Exercises — Frobenius method — regular singular points

Standard form. $x^2$ se divide karo: $y^{\prime \prime }+\frac{4}{x}{y}^{\prime }+\frac{2}{x^2}y=0.$ Toh $P=\frac{4}{x}$, $Q=\frac{2}{x^2}$.

Tamed pair kyun check karein? $P$ $1/x$ ki tarah blow up karta hai aur $Q$ $1/x^2$ ki tarah — worst allowed rates. Un bilkul powers se multiply karo: $xP=4(\text{analytic}),x^{2}Q=2(\text{analytic}).$ Dono constants hain, isliye $0$ par analytic hain, toh $x=0$ ek regular singular point hai.

Limits read karo: $p_0={\lim }_{x\to 0}xP=4$, $q_0={\lim }_{x\to 0}{x}^{2}Q=2$.

(Yeh actually ek Euler–Cauchy equation disguise mein hai — iska Frobenius series $x^r$ term ke baad terminate ho jaata hai kyunki $p,q$ mein koi higher powers nahi hain.)

Yahan $P=1/x^3$ hai, toh $xP=1/x^2$, jo $x=0$ par phir bhi blow up karta hai — analytic nahi. Kyunki $p(x)=xP$ analytic hone mein fail ho raha hai, $x=0$ ek irregular singular point hai. Consequence: Frobenius method yahan kaam karne ki guarantee nahi deta; indicial machinery cleanly apply bhi nahi hoti. Yeh woh "blow-up too strong for a single $x^r$ to balance" case hai jo parent note mein flag kiya gaya hai.

Root difference $r_1-r_2=3-1=2$, ek positive integer. Yeh Case 3 hai. Second solution ko shayad logarithm ki zaroorat ho sakti hai: $y_2=\kappa y_1\ln x+x^{r_2}{\sum }_{n\ge 0}{c}_n{x}^n,\kappa \text{ possibly }0.$ Abhi hum nahi bata sakte ki $\kappa =0$ hai ya nahi — uske liye actually step $N=2$ par recurrence run karni hogi.

Standard form: $2x^2$ se divide karo: $y^{\prime \prime }+\frac{1}{2x}{y}^{\prime }+\frac{x^2-3}{2x^2}y=0.$ Toh $xP=\frac{1}{2}$ aur $x^{2}Q=\frac{x^2-3}{2}$; dono analytic → RSP.

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

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 karo: $(2r-3)(r+1)=0$, toh $r_1=\frac{3}{2}$, $r_2=-1$. Difference: $\frac{3}{2}-(-1)=\frac{5}{2}\notin \mathbb{Z}$ → Case 1, do clean Frobenius series, koi log nahi.

$P=1/x$, $Q=1-\frac{1}{4x^2}$. Toh $xP=1$, $x^{2}Q=x^2-\frac{1}{4}$; dono analytic → RSP. $p_0=1$, $q_0=-\frac{1}{4}$. $r(r-1)+r-\frac{1}{4}=r^2-\frac{1}{4}=0\Rightarrow r=\pm \frac{1}{2}.$ Toh $r_1=\frac{1}{2}$, $r_2=-\frac{1}{2}$. Difference $=1$, ek positive integer → Case 3. Interesting: yahan log coefficient $\kappa =0$ nikalta hai (yeh exactly woh "integer difference but no log" wala surprise hai). Dekho Bessel's equation and Bessel functions.

$P=-1/x$, $Q=(1-x)/x^2$. $xP=-1$, $x^{2}Q=1-x$; dono 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) → $y_2$ mein logarithm forced hai.

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

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

$x^{n+r}$ ka coefficient $n\ge 1$ ke liye: $[2(n+r)(n+r-1)+3(n+r)-1]a_n-a_{n-1}=0.$ Bracket indicial polynomial $F(n+r)=2(n+r{)}^2+(n+r)-1=(2(n+r)-1)(n+r+1)$ hai. Toh $a_n=\frac{a_{n-1}}{(2(n+r)-1)(n+r+1)}.$

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

Branch $r=-1$: $2(n-1)-1=2n-3$, aur $n+r+1=n$, toh $a_n=\frac{a_{n-1}}{(2n-3)n}.$ $a_0=1$ ke saath: $a_1=\frac{1}{(-1)(1)}=-1$, $a_2=\frac{-1}{(1)(2)}=-\frac{1}{2}$, $a_3=\frac{-1/2}{(3)(3)}=-\frac{1}{18}$. $y_2=x^{-1}(1-x-\frac{1}{2}{x}^2-\frac{1}{18}{x}^3-\cdots ).$ Kyunki difference $\frac{5}{2}\notin \mathbb{Z}$ hai, yeh dono automatically independent hain — Wronskian and linear independence se verify karo agar chahte ho.

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

$r=0$ ke saath $y=\sum a_n{x}^{n+r}$ insert karo $x{y}^{\prime \prime }+y^{\prime }+xy=0$ mein:

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

Pehle do combine karte hain: $\sum a_n{n}^2{x}^{n-1}$. Teesre ko $n\to n-2$ se shift karo: $\sum a_{n-2}{x}^{n-1}$. $x^{n-1}$ ka coefficient: $a_n{n}^2+a_{n-2}=0\Rightarrow a_n=-\frac{a_{n-2}}{n^2}.$ $a_1=0$ (odd terms zero ho jaate hain). $a_0=1$ ke saath: $a_2=-\frac{1}{4},a_4=-\frac{a_2}{16}=\frac{1}{64}.$ $y_1=1-\frac{1}{4}{x}^2+\frac{1}{64}{x}^4-\cdots =J_0(x).$ Second solution mein $\ln x$ hoga (Case 2) — use Reduction of order se obtain karo.

Strategy. $y_2=y_1\ln x+w$ substitute karo, jahan $w={\sum }_{n\ge 1}{c}_n{x}^n$ hai, $x^2{y}^{\prime \prime }+x{y}^{\prime }+x^{2}y=0$ mein.

Log part handle karo. Kisi bhi smooth $u$ ke liye, agar $Y=u\ln x$ toh $Y^{\prime }=u^{\prime }\ln x+\frac{u}{x},Y^{\prime \prime }=u^{\prime \prime }\ln x+\frac{2u^{\prime }}{x}-\frac{u}{x^2}.$ $u=y_1$ ko $L[Y]:=x^2{Y}^{\prime \prime }+x{Y}^{\prime }+x^{2}Y$ mein plug karo: $L[y_1\ln x]=(\underset{=0}{\underbrace{x^2{y}_1^{\prime \prime }+x{y}_1^{\prime }+x^2{y}_1}})\ln x+2x{y}_1^{\prime }.$ Bracket zero hai kyunki $y_1$ ODE solve karta hai — yahi log ansatz ka poora point hai: $\ln x$ equation se nikal jaata hai aur ek clean forcing term $2x{y}_1^{\prime }$ chhod jaata hai.

Toh $w$ ki equation hai $L[w]=-2x{y}_1^{\prime }.$ $y_1=1-\frac{1}{4}{x}^2+\cdots$ ke saath, $y_1^{\prime }=-\frac{1}{2}x+\cdots$, toh $-2x{y}_1^{\prime }=x^2+\cdots$.

Left side $w=c_{1}x+c_2{x}^2+\cdots$ ke saath: $L[w]=\sum c_n(n(n-1)+n)x^n+\sum c_n{x}^{n+2}=\sum c_n{n}^2{x}^n+\sum c_n{x}^{n+2}.$ $x^1$ match karo: $c_1\cdot 1=0\Rightarrow c_1=0$ (hamare free choice ke saath consistent). $x^2$ match karo: $c_2\cdot 4=1\Rightarrow \boxed{c_2=\frac{1}{4}}.$ (Yeh normalization tak standard $Y_0$ ka leading term hai.)

Standard form: $y^{\prime \prime }-\frac{1}{x}y=0$, toh $P=0$, $Q=-1/x$. Phir $xP=0$, $x^{2}Q=-x$; dono 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. $y=\sum a_n{x}^{n+r}$ ko $x{y}^{\prime \prime }-y=0$ mein insert karo: $x{y}^{\prime \prime }=\sum a_n(n+r)(n+r-1)x^{n+r-1},-y=\sum -a_n{x}^{n+r}(\text{shift }n\to n-1).$ $x^{n+r-1}$ ka coefficient: $a_n(n+r)(n+r-1)-a_{n-1}=0.$

Chhota root $r_2=0$ try karo: $a_{n}n(n-1)=a_{n-1}$. $n=1$ par: $a_1\cdot (1)(0)=a_0$, matlab $0=a_0$. Lekin $a_0\ne 0$ — contradiction. Recurrence aage nahi chal sakti → log zaroor aayega, toh $\kappa \ne 0$.

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

Standard form: $4x^2$ se divide karo: $y^{\prime \prime }+\frac{1}{x}{y}^{\prime }-\frac{1+4x^2}{4x^2}y=0.$ $xP=1$ (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. ODE ko $4x^2{y}^{\prime \prime }+4x{y}^{\prime }-y-4x^{2}y=0$ likho, $y=\sum a_n{x}^{n+r}$ insert karo: $[4(n+r)(n+r-1)+4(n+r)-1]a_n-4a_{n-2}=0.$ Bracket $=4(n+r{)}^2-1=(2(n+r)-1\left)\right(2(n+r)+1)$ hai. Toh $a_n=\frac{4a_{n-2}}{(2(n+r)-1)(2(n+r)+1)}.$

Bada root $r=\frac{1}{2}$: $2(n+\frac{1}{2})-1=2n$, $2(n+\frac{1}{2})+1=2n+2$, toh $a_n=\frac{4a_{n-2}}{2n(2n+2)}=\frac{a_{n-2}}{n(n+1)}.$ $a_1=0$ (odd terms vanish, kyunki $n=1$ par: $4a_1=4a_{-1}=0$). $a_0=1$ ke saath: $a_2=\frac{1}{2\cdot 3}=\frac{1}{6},a_4=\frac{a_2}{4\cdot 5}=\frac{1}{120}.$ $y_1=x^{1/2}(1+\frac{1}{6}{x}^2+\frac{1}{120}{x}^4+\cdots )=x^{-1/2}(x+\frac{1}{6}{x}^3+\cdots )=x^{-1/2}\sinh x.$ (Pehchano $\sinh x=x+\frac{x^3}{6}+\frac{x^5}{120}+\cdots$.) Toh $y_1=x^{-1/2}\sinh x$ exactly hai.

Log ke liye critical-step check. $r_2=-\frac{1}{2}$ par, $n=N=1$: $a_1$ ko $4(n+r{)}^2-1=4(\frac{1}{2}{)}^2-1=0$ chahiye hoga, forcing $0\cdot a_1=4a_{-1}=0$, jo $0=0$ hai — forcing vanish ho gayi! Toh $\kappa =0$: chhota root ek clean second series deta hai $y_2=x^{-1/2}(1+\frac{1}{2}{x}^2+\cdots )=x^{-1/2}\cosh x.$ Independence: $y_1=x^{-1/2}\sinh x$ aur $y_2=x^{-1/2}\cosh x$ independent hain (inका Wronskian nonzero hai, bilkul $\sinh ,\cosh$ ki tarah). Integer difference ke bawajood koi logarithm nahi — Case 3 with $\kappa =0$.

Recurrence $a_n$ ko $F(n+r)$ se multiply karta hai. Chhote root $r_2$ ke liye, yeh factor $F(n+r_2)$ hai. Ab $F$ ke roots exactly $r_1$ aur $r_2$ hain, toh $F(s)=(s-r_1)(s-r_2)$. Isliye $F(n+r_2)=(n+r_2-r_1)(n+r_2-r_2)=(n-(r_1-r_2))\cdot n.$ Yeh zero hota hai exactly jab $n=r_1-r_2$ — possible tabhi jab $r_1-r_2$ ek non-negative integer $N$ ho.

• $r_1-r_2\notin \mathbb{Z}$: $F(n+r_2)$ kabhi zero nahi hota → dono series clean (Case 1).
• $r_1=r_2$ ($N=0$): roots collide hote hain; $F(s)=(s-r{)}^2$ ka double zero hai, toh $r$ ke respect mein differentiate karna (standard trick) ek $\ln x$ drop karta hai (Case 2).
• $r_1-r_2=N>0$: $F(N+r_2)=0$ → recurrence step $N$ par zero se divide karta hai. Agar wahan forcing nonzero hai, toh $\ln x$ second solution ko rescue karta hai; agar woh zero ho jaaye, toh hum $\kappa =0$ ke saath cleanly nikal jaate hain (Case 3).

Toh ek algebraic fact "$F(n+r_2)$ zero ke equal ho sakta hai" hi saari branching ka source hai. Upar wala figure dekho: $r_1-r_2$ ka axis position literally batata hai ki zero denominator aa sakta hai ya nahi.

Root difference $=2.4$, kaun sa case? ::: Case 1 (integer nahi) — do clean series, koi log nahi. Double indicial root, kya forced hai? ::: $y_2$ mein logarithm (Case 2). Case 3 mein step $n=N$ par kya check karna hai? ::: Kya forcing term vanish hoti hai; agar haan toh $\kappa =0$ (log nahi), agar nahi toh $\kappa \ne 0$ (log). $q_0=\lim x^{2}Q$ kyun hai aur $Q(0)$ kyun nahi? ::: Kyunki $Q$ singular point par blow up karta hai; sirf tamed $x^{2}Q$ finite hota hai.