02.12.E1 · analysis / ode

Qualitative theory of ODEs exercise pack (Arnold Ch. 2-3 supplement)

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Formal definition of the pack Intermediate

Arnold's chapters on the phase flow and on linear systems build the qualitative theory of ordinary differential equations: a system $\overset{x}{˙} = v (x)$ on a domain in $R^{n}$ is studied through its phase portrait — the partition of phase space into integral curves — rather than through closed-form solutions. The organizing objects are equilibria (zeros of $v$ ), the linearization $\dot{ξ} = A ξ$ with $A = D v (x_{0})$ at an equilibrium $x_{0}$ , the eigenvalue spectrum of $A$ that classifies the local picture, the phase flow $g^{t}$ as a one-parameter group of diffeomorphisms, the rectification theorem that straightens the flow near a regular point, and Lyapunov functions that certify stability without solving the system.

This pack collects nine exercises — three easy, four medium, two hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units. The exercises group loosely by Arnold's sections: drawing and reading phase portraits and computing equilibria (easy); classifying planar equilibria by linearization and applying the rectification theorem and the trace-determinant plane (medium); and proving stability by Lyapunov functions and reasoning about flows globally (hard).

Conventions follow Arnold: a vector field $v$ on phase space generates the phase flow $g^{t}$ with $\frac{d}{d t}_{t = 0} g^{t} (x) = v (x)$ ; an equilibrium is hyperbolic when the linearization has no eigenvalue on the imaginary axis; stability is in the sense of Lyapunov unless "asymptotic" is stated.

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Classify the equilibrium at the origin of the linear planar system $\overset{x}{˙} = A x$ by the eigenvalues of $A$ , and list every qualitative type.

Solution. For $\overset{x}{˙} = A x$ with $A$ a real $2 \times 2$ matrix and an isolated equilibrium at the origin ( $det A \neq = 0$ ), the local and global phase portrait is fixed by the eigenvalues $λ_{1}, λ_{2}$ of $A$ , equivalently by the trace $τ = λ_{1} + λ_{2}$ and the determinant $Δ = λ_{1} λ_{2}$ , through $λ_{1, 2} = \frac{1}{2} (τ \pm τ^{2} - 4Δ)$ .

Real distinct eigenvalues ( $τ^{2} - 4Δ > 0$ , $Δ \neq = 0$ ). If $λ_{1}, λ_{2} < 0$ (so $Δ > 0$ , $τ < 0$ ): a stable **node**, all trajectories approaching the origin tangent to the slow eigendirection. If $λ_{1}, λ_{2} > 0$ : an unstable node (time-reversed). If $λ_{1} < 0 < λ_{2}$ (so $Δ < 0$ ): a saddle, with a one-dimensional stable manifold (the $λ_{1}$ -eigenline) and a one-dimensional unstable manifold.

Complex eigenvalues ( $τ^{2} - 4Δ < 0$ , so $λ_{1, 2} = α \pm i β$ , $β \neq = 0$ ). If $α < 0$ : a stable **focus** (spiral) winding into the origin. If $α > 0$ : an unstable focus. If $α = 0$ (so $τ = 0$ , $Δ > 0$ ): a center, a one-parameter family of closed orbits.

Repeated real eigenvalue ( $τ^{2} - 4Δ = 0$ , $λ_{1} = λ_{2} = λ$ ). If $A$ is diagonalizable ( $A = λ I$ ): a star node, every ray a trajectory. If $A$ has a single eigenvector: a degenerate (improper) node.

These types fill the trace-determinant plane: $Δ < 0$ gives saddles; $Δ > 0$ with $τ^{2} > 4Δ$ gives nodes, $τ^{2} < 4Δ$ gives foci, and the parabola $τ^{2} = 4Δ$ carries the degenerate/star nodes; the half-line $τ = 0, Δ > 0$ carries the centers. Stability is governed by $τ$ : $τ < 0$ stable, $τ > 0$ unstable, the line $τ = 0$ marginal. $□$

This classification is the backbone of planar qualitative theory. By the Hartman-Grobman theorem, the phase portrait of a nonlinear system near a hyperbolic equilibrium is topologically conjugate to that of its linearization, so every exercise below that linearizes a nonlinear system reduces to reading off this table.

Exercises Intermediate

Exercise 3 (easy, proof). Phase flow is a one-parameter group.

Let $g^{t}$ be the phase flow of a complete vector field $v$ on $R^{n}$ . Show that $g^{t + s} = g^{t} \circ g^{s}$ for all $t, s \in R$ , and deduce that each $g^{t}$ is a diffeomorphism with inverse $g^{- t}$ .

Hint

Both $t \mapsto g^{t + s} (x)$ and $t \mapsto g^{t} (g^{s} (x))$ are integral curves of $v$ . Compare their values at $t = 0$ and invoke uniqueness of solutions.

Answer

Fix $s$ and $x$ . Define $γ (t) = g^{t + s} (x)$ and $η (t) = g^{t} (g^{s} (x))$ . Both are solutions of $\overset{u}{˙} = v (u)$ : indeed $\overset{γ}{˙} (t) = \frac{d}{d t} g^{t + s} (x) = v (g^{t + s} (x)) = v (γ (t))$ , and likewise $\overset{η}{˙} (t) = v (η (t))$ by definition of the flow. At $t = 0$ both equal $g^{s} (x)$ . By uniqueness of solutions of the initial-value problem, $γ \equiv η$ , i.e. $g^{t + s} = g^{t} \circ g^{s}$ .

Setting $s = - t$ gives $g^{0} = g^{t} \circ g^{- t}$ ; since $g^{0} = id$ , each $g^{t}$ is invertible with inverse $g^{- t}$ . Smooth dependence on initial conditions makes $g^{t}$ and $g^{- t}$ smooth, so $g^{t}$ is a diffeomorphism. The family ${g^{t}}_{t \in R}$ is thus a one-parameter group of diffeomorphisms — the group law $g^{t + s} = g^{t} g^{s}$ is exactly homomorphism of $(R, +)$ into $Diff (R^{n})$ .

Exercise 4 (medium, symbolic). Linearize and classify.

Classify the equilibrium $(0, 0)$ of the nonlinear system $\overset{x}{˙} = - x + y + x^{2}, \overset{y}{˙} = - x - y$ by linearization.

Hint

Compute the Jacobian $D v$ at $(0, 0)$ , drop the quadratic term, and read its eigenvalues off the trace and determinant.

Answer

The Jacobian is $$ Dv(x,y) = \begin{pmatrix} -1 + 2x & 1 \ -1 & -1 \end{pmatrix}, \qquad Dv(0,0) = \begin{pmatrix} -1 & 1 \ -1 & -1 \end{pmatrix}. $$ Trace $τ = - 2$ , determinant $Δ = (- 1) (- 1) - (1) (- 1) = 1 + 1 = 2$ . Then $τ^{2} - 4Δ = 4 - 8 = - 4 < 0$ , so the eigenvalues are complex: $λ = \frac{1}{2} (- 2 \pm - 4) = - 1 \pm i$ .

Since $Re λ = - 1 < 0$ and the eigenvalues are complex, the origin is a stable focus (spiral sink). The equilibrium is hyperbolic (no eigenvalue on the imaginary axis), so by Hartman-Grobman the nonlinear flow near the origin is topologically conjugate to the linear spiral: trajectories spiral into the origin.

Exercise 5 (medium, symbolic). Rectification away from an equilibrium.

The field $v (x, y) = (1, x)$ on $R^{2}$ has no zeros. Exhibit an explicit change of coordinates rectifying it to the constant field $(1, 0)$ near the origin.

Hint

The rectifying chart is built from the flow: $Φ (t, c) = g^{t} (σ (c))$ where $σ$ parametrizes a transverse section. Take the $y$ -axis ${x = 0}$ as the section.

Answer

Solve the flow. From $\overset{x}{˙} = 1$ , $x (t) = x_{0} + t$ ; from $\overset{y}{˙} = x = x_{0} + t$ , $y (t) = y_{0} + x_{0} t + \frac{1}{2} t^{2}$ . The $y$ -axis ${x = 0}$ is transverse to $v$ (there $v = (1, \cdot)$ has nonzero first component). Parametrize the section by $σ (c) = (0, c)$ and set $$ \Phi(t, c) = g^t(\sigma(c)) = \bigl(t,\ c + \tfrac{1}{2}t^2\bigr). $$ The map $Φ$ is a local diffeomorphism (Jacobian $det (1 t 01) = 1$ ). In the coordinates $(t, c)$ with $t = x$ and $c = y - \frac{1}{2} x^{2}$ , the field becomes $\partial_{t}$ , i.e. the constant field $(1, 0)$ .

Check directly: with new coordinate $c = y - \frac{1}{2} x^{2}$ , $\overset{c}{˙} = \overset{y}{˙} - x \overset{x}{˙} = x - x \cdot 1 = 0$ and $\dot{t} = \overset{x}{˙} = 1$ , so $\dot{(t, c)} = (1, 0)$ . The level sets $c = const$ are the integral curves (parabolas $y = \frac{1}{2} x^{2} + c$ ), straightened into horizontal lines by the rectifying chart.

Exercise 6 (medium, symbolic). A first integral and closed orbits.

For the system $\overset{x}{˙} = y, \overset{y}{˙} = - x - x^{3}$ , find a first integral and use it to show every orbit near the origin is closed.

Hint

Look for $H (x, y) = \frac{1}{2} y^{2} + V (x)$ conserved along trajectories; you need $V^{'} (x) = x + x^{3}$ . Show $H$ has a strict local minimum at the origin.

Answer

Try $H (x, y) = \frac{1}{2} y^{2} + V (x)$ with $V^{'} (x) = x + x^{3}$ , i.e. $V (x) = \frac{1}{2} x^{2} + \frac{1}{4} x^{4}$ . Then $$ \dot{H} = y\dot{y} + V'(x)\dot{x} = y(-x - x^3) + (x + x^3)y = 0, $$ so $H$ is a first integral: it is constant along every trajectory.

At the origin $H (0, 0) = 0$ , $\nabla H (0, 0) = 0$ , and the Hessian is $diag (V^{''} (0), 1) = diag (1, 1)$ , positive-definite. Hence $H$ has a strict local minimum at the origin, and for small $E > 0$ the level set ${H = E}$ is a small closed curve encircling the origin. Trajectories are confined to these level sets; since each level set near the origin is a single closed loop without equilibria on it, every such orbit is a periodic orbit. The origin is a (nonlinear) center, consistent with the linearization $\overset{x}{˙} = y, \overset{y}{˙} = - x$ having eigenvalues $\pm i$ .

Exercise 7 (medium, numeric). Read the trace-determinant plane.

For $A = (1243)$ , compute $τ$ and $Δ$ , locate the point $(τ, Δ)$ in the trace-determinant plane, and name the equilibrium type and its stability.

Hint

$τ = tr A$ , $Δ = det A$ . The sign of $Δ$ separates saddles from nodes/foci; then compare $τ^{2}$ with $4Δ$ .

Answer

$τ = 1 + 3 = 4$ , $Δ = (1) (3) - (4) (2) = 3 - 8 = - 5$ . Since $Δ = - 5 < 0$ , the point lies below the horizontal axis in the trace-determinant plane: the equilibrium is a saddle, which is always unstable.

Confirm via eigenvalues: $λ = \frac{1}{2} (4 \pm 16 + 20) = \frac{1}{2} (4 \pm 6) = 5$ or $- 1$ . One positive and one negative eigenvalue, exactly the saddle signature. The stable manifold is the $(- 1)$ -eigenline and the unstable manifold is the $5$ -eigenline. (The trace $τ = 4$ is irrelevant to stability once $Δ < 0$ : a saddle is unstable regardless of $τ$ .)

Exercise 8 (hard, proof). Lyapunov function proves asymptotic stability.

For the system $\overset{x}{˙} = - x + x y, \overset{y}{˙} = - y - x^{2}$ , use the candidate $L (x, y) = \frac{1}{2} (x^{2} + y^{2})$ to prove the origin is asymptotically stable.

Hint

Show $L > 0$ away from the origin and $L (0, 0) = 0$ (positive-definiteness), then compute $\dot{L}$ along trajectories and show it is negative-definite near the origin.

Answer

$L (x, y) = \frac{1}{2} (x^{2} + y^{2})$ is positive-definite: $L (0, 0) = 0$ and $L > 0$ for $(x, y) \neq = 0$ .

Compute the orbital derivative: $$ \dot{L} = x\dot{x} + y\dot{y} = x(-x + xy) + y(-y - x^2) = -x^2 + x^2 y - y^2 - x^2 y = -x^2 - y^2. $$ The cross terms $x^{2} y$ cancel exactly, leaving $\dot{L} = - (x^{2} + y^{2}) = - 2 L < 0$ for all $(x, y) \neq = 0$ . So $\dot{L}$ is negative-definite on all of $R^{2}$ .

By Lyapunov's direct method, a positive-definite $L$ with negative-definite $\dot{L}$ certifies asymptotic stability of the origin. (Here $\dot{L} = - 2 L$ even gives the explicit decay $L (t) = L (0) e^{- 2 t}$ , so $∣ (x (t), y (t)) ∣ \to 0$ exponentially.) No solving of the nonlinear system was required — the Lyapunov function delivers the stability conclusion directly.

Exercise 9 (hard, proof). No equilibrium inside a closed orbit forbidden in a gradient system.

Show that a planar gradient system $\overset{x}{˙} = - \nabla F (x)$ , with $F$ smooth, has no nonconstant periodic orbits.

Hint

$F$ decreases strictly along nonequilibrium trajectories. A periodic orbit would return $F$ to its starting value — a contradiction.

Answer

Along any trajectory $x (t)$ of $\overset{x}{˙} = - \nabla F (x)$ , $$ \frac{d}{dt} F(x(t)) = \nabla F(x(t)) \cdot \dot{x}(t) = \nabla F \cdot (-\nabla F) = -|\nabla F(x(t))|^2 \le 0, $$ with equality at time $t$ iff $\nabla F (x (t)) = 0$ , i.e. iff $x (t)$ is an equilibrium.

Suppose, for contradiction, a nonconstant periodic orbit of period $T > 0$ existed: $x (0) = x (T)$ with $x$ not identically an equilibrium. Then $F (x (T)) = F (x (0))$ , so $\int_{0}^{T} \frac{d}{d t} F (x (t)) d t = 0$ . But the integrand $- ∣\nabla F (x (t)) ∣^{2} \leq 0$ is strictly negative for at least one $t$ (since the orbit is nonconstant, it passes through points where $\nabla F \neq = 0$ ), forcing the integral to be strictly negative — a contradiction.

Therefore gradient flows have no nonconstant periodic orbits; their only recurrent behaviour is convergence to equilibria. This is the qualitative signature that distinguishes gradient systems from systems admitting limit cycles (such as the Van der Pol oscillator) and from Hamiltonian centers (Exercise 6), where $F = H$ is conserved rather than strictly decreasing.

Exercise pack — Arnold Ordinary Differential Equations Ch. 2-3 supplement: phase portraits, equilibria and stability, linearization, the phase flow, and the rectification theorem. Nine exercises, three easy / four medium / two hard.

Prerequisites

02.12.01 pending
02.12.02 pending
02.12.05 pending
02.12.08 pending
02.12.10 pending
02.12.12 pending
02.06.03 pending

Tier anchors

beginner
intermediate: Arnold *Ordinary Differential Equations* Ch. 2 §§7-8 (phase flow, rectification) and Ch. 3 §§12-13, §§19-23 (linear systems, equilibria, stability) exercises
master

References

Arnold — Ordinary Differential Equations (3rd ed., Springer 1992) · Ch. 2 §7 (rectification / flow-box), §8 (one-parameter groups); Ch. 3 §12 (phase plane), §17-19 (linear classification of equilibria), §22-23 (Lyapunov functions); end-of-section problems
Hirsch-Smale-Devaney — Differential Equations, Dynamical Systems, and an Introduction to Chaos (3rd ed., Academic Press 2013) · Ch. 3-4 (planar linear systems, the trace-determinant plane), Ch. 8-9 (equilibria, linearization, Lyapunov)
Perko — Differential Equations and Dynamical Systems (3rd ed., Springer 2001) · §1.5 (linear classification), §2.6-2.9 (Hartman-Grobman, stable manifold), §3.7-3.9 (Lyapunov, limit cycles)

Estimated time

intermediate: 5h