Determinism, and the one hypothesis it rests on

Idea. The recipe is mechanical: the state is the list of derivatives the equation does not determine, the first blocks of \(f\) are the tautologies \(\dot y_i = y_{i+1}\), and the last block is the equation itself.

1. Damped driven oscillator. \(y = (x, v) \in \mathbb{R}^2\), \[ f\big((x,v),t\big) = \Big(v,\ \tfrac{1}{m}\big(-bv - \kappa x + F_0\cos\Omega t\big)\Big), \qquad \Omega_{\text{dom}} = \mathbb{R}^2 \times \mathbb{R}. \] \(f\) is continuous everywhere, and Lipschitz in \(y\) with \(L = \sqrt{1 + \big((\kappa + b)/m\big)^2}\): indeed \(|\Delta f_1| = |\Delta v| \le \|\Delta y\|\) and \(|\Delta f_2| \le \frac{\kappa}{m}|\Delta x| + \frac{b}{m}|\Delta v| \le \frac{\kappa+b}{m}\|\Delta y\|\).

2. Radial fall. \(y = (x,v)\), \(f((x,v)) = \big(v, -GM/x^2\big)\), and the largest open set on which this is defined and continuous is \(\Omega_{\text{dom}} = \{(x,v) : x > 0\} \times \mathbb{R}\) (or \(x<0\); the two components are separate problems). The singularity at \(x = 0\) is not removable, and no extension of \(f\) to \(x = 0\) is continuous.

3. Third order. The equation determines \(\dddot x\), so the state is \(y = (x, \dot x, \ddot x) \in \mathbb{R}^3\) and \[ f(y) = (y_2,\ y_3,\ -y_1), \qquad \Omega_{\text{dom}} = \mathbb{R}^3\times\mathbb{R}. \] This is not of Newtonian form; nothing in the theory of this unit cares.

Which are global. (1) yes: \(\|f\| \le \frac{|F_0|}{m} + \frac{\kappa+b}{m}\|y\| \le C(1+\|y\|)\), so the global existence theorem applies. (3) yes: \(f\) is linear, \(\|f(y)\| = \|y\|\), again linear growth. (2) no: \(\Omega_{\text{dom}}\) is not all of \(\mathbb{R}^2\), and a mass released from rest at \(x = a\) reaches \(x = 0\) in finite time (P9 does the computation for \(n = 3\); \(n = 2\) is the same argument), so its maximal interval ends.

Check. Dimensions of the second block of (1): \(\mathrm{L\,T^{-2}}\), as an acceleration must be. Limit \(b = \kappa = F_0 = 0\): \(f = (v, 0)\), free motion, \(L = 1\), and the solution \(x_0 + v_0 t\) exists for all time. ✓

Idea. A Lipschitz bound is an inequality you exhibit; a failure is a sequence of pairs whose difference quotient is unbounded. Neither is ever established by looking at a graph.

Proofs.

\(|x|\): \(\big||a|-|b|\big| \le |a-b|\) is the reverse triangle inequality. Note it is not differentiable at \(0\) — Lipschitz does not require differentiability.

\(x^2\): \(|a^2-b^2| = |a+b|\,|a-b|\), so on \([-R,R]\) take \(L = 2R\); globally, the pairs \((n, n+1)\) give difference quotient \(2n+1 \to \infty\).

\(\sin x\): \(|\sin a - \sin b| = |\cos\xi||a-b| \le |a-b|\) by the mean value theorem.

\(\sqrt{|x|}\): the pairs \((1/n^2, 0)\) give quotient \(\frac{1/n}{1/n^2} = n \to \infty\), so no neighbourhood of \(0\) admits a constant. Away from \(0\), on \(|x| \ge \epsilon\), the mean value theorem gives \(L = 1/(2\sqrt\epsilon)\).

\(|x|^{4/3}\): for \(a, b\) of the same sign the mean value theorem gives \(L = \tfrac43 R^{1/3}\) on \([-R,R]\); for opposite signs split at \(0\) and add the two pieces, which at worst doubles the constant. At \(0\) itself the quotient is \(|x|^{4/3}/|x| = |x|^{1/3} \to 0\), so nothing goes wrong. Globally the quotient \(\tfrac43 R^{1/3}\) is unbounded, so it is not Lipschitz on \(\mathbb{R}\).

\(\phi\): the pairs \((1/n, 0)\) give quotient \(\ln n \to \infty\), so \(\phi\) is not Lipschitz near \(0\). Away from \(0\) it is smooth, hence locally Lipschitz there by the mean value theorem.

The Osgood integral. With \(\omega(u) = u\ln(1/u)\) on \((0,1/2]\), substitute \(w = \ln(1/u)\), \(\mathrm{d}w = -\mathrm{d}u/u\): \[ \int_{0^+}^{1/2}\frac{\mathrm{d}u}{u\ln(1/u)} = \int_{\ln 2}^{\infty}\frac{\mathrm{d}w}{w} = \infty . \] The criterion also wants the bound \(|\phi(a)-\phi(b)| \le \omega(|a-b|)\), which is worth one line rather than an assertion. Work on \((0,1/e]\), where \(\phi' = \ln(1/x)-1 \ge 0\) and \(\phi'' = -1/x < 0\): \(\phi\) is increasing and concave there, and a concave function with \(\phi(0)=0\) is subadditive: for \(0 \le b \le a\), \(\phi(a) = \phi\big(b + (a-b)\big) \le \phi(b) + \phi(a-b)\), i.e. \(|\phi(a)-\phi(b)| \le \omega(|a-b|)\). For \(a\) and \(b\) on opposite sides of \(0\), \(\phi\) being odd gives \(|\phi(a)-\phi(b)| = \phi(|a|)+\phi(|b|)\), and for \(|a|,|b| \le 1/(2e)\) — where \(\omega\) is increasing and \(|a|+|b| \le 1/e\) — each term is at most \(\omega(|a-b|)\), so \(2\omega\) does. Doubling \(\omega\) does not affect the divergence of the integral, and uniqueness is a local question, so this neighbourhood of \(0\) is all that is needed: away from \(0\), \(\phi\) is \(C^1\).

So Osgood's criterion applies and \(\dot x = \phi(x)\) has unique solutions through every point, although \(\phi\) is not Lipschitz anywhere near \(0\). The gap between Lipschitz and non-uniqueness is real, and \(\phi\) sits inside it.

Check. \(\sqrt{|x|}\) is Hölder-\(\tfrac12\) and fails Osgood (\(\int_0^\varepsilon u^{-1/2}\mathrm{d}u = 2\sqrt\varepsilon < \infty\)); \(\phi\) is "almost Lipschitz" and passes. The two examples bracket the line. ✓

Idea. Reversal is a substitution, and a substitution carries hypotheses with it. Nothing new is proved; a theorem is transported.

1. With \(z(t) = y(2t_0-t)\), the chain rule gives \(\dot z(t) = -\dot y(2t_0-t) = -f\big(y(2t_0-t), 2t_0-t\big) = -f\big(z(t), 2t_0-t\big) = g(z(t),t)\). For the Lipschitz constant, \[ \|g(z,t)-g(w,t)\| = \|f(z,2t_0-t) - f(w,2t_0-t)\| \le L\|z-w\|, \] the same \(L\), because the two arguments are compared at the same instant \(2t_0-t\) — which is exactly what "Lipschitz in \(y\), uniformly in \(t\)" allows. Continuity of \(g\) is likewise inherited.

2. Suppose two solutions of \(\dot y = f(y,t)\) agree at \(t_0\) and differ at some \(t < t_0\). Reflecting both gives two solutions of \(\dot z = g(z,t)\) agreeing at \(t_0\) and differing at \(2t_0 - t > t_0\), contradicting forward uniqueness for \(g\), which holds because \(g\) satisfies the same hypotheses. So backward uniqueness follows. (The theorem as stated already covers \([t_0-a,t_0]\); this argument is the reason one never has to prove anything twice.)

3. On the dome, take the state \(r = 1\) m, \(\dot r > 0\) at \(t = 0\). Energy gives \(\dot r = \sqrt{\tfrac43 k}\,r^{3/4}\) for the spontaneous solutions, so this state lies on the member with \(1 = \frac{k^2}{144}(0-T)^4\), i.e. \(T = -(144/k^2)^{1/4} = -3.464\) s. Going backwards from \(t = 0\), the right-hand side is locally Lipschitz everywhere the trajectory has been, since \(s > 0\) there, so the past is uniquely determined on \((-3.464, 0]\). At \(t = T = -3.464\) s the trajectory arrives at the apex with zero velocity, and before that instant nothing is determined: the mass may have been sitting there for any length of time, including forever. The past is determined exactly as far back as the apex, and not one instant further.

Check. The general formula is \(T = -\big(144\,r/k^2\big)^{1/4}\), and \([r/k^2] = \mathrm{L}/(\mathrm{L\,T^{-4}}) = \mathrm{T^4}\), so \(T\) is a time. ✓ With \(r = 1\) m and \(k = 1\,\mathrm{m^{1/2}s^{-2}}\), \(144^{1/4} = 3.4641\) s. Limit: a state further down the dome has a longer determined past, as it must. ✓

Idea. Completeness gives the limit; \(q < 1\) gives the Cauchy property. Remove either and the conclusion goes.

1. For \(u \ge 1\), \(Pu = u + 1/u \ge 1\), so \(P\) maps \([1,\infty)\) into itself. For \(1 \le u < w\), \[ Pw - Pu = (w-u) - \frac{w-u}{uw} = (w-u)\Big(1 - \frac{1}{uw}\Big) \in [0, w-u), \] strictly less than \(w - u\) because \(uw \ge 1\) and \(u \ne w\) forces \(uw > 1\). A fixed point would need \(1/u = 0\). None exists.

The reason strictness is not enough: the contraction factor here is \(1 - 1/(uw)\), which creeps up to \(1\). The Banach hypothesis demands a factor bounded away from \(1\) uniformly.

2. Let \(X\) be compact and \(d(Pu,Pw) < d(u,w)\) for \(u \ne w\). Then \(P\) is continuous (it is non-expansive), so \(\varphi(u) = d(u, Pu)\) is continuous, and on a compact space it attains its minimum at some \(u_\ast\). If \(Pu_\ast \ne u_\ast\) then \[ \varphi(Pu_\ast) = d\big(Pu_\ast, P(Pu_\ast)\big) < d(u_\ast, Pu_\ast) = \varphi(u_\ast), \] contradicting minimality. So \(Pu_\ast = u_\ast\). Uniqueness: two distinct fixed points \(u \ne w\) would give \(d(u,w) = d(Pu,Pw) < d(u,w)\).

3. \(X = \mathbb{R}\), \(Pu = u+1\). It is an isometry, so \(d(Pu,Pw) = d(u,w)\), and it has no fixed point. \(X\) is complete; only strictness has been dropped.

Check. Part 1 against part 2: \([1,\infty)\) is closed and complete but not compact, which is exactly the hypothesis part 2 adds. Part 3 against part 1: non-strict on a complete space, strict on a non-compact complete space, strict on a compact space — three points on one line, and only the last has a fixed point. ✓

Idea. The Picard operator on a linear system shifts a Taylor partial sum by one term per application, so the iteration is the power series, discovered rather than assumed.

1. The iterates. \(y = (x,v)\), \(f(x,v) = (v, x)\), \(y_0 = (1,0)\), \((Py)(t) = (1,0) + \int_0^t f(y(s))\,\mathrm{d}s\). Then \[ y^{(1)} = (1,\,t),\qquad y^{(2)} = \Big(1+\tfrac{t^2}{2},\ t\Big),\qquad y^{(3)} = \Big(1+\tfrac{t^2}{2},\ t+\tfrac{t^3}{6}\Big),\qquad y^{(4)} = \Big(1+\tfrac{t^2}{2}+\tfrac{t^4}{24},\ t+\tfrac{t^3}{6}\Big). \]

2. The limit. Claim: \(y^{(2m)} = \big(\sum_{j=0}^{m}\frac{t^{2j}}{(2j)!},\ \sum_{j=0}^{m-1}\frac{t^{2j+1}}{(2j+1)!}\big)\) and \(y^{(2m+1)}\) has the same first block with one more term in the second. Induction: applying \(P\) integrates the second block into the first and the first block into the second, and \(\int_0^t \frac{s^{2j}}{(2j)!}\mathrm{d}s = \frac{t^{2j+1}}{(2j+1)!}\), which is the shift. The limits are \(\cosh t\) and \(\sinh t\), whose series converge on all of \(\mathbb{R}\). By uniqueness (unit 01), \(x(t) = \cosh t\) is the solution.

3. Numbers at \(t = 1\). \(\cosh 1 = 1.543081\).

• \(y^{(2)}\): first component \(1.5\), error \(0.043081\); bound \(\frac{1}{4!}\cdot\frac{1}{1-\frac{1}{30}} = \frac{1}{24}\cdot\frac{30}{29} = 0.043103\). ✓
• \(y^{(4)}\): first component \(1.541667\), error \(0.001414\); bound \(\frac{1}{6!}\cdot\frac{1}{1-\frac{1}{56}} = \frac{1}{720}\cdot\frac{56}{55} = 0.001414\). ✓

The bound is the tail \(\sum_{j>m}1/(2j)!\) dominated by a geometric series with ratio \(1/((2m+3)(2m+4))\), and it is nearly attained because the first omitted term dominates the tail.

4. What Banach promises. On the cylinder \(b = 2\), \(c = 1\) about \((1,0)\): the largest \(\|y\|\) in it is \(\|y_0\| + b = 3\), and \(\|f(y)\| = \|y\|\), so \(M = 3\); \(L = 1\); \(a = \min(1, 2/3) = 2/3\). With \(\lambda = 2L = 2\), \(q = 1/2\), and \(\|y^{(1)}-y^{(0)}\|_\lambda \le \|y^{(1)}-y^{(0)}\|_\infty = \sup_{|t|\le a}|t| = a\), \[ \|y^{(n)}-y\|_\infty \le e^{\lambda a}\,\|y^{(n)}-y\|_\lambda \le e^{4/3}\cdot\frac{q^n}{1-q}\cdot a = 3.7937 \times 2^{1-n} \times \tfrac23 = 2.5291 \times 2^{1-n}. \] This is below \(10^{-3}\) first at \(n = 13\) (value \(6.17\times10^{-4}\)). The true sup-norm errors on \([-\tfrac23,\tfrac23]\) are \(2.36\times10^{-1}, 5.12\times10^{-2}, 8.43\times10^{-3}, 1.12\times10^{-3}, 1.23\times10^{-4}\) for \(n = 1,\dots,5\): five iterates suffice.

The gap. Banach's estimate is geometric, \(q^n\), because it uses only "each application shrinks distances by \(q\)". The truth is factorial, because iterating \(\|y^{(n+1)}-y\|(t) \le L\int_0^t \|y^{(n)}-y\|\) gives \(L^nt^n/n!\), and \(n!\) beats \(2^n\). The geometric bound is what a general theorem can promise; the factorial one needs the specific structure of the operator.

Check. Dimensions: \(t\) is dimensionless here, as it must be for \(\ddot x = x\) to be consistent. Limit: replacing \(f\) by \((v,0)\) — no force at all — leaves \(L = 1\), since the first block still reads the velocity, but the starting curve \(y^{(0)} \equiv (1,0)\) already satisfies \(f(y^{(0)}) = 0\), so \(y^{(1)} = y^{(0)}\) and the iteration terminates at the exact solution \(x \equiv 1\). Banach's bound is not what makes it stop: the start was already the fixed point. ✓

Idea. Two inequalities, then two theorems. Nothing else is needed, and nothing about pendulums in particular is used beyond \(|\sin a - \sin b| \le |a-b|\).

1. The system and \(L\). With \(y = (\theta, u)\), \[ f(y,t) = \Big(u,\ -\tfrac{g}{\ell}\sin\theta - \gamma u + A\cos\Omega t\Big), \qquad \Omega_{\text{dom}} = \mathbb{R}^2\times\mathbb{R}. \] For two states at the same \(t\) the drive cancels: \(|\Delta f_1| = |\Delta u| \le \|\Delta y\|\), and \[ \lvert\Delta f_2\rvert \le \tfrac{g}{\ell}|\sin\theta_1 - \sin\theta_2| + \gamma|\Delta u| \le \tfrac{g}{\ell}|\Delta\theta| + \gamma|\Delta u| \le \Big(\tfrac{g}{\ell}+\gamma\Big)\|\Delta y\|, \] using the mean value theorem for \(\sin\). Hence \[ \|\Delta f\| \le \sqrt{1 + \Big(\tfrac{g}{\ell}+\gamma\Big)^2}\;\|\Delta y\| =: L\,\|\Delta y\|, \] valid on all of \(\mathbb{R}^2\), for every \(t\). So \(f\) is (globally, hence locally) Lipschitz in \(y\).

2. Linear growth. \(\|f(y,t)\|^2 = u^2 + \big(\tfrac{g}{\ell}\sin\theta + \gamma u - A\cos\Omega t\big)^2\), so \[ \|f(y,t)\| \le |u| + \tfrac{g}{\ell} + \gamma|u| + |A| \le \Big(\tfrac{g}{\ell}+|A|\Big) + (1+\gamma)\|y\| \le C\big(1+\|y\|\big), \quad C = \tfrac{g}{\ell}+|A|+1+\gamma . \]

3. Conclusion. \(f\) is continuous on \(\mathbb{R}^2\times\mathbb{R}\) and locally Lipschitz in \(y\) (part 1), so Picard–Lindelöf gives a unique solution through every state on some interval, and global uniqueness extends that to any common interval. \(f\) has linear growth on every strip (part 2), so the global existence theorem gives \(T_\pm = \pm\infty\): Grönwall bounds \(\|y\|\) on any finite time interval, the solution therefore stays in a compact ball, and the escape lemma forbids the interval from ending. Every initial state has exactly one motion, defined for all time, forwards and backwards.

4. Dimensionless form. Put \(\tau = t\sqrt{g/\ell}\), \(\hat\gamma = \gamma\sqrt{\ell/g}\), \(\hat A = A\ell/g\), \(\hat\Omega = \Omega\sqrt{\ell/g}\). The equation becomes \(\theta'' = -\sin\theta - \hat\gamma\theta' + \hat A\cos(\hat\Omega\tau)\) and \[ L = \sqrt{1 + (1+\hat\gamma)^2}\,, \] a pure number. Undamped: \(L = \sqrt2\); the deck's sharper route — squaring the two components before bounding, rather than bounding each and adding — gives \(L = 1\) in that case, and any valid constant will do.

Check. Dimensions: in part 1, \(g/\ell\) has units \(\mathrm{T^{-2}}\) and the \(1\) under the square root has none — the mismatch the deck warned about, and the reason part 4 exists. Limit \(\gamma = A = 0\) and \(\ell \to \infty\): \(L \to 1\), \(C \to 1\), and the equation becomes \(\ddot\theta = 0\), free rotation, trivially deterministic. ✓

Idea. Grönwall is a bound over a class of equations. A conservation law is a statement about one. When the second is available it is usually enormously better.

1. The Grönwall bound. \(f(\theta,u) = (u,-\sin\theta)\) has \(L = 1\) (deck, "The Lipschitz condition"). Both pendulums are released from rest, so \(\Delta y(0) = (10^{-4}, 0)\) and \(\|\Delta y(0)\| = 10^{-4}\). Hence \[ \|\Delta y(\tau)\| \le 10^{-4}e^{\tau}, \qquad \text{and at } \tau = 20: \quad 10^{-4}\times 4.85165\times10^{8} = 4.852\times10^{4}\ \text{rad}. \]

2. The energy bound. For \(\theta'' = -\sin\theta\) released from rest at \(\theta_0\), the conserved energy per unit mass is \(E = \tfrac12 u^2 + (1-\cos\theta) = 1 - \cos\theta_0\). So \(1-\cos\theta \le 1-\cos\theta_0\), giving \(|\theta(\tau)| \le \theta_0\) for \(\theta_0 < \pi\), and \(u^2 = 2(\cos\theta - \cos\theta_0) \le 2(1-\cos\theta_0)\), giving \(|u(\tau)| \le 2\sin(\theta_0/2)\). Therefore, for all \(\tau\), \[ \lvert\Delta\theta\rvert \le \theta_0+\theta_0' = 0.4001, \qquad \lvert\Delta u\rvert \le 2\sin\tfrac{\theta_0}{2}+2\sin\tfrac{\theta_0'}{2} = 0.399433, \] \[ \|\Delta y(\tau)\| \le \sqrt{0.4001^2 + 0.399433^2} = 0.565356 . \] The two numbers differ by a factor \(4.852\times10^4 / 0.565356 = 8.6\times10^4\).

3. Where Grönwall stops saying anything. \(10^{-4}e^{\tau} = 0.565356\) at \[ \tau_\ast = \ln\big(0.565356\times10^{4}\big) = \boxed{8.640}. \] Beyond \(\tau_\ast\) the Grönwall bound exceeds a bound that holds for all time, so it carries no information. In seconds, for \(\ell = 1\) m, that is \(8.640\times 0.31933 = 2.759\) s, about \(1.4\) periods.

4. Why it is still sharp. Grönwall's bound is attained, with equality, by \(\dot y = Ly\): there \(\|\Delta y(\tau)\| = \|\Delta y(0)\|e^{L\tau}\) exactly, and that right-hand side has Lipschitz constant \(L\). So no function of \(L\) and \(\tau\) alone can do better. The pendulum is simply not the worst member of its class; it has a conserved energy and the exponential member does not.

Check. Limit \(\theta_0 \to 0\): both bounds scale linearly in the initial separation, so \(\tau_\ast\) moves only logarithmically — halving the measurement error buys \(\ln 2 = 0.69\) more units of dimensionless time. That is the shape of every such estimate, and it is why unit 17 measures exponents rather than horizons. ✓

Idea. The same subtraction as for continuous dependence, with one extra term that does not involve the unknown, so Grönwall still applies.

1. The simple bound. Both solve integral equations, so \[ y(t)-z(t) = \Delta_0 + \int_{t_0}^{t}\Big(f(y(s),s)-f(z(s),s)\Big)\mathrm{d}s + \int_{t_0}^{t}\Big(f(z(s),s)-\tilde f(z(s),s)\Big)\mathrm{d}s, \] where \(\Delta_0 = y(t_0)-z(t_0)\). Taking norms, using the Lipschitz condition on the first integral and \(\eta\) on the second, \[ u(t) := \|y(t)-z(t)\| \le \|\Delta_0\| + \eta\,(t-t_0) + L\int_{t_0}^{t}u(s)\,\mathrm{d}s . \] The term \(\eta(t-t_0)\) is not constant, so Grönwall as stated does not apply to the whole interval at once. Fix a \(t^\ast \in [t_0,t_1]\) and work on \([t_0,t^\ast]\): there the first two terms are at most the constant \(A := \|\Delta_0\| + \eta(t^\ast-t_0)\), Grönwall gives \(u(t) \le Ae^{L(t-t_0)}\) on \([t_0,t^\ast]\), and reading it at \(t = t^\ast\) — which is legitimate for every \(t^\ast\) — gives \(u(t^\ast) \le \big(\|\Delta_0\| + \eta(t^\ast-t_0)\big)e^{L(t^\ast-t_0)}\). ∎

2. The sharp bound. Do not freeze the inhomogeneous term. Put \(U(t) = \|\Delta_0\| + \eta(t-t_0) + L\int_{t_0}^t u\). Then \(u \le U\) and \(U' = \eta + Lu \le \eta + LU\), so \[ \frac{\mathrm{d}}{\mathrm{d}t}\Big(U e^{-L(t-t_0)}\Big) \le \eta\,e^{-L(t-t_0)} . \] Integrating from \(t_0\), where \(U(t_0) = \|\Delta_0\|\), \[ U(t)e^{-L(t-t_0)} \le \|\Delta_0\| + \frac{\eta}{L}\Big(1-e^{-L(t-t_0)}\Big) \ \Longrightarrow\ u(t) \le \|\Delta_0\|e^{L(t-t_0)} + \frac{\eta}{L}\Big(e^{L(t-t_0)}-1\Big). \qquad\square \]

Checks. \(\eta = 0\): the second term vanishes and this is exactly the continuous-dependence estimate. ✓ \(L \to 0\): expand \(\frac{\eta}{L}(e^{L\Delta t}-1) = \eta\Delta t + O(L)\), and the first term tends to \(\|\Delta_0\|\); the bound becomes \(\|\Delta_0\| + \eta\Delta t\), which is what you get by integrating \(\|\dot y - \dot z\| \le \eta\) directly. ✓ Dimensions: \(\eta\) has the units of \(f\), \(\eta/L\) those of \(y\). ✓ Sharpness: the two bounds agree to first order in \(L\Delta t\); the sharp one is smaller, since \(e^{x}-1 \le xe^{x}\).

Idea. Solve, then read the maximal interval off the formula, then say which mode of the escape lemma is at work. The theorem is never contradicted; it only ever promised something local.

1. \(\dot x = x^p\). Separating (unit 00), for \(p \ne 1\), \(\int x^{-p}\mathrm{d}x = \int\mathrm{d}t\) gives \(\frac{x^{1-p}}{1-p} = t + C\) and, with \(x(0) = x_0 > 0\), \[ x(t) = \Big(x_0^{1-p} - (p-1)t\Big)^{-1/(p-1)} \quad (p>1), \qquad x(t) = \Big(x_0^{1-p} + (1-p)t\Big)^{1/(1-p)} \quad (p<1). \] For \(p > 1\) the bracket vanishes at \(T_+ = \dfrac{x_0^{1-p}}{p-1}\) and \(x \to \infty\) there; for \(p < 1\) the bracket grows and the solution exists for all \(t \ge 0\); for \(p = 1\) it is \(x_0e^{t}\), global. Numbers: \[ (p,x_0) = (2,2): T_+ = \tfrac{1/2}{1} = 0.5; \quad (3,1): T_+ = \tfrac{1}{2} = 0.5; \quad (3/2,4): T_+ = \tfrac{4^{-1/2}}{1/2} = 1 . \]

2. \(m\ddot x = -k/x^n\), \(n > 1\). Unit 00's third case: \(F = -V'\) with \(V(x) = -\dfrac{k}{(n-1)x^{n-1}}\), and released from rest at \(a\), \(E = V(a)\). Then \[ \tfrac12 m\dot x^2 = E - V(x) = \frac{k}{n-1}\Big(\frac{1}{x^{n-1}} - \frac{1}{a^{n-1}}\Big), \qquad \dot x = -\sqrt{\frac{2k}{m(n-1)}}\,\sqrt{x^{1-n} - a^{1-n}} . \] Quadrature, then \(x = aw\): \[ t_{\text{fall}} = \sqrt{\frac{m(n-1)}{2k}}\int_0^a \frac{\mathrm{d}x}{\sqrt{x^{1-n}-a^{1-n}}} = \sqrt{\frac{m(n-1)}{2k}}\;a^{\frac{n+1}{2}} \underbrace{\int_0^1\frac{\mathrm{d}w}{\sqrt{w^{1-n}-1}}}_{=:I_n}. \] \(I_n\) is finite for every \(n > 1\): at \(w \to 0\) the integrand behaves like \(w^{(n-1)/2} \to 0\), and at \(w \to 1\) like \(\big((n-1)(1-w)\big)^{-1/2}\), whose integral converges — an integrable singularity at a turning point, exactly unit 00's case.

For \(n = 3\): \(I_3 = \int_0^1 \frac{\mathrm{d}w}{\sqrt{w^{-2}-1}} = \int_0^1\frac{w\,\mathrm{d}w}{\sqrt{1-w^2}} = \big[-\sqrt{1-w^2}\big]_0^1 = 1\), so \[ t_{\text{fall}} = \sqrt{\frac{2m}{2k}}\;a^{2} = a^2\sqrt{\frac{m}{k}} . \] With \(m = 2\) kg, \(a = 3\) m, \(k = 4\ \mathrm{kg\,m^4\,s^{-2}}\): \(t_{\text{fall}} = 9\sqrt{0.5} = 6.364\) s.

3. No contradiction. In (1), \(\Omega = \mathbb{R}\times\mathbb{R}\) and \(x^p\) is locally Lipschitz everywhere; the solution leaves every compact \(K \subseteq \mathbb{R}\), because \(x(t) \to \infty\) as \(t \to T_+^-\). First mode of the escape lemma: blow-up. In (2), \(\Omega = \{x>0\}\times\mathbb{R}\) because \(-k/x^n\) is undefined at \(x = 0\); the phase-space solution \((x,\dot x)\) leaves every compact subset of \(\Omega\) both by \(x \to 0\) (reaching \(\partial\Omega\)) and by \(|\dot x| \to \infty\). Second mode. In neither case is anything promised by Picard–Lindelöf violated: it guarantees a solution on \([t_0-a,t_0+a]\) with \(a = \min(c,b/M)\), and as the state approaches the bad set every admissible cylinder shrinks, so \(a \to 0\).

Check. Dimensions in (2): \([k] = \mathrm{M\,L^{n+1}\,T^{-2}}\) from \(F = k/x^n\), so \(\sqrt{m(n-1)/(2k)}\,a^{(n+1)/2}\) has units \(\sqrt{\mathrm{M}/(\mathrm{M\,L^{n+1}\,T^{-2}})}\cdot\mathrm{L}^{(n+1)/2} = \mathrm{T}\). ✓ For \(n = 3\), \([k] = \mathrm{M\,L^4\,T^{-2}} = \mathrm{kg\,m^4\,s^{-2}}\), as used. Limit \(a \to 0\): \(t_{\text{fall}} \to 0\), a mass released nearer the singularity falls sooner. ✓ Numerical check of the \(n=3\) quadrature against the closed form: \(6.363961\) both ways.

Statement. Let \(\Omega \subseteq \mathbb{R}^N\times\mathbb{R}\) be open, \(f : \Omega \to \mathbb{R}^N\) continuous and locally Lipschitz in \(y\), and \((y_0,t_0)\in\Omega\). Choose \(b,c>0\) with \(\mathcal{C} = \{(y,t): \|y-y_0\|\le b,\ |t-t_0|\le c\} \subseteq \Omega\), put \(M = \max_{\mathcal{C}}\|f\|\), let \(L\) be a Lipschitz constant for \(f\) in \(y\) on \(\mathcal{C}\), and set \(a = \min(c, b/M)\), \(J = [t_0-a, t_0+a]\). Then the initial value problem \(\dot y = f(y,t)\), \(y(t_0)=y_0\) has exactly one solution on \(J\).

Proof. Let \(X = \{y \in C(J,\mathbb{R}^N) : \|y(t)-y_0\|\le b\ \forall t\}\), and \((Py)(t) = y_0 + \int_{t_0}^{t}f(y(s),s)\,\mathrm{d}s\).

Step 0 (the integral equation). For continuous \(y\) with \((y(t),t)\in\Omega\), \(y\) is \(C^1\) and solves the IVP on \(J\) iff \(y = Py\). Forwards: the fundamental theorem of calculus, componentwise, applied to the continuous \(s\mapsto f(y(s),s)\). Backwards: the right-hand side is differentiable with derivative \(f(y(t),t)\), and at \(t = t_0\) it equals \(y_0\).

Step 1 (\(P\) maps \(X\) to \(X\)). For \(y\in X\), \((y(s),s)\in\mathcal{C}\), so the integrand is defined and continuous, and \(\|(Py)(t)-y_0\| \le \big|\int_{t_0}^t\|f\|\big| \le M|t-t_0| \le Ma \le b\).

Step 2 (\(X\) is complete). \(C(J,\mathbb{R}^N)\) with \(\|\cdot\|_\infty\) is complete: a uniformly Cauchy sequence converges pointwise because \(\mathbb{R}^N\) is complete, the convergence is uniform, and a uniform limit of continuous functions is continuous. \(X\) is closed in it, hence complete. The weighted norm \(\|y\|_\lambda = \sup_J e^{-\lambda|t-t_0|}\|y(t)\|\) satisfies \(e^{-\lambda a}\|y\|_\infty \le \|y\|_\lambda \le \|y\|_\infty\), so it is equivalent and \(X\) is complete in it too.

Step 3 (contraction). For \(t\ge t_0\), \[ \|(Py)(t)-(Pz)(t)\| \le \int_{t_0}^{t} L\|y-z\|\,\mathrm{d}s = \int_{t_0}^{t} L e^{\lambda(s-t_0)}\Big(e^{-\lambda(s-t_0)}\|y-z\|\Big)\mathrm{d}s \le \frac{L}{\lambda}e^{\lambda(t-t_0)}\|y-z\|_\lambda , \] and symmetrically for \(t \le t_0\). Multiplying by \(e^{-\lambda(t-t_0)}\) and taking the supremum, \(\|Py-Pz\|_\lambda \le \frac{L}{\lambda}\|y-z\|_\lambda\). Take \(\lambda = 2L\): \(q = 1/2\).

Step 4 (Banach, and back). \(P\) has exactly one fixed point in \(X\), which by Step 0 is exactly one solution in the tube. ∎

The five questions.

1. Completeness is used once, in Banach's theorem, and it is the completeness of \(X\) — a closed subset of \(C(J,\mathbb{R}^N)\) with the (weighted) sup norm. This is why the problem was moved to the integral equation: \(C^1\) with the sup norm is not complete.
2. Compactness of the cylinder \(\mathcal{C}\) is used twice: to know that \(M = \max_{\mathcal{C}}\|f\|\) is finite (continuous function on a compact set), and to convert "locally Lipschitz" into a single constant \(L\) valid on all of \(\mathcal{C}\).
3. The Lipschitz condition is used exactly once, in Step 3. Nowhere else. That is why its failure costs uniqueness and nothing else, and why Peano still gives existence without it.
4. The interval is \(\min(c, b/M)\) because two things can take the candidate out of the cylinder: time running past \(\pm c\), or the state moving more than \(b\); the state moves at speed at most \(M\), so it needs at least \(b/M\). \(L\) does not appear because Step 3 works for any \(L\) once \(\lambda\) is chosen large enough — that is the entire point of the weighted norm. With the plain sup norm one would need \(aL < 1\) as well.
5. Uniqueness in Step 4 is uniqueness in \(X\). Upgrade: let \(z\) be any solution on an interval \(I \ni t_0\) with \(z(t_0)=y_0\). Continuity gives \(\delta>0\) with \(\|z(t)-y_0\|\le b\) for \(|t-t_0|\le\delta\), so \(z\) lies in the tube there and equals \(y\) there. Then \(A = \{t\in I : y(t)=z(t)\}\) is non-empty, closed in \(I\) (preimage of \(\{0\}\) under the continuous \(y-z\)), and open in \(I\) (repeat the argument at each point of \(A\), building a fresh cylinder there). \(I\) is connected, so \(A = I\).

Checks. Dimensions: \(b/M\) is (length)/(length per time) = time. ✓ Special case \(f = f(t)\): \(L = 0\), \(\lambda\) may be anything, \(q = 0\), and one application of \(P\) gives the answer \(y_0 + \int f\) — unit 00's quadrature. ✓ Limit \(b \to 0\): \(a \to 0\), and the theorem says nothing, correctly, about a cylinder with no room in it. ✓

Idea. Non-uniqueness is exhibited, never argued for. Then the exponent \(\alpha\) is varied to see that \(1/2\) is not special and \(1\) is.

1. The family. For \(t > T\), \(r_T = \frac{k^2}{144}(t-T)^4\), \(\dot r_T = \frac{k^2}{36}(t-T)^3\), \(\ddot r_T = \frac{k^2}{12}(t-T)^2\), and \[ k\sqrt{r_T} = k\cdot\frac{k}{12}(t-T)^2 = \frac{k^2}{12}(t-T)^2 = \ddot r_T .\ ✓ \] For \(t<T\) both sides vanish. At \(t = T\) the one-sided limits of \(r_T\), \(\dot r_T\) and \(\ddot r_T\) are all \(0\) from both sides, so the glued function is \(C^2\) and solves the equation there too (both sides are \(0\)). The data: \(r_T(0)=\dot r_T(0)=0\) for every \(T\ge0\).

Exact smoothness. \(r_T''' = \frac{k^2}{6}(t-T) \to 0\), continuous, so \(r_T \in C^3\); \(r_T'''' = \frac{k^2}{6}\) for \(t>T\) and \(0\) for \(t<T\), a jump. So \(r_T\) is \(C^3\) and not \(C^4\).

2. The hypothesis. On \(s>0\), \(f(s,v) = (v, k\sqrt s)\). At the apex, the pairs \((s,0)\) and \((0,0)\) give \(\frac{\|f(s,0)-f(0,0)\|}{\|(s,0)\|} = \frac{k\sqrt s}{s} = \frac{k}{\sqrt s} \to \infty\), so no neighbourhood of the origin carries a constant. On \(\{\epsilon \le |s| \le g^2/k^2\}\), the mean value theorem gives \(|\sqrt{s_1}-\sqrt{s_2}| \le \frac{1}{2\sqrt\epsilon}|s_1-s_2|\) for \(s_1,s_2\) on the same side of the apex. A pair straddling it needs one more line, and a larger constant: for \(s_1 \ge \epsilon \ge -s_2\) the second components differ by \(k(\sqrt{s_1}+\sqrt{|s_2|})\), and \(\sqrt x \le x/\sqrt\epsilon\) for \(x \ge \epsilon\) bounds that by \(\frac{k}{\sqrt\epsilon}|s_1-s_2|\). So \[ L(\epsilon) = \sqrt{1 + \Big(\frac{k}{\sqrt\epsilon}\Big)^2} \xrightarrow[\epsilon\to0]{} \infty , \] and the factor \(2\) matters only in that it cannot be had: the set has two branches, and the estimate must cover pairs from both.

3. General \(\alpha\). Try \(x = At^p\), \(t>0\), \(A>0\), in \(\ddot x = kx^\alpha\): \[ A p(p-1)t^{p-2} = kA^\alpha t^{\alpha p} \ \Longrightarrow\ p-2 = \alpha p \ \text{ and } \ Ap(p-1) = kA^\alpha, \] so \[ p = \frac{2}{1-\alpha}, \qquad A^{1-\alpha} = \frac{k}{p(p-1)} = \frac{k(1-\alpha)^2}{2(1+\alpha)}, \qquad A = \left[\frac{k(1-\alpha)^2}{2(1+\alpha)}\right]^{\frac{1}{1-\alpha}} , \] using \(p-1 = \frac{1+\alpha}{1-\alpha}\) and \(p(p-1) = \frac{2(1+\alpha)}{(1-\alpha)^2}\). Since \(p = 2/(1-\alpha) > 2\) for every \(\alpha\in(0,1)\), the glued function is \(C^2\) and the same one-parameter family \(x_T(t) = A(t-T)^p\) exists. Osgood fails too: \(\omega(u) = ku^\alpha\) and \(\int_{0^+}u^{-\alpha}\mathrm{d}u < \infty\) for \(\alpha<1\).

Check \(\alpha = 1/2\): \(p = 4\) ✓ and \(A^{1/2} = \frac{k(1/2)^2}{2(3/2)} = \frac{k}{12}\), so \(A = k^2/144\) ✓.

Limit \(\alpha\to1^-\): \(p \to \infty\) and \(A \to 0\); the departure becomes infinitely flat and, in the limit, does not happen. At \(\alpha = 1\) the equation is \(\ddot x = kx\), the right-hand side is Lipschitz, the only solution with \(x(0)=\dot x(0)=0\) is \(x \equiv 0\), and a mass at an unstable Hookean apex takes infinite time to leave. The power-law ansatz has told us where it stops working, which is the right answer.

4. Numbers for \(k = 4\ \mathrm{m^{1/2}s^{-2}}\). \[ r_{\max} = \frac{g^2}{k^2} = \frac{96.1704}{16} = 6.0106\ \mathrm{m}, \qquad \text{depth} = \frac{2g^2}{3k^2} = 4.0071\ \mathrm{m}, \] \[ T_+ - T = \Big(\frac{144\,g^2}{k^4}\Big)^{1/4} = \frac{\sqrt{12\,g}}{k} = \frac{10.8480}{4} = 2.7120\ \mathrm{s}. \]

Check. Dimensions: \([\sqrt{g}/k] = (\mathrm{L\,T^{-2}})^{1/2}/(\mathrm{L^{1/2}T^{-2}}) = \mathrm{T}\). ✓ Scaling: multiplying \(k\) by 4 divides \(r_{\max}\) by 16 and the time by 4, which is what \(r \propto k^{-2}\) and \(t \propto k^{-1}\) demand — a steeper dome is smaller and is left sooner. ✓ The \(k=1\) values \(96.17\) m and \(10.85\) s are recovered by setting \(k=1\). ✓

Idea. The constant function is a solution. Uniqueness then does all the work: anything that touches an equilibrium is the equilibrium.

1. The proof. Since \(f(y_\ast) = 0\), the constant map \(c(t) \equiv y_\ast\) is a solution of \(\dot w = f(w)\) on all of \(\mathbb{R}\), with \(c(t_1) = y_\ast\). Suppose \(y\) solves the same equation on an interval \(I \ni t_1\) with \(y(t_1) = y_\ast\). Both \(y\) and \(c\) are solutions on \(I\) agreeing at \(t_1\), and \(f\) is continuous and locally Lipschitz, so global uniqueness gives \(y = c\) on \(I\): \(y \equiv y_\ast\). Contrapositive: a non-constant solution never takes the value \(y_\ast\), so it can only approach \(y_\ast\) in the limit \(t \to \pm\infty\). ∎

2. The pendulum. The system \(f(\theta,u) = (u,-\sin\theta)\) is Lipschitz with \(L = 1\) (P6 with \(\gamma = A = 0\)) and has equilibria at \((0,0)\) and \((\pm\pi, 0)\). The inverted state is the equilibrium \((\pi,0)\).

• Released from rest at \(\theta_0\in(-\pi,\pi)\) with \(\theta_0\ne0\), the motion is non-constant, so by part 1 it never equals \((\pi,0)\); and for \(\theta_0 = 0\) the motion is the equilibrium \((0,0)\), which is not \((\pi,0)\) either. In fact energy already forbids \(|\theta| > \theta_0\); part 1 is the argument that survives when no conservation law is at hand.
• On the separatrix, \(E = 2\) exactly, the motion tends to \((\pi,0)\). It is non-constant, so by part 1 it never arrives: the approach takes infinite time. (Quadrature confirms it — \(u = 2\cos(\theta/2)\) gives \(\int^{\pi}\frac{\mathrm{d}\theta}{2\cos(\theta/2)}\), logarithmically divergent at \(\theta = \pi\) — but the uniqueness argument needs no integral.)

3. Where it fails on the dome. The dome's system has \(f(0,0) = (0,0)\), so \((0,0)\) is an equilibrium and \(c \equiv (0,0)\) is a solution; yet every member of the family reaches it, backwards in time, at \(t = T\). The step that fails is the invocation of global uniqueness: it requires \(f\) to be locally Lipschitz at the point where the two solutions meet, and at \((0,0)\) it is not. Every other step survives — \(c\) really is a solution, the two really do agree at \(t = T\) — and the conclusion is false.

Check. Limit case: replace the dome's apex by a Hookean one, \(\ddot s = \kappa^2 s\). Now \(f\) is Lipschitz, part 1 applies, and the solutions through \((0,0)\) reduce to \(s\equiv0\); a mass approaching the apex does so like \(e^{-\kappa t}\), never arriving. The dome converts that exponential approach into a finite-time arrival, and part 1 says that is only possible where the hypothesis fails. ✓

Another way

Part 1 can be run without quoting global uniqueness, by a direct Grönwall argument that shows what is really going on. Let \(y\) be a solution with \(y(t_1) = y_\ast\), and work backwards on \([t_1-\delta, t_1]\) inside a neighbourhood where \(f\) is Lipschitz with constant \(L\). Then \(u(t) = \|y(t)-y_\ast\|\) satisfies, using \(f(y_\ast)=0\), \[ u(t) = \left\|\int_{t}^{t_1}f(y(s))\,\mathrm{d}s\right\| \le L\int_{t}^{t_1}u(s)\,\mathrm{d}s, \] which is Grönwall with \(A = 0\) (in reversed time), forcing \(u \equiv 0\). The dome's \(f\) admits no such \(L\), the integral inequality is unavailable, and \(u(t) = \frac{k^2}{144}(t_1-t)^4\) is free to be positive.