Laplace on disk — polar coordinates, Bessel functions connection

4.7.15 · Maths › Partial Differential Equations

1. KAUNSA problem solve kar rahe hain?

Bessel connection is note ka dil hai, isliye hum $\nabla^{2} u + λ^{2} u = 0$ par focus karte hain, aur plain Laplace ko $λ = 0$ special case ke roop mein recover karte hain.

2. KAISE: polar Laplacian scratch se derive karna

Hume $\nabla^{2} u = u_{xx} + u_{y y}$ chahiye $(r, θ)$ mein, jahan $x = r cos θ, y = r sin θ, r = x^{2} + y^{2}, θ = arctan \frac{y}{x} .$

Step 1 — basic derivatives. Kyon? Chain rule ko $r_{x}, r_{y}, θ_{x}, θ_{y}$ chahiye. $r_{x} = \frac{x}{r} = cos θ, r_{y} = sin θ, θ_{x} = - \frac{s i n θ}{r}, θ_{y} = \frac{c o s θ}{r} .$

Step 2 — chain rule se first derivatives. Kyon? $\partial_{x}$ ko $\partial_{r}, \partial_{θ}$ mein convert karo. $u_{x} = cos θ u_{r} - \frac{s i n θ}{r} u_{θ}, u_{y} = sin θ u_{r} + \frac{c o s θ}{r} u_{θ} .$

Step 3 — operator dobara lagao aur $u_{xx} + u_{y y}$ add karo. Cross terms cancel ho jaate hain (yeh orthogonal coordinates ka jadoo hai), aur standard result milta hai: $\nabla^{2} u = u_{r r} + \frac{1}{r} u_{r} + \frac{1}{r ^{2}} u_{θ θ}$

3. KAISE: variables ka separation

$\nabla^{2} u + λ^{2} u = 0$ ke liye $u (r, θ) = R (r) Θ (θ)$ dhundo.

Plug in karo aur $R Θ/ r^{2}$ se divide karo (carefully multiply through karo): $R^{''} Θ + \frac{1}{r} R^{'} Θ + \frac{1}{r ^{2}} R Θ^{''} + λ^{2} R Θ = 0.$ $R Θ$ se divide karo aur $r^{2}$ se multiply karo: $sirf r par depend karta hai \frac{r ^{2} R ^{''} + r R ^{'}}{R} + λ^{2} r^{2} = sirf θ par depend karta hai - \frac{Θ ^{''}}{Θ} = n^{2} .$ Constant kyon? LHS sirf $r$ par depend karta hai, RHS sirf $θ$ par; equal tabhi honge jab dono ek constant ke barabar hon. Hum ise $n^{2}$ kehte hain.

Angular ODE: $Θ^{''} + n^{2} Θ = 0$ . $n$ non-negative integer kyon hona chahiye? Kyunki $θ$ aur $θ + 2 π$ SAME point hain — periodicity $Θ (θ) = Θ (θ + 2 π)$ force karta hai ki $n = 0, 1, 2, \dots$ hoga, jahan $Θ_{n} (θ) = A cos n θ + B sin n θ .$

Radial ODE: $r^{2} R^{''} + r R^{'} + (λ^{2} r^{2} - n^{2}) R = 0.$

4. Bessel connection (is poore subtopic ka WHY)

Substitute karo $s = λ r$ (toh $\frac{d}{d r} = λ \frac{d}{d s}$ ). Kyon? $λ$ ko khatam karne ke liye aur ek universal form tak pahunchne ke liye. $s^{2} R^{''} (s) + s R^{'} (s) + (s^{2} - n^{2}) R = 0.$

Toh disk ka radial solution hai $R (r) = c_{1} J_{n} (λ r) + c_{2} Y_{n} (λ r) .$ Physical pruning: center $r = 0$ finite rehna chahiye ⇒ $c_{2} = 0$ ( $Y_{n}$ drop karo). Isliye $R (r) = J_{n} (λ r) .$

Series se $J_{0}$ derive karna (Derivation-from-scratch)

$n = 0$ ke liye: $s y^{''} + y^{'} + sy = 0$ . Try karo $y = \sum_{k \geq 0} a_{k} s^{2 k}$ (sirf even powers, symmetry ki wajah se). Plug in karo aur powers match karo toh milta hai $a_{k} = - \frac{a _{k - 1}}{( 2 k ) ^{2}}$ , toh $J_{0} (s) = \sum_{k = 0}^{\infty} \frac{( - 1 ) ^{k}}{( k ! ) ^{2}} (\frac{s}{2})^{2 k} = 1 - \frac{s ^{2}}{4} + \frac{s ^{4}}{64} - \dots$ Yeh kyon matter karta hai: tum literally ODE se $J_{0}$ generate kar sakte ho — kisi table ki zaroorat nahi.

5. Boundary condition eigenvalues fix karta hai

Vibrating drum ke liye jo rim par clamped hai ( $u (a, θ) = 0$ ): $J_{n} (λa) = 0 \Rightarrow λa = α_{n, m} \Rightarrow λ_{n, m} = \frac{α _{n, m}}{a},$ jahan $α_{n, m}$ , $J_{n}$ ka $m$ -va positive zero hai. Discrete kyon? $J_{n}$ fixed jagahon par zero cross karta hai; sirf wahi $λ$ clamp satisfy karte hain. Full modes: $u_{n, m} (r, θ) = J_{n} (\frac{α _{n, m}}{a} r) (A cos n θ + B sin n θ) .$

6. Pure Laplace ( $λ = 0$ ): koi Bessel nahi

$λ = 0$ ke saath radial ODE equidimensional hai: $r^{2} R^{''} + r R^{'} - n^{2} R = 0$ , solutions $R = r^{\pm n}$ (aur $ln r,$ const $n = 0$ ke liye). $0$ par finiteness $r^{∣ n ∣}$ rakhti hai: $u (r, θ) = \frac{a _{0}}{2} + \sum_{n \geq 1} r^{n} (a_{n} cos n θ + b_{n} sin n θ),$ boundary data $u (a, θ) = f (θ)$ se Fourier coefficients dwara match kiya jaata hai (yeh Poisson-kernel interior problem hai). Yahan $r^{n}$ kyon, pehle $J_{n}$ kyon? Koi $λ^{2} r^{2}$ term nahi ⇒ Bessel ki equation Euler ki equation mein degenerate ho jaati hai.

7. Worked examples

8. Common mistakes

Recall Feynman: 12-saal ke bachche ko samjhao

Ek round drum imagine karo. Jab tum use bajate ho, to skin rings aur pie-slices mein ripple karti hai. "Pie-slice" pattern sirf circle ke around nice waves hain (woh $cos n θ$ hai). "Ring" pattern — skin kitni uchi uchhalti hai jab tum center se edge ki taraf jaate ho — ek special wiggly curve hai jise Bessel function kehte hain. Yeh ek wave ki tarah wiggle karta hai lekin bahar jaane par thoda flat hota jaata hai, kyunki skin zyada space par spread ho jaati hai. Drum sirf kuch specific pitches par ring kar sakta hai: bilkul wahan jahan woh wiggly curve clamped rim par zero hit karta hai.

9. Connections

Separation of Variables — engine jo $r$ aur $θ$ split karta hai.
Bessel Functions — $J_{n}, Y_{n}$ , unki series, recurrences, zeros.
Fourier Series — boundary data $f (θ)$ ko $cos n θ, sin n θ$ mein expand karna.
Sturm-Liouville Theory — orthogonality $\int_{0}^{a} J_{n} (λ_{m} r) J_{n} (λ_{k} r) r d r = 0$ .
Poisson Kernel and Mean Value Property — pure-Laplace disk problem.
Vibrating Membrane Wave Equation — jahan drum eigenmodes rehte hain.
Euler-Cauchy Equation — $λ = 0$ radial reduction.

Flashcards

Polar Laplacian formula

u_{r r} + \frac{1}{r} u_{r} + \frac{1}{r ^{2}} u_{θ θ}

\frac{1}{r} u_{r}

kyon aata hai?

Area element

r d r d θ

; circles ki geometric spreading.

Angular part ke liye separation constant ko kya kehte hain?

n^{2}

, jahan periodicity

n = 0, 1, 2, \dots

force karti hai.

Angular equation aur uska solution

Θ^{''} + n^{2} Θ = 0 \Rightarrow A cos n θ + B sin n θ

Separation ke baad radial ODE (Helmholtz)

r^{2} R^{''} + r R^{'} + (λ^{2} r^{2} - n^{2}) R = 0

Ise Bessel mein convert karne wala substitution

s = λ r

, jis se

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

milta hai.

Bessel ki equation ke do solutions

J_{n}

(0 par finite) aur

Y_{n}

(0 par singular).

Full disk par kaunsa drop hota hai aur kyon?

Y_{n}

, kyunki yeh

r = 0

par blow up karta hai (unphysical center).

Clamped drum ke liye eigenvalue condition

J_{n} (λa) = 0 \Rightarrow λ_{n, m} = α_{n, m} / a

α

J_{n}

ke zeros.

J_{0}

ka pehla zero (fundamental drum)

\approx 2.405

J_{0}

series

\sum (- 1)^{k} / (k!)^{2} (s /2)^{2 k} = 1 - s^{2} /4 + s^{4} /64 - \dots

Pure Laplace (

λ = 0

) bounded radial solutions

r^{∣ n ∣}

(Euler equation), Bessel nahi.

Disk par general bounded Laplace solution

\frac{a _{0}}{2} + \sum r^{n} (a_{n} cos n θ + b_{n} sin n θ)

Bessel modes ke liye orthogonality weight

weight

r

\int_{0}^{a} J_{n} (λ_{m} r) J_{n} (λ_{k} r) r d r = 0

m \neq = k