Exercises — Bessel's equation and Bessel functions (intro, physical relevance)

KYA compare karte hain: standard form mein constant ${\nu }^2$ subtract hota hai $x^2$ se. Yahan woh constant $25$ hai. KYUN: $(x^2-{\nu }^2)$ mein number hamesha ${\nu }^2$ hota hai, kabhi $\nu$ nahi — yeh order ka square hai. Toh ${\nu }^2=25\Rightarrow \nu =5$ (convention se $\nu \ge 0$ lete hain). General solution: 2nd-order linear ODE ke liye do independent solutions chahiye, toh $y(x)=c_1{J}_5(x)+c_2{Y}_5(x).$

$J_3(0)=0$ (finite). $Y_3$ diverge karta hai. Integer order $n\ge 1$ ke liye second kind $0$ ke paas aise behave karta hai: $Y_n(x)\sim -\frac{(n-1)!}{\pi }{\left(\frac{2}{x}\right)}^n$, yani yahan $x^{-3}$ jaisa — yeh $-\infty$ ki taraf shoot karta hai. Kyun matter karta hai: kisi bhi region mein jo centre $r=0$ contain karta ho, hume $Y_3$ ka coefficient zero rakhna padega.

KYA karte hain: denominators clear karne ke liye har term ko $x^2$ se multiply karo. KYUN: standard form $x^2{y}^{\prime \prime }$ se shuru hoti hai; di gayi equation woh form hai $x^2$ se divide ki hui. $x^2{y}^{\prime \prime }+x{y}^{\prime }+(x^2-4)y=0\Rightarrow {\nu }^2=4\Rightarrow \nu =2.$ Solution: $y=c_1{J}_2(x)+c_2{Y}_2(x)$.

$\nu =1$ ke saath: $J_1(x)={\sum }_{m\ge 0}\frac{(-1{)}^m}{m!\Gamma (m+2)}{\left(\frac{x}{2}\right)}^{2m+1}$. Yahan $\Gamma (m+2)=(m+1)!$ (Gamma function integers par factorial hota hai).

• $m=0:\frac{1}{0!1!}{\left(\frac{x}{2}\right)}^1=\frac{x}{2}$.
• $m=1:\frac{-1}{1!2!}{\left(\frac{x}{2}\right)}^3=-\frac{x^3}{16}$. Toh $J_1(x)\approx \frac{x}{2}-\frac{x^3}{16}$. $x=0.5$ par: $0.25-\frac{0.125}{16}=0.25-0.0078125=0.2421875$. (True value $J_1(0.5)=0.24227$ — humare do terms already 4 decimals tak agree karte hain.)

KYUN yeh equation $\nu =\frac{1}{2}$ hai: constant $\frac{1}{4}=(\frac{1}{2}{)}^2$ hai. $J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin x$ proportional hai $\frac{\sin x}{\sqrt{x}}$ se, isliye use ODE satisfy karni chahiye. Direct check: maano $y=x^{-1/2}\sin x$. $y^{\prime }=-\frac{1}{2}{x}^{-3/2}\sin x+x^{-1/2}\cos x.$ $y^{\prime \prime }=\frac{3}{4}{x}^{-5/2}\sin x-x^{-3/2}\cos x-x^{-1/2}\sin x.$ Substitute karo: $x^2{y}^{\prime \prime }=\frac{3}{4}{x}^{-1/2}\sin x-x^{1/2}\cos x-x^{3/2}\sin x,$ $x{y}^{\prime }=-\frac{1}{2}{x}^{-1/2}\sin x+x^{1/2}\cos x,$ $(x^2-\frac{1}{4})y=x^{3/2}\sin x-\frac{1}{4}{x}^{-1/2}\sin x.$ $x^{1/2}\cos x$ terms ka sum: $-1+1=0$. $x^{3/2}\sin x$ ka sum: $-1+1=0$. $x^{-1/2}\sin x$ ka sum: $\frac{3}{4}-\frac{1}{2}-\frac{1}{4}=0$. Sab cancel ho jaate hain ⇒ yeh ek solution hai. ✓

General radial solution (physical variable $r$ mein wapas likha, kyunki $x=kr$): $R(r)=c_1{J}_0(kr)+c_2{Y}_0(kr)$. $r=0$ par finiteness: $Y_0\to -\infty$ wahan (upar definition callout dekho), isliye $c_2=0$ — yeh Sturm-Liouville theory wali boundedness condition hai singular endpoint par. Edge condition $R(a)=R(1)=0$: $J_0(k\cdot 1)=0\Rightarrow k={\alpha }_{0,j}$, yani $J_0$ ke zeros. Lowest two: $k_1=2.405,k_2=\mathrm{5.520.}$

$\frac{f_2}{f_1}=\frac{{\alpha }_{0,2}}{{\alpha }_{0,1}}=\frac{5.520}{2.405}=2.2952\dots$ String ke overtones $2,3,4,\dots$ (integers) hote hain, jo harmony dete hain. Yahan ratio $2.295$ integer nahi hai, toh overtones clash karte hain — awaaz inharmonic hoti hai, ek metallic "thud" na ki clear pitch.

(a) Pehle ek zaroori fact — KYUN $J_0^{\prime }(x)=-J_1(x)$. First-kind series ko term by term differentiate karo. General identity $\frac{d}{dx}(x^{-\nu }{J}_{\nu }(x))=-x^{-\nu }{J}_{\nu +1}(x)$, $\nu =0$ par lene se seedha milta hai $J_0^{\prime }(x)=-J_1(x)$ (yeh parent note ke Frobenius section mein series se prove hua hai). Ab reading: kyunki $J_0^{\prime }=-J_1$, $J_0$ ka slope exactly wahan zero hota hai jahan $J_1$ zero cross karta hai, toh $J_0$ ke turning points $J_1$ ke zeros ke saath line up karte hain. $x=0$ aur uske pehle zero $2.405$ ke beech, $J_0$ $1$ se girta hai (abhi koi turning point nahi), jabki $J_1$ $0$ se chadhta hai, apne pehle maximum ke paas $x\approx 1.84$ par pahunchta hai, phir wापas neeche jaata hai. Figure mein amber dot $J_1$ ka woh peak mark karta hai, aur woh $J_0$ ke pehle zero ($x=2.405$) par dotted cyan line ke left mein baitha hai. (b) Bade $x$ ke liye, $J_{\nu }(x)\approx \sqrt{\frac{2}{\pi x}}\cos (\cdots )$. Envelope $\sqrt{2/(\pi x)}$ hai (amber dashed curves), jo $1/\sqrt{x}$ ki tarah decay karti hai. Physically wave ki energy ek ever-larger ring (circumference $\propto x$) par spread hoti hai, isliye amplitude${}^2\propto 1/x$ ⇒ amplitude $\propto 1/\sqrt{x}$. Yahi shrinking tum dekh rahe ho.

(i) Time aur space ko separate karo. $u(r,\theta ,t)=v(r,\theta )T(t)$ likho. $u_{tt}=c^2{\nabla }^{2}u$ mein substitute karo aur $c^{2}vT$ se divide karo (yeh Separation of variables hai): $\frac{T^{\prime \prime }}{c^{2}T}=\frac{{\nabla }^{2}v}{v}=\text{constant}.$ KYUN constant negative hona chahiye (use $-k^2$ kaho): left side sirf $t$ par depend karti hai, right side sirf space par, toh dono ek single constant ke barabar hain. Agar woh constant $+k^2>0$ hota, toh $T^{\prime \prime }=c^2{k}^{2}T$ deta growing/decaying exponentials $e^{\pm ckt}$ — ek drum jo run away kare ya mar jaaye, bajne wala nahi. Ek vibrating membrane ko time mein oscillate karna hoga, jo chahiye $T^{\prime \prime }=-c^2{k}^{2}T\Rightarrow T=\cos (ckt),\sin (ckt)$. Yahi physical demand sign ko $-k^2$ force karta hai: ${\nabla }^{2}v+k^{2}v=0,T^{\prime \prime }+c^2{k}^{2}T=0.$ (ii) Radial modes. Radially-symmetric modes ke liye ($\nu =0$) Laplacian in polar and cylindrical coordinates deta hai $v_{rr}+\frac{1}{r}{v}_r+k^{2}v=0$; $x=kr$ ke saath yeh Bessel order $0$ hai: $v(r)=c_1{J}_0(kr)+c_2{Y}_0(kr).$ $r=0$ par boundedness $Y_0$ ko khatam karti hai ($c_2=0$); fixed rim $v(a)=0$ deta hai $J_0(ka)=0\Rightarrow k_j={\alpha }_{0,j}/a$. Mode shape: $v_j(r)=J_0({\alpha }_{0,j}r/a)$. (iii) Fundamental frequency. Angular frequency ${\omega }_1=c{k}_1=c{\alpha }_{0,1}/a$; frequency $f_1={\omega }_1/(2\pi )$: $f_1=\frac{c{\alpha }_{0,1}}{2\pi a}=\frac{100\times 2.405}{2\pi \times 0.15}=\frac{240.5}{0.94248}=255.2\text{Hz}.$

Centre $r=0$ ab region mein nahi hai, isliye $Y_0$ ko drop karne ki zaroorat nahi — dono survive karte hain: $R(r)=c_1{J}_0(kr)+c_2{Y}_0(kr).$ Do boundary conditions $R(b)=0$ aur $R(a)=0$: $c_1{J}_0(kb)+c_2{Y}_0(kb)=0,c_1{J}_0(ka)+c_2{Y}_0(ka)=0.$ Nonzero $(c_1,c_2)$ tabhi exist karta hai jab determinant vanish kare: $J_0(ka)Y_0(kb)-J_0(kb)Y_0(ka)=0.$ Is transcendental equation ke roots $k$ annular drum ki frequencies dete hain. L4.1 se contrast KYUN: survive karne wale constants ki sankhya is baat par depend karti hai ki singular point $r=0$ domain ke andar hai ya nahi.

KYA choose karte hain aur KYUN: coefficient $e^{2x}$ exponentially badhta hai; $t=e^x$ substitute karne par $x$ mein exponential growth $t$ mein polynomial behaviour mein badal jaati hai, jo Bessel's equation (ek polynomial-coefficient ODE) ko chahiye. $t=e^x$ ke saath: $\frac{dt}{dx}=e^x=t$, toh $\frac{dy}{dx}=t\frac{dy}{dt}$. $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(t\frac{dy}{dt}\right)=t\frac{d}{dt}\left(t\frac{dy}{dt}\right)=t\left(\frac{dy}{dt}+t\frac{d^{2}y}{d{t}^2}\right)=t^2{y}_{tt}+t{y}_t.$ Aur $e^{2x}=t^2$. ODE $y^{\prime \prime }+e^{2x}y=0$ ban jaata hai $t^2{y}_{tt}+t{y}_t+t^{2}y=0,$ jo exactly variable $t$ mein order $\nu =0$ ki Bessel's equation hai. Isliye $y=c_1{J}_0(t)+c_2{Y}_0(t)=c_1{J}_0(e^x)+c_2{Y}_0(e^x).$

KYUN recurrence: yeh hume known half-integer orders se agle tak bina series chhuye chadhne deta hai. $\nu =\frac{1}{2}$ set karo: $J_{-1/2}+J_{3/2}=\frac{2(\frac{1}{2})}{x}{J}_{1/2}=\frac{1}{x}{J}_{1/2}$. Toh $J_{3/2}=\frac{1}{x}{J}_{1/2}-J_{-1/2}=\sqrt{\frac{2}{\pi x}}\left(\frac{\sin x}{x}-\cos x\right).$ $x=\pi /2$ par evaluate karo: $\sin (\pi /2)=1$, $\cos (\pi /2)=0$, toh bracket hai $\frac{1}{\pi /2}-0=\frac{2}{\pi }$. Prefactor $\sqrt{\frac{2}{\pi \cdot \pi /2}}=\sqrt{\frac{4}{{\pi }^2}}=\frac{2}{\pi }$. $J_{3/2}(\pi /2)=\frac{2}{\pi }\cdot \frac{2}{\pi }=\frac{4}{{\pi }^2}=0.40528\dots$

KYUN orthogonality matter karti hai: yahi cheez tumhe ek arbitrary initial drum shape ko ${\sum }_j{c}_j{J}_0({\alpha }_{0,j}r)$ ke roop mein expand karne deti hai — Bessel analogue of Fourier series, Bessel's equation ke self-adjoint Sturm–Liouville form se guaranteed. Diagonal case $i=j=1$: $N_1=\frac{1}{2}{J}_1({\alpha }_{0,1}{)}^2=\frac{1}{2}(0.5191{)}^2=\frac{1}{2}(0.26946)=\mathrm{0.13473.}$

$\nu$ padhne se pehle standard form mein multiply karo ::: haan — leading term $x^2{y}^{\prime \prime }$ hona chahiye $(x^2-\square )$ ke slot mein hota hai ::: ${\nu }^2$, $\nu$ nahi Domain $r=0$ contain karta hai? drop karo ::: $Y_{\nu }$ (wahan diverge karta hai) Annulus (centre par hole)? rakho ::: dono $J_{\nu }$ aur $Y_{\nu }$ Disc par $J_{\nu }$ ke liye orthogonality weight ::: $w(r)=r$ Drum fundamental frequency ::: $f_1=c{\alpha }_{0,1}/(2\pi a)$ Vibrating drum ke liye separation constant ka sign ::: $-k^2$ (taaki time part oscillate kare, badhe nahi)