The Method of Frobenius

At a point where the coefficients blow up, an ordinary power series may fail entirely. The method of Frobenius rescues the situation by allowing an extra factor ${| x - x_{0} |}^{r}$ in front of the series, with $r$ possibly fractional, negative, or complex. The exponent $r$ is found from a quadratic called the indicial equation, and the relationship between its two roots determines everything that follows.

Quick Reference

Item	Statement
Regular singular form	y'' + \dfrac{P(x)}{x-x_0}y' + \dfrac{Q(x)}{\left(x-x_0\right)^{2}}y = 0 with $P$ , $Q$ analytic near $x_{0}$
Indicial equation	$r (r - 1) + p_{0} r + q_{0} = 0$ , with $p_{0} = P (x_{0})$ and $q_{0} = Q (x_{0})$
Frobenius series	$y = {\| x - x_{0} \|}^{r} \sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}$ , $a_{0} = 1$
Case 1: $r_{1} - r_{2}$ not an integer	Two Frobenius series, one for each root
Case 2: $r_{1} - r_{2}$ an integer, $r_{1} > r_{2}$	One Frobenius series for $r_{1}$ ; the second may need a logarithm
Equal roots $r_{1} = r_{2}$	The logarithm term is definitely present, $C \neq 0$
Complex roots $α \pm i β$	Take real and imaginary parts of ${\| x - x_{0} \|}^{α + i β} \sum a_{k} {(x - x_{0})}^{k}$
Radius	Both series converge for $\| x - x_{0} \| < ρ$ ; the equation holds for $0 < \| x - x_{0} \| < ρ$
Point at infinity	Substitute $z = 1 / x$ and classify $z = 0$

When Power Series Fail

When dealing with

y'' + P(x)y' + Q(x)y = 0 ,

if either $P (x)$ or $Q (x)$ is not analytic near $x_{0}$ , then power series solutions valid near $x_{0}$ may or may not exist.

Example 1.

Show that x^{2}y'' - 2y = 0 has no solution of the form $\sum a_{k} x^{k}$ with $a_{0} \neq 0$ other than the trivial one.

Solution

Substituting $y = \sum_{k \geq 0} a_{k} x^{k}$ ,

x^{2} \sum_{k = 2}^{\infty} k (k - 1) a_{k} x^{k - 2} - 2 \sum_{k = 0}^{\infty} a_{k} x^{k} = \sum_{k = 0}^{\infty} [k (k - 1) - 2] a_{k} x^{k} = 0,

where the $k = 0$ and $k = 1$ terms of the first sum are absent but $k (k - 1)$ vanishes there anyway, so the single formula covers all $k$ .

Setting each coefficient to zero requires $[k (k - 1) - 2] a_{k} = 0$ for every $k$ . Now $k^{2} - k - 2 = (k - 2) (k + 1)$ vanishes only at $k = 2$ among non-negative integers. So $a_{k} = 0$ for every $k \neq 2$ , leaving only $y = a_{2} x^{2}$ .

Only one solution of power series type exists, not the two that a second-order equation needs. The missing one is $y = 1 / x$ , which no ordinary power series can represent. Allowing the exponent $r = - 1$ is exactly what Frobenius does.

Regular Singular Points

Definition 1.

Consider a differential equation of the form

y'' + \frac{P(x)}{x-x_0}y' + \frac{Q(x)}{\left(x-x_0\right)^{2}}y = 0 .

If $P (x)$ and $Q (x)$ have power series expansions in some open interval $(x_{0} - ρ, x_{0} + ρ)$ , we say that $x_{0}$ is a regular singular point.

In words: the coefficient of y' may blow up no worse than $\frac{1}{x - x_{0}}$ , and the coefficient of $y$ no worse than $\frac{1}{{(x - x_{0})}^{2}}$ . A singular point that fails this test is called irregular, and the Frobenius method does not apply there.

Example 2.

Classify $x = 0$ for each equation.

x^{2}y'' + xy' + \left(x^{2}-\nu^{2}\right)y = 0
x^{3}y'' + y = 0
2xy'' + y' + y = 0

Solution

Dividing by $x^{2}$ gives y'' + \dfrac{1}{x}y' + \dfrac{x^{2}-\nu^{2}}{x^{2}}y = 0. Here $P (x) = 1$ and $Q (x) = x^{2} - ν^{2}$ , both analytic, so $x = 0$ is a regular singular point. This is Bessel's equation.
Dividing by $x^{3}$ gives y'' + \dfrac{1}{x^{3}}y = 0. Writing this as $\frac{Q (x)}{x^{2}} y$ requires $Q (x) = \frac{1}{x}$ , which is not analytic at $0$ . So $x = 0$ is an irregular singular point, and Frobenius does not apply.
Dividing by $2 x$ gives y'' + \dfrac{1}{2x}y' + \dfrac{1}{2x}y = 0. Here $P (x) = \frac{1}{2}$ and, writing the last coefficient as $\frac{Q (x)}{x^{2}}$ , we get $Q (x) = \frac{x}{2}$ . Both are analytic, so $x = 0$ is a regular singular point.

The Indicial Equation

Substituting the trial form $y = x^{r} \sum_{k \geq 0} a_{k} x^{k}$ into the equation and looking at the lowest power of $x$ produces a quadratic condition on $r$ .

The indicial equation is

r (r - 1) + P (x_{0}) r + Q (x_{0}) = 0,

where $P (x_{0}) = p_{0}$ and $Q (x_{0}) = q_{0}$ are the constant terms in the power series expansions of $P (x)$ and $Q (x)$ .

Note the resemblance to the Cauchy-Euler characteristic equation $a r (r - 1) + b r + c = 0$ of Section 4.5. That is no accident: keeping only the constant terms $p_{0}$ and $q_{0}$ replaces the equation near $x_{0}$ by the Cauchy-Euler equation it most resembles, and the leading behavior ${| x - x_{0} |}^{r}$ is inherited from that approximation.

The Frobenius Theorem

Theorem 1

Let the indicial equation have roots $r_{1}$ and $r_{2}$ .

1. If $r_{1} - r_{2}$ is not an integer, the differential equation has two linearly independent solutions

y_{1} (x) = {| x - x_{0} |}^{r_{1}} \sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k} with a_{0} = 1

and

y_{2} (x) = {| x - x_{0} |}^{r_{2}} \sum_{k = 0}^{\infty} b_{k} {(x - x_{0})}^{k} with b_{0} = 1 .

Both series converge on $| x - x_{0} | < ρ$ , and the differential equation is satisfied for $0 < | x - x_{0} | < ρ$ .

2. If $r_{1} - r_{2}$ is an integer, with $r_{1} > r_{2}$ , then one solution is

y_{1} (x) = {| x - x_{0} |}^{r_{1}} \sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k} with a_{0} = 1

and the other independent solution has the form

y_{2} (x) = {| x - x_{0} |}^{r_{2}} \sum_{k = 0}^{\infty} b_{k} {(x - x_{0})}^{k} + C y_{1} (x) \ln | x - x_{0} | with b_{0} = 1 .

If $r_{1} = r_{2}$ the constant $C$ is nonzero. If $r_{1} \neq r_{2}$ , $C$ may or may not be zero. As in case 1, both series converge on $| x - x_{0} | < ρ$ and the equation holds for $0 < | x - x_{0} | < ρ$ .

The equation is satisfied for $0 < | x - x_{0} | < ρ$ , with the center itself excluded. That is unavoidable: at a singular point the equation is not even well defined in standard form, and the solutions may be unbounded there.

Complex roots. If $r_{1} = α + i β$ and $r_{2} = α - i β$ with $β \neq 0$ , then two linearly independent real solutions are the real and imaginary parts of

y (x) = {| x - x_{0} |}^{α + i β} \sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k} with a_{0} = 1 .

This works exactly as in Section 4.5, where $x^{α + i β}$ expanded into $x^{α} \cos (β \ln x)$ and $x^{α} \sin (β \ln x)$ .

A Complete Worked Example

Example 3.

Solve 2xy'' + y' + y = 0 near the regular singular point $x = 0$ .

Solution

Step 1: check the point and find the indicial roots. From the classification example above, $p_{0} = P (0) = \frac{1}{2}$ and $q_{0} = Q (0) = 0$ . The indicial equation is

r (r - 1) + \frac{1}{2} r + 0 = r^{2} - \frac{1}{2} r = r (r - \frac{1}{2}) = 0,

so $r_{1} = \frac{1}{2}$ and $r_{2} = 0$ . Their difference is $\frac{1}{2}$ , not an integer, so case 1 applies and two clean Frobenius series exist.

Step 2: substitute the general Frobenius form. Let

y = \sum_{k = 0}^{\infty} a_{k} x^{k + r},

\begin{aligned} y' &= \sum_{k=0}^{\infty}(k+r)a_kx^{k+r-1}\\ y'' &= \sum_{k=0}^{\infty}(k+r)(k+r-1)a_kx^{k+r-2} . \end{aligned}

Then

2xy'' = \sum_{k=0}^{\infty}2(k+r)(k+r-1)a_kx^{k+r-1},y' = \sum_{k=0}^{\infty}(k+r)a_kx^{k+r-1} ,

while $y = \sum_{k = 0}^{\infty} a_{k} x^{k + r}$ must be shifted to match the power: putting $k \to k - 1$ gives

y = \sum_{k = 1}^{\infty} a_{k - 1} x^{k + r - 1} .

Step 3: collect coefficients. The coefficient of $x^{r - 1}$ , from $k = 0$ , is

[2 r (r - 1) + r] a_{0} = r (2 r - 1) a_{0} = 0,

which reproduces the indicial equation $2 r^{2} - r = 0$ once $a_{0} \neq 0$ is assumed. This is why $a_{0}$ is normalized to $1$ : if it were allowed to vanish, the indicial equation would carry no information.

For $k \geq 1$ , the coefficient of $x^{k + r - 1}$ gives

[2 (k + r) (k + r - 1) + (k + r)] a_{k} + a_{k - 1} = 0,

and factoring $(k + r)$ out of the bracket,

(k + r) (2 k + 2 r - 1) a_{k} = - a_{k - 1} ⟹ a_{k} = \frac{- a_{k - 1}}{(k + r) (2 k + 2 r - 1)} .

Step 4: the first root, $r = \frac{1}{2}$ . Here $(k + \frac{1}{2}) (2 k) = k (2 k + 1)$ , so

a_{k} = \frac{- a_{k - 1}}{k (2 k + 1)} .

With $a_{0} = 1$ ,

a_{1} = - \frac{1}{3}, a_{2} = - \frac{a_{1}}{2 \cdot 5} = \frac{1}{30}, a_{3} = - \frac{a_{2}}{3 \cdot 7} = - \frac{1}{630} .

Therefore

y_{1} = x^{1 / 2} (1 - \frac{x}{3} + \frac{x^{2}}{30} - \frac{x^{3}}{630} + \dots) .

Step 5: the second root, $r = 0$ . Here $(k) (2 k - 1)$ , so

a_{k} = \frac{- a_{k - 1}}{k (2 k - 1)} .

With $b_{0} = 1$ ,

b_{1} = - 1, b_{2} = - \frac{b_{1}}{2 \cdot 3} = \frac{1}{6}, b_{3} = - \frac{b_{2}}{3 \cdot 5} = - \frac{1}{90} .

Therefore

y_{2} = 1 - x + \frac{x^{2}}{6} - \frac{x^{3}}{90} + \dots .

Step 6: the general solution.

y = c_{1} y_{1} + c_{2} y_{2} .

The half-power in $y_{1}$ is what an ordinary power series could never have produced, and it also shows why the solution is defined only for $x > 0$ unless absolute values are used.

Further Examples

The worked example above was case 1, the simplest scenario. The next three examples show the other outcomes: equal roots, an integer difference with no logarithm, and complex roots.

Example 4.

Solve xy'' + y' - y = 0 near $x = 0$ .

Solution

Step 1: indicial roots. Dividing by $x$ gives y'' + \dfrac{1}{x}y' - \dfrac{1}{x}y = 0, so $P (x) = 1$ and $Q (x) = - x$ . Then $p_{0} = 1$ , $q_{0} = 0$ , and the indicial equation is

r (r - 1) + r = r^{2} = 0 .

The roots are equal, $r_{1} = r_{2} = 0$ . By the theorem, the second solution must contain a logarithm.

Step 2: recurrence. With $y = \sum a_{k} x^{k + r}$ , the terms xy'' and y' combine because $(k + r) (k + r - 1) + (k + r) = (k + r)^{2}$ :

xy'' + y' = \sum_{k=0}^{\infty}(k+r)^{2}a_kx^{k+r-1} ,

- y = - \sum_{k = 1}^{\infty} a_{k - 1} x^{k + r - 1} .

The $k = 0$ term gives $r^{2} a_{0} = 0$ , the indicial equation. For $k \geq 1$ ,

(k + r)^{2} a_{k} = a_{k - 1} .

Step 3: the first solution. With $r = 0$ and $a_{0} = 1$ ,

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

y_{1} = 1 + x + \frac{x^{2}}{4} + \frac{x^{3}}{36} + \dots .

Step 4: the logarithmic solution. Take $C = 1$ and try

y_{2} = y_{1} \ln x + \sum_{k = 0}^{\infty} b_{k} x^{k}, b_{0} = 1 .

Write L[y] = xy'' + y' - y. A short computation gives

L\left[y_1\ln x\right] = 2y_1' ,

because every term containing $\ln x$ collects into $L [y_{1}] \ln x = 0$ , and the $y_{1} / x$ terms cancel. So the series part must satisfy

\sum_{k=1}^{\infty}\left(k^{2}b_k - b_{k-1}\right)x^{k-1} = -2y_1' .

Since 2y_1' = \sum_{k\ge1}\dfrac{2k}{(k!)^{2}}x^{k-1}, matching coefficients gives

b_{k} = \frac{1}{k^{2}} (b_{k - 1} - \frac{2 k}{(k!)^{2}}) .

Then

b_{1} = - 1, b_{2} = - \frac{1}{2}, b_{3} = - \frac{2}{27},

and

y_{2} = y_{1} \ln x + 1 - x - \frac{x^{2}}{2} - \frac{2 x^{3}}{27} - \dots .

The general solution is $y = c_{1} y_{1} + c_{2} y_{2}$ for $x > 0$ . Notice that $y_{2} \to - \infty$ as $x \to 0^{+}$ , which is why the singular point itself is excluded.

Example 5.

Solve xy'' + 2y' + xy = 0 near $x = 0$ .

Solution

Step 1: indicial roots. Dividing by $x$ , $P (x) = 2$ and $Q (x) = x^{2}$ , so $p_{0} = 2$ and $q_{0} = 0$ . The indicial equation is

r (r - 1) + 2 r = r (r + 1) = 0,

with $r_{1} = 0$ and $r_{2} = - 1$ . The difference is $1$ , so case 2 applies and a logarithm may be needed.

Step 2: recurrence. Since $(k + r) (k + r - 1) + 2 (k + r) = (k + r) (k + r + 1)$ ,

xy'' + 2y' = \sum_{k=0}^{\infty}(k+r)(k+r+1)a_kx^{k+r-1} ,

x y = \sum_{k = 2}^{\infty} a_{k - 2} x^{k + r - 1} .

Collecting powers,

\begin{aligned} k = 0:&\quad r(r+1)a_0 = 0 ,\\ k = 1:&\quad (r+1)(r+2)a_1 = 0 ,\\ k \geq 2:&\quad (k+r)(k+r+1)a_k = -a_{k-2} . \end{aligned}

Step 3: start with the smaller root, $r = - 1$ . The $k = 1$ condition becomes $0 \cdot a_{1} = 0$ , so $a_{1}$ is free. This is exactly the step at which case 2 can break down, and here it does not. For $k \geq 2$ ,

a_{k} = - \frac{a_{k - 2}}{(k - 1) k} .

Taking $a_{0} = 1$ , $a_{1} = 0$ gives $a_{2} = - \frac{1}{2}$ , $a_{4} = \frac{1}{24}$ , and

y_{1} = x^{- 1} (1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} - \dots) = \frac{\cos x}{x} .

Taking $a_{0} = 0$ , $a_{1} = 1$ gives $a_{3} = - \frac{1}{6}$ , $a_{5} = \frac{1}{120}$ , and

y_{2} = x^{- 1} (x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \dots) = \frac{\sin x}{x} .

Step 4: conclusion. The smaller root alone produced both solutions, so $C = 0$ and no logarithm appears. The larger root $r = 0$ simply reproduces $y_{2}$ . When the roots differ by an integer it pays to try the smaller root first: if the recurrence survives, it delivers everything at once.

Example 6.

Find two real solutions of x^{2}y'' + xy' + (1+x)y = 0 near $x = 0$ .

Solution

Step 1: indicial roots. Here $P (x) = 1$ and $Q (x) = 1 + x$ , so $p_{0} = 1$ , $q_{0} = 1$ , and

r (r - 1) + r + 1 = r^{2} + 1 = 0 .

The roots are $r = \pm i$ , complex with $α = 0$ and $β = 1$ .

Step 2: recurrence. Substituting $y = \sum a_{k} x^{k + r}$ ,

x^{2}y'' + xy' + y = \sum_{k=0}^{\infty}\left[(k+r)^{2} + 1\right]a_kx^{k+r} ,

x y = \sum_{k = 1}^{\infty} a_{k - 1} x^{k + r},

so for $k \geq 1$

a_{k} = - \frac{a_{k - 1}}{(k + r)^{2} + 1} .

Step 3: use $r = i$ . Since $(k + i)^{2} + 1 = k^{2} + 2 i k = k (k + 2 i)$ ,

a_{k} = - \frac{a_{k - 1}}{k (k + 2 i)} .

With $a_{0} = 1$ ,

a_{1} = - \frac{1}{1 + 2 i} = \frac{- 1 + 2 i}{5}, a_{2} = - \frac{a_{1}}{4 (1 + i)} = \frac{- 1 - 3 i}{40} .

Step 4: separate real and imaginary parts. For $x > 0$ , $x^{i} = e^{i \ln x} = \cos (\ln x) + i \sin (\ln x)$ . So

y = [\cos (\ln x) + i \sin (\ln x)] (1 + a_{1} x + a_{2} x^{2} + \dots) .

Keeping terms through $x^{1}$ , the real and imaginary parts are

y_{1} = \cos (\ln x) - \frac{x}{5} [\cos (\ln x) + 2 \sin (\ln x)] + \dots,

y_{2} = \sin (\ln x) + \frac{x}{5} [2 \cos (\ln x) - \sin (\ln x)] + \dots .

The root $r = - i$ gives the complex conjugate of $y$ and nothing new. Both solutions oscillate infinitely often as $x \to 0^{+}$ , just like the Cauchy-Euler solutions $\cos (\ln x)$ and $\sin (\ln x)$ that they refine.

Singularities Off the Real Line

For the equation

y'' + \frac{1}{1+x^{2}}y' + \frac{2}{1+x^{2}}y = 0 ,

since $1 + x^{2} \neq 0$ for every real value of $x$ , one might think this equation has no singular points at all. However, we must consider the singularities at $x = \pm i$ in the complex plane. Since the distance from $x = 0$ to $x = \pm i$ is $1$ , the series about $x = 0$ converges for $| x | < 1$ and possibly for $| x | = 1$ .

This is the same warning that appeared in Section 6.2, and it explains an otherwise mysterious phenomenon. The function $\frac{1}{1 + x^{2}}$ is perfectly smooth on the whole real line, yet its Maclaurin series $1 - x^{2} + x^{4} - \dots$ diverges for $| x | > 1$ . The obstruction is invisible from the real axis and located at $\pm i$ .

The Point at Infinity as a Regular Singular Point

The behavior of solutions for large $x$ is often the question of real interest, and it can be studied by transporting infinity to the origin.

The transformation

z = \frac{1}{x}

carries the point at infinity to the origin. After rewriting the equation in the variable $z$ , the criteria for an ordinary point, a regular singular point, and an irregular singular point can be applied directly at $z = 0$ , and the classification obtained there is by definition the classification of the original equation "at infinity".

Under this change of variable the derivatives transform as

\frac{d y}{d x} = - z^{2} \frac{d y}{d z}, \frac{d^{2} y}{d x^{2}} = z^{4} \frac{d^{2} y}{d z^{2}} + 2 z^{3} \frac{d y}{d z},

which follow from the chain rule with $\frac{d z}{d x} = - z^{2}$ .

Example 7.

Classify the point at infinity for the Cauchy-Euler equation x^{2}y'' + 3xy' + y = 0.

Solution

Substituting $x = 1 / z$ and the derivative formulas,

\frac{1}{z^{2}} (z^{4} y_{z z} + 2 z^{3} y_{z}) + \frac{3}{z} (- z^{2} y_{z}) + y = 0,

that is,

z^{2} y_{z z} + 2 z y_{z} - 3 z y_{z} + y = 0 ⟹ z^{2} y_{z z} - z y_{z} + y = 0 .

Dividing by $z^{2}$ gives $y_{z z} - \frac{1}{z} y_{z} + \frac{1}{z^{2}} y = 0$ , in which $P (z) = - 1$ and $Q (z) = 1$ are analytic. So $z = 0$ is a regular singular point, meaning the original equation has a regular singular point at infinity.

That the transformed equation is again of Cauchy-Euler type is no coincidence: the equidimensional structure is preserved by $x \mapsto 1 / x$ .

Example 8.

Classify the point at infinity for y'' + y = 0.

Solution

Substituting the derivative formulas,

z^{4} y_{z z} + 2 z^{3} y_{z} + y = 0 .

Dividing by $z^{4}$ ,

y_{z z} + \frac{2}{z} y_{z} + \frac{1}{z^{4}} y = 0 .

The coefficient of $y_{z}$ is fine, with $P (z) = 2$ . But matching $\frac{1}{z^{4}}$ to the pattern $\frac{Q (z)}{z^{2}}$ forces $Q (z) = \frac{1}{z^{2}}$ , which is not analytic at $0$ . So infinity is an irregular singular point.

This agrees with what we know. The solutions $\cos x$ and $\sin x$ oscillate forever as $x \to \infty$ , and no expression of the form $x^{- r} \sum a_{k} x^{- k}$ can do that. Even the simplest constant-coefficient equation is beyond the reach of Frobenius at infinity.

Exercises

Exercise 1.

Find and solve the indicial equation for x^{2}y'' + xy' + \left(x^{2}-\tfrac{1}{4}\right)y = 0 at $x = 0$ , and state which case of the theorem applies.

Solution

Dividing by $x^{2}$ , $P (x) = 1$ and $Q (x) = x^{2} - \frac{1}{4}$ , so $p_{0} = 1$ and $q_{0} = - \frac{1}{4}$ . The indicial equation is

r (r - 1) + r - \frac{1}{4} = r^{2} - \frac{1}{4} = 0,

with roots $r = \pm \frac{1}{2}$ . Their difference is $1$ , an integer, so case 2 applies and a logarithm may be required.

In fact this is Bessel's equation of order $\frac{1}{2}$ , whose solutions are $\frac{\sin x}{\sqrt{x}}$ and $\frac{\cos x}{\sqrt{x}}$ as found in Section 4.1. No logarithm appears, illustrating that $C$ can be zero when the roots are distinct.

Exercise 2.

Classify $x = 0$ for x^{4}y'' + 2x^{2}y' + y = 0.

Solution

Dividing by $x^{4}$ gives y'' + \dfrac{2}{x^{2}}y' + \dfrac{1}{x^{4}}y = 0. Matching the pattern $\frac{P (x)}{x}$ requires $P (x) = \frac{2}{x}$ , which is not analytic at $0$ . So $x = 0$ is an irregular singular point and the Frobenius method does not apply.

Exercise 3.

Write down the indicial equation for the general Cauchy-Euler equation ax^{2}y'' + bxy' + cy = 0 and compare it with Section 4.5.

Solution

Dividing by $a x^{2}$ , we get $P (x) = \frac{b}{a}$ and $Q (x) = \frac{c}{a}$ , both constant. So $p_{0} = \frac{b}{a}$ and $q_{0} = \frac{c}{a}$ , and the indicial equation is

r (r - 1) + \frac{b}{a} r + \frac{c}{a} = 0 ⟺ a r (r - 1) + b r + c = 0,

exactly the Cauchy-Euler characteristic equation.

The three Frobenius cases match the three Cauchy-Euler cases term for term: distinct roots give $x^{r_{1}}$ and $x^{r_{2}}$ , equal roots force a logarithm, and complex roots give $x^{α} \cos (β \ln x)$ and $x^{α} \sin (β \ln x)$ . A Cauchy-Euler equation is the special case where the Frobenius series terminates after its first term.

Exercise 4.

For xy'' - y = 0, find the indicial roots, show that the smaller root does not produce a Frobenius series, and find the series for the larger root.

Solution

Dividing by $x$ gives $P (x) = 0$ and $Q (x) = - x$ , so $p_{0} = q_{0} = 0$ and the indicial equation is $r (r - 1) = 0$ , with $r_{1} = 1$ and $r_{2} = 0$ . The difference is $1$ : case 2.

Substituting $y = \sum a_{k} x^{k + r}$ ,

xy'' = \sum_{k=0}^{\infty}(k+r)(k+r-1)a_kx^{k+r-1} ,

y = \sum_{k = 1}^{\infty} a_{k - 1} x^{k + r - 1},

so for $k \geq 1$

(k + r) (k + r - 1) a_{k} = a_{k - 1} .

Smaller root $r = 0$ . At $k = 1$ the recurrence reads $1 \cdot 0 \cdot a_{1} = a_{0} = 1$ , which is impossible. The series breaks down at exactly the step $k = r_{1} - r_{2}$ , so the second solution must contain a logarithm: $C \neq 0$ . Contrast this with xy'' + 2y' + xy = 0 above, where the same step gave $0 = 0$ instead.

Larger root $r = 1$ . Now $(k + 1) k a_{k} = a_{k - 1}$ , so

a_{k} = \frac{a_{k - 1}}{k (k + 1)} = \frac{1}{k! (k + 1)!},

and

y_{1} = x + \frac{x^{2}}{2} + \frac{x^{3}}{12} + \frac{x^{4}}{144} + \dots .

Exercise 5.

Show that $x = 1$ is a regular singular point of Legendre's equation

\left(1-x^{2}\right)y'' - 2xy' + \nu(\nu+1)y = 0 ,

find the indicial roots there, and describe the two solutions near $x = 1$ .

Solution

Since $1 - x^{2} = - (x - 1) (x + 1)$ , dividing by $1 - x^{2}$ gives

y'' + \frac{2x}{(x-1)(x+1)}y' - \frac{\nu(\nu+1)}{(x-1)(x+1)}y = 0 .

Matching the pattern $\frac{P (x)}{x - 1}$ and $\frac{Q (x)}{(x - 1)^{2}}$ ,

P (x) = \frac{2 x}{x + 1}, Q (x) = - \frac{ν (ν + 1) (x - 1)}{x + 1} .

Both are analytic at $x = 1$ , where the denominator is $2$ , so $x = 1$ is a regular singular point. Here $p_{0} = P (1) = 1$ and $q_{0} = Q (1) = 0$ , so the indicial equation is

r (r - 1) + r = r^{2} = 0 .

The roots are equal, $r_{1} = r_{2} = 0$ . One solution is an ordinary power series in $x - 1$ , bounded at $x = 1$ . The other necessarily contains $\ln | x - 1 |$ and is unbounded there. By the symmetry $x \mapsto - x$ , the same holds at $x = - 1$ .

This is why, in Section 6.2, only one Legendre solution could be bounded on all of $[- 1, 1]$ : the polynomial $P_{n}$ . The companion functions $Q_{n}$ carry the logarithm.

Exercise 6.

Classify the point at infinity for Bessel's equation x^{2}y'' + xy' + \left(x^{2}-\nu^{2}\right)y = 0.

Solution

With $x = 1 / z$ ,

\frac{1}{z^{2}} (z^{4} y_{z z} + 2 z^{3} y_{z}) + \frac{1}{z} (- z^{2} y_{z}) + (\frac{1}{z^{2}} - ν^{2}) y = 0,

which simplifies to

z^{2} y_{z z} + z y_{z} + (\frac{1}{z^{2}} - ν^{2}) y = 0 .

Dividing by $z^{2}$ ,

y_{z z} + \frac{1}{z} y_{z} + (\frac{1}{z^{4}} - \frac{ν^{2}}{z^{2}}) y = 0 .

Then $P (z) = 1$ is fine, but $Q (z) = \frac{1}{z^{2}} - ν^{2}$ is not analytic at $0$ . Infinity is an irregular singular point.

The culprit is the term $x^{2} y$ , the same term that made y'' + y = 0 irregular at infinity in the example above. Bessel functions oscillate for large $x$ , roughly like $x^{- 1 / 2} \cos (x - const)$ , and such behavior cannot come from a Frobenius series in $1 / x$ .

Exercise 7.

For x^{2}y'' + 3xy' + (2+x)y = 0, find the indicial roots at $x = 0$ and state the form of two real solutions, including the first correction term.

Solution

Here $P (x) = 3$ and $Q (x) = 2 + x$ , so $p_{0} = 3$ , $q_{0} = 2$ , and

r (r - 1) + 3 r + 2 = r^{2} + 2 r + 2 = 0,

giving $r = - 1 \pm i$ , so $α = - 1$ and $β = 1$ .

Substituting $y = \sum a_{k} x^{k + r}$ , the non-shifted terms produce $[(k + r)^{2} + 2 (k + r) + 2] a_{k}$ and the term $x y$ produces $a_{k - 1}$ , so

a_{k} = - \frac{a_{k - 1}}{(k + r)^{2} + 2 (k + r) + 2} .

With $r = - 1 + i$ the denominator simplifies to $k (k + 2 i)$ , exactly as in the complex-roots example, so $a_{1} = \frac{- 1 + 2 i}{5}$ .

Since $x^{- 1 + i} = x^{- 1} [\cos (\ln x) + i \sin (\ln x)]$ , the two real solutions are

y_{1} = \frac{1}{x} {\cos (\ln x) - \frac{x}{5} [\cos (\ln x) + 2 \sin (\ln x)] + \dots},

y_{2} = \frac{1}{x} {\sin (\ln x) + \frac{x}{5} [2 \cos (\ln x) - \sin (\ln x)] + \dots} .

Both are unbounded and oscillate infinitely often as $x \to 0^{+}$ , because $α < 0$ .

Frequently Asked Questions

What is a regular singular point?

A singular point where the blow-up is mild enough to be controlled: writing the equation as y'' + \dfrac{P(x)}{x-x_0}y' + \dfrac{Q(x)}{(x-x_0)^{2}}y = 0, both $P$ and $Q$ must be analytic at $x_{0}$ . Equivalently, $(x - x_{0}) P_{std} (x)$ and $(x - x_{0})^{2} Q_{std} (x)$ are analytic, where the subscript denotes the standard-form coefficients.

Where does the indicial equation come from?

From the coefficient of the lowest power of $x - x_{0}$ after substituting $y = \sum a_{k} (x - x_{0})^{k + r}$ . That coefficient is $[r (r - 1) + p_{0} r + q_{0}] a_{0}$ , and since $a_{0} \neq 0$ by normalization, the bracket must vanish. Only the constant terms $p_{0}$ and $q_{0}$ of $P$ and $Q$ enter, because higher terms contribute to higher powers.

Why does the case

r_{1} - r_{2}

an integer cause trouble?

Because the recurrence for the smaller root eventually divides by $(k + r_{2}) (\dots)$ , and when $r_{1} - r_{2} = k$ that factor can vanish, making the recurrence unsolvable at that step. The logarithm term is what repairs the breakdown. Sometimes the numerator happens to vanish too, in which case $C = 0$ and no logarithm is needed.

Why set

a_{0} = 1

Because the indicial equation is derived by dividing out $a_{0}$ , so it must be nonzero, and scaling a solution by a constant changes nothing. Setting $a_{0} = 1$ is the simplest normalization. If your recurrence forces $a_{0} = 0$ , you have made an error, most likely by starting the series at the wrong power.

What if the point is irregular?

Frobenius offers no guarantee, and typically no Frobenius series solution exists. The equation x^{3}y'' + y = 0 is a standard example. Such equations require asymptotic methods, which describe the behavior of solutions near the singularity without producing a convergent series.

Why do the solutions use

{| x - x_{0} |}^{r}

rather than

{(x - x_{0})}^{r}

Because $r$ may be fractional or irrational, and a negative base raised to such a power is not real. The absolute value keeps the expression real on both sides of the singularity. In practice one usually works on $x > x_{0}$ and drops the bars.

How is Frobenius related to the Cauchy-Euler equation?

The indicial equation is the Cauchy-Euler characteristic equation of the "leading" equation obtained by replacing $P$ and $Q$ with their constant terms. Frobenius says the true solutions behave near $x_{0}$ like the Cauchy-Euler solutions of that approximation, corrected by a convergent power series. The three cases correspond exactly.

Quick Reference

When Power Series Fail

Regular Singular Points

The Indicial Equation

The Frobenius Theorem

A Complete Worked Example

Further Examples

Singularities Off the Real Line

The Point at Infinity as a Regular Singular Point

Exercises

Frequently Asked Questions

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