Frobenius method — regular singular points: Question bank

A rapid-fire self-quiz targeting the ideas behind the Frobenius method, not the arithmetic. Each item is a question ::: answer reveal — cover the right side, commit to an answer with a reason, then check. If your reason is wrong even when your verdict is right, count it as a miss.

Before you start, here is every symbol we lean on. Read this once so no answer surprises you:

$P, Q$ = the coefficients when the ODE reads $y^{''} + P y^{'} + Q y = 0$ .
$p (x) = x P (x)$ and $q (x) = x^{2} Q (x)$ = the tamed coefficients. At a regular singular point they are analytic, so they have power-series expansions $p (x) = \sum_{k \geq 0} p_{k} x^{k}$ and $q (x) = \sum_{k \geq 0} q_{k} x^{k}$ . Here $== p_{k}, q_{k} ==$ are just the Taylor coefficients of those expansions — $p_{0}, q_{0}$ are the constant terms, $p_{1}, q_{1}$ the $x^{1}$ coefficients, and so on.
$p_{0} = lim_{x \to 0} x P (x)$ and $q_{0} = lim_{x \to 0} x^{2} Q (x)$ = the leading tamed values (the $k = 0$ coefficients).
$r_{1} \geq r_{2}$ = the two roots of the indicial equation $r (r - 1) + p_{0} r + q_{0} = 0$ .
$== F (s) ==$ = the indicial polynomial $F (s) = s (s - 1) + p_{0} s + q_{0}$ . It is the exact left-hand bracket that multiplies $a_{n}$ in the recurrence: the coefficient of $x^{n + r}$ is $F (n + r) a_{n} = (stuff from earlier a_{n - k})$ . So $F (r) = 0$ is the indicial equation, and $F (n + r) \neq = 0$ is what lets us divide and solve for $a_{n}$ .
$== N ==$ = the positive integer root difference $N = r_{1} - r_{2}$ when the two roots differ by a whole number (Case 3). "Step $N$ " of the recurrence means the equation for $a_{N}$ .

The first figure below gives a picture of what these three cases look like on the number line of exponents — glance at it before the quiz.

Figure — Frobenius method — regular singular points

And the second figure shows the geometric meaning of the convergence radius — why a Frobenius series does not reach forever.

True or false — justify

The task is always: decide, then say why in one breath. A bare "true/false" scores zero.

Every singular point can be solved by the Frobenius method.

False. Only regular singular points are guaranteed; at an irregular point the singularity is stronger than a single

x^{r}

can balance, so the method may produce no valid series at all.

P

and

Q

are both analytic at

x_{0}

, then

x_{0}

is an ordinary point and you should NOT reach for Frobenius.

True. Analyticity of

P, Q

is exactly the ordinary-point condition; the plain power series about an ordinary point works, and Frobenius with

r = 0

would just reproduce it.

The Frobenius series

y = x^{r} \sum a_{n} x^{n}

always converges everywhere.

False. It converges at least up to the distance to the nearest other singular point of the ODE — its radius of convergence is finite when other singularities exist (see the second figure).

The indicial equation can have complex roots.

True. It is a quadratic in

r

; if

p_{0}, q_{0}

make the discriminant negative you get complex-conjugate exponents, giving oscillatory-plus-power behaviour (real solutions recovered by taking real/imaginary parts).

If the two indicial roots are equal, the two solutions are still linearly independent.

True. They must be — a 2nd-order ODE has a 2-dimensional solution space. The

ln x

term in

y_{2}

is precisely what makes the two solutions independent (their Wronskian is nonzero).

Multiplying the whole ODE by

x^{2}

changes its solutions.

False. Multiplying by

x^{2}

(for

x \neq = 0

) is just rescaling both sides by the same nonzero factor; it rearranges the algebra into

x^{2} y^{''} + x p y^{'} + q y = 0

but the solution set is untouched.

For the Euler–Cauchy equation

x^{2} y^{''} + b x y^{'} + cy = 0

, the Frobenius series terminates after one term.

True. Here the tamed coefficients are constants,

p (x) = b

and

q (x) = c

, so every higher Taylor coefficient

p_{k}, q_{k}

for

k \geq 1

is zero; the recurrence then forces

a_{n} = 0

for

n \geq 1

and you get pure powers

x^{r}

— exactly the Euler–Cauchy answer.

A larger indicial root

r_{1}

always gives a solution free of logarithms.

True. Recall

F (s)

is the bracket multiplying

a_{n}

. For the larger root,

F (r_{1} + n) \neq = 0

for every

n \geq 1

when

r_{1} \geq r_{2}

, so we never divide by zero; the trouble (if any) strikes only at the smaller root.

Spot the error

Each line states a plausible move. Find and name the flaw.

x = 0

is singular for

y^{''} + \frac{1}{x} y^{'} + \frac{1}{x ^{3}} y = 0

, and it's regular."

Wrong: it's irregular. Here

x^{2} Q = 1/ x

blows up as

x \to 0

, so

q

is not analytic; only

p = x P = 1

passes. Both

p

and

q

must be analytic.

"To find

p_{0}

, just evaluate

P (0)

Wrong.

P

itself blows up at a singular point, so

P (0)

is undefined. You need the tamed limit

p_{0} = lim_{x \to 0} x P (x)

, which multiplies away the

1/ x

before evaluating.

"The indicial equation for

x^{2} y^{''} + x y^{'} + x^{2} y = 0

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

Wrong. The indicial equation uses only the lowest-order pieces

p_{0}, q_{0}

, i.e.

q_{0} = lim x^{2} Q = 0

, not the full

q (x) = x^{2}

. Correct:

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

"Roots differ by an integer, so I immediately write

y_{2} = κ y_{1} ln x + x^{r_{2}} \sum c_{n} x^{n}

with

κ \neq = 0

Wrong. The coefficient

κ

may be zero. You must attempt the recurrence for

r_{2}

first; only if a coefficient becomes genuinely undefined is the log forced.

"Since

a_{0} \neq = 0

is required, I also set

a_{1} \neq = 0

Wrong. Only the lowest coefficient

a_{0}

is constrained to be nonzero (so

x^{r}

is truly the leading power). Every later

a_{n}

, including

a_{1}

, is whatever the recurrence dictates — often zero.

r = 1/2

came out, but that's not an integer, so this isn't a valid power series and Frobenius fails."

Wrong — that's the whole point. A non-integer

r

is exactly why we multiply by

x^{r}

: the factor

x^{1/2}

captures singular behaviour a plain Taylor series never could. It's a success, not a failure.

"I found

y_{1}

by Frobenius; for

y_{2}

I'll just plug in the other root and I'm guaranteed a clean second series."

Wrong in Cases 2 and 3. Equal roots (Case 2) give only one exponent, and integer-difference roots (Case 3) can clash; then you need Reduction of order or the log-form ansatz, not a second plain series.

Why questions

Explain the reason, not just the fact.

Why do we tame

P

with one power of

x

but

Q

with two?

Because

y^{'}

carries one fewer derivative than

y^{''}

: for

x^{2} y^{''}

to balance the

P

-term you need

x P

finite, and to balance the

Q

-term you need

x^{2} Q

finite. The powers match the derivative order so every term hits the same lowest power

x^{r}

Why does setting the coefficient of the lowest power

x^{r}

to zero give a quadratic in

r

The lowest power comes entirely from the

n = 0

term. Feeding

y = a_{0} x^{r}

into

x^{2} y^{''} + x p_{0} y^{'} + q_{0} y

produces

a_{0} [r (r - 1) + p_{0} r + q_{0}] x^{r} = a_{0} F (r) x^{r}

; with

a_{0} \neq = 0

that bracket

F (r)

— a quadratic — must vanish.

Why must

a_{0} \neq = 0

a_{0} = 0

the series would actually start at

x^{r + 1}

, meaning our claimed leading exponent

r

was wrong. Forcing

a_{0} \neq = 0

pins

r

to the genuine lowest power so the indicial equation is meaningful.

Why does a logarithm appear only in Cases 2 and 3, never Case 1?

A log is needed when the second exponent cannot supply an independent power series. In Case 1 the roots differ by a non-integer, so the bracket

F (r_{2} + n)

never hits zero and the recurrence runs cleanly; no rescue term is required.

Why is the log term structurally

y_{1} ln x

rather than some other function times

ln x

Reduction of order sets

y_{2} = v (x) y_{1}

; solving for

v

yields

\int \frac{d x}{x} = ln x

as its leading piece, so the log rides on top of

y_{1}

itself. The mathematics manufactures exactly the second independent solution.

Why classify the point before attempting any series?

Because Frobenius convergence and the whole three-case machinery are only guaranteed at a regular singular point. Skipping the check risks applying the method where it has no warranty (irregular points), wasting effort on a series that may not converge.

Why can the equal-root case never yield two power series?

Two distinct power-series solutions would need two distinct leading exponents. Equal roots supply only one exponent

r

, so the second independent solution is forced to differ qualitatively — hence the

ln x

Edge cases

The scenarios people forget until an exam supplies them.

What if

q_{0} = 0

(as in Bessel order 0)?

The indicial equation becomes

r (r - 1) + p_{0} r = 0 = r (r - 1 + p_{0})

, giving

r = 0

and

r = 1 - p_{0}

. A zero root is perfectly legal; here it happened to be a double root, forcing the log seen in $Y_0$.

What if

p_{0} = 0

AND

q_{0} = 0

Indicial equation is

r (r - 1) = 0

, roots

0

and

1

— differing by the integer

N = 1

(Case 3). This is the borderline where the point is technically singular but often behaves almost like an ordinary point; the log may or may not survive.

What happens at

x = 0

for Legendre's equation?

x = 0

is an ordinary point (its singularities sit at

x = \pm 1

), so you use a plain power series there — Frobenius is unnecessary. Frobenius is the tool at

x = \pm 1

instead.

If the discriminant of the indicial quadratic is zero, which case are we in?

Case 2 (equal roots

r_{1} = r_{2}

, so

N = 0

). A single exponent means the second solution must carry

ln x

; you can predict the log purely from the discriminant, before any recurrence.

What if two indicial roots differ by an integer

N

but the recurrence at step

N

has a zero right-hand side?

Then no division-by-zero crisis occurs; the equation for

a_{N}

leaves it free (often chosen

0

κ = 0

, and you get two clean Frobenius series with no logarithm — the "steel-manned" Case-3-without-log outcome.

What if the smaller root gives

x^{r_{2}}

with

r_{2}

negative and non-integer?

The solution

x^{r_{2}} \sum a_{n} x^{n}

blows up at

x = 0

but is still a genuine, independent solution on

x > 0

. Whether you keep it depends on physical boundedness requirements, not on any failure of the method.

Can a second-order ODE ever produce three independent Frobenius solutions?

No. The solution space of a 2nd-order linear ODE is exactly 2-dimensional; at most two independent solutions exist, and their independence is confirmed by a nonzero Wronskian.

Active Recall

Recall Fastest sanity checks

Tamed limits before roots: == $p_{0} = lim x P$ , $q_{0} = lim x^{2} Q$ ==, then the quadratic.
$F (s) = s (s - 1) + p_{0} s + q_{0}$ is the recurrence bracket; == $F (n + r) = 0$ is what triggers a log==.
Discriminant $= 0$ ⇒ ==equal roots ( $N = 0$ ) ⇒ guaranteed log==.
Integer difference $N \geq 1$ ⇒ maybe log; test the recurrence at step $N$ , don't assume.
Non-integer difference ⇒ two clean series, never a log.