Legendre's equation and Legendre polynomials (intro): Question bank

4.6.20 · D5 · Maths › Ordinary Differential Equations › Legendre's equation and Legendre polynomials (intro)

Setup: har symbol defined, ek picture ke saath

Variable $x$ , $[- 1, 1]$ pe chalta hai kyunki yeh $x = cos θ$ se aata hai, jahan $θ$ ek sphere par polar angle hai. Endpoints $x = \pm 1$ us sphere ke poles (north aur south) hain — neeche mapping figure dekho.

Numerator mein termination ki poori kahani hai: $m = n$ par yeh $n (n + 1) - n (n + 1) = 0$ ban jaata hai, isliye $a_{n + 2} = 0$ aur series ruk jaati hai, degree- $n$ polynomial mein collapse ho jaati hai. Figure mein dekho numerator ko zero hit karte hue.

Pehle chaar polynomials — interval par "building-block shapes" — yahan draw kiye hain taaki tum unki parity aur unki shared value $P_{n} (1) = 1$ dekh sako.

Formula Self-adjoint form aur orthogonality kahan se aati hai

ODE ko rewrite karke pehle do terms ko ek derivative mein fold karne par self-adjoint (Sturm–Liouville) form milti hai: $\frac{d}{d x} [(1 - x^{2}) y^{'}] + n (n + 1) y = 0,$ jise tum check kar sakte ho ki yeh original mein expand ho jaata hai kyunki $\frac{d}{d x} [(1 - x^{2}) y^{'}] = (1 - x^{2}) y^{''} - 2 x y^{'}$ . Yahan weight function $w (x) = 1$ hai aur brackets mein term $x = \pm 1$ par zero ho jaata hai. $P_{m}$ equation ko $P_{n}$ se multiply karke, $P_{n}$ equation ko $P_{m}$ se, subtract karke aur integration by parts se, boundary terms khatam ho jaate hain (kyunki $1 - x^{2} = 0$ ends par), aur bachta hai $[n (n + 1) - m (m + 1)] \int_{- 1}^{1} P_{m} P_{n} d x = 0.$ $m \neq = n$ ke liye bracket nonzero hai, isliye integral $0$ hona chahiye: yahi orthogonality hai, aur yeh ek structural consequence hai, luck nahi.

True or false — justify karo

Legendre's equation ke do independent solutions hain, lekin sirf ek polynomial hai.

True. Yeh second-order hai, isliye iske do independent solutions hain;

P_{n}

polynomial (terminating) wala hai, jabki doosra,

Q_{n}

, ek non-terminating series hai jo

x = \pm 1

par logarithmically blow up karti hai.

Legendre's equation ke dono solutions

[- 1, 1]

par bounded hain.

False. Sirf

P_{n}

poles

x = \pm 1

par bounded rehta hai; doosra solution

Q_{n}

wahan

lo g

singularity rakhta hai, aur yehi wajah hai ki physics use discard karta hai.

Point

x = 0

, Legendre's equation ka singular point hai.

False.

y^{''}

ka coefficient

1 - x^{2}

hai, jo

x = 0

par

1 \neq = 0

equal karta hai, isliye

x = 0

ek ordinary point hai — aur yehi wajah hai ki wahan plain Taylor series kaam karta hai.

Endpoints

x = \pm 1

ordinary points hain kyunki equation wahan bhi sense deta hai.

False.

x = \pm 1

par coefficient

1 - x^{2}

vanish ho jaata hai, isliye yeh regular singular points hain; equation "sense deta hai" lekin solutions misbehave kar sakte hain (ek blow up karta hai), jo poori selection story hai.

Non-integer

n

ke liye series solution phir bhi ek polynomial mein terminate hoti hai.

False. Termination ke liye numerator

m (m + 1) - n (n + 1)

ko kisi integer

m

par exactly

0

hit karna hoga; yeh tabhi hota hai jab

n

khud ek non-negative integer ho, warna series hamesha chalti rehti hai.

Orthogonality

\int_{- 1}^{1} P_{m} P_{n} d x = 0

for

m \neq = n

in specific polynomials ka ek lucky coincidence hai.

False. Yeh structurally self-adjoint (Sturm–Liouville) form se follow karta hai: alag eigenvalues

n (n + 1)

orthogonal eigenfunctions ko force karte hain, jaise upar integration-by-parts argument dikhata hai.

P_{n}

ki degree exactly

n

hai.

True. Recurrence series ko

m = n

par terminate karta hai, aur

x^{n}

coefficient nonzero hai (leading coefficient

\frac{( 2 n )!}{2 ^{n} ( n ! ) ^{2}}

), isliye degree exactly

n

hai, kam nahi.

P_{n} (1) = 1

ek theorem hai jo tum ODE se prove karte ho.

False. Yeh ek normalization convention hai — ek choice jo single free constant ko pin karti hai. ODE sirf solutions ko ek scalar multiple tak fix karta hai; hum choose karte hain

P_{n} (1) = 1

P_{5} (- 1) = - 1

True. Kyunki

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

aur

5

odd hai,

P_{5} (- 1) = (- 1)^{5} = - 1

; yeh yeh bhi reflect karta hai ki

P_{5}

ek odd function hai.

General solution ke do arbitrary constants

a_{0}

aur

a_{1}

ke correspond karte hain.

True. Recurrence coefficients ko do apart link karta hai, isliye saare even coefficients

a_{0}

se aate hain aur saare odd wale

a_{1}

se — exactly woh do free constants jo ek second-order ODE demand karta hai.

Error dhundho

"Recurrence numerator

m (m - 1) - n (n + 1)

hai, kyunki

y^{''}

m (m - 1)

contribute hota hai."

Error: tumne

- 2 x y^{'}

contribution chhor diya.

y^{''}

term

m (m - 1)

deta hai, lekin

- 2 x y^{'}

- 2 m

add hota hai; combine karne par,

m (m - 1) - 2 m = m (m + 1) - 2 \cdot (m + 1) + \dots

— cleanly,

m (m - 1) - 2 m = m^{2} - 3 m

, aur

n (n + 1)

term add karne par correct grouped numerator

m (m + 1) - n (n + 1)

hai.

- 2 m

miss karo aur

m = n

par termination toot jaati hai.

"Kyunki

P_{n}

ek polynomial hai, uska partner solution bhi symmetry se ek polynomial hona chahiye."

Error: form ki symmetry behaviour ki symmetry force nahi karti. Partner

Q_{n}

same-parity series hai jo kabhi terminate nahi hoti; yeh ek genuine non-polynomial hai jisme poles par singularity hai.

"Hum Legendre's equation solve karne ke liye

x = 0

par Frobenius/indicial analysis use karte hain."

Error: full Frobenius (indicial equation) singular points ke liye hai.

x = 0

par, jo ek ordinary point hai, plain power series

\sum a_{m} x^{m}

kaafi hai — koi indicial roots ki zaroorat nahi.

P_{2} (x) = 1 - 3 x^{2}

kyunki recurrence yahi deta hai."

Error: recurrence ek constant tak

y = a_{0} (1 - 3 x^{2})

deta hai.

P_{2} (1) = 1

apply karne par

a_{0} = - \frac{1}{2}

force hota hai, isliye

P_{2} (x) = \frac{1}{2} (3 x^{2} - 1)

hai,

1 - 3 x^{2}

nahi.

"Orthogonality ka matlab

\int_{- 1}^{1} P_{n}^{2} d x = 0

hai."

Error: orthogonality distinct indices ke liye hai. Same-index integral norm hai

\int_{- 1}^{1} P_{n}^{2} d x = \frac{2}{2 n + 1} \neq = 0

, jo expansion coefficients ko well-defined banata hai.

"Rodrigues' formula

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

bina

1/ (2^{n} n!)

ke bhi kaam karta hai."

Error: woh factor drop karne par bhi ek degree-

n

ODE solution milta hai, lekin woh

P_{n} (1) = 1

satisfy nahi karega. Constant exactly woh normalizer hai jo convention enforce karta hai.

Why questions

x = cos θ

substitution aata hi kyun hai?

Laplace's Equation in Spherical Coordinates ka angular part naturally

θ

mein hota hai;

x = cos θ

set karne par yeh Legendre's form mein convert ho jaata hai

x \in [- 1, 1]

ke saath, aur poles

θ = 0, π

endpoints

x = \pm 1

ban jaate hain.

Hum self-adjoint form par insist kyun karte hain jab dono forms same equation hain?

Self-adjoint (Sturm–Liouville) shape integration-by-parts mein boundary terms ko

x = \pm 1

par vanish karati hai, jo proves orthogonality via Sturm-Liouville Theory; us structure ke bina tum kabhi clean

δ_{mn}

integral expect nahi karte.

Poles par boundedness require karne se

n

non-negative integer kyun force hota hai?

Sirf integer

n

series ko terminate karta hai; kisi bhi aur

n

ke liye surviving infinite series

x = \pm 1

par diverge karti hai. Isliye "poles par bounded" woh physical rule hai jo integers select karti hai.

Recurrence coefficients ko ek step ki bajaye do steps apart kyun link karta hai?

Equation mein sirf parity-preserving pieces hain har ek family par —

x^{2}

aur

2 x y^{'}

parity back shift karte hain,

n (n + 1) y

preserve karta hai — isliye even aur odd powers kabhi mix nahi karte. Isliye

a_{m}

a_{m + 2}

determine hota hai, do parity families ko alag rakhta hai.

Hum

[- 1, 1]

par kisi bhi nice

f (x)

\sum c_{n} P_{n} (x)

ke roop mein expand kyun kar sakte hain?

Kyunki

P_{n}

ek complete orthogonal set (Sturm–Liouville eigenbasis) form karte hain; orthogonality se hum har

c_{n} = \frac{2 n + 1}{2} \int_{- 1}^{1} f P_{n} d x

isolate kar sakte hain, bilkul jaise Fourier Series orthogonal sines aur cosines use karta hai.

Terminating solution ko "physical" kyun kaha jaata hai aur doosra reject kyun hota hai?

Physical potentials, temperatures, aur fields sphere par hamesha finite rehne chahiye, poles par bhi;

Q_{n}

wahan blow up karta hai, isliye yeh ek real bounded field describe nahi kar sakta aur discard ho jaata hai.

Edge cases

n = 0

ke liye, recurrence kya karta hai aur

P_{0} = 1

kyun hai?

n = 0

ke saath numerator

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

sirf

m = 0

par zero hai,

a_{2}

ko turant khatam kar deta hai, isliye sirf

a_{0}

bachta hai — ek constant,

P_{0} (x) = 1

tak normalize kiya gaya.

Kya

P_{0} (x) = 1

P_{2} (x) = \frac{1}{2} (3 x^{2} - 1)

ke saath orthogonal hai, aur geometrically iska kya matlab hai?

Haan;

\int_{- 1}^{1} 1 \cdot \frac{1}{2} (3 x^{2} - 1) d x = 0

. Iska matlab hai "flat" pattern aur "ring" pattern sphere par temperature ke baare mein independent information carry karte hain.

Jab

n

even ho (maan lo

n = 2

) toh odd-series solution ka kya hota hai?

Even

n

ke liye even series

P_{n}

mein terminate hoti hai, lekin odd series (

a_{1} \neq = 0

) terminate nahi hoti — yeh unbounded

Q_{n}

-type solution hai aur physics ke liye reject ho jaati hai.

Kya

P_{n}

kabhi degree

n

se kam hota hai (jaise leading term cancel ho jaaye)?

Nahi. Leading coefficient

\frac{( 2 n )!}{2 ^{n} ( n ! ) ^{2}}

har

n

ke liye strictly positive hai, isliye degree hamesha exactly

n

hoti hai; polynomial kabhi lower degree mein collapse nahi ho sakti.

Exact points

x = \pm 1

par, kya

P_{n}

abhi bhi ODE satisfy karta hai?

P_{n}

ek polynomial hai, isliye yeh

\pm 1

par defined aur finite hai, lekin yeh equation ke singular points hain — ODE ka leading coefficient wahan vanish ho jaata hai, isliye yeh validity ki boundary hai, interior check nahi.

Odd

n

ke liye

P_{n} (0)

kya hai, aur kyun?

Yeh har odd

n

ke liye $0$ hai, kyunki odd

P_{n}

odd functions hain (

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

), aur koi bhi odd function origin se guzarni chahiye.

Kya negative degree

n < 0

genuinely naya hai, ya koi existing polynomial repeat karta hai?

Yeh repeat karta hai: kyunki

n (n + 1)

n \mapsto - (n + 1)

ke under symmetric hai, degree

n = - 1

same eigenvalue deta hai jaise

n = 0

n = - 2

same jaise

n = 1

, aur aise hi — isliye negative integers

n = 0, 1, 2, \dots

se aage kuch naya nahi laate.

Half-integer

n

ke baare mein kya, maan lo

n = \frac{1}{2}

Numerator

m (m + 1) - \frac{1}{2} \cdot \frac{3}{2}

kisi bhi whole

m

ke liye zero nahi hit karta, isliye series kabhi terminate nahi hoti aur

x = \pm 1

par diverge karti hai; koi bounded polynomial nahi hai, isliye sirf non-negative integers allow hote hain.

Jab

n \to \infty

[- 1, 1]

par

P_{n}

ke graphs ka kya hota hai?

Woh zyada zeros ke saath tezi se oscillate karte hain (exactly

n

of them in

(- 1, 1)

), phir bhi interior mein hamesha

[- 1, 1]

ke andar rehte hain aur

P_{n} (1) = 1

par pin hote hain; zyada oscillation polynomial analogue hai higher-frequency modes ka Fourier Series mein.