Part C · the analytic toolkit · spine

Fourier and Sturm–Liouville: the modes are a basis, and that is a theorem

Separation of variables produces modes and then asks you to believe that every initial shape is a sum of them. It is true, it is not obvious, and the reason is that one differential operator is symmetric with respect to one inner product.

needs 18, 09, 10, 11~3½ h slides + 7 h problemsproblems: ./mr problems 19

Unit 18 gave the wave equation and solved it on the infinite line with d'Alembert's formula. On a finite string d'Alembert still works, by reflecting the data forever, but it tells you nothing about pitch. Separation of variables answers the other question — which frequencies can this object have — and immediately runs up against a debt it cannot pay by itself: it produces a family of special solutions and then claims that superpositions of them reach every initial condition.

That claim is completeness, and this unit is about why it holds. The short answer: the modes are eigenfunctions of a symmetric operator, and a symmetric operator with compact resolvent has an orthonormal basis of eigenfunctions. That is the spectral theorem of unit 07 — the one that gave the normal modes of a chain of masses — moved from $\mathbb R^n$ to a space of functions. Sturm–Liouville theory is the machinery that makes the move legitimate.

How to read this deck

One object runs through it: a circular drum, radius $a$, clamped at the rim. Section 1 separates it and the string beside it; section 2 builds the inner product that makes coefficients unique; section 3 asks in what sense the series converges and how fast; section 4 proves the Sturm–Liouville theorems that the drum, the string, Bessel and Legendre are all instances of; section 5 bounds the drum's lowest frequency without solving anything; section 6 puts the three geometries side by side and explains why a drum has no pitch.

Every number quoted was computed. Proofs are on the slides where they are short; the longer ones are in these notes under Full proof, and anything imported is labelled cited with where its proof lives. Terms with a dotted underline are glossary entries — click one, or press G. N opens the notes, R stacks the deck into one readable page.

19 · overview

One drum, six sections

A drum of radius 15 cm. Strike it and it makes a noise, not a note; a string of the same fundamental makes a note. The difference is a list of numbers, and this unit computes it — by separating the equation, getting an eigenvalue problem for an ordinary differential operator, finding that the operator is symmetric for one inner product, and harvesting reality, orthogonality, discreteness, oscillation and a variational principle from that one fact.

1 · separation

String and drum split into a time factor and a shape; the debt that leaves.

2 · orthogonality

Projection, Bessel, Parseval, completeness.

3 · convergence

Pointwise, uniform, mean-square; Gibbs; smoothness against decay.

4 · Sturm–Liouville

$-(pu')' + qu = \lambda wu$ for given coefficient functions $p, q, w$: one identity, five theorems.

5 · Rayleigh

The lowest eigenvalue is a minimum, so a guess bounds it.

6 · three geometries

Sines, Bessel, Legendre — why drums have no pitch.

getting around

A dotted underline is a glossary term — click it, or press G. A solid underline like the drum is the running example: hover it. N shows the notes, where the long proofs are.

The problem sheet follows the same order. P1–P5 are Fourier series and their convergence; P6–P10 are the Sturm–Liouville theorems, most of them proofs; P11–P13 use the theory — a Rayleigh bound, a system that is orthogonal but not complete, and Fourier's own heat problem. Seven end in a number.

A warning about what this unit is not. It is not a catalogue of special functions. Bessel functions arrived in unit 09 and Legendre polynomials in unit 10, each for its own reason, and neither will be re-derived. What is new is that both are eigenfunctions of the same kind of operator, so that everything one knows about sines — orthogonality, expansions, Parseval, a variational principle for the lowest frequency — is true of them too, for one reason and not three.

It is also not the Fourier transform. That is unit 20, and it needs complex analysis. Here the interval is finite, the spectrum is a sequence, and no contour integral is required.

section 1

Separation of variables, and the question it leaves

A method that produces infinitely many special solutions, and then needs a theorem to turn them into every solution.

19 · separation

Splitting the string

Unit 18's wave equation on $0 \le x \le L$ with clamped ends:

$$u_{tt} = c^2u_{xx},\qquad u(0,t) = u(L,t) = 0 .$$

Look for solutions of the special form $u(x,t) = X(x)\,T(t)$ — not because every solution has it, but because for these the equation becomes two ordinary ones. Substituting and dividing by $XT$ wherever it is nonzero:

$$\frac{T''(t)}{c^2T(t)} = \frac{X''(x)}{X(x)} .$$

The left side does not depend on $x$; the right side does not depend on $t$; they are equal. So both are one constant, written $-\lambda$.

separation of variables

Seeking solutions of a linear PDE as a product of functions of one variable each. It converts the PDE into a family of ODEs indexed by the separation constant, which then turns out to be an eigenvalue.

two things are being assumed

That a product solution exists at all, and — much more seriously — that superposing them reaches every initial condition. The first is checked by substitution, in a line. The second is the content of the unit.

The division by $XT$ is legitimate on any open set where neither vanishes; the resulting ODEs then hold everywhere by continuity.

Why "both sides are constant" is a proof and not a slogan. Write $F(t) = T''/c^2T$ and $G(x) = X''/X$ on an open rectangle where $XT \ne 0$. The identity $F(t) = G(x)$ holds for all $(x,t)$ in it. Fix $t_0$: then $G(x) = F(t_0)$ for all $x$, a constant. Fix $x_0$: then $F(t) = G(x_0)$, the same constant. No differentiation is needed, though differentiating $F(t) - G(x) = 0$ in $x$ gives $G' = 0$ just as fast.

Why $-\lambda$ and not $+\lambda$. Pure convention, chosen so that the physically relevant case comes out with $\lambda > 0$. Nothing is assumed: all three signs are tested on the next slide and two of them die on the boundary conditions.

Where the method fails. Separation needs the domain to be a product of coordinate ranges in a coordinate system in which the operator separates. A rectangle and a disc are fine; an L-shaped membrane is not, and its eigenfunctions have no closed form. That is not a defect of this unit's theorems — Sturm–Liouville theory is one-dimensional, but the spectral theory it is a model of applies to the L-shape too. What fails is only the computation, and unit 21's asymptotics and numerical eigenvalue methods take over.

History

D'Alembert published the travelling-wave solution in 1747; Daniel Bernoulli replied in 1753 that the general solution was instead a sum of sinusoidal modes. Euler and d'Alembert both rejected Bernoulli's claim, on the grounds that a sum of smooth periodic functions could not possibly equal an arbitrary hand-drawn initial shape. They were wrong, but not stupidly wrong: what they were denying is exactly the completeness theorem of section 2, which nobody could prove for another seventy years. Fourier asserted it for heat conduction in 1807, was refereed harshly by Lagrange for exactly that assertion, and was right.

19 · separation

The boundary conditions do the selecting

The spatial half is $X'' = -\lambda X$ with $X(0) = X(L) = 0$. Take the three cases in turn — the general solution of a constant-coefficient linear ODE is unit 06's.

$\lambda$	general $X$	after $X(0)=X(L)=0$
$\lambda = -\mu^2 < 0$	$A\mathrm e^{\mu x} + B\mathrm e^{-\mu x}$	$A = B = 0$
$\lambda = 0$	$A + Bx$	$A = B = 0$
$\lambda = k^2 > 0$	$A\cos kx + B\sin kx$	$A = 0$, $\sin kL = 0$

The eigenvalues $\lambda_n$ are the separation constants the boundary conditions permit; the eigenfunctions $X_n$ are the shapes that go with them. A normal mode is one $u_n$: every point moving at one frequency, in fixed proportion.

Start every solution at $X(0) = 0$ with the same slope and see where it is at $x = L$. For $\lambda \le 0$ it climbs away and never comes back; for $\lambda = k^2$ it oscillates, and it lands on zero only when $kL$ is a multiple of $\pi$. The second boundary condition picks out a discrete set.

Reading the figure. Fixing $X(0) = 0$ leaves a one-parameter family for each $\lambda$, all multiples of one solution; normalising the slope picks one representative. The right-hand condition $X(L) = 0$ is then a single equation in $\lambda$. Below $k = \pi/L$ the sine has not yet turned round by $x = L$; above it, it has crossed zero and gone negative; only at $k = \pi/L$ exactly (and at $2\pi/L$, $3\pi/L$, …, which would cross zero $n - 1$ times first) does it land. This "shooting" picture is how eigenvalues of problems with no closed form are found numerically, and it is the picture behind the Prüfer angle in section 4.

The $\lambda < 0$ row in full. $X = A\mathrm e^{\mu x} + B\mathrm e^{-\mu x}$ with $\mu > 0$. $X(0) = 0$ gives $B = -A$, so $X = 2A\sinh\mu x$; $X(L) = 0$ then needs $A\sinh\mu L = 0$, and $\sinh$ has no positive zero, so $A = 0$. The $\lambda = 0$ row: $A = 0$ from the left end, $BL = 0$ from the right. Both cases give only $X \equiv 0$, which is not an eigenfunction by definition.

Why that is worth doing rather than skipping. Section 4 proves, for every Sturm–Liouville problem of this type, that the eigenvalues are real and bounded below — so the case analysis above will never again have to be done by hand. The general theorem also says something the case analysis cannot: that there are infinitely many eigenvalues and they run off to $+\infty$. Here we can see that; on the drum we could not.

19 · separation

The string's spectrum, and three checks

the string's spectrum

$\lambda_n = (n\pi/L)^2$ and $X_n(x) = \sin(n\pi x/L)$, $n = 1, 2, 3,\dots$, and nothing else. The time factor is then $T'' = -c^2\lambda_nT$, so

$$u_n = \sin\tfrac{n\pi x}{L}\,\big(\alpha_n\cos\omega_nt + \beta_n\sin\omega_nt\big),\quad \omega_n = \tfrac{n\pi c}{L}.$$

The first three modes, each at both extremes of its swing. Mode $n$ has $n - 1$ interior nodes and frequency $n\omega_1$: the shapes and the frequencies come in the same order.

checks

Dimensions. $[\lambda] = \mathrm L^{-2}$, and $\omega_n = n\pi c/L$ has $[\mathrm{L\,T^{-1}}]/[\mathrm L] = \mathrm T^{-1}$ ✓.

Special case. $N$ masses on a chain had $\omega_n = 2\sqrt{k/m}\sin\big(n\pi/2(N+1)\big)$ (unit 07). For $n \ll N$ this is $\approx n\pi c/L$ with $c = a\sqrt{k/m}$ — the same formula, as unit 18's continuum limit requires ✓.

Limit. $L \to \infty$ at fixed $n$ sends $\omega_n \to 0$: an infinite string has no lowest note ✓.

Completeness of the list. $\sin kL = 0$ with $k > 0$ forces $kL \in \pi\mathbb Z_{>0}$, and $k$ and $-k$ give the same $\lambda$ and proportional $X$, so $n \ge 1$ counts every eigenvalue exactly once. Each is simple: one eigenfunction up to scale. Section 4 proves that simplicity in general, for separated boundary conditions, and exhibits the periodic case where it fails.

What to look for in the figure. The nodes sit at $x = kL/n$, evenly spaced, and the count $n - 1$ goes up by one with each mode. That is not special to sines: section 4's oscillation theorem says the $n$-th eigenfunction of every regular Sturm–Liouville problem with separated conditions has exactly $n - 1$ interior zeros, and the drum's radial factors obey the same count in nodal circles. What is special to the uniform string is that the frequencies are equally spaced too — the subject of the unit's last slide.

19 · separation

Superposition, and the debt it runs up

The equation is linear and the boundary conditions are homogeneous, so any finite sum of the $u_n$ solves the problem. Allow infinite sums and the general candidate is

$$u(x,t) = \sum_{n\ge1}\sin\tfrac{n\pi x}{L}\,\big(\alpha_n\cos\omega_nt + \beta_n\sin\omega_nt\big).$$

Matching $u(\cdot,0)$ and $u_t(\cdot,0)$ to the initial data $f$ and $g$:

$$f = \sum_{n\ge1}\alpha_n\sin\tfrac{n\pi x}{L},\qquad g = \sum_{n\ge1}\omega_n\beta_n\sin\tfrac{n\pi x}{L}.$$

the debt

Everything rests on this: can every admissible $f$ be written that way, and with which coefficients? Separation cannot answer it — it is a question about the sines, not about the wave equation.

the same debt, in heat

Fourier's own problem (1807). A rod, $\theta_t = \kappa\,\theta_{xx}$, ends held at zero. Separation gives the same $X_n$ and $T_n = \mathrm e^{-\kappa\lambda_nt}$, so

$$\theta = \sum_n \alpha_n\sin\tfrac{n\pi x}{L}\,\mathrm e^{-t/\tau_n},\ \ \tau_n = \frac{L^2}{\kappa n^2\pi^2}.$$

Copper, $\kappa = 1.11\times10^{-4}\ \mathrm{m^2\,s^{-1}}$, $L = 1\ \mathrm m$: $\tau_1 = 912.8\ \mathrm s = 15.2$ min, and $\tau_{10} = 9.13\ \mathrm s$. The tenth harmonic is gone a hundred times sooner than the first.

That is why heat conduction smooths: the high modes, which carry the detail, die as $\mathrm e^{-n^2t/\tau_1}$. It is also why running it backwards is hopeless, which is unit 18's ill-posedness with a rate attached.

The three loose ends in "allow infinite sums". Differentiating the series twice term by term to check it still solves the equation is not free: it multiplies the $n$-th coefficient by $n^2$, and a convergent series need not survive that. Section 3's slide on term-by-term differentiation says exactly what it costs. For the heat equation the factor $\mathrm e^{-\kappa\lambda_nt}$ beats every power of $n$ for $t > 0$, so the sum is smooth for positive time no matter how rough $f$ is — a property the wave equation does not share, and the cleanest statement of the difference between parabolic and hyperbolic (unit 18's classification).

The numbers. $\tau_1 = L^2/(\kappa\pi^2) = 1/(1.11\times10^{-4}\times9.8696) = 912.8\ \mathrm s$, which is $15.21$ minutes; $\tau_{10} = \tau_1/100 = 9.128\ \mathrm s$. Dimensions: $[\kappa] = \mathrm{L^2T^{-1}}$, so $L^2/\kappa$ is a time ✓.

Why the sine series and not some other. Because the boundary conditions were $u = 0$ at both ends. Change them and the eigenfunctions change: free ends ($u_x = 0$) give cosines, a periodic ring gives both, an elastic support gives a transcendental condition and eigenfunctions that are neither. The dictionary is on "Sine, cosine, complex: one system per boundary condition". The unifying statement is section 4's: the boundary conditions are not decoration, they are half of the operator.

19 · separation · example

Worked: what a pluck puts into each mode

the plucked string

Pull the string aside by $h$ at $x = a$ and release from rest: $g = 0$, and $f$ is the triangle $hx/a$ on $[0,a]$, $h(L-x)/(L-a)$ on $[a,L]$. Use section 2's formula $\alpha_n = \frac2L\int_0^Lf\sin\frac{n\pi x}{L}\mathrm dx$ and integrate each leg by parts. With $k = n\pi/L$, the legs give $\frac{2h}{La}\big[\frac{\sin ka}{k^2} - \frac{a\cos ka}{k}\big]$ and $\frac{2h}{L(L-a)}\big[\frac{\sin ka}{k^2} + \frac{(L-a)\cos ka}{k}\big]$: the $\cos ka$ terms are $\mp2h/Lk$ and cancel, leaving

$$\alpha_n = \frac{2hL^2}{\pi^2a(L-a)}\cdot\frac{1}{n^2}\,\sin\frac{n\pi a}{L}.$$

$1/n^2$ is the kink; $\sin(n\pi a/L)$ is where it is.

checks

Dimensions. $hL^2/(a(L-a))$ is a length ✓.

Special case. $a = L/2$: $\sin(n\pi/2)$ kills every even $n$ — a symmetric shape cannot excite an antisymmetric mode ✓.

Limit. $a \to 0$: for $n \ll L/a$ the sine is $\approx n\pi a/L$, so $\alpha_n \approx 2h/(n\pi)$ — decay $1/n$, not $1/n^2$, over the first $\sim L/a$ modes. A sharper corner is brighter ✓.

Full computation of the cancellation

Write $k = n\pi/L$. First leg: $\int_0^a x\sin kx\,\mathrm dx = \big[-\tfrac{x}{k}\cos kx\big]_0^a + \tfrac1k\int_0^a\cos kx\,\mathrm dx = -\tfrac ak\cos ka + \tfrac1{k^2}\sin ka$. Second leg, substituting $s = L - x$ so that $\sin kx = \sin(kL - ks) = -\cos kL\sin ks + \sin kL\cos ks$, and $\sin kL = \sin n\pi = 0$, $\cos kL = (-1)^n$: $\int_a^L (L-x)\sin kx\,\mathrm dx = -(-1)^n\int_0^{L-a}s\sin ks\,\mathrm ds = -(-1)^n\big[-\tfrac{L-a}{k}\cos k(L-a) + \tfrac1{k^2}\sin k(L-a)\big]$. Now $\cos k(L-a) = (-1)^n\cos ka$ and $\sin k(L-a) = -(-1)^n\sin ka$, so this is $\tfrac{L-a}{k}\cos ka + \tfrac1{k^2}\sin ka$. Multiply the first leg by $2h/(La)$ and the second by $2h/(L(L-a))$ and add: the $\cos ka$ terms are $\tfrac{2h}{La}\big(-\tfrac ak\big)\cos ka + \tfrac{2h}{L(L-a)}\tfrac{L-a}{k}\cos ka = \big(-\tfrac{2h}{Lk} + \tfrac{2h}{Lk}\big)\cos ka = 0$, and the $\sin ka$ terms combine to $\tfrac{2h}{Lk^2}\big(\tfrac1a + \tfrac1{L-a}\big)\sin ka = \tfrac{2hL}{k^2a(L-a)}\sin ka$. With $k^2 = n^2\pi^2/L^2$ that is the boxed formula.

Does it actually work? Yes, and it is the standard demonstration. Pluck a string at its midpoint and every even partial vanishes, which is why a centre-plucked string sounds hollow and clarinet-like. Pluck very near the bridge, $a \ll L$, and the decay is $1/n$ over the first $L/a$ modes instead of $1/n^2$: that is the thin, bright sul ponticello sound. The whole of timbre, for a plucked string, is in this one formula.

Honesty about one step. The coefficient formula was used before it was derived. Section 2 derives it in two lines from orthogonality, and proves that it is the only possible choice; nothing above depends on the order.

19 · separation · example

The pluck at L/7, mode by mode

Top: the triangle plucked at $L/7$ and its first eight modes summed — already close everywhere but the corner. Bottom: $|\alpha_n/\alpha_1|$ on a log scale. The amplitudes ride down the $1/n^2$ envelope, modulated by $\sin(n\pi/7)$, which kills modes 7 and 14 exactly.

pluck at $a = L/7$ and mode 7 is silent

$\sin(7\pi/7) = 0$ exactly. Amplitudes $\alpha_n/\alpha_1$ for $n = 1,\dots,8$: $1$, $.450$, $.250$, $.140$, $.072$, $.028$, $\mathbf 0$, $-.016$. Helmholtz used this to explain why piano hammers strike near $L/7$ to $L/9$: the seventh partial is the first that lies well off the tempered scale.

Read the formula as two factors: $1/n^2$ says what kind of shape it is (one kink), and $\sin(n\pi a/L)$ says where the kink sits. A mode with a node at the plucking point cannot be excited by it.

The table, computed. $\alpha_n/\alpha_1 = n^{-2}\sin(n\pi/7)/\sin(\pi/7)$ with $\sin(\pi/7) = 0.433884$: $1$, $0.45048$, $0.24966$, $0.14044$, $0.07208$, $0.02778$, $0$, $-0.01562$. The sign flip at $n = 8$ is $\sin(8\pi/7) < 0$; it is inaudible, since the ear does not hear the relative phase of partials for a steady tone, but it is real and it matters for the waveform.

What to look for in the figure. In the top panel the eight-mode sum is already within a few per cent of the triangle except at the corner, where it rounds it off — the missing high modes are exactly what a corner is made of. In the bottom panel the open circles (modes 8 to 13) are the negative amplitudes; on a log scale only the size is shown, and the dashed envelope $1/(n^2\sin(\pi/7))$ is what the amplitudes would be if every $\sin(n\pi/7)$ were $\pm1$. Mode 7 sits at a node of the pluck and is absent; so is mode 14, for the same reason.

19 · separation · the drum

The drum, separated

The membrane obeys $u_{tt} = c^2\nabla^2u$ on the disc $r < a$, with $u = 0$ on $r = a$. In polar coordinates (unit 05) the Laplacian is

$$\nabla^2u = u_{rr} + \frac1r u_r + \frac1{r^2}u_{\theta\theta}.$$

Try $u = R(r)\Theta(\theta)T(t)$; the time factor separates as before, with constant $-c^2\lambda$. Multiplying what is left by $r^2/(R\Theta)$,

$$\frac{r^2R'' + rR'}{R} + \lambda r^2 \;=\; -\frac{\Theta''}{\Theta} \;=\; m^2,$$

a function of $r$ equal to a function of $\theta$, hence constant.

$\Theta$ must be $2\pi$-periodic — not imposed by hand, but required for $u$ to be a function on the disc at all — so $\Theta = \cos m\theta$ or $\sin m\theta$, $m \in \{0,1,2,\dots\}$.

what is left is Bessel's equation

With $\kappa = \sqrt\lambda$ and $s = \kappa r$,

$$r^2R'' + rR' + (\lambda r^2 - m^2)R = 0,$$

i.e. $s^2\ddot R + s\dot R + (s^2 - m^2)R = 0$,

which is exactly the equation unit 09 solved by Frobenius, with solutions $J_m$ (regular at $0$) and $Y_m$ (logarithmically singular there).

the centre is not a boundary

$r = 0$ is a coordinate singularity, not an edge of the drum (unit 05). Nothing physical can be imposed there — only that $u$ be bounded, which discards $Y_m$. Section 4 shows this is what a singular endpoint is.

Full derivation: the membrane equation

A membrane under uniform tension $T$ (force per unit length across any cut, $\mathrm{N\,m^{-1}}$) with surface density $\sigma$ ($\mathrm{kg\,m^{-2}}$). Take a small rectangle $[x, x+\Delta x]\times[y, y+\Delta y]$ of the membrane, displaced by $u(x,y,t)$. The two edges normal to $x$ pull with vertical components $T\Delta y\sin\vartheta \approx T\Delta y\,u_x$ evaluated at the two ends, so their net vertical force is $T\Delta y[u_x(x+\Delta x) - u_x(x)] \approx T\,u_{xx}\Delta x\Delta y$; likewise for $y$. Newton's second law on the patch, mass $\sigma\Delta x\Delta y$, gives $\sigma u_{tt} = T(u_{xx} + u_{yy}) = T\nabla^2u$, and $c^2 = T/\sigma$ has dimensions $(\mathrm{N\,m^{-1}})/(\mathrm{kg\,m^{-2}}) = \mathrm{m^2\,s^{-2}}$ ✓. The small-slope approximation is the same one unit 18 made for the string, with the same error: the dropped terms are $O(|\nabla u|^2)$ relative to those kept.

Full derivation: the Laplacian in polar coordinates

With $x = r\cos\theta$, $y = r\sin\theta$, the chain rule (unit 03) gives $\partial_r = \cos\theta\,\partial_x + \sin\theta\,\partial_y$ and $\partial_\theta = -r\sin\theta\,\partial_x + r\cos\theta\,\partial_y$. Hence $\partial_{rr} = \cos^2\!\theta\,\partial_{xx} + 2\sin\theta\cos\theta\,\partial_{xy} + \sin^2\!\theta\,\partial_{yy}$, and $\partial_{\theta\theta} = r^2(\sin^2\!\theta\,\partial_{xx} - 2\sin\theta\cos\theta\,\partial_{xy} + \cos^2\!\theta\,\partial_{yy}) - r\partial_r$. Adding $\partial_{rr} + r^{-2}\partial_{\theta\theta} + r^{-1}\partial_r$, the cross terms cancel, the $\sin^2 + \cos^2$ pairs give $\partial_{xx} + \partial_{yy}$, and the stray $-r^{-1}\partial_r$ cancels the one added. So $\nabla^2 = \partial_{rr} + r^{-1}\partial_r + r^{-2}\partial_{\theta\theta}$, valid on $r > 0$. The three-dimensional spherical version is the same computation with one more angle; it is cited here and used on "The sphere, separated", where it is written out.

Why $m$ must be an integer. $\Theta'' = -m^2\Theta$ with $\Theta(\theta + 2\pi) = \Theta(\theta)$. For $m^2 < 0$ the solutions are real exponentials and never periodic; for $m^2 = 0$, $\Theta = A + B\theta$ is periodic only if $B = 0$; for $m^2 = k^2 > 0$, periodicity needs $k \in \mathbb Z$. This is the first periodic Sturm–Liouville problem in the course, and note what is different: $m^2 = k^2$ has a two-dimensional eigenspace, spanned by $\cos k\theta$ and $\sin k\theta$. Section 4 explains why — periodic boundary conditions are not separated, and simplicity is exactly the conclusion that fails.

19 · separation · the drum

The drum's spectrum, in numbers

Boundedness leaves $R(r) = J_m(\kappa r)$; the clamped rim needs $J_m(\kappa a) = 0$. So $\kappa a$ must be a zero of $J_m$, and with $j_{mk}$ the $k$-th positive one:

the drum's frequencies $$\omega_{mk} = \frac{c\,j_{mk}}{a},\qquad u_{mk} = J_m\!\Big(\frac{j_{mk}r}{a}\Big)\begin{Bmatrix}\cos m\theta\\ \sin m\theta\end{Bmatrix}.$$

$(m,k)$	$j_{mk}$	$j_{mk}/j_{01}$	$f$ / Hz	nodes inside
$(0,1)$	2.404826	1.0000	280.39	none
$(1,1)$	3.831706	1.5933	446.76	1 diameter
$(2,1)$	5.135622	2.1355	598.79	2 diameters
$(0,2)$	5.520078	2.2954	643.62	1 circle

The four modes, seen from above, nodal lines to scale. Shaded and unshaded regions move in opposite directions. $m$ counts nodal diameters, $k - 1$ counts nodal circles: the $(0,2)$ circle sits at $r = (j_{01}/j_{02})\,a = 0.436\,a$.

Reading the mode shapes. $u_{mk}$ vanishes on $m$ diameters (the zeros of $\cos m\theta$ in $[0,2\pi)$, which come in $m$ collinear pairs) and on $k-1$ interior circles (the zeros $j_{m1},\dots,j_{m,k-1}$ pulled back by $r = j_{mk}^{-1}a\,j_{mi}$). Those are Chladni's figures, which he produced in 1787 by bowing plates and watching sand collect on the still lines — the first direct sight of an eigenfunction, more than a century before there was a word for one.

Two eigenfunctions per $(m,k)$ when $m \ge 1$. $\cos m\theta$ and $\sin m\theta$ give the same frequency, so the eigenvalue $\lambda_{mk}$ has multiplicity 2 for $m \ge 1$ and 1 for $m = 0$. The degeneracy is forced by the rotational symmetry of the disc: rotating an eigenfunction gives an eigenfunction of the same eigenvalue, and for $m \ge 1$ the rotate is genuinely new. Break the symmetry — a drum with a coin stuck to it — and the pair splits, which is audible as a beat (unit 07's beats, with the same mechanism). The figure draws the $\cos m\theta$ member of each pair; its $\sin m\theta$ partner is the same picture rotated by $\pi/2m$.

19 · separation · the drum

The drum's radial factors, and three checks

The radial factors $J_m(j_{mk}r/a)$ of the same four modes. Every one vanishes at the rim — that is how $j_{mk}$ was chosen. Only the $m = 0$ modes move at the centre; for $m \ge 1$ the factor starts at zero, as a function of $r$ and $\theta$ must if it is to be continuous there.

checks

Dimensions. $j_{mk}$ is a pure number, so $cj_{mk}/a$ is $\mathrm{T^{-1}}$ ✓.

Asymptotic limit. Unit 09's large-argument form $J_m(s) \sim \sqrt{2/\pi s}\cos(s - m\pi/2 - \pi/4)$ predicts $j_{mk} \approx (k + \tfrac m2 - \tfrac14)\pi$. For $(0,3)$: $2.75\pi = 8.6394$ against $8.6537$, out by $0.17\%$; for $(0,1)$, $2.0\%$ ✓ — high modes become string-like, as they must.

The number. $c = \sqrt{3200/0.265} = 109.89\ \mathrm{m\,s^{-1}}$ and $f_{01} = 109.89\times2.404826/(2\pi\times0.150) = 280.39\ \mathrm{Hz}$.

A string's frequencies are $1 : 2 : 3 : 4$. The drum's are $1 : 1.5933 : 2.1355 : 2.2954$ — ratios of Bessel zeros, and there is no reason for them to be rational. The last slide of the unit cashes that observation.

Why the $m \ge 1$ factors must vanish at the centre. At $r = 0$ every value of $\theta$ names the same point, so a continuous $u$ cannot depend on $\theta$ there; a factor $\cos m\theta$ with $m \ge 1$ forces $R(0) = 0$. The Frobenius exponent $m$ of unit 09, $J_m(s) \sim (s/2)^m/m!$, delivers exactly that, and the higher $m$, the flatter the start — visible in the figure as $(2,1)$ leaving the axis more slowly than $(1,1)$.

Where the real drum differs. A real timpano is loaded by the air inside the kettle and by the air it radiates into, which lowers $(1,1)$, $(2,1)$, $(3,1)$ and $(4,1)$ towards the ratios $1 : 1.5 : 2 : 2.5$ relative to $(1,1)$ — approximately harmonic, which is precisely why a timpano can be tuned and a bare membrane cannot. The ideal membrane computed here is the baseline that the kettle's design is fighting against, and it is the right baseline for this unit because it is the one with a Sturm–Liouville problem behind it.

Zeros used: $j_{01} = 2.4048255577$, $j_{11} = 3.8317059702$, $j_{21} = 5.1356223018$, $j_{02} = 5.5200781103$, $j_{03} = 8.6537279129$, all computed to twelve digits.

section 2

Orthogonality, and the best approximation

One inner product turns "find the coefficients" into a projection, and "is every shape reachable" into a theorem.

19 · orthogonality

An inner product on a space of functions

definition

On the real vector space $PC[a,b]$ of piecewise continuous functions on $[a,b]$, with a fixed weight $w$ continuous and $w > 0$, put

$$\langle f,g\rangle_w := \int_a^b\! fg\,w\,\mathrm dx,\ \ \lVert f\rVert_w := \langle f,f\rangle_w^{1/2}.$$

Bilinear and symmetric by linearity of the integral; positive because $w > 0$ — and definite once functions agreeing off a finite set are identified.

Everything from unit 07 applies verbatim: Cauchy–Schwarz, the triangle inequality, orthogonal projection, Pythagoras. The space is infinite-dimensional, which changes nothing here and everything in section 3.

the sines are orthogonal

On $[0,L]$ with $w = 1$, write $S_n(x) = \sin(n\pi x/L)$. Then for $m \ne n$

$$\langle S_m, S_n\rangle = 0,\qquad \lVert S_n\rVert^2 = L/2 .$$

proof

$\sin A\sin B = \tfrac12[\cos(A-B) - \cos(A+B)]$, so the integral is

$$\tfrac12\int_0^L\!\Big[\cos\tfrac{(m-n)\pi x}{L} - \cos\tfrac{(m+n)\pi x}{L}\Big]\mathrm dx$$

and $\int_0^L\cos(k\pi x/L)\mathrm dx = (L/k\pi)\sin k\pi = 0$ for nonzero integer $k$. For $m = n$ the first cosine is $1$: contribution $L/2$. $\square$

A computation — but section 4 proves it without computing, and that proof also covers $J_m$ and $P_n$, which this one does not.

Why "piecewise continuous" and not "continuous". Because the initial data we care about — the plucked string's kink, a struck string's step, Fourier's rod at two temperatures — are not smooth, and the whole point is to expand them anyway. A piecewise continuous function on $[a,b]$ is continuous except at finitely many points, with one-sided limits there. It is Riemann integrable, and that is all the integration theory this unit needs.

Why definiteness needs the identification. $\lVert f\rVert_w = 0$ forces $f = 0$ at every point of continuity — if $f(x_0) \ne 0$ with $f$ continuous at $x_0$, then $f^2w > \varepsilon$ on a neighbourhood and the integral is positive. So $f$ can be nonzero only at its finitely many jump points. Declaring two functions equal when they agree off a finite set makes $\langle\cdot,\cdot\rangle_w$ an honest inner product. This is the elementary shadow of the measure-theoretic identification "equal almost everywhere" that defines $L^2$; we do not need the general version.

What is genuinely missing. $PC[a,b]$ with this norm is not complete as a metric space: Cauchy sequences of piecewise continuous functions can converge to things that are not. The completion is $L^2_w(a,b)$ and constructing it needs the Lebesgue integral. Everything proved in this unit is proved for piecewise continuous $f$ with the Riemann integral, so nothing here rests on that gap; but the clean statement "the eigenfunctions are an orthonormal basis of a Hilbert space" does, and it is flagged as cited when it is used.

Cosines and the constant. The same product formula gives $\langle\cos(m\pi x/L),\cos(n\pi x/L)\rangle = 0$ for $m \ne n \ge 0$, with norm$^2$ equal to $L/2$ for $n \ge 1$ and $L$ for $n = 0$. The exceptional norm of the constant is why the constant term of a cosine series is written $a_0/2$: it makes one formula cover every coefficient.

19 · orthogonality

Coefficients are a projection, and that is optimal

Let $\{e_n\}$ be orthogonal and nonzero in an inner-product space $V$, $f \in V$, and $S = \sum_{n\le N}c_ne_n$.

best approximation

$\lVert f - S\rVert$ is smallest exactly when $c_n = \hat f_n := \langle f, e_n\rangle/\lVert e_n\rVert^2$ for $n \le N$, and then $f - S \perp e_1,\dots,e_N$.

proof

Expand, using orthogonality, then complete the square in each $c_n$ separately:

$$\lVert f - S\rVert^2 = \lVert f\rVert^2 - 2\sum_{n\le N} c_n\langle f,e_n\rangle + \sum_{n\le N}c_n^2\lVert e_n\rVert^2$$ $$= \lVert f\rVert^2 - \sum_{n\le N}\hat f_n^2\lVert e_n\rVert^2 + \sum_{n\le N}\lVert e_n\rVert^2(c_n - \hat f_n)^2 .$$

Only the last sum depends on $c$; it is nonnegative and zero exactly at $c = \hat f$. $\square$

Schematic. The best combination $S_Nf$ is the foot of the perpendicular from $f$ to the span, and its coordinates along the $e_n$ are the $\hat f_n$. Every other point $S$ of the span is farther, by Pythagoras: $\lVert f - S\rVert^2 = \lVert f - S_Nf\rVert^2 + \lVert S_Nf - S\rVert^2$.

The completion of the square, step by step. For one index, $-2c_n\langle f,e_n\rangle + c_n^2\lVert e_n\rVert^2 = \lVert e_n\rVert^2\big(c_n^2 - 2c_n\hat f_n\big) = \lVert e_n\rVert^2\big((c_n - \hat f_n)^2 - \hat f_n^2\big)$, using $\langle f,e_n\rangle = \hat f_n\lVert e_n\rVert^2$. Summing over $n \le N$ gives the displayed line. Note that the cross terms between different $e_n$ never appeared — that is exactly where orthogonality is spent, and it is the only place.

Why this is the important slide of the section. It says the truncated series is not merely a approximation but the best one in the mean-square sense, among all combinations of those $N$ modes. So the question "does the series converge to $f$?" becomes "do the spans of the $e_n$ eventually come arbitrarily close to $f$?" — a question about the system, not about $f$'s coefficients. That is completeness, and separating the two questions is what makes it provable.

The picture and the algebra. The figure is the proof read geometrically. The last sum in the displayed identity is $\lVert S - S_Nf\rVert^2$ for $S = \sum c_ne_n$ — the squared distance, inside the span, from the trial point to the foot of the perpendicular. So the identity says $\lVert f - S\rVert^2 = \lVert f - S_Nf\rVert^2 + \lVert S - S_Nf\rVert^2$, Pythagoras in the right triangle with vertices $f$, $S_Nf$, $S$, and the minimum is at $S = S_Nf$. Write $S_Nf := \sum_{n\le N}\hat f_ne_n$ from here on: it is the partial sum of the Fourier series.

19 · orthogonality

Fourier coefficients: coordinates that stay put

Fourier coefficient

$\hat f_n = \langle f,e_n\rangle/\lVert e_n\rVert^2$ — the coordinate of $f$ along $e_n$, for any orthogonal system in any inner-product space. The best approximation $S_Nf := \sum_{n\le N}\hat f_ne_n$ is the partial sum. For the sines on $[0,L]$: $\alpha_n = \frac2L\int_0^Lf\sin\frac{n\pi x}{L}\mathrm dx$, the formula the plucked string used.

three things follow at once

Uniqueness. If $f = \sum c_ne_n$ in norm then $c_n = \hat f_n$: take $\langle\cdot,e_m\rangle$ through the limit by Cauchy–Schwarz.

Independence of $N$. $\hat f_n$ does not move when more terms are allowed. Fit a non-orthogonal family and every coefficient moves.

The error is orthogonal to the span — Pythagoras, and the next slide.

Schematic. Fit $f$ with $e_1$ alone, then with $e_1$ and $e_2$, and watch the $e_1$ coefficient. With $e_2 \perp e_1$ the two fits agree (left). With $e_2$ at $60^\circ$ the best $e_1$ coefficient jumps when $e_2$ is admitted (right) — and every coefficient would move again with $e_3$.

Reading the figure. Fitting with $e_1$ alone drops a perpendicular onto the $e_1$ axis (orange dashes). Fitting with both writes $f$ in the oblique coordinates of the $e_1, e_2$ plane, so the $e_1$ coordinate is read off by sliding parallel to $e_2$ (teal dashes). The two routes agree exactly when sliding parallel to $e_2$ is dropping a perpendicular to $e_1$ — that is, when $e_2 \perp e_1$. That is the whole content of "independence of $N$", and it is why a synthesiser, a spectrum analyser and a normal-mode calculation all use orthogonal building blocks.

The physics reading. $\hat f_n$ is how much of mode $n$ the initial shape contains, and the best-approximation property says a mode-$N$ truncation of a synthesiser is doing the best any $N$-mode instrument could. It also says the modes do not interfere in the accounting: adding a higher partial never changes how much of the fundamental is there. That is false for non-orthogonal building blocks, and it is why physicists reach for orthogonal ones every time.

The complex case. With $\langle f,g\rangle = \int f\bar g\,w$, the same proof runs with $|c_n - \hat f_n|^2$ in place of the square and $\hat f_n = \langle f,e_n\rangle/\lVert e_n\rVert^2$ unchanged. Only the bookkeeping of conjugates differs.

19 · orthogonality

Bessel's inequality, and what it hands you free

Bessel's inequality

For any orthogonal system $\{e_n\}$ and any $f$,

$$\sum_{n=1}^{\infty}\hat f_n^2\,\lVert e_n\rVert^2 \;\le\; \lVert f\rVert^2 .$$

proof

The best-approximation identity with $c = \hat f$ reads

$$0 \le \lVert f - S_Nf\rVert^2 = \lVert f\rVert^2 - \sum_{n\le N}\hat f_n^2\lVert e_n\rVert^2 ,$$

so the partial sums are bounded by $\lVert f\rVert^2$; nondecreasing, they converge. $\square$

Free: no hypothesis on $f$, none on the system beyond orthogonality.

corollary: Riemann–Lebesgue

A convergent series has terms tending to zero, so $\hat f_n\lVert e_n\rVert \to 0$. For the normalised trigonometric system on $[-\pi,\pi]$:

$$\int_{-\pi}^{\pi}\!\!f\sin nx\,\mathrm dx \to 0,\quad \int_{-\pi}^{\pi}\!\!f\cos nx\,\mathrm dx \to 0$$

for every piecewise continuous $f$. Oscillate fast enough against anything and the average wins.

check

Dimensions. Both sides carry $[f]^2\,[w]\,\mathrm L$ ✓. Special case. $f = e_1$: the left side is $\lVert e_1\rVert^2$ and the right is the same — equality, as it must be for $f$ in the span ✓.

Riemann–Lebesgue is the engine of Dirichlet's pointwise theorem, and the reason it works for merely piecewise smooth data.

Riemann–Lebesgue in the form actually used. Dirichlet's theorem needs $\int_0^{\pi}g(t)\sin\big((N+\tfrac12)t\big)\,\mathrm dt \to 0$ for piecewise continuous $g$, with the half-integer frequency. Expand $\sin\big((N+\tfrac12)t\big) = \sin Nt\cos\tfrac t2 + \cos Nt\sin\tfrac t2$ and apply the boxed statement to the two piecewise continuous functions $g(t)\cos(t/2)$ and $g(t)\sin(t/2)$, extended by zero to $[-\pi,\pi]$. Both integrals tend to zero, so their sum does.

What equality would mean. Bessel's inequality is an equality for every $f$ exactly when the system is complete — that equality is Parseval's identity, next slide. A strict inequality for some $f$ means the system has missed a direction: there is something in the space that the $e_n$ cannot reach. Problem P12 builds such a system on purpose, from perfectly good orthogonal functions, so that "orthogonal" and "complete" stop being able to blur into each other.

A caution about the name. This Bessel is the same Bessel as the functions of unit 09 — Friedrich Wilhelm Bessel, astronomer, who met both objects in the same line of work. The inequality has nothing to do with the functions.

19 · orthogonality

Completeness, and Parseval

completeness

$\{e_n\}$ is complete in $V$ if $\lVert f - S_Nf\rVert \to 0$ for every $f \in V$. Then the only $f$ orthogonal to every $e_n$ is $0$; the converse needs more (notes).

Parseval's identity

Complete $\iff$ Bessel's inequality is an equality for every $f$:

$$\lVert f\rVert^2 = \sum_{n\ge1}\hat f_n^2\,\lVert e_n\rVert^2 .$$

Proof. Immediate from $\lVert f - S_Nf\rVert^2 = \lVert f\rVert^2 - \sum_{n\le N}\hat f_n^2\lVert e_n\rVert^2$, which is the best-approximation identity. $\square$

Physically: the energy is the sum of the modes' energies, no cross terms.

why not just use $S_N$?

Because $S_Nf$ need not converge pointwise for continuous $f$: du Bois-Reymond, 1876 (cited). Their averages $\sigma_N := \frac1{N+1}\sum_{k\le N}S_kf$ do, being $f$ convolved with a nonnegative kernel — which $S_Nf$ is not.

the trigonometric system is complete

$\{1,\cos nx,\sin nx\}$ is complete in $PC[-\pi,\pi]$; so are $\{\sin(n\pi x/L)\}$ and $\{\cos(n\pi x/L)\}$ in $PC[0,L]$.

full proof in the notes (N) — the Fejér route

Three steps: (i) Fejér's theorem — the averages $\sigma_N$ converge uniformly to $f$ for continuous periodic $f$; (ii) any piecewise continuous $f$ is within $\varepsilon$ in norm of a continuous periodic one; (iii) best approximation: once a trigonometric polynomial of degree $D$ is within $\varepsilon$, so is $S_Nf$ for $N \ge D$.

Full proof, step (i): Fejér's theorem

Write $S_kf(x) = \frac1{2\pi}\int_{-\pi}^{\pi}f(x-t)D_k(t)\,\mathrm dt$ with the Dirichlet kernel $D_k(t) = \sum_{|n|\le k}\mathrm e^{\mathrm int} = \sin\big((k+\tfrac12)t\big)/\sin(t/2)$. Averaging, $\sigma_Nf = \frac1{2\pi}\int f(x-t)F_N(t)\,\mathrm dt$ with the Fejér kernel $F_N = \frac1{N+1}\sum_{k\le N}D_k$. Summing the geometric-type series gives $$F_N(t) = \frac{1}{N+1}\left(\frac{\sin\frac{(N+1)t}{2}}{\sin\frac t2}\right)^{\!2},$$ from which three facts: $F_N \ge 0$; $\frac1{2\pi}\int_{-\pi}^{\pi}F_N = 1$ (each $D_k$ integrates to $2\pi$); and for any $\delta > 0$, $\sup_{\delta\le|t|\le\pi}F_N \le \frac{1}{(N+1)\sin^2(\delta/2)} \to 0$. A family with those three properties is an approximate identity. Then, for $f$ continuous and $2\pi$-periodic, hence uniformly continuous, choose $\delta$ with $|f(x-t) - f(x)| < \varepsilon$ for $|t| < \delta$ and all $x$; splitting the integral, $$|\sigma_Nf(x) - f(x)| \le \tfrac1{2\pi}\!\int_{|t|<\delta}\!\!|f(x-t)-f(x)|F_N + \tfrac1{2\pi}\!\int_{\delta\le|t|\le\pi}\!\!2\lVert f\rVert_\infty F_N \le \varepsilon + \frac{2\lVert f\rVert_\infty}{(N+1)\sin^2(\delta/2)},$$ uniformly in $x$. Let $N \to \infty$, then $\varepsilon \to 0$. $\square$

Full proof, steps (ii) and (iii)

(ii) Let $f$ be piecewise continuous on $[-\pi,\pi]$, bounded by $M$, with jumps at $x_1,\dots,x_J$ (count $\pm\pi$ as one, for periodicity). Given $\varepsilon > 0$, let $g$ agree with $f$ outside intervals of half-width $\eta$ about each jump and interpolate linearly across them. Then $g$ is continuous and periodic and $\lVert f-g\rVert^2 \le (2M)^2\cdot 2\eta J$, which is below $\varepsilon^2$ for small $\eta$.

(iii) Fejér gives a trigonometric polynomial $T = \sigma_Ng$ with $\lVert g-T\rVert_\infty < \varepsilon$, hence $\lVert g-T\rVert_2 \le \sqrt{2\pi}\,\varepsilon$. Say $T$ has degree $D$. For $N \ge D$, $T$ lies in the span of $e_1,\dots$ up to degree $N$, so best approximation gives $\lVert f - S_Nf\rVert \le \lVert f - T\rVert \le \lVert f-g\rVert + \lVert g-T\rVert \le \varepsilon(1 + \sqrt{2\pi})$. Since $\varepsilon$ was arbitrary, $\lVert f - S_Nf\rVert \to 0$. $\square$ (du Bois-Reymond's continuous function whose Fourier series diverges at a point is cited: Stein & Shakarchi, Fourier Analysis, ch. 3 §2.2.) For the sine system on $[0,L]$, apply this to the odd $2L$-periodic extension of $f$, whose full Fourier series is its sine series.

Complete, and "nothing is orthogonal to all of them". If the system is complete and $\langle f,e_n\rangle = 0$ for every $n$, then every $S_Nf = 0$ and $\lVert f\rVert = \lim\lVert f - S_Nf\rVert = 0$. The converse — no nonzero $f$ orthogonal to every $e_n$, hence complete — is true in a metrically complete space such as $L^2_w$ (cited: it is the projection theorem for Hilbert spaces), but it can fail in $PC$, which is not metrically complete. So in this unit completeness is always checked through the definition, $\lVert f - S_Nf\rVert \to 0$, and the orthogonality test is used only in the safe direction: to refute completeness, as P12 does.

The two senses of "complete". This unit's sense — the span is dense — is a property of a system. The metric sense — every Cauchy sequence converges — is a property of a space, and $PC$ does not have it. The words collide and always have; the context always disambiguates, because a system is complete in a space.

19 · orthogonality

Sine, cosine, complex: one system per boundary condition

problem on $[0,L]$	eigenfunctions	expansion of $f$	extension
$u(0) = u(L) = 0$	$\sin\frac{n\pi x}{L}$, $n\ge1$	sine series	odd, $2L$-periodic
$u'(0) = u'(L) = 0$	$\cos\frac{n\pi x}{L}$, $n\ge0$	cosine series	even, $2L$-periodic
$u(0)=u(L),\ u'(0)=u'(L)$	$\mathrm e^{2\pi\mathrm inx/L}$, $n\in\mathbb Z$	complex series	$L$-periodic

One function on one interval has all three expansions, with different coefficients and different behaviour at the ends, because the extensions differ there.

the complex Fourier series

With $\langle f,g\rangle = \int_0^L f\bar g$, the $\mathrm e^{2\pi\mathrm inx/L}$ are orthogonal with norm$^2$ $L$, so

$$f \sim \sum_{n\in\mathbb Z}c_n\mathrm e^{2\pi\mathrm inx/L},\quad c_n = \frac1L\int_0^Lf(x)\mathrm e^{-2\pi\mathrm inx/L}\mathrm dx .$$

$f$ real $\iff c_{-n} = \overline{c_n}$.

Choosing an expansion is choosing a boundary condition. Section 4 says why: the boundary condition is part of the operator.

19 · orthogonality · example

Worked: one function, three expansions

$f(x) = x$ on $[0,L]$ (solid) and the extension each system implicitly makes (dashed). The odd extension jumps at $x = L$, the even one only has corners, the $L$-periodic one jumps at $0$ and $L$. Open circles: the jump midpoints, where Dirichlet's theorem (next section) says the series land.

$f(x) = x$ on $[0,L]$, three ways

Sine: $\alpha_n = \dfrac{2L(-1)^{n+1}}{n\pi}$ — decay $1/n$, because the odd extension jumps from $L$ to $-L$ at $x = L$.

Cosine: $a_0/2 = L/2$, $a_n = \dfrac{-4L}{n^2\pi^2}$ for odd $n$ and $0$ for even — decay $1/n^2$, because the even extension is continuous with a kink.

Complex: $c_0 = L/2$, $c_n = \dfrac{\mathrm iL}{2\pi n}$ — decay $1/n$, jump at $x = 0$.

At $x = L$ the three converge to $0$, to $L$, and to $L/2$. All three are right; they are answering different questions.

The three coefficient computations. Sine: $\alpha_n = \frac2L\int_0^Lx\sin\frac{n\pi x}L\mathrm dx = \frac2L\big[-\frac{xL}{n\pi}\cos\frac{n\pi x}L\big]_0^L + \frac{2}{n\pi}\int_0^L\cos\frac{n\pi x}L\mathrm dx = -\frac{2L}{n\pi}(-1)^n + 0$. Cosine: $a_n = \frac2L\int_0^Lx\cos\frac{n\pi x}L\mathrm dx = \frac2L\big(\big[\frac{xL}{n\pi}\sin\big]_0^L - \frac{L}{n\pi}\int_0^L\sin\big) = \frac{2L}{n^2\pi^2}\big((-1)^n - 1\big)$, which is $-4L/n^2\pi^2$ for odd $n$; and $a_0 = \frac2L\int_0^Lx\,\mathrm dx = L$. Complex: $c_n = \frac1L\int_0^Lx\mathrm e^{-2\pi\mathrm inx/L}\mathrm dx$; integrating by parts, the boundary term is $\frac1L\cdot\frac{L}{-2\pi\mathrm in}\big[x\mathrm e^{-2\pi\mathrm inx/L}\big]_0^L = \frac{L}{-2\pi\mathrm in} = \frac{\mathrm iL}{2\pi n}$ and the remaining integral vanishes.

The end values, from Dirichlet's theorem ("Dirichlet: what the series converges to at a jump"). At $x = L$ the odd $2L$-periodic extension has left limit $L$ and right limit $-L$, midpoint $0$. The even extension is continuous there, value $L$. The $L$-periodic extension jumps from $L$ to $0$, midpoint $L/2$. All three predictions are confirmed by the formulas: the sine series is a sum of $\sin n\pi = 0$; the cosine series at $x = L$ is $L/2 + \sum_{\text{odd}}\frac{4L}{n^2\pi^2} = L/2 + \frac{4L}{\pi^2}\cdot\frac{\pi^2}{8} = L$ ✓, using $\sum_{\text{odd}}n^{-2} = \pi^2/8$; the complex series at $x = L$ is $c_0 = L/2$, since every other term is $c_n\mathrm e^{-2\pi\mathrm in} = c_n$ and $\sum_{n\ne0}c_n = 0$ by the cancellation of $\pm n$ ✓.

Why the decay rates differ, in one line. A jump in the periodic extension gives $1/n$; a jump in its first derivative gives $1/n^2$. That is the dictionary on "Smoothness against decay", and it is the practical reason to prefer a cosine series when either will do: the same accuracy for the square root of the terms.

19 · orthogonality · example

Worked: Parseval, twice, with the drum's weight

a sum from a square wave

$f \equiv 1$ on $[0,L]$ in sines: $\alpha_n = \frac2L\int_0^L\sin\frac{n\pi x}L\mathrm dx = \frac{2}{n\pi}(1 - (-1)^n)$, so $4/n\pi$ for odd $n$, $0$ for even. Parseval, with $\lVert S_n\rVert^2 = L/2$:

$$L = \frac L2\sum_{n\ \text{odd}}\frac{16}{n^2\pi^2} \ \Longrightarrow\ \sum_{k\ge0}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8}$$

Summing the left side to $10^6$ terms gives $1.2337003$, against $\pi^2/8 = 1.2337006$ ✓. The same manoeuvre on $f(x) = x$, with $\alpha_n = 2L(-1)^{n+1}/n\pi$, gives $\sum n^{-2} = \pi^2/6$.

checks

Dimensions. Both sides of each identity carry $[f]^2$ times a length (a length$^2$ for the drum) ✓.

Special case. $f = J_0(j_{01}r/a)$ gives $\hat f_1 = 1$ and all other $\hat f_k = 0$, and Parseval reduces to the norm formula ✓.

the drum's version, with $w = r$

A radial initial shape on the drum expands in $\{J_0(j_{0k}r/a)\}$, orthogonal with weight $w = r$ — the polar area element (section 4). So

$$\hat f_k = \frac{\int_0^a f(r)J_0\!\big(\tfrac{j_{0k}r}{a}\big)\,r\,\mathrm dr}{\int_0^a J_0\!\big(\tfrac{j_{0k}r}{a}\big)^2\,r\,\mathrm dr},$$

with $\int_0^aJ_0(j_{0k}r/a)^2r\,\mathrm dr = \tfrac12a^2J_1(j_{0k})^2$, and Parseval reads

$$\int_0^a f^2\,r\,\mathrm dr = \tfrac12a^2\sum_{k\ge1}\hat f_k^2\,J_1(j_{0k})^2 .$$

The weight is not a decoration. Without the $r$, the $J_0(j_{0k}r/a)$ are not orthogonal: $\int_0^aJ_0(j_{01}r/a)J_0(j_{02}r/a)\,\mathrm dr = 0.19608\,a$ for the first pair, while with the weight it is zero to every digit computed. The $r$ is there because $\mathrm dA = r\,\mathrm dr\,\mathrm d\theta$ in polar coordinates (unit 05), so $\int_{\text{disc}}fg\,\mathrm dA = 2\pi\int_0^afg\,r\,\mathrm dr$ for radial $f,g$. The inner product that makes the operator symmetric is the one the geometry hands you, every time.

Full derivation of the norm $\tfrac12a^2J_1(j_{0k})^2$

Let $R(r) = J_0(\kappa r)$ with $\kappa = j_{0k}/a$, so $(rR')' = -\kappa^2rR$. Multiply by $2r R'$ and integrate: $\int_0^a 2rR'(rR')'\,\mathrm dr = \big[(rR')^2\big]_0^a$, while the right side is $-2\kappa^2\int_0^ar^2RR'\,\mathrm dr = -\kappa^2\big[r^2R^2\big]_0^a + 2\kappa^2\int_0^arR^2\mathrm dr$. With $R(a) = J_0(j_{0k}) = 0$ the term $[r^2R^2]_0^a$ vanishes, leaving $\int_0^arR^2\mathrm dr = \frac{a^2R'(a)^2}{2\kappa^2}$. Since $J_0' = -J_1$ (unit 09's recurrence), $R'(a) = -\kappa J_1(j_{0k})$, so the integral is $\tfrac12a^2J_1(j_{0k})^2$. Dimensions: length$^2$ ✓. Numerically $J_1(j_{01}) = 0.5191474973$, so the norm$^2$ is $0.1347571\,a^2$.

Euler and the Basel problem. $\sum n^{-2} = \pi^2/6$ is what Euler found in 1735 by a formal factorisation of $\sin x$; Parseval gives it in three lines and with no leap of faith. Parseval himself stated the identity in 1799, also formally — the proof above, via best approximation and Fejér, is a nineteenth- and twentieth-century assembly.

section 3

Convergence: which kind, and how fast

Mean-square convergence is free and says nothing about any single point. Pointwise and uniform cost hypotheses, and the price is exactly the smoothness of the data.

19 · convergence

Dirichlet: what the series converges to at a jump

Dirichlet's theorem (1829)

Let $f$ be $2\pi$-periodic and piecewise $C^1$: continuous but for finitely many points, $f'$ piecewise continuous, with one-sided limits $f(x\pm)$, $f'(x\pm)$ everywhere. Then for every $x$,

$$S_Nf(x) \longrightarrow \tfrac12\big(f(x+) + f(x-)\big).$$

In particular $S_Nf(x) \to f(x)$ at every point of continuity, and to the midpoint of the jump elsewhere.

full proof in the notes (N)

It cannot do better: the coefficients are integrals of $f$, which cannot see the value at one point. Changing $f$ at a single point changes no coefficient at all.

the hypothesis is not decoration

Continuity alone is not enough — du Bois-Reymond, 1876. What piecewise $C^1$ buys is a bounded difference quotient, which is exactly what the proof feeds to Riemann–Lebesgue.

the square wave

$f = +1$ on $(0,\pi)$, $-1$ on $(-\pi,0)$, extended $2\pi$-periodically. It is odd, so only sines appear, with coefficients $b_n = \frac1\pi\int_{-\pi}^{\pi}f\sin nx\,\mathrm dx = \frac{2}{n\pi}(1-(-1)^n)$, and

$$S_N(x) = \frac4\pi\sum_{k\ \text{odd}\le N}\frac{\sin kx}{k}.$$

At $x = 0$ every term is zero, so $S_N(0) = 0 = \tfrac12(1 + (-1))$ ✓ — the theorem, visible without any analysis.

At $x = \pi/2$: $\frac4\pi(1 - \tfrac13 + \tfrac15 - \cdots) = \frac4\pi\cdot\frac\pi4 = 1$ ✓, Leibniz's series recovered.

Full proof of Dirichlet's theorem

With the Dirichlet kernel $D_N(t) = \sum_{|n|\le N}\mathrm e^{\mathrm int} = \dfrac{\sin\big((N+\frac12)t\big)}{\sin(t/2)}$, convolution gives $S_Nf(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-t)D_N(t)\,\mathrm dt$, and $D_N$ is even with $\frac1{2\pi}\int_{-\pi}^{\pi}D_N = 1$, hence $\frac1{2\pi}\int_0^{\pi}D_N = \tfrac12$. Therefore $$S_Nf(x) - \tfrac12\big(f(x+)+f(x-)\big) = \frac1{2\pi}\int_0^{\pi}\Big[\big(f(x-t)-f(x-)\big) + \big(f(x+t)-f(x+)\big)\Big]D_N(t)\,\mathrm dt .$$ Write the bracket as $\Delta(t)$ and set $g(t) := \Delta(t)/\sin(t/2)$, so that the integral is $\frac1{2\pi}\int_0^\pi g(t)\sin\big((N+\tfrac12)t\big)\mathrm dt$. The only question is whether $g$ is piecewise continuous on $(0,\pi]$ including at $t = 0^+$, since $\sin(t/2)$ vanishes there. It is: by the mean value theorem, for small $t > 0$, $f(x+t) - f(x+) = t\,f'(\xi)$ for some $\xi \in (x, x+t)$, and $f'$ is bounded near $x$ on that side, so $|\Delta(t)| \le Ct$; and $\sin(t/2) \ge t/\pi$ on $[0,\pi]$, giving $|g(t)| \le \pi C$. Better, $f'(\xi) \to f'(x+)$ as $t \to 0^+$ (and likewise on the left), so $g$ has a one-sided limit at $0$. Hence $g$ is piecewise continuous on $[0,\pi]$, and the Riemann–Lebesgue corollary of Bessel's inequality — in the half-integer form spelt out in the notes of "Bessel's inequality" — sends the integral to zero. $\square$

Where each hypothesis is spent. Piecewise continuity of $f$ makes the coefficients exist. Existence of the one-sided limits $f(x\pm)$ makes the statement meaningful. Boundedness of $f'$ on each side near $x$ — the piecewise-$C^1$ part — is what bounds $g$; it can be weakened to a Dini or Hölder condition at the single point $x$, and the same proof runs. Nothing about $f$ away from $x$ matters beyond integrability: this is Riemann's localisation principle, and it is surprising, because the coefficients depend on all of $f$ while the convergence at $x$ does not.

The geometric series behind $D_N$

$\sum_{n=-N}^{N}z^n = \frac{z^{-N} - z^{N+1}}{1-z}$ with $z = \mathrm e^{\mathrm it}$; multiplying numerator and denominator by $\mathrm e^{-\mathrm it/2}$ turns it into $\frac{\mathrm e^{-\mathrm i(N+1/2)t} - \mathrm e^{\mathrm i(N+1/2)t}}{\mathrm e^{-\mathrm it/2} - \mathrm e^{\mathrm it/2}} = \frac{\sin((N+\frac12)t)}{\sin(t/2)}$. Only unit 06's complex exponential algebra is used; nothing here needs complex analysis.

19 · convergence

Uniform convergence, and exactly when it fails

when the series converges uniformly

If $f$ is continuous, $2\pi$-periodic and piecewise $C^1$, then $\sum_n|c_n| < \infty$ and $S_Nf \to f$ uniformly and absolutely.

proof

Integrating by parts, with continuity and periodicity killing the boundary term, $c_n(f') = \mathrm in\,c_n(f)$; Bessel for $f'$ gives $\sum n^2|c_n|^2 \lt \infty$. Then by Cauchy–Schwarz

$$\sum_{n\ne0}|c_n| = \sum_{n\ne0}\big(n|c_n|\big)\tfrac1{|n|}$$ $$\le \Big(\sum n^2|c_n|^2\Big)^{\!1/2}\Big(\sum_{n\ne0}\tfrac1{n^2}\Big)^{\!1/2} \lt \infty .$$

The $M$-test then gives uniform convergence to some continuous limit, and Dirichlet identifies it as $f$. $\square$

and when it cannot

A uniform limit of continuous functions is continuous, and each $S_Nf$ is a trigonometric polynomial. So if $f$ jumps, the convergence is not uniform on any interval containing the jump — no hypothesis can rescue it. The series still converges in norm and pointwise to the midpoint; the next slide measures by how much uniformity fails.

the three modes, ranked

Mean-square: free, for every piecewise continuous $f$.
Pointwise: guaranteed by piecewise $C^1$; midpoints at jumps.
Uniform: guaranteed by continuity as well, through $\sum|c_n| < \infty$.

Each hypothesis is sufficient, not necessary; only the first mode is automatic.

The integration by parts, in full. $c_n(f') = \frac1{2\pi}\int_{-\pi}^{\pi}f'(x)\mathrm e^{-\mathrm inx}\mathrm dx = \frac1{2\pi}\big[f\mathrm e^{-\mathrm inx}\big]_{-\pi}^{\pi} + \frac{\mathrm in}{2\pi}\int_{-\pi}^{\pi}f\mathrm e^{-\mathrm inx}\mathrm dx$. The bracket is $f(\pi)\mathrm e^{-\mathrm in\pi} - f(-\pi)\mathrm e^{\mathrm in\pi} = (-1)^n\big(f(\pi)-f(-\pi)\big) = 0$ by periodicity and continuity — drop continuity and this is exactly the term that survives and destroys the whole argument. For piecewise $C^1$ $f$ with $f$ continuous, do the same on each subinterval where $f'$ is continuous and note the interior boundary terms cancel in pairs.

Why $\sum|c_n| < \infty$ is the right criterion. It is sufficient for uniform convergence by the $M$-test, and it is not necessary in general — there exist continuous functions with uniformly convergent Fourier series and $\sum|c_n| = \infty$. What is true and clean: $\sum|c_n| < \infty$ $\implies$ uniform, and $f$ merely continuous $\implies$ nothing. The gap between them is the subject of a century of hard analysis (Carleson's 1966 theorem: for $f \in L^2$, $S_Nf \to f$ almost everywhere. Cited; the proof is far outside this course).

What this means for term-by-term integration, which is easy. If $\lVert f - S_Nf\rVert_2 \to 0$ then $\int_0^xS_Nf \to \int_0^xf$ uniformly in $x$, by Cauchy–Schwarz against the indicator of $[0,x]$. So a Fourier series may always be integrated term by term, even when it does not converge. Integration divides the $n$-th coefficient by $n$; differentiation multiplies by $n$. That asymmetry is the whole story of "Smoothness against decay" and "Differentiating term by term".

19 · convergence · example

Worked: Gibbs, and the number 8.9490 %

the overshoot at a jump

Square wave, $S_{2M-1} = \frac4\pi\sum_{j<M}\frac{\sin(2j+1)x}{2j+1}$. Differentiate and sum the cosines (the geometric sum is in the notes):

$$S'(x) = \frac4\pi\sum_{j=0}^{M-1}\cos(2j+1)x = \frac{2}{\pi}\,\frac{\sin 2Mx}{\sin x},$$

so the first maximum right of the jump is at $x_M = \pi/2M$. Integrate from $0$, where $S = 0$, and put $u = 2Mt$:

$$S(x_M) = \frac2\pi\int_0^{\pi/2M}\!\!\frac{\sin 2Mt}{\sin t}\mathrm dt = \frac{2}{\pi}\int_0^{\pi}\!\frac{\sin u\,\mathrm du}{2M\sin\frac{u}{2M}} .$$

As $M\to\infty$, $2M\sin(u/2M) \to u$ uniformly on $[0,\pi]$, so

$$S(x_M) \to \frac2\pi\int_0^{\pi}\frac{\sin u}{u}\mathrm du = \frac2\pi\mathrm{Si}(\pi) = 1.17898 .$$

the Gibbs phenomenon

Near a jump of size $J$ the partial sums overshoot by $\big(\tfrac2\pi\mathrm{Si}(\pi) - 1\big)\tfrac J2 = 0.08948987\,J$ — 8.9490 % of the jump, on each side. It does not shrink with $N$; it only moves closer to the jump, at distance $\pi/2M$.

checks

Dimensions. A fraction of a jump; dimensionless ✓.

Numbers. Summing directly at $x = \pi/2M$: $9.116\%$ at $M=5$, $8.959\%$ at $M=20$, $8.9494\%$ at $M=100$, $8.94899\%$ at $M=1000$ ✓.

Consistency. The width $\pi/2M \to 0$, so the overshoot occupies a vanishing interval and contributes nothing to $\lVert f - S_Nf\rVert_2$ ✓.

The cosine sum. $\sum_{j=0}^{M-1}\cos(2j+1)x = \operatorname{Re}\sum_{j=0}^{M-1}\mathrm e^{\mathrm i(2j+1)x} = \operatorname{Re}\Big(\mathrm e^{\mathrm ix}\dfrac{\mathrm e^{2\mathrm iMx}-1}{\mathrm e^{2\mathrm ix}-1}\Big)$. Pull a half-angle out of each difference of exponentials: $\mathrm e^{2\mathrm iMx}-1 = \mathrm e^{\mathrm iMx}\cdot2\mathrm i\sin Mx$ and $\mathrm e^{2\mathrm ix}-1 = \mathrm e^{\mathrm ix}\cdot2\mathrm i\sin x$, so the quotient is $\mathrm e^{\mathrm i(M-1)x}\sin Mx/\sin x$ and the whole bracket is $\mathrm e^{\mathrm iMx}\sin Mx/\sin x$. Its real part is $\cos Mx\,\sin Mx/\sin x = \sin(2Mx)/(2\sin x)$. Multiplying by $4/\pi$ gives the displayed $S'$.

Why $x_M = \pi/2M$ is the first maximum. $S'$ has the sign of $\sin 2Mx$ for $0 < x < \pi$, so $S$ increases on $(0,\pi/2M)$ and decreases after it. The subsequent stationary points give the decaying ripples, at $x = k\pi/2M$.

The uniform limit, justified. For $u \in [0,\pi]$ and $M \ge 1$, $\big|\frac{1}{2M\sin(u/2M)} - \frac1u\big| = \frac{|u - 2M\sin(u/2M)|}{2Mu\sin(u/2M)} \le \frac{u^3/24M^2}{u\cdot 2u/\pi} = \frac{\pi u}{48M^2} \le \frac{\pi^2}{48M^2}$. The numerator bound is $|s - \sin s| \le s^3/6$ with $s = u/2M$, so $|u - 2M\sin s| \le 2M s^3/6 = u^3/24M^2$; the denominator bound is $\sin s \ge 2s/\pi$ on $[0,\pi/2]$, so $2M\sin s \ge 2u/\pi$. Multiplying by $|\sin u| \le 1$ and integrating over $[0,\pi]$, the two integrals differ by at most $\pi^3/48M^2 \to 0$, so the limit may be taken inside.

History, and a correction. Henry Wilbraham found this in 1848 and was ignored. J. Willard Gibbs described it in a letter to Nature in 1899, after Michelson's mechanical harmonic analyser produced the ripples and Michelson suspected his machine. It was not the machine. The name is Bôcher's, from 1906.

19 · convergence · example

Gibbs, seen and repaired

$S_9$ ($M = 5$) and $S_{49}$ ($M = 25$) at the jump. Both pass through the midpoint $0$; both peak near $1.18$ ($1.182$ and $1.179$); more terms only squeeze the peak towards the jump.

The same jump with $N = 25$: the partial sum $S_{25}$ overshoots, the Fejér average $\sigma_{25}$ of $S_0, \dots, S_{25}$ does not — it climbs monotonically to the jump height and stays below it, at the price of a softer edge.

the repair

$\sigma_Nf$ is $f$ averaged against the Fejér kernel, which is nonnegative with total weight $1$ ("Completeness, and Parseval", notes). An average of values in $[-1,1]$ lies in $[-1,1]$: $\sigma_N$ cannot overshoot the range of $f$.

The left figure. It is the worked example of the previous slide, drawn: the first maximum of $S_9$ is at $x = \pi/10$ and that of $S_{49}$ at $\pi/50$, five times closer to the jump, at almost the same height. Every partial sum passes through the midpoint $0$ at the jump itself, as Dirichlet's theorem requires.

What is actually done about it. Nothing, if you are computing $\lVert\cdot\rVert_2$. If you need pointwise accuracy near a jump, average the partial sums — Fejér's $\sigma_N$, whose kernel is nonnegative and which therefore cannot overshoot the range of $f$ at all. That is the same repair that made the completeness proof work, used for a second reason, and it costs a slower convergence rate away from the jump.

What to look for in the right figure. Averaging $S_0,\dots,S_N$ multiplies the $n$-th coefficient by $1 - n/(N+1)$, tapering the high modes linearly to zero instead of cutting them off. The ripples of $S_{25}$ are those high modes beating against the cut-off; tapered, they are gone, and so is the overshoot. The cost is visible too: $\sigma_{25}$ reaches the plateau more slowly than $S_{25}$, and away from jumps it converges at rate $1/N$ even for smooth $f$, where $S_Nf$ would do much better.

19 · convergence

Smoothness against decay: a dictionary, and an error bar

smoothness controls decay

If $f$ is $2\pi$-periodic and $C^{k-1}$ with $f^{(k)}$ piecewise continuous, integrating by parts $k$ times gives $c_n(f^{(k)}) = (\mathrm in)^kc_n(f)$; Bessel applied to $f^{(k)}$ then gives $\sum_nn^{2k}|c_n|^2 \lt \infty$, so $|c_n| = o(|n|^{-k})$, and by Parseval

$$\lVert f - S_Nf\rVert_2^2 = 2\pi\!\!\sum_{|n|>N}\!\!|c_n|^2 \le \frac{2\pi}{N^{2k}}\!\!\sum_{|n|>N}\!\!n^{2k}|c_n|^2 .$$

lowest derivative that jumps	$\|c_n\|$	$\lVert f-S_Nf\rVert_2$	example
$f$ itself	$\sim1/n$	$\sim N^{-1/2}$	square wave
$f'$	$\sim1/n^2$	$\sim N^{-3/2}$	plucked string
$f''$	$\sim1/n^3$	$\sim N^{-5/2}$	a parabolic arch
none ($C^\infty$)	faster than every power	same	$\mathrm e^{\cos x}$

The same dictionary governs the drum, with $J_0(j_{0k}r/a)$ for sines: a finger-pressed dimple has a kink and needs many modes; a smooth bulge needs few.

$|c_n|$ against $n$, both on log scales, for the table's four rows (odd $n$; the even coefficients vanish for the first three). A power law is a straight line whose slope is the power — $-1$, $-2$, $-3$, one step for each extra continuous derivative — and $\mathrm e^{\cos x}$, with no derivative that jumps, falls off a cliff.

Why the table's "lowest derivative that jumps" is the right index. If $f^{(m)}$ has a jump of size $J$ at $x_0$ and $f^{(m-1)}$ is continuous, integrating by parts $m+1$ times leaves a surviving boundary term $\propto J\mathrm e^{-\mathrm inx_0}/n^{m+1}$ together with terms that are $o(n^{-(m+1)})$. So $c_n \sim CJ/n^{m+1}$ exactly — the decay rate is not merely bounded but determined, and the constant carries the size of the jump. That is why the table's third column is an asymptotic and not only an inequality, and it is the practical version of the theorem: to find out how smooth a signal is, look at how its coefficients fall.

The truncation bound made usable. The displayed inequality holds for each fixed $N$ with an explicit constant if $f^{(k)}$ has a known $L^2$ norm: by Parseval, $\sum n^{2k}|c_n|^2 = \lVert f^{(k)}\rVert_2^2/2\pi$, so $\lVert f - S_Nf\rVert_2 \le \lVert f^{(k)}\rVert_2/N^k$. That is a genuine a priori error bar: it says how many modes are needed, before any of them is computed. Problem P3 computes both sides for the plucked and the struck string and reads the answer as a statement about brightness.

The four curves, exactly. On $[-\pi,\pi]$: the square wave ($\pm1$) has $|c_n| = 2/\pi n$; the triangle wave $|x|$ — the plucked string's shape, a kink — has $|c_n| = 2/\pi n^2$; the parabolic arch $x(\pi - x)$ on $[0,\pi]$, extended oddly, has $|c_n| = 4/\pi n^3$; all three for odd $n$ only. And $\mathrm e^{\cos x}$ has $c_n = I_n(1)$, the modified Bessel function, which falls like $1/(2^nn!)$ — faster than any power, so on log-log axes it bends ever more steeply down.

19 · convergence · example

Smoothness in decibels: pluck against strike

what it is worth, in decibels

Plucked: $\alpha_n \propto n^{-2}$, so partial $n$ is $40\log_{10}n$ dB below the first — partial 10 at $-40$ dB. Struck by a hammer of width $\varepsilon \ll L$: $\alpha_n \propto n^{-1}$ while $n \ll L/\pi\varepsilon$ — partial 10 at $-20$ dB.

Twenty decibels at the tenth partial, idealised pluck against idealised strike — and it is a statement about smoothness.

check

Special case. $f = \cos x$ exactly: $k$ is unbounded and all but two coefficients are zero, so every bound is satisfied trivially ✓. Dimensions. $|c_n|$ carries $[f]$, and $n$ is a pure number ✓.

Partial levels relative to the first. The pluck's $1/n^2$ is a straight $-40$ dB per decade. A hammer of width $\varepsilon = L/50$ gives $1/n$ — partial 10 at $-20.6$ dB — up to a knee near $n \approx L/\pi\varepsilon \approx 16$, beyond which its smoothness on the scale $\varepsilon$ shows and the slope steepens.

The decibel arithmetic. An amplitude ratio $\rho$ is $20\log_{10}\rho$ decibels. For $\alpha_n/\alpha_1 = n^{-2}$ and $n = 10$ that is $-20\log_{10}100 = -40$ dB; for $n^{-1}$ it is $-20$ dB. Both refer to amplitude, not energy; the energy ratios are the same numbers because the decibel already accounts for the square.

Where the hammer width cuts in. A hammer of half-width $\varepsilon$ striking at $x = a$ gives velocity coefficients $\propto \sin(n\pi a/L)\sin(n\pi\varepsilon/L)/n$, hence displacement coefficients $\propto \sin(n\pi a/L)\sin(n\pi\varepsilon/L)/n^2$. While $n\pi\varepsilon/L \ll 1$ the second sine is $\approx n\pi\varepsilon/L$ and the net decay is $1/n$; beyond $n \approx L/\pi\varepsilon$ it saturates and the true $1/n^2$ resumes. A real piano hammer is soft and wide, which is exactly a designer's choice about where that knee sits. Real instruments blur the idealised contrast in both directions: a harpsichord plucks close to the end, and by the $a \to 0$ limit of the plucked-string slide that is itself $1/n$-bright over the first $L/a$ partials, while a felt hammer's width pulls the piano's knee down. The dictionary is exact; the instruments are engineered compromises on it.

19 · convergence

Differentiating term by term, and what it costs

when it is legal

Let $f$ be $2\pi$-periodic, continuous, and piecewise $C^1$. Then the Fourier series of $f'$ is obtained by differentiating that of $f$ term by term, and it converges to $f'$ in mean square — and to midpoints pointwise if $f'$ is itself piecewise $C^1$.

Proof. $c_n(f') = \mathrm inc_n(f)$, proved on the uniform-convergence slide; then completeness for the piecewise continuous $f'$, and Dirichlet for $f'$ under the extra hypothesis. $\square$

the price

Differentiation multiplies $c_n$ by $\mathrm in$: one power of decay is lost each time. A series with $c_n \sim n^{-2}$ that converged uniformly differentiates to one with $c_n \sim n^{-1}$, which converges in mean square and not uniformly. Differentiate once more and it need not converge at all.

two failures, for two different reasons

Continuity fails. The square wave has $S(x) = \frac4\pi\sum_{\text{odd}}\frac{\sin kx}{k}$; differentiating termwise gives $\frac4\pi\sum_{\text{odd}}\cos kx$, whose terms do not tend to zero, so it diverges at every $x$ except $\pm\pi/2$. The theorem's hypothesis is exactly what is missing: $f$ jumps.

Continuity holds and it still fails. $\displaystyle f(x) = \sum_{n\ge1}\frac{\sin n^2x}{n^2}$ converges uniformly by the $M$-test, so $f$ is continuous. Termwise, $\sum\cos n^2x$ — terms not tending to zero, divergent everywhere. Here $f$ is continuous but not piecewise $C^1$; the second hypothesis is the one being used.

This is why the wave equation's series solution needs care and the heat equation's does not: $\mathrm e^{-\kappa\lambda_nt}$ restores every power of $n$ lost, for any $t > 0$.

The rule, in one sentence. Integration is always safe and gains a power of $n$; differentiation needs continuity of the function being differentiated and costs a power of $n$. Both facts come from the same integration by parts, read in opposite directions.

Where the square wave's derived series converges. If $\cos kx \to 0$ along odd $k$, then $\cos 2kx = 2\cos^2kx - 1 \to -1$ along odd $k$, so both $2kx$ and $2(k+2)x = 2kx + 4x$ tend to $\pi$ modulo $2\pi$, forcing $4x \in 2\pi\mathbb Z$. Of the four candidates $x = 0, \pm\pi/2, \pi$, the terms are $\pm1$ at $0$ and $\pi$ and exactly $0$ at $\pm\pi/2$. So the derived series converges at $\pm\pi/2$ and nowhere else. For $\sum\cos n^2x$ the analogous argument (P4) leaves no exceptional point at all.

Why the second counterexample is worth having. Without it a reader concludes that "continuous" is the real hypothesis and "piecewise $C^1$" is bookkeeping. It is not: $\sum n^{-2}\sin(n^2x)$ is continuous everywhere, and it is differentiable at no irrational multiple of $\pi$ — a dense set (Hardy, 1916; cited, with the exact set of points of differentiability found by Gerver in 1970–71). A continuous function can be far too rough to differentiate at all, and its Fourier series knows.

Back to the wave equation. The series $u(x,t) = \sum\alpha_n\sin(n\pi x/L)\cos(\omega_nt)$ must be differentiated twice in each variable for the equation to hold classically, which $\sum n^2|\alpha_n| < \infty$ guarantees — true, for instance, when the odd periodic extension of $f$ is $C^2$ with a piecewise continuous third derivative (the uniform-convergence argument, two derivatives up). The plucked string has $\alpha_n \sim n^{-2}$ and does not qualify: the series converges uniformly to a continuous $u$, but that $u$ is not twice differentiable at the travelling kinks, as unit 18's d'Alembert picture already showed. The honest statement is that $u$ is a weak solution, and the machinery that makes "weak" precise is unit 20's distributions. Until then, the physical claim stands on unit 18's d'Alembert formula, which needs no differentiation of a series at all — and it agrees with the mode sum wherever both make sense, by uniqueness.

section 4

Sturm–Liouville: why any of this was inevitable

One operator, one identity, and five theorems that were proved separately for sines, Bessel functions and Legendre polynomials before anyone noticed they were the same proof.

19 · Sturm–Liouville

The operator, and the weight that comes with it

the Sturm–Liouville operator

On $[a,b]$ let $p \in C^1$ with $p > 0$, $q \in C^0$ real, $w \in C^0$ with $w > 0$. Define a linear map on $C^2[a,b]$ by

$$\mathcal Lu := -(p\,u')' + q\,u .$$

The Sturm–Liouville problem is: find $\lambda \in \mathbb C$ and $u \not\equiv 0$ with

$$\mathcal Lu = \lambda\,w\,u \quad\text{on }(a,b),$$

together with boundary conditions.

Divide by $w$ and it is an eigenvalue problem for the operator $\mathcal A := w^{-1}\mathcal L$ on the inner-product space $\big(PC[a,b],\ \langle\cdot,\cdot\rangle_w\big)$ of the last section. The weight in the equation and the weight in the inner product are the same $w$. That is not a coincidence; it is the next slide.

every second-order equation has this form

Given $A(x)u'' + B(x)u' + C(x)u + \lambda D(x)u = 0$ with $A > 0$, multiply through by the integrating factor $\mu = A^{-1}\exp\!\int\! (B/A)$. Then $\mu A = \exp\!\int\!(B/A) =: p$ and $\mu(Au'' + Bu') = (pu')'$ exactly, giving $-(pu')' - \mu Cu = \lambda\mu Du$. $\square$

equation	$p$	$q$	$w$	on
string, $u''=-\lambda u$	$1$	$0$	$1$	$[0,L]$
Bessel, order $m$	$r$	$m^2/r$	$r$	$(0,a]$
Legendre (unit 10)	$1-x^2$	$0$	$1$	$[-1,1]$
non-uniform string	$T$	$0$	$\rho(x)$	$[0,L]$

The weight is where the physics hides: for a string it is the mass density, for the drum it is the area element. Orthogonality is always with respect to the thing that measures how much stuff is there.

Sign convention. Many texts write the problem as $(pu')' + qu + \lambda wu = 0$, which is the same equation with $q \to -q$ and the operator negated. The convention used here, $\mathcal L = -(pu')' + q$, is chosen because it makes $\mathcal L$ bounded below and the eigenvalues increase to $+\infty$, matching the physical reading "$\lambda = \omega^2/c^2$" and matching unit 07's positive semidefinite $K$. When comparing with a book, check the sign of $q$ and the sign of $\lambda$ before comparing anything else; the two conventions differ in exactly those two places.

The integrating factor, checked. With $p = \exp\int^x(B/A)$, we have $p' = (B/A)p$, so $(pu')' = pu'' + p'u' = p\big(u'' + (B/A)u'\big) = (p/A)(Au'' + Bu')$, and $p/A = \mu$ ✓. The construction needs $A > 0$ on the interval — where $A$ vanishes the endpoint is singular, which is exactly Bessel at $r = 0$ and Legendre at $x = \pm1$, and the reason those two need separate handling.

Why the domain is not yet specified. $\mathcal A$ is not defined on all of $PC[a,b]$ — it needs two derivatives and it needs the boundary conditions. The honest object is an unbounded operator with a domain $D(\mathcal A) = \{u \in C^2[a,b] : u \text{ satisfies the boundary conditions}\}$, dense in the space. Everything proved below is proved for $u, v \in D(\mathcal A)$ and is elementary; the functional-analytic theory of unbounded self-adjoint operators, which distinguishes "symmetric" from "self-adjoint" and matters enormously in quantum mechanics, is not needed here and is not developed. What is called self-adjointness below is symmetry on this domain.

Where the name comes from. Charles Sturm and Joseph Liouville, in a sequence of joint and separate papers in 1836–1837, in the Journal de mathématiques pures et appliquées that Liouville had just founded. They were the first to study an eigenvalue problem for a differential operator as an object in its own right rather than as a step in solving one equation, and the first to prove things about eigenfunctions they could not write down.

19 · Sturm–Liouville

Lagrange's identity: one line, everything follows

Lagrange's identity

For any $u, v \in C^2[a,b]$ (complex-valued, $p$ and $q$ real),

$$\bar v\,\mathcal Lu - u\,\overline{\mathcal Lv} = \big[\,p\,(u\bar v{}' - u'\bar v)\,\big]' .$$

proof

The $q$ terms cancel ($q$ real), leaving $u(p\bar v{}')' - \bar v(pu')'$. With $W := u\bar v{}' - u'\bar v$, expand the right-hand side:

$$(pW)' = p'W + p\,(u\bar v{}'' - u''\bar v)$$ $$= (p\bar v{}')'u - (pu')'\bar v .\ \square$$

Integrate; the $w$ in $\mathcal A = w^{-1}\mathcal L$ cancels the $w$ in $\langle\cdot,\cdot\rangle_w$:

$$\langle \mathcal Au, v\rangle_w - \langle u, \mathcal Av\rangle_w = \big[\,p(u\bar v{}' - u'\bar v)\,\big]_a^b .$$

the bilinear concomitant

$\;[u,v](x) := p(x)\big(u\bar v{}' - u'\bar v\big)(x)$ — a weighted Wronskian, and the only obstruction to symmetry.

what has just happened

$\mathcal A$ is symmetric up to a boundary term, for every $p$, $q$, $w$, with no conditions at all. The differential operator is formally self-adjoint by construction.

So self-adjointness reduces to: which boundary conditions kill $[u,v]$ at both ends? A finite-dimensional question, settled next slide.

Notice where the weight went: $\langle\mathcal Au,v\rangle_w = \int(w^{-1}\mathcal Lu)\bar vw = \int(\mathcal Lu)\bar v$. Choose the wrong inner product and there is no identity to have.

Why "Lagrange". Lagrange used the identity in the 1760s to construct adjoint equations, long before there was an inner product to interpret it in. The modern reading — a differential operator is symmetric precisely when an exact derivative integrates to zero — is Sturm's and Liouville's, and it is the template for every self-adjointness argument that follows in this course: the same manoeuvre proves the Laplacian symmetric in unit 23 (Green's identities), and the same boundary-term bookkeeping produces the Gibbons–Hawking–York term in unit 35.

The real case, for orientation. For real $u, v$ the identity reads $v\mathcal Lu - u\mathcal Lv = [p(uv' - u'v)]'$, and $uv' - u'v$ is the Wronskian of $u$ and $v$ (unit 06). So the concomitant is $p$ times the Wronskian, and Abel's identity from unit 06 — that $pW$ is constant for two solutions of the same equation — is the special case $\mathcal Lu = \lambda wu$, $\mathcal Lv = \lambda wv$, where the left side vanishes identically. Two facts that looked unrelated are one line apart.

A check of the identity. Take $p = 1$, $q = 0$, so $\mathcal L = -\mathrm d^2/\mathrm dx^2$. The identity says $-v u'' + u v'' = (uv' - u'v)'$, and differentiating the right side gives $u'v' + uv'' - u''v - u'v' = uv'' - u''v$ ✓. Take $u = \sin x$, $v = \cos x$ on $[0,\pi]$: the concomitant is $\sin x\cdot(-\sin x) - \cos x\cos x = -1$, constant, so the right side is zero; and since $\mathcal L\sin = \sin$ and $\mathcal L\cos = \cos$, the left side is $\cos x\sin x - \sin x\cos x = 0$. Both sides vanish, as Abel promised, since both are solutions of the same equation with $\lambda = 1$ ✓.

19 · Sturm–Liouville

Three families of boundary conditions, and one that fails

1 · separated

$\alpha_1u(a) + \alpha_2u'(a) = 0$ and $\beta_1u(b) + \beta_2u'(b) = 0$, with $\alpha,\beta$ real and each pair not both zero.

Why it works. At $a$, both $(u,u')$ and $(\bar v,\bar v{}')$ lie in the complex line $\alpha^\perp$ ($\alpha$ real), so the determinant $u\bar v{}' - u'\bar v$ vanishes. Same at $b$. Dirichlet ($\alpha_2 = 0$), Neumann ($\alpha_1 = 0$), Robin (neither) are the cases.

2 · periodic

$p(a) = p(b)$, $u(a) = u(b)$, $u'(a) = u'(b)$.

Why it works. $[u,v](a)$ and $[u,v](b)$ are then literally the same number, so the difference is zero.

The drum's angular problem is this one, on $[0,2\pi]$ with $p = 1$. Each eigenvalue $m^2$, $m \ge 1$, carries a two-dimensional eigenspace — the exception this family is responsible for.

3 · singular

$p \to 0$ at an endpoint. Then $[u,v] \to 0$ there provided $u, v$ and their derivatives do not blow up faster than $1/p$ — in practice, provided both are bounded with bounded derivative.

Bessel at $r = 0$: $p = r \to 0$, and boundedness discards $Y_m$. Legendre at $x = \pm1$: $p = 1 - x^2 \to 0$, and boundedness discards $Q_n$ and forces $\lambda = n(n+1)$.

and one that fails

$\mathcal L = -\mathrm d^2/\mathrm dx^2$ on $[0,1]$ with $u(0) = 0$, $u'(1) = \mathrm i\,u(1)$ — a Robin condition with a complex coefficient, which is what an absorbing boundary looks like. Then $u = \sin kx$ needs $\tan k = -\mathrm ik$, which no real $k \ne 0$ satisfies: the first root is $k = 1.77927 - 0.54554\,\mathrm i$, so $\lambda = 2.86819 - 1.94131\,\mathrm i$. Reality goes first, and orthogonality with it. Problem P6 runs it through.

The determinant argument in full. Let $\alpha = (\alpha_1,\alpha_2) \in \mathbb R^2\setminus\{0\}$. The condition at $a$ says $\alpha\cdot(u(a),u'(a)) = 0$ and $\alpha\cdot(v(a),v'(a)) = 0$; conjugating the second (legitimate because $\alpha$ is real) gives $\alpha\cdot(\bar v(a),\bar v{}'(a)) = 0$. Two vectors of $\mathbb C^2$ both annihilated by the same nonzero real functional are linearly dependent, so $\det\begin{pmatrix}u & \bar v\\ u' & \bar v{}'\end{pmatrix}(a) = u\bar v{}' - u'\bar v = 0$. Realness of $\alpha$ is doing genuine work: it is the hypothesis that fails in the warn box, and the failure is not cosmetic.

The singular case, stated honestly. "Bounded with bounded derivative" is sufficient here and is what every example in this course satisfies, but it is not the general criterion. The general theory is Weyl's limit-point/limit-circle classification: at a singular endpoint either every solution is square-integrable near it (limit circle, and a boundary condition is still needed) or only a one-dimensional family is (limit point, and no condition is needed — square-integrability selects the solution by itself). Bessel of order $m \ge 1$ is limit-point at $r = 0$: $Y_m \sim r^{-m}$ is not square-integrable against $r\,\mathrm dr$, so boundedness only confirms a choice already forced. Bessel of order $0$ and Legendre at $x = \pm1$ are limit-circle: $Y_0$ and $Q_\ell$ grow only logarithmically and are square-integrable, so a condition is genuinely needed there, and boundedness is that condition. Cited: Coddington & Levinson, Theory of Ordinary Differential Equations, ch. 9; Weidmann, Spectral Theory of Ordinary Differential Operators, ch. 5.

Check on Bessel's concomitant. Near $r = 0$, $J_m(\kappa r) \sim (\kappa r/2)^m/m!$ so $p(u\bar v{}' - u'\bar v) = r\cdot O(r^{2m-1}) = O(r^{2m}) \to 0$ for $m \ge 1$, and for $m = 0$, $J_0 \to 1$ with $J_0' \sim -\kappa^2r/2$, giving $r\cdot O(r) \to 0$ ✓. With $Y_0 \sim (2/\pi)\log r$ instead, $rY_0' \sim 2/\pi$ does not tend to zero, and the symmetry is genuinely lost — boundedness is not a convenience.

19 · Sturm–Liouville

Real eigenvalues, orthogonal eigenfunctions

theorem

Let the boundary conditions be self-adjoint, so $\langle\mathcal Au,v\rangle_w = \langle u,\mathcal Av\rangle_w$ on the domain. Then:

(i) every eigenvalue is real, and every eigenfunction may be taken real;
(ii) eigenfunctions with different eigenvalues are orthogonal in $\langle\cdot,\cdot\rangle_w$.

proof of (i)

$\mathcal Au = \lambda u$ with $u \ne 0$; drop the subscript $w$ on every product. Then

$$\lambda\lVert u\rVert^2 = \langle\mathcal Au,u\rangle = \langle u,\mathcal Au\rangle = \bar\lambda\lVert u\rVert^2 ,$$

and $\lVert u\rVert_w^2 > 0$ because $w > 0$, so $\lambda = \bar\lambda$. For the eigenfunction: $\mathcal L$ and the boundary conditions have real coefficients, so $\bar u$ is an eigenfunction for the same $\lambda$; hence so are $\operatorname{Re}u$ and $\operatorname{Im}u$, and one is nonzero. $\square$

proof of (ii)

$\mathcal Au = \lambda u$, $\mathcal Av = \mu v$ with $\lambda \ne \mu$, both real by (i). Then

$$\lambda\langle u,v\rangle = \langle\mathcal Au,v\rangle = \langle u,\mathcal Av\rangle = \mu\langle u,v\rangle ,$$

so $(\lambda - \mu)\langle u,v\rangle_w = 0$, hence $\langle u,v\rangle_w = 0$. $\square$

what this has just replaced

The trigonometric orthogonality computed by a product formula in section 2 — now a corollary with $p = w = 1$, $q = 0$. And the Bessel and Legendre relations, and every case nobody has computed yet.

Unit 07's $M$-orthogonality of normal modes is the same statement for $\mathbb R^n$: $M$ for $w$, the spectral theorem for Lagrange's identity.

Both proofs used only symmetry. Neither used $p > 0$, the form of $\mathcal L$, or anything about the interval.

Where positivity of $w$ is genuinely needed in (i). If $w$ changed sign, then $\langle u,u\rangle_w$ could vanish for $u \ne 0$ and the division would be illegal — and indeed indefinite-weight problems (which occur physically, in stability analysis of stratified fluids) do have complex eigenvalues. This is the exact analogue of unit 07's insistence that the mass matrix $M$ be positive definite: the same hypothesis, in the same role, one dimension count later.

Orthogonality when eigenvalues repeat. Conclusion (ii) says nothing about two eigenfunctions of the same eigenvalue. For separated conditions this is empty, since the next slide proves eigenvalues simple. For periodic conditions it is not: the drum's $\cos m\theta$ and $\sin m\theta$ share an eigenvalue, and they happen to be orthogonal for a different reason (parity). In general one applies Gram–Schmidt inside each eigenspace, which is finite-dimensional — at most two-dimensional here, since the ODE is second order.

The physical reading of reality. $\lambda = \omega^2/c^2$, so a complex $\lambda$ would mean a complex frequency, which means exponential growth or decay. A self-adjoint boundary condition is precisely one through which no energy flows — clamped, free, or periodic — and the theorem says energy conservation forces purely oscillatory modes. Break it by letting the boundary absorb, as the warn box on the previous slide does, and the eigenvalues go complex immediately, with the imaginary part measuring the damping rate. The mathematics and the physics are the same statement.

19 · Sturm–Liouville

Simple, discrete, unbounded above

simplicity — separated conditions only

Each eigenvalue has a one-dimensional eigenspace.

proof

Let $u, v$ satisfy $\mathcal Ly = \lambda wy$ and $\alpha_1y(a) + \alpha_2y'(a) = 0$, both real (previous slide). The pairs $(u,u')$ and $(v,v')$ at $a$ lie in the one-dimensional $\alpha^\perp \subset \mathbb R^2$, so $(v,v')(a) = c(u,u')(a)$; then $v$ and $cu$ solve the same ODE with the same data at $a$, so $v = cu$ — Picard–Lindelöf, unit 01. $\square$

why periodic is different

It constrains the two ends jointly, so the argument has nothing to start from: $-y'' = \lambda y$ on $[0,2\pi]$, periodic, has eigenspace $\operatorname{span}(\cos m\theta,\sin m\theta)$ at $\lambda = m^2$, $m \ge 1$.

bounded below

With Dirichlet conditions, one integration by parts — boundary term $-[pu'\bar u]_a^b = 0$ — gives $\lambda\lVert u\rVert_w^2 = \int_a^b(p|u'|^2 + q|u|^2)$, which is at least $\inf(q/w)\lVert u\rVert_w^2$. So $\lambda \ge \inf(q/w)$. $\square$

discrete and unbounded — cited

The eigenvalues form a strictly increasing sequence $\lambda_1 \lt \lambda_2 \lt \cdots \to +\infty$, with no accumulation point.

By the Prüfer angle: $pu' = \rho\cos\vartheta$, $u = \rho\sin\vartheta$, and $\vartheta$ obeys a first-order ODE, increasing through multiples of $\pi$, monotone in $\lambda$. Cited: Coddington & Levinson ch. 8; Al-Gwaiz §3.3.

The integration by parts, in full. $\langle\mathcal Au,u\rangle_w = \int_a^b(\mathcal Lu)\bar u = \int_a^b\big[-(pu')'\bar u + q|u|^2\big] = \big[-pu'\bar u\big]_a^b + \int_a^b\big[p|u'|^2 + q|u|^2\big]$. The boundary term vanishes for Dirichlet ($\bar u = 0$ at both ends) and for Neumann ($u' = 0$); for Robin it becomes $(\beta_1/\beta_2)p(b)|u(b)|^2 - (\alpha_1/\alpha_2)p(a)|u(a)|^2$, which is $\ge 0$ for the usual sign choice $\alpha_1/\alpha_2 \le 0 \le \beta_1/\beta_2$ and only improves the lower bound. Dimensions: every term carries $[p][u]^2\mathrm L^{-1}$ against $[w][u]^2\mathrm L$, and their ratio is $[\lambda]$ ✓ — for $p = w = 1$ that is $\mathrm L^{-2}$, as it must be.

A check of the lower bound. String, $p = w = 1$, $q = 0$, Dirichlet: the bound says $\lambda \ge 0$. The true $\lambda_1 = \pi^2/L^2 > 0$ ✓. Bessel of order $m$, $q = m^2/r > 0$, $w = r$: the bound says $\lambda \ge \inf m^2/r^2 = m^2/a^2$, and the true $\lambda_{m1} = j_{m1}^2/a^2$ with $j_{m1} > m$ ✓ (indeed $j_{11} = 3.83 > 1$, $j_{21} = 5.14 > 2$). The bound is crude and correct, which is what a bound is for.

Why discreteness is not obvious. Nothing said so far forbids a continuum of eigenvalues; the Laplacian on the whole line has one. What rules it out is the compactness of $[a,b]$ together with $p > 0$ on it. In the operator-theoretic language: the resolvent $(\mathcal A - \mu)^{-1}$ is an integral operator whose kernel is the Green's function, continuous on the compact square $[a,b]^2$, hence a compact operator; a compact self-adjoint operator has a discrete spectrum accumulating only at zero, and inverting sends that to a discrete spectrum running off to infinity. The Prüfer route proves the same thing by staying with the ODE and is the one to read first, because it also delivers the oscillation theorem of "Counting nodes: the oscillation theorem".

19 · Sturm–Liouville

Completeness — the theorem the whole method was waiting for

the expansion theorem — cited

For a regular problem ($[a,b]$ bounded, $p, w > 0$ on all of it: no singular end) with self-adjoint boundary conditions, the eigenfunctions $\{u_n\}$ are complete in $L^2_w(a,b)$, the completion of $PC[a,b]$ in $\lVert\cdot\rVert_w$: for every $f$ there,

$$\Big\lVert f - \sum_{n\le N}\hat f_nu_n\Big\rVert_w \to 0,\quad \hat f_n = \frac{\langle f,u_n\rangle_w}{\lVert u_n\rVert_w^2},$$

and Parseval holds. If $f$ is also $C^1$ and satisfies the boundary conditions, the convergence is uniform and absolute.

Why, in one sentence. $(\mathcal A - \mu)^{-1}$ is an integral operator whose kernel — a Green's function for $\mathcal L$, built by unit 06's jump-matching — is continuous on a compact square, hence compact and self-adjoint; and a compact self-adjoint operator has an orthonormal basis of eigenfunctions.

what this unit is not proving, and why

That needs three things this course has not built: the Lebesgue integral, metric completeness of $L^2_w$, and the spectral theorem for compact operators. None is hard; all are long.

What is proved in full here is the trigonometric case, by Fejér — the case every other is modelled on, and enough for the string, the rod, and units 20, 23 and 38.

Cited: Coddington & Levinson ch. 7 §§3–4; Courant & Hilbert vol. I ch. V–VI; Al-Gwaiz ch. 2.

The singular cases. The theorem as stated is for a regular problem: $[a,b]$ compact, $p > 0$ and $w > 0$ on the closed interval. The drum ($p = w = r$, vanishing at $r = 0$) and Legendre ($p = 1 - x^2$, vanishing at both ends) are singular, and the corresponding completeness statements — the Fourier–Bessel expansion theorem and the Legendre expansion theorem — are true and are proved separately, by Weyl's theory or by direct kernel estimates. Cited: Watson, A Treatise on the Theory of Bessel Functions, ch. XVIII, for Fourier–Bessel; Sansone, Orthogonal Functions, ch. III, for Legendre. Every use of them in this course is a use of the statement, never of the proof.

Why the uniform-convergence clause matters in practice. Mean-square convergence permits the truncation to be wrong by a lot at any given point, as Gibbs showed. The extra hypothesis — $f$ is $C^1$ and satisfies the boundary conditions — buys pointwise control everywhere at once, and it is a hypothesis one can arrange: for the drum it says the initial shape must actually vanish at the rim, which a physical membrane does.

What "the modes are a basis" does and does not mean. It means: the finite linear combinations are dense, so every element is a limit of them. It does not mean every element is a finite combination, and it does not mean the infinite sum converges pointwise. An orthonormal basis of a Hilbert space is a topological notion, not the linear-algebraic one of unit 07 — an infinite-dimensional space has no countable algebraic basis at all. The title of the unit says "the modes are a basis, and that is a theorem"; the theorem is this, with that meaning.

19 · Sturm–Liouville · the drum

The debt from section 1, paid on the drum

A finger-pressed dimple, $f(r) = 1 - r/0.4a$ inside $r = 0.4a$ and $0$ outside, expanded in the drum's radial modes $J_0(j_{0k}r/a)$ with weight $r$. One mode is a poor caricature; ten are within $2.6\%$ in the weighted norm. The corners at $r = 0$ and $r = 0.4a$ are the last thing to be reached.

the debt from section 1, paid

"Is every initial shape a superposition of modes?" — yes: in mean square always, uniformly when the shape is smooth and respects the boundary conditions.

The coefficients are the projection formula of section 2 with the drum's weight, $\hat f_k = \int_0^af J_0(j_{0k}r/a)\,r\,\mathrm dr\big/\tfrac12a^2J_1(j_{0k})^2$. The shape has kinks, so the uniform clause of the theorem does not apply — but mean-square convergence is unconditional, and it is what the figure's percentages measure.

The figure's numbers. The relative weighted error $\lVert f - S_Nf\rVert_w/\lVert f\rVert_w$, with $w = r$ on $[0,a]$, is $0.810$, $0.176$, $0.026$ for $N = 1$, $3$, $10$, and $0.003$ for $N = 40$. The first coefficients are $\hat f_1, \dots, \hat f_4 = 0.185$, $0.315$, $0.265$, $0.114$: the lowest mode is not the largest component, because the dimple is narrow and the low modes are broad. That is the drum's version of the smoothness dictionary — a narrow, kinked shape needs many modes — and it is why a hard stick, which dents only a small patch of the membrane, excites so many partials.

Which theorem is being used. The drum's radial problem is singular at $r = 0$, so the regular expansion theorem does not literally apply; the Fourier–Bessel expansion theorem does (cited, Watson ch. XVIII, as in the notes of the previous slide). Its conclusions are the same: mean-square convergence for every square-integrable shape, with Parseval, and pointwise convergence to the midpoint at a jump.

19 · Sturm–Liouville

Sturm's comparison theorem, and counting nodes

Sturm comparison

Let $u'' + Qu = 0$ and $v'' + \tilde Qv = 0$ on $[x_1,x_2]$ with $\tilde Q \ge Q$ and $\tilde Q \not\equiv Q$. If $x_1 < x_2$ are consecutive zeros of $u$, then $v$ vanishes somewhere in $(x_1,x_2)$.

proof

Suppose not; normalise $u > 0$ and $v > 0$ on $(x_1,x_2)$. Then

$$(u'v - uv')' = (\tilde Q - Q)\,uv \ \ge\ 0 ,$$

since $u''v - uv'' = (\tilde Q - Q)uv$. So $u'v - uv'$ is nondecreasing. But $u(x_1) = u(x_2) = 0$ with $u'(x_1) > 0$, $u'(x_2) < 0$, so it is $u'(x_1)v(x_1) \ge 0$ at the left end and $u'(x_2)v(x_2) \le 0$ at the right. Nondecreasing from $\ge0$ to $\le0$ forces both to vanish and $(\tilde Q-Q)uv \equiv 0$ — impossible on the interior. $\square$

Drawn with $Q = 1$ and $\tilde Q = 2.25$. $u$ has consecutive zeros $x_1, x_2$; $v$, under the larger $\tilde Q$, bends back towards the axis harder and cannot stay positive across $(x_1, x_2)$, whatever its starting value.

Getting a general $\mathcal L$ into the form $u'' + Qu = 0$. Substitute the Liouville transformation $t = \int_a^x\sqrt{w/p}$, $y = (pw)^{1/4}u$; then $-(pu')' + qu = \lambda wu$ becomes $\ddot y + (\lambda - \tilde q(t))y = 0$ for a computable $\tilde q$. So the comparison theorem applies to every Sturm–Liouville problem, with $Q = \lambda - \tilde q$, and "larger $\lambda$" is literally "larger $Q$". Zeros are preserved because $(pw)^{1/4} > 0$ and $t(x)$ is strictly increasing.

The intuition, worth having. $u'' = -Qu$ says the curve bends towards the axis at a rate $Q$. Bend harder and you cross sooner. Every statement in this slide is that sentence, made precise: higher eigenvalue, more bending, more crossings; heavier string (larger $w$), more bending per unit length, lower frequency for the same number of crossings.

19 · Sturm–Liouville

Counting nodes: the oscillation theorem

Sturm oscillation theorem — cited

For a regular problem with separated conditions, ordering $\lambda_1 < \lambda_2 < \cdots$, the eigenfunction $u_n$ has exactly $n-1$ zeros in the open interval $(a,b)$.

Comparison gives the monotone half — larger $\lambda$ means at least as many zeros — and the Prüfer angle supplies exactness. Cited: Coddington & Levinson ch. 8 §2.

checks, on things already known

String. $u_n = \sin(n\pi x/L)$ has zeros at $x = kL/n$, $k = 1,\dots,n-1$: exactly $n-1$ ✓.

Legendre. $P_n$ has $n$ zeros in $(-1,1)$, and $\lambda_n = n(n+1)$ is the $(n+1)$-st eigenvalue counting from $\lambda_0 = 0$ ✓.

The drum. $J_0(j_{0k}r/a)$ vanishes at $r = aj_{0i}/j_{0k}$, $i \lt k$: $k-1$ nodal circles ✓.

A problem that is not a plain sine series: $-u'' = \lambda u$ on $[0,1]$ with $u(0) = 0$ and the Robin condition $u(1) + u'(1) = 0$. The eigenfunctions are $\sin k_nx$ with $\tan k = -k$; they do not vanish at $x = 1$, yet $u_n$ still has exactly $n - 1$ interior zeros.

The figure's numbers. The roots of $\tan k = -k$ are $k_1 = 2.0288$, $k_2 = 4.9132$, $k_3 = 7.9787$, one in each interval $((n - \tfrac12)\pi, n\pi)$; $\sin k_nx$ vanishes at $x = j\pi/k_n$, and the zeros inside $(0,1)$ are $0.639$ for $n = 2$ and $0.394$, $0.787$ for $n = 3$. This is the same problem as the checkpoint's practice problem, so the picture is also its answer to "how many zeros has the third eigenfunction".

Two corollaries worth naming. (1) If $Q \le 0$ on an interval, a nontrivial solution has at most one zero there — compare with $v'' = 0$, whose solutions are straight lines. For $p = w = 1$ that is another proof that a problem with $q \ge 0$ and Dirichlet conditions has no eigenvalue $\le 0$; P10 gives the direct argument for any $p$. (2) Zeros of consecutive eigenfunctions interlace: between two consecutive zeros of $u_n$ — counting the endpoints, for Dirichlet conditions — there is at least one zero of $u_{n+1}$. Both are one application of the boxed proof.

What it is used for later. Counting nodes is how one identifies which mode a measured resonance is, without computing eigenvalues: count the still lines. It is also how unit 21's WKB approximation is normalised — the Bohr–Sommerfeld counting rule is the oscillation theorem with an asymptotic estimate of the number of zeros.

19 · Sturm–Liouville · example

Worked: the drum is a singular Sturm–Liouville problem

put Bessel's equation into the form

Section 1's radial equation is $r^2R'' + rR' + (\lambda r^2 - m^2)R = 0$ on $(0,a)$. Divide by $r$ — the integrating-factor recipe, since $A = r^2$, $B = r$, $\exp\int(B/A) = r$:

$$rR'' + R' + \Big(\lambda r - \frac{m^2}{r}\Big)R = 0,$$

i.e. $(rR')' - \frac{m^2}{r}R = -\lambda rR$,

since $(rR')' = rR'' + R'$. Negated, this is $-(pR')' + qR = \lambda wR$ with

$$p = r,\qquad q = m^2/r,\qquad w = r .$$

Endpoints. $r = a$ is regular with $R(a) = 0$; at $r = 0$, $p(0) = 0$ and $q$ blows up — singular, and boundedness is the condition.

what the theorems now say, for free

Real. Every $\lambda_{mk}$ is real, so $\omega_{mk} = c\sqrt{\lambda_{mk}}$ is a genuine frequency and no mode grows ✓.

Orthogonal. $\int_0^aJ_m(\tfrac{j_{mk}r}{a})J_m(\tfrac{j_{ml}r}{a})r\,\mathrm dr = 0$ for $k \ne l$, proved without an integral ✓.

Bounded below. $\lambda \ge \inf(q/w) = m^2/a^2$, true since $j_{mk} > m$ ✓. Nodes. The $k$-th radial eigenfunction has $k-1$ interior zeros ✓. Complete (cited: Watson ch. XVIII).

check the concomitant at the centre

As $r \to 0^+$, $p(u\bar v{}' - u'\bar v) = r\cdot O(r^{2m-1}) \to 0$ for $J_m$; but $Y_0 \sim \frac2\pi\log r$ gives $rY_0' \to \frac2\pi \ne 0$. Without boundedness the operator is not symmetric and every line above fails.

Why this example is the point of the section. Everything in the right-hand column was known for Bessel functions by 1850, each fact proved by its own manipulation of the series or the integral representation. Sturm–Liouville theory says: none of that was about Bessel functions. It was about $p = w = r$ and a symmetric operator, and any other $(p,q,w)$ gives the same list with no new work. That is the difference between a catalogue of special functions and a theory.

The orthogonality, if one insists on seeing the integral. Let $u = J_m(\kappa r)$, $v = J_m(\kappa'r)$ with $J_m(\kappa a) = J_m(\kappa'a) = 0$. Lagrange's identity integrated over $[\varepsilon,a]$ gives $(\kappa^2 - \kappa'^2)\int_\varepsilon^a uv\,r\,\mathrm dr = \big[r(uv' - u'v)\big]_\varepsilon^a$. At $r = a$ both $u$ and $v$ vanish; at $r = \varepsilon$ the bracket is $O(\varepsilon^{2m})$ or, for $m = 0$, $O(\varepsilon^2)$. Let $\varepsilon \to 0$ ✓. That is the classical computation, and it is Lagrange's identity — which is the whole thesis of the section restated.

A caution about $q$. Here $q = m^2/r$ is unbounded at the singular endpoint, so even the standing hypothesis "$q \in C^0[a,b]$" is violated. That is normal for singular problems and is exactly what Weyl's theory is for. In the course's later use — separating the Laplacian in units 23 and 39 — the same structure recurs with the same resolution, and it is always worth naming which endpoint is singular before trusting a boundary condition there.

The angular factor, for completeness of the picture. $\Theta'' = -m^2\Theta$ on $[0,2\pi]$ with periodic conditions is a second, independent Sturm–Liouville problem, regular, with $p = w = 1$ and $q = 0$. The full two-dimensional eigenfunctions are products of the two problems' eigenfunctions, and their orthogonality on the disc is the product of the two orthogonality relations, with the measure $r\,\mathrm dr\,\mathrm d\theta$ factorising to match.

section 5

The Rayleigh quotient

The lowest eigenvalue is the minimum of a functional, so any admissible guess is an upper bound — and unit 11's machinery is what proves it.

19 · Rayleigh

The quotient, and the identity behind it

the Rayleigh quotient

For $u$ in the domain, $u \not\equiv 0$,

$$R[u] := \frac{\displaystyle\int_a^b\big(p\,u'^2 + q\,u^2\big)\mathrm dx}{\displaystyle\int_a^b w\,u^2\,\mathrm dx} \;=\; \frac{\langle\mathcal Au,u\rangle_w}{\langle u,u\rangle_w}.$$

The second equality is one integration by parts, with the boundary term killed by the boundary conditions — the same computation as the lower bound on "Simple, discrete, unbounded above".

For a string, the numerator is twice the elastic potential energy stored in the shape $u$, and the denominator is twice the kinetic energy of the same profile read as a velocity: $R$ is stiffness over inertia, as $\omega^2 = k/m$ is for one oscillator and $v^{\!\top}\!Kv/v^{\!\top}\!Mv$ was in unit 07.

the variational principle

$\displaystyle \lambda_1 = \min_{u \ne 0} R[u]$, attained exactly on the multiples of $u_1$. In particular every admissible $u$ gives $\lambda_1 \le R[u]$ — admissible meaning continuous, piecewise $C^1$, meeting the Dirichlet conditions (notes).

proof (granting completeness)

Expand $u = \sum_nc_nu_n$ with $\{u_n\}$ orthonormal in $\langle\cdot,\cdot\rangle_w$. Then $\langle u,u\rangle_w = \sum c_n^2$ and, by symmetry and $\mathcal Au_n = \lambda_nu_n$,

$$\langle\mathcal Au,u\rangle_w = \sum_n\lambda_nc_n^2 \ \ge\ \lambda_1\sum_nc_n^2 ,$$

so $R[u] \ge \lambda_1$, with equality iff $c_n = 0$ whenever $\lambda_n > \lambda_1$, i.e. iff $u \in \operatorname{span}(u_1)$. Taking $u = u_1$ gives $R = \lambda_1$. $\square$

One-sided by construction: a trial function bounds $\lambda_1$ from above, never below. Lower bounds are genuinely harder and need a different idea.

The integration by parts once more, with the boundary term retained. $\int_a^b(\mathcal Lu)u = [-pu'u]_a^b + \int_a^b(pu'^2 + qu^2)$. For Dirichlet or Neumann the bracket is zero. For Robin, $u'(a) = -(\alpha_1/\alpha_2)u(a)$ makes it $-p(a)(\alpha_1/\alpha_2)u(a)^2$ at the left end, so the correct Rayleigh quotient for a Robin problem carries extra boundary terms in the numerator: $R[u] = \big(\int(pu'^2 + qu^2) + \sigma_ap(a)u(a)^2 + \sigma_bp(b)u(b)^2\big)/\int wu^2$ with $\sigma$ the appropriate ratios. Forgetting them is the single commonest error in using this method, and the practice problem in the checkpoint's notes makes the reader meet it.

Why an upper bound is the useful direction. $\lambda_1$ is the lowest frequency: the pitch of the string, the resonance of the bridge, the buckling load of the column. An engineer needs to know the structure is stiff enough, which is a lower bound on $\lambda_1$ — so the Rayleigh quotient is on the wrong side. The reason it is used anyway is that it is extraordinarily accurate: the error in $R$ is quadratic in the error of the trial function, because $\lambda_1$ is a stationary point. Guess the shape to 10 % and get the eigenvalue to 1 %, which the drum's worked example demonstrates and problem P11 repeats.

Full proof: trial functions outside the domain

The proof on the slide takes $u$ in the domain, $C^2$ and meeting every boundary condition. The trial functions actually used are rougher: the drum's cone has a corner at the centre, and $1 - x^{\sqrt2}$ has an unbounded second derivative there. The bound survives. Take Dirichlet conditions and $q \ge 0$, so every $\lambda_n > 0$, and the energy form $E(u,v) = \int_a^b(pu'v' + quv)$, an inner product on admissible functions. For an eigenfunction $u_n$, orthonormal in $\langle\cdot,\cdot\rangle_w$, one integration by parts moves the derivative onto the smooth $u_n$: $E(u,u_n) = \int_a^bu\,\mathcal Lu_n = \lambda_n\langle u,u_n\rangle_w = \lambda_nc_n$, the boundary term $[pu_n'u]$ vanishing because $u = 0$ at a regular end and $p \to 0$ with $u$, $u_n'$ bounded at a singular one. So the $u_n/\sqrt{\lambda_n}$ are orthonormal for $E$, and Bessel's inequality in the inner product $E$ gives $E(u,u) \ge \sum_nE(u,u_n)^2/\lambda_n = \sum_n\lambda_nc_n^2 \ge \lambda_1\sum_nc_n^2 = \lambda_1\lVert u\rVert_w^2$, the last step by Parseval (completeness, cited). That is $R[u] \ge \lambda_1$ for every admissible $u$. Neumann conditions need not be imposed on a trial function at all — they are natural, next slide — and Robin ones enter through the boundary terms of $E$.

Quadratic accuracy, proved. If $u = u_1 + \varepsilon\eta$ with $\eta \perp_w u_1$ and $\lVert u_1\rVert_w = 1$, then $\langle\mathcal Au,u\rangle_w = \lambda_1 + \varepsilon^2\langle\mathcal A\eta,\eta\rangle_w$ and $\langle u,u\rangle_w = 1 + \varepsilon^2\lVert\eta\rVert_w^2$, so $R[u] = \lambda_1 + \varepsilon^2\big(\langle\mathcal A\eta,\eta\rangle_w - \lambda_1\lVert\eta\rVert_w^2\big) + O(\varepsilon^4)$. The first-order term is absent — which is precisely the statement that $u_1$ is a stationary point of $R$, the next slide's content.

19 · Rayleigh

Unit 11's second job: the eigenvalue problem is an Euler–Lagrange equation

Run it the other way: minimise $J[u] = \int_a^b(pu'^2 + qu^2)\,\mathrm dx$ subject to $G[u] = \int_a^b wu^2\,\mathrm dx = 1$ — unit 11's isoperimetric problem.

the Euler–Lagrange equation is $\mathcal Lu = \lambda wu$

With a multiplier $\lambda$, the first variation along $\eta$ is

$$\delta(J - \lambda G)[\eta] = 2\!\int_a^b\!\big(pu'\eta' + (q{-}\lambda w)u\eta\big)$$ $$= 2\big[pu'\eta\big]_a^b + 2\!\int_a^b\!\big(\mathcal Lu - \lambda wu\big)\eta .$$

For $\eta$ vanishing at the ends the bracket dies and the fundamental lemma (unit 11) forces $\mathcal Lu = \lambda wu$. $\square$

No ansatz, then: the equation is what "least stiffness per inertia" means, and the multiplier is the eigenvalue.

natural boundary conditions, free

Drop $\eta(a) = \eta(b) = 0$ and the bracket must vanish by itself, forcing $pu' = 0$ at each free end — the Neumann condition, arriving as unit 11's natural boundary condition rather than imposed.

higher eigenvalues: min–max — cited

$\lambda_n = \min_{\dim V = n}\max_{u \in V\setminus0}R[u]$; equivalently, $\lambda_n = \min\{R[u] : u \perp_w u_1,\dots,u_{n-1}\}$.

Courant–Fischer, proved for matrices in unit 07; the proof transfers given completeness. Cited: Courant & Hilbert vol. I ch. VI §1.

The integration by parts in the first variation. $\int_a^b pu'\eta' = [pu'\eta]_a^b - \int_a^b(pu')'\eta$. That is the only step; the rest is collecting terms. Note that the variation was taken of $J - \lambda G$ with $\lambda$ treated as a constant, which is legitimate because unit 11's multiplier theorem for isoperimetric constraints says exactly that: at a constrained extremum there exists a $\lambda$ making the unconstrained variation of $J - \lambda G$ vanish, provided $u$ is not a critical point of $G$ alone — and $\delta G[\eta] = 2\int wu\eta$ vanishes for all $\eta$ only if $u \equiv 0$, which is excluded.

Rayleigh's own use of it. Rayleigh's Theory of Sound (1877) introduces the quotient to estimate the pitch of bells, plates and air columns whose exact eigenfunctions were hopeless. Ritz's 1909 refinement — minimise $R$ over a finite-dimensional family of trial functions, which reduces the problem to a matrix generalised eigenvalue problem $Kc = \lambda Mc$, unit 07's — is the Rayleigh–Ritz method, and it is the direct ancestor of the finite element method. The whole of structural engineering computation is this slide, iterated.

What "unit 11's second job" means. Unit 11 built the calculus of variations to get equations of motion out of an action, and unit 12 spent it on Lagrangian mechanics. This is its other, independent use: characterising eigenvalues. The two look unrelated and share every theorem — first variation, fundamental lemma, natural boundary conditions, multipliers.

A warning about "min" versus "inf". The variational principle as stated asserts the minimum is attained. On a compact interval with $p, w > 0$ it is, and that is a theorem, not a formality — Weierstrass's critique of Dirichlet's principle (unit 11) is exactly the observation that a bounded-below functional need not attain its infimum. Here attainment comes from the cited spectral theorem: $u_1$ exists, and $R[u_1] = \lambda_1$. The direct method of unit 11 is the other route, and the one that generalises.

19 · Rayleigh · example

Worked: bounding the drum's lowest note, one parameter

a family of guesses

The drum's radial modes have $m = 0$, so $p = w = r$, $q = 0$:

$$R[u] = \int_0^a\! u'^2 r\,\mathrm dr \;\Big/\; \int_0^a\! u^2 r\,\mathrm dr .$$

Take the family $u_p = 1 - x^p$, $x = r/a$, $p > 0$: bounded at the centre, zero at the rim. With $u_p' = -(p/a)x^{p-1}$,

$$\int_0^a\!u_p'^2\,r\,\mathrm dr = p^2\!\int_0^1\!\!x^{2p-1}\,\mathrm dx = \tfrac p2 .$$

And $\int_0^au_p^2r\,\mathrm dr = a^2(\tfrac12 - \tfrac{2}{p+2} + \tfrac1{2p+2}) = a^2p^2/[2(p+1)(p+2)]$, the numerator collapsing to $p^2$ (notes). So

$$a^2R[u_p] = \frac{(p+1)(p+2)}{p} = p + 3 + \frac2p .$$

minimise, and compare

$1 - 2/p^2 = 0$ at $p = \sqrt2$, a minimum as $4/p^3 > 0$. There

$$a^2\lambda_1 \le 3 + 2\sqrt2 = 5.8284 .$$

So $\sqrt{a^2\lambda_1} \le 1 + \sqrt2 = 2.414214$; $j_{01} = 2.404826$.

trial	$a^2R$	error
$p = 1$ (a cone)	6.000000	$+3.75\%$
$p = 2$ (a paraboloid)	6.000000	$+3.75\%$
$p = \sqrt2$ (best)	5.828427	$+0.78\%$
exact $J_0$	5.783186	—

Checks. $a^2R$ dimensionless ✓; every entry above the exact value ✓; the $p = 1$, $2$ tie is the symmetry $p \mapsto 2/p$ ✓.

The algebraic collapse, in full. $\tfrac12 - \tfrac2{p+2} + \tfrac1{2(p+1)}$ over $2(p+1)(p+2)$ has numerator $(p+1)(p+2) - 4(p+1) + (p+2) = (p^2+3p+2) - (4p+4) + (p+2) = p^2$. So the denominator integral is $a^2p^2/[2(p+1)(p+2)]$, and $a^2R = (p/2)\cdot 2(p+1)(p+2)/p^2 = (p+1)(p+2)/p$ ✓. At $p = \sqrt2$: $(\sqrt2+1)(\sqrt2+2)/\sqrt2 = (2 + 3\sqrt2 + 2)/\sqrt2 = 4/\sqrt2 + 3 = 2\sqrt2 + 3$ — and $3 + 2\sqrt2 = (1+\sqrt2)^2$, so the bound on $\sqrt\lambda$ is exactly $1 + \sqrt2$. A one-parameter guess, in closed form, within $0.4\%$ of a Bessel zero.

In frequency. For the deck's numbers, $a = 0.150\ \mathrm m$ and $c = 109.89\ \mathrm{m\,s^{-1}}$: the exact fundamental is $280.39$ Hz and the bound is $c(1+\sqrt2)/(2\pi a) = 281.49$ Hz — one hertz high, from a guess that did not know Bessel functions exist.

Why $p = 1$ and $p = 2$ tie. $a^2R = p + 3 + 2/p$ is symmetric under $p \mapsto 2/p$, which swaps $1$ and $2$. The symmetry is an accident of this family and proves nothing. Quadratic accuracy shows instead in the sizes: in the weighted norm the cone's shape is $6.9\%$ away from $J_0(j_{01}r/a)$, the paraboloid's $8.6\%$ and the best trial's $2.5\%$, while their eigenvalues are off by $3.75\%$, $3.75\%$ and $0.78\%$ — each a modest multiple of the square of the shape error, never of the shape error itself.

The obvious improvement. Use two parameters — Rayleigh–Ritz with $u = (1-x^2) + \beta(1-x^4)$, say — and the bound drops to within a few parts in $10^4$. That computation is a generalised eigenvalue problem $Kc = \lambda Mc$ for a $2\times2$ pencil, which is unit 07's slide "The generalised eigenvalue problem" verbatim. Problem P11 does a two-term version for a string of varying density and lands within one part in $10^5$ of the exact eigenvalue.

19 · Rayleigh · example

The drum's bound, plotted

$a^2R[u_p] = p + 3 + 2/p$ over the whole family. Every trial lies above the exact $j_{01}^2 = 5.7832$, as the variational principle demands; the best, at $p = \sqrt2$, misses it by $0.78\%$.

The trial shapes themselves, against the true fundamental $J_0(j_{01}r/a)$. The best trial is visibly not the eigenfunction — about $2.5\%$ off in the weighted norm — yet its eigenvalue is good to $0.78\%$: the error in $R$ is quadratic in the error of the shape.

What to look for. On the left, the curve is convex with a single minimum, so one derivative found it; and the gap between the minimum and the dashed line is the price of the one-parameter family, not of any arithmetic. On the right, the cone has a corner at the centre where the drum is flat, and sags below $J_0$; the paraboloid is flat at the centre but bulges above $J_0$ and meets the rim too steeply (slope $-2/a$ against $-j_{01}J_1(j_{01})/a = -1.248/a$); the best trial compromises between them and hugs $J_0$ to within a few per cent everywhere. Its shape error of $2.5\%$ squared is $0.06\%$, and the eigenvalue error of $0.78\%$ is a modest multiple of that — the quadratic accuracy proved in the notes of "The quotient, and the identity behind it".

section 6

Three geometries, one theorem

Interval, disc and sphere. The equations look unrelated and the proofs are identical, because the operator is.

19 · geometries

One table, three special functions

domain	separated equation	$p$	$q$	$w$	endpoints	eigenfunctions	eigenvalue condition
interval $[0,L]$	$-u'' = \lambda u$	$1$	$0$	$1$	both regular	$\sin(n\pi x/L)$	$\sin\sqrt\lambda L = 0$
disc, radius $a$	$-(rR')' + \tfrac{m^2}{r}R = \lambda rR$	$r$	$m^2/r$	$r$	$r=0$ singular	$J_m(\sqrt\lambda r)$	$J_m(\sqrt\lambda a) = 0$
sphere, in $x=\cos\theta$	$-\big((1-x^2)\Theta'\big)' + \tfrac{m^2}{1-x^2}\Theta = \lambda\Theta$	$1-x^2$	$\tfrac{m^2}{1-x^2}$	$1$	both singular	$P_\ell^m(x)$	$\lambda = \ell(\ell+1)$, $\ell \ge m$

what is the same

Real eigenvalues. Orthogonality in $\langle\cdot,\cdot\rangle_w$ — $\mathrm dx$, $r\,\mathrm dr$, $\mathrm dx = \sin\theta\,\mathrm d\theta$. Simplicity for the separated cases. Discrete spectrum running to $+\infty$. Node counting. The Rayleigh quotient with the same $p$, $q$, $w$. Completeness.

what is different, and only this

Which endpoints are singular, and therefore what plays the role of a boundary condition there — and hence which solutions survive. On the sphere, boundedness at $x = \pm1$ is what quantises $\lambda$ to $\ell(\ell+1)$: the integers of spherical harmonics come from a boundary condition, not from anywhere else.

The weight is the geometry, every time. $\mathrm d\ell = \mathrm dx$ on the interval; $\mathrm dA = r\,\mathrm dr\,\mathrm d\theta$ on the disc; $\mathrm d\Omega = \sin\theta\,\mathrm d\theta\,\mathrm d\varphi = \mathrm dx\,\mathrm d\varphi$ on the sphere, which is why the Legendre weight is $1$ in the variable $x = \cos\theta$ and $\sin\theta$ in the variable $\theta$. Both statements are the same statement, and the change of variable is why $x = \cos\theta$ is universally used. Unit 10 chose it for the generating function and got the measure right by luck; here one can see it was never luck.

Why $\ell \ge m$. $P_\ell^m \equiv 0$ for $m > \ell$, since its definition differentiates the degree-$\ell$ polynomial $P_\ell$ $m$ times. So for each $m$ the eigenvalues are $\ell(\ell+1)$ with $\ell = m, m+1,\dots$ — an infinite increasing sequence, exactly as the general theory demands, but with the first few eigenvalues of the $m = 0$ problem missing from the $m = 3$ problem. The pattern is the same on the disc, where $j_{mk}$ increases with $m$ at fixed $k$.

A fourth row worth knowing about. Hermite's equation ($p = \mathrm e^{-x^2}$, $w = \mathrm e^{-x^2}$, on all of $\mathbb R$) and Laguerre's ($p = x\mathrm e^{-x}$, $w = \mathrm e^{-x}$, on $(0,\infty)$) are singular Sturm–Liouville problems on unbounded intervals, and they are the quantum harmonic oscillator and the hydrogen atom's radial equation. This course does not go there, but the row is the same row, and the reason those polynomial families are orthogonal is this section's reason and no other.

19 · geometries

The sphere, separated

In spherical coordinates (unit 05, cited — the derivation is the polar computation with one more angle):

$$\nabla^2\Phi = \frac1{r^2}\partial_r\big(r^2\partial_r\Phi\big) + \frac{1}{r^2}\Delta_{S^2}\Phi,$$ $$\Delta_{S^2} = \frac{1}{\sin\theta}\partial_\theta\big(\sin\theta\,\partial_\theta\big) + \frac{1}{\sin^2\theta}\partial_\varphi^2 .$$

Separate $\Phi = R(r)Y(\theta,\varphi)$: the two pieces depend on different variables, so both equal one constant, written $\ell(\ell+1)$, and $-\Delta_{S^2}Y = \ell(\ell+1)Y$. Separate again, $Y = \Theta(\theta)\mathrm e^{\mathrm im\varphi}$ with $m \in \mathbb Z$ by periodicity, and put $x = \cos\theta$, so that $\sin\theta\,\partial_\theta = -(1-x^2)\partial_x$:

$$\big((1-x^2)\Theta'\big)' + \Big(\ell(\ell+1) - \frac{m^2}{1-x^2}\Big)\Theta = 0,$$

the associated Legendre equation, whose $m = 0$ case is unit 10's.

Unit 10 got the $P_\ell$ from a generating function and their orthogonality from Rodrigues' formula; both are now corollaries.

spherical harmonics

$Y_\ell^m = N_{\ell m}P_\ell^{|m|}(\cos\theta)\mathrm e^{\mathrm im\varphi}$, $|m| \le \ell$, with $P_\ell^m := (1-x^2)^{m/2}P_\ell^{(m)}$ (notes): $2\ell+1$ per $\ell$, orthonormal for the right $N_{\ell m}$, complete in $L^2(S^2)$ (cited: Sansone ch. III).

what the singular endpoints buy

$p = 1-x^2$ vanishes at the poles $x = \pm1$, coordinate singularities, not edges. Boundedness forces $\lambda = \ell(\ell+1)$ with $\ell \in \{0, 1, \dots\}$ (P7 for $m = 0$, cited for $m \ge 1$), and discards $Q_\ell$. The integer $\ell$ is a consequence of the geometry being closed; nothing was quantised by hand.

The substitution, checked. With $x = \cos\theta$, $\mathrm dx = -\sin\theta\,\mathrm d\theta$, so $\partial_\theta = -\sin\theta\,\partial_x$ and $\frac1{\sin\theta}\partial_\theta(\sin\theta\,\partial_\theta\Theta) = \frac1{\sin\theta}\cdot(-\sin\theta\,\partial_x)\big(\sin\theta\cdot(-\sin\theta\,\partial_x\Theta)\big) = \partial_x\big(\sin^2\theta\,\partial_x\Theta\big) = \big((1-x^2)\Theta'\big)'$ ✓. And $1/\sin^2\theta = 1/(1-x^2)$ ✓. This is why $x = \cos\theta$ is not a trick: it is the substitution that turns the angular Laplacian into Sturm–Liouville form, and the measure $\sin\theta\,\mathrm d\theta$ into $\mathrm dx$, at one stroke.

Full proof: $P_\ell^m$ solves the associated equation

Differentiate Legendre's equation $(1-x^2)y'' - 2xy' + \ell(\ell+1)y = 0$ $m$ times by Leibniz's rule. The first term gives $(1-x^2)v'' - 2mxv' - m(m-1)v$ with $v = y^{(m)}$, the second $-2xv' - 2mv$, so $(1-x^2)v'' - 2(m+1)xv' + \big(\ell(\ell+1) - m(m+1)\big)v = 0$. Now put $\Theta = (1-x^2)^{m/2}v$. Then $(1-x^2)\Theta' = (1-x^2)^{m/2}\big((1-x^2)v' - mxv\big)$, and differentiating once more and collecting, $\big((1-x^2)\Theta'\big)' + \big(\ell(\ell+1) - \tfrac{m^2}{1-x^2}\big)\Theta = (1-x^2)^{m/2}\big[(1-x^2)v'' - 2(m+1)xv' + (\ell(\ell+1) - m(m+1))v\big] = 0$. With $y = P_\ell$ this is $P_\ell^m$: a polynomial times $(1-x^2)^{m/2}$, bounded on $[-1,1]$, nonzero exactly when $m \le \ell$. The converse — that for $m \ge 1$ boundedness at both poles forces $\lambda = \ell(\ell+1)$ with $\ell \ge m$ — is cited: Sansone, Orthogonal Functions, ch. III. The $m = 0$ case is proved in P7.

Where this is spent. Unit 23 solves Laplace's equation outside a charged body by expanding the boundary data in $Y_\ell^m$ and matching term by term — the multipole expansion of unit 10, now with a completeness theorem behind it, so that "the expansion exists" is no longer an assumption. Unit 39 uses the same harmonics on the sphere of directions to classify perturbations of a cosmological model, and unit 38 uses the $\ell = 2$ subspace to explain why gravitational radiation is quadrupolar.

The count $2\ell+1$, and why it is a degeneracy. All $2\ell+1$ functions $Y_\ell^m$ share the eigenvalue $\ell(\ell+1)$. On the drum the analogous degeneracy was $2$ and came from rotations of the disc; here it is $2\ell+1$ and comes from rotations of the sphere, a three-parameter group. Degeneracy usually has a symmetry behind it — which is Noether's theorem's near neighbour (unit 13), and exactly the mechanism unit 07 saw for zero modes. Not always: a square membrane has eigenvalues proportional to $j^2 + k^2$, and $1^2 + 7^2 = 5^2 + 5^2$ gives a degeneracy between the $(1,7)$ and $(5,5)$ modes that no symmetry of the square explains. Such coincidences are called accidental.

What the radial half becomes. $\big(r^2R'\big)' = \ell(\ell+1)R$ is an Euler equation with solutions $r^\ell$ and $r^{-\ell-1}$ — the growing and decaying multipoles of unit 10, recovered as the radial companion of each harmonic. For Laplace's equation the radial problem has no eigenvalue; for the wave equation in a ball it becomes a Bessel problem of half-integer order, and the spherical Bessel functions appear.

19 · geometries

Legendre's orthogonality, with no generating function

two lines, where unit 10 needed a page

Legendre's equation is $\big((1-x^2)P'\big)' = -\ell(\ell+1)P$ on $[-1,1]$: Sturm–Liouville with $p = 1-x^2$, $q = 0$, $w = 1$, both endpoints singular because $p(\pm1) = 0$.

Lagrange's identity for $P_\ell$ and $P_n$, integrated:

$$(\lambda_\ell - \lambda_n)\!\int_{-1}^{1}\!\!P_\ell P_n\,\mathrm dx = \big[(1{-}x^2)W\big]_{-1}^{1} = 0,$$

writing $\lambda_\ell = \ell(\ell+1)$, $W = P_\ell P_n' - P_\ell'P_n$,

because $1 - x^2$ vanishes at both ends while $P_\ell$, $P_n$ and their derivatives are polynomials, hence finite there. Since $\ell \ne n$ gives $\ell(\ell+1) \ne n(n+1)$ for non-negative integers, the integral is zero. $\square$

The whole content is that $p$ vanishes at the endpoints and the eigenfunctions do not blow up. Rodrigues' formula was never needed.

checks

Special case. $P_0 = 1$, $P_1 = x$: $\int_{-1}^1x\,\mathrm dx = 0$ ✓. $P_2 = \tfrac12(3x^2-1)$ against $P_0$: $\int_{-1}^1\tfrac12(3x^2-1) = \tfrac12(2 - 2) = 0$ ✓.

Monotonicity of eigenvalues. $\ell(\ell+1)$ is strictly increasing in $\ell \ge 0$: $0, 2, 6, 12, 20$ — discrete, bounded below, unbounded above, as section 4 requires ✓.

Nodes. $P_\ell$ has $\ell$ zeros in $(-1,1)$, and $\lambda_\ell = \ell(\ell+1)$ is the $(\ell+1)$-st eigenvalue: "index minus one" ✓.

where $Q_\ell$ went

The second solution is $Q_0 = \tfrac12\log\frac{1+x}{1-x}$ and its relatives, infinite at $x = \pm1$. Keep it and the concomitant $\big[(1-x^2)(\cdot)\big]_{-1}^1$ no longer vanishes — $(1-x^2)Q_0' = 1$, a constant — so the operator stops being symmetric and the proof above collapses. Boundedness at a singular endpoint is load-bearing, once again.

Why $\lambda = \ell(\ell+1)$ and not something else. Nothing above assumed it. Feeding a general $\lambda$ into Legendre's equation and demanding boundedness at both $x = 1$ and $x = -1$ is what forces $\lambda = \ell(\ell+1)$: the Frobenius series about $x = 1$ (unit 09's method, with a regular singular point) terminates — and so stays bounded at the other end — only for those $\lambda$. Unit 10 derived this; the point here is that "the series must terminate" is the analytic face of "the endpoint is singular and the eigenfunction must be bounded", which is the operator-theoretic face. One statement, two languages.

$(1-x^2)Q_0' = 1$, checked. $Q_0 = \tfrac12\log\frac{1+x}{1-x}$ gives $Q_0' = \tfrac12\big(\tfrac1{1+x} + \tfrac1{1-x}\big) = \tfrac1{1-x^2}$, so $(1-x^2)Q_0' = 1$ exactly ✓. The concomitant of $P_0 = 1$ and $Q_0$ is then $(1-x^2)(P_0Q_0' - P_0'Q_0) = 1$ at every $x$, and $[1]_{-1}^{1} = 0$ — so this particular pair happens not to break it, but $P_1$ and $Q_0$ do: $(1-x^2)(xQ_0' - Q_0) = x - (1-x^2)Q_0$, which tends to $1$ at $x = 1$ and $-1$ at $x = -1$, giving a concomitant difference of $2 \ne 0$.

The norm, for completeness of the record. $\int_{-1}^1P_\ell^2\,\mathrm dx = 2/(2\ell+1)$, proved in unit 10 from Rodrigues' formula. Sturm–Liouville theory gives orthogonality for free but says nothing about norms — those always need a computation specific to the family, exactly as the drum's $\tfrac12a^2J_1(j_{0k})^2$ did.

19 · geometries

Why a string has a pitch and a drum has not

	string, $f_n/f_1$	cents	drum, $j/j_{01}$	cents
2nd partial	2.0000	1200.0	1.5933	806.5
3rd	3.0000	1902.0	2.1355	1313.5
4th	4.0000	2400.0	2.2954	1438.5
5th	5.0000	2786.3	2.6531	1689.2

The string's partials are exact integer multiples, because $\lambda_n = (n\pi/L)^2$ makes $\sqrt{\lambda_n}$ linear in $n$. The drum's are ratios of Bessel zeros, with no reason to be rational: its second partial sits $806.5$ cents up, where a harmonic series puts the octave, and no low harmonic series fits the four together.

the mechanism, stated once

Pitch is the brain's estimate of the fundamental of a harmonic series. Feed it $f, 2f, 3f,\dots$ and it reports $f$, even if $f$ itself is missing. Feed it $f, 1.5933f, 2.1355f,\dots$ and there is no harmonic series to fit: the report is a thud. "Has a pitch" is a statement about the eigenvalues of a Sturm–Liouville problem.

The cents. An interval of ratio $\rho$ is $1200\log_2\rho$ cents; $100$ cents is a semitone. $1200\log_2 1.59334 = 806.5$, $1200\log_2 2.13555 = 1313.5$, $1200\log_2 2.29542 = 1438.5$, $1200\log_2 2.65307 = 1689.2$. For comparison, the just intervals: octave $1200$, fifth $702$, twelfth $1902$, double octave $2400$. The drum's second partial is $104.5$ cents above a fifth — beyond a semitone — and its third is $113.5$ cents above an octave. Taken one at a time, any ratio lies within $50$ cents of some tempered interval (the second partial is $6.5$ cents from a tempered minor sixth), so "is it near an interval?" is not the test. The test is whether the partials sit near multiples of one frequency, where a harmonic series puts them at $1200$, $1902$ and $2400$ cents; the drum's sit at $806.5$, $1313.5$ and $1438.5$.

19 · geometries

Partials on one axis: why only one dimension rings

Every partial within two octaves (2400 cents) of the lowest, string above and drum below, on one axis; the dotted lines are the string's harmonics. The string's four partials sit on them by construction. The drum's ten — counting each degenerate pair once — fall between, and crowd closer together as they rise.

why one dimension is special

$\sqrt{\lambda_n}$ linear in $n$ needs $\sqrt{\lambda_n}$ to be an arithmetic progression, and Weyl's law (cited, notes) says $\lambda_n \sim Cn^{2/d}$ in $d$ dimensions — so $\sqrt{\lambda_n} \sim n^{1/d}$, linear only when $d = 1$. Tuned percussion is the craft of deforming an object until a few modes take ratios they had no reason to take.

The crowding in the figure is Weyl's law made visible: in two dimensions the number of modes below frequency $\omega$ grows like $\omega^2$, not $\omega$, so the gaps shrink as the pitch rises.

The figure's drum partials, in cents above $(0,1)$: $806.5$ $(1,1)$, $1313.5$ $(2,1)$, $1438.5$ $(0,2)$, $1689.2$ $(3,1)$, $1853.6$ $(1,2)$, $1989.4$ $(4,1)$, $2168.9$ $(2,2)$, $2216.9$ $(0,3)$ and $2240.3$ $(5,1)$ — every $j_{mk}$ with $1200\log_2(j_{mk}/j_{01}) \le 2400$, each counted once although the $m \ge 1$ ones are doubly degenerate. In the same two octaves the string has four partials. The next drum partials, $(3,2)$ at $2425.3$ and $(6,1)$ at $2456.1$, crowd in just above.

Weyl's law is Hermann Weyl's theorem of 1911: for the Dirichlet Laplacian on a bounded domain in $\mathbb R^d$, $\lambda_n \sim C_d\,(n/\mathrm{vol})^{2/d}$. Cited: Courant & Hilbert vol. I ch. VI §4, where it is proved by the min–max principle of section 5 and a comparison with boxes.

How a timpano is fixed. Three things push the ideal membrane's ratios towards $1 : 1.5 : 2 : 2.5$ relative to the $(1,1)$ mode, which is then heard as the pitch: the air in the kettle stiffens the low modes selectively; radiation loading adds effective mass, most to the modes that move the most air; and the strike point, about a quarter of the way in from the rim, is where the $(0,1)$ mode moves with only $0.34$ of its centre amplitude, since $J_0(0.75\,j_{01}) = 0.338$ — so the mode that fits worst is weakly excited. It has no interior node to strike at; the kettle designer settles for small. The result is a genuine pitch, and every one of those three is an engineered departure from the Sturm–Liouville problem computed here.

"Can one hear the shape of a drum?" Mark Kac asked it in 1966 — does the spectrum determine the domain? Weyl's law makes the area audible, and its two-term refinement the perimeter. The full answer is no: Gordon, Webb and Wolpert exhibited two different plane domains with identical spectra in 1992. Cited; the construction is elementary but outside this course. The question is the exact converse of this unit, which has spent fifty slides going from a shape to its spectrum.

The running example, closed. A drum of radius $15$ cm under $3200\ \mathrm{N\,m^{-1}}$ of tension has modes at $280.39$, $446.76$, $598.79$ and $643.62$ Hz; those numbers are zeros of Bessel functions, which are eigenvalues of the Sturm–Liouville problem $p = w = r$, $q = m^2/r$, singular at the centre, Dirichlet at the rim; their eigenfunctions are real, orthogonal with weight $r$, node-counted, complete, and bounded from below by a one-parameter guess to within $0.78\%$. Every one of those adjectives came from one identity on one slide.

19 · where it is spent

What this unit bought

20 Fourier transform	the same idea with the interval sent to $\mathbb R$: the sequence of eigenvalues becomes a continuum, and Parseval becomes Plancherel.
23 static fields	boundary-value problems for Laplace's equation solved by expanding the data in $Y_\ell^m$ — legitimate because of this unit's completeness.
24 Maxwell	mode expansions in a cavity and a waveguide; the cut-off frequency is a lowest eigenvalue.
32 geodesics	the Rayleigh quotient's parent idea — a differential equation as the stationarity condition of a quotient — reappears as the geodesic equation.
38 gravitational waves	the $\ell = 2$ subspace of spherical harmonics, and why radiation starts at the quadrupole.
39 cosmology	harmonics on a maximally symmetric space; the same separation, one curvature term further on.

the three sentences to keep

1. The eigenfunctions of a symmetric operator are orthogonal, and that one fact is what makes an expansion's coefficients computable and unique.

2. The boundary conditions are half of the operator. Change them and you change the eigenfunctions, the expansion, and what it converges to at the ends.

3. How fast the coefficients fall is how smooth the function is — a dictionary, not a heuristic, and it prices every truncation in advance.

And one habit: when a new orthogonal family turns up — Hermite, Laguerre, Chebyshev, Mathieu — the first question is never "what are its identities?" but "what are $p$, $q$ and $w$, and which endpoints are singular?". Everything else is on this deck already.

Reading. For the Fourier half: Stein & Shakarchi, Fourier Analysis, ch. 2–3 — ch. 2 is Dirichlet and the kernels, ch. 3 is mean-square convergence; skip §3.3 onwards, which needs the Lebesgue integral this course has not built. Körner's Fourier Analysis is a collection of short chapters: 1–8, 15–18 are this unit, and each is self-contained. For the Sturm–Liouville half: Al-Gwaiz, Sturm–Liouville Theory and its Applications, ch. 2–3 is the shortest complete account, with the drum in ch. 5. Coddington & Levinson ch. 7–8 is the reference for the two theorems cited here, and is harder going. Courant & Hilbert vol. I ch. V–VI is where the variational characterisation is done properly, and is still the best written account of why it is the right way to think.

What was deliberately left out. The Fourier transform and everything needing contour integration (unit 20). Distributions, so no delta functions and no Green's function for $\mathcal L$ written as $\mathcal LG = \delta$ — the Green's function was named only inside a citation. Weyl's limit-point/limit-circle theory, beyond a mention. Convergence of Fourier series in $L^p$ for $p \ne 2$, and Carleson's theorem. Numerical eigenvalue methods. Each is a real subject and none of them is needed to get to the field equations.

19 · checkpoint

Done when

checkpoint

Given a Sturm–Liouville problem you can, on paper and without notes:

prove that its eigenvalues are real and its eigenfunctions $w$-orthogonal, starting from Lagrange's identity and naming where the boundary conditions are spent;
say how many zeros the third eigenfunction has in the open interval, and be right;
bound the lowest eigenvalue from above, to within a few per cent, with a trial function of your own choosing — and say which side of the truth the bound is on and why;
expand a given initial profile in its eigenfunctions, and state which kind of convergence you have — mean-square, pointwise, or uniform — together with the hypothesis that buys it.

Failing it is unambiguous

Two eigenfunctions declared orthogonal without a weight; a third eigenfunction with the wrong number of nodes; a Rayleigh bound below the true eigenvalue; or "the series converges to $f$" at a jump.

20 the Fourier transform and complex analysis · 21 asymptotics · 23 static fields, which is the first place this unit's completeness is cashed for a physical answer.

Try the checkpoint on a problem not in the deck before opening the sheet. A good one: $-u'' = \lambda u$ on $[0,1]$ with $u(0) = 0$ and $u(1) + u'(1) = 0$. The eigenvalue condition is $\tan k = -k$; the first three roots are $k = 2.0288$, $4.9132$, $7.9787$, so $\lambda_3 = 63.66$ and $u_3 = \sin(7.9787x)$ has zeros at $x = \pi/7.9787 = 0.3938$ and $2\pi/7.9787 = 0.7875$ — two of them, as the oscillation theorem demands. Then bound $\lambda_1 = 4.1159$ from above with any trial function vanishing at $0$, remembering the Robin boundary term $+u(1)^2$ in the numerator: $u = x$ gives $(1 + 1)/(1/3) = 6$, which is high by $46\%$ and correctly on the high side; $u = \sin(\pi x/2)$ gives $(\pi^2/8 + 1)/(1/2) = 4.4674$, high by $8.5\%$. Then expand $f(x) = 1$ in those eigenfunctions and say what kind of convergence you have: mean-square, by completeness (cited); pointwise to $f$ on $(0,1]$, by the Sturm–Liouville analogue of Dirichlet's theorem (cited: Titchmarsh, Eigenfunction Expansions, Part I, ch. I, the equiconvergence theorem), with the value $0$ at $x = 0$, where every $\sin k_nx$ vanishes; and not uniform, provably: each partial sum $S_N$ is continuous and vanishes at $0$, so $\sup_{(0,1]}|S_N - 1| \ge 1$ for every $N$. Saying “uniform” is the fourth way to fail the checkpoint.

Every number in that paragraph was computed. If yours disagree, the deck is not the thing to re-read first — the arithmetic is.

Fourier and Sturm–Liouville: the modes are a basis, and that is a theorem

One drum, six sections

Separation of variables, and the question it leaves

Splitting the string

The boundary conditions do the selecting

The string's spectrum, and three checks

Superposition, and the debt it runs up

Worked: what a pluck puts into each mode

The pluck at L/7, mode by mode

The drum, separated

The drum's spectrum, in numbers

The drum's radial factors, and three checks

Orthogonality, and the best approximation

An inner product on a space of functions

Coefficients are a projection, and that is optimal

Fourier coefficients: coordinates that stay put

Bessel's inequality, and what it hands you free

Completeness, and Parseval

Sine, cosine, complex: one system per boundary condition

Worked: one function, three expansions

Worked: Parseval, twice, with the drum's weight

Convergence: which kind, and how fast

Dirichlet: what the series converges to at a jump

Uniform convergence, and exactly when it fails

Worked: Gibbs, and the number 8.9490 %

Gibbs, seen and repaired

Smoothness against decay: a dictionary, and an error bar

Smoothness in decibels: pluck against strike

Differentiating term by term, and what it costs

Sturm–Liouville: why any of this was inevitable

The operator, and the weight that comes with it

Lagrange's identity: one line, everything follows

Three families of boundary conditions, and one that fails

Real eigenvalues, orthogonal eigenfunctions

Simple, discrete, unbounded above

Completeness — the theorem the whole method was waiting for

The debt from section 1, paid on the drum

Sturm's comparison theorem, and counting nodes

Counting nodes: the oscillation theorem

Worked: the drum is a singular Sturm–Liouville problem

The Rayleigh quotient

The quotient, and the identity behind it

Unit 11's second job: the eigenvalue problem is an Euler–Lagrange equation

Worked: bounding the drum's lowest note, one parameter

The drum's bound, plotted

Three geometries, one theorem

One table, three special functions

The sphere, separated

Legendre's orthogonality, with no generating function

Why a string has a pitch and a drum has not

Partials on one axis: why only one dimension rings

What this unit bought

Done when

Failing it is unambiguous

Next