Frobenius method — regular singular points

4.6.18 · Maths › Ordinary Differential Equations

Ek tarika hai $y^{''} + P (x) y^{'} + Q (x) y = 0$ ko solve karne ka, uss point ke paas jahan simple power-series method fail ho jaati hai.

Hume yeh kyun chahiye?

Toh KYA hai: hum $y = x^{r} \sum_{n \geq 0} a_{n} x^{n}$ dhundh rahe hain jahan $r$ determine karna hai, integer assume nahi karna.

Singular point ko classify karna (the gatekeeper)

ODE ko standard form mein likho: $y^{''} + P (x) y^{'} + Q (x) y = 0.$

YEH EXACT POWERS KYUN? Kyunki agar $P$ se zyada $1/ (x - x_{0})$ nahi badha aur $Q$ se zyada $1/ (x - x_{0})^{2}$ nahi badha, toh $x^{r}$ se multiply karna equation mein worst singularity ko exactly balance kar deta hai. Zyada strong blow-ups ko ek single power se balance nahi kiya ja sakta — isliye irregular points method ko tod dete hain.

Figure — Frobenius method — regular singular points

Indicial equation SCRATCH se derive karna

$x_{0} = 0$ lo aur RSP equation ko $x^{2}$ se multiply karke sabse clean form mein likho: $x^{2} y^{''} + x p (x) y^{'} + q (x) y = 0,$ jahan $p (x) = \sum_{k \geq 0} p_{k} x^{k}$ aur $q (x) = \sum_{k \geq 0} q_{k} x^{k}$ analytic hain.

Step 1 — Frobenius ansatz assume karo. $y = \sum_{n = 0}^{\infty} a_{n} x^{n + r}, a_{0} \neq = 0.$ $a_{0} \neq = 0$ kyun? Hum demand karte hain ki present sabse chhoti power $x^{r}$ ho; agar $a_{0} = 0$ ho toh hum $r$ ko relabel kar dete hain, isliye insist karo ki yeh nonzero ho.

Step 2 — Term by term differentiate karo. $y^{'} = \sum_{n} a_{n} (n + r) x^{n + r - 1}, y^{''} = \sum_{n} a_{n} (n + r) (n + r - 1) x^{n + r - 2} .$

Step 3 — Plug in karo aur SIRF lowest power $x^{r}$ dekho (yaani $n = 0$ , $p \to p_{0}$ , $q \to q_{0}$ ): $x^{2} y^{''} \to a_{0} r (r - 1) x^{r}, x p y^{'} \to a_{0} p_{0} r x^{r}, q y \to a_{0} q_{0} x^{r} .$ Jodh ke aur $x^{r}$ ka coefficient zero set karke (with $a_{0} \neq = 0$ ):

Step 4 — Baaki ke liye Recurrence. $x^{n + r}$ ka coefficient zero set karne se milta hai: $[(n + r) (n + r - 1) + p_{0} (n + r) + q_{0}] a_{n} = - \sum_{k = 1}^{n} [p_{k} (n - k + r) + q_{k}] a_{n - k} .$ Left side ka bracket $F (n + r)$ hai jahan $F (s) = s (s - 1) + p_{0} s + q_{0}$ indicial polynomial hai. Toh $a_{n}$ solve ho sakta hai jab tak $F (n + r) \neq = 0$ .

Teen cases (WHY second solution ka form badalta hai)

$r_{1} - r_{2} =$ root difference maano.

Cases 2 & 3 mein log kyun? Jab $F (n + r_{2}) = 0$ kisi $n$ par ho, toh recurrence zero se division demand karta hai. Reduction of order $y_{2} = v (x) y_{1}$ phir ek $\int d x / x = ln x$ term produce karta hai. Log mathematics ka tarika hai ek second independent solution supply karne ka jab power series mein room khatam ho gaya.

Worked Example 1 — $p_{0}, q_{0}$ nikalo aur indicial equation solve karo

ODE: $2 x^{2} y^{''} + 3 x y^{'} - (1 + x) y = 0.$

Step 1: Standard form. $2 x^{2}$ se divide karo: $y^{''} + \frac{3}{2 x} y^{'} - \frac{1 + x}{2 x ^{2}} y = 0.$ Kyun? Hume $P, Q$ chahiye singularity read off karne ke liye.

Step 2: RSP check karo. $x P = \frac{3}{2}$ (analytic), $x^{2} Q = - \frac{1 + x}{2}$ (analytic). ✔ Toh $x = 0$ ek regular singular point hai. Yeh step kyun? Frobenius sirf RSP par guaranteed hai.

Step 3: $p_{0}, q_{0}$ padho. $p_{0} = \frac{3}{2}$ , $q_{0} = x^{2} Q ∣_{x = 0} = - \frac{1}{2}$ .

Step 4: Indicial equation. $r (r - 1) + \frac{3}{2} r - \frac{1}{2} = r^{2} + \frac{1}{2} r - \frac{1}{2} = 0 \Rightarrow 2 r^{2} + r - 1 = 0 \Rightarrow (2 r - 1) (r + 1) = 0.$ Roots $r_{1} = \frac{1}{2}, r_{2} = - 1$ . Difference $= \frac{3}{2} \in / Z$ → Case 1, do clean Frobenius series. Yeh kyun matter karta hai? Hum instantly jaante hain ki koi logarithm nahi aayega.

Worked Example 2 — full recurrence (Bessel-type, equal roots)

ODE: $x^{2} y^{''} + x y^{'} + x^{2} y = 0$ (Bessel order 0).

Step 1: $P = 1/ x, Q = 1$ . $x P = 1$ , $x^{2} Q = x^{2}$ , dono analytic → RSP. $p_{0} = 1, q_{0} = 0$ .

Step 2: Indicial. $r (r - 1) + r + 0 = r^{2} = 0 \Rightarrow r_{1} = r_{2} = 0$ (Case 2 — $y_{2}$ mein log expect karo). Yeh step kyun? Equal roots kuch bhi compute karne se pehle structure predict kar dete hain.

Step 3: $y = \sum a_{n} x^{n}$ plug karo (kyunki $r = 0$ ). Tab $\sum a_{n} n (n - 1) x^{n} + \sum a_{n} n x^{n} + \sum a_{n} x^{n + 2} = 0.$ Pehle do combine karo: $\sum a_{n} n^{2} x^{n} + \sum a_{n} x^{n + 2} = 0$ .

Step 4: Powers $x^{m}$ match karo. $a_{m} m^{2} + a_{m - 2} = 0 \Rightarrow a_{m} = - \frac{a _{m - 2}}{m ^{2}} .$ Yeh step kyun? Index $n \to m - 2$ shift karne se powers align ho jaati hain taaki hum coefficients equate kar sakein.

Step 5: Iterate karo. $a_{1} = 0$ (kyunki $1^{2} a_{1} = 0$ ), toh saare odd terms vanish ho jaate hain. $a_{2} = - \frac{a _{0}}{4}, a_{4} = - \frac{a _{2}}{16} = \frac{a _{0}}{64}, \dots, a_{2 k} = \frac{( - 1 ) ^{k} a _{0}}{2 ^{2 k} ( k ! ) ^{2}} .$ $a_{0} = 1$ ke saath yeh hai $J_{0} (x) = \sum_{k} \frac{( - 1 ) ^{k}}{( k ! ) ^{2}} (x /2)^{2 k}$ — pehla Bessel function. Doosra solution $Y_{0}$ predicted $ln x$ carry karta hai.

Common mistakes (Steel-manned)

Active Recall

Recall Quick self-test

$x_{0}$ ko RSP hone ke liye kaun se do functions analytic hone chahiye?
Indicial equation likho aur $p_{0}, q_{0}$ define karo.
Kaun sa case log ko certainty ke saath force karta hai?
Example 1 mein log kyun nahi aaya?

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho ek jhule ki chain uske upar ke pivot ( $x = 0$ ) par ghisi hui hai. Ek normal smooth wiggle-formula ek ghisi hui jagah par motion describe nahi kar sakta. Toh hum cheat karte hain: hum kehte hain "motion $x^{r}$ times ek smooth wiggle jaisi lagti hai," aur hum equation ko khud magic exponent $r$ batane dete hain ek chhoti quadratic solve karke (indicial equation). Kabhi kabhi dono magic numbers achhe se kaam karte hain; kabhi kabhi woh same hote hain ya ek poore step se differ karte hain, aur phir ek " $ln x$ " ghus aata hai taaki hume genuinely alag second motion mile.

Flashcards

x_{0}

ko regular singular point kya banata hai?

p (x) = (x - x_{0}) P

aur

q (x) = (x - x_{0})^{2} Q

dono

x_{0}

par analytic hain.

x = 0

ke baare mein Frobenius ansatz?

y = \sum_{n \geq 0} a_{n} x^{n + r}

jahan

a_{0} \neq = 0

Indicial equation?

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

, jahan

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

q_{0} = lim_{x \to 0} x^{2} Q

a_{0} \neq = 0

kyun hona chahiye?

Taaki

x^{r}

genuinely lowest power ho; warna hum

r

redefine kar dete.

Equal indicial roots wala case?

y_{2} = y_{1} ln x + x^{r} \sum_{n \geq 1} c_{n} x^{n}

— log forced hai.

Roots positive integer

N

se differ karte hain — general

y_{2}

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

κ

possibly 0.

Log kyun appear ho sakta hai?

Recurrence

F (n + r) = 0

hit karta hai (zero se division); reduction of order

\int d x / x = ln x

yield karta hai.

Distinct roots ke saath log kab nahi aata?

Jab

r_{1} - r_{2} \in / Z

(Case 1).

Irregular singular point ka Frobenius ke liye kya matlab hai?

Method guaranteed nahi; standard series poori tarah fail ho sakti hai.

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

ke indicial roots?

r = 0

(double) —

J_{0}

aur

ln

-carrying

Y_{0}

deta hai.

Connections

Power series solutions — ordinary points
Bessel's equation and Bessel functions
Legendre's equation and polynomials
Reduction of order
Wronskian and linear independence
Analytic functions and radius of convergence
Euler–Cauchy equation (indicial equation literally iska characteristic equation hai)