Phase portraits: what you know when there is no formula

Idea. One separable equation (unit 00), then the group property is pure algebra, and the maximal interval is read off the formula's pole.

Working.

1. Away from \(x = 0\), \(\mathrm{d}x/x^2 = \mathrm{d}t\) gives \(-1/x = t + C\), so \(x(t) = -1/(t+C)\); imposing \(x(0) = x_0 \ne 0\) gives \(C = -1/x_0\) and \[\varphi_t(x_0) = \frac{-1}{t - 1/x_0} = \frac{x_0}{1 - x_0 t}.\] For \(x_0 = 0\) the constant solution \(x \equiv 0\) is the unique one (Picard–Lindelöf; \(f(x) = x^2\) is \(C^\infty\)), and the formula gives it too.
2. \[\varphi_t(\varphi_s(x_0)) = \frac{\dfrac{x_0}{1-x_0 s}}{1 - t\,\dfrac{x_0}{1-x_0 s}} = \frac{x_0}{1 - x_0 s - x_0 t} = \varphi_{t+s}(x_0),\] multiplying numerator and denominator by \(1 - x_0 s\). Both sides are defined exactly when \(1 - x_0 s \ne 0\) and \(1 - x_0(s+t) \ne 0\); the second condition alone does not suffice, because the intermediate point must exist.
3. \(x_0 > 0\): the denominator vanishes at \(t = 1/x_0 > 0\), and \(x(t) \to +\infty\) there, so the maximal interval is \((-\infty,\ 1/x_0)\). \(x_0 = 0\): all of \(\mathbb{R}\). \(x_0 < 0\): the pole is at \(t = 1/x_0 < 0\), so the interval is \((1/x_0,\ \infty)\). Hence no single \(t > 0\) gives a map defined on all of \(\mathbb{R}\): for fixed \(t > 0\), \(\varphi_t\) is undefined at \(x_0 = 1/t\). This is a local flow, not a one-parameter group of maps of \(\mathbb{R}\).

Check. Dimensions: if \([x] = \mathsf{L}\) then \([\dot x] = \mathsf{L}\mathsf{T}^{-1}\) forces \(x^2\) to be read as \(x^2/\ell\) for some length \(\ell\); with \(\ell = 1\) the product \(x_0 t\) is dimensionless, as the formula requires. Limit: for small \(t\), \(\varphi_t(x_0) = x_0(1 + x_0t + \dots) = x_0 + x_0^2 t + O(t^2)\), and \(\dot x(0) = x_0^2\) — the equation, recovered. Number: from \(x_0 = 2\), blow-up at \(t = 0.5\); at \(t = 0.4\), \(x = 2/(1-0.8) = 10\).

Idea. Compute \(\tau\) and \(\Delta\), locate the point in the trace–determinant plane, and only then solve the quadratic.

Working.

\(A_5\) is not \(2I\), so it is defective: the solution is \(e^{2t}(c_1\mathbf v + c_2(\mathbf w + t\mathbf v))\) with \(\mathbf v = (1,0)\). \(A_6\) gives \(\dot x = y\), \(\dot y = 0\), so \(x(t) = x_0 + y_0 t\), \(y \equiv y_0\): every point of the \(x\)-axis is an equilibrium and every other orbit is a horizontal line traversed at constant speed.

The two non-hyperbolic ones are \(A_3\) (\(\operatorname{Re}\lambda = 0\)) and \(A_6\) (\(\lambda = 0\)). Using them as a guide to a nonlinear system is illegitimate: Hartman–Grobman does not apply, and the deck's family \(\dot r = ar^3\) shows that a linear centre is compatible with a nonlinear sink, source or centre. For \(A_6\) the failure is worse still — an arbitrarily small nonlinear term can destroy the line of equilibria entirely, leaving none.

Check. For each row, the eigenvalues multiply to \(\Delta\) and add to \(\tau\): \(4\times(-1) = -4\) and \(4-1 = 3\); \((-2)(-4) = 8\), \(-6\); \((2i)(-2i) = 4\), \(0\); \((-1)^2 + 2^2 = 5\), \(-2\); \(2\times2 = 4\), \(4\); and \(0\times0 = 0\), \(0+0 = 0\). All six check.

Idea. Both Jacobians have \(\tau = -\gamma\); only \(\Delta = \cos\theta^*\) differs. One parameter moves both points horizontally in the \(\tau\)–\(\Delta\) plane.

Working.

1. At \((0,0)\): \(\tau = -\gamma\), \(\Delta = 1\), so \(\lambda_\pm = \tfrac12(-\gamma \pm \sqrt{\gamma^2-4})\).

   • \(0 < \gamma < 2\): complex with real part \(-\gamma/2 < 0\) — a *stable spiral*.
   • \(\gamma = 2\): double root \(\lambda = -1\), and the matrix \(\begin{pmatrix}0&1\\-1&-2\end{pmatrix}\) is not \(-I\), so a *degenerate stable node*.
   • \(\gamma > 2\): two distinct negative real roots — a stable node (\(\lambda_+\lambda_- = 1 > 0\) and \(\lambda_++\lambda_- = -\gamma < 0\)).

   At \((\pi, 0)\): \(\Delta = -1 < 0\) for every \(\gamma\), so the roots are real of opposite sign: a saddle, always.

2. \(\gamma = 2\): critical damping, \(b = 2m\sqrt{gl^3}\) in physical units. It is the least damping that returns the pendulum without overshoot, and the boundary between spiral and node in the \(\tau\)–\(\Delta\) plane.
3. \(\gamma = 0.1\): down, \(\lambda = -0.0500 \pm 0.9987\,i\) (sum \(-0.1 = \tau\), modulus squared \(0.0025 + 0.9975 = 1.0000 = \Delta\)); up, \(\lambda = 0.9512,\ -1.0512\) (sum \(-0.1000\), product \(-1.0000\)). \(\gamma = 2.5\): down, \(\lambda = -0.5000,\ -2.0000\) (sum \(-2.5\), product \(1.0000\)); up, \(\lambda = 0.3508,\ -2.8508\) (sum \(-2.5000\), product \(-1.0000\)).
4. No. At the hanging position \(\Delta = 1 > 0\) and \(\tau = -\gamma < 0\), so both roots have real part \(-\gamma/2 \ne 0\); at the upright position \(\Delta = -1 \ne 0\), so neither root is zero and both are real. Hence Hartman–Grobman applies at both for every \(\gamma > 0\): the linear picture is the true local picture, and no bifurcation can occur as \(\gamma\) varies over \((0,\infty)\) — the transition at \(\gamma = 2\) changes the shape of the approach but not the topology, which is why it is not a bifurcation.

Check. Limit \(\gamma \to 0^+\): the down eigenvalues tend to \(\pm i\), the linearised small-amplitude frequency, and the period \(2\pi\) in units of \(\sqrt{l/g}\) — unit 00's \(2\pi\sqrt{l/g}\). Limit \(\gamma \to \infty\): \(\lambda_+ \to -1/\gamma \to 0^-\) and \(\lambda_- \to -\gamma\), the overdamped creep, slower and slower as the fluid thickens, as it should be.

Idea. The linearisation is a centre for both systems, so it decides nothing. The radial derivative decides everything, and \(V = x^2+y^2\) computes it.

Working.

1. Equilibria: \(-y = x^3\) and \(x = y^3\) give \(x = (-x^3)^3 = -x^9\), i.e. \(x(1 + x^8) = 0\), so \(x = 0\) and then \(y = 0\): the origin only. The Jacobian is \(\begin{pmatrix}-3x^2 & -1\\ 1 & -3y^2\end{pmatrix}\), which at the origin is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\): \(\tau = 0\), \(\Delta = 1\), a centre. Both eigenvalues are \(\pm i\), purely imaginary, so the equilibrium is not hyperbolic and Hartman–Grobman is silent.
2. Take \(V(x,y) = x^2 + y^2\) on \(N = \mathbb{R}^2\). Hypotheses, one at a time: \(V\) is \(C^1\); \(V(0,0) = 0\); \(V > 0\) elsewhere; and \[\dot V = 2x(-y - x^3) + 2y(x - y^3) = -2xy + 2xy - 2x^4 - 2y^4 = -2(x^4+y^4),\] which is \(< 0\) for every \((x,y) \ne (0,0)\). Lyapunov's theorem therefore gives asymptotic stability. Since \(\dot V < 0\) off the origin on the whole plane and every sublevel set \(\{V \le c\}\) is compact, the argument runs from any initial point: the origin is globally asymptotically stable.

3. For \(\dot x = -y + x^3\), \(\dot y = x + y^3\) the Jacobian at the origin is again \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). Now \(\dot V = +2(x^4+y^4) > 0\) off the origin. Instability: fix a small \(\varepsilon\) and a starting point with \(a := V(\mathbf x_0) > 0\) inside \(B_\varepsilon\). While the orbit remains in \(B_\varepsilon\) it stays in the compact set \(\{a \le V \le \varepsilon^2\}\), on which \(\dot V \ge k > 0\) by continuity and compactness; so \(V \ge a + kt\), which exceeds \(\varepsilon^2\) in finite time. The orbit must leave \(B_\varepsilon\), and it does so however small \(a\) was: the origin is unstable.

   What the pair proves. Two systems with identical linearisations, one with a globally asymptotically stable origin and one with an unstable origin. So at a non-hyperbolic equilibrium the linearisation carries no information about stability whatsoever — not "weak" information, none. The hypothesis in Hartman–Grobman is not an artefact of its proof.

Check. Dimensions: \(V\) has the dimensions of a squared length, \(\dot V\) of (length\(^4\))/time, so the cubic terms carry the compensating length\(^{-2}\)time\(^{-1}\) — the systems are already written in dimensionless variables. Limit: drop the cubic terms and \(\dot V = 0\), the linear centre, consistent. Number: from \(\mathbf x_0 = (1,0)\) the first system has \(\dot V = -2\) initially, so \(V\) falls from \(1\) at rate \(2\) — a fast start and, since \(\dot V \sim -2V^2\) near the axes, a slow \(t^{-1}\)-like tail, which is not exponential and so not something a linearisation could have predicted.

Idea. Energy again, but \(\dot V\) vanishes on a whole line, so Lyapunov's asymptotic test fails and LaSalle is the right instrument. The potential grows without bound, so the sublevel sets are compact and the basin is everything.

Working.

1. \(\dot x = v\), \(\dot v = -v|v| - x^3\). The right-hand side is \(C^1\) (the derivative of \(v|v|\) is \(2|v|\), continuous), so Picard–Lindelöf applies. Equilibria need \(v = 0\) and \(x^3 = 0\): the origin alone.
2. \(\dot V = v\dot v + x^3\dot x = v(-v|v| - x^3) + x^3 v = -v^2|v| = -|v|^3 \le 0\). Lyapunov's asymptotic half requires \(\dot V < 0\) on a punctured neighbourhood, but \(\dot V = 0\) on the entire line \(\{v = 0\}\), which meets every neighbourhood of the origin. So that test gives stability only.
3. Fix \(c > 0\) and let \(\Omega = \{V \le c\}\). It is closed, and bounded because \(V \ge \tfrac14x^4\) and \(V \ge \tfrac12 v^2\) force \(|x| \le (4c)^{1/4}\), \(|v| \le \sqrt{2c}\); hence compact. It is forward invariant since \(\dot V \le 0\). Now \(E = \{v = 0\} \cap \Omega\). If a solution lies in \(E\) for all \(t\), then \(v \equiv 0\), so \(\dot v \equiv 0\), and \(\dot v = -0 - x^3\) forces \(x \equiv 0\). So \(M = \{(0,0)\}\), and LaSalle gives \(\varphi_t(\mathbf x_0) \to (0,0)\) for every \(\mathbf x_0 \in \Omega\). Since \(c\) was arbitrary and every point of the plane lies in some \(\Omega\), the basin is all of \(\mathbb{R}^2\): the origin is globally asymptotically stable.
4. Because the pendulum's \(M\) is bigger. Running the same argument on the cylinder with \(\Omega = \{V \le c\}\), \(c > 2\), the set is still compact and forward invariant and \(E\) is still \(\{\omega = 0\}\); but now \(\dot\omega = -\sin\theta\) vanishes at \(\theta = 0\) and \(\theta = \pi\), so \(M = \{(0,0), (\pi,0)\}\) and the conclusion is only that every orbit approaches the pair. That is the truth: the saddle's stable manifold really does end upright, so the hanging position's basin is not the whole cylinder. Taking \(c < 2\) excludes the saddle and recovers the deck's result.

Check. Dimensions: \(v|v|\) is an acceleration, so the damping coefficient hidden in it carries \(\mathsf{L}^{-1}\); \(V\) is then an energy per unit mass, and \(\dot V = -|v|^3\) is a power per unit mass. Limit: replace \(v|v|\) by \(\gamma v\) and \(\dot V = -\gamma v^2\), the linear case, with the same conclusion by the same argument. Number: starting from rest at \(x = 2\), \(V = \tfrac14\cdot16 = 4\), so the motion is forever confined to \(|x| \le (16)^{1/4} = 2\) and \(|v| \le \sqrt 8 = 2.8284\).

Idea. An inner boundary from the energy \(W\), an outer boundary from three straight edges chosen so that the field crosses each inwards, and Poincaré–Bendixson applied to the annulus between them.

Working.

1. Differentiating \(\dot x = y - F(x)\) and using \(\dot y = -x\): \(\ddot x = \dot y - F'(x)\dot x = -x - (x^2-1)\dot x\), i.e. \(\ddot x - (1-x^2)\dot x + x = 0\). Equilibria: \(\dot y = 0\) gives \(x = 0\), and then \(\dot x = 0\) gives \(y = F(0) = 0\) — the origin only. Its Jacobian is \(\begin{pmatrix}-F'(x) & 1\\ -1 & 0\end{pmatrix}_{(0,0)} = \begin{pmatrix}1&1\\-1&0\end{pmatrix}\), so \(\tau = 1\), \(\Delta = 1\), \(\tau^2 - 4\Delta = -3 < 0\): an unstable spiral, eigenvalues \(0.5 \pm 0.8660\,i\).
2. \(\dot W = x\dot x + y\dot y = x\big(y - F(x)\big) - xy = -xF(x) = -x\big(\tfrac13x^3 - x\big) = x^2\big(1 - \tfrac13x^2\big)\). On the closed unit disc \(|x| \le 1 < \sqrt3\), so \(\dot W \ge 0\) there. Suppose an orbit had \(W(t_0) \ge \tfrac12\) and \(W(t_1) < \tfrac12\) for some \(t_1 > t_0\). Let \(t_2\) be the last time before \(t_1\) with \(W = \tfrac12\); on \([t_2, t_1]\) the orbit lies in the disc \(\{W \le \tfrac12\}\), where \(\dot W \ge 0\), so \(W(t_1) \ge W(t_2) = \tfrac12\) — a contradiction. Hence \(\{x^2+y^2 \ge 1\}\) is forward invariant.

3. Write \(c = F(4) = \tfrac{64}{3} - 4 = \tfrac{52}{3}\) and \(s = c/4 = \tfrac{13}{3}\). The field satisfies \(f(-x,-y) = -f(x,y)\) (because \(F\) is odd), so the symmetry \(\sigma(x,y) = (-x,-y)\) maps orbits to orbits, and \(K\) is \(\sigma\)-invariant: it suffices to check three edges.

   • Edge A, \(y = c\), \(0 \le x \le 4\). With \(g = y - c\), \(\dot g = \dot y = -x \le 0\), strictly for \(x > 0\).
   • Edge B, \(x = 4\), \(0 \le y \le c\). With \(g = x - 4\), \(\dot g = \dot x = y - F(4) = y - c \le 0\), strictly for \(y < c\).
   • Edge C, the segment from \((4,0)\) to \((0,-c)\), i.e. \(y = sx - c\) with \(0 \le x \le 4\); the region lies above it, \(g = sx - c - y\). Then \[\dot g = s\dot x - \dot y = s\big(y - F(x)\big) + x = \underbrace{s\big(sx - c - \tfrac13x^3 + x\big) + x}_{=:h(x)} .\] \(h'(x) = s^2 + s + 1 - sx^2\) vanishes at \(x^\ast = \sqrt{(s^2+s+1)/s} = \sqrt{217/39} = 2.3588\), the only maximum on \([0,4]\), and \(h(x^\ast) = -sc + x^\ast(s^2+s+1) - \tfrac13 s (x^\ast)^3 = -75.111 + 56.874 - 18.958 = -37.195 < 0\). (Endpoints, for reassurance: \(h(0) = -sc = -75.111\) and \(h(4) = -71.111\).) So \(\dot g < 0\) strictly along the whole edge.

   Tangential points. Two boundary points have \(\dot g = 0\) for one of their active constraints, and by the symmetry it is enough to treat one of each kind. At \((0, c)\), edge A is tangent; but there \(\dot x = c - F(0) = c > 0\), so \(x(t) > 0\) immediately afterwards and hence \(\dot y = -x < 0\): the orbit moves strictly below \(y = c\). At the corner \((4, c)\), edge B is tangent (\(\dot x = c - c = 0\)); there \(\ddot x = \dot y - F'(x)\dot x = -4 - 0 < 0\), so \(x\) strictly decreases, and edge A is strictly inward (\(\dot y = -4 < 0\)). In both cases the orbit enters the interior. Every other boundary point has every active constraint strictly decreasing, so a first exit time cannot exist: at such a time some \(g\) would have to be non-decreasing. Hence \(K\) is forward invariant.

4. Let \(A = K \cap \{x^2+y^2 \ge 1\}\). It is compact (closed subset of the compact \(K\)), forward invariant (an intersection of two forward invariant sets), non-empty, and contains no equilibrium, since the only equilibrium is the origin and the unit disc has been removed — note the disc is inside \(K\), because \(K\) is convex and the origin's distances to the three edge lines are \(4\), \(c = 17.3333\) and \(c/\sqrt{s^2+1} = 3.8976\), all greater than \(1\). Any orbit starting in \(A\) therefore stays in \(A\) forever, and by Poincaré–Bendixson its \(\omega\)-limit set is a periodic orbit contained in \(A\). So the Van der Pol equation has a periodic solution, and we have not computed it.

   Where the plane was used: only in Poincaré–Bendixson, whose proof needs the Jordan curve theorem. Steps 1–3 are valid in any dimension and prove only that the orbits are trapped — which in \(\mathbb{R}^3\) would allow chaos instead of a cycle.

Check. Dimensions: the Liénard variables are already dimensionless; \(W\) has the dimensions of \(x^2\) and \(\dot W\) of \(x^2\) per unit time, matching \(x^2(1-x^2/3)\) once the \(1\) is read as an inverse time. Limit: the sign change of \(\dot W\) at \(|x| = \sqrt3 = 1.7321\) predicts a cycle crossing that value; the true Van der Pol cycle at \(\mu = 1\) has maximum \(|x| = 2.0086\), comfortably outside, as the inequality only requires. Number: the region \(A\) is enormous compared with the cycle — \(K\) reaches \(|y| = 17.33\) — which is the usual price of an explicit trapping region: it proves existence, not location.

Idea. The index is an angle count: write the field in polar form and read off how fast its angle turns.

Working.

1. Put \(x = \rho\cos\phi\), \(y = \rho\sin\phi\). Then \(x^2 - y^2 = \rho^2\cos 2\phi\) and \(2xy = \rho^2\sin 2\phi\), so \(f = \rho^2(\cos2\phi, \sin2\phi)\): the angle of \(f\) is \(\vartheta = 2\phi\). As \(\phi\) runs from \(0\) to \(2\pi\), \(\vartheta\) increases by \(4\pi\), so \(I_f = 4\pi/2\pi = 2\), for every \(\rho > 0\). (\(f\) is the map \(z \mapsto z^2\) in complex notation, and its index is its degree.)
2. A hyperbolic equilibrium has \(\operatorname{ind} = \operatorname{sign}\det Df \in \{+1,-1\}\), and \(2\) is neither; so the origin is not hyperbolic. Indeed every entry of \(Df = \begin{pmatrix}2x & -2y\\ 2y & 2x\end{pmatrix}\) vanishes at the origin, so \(Df(0,0) = 0\), which is as non-hyperbolic as a matrix can be.
3. By the Poincaré index theorem, \(I_f(\Gamma) = (+1) + (+1) + (-1) = +1\). That is consistent with \(\Gamma\) being a periodic orbit, since the Umlaufsatz requires index \(+1\) — so the index test does not rule it out. It does not prove one exists either: the index is a necessary condition, never a sufficient one.
4. Suppose \(\Gamma\) is a periodic orbit enclosing exactly one equilibrium, a saddle. The Umlaufsatz gives \(I_f(\Gamma) = +1\); the index theorem gives \(I_f(\Gamma) = \operatorname{ind}(\text{saddle}) = -1\). Since \(+1 \ne -1\), no such \(\Gamma\) exists. \(\square\)

Check. Consistency: a field whose index is \(2\) must, under any small perturbation that makes the equilibria hyperbolic, break into equilibria whose indices sum to \(2\) — for instance two sinks, or two spirals. Trying \(f_\epsilon = (x^2-y^2-\epsilon,\ 2xy)\) gives equilibria at \((\pm\sqrt\epsilon, 0)\); their determinants are \(4x^2 = 4\epsilon > 0\), so each has index \(+1\) and the total is \(2\), as the homotopy invariance of the index on a large circle demands.

Idea. The system is \(\dot x = -V_x\), \(\dot y = -V_y\), and here it decouples, so everything reduces to a one-dimensional picture times a decaying \(y\).

Working.

1. \(V_x = 2x(x-1)^2 + 2x^2(x-1) = 2x(x-1)(2x-1)\) and \(V_y = 2y\). So \[\dot x = -2x(x-1)(2x-1), \qquad \dot y = -2y,\] with equilibria \((0,0)\), \((\tfrac12, 0)\), \((1,0)\). The Jacobian is the diagonal matrix \(\operatorname{diag}(-(12x^2-12x+2),\ -2)\), giving

   • \((0,0)\): \(\operatorname{diag}(-2,-2)\) — a stable star node;
   • \((\tfrac12,0)\): \(\operatorname{diag}(+1,-2)\) — a saddle;
   • \((1,0)\): \(\operatorname{diag}(-2,-2)\) — a stable star node.
2. Indices \(\operatorname{sign}\det\): \(+1\), \(-1\), \(+1\), total \(+1\). A large circle must give \(+1\), and it does: for large \(|x|\) the field points inward along \(x\) (\(\dot x \approx -4x^3\)) and along \(y\), and tracing the angle of \(f\) once round a circle of radius \(R\) gives one full turn. No periodic orbit: (i) a cycle would need to enclose equilibria of total index \(+1\), which is possible only by enclosing a sink alone or all three — but (ii) along any orbit \(\dot V = -\big(V_x^2 + V_y^2\big) \le 0\) with equality only at equilibria, so \(0 = \oint\dot V\,\mathrm{d}t < 0\) for a non-constant periodic orbit, a contradiction. The second argument is decisive on its own; the first shows how far index bookkeeping gets you without it.
3. The portrait: \(y\) decays to \(0\) exponentially everywhere, and the \(x\)-dynamics is one-dimensional with stable points \(0\) and \(1\) and an unstable point \(\tfrac12\). The separatrix is the vertical line \(x = \tfrac12\) — the saddle's stable manifold, which here is exact rather than approximate because the equations decouple. Basins: \(\{x < \tfrac12\}\) drains to \((0,0)\), \(\{x > \tfrac12\}\) to \((1,0)\). Orbits cross the level curves of \(V\) at right angles, as in any gradient system.

Check. \(V\) has a double well with minima \(V = 0\) at \(x = 0, 1\) and a local maximum \(V(\tfrac12, 0) = \tfrac1{16} = 0.0625\) between them, and the equilibria of the flow are exactly the critical points of \(V\) — as they must be, since \(\nabla V = 0\) is the equilibrium condition. Limit: drop the \(y\) equation and the remaining one-dimensional flow has the stability pattern stable–unstable–stable that any smooth double well forces.

Idea. Solve \(\mu x = \sin x\) by comparing \(\sin x / x\) with \(\mu\); the symmetry \(x \mapsto -x\) is what makes the bifurcation a pitchfork rather than a saddle-node.

Working.

1. \(f(x,\mu) = \mu x - \sin x\) vanishes at \(x = 0\) for every \(\mu\). \(f'(x) = \mu - \cos x\), so \(f'(0) = \mu - 1\): the origin is stable for \(\mu < 1\) and unstable for \(\mu > 1\).
2. Non-zero equilibria satisfy \(\mu = \sin x / x\). On \((0,\pi)\) the function \(\sin x/x\) is strictly decreasing from \(1\) (as \(x \to 0^+\)) to \(0\), so there is exactly one positive root when \(0 < \mu < 1\) and none when \(\mu \ge 1\); the odd symmetry of \(f\) gives the mirror root. So the pair exists below the threshold, on the side where the origin is stable: a subcritical pitchfork at \(\mu = 1\). The outer branches are unstable: at a root \(x^\ast \in (0,\pi)\), \(f'(x^\ast) = \mu - \cos x^\ast = \frac{\sin x^\ast - x^\ast\cos x^\ast}{x^\ast} > 0\), since \(\tan x^\ast > x^\ast\) on \((0,\pi/2)\) and \(\cos x^\ast < 0\) on \([\pi/2, \pi)\).
3. \(\mu x = x - x^3/6 + O(x^5)\) gives \(x^2 = 6(1-\mu)\), i.e. \(x \approx \pm\sqrt{6(1-\mu)}\). At \(\mu = 0.9\): the approximation is \(\sqrt{0.6} = 0.7746\), the exact root of \(0.9x = \sin x\) is \(0.7867\). The error is \(0.0121\), about \(1.5\%\). Keeping the next term, \(\mu x = x - x^3/6 + x^5/120\) gives \(x^2 = 6(1-\mu) + x^4/20\), a relative correction of \(x^2/40 = 0.0155\) — which is the \(1.5\%\) observed, so the discrepancy is exactly the \(O(x^5)\) truncation and nothing else.

Check. Limit \(\mu \to 1^-\): both branches tend to \(0\) like \(\sqrt{6(1-\mu)}\), the square-root law of every pitchfork. Limit \(\mu \to 0^+\): the roots tend to \(\pm\pi\), where \(\sin x/x \to 0\) — correct, since \(\dot x = -\sin x\) has equilibria at \(0\) and \(\pm\pi\). Sanity: \(\sin(0.7867)/0.7867 = 0.9000\), so the quoted root really does solve the equation at \(\mu = 0.9\).

Idea. The system is the Hopf normal form in disguise; polar coordinates separate it completely.

Working.

1. \(r\dot r = x\dot x + y\dot y = \mu(x^2+y^2) - (x^2+y^2)^2 = \mu r^2 - r^4\), so \(\dot r = \mu r - r^3\); and \(r^2\dot\theta = x\dot y - y\dot x = x^2 + y^2 = r^2\), so \(\dot\theta = 1\) (the cross terms cancel in both computations). At the origin the Jacobian is \(\begin{pmatrix}\mu & -1\\ 1 & \mu\end{pmatrix}\), eigenvalues \(\mu \pm i\): a stable spiral for \(\mu < 0\), a centre in the linearisation for \(\mu = 0\) — where the cubic decides, and \(\dot r = -r^3 < 0\) makes it an asymptotically stable non-hyperbolic sink — and an unstable spiral for \(\mu > 0\).
2. For \(\mu > 0\), \(\dot r = 0\) with \(r > 0\) forces \(r = \sqrt\mu\), so the circle \(r = \sqrt\mu\) is an orbit and it is the only periodic one: any periodic orbit must have \(r\) returning to its initial value, and \(r\) is strictly monotone off \(r = \sqrt\mu\). The period is \(2\pi\), since \(\dot\theta = 1\). It attracts because \(\partial_r(\mu r - r^3)\big|_{r=\sqrt\mu} = \mu - 3\mu = -2\mu < 0\), so \(\rho = r - \sqrt\mu\) obeys \(\dot\rho = -2\mu\rho + O(\rho^2)\). At \(\mu = 0.36\): radius \(\sqrt{0.36} = 0.6\) exactly, period \(2\pi = 6.2832\).
3. With the signs flipped, \(\dot r = \mu r + r^3\). For \(\mu < 0\) the cycle is at \(r = \sqrt{-\mu}\), and \(\partial_r(\mu r + r^3) = \mu + 3r^2 = -2\mu > 0\): repelling. It is the boundary of the basin of the (still stable) origin, and it shrinks to nothing as \(\mu \to 0^-\). What the experimenter sees. Supercritical: nothing until \(\mu = 0\), then a smooth oscillation whose amplitude grows as \(\sqrt\mu\) — reversible, no hysteresis. Subcritical: nothing until \(\mu = 0\), then a sudden jump to whatever large-amplitude behaviour exists elsewhere in the system, with no small oscillation in between, and usually hysteresis on the way back down.

Check. Dimensions: \(\mu\) is an inverse time (it is a growth rate), so the cubic term must be \(k r^3\) with \([k] = \mathsf{T}^{-1}\mathsf{L}^{-2}\), and the cycle radius \(\sqrt{\mu/k}\) is then a length, as an amplitude must be. Writing \(k = 1\), as the statement does, is a choice of length unit. Limit \(\mu \to 0^+\): the cycle radius \(\to 0\) and its period stays \(2\pi\) — the cycle is born from the equilibrium, not from infinity. Number: at \(\mu = 0.36\) an orbit starting at \(r_0 = 0.1\) has \(\dot r = 0.036 - 0.001 = 0.035 > 0\), heading outward toward \(0.6\), as claimed.

Idea. Everything follows from one identity, \(\dot V = -|\nabla V|^2\), except the last part, which is the spectral theorem.

Working.

1. Along a solution, \(\dot V = \sum_i \partial_i V \cdot \dot x_i = -\sum_i (\partial_i V)^2 = -|\nabla V|^2 \le 0\), with equality exactly when \(\nabla V(\mathbf x) = 0\), which is exactly the equilibrium condition.
2. If \(\mathbf x(\cdot)\) is \(T\)-periodic, then \(0 = V(\mathbf x(T)) - V(\mathbf x(0)) = \int_0^T \dot V\,\mathrm{d}t = -\int_0^T|\nabla V|^2\,\mathrm{d}t \le 0\), so the integrand vanishes identically, so \(\nabla V \equiv 0\) along the orbit and the orbit is a single equilibrium. \(\square\)
3. Suppose an orbit is homoclinic to \(\mathbf p\): \(\mathbf x(t) \to \mathbf p\) as \(t \to \pm\infty\), with \(\mathbf x(t) \ne \mathbf p\). Then \(V(\mathbf x(t))\) is strictly decreasing (part 1, since \(\nabla V \ne 0\) along a non-constant orbit), while continuity gives \(\lim_{t\to-\infty}V(\mathbf x(t)) = V(\mathbf p) = \lim_{t\to+\infty}V(\mathbf x(t))\). A strictly decreasing function cannot have equal limits at \(\pm\infty\). For a cycle of heteroclinic orbits \(\mathbf p_1 \to \mathbf p_2 \to \dots \to \mathbf p_k \to \mathbf p_1\) the same monotonicity gives \(V(\mathbf p_1) > V(\mathbf p_2) > \dots > V(\mathbf p_k) > V(\mathbf p_1)\), which is impossible. \(\square\)
4. The Jacobian of \(-\nabla V\) is \(-\big(\partial_i\partial_j V\big)\), the negative Hessian. For \(V \in C^2\) the mixed partials commute — Clairaut's theorem, proved in unit 03 — so the matrix is symmetric, and the spectral theorem (linear algebra) gives an orthonormal eigenbasis with real eigenvalues. A spiral needs a non-real pair and a centre needs a purely imaginary one, so neither can occur. In the plane a hyperbolic equilibrium then has two real non-zero eigenvalues: same sign gives a node, opposite signs a saddle. \(\square\)

Check. Test on P8: \(V = x^2(x-1)^2 + y^2\) gave two star nodes and a saddle, with no spiral and no cycle — exactly the menu this problem says is available. Test on the pendulum: the damped pendulum has a spiral, so it is not a gradient system, however much it behaves like one; the deck called it gradient-like for precisely this reason. Limit: for \(V\) a positive definite quadratic form \(\tfrac12\mathbf x^\top P\mathbf x\), the flow is \(\dot{\mathbf x} = -P\mathbf x\) with \(P\) symmetric positive definite — a stable node in every eigendirection, and never a spiral.

Idea. Successive maxima of a decaying oscillation are one full period apart, so their ratio is \(e^{-(\gamma/2)T}\) with \(T = 2\pi/\beta\); inverting gives \(\gamma\).

Working.

1. Small oscillations obey the linearisation \(\ddot\theta + \gamma\dot\theta + \theta = 0\), whose solutions are (deck, “Solving \(\dot{\boldsymbol\xi} = A\boldsymbol\xi\), all three cases”, with \(\alpha = -\gamma/2\) and \(\beta = \sqrt{1-\gamma^2/4}\)) \(\theta(t) = Ae^{-\gamma t/2}\cos(\beta t + \phi)\). Maxima of \(\theta\) occur where \(\dot\theta = 0\) with \(\ddot\theta < 0\); differentiating, \(\dot\theta = 0\) requires \(\tan(\beta t + \phi) = -\gamma/(2\beta)\), whose solutions are spaced by \(\pi/\beta\) and alternate between maxima and minima. Successive maxima are therefore \(T = 2\pi/\beta\) apart, and at each the cosine factor repeats, so \[\frac{\theta_{n+1}}{\theta_n} = e^{-\gamma T/2} = e^{-\Lambda}, \qquad \Lambda = \frac{\gamma}{2}\cdot\frac{2\pi}{\beta} = \frac{\gamma\pi}{\sqrt{1-\gamma^2/4}} .\]
2. Squaring, \(\Lambda^2(1 - \gamma^2/4) = \gamma^2\pi^2\), so \(\Lambda^2 = \gamma^2(\pi^2 + \Lambda^2/4)\) and \(\gamma = \Lambda/\sqrt{\pi^2 + \Lambda^2/4} = 2\Lambda/\sqrt{4\pi^2+\Lambda^2}\). With the measured ratio \(0.1974\), \(\Lambda = \ln(1/0.1974) = 1.6225\), so \[\gamma = \frac{2(1.6225)}{\sqrt{4\pi^2 + 1.6225^2}} = \frac{3.2450}{6.4893} = 0.5001 .\]
3. The scaling of the deck's overview note is \(\gamma = b/(m\sqrt{gl^3})\), so \[b = \gamma\, m\sqrt{gl^3} = 0.5001 \times 0.2 \times \sqrt{9.81} = 0.3132 \ \text{N m s}.\] The period in dimensionless time is \(2\pi/\beta\) with \(\beta = \sqrt{1-\gamma^2/4} = 0.9682\), giving \(6.4893\) — which is \(\sqrt{4\pi^2+\Lambda^2}\) again, and not by accident: squaring the definition of \(\Lambda\) gives \(1-\gamma^2/4 = 4\pi^2/(4\pi^2+\Lambda^2)\), so \(2\pi/\beta = \sqrt{4\pi^2+\Lambda^2}\) exactly. Multiplying by the time unit \(\sqrt{l/g} = 0.3193\) s gives \(T = 2.0719\) s.
4. \(\gamma = 0.5001 < 2\), so the equilibrium is a stable spiral and the motion oscillates — consistent with there being successive maxima to measure at all. Energy is quadratic in the amplitude for small oscillations, so it falls by \(0.1974^2 = 0.0390\) per period, a loss of \(96.1\%\) — a heavily damped pendulum, which is why only a handful of swings are visible.

Check. Dimensions: \(b\) must be a torque per angular velocity, \(\mathsf{M}\mathsf{L}^2\mathsf{T}^{-1}\); \(m\sqrt{gl^3}\) is \(\mathsf{M}\sqrt{\mathsf{L}\mathsf{T}^{-2}\cdot\mathsf{L}^3} = \mathsf{M}\mathsf{L}^2\mathsf{T}^{-1}\) — and N m s is exactly that. Limit \(\gamma \to 0\): \(\Lambda \to 0\) and the ratio \(\to 1\), an undamped pendulum whose maxima never shrink. Cross-check on the period: the undamped period would be \(2\pi\sqrt{l/g} = 2.0061\) s, so damping has lengthened it by \(3.3\%\), the figure quoted on the deck's “The pendulum's two equilibria, classified” for the same \(\gamma\).