the size found at a sharp metal tip) $\ell = 0.336$ nm. The probability density $\lvert\psi\rvert^2 \propto \operatorname{Ai}(z)^2$ is $14.5\%$ of its value at the turning point one length $\ell$ (0.34 nm) beyond it, and $0.97\%$ two lengths (0.67 nm) beyond: the electron is found where classically it has negative kinetic energy. This is tunnelling in its simplest form.

A neutron bouncing on a mirror. A neutron above a horizontal mirror has $V = mgz$ and $\psi(0) = 0$ . The same substitution gives $\psi \propto \operatorname{Ai}\left(\frac{z - E/mg}{z_0}\right)$ with $z_0 = \left(\frac{\hbar^2}{2m^2g}\right)^{1/3} = 5.87\ \mu\text{m}$ , and $\psi(0) = 0$ requires $\operatorname{Ai}(-E/(mgz_0)) = 0$ . So the allowed energies are $E_n = mgz_0\,\lvert a_n\rvert$ , where $a_n$ is the $n$ -th zero of Ai counted from $0$ downwards. With $\lvert a_n\rvert = 2.338, 4.088, 5.521$ this gives $E_1 = 1.41$ peV, $E_2 = 2.46$ peV, $E_3 = 3.32$ peV (1 peV $= 10^{-12}$ eV), the neutron’s classical heights being $13.7, 24.0, 32.4\ \mu\text{m}$ . Nesvizhevsky and his colleagues reported neutrons in such quantum states, with a ground state of about $1.4$ peV, in Nature in 2002.

The rainbow. Geometric optics says the brightness at the edge of a rainbow is infinite and beyond it is zero. Airy’s paper of 1838 replaced this by a wave calculation in which the intensity near the edge is proportional to $\operatorname{Ai}(-z)^2$ , with $z$ the angular distance from the geometric edge in a natural unit. That unit scales like $(\lambda/r)^{2/3}$ for light of wavelength $\lambda$ and drops of radius $r$ . On the dark side the intensity decays like $e^{-\frac43z^{3/2}}$ (a faint leakage), so it is finite everywhere. On the bright side it oscillates, with maxima where $\operatorname{Ai}' = 0$ , at $z = 1.019, 3.248, 4.820, \dots$ The first is the bright primary bow, set back from the geometric edge by $1.019$ units, and the later ones are the faint supernumerary bows, which wave optics predicts and geometric optics cannot. Because the unit grows when the drops are smaller, the extra bands are widest and most visible in fine, uniform drops: reducing the drop radius by a factor $10$ widens the pattern by $10^{2/3} \approx 4.6$ .

Used in the real world· Acoustics: why a drum has no clear pitch (Bessel functions)

A circular membrane of radius $a$ and wave speed $c$ vibrates as $u(r, \theta, t) = R(r)\cos(m\theta)\cos(\omega t)$ . Substituting into the wave equation in polar coordinates gives Bessel’s equation for $R$ in the variable $\omega r/c$ , with order $m$ . A membrane fixed at the rim needs $R(a) = 0$ , so $J_m(\omega a/c) = 0$ , and the frequencies are $f = \frac{c\,j_{m,k}}{2\pi a}$ , where $j_{m,k}$ is the $k$ -th positive zero of $J_m$ . Take $a = 0.30$ m and $c = 100$ m/s (illustrative values for a stretched drumhead), so $\frac{c}{2\pi a} = 53.05$ Hz. Then the modes with no angular variation ( $m = 0$ ) are at $2.405 \times 53.05 = 127.6$ Hz, $292.8$ Hz, $459.1$ Hz, $625.6$ Hz, with ratios $1 : 2.295 : 3.598 : 4.903$ . The mode $m = 1$ appears at $3.832 \times 53.05 = 203.3$ Hz and the mode $m = 2$ at $5.136 \times 53.05 = 272.5$ Hz. A string would give ratios $1 : 2 : 3 : 4$ , which the ear hears as one pitch. The drum’s overtones are not whole-number multiples of the lowest frequency, so an ideal drumhead has no definite pitch, only a thump. (Real timpani are shaped and loaded with air so that some overtones line up.)

Used in the real world· Satellites: the Legendre polynomial P2 and the sun-synchronous orbit

Outside the Earth the gravitational potential is $\Phi = -\frac{GM}{r}\Big[1 - \sum_{n \ge 2}J_n\left(\frac Rr\right)^nP_n(\sin\varphi)\Big]$ , where $\varphi$ is latitude and $R$ the equatorial radius. This is an expansion in Legendre polynomials, justified by the orthogonality above. The dominant correction comes from $n = 2$ , with $J_2 = 1.08263 \times 10^{-3}$ (the Earth’s equatorial bulge), and the angular pattern is $P_2(\sin\varphi) = \frac{3\sin^2\varphi - 1}{2}$ . The bulge makes the plane of a circular orbit of radius $a$ and inclination $i$ turn about the Earth’s axis at the rate $\dot\Omega = -\frac32J_2\,n\left(\frac Ra\right)^2\cos i$ , where $n = \sqrt{GM/a^3}$ is the orbital angular speed. For an orbit at height $800$ km ( $a = 7178$ km, $n = 1.038\times10^{-3}$ rad/s, period $100.9$ min), choosing $i = 98.6^\circ$ (so $\cos i = -0.1496$ ) gives $\dot\Omega = +0.985^\circ$ per day, which is $360^\circ$ per year (the Sun’s mean motion is $0.9856^\circ$ per day, and a change of about $0.003^\circ$ in $i$ closes the small gap). The plane follows the Sun around once a year, so the satellite passes over a given latitude at the same local solar time each day. Earth-observation satellites use these sun-synchronous orbits.

Traps

Watch out· Index shifts: the sums must start where the terms start

In $y'' = \sum_{n \ge 2}n(n-1)a_nx^{n-2}$ the shift $n \to n + 2$ makes the sum start at $0$ . In $xy = \sum_{n \ge 0}a_nx^{n+1}$ the shift $n \to n - 1$ makes it start at $1$ , so the coefficient of $x^0$ gets nothing from $xy$ (this is why $a_2 = 0$ for the Airy equation). After a shift, write the first few terms out in full and check that the lowest power is the same on both sides. An equation for the coefficient of $x^0$ must be written separately when the sums start at different places.

Watch out· Two constants are free, so do not set one to zero by habit

At an ordinary point $a_0 = y(x_0)$ and $a_1 = y'(x_0)$ are both free. The general solution is $a_0u_1 + a_1u_2$ . Worksheet 26 #5 gives both starting values, which is why a single series came out. When they are not given, keep both letters, or give both solutions. The even and the odd chains of Legendre and Airy never mix, which is why the two solutions split cleanly.

Watch out· A real equation with no real trouble can still have a small radius

$(1 + x^2)y'' + 2xy' = 0$ has coefficients with no singularity on the real line, but the series of its solution $\arctan x$ about $0$ has radius $1$ . The reason is the complex points $\pm i$ . So “the equation is smooth at $x = 1$ ” does not prove that the series converges at $x = 1$ . Use the theorem as stated: find the radius of convergence of $p$ and $q$ as power series about the centre. That radius is the guarantee for the solution.

Watch out· Legendre: the sign of n − ν, and which chain stops

The recurrence has $(n - \nu)(n + \nu + 1)$ , with a minus sign in the first factor. The factor $(n - \nu)$ vanishes at $n = \nu$ and only in the chain of that parity: for $\nu = 2$ the even chain stops after $a_2$ , while the odd chain ( $a_1, a_3, \dots$ ) never stops. And $P_n$ is normalised by $P_n(1) = 1$ . If you write $P_2 = x^2 - \frac13$ (the monic version from Gram–Schmidt), the equation still holds but the value at $1$ is $\frac23$ , not $1$ .

Watch out· Frobenius: standard form first, then the indicial equation

The indicial equation needs the equation in the form $x^2y'' + xPy' + Qy = 0$ . For $2xy'' + y' - y = 0$ the coefficient of $y'$ is $1$ , but $p_0$ is $\frac12$ , because you must divide by $2$ and multiply by $x$ to reach $x^2y'' + x\cdot\frac12y' - \frac x2y = 0$ . Reading the roots off the original coefficients gives wrong exponents. Also: take the LARGER root first; it never fails. If the roots differ by a whole number do not conclude “no second solution”: try a logarithm.

Watch out· At an irregular singular point the formal series can have radius zero

$x^2y'' - y' - y = 0$ is irregular at $0$ (the coefficient of $y'$ is $-1$ , so $P = -\frac1x$ is not analytic). Substituting $\sum a_nx^n$ gives $a_{n+1} = \frac{n^2 - n - 1}{n+1}a_n$ , and the ratio $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert$ grows like $n$ , so the series converges only at $x = 0$ (Apostol 6.22). No arithmetic error caused this: the method does not apply at irregular points.

Watch out· x > 0 or x < 0

Frobenius solutions contain $x^s$ with $s$ possibly not a whole number, such as $x^{1/2}$ or $x^{-1/2}$ . They are defined for $x > 0$ (or use $\lvert x\rvert^s$ for $x < 0$ ). Write the interval with your answer. $\ln x$ in the second solution is likewise $\ln\lvert x\rvert$ on the left.

Problems

checkApostol 6.22; new

Classify the point $x = 0$ (ordinary, regular singular, irregular) for each equation, and give the indicial equation when it is regular singular. (a) $y'' + 3y' + 2y = 0$ . (b) $x^2y'' + xy' + (x^2 - 4)y = 0$ . (c) $x^3y'' + y = 0$ . (d) $x^2y'' + 4xy' + 2y = 0$ .

Solution

(a) Ordinary: $p = 3$ and $q = 2$ are constants, analytic. (b) Regular singular, since $P = 1$ and $Q = x^2 - 4$ are analytic. $F(s) = s(s-1) + s - 4 = s^2 - 4$ , roots $\pm2$ . (c) Dividing by $x^3$ , $p = 0$ and $q = x^{-3}$ . Then $x^2q = x^{-1}$ , which is not analytic at $0$ : irregular. (d) Regular singular with $P = 4$ , $Q = 2$ , so $F(s) = s(s-1) + 4s + 2 = s^2 + 3s + 2 = (s+1)(s+2)$ , roots $-1$ and $-2$ . It is an Euler equation, with solutions $x^{-1}$ and $x^{-2}$ and no logarithm.

checkMATH 376 worksheet 26 #6 style

Find the recurrence for the power series solutions of $y'' + xy = 0$ and write the first three non-zero terms of the solution with $y(0) = 1$ , $y'(0) = 0$ .

Solution

With $y = \sum a_nx^n$ : $\sum(n+2)(n+1)a_{n+2}x^n + \sum_{n \ge 1}a_{n-1}x^n = 0$ . The constant term gives $a_2 = 0$ , and for $n \ge 1$ , $(n+2)(n+1)a_{n+2} = -a_{n-1}$ , that is $a_{n+3} = -\frac{a_n}{(n+3)(n+2)}$ . With $a_0 = 1$ and $a_1 = 0$ : $a_3 = -\frac16$ , $a_6 = \frac{1}{180}$ . So $y = 1 - \frac{x^3}{6} + \frac{x^6}{180} - \cdots$ . The sign alternation compared with the Airy equation turns growth into oscillation for $x > 0$ .

checkApostol 6.22; new

Find the indicial roots at $x = 0$ of (a) $x^2y'' + xy' + \left(x - \frac19\right)y = 0$ and (b) $2x^2y'' + xy' - y = 0$ , and say which case of the Frobenius theorem applies.

Solution

(a) $P = 1$ , $Q = x - \frac19$ , so $p_0 = 1$ , $q_0 = -\frac19$ and $F(s) = s^2 - \frac19$ , roots $\pm\frac13$ . The difference $\frac23$ is not a whole number: case 1, two Frobenius series. (b) Divide by $2$ : $x^2y'' + x\cdot\frac12y' - \frac12y = 0$ , so $p_0 = \frac12$ , $q_0 = -\frac12$ and $F(s) = s^2 - \frac s2 - \frac12 = (s - 1)\left(s + \frac12\right)$ . The roots are $1$ and $-\frac12$ ; the difference $\frac32$ is not a whole number: case 1 (and in fact an Euler equation with solutions $x$ and $x^{-1/2}$ ).

checkApostol 6.19; new

Find $P_n(1)$ , $P_n(-1)$ , $P_2(0)$ and $P_4(0)$ using the facts of this chapter.

Solution

$P_n(1) = 1$ for all $n$ . By the parity result of problem P1, $P_n(-1) = (-1)^nP_n(1) = (-1)^n$ . From $P_2 = \frac{3x^2-1}{2}$ : $P_2(0) = -\frac12$ . From $P_4 = \frac{35x^4 - 30x^2 + 3}{8}$ : $P_4(0) = \frac38$ .

useApostol 6.21 #4

Find the recurrence for power series solutions of the Hermite equation $y'' - 2xy' + 2\alpha y = 0$ . For which $\alpha$ is there a polynomial solution? Find it for $\alpha = 3$ , and say where the series solutions converge.

Solution

With $y = \sum a_nx^n$ : $(n+2)(n+1)a_{n+2} - 2na_n + 2\alpha a_n = 0$ , so $a_{n+2} = \dfrac{2(n - \alpha)}{(n+1)(n+2)}a_n$ . If $\alpha = m$ is a non-negative integer the chain of the same parity stops after $a_m$ , giving a polynomial of degree $m$ . For $\alpha = 3$ take $a_1 = 1$ ( $a_0 = 0$ ): $a_3 = \frac{2(1-3)}{2\cdot3} = -\frac23$ , and $a_5 = \frac{2(3-3)}{4\cdot5}a_3 = 0$ . So $y = x - \frac23x^3$ ; multiplied by $-6$ this is $4x^3 - 6x$ , a multiple of the Hermite polynomial $H_3 = 8x^3 - 12x$ . The coefficients $p = -2x$ and $q = 2\alpha$ are polynomials, analytic on every interval, so the series converge for all $x$ .

useApostol 6.22 style; new

Solve $4xy'' + 2y' + y = 0$ for $x > 0$ by the Frobenius method, and recognise the solutions.

Solution

Multiply by $\frac x4$ : $x^2y'' + x\cdot\frac12y' + \frac x4y = 0$ . So $p_0 = \frac12$ , $q_0 = 0$ , $q_1 = \frac14$ and $F(s) = s\left(s - \frac12\right)$ , roots $0$ and $\frac12$ (case 1). The recurrence (F) reads $F(s+n)a_n = -\frac14a_{n-1}$ . For $s = 0$ : $n\left(n - \frac12\right)a_n = -\frac14a_{n-1}$ , so $a_n = -\frac{a_{n-1}}{2n(2n-1)}$ and $a_n = \frac{(-1)^n}{(2n)!}$ : $u_1 = \sum\frac{(-1)^nx^n}{(2n)!} = \cos\sqrt x$ . For $s = \frac12$ : $\left(n + \frac12\right)na_n = -\frac14a_{n-1}$ , so $a_n = -\frac{a_{n-1}}{2n(2n+1)}$ and $a_n = \frac{(-1)^n}{(2n+1)!}$ : $u_2 = x^{1/2}\sum\frac{(-1)^nx^n}{(2n+1)!} = \sin\sqrt x$ . General solution $y = A\cos\sqrt x + B\sin\sqrt x$ . Check: for $y = \cos\sqrt x$ , $4xy'' + 2y' + y = (-\cos\sqrt x + \frac{\sin\sqrt x}{\sqrt x}) - \frac{\sin\sqrt x}{\sqrt x} + \cos\sqrt x = 0$ .

useApostol 6.20; new

Expand $f(x) = x^4$ in Legendre polynomials.

Solution

Only even $k \le 4$ contribute. $c_0 = \frac12\int x^4 = \frac12\cdot\frac25 = \frac15$ . $c_2 = \frac52\int x^4\cdot\frac{3x^2-1}{2} = \frac54\left(\frac67 - \frac25\right) = \frac54\cdot\frac{16}{35} = \frac47$ . $c_4 = \frac92\int x^4\cdot\frac{35x^4 - 30x^2 + 3}{8} = \frac9{16}\left(\frac{70}{9} - \frac{60}{7} + \frac65\right) = \frac{9}{16}\cdot\frac{128}{315} = \frac{8}{35}$ . So $x^4 = \frac15P_0 + \frac47P_2 + \frac8{35}P_4$ . Check at $x = 1$ : $\frac15 + \frac47 + \frac8{35} = \frac{7 + 20 + 8}{35} = 1$ .

useApostol 6.23; new

Use the first four terms of the series for $J_0(1)$ and bound the error.

Solution

$J_0(1) = 1 - \frac14 + \frac1{64} - \frac1{2304} + \cdots = 1 - 0.25 + 0.015625 - 0.000434 = 0.765191$ . The series alternates with terms decreasing in size, so the error is less than the first omitted term, $\frac{1}{4^4(4!)^2} = \frac{1}{147456} = 6.8 \times 10^{-6}$ . The true value is $0.765198$ , within that bound.

proveApostol 6.20; new

Prove that $P_n(-x) = (-1)^nP_n(x)$ for every $n$ .

Solution

Plan. Rodrigues’ formula applies $n$ derivatives to an even function.

The function $w(x) = (x^2-1)^n$ is even: $w(-x) = w(x)$ . If $g$ is even then $g'$ is odd (differentiate $g(-x) = g(x)$ : $-g'(-x) = g'(x)$ ), and if $g$ is odd then $g'$ is even. So each derivative reverses the parity, and $w^{(n)}$ has parity $(-1)^n$ : $w^{(n)}(-x) = (-1)^nw^{(n)}(x)$ . Dividing by $2^nn!$ gives the claim. $\blacksquare$

proveApostol 6.20 (last paragraph)

Prove that $P_n$ has $n$ distinct real roots, all in the interval $(-1, 1)$ .

Solution

Plan. If $P_n$ changed sign in $(-1,1)$ fewer than $n$ times, multiply it by a polynomial that changes sign at the same places. The product has constant sign, so its integral is not $0$ ; but orthogonality says it is $0$ .

Let $x_1, \dots, x_m$ be the points in $(-1, 1)$ where $P_n$ changes sign (roots of odd multiplicity), $m \ge 0$ . Suppose $m < n$ . Put $q(x) = (x - x_1)\cdots(x - x_m)$ , with $q = 1$ if $m = 0$ . Then $P_nq$ does not change sign on $(-1, 1)$ : each sign change of $P_n$ at $x_i$ is cancelled by the sign change of the factor $x - x_i$ , and no other sign changes exist. It is not identically zero, so $\int_{-1}^{1}P_nq\,dx \ne 0$ . But $\deg q = m < n$ , so $\int P_nq = 0$ by the corollary on Legendre polynomials, a contradiction. Hence $m \ge n$ . A polynomial of degree $n$ has at most $n$ roots, so $m = n$ , and the $n$ sign-change points are all of its roots, each simple. $\blacksquare$

proveApostol 6.23; new

Let $J_1(x) = \sum_{k \ge 0}\frac{(-1)^k}{k!\,(k+1)!}\left(\frac x2\right)^{2k+1}$ . Prove that $J_0'(x) = -J_1(x)$ .

Solution

Plan. Differentiate the series term by term and shift the index.

Power series can be differentiated term by term inside the radius of convergence, here infinite. $\frac{d}{dx}\left(\frac x2\right)^{2k} = k\left(\frac x2\right)^{2k-1}$ , so $J_0'(x) = \sum_{k \ge 1}\frac{(-1)^kk}{(k!)^2}\left(\frac x2\right)^{2k-1}$ (the term $k = 0$ is constant). Since $\frac{k}{(k!)^2} = \frac{1}{k!\,(k-1)!}$ , put $j = k - 1$ : $J_0'(x) = \sum_{j \ge 0}\frac{(-1)^{j+1}}{(j+1)!\,j!}\left(\frac x2\right)^{2j+1} = -J_1(x)$ . $\blacksquare$

proveuses the corollary on Legendre polynomials

Prove that the two-point rule $\int_{-1}^{1}f\,dx \approx f\left(-\frac1{\sqrt3}\right) + f\left(\frac1{\sqrt3}\right)$ is exact for every cubic polynomial, and show it is not exact for $x^4$ .

Solution

Plan. By linearity it is enough to check $1, x, x^2, x^3$ .

The rule is linear in $f$ , and so is the integral, so it suffices to check them on $1, x, x^2, x^3$ . For $f = 1$ : $2 = 1 + 1$ . For $f = x$ : $0 = -\frac1{\sqrt3} + \frac1{\sqrt3}$ . For $f = x^2$ : $\frac23 = \frac13 + \frac13$ . For $f = x^3$ : $0 = -\frac{1}{3\sqrt3} + \frac{1}{3\sqrt3}$ . All four agree, so it is exact for every cubic. For $x^4$ the integral is $\frac25 = 0.4$ and the rule gives $\frac19 + \frac19 = \frac29 = 0.22$ . $\blacksquare$ (The nodes are the roots of $P_2$ , and this is the case $n = 2$ of the quadrature box.)

stretchApostol 6.18 (remark on convergence); new

Show that for $\nu$ not an integer the even series solution of Legendre’s equation diverges at $x = 1$ .

Solution

Plan. Show that the coefficients behave like $\frac1k$ , so the series compares with the harmonic series.

For $\nu$ not an integer, no coefficient of the even chain vanishes (the factor $2k - \nu$ is never $0$ ). For $2k > \max(\nu, -\nu - 1)$ the factor $(2k - \nu)(2k + \nu + 1)$ is positive, so from some index on all $a_{2k}$ have the same sign. The ratio is

r_k = \frac{a_{2k+2}}{a_{2k}} = \frac{4k^2 + 2k - \mu}{4k^2 + 6k + 2}, \qquad \mu = \nu(\nu+1),

so $1 - r_k = \frac{4k + 2 + \mu}{4k^2 + 6k + 2} = \frac1k + O\left(\frac1{k^2}\right)$ . Then $\ln r_k = -(1 - r_k) + O\left((1-r_k)^2\right) = -\frac1k + O\left(\frac1{k^2}\right)$ , and $\sum_k\left(\ln r_k + \frac1k\right)$ converges (its terms are $O(k^{-2})$ ). Hence $\ln\lvert a_{2K}\rvert = \ln\lvert a_{2k_0}\rvert + \sum_{k=k_0}^{K-1}\ln r_k = -\ln K + O(1)$ , that is $\lvert a_{2K}\rvert \ge \frac cK$ for some $c > 0$ and large $K$ . Comparison with the harmonic series shows $\sum a_{2k}$ diverges. Because the terms eventually have one sign, the sum $\sum a_{2k}x^{2k} \to \pm\infty$ as $x \to 1^-$ : the solution is unbounded. $\blacksquare$

stretchAbel's formula of 7.3

Let $u$ be any solution of Bessel’s equation of order $0$ that is not a multiple of $J_0$ . Show that $u = J_0\left(c\int\frac{dx}{xJ_0^2} + d\right)$ near $0$ for constants $c \ne 0$ and $d$ , and that $\lvert u(x) \rvert \to \infty$ as $x \to 0^+$ .

Solution

Plan. Use Abel’s formula for the Wronskian, then divide.

The equation in standard form is $y'' + \frac1xy' + y = 0$ , so by Abel’s formula (7.3) $W(J_0, u)(x) = \frac cx$ for a constant $c$ ( $\int\frac1x = \ln x$ , so $e^{-\int p} = \frac1x$ ). The Wronskian of two independent solutions is not $0$ anywhere, so $c \ne 0$ . Near $0$ , $J_0 \ne 0$ , and $W = J_0u' - J_0'u = J_0^2\left(\frac{u}{J_0}\right)'$ . So $\left(\frac{u}{J_0}\right)' = \frac{c}{xJ_0^2}$ , and integrating gives $u = J_0\left(c\int\frac{dx}{xJ_0^2} + d\right)$ . Since $J_0^2 \to 1$ , $\frac{1}{xJ_0^2} = \frac1x + O(x)$ , so $\int\frac{dx}{xJ_0^2} = \ln x + O(x^2)$ , and $u = J_0\,(c\ln x + d + O(x^2)) \to \pm\infty$ as $x \to 0^+$ because $c \ne 0$ and $J_0 \to 1$ . $\blacksquare$ (This is why only $J_0$ describes a drumhead that is finite at its centre.)

Summary

A power series method assumes $y = \sum a_nx^n$ , substitutes, and sets each coefficient to zero (matching coefficients is justified by $f^{(k)}(0) = k!\,c_k$ ). The result is a recurrence; the starting values $a_0, a_1$ are free.
At an ordinary point ( $p$ and $q$ analytic) there are two independent power-series solutions, and they converge at least as far as $p$ and $q$ do. The proof bounds $\lvert a_n\rvert$ by $K\rho^{-n}$ . The radius is set by the nearest singularity in the complex plane ( $\arctan$ : radius $1$ ).
Legendre’s equation $(1-x^2)y'' - 2xy' + \nu(\nu+1)y = 0$ has the recurrence $a_{n+2} = \frac{(n-\nu)(n+\nu+1)}{(n+1)(n+2)}a_n$ . For $\nu = n$ a whole number the series stops: the Legendre polynomial $P_n = \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$ (Rodrigues), with $P_n(1) = 1$ .
The $P_n$ are orthogonal: $\int_{-1}^{1}P_mP_n = 0$ for $m \ne n$ and $\frac{2}{2n+1}$ for $m = n$ , because they are eigenvectors of a symmetric operator. This gives expansions $f = \sum c_kP_k$ , Gauss–Legendre quadrature, and multipole expansions.
At a regular singular point ( $(x-x_0)p$ and $(x-x_0)^2q$ analytic) try the Frobenius series $x^s\sum a_nx^n$ . The indicial equation $s(s-1) + p_0s + q_0 = 0$ gives $s$ , and the recurrence (F) gives the coefficients. The larger root always works. Equal roots, or roots differing by a whole number, may force $\ln x$ in the second solution, as Abel’s formula explains.
Bessel’s equation of order $0$ gives $J_0(x) = \sum\frac{(-1)^k}{(k!)^2}(x/2)^{2k}$ , finite at $0$ and with infinitely many zeros ( $2.405, 5.520, \dots$ ); the other solution has $\ln x$ . Airy’s equation $y'' = xy$ has solutions that oscillate for $x < 0$ and are exponential for $x > 0$ : the model of a turning point, for tunnelling and rainbows.
In practice: shift indices carefully, keep both free constants, standardise the equation before reading $p_0$ and $q_0$ , and remember that a convergent series is not always a good way to compute (the lab).

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

7.5 Power series solutions, Legendre and Bessel

The picture

The idea, in words

Definitions and theorems

Matching coefficients

Matching coefficients

Analytic functions and ordinary points

Analytic function

Ordinary point

Singular point

The main theorem at an ordinary point

Series solutions at an ordinary point

Example: the Airy equation

The Legendre equation

Legendre polynomial

Recurrence for Legendre's equation

Legendre polynomials solve Legendre's equation

Orthogonality

Orthogonality of the Legendre polynomials

Expansion in Legendre polynomials

Regular singular points and the Frobenius method

Regular singular point

Indicial equation

Frobenius series

The Frobenius recurrence

The Frobenius theorem

Bessel’s equation

Bessel function

J0 solves Bessel's equation of order zero

Where it is used

Traps

Problems

Summary

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Where this sits

Comes from

Leads to

Same idea elsewhere