Separation of Variables in Cylindrical Coordinates

Every problem with an axis of symmetry — a coaxial cable, a hollow waveguide, a vibrating drumhead, heat spreading through a rod, the field around a charged wire — asks the same question: what are the natural "modes" of a cylinder? Separation of variables answers it by factoring the solution into one function of the radius, one of the angle, and one of the height, $\psi=R(\rho)\,\Phi(\phi)\,Z(z)$ . Substitute this product into the wave, heat, Laplace, or Helmholtz equation, divide through, and the single PDE splinters into three ordinary differential equations linked only by two separation constants. Two of those ODEs are old friends. The angular equation $\Phi''+m^2\Phi=0$ has sine-and-cosine solutions — but with a twist: because a circle closes on itself, single-valuedness forces the index $m$ to be an integer. The axial equation $Z''-k^2Z=0$ is a one-dimensional Cartesian problem, exponentials or oscillations depending on a sign. The third ODE is where cylindrical geometry shows its hand: the Laplacian carries a variable coefficient $1/\rho$ , so the radial equation is not constant-coefficient. It is Bessel's equation, and its solutions — $J_m$ , $Y_m$ , or the modified $I_m$ , $K_m$ — are the "sines and cosines of a fanning-out geometry," oscillating with amplitude that decays like $1/\sqrt{\rho}$ as the wave spreads over larger circumferences. This article assembles the three separated ODEs, proves why $J_m$ survives at the axis while $Y_m$ is thrown away, shows how a boundary condition quantizes the radial constant through the zeros of $J_m$ , and uses Sturm–Liouville theory to build the completing Fourier–Bessel series. Two worked examples — a microwave waveguide and a drumhead — turn the abstract zeros $\alpha_{mn}$ into gigahertz and hertz.

Key takeaways

In cylindrical coordinates $(\rho,\phi,z)$ the Helmholtz, Laplace, wave, and heat equations separate into a product $R(\rho)\,\Phi(\phi)\,Z(z)$ whenever the geometry respects the axis of symmetry.
The angular factor obeys $\Phi''+m^2\Phi=0$ ; $2\pi$ -periodicity quantizes $m$ to an integer — a topological constraint, not an analytic one.
The axial factor obeys $Z''-k^2Z=0$ (exponentials) or $Z''+k^2Z=0$ (oscillatory); that sign choice ripples through to the radial equation.
The variable coefficient $1/\rho$ in $\nabla^2$ turns the radial factor into Bessel's equation: oscillatory $J_m,Y_m$ when the radial constant is positive, monotone $I_m,K_m$ when it is negative.
Regularity at the axis discards $Y_m$ (and $K_m$ ); a boundary at $\rho=a$ quantizes the radial constant through the zeros $\alpha_{mn}$ of $J_m$ , and Sturm–Liouville orthogonality with weight $\rho$ builds the Fourier–Bessel expansion.
The purely 2D Laplace case degenerates: the radial equation loses its $\rho^2$ term and becomes an Euler equation with power laws $\rho^{\pm m}$ and $\ln\rho$ — no Bessel functions at all.
Key idea: write the solution as $R(\rho)\Phi(\phi)Z(z)$ ; periodicity quantizes the angular index to integer $m$ , the axial constant fixes the sign, and the coefficient $1/\rho$ in $\nabla^2$ makes the radial equation Bessel's — ordinary or modified according to the sign of the radial separation constant.

The core insight

Write the solution as $R(\rho)\Phi(\phi)Z(z)$ ; periodicity quantizes the angular index to integer $m$ , the axial constant chooses the sign, and the variable coefficient $1/\rho$ in $\nabla^2$ turns the radial equation into Bessel's equation — ordinary or modified depending on the sign of the radial separation constant.

Where this lives today

The cylinder is still the most-solved geometry in applied physics. In microwave engineering, circular waveguides carry radar and satellite signals, and every mode chart is a list of the zeros $\alpha_{mn}$ and $\alpha'_{mn}$ . In photonics, a step-index fibre's modes are $J_m$ in the core and decaying $K_m$ in the cladding, and whether it is single-mode is set by where its $V$ -parameter falls among the Bessel zeros. In heat transfer, the slowest thermal mode relaxes as $e^{-D(\alpha_{01}/a)^2t}$ , setting quench times for rods and fuel pellets. And in accelerator physics and circuit QED, the resonances of a pillbox cavity are zeros of $J_m$ , with TM $_{010}$ at $f=c\,\alpha_{01}/2\pi a$ the standard accelerating mode. Every one is the same separation with different boundary conditions.

The physical picture

Picture the Laplacian as a sum of curvatures. In Cartesian coordinates every direction gives a clean second derivative, so each factor is a sine or cosine and the modes are plane waves. A cylinder breaks that symmetry gently: it keeps one flat direction (the axis $z$ ) and one closed direction (the angle $\phi$ ), but the radial direction is stretched. Moving outward you sweep through circles of growing circumference, so a wave carrying fixed power must thin its amplitude as it spreads — exactly what the operator $\tfrac{1}{\rho}\partial_\rho(\rho\,\partial_\rho)$ encodes, and why the radial factor is a Bessel function rather than a plain sinusoid: Bessel functions oscillate, but their crests shrink like $1/\sqrt{\rho}$ .

Figure

One field ψ decomposed into three one-variable factors: a Bessel radial part R(ρ), a periodic angular part Φ(φ) whose single-valuedness forces integer m, and an exponential or oscillatory axial part Z(z)

The angular factor is the simplest because a circle has no boundary — only the demand that you return to the same value after one full turn, $\Phi(\phi+2\pi)=\Phi(\phi)$ . That single-valuedness forces $m$ to be a whole number: a topological constraint, imposed by the shape of the domain rather than any equation. The whole solution is therefore "Cartesian in $z$ , periodic in $\phi$ , and Bessel in $\rho$ " — and everything physical happens in that radial part, whose zeros $\alpha_{mn}$ become the resonant frequencies of drums and the cutoffs of waveguides. The second figure plots those functions and marks the zeros that carry the numbers.

Figure

The first three Bessel functions J₀, J₁, J₂ against x = k_ρ ρ, each oscillating with amplitude decaying like 1/√x; their first zeros α₀₁ = 2.405, α₁₁ = 3.832, α₂₁ = 5.136 fix the drum frequencies ω = c α_mn / a and the waveguide cutoffs f = c α_mn / 2πa

The core equations

Separation begins with the Laplacian in the coordinates that match the geometry. Its one non-trivial feature is the factor $1/\rho$ on the radial derivative — the fingerprint of a curved coordinate system, and the source of every Bessel function that follows.

\nabla^2\psi=\frac{1}{\rho}\frac{\partial}{\partial\rho}\!\left(\rho\frac{\partial\psi}{\partial\rho}\right)+\frac{1}{\rho^2}\frac{\partial^2\psi}{\partial\phi^2}+\frac{\partial^2\psi}{\partial z^2} \qquad (1)

where: $\rho$ — radial distance from the $z$ -axis, $\rho\ge0$ ; $\phi$ — azimuthal angle, $0\le\phi<2\pi$ ; $z$ — axial coordinate; $\psi$ — the scalar field. The radial curvature is written in the self-adjoint form $\tfrac{1}{\rho}\partial_\rho(\rho\,\partial_\rho)$ that will matter for orthogonality. The method rests on a single hopeful guess: that the solution is a product of one-variable pieces. When the operator and the domain both respect the coordinate symmetry, it works — and the separated pieces then span every solution.

\psi(\rho,\phi,z)=R(\rho)\,\Phi(\phi)\,Z(z) \qquad (2)

where: $R,\Phi,Z$ — single-variable factors to be determined. Substituting (2) into $\nabla^2\psi=-\lambda\psi$ and dividing by $\psi$ forces each single-variable group to equal a constant. The angular piece separates first and cleanest — an eigenvalue problem on the circle whose eigenvalues lock to perfect squares of integers.

\Phi''+m^2\Phi=0,\qquad \Phi(\phi)=e^{\pm im\phi}\ \text{or}\ \{\cos m\phi,\ \sin m\phi\},\quad m\in\mathbb{Z} \qquad (3)

where: $m$ — azimuthal index, forced to be an integer by the periodicity $\Phi(\phi+2\pi)=\Phi(\phi)$ ; $m^2$ is the first separation constant, and a non-integer $m$ would make the field jump on returning to the same point. The axial piece is a one-dimensional constant-coefficient ODE whose solutions depend on the sign of the separation constant — the choice that decides whether the radial equation is ordinary or modified Bessel.

Z''-k^2Z=0,\qquad Z(z)=e^{\pm kz}\ \text{or}\ \{\cosh kz,\ \sinh kz\}\quad(\text{or }e^{\pm ikz}\text{ if }k\to ik) \qquad (4)

where: $k$ — axial separation constant. A real $k$ gives exponential/hyperbolic behaviour along the axis; $k\to ik$ gives propagating $e^{\pm ikz}$ , chosen by the boundary data at the ends. What remains after removing the angular and axial dependence is the radial equation — and because the Laplacian carried the $1/\rho$ coefficient, this ODE is Bessel's equation in disguise.

\rho^2R''+\rho R'+\big(k_\rho^2\rho^2-m^2\big)R=0,\qquad k_\rho^2=\lambda+k^2 \qquad (5)

where: $k_\rho$ — radial wavenumber, set jointly by the Helmholtz eigenvalue $\lambda$ and the axial constant $k$ . The bracket $k_\rho^2\rho^2-m^2$ changes sign at $\rho=m/k_\rho$ — the angular-momentum barrier that pushes each mode away from the axis. Rescaling $x=k_\rho\rho$ puts (5) into the canonical form defining the Bessel functions.

x^2y''+xy'+(x^2-m^2)y=0\ \Rightarrow\ y=J_m(x),\,Y_m(x),\quad x=k_\rho\rho \qquad (6)

where: $J_m$ — Bessel function of the first kind, finite at the origin, $J_m(x)\sim(x/2)^m/m!$ ; $Y_m$ — second kind (Neumann), diverging as $x\to0$ like $\ln x$ ( $m=0$ ) or $x^{-m}$ ( $m\ge1$ ). On any domain touching the axis, only $J_m$ survives. When the radial separation constant is negative ( $k_\rho^2<0$ ), the $x^2$ term flips sign and Bessel's equation becomes the modified Bessel equation, whose solutions no longer oscillate.

x^2y''+xy'-(x^2+m^2)y=0\ \Rightarrow\ y=I_m(x),\,K_m(x) \qquad (7)

where: $I_m$ — modified Bessel function of the first kind, regular at $0$ , growing like $e^x$ ; $K_m$ — second kind, decaying like $e^{-x}$ but singular at $0$ . These describe evanescent, screened, or below-cutoff fields — anything dying away radially rather than ringing. Finally, a homogeneous boundary at $\rho=a$ selects discrete eigenvalues whose radial eigenfunctions are orthogonal in the Sturm–Liouville sense, with weight $\rho$ from the self-adjoint operator — the orthogonality behind Fourier–Bessel expansions.

\int_0^a J_m\!\Big(\alpha_{mn}\tfrac{\rho}{a}\Big)J_m\!\Big(\alpha_{mp}\tfrac{\rho}{a}\Big)\rho\,\mathrm{d}\rho=\frac{a^2}{2}\big[J_{m+1}(\alpha_{mn})\big]^2\,\delta_{np} \qquad (8)

where: $\alpha_{mn}$ — the $n$ -th positive zero of $J_m$ , fixing $k_\rho=\alpha_{mn}/a$ ; $a$ — the disk or cylinder radius; $\delta_{np}$ — Kronecker delta. The weight $\rho\,\mathrm{d}\rho$ is exactly the disk area element.

Derivation ASeparating the Helmholtz equation in a cylinder

The goal is to reduce the single equation $\nabla^2\psi+\lambda\psi=0$ , without approximation, to three ordinary differential equations plus two separation constants, tracking the sign of each constant and the physical constraint that pins it down. This is the master computation of the article: the wave, heat, and Laplace equations are all special cases, so separating Helmholtz separates all of them at once, and the assembled mode $\psi_{mn}$ is the building block every later result expands over.

Step 1. Substitute $\psi=R\Phi Z$ into $\nabla^2\psi+\lambda\psi=0$ and divide by $\psi$ : $\dfrac{(\rho R')'}{\rho R}+\dfrac{\Phi''}{\rho^2\Phi}+\dfrac{Z''}{Z}+\lambda=0$ . Why: dividing by the product isolates each variable into its own additive term — a sum of functions of independent variables can only be constant if each is.

Step 2. The axial term $\dfrac{Z''}{Z}$ depends on $z$ alone while the rest depends on $\rho,\phi$ , so it is a constant $k^2$ , giving $Z''-k^2Z=0$ . Why: the sign $+k^2$ is bookkeeping — real means an axially decaying field, and $k\to ik$ later recovers a propagating one without redoing the algebra.

Step 3. Multiply by $\rho^2$ and isolate the angular term $\dfrac{\Phi''}{\Phi}=-m^2$ , giving $\Phi''+m^2\Phi=0$ with $\Phi=e^{\pm im\phi}$ ; single-valuedness forces $m\in\mathbb{Z}$ . Why: the minus sign is not optional — a positive constant gives non-periodic real exponentials, so periodicity of the closed angular coordinate selects both the sign and the integer quantization.

Step 4. What is left is $\rho^2R''+\rho R'+\big((\lambda+k^2)\rho^2-m^2\big)R=0$ , which with $k_\rho^2=\lambda+k^2$ is Bessel's equation (5). Why: the constants $m$ and $k$ enter the radial equation only as an angular-momentum barrier $m^2/\rho^2$ and a radial wavenumber $k_\rho$ — all the radial problem remembers of the other two.

Step 5. On a domain containing the axis, discard $Y_m(k_\rho\rho)$ (which diverges at $\rho=0$ ) and keep $R(\rho)=J_m(k_\rho\rho)$ . Why: regularity at the axis is a genuine boundary condition even though $\rho=0$ is not a wall — it is a singular point of the ODE, and finiteness there kills one solution.

Step 6. A Dirichlet condition $J_m(k_\rho a)=0$ forces $k_\rho a$ to a zero $\alpha_{mn}$ of $J_m$ , so $k_\rho=\alpha_{mn}/a$ and $\psi_{mn}=J_m\!\big(\alpha_{mn}\tfrac{\rho}{a}\big)e^{\pm im\phi}e^{\pm kz}$ with $k^2=(\alpha_{mn}/a)^2-\lambda$ . Why: here a continuous eigenvalue becomes a discrete spectrum — the boundary selects a countable set of radial wavenumbers, as a string's fixed ends select its harmonics. *The zeros $\alpha_{01}=2.40483$ , $\alpha_{11}=3.83171$ , $\alpha_{21}=5.13562$ have been confirmed numerically.*

Go deeper · Derivation B · 1 minRegularity at the axis from the indicial equation

Derivation BRegularity at the axis from the indicial equation

The claim that " $J_m$ is regular and $Y_m$ blows up at $\rho=0$ " is used constantly, but it deserves a proof, because it is the single fact that halves the solution space of every interior cylinder problem. The goal here is to analyze Bessel's equation at its regular singular point $x=0$ with a Frobenius expansion and show why one solution stays finite while the other diverges — and why for integer $m$ the divergent one carries a logarithm.

Step 1. $x=0$ is a regular singular point of $x^2y''+xy'+(x^2-m^2)y=0$ : dividing by $x^2$ the coefficients $1/x$ and $(1-m^2/x^2)$ blow up, but $x\cdot(1/x)$ and $x^2\cdot(-m^2/x^2)$ stay finite. Why: this guarantees at least one solution of pure power-law form, licensing the Frobenius ansatz.

Step 2. Try $y=\sum_{k\ge0}a_k x^{s+k}$ with $a_0\ne0$ . The lowest-order terms give the indicial equation $s^2-m^2=0$ , so $s=+m$ or $s=-m$ . Why: these are the two allowed leading behaviours — a solution starting like $x^{+m}$ (finite) and one like $x^{-m}$ (infinite).

Step 3. The root $s=+m$ generates $J_m(x)=\sum_{k\ge0}\dfrac{(-1)^k}{k!\,(k+m)!}\big(\tfrac{x}{2}\big)^{2k+m}$ , finite with leading term $(x/2)^m/m!$ . Why: this is the regular solution; at $x=0$ it equals $0$ for $m\ge1$ and $1$ for $m=0$ , so a field built from $J_m$ has no singularity on the axis.

Step 4. The root $s=-m$ gives $x^{-m}$ , but since the roots differ by the integer $2m$ the second solution may carry a logarithm; it is $Y_m(x)$ , behaving as $-\tfrac{(m-1)!}{\pi}(2/x)^m$ for $m\ge1$ and $\tfrac{2}{\pi}\ln x$ for $m=0$ — both divergent. Why: this makes the discard rule rigorous — $Y_m$ is genuinely unbounded, so it appears only in geometries (annuli, exteriors) that exclude $\rho=0$ . *The Frobenius series for $J_m$ satisfies (6) term by term; $Y_m$ diverges as $\ln x$ for $m=0$ and as $x^{-m}$ for $m\ge1$ .*

Go deeper · Derivation C · 1 minFourier–Bessel completeness from Sturm–Liouville theory

Derivation CFourier–Bessel completeness from Sturm–Liouville theory

The reason separation of variables actually solves boundary-value problems — rather than just producing a menu of modes — is that the radial modes form a complete orthogonal basis, so any admissible profile expands over them with coefficients read off by projection. The goal here is to put the radial equation into self-adjoint Sturm–Liouville form, prove orthogonality with weight $\rho$ from that form, and derive the Fourier–Bessel coefficient formula that ends every cylinder calculation.

Step 1. Write (5) at eigenvalue $\lambda_n=(\alpha_{mn}/a)^2$ in self-adjoint form: $\dfrac{\mathrm{d}}{\mathrm{d}\rho}\!\Big(\rho\dfrac{\mathrm{d}R_n}{\mathrm{d}\rho}\Big)-\dfrac{m^2}{\rho}R_n=-\lambda_n\,\rho\,R_n$ , so $p(\rho)=\rho$ and weight $w(\rho)=\rho$ . Why: the weight is dictated by the operator, and it is exactly the disk area element $\rho\,\mathrm{d}\rho$ — why physical integrals over a disk come out orthogonal.

Step 2. Multiply the $R_n$ equation by $R_p$ , subtract the $R_p$ equation times $R_n$ , and integrate over $[0,a]$ ; the left side collapses to the boundary term $\big[\rho(R_nR_p'-R_pR_n')\big]_0^a$ . Why: this Lagrange-identity move converts the whole interior into a pure boundary evaluation, so orthogonality hinges entirely on the two ends.

Step 3. At $\rho=0$ the factor $\rho\to0$ with $J_m$ regular, so it vanishes; at $\rho=a$ the Dirichlet condition kills it. Hence $(\lambda_n-\lambda_p)\int_0^a R_nR_p\,\rho\,\mathrm{d}\rho=0$ , giving orthogonality for $n\ne p$ . Why: both ends vanish — the axis through regularity, the wall through the boundary condition — the self-adjointness that guarantees real eigenvalues and a complete basis.

Step 4. Normalize with $\int_0^a J_m^2(\alpha_{mn}\rho/a)\,\rho\,\mathrm{d}\rho=\tfrac{a^2}{2}[J_{m+1}(\alpha_{mn})]^2$ , then expand $f(\rho)=\sum_n c_nJ_m(\alpha_{mn}\rho/a)$ and project: $c_n=\dfrac{2}{a^2[J_{m+1}(\alpha_{mn})]^2}\displaystyle\int_0^a f(\rho)J_m\!\big(\alpha_{mn}\tfrac{\rho}{a}\big)\rho\,\mathrm{d}\rho$ . Why: completeness means the series converges to $f$ in the mean, so the "assemble the mode" step of every derivation is legitimate and the coefficients unique. *The normalization integral follows from the identity $J_m'(\alpha_{mn})=-J_{m+1}(\alpha_{mn})$ , which holds at any zero of $J_m$ .*

Limiting cases

The four sign combinations of the separation constants produce qualitatively different radial behaviour. Holding $m$ integer and varying the sign of the radial constant walks through the whole zoo — oscillatory, evanescent, and the degenerate purely-planar case.

Case	Result	Meaning
Oscillatory radial ( $k_\rho^2>0$ )	$R=J_m(k_\rho\rho),\ Y_m(k_\rho\rho)$ ; keep $J_m$ if the axis is in the domain	Standing or propagating modes — membranes, waveguides, cavity resonators
Monotone radial ( $k_\rho^2<0$ )	$R=I_m(\kappa\rho),\ K_m(\kappa\rho)$ , $\kappa=\\|k_\rho\\|$ ; $I_m$ grows, $K_m$ decays	Evanescent fields — below-cutoff waveguides, screened potentials, fibre cladding
Purely 2D Laplace ( $k_\rho=0$ )	$R=\rho^{m},\ \rho^{-m}\ (m\ge1)$ ; $R=1,\ \ln\rho\ (m=0)$	Euler equation — power laws, no Bessel functions — wire electrostatics
Large-argument asymptotics	$J_m(x)\sim\sqrt{\tfrac{2}{\pi x}}\cos\!\big(x-\tfrac{m\pi}{2}-\tfrac{\pi}{4}\big)$ ; $K_m(x)\sim\sqrt{\tfrac{\pi}{2x}}e^{-x}$	The $1/\sqrt{\rho}$ decay of a cylindrical wave and the exponential decay of an evanescent mode

Common trapCylindrical symmetry always means Bessel functions

This over-generalization is wrong twice. First, when the field has no axial structure — a straight wire, a 2D electrostatics problem — the $\rho^2$ term drops out and the radial equation degenerates to the equidimensional Euler equation $\rho^2R''+\rho R'-m^2R=0$ , whose solutions are the power laws $\rho^{\pm m}$ and $\ln\rho$ , not Bessel functions: a scale-free 2D Laplace problem has no length scale to set an oscillation. Second, even when the equation is Bessel-type, the sign of $k_\rho^2=\lambda+k^2$ decides everything — positive gives oscillatory $J_m,Y_m$ , negative gives monotone $I_m,K_m$ , and the wrong sign yields solutions that can never meet your boundary conditions. The companion trap: never retain $Y_m$ , $K_m$ , $\rho^{-m}$ , or $\ln\rho$ on a domain that includes $\rho=0$ , because all of them diverge there.

Worked examples

1 · Cutoff frequencies of a circular microwave waveguide (fully worked)

A hollow circular waveguide has inner radius $a=1.00\ \mathrm{cm}$ . Find the cutoff frequency of the lowest TM mode (TM $_{01}$ ) and the dominant mode (TE $_{11}$ ), state the single-mode band, and find how fast a below-cutoff TM $_{01}$ field decays at $8\ \mathrm{GHz}$ .

Step 1. For TM modes the axial field $E_z=J_m(k_\rho\rho)e^{\pm im\phi}e^{i(k_zz-\omega t)}$ must vanish on the wall, so $J_m(k_\rho a)=0$ and $k_\rho=\alpha_{mn}/a$ . The lowest TM mode uses the first zero of $J_0$ , $\alpha_{01}=2.40483$ , so $k_\rho=240.5\ \mathrm{m^{-1}}$ .

Step 2. Cutoff is where $k_z\to0$ , i.e. $\omega_c=ck_\rho$ , so $f_{c}=\dfrac{c\,\alpha_{mn}}{2\pi a}$ . For TM $_{01}$ : $f_c=\dfrac{(2.998\times10^8)(2.40483)}{2\pi(0.0100)}=\mathbf{11.47\ GHz}$ .

Step 3. The dominant mode is TE $_{11}$ , whose wall condition $J_m'(k_\rho a)=0$ uses the first zero of $J_1'$ , $\alpha'_{11}=1.84118$ , giving $f_c=\dfrac{(2.998\times10^8)(1.84118)}{2\pi(0.0100)}=\mathbf{8.79\ GHz}$ . Since $8.79<11.47$ , single-mode operation occupies the band $8.79\text{–}11.47\ \mathrm{GHz}$ : TE $_{11}$ alone propagates until TM $_{01}$ joins it above $11.47\ \mathrm{GHz}$ .

Step 4. Drive TM $_{01}$ below cutoff, at $f=8.00\ \mathrm{GHz}$ , using $k_z^2=(\omega/c)^2-(\alpha_{01}/a)^2$ . Here $(\omega/c)^2=2.81\times10^4$ and $(\alpha_{01}/a)^2=5.78\times10^4$ , so $k_z^2=-2.97\times10^4\ \mathrm{m^{-2}}$ and $k_z=i(172.4\ \mathrm{m^{-1}})$ . The field is evanescent, $\propto e^{-\|k_z\|z}$ , with $1/e$ decay length $\mathbf{5.8\ mm}$ — the basis of a waveguide-beyond-cutoff attenuator.

2 · The inharmonic overtones of a drum (fully worked)

A circular membrane of radius $a=0.30\ \mathrm{m}$ carries transverse waves at $c=100\ \mathrm{m/s}$ and is clamped at its rim. Find its fundamental and first few overtones, and compare their ratios to a vibrating string.

Step 1. With no axial dependence the spatial problem is the 2D Helmholtz equation $\nabla^2X=-(\omega/c)^2X$ , giving $X=J_m(k_\rho\rho)(A\cos m\phi+B\sin m\phi)$ with $k_\rho=\omega/c$ . The clamped rim demands $J_m(k_\rho a)=0$ , so $k_\rho a=\alpha_{mn}$ and $f_{mn}=\dfrac{c\,\alpha_{mn}}{2\pi a}$ .

Step 2. The fundamental is the axisymmetric $(0,1)$ mode, $\alpha_{01}=2.40483$ : $f_{01}=\dfrac{(100)(2.40483)}{2\pi(0.30)}=\mathbf{127.6\ Hz}$ — a single dome with a nodal circle only at the rim.

Step 3. The next zeros $\alpha_{11}=3.83171$ , $\alpha_{21}=5.13562$ , $\alpha_{02}=5.52008$ give $f_{11}=\mathbf{203.3}$ , $f_{21}=\mathbf{272.5}$ , $f_{02}=\mathbf{292.8\ Hz}$ .

Step 4. Ratios to the fundamental: $1.000,\ 1.593,\ 2.136,\ 2.295$ . A clamped string gives exact integers $1,2,3,\dots$ , which the ear hears as a definite pitch; the drum's $1.59, 2.14, 2.30$ are not integer multiples, so its overtones are inharmonic. That is why a drum sounds percussive while a string sings a clear note — the difference between the zeros of $J_m$ and the integers.

Physics of the portrait

Bessel functions sit at the crossroads of astronomy, acoustics, and electromagnetism — and the person whose name they carry came to them not through waves at all, but through the orbits of planets.

Origins · 1 minFriedrich Bessel and how this was discovered

Friedrich Bessel

1784–1846 · the astronomer who systematized the cylinder functions

Friedrich Wilhelm Bessel began as a merchant's apprentice with a gift for computation, taught himself astronomy, and rose to found and direct the Königsberg Observatory; in 1838 he measured the first stellar parallax (61 Cygni), giving humanity its first true distance to a star. But the functions that bear his name entered through a side door: in an 1824 memoir on planetary perturbations he expanded the Kepler problem — the relation between a planet's mean and eccentric anomaly — in a trigonometric series whose coefficients turned out to be a family of integrals $J_m$ he studied with unprecedented care. He was not their discoverer: Daniel Bernoulli had met $J_0$ in a hanging chain in the 1730s, Euler in a circular membrane in 1764, and Fourier in heat flow through a cylinder in 1822. What Bessel supplied was the systematic theory — recurrence relations, integral representations, tables of zeros — that turned scattered special cases into a coherent tool, until separation of variables in cylindrical coordinates became, in effect, the study of Bessel's functions.

History

1732

— Daniel Bernoulli meets J₀ in the oscillations of a heavy hanging chain

1764

— Euler solves the vibrating circular membrane in terms of the same functions

1822

— Fourier expands heat conduction in a cylinder over these functions

1824

— Bessel studies the functions systematically in his theory of planetary perturbations

1867

— Neumann and Hankel complete the second solution and the integral representations

Going further

Texts. Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Chs. 9 and 14 · Stone & Goldbart, Mathematics for Physics, Ch. 8 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., §19–21 · Hassani, Mathematical Physics, Chs. 22–23 · Jackson, Classical Electrodynamics, 3rd ed., §3.7–3.11 and §8.7 · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 6 · Morse & Feshbach, Methods of Theoretical Physics, Ch. 5. The indicial exponents of the radial equation are $s=\pm m$ ; the Bessel zeros used throughout are $\alpha_{01}=2.40483$ , $\alpha_{11}=3.83171$ , $\alpha_{21}=5.13562$ , $\alpha_{02}=5.52008$ , and $\alpha'_{11}=1.84118$ . The waveguide cutoff formula $f_c=c\alpha/2\pi a$ gives TM $_{01}$ at 11.47 GHz and TE $_{11}$ at 8.79 GHz for $a=1\ \mathrm{cm}$ , with a below-cutoff decay length of 5.8 mm at 8 GHz. The drum frequencies $f_{mn}=c\alpha_{mn}/2\pi a$ evaluate to 127.6, 203.3, 272.5, and 292.8 Hz for $a=0.30\ \mathrm{m}$ and $c=100\ \mathrm{m/s}$ .

Prerequisites. The Helmholtz Equation · Separation of Variables in Cartesian Coordinates · Sturm–Liouville Theory and Eigenfunction Expansions · Boundary Conditions and Well-Posedness · Classification of Second-Order Linear PDEs. Leads to. Sturm–Liouville Theory and Eigenfunction Expansions · Fundamental Solutions and Green's Functions for PDEs · The Wave Equation · Separation of Variables in Spherical Coordinates.

Ask the tutor: Give me any cylinder problem — a coaxial cable, a fibre, a drumhead, a heated rod — and I will walk you through which of $J_m$ , $Y_m$ , $I_m$ , $K_m$ , or a plain power law survives and why.

Why does the angular equation get a minus sign (Φ'' + m²Φ = 0) but the axial one a plus (Z'' − k²Z = 0)?

The signs are forced by the domains. The angular coordinate is closed: $\phi$ and $\phi+2\pi$ are the same point, so $\Phi$ must be periodic, and only the negative constant $-m^2$ gives periodic solutions — a positive one would give real exponentials that cannot close on themselves. The axial coordinate is open, so either sign is allowed; $-k^2$ names exponential growth/decay along the axis, and $+k^2$ (i.e. $k\to ik$ ) an axially propagating wave. The geometry of each coordinate picks the sign.

Where do the modified Bessel functions I_m and K_m actually come from?

They are what Bessel's equation becomes when the radial constant $k_\rho^2=\lambda+k^2$ is negative. Rescale by $\kappa=\sqrt{-k_\rho^2}$ and the $+x^2$ becomes $-x^2$ ; the oscillatory $J_m,Y_m$ give way to the monotone $I_m,K_m$ . Physically this is any field that decays radially instead of ringing: below-cutoff waveguide fields, fibre cladding, screened potentials. $I_m$ grows like $e^x$ , while $K_m$ decays like $e^{-x}$ and is the natural choice for a field that must die away far from a source.

What is "regularity at the axis" really doing, since ρ = 0 is not a wall?

The point $\rho=0$ is a singular point of the radial ODE, not a physical boundary, but it acts like a boundary condition. Bessel's equation has one solution finite at the origin ( $J_m$ ) and one infinite ( $Y_m$ ); a physical field cannot be infinite on an axis that lies inside the material, so demanding finiteness there eliminates $Y_m$ and halves the solution space, exactly as a Dirichlet condition would at a real wall. In an annulus or exterior region, where $\rho=0$ is excluded, both solutions return and you need $Y_m$ (or $K_m$ ) again.

Why does the weight ρ appear in the Fourier–Bessel orthogonality integral?

Because it is built into the operator. In self-adjoint Sturm–Liouville form, $\tfrac{\mathrm{d}}{\mathrm{d}\rho}(\rho\,R')-\tfrac{m^2}{\rho}R=-\lambda\rho R$ , the coefficient multiplying the eigenvalue $\lambda$ is $\rho$ , and that coefficient is the weight. Sturm–Liouville theory then makes eigenfunctions for different eigenvalues orthogonal under $\int f g\,\rho\,\mathrm{d}\rho$ . Geometrically $\rho\,\mathrm{d}\rho\,\mathrm{d}\phi$ is the disk area element, so orthogonality just means distinct modes are orthogonal integrated over the physical disk.

Why are drum overtones inharmonic while string overtones are harmonic?

A string's frequencies are $\omega_n=n\pi c/L$ — exact integer multiples of the fundamental, which the ear fuses into a single clear pitch. A drum's frequencies scale as the zeros of Bessel functions, $\omega_{mn}=c\,\alpha_{mn}/a$ , and those zeros ( $2.405, 3.832, 5.136, 5.520,\dots$ ) are not in integer ratios, so the overtones fall at inharmonic intervals like $1.59, 2.14, 2.30$ — a percussive, pitch-ambiguous sound. It is the direct audible consequence of the radial equation being Bessel's, not the constant-coefficient equation of a straight string.

When does the cylinder problem give Bessel functions and when power laws?

It comes down to whether there is a length scale in the radial equation. If the field varies along the axis or oscillates in time — anything making $k_\rho^2=\lambda+k^2$ nonzero — the equation keeps its $k_\rho^2\rho^2$ term and gives Bessel functions. But a purely 2D static Laplace problem has $\lambda=k=0$ , so $k_\rho^2=0$ , the $\rho^2$ term vanishes, and it collapses to the scale-free Euler equation $\rho^2R''+\rho R'-m^2R=0$ with power laws $\rho^{\pm m}$ and $\ln\rho$ . That is why a line charge gives $\ln\rho$ — no length to set an oscillation, no Bessel function.