Legendre's equation and Legendre polynomials (intro)

4.6.20 · Maths › Ordinary Differential Equations

Legendre's equation HAI KYA?

$x = cos θ$ KYU aata hai? Spherical coordinates mein angular equation mein $θ$ variable hota hai; $x = cos θ$ substitute karne par woh upar wali form mein aa jaata hai, jahan $x \in [- 1, 1]$ . Endpoints $x = \pm 1$ sphere ke poles hain — aur woh ODE ke singular points hain.

Isse KAISE derive karein (pehle self-adjoint form)

Solve KAISE karein: power-series (Frobenius) method

Series KYU? $x = 0$ ek ordinary point hai ( $y^{''}$ ka coefficient $1 - x^{2}$ wahan nonzero hai), toh ek plain Taylor series $y = \sum a_{m} x^{m}$ converge karti hai aur kaam karti hai.

Assume karo $y = \sum_{m = 0}^{\infty} a_{m} x^{m} .$

Toh $y^{'} = \sum m a_{m} x^{m - 1}$ aur $y^{''} = \sum m (m - 1) a_{m} x^{m - 2}$ . Plug in karo:

$(1 - x^{2}) \sum m (m - 1) a_{m} x^{m - 2} - 2 x \sum m a_{m} x^{m - 1} + n (n + 1) \sum a_{m} x^{m} = 0.$

Yeh step kyun? Hum har sum ko same power $x^{m}$ par re-index karte hain:

$\sum m (m - 1) a_{m} x^{m - 2} \to \sum (m + 2) (m + 1) a_{m + 2} x^{m}$ (shift $m \to m + 2$ ).
Baaki already $x^{m}$ carry kar rahe hain.

$x^{m}$ ka coefficient collect karo aur $0$ set karo: $(m + 2) (m + 1) a_{m + 2} - m (m - 1) a_{m} - 2 m a_{m} + n (n + 1) a_{m} = 0.$

$a_{m}$ terms group karo: $- m (m - 1) - 2 m + n (n + 1) = n (n + 1) - m (m + 1)$ . Toh

Pehle kuch Legendre polynomials banao

Hum $P_{n} (1) = 1$ convention se normalize karte hain.

Figure — Legendre's equation and Legendre polynomials (intro)

Do shortcut formulas (derive kiye, dump nahi kiye)

Quick parity aur value facts

$P_{n} (- x) = (- 1)^{n} P_{n} (x)$ — even $n$ even function, odd $n$ odd.
$P_{n} (1) = 1$ , $P_{n} (- 1) = (- 1)^{n}$ .
$P_{n}$ ki degree exactly $n$ hai; leading coefficient $\frac{( 2 n )!}{2 ^{n} ( n ! ) ^{2}}$ .

Common mistake Common errors ko steel-man karo

Mistake 1: "Legendre's equation ke dono solutions polynomials hain." Kyun sahi lagta hai: tumne ek polynomial $P_{n}$ dhundha, toh symmetry se lagta hai doosra bhi hoga. Sach yeh hai: yeh 2nd-order ODE hai, toh ek doosra, independent solution $Q_{n} (x)$ hota hai — Legendre function of the second kind — jo woh series hai jo terminate nahi karti aur $x = \pm 1$ par logarithmically blow up ho jaati hai. Hum ise discard karte hain kyunki yeh poles par unbounded hai.

Mistake 2: $x = 0$ ordinary hai yeh dekhke $x = \pm 1$ ke singular hone ko bhool jaana. Fix: equation interior mein theek hai lekin $(1 - x^{2})$ $\pm 1$ par vanish hota hai, jo unhe regular singular points banata hai. Poles par bounded hona physical selection rule hai jo force karta hai $n \in {0, 1, 2, \dots}$ .

Mistake 3: Recurrence numerator ko $m (m - 1) - n (n + 1)$ padhna. Kyun sahi lagta hai: $y^{''}$ term akela $m (m - 1)$ deta hai. Fix: tumhe $- 2 x y^{'}$ term se $- 2 m$ bhi add karna hai, jo $m (m + 1) - n (n + 1)$ deta hai. Isse bhool jaana termination tod deta hai.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho tum ek beach ball par temperature ko sirf smooth "shape patterns" use karke describe kar rahe ho. Ek flat pattern hai (har jagah same), ek top-vs-bottom pattern hai, ek "ring" pattern hai, aur aise hi aur. Legendre polynomials exactly woh building-block patterns hain. Inhe dhundhne ka rule ek clean equation hai, aur cool trick yeh hai: sirf woh patterns allowed hain jo ball ke dono poles par crazy nahi jaate. Inhe sahi matra mein add karo aur tum ball par koi bhi smooth temperature describe kar sakte ho.

Flashcards

Legendre's equation (standard form)

(1 - x^{2}) y^{''} - 2 x y^{'} + n (n + 1) y = 0

Legendre's equation self-adjoint form mein

\frac{d}{d x} [(1 - x^{2}) y^{'}] + n (n + 1) y = 0

Series coefficients ke liye recurrence relation

a_{m + 2} = \frac{m ( m + 1 ) - n ( n + 1 )}{( m + 2 ) ( m + 1 )} a_{m}

Integer

n

ke liye series terminate kyun hoti hai?

m = n

par numerator

m (m + 1) - n (n + 1) = 0

ho jaata hai, toh

a_{n + 2} = 0

aur saare aage ke same-parity coefficients khatam, degree-

n

polynomial milti hai.

P_{0}, P_{1}, P_{2}, P_{3}

1, x, \frac{1}{2} (3 x^{2} - 1), \frac{1}{2} (5 x^{3} - 3 x)

Rodrigues' formula

P_{n} (x) = \frac{1}{2 ^{n} n !} \frac{d ^{n}}{d x ^{n}} (x^{2} - 1)^{n}

Orthogonality relation

\int_{- 1}^{1} P_{m} P_{n} d x = \frac{2}{2 n + 1} δ_{mn}

Normalization convention

P_{n} (1) = 1

(aur

P_{n} (- 1) = (- 1)^{n}

)

P_{n}

ki parity

P_{n} (- x) = (- 1)^{n} P_{n} (x)

Doosra solution kya hai aur ise discard kyun karte hain?

Q_{n} (x)

, Legendre function of the second kind; yeh

x = \pm 1

par unbounded hai (log singularity), toh physically reject kiya jaata hai.

x = 0

ordinary point kyun hai lekin

x = \pm 1

singular?

y^{''}

ka coefficient

(1 - x^{2})

0

par nonzero hai lekin

\pm 1

par vanish karta hai (regular singular points = sphere ke poles).

Connections

Power Series / Frobenius Method — jaise humne $y = \sum a_{m} x^{m}$ banaya.
Sturm-Liouville Theory — self-adjoint form ⇒ orthogonality aur real eigenvalues.
Laplace's Equation in Spherical Coordinates — jahan $x = cos θ$ aur Legendre's equation paida hoti hai.
Fourier Series — same idea, orthogonal basis expansion, sines ki jagah polynomials.
Associated Legendre Functions & Spherical Harmonics — agla step ( $m \neq = 0$ angular modes).
Generating Functions — $\frac{1}{1 - 2 x t + t ^{2}} = \sum P_{n} (x) t^{n}$ multipole expansion se connect karta hai.