Exercises — Frobenius method — regular singular points

Standard form. Divide by $x^2$: $y^{\prime \prime }+\frac{4}{x}{y}^{\prime }+\frac{2}{x^2}y=0.$ So $P=\frac{4}{x}$, $Q=\frac{2}{x^2}$.

Why check the tamed pair? $P$ blows up like $1/x$ and $Q$ like $1/x^2$ — the worst allowed rates. Multiply by exactly those powers: $xP=4(\text{analytic}),x^{2}Q=2(\text{analytic}).$ Both are constants, hence analytic at $0$, so $x=0$ is a regular singular point.

Read off the limits: $p_0={\lim }_{x\to 0}xP=4$, $q_0={\lim }_{x\to 0}{x}^{2}Q=2$.

(This is actually an Euler–Cauchy equation in disguise — its Frobenius series terminates after the $x^r$ term because $p,q$ have no higher powers.)

Here $P=1/x^3$, so $xP=1/x^2$, which still blows up at $x=0$ — not analytic. Because $p(x)=xP$ fails to be analytic, $x=0$ is an irregular singular point. Consequence: the Frobenius method is not guaranteed to work here; the indicial machinery doesn't even apply cleanly. This is the "blow-up too strong for a single $x^r$ to balance" case flagged in the parent note.

Root difference $r_1-r_2=3-1=2$, a positive integer. This is Case 3. The second solution may need a logarithm: $y_2=\kappa y_1\ln x+x^{r_2}{\sum }_{n\ge 0}{c}_n{x}^n,\kappa \text{ possibly }0.$ We cannot tell yet whether $\kappa =0$ — that requires actually running the recurrence at step $N=2$.

Standard form: divide by $2x^2$: $y^{\prime \prime }+\frac{1}{2x}{y}^{\prime }+\frac{x^2-3}{2x^2}y=0.$ So $xP=\frac{1}{2}$ and $x^{2}Q=\frac{x^2-3}{2}$; both analytic → RSP.

Tamed limits: $p_0=\frac{1}{2}$, and $q_0=x^{2}Q{|}_{x=0}=\frac{0-3}{2}=-\frac{3}{2}$.

Indicial equation $r(r-1)+p_{0}r+q_0=0$: $r^2-r+\frac{1}{2}r-\frac{3}{2}=r^2-\frac{1}{2}r-\frac{3}{2}=0\Rightarrow 2r^2-r-3=0.$ Factor: $(2r-3)(r+1)=0$, so $r_1=\frac{3}{2}$, $r_2=-1$. Difference: $\frac{3}{2}-(-1)=\frac{5}{2}\notin \mathbb{Z}$ → Case 1, two clean Frobenius series, no log.

$P=1/x$, $Q=1-\frac{1}{4x^2}$. Then $xP=1$, $x^{2}Q=x^2-\frac{1}{4}$; both analytic → RSP. $p_0=1$, $q_0=-\frac{1}{4}$. $r(r-1)+r-\frac{1}{4}=r^2-\frac{1}{4}=0\Rightarrow r=\pm \frac{1}{2}.$ So $r_1=\frac{1}{2}$, $r_2=-\frac{1}{2}$. Difference $=1$, a positive integer → Case 3. Interesting: here the log coefficient turns out $\kappa =0$ (this is exactly the "integer difference but no log" surprise). See Bessel's equation and Bessel functions.

$P=-1/x$, $Q=(1-x)/x^2$. $xP=-1$, $x^{2}Q=1-x$; both analytic → RSP. $p_0=-1$, $q_0=1$. $r(r-1)-r+1=r^2-2r+1=(r-1{)}^2=0.$ Double root $r_1=r_2=1$ → Case 2 (equal roots) → a logarithm is forced in $y_2$.

Setup. Multiply-free form: $2x^2{y}^{\prime \prime }+3x{y}^{\prime }-y-xy=0$. Insert $y={\sum }_{n\ge 0}{a}_n{x}^{n+r}$.

• $2x^2{y}^{\prime \prime }=\sum 2a_n(n+r)(n+r-1)x^{n+r}$
• $3x{y}^{\prime }=\sum 3a_n(n+r)x^{n+r}$
• $-y=\sum -a_n{x}^{n+r}$
• $-xy=\sum -a_n{x}^{n+r+1}$ (shift index $n\to n-1$: $\sum -a_{n-1}{x}^{n+r}$)

Coefficient of $x^{n+r}$ for $n\ge 1$: $[2(n+r)(n+r-1)+3(n+r)-1]a_n-a_{n-1}=0.$ The bracket is the indicial polynomial $F(n+r)=2(n+r{)}^2+(n+r)-1=(2(n+r)-1)(n+r+1)$. So $a_n=\frac{a_{n-1}}{(2(n+r)-1)(n+r+1)}.$

Branch $r=\frac{1}{2}$: $2(n+\frac{1}{2})-1=2n$, and $n+r+1=n+\frac{3}{2}$, so $a_n=\frac{a_{n-1}}{2n(n+\frac{3}{2})}=\frac{a_{n-1}}{n(2n+3)}.$ With $a_0=1$: $a_1=\frac{1}{1\cdot 5}=\frac{1}{5}$, $a_2=\frac{1/5}{2\cdot 7}=\frac{1}{70}$, $a_3=\frac{1/70}{3\cdot 9}=\frac{1}{1890}$. $y_1=x^{1/2}(1+\frac{1}{5}x+\frac{1}{70}{x}^2+\frac{1}{1890}{x}^3+\cdots ).$

Branch $r=-1$: $2(n-1)-1=2n-3$, and $n+r+1=n$, so $a_n=\frac{a_{n-1}}{(2n-3)n}.$ With $a_0=1$: $a_1=\frac{1}{(-1)(1)}=-1$, $a_2=\frac{-1}{(1)(2)}=-\frac{1}{2}$, $a_3=\frac{-1/2}{(3)(3)}=-\frac{1}{18}$. $y_2=x^{-1}(1-x-\frac{1}{2}{x}^2-\frac{1}{18}{x}^3-\cdots ).$ Since the difference $\frac{5}{2}\notin \mathbb{Z}$, these two are automatically independent — check via the Wronskian and linear independence if you like.

Standard form: $y^{\prime \prime }+\frac{1}{x}{y}^{\prime }+y=0$. So $xP=1$, $x^{2}Q=x^2$ → RSP, $p_0=1,q_0=0$. Indicial: $r(r-1)+r=r^2=0$ → $r_1=r_2=0$ (Case 2).

Insert $y=\sum a_n{x}^{n+r}$ with $r=0$ into $x{y}^{\prime \prime }+y^{\prime }+xy=0$:

• $x{y}^{\prime \prime }=\sum a_{n}n(n-1)x^{n-1}$
• $y^{\prime }=\sum a_{n}n{x}^{n-1}$
• $xy=\sum a_n{x}^{n+1}$

First two combine: $\sum a_n{n}^2{x}^{n-1}$. Shift the third by $n\to n-2$: $\sum a_{n-2}{x}^{n-1}$. Coefficient of $x^{n-1}$: $a_n{n}^2+a_{n-2}=0\Rightarrow a_n=-\frac{a_{n-2}}{n^2}.$ $a_1=0$ (odd terms vanish). With $a_0=1$: $a_2=-\frac{1}{4},a_4=-\frac{a_2}{16}=\frac{1}{64}.$ $y_1=1-\frac{1}{4}{x}^2+\frac{1}{64}{x}^4-\cdots =J_0(x).$ The second solution carries $\ln x$ (Case 2) — obtain it by Reduction of order.

Strategy. Substitute $y_2=y_1\ln x+w$, where $w={\sum }_{n\ge 1}{c}_n{x}^n$, into $x^2{y}^{\prime \prime }+x{y}^{\prime }+x^{2}y=0$.

Handle the log part. For any smooth $u$, if $Y=u\ln x$ then $Y^{\prime }=u^{\prime }\ln x+\frac{u}{x},Y^{\prime \prime }=u^{\prime \prime }\ln x+\frac{2u^{\prime }}{x}-\frac{u}{x^2}.$ Plug $u=y_1$ into $L[Y]:=x^2{Y}^{\prime \prime }+x{Y}^{\prime }+x^{2}Y$: $L[y_1\ln x]=(\underset{=0}{\underbrace{x^2{y}_1^{\prime \prime }+x{y}_1^{\prime }+x^2{y}_1}})\ln x+2x{y}_1^{\prime }.$ The bracket is zero because $y_1$ solves the ODE — that is the whole point of the log ansatz: the $\ln x$ dies out of the equation and leaves a clean forcing term $2x{y}_1^{\prime }$.

So the equation for $w$ is $L[w]=-2x{y}_1^{\prime }.$ With $y_1=1-\frac{1}{4}{x}^2+\cdots$, $y_1^{\prime }=-\frac{1}{2}x+\cdots$, so $-2x{y}_1^{\prime }=x^2+\cdots$.

Left side with $w=c_{1}x+c_2{x}^2+\cdots$: $L[w]=\sum c_n(n(n-1)+n)x^n+\sum c_n{x}^{n+2}=\sum c_n{n}^2{x}^n+\sum c_n{x}^{n+2}.$ Match $x^1$: $c_1\cdot 1=0\Rightarrow c_1=0$ (consistent with our free choice). Match $x^2$: $c_2\cdot 4=1\Rightarrow \boxed{c_2=\frac{1}{4}}.$ (This is the leading term of the standard $Y_0$ up to normalization.)

Standard form: $y^{\prime \prime }-\frac{1}{x}y=0$, so $P=0$, $Q=-1/x$. Then $xP=0$, $x^{2}Q=-x$; both analytic → RSP. $p_0=0$, $q_0=0$. Indicial: $r(r-1)=0$ → $r_1=1$, $r_2=0$. Difference $N=1$ → Case 3.

Recurrence. Insert $y=\sum a_n{x}^{n+r}$ into $x{y}^{\prime \prime }-y=0$: $x{y}^{\prime \prime }=\sum a_n(n+r)(n+r-1)x^{n+r-1},-y=\sum -a_n{x}^{n+r}(\text{shift }n\to n-1).$ Coefficient of $x^{n+r-1}$: $a_n(n+r)(n+r-1)-a_{n-1}=0.$

Try the smaller root $r_2=0$: $a_{n}n(n-1)=a_{n-1}$. At $n=1$: $a_1\cdot (1)(0)=a_0$, i.e. $0=a_0$. But $a_0\ne 0$ — contradiction. The recurrence cannot proceed → the log must appear, so $\kappa \ne 0$.

Larger root $r_1=1$ gives the clean series: $a_n(n+1)(n)=a_{n-1}$, so $a_1=\frac{a_0}{2}$, $a_2=\frac{a_1}{6}=\frac{a_0}{12}$. $y_1=x(1+\frac{1}{2}x+\frac{1}{12}{x}^2+\cdots ),y_2=\kappa y_1\ln x+{\sum }_{n\ge 0}{c}_n{x}^n,\kappa \ne 0.$

Standard form: divide by $4x^2$: $y^{\prime \prime }+\frac{1}{x}{y}^{\prime }-\frac{1+4x^2}{4x^2}y=0.$ $xP=1$ (analytic), $x^{2}Q=-\frac{1+4x^2}{4}$ (analytic) → RSP. $p_0=1$, $q_0=-\frac{1}{4}$.

Indicial: $r(r-1)+r-\frac{1}{4}=r^2-\frac{1}{4}=0\Rightarrow r=\pm \frac{1}{2}$. Difference $=1$ → Case 3.

Recurrence. Write ODE as $4x^2{y}^{\prime \prime }+4x{y}^{\prime }-y-4x^{2}y=0$, insert $y=\sum a_n{x}^{n+r}$: $[4(n+r)(n+r-1)+4(n+r)-1]a_n-4a_{n-2}=0.$ Bracket $=4(n+r{)}^2-1=(2(n+r)-1\left)\right(2(n+r)+1)$. So $a_n=\frac{4a_{n-2}}{(2(n+r)-1)(2(n+r)+1)}.$

Larger root $r=\frac{1}{2}$: $2(n+\frac{1}{2})-1=2n$, $2(n+\frac{1}{2})+1=2n+2$, so $a_n=\frac{4a_{n-2}}{2n(2n+2)}=\frac{a_{n-2}}{n(n+1)}.$ $a_1=0$ (odd terms vanish, since at $n=1$: $4a_1=4a_{-1}=0$). With $a_0=1$: $a_2=\frac{1}{2\cdot 3}=\frac{1}{6},a_4=\frac{a_2}{4\cdot 5}=\frac{1}{120}.$ $y_1=x^{1/2}(1+\frac{1}{6}{x}^2+\frac{1}{120}{x}^4+\cdots )=x^{-1/2}(x+\frac{1}{6}{x}^3+\cdots )=x^{-1/2}\sinh x.$ (Recognize $\sinh x=x+\frac{x^3}{6}+\frac{x^5}{120}+\cdots$.) So $y_1=x^{-1/2}\sinh x$ exactly.

Critical-step check for the log. At $r_2=-\frac{1}{2}$, $n=N=1$: $a_1$ would need $4(n+r{)}^2-1=4(\frac{1}{2}{)}^2-1=0$, forcing $0\cdot a_1=4a_{-1}=0$, which is $0=0$ — the forcing vanishes! So $\kappa =0$: the smaller root gives a clean second series $y_2=x^{-1/2}(1+\frac{1}{2}{x}^2+\cdots )=x^{-1/2}\cosh x.$ Independence: $y_1=x^{-1/2}\sinh x$ and $y_2=x^{-1/2}\cosh x$ are independent (their Wronskian is nonzero, just like $\sinh ,\cosh$). No logarithm despite integer difference — Case 3 with $\kappa =0$.

The recurrence multiplies $a_n$ by $F(n+r)$. For the smaller root $r_2$, this factor is $F(n+r_2)$. Now $F$ has roots exactly $r_1$ and $r_2$, so $F(s)=(s-r_1)(s-r_2)$. Thus $F(n+r_2)=(n+r_2-r_1)(n+r_2-r_2)=(n-(r_1-r_2))\cdot n.$ This is zero precisely when $n=r_1-r_2$ — possible only if $r_1-r_2$ is a non-negative integer $N$.

• $r_1-r_2\notin \mathbb{Z}$: $F(n+r_2)$ never hits zero → both series clean (Case 1).
• $r_1=r_2$ ($N=0$): the roots collide; $F(s)=(s-r{)}^2$ has a double zero, so differentiating with respect to $r$ (the standard trick) drops a $\ln x$ (Case 2).
• $r_1-r_2=N>0$: $F(N+r_2)=0$ → the recurrence divides by zero at step $N$. If the forcing there is nonzero, a $\ln x$ rescues the second solution; if it happens to vanish, we sail through with $\kappa =0$ (Case 3).

So the single algebraic fact "$F(n+r_2)$ can equal zero" is the source of all the branching. See the figure at the top: the axis position of $r_1-r_2$ literally tells you whether a zero denominator can occur.

Root difference $=2.4$, which case? ::: Case 1 (not an integer) — two clean series, no log. Double indicial root, what's forced? ::: A logarithm in $y_2$ (Case 2). What must you check at step $n=N$ in Case 3? ::: Whether the forcing term vanishes; if yes $\kappa =0$ (no log), if no $\kappa \ne 0$ (log). Why is $q_0=\lim x^{2}Q$ and not $Q(0)$? ::: Because $Q$ blows up at the singular point; only the tamed $x^{2}Q$ is finite.