Power series solutions — ordinary points: Exercises

4.6.17 · D4 · Maths › Ordinary Differential Equations › Power series solutions — ordinary points

Shuru karne se pehle, ek common picture — "power series solution" ka matlab kya hota hai — ek curve jo infinite polynomial se, coefficient by coefficient, rebuild hoti hai:

Figure — Power series solutions — ordinary points

Level 1 — Recognition

Exercise 1.1

Har ODE ko standard form $y^{''} + P (x) y^{'} + Q (x) y = 0$ mein likho aur batao ki $x_{0} = 0$ ordinary point hai ya singular point.

(a) $y^{''} + x y^{'} + y = 0$ (b) $(1 - x^{2}) y^{''} - 2 x y^{'} + 12 y = 0$ (c) $x y^{''} + y^{'} + x y = 0$

Recall Solution 1.1

KYA karna hai: jo bhi $y^{''}$ ko multiply kar raha hai ussse divide karo, phir $P$ aur $Q$ padho aur pucho — "kya dono ke convergent power series hain (kya woh analytic hain) $x = 0$ par?" Polynomials ka ratio wahan analytic hota hai jahan denominator non-zero ho. Bas yehi test hai.

(a) Pehle se hi standard form mein hai: $P (x) = x$ , $Q (x) = 1$ . Dono polynomials hain — har jagah analytic. To $x_{0} = 0$ ek ordinary point hai. ✔ (Yeh Hermite-type shape hai.)

(b) $(1 - x^{2})$ se divide karo: $P = \frac{- 2 x}{1 - x ^{2}}, Q = \frac{12}{1 - x ^{2}} .$ Denominator $1 - x^{2} = 0$ sirf $x = \pm 1$ par. $x = 0$ par denominator $1 \neq = 0$ hai, to dono wahan analytic hain. $x_{0} = 0$ ek ordinary point hai; $x = \pm 1$ singular hain.

(c) $x$ se divide karo: $P = \frac{1}{x}, Q = 1.$ $P$ $x = 0$ par blow up karta hai (zero se division). To $x_{0} = 0$ ek singular point hai — ordinary point wala power series method apply nahi hoga; tumhe Singular points and Frobenius method chahiye hoga.

Level 2 — Application

Exercise 2.1

$y^{''} - x y = 0$ (Airy equation) ka recurrence relation $x_{0} = 0$ ke baare mein nikalo, aur $a_{0}, a_{1}$ ke terms mein $a_{2}, a_{3}, a_{4}, a_{5}$ compute karo.

Recall Solution 2.1

Step 1 (assume karo). $y = n = 0 \sum \infty a_{n} x^{n}$ , to $y^{''} = n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2}$ .

Step 2 (substitute karo). Term $- x y$ har $x^{n}$ ko $x$ se multiply karta hai, jo $x^{n + 1}$ deta hai: $\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - \sum_{n = 0}^{\infty} a_{n} x^{n + 1} = 0.$

Step 3 (common power $x^{k}$ ke liye re-index karo). Kyun? Hum sirf same power ke coefficients add kar sakte hain. Pehle sum mein $k = n - 2$ lo (to $n = k + 2$ , aur $n = 2 \Rightarrow k = 0$ ). Doosre mein $k = n + 1$ lo (to $n = k - 1$ , aur $n = 0 \Rightarrow k = 1$ ): $\sum_{k = 0}^{\infty} (k + 2) (k + 1) a_{k + 2} x^{k} - \sum_{k = 1}^{\infty} a_{k - 1} x^{k} = 0.$

Step 4 ( $k = 0$ wale odd-man-out ko handle karo). Doosra sum $k = 1$ par start hota hai, to $k = 0$ sirf pehle mein aata hai: $k = 0 : (2) (1) a_{2} = 0 \Rightarrow a_{2} = 0 .$ $k \geq 1$ ke liye, $x^{k}$ ka coefficient zero karo: $(k + 2) (k + 1) a_{k + 2} - a_{k - 1} = 0 \Rightarrow a_{k + 2} = \frac{a _{k - 1}}{( k + 2 ) ( k + 1 )}, k \geq 1.$

Step 5 (calculate karo).

$k = 1 : a_{3} = \frac{a _{0}}{3 \cdot 2} = \frac{a _{0}}{6}$
$k = 2 : a_{4} = \frac{a _{1}}{4 \cdot 3} = \frac{a _{1}}{12}$
$k = 3 : a_{5} = \frac{a _{2}}{5 \cdot 4} = 0 (kyunki a_{2} = 0)$

To $a_{2} = 0, a_{3} = \frac{a _{0}}{6}, a_{4} = \frac{a _{1}}{12}, a_{5} = 0$ . Aage ka pattern: $a_{2}$ se shuru hoke har teesra coefficient vanish hota hai ( $a_{2}, a_{5}, a_{8}, \dots = 0$ ), aur solution ek $a_{0}$ -branch aur ek $a_{1}$ -branch mein split hoti hai, exactly jaisa Fuchs promise karta hai.

Level 3 — Analysis

Exercise 3.1

Hermite-type equation $y^{''} - 2 x y^{'} + 4 y = 0$ ko $x_{0} = 0$ ke baare mein solve karo. Dikhao ki do independent solutions mein se ek polynomial hai, aur use identify karo.

Recall Solution 3.1

Substitute karo $y = \sum a_{n} x^{n}$ . Yahan $y^{'} = \sum n a_{n} x^{n - 1}$ hai, to $- 2 x y^{'} = - 2 \sum n a_{n} x^{n}$ (woh $x$ aur $x^{n - 1}$ milke $x^{n}$ de deta hai): $\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - 2 \sum_{n = 1}^{\infty} n a_{n} x^{n} + 4 \sum_{n = 0}^{\infty} a_{n} x^{n} = 0.$

Sirf pehle sum ko re-index karo ( $k = n - 2$ ); baaki dono mein pehle se $x^{k}$ hai (rename $n \to k$ , aur note karo ki $- 2 n a_{n}$ at $n = 0$ zero hai to sab $k = 0$ se start kar sakte hain): $\sum_{k = 0}^{\infty} [(k + 2) (k + 1) a_{k + 2} - 2 k a_{k} + 4 a_{k}] x^{k} = 0.$

$a_{k}$ terms combine karo (woh $x^{k}$ share karte hain): $- 2 k + 4 = - 2 (k - 2)$ . Coefficient zero karo: $(k + 2) (k + 1) a_{k + 2} = 2 (k - 2) a_{k} \Rightarrow a_{k + 2} = \frac{2 ( k - 2 )}{( k + 2 ) ( k + 1 )} a_{k} .$

Key observation: numerator $2 (k - 2)$ $k = 2$ par zero hai. To:

$a_{2} = \frac{2 ( 0 - 2 )}{2 \cdot 1} a_{0} = \frac{- 4}{2} a_{0} = - 2 a_{0}$ .
$a_{4} = \frac{2 ( 2 - 2 )}{4 \cdot 3} a_{2} = 0$ , aur uske baad ke har even term $0$ hai.

Even branch terminate ho jaati hai: $y_{even} = a_{0} (1 - 2 x^{2}) .$ Yeh (scaling tak) Hermite polynomial $H_{2} (x) = 4 x^{2} - 2$ hai: indeed $1 - 2 x^{2} = - \frac{1}{2} H_{2} (x)$ . Dekho Hermite and Airy equations.

Odd branch terminate nahi hoti (numerator $2 (k - 2)$ odd $k \geq 1$ ke liye kabhi $0$ nahi aata):

$k = 1 : a_{3} = \frac{2 ( - 1 )}{3 \cdot 2} a_{1} = - \frac{1}{3} a_{1}$
$k = 3 : a_{5} = \frac{2 ( 1 )}{5 \cdot 4} a_{3} = \frac{2}{20} a_{3} = \frac{1}{10} (- \frac{1}{3}) a_{1} = - \frac{1}{30} a_{1}$

$y = a_{0} (1 - 2 x^{2}) + a_{1} (x - \frac{1}{3} x^{3} - \frac{1}{30} x^{5} - \dots) .$

Level 4 — Synthesis

Exercise 4.1

Initial value problem solve karo $y^{''} + x y^{'} + y = 0, y (0) = 1, y^{'} (0) = 0,$ $x_{0} = 0$ ke baare mein. Recurrence nikalo, initial conditions se $a_{0}, a_{1}$ pin karo, aur solution ke pehle chaar non-zero terms likho.

Recall Solution 4.1

Kyun initial data constants pin karta hai: power series $y = a_{0} + a_{1} x + a_{2} x^{2} + \dots$ mein, $x = 0$ plug karne par $y (0) = a_{0}$ milta hai, aur differentiate karke $x = 0$ set karne par $y^{'} (0) = a_{1}$ milta hai. To do free constants wahi hain jo do pieces of initial data hain: $a_{0} = y (0) = 1, a_{1} = y^{'} (0) = 0.$

Recurrence. $- 2 x$ ki jagah $+ x$ hai: $x y^{'} = \sum n a_{n} x^{n}$ , to $\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} + \sum_{n = 1}^{\infty} n a_{n} x^{n} + \sum_{n = 0}^{\infty} a_{n} x^{n} = 0.$ Pehle ko re-index karo ( $k = n - 2$ ), baaki rename karo ( $n \to k$ , $n = 1, 0$ wale starts absorb ho jaate hain kyunki $k a_{k} + a_{k}$ at $k = 0$ bas $a_{0}$ hai): $\sum_{k = 0}^{\infty} [(k + 2) (k + 1) a_{k + 2} + k a_{k} + a_{k}] x^{k} = 0.$ Combine karo: $k a_{k} + a_{k} = (k + 1) a_{k}$ , to $(k + 2) (k + 1) a_{k + 2} = - (k + 1) a_{k} \Rightarrow a_{k + 2} = - \frac{a _{k}}{k + 2} .$ (Ek factor $(k + 1)$ cancel ho gaya — ek choti si gift.)

$a_{0} = 1, a_{1} = 0$ se calculate karo. Kyunki $a_{1} = 0$ hai, har odd coefficient $0$ hai (har odd term $a_{1}$ tak trace back karta hai).

$a_{2} = - \frac{a _{0}}{2} = - \frac{1}{2}$
$a_{4} = - \frac{a _{2}}{4} = - \frac{1}{4} (- \frac{1}{2}) = \frac{1}{8}$
$a_{6} = - \frac{a _{4}}{6} = - \frac{1}{6} \cdot \frac{1}{8} = - \frac{1}{48}$

Pehle chaar non-zero terms: $y = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{8} - \frac{x ^{6}}{48} + \dots$

Bonus recognition: pattern $a_{2 m} = \frac{( - 1 ) ^{m}}{2 ^{m} m !}$ se $y = \sum_{m = 0}^{\infty} \frac{( - 1 ) ^{m}}{2 ^{m} m !} x^{2 m} = \sum_{m = 0}^{\infty} \frac{( - x ^{2} /2 ) ^{m}}{m !} = e^{- x^{2} /2} .$ Check: $\frac{1}{2 ^{1} 1 !} = \frac{1}{2}$ ✔, $\frac{1}{2 ^{2} 2 !} = \frac{1}{8}$ ✔, $\frac{1}{2 ^{3} 3 !} = \frac{1}{48}$ ✔. Ek closed form nikal aya — acha reward hai.

Level 5 — Mastery

Exercise 5.1

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

(a) Dikhao ki $x_{0} = 0$ ek ordinary point hai aur Fuchs' theorem se series solution ke radius of convergence ka guaranteed lower bound batao. (b) $x_{0} = 0$ ke baare mein recurrence derive karo. (c) Dikhao ki ek solution exact polynomial $y_{1} = x$ hai, aur $a_{0} = 1$ se shuru hone wale doosre (even) solution ke liye $a_{2}, a_{4}, a_{6}$ nikalo.

Recall Solution 5.1

(a) Ordinary point + radius. Standard form: $x^{2} + 1$ se divide karo, $P = \frac{x}{x ^{2} + 1}, Q = \frac{- 1}{x ^{2} + 1} .$ Denominator $x^{2} + 1 = 0$ sirf $x = \pm i$ par (complex mein). $x = 0$ par yeh $1 \neq = 0$ hai, to dono $P, Q$ analytic hain — $x_{0} = 0$ ek ordinary point hai.

Fuchs' theorem se, radius of convergence kam se kam $x_{0} = 0$ se nearest singularity ki distance ke barabar hai. Singularities $x = \pm i$ par hain, har ek ki distance $∣ \pm i - 0∣ = 1$ hai. To $R \geq 1.$ Kyun imaginary singularities count karti hain: analyticity aur radius of convergence complex plane mein rehte hain; ek real series phir bhi nearest complex singularity ko "feel" karti hai. Dekho Taylor series and analyticity.

(b) Recurrence. $y = \sum a_{n} x^{n}$ substitute karo. Pieces: $x^{2} y^{''} \sum n (n - 1) a_{n} x^{n} + y^{''} \sum n (n - 1) a_{n} x^{n - 2} + x y^{'} \sum n a_{n} x^{n} - y \sum a_{n} x^{n} = 0.$ Sirf $y^{''}$ (bare) sum ko re-index karna hai ( $k = n - 2$ ); baaki mein pehle se $x^{k}$ hai: $\sum_{k} [(k + 2) (k + 1) a_{k + 2} + k (k - 1) a_{k} + k a_{k} - a_{k}] x^{k} = 0.$ $a_{k}$ terms combine karo: $k (k - 1) + k - 1 = k^{2} - k + k - 1 = k^{2} - 1 = (k - 1) (k + 1)$ . To $(k + 2) (k + 1) a_{k + 2} = - (k - 1) (k + 1) a_{k} \Rightarrow a_{k + 2} = - \frac{( k - 1 )}{( k + 2 )} a_{k} .$ (Common factor $(k + 1)$ cancel ho gaya.)

(c) Polynomial solution. Numerator $- (k - 1)$ $k = 1$ par zero hai. Odd branch $a_{1}$ se shuru karte hain: $a_{3} = - \frac{( 1 - 1 )}{3} a_{1} = 0,$ to saare higher odd terms vanish ho jaate hain. Poori odd branch bas $y_{1} = a_{1} x$ hai. $a_{1} = 1$ lete hain to exact solution $y_{1} = x$ milti hai. (Quick sanity check: $y_{1} = x \Rightarrow y_{1}^{'} = 1, y_{1}^{''} = 0$ , aur $(x^{2} + 1) (0) + x (1) - x = 0$ ✔.)

Even branch $a_{0} = 1$ ke saath:

$k = 0 : a_{2} = - \frac{( 0 - 1 )}{2} a_{0} = - \frac{- 1}{2} = \frac{1}{2}$
$k = 2 : a_{4} = - \frac{( 2 - 1 )}{4} a_{2} = - \frac{1}{4} \cdot \frac{1}{2} = - \frac{1}{8}$
$k = 4 : a_{6} = - \frac{( 4 - 1 )}{6} a_{4} = - \frac{3}{6} (- \frac{1}{8}) = \frac{1}{16}$

To doosri solution shuru hoti hai $y_{2} = 1 + \frac{x ^{2}}{2} - \frac{x ^{4}}{8} + \frac{x ^{6}}{16} - \dots (a_{2}, a_{4}, a_{6}) = (\frac{1}{2}, - \frac{1}{8}, \frac{1}{16}) .$ (Actually $y_{2} = 1 + x^{2}$ hai — uski binomial series term for term match karti hai, aur uska radius of convergence exactly $1$ hai, jo Fuchs bound ko equality ke saath meet karta hai.)

Wrap-up recall

Recall Ek-line takeaways

Ordinary point test mein kaunse coefficients use hote hain? ::: $P, Q$ standard form $y^{''} + P y^{'} + Q y = 0$ se — dono $x_{0}$ par analytic hone chahiye. Initial conditions $a_{0}, a_{1}$ kaise fix karte hain? ::: $a_{0} = y (x_{0})$ , $a_{1} = y^{'} (x_{0})$ . Radius of convergence ko kya cap karta hai? ::: Nearest singular point ki distance, complex plane mein measure hoti hai (Fuchs). Series branch polynomial mein kab terminate hoti hai? ::: Jab recurrence ka numerator kisi index par zero ho jaata hai, us coefficient aur branch ke saare baad waale coefficients ko zero kar deta hai.