Perturbation theory: the series that diverges and is still right

Idea. Scale so that the uniform-gravity problem has everything of size one; what is left over is the ratio of the height reached to the Earth's radius.

1. With \(z = (V^2/g)y\), \(t = (V/g)s\), the chain rule gives \(\dot z = V y'\) and \(\ddot z = g\,y''\). So \(g y'' = -g/(1 + \varepsilon y)^2\) with \[ \varepsilon = \frac{V^2}{gR} . \] Data: \(z(0) = 0\) gives \(y(0) = 0\); \(\dot z(0) = V\) gives \(y'(0) = 1\). Since \(V^2/2g\) is the uniform-gravity height, \(\varepsilon\) is twice that height over the Earth's radius: how far up the projectile goes, measured against the scale on which gravity weakens.

2. \((1 + \varepsilon y)^{-2} = 1 - 2\varepsilon y + O(\varepsilon^2)\). Order 1: \(y_0'' = -1\), \(y_0(0) = 0\), \(y_0'(0) = 1\), so \(y_0 = s - \tfrac12 s^2\). Order \(\varepsilon\): \(y_1'' = 2y_0 = 2s - s^2\), zero data, so \(y_1 = \tfrac13 s^3 - \tfrac1{12}s^4\).

3. \(y' = 1 - s + \varepsilon(s^2 - \tfrac13 s^3) + O(\varepsilon^2)\) vanishes at \(s^* = 1 + \tfrac23\varepsilon + O(\varepsilon^2)\). Then \(y(s^*) = y_0(s^*) + \varepsilon y_1(1) + O(\varepsilon^2)\), and \(y_0(s^*) = \tfrac12 - \tfrac12(s^* - 1)^2 = \tfrac12 + O(\varepsilon^2)\) because \(y_0\) is stationary at \(s = 1\). With \(y_1(1) = \tfrac13 - \tfrac1{12} = \tfrac14\): \[ y_{\max} = \tfrac12 + \tfrac14\varepsilon + O(\varepsilon^2),\qquad h = \frac{V^2}{2g}\Big(1 + \frac{V^2}{2gR}\Big) + O(\varepsilon^2) . \]

4. \(\varepsilon = 10^6/(9.81\times6.371\times10^6) = 0.016000\). Zeroth order: \(V^2/2g = 50\,968\) m. First order: \(50\,968\times1.0080 = 51\,376\) m. Exact, from energy, \(\tfrac12V^2 = gR^2\big(\tfrac1R - \tfrac1{R+h}\big)\), so \(h = \frac{V^2/2g}{1 - \varepsilon/2} = 51\,379\) m. The zeroth order is \(0.80\%\) low; the first order is \(6.4\times10^{-5}\) low, which is \((\varepsilon/2)^2 = 6.4\times10^{-5}\) — the next term of the exact expansion \((1 - \varepsilon/2)^{-1} = 1 + \varepsilon/2 + \varepsilon^2/4 + \cdots\).

Check. Dimensions: \(V^2/gR\) is (m²s⁻²)/(m s⁻² · m), dimensionless. Limit \(R \to \infty\) gives \(\varepsilon \to 0\) and the uniform-gravity \(V^2/2g\). The sign is right: weaker gravity higher up lets it go higher.

Idea. At \(\varepsilon = 0\) the cubic drops to degree one and two roots are lost: a singular perturbation in the equation. They are lost to infinity, and the scaling that finds them is the one that makes the cubic term compete again.

1. Regular root. \(x = 1 + \varepsilon x_1 + \varepsilon^2x_2\): \(\varepsilon(1 + 3\varepsilon x_1) - 1 - \varepsilon x_1 - \varepsilon^2x_2 + 1 + O(\varepsilon^3) = 0\). Order \(\varepsilon\): \(1 - x_1 = 0\). Order \(\varepsilon^2\): \(3x_1 - x_2 = 0\). So \(x = 1 + \varepsilon + 3\varepsilon^2 + O(\varepsilon^3)\).

2. Singular roots. Balances: \(\varepsilon^{1-3\alpha} = \varepsilon^{-\alpha}\) gives \(\alpha = \tfrac12\), with the constant \(1 \ll \varepsilon^{-1/2}\): consistent. (\(\alpha = 0\) is the regular root; \(\varepsilon^{1-3\alpha} = 1\) gives \(\alpha = \tfrac13\) with the middle term \(\varepsilon^{-1/3}\) dominating both: inconsistent.) So \(x = X/\delta\), \(\delta = \sqrt\varepsilon\), and \(X^3 - X + \delta = 0\). Put \(X = X_0 + a_1\delta + a_2\delta^2\):

• order 1: \(X_0^3 - X_0 = 0\), so \(X_0 = \pm1\) (the root \(X_0 = 0\) is the regular one again);
• order \(\delta\): \(3X_0^2a_1 - a_1 + 1 = 2a_1 + 1 = 0\), so \(a_1 = -\tfrac12\);
• order \(\delta^2\): \(3X_0^2a_2 + 3X_0a_1^2 - a_2 = 2a_2 + \tfrac34X_0 = 0\), so \(a_2 = -\tfrac38X_0\).

Dividing by \(\delta\): \[ x = \pm\frac1{\sqrt\varepsilon} - \frac12 \mp \frac38\sqrt\varepsilon + O(\varepsilon) . \]

3. At \(\varepsilon = 0.01\): regular \(1 + 0.01 + 0.0003 = 1.0103\) against \(1.0103126\) — error \(1.3\times10^{-5}\), order \(\varepsilon^3\). Singular: \(10 - 0.5 - 0.0375 = 9.4625\) against \(9.4564927\), and \(-10 - 0.5 + 0.0375 = -10.4625\) against \(-10.4668053\) — errors \(6\times10^{-3}\) and \(4\times10^{-3}\), order \(\delta^2 = \varepsilon\), the next term in \(x\).

Check. The roots of \(\varepsilon x^3 + 0\cdot x^2 - x + 1\) sum to zero. The expansions give \(1 + (-\tfrac12 - \tfrac12) + (\mp\tfrac38\sqrt\varepsilon\ \text{cancelling}) = 0 + O(\varepsilon)\), and \(O(\varepsilon)\) is the first order not kept in the singular roots. Their product is \(-1/\varepsilon\): \(1\cdot(1/\delta)(-1/\delta) = -1/\varepsilon\) at leading order.

Idea. Every claim is about a limit of a ratio; write the ratio down and bound it.

1. True. Taylor with Lagrange remainder: \(\sin x = x - \tfrac{\cos\xi}{6}x^3\) for some \(\xi\) between 0 and \(x\), so \(|\sin x - x|/|x|^3 \le \tfrac16\).

2. True. With \(s = 1/x\): \(|x\ln x| = \ln s/s\). Since \(e^u \ge u^2/2\) for \(u \ge 0\), \(s \ge (\ln s)^2/2\), so \(\ln s/s \le 2/\ln s \to 0\).

3. True. \(e^{-1/x}/x^{100} = s^{100}e^{-s} \le s^{100}\cdot 101!/s^{101} = 101!/s \to 0\), using \(e^s \ge s^{101}/101!\).

4. True. \(\ln(1+x) = x - \tfrac12x^2 + O(x^3)\), so \(\tfrac1x\ln(1+x) = 1 - \tfrac12x + O(x^2)\) and \((1+x)^{1/x} = e\cdot\exp(-\tfrac12x + O(x^2)) = e(1 - \tfrac12x + O(x^2))\). Hence \((1+x)^{1/x} - e = -\tfrac12ex + O(x^2)\), and dividing by \(-\tfrac12ex\) gives \(1 + O(x) \to 1\). At \(x = 0.01\): \(2.7048138 - 2.7182818 = -0.0134680\), against \(-\tfrac12e(0.01) = -0.0135914\): \(0.9\%\) apart, of order \(x\), as the \(O(x^2)\) remainder predicts.

5. False. \(f = x^2 + x\), \(g = x^2\): \(f/g \to 1\), but \(e^f/e^g = e^x \to \infty\). Exponentiating converts an absolute error into a relative one, so \(\sim\), which controls only the relative error, is not preserved. This is the deck's phase lesson: inside a cosine or an exponential you need \(f - g \to 0\), not \(f/g \to 1\).

Check. Item 4 is the kind of statement the whole unit makes: an equivalence proved from a limit, then confirmed with a number whose error has the predicted size.

Idea. Uniqueness comes from dividing by the last term kept; non-uniqueness comes from functions smaller than every term of the sequence.

1. Subtracting, \(\sum_{n=0}^N c_n\phi_n = o(\phi_N)\) for every \(N\), with \(c_n = a_n - b_n\). Suppose some \(c_n \ne 0\) and let \(k\) be the least such. Then at \(N = k\) the sum is \(c_k\phi_k\), so \(c_k\phi_k/\phi_k = c_k \to 0\), that is, \(c_k = 0\): contradiction. Hence all \(c_n = 0\). \(\square\) (Equivalently, the slide's formula \(a_N = \lim(f - \sum_{n\lt N}a_n\phi_n)/\phi_N\) determines each coefficient from the earlier ones.)

2. \(\phi_{n+1}/\phi_n = x^2 \to 0\), so the sequence is asymptotic. By the rung-3 identity, \[ \frac1{1-x} - \sum_{n=0}^N\phi_n = \frac{x^{2N+2}}{1-x}, \qquad \frac{x^{2N+2}/(1-x)}{(1+x)x^{2N}} = \frac{x^2}{1 - x^2} \to 0 , \] which is \(o(\phi_N)\) for every \(N\). \(\square\) All coefficients are 1 here, while in powers of \(x\) they are also all 1 but attached to different functions: the coefficients are unique only relative to the sequence.

3. We need \(e^{-1/x} = o(x^N)\) for every \(N\): P3 part 3 with \(100\) replaced by \(N\), via \(e^s \ge s^{N+1}/(N+1)!\). So all coefficients are zero. Then \(g\) and \(g + e^{-1/x}\) have the same expansion for any \(g\) that has one, and they are different functions. \(\square\)

Check. At \(x = 0.1\), \(e^{-1/x} = 4.5\times10^{-5}\), already smaller than \(x^4\); at \(x = 0.05\) it is \(2.1\times10^{-9}\), smaller than \(x^6 = 1.6\times10^{-8}\). "Beyond all orders" is visible in numbers only for small \(x\), but it is always there.

Idea. The remainder formula of the deck, \(S - \sum_{n\lt N}t_n = R_N\) with \(R_N\) of the sign of \((-1)^N\) and \(|R_N| \lt N!\,x^N\), turns truncation into a controlled estimate.

1. \(|t_{n+1}/t_n| = (n+1)/10\), which is below 1 for \(n \le 8\), equal to 1 at \(n = 9\), above after. So \(|t_9| = |t_{10}| = 9!/10^9 = 3.6288\times10^{-4}\) are the smallest.

2. The partial sums: \(1,\ 0.9,\ 0.92,\ 0.914,\ 0.9164,\ 0.9152,\ 0.91592,\ 0.915416,\ 0.9158192\) (through \(n = 8\)). So \(S_9 = 0.9158192\).

3. \(S(0.1) - S_9 = 0.9156333 - 0.9158192 = -1.859\times10^{-4}\). The first omitted term is \(t_9 = -3.629\times10^{-4}\): same sign ✓, and the error is \(0.51\) of it in size ✓. Keeping \(t_9\) too gives \(S_{10} = 0.9154563\), error \(+1.770\times10^{-4}\), again of the sign of the next term \(t_{10} \gt 0\) and smaller than it. Either way the accuracy is about \(1.8\times10^{-4}\) — about four digits, from a series that diverges.

4. No. From \(n = 10\) on the terms grow: \(t_{19} = -19!/10^{19} = -0.0122\). The error bound \(|R_{20}| \lt |t_{20}| = 20!/10^{20} = 0.024\) is more than a hundred times the optimal error, and the true error after twenty terms, \(S_{20} - S = -0.0080\), is of that size: about forty-five times worse than stopping at the smallest term.

Check. The deck's bound \((e^2/x)e^{-1/x} = 73.9\times4.54\times10^{-5} = 3.4\times10^{-3}\) holds, crudely. The Stirling estimate of half the smallest term, \(\sqrt{\pi/2x}\,e^{-1/x} = 1.80\times10^{-4}\), sits between the two optimal errors. Dimensions: all dimensionless.

Idea. The general solution is one particular solution plus \(Ce^{1/x}\). For \(x \to 0^-\), \(e^{1/x}\) is beyond all orders, so the series is blind to \(C\): an asymptotic series cannot choose among solutions that differ by an exponentially small amount.

1. Coefficient of \(x^0\): \(c_0 = 0\). Coefficient of \(x^{n+1}\), \(n \ge 0\): \(nc_n + c_{n+1} = [n = 0]\). So \(c_1 = 1\) and \(c_{n+1} = -nc_n\) for \(n \ge 1\): \(c_2 = -1\), \(c_3 = 2\), \(c_4 = -6\), and by induction \(c_{n+1} = (-1)^nn!\). The recurrence forces every coefficient, so the formal solution is unique: \[ y \sim \sum_{n\ge0}(-1)^nn!\,x^{n+1} = x\sum_{n\ge0}(-1)^nn!\,x^n , \] which is \(x\) times the Stieltjes series, divergent for \(x \ne 0\) by the ratio test.

2. For \(x \gt 0\), \((e^{-1/x}y)' = e^{-1/x}(y' + y/x^2) = e^{-1/x}/x\). The integrand \(e^{-1/u}/u \to 0\) as \(u \to 0^+\), so it is integrable at 0, and \[ y = e^{1/x}\int_0^x \frac{e^{-1/u}}u\,\mathrm du + Ce^{1/x} . \] Substitute \(1/u = 1/x + t\), \(t\) from \(\infty\) (at \(u = 0\)) to \(0\) (at \(u = x\)): \(u = x/(1 + xt)\), \(\mathrm du = -x^2(1 + xt)^{-2}\mathrm dt\), \(e^{-1/u} = e^{-1/x}e^{-t}\), \(1/u = (1 + xt)/x\). The integral becomes \(e^{-1/x}\int_0^\infty \frac{x\,e^{-t}}{1 + xt}\,\mathrm dt\), so the first term is exactly \(xS(x)\). Since \(0 \lt S \le 1\), \(xS(x) \to 0\); while \(Ce^{1/x} \to \pm\infty\) unless \(C = 0\). Exactly one solution tends to 0, and it is \(xS(x)\). \(\square\)

3. The equation is linear, so the difference of two solutions solves the homogeneous equation \(x^2y' + y = 0\), whose solutions are \(Ce^{1/x}\). For \(x \lt 0\) write \(x = -\xi\), \(\xi \gt 0\): \(e^{1/x} = e^{-1/\xi} = o(\xi^N)\) for every \(N\) (P4). So if \(y_1 \sim \sum c_nx^n\) then \(y_1 + Ce^{1/x} \sim \sum c_nx^n\) too: at every \(N\) the extra term is \(o(|x|^N)\), so the defining condition of the expansion holds for the sum exactly when it holds for \(y_1\). \(\square\)

Which expansion it would be is suggested by part 1: an expansion that could be substituted into the equation term by term would have to satisfy the same recurrence, so it can only be part 1's series. Proving that the solutions really do have an expansion as \(x \to 0^-\) needs a remainder estimate that this problem does not attempt, and nothing above depends on it.

The moral: as \(x \to 0^-\) the divergent series fits a whole one-parameter family of solutions equally well, and so determines none of them. As \(x \to 0^+\) it is the other way round: \(e^{1/x}\) blows up, only one solution has an expansion at all, and the series names it. Whether a divergent series can pick out a function depends on what exponentially small terms are allowed — which is the question Borel summation answers.

Check. At \(x = 0.1\), \(xS(x) = 0.0915633\) (from P5's value), and the optimally truncated series times \(x\) gives \(0.0915819\): the same accuracy as P5, scaled by \(x\). Limit: as \(x \to 0^+\), \(xS(x) = x + O(x^2)\), matching \(c_1 = 1\). Direct check of part 2 without the substitution: \(y = xS\) and \(y = \sum(-1)^nn!x^{n+1}\) agree to the optimal-truncation accuracy at every small \(x\) tried, as they must if both describe the same solution.

Idea. Exactly the Duffing calculation, but the first-order forcing has no resonant part, so the frequency moves only at second order, and \(x_1\) (which now contains a steady offset) feeds the resonance at the next order.

1. With \(\tau = \omega t\), \(\omega^2x'' + x + \varepsilon x^2 = 0\), \(x = x_0 + \varepsilon x_1 + \varepsilon^2x_2\), \(\omega = 1 + \varepsilon\omega_1 + \varepsilon^2\omega_2\), so \(\omega^2 = 1 + 2\varepsilon\omega_1 + \varepsilon^2(\omega_1^2 + 2\omega_2)\). Order 1: \(x_0 = A\cos\tau\). Order \(\varepsilon\): \(x_1'' + x_1 = -2\omega_1x_0'' - x_0^2 = 2\omega_1A\cos\tau - \tfrac12A^2 - \tfrac12A^2\cos2\tau\). Solvability: \(\omega_1 = 0\). Particular solution: the constant gives \(-\tfrac12A^2\); \(c\cos2\tau\) with \(-3c = -\tfrac12A^2\) gives \(c = \tfrac16A^2\). With \(x_1(0) = x_1'(0) = 0\): \[ x_1 = -\tfrac12A^2 + \tfrac16A^2\cos2\tau + \tfrac13A^2\cos\tau . \]

2. Order \(\varepsilon^2\) (with \(\omega_1 = 0\)): \(x_2'' + x_2 = -2\omega_2x_0'' - 2x_0x_1\). The \(\cos\tau\) coefficient: from \(-2\omega_2x_0''\), \(2\omega_2A\); from \(-2x_0x_1 = -2A\cos\tau\,x_1\): the constant of \(x_1\) gives \(-2A(-\tfrac12A^2) = A^3\); the \(\cos2\tau\) term gives \(-2A\cdot\tfrac16A^2\cdot\tfrac12 = -\tfrac16A^3\); the \(\cos\tau\) term gives \(\cos^2\tau\), which has no \(\cos\tau\) part. Solvability: \[ 2\omega_2A + A^3 - \tfrac16A^3 = 0 \ \Rightarrow\ \omega_2 = -\tfrac5{12}A^2,\qquad \omega = 1 - \tfrac5{12}\varepsilon^2A^2 + O(\varepsilon^3) . \]

3. The first-order shift is decided by the \(\cos\tau\) component of the first-order forcing, \(-x_0^2\) here and \(-x_0^3\) for Duffing. An even power of \(x_0 = A\cos\tau\) contains only even harmonics (\(1, \cos2\tau, \dots\)); an odd power contains \(\cos\tau\). In terms of the potential: the perturbation \(\tfrac13\varepsilon x^3\) is odd, stiffening one side of the well exactly as much as it softens the other, and the unperturbed orbit visits \(x\) and \(-x\) equally, so the effects cancel at first order. What survives is second order, through the distortion \(x_1\) that the odd term produces. Duffing's \(\tfrac14\varepsilon x^4\) is even and stiffens both sides, so it shifts the frequency at first order.

4. \(1 - \tfrac5{12}(0.01) = 0.9958333\) against \(0.9954982\): error \(3.4\times10^{-4}\). At \(\varepsilon A = 0.05\) the error is \(3.8\times10^{-5}\), and at \(0.2\) it is \(3.3\times10^{-3}\): a factor near 9–10 per halving, so the next correction is third order. (It can be: the frequency depends on \(\varepsilon A\), and releasing at \(+A\) differs from releasing at \(-A\), so odd powers are allowed.)

Check. Dimensions: in physical units \(\varepsilon\) has units of inverse length, so \(\varepsilon A\) is the dimensionless group, as the answer uses. Sign: negative, and the net effect of an asymmetric well is to soften it — the soft side is visited with more amplitude.

Idea. One new unknown, \(\omega_2\), one condition, the solvability of the order-\(\varepsilon^2\) equation. The only labour is collecting the \(\cos\tau\) coefficient.

Working. Order \(\varepsilon^2\) of \(\omega^2x'' + x + \varepsilon x^3 = 0\), with \(\omega^2 = 1 + 2\varepsilon\omega_1 + \varepsilon^2(\omega_1^2 + 2\omega_2)\): \[ x_2'' + x_2 = -(\omega_1^2 + 2\omega_2)x_0'' - 2\omega_1x_1'' - 3x_0^2x_1 . \] The \(\cos\tau\) coefficients:

• \(-(\omega_1^2 + 2\omega_2)x_0'' = (\omega_1^2 + 2\omega_2)A\cos\tau\): coefficient \(\tfrac9{64}A^5 + 2\omega_2A\);
• \(-2\omega_1x_1''\): \(-\tfrac3{128}A^5\) (rung 3);
• \(-3x_0^2x_1 = -\tfrac3{32}A^5\cos^2\tau(\cos3\tau - \cos\tau)\): the \(\cos\tau\) part of the bracket product is \(\tfrac14 - \tfrac34 = -\tfrac12\), giving \(+\tfrac3{64}A^5\).

Setting the sum to zero: \(2\omega_2A = -\tfrac9{64}A^5 + \tfrac3{128}A^5 - \tfrac3{64}A^5 = -\tfrac{18 - 3 + 6}{128}A^5 = -\tfrac{21}{128}A^5\), so \(\omega_2 = -\tfrac{21}{256}A^4\). \(\square\)

The \(\cos^2\tau\cos3\tau\) identity: \(\cos^2\tau = \tfrac12(1 + \cos2\tau)\) and \(\cos2\tau\cos3\tau = \tfrac12(\cos\tau + \cos5\tau)\), so the \(\cos\tau\) part is \(\tfrac14\).

Check. At \(\lambda = \varepsilon A^2 = 0.1\): \(1 + 0.0375 - 0.00082031 = 1.0366797\) against \(1.0367169\), error \(3.7\times10^{-5}\) (first order alone: \(7.8\times10^{-4}\)). At \(\lambda = 0.01\): \(1.0037418\) against \(1.0037418\), error \(3.9\times10^{-8}\). The error falls by about \(10^3\) for a tenfold drop in \(\lambda\): third order, as it must be once \(\omega_2\) is right. The deck's table of \((\omega - 1 - \tfrac38\lambda)/\lambda^2\), tending to \(-0.082\), is the same check from the other side. Dimensions: \(\varepsilon A^2\) is dimensionless, so \(\omega_2\varepsilon^2 \propto (\varepsilon A^2)^2\) ✓.

Idea. The slow drift of the amplitude is what multiple scales exists to capture; the limit cycle is a fixed point of that drift.

1. Order \(\varepsilon\): \(\partial_t^2X_1 + X_1 = -2\partial_t\partial_TX_0 + (1 - X_0^2)\partial_tX_0\). With \(\psi = t + \phi\), the forcing is \[ 2a'\sin\psi + 2a\phi'\cos\psi - a\sin\psi + \tfrac14a^3(\sin\psi + \sin3\psi) . \] Solvability (no \(\sin\psi\), no \(\cos\psi\)): \[ a' = \tfrac12a\big(1 - \tfrac14a^2\big),\qquad \phi' = 0 . \]

2. Put \(u = a^2\): \(u' = 2aa' = u(1 - u/4)\), logistic. So \(u = 4/(1 + Ce^{-T})\) with \(C = 4/a_0^2 - 1\): \[ a(T) = \frac{2}{\sqrt{1 + (4/a_0^2 - 1)e^{-T}}} \to 2 . \] The limit cycle has amplitude \(2\), whatever \(a_0 \gt 0\), and the equilibrium \(a = 2\) is stable (\(\mathrm d a'/\mathrm da = \tfrac12 - \tfrac38a^2 = -1\) there), while \(a = 0\) is unstable. The phase does not drift: the frequency is \(1 + O(\varepsilon^2)\).

3. Poincaré–Lindstedt assumes a periodic solution from the start. It can find the cycle itself (its amplitude comes out as 2 from the solvability condition), but the approach from \(a_0 \ne 2\) is not periodic, so there is nothing for it to strain.

4. \(2.00010\) and \(2.000001\) against the prediction \(2 + O(\varepsilon)\): the deviations, \(1.0\times10^{-4}\) and \(1.0\times10^{-6}\), are well inside \(O(\varepsilon)\), and they fall a hundredfold for a tenfold drop in \(\varepsilon\) — so, numerically, the first correction to the amplitude is of order \(\varepsilon^2\), better than the first-order theory promises. From \(a_0 = 0.5\), the prediction reaches \(a = 1.9\) at \(T = \ln\big(15/(4/3.61 - 1)\big) = 4.93\), that is \(t = 49\) at \(\varepsilon = 0.1\).

Check. Limit \(\varepsilon \to 0\): the cycle becomes one of the circles of the harmonic oscillator, radius 2 — the one on which the energy fed in by negative damping at \(|x| \lt 1\) balances the energy removed at \(|x| \gt 1\). Averaging \(\varepsilon(1 - x^2)\dot x^2\) over \(x = a\cos t\) gives \(\varepsilon a^2(\tfrac12 - \tfrac18a^2)\), zero at \(a = 2\) ✓ — an independent energy-balance check.

Idea. \(\varepsilon\) multiplies the highest derivative: singular. The reduced equation is first order and can keep one condition; the other is met in a layer of width \(\varepsilon\) at the end where the inner solution decays into the interior.

1. At \(x = 0\), \(X = x/\varepsilon\): \(Y'' + Y' = 0\), solutions \(1, e^{-X}\), decaying inward ✓. At \(x = 1\), \(X = (1 - x)/\varepsilon\): \(Y'' - Y' = 0\), solutions \(1, e^{X}\), growing inward ✗. The layer is at \(x = 0\).

2. Outer: \(y' + y = 0\), \(y(1) = 1\), so \(y_{\rm out} = e^{1-x}\). Inner: \(Y_0 = C(1 - e^{-X})\) from \(Y_0(0) = 0\). Matching (Van Dyke, one term each): the outer as \(x \to 0\) is \(e\); the inner as \(X \to \infty\) is \(C\). So \(C = e\). Composite, outer plus inner minus common part: \[ y_c = e^{1-x} - e^{1 - x/\varepsilon} . \]

3. Characteristic equation \(\varepsilon r^2 + r + 1 = 0\): \(r_\pm = \frac{-1 \pm \sqrt{1 - 4\varepsilon}}{2\varepsilon}\). At \(\varepsilon = 0.01\), \(r_+ = -1.0102051\), \(r_- = -98.989795\). With \(y(0) = 0\), \[ y = \frac{e^{r_+x} - e^{r_-x}}{e^{r_+} - e^{r_-}} . \] Comparison:

The largest error, \(0.024\), is at \(x \approx 0.058\), just outside the layer: it is the \(O(\varepsilon)\) error of the outer solution, since \(r_+ = -1 - \varepsilon + O(\varepsilon^2)\) and so the true outer decay is \(e^{(1+\varepsilon)(1 - x)}\), not \(e^{1-x}\). The exact solution peaks at \(x = 0.0468\) with value \(2.5926\), almost \(e = 2.718\).

Check. Order: the maximum error \(0.024 \approx 2.4\varepsilon\) is \(O(\varepsilon)\), and at \(\varepsilon = 0.05\) it is \(0.086 \approx 1.7\varepsilon\). Limit: as \(\varepsilon \to 0\), \(r_+ \to -1\) and \(r_- \approx -1/\varepsilon\), so the exact solution itself tends to the composite. The roots are exactly the outer and inner exponents — this is why the method works.

Idea. The string is locally uniform on the scale of a wavelength when \(\lambda\) is large, so the wavenumber is \(\lambda(1 + x)\) at each point; the eigenvalue condition is that a whole number of half-wavelengths fits.

1. \(q = (1 + x)^2\), \(\sqrt q = 1 + x\), \(q^{-1/4} = (1 + x)^{-1/2}\). WKB: \(y \approx (1 + x)^{-1/2}\big[C_1\cos\Phi + C_2\sin\Phi\big]\) with \(\Phi = \lambda(x + \tfrac12x^2)\). \(y(0) = 0\) needs \(C_1 = 0\). \(y(1) = 0\) needs \(\sin\Phi(1) = 0\), \(\Phi(1) = \tfrac32\lambda = n\pi\): \[ \lambda_n \approx \frac{2n\pi}{3},\qquad y_n \approx (1 + x)^{-1/2}\sin\big(\tfrac{2n\pi}{3}(x + \tfrac12x^2)\big). \] The amplitude falls by \(1/\sqrt2\) from the light end to the heavy end, and the wavelength shortens there.

2. \(q' = 2(1 + x)\), so \(\varepsilon|q'|/4q^{3/2} = \frac{1}{2\lambda(1 + x)^2} \le \frac1{2\lambda} = \frac{3}{4n\pi}\). For \(n = 1\) this is \(0.24\): marginal. For \(n = 6\), \(0.04\): good.

3. WKB: \(2.09440\), \(6.28319\), \(12.56637\). Relative errors: \(1.65\%\), \(0.23\%\), \(0.058\%\). From \(n = 3\) to \(n = 6\) the error falls by a factor \(3.9\), close to \(2^2 = 4\): the relative error scales as \(1/n^2\), that is as \(\varepsilon^2\). (From \(n = 1\) to \(n = 6\) the factor is \(28\) rather than \(36\): at \(n = 1\), where the validity ratio is \(0.24\), the higher-order terms are not yet negligible.) The \(\varepsilon^2\) is the size of the first neglected term of the phase, which enters \(\lambda\) at relative order \(\varepsilon^2\).

Check. Limit of a uniform string, \(q = 1\): WKB gives \(\lambda_n = n\pi\), exact ✓. The WKB estimate lies above the true value in all three cases, and the error is largest where the validity ratio of part 2 is largest ✓. Dimensions: with \(x\) in units of the string length, \(\lambda\) is dimensionless ✓.

Idea. Unit 00's Buckingham theorem with three quantities and two dimensions leaves one group, so the spectrum is a power law with a pure-number prefactor. The Kolmogorov scale is where the neglected quantity, \(\nu\), stops being negligible — the same logic as a boundary layer.

1. \(a = \tfrac23\), \(b = 2a - 3 = -\tfrac53\): \(E(k) = C\epsilon^{2/3}k^{-5/3}\). The dimension matrix of \((E, \epsilon, k)\) in \((\mathsf L, \mathsf T)\) has rank 2, so there is exactly one dimensionless group, \(E\,\epsilon^{-2/3}k^{5/3}\), and it must be a constant.

2. From \(\nu\) and \(\epsilon\): \(\eta = (\nu^3/\epsilon)^{1/4}\) (check: \(\mathsf{L^6T^{-3}}/\mathsf{L^2T^{-3}} = \mathsf{L^4}\)), \(\tau_\eta = (\nu/\epsilon)^{1/2}\), \(u_\eta = (\nu\epsilon)^{1/4}\). Then \(u_\eta\eta/\nu = (\nu\epsilon)^{1/4}(\nu^3/\epsilon)^{1/4}/\nu = \nu/\nu = 1\). At that scale inertia and viscosity are of the same size; below it, viscosity wins and the cascade ends.

3. \(\eta = (3.375\times10^{-15}/10^{-2})^{1/4} = 7.6\times10^{-4}\) m \(= 0.76\) mm; \(\tau_\eta = 0.039\) s; \(u_\eta = 0.020\ \mathrm{m\,s^{-1}}\). Eddies smaller than about a millimetre do not exist in that flow.

4. On a log–log plot of a measured spectrum, the inertial range is a straight line whose slope is compared with \(-5/3 = -1.667\). The first convincing confirmation was in a tidal channel (Grant, Stewart and Moilliet, J. Fluid Mech. 12, 1962); high-Reynolds-number wind-tunnel boundary layers show it over more than a decade of \(k\) (Saddoughi and Veeravalli, J. Fluid Mech. 268, 1994) — cited, since the data are not reproduced here. A small deviation would mean the assumption fails: that \(E\) depends on more than \(\epsilon\) and \(k\), for instance on the large-scale fluctuations of \(\epsilon\) itself (intermittency). The She–Leveque model, for example, predicts a slope of \(-1 - \zeta_2\) with \(\zeta_2 = \tfrac29 + 2\big(1 - (\tfrac23)^{2/3}\big) = 0.696\), that is \(-1.696\), a correction of \(0.03\) that a single spectrum over one or two decades can barely resolve.

Check. Dimensions of \(C\epsilon^{2/3}k^{-5/3}\): \(\mathsf{L^{4/3}T^{-2}}\cdot\mathsf{L^{5/3}} = \mathsf{L^3T^{-2}}\) ✓. Limit: larger \(\epsilon\) (a more violent flow) gives a smaller \(\eta\) — finer eddies ✓. This is the unit's first lesson in its working form: the dimensional argument says which terms you may drop, and where they stop being droppable.

Idea. The secular term is forced by the equation — it cannot be removed without changing the radial frequency — and that frequency change is the precession, which accumulates and is measured.

1. Order \(\varepsilon\): \(w_1'' + w_1 = w_0^2 = (1 + \tfrac12e^2) + 2e\cos\theta + \tfrac12e^2\cos2\theta\). By the solvability condition, a \(2\pi\)-periodic \(w_1\) exists iff \(\int_0^{2\pi}w_0^2\cos\theta = 0\) and \(\int_0^{2\pi}w_0^2\sin\theta = 0\). The first is \(2e\pi\), nonzero unless \(e = 0\). So for \(e \ne 0\) no first-order periodic solution exists: the orbit does not close. \(\square\)

2. The resonant part gives \(w_1 \supset e\theta\sin\theta\) (since \(L(\theta\sin\theta) = 2\cos\theta\)). Then \(w \approx 1 + e\cos\theta + \varepsilon e\theta\sin\theta = 1 + e\cos((1 - \varepsilon)\theta) + O(\varepsilon^2\theta^2)\), plus bounded \(O(\varepsilon)\) terms. Perihelion, the maximum of \(w\), is at \((1 - \varepsilon)\theta = 2\pi k\), so successive perihelia are \(2\pi/(1 - \varepsilon) = 2\pi(1 + \varepsilon) + O(\varepsilon^2)\) apart: an advance of \(2\pi\varepsilon\).

3. \(\varepsilon = 7.9875\times10^{-8}\); \(2\pi\varepsilon = 5.0187\times10^{-7}\) rad \(= 0.10352''\) per orbit; \(36525/87.9691 = 415.20\) orbits per century; total \(\mathbf{42.98''}\) per century.

4. \(P = 0.322997449\times86400 = 27\,906.98\) s; \(GM = 2.828378\times1.32712\times10^{20} = 3.7536\times10^{20}\ \mathrm{m^3\,s^{-2}}\); \(a = (GMP^2/4\pi^2)^{1/3} = 1.9491\times10^9\) m. Then \(\varepsilon = 3GM/c^2a(1 - e^2) = 1.038\times10^{-5}\), 130 times Mercury's. Advance per orbit \(2\pi\varepsilon = 6.52\times10^{-5}\) rad \(= 0.0037377^\circ\); orbits per year \(365.25/0.322997 = 1130.8\); advance \(\mathbf{4.2266^\circ}\) per year — in one year, roughly what Mercury's perihelion does in thirty-five thousand.

A caution on the pulsar. The measured advance is \(4.2266^\circ\) per year, but this is not an independent test: the total mass quoted was itself inferred from the measured advance, assuming this formula. The independent tests of general relativity in that system use other effects (the orbital decay by gravitational radiation, unit 38). The calculation shows the size of the effect and that the formula is self-consistent.

Check. Dimensions: \(GM/c^2p\) is length over length ✓. Limit \(c \to \infty\): zero advance ✓. \(e = 0\): no perihelion and no secular term ✓. And the answer to the checkpoint's last question: the secular term must be kept because the equation forbids a closed orbit (part 1) and the phase it encodes accumulates into the observed precession; what Poincaré–Lindstedt removes is only the false growth, and it keeps the frequency \(1 - \varepsilon\) that the term encodes.