Worked examples — Frobenius method — regular singular points

One page, every cell. We march through each way the indicial roots can behave, plus the traps (a fake singular point, a huge blow-up that kills the method, a degenerate root, a limiting case). If a scenario can happen at a regular singular point, it appears below with its label.

Everything here builds on the parent note. If a symbol shows up, we re-earn it here so you never have to scroll back.

---

The scenario matrix

The whole game is: compute the tamed limits $p_0,q_0$, solve the tiny quadratic (the indicial equation), look at the difference of its two roots. That difference sorts every problem into a cell.

Recall the two tamed coefficients (re-stated so nothing is assumed):

Sometimes we need more than just the leading numbers $p_0,q_0$. The tamed coefficients $p(x)=xP(x)$ and $q(x)=x^{2}Q(x)$ are analytic, so each has its own Taylor series:

For the record, here is the recurrence (the parent note's Step 4), re-stated so this page stands alone. After plugging the ansatz $y=\sum a_n{x}^{n+r}$ into the equation, matching the coefficient of $x^{n+r}$ gives

The indicial equation is $F(r)=r(r-1)+p_{0}r+q_0=0$, two roots $r_1\ge r_2$. Here is every cell:

Cell	What triggers it	Root behaviour	Second solution shape
A — ordinary point in disguise	$P,Q$ actually analytic ($x=0$ not singular)	roots $0,1$	plain Taylor, no $x^r$ needed
B — distinct roots, non-integer gap	$r_1-r_2\notin \mathbb{Z}$	two clean $x^r$ series	no log
C — fractional roots	roots like $1,\frac{1}{2}$	still Case-B mechanics	no log, non-integer powers
D — equal roots	$r_1=r_2$	one exponent only	log forced
E — integer gap with log	$r_1-r_2=N$, coefficient blows up	recurrence hits $0$	log present ($\kappa \ne 0$)
F — integer gap without log	$r_1-r_2=N$, but it survives	recurrence stays finite	two clean series ($\kappa =0$)
G — irregular point (method fails)	$xP$ or $x^{2}Q$ not analytic	no valid indicial eq.	Frobenius not guaranteed
H — word problem / limiting case	physical setup, degenerate parameter	any of above	interpret the exponent

We now hit every cell with a worked example.

---

Example A — the point wasn't even singular (Cell A)

Forecast: Guess now — is $x=0$ regular singular, irregular, or ordinary?

1. Read off $P,Q$. Already standard form: $P(x)=x$, $Q(x)=1+x^2$. Why this step? The classification is a statement about $P,Q$; we must see them bare.
2. Test analyticity at $0$. Both $x$ and $1+x^2$ are polynomials — analytic everywhere. Why this step? An analytic $P$ and $Q$ means no blow-up, so $x=0$ is an ordinary point.
3. Conclusion. No $x^r$ needed. Use the plain power-series method: $y=\sum a_n{x}^n$.

Verify: The tamed limits are $p_0=\lim x\cdot x=0$ and $q_0=\lim x^2(1+x^2)=0$. Indicial equation $r(r-1)=0$ gives roots $0,1$ — exactly the two lowest integer powers of a Taylor series. Frobenius degenerates back into the ordinary method, confirming there was nothing singular. ✔

---

Example B — distinct roots, non-integer gap (Cell B)

Forecast: From the parent note the roots are $\frac{1}{2},-1$. Will you need to divide by zero anywhere in the recurrence?

1. Confirm RSP and $p_0,q_0$. Standard form $y^{\prime \prime }+\frac{3}{2x}{y}^{\prime }-\frac{1+x}{2x^2}y=0$, so $xP=\frac{3}{2}\to p_0=\frac{3}{2}$ and $x^{2}Q=-\frac{1+x}{2}\to q_0=-\frac{1}{2}$. Why this step? These feed both the indicial equation and the recurrence.
2. Indicial equation. $r(r-1)+\frac{3}{2}r-\frac{1}{2}=0\Rightarrow 2r^2+r-1=0\Rightarrow (2r-1)(r+1)=0$, roots $r_1=\frac{1}{2},r_2=-1$. Gap $=\frac{3}{2}\notin \mathbb{Z}$ → Cell B, no log. Why this step? The gap tells us both series are independent and clean.
3. Read off the coefficient series $p_k,q_k$. Using the definitions above, expand the tamed coefficients: $p(x)=xP=\frac{3}{2}$, a constant, so $p_0=\frac{3}{2}$ and $p_k=0$ for $k\ge 1$. And $q(x)=x^{2}Q=-\frac{1+x}{2}=-\frac{1}{2}-\frac{1}{2}x$, so $q_0=-\frac{1}{2}$ and $q_1=-\frac{1}{2}$ (higher $q_k=0$). Why this step? The recurrence needs these Taylor entries; $q_1\ne 0$ is precisely what couples $a_n$ to $a_{n-1}$.
4. Build the recurrence for $r_1=\frac{1}{2}$. Plug the ansatz into the restated recurrence above. Since only $q_1$ survives among the higher coefficients, the sum on the right has just its $k=1$ term ($p_1=0$, and $q_1=-\frac{1}{2}$), so $a_n$ depends only on $a_{n-1}$. Multiplying through by $-2$ to clear fractions, the coefficient of $x^{n+r}$ gives $[2(n+r)(n+r-1)+3(n+r)-1]a_n=a_{n-1}.$ Why this step? Only the $-xy$ term (carried by $q_1$) connects neighbouring coefficients, so each $a_n$ depends on just $a_{n-1}$.
5. Evaluate at $r=\frac{1}{2}$. Bracket $=2(n+\frac{1}{2})(n-\frac{1}{2})+3(n+\frac{1}{2})-1=2n^2+3n$. So $a_n=\frac{a_{n-1}}{2n^2+3n}=\frac{a_{n-1}}{n(2n+3)}.$ With $a_0=1$: $a_1=\frac{1}{1\cdot 5}=\frac{1}{5}$, $a_2=\frac{a_1}{2\cdot 7}=\frac{1}{70}$. Why this step? Concrete numbers let us verify against a machine.

Verify: The bracket $2n^2+3n$ never hits zero for $n\ge 1$, so no division by zero — consistent with "no log". Numerically $a_1=0.2$, $a_2=1/70\approx 0.014286$. ✔

---

Example C — a fractional root living beside an integer one (Cell C)

Forecast: One root is a whole number, the other is a half. What does the exponent $\frac{1}{2}$ look like for the solution as $x\to 0^{+}$?

1. Standard form. Divide by $x^2$: $P=-\frac{1}{2x}$, $Q=\frac{1-x}{2x^2}$. Why this step? Need bare coefficients for the limits.
2. Tamed limits. $p_0=\lim xP=-\frac{1}{2}$. $q_0=\lim x^{2}Q=\frac{1}{2}$. Why this step? These are the indicial inputs.
3. Indicial equation. $r(r-1)-\frac{1}{2}r+\frac{1}{2}=0\Rightarrow r^2-\frac{3}{2}r+\frac{1}{2}=0\Rightarrow 2r^2-3r+1=0\Rightarrow (2r-1)(r-1)=0$. Roots $r_1=1,r_2=\frac{1}{2}$ (one integer, one half). Gap $=\frac{1}{2}\notin \mathbb{Z}$ → Cell C, clean, two independent series. Why this step? Confirms a fractional exponent is genuinely present and that the non-integer gap rules out any logarithm.
4. Interpret the smaller exponent $r_2=\frac{1}{2}$. The solution $y_2\sim x^{1/2}$ near $0$ is finite (it $\to 0$) but has infinite slope: its derivative $\sim \frac{1}{2}{x}^{-1/2}\to \infty$. Look at the figure — the pale-yellow $x^{1/2}$ curve rises with a vertical tangent at the origin, while the blue $x^1$ curve (the larger root) enters flat. Why this step? Reading the geometry off the exponent is the whole payoff of finding $r$ — the number $\frac{1}{2}$ literally predicts a vertical tangent, so we can picture the solution without solving further.

Verify: $(2r-1)(r-1)=2r^2-3r+1$ matches. Roots $1$ and $0.5$; both positive, difference $0.5$. ✔ (Neither root is negative here, so nothing blows up; a genuinely negative root — as in Example B's $r_2=-1$ — is what would drive $y\to \infty$ at the origin.)

---

Example D — equal roots, log guaranteed (Cell D)

First we re-earn the symbol $J_0$ so nothing is borrowed unexplained:

Forecast: The parent showed $y_1=J_0$. Guess the exact template for $y_2$.

1. Indicial. $p_0=1,q_0=0$ so $r(r-1)+r=r^2=0$, double root $r=0$. → Cell D. Why this step? A repeated root means only one exponent, so the two series can't both be pure powers.
2. Why a log must appear. With one exponent, the recurrence produces exactly one power series (the sum we named $J_0$). A second, independent solution cannot be another power series about the same exponent — independence would fail. Reduction of order with $y_2=v(x)y_1$ then forces $v^{\prime }=\frac{1}{y_1^2}\exp (-\int P)$; here $\int P=\int \frac{1}{x}=\ln x$, and unwinding gives a $\int \frac{dx}{x}=\ln x$ term. Why this step? This is the mechanism the parent asserted; we make it explicit.
3. Write the template. $y_2=J_0(x)\ln x+{\sum }_{n\ge 1}{c}_n{x}^n.$ Why this step? Cell D always has this shape; the $c_n$ come from re-substituting.

Verify: Compute the Wronskian shape: $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$. Since $y_2$ contains $\ln x$, $y_2^{\prime }$ contains $y_1/x$, so $W\propto 1/x\ne 0$ — genuinely independent. Abel's formula gives $W=C\exp (-\int P)=C/x$, matching. ✔

---

Example E — integer gap WITH a log (Cell E)

Forecast: Roots differ by an integer. Will step $N$ demand division by zero?

1. Indicial. Here $p_0=xP{|}_0=1$, and $q_0=x^{2}Q{|}_0=(x^2-1){|}_0=-1$. So $r(r-1)+r-1=r^2-1=0$, roots $r_1=1,r_2=-1$. Gap $N=2$ → integer, Cell E or F. Why this step? Integer gap warns us of a possible clash; only the recurrence tells us which cell.
2. Set up the recurrence and see why the shift is by $2$. Plug $y=\sum a_n{x}^{n+r}$. The tamed coefficient $q(x)=x^{2}Q=x^2-1$ has entries $q_0=-1$ and $q_2=+1$ (all other $q_k=0$, and $p(x)=1$ so only $p_0$ survives). The $-1$ piece pairs $a_n$ with itself and builds the left-hand indicial factor $F(n+r)$; the $+x^2$ piece multiplies the series by $x^2$, which shifts every term up by two powers, so $a_{n-2}$ lands on the same power $x^{n+r}$ as $a_n$. Matching the coefficient of $x^{m+r}$ therefore gives $F(m+r)a_m=-a_{m-2},F(s)=s^2-1.$ Why this step? The "shift by $m-2$" is not a trick — it is exactly because the only surviving higher coefficient is $q_2$, attached to $x^2$; that single term is why neighbours two apart couple and nothing else survives.
3. Watch the break at $r_2=-1$. With $r=-1$, $F(m-1)=(m-1{)}^2-1=m(m-2)$. At $m=N=2$: $F(2-1)=F(1)=1^2-1=0$ — the left side vanishes while the right side $-a_0\ne 0$. No finite $a_2$ can satisfy $0\cdot a_2=-a_0$. Why this step? This is exactly $F(r_2+N)=F(r_1)=0$: the recurrence tries to divide by zero at step $N=2$.
4. Consequence. The pure $x^{-1}$ series cannot be completed. The rescue is the log: $y_2=\kappa y_1\ln x+x^{-1}{\sum }_{n\ge 0}{c}_n{x}^n,\kappa \ne 0.$ Why this step? The log term contributes the missing piece at order $x^1$, filling the gap.

The figure below plots $F(s)=s^2-1$ so you can see the collision: the two roots sit at $s=\pm 1$, and the step $N=2$ starting from $r_2=-1$ lands the argument $s=m+r_2=2+(-1)=1$ exactly on the other root, where $F=0$. Whenever a step lands on a zero of $F$, the recurrence stalls — that is the geometric fingerprint of a Cell-E logarithm.

Verify: $F(1)=1-1=0$ confirms the breakdown at exactly $m=N=2$, and $r_1=1$ is where $F$ vanishes — the log is unavoidable here ($\kappa \ne 0$). ✔

---

Example F — integer gap WITHOUT a log (Cell F)

Forecast: The parent's third "mistake" warns integer-gap $\ne$ automatic log. Watch $\kappa$ turn out to be $0$.

1. Standard form. Divide by $x$: $y^{\prime \prime }+\frac{2}{x}{y}^{\prime }+y=0$, so $P=2/x$, $Q=1$. $p_0=\lim x\cdot \frac{2}{x}=2$, $q_0=\lim x^2\cdot 1=0$. Why this step? Indicial inputs.
2. Indicial. $r(r-1)+2r+0=r^2+r=r(r+1)=0$, roots $r_1=0,r_2=-1$. Gap $N=1$ → integer, Cell E or F. Why this step? Another integer gap; must test the recurrence.
3. Recurrence. Here $p(x)=xP=2$ (only $p_0$) and $q(x)=x^{2}Q=x^2$ (only $q_2=1$), so again the surviving $+x^2$ couples terms two apart. Plug $y=\sum a_n{x}^{n+r}$; matching $x^{n+r}$ gives $(n+r)(n+r+1)a_n=-a_{n-2}.$ For $r_2=-1$: $(n-1)n{a}_n=-a_{n-2}$. At the dangerous step $n=N=1$: $(0)(1)a_1=-a_{-1}=0$ (there is no $a_{-1}$, so the right side is $0$). Thus $0\cdot a_1=0$ is satisfied by any $a_1$ — no contradiction! Why this step? The right side vanished too, so there is no forced division by zero — this is precisely the Cell-F escape.
4. Result and identification. Choose $a_1=0$; the $x^{-1}$ series completes freely, $\kappa =0$, no log. To name the solutions, notice the ODE is exactly $(xy{)}^{\prime \prime }+(xy)=0$ in disguise: let $u=xy$, then substituting gives $u^{\prime \prime }+u=0$, whose solutions are $u=\sin x$ and $u=\cos x$. Undoing $u=xy$ gives $y=\frac{\sin x}{x}$ and $y=\frac{\cos x}{x}$ — two clean, independent solutions, the first a Frobenius series with $r_1=0$, the second the singular $x^{-1}$ series with no logarithm. Why this step? The substitution $u=xy$ is a legitimate change of variable that turns the equation into one we already know; it proves the two solutions are elementary and confirms $\kappa =0$.

Verify: $r(r+1)$ has roots $0,-1$, gap $1$. At $n=1$ both sides zero, so $\kappa =0$. Substituting $y=\cos x/x$ (the $r=-1$ solution) into the original ODE gives $0$, and there is no logarithm. ✔

---

Example G — irregular point, method NOT guaranteed (Cell G)

Forecast: Guess before computing — regular or irregular?

1. Standard form. Divide by $x^3$: $P=\frac{1}{x^2}$, $Q=\frac{1}{x^3}$. Why this step? Need bare coefficients.
2. Tame them. $xP=\frac{1}{x}\to \infty$ as $x\to 0$ — not analytic. Already fails; $x^{2}Q=\frac{1}{x}\to \infty$ too. Why this step? The RSP test requires both $xP$ and $x^{2}Q$ analytic. One failure is enough.
3. Conclusion. $x=0$ is an irregular singular point. There is no valid indicial equation (the "$p_0$" limit diverges). Frobenius is not guaranteed; solutions may involve essential singularities like $e^{1/x}$. Why this step? This is the boundary of the method — you must recognise it and stop.

Verify: ${\lim }_{x\to 0}xP=\lim 1/x$ diverges ⇒ no finite $p_0$ ⇒ Cell G. The parent's "WHY these exact powers?" explains it: a $1/x^2$ in $P$ is a stronger blow-up than a single $x^r$ can balance. ✔

---

Example H — word problem / limiting case (Cell H)

Forecast: Which mode do you expect to be nonzero at the drum's centre — the symmetric one or the asymmetric one?

1. Indicial equation (general $m$). $p_0=1$, $q_0=(x^2-m^2){|}_0=-m^2$, so $r(r-1)+r-m^2=r^2-m^2=0$, roots $r=\pm m$. Why this step? One quadratic handles every mode at once.
2. Case $m=0$. Roots $0,0$ — Cell D, equal roots. Leading behaviour $R\sim x^0=$ constant. So the symmetric mode is finite and nonzero at the centre (this is $J_0$, peak at $x=0$). Why this step? The exponent $0$ literally says "finite at the origin".
3. Case $m=1$. Roots $+1,-1$, gap $2$ — Cell E. Physically we keep the regular root $r_1=+1$: $R\sim x^1\to 0$ at the centre. The other root $r_2=-1$ gives $R\sim x^{-1}\to \infty$, which is unphysical for a drum (infinite displacement at the centre) and is discarded. Why this step? The exponent sign selects the physically admissible solution.
4. Answer. $m=0$: $R\sim \text{const}$, finite at centre. $m=1$: physical $R\sim x$, zero at centre; the singular partner is thrown away.

The figure contrasts the two admissible modes. The pale-yellow curve is the $m=0$ mode ($J_0$): it starts at a finite peak at the centre — that is the exponent $r=0$ made visible. The chalk-blue curve is the physical $m=1$ mode ($J_1$): it starts at zero with a linear ramp — the exponent $r=+1$ made visible. The discarded $x^{-1}$ partner would shoot to infinity at $x=0$ and is not drawn, exactly because a real drum cannot have infinite centre displacement.

Verify: Roots of $r^2-m^2$ are $\pm m$: for $m=0$ double root $0$; for $m=1$ roots $\pm 1$ gap $2$. The finite-at-centre modes correspond to $r\ge 0$. This is precisely why Bessel functions $J_m$ (regular) appear in drum physics while $Y_m$ (singular) are excluded. ✔

---

Active Recall

Recall Which cell, at a glance?

Compute $p_0,q_0$ → solve $r^2+(p_0-1)r+q_0=0$ → look at the gap $r_1-r_2$. Gap not real integer? ::: Cell B/C — two clean series. Gap $=0$? ::: Cell D — log guaranteed. Gap $=N\in {\mathbb{Z}}^{+}$ and recurrence divides by zero at step $N$? ::: Cell E — log. Gap $=N$ but both sides vanish at step $N$? ::: Cell F — no log ($\kappa =0$). $xP$ or $x^{2}Q$ not analytic? ::: Cell G — irregular, method not guaranteed.