Boundary Values & Fourier Series

$y'' + \lambda y = 0$ with $y(0) = y(\pi) = 0$: the general solution $\sin(\sqrt\lambda\,x)$ meets the far boundary only when $\sqrt\lambda$ is an integer — eigenvalues $\lambda_n = n^2$ with eigenfunctions $\sin nx$. A guitar string of length $L$ can vibrate only at frequencies $\tfrac{n\pi c}{L}$: harmonics are a boundary-value theorem.

The flat function $f = 1$ becomes the square-wave series $\tfrac{4}{\pi}\left(\sin x + \tfrac{\sin 3x}{3} + \tfrac{\sin 5x}{5} + \cdots\right)$. Why: multiply $f = \sum b_m\sin mx$ by $\sin nx$ and integrate over $[0, \pi]$; orthogonality kills every term except $m = n$, which contributes $b_n\cdot\tfrac\pi2$. For $f = 1$: $b_n = \tfrac2\pi\int_0^\pi\sin nx\,dx = \tfrac{2(1 - \cos n\pi)}{n\pi}$, which is $\tfrac{4}{n\pi}$ for odd $n$ and $0$ for even $n$. Each partial sum is a smooth wave that must be $0$ at the ends, so near $x = 0$ and $x = \pi$ it overshoots before settling.

Wave ($u_{tt} = c^2u_{xx}$): each mode oscillates at frequency $nc$, or equivalently d'Alembert's $f(x - ct) + g(x + ct)$ — disturbances travel at speed $c$ unchanged. Expand the initial data, evolve each mode by its own simple rule, superpose: that is the whole method.

Why these formulas: on $[-L, L]$ the functions $1, \cos\tfrac{n\pi x}{L}, \sin\tfrac{n\pi x}{L}$ are mutually orthogonal, each cosine and sine has $\int_{-L}^L(\cdot)^2 = L$, but the constant has $\int_{-L}^L 1^2 = 2L$. Projecting $f$ onto each function therefore divides by $L$ — except the constant, which divides by $2L$; writing the constant term as $\tfrac{a_0}{2}$ lets one formula serve all $a_n$. Why the even/odd shortcut: for even $f$, $f\sin$ is odd and integrates to $0$ over a symmetric interval.

Step 1 — the constant: $a_0 = \dfrac2\pi\int_0^\pi x\,dx = \pi$, so the series starts $\dfrac{a_0}{2} = \dfrac\pi2$, the average value of $|x|$.

Step 2 — by parts, $\int_0^\pi x\cos nx\,dx = \left[\dfrac{x\sin nx}{n}\right]_0^\pi - \dfrac1n\int_0^\pi\sin nx\,dx = \dfrac{(-1)^n - 1}{n^2}$. So $a_n = \dfrac{2\big((-1)^n - 1\big)}{\pi n^2}$: zero for even $n$, $-\dfrac{4}{\pi n^2}$ for odd $n$.

Step 3 — $|x| = \dfrac\pi2 - \dfrac4\pi\left(\cos x + \dfrac{\cos 3x}{9} + \dfrac{\cos 5x}{25} + \cdots\right)$. The coefficients fall like $\tfrac1{n^2}$ because $|x|$ has corners but no jumps; two terms are already close.

Step 4 — a payoff: at $x = 0$, $0 = \dfrac\pi2 - \dfrac4\pi\sum_{k\ge0}\dfrac{1}{(2k+1)^2}$, so $1 + \dfrac19 + \dfrac1{25} + \cdots = \dfrac{\pi^2}{8}$.

Step 1 — $a_0 = \dfrac1\pi\int_0^\pi dx = 1$, so the constant term is $\tfrac12$. For $n \ge 1$, $a_n = \dfrac1\pi\int_0^\pi\cos nx\,dx = 0$.

Step 2 — $b_n = \dfrac1\pi\int_0^\pi\sin nx\,dx = \dfrac{1 - (-1)^n}{n\pi}$: $\dfrac{2}{n\pi}$ for odd $n$, $0$ for even. So $f \sim \dfrac12 + \dfrac2\pi\left(\sin x + \dfrac{\sin 3x}{3} + \dfrac{\sin 5x}{5} + \cdots\right)$.

Step 3 — at $x = 0$ every sine vanishes and the series gives $\tfrac12$: not $0$, not $1$, but the midpoint of the jump. This is the general rule for piecewise smooth $f$: the series converges to $\tfrac12\big(f(x^-) + f(x^+)\big)$.

Step 4 — near the jump every partial sum overshoots, and the overshoot does not shrink as terms are added: the peaks approach $\tfrac12 + \tfrac1\pi\mathrm{Si}(\pi) \approx 1.0895$, about $9\%$ of the jump (the Gibbs phenomenon). The peaks only move closer to the jump.

Why it works: each $\sin\tfrac{n\pi x}{L}$ times its time factor is an exact solution that meets the boundary conditions; the equations are linear, so sums of solutions are solutions; and orthogonality makes the expansion of the initial data unique, so the sum matches it. The steady state must be removed first because superposition only preserves boundary values of zero: adding two solutions with $u(\pi) = \pi$ would give $u(\pi) = 2\pi$.

Step 1 — attach each mode's time factor: $u(x, t) = \dfrac4\pi\displaystyle\sum_{n\text{ odd}}\frac{e^{-n^2t}\sin nx}{n}$.

Step 2 — at the centre $x = \tfrac\pi2$, $\sin\tfrac{n\pi}{2} = \pm1$ alternately for odd $n$: $u = \dfrac4\pi\left(e^{-t} - \dfrac{e^{-9t}}{3} + \dfrac{e^{-25t}}{5} - \cdots\right)$.

Step 3 — at $t = 1$: $\dfrac4\pi e^{-1} \approx 0.4684$, and the next term is smaller by the factor $\tfrac{e^{-8}}{3} \approx 10^{-4}$, so $u(\tfrac\pi2, 1) \approx 0.468$.

Step 4 — the lesson: at $t = 0$ the series needs infinitely many terms to make the flat profile (and still misses the corners), but after a short time one term describes the bar. Heat flow turns any initial shape into the fundamental mode $\sin x$, decaying like $e^{-t}$.

Step 1 — the initial shape fixes the cosine parts: at $t = 0$, $u = \sum A_n\sin nx = \sin 2x$, so $A_2 = 1$ and all other $A_n = 0$.

Step 2 — the initial velocity fixes the sine parts: $u_t(x, 0) = \sum 2nB_n\sin nx = 3\sin x$, so $2B_1 = 3$, $B_1 = \tfrac32$.

Step 3 — $u = \sin 2x\cos 4t + \tfrac32\sin x\sin 2t$. Check: each term satisfies $u_{tt} = 4u_{xx}$ (for the first, $-16 = 4\cdot(-4)$; for the second, $-4 = 4\cdot(-1)$), and both vanish at $x = 0, \pi$.

Step 4 — the motion repeats with period $\pi$, the common period of $\cos 4t$ and $\sin 2t$. At $t = \tfrac\pi8$ the $\sin 2x$ part is momentarily flat and the string is $\tfrac{3}{2\sqrt2}\sin x$; at $t = \tfrac\pi4$ it is $-\sin 2x + \tfrac32\sin x$. Unlike heat, nothing decays: each mode keeps its energy forever.

Step 1 — the difference $w = u - v$ has zero boundary values and satisfies the same equation (since $v_t = v_{xx} = 0$), with $w(x, 0) = 0 - x = -x$.

Step 2 — expand $-x$ in sines: from the worked problem below, $x = 2\sum(-1)^{n+1}\tfrac{\sin nx}{n}$, so $w(x, 0) = -2\sum(-1)^{n+1}\tfrac{\sin nx}{n}$.

Step 3 — attach decay factors and add back the steady state: $u = x - 2\displaystyle\sum_{n\ge1}(-1)^{n+1}\frac{e^{-n^2t}\sin nx}{n}$.

Step 4 — as $t \to \infty$ every mode dies and $u \to x$, the straight-line profile that conducts heat steadily from the hot end to the cold end. The slowest mode, $n = 1$, sets the time scale: the gap at the centre shrinks like $2e^{-t}$.

Hence $\int_0^\pi f\sin nx\,dx = b_n\cdot\tfrac\pi2$ when $f = \sum b_m\sin mx$ — every other term integrates away, isolating $b_n$. $\blacksquare$

This splits the PDE into $X'' + \lambda X = 0$ (the boundary-value problem, quantizing $\lambda = n^2$) and $T' = -k\lambda T$ (exponential decay $e^{-kn^2t}$). Superposition assembles the general solution because the heat equation is linear. $\blacksquare$

Step 1 — separation gives modes $\sin nx$ (the boundary conditions quantize to integer $n$), each decaying like $e^{-2n^2t}$ (the $2$ is the diffusivity).

Step 2 — the initial data is already a finite Fourier sine series: $b_1 = 3$, $b_3 = -1$, all others $0$ — no coefficient integrals needed.

Step 3 — evolve each mode: $u(x,t) = 3e^{-2t}\sin x - e^{-18t}\sin 3x$.

Step 4 — check: $u_t = -6e^{-2t}\sin x + 18e^{-18t}\sin 3x$ and $2u_{xx} = 2(-3e^{-2t}\sin x + 9e^{-18t}\sin 3x)$ — equal ✓; ends: $\sin 0 = \sin 3\pi = 0$ ✓.

Step 5 — by $t = 0.5$ the $\sin 3x$ ripple has shrunk by $e^{-9} \approx 0.01\%$ while the fundamental keeps $e^{-1} \approx 37\%$: the bar forgets its fine structure almost instantly.

Step 1 — $b_n = \dfrac2\pi\displaystyle\int_0^\pi x\sin nx\,dx$.

Step 2 — integrate by parts with $u = x$, $dv = \sin nx\,dx$: $\left[-\dfrac{x\cos nx}{n}\right]_0^\pi + \dfrac1n\int_0^\pi\cos nx\,dx = -\dfrac{\pi\cos n\pi}{n} + 0 = -\dfrac{\pi(-1)^n}{n}$.

Step 3 — so $b_n = \dfrac2\pi\cdot\dfrac{\pi(-1)^{n+1}}{n} = \dfrac{2(-1)^{n+1}}{n}$: $x = 2\left(\sin x - \dfrac{\sin 2x}{2} + \dfrac{\sin 3x}{3} - \cdots\right)$.

Step 4 — a famous payoff: at $x = \tfrac\pi2$ the series gives $\dfrac\pi2 = 2\left(1 - \dfrac13 + \dfrac15 - \cdots\right)$ — the Leibniz formula for $\pi$, falling out of a heat-flow calculation.