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

Idea. One function, three different operators — Dirichlet, Neumann and periodic boundary conditions — and therefore three different eigenfunction systems. Everything that differs between the three answers is a fact about the extension, not about \(f\).

1. The coefficients. Write \(k_n = n\pi/L\).

• Sine. \(\alpha_n = \frac2L\int_0^Lx\sin k_nx\,\mathrm dx = \frac2L\Big(\big[-\tfrac{x}{k_n}\cos k_nx\big]_0^L + \tfrac1{k_n}\int_0^L\cos k_nx\,\mathrm dx\Big) = \frac2L\cdot\big(-\tfrac{L}{k_n}(-1)^n\big) = \dfrac{2L(-1)^{n+1}}{n\pi}.\)
• Cosine. \(a_0 = \frac2L\int_0^Lx\,\mathrm dx = L\). For \(n \ge 1\), \(a_n = \frac2L\int_0^Lx\cos k_nx\,\mathrm dx = \frac2L\Big(\big[\tfrac{x}{k_n}\sin k_nx\big]_0^L - \tfrac1{k_n}\int_0^L\sin k_nx\,\mathrm dx\Big) = \frac{2L\big((-1)^n - 1\big)}{n^2\pi^2},\) which is \(-4L/n^2\pi^2\) for odd \(n\) and \(0\) for even \(n\).
• Full, period \(L\). \(A_0 = \frac2L\int_0^Lx\,\mathrm dx = L\); \(A_n = \frac2L\int_0^Lx\cos\frac{2\pi nx}{L}\mathrm dx = 0\) (the boundary term is \(\frac2L\cdot\frac{L}{2\pi n}\big[x\sin\big]_0^L = 0\) and the remaining integral vanishes); \(B_n = \frac2L\int_0^Lx\sin\frac{2\pi nx}{L}\mathrm dx = \frac2L\big[-\tfrac{Lx}{2\pi n}\cos\big]_0^L = -\dfrac{L}{\pi n}.\)

2. The endpoint values. Draw the extensions.

• Sine \(\to\) odd \(2L\)-periodic extension. It is continuous at \(0\) with value \(0\), so the series gives \(\mathbf 0\) there — as it must, every term being \(\sin 0\). At \(x = L\) the extension jumps from \(L\) (approaching from the left) to \(-L\) (the value just right of \(L\), by oddness and \(2L\)-periodicity), so Dirichlet gives the midpoint \(\mathbf 0\) — again visible, since \(\sin n\pi = 0\).
• Cosine \(\to\) even \(2L\)-periodic extension, which is continuous everywhere (it is the triangular wave). So the series converges to \(f\): \(\mathbf 0\) at \(x=0\) and \(\mathbf L\) at \(x = L\).
• Full \(\to\) \(L\)-periodic extension, which jumps from \(L\) to \(0\) at every multiple of \(L\). Dirichlet gives the midpoint \(\mathbf{L/2}\) at both ends.

3. Decay. Sine: \(|\alpha_n| \sim 2L/\pi n\), rate \(1/n\), because the odd extension itself jumps (lowest jumping derivative: the zeroth). Cosine: \(|a_n| \sim 4L/\pi^2n^2\), rate \(1/n^2\), because the even extension is continuous but its derivative jumps from \(+1\) to \(-1\) at \(x = L\) (lowest jumping derivative: the first). Full: \(|B_n| = L/\pi n\), rate \(1/n\), jump in the function. So the cosine series is one power better than the other two, and this is the practical reason to prefer it.

4. Uniform convergence. Only the cosine series. Its limit is continuous and \(\sum|a_n| < \infty\), so the \(M\)-test applies. The other two cannot converge uniformly on any interval containing \(x = L\) (sine) or \(x = 0\) (full), because a uniform limit of continuous functions is continuous while their limits jump there. No hypothesis on \(f\) could rescue them: the obstruction is in the extension, which the boundary conditions chose.

Check. Dimensions: \(\alpha_n\), \(a_n\), \(B_n\) all carry a length, as they must, \(f\) being a length ✓. Limit: as \(n \to \infty\) all three tend to \(0\), which is Riemann–Lebesgue ✓. A number: at \(x = L/2\) the cosine series gives \(L/2 - \frac{4L}{\pi^2}\sum_{\text{odd}}\frac{\cos(n\pi/2)}{n^2} = L/2 - 0 = L/2\), since \(\cos(n\pi/2) = 0\) for every odd \(n\) — and \(f(L/2) = L/2\) ✓.

Another way

Get the cosine series from the sine series without integrating. The even extension of \(x\) on \([0,L]\) is the triangular wave; the odd extension is the sawtooth; and the triangular wave is the integral of a square wave. Term-by-term integration is always legal (the uniform-convergence slide's note), so integrating the square wave's sine series once produces the triangular wave's cosine series, complete with the extra factor of \(n\) in the denominator. The gain of one power of \(n\) under integration is exactly the gain of one power in the decay rate.

Idea. Parseval turns "how big is \(f\)?", answered by a one-line integral, into "how big are its coefficients?", answered by a series. Every classical sum of this kind is that trade, done with a well-chosen \(f\).

1. With \(\alpha_n = 4/n\pi\) for odd \(n\) and \(\lVert f\rVert^2 = \int_0^L1 = L\), \[ L = \frac L2\sum_{n\ \text{odd}}\frac{16}{n^2\pi^2} \;\Longrightarrow\; \sum_{n\ \text{odd}}\frac1{n^2} = \frac{\pi^2}{8} = 1.2337006 . \]

2. \(\alpha_n = 2L(-1)^{n+1}/n\pi\) and \(\lVert f\rVert^2 = \int_0^Lx^2 = L^3/3\): \[ \frac{L^3}{3} = \frac L2\sum_{n\ge1}\frac{4L^2}{n^2\pi^2} \;\Longrightarrow\; \sum_{n\ge1}\frac1{n^2} = \frac{\pi^2}{6} = 1.6449341 . \]

3. The cosine system is \(\{1, \cos(n\pi x/L)\}\) with \(\lVert1\rVert^2 = L\) and \(\lVert\cos\rVert^2 = L/2\). The expansion is \(x = \frac{a_0}2 + \sum a_n\cos\), so the coefficient along the eigenfunction \(1\) is \(a_0/2 = L/2\). Parseval: \[ \frac{L^3}{3} = \Big(\frac L2\Big)^2\! L + \frac L2\sum_{n\ \text{odd}}\frac{16L^2}{n^4\pi^4} \;\Longrightarrow\; \frac{L^3}{3} - \frac{L^3}{4} = \frac{8L^3}{\pi^4}\sum_{n\ \text{odd}}\frac1{n^4} , \] and \(\frac13 - \frac14 = \frac1{12}\), so \(\sum_{\text{odd}}n^{-4} = \pi^4/96 = 1.0146780\).

4. For \(s > 1\), \(\sum_{n\ge1}n^{-s} = \sum_{\text{odd}} + \sum_{\text{even}} = \sum_{\text{odd}} + 2^{-s}\sum_{n\ge1}n^{-s}\), so \(\sum_{\text{odd}}n^{-s} = (1 - 2^{-s})\sum_{n\ge1}n^{-s}\). At \(s = 2\): factor \(3/4\), so \(\sum n^{-2} = \frac43\sum_{\text{odd}}n^{-2} = \frac43\cdot\frac{\pi^2}{8} = \frac{\pi^2}{6}\) ✓, consistent with part 2. At \(s = 4\): factor \(15/16\), so \(\sum n^{-4} = \frac{16}{15}\cdot\frac{\pi^4}{96} = \frac{\pi^4}{90} = 1.0823232\).

Check. Dimensions: every Parseval identity above has \(L^3\) or \(L\) on both sides ✓. Numbers: \(\pi^2/8 = 1.2337006\) against the partial sum \(1 + \frac19 + \frac1{25} + \frac1{49} + \frac1{81} = 1.1839\), converging slowly from below as Bessel's inequality demands ✓. \(\pi^4/90 = 1.0823232\) against \(1 + \frac1{16} + \frac1{81} = 1.0748\) ✓. Limit: both sums exceed their first term and are below it plus \(\int_1^\infty x^{-s}\), i.e. \(\pi^2/6 \in (1,2)\) and \(\pi^4/90 \in (1, 4/3)\) ✓.

Another way

Part 3 without the cosine series: apply Parseval to the derivative instead. The triangular wave's derivative is the square wave, and the coefficients differ by a factor \(n\), so \(\sum_{\text{odd}}n^{-4}\) and \(\sum_{\text{odd}}n^{-2}\) are the Parseval sums of a function and its derivative. That is the smoothness–decay dictionary being read as an arithmetic identity, and it generalises: integrating the square wave \(k\) times evaluates \(\sum_{\text{odd}}n^{-2k}\).

Idea. Timbre is a decay rate. A pluck puts a kink into the string, a strike puts a step into its velocity, and one integration by parts separates the two: the kink gives \(n^{-2}\), the step-in-velocity gives \(n^{-1}\) after dividing by \(\omega_n \propto n\).

1. The initial shape is \(f = hx/a\) on \([0,a]\) and \(h(L-x)/(L-a)\) on \([a,L]\), and \(\alpha_n = \frac2L\int_0^Lf\sin k_nx\,\mathrm dx\) with \(k_n = n\pi/L\). On the first leg, \(\int_0^ax\sin k_nx\,\mathrm dx = \frac{\sin k_na}{k_n^2} - \frac{a\cos k_na}{k_n}\). On the second, substituting \(s = L - x\) and using \(\sin k_nL = 0\), \(\cos k_nL = (-1)^n\), \(\int_a^L(L-x)\sin k_nx\,\mathrm dx = \frac{\sin k_na}{k_n^2} + \frac{(L-a)\cos k_na}{k_n}\). Multiply the first by \(2h/La\) and the second by \(2h/L(L-a)\) and add: the \(\cos k_na\) terms are \(-\frac{2h}{Lk_n}\) and \(+\frac{2h}{Lk_n}\) and cancel, while the \(\sin k_na\) terms give \(\frac{2h}{Lk_n^2}\big(\frac1a + \frac1{L-a}\big)\sin k_na = \frac{2hL}{k_n^2a(L-a)}\sin k_na\). With \(k_n^2 = n^2\pi^2/L^2\) that is the stated formula. ∎

2. \(\sin(7\pi/7) = \sin\pi = 0\), so \(\alpha_7 = 0\) exactly. With \(\alpha_n/\alpha_1 = n^{-2}\sin(n\pi/7)/\sin(\pi/7)\) and \(\sin(\pi/7) = 0.4338837\):

3. A strike sets \(u(x,0) = 0\) and \(u_t(x,0) = g\), so \(\alpha_n = 0\) and \(\omega_n\beta_n = \frac2L\int_0^Lg\sin k_nx\,\mathrm dx\). With \(g = V\) on \((a-\varepsilon, a+\varepsilon)\), \[ \omega_n\beta_n = \frac{2V}{L}\cdot\frac{L}{n\pi}\Big[\cos k_n(a-\varepsilon) - \cos k_n(a+\varepsilon)\Big] = \frac{4V}{n\pi}\sin\frac{n\pi a}{L}\sin\frac{n\pi\varepsilon}{L}. \] Divide by \(\omega_n = n\pi c/L\): \(\beta_n = \dfrac{4VL}{\pi^2c}\cdot\dfrac{1}{n^2}\sin\dfrac{n\pi a}{L}\sin\dfrac{n\pi\varepsilon}{L}\), which is the stated proportionality. For \(n\pi\varepsilon/L \ll 1\), \(\sin(n\pi\varepsilon/L) \approx n\pi\varepsilon/L\), and one power of \(n\) is eaten: \(\beta_n \approx \frac{4V\varepsilon}{\pi c}\cdot\frac1n\sin\frac{n\pi a}{L}\). Beyond \(n \approx L/\pi\varepsilon\) the second sine stops growing and the true \(n^{-2}\) resumes — a wide, soft hammer is a dull one, and that is the design parameter.

4. By Parseval the relative mean-square truncation error is \(E(N) = \sum_{n>N}\alpha_n^2 \big/ \sum_{n\ge1}\alpha_n^2\). For the pluck, \(\alpha_n^2 \propto n^{-4}\sin^2(n\pi a/L)\) and \(\sum_{n>N}n^{-4} \approx \frac1{3N^3}\); for the strike in its \(n^{-1}\) regime, \(\beta_n^2 \propto n^{-2}\sin^2\) and \(\sum_{n>N}n^{-2} \approx \frac1N\). So \(E \sim N^{-3}\) and \(E \sim N^{-1}\) respectively.

Evaluated exactly at \(N = 10\), \(a = L/7\), by summing the series:

The struck string is brighter, by a wide margin: ten modes capture 99.94 % of the plucked string's mean square but only 92.0 % of the struck string's, so far more of the struck string's content lies above the tenth partial. Equivalently, the envelope \(n^{-2}\) puts partial 10 at \(-40\) dB and the envelope \(n^{-1}\) puts it at \(-20\) dB. The cause is one power of \(n\), and that power came from dividing by \(\omega_n\) in part 3 — i.e. from the fact that a strike specifies a velocity and a pluck specifies a displacement.

Check. Dimensions: \(\alpha_n\) carries \(h\), a length ✓; \(\beta_n \sim VL/c\) also a length, since \(V/c\) is dimensionless ✓. Special case: \(a = L/2\) in part 1 kills every even \(n\), as symmetry demands ✓. Number: \(E(20)/E(10)\) for the pluck is \(0.1105\), against the predicted \((10/20)^3 = 0.125\) ✓; for the strike it is \(0.488\) against \((10/20)^1 = 0.5\) ✓ — both rates confirmed.

Another way

Part 2 without any arithmetic: a mode with a node at \(x = a\) cannot be excited by anything done at \(x = a\), because its amplitude is a projection against \(\sin(n\pi a/L)\), which is the mode's own displacement there. Mode 7 has nodes at \(L/7, 2L/7, \dots\); plucking at \(L/7\) therefore misses it entirely. The same sentence is Helmholtz's explanation of why piano hammers strike near \(L/7\) to \(L/9\): a strike at or near a node of the seventh partial weakens it, and the seventh is the first partial that lies well off the tempered scale.

Idea. One integration by parts trades a derivative for a factor of \(\mathrm in\). Every statement in this problem is that trade, run forwards (smoothness buys decay) or backwards (differentiating a series costs decay), with the boundary term deciding whether the trade is legal.

1. One step: for \(f\) continuous, \(2\pi\)-periodic with \(f'\) piecewise continuous, \[ 2\pi c_n(f') = \int_{-\pi}^{\pi}\!f'\mathrm e^{-\mathrm inx}\mathrm dx = \big[f\mathrm e^{-\mathrm inx}\big]_{-\pi}^{\pi} + \mathrm in\!\int_{-\pi}^{\pi}\!f\mathrm e^{-\mathrm inx}\mathrm dx = 0 + 2\pi\,\mathrm in\,c_n(f), \] the bracket vanishing because \(\mathrm e^{-\mathrm in\pi} = \mathrm e^{\mathrm in\pi}\) and \(f(\pi) = f(-\pi)\). (If \(f'\) has finitely many jumps, split the integral at them; the interior boundary terms cancel in pairs because \(f\) is continuous.) Now induct: \(f \in C^{k-1}\) means \(f, f', \dots, f^{(k-2)}\) are all continuous and periodic, so the step applies \(k\) times and \(c_n(f^{(k)}) = (\mathrm in)^kc_n(f)\).

Bessel's inequality for \(f^{(k)}\) gives \(\sum_n n^{2k}|c_n(f)|^2 = \sum_n|c_n(f^{(k)})|^2 < \infty\); the terms of a convergent series tend to zero, so \(n^k|c_n| \to 0\), i.e. \(|c_n| = o(|n|^{-k})\). For the error bound, Parseval for \(f^{(k)}\) reads \(\sum_n|c_n(f^{(k)})|^2 = \lVert f^{(k)}\rVert_2^2/2\pi\), and \[ \lVert f - S_Nf\rVert_2^2 = 2\pi\!\!\sum_{|n|>N}\!|c_n|^2 \;\le\; \frac{2\pi}{N^{2k}}\!\!\sum_{|n|>N}\! n^{2k}|c_n|^2 \;\le\; \frac{2\pi}{N^{2k}}\cdot\frac{\lVert f^{(k)}\rVert_2^2}{2\pi} = \frac{\lVert f^{(k)}\rVert_2^2}{N^{2k}} . \ \blacksquare\]

2. It enters only in killing the boundary term — that is its entire job, and it is not a formality. Take the sawtooth \(f(x) = x\) on \((-\pi,\pi)\), extended \(2\pi\)-periodically. Off the jumps \(f' \equiv 1\), which is as piecewise continuous as anything could be, so the naive reading of the hypothesis holds with \(k = 1\). But \(c_n(f') = c_n(1) = 0\) for \(n \ne 0\), whereas \(\mathrm inc_n(f) = \mathrm in\cdot\frac{\mathrm i(-1)^n}{n} = -(-1)^n \ne 0\). The identity fails, and with it the decay: \(|c_n| = 1/|n|\) is not \(o(n^{-1})\). The missing piece is exactly \(\big[f\mathrm e^{-\mathrm inx}\big]_{-\pi}^{\pi} = (-1)^n(\pi - (-\pi)) = 2\pi(-1)^n\).

3. \(|n^{-2}\sin(n^2x)| \le n^{-2} =: M_n\) for all \(x\), and \(\sum M_n = \pi^2/6 < \infty\). By the Weierstrass \(M\)-test the series converges uniformly on \(\mathbb R\); a uniform limit of continuous functions is continuous, and each partial sum is \(2\pi\)-periodic, so \(g\) is continuous and \(2\pi\)-periodic. ∎

4. Differentiating term by term gives \(\sum_{n\ge1}\cos(n^2x)\). Fix \(x\) and suppose the series converges there. Then its terms tend to zero: \(\cos(n^2x) \to 0\). Apply the double-angle formula twice: \[ \cos(2n^2x) = 2\cos^2(n^2x) - 1 \to -1, \qquad \cos(4n^2x) = 2\cos^2(2n^2x) - 1 \to 2(-1)^2 - 1 = 1 . \] But \(4n^2 = (2n)^2\), so \(\cos(4n^2x) = \cos\big((2n)^2x\big)\) is a subsequence of \(\cos(m^2x)\), which we assumed tends to \(0\). Hence \(1 = 0\) — a contradiction. So \(\cos(n^2x)\not\to0\) and the differentiated series diverges, at every real \(x\). ∎

\(g\) satisfies the continuity hypothesis and fails piecewise \(C^1\): it has no piecewise continuous derivative, because the formal derivative does not exist as a function anywhere. So the second hypothesis is the one doing the work, which is precisely what this counterexample is for.

Check. Dimensions: all quantities are pure numbers ✓. Special case of part 1: \(f = \cos x\), \(k\) arbitrary, \(c_{\pm1} = 1/2\) and the bound reads \(\lVert f - S_Nf\rVert_2 = 0 \le \lVert f^{(k)}\rVert_2/N^k\) for \(N \ge 1\) ✓ (trivially true). Number: for the sawtooth of part 2, \(\lVert f - S_Nf\rVert_2^2 = 2\pi\sum_{|n|>N}n^{-2} \approx 4\pi/N\), so the error falls only as \(N^{-1/2}\) — the slowest rate in the dictionary, and the right one for a function that jumps ✓.

Another way

Part 4 also follows from a classical fact, if one is willing to import it: for \(x/\pi\) irrational the sequence \(n^2x \bmod 2\pi\) is equidistributed (Weyl, 1916), so \(\cos(n^2x)\) visits every value in \([-1,1]\) and certainly does not tend to zero; for \(x/\pi\) rational the sequence is eventually periodic and again does not. That is cited, not proved — Weyl's equidistribution theorem is outside this course — and the double-angle argument above needs nothing at all, which is why it is the one given.

Idea. The partial sum has a closed-form derivative, so its first peak can be located exactly; at that peak the sum is a Riemann sum for \(\int_0^\pi\frac{\sin u}{u}\mathrm du\), and that integral exceeds \(\pi/2\) by a fixed amount. The overshoot never shrinks because the peak tracks the jump.

1. With \(z = \mathrm e^{2\mathrm ix}\), \[ \sum_{j=0}^{M-1}\cos(2j+1)x = \operatorname{Re}\Big(\mathrm e^{\mathrm ix}\frac{z^M-1}{z-1}\Big) = \operatorname{Re}\Big(\mathrm e^{\mathrm ix}\cdot\frac{\mathrm e^{\mathrm iMx}\,2\mathrm i\sin Mx}{\mathrm e^{\mathrm ix}\,2\mathrm i\sin x}\Big) = \operatorname{Re}\Big(\mathrm e^{\mathrm iMx}\frac{\sin Mx}{\sin x}\Big) = \frac{\cos Mx\,\sin Mx}{\sin x} = \frac{\sin 2Mx}{2\sin x}. \] Multiplying by \(4/\pi\) gives \(S_{2M-1}'(x) = \frac{2}{\pi}\frac{\sin 2Mx}{\sin x}\). On \((0,\pi)\) the denominator is positive, so the sign of \(S'\) is the sign of \(\sin 2Mx\): positive on \((0,\pi/2M)\), negative just after. Hence \(x_M = \pi/2M\) is the first maximum. ∎

2. \(S_{2M-1}\) is odd, so \(S_{2M-1}(0) = 0\), and \[ S_{2M-1}(x_M) = \int_0^{\pi/2M}\!\!\frac{2}{\pi}\frac{\sin 2Mt}{\sin t}\,\mathrm dt \;\overset{u = 2Mt}{=}\; \frac2\pi\int_0^{\pi}\frac{\sin u}{\sin(u/2M)}\cdot\frac{\mathrm du}{2M} = \frac2\pi\int_0^{\pi}\frac{\sin u}{2M\sin(u/2M)}\,\mathrm du . \] For the limit, put \(s = u/2M \in (0, \pi/2M] \subset (0,\pi/2]\). Then \[ \Big|\frac{1}{2M\sin s} - \frac{1}{2Ms}\Big| = \frac{s - \sin s}{2Ms\sin s} \le \frac{s^3/6}{2Ms\cdot(2s/\pi)} = \frac{\pi s}{24M} = \frac{\pi u}{48M^2} \le \frac{\pi^2}{48M^2}, \] using \(0 \le s - \sin s \le s^3/6\) and \(\sin s \ge 2s/\pi\) on \([0,\pi/2]\). Since \(\frac1{2Ms} = \frac1u\), the integrand differs from \(\frac{\sin u}{u}\) by at most \(\pi^2/48M^2\) uniformly on \((0,\pi]\), so the integrals differ by at most \(\pi^3/48M^2 \to 0\). ∎

3. Therefore \[ S_{2M-1}(x_M) \longrightarrow \frac2\pi\int_0^\pi\frac{\sin u}{u}\,\mathrm du = \frac2\pi\,\mathrm{Si}(\pi) = \frac{2\times1.8519371}{\pi} = 1.1789797 . \] The limiting value of \(f\) just right of the jump is \(+1\), so the overshoot in absolute terms is \(0.1789797\); the jump itself is \(1 - (-1) = 2\); so as a fraction of the jump the overshoot is \(0.1789797/2 = 0.0894899\), i.e. 8.9490 % (to four decimals, \(8.9490\%\); the next digits are \(8.948987\%\)). The same number occurs at every jump of every piecewise-\(C^1\) function: subtract \(\tfrac J2\) times a square wave with its jump at the same point, and what is left is continuous there, so its series converges uniformly near that point — by the uniform-convergence theorem of the slides when there is one jump per period, and in general by a uniform form of Riemann's localisation principle (cited: Zygmund, Trigonometric Series, vol. I, ch. II).

4. No contradiction, because the overshoot lives on a shrinking interval. The ripple occupies \(x \in (0, C/M)\) for a fixed \(C\) of order \(\pi\), on which \(|f - S_N| \le 0.18 + 2 = O(1)\); so its contribution to \(\lVert f - S_Nf\rVert_2^2\) is \(O(1)\times O(1/M) \to 0\). Mean-square convergence cares about area, and the overshoot has none in the limit.

Check. Dimensions: everything is a pure number ✓. Numbers: summing \(S_{2M-1}(\pi/2M)\) directly gives \(9.1164\%\) at \(M = 5\), \(8.9594\%\) at \(M = 20\), \(8.9494\%\) at \(M = 100\) and \(8.94899\%\) at \(M = 1000\) — approaching \(8.948987\%\) from above ✓. Special case: at \(x = \pi/2\), \(S_{2M-1}(\pi/2) = \frac4\pi(1 - \frac13 + \frac15 - \cdots) \to \frac4\pi\cdot\frac\pi4 = 1\) with no overshoot, because \(\pi/2\) is far from the jump ✓.

Another way

The whole phenomenon can be read off the Dirichlet kernel without any of this computation. \(S_Nf = f * D_N/2\pi\), and \(D_N\) is not nonnegative: it has a central spike of area about \(4\,\mathrm{Si}(\pi) = 7.408\), some \(18\) per cent more than \(2\pi\), surrounded by negative lobes. Convolving a step with a kernel whose positive part has area exceeding the total is exactly an overshoot, and its size is \(\frac2\pi\int_0^\pi\frac{\sin u}{u}\mathrm du - 1\), the relative excess area of that central lobe. The Fejér kernel, being nonnegative, has no excess and \(\sigma_N\) never overshoots — which is the practical cure.

Idea. All four Sturm–Liouville theorems are one identity plus one vanishing boundary term. Part 3 keeps the identity and destroys the boundary term, and part 4 watches the theorems fall in the order the proof predicts.

1. Since \(q\) is real, \(\bar v(qu) - u\overline{(qv)} = q u\bar v - q u\bar v = 0\), so the left-hand side is \(u(p\bar v{}')' - \bar v(pu')'\) — this is the first place reality of \(q\) is spent, and reality of \(p\) is spent in writing \(\overline{(pv')'} = (p\bar v{}')'\). Differentiating the right-hand side, \[ \big[p(u\bar v{}' - u'\bar v)\big]' = p'(u\bar v{}' - u'\bar v) + p\big(u'\bar v{}' + u\bar v{}'' - u''\bar v - u'\bar v{}'\big) = (p'\bar v{}' + p\bar v{}'')u - (p'u' + pu'')\bar v = (p\bar v{}')'u - (pu')'\bar v, \] which is the left-hand side. ∎

Integrating over \([a,b]\): \(\int_a^b(\mathcal Lu)\bar v - \int_a^bu\overline{(\mathcal Lv)} = \big[p(u\bar v{}' - u'\bar v)\big]_a^b\). And \(\langle\mathcal Au,v\rangle_w = \int_a^b(w^{-1}\mathcal Lu)\bar v\,w\,\mathrm dx = \int_a^b(\mathcal Lu)\bar v\,\mathrm dx\): the weight in the operator cancels the weight in the inner product exactly, which is why the eigenvalue problem must be written with the same \(w\) in both places.

2. At \(a\), both \((u(a),u'(a))\) and \((v(a),v'(a))\) satisfy \(\alpha\cdot(y,y') = 0\). Conjugating the second — legitimate only because \(\alpha\) is real — gives \(\alpha\cdot(\bar v(a),\bar v{}'(a)) = 0\). Two vectors of \(\mathbb C^2\) annihilated by the same nonzero real functional span at most a line, so \(\det\begin{pmatrix}u & \bar v\\ u' & \bar v{}'\end{pmatrix}(a) = u\bar v{}' - u'\bar v = 0\). Same at \(b\). Hence \(\langle\mathcal Au,v\rangle_w = \langle u,\mathcal Av\rangle_w\).

(i) If \(\mathcal Au = \lambda u\) with \(u \not\equiv 0\), then \(\lambda\lVert u\rVert_w^2 = \langle\mathcal Au,u\rangle_w = \langle u,\mathcal Au\rangle_w = \bar\lambda\lVert u\rVert_w^2\), and \(\lVert u\rVert_w^2 > 0\) because \(w > 0\), so \(\lambda = \bar\lambda\). (ii) If also \(\mathcal Av = \mu v\) with \(\mu \ne \lambda\), both real by (i), then \(\lambda\langle u,v\rangle_w = \langle\mathcal Au,v\rangle_w = \langle u,\mathcal Av\rangle_w = \bar\mu\langle u,v\rangle_w = \mu\langle u,v\rangle_w\), so \((\lambda - \mu)\langle u,v\rangle_w = 0\) and \(\langle u,v\rangle_w = 0\). ∎

3. For \(\lambda = 0\), \(u(0) = 0\) forces \(u = x\), and \(u'(1) = 1 \ne \mathrm i = \mathrm iu(1)\), so \(0\) is not an eigenvalue. Otherwise \(u(0) = 0\) forces \(u = \sin kx\) up to scale, with \(\lambda = k^2\), \(k \ne 0\). Then \(u'(1) = k\cos k\) and \(u(1) = \sin k\), so the condition \(u'(1) = \mathrm iu(1)\) reads \(k\cos k = \mathrm i\sin k\), i.e. \(\tan k = k/\mathrm i = -\mathrm ik\). If \(k\) is real and nonzero, \(\tan k\) is real while \(-\mathrm ik\) is purely imaginary and nonzero: impossible. If \(k = \mathrm is\) with \(s\) real and nonzero, \(\tan(\mathrm is) = \mathrm i\tanh s\) while \(-\mathrm ik = s\), and \(\mathrm i\tanh s = s\) needs both sides real and imaginary at once, so \(s = 0\). Hence no \(\lambda\) is real. Numerically, at \(k = 1.7792684 - 0.5455364\,\mathrm i\) one finds \(\tan k + \mathrm ik = 0\) to ten digits, and \(\lambda = k^2 = 2.8681861 - 1.9413113\,\mathrm i\).

4. Reality fails first, and orthogonality fails as a consequence — but only partly. The line that breaks is the conjugation step in part 2: the boundary functional at \(x = 1\) is \(\gamma = (-\mathrm i, 1)\), which is not real, so \(\gamma\cdot(v,v') = 0\) does not give \(\gamma\cdot(\bar v,\bar v{}') = 0\), the determinant does not vanish, and \(\mathcal A\) is not symmetric. With symmetry gone, step (i) has nothing to stand on and \(\lambda\) may be complex; and step (ii) used (i) in replacing \(\bar\mu\) by \(\mu\), so it fails too. Orthogonality fails because reality did: what survives for a non-self-adjoint problem is a weaker "bi-orthogonality" between the eigenfunctions of \(\mathcal A\) and those of its adjoint, which is not the same statement.

Check. Dimensions: with \(p = w = 1\), \(\lambda\) carries \(\mathrm L^{-2}\) and \(k\) carries \(\mathrm L^{-1}\), so \(\tan k = -\mathrm ik\) is consistent on the unit interval where lengths are pure numbers ✓. Limit: replace \(\mathrm i\) by a real \(\sigma\) in the boundary condition \(u'(1) = \sigma u(1)\). Then the condition is \(\tan k = k/\sigma\), every root of which gives a real \(\lambda\) (\(k\) real, or purely imaginary when \(\sigma > 1\)), and reality is restored ✓ — the failure is caused precisely by the \(\mathrm i\), not by the Robin form. Physical reading: \(\operatorname{Im}\lambda \ne 0\) means a complex frequency, i.e. energy leaving through the boundary, which is exactly what an absorbing boundary does ✓.

Another way

Part 4 can be seen without any algebra by an energy argument. For a real separated condition the boundary term \(p(u\bar u{}' - u'\bar u) = 2\mathrm i\,p\operatorname{Im}(u\bar u{}')\) vanishes, which says no energy flux crosses the ends. With \(u'(1) = \mathrm iu(1)\) the concomitant at \(x = 1\) is \(-2\mathrm i\,|u(1)|^2 \ne 0\): energy is being removed at a definite rate. A mode losing energy must decay, so its frequency must be complex — and the imaginary part of \(\lambda\) is that decay rate.

Idea. Legendre's equation is already self-adjoint; nothing has to be forced on it. The orthogonality that unit 10 obtained from Rodrigues' formula is really a statement that \(p = 1-x^2\) vanishes at both ends while the eigenfunctions do not blow up — and that is all.

1. Note \(\big((1-x^2)y'\big)' = (1-x^2)y'' - 2xy'\) exactly. So Legendre's equation is \(\big((1-x^2)y'\big)' + \lambda y = 0\), i.e. \[ -\big((1-x^2)y'\big)' = \lambda y : \qquad p(x) = 1-x^2,\quad q \equiv 0,\quad w \equiv 1 . \] Both endpoints are singular, because \(p(\pm1) = 0\). At each, no boundary condition is imposed; what replaces one is the demand that \(y\) stay bounded (with bounded derivative), which is enough to make \(p(u v' - u'v) \to 0\) there.

2. Write \(\lambda_\ell = \ell(\ell+1)\) and \(\lambda_n = n(n+1)\). Lagrange's identity for real solutions reads \(P_n\mathcal LP_\ell - P_\ell\mathcal LP_n = \big[p(P_\ell P_n' - P_\ell'P_n)\big]'\). The left side is \((\lambda_\ell - \lambda_n)P_\ell P_n\). Integrating over \([-1,1]\), \[ (\lambda_\ell - \lambda_n)\!\int_{-1}^{1}\!P_\ell P_n\,\mathrm dx = \Big[(1-x^2)\big(P_\ell P_n' - P_\ell'P_n\big)\Big]_{-1}^{1} = 0, \] because \(1-x^2\) vanishes at \(x = \pm1\) while \(P_\ell, P_n, P_\ell', P_n'\) are polynomials and hence finite there. Finally \(\lambda_\ell \ne \lambda_n\) when \(\ell \ne n\), since \(t \mapsto t(t+1)\) is strictly increasing on \([0,\infty)\). Divide. ∎

3. \(Q_0' = \tfrac12\big(\tfrac1{1+x} + \tfrac1{1-x}\big) = \tfrac{1}{1-x^2}\), so \((1-x^2)Q_0' = 1\) identically. With \(P_1 = x\) (and \(\lambda = 2\)) against \(Q_0\) (\(\lambda = 0\)), the concomitant is \[ (1-x^2)\big(P_1Q_0' - P_1'Q_0\big) = x\cdot 1 - (1-x^2)Q_0 = x - \tfrac{1-x^2}{2}\log\tfrac{1+x}{1-x}. \] As \(x \to 1^-\), \((1-x^2)\log\frac{1+x}{1-x} \to 0\) (the logarithm beats no power), so the concomitant \(\to 1\); as \(x \to -1^+\) it \(\to -1\). The bracket is therefore \(1 - (-1) = 2 \ne 0\). So admitting \(Q_0\) destroys symmetry: Lagrange's identity would then give \((\lambda_1 - 0)\int_{-1}^1 P_1Q_0\,\mathrm dx = 2\), and \(P_1\) and \(Q_0\) are not orthogonal despite having different eigenvalues. Boundedness at a singular endpoint is not tidiness; it is the hypothesis.

4. Feed a general \(\lambda\) into Legendre's equation and solve by power series about the ordinary point \(x = 0\) — unit 09's series method, and unit 10's derivation. Substituting \(y = \sum a_mx^m\) gives the two-term recurrence \(a_{m+2} = \frac{m(m+1) - \lambda}{(m+1)(m+2)}a_m\), so every solution is an even series plus an odd series, each converging on \((-1,1)\). A bounded solution has bounded even and odd parts \(\tfrac12(y(x) \pm y(-x))\), so it is enough to show that a non-terminating one of these series is unbounded near \(x = 1\). Write the ratio as \(\frac{m}{m+2}\big(1 - \frac{\lambda}{m(m+1)}\big)\). Once \(m(m+1) > \lambda\) every factor is positive, the coefficients keep one sign, and multiplying the ratios from some \(m_0\) on gives \(a_{m} = a_{m_0}\frac{m_0}{m}\prod\big(1 - \frac{\lambda}{k(k+1)}\big)\), the product over \(k = m_0, m_0 + 2, \dots, m - 2\). That product converges to a nonzero limit, because \(\sum\lambda/k(k+1) < \infty\), so \(|a_m| \ge c/m\) for some \(c > 0\). Then \(|y_{\text{part}}(x)| \ge c\sum_{m \ge m_0}x^m/m - (\text{a polynomial})\), which tends to \(\infty\) as \(x \to 1^-\), because \(\sum x^m/m\) over every other \(m\) does. So boundedness at \(x = \pm1\) forces each series to terminate or vanish, and termination needs \(m(m+1) = \lambda\) for some \(m\): \(\lambda = \ell(\ell+1)\) with \(\ell\) a non-negative integer, the solution then a polynomial of degree \(\ell\). The quantisation is the boundedness condition at the singular endpoints, read through the series.

Check. Dimensions: \(x\) is dimensionless and \(\lambda\) is a pure number ✓. Special cases: \(\int_{-1}^1P_0P_1 = \int_{-1}^1x\,\mathrm dx = 0\) ✓ and \(\int_{-1}^1P_0P_2 = \int_{-1}^1\tfrac12(3x^2-1)\mathrm dx = \tfrac12(2-2) = 0\) ✓. Nodes: \(P_3 = \tfrac12(5x^3-3x)\) has zeros at \(0, \pm\sqrt{3/5} = \pm0.7746\) — three of them, and \(\lambda_3 = 12\) is the fourth eigenvalue counting from \(\lambda_0 = 0\), so "index minus one" holds ✓.

Another way

For polynomials only, orthogonality also follows from degree counting plus one integration by parts: \(\int_{-1}^1 P_\ell\,x^m\mathrm dx = 0\) for \(m < \ell\) can be proved by integrating \(\big((1-x^2)P_\ell'\big)' = -\lambda_\ell P_\ell\) against \(x^m\) twice and inducting on \(m\). That gives orthogonality to every polynomial of lower degree, hence to \(P_n\) for \(n < \ell\). It is shorter for this family and it does not generalise — the Lagrange argument works verbatim for Bessel, Hermite, Laguerre and everything else, which is the point of having it.

Idea. Bessel's equation is a Sturm–Liouville problem whose weight is the polar area element and whose left endpoint is singular. Once that is said, every property of the drum's modes is a theorem already proved, and the only work left is arithmetic.

1. Divide by \(r\): \(rR'' + R' + (\lambda r - m^2/r)R = 0\). Since \(rR'' + R' = (rR')'\), this is \((rR')' - \frac{m^2}{r}R = -\lambda rR\), i.e. \[ -\big(rR'\big)' + \frac{m^2}{r}R = \lambda\,r\,R : \qquad p(r) = r,\quad q(r) = \frac{m^2}{r},\quad w(r) = r . \] The recipe agrees: \(A = r^2\), \(B = r\), so \(\exp\int(B/A)\mathrm dr = \exp\int\frac{\mathrm dr}r = r = p\) and \(\mu = A^{-1}p = 1/r\), which is exactly the division performed. The endpoint \(r = a\) is regular with the Dirichlet condition \(R(a) = 0\); \(r = 0\) is singular (\(p(0) = 0\), and \(q\) unbounded for \(m \ge 1\)), where boundedness discards \(Y_m\).

2. Because the area element in polar coordinates is \(\mathrm dA = r\,\mathrm dr\,\mathrm d\theta\), so the honest inner product of two radial functions on the disc is \(2\pi\int_0^a fg\,r\,\mathrm dr\) — the weight is the measure the geometry supplies, not a choice. Numerically, with \(a = 1\): \(\int_0^1J_0(j_{01}r)J_0(j_{02}r)\,r\,\mathrm dr = 0\) (to all thirty digits of a thirty-digit quadrature), while \(\int_0^1J_0(j_{01}r)J_0(j_{02}r)\,\mathrm dr = 0.19608\) — not remotely zero. Drop the weight and orthogonality is simply false.

3. Let \(u = J_m(\kappa r)\) and \(v = J_m(\kappa'r)\) with \(\kappa = j_{mk}/a \ne \kappa' = j_{ml}/a\). Both satisfy \(-(ru')' + \frac{m^2}{r}u = \kappa^2 ru\) and the same with \(\kappa'\), \(v\). Lagrange's identity integrated over \([\varepsilon, a]\) gives \[ (\kappa^2 - \kappa'^2)\!\int_\varepsilon^a\! uv\,r\,\mathrm dr = \Big[r\big(uv' - u'v\big)\Big]_\varepsilon^a , \] the sign from \(v\,\mathcal Lu - u\,\mathcal Lv = (\kappa^2 - \kappa'^2)\,r\,uv\). At \(r = a\): \(u(a) = J_m(j_{mk}) = 0\) and \(v(a) = J_m(j_{ml}) = 0\), so the bracket vanishes there. At \(r = \varepsilon\): unit 09's series gives \(J_m(s) = \frac{(s/2)^m}{m!}\big(1 + O(s^2)\big)\), so \(u, v = O(\varepsilon^m)\) and \(u', v' = O(\varepsilon^{m-1})\), whence \(\varepsilon(uv' - u'v) = O(\varepsilon^{2m}) \to 0\) for \(m \ge 1\); and for \(m = 0\), \(J_0 \to 1\) with \(J_0'(\kappa\varepsilon) = -\kappa J_1(\kappa\varepsilon) = O(\varepsilon)\), so the bracket is \(O(\varepsilon^2) \to 0\). Letting \(\varepsilon \to 0\), \((\kappa^2 - \kappa'^2)\int_0^a uv\,r\,\mathrm dr = 0\), and \(\kappa \ne \kappa'\). ∎

4. \(c = \sqrt{T/\sigma} = \sqrt{3200/0.265} = \sqrt{12075.47} = 109.888\ \mathrm{m\,s^{-1}}\), and \(f_{mk} = cj_{mk}/2\pi a\) with \(2\pi a = 0.94248\ \mathrm m\), so \(f_{mk} = 116.5953\,j_{mk}\ \mathrm{Hz}\).

The second partial is \(806.5 - 702 = \mathbf{104.5}\) cents above a perfect fifth — more than a semitone sharp of it. (It happens to lie \(6.5\) cents from a tempered minor sixth; but any single ratio lies within \(50\) cents of some tempered interval, so that is not the test.) The test is whether the partials sit near multiples of one frequency — a harmonic series would put the next three at \(1200\), \(1902\) and \(2400\) cents, and the drum's are at \(806.5\), \(1313.5\) and \(1438.5\). That is why the drum has no pitch: the ear finds no harmonic series to fit.

Check. Dimensions: \(c^2 = T/\sigma\) is \((\mathrm{N\,m^{-1}})/(\mathrm{kg\,m^{-2}}) = \mathrm{m^2\,s^{-2}}\) ✓, and \(cj/2\pi a\) is \(\mathrm{s^{-1}}\) ✓. Limit: unit 09's asymptotic \(j_{mk} \approx (k + \frac m2 - \frac14)\pi\) gives \(8.6394\) for \((0,3)\) against the true \(8.6537\), out by \(0.17\%\) — the high modes are string-like ✓. Special case: for \(m = 0\), \(q \equiv 0\) and the lower bound \(\lambda \ge \inf(q/w) = 0\) is satisfied by \(\lambda_{01} = (j_{01}/a)^2 > 0\) ✓.

Another way

Part 3 in one line, if one is willing to quote the theorem instead of re-deriving it: the problem is a singular Sturm–Liouville problem with \(p = w = r\), the concomitant vanishes at both ends (boundedness at \(0\), Dirichlet at \(a\)), and the theorem of the deck's "Real eigenvalues, orthogonal eigenfunctions" applies verbatim. The \(\varepsilon\)-argument above is what that theorem's singular case is, written out once so it never has to be again.

Idea. The solution space at a fixed \(\lambda\) is two-dimensional; a separated boundary condition at one end cuts it to one dimension; and one dimension cannot contain two independent functions. Periodic conditions do not cut at one end, so they cut nothing until both ends are used, and then they may cut nothing at all.

1. Let \(u\) and \(v\) both satisfy \(\mathcal Ly = \lambda wy\) and \(\alpha_1y(a) + \alpha_2y'(a) = 0\) with \(\alpha \ne 0\). The vectors \((u(a),u'(a))\) and \((v(a),v'(a))\) lie in \(\alpha^\perp\), a one-dimensional subspace of \(\mathbb R^2\). If \((u(a),u'(a)) = (0,0)\) then \(u \equiv 0\) by uniqueness and \(u\) is not an eigenfunction; so it is nonzero and spans \(\alpha^\perp\), giving \((v(a),v'(a)) = c\,(u(a),u'(a))\) for some scalar \(c\). Now \(v\) and \(cu\) solve the same linear second-order ODE with the same value and derivative at \(a\), so \(v = cu\) everywhere by Picard–Lindelöf (unit 01, applied to the equivalent first-order system; unit 06 states the linear case). Hence the eigenspace is spanned by \(u\). ∎

2. For \(\lambda = m^2\) with \(m \ge 1\), both \(\cos mx\) and \(\sin mx\) satisfy \(-y'' = m^2y\) and are \(2\pi\)-periodic with \(2\pi\)-periodic derivative, so both meet the boundary conditions; they are linearly independent; so the eigenspace contains a two-dimensional subspace, and by part 3 it is exactly two-dimensional. The proof of part 1 has nothing to start from: the periodic conditions constrain the pair \(\big((y(0),y'(0)),(y(2\pi),y'(2\pi))\big)\) jointly and place no restriction on \((y(0),y'(0))\) by itself — indeed for \(\lambda = m^2\) every initial vector works.

For \(\lambda = 0\) the general solution is \(y = A + Bx\); periodicity of \(y\) forces \(B = 0\), so the eigenspace is the constants, one-dimensional. The periodic problem therefore has a simple lowest eigenvalue and doubled ones thereafter, which is exactly the drum's angular spectrum.

3. Fix \(\lambda\). The set of solutions of \(\mathcal Ly = \lambda wy\) on \([a,b]\) is a vector space of dimension exactly two: the map \(y \mapsto (y(a), y'(a))\) into \(\mathbb R^2\) is linear, injective by uniqueness of the initial value problem, and surjective by existence (Picard–Lindelöf, unit 01; the equation is linear with continuous coefficients, so the solution exists on all of \([a,b]\)). An eigenspace is a subspace of that solution space that additionally satisfies the boundary conditions, hence has dimension at most two. ∎ (The same counting says an \(n\)-th-order problem has multiplicity at most \(n\).)

4. \((pW)' = (p(uv' - u'v))' = u(pv')' - v(pu')'\), by the expansion used in Lagrange's identity. Both \(u\) and \(v\) satisfy \((py')' = (q - \lambda w)y\), so the right side is \(u(q-\lambda w)v - v(q-\lambda w)u = 0\). Hence \(pW\) is constant — which is Abel's identity of unit 06, specialised to this form. Evaluate it at \(x = a\): the separated condition makes \((u,u')(a)\) and \((v,v')(a)\) parallel, so \(W(a) = 0\) and the constant is \(0\). Then \(W \equiv 0\) (since \(p > 0\)), so \(u\) and \(v\) are linearly dependent — which is part 1 again.

That the constant is zero is the vanishing of the bilinear concomitant \([u,v]\) in the real case with \(u, v\) sharing an eigenvalue. So the three facts — simplicity, self-adjointness, and Abel's identity — are one computation seen from three sides.

Check. Dimensions: \(pW\) carries \([p][u]^2\mathrm L^{-1}\) throughout, and a constant is dimensionally consistent ✓. Special case: for \(p = w = 1\), \(q = 0\), Dirichlet on \([0,L]\), the eigenspace of \(\lambda_n = (n\pi/L)^2\) is spanned by \(\sin(n\pi x/L)\) alone — \(\cos(n\pi x/L)\) has the same \(\lambda\) but fails \(y(0) = 0\), so it is excluded, exactly as part 1 predicts ✓. Number: for the periodic problem the eigenvalue list is \(0, 1, 1, 4, 4, 9, 9, \dots\), so the \(n\)-th distinct eigenvalue is \((n-1)^2\) and the count of eigenfunctions up to \(\lambda = 9\) is \(7\) ✓.

Another way

Part 2 is also a symmetry statement, and that is the version worth remembering. The periodic problem is invariant under the rotation \(x \mapsto x + \theta\), which maps eigenfunctions to eigenfunctions of the same eigenvalue; for \(m \ge 1\) it maps \(\cos mx\) to \(\cos m(x+\theta) = \cos m\theta\cos mx - \sin m\theta\sin mx\), genuinely new, so the eigenspace must be at least two-dimensional. For \(\lambda = 0\) the eigenfunction is constant and the rotation does nothing, so nothing is forced. Degeneracy usually has a symmetry behind it (the exceptions are called accidental) — which is why perturbing the symmetry splits the pair, and why a coin stuck to a drum makes it beat.

Idea. The oscillation theorem makes the node count a label rather than a calculation: the \(n\)-th eigenfunction has \(n-1\) interior zeros, always. Everything else here is the same statement seen in a different geometry or used to compare two problems.

1. \(u(0) = 0\) forces \(u = \sin kx\), \(\lambda = k^2\). The condition at \(x=1\) is \(\sin k + k\cos k = 0\), i.e. \(\tan k = -k\). Solving numerically: \(k_1 = 2.028758\), \(k_2 = 4.913180\), \(k_3 = 7.978666\), so \[ \lambda_1 = 4.115858,\qquad \lambda_2 = 24.139342,\qquad \lambda_3 = 63.659107 . \] \(u_3 = \sin(7.978666\,x)\) vanishes where \(7.978666\,x = \pi\) or \(2\pi\), i.e. at \(x = 0.393749\) and \(x = 0.787498\); the next would be at \(x = 1.18\), outside. So two interior zeros — exactly \(n-1 = 2\), as the oscillation theorem demands ✓.

2. \(J_2(j_{23}r/a)\) vanishes inside the disc where \(j_{23}r/a\) equals \(j_{21}\) or \(j_{22}\), i.e. at two radii: two nodal circles (the rim is a third zero but is the boundary, not an interior node). That count is the oscillation theorem applied to the radial Sturm–Liouville problem, whose third eigenfunction has \(3-1 = 2\) interior zeros. \(\cos2\theta\) vanishes at \(\theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4\) — four rays, forming two nodal diameters, which is the angular problem's own count.

3. Suppose \(u\) had two zeros \(x_1 < x_2\) in \([a,b]\); take them consecutive (zeros of a nontrivial solution are isolated, since a double zero would force \(u \equiv 0\) by uniqueness), so \(u \ne 0\) on \((x_1,x_2)\), and normalise \(u > 0\) there. Multiply \(-(pu')' + qu = 0\) by \(u\) and integrate over \([x_1,x_2]\): \[ 0 = \int_{x_1}^{x_2}\!\big(-(pu')'u + qu^2\big) = \big[-pu'u\big]_{x_1}^{x_2} + \int_{x_1}^{x_2}\!\big(pu'^2 + qu^2\big) = \int_{x_1}^{x_2}\!\big(pu'^2 + qu^2\big), \] the boundary term vanishing because \(u(x_1) = u(x_2) = 0\). But then \(\int pu'^2 = -\int qu^2 \le 0\) since \(q \ge 0\), while \(p > 0\) forces \(\int pu'^2 \ge 0\); so both are zero, hence \(u' \equiv 0\) on \((x_1,x_2)\), hence \(u\) is constant there and equal to \(u(x_1) = 0\), contradicting \(u > 0\). ∎

When \(p \equiv 1\) the equation is \(u'' + Qu = 0\) with \(Q = -q \le 0\), and this is Sturm comparison against \(v'' + 0\cdot v = 0\): if \(Q \not\equiv 0\) on \([x_1,x_2]\), the solution \(v \equiv 1\) would have to vanish between two consecutive zeros of \(u\), which it does not; and if \(Q \equiv 0\) there, \(u\) is a straight line with two zeros, so \(u \equiv 0\). The sign matters: with \(q \le 0\) the statement is false — \(q = -1\), \(p = 1\) gives \(u = \sin x\), with a zero every \(\pi\).

4. Fundamental. The Rayleigh quotient is \(R[u] = \int_0^LTu'^2\big/\int_0^L\rho u^2\). Doubling \(\rho\) halves \(R[u]\) for every trial \(u\), so \(\lambda_1\) is halved and \(\omega_1 = \sqrt{\lambda_1}\) falls by \(\sqrt2\): the heavier string has the lower fundamental. (It also follows from \(\omega_1 = \pi\sqrt{T/\rho}/L\), but the Rayleigh argument does not need the spectrum.)

Zeros. Neither: both third eigenfunctions have exactly two interior zeros. The oscillation theorem counts by index, and the index does not know about \(\rho\). Changing the density moves the eigenvalues and the shapes; it cannot move the node count, which is why counting nodes is how one identifies a measured resonance.

Check. Dimensions: lengths in part 1 are scaled to the unit interval, so \(k\) and \(\lambda\) are pure numbers ✓. Limit: a Robin condition sits between Neumann, whose roots are \((n-\tfrac12)\pi\), and Dirichlet, whose roots are \(n\pi\), so \(k_n\) should lie in \(\big((n-\tfrac12)\pi,\ n\pi\big)\) and approach the lower end as \(n\) grows. It does: \(k_1 = 2.0288 \in (1.571, 3.142)\), \(k_2 = 4.9132 \in (4.712, 6.283)\), \(k_3 = 7.9787 \in (7.854, 9.425)\), with gaps \(0.458\), \(0.201\), \(0.125\) shrinking ✓. Number: \(\lambda_3/\lambda_1 = 63.659/4.1159 = 15.47\), against \((k_3/k_1)^2 = (3.9328)^2 = 15.47\) ✓.

Another way

Part 1's node count without solving anything. The eigenfunction is \(\sin k_3x\) whatever \(k_3\) is, and \(k_3\) lies strictly between the third Neumann-type root \(\tfrac52\pi = 7.854\) and the third Dirichlet root \(3\pi = 9.425\). On \((0,1)\), \(\sin kx\) has \(\lfloor k/\pi\rfloor\) zeros, and \(k_3/\pi \in (2.5, 3)\) gives \(\lfloor\cdot\rfloor = 2\). The interval is enough; the value is not needed.

Idea. The lowest eigenvalue is the minimum of a quotient, so any admissible guess overestimates it, and a one-parameter family lets you minimise within the guess. The accuracy is spectacular because \(\lambda_1\) is a stationary value.

1. The equation is \(-(1\cdot u')' + 0\cdot u = \lambda(1+x)u\), so \(p \equiv 1\), \(q \equiv 0\), \(w(x) = 1+x\). Both endpoints are regular with Dirichlet conditions, so the boundary term in the Rayleigh identity vanishes and \[ R[u] = \frac{\int_0^1 u'^2\,\mathrm dx}{\int_0^1 (1+x)\,u^2\,\mathrm dx}. \]

2. Numerator. \(u' = \pi\cos\pi x + 2\pi\beta\cos2\pi x\), and \(\int_0^1\cos^2\pi x = \int_0^1\cos^22\pi x = \tfrac12\) while \(\int_0^1\cos\pi x\cos2\pi x = \tfrac12\int_0^1[\cos\pi x + \cos3\pi x] = 0\). So \(\int_0^1u'^2 = \tfrac{\pi^2}{2}\big(1 + 4\beta^2\big)\).

Denominator. Three integrals, using \(\int_0^1\sin^2k\pi x = \tfrac12\), \(\int_0^1 x\sin^2k\pi x = \tfrac14\) and \(\int_0^1x\cos k\pi x\,\mathrm dx = \big((-1)^k - 1\big)/(k\pi)^2\):

• \(\int_0^1(1+x)\sin^2\pi x = \tfrac12 + \tfrac14 = \tfrac34\);
• \(\int_0^1(1+x)\sin^22\pi x = \tfrac12 + \tfrac14 = \tfrac34\);
• \(\int_0^1(1+x)\sin\pi x\sin2\pi x = \int_0^1 x\cdot\tfrac12[\cos\pi x - \cos3\pi x]\,\mathrm dx = \tfrac12\Big(\tfrac{-2}{\pi^2} - \tfrac{-2}{9\pi^2}\Big) = -\tfrac{8}{9\pi^2}\) (the \(1\) contributes nothing, since \(\int_0^1\cos k\pi x = 0\)).

Hence \(\int_0^1(1+x)u^2 = \tfrac34(1+\beta^2) + 2\beta\big(-\tfrac{8}{9\pi^2}\big) = \tfrac34(1+\beta^2) - \tfrac{16\beta}{9\pi^2}\), which is the stated \(R(\beta)\). ∎

3. \(R(0) = \dfrac{\pi^2/2}{3/4} = \dfrac{2\pi^2}{3} = 6.579736\), against the true \(\lambda_1 = 6.548395\): high by \(0.0313\), i.e. 0.479 %.

Minimising \(R\) over \(\beta\) numerically gives \(\beta_{\min} = -0.0397748\) and \(R(\beta_{\min}) = 6.548459\), high by \(6.3\times10^{-5}\), i.e. 0.00097 % — one part in a hundred thousand, from a two-term guess.

4. Guaranteed above, because \(\lambda_1 = \min_{u\ne0}R[u]\) over admissible \(u\) and our \(u\) is admissible (it is smooth and vanishes at both ends). So \(R[u] \ge \lambda_1\) for every choice of \(\beta\); no trial function can ever undershoot.

The error is small because \(u_1\) is a stationary point of \(R\), not merely a minimum-value point: writing \(u = u_1 + \varepsilon\eta\) with \(\eta \perp_w u_1\) and \(\lVert u_1\rVert_w = 1\), \(R[u] = \lambda_1 + \varepsilon^2\big(\langle\mathcal A\eta,\eta\rangle_w - \lambda_1\lVert\eta\rVert_w^2\big) + O(\varepsilon^4)\), with no first-order term. A shape wrong by \(\varepsilon\) gives an eigenvalue wrong by \(\varepsilon^2\). Here the one-term trial differs from the true \(u_1\) by about \(4\%\) in shape (the size of \(\beta_{\min}\)), and \(0.04^2 \approx 1.6\times10^{-3}\) — the right order for the \(0.48\%\) observed.

Check. Dimensions: with \(T = L = \rho_0 = 1\) everything is a pure number; restoring units, the problem is \(-(Tu')' = \lambda\rho u\), so \(R = \int Tu'^2/\int\rho u^2\) carries \(\mathrm N/(\mathrm{kg\,m^{-1}}\cdot\mathrm m^2) = \mathrm s^{-2}\): here \(\lambda\) is \(\omega^2\) itself ✓. Special case: set \(w \equiv 1\) (uniform string) and the same one-term trial gives \(R = (\pi^2/2)/(1/2) = \pi^2\), the exact answer, because \(\sin\pi x\) is then the exact eigenfunction ✓. Limit: replacing \(w = 1+x\) by its mean value \(3/2\) gives \(\lambda_1 \approx \pi^2/1.5 = 6.5797\) — which is exactly \(R(0)\), so the one-term bound is "the uniform string with the average density", a reading worth keeping ✓.

Another way

Minimising \(R(\beta)\) over a two-dimensional trial space is the Rayleigh–Ritz method, and it need not be done by calculus on a quotient: setting \(K_{ij} = \int_0^1\phi_i'\phi_j'\) and \(M_{ij} = \int_0^1(1+x)\phi_i\phi_j\) for \(\phi_1 = \sin\pi x\), \(\phi_2 = \sin2\pi x\) turns the problem into the generalised eigenvalue problem \(Kc = \lambda Mc\) of unit 07, with \(K = \tfrac{\pi^2}{2}\mathrm{diag}(1,4)\) and \(M = \begin{pmatrix}3/4 & -8/9\pi^2\\ -8/9\pi^2 & 3/4\end{pmatrix}\). Its smaller root is \(6.548459\), the same number, and the eigenvector is \((1, -0.03977)\). Adding more trial functions enlarges the pencil and the bound falls monotonically — which is the finite element method in miniature.

Idea. Take a perfectly good complete system, delete half of it, and watch every theorem that depended on completeness fail while every theorem that depended only on orthogonality survives. It is the cleanest way to see which is which.

1. For \(m \ne n\), \(\int_0^\pi\sin2mx\sin2nx\,\mathrm dx = \tfrac12\int_0^\pi\big[\cos2(m-n)x - \cos2(m+n)x\big]\mathrm dx\), and \(\int_0^\pi\cos2kx\,\mathrm dx = \sin(2k\pi)/2k = 0\) for every nonzero integer \(k\). So the system is orthogonal. For \(m = n\) the first cosine is \(1\), giving \(\lVert\sin2nx\rVert^2 = \tfrac12\cdot\pi = \pi/2\). ✓

2. \(\int_0^\pi\sin x\sin2nx\,\mathrm dx = \tfrac12\int_0^\pi\big[\cos(2n-1)x - \cos(2n+1)x\big]\mathrm dx = \tfrac12\Big[\frac{\sin(2n-1)\pi}{2n-1} - \frac{\sin(2n+1)\pi}{2n+1}\Big] = 0\), since \(2n\pm1\) are integers and \(\sin(k\pi) = 0\). ∎

So \(f = \sin x\) is a nonzero element of \(PC[0,\pi]\) — indeed a smooth one, vanishing at both endpoints — orthogonal to every member of \(\mathcal E\). Every coefficient \(\hat f_n\) is then zero, so every partial sum \(S_Nf\) is the zero function and \(\lVert f - S_Nf\rVert = \lVert f\rVert > 0\) for all \(N\): the definition of completeness fails at \(f\), and \(\mathcal E\) is not complete. (This is the safe direction of the orthogonality test; the slides' notes explain why the other direction needs care in \(PC\).)

3. \(\hat f_n = \langle f, e_n\rangle/\lVert e_n\rVert^2 = 0/(\pi/2) = 0\) for every \(n\). So the left side of Bessel's inequality is \(\sum_n\hat f_n^2\lVert e_n\rVert^2 = 0\), while \(\lVert f\rVert^2 = \int_0^\pi\sin^2x\,\mathrm dx = \pi/2 = 1.5707963\). Bessel's inequality holds — \(0 \le \pi/2\) — and Parseval fails by the whole of \(\lVert f\rVert^2\), a gap of \(\pi/2\). The "expansion" of \(\sin x\) in this system is the zero function, and \(\lVert f - S_Nf\rVert = \lVert f\rVert = \sqrt{\pi/2} = 1.2533141\) for every \(N\): it does not converge to \(f\) at all.

4. \(\{\sin2nx\}\) is the Dirichlet eigenfunction system of \(-u'' = \lambda u\) on \(\big[0, \tfrac\pi2\big]\), with eigenvalues \(\lambda_n = (2n)^2\) — set \(L = \pi/2\) in \(\sin(n\pi x/L)\). On that interval it is complete, and every theorem of the unit applies.

On \([0,\pi]\) it is not the full eigenfunction system of any regular Sturm–Liouville problem — if it were, the cited expansion theorem would make it complete, contradicting part 2 — and the hypothesis being violated is the one word "all": the expansion theorem asserts completeness for the full set of eigenfunctions of one problem on one interval, and \(\{\sin2nx\}\) is only the even-indexed half of \(\{\sin nx\}\), the Dirichlet system on \([0,\pi]\). Deleting the odd-indexed members deletes exactly the directions that \(\sin x\) lives in. Nothing about the surviving members changed — they are still orthogonal, still eigenfunctions of \(-\mathrm d^2/\mathrm dx^2\) with Dirichlet conditions at \(0\) and \(\pi\) — which is why orthogonality is no evidence at all for completeness.

Check. Dimensions: all pure numbers ✓. Special case: restore the missing half. Then \(\{\sin nx\}_{n\ge1}\) is complete on \([0,\pi]\), \(\sin x\) has coefficients \((1,0,0,\dots)\), and Parseval reads \(\pi/2 = 1^2\cdot\pi/2\) ✓ — the gap closes exactly, and it closes on the one term that was deleted. Number: the partial sums of the deficient expansion give mean-square error \(\pi/2 = 1.5708\) at every \(N\), against the complete system's \(0\) for \(N \ge 1\) ✓.

Another way

A faster route to part 2, and one that explains rather than computes: on \([0,\pi]\), every \(\sin2nx\) is antisymmetric about the midpoint \(x = \pi/2\), because \(\sin2n(\pi - x) = -\sin2nx\). And \(\sin x\) is symmetric there: \(\sin(\pi - x) = \sin x\). A symmetric function and an antisymmetric one are automatically orthogonal on an interval symmetric about that midpoint, so the integral vanishes without any trigonometry. The deleted half of the system is exactly the symmetric half, and \(\sin x\) is the cheapest witness to its absence.

Idea. The same eigenfunctions as the string, because the same operator with the same boundary conditions; only the time factor differs, and the exponential in \(n^2\) is what makes diffusion smooth and irreversible.

1. With \(\theta = X(x)T(t)\), \(T'/\kappa T = X''/X = -\lambda\). The spatial problem with \(X(0) = X(L) = 0\) gives \(\lambda_n = (n\pi/L)^2\) and \(X_n = \sin(n\pi x/L)\), and \(T_n = \mathrm e^{-\kappa\lambda_nt}\). So \[ \theta(x,t) = \sum_{n\ \text{odd}}\frac{4\theta_0}{n\pi}\sin\frac{n\pi x}{L}\,\mathrm e^{-t/\tau_n}, \qquad \tau_n = \frac{1}{\kappa\lambda_n} = \frac{L^2}{\kappa n^2\pi^2} = \frac{\tau_1}{n^2}, \] with \(\alpha_n = 4\theta_0/n\pi\) for odd \(n\) and \(0\) for even (a uniform profile is symmetric about the midpoint, so it excites no antisymmetric mode).

2. \(\tau_1 = \dfrac{L^2}{\kappa\pi^2} = \dfrac{1}{1.11\times10^{-4}\times9.8696} = 912.8\ \mathrm s = \mathbf{15.21\ minutes}\), and \(\tau_{10} = \tau_1/100 = \mathbf{9.128\ s}\). The tenth harmonic is gone a hundred times sooner than the first.

3. At \(x = L/2\), \(\sin(n\pi/2) = (-1)^{(n-1)/2}\) for odd \(n\), so \[ \theta(L/2, t) = \frac{4\theta_0}{\pi}\sum_{k\ge0}\frac{(-1)^k}{2k+1}\,\mathrm e^{-(2k+1)^2t/\tau_1}. \]

• \(t = 0\): the series is \(\frac{4\theta_0}{\pi}(1 - \tfrac13 + \tfrac15 - \cdots) = \theta_0 = \mathbf{100.0}\,{}^{\circ}\mathrm{C}\), by Leibniz — but it converges very slowly. For an alternating series with decreasing terms the error after \(N\) terms is below the first omitted term, \(\frac{400}{\pi(2N+1)}\,{}^{\circ}\mathrm{C}\), so being sure of three figures (error under \(0.5\,{}^{\circ}\mathrm{C}\)) takes 127 terms, and being sure of \(100.0\) to the first decimal takes \(1273\). This is the discontinuity at the ends making itself felt, and the right answer here is to sum the series in closed form rather than numerically.
• \(t = 100\ \mathrm s\): \(t/\tau_1 = 0.10955\), so the factors are \(\mathrm e^{-0.1096} = 0.8962\), \(\mathrm e^{-0.9860} = 0.3731\), \(\mathrm e^{-2.739} = 0.0646\), \(\mathrm e^{-5.368} = 0.00466\), \(\mathrm e^{-8.874} = 1.4\times10^{-4}\). Summing, \(\frac{400}{\pi}\big(0.8962 - \tfrac{0.3731}{3} + \tfrac{0.0646}{5} - \tfrac{0.00466}{7} + \tfrac{0.00014}{9}\big) = \frac{400}{\pi}(0.784155) = \mathbf{99.8}\,{}^{\circ}\mathrm{C}\) — five terms suffice, and four give the same three figures.
• \(t = \tau_1 = 912.8\ \mathrm s\): factors \(\mathrm e^{-1} = 0.36788\), \(\mathrm e^{-9} = 1.234\times10^{-4}\), \(\mathrm e^{-25} = 1.4\times10^{-11}\). So \(\theta(L/2,\tau_1) = \frac{400}{\pi}\big(0.36788 - \tfrac{1.234\times10^{-4}}{3}\big) = \mathbf{46.8}\,{}^{\circ}\mathrm{C}\) — one term already gives three figures (\(46.840\) against the two-term \(46.835\)), and the second is below the last figure kept.

The number of terms needed collapses from "infinitely many" to two within one decay time. That collapse is the smoothing.

4. The coefficient of mode \(n\) at time \(t\) is \(\frac{4\theta_0}{n\pi}\mathrm e^{-n^2t/\tau_1}\). For any fixed \(t > 0\), and any \(k\), \(n^k\mathrm e^{-n^2t/\tau_1} \to 0\) faster than any power of \(1/n\) — the exponential beats every polynomial — so \(\sum_n n^k|\alpha_n|\mathrm e^{-n^2t/\tau_1} < \infty\) for every \(k\). Each \(x\)-derivative multiplies term \(n\) by at most \((n\pi/L)^k\), so the \(k\)-times differentiated series converges uniformly by the \(M\)-test, and the real-analysis theorem on uniformly convergent derivatives lets it be differentiated term by term, for every \(k\). So \(\theta(\cdot,t)\) has continuous derivatives of every order — the smoothness–decay dictionary read backwards: the profile is \(C^\infty\) at once, however rough the data.

Running backwards multiplies mode \(n\) by \(\mathrm e^{+n^2|t|/\tau_1}\), which amplifies a coefficient of size \(\epsilon\) in mode \(n\) by a factor \(\mathrm e^{n^2|t|/\tau_1}\). At \(|t| = \tau_1\) and \(n = 10\) that factor is \(\mathrm e^{100} = 2.7\times10^{43}\): an error in the data of one part in \(10^{40}\) becomes an error of thousands of degrees. No continuous dependence on the data means ill-posed in the Hadamard sense of unit 18 — and now with a rate attached.

Check. Dimensions: \([\kappa] = \mathrm{L^2T^{-1}}\), so \(L^2/\kappa\) is a time ✓; \(\alpha_n\) carries \(\theta_0\), a temperature ✓. Limit: as \(t \to \infty\), every term dies and \(\theta \to 0\), which is right for a rod with both ends held at zero ✓. Special case: at \(t = 0\) the series must return the constant \(100\), and the Leibniz sum \(\frac4\pi\cdot\frac\pi4 = 1\) says it does, at every interior point — while at \(x = 0\) and \(x = L\) it returns \(0\), the midpoint of the jump, exactly as Dirichlet's theorem predicts ✓.

Another way

For very short times the mode sum is useless and the right picture is the other one: on a timescale \(t \ll \tau_1\) the ends have only influenced a layer of thickness \(\sim\sqrt{\kappa t}\), so \(\theta(L/2,t) = \theta_0\) to any accuracy you like. Setting \(\sqrt{\kappa t} = L/2\) gives \(t \approx 2250\ \mathrm s\) as the time at which the ends are first felt at the middle — the same order as \(\tau_1 = 913\ \mathrm s\), which is the consistency one wants between a similarity argument and a mode sum. The two descriptions swap over at \(t \sim \tau_1\), and unit 20's Fourier transform is what makes the short-time one exact.