Worked examples — Frobenius method — regular singular points

4.6.18 · D3 · Maths › Ordinary Differential Equations › Frobenius method — regular singular points

Ek hi page, har cell. Hum har us tarike se guzarte hain jis tarah indicial roots behave kar sakti hain, saath mein traps bhi (ek fake singular point, ek bada blow-up jo method ko kill kar de, ek degenerate root, ek limiting case). Agar koi scenario regular singular point pe ho sakta hai, toh woh neeche apne label ke saath diya gaya hai.

Yahan sab kuch parent note pe build karta hai. Agar koi symbol aata hai, toh hum use yahan re-earn karte hain taaki tumhe kabhi scroll back na karna pade.

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The scenario matrix

Poora game yeh hai: tamed limits $p_0,q_0$ compute karo, chhota sa quadratic (indicial equation) solve karo, aur dono roots ka difference dekho. Woh difference har problem ko ek cell mein sort karta hai.

Recall karo dono tamed coefficients (re-stated taaki kuch assume na ho):

Kabhi kabhi hume sirf leading numbers $p_0,q_0$ se zyada chahiye. Tamed coefficients $p(x)=xP(x)$ aur $q(x)=x^{2}Q(x)$ analytic hote hain, isliye inke apne Taylor series hote hain:

Record ke liye, yahan recurrence hai (parent note ka Step 4), re-stated taaki yeh page akele khada rahe. Ansatz $y=\sum a_n{x}^{n+r}$ ko equation mein plug karne aur $x^{n+r}$ ka coefficient match karne ke baad

Indicial equation hai $F(r)=r(r-1)+p_{0}r+q_0=0$, do roots $r_1\ge r_2$. Yahan har cell hai:

Cell	Kya trigger karta hai	Root behaviour	Second solution shape
A — disguise mein ordinary point	$P,Q$ actually analytic hain ($x=0$ singular nahi)	roots $0,1$	plain Taylor, koi $x^r$ nahi chahiye
B — distinct roots, non-integer gap	$r_1-r_2\notin \mathbb{Z}$	do clean $x^r$ series	koi log nahi
C — fractional roots	roots jaise $1,\frac{1}{2}$	phir bhi Case-B mechanics	koi log nahi, non-integer powers
D — equal roots	$r_1=r_2$	sirf ek exponent	log forced
E — integer gap with log	$r_1-r_2=N$, coefficient blow up karta hai	recurrence $0$ hit karta hai	log present ($\kappa \ne 0$)
F — integer gap without log	$r_1-r_2=N$, lekin survive karta hai	recurrence finite rehti hai	do clean series ($\kappa =0$)
G — irregular point (method fail)	$xP$ ya $x^{2}Q$ analytic nahi	koi valid indicial eq. nahi	Frobenius guarantee nahi
H — word problem / limiting case	physical setup, degenerate parameter	upar mein se koi bhi	exponent interpret karo

Ab hum har cell ko ek worked example ke saath hit karte hain.

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Example A — point singular tha hi nahi (Cell A)

Forecast: Abhi guess karo — kya $x=0$ regular singular hai, irregular hai, ya ordinary hai?

1. $P,Q$ padho. Pehle se standard form hai: $P(x)=x$, $Q(x)=1+x^2$. Yeh step kyun? Classification $P,Q$ ke baare mein statement hai; inhe bare dekhna zaroori hai.
2. $0$ pe analyticity test karo. $x$ aur $1+x^2$ dono polynomials hain — har jagah analytic. Yeh step kyun? Analytic $P$ aur $Q$ ka matlab hai koi blow-up nahi, toh $x=0$ ek ordinary point hai.
3. Conclusion. Koi $x^r$ nahi chahiye. Plain power-series method use karo: $y=\sum a_n{x}^n$.

Verify: Tamed limits hain $p_0=\lim x\cdot x=0$ aur $q_0=\lim x^2(1+x^2)=0$. Indicial equation $r(r-1)=0$ roots $0,1$ deta hai — exactly woh do lowest integer powers jo Taylor series mein hote hain. Frobenius ordinary method mein wapis degenerate ho jaata hai, confirm karta hai ki kuch singular nahi tha. ✔

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

Forecast: Parent note se roots $\frac{1}{2},-1$ hain. Kya recurrence mein kahin zero se divide karna padega?

1. RSP aur $p_0,q_0$ confirm karo. Standard form $y^{\prime \prime }+\frac{3}{2x}{y}^{\prime }-\frac{1+x}{2x^2}y=0$, toh $xP=\frac{3}{2}\to p_0=\frac{3}{2}$ aur $x^{2}Q=-\frac{1+x}{2}\to q_0=-\frac{1}{2}$. Yeh step kyun? Yeh dono indicial equation aur recurrence dono ko feed karte hain.
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, koi log nahi. Yeh step kyun? Gap batata hai ki dono series independent aur clean hain.
3. Coefficient series $p_k,q_k$ padho. Upar di gayi definitions use karke tamed coefficients expand karo: $p(x)=xP=\frac{3}{2}$, ek constant hai, toh $p_0=\frac{3}{2}$ aur $p_k=0$ for $k\ge 1$. Aur $q(x)=x^{2}Q=-\frac{1+x}{2}=-\frac{1}{2}-\frac{1}{2}x$, toh $q_0=-\frac{1}{2}$ aur $q_1=-\frac{1}{2}$ (higher $q_k=0$). Yeh step kyun? Recurrence ko yeh Taylor entries chahiye; $q_1\ne 0$ exactly woh hai jo $a_n$ ko $a_{n-1}$ se couple karta hai.
4. $r_1=\frac{1}{2}$ ke liye recurrence build karo. Ansatz ko upar re-stated recurrence mein plug karo. Kyunki higher coefficients mein sirf $q_1$ survive karta hai, right side pe sum mein sirf uska $k=1$ term hai ($p_1=0$, aur $q_1=-\frac{1}{2}$), toh $a_n$ sirf $a_{n-1}$ pe depend karta hai. Fractions clear karne ke liye $-2$ se multiply karo, $x^{n+r}$ ka coefficient deta hai $[2(n+r)(n+r-1)+3(n+r)-1]a_n=a_{n-1}.$ Yeh step kyun? Sirf $-xy$ term (jo $q_1$ carry karta hai) neighboring coefficients ko connect karta hai, toh har $a_n$ sirf $a_{n-1}$ pe depend karta hai.
5. $r=\frac{1}{2}$ pe evaluate karo. Bracket $=2(n+\frac{1}{2})(n-\frac{1}{2})+3(n+\frac{1}{2})-1=2n^2+3n$. Toh $a_n=\frac{a_{n-1}}{2n^2+3n}=\frac{a_{n-1}}{n(2n+3)}.$ $a_0=1$ ke saath: $a_1=\frac{1}{1\cdot 5}=\frac{1}{5}$, $a_2=\frac{a_1}{2\cdot 7}=\frac{1}{70}$. Yeh step kyun? Concrete numbers hume machine ke saath verify karne dete hain.

Verify: Bracket $2n^2+3n$ $n\ge 1$ ke liye kabhi zero nahi hota, toh koi division by zero nahi — "no log" ke consistent hai. Numerically $a_1=0.2$, $a_2=1/70\approx 0.014286$. ✔

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Example C — ek integer ke saath reh raha ek fractional root (Cell C)

Forecast: Ek root whole number hai, doosra half hai. $x\to 0^{+}$ par solution ke liye exponent $\frac{1}{2}$ kaisa dikhta hai?

1. Standard form. $x^2$ se divide karo: $P=-\frac{1}{2x}$, $Q=\frac{1-x}{2x^2}$. Yeh step kyun? Limits ke liye bare coefficients chahiye.
2. Tamed limits. $p_0=\lim xP=-\frac{1}{2}$. $q_0=\lim x^{2}Q=\frac{1}{2}$. Yeh step kyun? Yeh indicial inputs hain.
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 $r_1=1,r_2=\frac{1}{2}$ (ek integer, ek half). Gap $=\frac{1}{2}\notin \mathbb{Z}$ → Cell C, clean, do independent series. Yeh step kyun? Confirm karta hai ki fractional exponent genuinely present hai aur non-integer gap kisi logarithm ko rule out karta hai.
4. Chhote exponent $r_2=\frac{1}{2}$ ko interpret karo. Solution $y_2\sim x^{1/2}$ near $0$ finite hai (yeh $\to 0$ jaata hai) lekin infinite slope hai: iska derivative $\sim \frac{1}{2}{x}^{-1/2}\to \infty$. Figure dekho — pale-yellow $x^{1/2}$ curve origin par vertical tangent ke saath utha hai, jabki blue $x^1$ curve (larger root) flat enter karta hai. Yeh step kyun? $r$ nikalna ka poora fayeda geometry ko exponent se padhna hai — number $\frac{1}{2}$ literally ek vertical tangent predict karta hai, toh hum aage solve kiye bina solution picture kar sakte hain.

Verify: $(2r-1)(r-1)=2r^2-3r+1$ match karta hai. Roots $1$ aur $0.5$; dono positive, difference $0.5$. ✔ (Yahan koi root negative nahi hai, toh kuch blow up nahi hota; genuinely negative root — jaise Example B ka $r_2=-1$ — woh hai jo $y\to \infty$ at the origin drive karta.)

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

Pehle hum symbol $J_0$ re-earn karte hain taaki kuch borrowed na ho:

Forecast: Parent ne dikhaya tha $y_1=J_0$. $y_2$ ke liye exact template guess karo.

1. Indicial. $p_0=1,q_0=0$ toh $r(r-1)+r=r^2=0$, double root $r=0$. → Cell D. Yeh step kyun? Repeated root ka matlab hai sirf ek exponent, toh dono series pure powers nahi ho sakti.
2. Log kyun aana chahiye. Ek exponent ke saath, recurrence exactly ek power series produce karta hai (woh sum jise hum ne $J_0$ naam diya). Ek doosra, independent solution same exponent ke aas paas ek aur power series nahi ho sakta — independence fail ho jaati. Reduction of order with $y_2=v(x)y_1$ tab $v^{\prime }=\frac{1}{y_1^2}\exp (-\int P)$ force karta hai; yahan $\int P=\int \frac{1}{x}=\ln x$, aur unwinding ek $\int \frac{dx}{x}=\ln x$ term deta hai. Yeh step kyun? Yeh woh mechanism hai jo parent ne assert kiya tha; hum ise explicit bana rahe hain.
3. Template likho. $y_2=J_0(x)\ln x+{\sum }_{n\ge 1}{c}_n{x}^n.$ Yeh step kyun? Cell D mein hamesha yahi shape hoti hai; $c_n$ re-substituting se aate hain.

Verify: Wronskian shape compute karo: $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$. Kyunki $y_2$ mein $\ln x$ hai, $y_2^{\prime }$ mein $y_1/x$ hai, toh $W\propto 1/x\ne 0$ — genuinely independent. Abel's formula deta hai $W=C\exp (-\int P)=C/x$, match karta hai. ✔

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

Forecast: Roots ek integer se differ karte hain. Kya step $N$ zero se division maangega?

1. Indicial. Yahan $p_0=xP{|}_0=1$, aur $q_0=x^{2}Q{|}_0=(x^2-1){|}_0=-1$. Toh $r(r-1)+r-1=r^2-1=0$, roots $r_1=1,r_2=-1$. Gap $N=2$ → integer, Cell E ya F. Yeh step kyun? Integer gap possible clash ki warning deta hai; sirf recurrence batati hai kaunsa cell hai.
2. Recurrence setup karo aur dekho shift $2$ kyun hai. $y=\sum a_n{x}^{n+r}$ plug karo. Tamed coefficient $q(x)=x^{2}Q=x^2-1$ mein entries hain $q_0=-1$ aur $q_2=+1$ (baaki sab $q_k=0$, aur $p(x)=1$ toh sirf $p_0$ survive karta hai). $-1$ piece $a_n$ ko khud se pair karta hai aur left-hand indicial factor $F(n+r)$ build karta hai; $+x^2$ piece series ko $x^2$ se multiply karta hai, jo har term ko do powers upar shift karta hai, toh $a_{n-2}$ us power $x^{n+r}$ pe land karta hai jahan $a_n$ hai. Toh $x^{m+r}$ ka coefficient match karna deta hai $F(m+r)a_m=-a_{m-2},F(s)=s^2-1.$ Yeh step kyun? "Shift by $m-2$" koi trick nahi hai — yeh exactly isliye hai kyunki sirf surviving higher coefficient $q_2$ hai, $x^2$ se attached; woh ek term hi wajah hai ki do apart ke neighbors couple hote hain aur kuch aur survive nahi karta.
3. $r_2=-1$ pe break watch karo. $r=-1$ ke saath, $F(m-1)=(m-1{)}^2-1=m(m-2)$. $m=N=2$ pe: $F(2-1)=F(1)=1^2-1=0$ — left side vanish ho jaati hai jabki right side $-a_0\ne 0$. Koi finite $a_2$ satisfy nahi kar sakta $0\cdot a_2=-a_0$. Yeh step kyun? Yeh exactly $F(r_2+N)=F(r_1)=0$ hai: recurrence step $N=2$ pe zero se divide karne ki koshish karta hai.
4. Consequence. Pure $x^{-1}$ series complete nahi ho sakti. Rescue log hai: $y_2=\kappa y_1\ln x+x^{-1}{\sum }_{n\ge 0}{c}_n{x}^n,\kappa \ne 0.$ Yeh step kyun? Log term missing piece $x^1$ order pe contribute karta hai, gap fill karta hai.

Neeche figure $F(s)=s^2-1$ plot karta hai taaki tum dekh sako collision: do roots $s=\pm 1$ pe hain, aur step $N=2$ starting from $r_2=-1$ argument $s=m+r_2=2+(-1)=1$ ko exactly doosre root pe land karta hai, jahan $F=0$ hai. Jab bhi koi step $F$ ke zero pe land karta hai, recurrence ruk jaati hai — yeh Cell-E logarithm ka geometric fingerprint hai.

Verify: $F(1)=1-1=0$ breakdown exactly $m=N=2$ pe confirm karta hai, aur $r_1=1$ wahan hai jahan $F$ vanish karta hai — log yahan unavoidable hai ($\kappa \ne 0$). ✔

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

Forecast: Parent ka teesra "mistake" warn karta hai ki integer-gap automatic log nahi hai. Dekho $\kappa$ zero nikle.

1. Standard form. $x$ se divide karo: $y^{\prime \prime }+\frac{2}{x}{y}^{\prime }+y=0$, toh $P=2/x$, $Q=1$. $p_0=\lim x\cdot \frac{2}{x}=2$, $q_0=\lim x^2\cdot 1=0$. Yeh step kyun? 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 ya F. Yeh step kyun? Ek aur integer gap; recurrence test karna zaroori.
3. Recurrence. Yahan $p(x)=xP=2$ (sirf $p_0$) aur $q(x)=x^{2}Q=x^2$ (sirf $q_2=1$), toh phir surviving $+x^2$ terms two apart couple karta hai. $y=\sum a_n{x}^{n+r}$ plug karo; $x^{n+r}$ match karta hai $(n+r)(n+r+1)a_n=-a_{n-2}.$ $r_2=-1$ ke liye: $(n-1)n{a}_n=-a_{n-2}$. Dangerous step $n=N=1$ pe: $(0)(1)a_1=-a_{-1}=0$ (koi $a_{-1}$ nahi hai, toh right side $0$ hai). Toh $0\cdot a_1=0$ kisi bhi $a_1$ se satisfy hota hai — koi contradiction nahi! Yeh step kyun? Right side bhi vanish ho gayi, toh koi forced division by zero nahi hai — yeh exactly Cell-F escape hai.
4. Result aur identification. $a_1=0$ choose karo; $x^{-1}$ series freely complete hoti hai, $\kappa =0$, koi log nahi. Solutions ko naam dene ke liye, notice karo ki ODE exactly $(xy{)}^{\prime \prime }+(xy)=0$ hai disguise mein: $u=xy$ rakho, tab substituting deta hai $u^{\prime \prime }+u=0$, jiske solutions hain $u=\sin x$ aur $u=\cos x$. $u=xy$ undo karo: $y=\frac{\sin x}{x}$ aur $y=\frac{\cos x}{x}$ milte hain — do clean, independent solutions, pehla ek Frobenius series $r_1=0$ ke saath, doosra singular $x^{-1}$ series bina logarithm ke. Yeh step kyun? Substitution $u=xy$ ek legitimate change of variable hai jo equation ko ek aise equation mein turn karta hai jo hum already jaante hain; yeh prove karta hai ki do solutions elementary hain aur $\kappa =0$ confirm karta hai.

Verify: $r(r+1)$ ke roots $0,-1$ hain, gap $1$. $n=1$ pe dono sides zero hain, toh $\kappa =0$. $y=\cos x/x$ ($r=-1$ solution) ko original ODE mein substitute karne par $0$ milta hai, aur koi logarithm nahi hai. ✔

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

Forecast: Compute karne se pehle guess karo — regular hai ya irregular?

1. Standard form. $x^3$ se divide karo: $P=\frac{1}{x^2}$, $Q=\frac{1}{x^3}$. Yeh step kyun? Bare coefficients chahiye.
2. Inhe tame karo. $xP=\frac{1}{x}\to \infty$ as $x\to 0$ — analytic nahi. Pehle se hi fail; $x^{2}Q=\frac{1}{x}\to \infty$ bhi. Yeh step kyun? RSP test require karta hai ki dono $xP$ aur $x^{2}Q$ analytic hon. Ek failure kaafi hai.
3. Conclusion. $x=0$ ek irregular singular point hai. Koi valid indicial equation nahi hai ("$p_0$" limit diverge karta hai). Frobenius guaranteed nahi hai; solutions mein $e^{1/x}$ jaisi essential singularities ho sakti hain. Yeh step kyun? Yeh method ki boundary hai — ise recognize karo aur ruk jao.

Verify: ${\lim }_{x\to 0}xP=\lim 1/x$ diverge karta hai ⇒ koi finite $p_0$ nahi ⇒ Cell G. Parent ka "WHY these exact powers?" explain karta hai: $P$ mein $1/x^2$ ek aisa zyada strong blow-up hai jo koi single $x^r$ balance nahi kar sakta. ✔

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

Forecast: Kaunse mode ko tum expect karte ho ki drum ke centre pe nonzero ho — symmetric ya asymmetric?

1. Indicial equation (general $m$). $p_0=1$, $q_0=(x^2-m^2){|}_0=-m^2$, toh $r(r-1)+r-m^2=r^2-m^2=0$, roots $r=\pm m$. Yeh step kyun? Ek quadratic har mode ko ek saath handle karta hai.
2. Case $m=0$. Roots $0,0$ — Cell D, equal roots. Leading behaviour $R\sim x^0=$ constant. Toh symmetric mode centre pe finite aur nonzero hai (yeh $J_0$ hai, peak at $x=0$). Yeh step kyun? Exponent $0$ literally kehta hai "origin pe finite".
3. Case $m=1$. Roots $+1,-1$, gap $2$ — Cell E. Physically hum regular root $r_1=+1$ rakhte hain: $R\sim x^1\to 0$ at the centre. Doosra root $r_2=-1$ deta hai $R\sim x^{-1}\to \infty$, jo drum ke liye unphysical hai (centre pe infinite displacement) aur discard kar diya jaata hai. Yeh step kyun? Exponent ka sign physically admissible solution select karta hai.
4. Answer. $m=0$: $R\sim \text{const}$, centre pe finite. $m=1$: physical $R\sim x$, centre pe zero; singular partner throw away kar diya jaata hai.

Figure dono admissible modes contrast karta hai. Pale-yellow curve $m=0$ mode ($J_0$) hai: yeh centre pe finite peak se shuru hota hai — woh exponent $r=0$ visible hua. Chalk-blue curve physical $m=1$ mode ($J_1$) hai: yeh zero se linear ramp ke saath shuru hota hai — exponent $r=+1$ visible hua. Discarded $x^{-1}$ partner $x=0$ pe infinity tak shoot karta aur draw nahi kiya gaya, exactly isliye kyunki ek real drum ka centre displacement infinite nahi ho sakta.

Verify: $r^2-m^2$ ke roots $\pm m$ hain: $m=0$ ke liye double root $0$; $m=1$ ke liye roots $\pm 1$ gap $2$. Centre-pe-finite modes $r\ge 0$ ke correspond karte hain. Yahi exact wajah hai ki Bessel functions $J_m$ (regular) drum physics mein appear karte hain jabki $Y_m$ (singular) exclude hote hain. ✔

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Active Recall

Recall Ek nazar mein kaunsa cell?

$p_0,q_0$ compute karo → $r^2+(p_0-1)r+q_0=0$ solve karo → gap $r_1-r_2$ dekho. Gap real integer nahi? ::: Cell B/C — do clean series. Gap $=0$? ::: Cell D — log guaranteed. Gap $=N\in {\mathbb{Z}}^{+}$ aur recurrence step $N$ pe zero se divide karta hai? ::: Cell E — log. Gap $=N$ lekin dono sides step $N$ pe vanish ho jaati hain? ::: Cell F — koi log nahi ($\kappa =0$). $xP$ ya $x^{2}Q$ analytic nahi? ::: Cell G — irregular, method guaranteed nahi.