Bessel's equation and Bessel functions (intro, physical relevance): Worked examples

The scenario matrix

Read this table first. Each cell is a distinct thing an exam can test. The examples below are tagged with the cell they cover.

Cell	What changes	The question it asks	Example
A Integer order	$ν = n$ (whole number)	Recognise standard form, name the solution	Ex 1
B Half-integer order	$ν = \frac{1}{2}, \frac{3}{2}$	Does it collapse to sin/cos?	Ex 2
C Centre included ( $r = 0$ )	region is a solid disc	Which solution survives boundedness?	Ex 3
D Centre excluded (annulus)	hole in the middle	Do we keep both solutions?	Ex 4
E Disguised equation	not yet in $x^{2} y^{''} + x y^{'} + \dots$ form	Rescale/multiply to reveal $ν$	Ex 5
F Rescaled argument	$x = λ r$	Where do the $λ$ 's go?	Ex 6
G Degenerate: $ν = 0$	order vanishes	$J_{0} (0) = 1$ , frequencies	Ex 7
H Real-world word problem	physical drum, units	Turn physics into a number	Ex 7
I Limiting behaviour	$x \to 0$ and $x \to \infty$	Small/large- $x$ shapes	Ex 8
J Exam twist	a " $- ν^{2}$ " sign trap / wrong-order trap	Spot the misread	Ex 9

Definition Symbols used on this page (each earned before use)

$x$ — the independent variable (a number you feed in). It runs along the horizontal axis of every graph.
$r$ — the radial coordinate: the straight-line distance from the centre of a disc or cylinder out to the point you care about. In every physical example it is the "how far from the middle" number, and it links to $x$ through a rescaling like $x = k r$ or $x = λ r$ .
$ν$ (Greek "nu") — the order, a fixed constant chosen once per problem. Never varies.
$J_{ν} (x)$ — the Bessel function of the first kind: the wiggly, decaying "sine for circles" that stays finite at $x = 0$ .
$Y_{ν} (x)$ — the Bessel function of the second kind: the partner solution that shoots to $- \infty$ at $x = 0$ .
$c_{1}, c_{2}$ — the two arbitrary constants (integration constants) that any second-order linear ODE carries. They are fixed later by boundary/finiteness conditions; on a solid disc one of them is forced to zero.
$α_{ν, k}$ — the $k$ -th positive zero of $J_{ν}$ (a value of $x$ where the curve crosses the axis).
$Γ$ — the Gamma function, the smooth version of factorial: $Γ (n + 1) = n!$ .

Ex 1 — Integer order (Cell A)

Worked example Statement

Solve $x^{2} y^{''} + x y^{'} + (x^{2} - 25) y = 0$ .

Forecast: Guess the value of $ν$ before reading on. What number, squared, is $25$ ?

Step 1 — Line up against the standard form. $x^{2} y^{''} + x y^{'} + (x^{2} - ν^{2}) y = 0.$ Why this step? Bessel's equation has a rigid skeleton — coefficients $x^{2}, x, (x^{2} - ν^{2})$ . Matching term by term is the fastest legal move; no integration is needed once we recognise the pattern.

Step 2 — Read off the order. The constant subtracted from $x^{2}$ is $ν^{2}$ , so $ν^{2} = 25 \Rightarrow ν = 5$ (we take $ν \geq 0$ ). Why this step? $ν$ is defined non-negative; the sign is fixed by convention, and $J_{- 5} = J_{5}$ up to a sign anyway.

Step 3 — Write the general solution. $y (x) = c_{1} J_{5} (x) + c_{2} Y_{5} (x) .$ Why this step? A second-order linear ODE always has two independent solutions; $J_{5}$ and $Y_{5}$ are the standard pair.

Verify: $ν^{2} = 5^{2} = 25$ ✓ matches the " $- 25$ ". The equation has no extra multiplier on $y^{''}$ , so no rescaling was hidden.

Ex 2 — Half-integer order collapses to sin/cos (Cell B)

Worked example Statement

Show that $J_{1/2} (x) = \frac{2}{π x} sin x$ using the series, and evaluate it at $x = \frac{π}{2}$ .

Forecast: Half-integer orders are secretly elementary. Do you expect a clean number at $x = π /2$ ?

Step 1 — Write the series with $ν = \frac{1}{2}$ . $J_{1/2} (x) = \sum_{m = 0}^{\infty} \frac{( - 1 ) ^{m}}{m ! Γ ( m + \frac{3}{2} )} (\frac{x}{2})^{2 m + 1/2} .$ Why this step? The Frobenius method gives $J_{ν}$ as this power series; plugging $ν = \frac{1}{2}$ is a direct substitution.

Step 2 — Use the Gamma function half-integer values. With $Γ (m + \frac{3}{2}) = \frac{( 2 m + 1 )!}{2 ^{2 m + 1} m !} π$ , the $m!$ cancels and the sum rearranges into the Taylor series of $sin x$ divided by $x$ . Why this step? The half-integer Gamma is the only bridge from the abstract series back to a function you already know. It is what makes the collapse happen.

Step 3 — Read the closed form. $J_{1/2} (x) = \frac{2}{π x} sin x .$ Why this step? Once the rearranged sum is literally $2/ (π x)$ times the $sin x$ Taylor series, we are entitled to replace that series by its named function — that recognition is the whole payoff, turning an infinite sum into one elementary expression we can graph and differentiate.

Step 4 — Evaluate at $x = π /2$ . $sin (π /2) = 1$ , so $J_{1/2} (\frac{π}{2}) = \frac{2}{π \cdot π /2} \cdot 1 = \frac{4}{π ^{2}} = \frac{2}{π} .$ Why this step? A concrete number is the best proof the closed form is right.

Verify: Numerically $2/ π \approx 0.6366$ . You can reproduce this on any calculator, and the automated SymPy snippet confirms the symbolic identity $2/ (π \cdot π /2) sin (π /2) = 2/ π$ .

Ex 3 — Centre included: boundedness kills $Y_{ν}$ (Cell C)

Worked example Statement

Heat in a solid cylinder ( $0 \leq r \leq a$ ). The radial part obeys $x^{2} R^{''} + x R^{'} + (x^{2} - 4) R = 0$ with $x = k r$ . Which solution survives, and why?

Forecast: Two constants $c_{1}, c_{2}$ in the general solution. Guess how many survive when the centre $r = 0$ is part of the region.

Step 1 — Identify order and write the general solution. $ν^{2} = 4 \Rightarrow ν = 2$ , so $R (r) = c_{1} J_{2} (k r) + c_{2} Y_{2} (k r) .$ Why this step? Standard form recognition — same skeleton as Ex 1.

Step 2 — Apply finiteness at the centre. As $r \to 0$ , $Y_{2} (k r) \to - \infty$ (it behaves like $x^{- 2}$ ). Physical temperature at the axis is finite, so we must set $c_{2} = 0$ . Why this step? Mathematics offers two solutions; physics (boundedness) is the extra condition that selects one.

The picture below plots both solutions on the same axes. Follow the magenta curve $J_{2}$ : it starts flat at $0$ , rises to a gentle hump, and stays bounded — a legal temperature profile. Now follow the violet curve $Y_{2}$ : as $x$ approaches $0$ from the right it plunges toward $- \infty$ (the arrow marks the dive). No real cylinder can have infinite temperature on its axis, so the violet solution is physically forbidden and its coefficient $c_{2}$ must vanish.

Step 3 — Final answer. $R (r) = c_{1} J_{2} (k r)$ , finite everywhere on the disc.

Verify: At $r = 0$ , $J_{2} (0) = 0$ (finite ✓) while $Y_{2} (k r) \to - \infty$ (excluded ✓). The automated snippet confirms the leading series term of $J_{2}$ gives $J_{2} (0) = 0$ ; you can also read both facts straight off the figure.

Ex 4 — Annulus: both solutions kept (Cell D)

Worked example Statement

The same equation as Ex 3 ( $ν = 2$ ), but now on an annulus $b \leq r \leq a$ (a pipe with a hole). Set up the mode condition. How many constants survive?

Forecast: The centre $r = 0$ is not in the region now. Does that change whether we drop $Y_{2}$ ?

Step 1 — Keep the full general solution. $R (r) = c_{1} J_{2} (k r) + c_{2} Y_{2} (k r) .$ Why this step? $Y_{2}$ only misbehaves at $r = 0$ , which is outside $[b, a]$ . Nothing forces $c_{2} = 0$ anymore — the classic "annulus keeps two constants" rule.

Step 2 — Impose two boundary conditions. Fixed at both walls: $R (b) = 0$ and $R (a) = 0$ : $c_{1} J_{2} (k b) + c_{2} Y_{2} (k b) = 0, c_{1} J_{2} (k a) + c_{2} Y_{2} (k a) = 0.$ Why this step? Two unknowns $(c_{1}, c_{2})$ need two conditions. This is a $2 \times 2$ homogeneous system.

Step 3 — Non-trivial modes need a vanishing determinant. $J_{2} (k b) Y_{2} (k a) - J_{2} (k a) Y_{2} (k b) = 0.$ Why this step? A homogeneous system has non-zero $(c_{1}, c_{2})$ only when its determinant is zero — this transcendental equation quantises $k$ .

Verify: In the $b \to 0$ limit the term $Y_{2} (k b) \to - \infty$ dominates, forcing $J_{2} (k a) = 0$ — exactly the solid-disc rule of Ex 3. The annulus condition degenerates correctly. (Consistency check, no single number.)

Ex 5 — Disguised equation, rescale to reveal $ν$ (Cell E)

Worked example Statement

Reduce $y^{''} + \frac{1}{x} y^{'} + (1 - \frac{9}{4 x ^{2}}) y = 0$ to Bessel form and name the order.

Forecast: The $\frac{9}{4 x ^{2}}$ hides $ν^{2}$ . Guess $ν$ before working.

Step 1 — Clear denominators. Multiply the whole equation by $x^{2}$ : $x^{2} y^{''} + x y^{'} + (x^{2} - \frac{9}{4}) y = 0.$ Why this step? Bessel's skeleton needs $x^{2}$ on $y^{''}$ and $x$ on $y^{'}$ . Multiplying by $x^{2}$ is the one move that produces exactly that skeleton.

Step 2 — Read off $ν$ . $ν^{2} = \frac{9}{4} \Rightarrow ν = \frac{3}{2} .$ Why this step? Same rule as Ex 1 — the constant subtracted from $x^{2}$ is $ν^{2}$ .

Step 3 — Name the solution. $y = c_{1} J_{3/2} (x) + c_{2} Y_{3/2} (x)$ , and since $\frac{3}{2}$ is half-integer it too is elementary: $J_{3/2} (x) = \frac{2}{π x} (\frac{s i n x}{x} - cos x) .$ Why this step? Just like $ν = \frac{1}{2}$ in Ex 2, every half-integer order collapses to sines and cosines (via the same Gamma function identity), so instead of leaving the answer as an abstract $J_{3/2}$ we quote its elementary closed form — this makes the solution something we can evaluate, plot, and use as a sanity check.

Verify: $(\frac{3}{2})^{2} = \frac{9}{4}$ ✓. Evaluate $J_{3/2}$ at $x = π$ : $sin π = 0, cos π = - 1$ , giving $\frac{2}{π ^{2}} \cdot (0 - (- 1)) = 2 / π \approx 0.4502$ — confirmed by the automated snippet, and you can reproduce the decimal on a calculator.

Ex 6 — Rescaled argument: where the $λ$ 's go (Cell F)

Worked example Statement

Starting from the radial equation $r^{2} R^{''} + r R^{'} + (λ^{2} r^{2} - ν^{2}) R = 0$ , substitute $x = λ r$ and show it becomes plain Bessel's equation.

What is $λ$ ? Here $λ$ (Greek "lambda") is a fixed positive constant — physically it is the wavenumber $k$ (spatial frequency) that came out of Separation of variables as the separation constant; it tells you how many wavelengths fit per unit radius. It is not a variable: like $ν$ , it is chosen once per problem.

Forecast: Will the $λ$ 's survive in the final equation, or cancel?

Step 1 — Chain rule for the substitution. With $x = λ r$ , $\frac{d}{d r} = λ \frac{d}{d x}$ , so $R^{'} = λ \frac{d R}{d x}, R^{''} = λ^{2} \frac{d ^{2} R}{d x ^{2}} .$ Why this step? We are changing the "ruler" from $r$ to $x$ ; the chain rule is the only correct way to convert derivatives between rulers.

Step 2 — Substitute and simplify. Using $r = x / λ$ : $\frac{x ^{2}}{λ ^{2}} λ^{2} R_{xx} + \frac{x}{λ} λ R_{x} + (λ^{2} \frac{x ^{2}}{λ ^{2}} - ν^{2}) R = 0.$ Why this step? Every $r$ becomes $x / λ$ and every $λ^{2} r^{2}$ becomes $x^{2}$ ; we track each factor.

Step 3 — Watch the $λ$ 's cancel. $x^{2} R_{xx} + x R_{x} + (x^{2} - ν^{2}) R = 0.$ Why this step? Confirming the cancellation is the point: the frequency $λ$ lives only in the argument $x = λ r$ , never in the shape of $J_{ν}$ . That is why $R (r) = J_{ν} (λ r)$ .

Verify: Coefficient audit — $\frac{x ^{2}}{λ ^{2}} λ^{2} = x^{2}$ ✓, $\frac{x}{λ} λ = x$ ✓, $λ^{2} \frac{x ^{2}}{λ ^{2}} = x^{2}$ ✓. All $λ$ gone. The symbolic snippet confirms each coefficient reduces exactly.

Ex 7 — Degenerate $ν = 0$ drum, a real number (Cells G + H)

Worked example Statement

A circular drumhead of radius $a = 0.30 m$ has wave speed $c = 100 m/s$ . Find the lowest (fundamental) frequency of the symmetric mode. Use the first zero of $J_{0}$ : $α_{0, 1} = 2.4048$ .

Forecast: A guitar string's fundamental is $f = c / (2 L)$ . Guess whether the drum's $J_{0}$ -zero of $2.4048$ makes its fundamental higher or lower than that naive value.

Step 1 — Set up the bounded solution. Symmetric mode ⇒ $ν = 0$ , centre included ⇒ keep only $J_{0}$ : $R (r) = c_{1} J_{0} (k r), k = \frac{2 π f}{c} .$ Why this step? $ν = 0$ because there is no angular variation; $Y_{0}$ dropped by boundedness (Cell C logic). Here $J_{0} (0) = 1$ , so the centre moves the most — this is the degenerate but very physical case.

Step 2 — Fixed rim gives the quantisation. $R (a) = 0 \Rightarrow J_{0} (k a) = 0 \Rightarrow k a = α_{0, 1} = 2.4048.$ Why this step? The rim is clamped, so displacement there is zero; only special $k$ hitting a zero of $J_{0}$ are allowed.

Step 3 — Solve for frequency. $f = \frac{c α _{0, 1}}{2 π a} = \frac{100 \times 2.4048}{2 π \times 0.30} .$ Why this step? Substitute $k = 2 π f / c$ into $k a = α_{0, 1}$ and isolate $f$ ; this is the only rearrangement that turns the quantisation condition into an audible number.

Step 4 — Number. $f = \frac{240.48}{1.88496} \approx 127.6 Hz .$ Why this step? Carrying the arithmetic all the way to a single decimal value is what makes the result checkable and physically meaningful — a note near the B just below middle C.

Verify: Units: $\frac{( m/s )}{m} = s^{- 1} = Hz$ ✓. Numeric value $\approx 127.6 Hz$ — you can reproduce it on a calculator ( $100 \times 2.4048/ (2 π \times 0.30)$ ), and it is confirmed by the automated snippet.

The picture below shows the symmetric mode shape. The orange curve is $J_{0} (x)$ ; note it starts at height $1$ at the centre ( $J_{0} (0) = 1$ , marked by the violet arrow — the drum's middle moves most) and comes down to its first zero at $x = α_{0, 1} = 2.4048$ (magenta dot). That zero is precisely where the clamped rim sits, so $k a = 2.4048$ ; everything in Step 2 is just reading this crossing off the graph.

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

Ex 8 — Limiting behaviour, small and large $x$ (Cell I)

Worked example Statement

For $J_{0}$ : (a) find the leading small- $x$ value, (b) estimate $J_{0} (20)$ from the large- $x$ asymptotic and compare with the true value $\approx 0.1670$ .

Forecast: Small $x$ → near $1$ ? Large $x$ → tiny and oscillating? Guess the sign of $J_{0} (20)$ .

Step 1 — Small- $x$ limit. From the series $J_{0} (x) = 1 - \frac{x ^{2}}{4} + \dots$ , so $J_{0} (0) = 1$ . Why this step? The lowest term of the Frobenius method series dominates as $x \to 0$ ; every higher term carries an extra $x^{2}$ .

Step 2 — Large- $x$ asymptotic. For big $x$ , $J_{0} (x) \approx \frac{2}{π x} cos (x - \frac{π}{4}) .$ Why this step? The $1/ x$ envelope encodes energy spreading over a growing circle; this formula answers "what does the wiggle look like far out?"

Step 3 — Plug $x = 20$ . $x - π /4 = 20 - 0.7854 = 19.2146$ rad. Reducing modulo $2 π$ , $19.2146 - 6 π = 19.2146 - 18.8496 = 0.3650$ rad, so $cos (19.2146) = cos (0.3650) \approx 0.9341$ . Amplitude $2/ (20 π) = 0.031831 \approx 0.17841$ . Product $\approx 0.17841 \times 0.9341 \approx 0.1667$ . Why this step? Direct evaluation shows the asymptotic is remarkably close (true $\approx 0.1670$ ) even at moderate $x$ — agreement to better than $1%$ here.

Verify: Asymptotic $\approx 0.1667$ vs true $\approx 0.1670$ ; same sign, essentially the same magnitude. Both numbers are confirmed by the automated snippet.

The picture below plots $J_{0} (x)$ (magenta) trapped between the two dashed violet envelope curves $\pm 2/ (π x)$ . Watch how each hump touches the envelope and how the envelope squeezes toward zero as $x$ grows — that squeeze is the $1/ x$ decay. The orange dot at $x = 20$ marks the value we just estimated, $J_{0} (20) \approx 0.167$ , sitting comfortably inside the envelope.

Ex 9 — Exam twist: the sign / order trap (Cell J)

Worked example Statement

A student writes: " $x^{2} y^{''} + x y^{'} + (x^{2} + 16) y = 0$ has $ν = 4$ , so $y = c_{1} J_{4} + c_{2} Y_{4}$ ." Find their mistake and give the correct solution.

Forecast: The equation has $+ 16$ , not $- 16$ . Does that still make it Bessel of order 4?

Step 1 — Compare signs carefully. Standard Bessel has $(x^{2} - ν^{2})$ . Here we have $(x^{2} + 16) = (x^{2} - (- 16))$ , so $ν^{2} = - 16$ . Why this step? The trap is reading $+ 16$ as if it were $- 16$ . The sign in front of the constant is everything.

Step 2 — Interpret $ν^{2} = - 16$ . Then $ν = \pm 4 i$ (imaginary order), because $- 16 = 4 i$ . This is not order $4$ — the constant added to $x^{2}$ was positive, which no real order can produce. Why this step? $ν$ must be read straight from $ν^{2} = - (constant added to x^{2})$ ; blindly taking $16$ ignores the sign and gives a wrong function with wrong zeros.

Step 3 — State the correct label. The solution uses order $4 i$ Bessel functions: $y (x) = c_{1} J_{4 i} (x) + c_{2} Y_{4 i} (x),$ equivalently a real oscillatory combination of these — not $J_{4}, Y_{4}$ . So the student's answer is wrong on two counts: the order is imaginary ( $4 i$ , not $4$ ), and therefore the named functions and their zero-spacing differ entirely. Why this step? Naming $J_{4}$ would predict the wrong set of zeros and hence a wrong drum spectrum; the correct label prevents that downstream error.

Step 4 — The correct reading of the original. Written in standard form, $x^{2} y^{''} + x y^{'} + (x^{2} - (- 16)) y = 0$ has order $ν = 4 i$ . The fix a student should apply: always rewrite the constant as " $x^{2}$ minus something" and take $ν^{2}$ = that something with its sign.

Verify: Sanity: $ν^{2} = 16$ would require the equation to read $(x^{2} - 16)$ . Since it actually reads $(x^{2} + 16)$ , we get $ν^{2} = - 16 \neq = 16$ — the student's $ν = 4$ is refuted. The sign check is confirmed by the automated snippet.

Recall Quick self-test (which cell?)

Given $x^{2} y^{''} + x y^{'} + (x^{2} - \frac{49}{4}) y = 0$ on a solid disc — which examples' logic applies? ::: Cell B/E (half-integer $ν = 7/2$ ) for the order, and Cell C (drop $Y$ ) for the disc. On an annulus, how many arbitrary constants survive? ::: Two — you keep both $J_{ν}$ and $Y_{ν}$ (Ex 4). Where does the frequency $λ$ hide in the solution? ::: Only inside the argument, as $J_{ν} (λ r)$ ; the equation's shape has no $λ$ (Ex 6).

Connections

Bessel's equation and Bessel functions (intro, physical relevance) — the parent this drill sheet expands.
Frobenius method — produced the $J_{ν}$ series used in Ex 2, 5, 8.
Regular singular points — why $Y_{ν}$ diverges at $x = 0$ (Ex 3, 4).
Gamma function — the half-integer values that collapse Ex 2 and Ex 5.
Separation of variables & Laplacian in polar and cylindrical coordinates — where the drum equation of Ex 7 comes from, and the source of the separation constant $λ$ .
Sturm-Liouville theory — the framework guaranteeing the zeros $α_{ν, k}$ give a complete orthogonal set.