Frobenius method — regular singular points: Question bank

4.6.18 · D5 · Maths › Ordinary Differential Equations › Frobenius method — regular singular points

Yeh ek rapid-fire self-quiz hai jo Frobenius method ke ideas ko target karta hai, arithmetic ko nahi. Har item ek question ::: answer reveal hai — right side cover karo, pehle ek answer reason ke saath commit karo, phir check karo. Agar tumhara reason galat tha chahe verdict sahi tha, toh use miss count karo.

Shuru karne se pehle, yeh hain saare symbols jinpar hum rely karte hain. Isse ek baar padh lo taaki koi answer surprise na kare:

$P, Q$ = coefficients jab ODE $y^{''} + P y^{'} + Q y = 0$ likhte hain.
$p (x) = x P (x)$ aur $q (x) = x^{2} Q (x)$ = tamed coefficients. Ek regular singular point par yeh analytic hote hain, isliye inke power-series expansions hote hain $p (x) = \sum_{k \geq 0} p_{k} x^{k}$ aur $q (x) = \sum_{k \geq 0} q_{k} x^{k}$ . Yahan $== p_{k}, q_{k} ==$ sirf un expansions ke Taylor coefficients hain — $p_{0}, q_{0}$ constant terms hain, $p_{1}, q_{1}$ the $x^{1}$ coefficients, aur aise hi aage.
$p_{0} = lim_{x \to 0} x P (x)$ aur $q_{0} = lim_{x \to 0} x^{2} Q (x)$ = leading tamed values (the $k = 0$ coefficients).
$r_{1} \geq r_{2}$ = indicial equation $r (r - 1) + p_{0} r + q_{0} = 0$ ke do roots.
$== F (s) ==$ = indicial polynomial $F (s) = s (s - 1) + p_{0} s + q_{0}$ . Yeh wahi exact left-hand bracket hai jo recurrence mein $a_{n}$ ko multiply karta hai: $x^{n + r}$ ka coefficient hai $F (n + r) a_{n} = (stuff from earlier a_{n - k})$ . Toh $F (r) = 0$ hi indicial equation hai, aur $F (n + r) \neq = 0$ hi woh condition hai jo humein $a_{n}$ ke liye divide karke solve karne deti hai.
$== N ==$ = positive integer root difference $N = r_{1} - r_{2}$ jab do roots ek puri number se differ karte hain (Case 3). Recurrence ka "Step $N$ " matlab hai $a_{N}$ ke liye equation.

Pehli figure neeche exponents ki number line par in teen cases ka picture dikhati hai — quiz se pehle ek nazar daal lo.

Figure — Frobenius method — regular singular points

Aur doosri figure convergence radius ka geometric meaning dikhati hai — kyun ek Frobenius series hamesha ke liye nahi jaati.

True or false — justify

Task hamesha yeh hai: decide karo, phir ek breath mein kyun bolo. Sirf "true/false" ka score zero hai.

Every singular point ko Frobenius method se solve kiya ja sakta hai.

False. Sirf regular singular points ki guarantee hai; ek irregular point par singularity itni strong hoti hai ki ek akela

x^{r}

balance nahi kar sakta, isliye method koi valid series produce hi nahi kar sakta.

Agar

P

aur

Q

dono

x_{0}

par analytic hain, toh

x_{0}

ek ordinary point hai aur tumhe Frobenius nahi pakadna chahiye.

True.

P, Q

ki analyticity exactly ordinary-point condition hai; ek ordinary point ke baare mein plain power series kaam karti hai, aur

r = 0

ke saath Frobenius use hi reproduce kar deta.

Frobenius series

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

hamesha everywhere converge karti hai.

False. Yeh kam se kam nearest other singular point tak ki doori tak converge karti hai — jab doosri singularities exist karti hain toh radius of convergence finite hota hai (doosri figure dekho).

Indicial equation ke complex roots ho sakte hain.

True. Yeh

r

mein ek quadratic hai; agar

p_{0}, q_{0}

discriminant ko negative banate hain toh complex-conjugate exponents milte hain, jo oscillatory-plus-power behaviour dete hain (real solutions real/imaginary parts lekar recover hote hain).

Agar do indicial roots equal hain, toh bhi do solutions linearly independent hote hain.

True. Yeh hone hi chahiye — ek 2nd-order ODE ka 2-dimensional solution space hota hai.

y_{2}

mein

ln x

term precisely wohi cheez hai jo do solutions ko independent banati hai (unka Wronskian nonzero hota hai).

Poore ODE ko

x^{2}

se multiply karna uske solutions ko change kar deta hai.

False.

x^{2}

se multiply karna (

x \neq = 0

ke liye) dono sides ko same nonzero factor se rescale karna hai; yeh algebra ko

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

mein rearrange karta hai lekin solution set untouched rehta hai.

Euler–Cauchy equation

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

ke liye, Frobenius series ek term ke baad terminate ho jaati hai.

True. Yahan tamed coefficients constants hain,

p (x) = b

aur

q (x) = c

, isliye

k \geq 1

ke liye har higher Taylor coefficient

p_{k}, q_{k}

zero hai; recurrence phir

n \geq 1

ke liye

a_{n} = 0

force karta hai aur pure powers

x^{r}

milte hain — exactly Euler–Cauchy ka answer.

Bada indicial root

r_{1}

hamesha logarithm-free solution deta hai.

True. Yaad karo

F (s)

woh bracket hai jo

a_{n}

ko multiply karta hai. Bade root ke liye,

r_{1} \geq r_{2}

hone par har

n \geq 1

ke liye

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

hota hai, isliye hum kabhi zero se divide nahi karte; trouble (agar koi ho) sirf chhote root par aata hai.

Spot the error

Har line ek plausible move state karti hai. Flaw dhundho aur name karo.

x = 0

is singular for

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

, aur yeh regular hai."

Galat: yeh irregular hai. Yahan

x^{2} Q = 1/ x

x \to 0

blow up karta hai, isliye

q

analytic nahi hai; sirf

p = x P = 1

pass karta hai. Dono

p

aur

q

analytic hone chahiye.

p_{0}

find karne ke liye, bas

P (0)

evaluate karo."

Galat.

P

khud ek singular point par blow up karta hai, isliye

P (0)

undefined hai. Tumhe tamed limit

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

chahiye, jo evaluate karne se pehle

1/ x

ko multiply karke hata deta hai.

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

ke liye indicial equation

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

hai."

Galat. Indicial equation sirf lowest-order pieces

p_{0}, q_{0}

use karta hai, yaani

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

, na ki poora

q (x) = x^{2}

. Correct:

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

"Roots ek integer se differ karte hain, isliye main immediately likhta hoon

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

with

κ \neq = 0

Galat. Coefficient

κ

zero bhi ho sakta hai. Tumhe pehle

r_{2}

ke liye recurrence try karna hoga; log tabhi forced hai jab koi coefficient genuinely undefined ho jaye.

"Kyunki

a_{0} \neq = 0

required hai, main

a_{1} \neq = 0

bhi set karta hoon."

Galat. Sirf lowest coefficient

a_{0}

ko nonzero hone ki constraint hai (taaki

x^{r}

sach mein leading power ho). Har baad wala

a_{n}

, including

a_{1}

, jo recurrence dictate kare woh hoga — aksar zero hota hai.

r = 1/2

aaya, lekin yeh integer nahi hai, isliye yeh valid power series nahi hai aur Frobenius fail karta hai."

Galat — yahi toh poora point hai. Non-integer

r

precisely isliye hai ki hum

x^{r}

se multiply karte hain: factor

x^{1/2}

singular behaviour capture karta hai jo plain Taylor series kabhi nahi kar sakti. Yeh success hai, failure nahi.

"Maine Frobenius se

y_{1}

find kiya;

y_{2}

ke liye main doosra root plug in karunga aur mujhe guaranteed clean second series milegi."

Cases 2 aur 3 mein galat. Equal roots (Case 2) sirf ek exponent dete hain, aur integer-difference roots (Case 3) clash kar sakte hain; tab tumhe Reduction of order ya log-form ansatz chahiye, doosri plain series nahi.

Why questions

Reason explain karo, sirf fact nahi.

Hum

P

ko ek power of

x

se aur

Q

ko do se kyun tame karte hain?

Kyunki

y^{'}

mein

y^{''}

se ek kam derivative hai:

x^{2} y^{''}

ko balance karne ke liye

P

-term ke saath

x P

finite chahiye, aur

Q

-term ke saath

x^{2} Q

finite chahiye. Powers derivative order se match karti hain taaki har term same lowest power

x^{r}

hit kare.

Lowest power

x^{r}

ke coefficient ko zero set karne se

r

mein quadratic kyun milti hai?

Lowest power entirely

n = 0

term se aati hai.

y = a_{0} x^{r}

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

mein daalne par

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

milta hai;

a_{0} \neq = 0

ke saath woh bracket

F (r)

— ek quadratic — zero hona chahiye.

a_{0} \neq = 0

kyun hona chahiye?

Agar

a_{0} = 0

hota toh series actually

x^{r + 1}

par start hoti, matlab hamara claimed leading exponent

r

galat tha.

a_{0} \neq = 0

force karna

r

ko genuine lowest power par pin karta hai taaki indicial equation meaningful ho.

Logarithm sirf Cases 2 aur 3 mein kyun appear hota hai, Case 1 mein kabhi nahi?

Log tab chahiye jab doosra exponent independent power series supply nahi kar sakta. Case 1 mein roots non-integer se differ karte hain, isliye bracket

F (r_{2} + n)

kabhi zero nahi hota aur recurrence cleanly chalta hai; koi rescue term required nahi.

Log term structurally

y_{1} ln x

kyun hai, kisi aur function ke saath nahi?

Reduction of order

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

set karta hai;

v

ke liye solve karne par

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

leading piece ke roop mein milta hai, isliye log

y_{1}

ke upar hi ride karta hai. Mathematics exactly doosra independent solution manufacture karta hai.

Point ko koi bhi series try karne se pehle classify kyun karein?

Kyunki Frobenius convergence aur poora three-case machinery sirf ek regular singular point par guaranteed hai. Check skip karna method ko wahan apply karne ka risk hai jahan uski koi warranty nahi (irregular points), ek aisi series par effort waste karna jo converge hi nahi kar sakti.

Equal-root case mein kabhi do power series kyun nahi milti?

Do distinct power-series solutions ko do distinct leading exponents chahiye. Equal roots sirf ek exponent

r

supply karte hain, isliye doosra independent solution qualitatively differ karne par majboor hai — isliye

ln x

aata hai.

Edge cases

Woh scenarios jo log exam tak bhool jaate hain.

Kya hoga agar

q_{0} = 0

ho (jaise Bessel order 0 mein)?

Indicial equation ban jaati hai

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

, jo

r = 0

aur

r = 1 - p_{0}

deta hai. Zero root bilkul legal hai; yahan yeh double root bhi tha, jo $Y_0$ mein dikhne wala log force karta hai.

Kya hoga agar

p_{0} = 0

AUR

q_{0} = 0

dono ho?

Indicial equation hai

r (r - 1) = 0

, roots

0

aur

1

— integer

N = 1

se differ karte hain (Case 3). Yeh borderline hai jahan point technically singular hai lekin aksar almost ordinary point jaisa behave karta hai; log survive kare ya na kare.

Legendre's equation ke liye

x = 0

par kya hota hai?

x = 0

ek ordinary point hai (iske singularities

x = \pm 1

par hain), isliye wahan plain power series use karo — Frobenius unnecessary hai. Frobenius wahan

x = \pm 1

par tool hai.

Agar indicial quadratic ka discriminant zero ho, toh hum kis case mein hain?

Case 2 (equal roots

r_{1} = r_{2}

, toh

N = 0

). Ek single exponent matlab doosra solution zaroor

ln x

carry karega; log sirf discriminant se predict kar sakte ho, kisi recurrence se pehle.

Agar do indicial roots integer

N

se differ karein lekin step

N

par recurrence ka right-hand side zero ho?

Tab koi division-by-zero crisis nahi aati;

a_{N}

ke liye equation use free chhod deta hai (aksar

0

choose kiya jaata hai),

κ = 0

, aur do clean Frobenius series milti hain bina logarithm ke — yeh "steel-manned" Case-3-without-log outcome hai.

Agar chhota root

x^{r_{2}}

deta hai jahan

r_{2}

negative aur non-integer ho?

Solution

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

x = 0

par blow up karta hai lekin

x > 0

par ek genuine, independent solution hai. Ise rakhna hai ya nahi yeh physical boundedness requirements par depend karta hai, method ke kisi failure par nahi.

Kya ek second-order ODE kabhi teen independent Frobenius solutions produce kar sakti hai?

Nahi. Ek 2nd-order linear ODE ka solution space exactly 2-dimensional hota hai; zyada se zyada do independent solutions exist karti hain, aur unki independence nonzero Wronskian se confirm hoti hai.

Active Recall

Recall Sabse fast sanity checks

Tamed limits before roots: == $p_{0} = lim x P$ , $q_{0} = lim x^{2} Q$ ==, phir quadratic.
$F (s) = s (s - 1) + p_{0} s + q_{0}$ recurrence bracket hai; == $F (n + r) = 0$ woh hai jo log trigger karta hai==.
Discriminant $= 0$ ⇒ ==equal roots ( $N = 0$ ) ⇒ guaranteed log==.
Integer difference $N \geq 1$ ⇒ maybe log; step $N$ par recurrence test karo, assume mat karo.
Non-integer difference ⇒ do clean series, kabhi log nahi.