Intuition The big pictureA disk has no corners, only a circular edge. Cartesian x , y x,y x , y polar coordinates ( r , θ ) (r,\theta) ( r , θ ) radial part is no longer a sine/cosine — it becomes a Bessel equation , and its solutions are Bessel functions J n ( λ r ) J_n(\lambda r) J n ( λ r ) WHY Bessel appears: the 1 r \tfrac1r r 1 1 r 2 \tfrac1{r^2} r 2 1 variable coefficient , and that ODE is precisely Bessel's equation.
Definition Laplace's / Helmholtz problem on the diskOn the disk r ≤ a r\le a r ≤ a Laplace ∇ 2 u = 0 \nabla^2 u=0 ∇ 2 u = 0 eigenvalue (Helmholtz) problem ∇ 2 u + λ 2 u = 0 \nabla^2 u + \lambda^2 u = 0 ∇ 2 u + λ 2 u = 0
Pure Laplace → solution built from r ∣ n ∣ r^{|n|} r ∣ n ∣ Helmholtz / wave / heat with circular symmetry → ==Bessel functions J n ( λ r ) J_n(\lambda r) J n ( λ r )
The Bessel connection is the heart of this note, so we focus on ∇ 2 u + λ 2 u = 0 \nabla^2 u+\lambda^2 u=0 ∇ 2 u + λ 2 u = 0 λ = 0 \lambda=0 λ = 0
We need ∇ 2 u = u x x + u y y \nabla^2 u = u_{xx}+u_{yy} ∇ 2 u = u xx + u y y ( r , θ ) (r,\theta) ( r , θ ) x = r cos θ , y = r sin θ , r = x 2 + y 2 , θ = arctan y x . x=r\cos\theta,\quad y=r\sin\theta,\quad r=\sqrt{x^2+y^2},\quad \theta=\arctan\tfrac yx. x = r cos θ , y = r sin θ , r = x 2 + y 2 , θ = arctan x y .
Step 1 — basic derivatives. Why? The chain rule needs r x , r y , θ x , θ y r_x,r_y,\theta_x,\theta_y r x , r y , θ x , θ y r x = x r = cos θ , r y = sin θ , θ x = − sin θ r , θ y = cos θ r . r_x=\frac{x}{r}=\cos\theta,\quad r_y=\sin\theta,\quad \theta_x=-\frac{\sin\theta}{r},\quad \theta_y=\frac{\cos\theta}{r}. r x = r x = cos θ , r y = sin θ , θ x = − r s i n θ , θ y = r c o s θ .
Step 2 — first derivatives via chain rule. Why? Convert ∂ x \partial_x ∂ x ∂ r , ∂ θ \partial_r,\partial_\theta ∂ r , ∂ θ u x = cos θ u r − sin θ r u θ , u y = sin θ u r + cos θ r u θ . u_x=\cos\theta\,u_r-\frac{\sin\theta}{r}u_\theta,\qquad u_y=\sin\theta\,u_r+\frac{\cos\theta}{r}u_\theta. u x = cos θ u r − r s i n θ u θ , u y = sin θ u r + r c o s θ u θ .
Step 3 — apply the operator again and add u x x + u y y u_{xx}+u_{yy} u xx + u y y ∇ 2 u = u r r + 1 r u r + 1 r 2 u θ θ \boxed{\nabla^2 u=u_{rr}+\frac1r u_r+\frac1{r^2}u_{\theta\theta}} ∇ 2 u = u r r + r 1 u r + r 2 1 u θ θ
Seek u ( r , θ ) = R ( r ) Θ ( θ ) u(r,\theta)=R(r)\,\Theta(\theta) u ( r , θ ) = R ( r ) Θ ( θ ) ∇ 2 u + λ 2 u = 0 \nabla^2 u+\lambda^2 u=0 ∇ 2 u + λ 2 u = 0
Plug in and divide by R Θ / r 2 R\Theta/r^2 R Θ/ r 2 R ′ ′ Θ + 1 r R ′ Θ + 1 r 2 R Θ ′ ′ + λ 2 R Θ = 0. R''\Theta+\frac1r R'\Theta+\frac1{r^2}R\Theta''+\lambda^2 R\Theta=0. R ′′ Θ + r 1 R ′ Θ + r 2 1 R Θ ′′ + λ 2 R Θ = 0. R Θ R\Theta R Θ r 2 r^2 r 2 r 2 R ′ ′ + r R ′ R + λ 2 r 2 ⏟ depends on r = − Θ ′ ′ Θ ⏟ depends on θ = n 2 . \underbrace{\frac{r^2R''+rR'}{R}+\lambda^2 r^2}_{\text{depends on }r}=\underbrace{-\frac{\Theta''}{\Theta}}_{\text{depends on }\theta}=n^2. depends on r R r 2 R ′′ + r R ′ + λ 2 r 2 = depends on θ − Θ Θ ′′ = n 2 . Why a constant? LHS depends only on r r r θ \theta θ n 2 n^2 n 2
Angular ODE: Θ ′ ′ + n 2 Θ = 0 \Theta''+n^2\Theta=0 Θ ′′ + n 2 Θ = 0 Why must n n n Because θ \theta θ θ + 2 π \theta+2\pi θ + 2 π Θ ( θ ) = Θ ( θ + 2 π ) \Theta(\theta)=\Theta(\theta+2\pi) Θ ( θ ) = Θ ( θ + 2 π ) n = 0 , 1 , 2 , … n=0,1,2,\dots n = 0 , 1 , 2 , … Θ n ( θ ) = A cos n θ + B sin n θ . \Theta_n(\theta)=A\cos n\theta+B\sin n\theta. 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. r^2R''+rR'+(\lambda^2 r^2-n^2)R=0. r 2 R ′′ + r R ′ + ( λ 2 r 2 − n 2 ) R = 0.
Substitute s = λ r s=\lambda r s = λ r d d r = λ d d s \frac{d}{dr}=\lambda\frac{d}{ds} d r d = λ d s d Why? To kill the λ \lambda λ s 2 R ′ ′ ( s ) + s R ′ ( s ) + ( s 2 − n 2 ) R = 0. s^2R''(s)+sR'(s)+(s^2-n^2)R=0. s 2 R ′′ ( s ) + s R ′ ( s ) + ( s 2 − n 2 ) R = 0.
Definition Bessel's equation of ordern n n s 2 y ′ ′ + s y ′ + ( s 2 − n 2 ) y = 0 s^2 y''+s y'+(s^2-n^2)y=0 s 2 y ′′ + s y ′ + ( s 2 − n 2 ) y = 0 Bessel function of the first kind J n ( s ) J_n(s) J n ( s ) Y n ( s ) Y_n(s) Y n ( s ) blows up as s → 0 s\to0 s → 0
So the disk radial solution is
R ( r ) = c 1 J n ( λ r ) + c 2 Y n ( λ r ) . R(r)=c_1 J_n(\lambda r)+c_2 Y_n(\lambda r). R ( r ) = c 1 J n ( λ r ) + c 2 Y n ( λ r ) . Physical pruning: the center r = 0 r=0 r = 0 c 2 = 0 c_2=0 c 2 = 0 Y n Y_n Y n R ( r ) = J n ( λ r ) . \boxed{R(r)=J_n(\lambda r)}. R ( r ) = J n ( λ r ) .
J n J_n J n J n J_n J n sin ( n π x / L ) \sin(n\pi x/L) sin ( nπ x / L ) 1 r u r \tfrac1r u_r r 1 u r is J n J_n J n
For n = 0 n=0 n = 0 s y ′ ′ + y ′ + s y = 0 s y''+y'+s y=0 s y ′′ + y ′ + sy = 0 y = ∑ k ≥ 0 a k s 2 k y=\sum_{k\ge0}a_k s^{2k} y = ∑ k ≥ 0 a k s 2 k a k = − a k − 1 ( 2 k ) 2 a_k=-\dfrac{a_{k-1}}{(2k)^2} a k = − ( 2 k ) 2 a k − 1 J 0 ( s ) = ∑ k = 0 ∞ ( − 1 ) k ( k ! ) 2 ( s 2 ) 2 k = 1 − s 2 4 + s 4 64 − ⋯ J_0(s)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(k!)^2}\Big(\frac s2\Big)^{2k}=1-\frac{s^2}{4}+\frac{s^4}{64}-\cdots J 0 ( s ) = ∑ k = 0 ∞ ( k ! ) 2 ( − 1 ) k ( 2 s ) 2 k = 1 − 4 s 2 + 64 s 4 − ⋯ Why this matters: you can literally generate J 0 J_0 J 0
For a vibrating drum clamped at the rim (u ( a , θ ) = 0 u(a,\theta)=0 u ( a , θ ) = 0 J n ( λ a ) = 0 ⇒ λ a = α n , m ⇒ λ n , m = α n , m a , J_n(\lambda a)=0\ \Rightarrow\ \lambda a=\alpha_{n,m}\ \Rightarrow\ \lambda_{n,m}=\frac{\alpha_{n,m}}{a}, J n ( λa ) = 0 ⇒ λa = α n , m ⇒ λ n , m = a α n , m , α n , m \alpha_{n,m} α n , m m m m zero of J n J_n J n Why discrete? J n J_n J n λ \lambda λ u n , m ( r , θ ) = J n ( α n , m a r ) ( A cos n θ + B sin n θ ) . u_{n,m}(r,\theta)=J_n\!\Big(\frac{\alpha_{n,m}}{a}r\Big)(A\cos n\theta+B\sin n\theta). u n , m ( r , θ ) = J n ( a α n , m r ) ( A cos n θ + B sin n θ ) .
With λ = 0 \lambda=0 λ = 0 equidimensional : r 2 R ′ ′ + r R ′ − n 2 R = 0 r^2R''+rR'-n^2R=0 r 2 R ′′ + r R ′ − n 2 R = 0 R = r ± n R=r^{\pm n} R = r ± n ln r , \ln r,\; ln r , n = 0 n=0 n = 0 0 0 0 r ∣ n ∣ r^{|n|} r ∣ n ∣ u ( r , θ ) = a 0 2 + ∑ n ≥ 1 r n ( a n cos n θ + b n sin n θ ) , u(r,\theta)=\frac{a_0}{2}+\sum_{n\ge1}r^n(a_n\cos n\theta+b_n\sin n\theta), u ( r , θ ) = 2 a 0 + ∑ n ≥ 1 r n ( a n cos n θ + b n sin n θ ) , u ( a , θ ) = f ( θ ) u(a,\theta)=f(\theta) u ( a , θ ) = f ( θ ) Poisson-kernel interior problem). Why r n r^n r n J n J_n J n No λ 2 r 2 \lambda^2 r^2 λ 2 r 2
∇ 2 u = 0 \nabla^2 u=0 ∇ 2 u = 0 r ≤ 1 r\le1 r ≤ 1 u ( 1 , θ ) = 3 + 2 cos θ u(1,\theta)=3+2\cos\theta u ( 1 , θ ) = 3 + 2 cos θ Step 1. Match form u = a 0 2 + ∑ r n ( a n cos n θ + b n sin n θ ) u=\frac{a_0}2+\sum r^n(a_n\cos n\theta+b_n\sin n\theta) u = 2 a 0 + ∑ r n ( a n cos n θ + b n sin n θ ) Why? It's the general bounded Laplace solution.
Step 2. At r = 1 r=1 r = 1 a 0 2 + a 1 cos θ + ⋯ = 3 + 2 cos θ \frac{a_0}{2}+a_1\cos\theta+\dots=3+2\cos\theta 2 a 0 + a 1 cos θ + ⋯ = 3 + 2 cos θ a 0 / 2 = 3 a_0/2=3 a 0 /2 = 3 a 1 = 2 a_1=2 a 1 = 2 0 0 0 Step 3. Restore r r r u = 3 + 2 r cos θ = 3 + 2 x \boxed{u=3+2r\cos\theta=3+2x} u = 3 + 2 r cos θ = 3 + 2 x Why this step? r cos θ = x r\cos\theta=x r cos θ = x
Worked example (b) Fundamental drum mode, radiusa a a Step 1. Symmetric mode ⇒ n = 0 n=0 n = 0 R = J 0 ( λ r ) R=J_0(\lambda r) R = J 0 ( λ r ) Why n = 0 n=0 n = 0 Lowest mode has no angular nodes.
Step 2. Clamp: J 0 ( λ a ) = 0 J_0(\lambda a)=0 J 0 ( λa ) = 0 α 0 , 1 ≈ 2.405 \alpha_{0,1}\approx2.405 α 0 , 1 ≈ 2.405 Step 3. λ 0 , 1 = 2.405 / a \lambda_{0,1}=2.405/a λ 0 , 1 = 2.405/ a u = J 0 ( 2.405 r / a ) u=J_0(2.405\,r/a) u = J 0 ( 2.405 r / a ) Why care? This λ \lambda λ ω = c λ \omega=c\lambda ω = c λ
Y n Y_n Y n Y 0 ( s ) ∼ 2 π ln s → − ∞ Y_0(s)\sim\frac2\pi\ln s\to-\infty Y 0 ( s ) ∼ π 2 ln s → − ∞ s → 0 s\to0 s → 0 c 2 = 0 c_2=0 c 2 = 0 Why feel finite? Infinite displacement at the center is unphysical.
Common mistake "The radial part should besin / cos \sin/\cos sin / cos Why it feels right: the angular part gave clean trig, so we expect symmetry.
Fix: the radial ODE has variable coefficients (1 r , 1 r 2 \frac1r,\frac1{r^2} r 1 , r 2 1 J n J_n J n n 2 n^2 n 2
Y n Y_n Y n Why it feels right: it's a valid mathematical solution of Bessel's equation.
Fix: Y n Y_n Y n r = 0 r=0 r = 0 includes the center, so physical/bounded solutions require c 2 = 0 c_2=0 c 2 = 0 Y n Y_n Y n annulus 0 < b ≤ r ≤ a 0<b\le r\le a 0 < b ≤ r ≤ a
λ 2 \lambda^2 λ 2 J n J_n J n Why it feels right: for a string you used λ = n π / L \lambda=n\pi/L λ = nπ / L Fix: here λ a \lambda a λa zero of J n J_n J n , α n , m \alpha_{n,m} α n , m λ n , m = α n , m / a \lambda_{n,m}=\alpha_{n,m}/a λ n , m = α n , m / a
Common mistake Forgetting the1 r u r \frac1r u_r r 1 u r Why it feels right: by analogy with Cartesian u x x + u y y u_{xx}+u_{yy} u xx + u y y Fix: area element is r d r d θ r\,dr\,d\theta r d r d θ 1 r u r \frac1r u_r r 1 u r
Recall Feynman: explain to a 12-year-oldImagine a round drum. When you hit it, the skin ripples in rings and pie-slices. The "pie-slice" pattern is just nice waves around the circle (that's the cos n θ \cos n\theta cos n θ Bessel function . It wiggles like a wave but gets a bit flatter as you go out, because the skin spreads over more space. The drum can only ring at certain pitches: exactly where that wiggly curve hits zero at the clamped rim.
Mnemonic Remember the setup"Round Things Bring Bessel." R R R T T T s = λ r s=\lambda r s = λ r B B B B B B J n J_n J n Y n Y_n Y n "J J J Y Y Y
Separation of Variables — engine that splits r r r θ \theta θ Bessel Functions — J n , Y n J_n, Y_n J n , Y n Fourier Series — expanding boundary data f ( θ ) f(\theta) f ( θ ) cos n θ , sin n θ \cos n\theta,\sin n\theta cos n θ , sin n θ Sturm-Liouville Theory — orthogonality ∫ 0 a J n ( λ m r ) J n ( λ k r ) r d r = 0 \int_0^a J_n(\lambda_m r)J_n(\lambda_k r)\,r\,dr=0 ∫ 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 — where drum eigenmodes live.Euler-Cauchy Equation — the λ = 0 \lambda=0 λ = 0
Polar Laplacian formula u r r + 1 r u r + 1 r 2 u θ θ u_{rr}+\frac1r u_r+\frac1{r^2}u_{\theta\theta} u r r + r 1 u r + r 2 1 u θ θ Why does 1 r u r \frac1r u_r r 1 u r Area element
r d r d θ r\,dr\,d\theta r d r d θ ; geometric spreading of circles.
Separation constant for the angular part is called? n 2 n^2 n 2 , with periodicity forcing
n = 0 , 1 , 2 , … n=0,1,2,\dots n = 0 , 1 , 2 , … Angular equation and its solution Θ ′ ′ + n 2 Θ = 0 ⇒ A cos n θ + B sin n θ \Theta''+n^2\Theta=0\Rightarrow A\cos n\theta+B\sin n\theta Θ ′′ + n 2 Θ = 0 ⇒ A cos n θ + B sin n θ .
Radial ODE after separation (Helmholtz) r 2 R ′ ′ + r R ′ + ( λ 2 r 2 − n 2 ) R = 0 r^2R''+rR'+(\lambda^2r^2-n^2)R=0 r 2 R ′′ + r R ′ + ( λ 2 r 2 − n 2 ) R = 0 .
Substitution turning it into Bessel s = λ r s=\lambda r s = λ r , giving
s 2 y ′ ′ + s y ′ + ( s 2 − n 2 ) y = 0 s^2y''+sy'+(s^2-n^2)y=0 s 2 y ′′ + s y ′ + ( s 2 − n 2 ) y = 0 .
Two solutions of Bessel's equation J n J_n J n (finite at 0) and
Y n Y_n Y n (singular at 0).
Which is dropped on a full disk and why? Y n Y_n Y n , since it blows up at
r = 0 r=0 r = 0 (unphysical center).
Eigenvalue condition for clamped drum J n ( λ a ) = 0 ⇒ λ n , m = α n , m / a J_n(\lambda a)=0\Rightarrow \lambda_{n,m}=\alpha_{n,m}/a J n ( λa ) = 0 ⇒ λ n , m = α n , m / a ,
α \alpha α = zeros of
J n J_n J n .
First zero of J 0 J_0 J 0 ≈ 2.405 \approx2.405 ≈ 2.405 .
J 0 J_0 J 0 ∑ ( − 1 ) k / ( k ! ) 2 ( s / 2 ) 2 k = 1 − s 2 / 4 + s 4 / 64 − ⋯ \sum (-1)^k/(k!)^2 (s/2)^{2k}=1-s^2/4+s^4/64-\cdots ∑ ( − 1 ) k / ( k ! ) 2 ( s /2 ) 2 k = 1 − s 2 /4 + s 4 /64 − ⋯ Pure Laplace (λ = 0 \lambda=0 λ = 0 r ∣ n ∣ r^{|n|} r ∣ n ∣ (Euler equation), not Bessel.
General bounded Laplace solution on disk a 0 2 + ∑ r n ( a n cos n θ + b n sin n θ ) \frac{a_0}2+\sum r^n(a_n\cos n\theta+b_n\sin n\theta) 2 a 0 + ∑ r n ( a n cos n θ + b n sin n θ ) .
Orthogonality weight for Bessel modes weight
r r r :
∫ 0 a J n ( λ m r ) J n ( λ k r ) r d r = 0 \int_0^a J_n(\lambda_m r)J_n(\lambda_k r)r\,dr=0 ∫ 0 a J n ( λ m r ) J n ( λ k r ) r d r = 0 ,
m ≠ k m\ne k m = k .
special case lambda equals 0
Polar coordinates r theta
Variable coefficient radial ODE
Helmholtz problem lambda^2 u
Bessel functions Jn lambda r
Intuition Hinglish mein samjhoDekho, jab domain ek disk ho (gol plate ya drum), to Cartesian x , y x,y x , y polar coordinates ( r , θ ) (r,\theta) ( r , θ ) u r r + 1 r u r + 1 r 2 u θ θ u_{rr}+\frac1r u_r+\frac1{r^2}u_{\theta\theta} u r r + r 1 u r + r 2 1 u θ θ 1 r \frac1r r 1 r d r d θ r\,dr\,d\theta r d r d θ
Ab separation of variables karo: u = R ( r ) Θ ( θ ) u=R(r)\Theta(\theta) u = R ( r ) Θ ( θ ) cos n θ , sin n θ \cos n\theta,\sin n\theta cos n θ , sin n θ n n n s = λ r s=\lambda r s = λ r Bessel's equation . Iski solution hai Bessel function J n ( λ r ) J_n(\lambda r) J n ( λ r )
Center r = 0 r=0 r = 0 Y n Y_n Y n J n ( λ a ) = 0 J_n(\lambda a)=0 J n ( λa ) = 0 λ a \lambda a λa J n J_n J n zeros ke barabar hona padta hai — yeh zeros equally spaced nahi hote, isliye drum ki frequencies string jaisi simple nahi hotin. Aur agar pure Laplace ho (λ = 0 \lambda=0 λ = 0 r n r^n r n