Part C · unit 20 · problems

The Fourier transform, complex analysis, and why causality is a statement about poles

the slides · the curriculum

Checkpoint. On paper and without notes, you compute a retarded Green's function by contour integration from its frequency response and prove it vanishes for \(t \lt 0\) by closing the contour in the half-plane with no poles and bounding the arc, not by inspecting the answer; you evaluate a real integral by steepest descent, justifying every arc and connecting segment by an explicit estimate; and you settle unit 09's Laplace limit — Lagrange's series for Kepler's equation converges for every \(M\) when \(e \lt 0.6627434\) and diverges at \(M = \pi/2\) when \(e \gt 0.6627434\).

Work each problem on paper first. Under it, Hints opens a ladder one rung at a time — each rung appears only once the one before it is open, and the last gives the first line of the working — and Worked solution has the whole thing: the idea, the working, then a check. Read the solution after your attempt, including when you got it right. F folds everything back; D switches dark and light.

Twelve problems, about seven hours. Work them on paper first. Conventions as in the deck: \(\hat f(\omega) = \int f(t)e^{i\omega t}\,\mathrm dt\), \(f(t) = \frac1{2\pi}\int\hat f(\omega)e^{-i\omega t}\,\mathrm d\omega\), a single frequency is \(e^{-i\omega t}\), and the residue theorem carries \(+2\pi i\) anticlockwise.

The running example: unit 06's galvanometer, \(\ddot\theta + \gamma\dot\theta + \omega_0^2\theta = \tau(t)/I\), with \(\omega_0 = 8\ \mathrm{s^{-1}}\), \(\gamma = 0.4\ \mathrm{s^{-1}}\), \(\omega_\mathrm d = \sqrt{\omega_0^2 - \gamma^2/4} = 7.9974996\ \mathrm{s^{-1}}\), susceptibility \(\chi(\omega) = 1/(\omega_0^2 - \omega^2 - i\gamma\omega)\) with poles \(\omega_\pm = \pm\omega_\mathrm d - i\gamma/2\), and Green's function \(G(t) = e^{-\gamma t/2}\sin(\omega_\mathrm dt)/\omega_\mathrm d\) for \(t \gt 0\), zero before. Its coil has moment of inertia \(I = 1.0\times10^{-8}\ \mathrm{kg\,m^2}\) and turns a current into a torque \(\alpha i\) with \(\alpha = 2.0\times10^{-3}\ \mathrm{N\,m\,A^{-1}}\).

The transform and distributions

P1 Three transforms and a sum routine

exercises: transforms from the definition, the inversion theorem, the convolution theorem, a number

  1. Compute the transforms of \(f_1(t) = e^{-a|t|}\) (\(a \gt 0\)) and of the box \(f_2 = \mathbf 1_{[-1,1]}\).
  2. Check the inversion theorem for \(f_1\) at \(t = 0\), evaluating \(\frac1{2\pi}\int\hat f_1\,\mathrm d\omega\) directly.
  3. Show that \(f_2 * f_2\) is the triangle \(\Lambda(t) = \max(2 - |t|, 0)\), and write down \(\hat\Lambda\) without integrating.
  4. Apply the inversion theorem to \(\Lambda\) at \(t = 0\) to evaluate \(\displaystyle\int_{-\infty}^\infty\frac{\sin^2x}{x^2}\,\mathrm dx\). Why may the theorem be applied to \(\Lambda\) but not to \(f_2\)?
Hints · 3 rungs
Rung 1

Two functions of \(\omega\), then a number, then a function of \(t\) and its transform, then a number that should look familiar. Only parts 1 and 3 need an integral computed from a definition.

Rung 2

Split \(f_1\)'s integral at \(0\) and use \(\int_0^\infty e^{-ct}\mathrm dt = 1/c\) for \(\operatorname{Re}c \gt 0\). For part 3, the convolution theorem says \(\hat\Lambda = \hat f_2^{\,2}\). The inversion theorem wants \(f\) continuous and both \(f\) and \(\hat f\) absolutely integrable.

Rung 3

First line: \(\hat f_1(\omega) = \int_0^\infty e^{-(a - i\omega)t}\mathrm dt + \int_{-\infty}^0e^{(a + i\omega)t}\mathrm dt = \frac{1}{a - i\omega} + \frac{1}{a + i\omega}\). For part 4, \(\Lambda(0) = 2\) and \(\hat\Lambda(\omega) = 4\sin^2\omega/\omega^2\).

Worked solution

Idea. Compute two transforms from the definition; then let the convolution theorem build a third, smoother one, for which the inversion theorem is legitimate at every point.

1. \(\hat f_1 = \frac{1}{a - i\omega} + \frac{1}{a + i\omega} = \frac{2a}{a^2 + \omega^2}\). \(\hat f_2 = \int_{-1}^1e^{i\omega t}\mathrm dt = \frac{e^{i\omega} - e^{-i\omega}}{i\omega} = \frac{2\sin\omega}{\omega}\) (value \(2\) at \(\omega = 0\)).

2. \(\frac1{2\pi}\int\frac{2a\,\mathrm d\omega}{a^2 + \omega^2} = \frac{a}{\pi}\cdot\frac{1}{a}\big[\arctan\frac\omega a\big]_{-\infty}^\infty = \frac a\pi\cdot\frac\pi a = 1 = f_1(0)\) ✓. The theorem applies: \(f_1\) is continuous (with a kink), integrable, and \(\hat f_1\sim2a/\omega^2\) is integrable.

3. \((f_2*f_2)(t) = \int\mathbf 1_{[-1,1]}(s)\mathbf 1_{[-1,1]}(t - s)\,\mathrm ds\) is the length of \([-1,1]\cap[t-1, t+1]\), which is \(2 - |t|\) for \(|t|\le2\) and \(0\) otherwise: \(\Lambda\). By the convolution theorem (both factors bounded and integrable), \(\hat\Lambda = \hat f_2^{\,2} = 4\sin^2\omega/\omega^2\).

4. \(\Lambda\) is continuous and integrable, and \(\hat\Lambda\le\min(4, 4/\omega^2)\) is integrable. So inversion at \(t = 0\): \[ 2 = \Lambda(0) = \frac1{2\pi}\int\frac{4\sin^2\omega}{\omega^2}\,\mathrm d\omega \quad\Longrightarrow\quad \int_{-\infty}^\infty\frac{\sin^2x}{x^2}\,\mathrm dx = \pi = 3.14159 . \] For \(f_2\) the theorem does not apply: \(f_2\) is discontinuous and \(\hat f_2 = 2\sin\omega/\omega\) is not absolutely integrable (the deck's inversion slide). Convolving with itself smoothed one jump into a kink, and squared the decay from \(1/\omega\) to \(1/\omega^2\) — the dictionary between smoothness and decay.

Check. Dimensions: with \(t\) a time and \(f\) dimensionless, \(\hat f_1 = 2a/(a^2 + \omega^2)\) has dimension \(\mathrm T\) ✓. Limit: \(a\to0\) makes \(f_1\to1\) and \(\hat f_1\to\) a spike of area \(\int2a/(a^2 + \omega^2) = 2\pi\) — the table's \(\hat1 = 2\pi\delta\) ✓. Number: Plancherel on \(f_2\) (by the cited \(L^2\) extension) gives \(\int|f_2|^2 = 2 = \frac1{2\pi}\int4\sin^2\omega/\omega^2\), the same identity ✓.

Another way

\(\int\sin^2x/x^2 = \int\frac{1 - \cos2x}{2x^2}\); integrate by parts once to get \(\int\frac{\sin2x}{x}\mathrm dx = \pi\), which is worked example 2 of the deck.

P2 The uncertainty relation, sharp and not medium

exercises: the uncertainty relation's equality case, Plancherel for a derivative, moments by complex exponentials, a number

  1. For \(f(t) = Ce^{-at^2}\), \(a \gt 0\), normalised so \(\int|f|^2 = 1\), compute \(\Delta t\) and \(\Delta\omega\) and show \(\Delta t\,\Delta\omega = \frac12\).
  2. For the galvanometer's \(G\), show that \(\bar\omega = 0\) and \(\Delta\omega = \omega_0\) exactly, using \(\int\dot G^2 = \frac1{2\pi}\int\omega^2|\chi|^2\) and the equation \(G\) satisfies for \(t \gt 0\).
  3. Given \(\Delta t^2 = \frac1{\gamma^2} - \frac{1}{2\omega_0^2} + \frac{\gamma^2}{4\omega_0^4}\) (derive it if you have time; the solution does), evaluate \(\Delta t\,\Delta\omega\) for the galvanometer and compare with \(Q = \omega_0/\gamma\).
  4. Explain in two sentences why a sharply resonant real signal is so far from the bound.
Hints · 3 rungs
Rung 1

Part 1 is two Gaussian integrals. Part 2 is an integration by parts, no transform needed. Part 3 is arithmetic. Part 4 is about where \(|\chi|^2\) puts its weight on the frequency axis.

Rung 2

For part 1 use worked example 1 with scaling: \(\hat f\) is a Gaussian in \(\omega\), and \(\int u^2e^{-cu^2}\mathrm du = \frac{1}{2c}\int e^{-cu^2}\mathrm du\). For part 2, \(|\chi(\omega)|^2\) is even. Compute \(\int_0^\infty\dot G^2\) by parts, then use \(\ddot G = -\gamma\dot G - \omega_0^2G\).

Rung 3

First line of part 2: \(\int_0^\infty\dot G^2\,\mathrm dt = [G\dot G]_0^\infty - \int_0^\infty G\ddot G\,\mathrm dt\), and the boundary term vanishes at both ends.

Worked solution

Idea. The Gaussian meets the bound; a damped ringing signal carries most of its frequency spread in the distance between its two spectral peaks, not in their width.

1. \(|f|^2 = C^2e^{-2at^2}\), so \(\Delta t^2 = \int t^2e^{-2at^2}/\int e^{-2at^2} = \frac{1}{4a}\). By worked example 1 and scaling, \(\hat f\propto e^{-\omega^2/4a}\), so \(|\hat f|^2\propto e^{-\omega^2/2a}\) and \(\Delta\omega^2 = \frac{1}{2\cdot(1/2a)} = a\). Then \(\Delta t\,\Delta\omega = \sqrt{a/4a} = \frac12\) ✓. Both means are \(0\) by symmetry.

2. \(G\) is real, so \(\chi(-\omega) = \overline{\chi(\omega)}\) and \(|\chi|^2\) is even: \(\bar\omega = 0\). \(G\) is continuous and piecewise \(C^1\), so \(\widehat{\dot G} = -i\omega\chi\) (the jump formula adds no delta because \(G\) does not jump), and Plancherel gives \(\frac1{2\pi}\int\omega^2|\chi|^2 = \int\dot G^2\). Now \[ \int_0^\infty\dot G^2 = [G\dot G]_0^\infty - \int_0^\infty G\ddot G = \int_0^\infty G(\gamma\dot G + \omega_0^2G) = \gamma\big[\tfrac12G^2\big]_0^\infty + \omega_0^2\int_0^\infty G^2 = \omega_0^2\int_0^\infty G^2, \] using \(G(0) = 0\) and decay at infinity. So \(\Delta\omega^2 = \int\dot G^2/\int G^2 = \omega_0^2\): \(\Delta\omega = 8\ \mathrm{s^{-1}}\), for every \(\gamma\).

3. \(\Delta t^2 = 6.25 - 0.0078125 + 0.0000098 = 6.2421973\ \mathrm{s^2}\), \(\Delta t = 2.49844\) s, and \[ \Delta t\,\Delta\omega = 19.9875 = \sqrt{Q^2 - \tfrac12 + \tfrac1{4Q^2}}\quad(Q = 20). \] Derivation of \(\Delta t^2\). Write \(G^2 = \frac{e^{-\gamma t}}{2\omega_\mathrm d^2}(1 - \cos2\omega_\mathrm dt)\) and use \(\int_0^\infty t^ne^{-\gamma t}\mathrm dt = n!/\gamma^{n+1}\) and \(\int_0^\infty t^ne^{-\gamma t}\cos2\omega_\mathrm dt\,\mathrm dt = \operatorname{Re}\,n!/w^{n+1}\) with \(w = \gamma - 2i\omega_\mathrm d\), \(|w|^2 = \gamma^2 + 4\omega_\mathrm d^2 = 4\omega_0^2\). The moments \(m_n = \int t^nG^2\) are then \(m_0 = \frac{1}{2\omega_\mathrm d^2}\big(\frac1\gamma - \frac{\gamma}{4\omega_0^2}\big) = \frac{1}{2\gamma\omega_0^2}\), \(m_1 = \frac{1}{2\omega_\mathrm d^2}\big(\frac1{\gamma^2} - \frac{\gamma^2 - 4\omega_\mathrm d^2}{16\omega_0^4}\big)\), \(m_2 = \frac{1}{\omega_\mathrm d^2}\big(\frac1{\gamma^3} - \frac{\gamma^3 - 12\gamma\omega_\mathrm d^2}{64\omega_0^6}\big)\), and simplifying with \(4\omega_\mathrm d^2 = 4\omega_0^2 - \gamma^2\): \(\bar t = m_1/m_0 = \frac1\gamma + \frac{\gamma}{2\omega_0^2}\) and \(\Delta t^2 = m_2/m_0 - \bar t^2\) as stated. (Numerically \(\bar t = 2.503125\) s.)

4. \(|\chi|^2\) has two narrow peaks, at \(\pm\omega_\mathrm d\), each of width about \(\gamma\). The spread about \(\bar\omega = 0\) is dominated by the distance \(\omega_\mathrm d\) of the peaks from the origin, while \(\Delta t\approx1/\gamma\) is the ringing time: the product is about \(\omega_0/\gamma = Q\). The uncertainty relation constrains the width of a spectral line only when the spread is measured about the line, which for a real signal means working with one of the two peaks (the analytic signal), not both.

Check. Dimensions: \(\Delta t\) in s, \(\Delta\omega\) in s\(^{-1}\), product a number ✓. The bound holds for every underdamped \(Q \gt \frac12\): \(Q^2 - \frac12 + \frac1{4Q^2}\) is smallest at \(Q = 1/\sqrt2\), where it equals \(\frac12\), so the product is at least \(0.707 \gt 0.5\) ✓. The hypotheses: \(G\) is only piecewise \(C^1\), but the deck's proof uses only one integration by parts and Plancherel for \(\dot G\), both valid here.

P3 Distributions by hand routine

exercises: distributional derivatives, the jump formula, a proof, the ballistic galvanometer as \(L\theta = \)const\(\cdot\delta\), a number

  1. Compute the first and second distributional derivatives of \(|t|\).
  2. Prove that \(t\,\delta' = -\delta\), where \(\langle t\,T,\varphi\rangle := \langle T, t\varphi\rangle\).
  3. For \(a\ne0\) define \(\delta(at)\) so that it agrees with the change of variable for functions, and show \(\delta(at) = \delta/|a|\).
  4. A capacitor discharges a charge \(q = 1\ \mu\mathrm C\) through the galvanometer in a time far shorter than its period. Model the torque as \(\alpha q\,\delta(t)\), solve for \(\theta(t)\) using \(LG = \delta\) and the uniqueness of the causal solution, and give the first maximum of \(\theta\).
Hints · 3 rungs
Rung 1

Two distributions, one identity between distributions, one definition-plus-identity, one function of time and a number in radians.

Rung 2

The jump formula does part 1 in two lines. For parts 2 and 3, test against an arbitrary \(\varphi\) and use only the definitions. For part 3, compute \(\int f(at)\varphi(t)\mathrm dt\) for a function \(f\) by substitution first.

Rung 3

First line of part 2: \(\langle t\,\delta',\varphi\rangle = \langle\delta', t\varphi\rangle = -\langle\delta,(t\varphi)'\rangle = -(t\varphi)'(0)\). For part 4, divide the equation by \(I\): \(L\theta = (\alpha q/I)\,\delta\).

Worked solution

Idea. Every statement about a distribution is a statement about what it does to test functions; write that down and the proof follows.

1. \(|t|\) is continuous and \(C^1\) off \(0\) with derivative \(\operatorname{sgn}t\): no jump, so \(|t|' = \operatorname{sgn}t\). \(\operatorname{sgn}\) jumps by \(2\) at \(0\) and has classical derivative \(0\) elsewhere, so \(|t|'' = 2\delta\).

2. \(-(t\varphi)'(0) = -(\varphi(0) + 0\cdot\varphi'(0)) = -\varphi(0) = \langle-\delta,\varphi\rangle\). \(\square\)

3. For a function, \(\int f(at)\varphi(t)\,\mathrm dt = \frac1{|a|}\int f(u)\varphi(u/a)\,\mathrm du\) (\(u = at\); the modulus because a negative \(a\) reverses the limits). So define \(\langle T(a\,\cdot),\varphi\rangle := \frac1{|a|}\langle T,\varphi(\cdot/a)\rangle\). For \(T = \delta\): \(\frac1{|a|}\varphi(0) = \langle\delta/|a|,\varphi\rangle\). \(\square\)

4. The equation is \(\ddot\theta + \gamma\dot\theta + \omega_0^2\theta = (\alpha q/I)\,\delta(t)\) with the coil at rest before \(t = 0\). \(\theta = (\alpha q/I)G\) satisfies it, since \(L\) is linear and \(LG = \delta\); it is causal; and the causal solution is unique (deck, notes to the \(LG = \delta\) slide). With \(\alpha q/I = (2.0\times10^{-3})(10^{-6})/10^{-8} = 0.2\ \mathrm{rad\,s^{-1}}\), \[ \theta(t) = 0.2\,e^{-0.2t}\,\frac{\sin(7.9975\,t)}{7.9975}\ \mathrm{rad},\qquad t \gt 0 . \] The first maximum is where \(\dot G = 0\), \(\tan\omega_\mathrm dt = 2\omega_\mathrm d/\gamma = 39.987\), at \(t = 0.193285\) s, where \(G = 0.120260\) s: \(\theta_{\max} = 0.2\times0.120260 = 0.024052\) rad \(= 1.378^\circ\), unit 06's number ✓.

Check. Dimensions: \([\alpha q/I] = \mathrm{N\,m\,A^{-1}\,C}/\mathrm{kg\,m^2} = \mathrm{s^{-1}}\) (an angular velocity), times \([G] = \mathrm s\): radians ✓. Limit: \(\gamma\to0\) gives \(\theta_{\max}\to0.2/8 = 0.025\) rad, the undamped amplitude \(v_0/\omega_0\), slightly above the damped \(0.02405\) ✓. Part 1 check: pair \(|t|'' = 2\delta\) with a \(\varphi\) that equals \(1\) on \([-1,1]\): \(\int|t|\varphi''\) — integrate by parts twice on each half — gives \(2\varphi(0) = 2\) ✓.

Holomorphy and contour integrals

P4 Holomorphic or not routine

exercises: Cauchy–Riemann, building a holomorphic function from its real part, Cauchy's integral formula, a number

  1. Where is each of these complex-differentiable: \(z^2\), \(\bar z\), \(|z|^2\), \(e^{\bar z}\), \(\operatorname{Re}z\)?
  2. Find all real \(v(x,y)\) such that \(u + iv\) is holomorphic on \(\mathbb C\), where \(u = x^3 - 3xy^2\). Recognise \(u + iv\).
  3. Evaluate \(\oint\chi(\omega)\,\mathrm d\omega\) anticlockwise round the circle \(|\omega - 8| = 1\), for the galvanometer's \(\chi\), using only Cauchy's integral formula. Then round \(|\omega| = 10\).
Hints · 3 rungs
Rung 1

Part 1 is five verdicts, each a set of points. Part 2 is a family of functions differing by a constant. Part 3 is two complex numbers.

Rung 2

Write each function as \(u + iv\) and test Cauchy–Riemann; the converse theorem (continuous partials) settles the points where it holds. For part 2, \(v_y = u_x\) determines \(v\) up to a function of \(x\) alone; \(v_x = -u_y\) pins that down. For part 3, which poles lie inside each circle?

Rung 3

Part 3, first line: \(\chi(\omega) = \frac{g(\omega)}{\omega - \omega_+}\) with \(g(\omega) = \frac{-1}{\omega - \omega_-}\), holomorphic on and inside \(|\omega - 8| = 1\) because \(|\omega_- - 8|\approx16\).

Worked solution

Idea. Cauchy–Riemann is a local test; Cauchy's formula turns a contour integral round an isolated pole into a function value.

1. \(z^2 = x^2 - y^2 + 2ixy\): \(u_x = 2x = v_y\), \(u_y = -2y = -v_x\) — everywhere (it is a polynomial). \(\bar z\): \(u_x = 1\ne-1 = v_y\) — nowhere. \(|z|^2 = x^2 + y^2\), \(v = 0\): CR needs \(2x = 0\) and \(2y = 0\) — only at \(0\), and there it is differentiable (\(|h|^2/h\to0\)). \(e^{\bar z} = e^x(\cos y - i\sin y)\): \(u_x = e^x\cos y\), \(v_y = -e^x\cos y\), equal only where \(\cos y = 0\); \(u_y = -e^x\sin y\), \(-v_x = e^x\sin y\), equal only where \(\sin y = 0\) — never both: nowhere. \(\operatorname{Re}z = x\): \(u_x = 1\ne0 = v_y\) — nowhere.

2. \(u_x = 3x^2 - 3y^2 = v_y\) gives \(v = 3x^2y - y^3 + c(x)\). Then \(v_x = 6xy + c'(x)\) must equal \(-u_y = 6xy\), so \(c' = 0\): \(v = 3x^2y - y^3 + C\). And \(u + iv = (x + iy)^3 + iC = z^3 + iC\).

3. \(|\omega_+ - 8| = |{-0.0025} - 0.2i| = 0.2000\lt1\), and \(\omega_-\) is far outside. So by Cauchy's formula \[ \oint_{|\omega - 8| = 1}\frac{g(\omega)}{\omega - \omega_+}\,\mathrm d\omega = 2\pi i\,g(\omega_+) = \frac{-2\pi i}{\omega_+ - \omega_-} = \frac{-2\pi i}{2\omega_\mathrm d} = -\frac{\pi i}{\omega_\mathrm d} = -0.392822\,i\ \mathrm{s}. \] Round \(|\omega| = 10\) both poles are inside. Partial fractions, \(\chi = \frac{-1}{2\omega_\mathrm d}\big(\frac1{\omega - \omega_+} - \frac1{\omega - \omega_-}\big)\), give \(\frac{-2\pi i}{2\omega_\mathrm d}(1 - 1) = 0\). Equivalently: \(|\chi|\le1/(R^2 - \gamma R - \omega_0^2)\) on \(|\omega| = R\), so the integral is \(O(1/R)\); it is the same for every \(R \gt 8\) (the annulus between two such circles contains no pole), so it is \(0\).

Check. Dimensions: \([\chi][\omega] = \mathrm{s^2\,s^{-1}} = \mathrm s\) ✓. Limit \(\gamma\to0\): \(\omega_+\to8\), and the integral tends to \(-\pi i/8 = -0.392699i\), close to the damped value ✓. Part 2 check: \(z^3\) at \(z = 1 + i\) is \(-2 + 2i\); \(u(1,1) = 1 - 3 = -2\), \(v(1,1) = 3 - 1 = 2\) ✓.

P5 Plancherel for the galvanometer, by residues medium

exercises: the residue theorem on a semicircle, residues at simple poles, an ML arc bound, Plancherel checked, a number

Let \(D(\omega) = (\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2 = |\omega_0^2 - \omega^2 - i\gamma\omega|^2\) for real \(\omega\).

  1. Prove \(\displaystyle\int_{-\infty}^\infty\frac{\mathrm d\omega}{D(\omega)} = \frac{\pi}{\gamma\omega_0^2}\) by closing in the upper half-plane.
  2. Prove \(\displaystyle\int_{-\infty}^\infty\frac{\omega^2\,\mathrm d\omega}{D(\omega)} = \frac\pi\gamma\) the same way.
  3. Evaluate the first for the galvanometer and confirm Plancherel: \(\frac1{2\pi}\int|\chi|^2 = \int G^2 = 1/(2\gamma\omega_0^2)\).
Hints · 3 rungs
Rung 1

Two closed-form real integrals and one number. The integrands are rational with four simple poles, two above the axis and two below.

Rung 2

For real \(\omega\), \(D = h(\omega)h(-\omega)\) with \(h(\omega) = \omega_0^2 - \omega^2 - i\gamma\omega\). The zeros of \(h(-\omega)\) are \(-\omega_\pm\), in the upper half-plane. At a zero \(p\) of \(h(-\omega)\), \(h(p) = h(p) - h(-p)\), and the even terms cancel; or use \(\omega_0^2 - \omega_\pm^2 = i\gamma\omega_\pm\) to simplify.

Rung 3

First line: \(h(-\omega) = -(\omega + \omega_+)(\omega + \omega_-)\), so at \(\omega = -\omega_+\) the residue of \(1/(h(\omega)h(-\omega))\) is \(\frac{1}{h(-\omega_+)}\cdot\frac{-1}{\omega_- - \omega_+}\), and \(h(-\omega_+) = \omega_0^2 - \omega_+^2 + i\gamma\omega_+ = 2i\gamma\omega_+\).

Worked solution

Idea. Factor \(D\) into a function with poles below and its mirror with poles above; close upwards and collect the two mirror poles.

The factorisation. For real \(\omega\), \(\overline{h(\omega)} = \omega_0^2 - \omega^2 + i\gamma\omega = h(-\omega)\), so \(D = h(\omega)h(-\omega)\), and \(D(\omega) = h(\omega)h(-\omega)\) continues as a polynomial to all of \(\mathbb C\). \(h(\nu) = -(\nu - \omega_+)(\nu - \omega_-)\), so \(h(-\omega) = -(\omega + \omega_+)(\omega + \omega_-)\), zero at \(-\omega_\pm = \mp\omega_\mathrm d + i\gamma/2\) (upper half-plane).

The residues. Since \(h(\omega_\pm) = 0\), \(\omega_0^2 - \omega_\pm^2 = i\gamma\omega_\pm\), hence \(h(-\omega_\pm) = \omega_0^2 - \omega_\pm^2 + i\gamma\omega_\pm = 2i\gamma\omega_\pm\). At \(-\omega_+\): \(\operatorname{Res}\frac1D = \frac{1}{h(-\omega_+)}\cdot\frac{-1}{-\omega_+ + \omega_-} = \frac{1}{2i\gamma\omega_+\cdot2\omega_\mathrm d}\), using \(\omega_+ - \omega_- = 2\omega_\mathrm d\). At \(-\omega_-\): \(\frac{1}{h(-\omega_-)}\cdot\frac{-1}{-\omega_- + \omega_+} = \frac{-1}{2i\gamma\omega_-\cdot2\omega_\mathrm d}\). Sum: \[ \frac{1}{4i\gamma\omega_\mathrm d}\Big(\frac1{\omega_+} - \frac1{\omega_-}\Big) = \frac{1}{4i\gamma\omega_\mathrm d}\cdot\frac{\omega_- - \omega_+}{\omega_+\omega_-} = \frac{1}{4i\gamma\omega_\mathrm d}\cdot\frac{-2\omega_\mathrm d}{-\omega_0^2} = \frac{1}{2i\gamma\omega_0^2}, \] with \(\omega_+\omega_- = (\omega_\mathrm d - i\gamma/2)(-\omega_\mathrm d - i\gamma/2) = -\omega_\mathrm d^2 - \gamma^2/4 = -\omega_0^2\).

1. On the upper semicircle of radius \(R\), \(|D|\ge(R^2 - \gamma R - \omega_0^2)^2\ge R^4/4\) once \(R\ge12\), so the arc contributes at most \(\pi R\cdot4/R^4\to0\). The residue theorem: \(\int\frac{\mathrm d\omega}{D} = 2\pi i\cdot\frac{1}{2i\gamma\omega_0^2} = \frac{\pi}{\gamma\omega_0^2}\). \(\square\)

2. The numerator \(\omega^2\) multiplies the two residues by \(\omega_+^2\) and \(\omega_-^2\) (evaluated at \(-\omega_\pm\), the square is the same): sum \(\frac{1}{4i\gamma\omega_\mathrm d}(\omega_+ - \omega_-) = \frac{2\omega_\mathrm d}{4i\gamma\omega_\mathrm d} = \frac{1}{2i\gamma}\). The arc is \(O(R\cdot R^2/R^4)\to0\). So \(\int\omega^2/D = \pi/\gamma\). \(\square\)

3. \(\pi/(0.4\times64) = 0.1227185\ \mathrm{s^3}\), and \(0.1227185/2\pi = 0.01953125\ \mathrm{s^3} = 1/(2\times0.4\times64)\) ✓ — Plancherel, the two sides computed by unrelated means.

Check. Dimensions: \([D] = \mathrm s^{-4}\), so \(\int\mathrm d\omega/D\) is \(\mathrm s^3\), as is \(\pi/\gamma\omega_0^2\) ✓; and \(\int\omega^2/D\) is \(\mathrm s\), as is \(\pi/\gamma\) ✓. Limit: \(\gamma\to0\) makes both blow up like \(1/\gamma\) — the resonance peak of height \(1/\gamma^2\omega_0^2\) and width \(\gamma\) ✓. Part 2 is the deck's \(f\)-sum rule: \(\gamma\int_0^\infty\omega^2/D = \pi/2\) ✓.

P6 Fresnel by octant medium

exercises: a sector contour, Jordan's inequality as an honest arc estimate, the constant in stationary phase, a number

  1. Show that \(\int_0^\infty\cos(x^2)\,\mathrm dx\) and \(\int_0^\infty\sin(x^2)\,\mathrm dx\) converge.
  2. Integrate \(e^{iz^2}\) round the boundary of the sector \(0\le\arg z\le\pi/4\), \(|z|\le R\), bound the arc explicitly, and deduce \[ \int_0^\infty\cos(x^2)\,\mathrm dx = \int_0^\infty\sin(x^2)\,\mathrm dx = \sqrt{\frac\pi8}. \]
  3. Deduce \(\int_{-\infty}^\infty e^{i\lambda u^2/2}\,\mathrm du = \sqrt{2\pi/\lambda}\,e^{i\pi/4}\) for \(\lambda \gt 0\): the constant the deck's stationary-phase theorem uses.
Hints · 3 rungs
Rung 1

One convergence proof, one number (the same for both integrals), one formula. The contour is a slice of pizza with its point at the origin.

Rung 2

For convergence substitute \(u = x^2\) and integrate by parts once. On the ray \(z = re^{i\pi/4}\), \(iz^2 = -r^2\): a Gaussian. On the arc \(z = Re^{i\theta}\), \(|e^{iz^2}| = e^{-R^2\sin2\theta}\).

Rung 3

First line of the arc estimate: \(\big|\int_\text{arc}\big|\le\int_0^{\pi/4}e^{-R^2\sin2\theta}R\,\mathrm d\theta\le\int_0^{\pi/4}e^{-4R^2\theta/\pi}R\,\mathrm d\theta\), by Jordan's inequality \(\sin2\theta\ge\frac{2}{\pi}\cdot2\theta\) for \(2\theta\in[0,\pi/2]\).

Worked solution

Idea. Rotate the line of integration by \(45^\circ\), where the oscillating exponential becomes a decaying Gaussian; the only cost is an arc, and Jordan's inequality pays it.

1. With \(u = x^2\), \(\int_1^X\cos(x^2)\,\mathrm dx = \int_1^{X^2}\frac{\cos u}{2\sqrt u}\,\mathrm du = \big[\frac{\sin u}{2\sqrt u}\big]_1^{X^2} + \int_1^{X^2}\frac{\sin u}{4u^{3/2}}\,\mathrm du\); the boundary term converges and the integral converges absolutely. Same for \(\sin\).

2. \(e^{iz^2}\) is entire and the sector is convex, so the boundary integral is \(0\): \[ \int_0^Re^{ix^2}\mathrm dx + \int_\text{arc}e^{iz^2}\mathrm dz - \int_0^Re^{-r^2}e^{i\pi/4}\,\mathrm dr = 0, \] the ray traversed inwards, with \(z = re^{i\pi/4}\), \(z^2 = ir^2\), \(\mathrm dz = e^{i\pi/4}\mathrm dr\). The arc: \(z = Re^{i\theta}\), \(\mathrm dz = iRe^{i\theta}\mathrm d\theta\), \(|e^{iz^2}| = |e^{iR^2(\cos2\theta + i\sin2\theta)}| = e^{-R^2\sin2\theta}\). By concavity of \(\sin\) on \([0,\pi/2]\), \(\sin2\theta\ge4\theta/\pi\) for \(0\le\theta\le\pi/4\), so \[ \Big|\int_\text{arc}\Big|\le R\int_0^{\pi/4}e^{-4R^2\theta/\pi}\,\mathrm d\theta\le R\cdot\frac{\pi}{4R^2} = \frac{\pi}{4R}\to0 . \] Let \(R\to\infty\): \(\int_0^\infty e^{ix^2}\mathrm dx = e^{i\pi/4}\int_0^\infty e^{-r^2}\mathrm dr = \frac{\sqrt\pi}{2}\cdot\frac{1 + i}{\sqrt2}\). Real and imaginary parts: both \(\frac{\sqrt\pi}{2\sqrt2} = \sqrt{\pi/8} = 0.626657\).

3. Evenness gives \(\int_{-\infty}^\infty e^{ix^2}\mathrm dx = \sqrt\pi\,e^{i\pi/4}\). Substitute \(x = u\sqrt{\lambda/2}\): \(\int e^{i\lambda u^2/2}\mathrm du = \sqrt{2/\lambda}\,\sqrt\pi\,e^{i\pi/4} = \sqrt{2\pi/\lambda}\,e^{i\pi/4}\). \(\square\)

Check. Part 3 is the analytic continuation slide's \(\sqrt{2\pi/a}\) at \(a = -i\lambda\), on the boundary of its half-plane: \(\sqrt{2\pi/(-i\lambda)} = \sqrt{2\pi/\lambda}\,e^{i\pi/4}\) with the principal root ✓ — which is why the limit needed an argument. Numerically \(\int_0^{30}\cos(x^2)\,\mathrm dx\) is within \(0.02\) of \(0.6267\), the boundary term \(\sin(900)/60\) being the error's size ✓. Dimensions: if \(x\) carries a unit, \(x^2\) must be made dimensionless, and \(\sqrt{2\pi/\lambda}\) carries the unit of \(u\) ✓.

Another way

Continue \(\int e^{-au^2/2}\mathrm du = \sqrt{2\pi/a}\) to \(a = \varepsilon - i\lambda\) and let \(\varepsilon\to0^+\), justifying the limit by the same substitution-and-parts bound as in part 1, uniform in \(\varepsilon\).

Causality

P7 Kramers–Kronig hard

exercises: dispersion relations from analyticity by an indented contour, exact checks on the oscillator, a proof that refraction implies absorption

  1. Let \(\chi\) be holomorphic on \(\operatorname{Im}\omega \gt 0\), continuous and \(C^1\) up to the real axis, with \(\max_{|\omega| = R,\,\operatorname{Im}\omega\ge0}|\chi|\to0\). Prove, for real \(\omega\), that \(\chi(\omega) = \frac{1}{i\pi}\,\mathrm{PV}\!\int\frac{\chi(\omega')}{\omega' - \omega}\,\mathrm d\omega'\), by integrating \(\chi(\omega')/(\omega' - \omega)\) round an upper semicircle indented at \(\omega\). Deduce the two real relations.
  2. For the galvanometer, verify the relation for \(\chi'\) exactly at \(\omega = 0\), and verify the sum rule \(\int_0^\infty\omega\chi''\,\mathrm d\omega = \pi/2\) exactly. (P5's integrals may be used.)
  3. Prove: if a medium's \(\varepsilon(\omega) - 1\) satisfies the hypotheses of part 1 and \(\operatorname{Im}\varepsilon(\omega) = 0\) for every real \(\omega\), then \(n = \sqrt\varepsilon = 1\) at every frequency. State the physical conclusion.
Hints · 4 rungs
Rung 1

Part 1 is a proof with one contour; the answer is a principal-value identity. Part 2 is two exact numbers, \(1/64\) and \(\pi/2\). Part 3 is two lines once part 1 is in hand.

Rung 2

Inside the indented semicircle there are no singularities. Four pieces: two straight segments (together a principal-value integral in the limit), the big arc (Jordan is not even needed — why?), and the small arc (the indentation lemma, clockwise).

Rung 3

On the big arc \(|\chi(\omega')/(\omega' - \omega)|\le M_R/(R - |\omega|)\), and the arc has length \(\pi R\): that is \(O(M_R)\to0\). The small arc, clockwise over angle \(\pi\), contributes \(-i\pi\chi(\omega)\) in the limit.

Rung 4

First line: \(0 = \int_{|\omega' - \omega| \gt \varepsilon,\,|\omega'|\lt R}\frac{\chi(\omega')}{\omega' - \omega}\mathrm d\omega' + \int_{c_\varepsilon} + \int_{C_R}\). For part 2, \(\chi''(\omega')/\omega' = \gamma/D(\omega')\).

Worked solution

Idea. Cauchy's theorem applied to \(\chi(\omega')/(\omega' - \omega)\) says the whole real-axis integral is carried by the pole at \(\omega' = \omega\), and a pole on the contour gives half a residue.

1. The integrand is holomorphic inside the indented half-disc (the only singularity, \(\omega' = \omega\), is cut out), continuous up to its boundary, so its boundary integral is \(0\) (Cauchy, applied to the region raised by \(\eta\) and \(\eta\to0\), as in the deck's (b) ⇒ (a)). Big arc: at most \(\pi R\,M_R/(R - |\omega|)\to0\), since \(M_R\to0\). Small arc \(c_\varepsilon\), clockwise over angle \(\pi\): \(\chi\) is \(C^1\), so \(\chi(\omega') = \chi(\omega) + O(\omega' - \omega)\), and \(\int_{c_\varepsilon}\frac{\chi(\omega)}{\omega' - \omega}\mathrm d\omega' = -i\pi\chi(\omega)\) exactly while the remainder is \(O(\varepsilon)\). Letting \(\varepsilon\to0\), \(R\to\infty\): \[ \mathrm{PV}\!\int\frac{\chi(\omega')}{\omega' - \omega}\,\mathrm d\omega' - i\pi\chi(\omega) = 0 . \] That is the claim. Writing \(\chi = \chi' + i\chi''\) and taking real and imaginary parts of \(i\pi(\chi' + i\chi'') = \mathrm{PV}\int\frac{\chi' + i\chi''}{\omega' - \omega}\): \(\chi' = \frac1\pi\mathrm{PV}\int\frac{\chi''}{\omega' - \omega}\) and \(\chi'' = -\frac1\pi\mathrm{PV}\int\frac{\chi'}{\omega' - \omega}\). \(\square\)

The galvanometer meets the hypotheses. \(\chi\) is rational with poles only at \(\omega_\pm\), below the axis; it is holomorphic on a neighbourhood of the closed upper half-plane and \(|\chi|\le2/R^2\) there for \(R\ge12\).

2. \(\chi'' = \gamma\omega/D\) is odd and \(\chi'\) even, so at \(\omega = 0\) the first relation reads \(\chi'(0) = \frac1\pi\int\frac{\chi''(\omega')}{\omega'}\mathrm d\omega'\) (no principal value is needed: \(\chi''/\omega' = \gamma/D\) is smooth). Then \(\frac\gamma\pi\int\frac{\mathrm d\omega'}{D} = \frac\gamma\pi\cdot\frac{\pi}{\gamma\omega_0^2} = \frac1{\omega_0^2} = \frac1{64} = 0.015625\ \mathrm{s^2}\), which is \(\chi(0)\) ✓. Sum rule: \(\int_0^\infty\omega\chi''\,\mathrm d\omega = \gamma\int_0^\infty\frac{\omega^2}{D}\mathrm d\omega = \frac\gamma2\cdot\frac\pi\gamma = \frac\pi2\) ✓, for every \(\gamma\) and \(\omega_0\).

3. Apply part 1 to \(\chi_\varepsilon = \varepsilon - 1\). If \(\operatorname{Im}\chi_\varepsilon\equiv0\) on the real axis, the first real relation gives \(\operatorname{Re}\chi_\varepsilon(\omega) = \frac1\pi\mathrm{PV}\int\frac{0}{\omega' - \omega} = 0\) for every \(\omega\). So \(\varepsilon\equiv1\) and \(n\equiv1\). \(\square\) Physically: a medium that refracts at even one frequency (\(n\ne1\) there) must absorb at some frequency. Glass is transparent in the visible because its absorption is elsewhere — in the ultraviolet and the infrared — and the static sum rule \(n(0)^2 - 1 = \frac2\pi\int_0^\infty\operatorname{Im}\varepsilon\,\frac{\mathrm d\omega}{\omega}\) says how much.

Check. Dimensions: \([\chi''\,\mathrm d\omega'/(\omega' - \omega)] = [\chi]\) ✓. Limit: in part 2, \(\gamma\to0\) leaves \(\chi(0) = 1/\omega_0^2\) unchanged while the absorption collapses to delta functions at \(\pm\omega_0\) of weight \(\pi/2\omega_0\): \(\frac2\pi\cdot\frac{\pi}{2\omega_0}\cdot\frac{1}{\omega_0} = \frac1{\omega_0^2}\) ✓. Numbers at \(\omega = 7\): \(\chi(7) = 0.06442192 + 0.01202542i\) and the two principal values reproduce both parts to seven figures (deck, worked example 4).

P8 A keyhole medium

exercises: branch cuts, choosing a branch and stating it, the keyhole contour, both circle estimates, a number

For \(0 \lt a \lt 1\) prove \[ \int_0^\infty\frac{x^{a-1}}{1 + x}\,\mathrm dx = \frac{\pi}{\sin\pi a}, \] checking the case \(a = \frac12\) by an elementary substitution, and evaluate the integral at \(a = \frac13\).

Hints · 3 rungs
Rung 1

A function of \(a\), then a check, then a number. The integrand is multivalued in the complex plane; the only pole is at \(z = -1\).

Rung 2

Cut along \([0,\infty)\) and take \(z^{a-1} = |z|^{a-1}e^{i(a-1)\arg z}\) with \(0 \lt \arg z \lt 2\pi\). The keyhole: a large circle \(C_R\), a small circle \(c_\varepsilon\), and the two edges of the cut. On the lower edge \(\arg z = 2\pi\).

Rung 3

First line: the two edges give \((1 - e^{2\pi i(a-1)})\int_\varepsilon^R\frac{x^{a-1}}{1+x}\mathrm dx = (1 - e^{2\pi ia})\int_\varepsilon^R\dots\), and the residue at \(z = -1 = e^{i\pi}\) is \(e^{i\pi(a-1)} = -e^{i\pi a}\).

Worked solution

Idea. The two edges of a branch cut carry the same integral multiplied by different phases; their difference is a nonzero multiple of the integral you want, and the residue theorem supplies it.

The branch. \(z^{a-1} := e^{(a-1)\log z}\) with \(\log z = \ln|z| + i\arg z\), \(\arg z\in(0,2\pi)\), holomorphic on \(\mathbb C\setminus[0,\infty)\). The keyhole runs along the upper edge from \(\varepsilon\) to \(R\) (\(\arg = 0\)), round \(C_R\) anticlockwise, back along the lower edge from \(R\) to \(\varepsilon\) (\(\arg = 2\pi\)), and round \(c_\varepsilon\) clockwise.

Edges. Upper: \(\int_\varepsilon^R\frac{x^{a-1}}{1+x}\). Lower: \(-\int_\varepsilon^R\frac{x^{a-1}e^{2\pi i(a-1)}}{1+x} = -e^{2\pi ia}\int_\varepsilon^R\frac{x^{a-1}}{1+x}\).

Circles. On \(C_R\): \(|z^{a-1}/(1+z)|\le R^{a-1}/(R - 1)\), length \(2\pi R\): total \(O(R^{a-1})\to0\) as \(a \lt 1\). On \(c_\varepsilon\): \(\le\varepsilon^{a-1}/(1 - \varepsilon)\cdot2\pi\varepsilon = O(\varepsilon^a)\to0\) as \(a \gt 0\).

Residue at \(z = -1\), where \(\arg z = \pi\): \((e^{i\pi})^{a-1} = e^{i\pi a}e^{-i\pi} = -e^{i\pi a}\), a simple pole of \(1/(1+z)\). So \[ (1 - e^{2\pi ia})\,I = 2\pi i\,(-e^{i\pi a})\quad\Longrightarrow\quad I = \frac{-2\pi i}{e^{-i\pi a} - e^{i\pi a}} = \frac{-2\pi i}{-2i\sin\pi a} = \frac{\pi}{\sin\pi a}. \] \(\square\)

Check \(a = \frac12\). \(x = u^2\): \(\int_0^\infty\frac{u^{-1}\cdot2u\,\mathrm du}{1 + u^2} = 2\cdot\frac\pi2 = \pi = \pi/\sin\frac\pi2\) ✓. Number: \(a = \frac13\) gives \(\pi/\sin60^\circ = 2\pi/\sqrt3 = 3.62760\); direct quadrature agrees to six figures ✓. Limits: as \(a\to0^+\) or \(a\to1^-\) the integral diverges (at \(0\), respectively at \(\infty\)), and so does \(\pi/\sin\pi a\) ✓. Symmetry: \(x\to1/x\) maps the \(a\) integral to the \(1 - a\) integral, and \(\sin\pi a = \sin\pi(1 - a)\) ✓.

Another way

This is \(\Gamma(a)\Gamma(1 - a)\) (write \(\frac1{1+x} = \int_0^\infty e^{-(1+x)s}\mathrm ds\) and swap), so the result is Euler's reflection formula \(\Gamma(a)\Gamma(1-a) = \pi/\sin\pi a\).

P9 Switching on: the Bromwich contour medium

exercises: the Laplace transform of an initial value problem, the Bromwich inversion by residues, closing left and right, a number

At \(t = 0\) a steady current is switched on through the galvanometer, which was at rest, so that \(\ddot\theta + \gamma\dot\theta + \omega_0^2\theta = u_0H(t)\) with \(\theta(0) = \dot\theta(0) = 0\).

  1. Find \(\Theta(s) = \mathcal L[\theta](s)\) and locate its poles.
  2. Invert by the Bromwich integral, closing the contour to the left for \(t \gt 0\) and to the right for \(t \lt 0\); justify both arcs.
  3. Find the time and size of the first maximum, as a multiple of the final deflection \(u_0/\omega_0^2\).
Hints · 3 rungs
Rung 1

A rational function of \(s\) with three simple poles; then a function of \(t\), zero for \(t \lt 0\); then a time and a number close to \(2\).

Rung 2

\(\mathcal L[\ddot\theta] = s^2\Theta - s\theta(0) - \dot\theta(0)\) and \(\mathcal L[H] = 1/s\). The poles are \(0\) and \(s_\pm = -\gamma/2\pm i\omega_\mathrm d\). \(|e^{st}| = e^{t\operatorname{Re}s}\) decides which side to close on.

Rung 3

First line: \(\Theta(s) = \dfrac{u_0}{s(s^2 + \gamma s + \omega_0^2)}\), and the residue at \(s_+\) of \(\Theta e^{st}\) is \(\dfrac{u_0e^{s_+t}}{s_+(s_+ - s_-)} = \dfrac{u_0e^{s_+t}}{2i\omega_\mathrm d s_+}\).

Worked solution

Idea. The Laplace transform turns the switched-on problem into algebra; the Bromwich line lies to the right of every pole, which makes the answer causal automatically.

1. Transforming, \((s^2 + \gamma s + \omega_0^2)\Theta = u_0/s\), so \(\Theta = u_0/[s(s^2 + \gamma s + \omega_0^2)]\): poles at \(0\) and \(s_\pm = -0.2\pm7.9975i\). Take the Bromwich line at \(c = 1 \gt 0\).

2. Arcs. On a semicircle of radius \(R\) centred at \(c\), \(|\Theta|\le2u_0/R^3\) for large \(R\). For \(t \gt 0\), on the left semicircle \(\operatorname{Re}s\le c\), so \(|e^{st}|\le e^{ct}\): the arc is at most \(\pi R\cdot e^{ct}\cdot2u_0/R^3\to0\). For \(t \lt 0\), on the right semicircle \(\operatorname{Re}s\ge c\), so \(|e^{st}|\le e^{ct}\) again (now \(t \lt 0\)): the arc vanishes, and there are no poles to the right of the line, so \(\theta(t) = 0\) for \(t \lt 0\) — causality again, from the position of the poles.

Residues, \(t \gt 0\) (anticlockwise, since closing left keeps the inside on the left): at \(0\), \(u_0/\omega_0^2\). At \(s_\pm\), \(\frac{u_0e^{s_\pm t}}{s_\pm(\pm2i\omega_\mathrm d)}\) — complex conjugates, summing to \(2\operatorname{Re}\frac{u_0e^{s_+t}}{2i\omega_\mathrm ds_+}\). With \(\frac{1}{s_+} = \frac{\bar s_+}{\omega_0^2}\) (since \(|s_+|^2 = \omega_0^2\)) and \(\bar s_+ = -\gamma/2 - i\omega_\mathrm d\): \[ 2\operatorname{Re}\frac{u_0e^{-\gamma t/2}e^{i\omega_\mathrm dt}(-\gamma/2 - i\omega_\mathrm d)}{2i\omega_\mathrm d\omega_0^2} = -\frac{u_0}{\omega_0^2}e^{-\gamma t/2}\Big(\cos\omega_\mathrm dt + \frac{\gamma}{2\omega_\mathrm d}\sin\omega_\mathrm dt\Big). \] So \[ \theta(t) = \frac{u_0}{\omega_0^2}\Big[1 - e^{-\gamma t/2}\Big(\cos\omega_\mathrm dt + \frac{\gamma}{2\omega_\mathrm d}\sin\omega_\mathrm dt\Big)\Big],\qquad t \gt 0 . \]

3. \(\dot\theta = \frac{u_0}{\omega_\mathrm d}e^{-\gamma t/2}\sin\omega_\mathrm dt\) (\(= u_0G\), as it must be: the step response is the integral of the impulse response). First zero at \(t = \pi/\omega_\mathrm d = 0.392822\) s, where \(\theta = \frac{u_0}{\omega_0^2}\big(1 + e^{-\pi\gamma/2\omega_\mathrm d}\big) = 1.92444\,\frac{u_0}{\omega_0^2}\): an overshoot of \(92.4\%\).

Check. \(\theta(0) = \frac{u_0}{\omega_0^2}(1 - 1) = 0\) and \(\dot\theta(0) = 0\) ✓. \(t\to\infty\): \(u_0/\omega_0^2\), the static deflection, which is also \(\lim_{s\to0}s\Theta(s)\) ✓. Dimensions: \([u_0] = \mathrm{rad\,s^{-2}}\), so \(u_0/\omega_0^2\) is in radians ✓. Critical damping \(\gamma\to2\omega_0\) sends the overshoot factor \(e^{-\pi\gamma/2\omega_\mathrm d}\to0\) ✓.

Unit 09's debts, and asymptotics

P10 The Laplace limit, finally hard

exercises: Rouché's theorem, holomorphy of a function defined by a contour integral, Lagrange inversion by residues, radius of convergence as distance to a singularity, a proof and a number

Kepler's equation \(E - e\sin E = M\) (unit 09) has, for real \(M\) and \(0\le e \lt 1\), a unique real solution \(E(e)\), and Lagrange's series writes it as a power series in \(e\): \(E = M + \sum_{n\ge1}\frac{e^n}{n!}\frac{\mathrm d^{n-1}}{\mathrm dM^{n-1}}\sin^nM\). Let \(x_* \gt 0\) solve \(x\tanh x = 1\) and \(\lambda = 1/\sinh x_*\).

  1. Show \(|\sin(x + iy)|\le\cosh y\), and that \(\max_{r \gt 0}r/\cosh r = \lambda\), attained at \(r = x_*\).
  2. For real \(M\) and complex \(|e| \lt \lambda\), show that \(z - M - e\sin z\) has exactly one zero \(E(e)\) in the disc \(|z - M| \lt x_*\).
  3. Show that \(E(e) = \frac1{2\pi i}\oint_{|z - M| = x_*}\frac{z(1 - e\cos z)}{z - M - e\sin z}\,\mathrm dz\), that it is holomorphic in \(|e| \lt \lambda\), and that it agrees with Kepler's real solution for real \(e\). Conclude that Lagrange's series converges for every \(M\) when \(|e| \lt \lambda\).
  4. (Lagrange by residues.) Expand the integrand of part 3 to recover Lagrange's coefficients.
  5. At \(M = \pi/2\), follow \(E(is)\) for \(0\le s \lt \lambda\) and show \(|\mathrm dE/\mathrm de|\to\infty\) as \(s\to\lambda\). Conclude that the series diverges at \(M = \pi/2\) for every real \(e \gt \lambda\). Evaluate \(x_*\) and \(\lambda\).
Hints · 4 rungs
Rung 1

The answer is a number, \(\lambda\approx0.66\), with two halves to its proof: a disc of radius \(\lambda\) on which the solution is holomorphic in \(e\) (so the series converges there), and a singular point on its boundary (so it converges nowhere beyond). Rouché gives the first, a blow-up of \(\mathrm dE/\mathrm de\) the second.

Rung 2

\(|\sin(x + iy)|^2 = \sin^2x + \sinh^2y\). For Rouché compare \(f = z - M\) (one zero, at \(M\)) with \(g = -e\sin z\) on the circle of radius \(r\); you need \(|e|\cosh r \lt r\), and the best \(r\) is the one maximising \(r/\cosh r\). Part 3 is the argument principle weighted by \(z\): the residue of \(zf'/f\) at a simple zero \(z_0\) of \(f\) is \(z_0\).

Rung 3

For part 4 write \(z - M - e\sin z = (z - M)(1 - w)\) with \(w = \frac{e\sin z}{z - M}\), \(|w| \lt 1\) on the circle, so \(\frac{f'}{f} = \frac{1}{z - M} + \frac{\mathrm d}{\mathrm dz}\log(1 - w)\) with a single-valued logarithm; integrate by parts. For part 5, put \(E = \frac\pi2 + iy\), \(e = is\) and use \(\sin(\frac\pi2 + iy) = \cosh y\), \(\cos(\frac\pi2 + iy) = -i\sinh y\).

Rung 4

First line of part 5: \(E - e\sin E = \frac\pi2 + iy - is\cosh y\), so \(E(is) = \frac\pi2 + iy(s)\) where \(y = s\cosh y\); and \(\frac{\mathrm dE}{\mathrm de} = \frac{\sin E}{1 - e\cos E} = \frac{\cosh y}{1 - s\sinh y}\).

Worked solution

Idea. A power series converges on the largest disc on which its sum is holomorphic. Rouché builds that disc — radius \(\lambda\), for every \(M\) at once — and at \(M = \pi/2\) two roots of Kepler's equation collide on its edge, at \(e = \pm i\lambda\), so no larger disc exists.

1. \(\sin(x + iy) = \sin x\cosh y + i\cos x\sinh y\), so \(|\sin|^2 = \sin^2x(1 + \sinh^2y) + \cos^2x\sinh^2y = \sin^2x + \sinh^2y\le\cosh^2y\). \(\phi(r) = r/\cosh r\) has \(\phi' = (\cosh r - r\sinh r)/\cosh^2r\), positive for \(r\tanh r \lt 1\) and negative after (\(r\tanh r\) is increasing), so the maximum is at \(x_*\), with value \(x_*/\cosh x_* = 1/\sinh x_*\) because \(x_* = \coth x_*\).

2. On \(|z - M| = x_*\), \(|\operatorname{Im}z|\le x_*\), so \(|e\sin z|\le|e|\cosh x_* \lt \lambda\cosh x_* = x_* = |z - M|\). By Rouché, \(z - M - e\sin z\) has as many zeros inside as \(z - M\): exactly one. \(\square\)

3. With \(f_e(z) = z - M - e\sin z\), \(f_e' = 1 - e\cos z\), and \(zf_e'/f_e\) has, at the simple zero \(E(e)\) (simple: a double zero would count twice), residue \(E(e)\cdot1\); no other singularity inside. So the formula holds. The integrand is jointly continuous in \((e, z)\) for \(|e| \lt \lambda\), \(z\) on the circle (its denominator does not vanish there, by part 2's inequality), and holomorphic in \(e\); swapping a triangle integral in \(e\) with the \(z\)-integral (both over compact sets) and using Goursat shows, by Morera, that \(E(e)\) is holomorphic on \(|e| \lt \lambda\). For real \(e\in[0,\lambda)\), Kepler's real solution satisfies \(|E - M| = |e\sin E|\le e \lt x_*\), so it lies in the disc and is the unique root there: the two agree. A function holomorphic on \(|e| \lt \lambda\) equals its Taylor series there (deck, "What the integral formula buys"), and the Taylor series of \(E(e)\) at \(e = 0\) is Lagrange's series (part 4 computes its coefficients from the contour formula and finds Lagrange's). So the series converges for \(|e| \lt \lambda\), for every real \(M\). \(\square\)

4. On the circle \(|w| = |e\sin z|/|z - M| \le q \lt 1\), so \(\log(1 - w) = -\sum_{n\ge1}w^n/n\) converges uniformly and is single-valued. Then \(\frac{f'}{f} = \frac{1}{z - M} - \frac{\mathrm d}{\mathrm dz}\sum_n\frac{w^n}{n}\), and \[ E = \frac{1}{2\pi i}\oint\frac{z\,\mathrm dz}{z - M} - \frac{1}{2\pi i}\oint z\,\frac{\mathrm d}{\mathrm dz}\sum_n\frac{w^n}{n}\,\mathrm dz = M + \sum_{n\ge1}\frac1n\cdot\frac{1}{2\pi i}\oint\frac{e^n\sin^nz}{(z - M)^n}\,\mathrm dz, \] integrating by parts round the closed curve (the boundary term vanishes because \(w^n\) is single-valued). Cauchy's formula for the \((n-1)\)-th derivative turns the last integral into \(\frac{e^n}{(n-1)!}\frac{\mathrm d^{n-1}}{\mathrm dM^{n-1}}\sin^nM\). With the \(\frac1n\): Lagrange's \(\frac{e^n}{n!}\frac{\mathrm d^{n-1}}{\mathrm dM^{n-1}}\sin^nM\). \(\square\)

5. At \(M = \frac\pi2\), \(e = is\): \(E = \frac\pi2 + iy\) solves Kepler's equation iff \(y = s\cosh y\). The function \(y/\cosh y\) increases from \(0\) to \(\lambda\) on \([0, x_*]\) (part 1), so it has a continuous increasing inverse \(y(s)\) on \([0,\lambda]\) with \(y(\lambda) = x_*\). For \(s \lt \lambda\), \(\frac\pi2 + iy(s)\) lies in the disc \(|z - \frac\pi2| \lt x_*\), so it is \(E(is)\) (uniqueness, part 2). Differentiating Kepler's equation, \(\frac{\mathrm dE}{\mathrm de} = \frac{\sin E}{1 - e\cos E} = \frac{\cosh y}{1 - s\sinh y}\), and as \(s\to\lambda\), \(s\sinh y(s)\to\lambda\sinh x_* = 1\): the derivative blows up. If Lagrange's series at \(M = \frac\pi2\) had radius \(\rho \gt \lambda\), its sum would be holomorphic on \(|e| \lt \rho\), equal to \(E(e)\) on \(|e| \lt \lambda\) (identity theorem), with bounded derivative near \(e = i\lambda\) — a contradiction. So the radius is exactly \(\lambda\), and a power series diverges at every point outside its disc of convergence, in particular at every real \(e \gt \lambda\). \(\square\)

Numbers. Newton on \(x\tanh x = 1\) from \(x = 1.2\): \(x_* = 1.1996786\), \(\sinh x_* = 1.5088796\), so \[ \lambda = 1/\sinh x_* = \sqrt{x_*^2 - 1} = 0.6627434 . \]

Check. \(\lambda^2 = x_*^2 - 1\) because \(\cosh^2 - \sinh^2 = 1\) with \(\cosh x_* = x_*\sinh x_*\), agreeing with unit 09's form ✓. The singular point \(E = \frac\pi2 + ix_*\), \(e = i\lambda\) satisfies \(1 - e\cos E = 1 - \lambda\sinh x_* = 0\): exactly unit 09's "implicit function theorem fails" point ✓. Unit 09's cruder real bound gave convergence to \(0.6599\), just inside \(0.6627\) ✓. Mars (\(e = 0.0934\)) is deep inside; Halley (\(e = 0.967\)) far outside, where unit 09 saw the terms grow by \((0.967/0.6627)^2\approx2.1\) per two orders ✓.

P11 Bessel's constants by stationary phase medium

exercises: stationary phase with two stationary points, endpoint contributions and their cancellation, unit 09's debt, numbers

Bessel's integral (unit 09) is \(J_n(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(n\theta - x\sin\theta)}\,\mathrm d\theta\) for integer \(n\).

  1. Apply the deck's stationary-phase theorem, with \(\lambda = x\), to show \(J_n(x) = \sqrt{\frac{2}{\pi x}}\cos\big(x - \frac{n\pi}{2} - \frac\pi4\big) + O(1/x)\).
  2. Compare with \(J_0(10) = -0.245936\) and \(J_0(20) = 0.167025\), and predict the first two positive zeros of \(J_0\) (true values \(2.40483\), \(5.52008\)).
Hints · 3 rungs
Rung 1

A formula with two constants, \(\sqrt{2/\pi}\) and the phase \(n\pi/2 + \pi/4\); then three numbers.

Rung 2

The phase is \(\psi(\theta) = -\sin\theta\), stationary at \(\theta = \pm\frac\pi2\). The theorem as stated handles one stationary point on an interval, so split \([-\pi, \pi]\) at \(0\) and at \(\pm\pi\); ask what the endpoint terms of the integration by parts do.

Rung 3

First line: at \(\theta = \frac\pi2\), \(\psi = -1\), \(\psi'' = \sin\frac\pi2 = 1 \gt 0\), and \(g = \frac{e^{in\pi/2}}{2\pi}\): contribution \(\frac{e^{in\pi/2}}{2\pi}\sqrt{\frac{2\pi}{x}}\,e^{-ix + i\pi/4}\).

Worked solution

Idea. The integrand oscillates fast everywhere except where \(\sin\theta\) is stationary, at the top and bottom of the circle; each of those two points contributes a Fresnel integral, and they are complex conjugates.

1. On \([0,\pi]\) the only stationary point is \(\frac\pi2\); on \([-\pi,0]\) it is \(-\frac\pi2\); \(\psi'' = \sin\theta = \pm1\ne0\) there. Endpoints: the \(O(1/x)\) boundary terms \(\big[\frac{g e^{ix\psi}}{ix\psi'}\big]\) at \(\theta = 0\) from the two halves are equal and opposite, and those at \(\pm\pi\) cancel by periodicity — so they cancel, and in any case they are \(O(1/x)\). Stationary points: at \(\frac\pi2\), \(\psi = -1\), \(\psi'' = 1\): \(\frac{e^{in\pi/2}}{2\pi}\sqrt{\frac{2\pi}{x}}e^{-ix + i\pi/4}\). At \(-\frac\pi2\), \(\psi = 1\), \(\psi'' = -1\): \(\frac{e^{-in\pi/2}}{2\pi}\sqrt{\frac{2\pi}{x}}e^{ix - i\pi/4}\), the conjugate. Sum: \[ J_n(x) = \frac{1}{2\pi}\sqrt{\frac{2\pi}{x}}\cdot2\cos\Big(x - \frac{n\pi}{2} - \frac\pi4\Big) + O(1/x) = \sqrt{\frac{2}{\pi x}}\cos\Big(x - \frac{n\pi}2 - \frac\pi4\Big) + O(1/x). \]

2. \(x = 10\): \(\sqrt{2/10\pi} = 0.252313\), \(\cos(10 - \pi/4) = -0.977994\), product \(-0.246761\) against \(-0.245936\) (error \(8.3\times10^{-4}\)). \(x = 20\): \(0.166656\) against \(0.167025\) (error \(3.7\times10^{-4}\)). Zeros of the cosine: \(x - \frac\pi4 = (k - \frac12)\pi\), \(x = (k - \frac14)\pi\): \(2.35619\) and \(5.49779\), against \(2.40483\) and \(5.52008\) — low by \(2.0\%\) and \(0.40\%\), improving as \(x\) grows ✓.

Check. The errors are smaller than the \(O(1/x)\) allowed: because the endpoint terms cancel exactly, the true error is \(O(x^{-3/2})\) (cited in the deck: Watson's Treatise, §7.21), and the two errors above fall by \(2.2\) when \(x\) doubles, between \(2\) and \(2^{3/2}\). Dimensions: \(x\) is a pure number; so is \(J_n\) ✓. Unit 09 proved the \(x^{-1/2}\) decay and unit frequency but could not fix \(\sqrt{2/\pi}\) or \(\frac{n\pi}2 + \frac\pi4\); both are now proved.

P12 Steepest descent: the transform of \(e^{-t^4}\) hard

exercises: locating saddles, choosing the admissible ones, a contour shift with every side estimated, Laplace's method with a complex exponent, numbers

Let \(F(\lambda) = \int_{-\infty}^\infty e^{-t^4}\cos(\lambda t)\,\mathrm dt\). It decays faster than any power of \(\lambda\) (why?), but how fast?

  1. Write \(F\) as \(\int e^{-t^4 + i\lambda t}\,\mathrm dt\), rescale \(t = a\tau\) with \(a = (\lambda/4)^{1/3}\), and find the saddles of the exponent.
  2. Show that the two saddles in the upper half-plane lie on one horizontal line, and move the contour onto it. Bound the two vertical sides explicitly.
  3. Show that on the new line the real part of the exponent has exactly two maxima, at the saddles, and deduce \[ F(\lambda)\sim\frac{2}{a}\sqrt{\frac\pi6}\;e^{-3a^4/2}\cos\Big(\frac{3\sqrt3}{2}a^4 - \frac\pi6\Big). \]
  4. Compare with the exact values \(F(10) = -2.6942\times10^{-3}\) and \(F(20) = -2.1669\times10^{-6}\).
Hints · 4 rungs
Rung 1

The answer is exponentially small in \(\lambda^{4/3}\) and oscillates in sign. Steepest descent will give it as the real part of one saddle's contribution, doubled.

Rung 2

With \(t = a\tau\) and \(\lambda = 4a^3\), the exponent is \(a^4\varphi(\tau)\), \(\varphi(\tau) = -\tau^4 + 4i\tau\). Saddles: \(\tau^3 = i\). Two of the three cube roots have imaginary part \(\frac12\).

Rung 3

On the side \(\tau = \pm X + iy\), \(0\le y\le\frac12\): \(\operatorname{Re}\varphi = -(X^4 - 6X^2y^2 + y^4) - 4y\le-X^4 + \frac32X^2\). On the line \(\tau = \sigma + \frac i2\): \(\operatorname{Re}\varphi = -\sigma^4 + \frac32\sigma^2 - \frac{33}{16}\).

Rung 4

First line of part 3: at \(\tau_0 = e^{i\pi/6}\), \(\varphi(\tau_0) = \tau_0(-\tau_0^3 + 4i) = 3i\tau_0 = 3e^{2\pi i/3}\) and \(\varphi''(\tau_0) = -12\tau_0^2 = -12e^{i\pi/3}\); the contribution is \(a\,e^{a^4\varphi(\tau_0)}\sqrt{2\pi/(-a^4\varphi''(\tau_0))}\).

Worked solution

Idea. On the real axis the integrand oscillates and cancels itself almost completely; lifted to the horizontal line through the two saddles it has two honest peaks, and Laplace's method reads the answer off them.

Why faster than any power. \(e^{-t^4}\) is Schwartz, so the dictionary gives \(\widehat{f^{(k)}} = (-i\lambda)^k\hat f\) with the left side bounded: \(\hat f = O(\lambda^{-k})\) for every \(k\). The sine part vanishes by oddness, so \(F = \int e^{-t^4 + i\lambda t}\mathrm dt\).

1. \(t = a\tau\): \(-t^4 + i\lambda t = -a^4\tau^4 + 4ia^4\tau = a^4\varphi(\tau)\), and \(F = a\int e^{a^4\varphi(\tau)}\mathrm d\tau\), a large-parameter integral with \(\Lambda = a^4\). \(\varphi' = -4\tau^3 + 4i = 0\) at \(\tau^3 = i\): \(\tau_0 = e^{i\pi/6}\), \(\tau_1 = e^{5i\pi/6}\), \(\tau_2 = -i\).

2. \(\operatorname{Im}\tau_0 = \operatorname{Im}\tau_1 = \frac12\): both on the line \(\operatorname{Im}\tau = \frac12\). (\(\tau_2\) is below the axis, where \(\operatorname{Re}\varphi\) at the saddle is \(+3\): a huge contribution that must cancel — not admissible, as in the deck's worked example 5.) Rectangle \(-X, X, X + \frac i2, -X + \frac i2\), integrand entire. On the sides, from rung 3, the integrand is at most \(e^{a^4(-X^4 + 3X^2/2)}\), over a length \(\frac12\): \(\to0\) as \(X\to\infty\). So \(F = a\int_{-\infty}^\infty e^{a^4\varphi(\sigma + i/2)}\,\mathrm d\sigma\). (Expanding: \((\sigma + \frac i2)^4 = \sigma^4 + 2i\sigma^3 - \frac32\sigma^2 - \frac i2\sigma + \frac1{16}\), and \(4i(\sigma + \frac i2) = 4i\sigma - 2\), giving the real part in rung 3.)

3. \(\frac{\mathrm d}{\mathrm d\sigma}(-\sigma^4 + \frac32\sigma^2) = -4\sigma^3 + 3\sigma = 0\) at \(\sigma = 0, \pm\frac{\sqrt3}2\); the maxima are at \(\pm\frac{\sqrt3}{2}\), i.e. at \(\tau_0\) and \(\tau_1\), with \(\operatorname{Re}\varphi = -\frac{9}{16} + \frac98 - \frac{33}{16} = -\frac32\), while at \(\sigma = 0\) it is \(-\frac{33}{16}\). Near each maximum \(\varphi(\sigma + \frac i2) = \varphi(\tau_k) + \frac12\varphi''(\tau_k)(\sigma - \sigma_k)^2 + O(|\sigma - \sigma_k|^3)\), and \(\operatorname{Re}(-\varphi''(\tau_0)) = 12\cos\frac\pi3 = 6 \gt 0\): Laplace's method with a complex exponent (proof as in the deck, with \(\sqrt{2\pi/(-\Lambda\varphi'')}\) the principal root, since \(\int e^{-\Lambda c u^2/2}\mathrm du = \sqrt{2\pi/\Lambda c}\) for \(\operatorname{Re}c \gt 0\) by analytic continuation). Contribution of \(\tau_0\): \[ a\,e^{3a^4e^{2\pi i/3}}\sqrt{\frac{2\pi}{12a^4e^{i\pi/3}}} = \frac1a\sqrt{\frac\pi6}\,e^{-\frac32a^4}e^{i\frac{3\sqrt3}{2}a^4 - i\frac\pi6}. \] Since \(\tau_1 = -\bar\tau_0\) and \(\varphi(-\bar\tau) = \overline{\varphi(\tau)}\), the other saddle gives the complex conjugate; the sum is twice the real part — the stated formula. \(\square\)

4. \(\lambda = 10\): \(a = 1.357209\), \(a^4 = 3.3930\), formula \(-2.7848\times10^{-3}\) against \(-2.6942\times10^{-3}\) (\(3.4\%\)). \(\lambda = 20\): \(a = 1.709976\), \(a^4 = 8.5499\), formula \(-2.1764\times10^{-6}\) against \(-2.1669\times10^{-6}\) (\(0.44\%\)). The relative error falls roughly like \(1/a^4\), as the next Laplace term would.

Check. Decay rate \(e^{-\frac32(\lambda/4)^{4/3}}\): faster than any power, slower than a Gaussian's \(e^{-\lambda^2/4}\) — right, because the rate is set by how far the contour can be lifted: here the saddles sit at height \(a/2\propto\lambda^{1/3}\), while for \(e^{-t^2}\) the single saddle is at \(i\lambda/2\), and lifting by \(\lambda/2\) buys \(e^{-\lambda^2/4}\) ✓. Dimensions: all quantities dimensionless ✓. Sign changes: zeros where \(\frac{3\sqrt3}2a^4 - \frac\pi6 = (k + \frac12)\pi\), infinitely many, as a real transform of an even function may have ✓.