---
title: "Ordinary Differential Equations"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 19
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/19-ordinary-differential-equations
---

# Chapter 19 — Ordinary Differential Equations

Year 2 solved linear differential equations and stated the Cauchy–Lipschitz theorem; this chapter proves it — twice over: existence and uniqueness by the Banach fixed point, global structure by the theory of [maximal solutions](#thm-b3-ode-maximal) and the escape-from-compacts theorem. The linear theory is then rebuilt on honest foundations ([resolvent](#thm-b3-ode-linearstructure), [Wronskian](#thm-b3-ode-linearstructure), [matrix exponential](#thm-b3-ode-matrixexp), Duhamel), and the chapter’s second half opens the *qualitative* theory — [flows](#ex-b3-ode-planeclassification), equilibria, [Lyapunov functions](#thm-b3-ode-lyapunov), and [stability by linearization](#thm-b3-ode-linearization): how to understand solutions one will never compute. The pendulum, in the weekend problem, is the eternal case study. Throughout, $U \subseteq \R\times\R^d$ is open and $f \colon U \to \R^d$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity); a *solution* of $x' = f(t, x)$ is a $\mathcal C^1$ map $x \colon I \to \R^d$ ($I$ an interval) with graph in $U$ satisfying the equation.

## 19.1 Cauchy–Lipschitz

**Definition 19.1.**

$f$ is *locally Lipschitz in $x$* if every point of $U$ has a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $V$ and a constant $L$ with $\norm{f(t, x_1) - f(t, x_2)} \leq L\norm{x_1 -
x_2}$ for $(t, x_1), (t, x_2) \in V$. If $f$ is $\mathcal
C^1$ (or merely $\partial_xf$ exists and is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)), it is locally Lipschitz in $x$: on a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) convex [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology), the mean value inequality with $L =
\sup\vertiii{\partial_xf}$.

**Theorem 19.2 (Cauchy–Lipschitz, local).**

Let $f$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and [locally Lipschitz](#def-b3-ode-lipschitz) in $x$, and $(t_0,
x_0) \in U$. There is $T > 0$ such that the Cauchy problem

$$
x' = f(t, x), \qquad x(t_0) = x_0
$$

has exactly one solution on $[t_0 - T, t_0 +
T]$.

**Proof.** Choose $a, b > 0$ with $Q = [t_0 - a, t_0 + a]\times\bar
B(x_0, b) \subseteq U$, on which $\norm f \leq M$ and $f$ is $L$-Lipschitz in $x$. A $\mathcal C^1$ function is a solution iff it satisfies the integral equation

$$
x(t) = x_0 + \int_{t_0}^{t}f\bigl(s, x(s)\bigr)\,\dd s
$$

(fundamental theorem of calculus, both ways). Let $T =
\min\bigl(a, \frac bM, \frac1{2L}\bigr)$, $I = [t_0 - T, t_0
+ T]$, and

$$
\mathcal E = \{x \in \mathcal C(I, \R^d) : \norm{x(t) - x_0}
\leq b\ \text{on } I\},
$$

a closed subset of the Banach space $(\mathcal C(I, \R^d),
\norm\cdot_\infty)$: [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) ([Definition 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete)). Define $\Phi(x)(t) = x_0 +
\int_{t_0}^tf(s, x(s))\dd s$: for $x \in \mathcal E$, $\norm{\Phi(x)(t) - x_0} \leq M\abs{t - t_0} \leq MT \leq b$ — $\Phi$ maps $\mathcal E$ to itself — and for $x, y \in
\mathcal E$:

$$
\norm{\Phi(x)(t) - \Phi(y)(t)} \leq \Bigl|\int_{t_0}^t
L\,\norm{x(s) - y(s)}\,\dd s\Bigr| \leq LT\,\norm{x -
y}_\infty \leq \tfrac12\norm{x - y}_\infty :
$$

a contraction. The Banach fixed point ([Theorem 7.4](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-banach)) gives a unique fixed point in $\mathcal E$: existence, and uniqueness among solutions staying in $\bar B(x_0, b)$ — but any solution on $I$ stays there ($\norm{x(t) - x_0} \leq M\abs{t - t_0} \leq b$ as long as the graph remains in $Q$, a [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) argument): uniqueness on $I$. ∎

**Lemma 19.3 (Grönwall).**

Let $u \colon I \to \intco0\infty$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), $t_0 \in
I$, and suppose

$$
u(t) \leq \alpha + \beta\,\Bigl|\int_{t_0}^{t}u(s)\,\dd
s\Bigr| \qquad (t \in I)
$$

with $\alpha \geq 0$, $\beta > 0$. Then $u(t) \leq
\alpha\,\eu^{\beta\abs{t - t_0}}$ on $I$.

**Proof.** For $t \geq t_0$: let $v(t) = \alpha +
\beta\int_{t_0}^tu(s)\dd s$, so $u \leq v$, $v' = \beta u
\leq \beta v$, and $(v\eu^{-\beta(t - t_0)})' \leq 0$: $v(t)
\leq v(t_0)\eu^{\beta(t-t_0)} = \alpha\eu^{\beta(t - t_0)}$. For $t \leq t_0$, apply the same to $\tilde u(t) = u(2t_0 -
t)$. ∎

**Corollary 19.4 (Uniqueness and continuous dependence).**

Under the hypotheses of [Theorem 19.2](#thm-b3-ode-cauchylipschitz), two solutions of $x' = f(t,x)$ that agree at one point agree on their common interval of definition. Quantitatively, if $x, y$ are two solutions with graphs in a region where $f$ is $L$-Lipschitz in $x$, then

$$
\norm{x(t) - y(t)} \leq \norm{x(t_0) -
y(t_0)}\,\eu^{L\abs{t - t_0}} .
$$

**Proof.** The estimate: $u = \norm{x - y}$ satisfies $u(t) \leq u(t_0)
+ L\abs{\int_{t_0}^tu}$ (subtract the integral equations); Grönwall. Global uniqueness: the agreement set $\{t : x(t)
= y(t)\}$ is closed in the common interval, nonempty, and open — around any agreement point, cover a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) piece of the common graph by finitely many Lipschitz boxes and apply the estimate with $u(t_1) = 0$ on each: locally $x \equiv y$. A nonempty open closed subset of an interval is everything. ∎

## 19.2 Maximal solutions

**Theorem 19.5 (Maximal solutions; escape from compacts).**

Assume $f$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), [locally Lipschitz](#def-b3-ode-lipschitz) in $x$.

1. Every Cauchy problem has a unique *maximal* solution $x \colon \intoo{T_-}{T_+} \to \R^d$ : every other solution through $(t_0, x_0)$ is its restriction. The interval is open.
2. ( *Escape* ) For every [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K \subseteq U$ there is $\varepsilon > 0$ such that $(t, x(t)) \notin  K$ for all $t \in \intoo{T_+ - \varepsilon}{T_+}$ (and symmetrically at $T_-$ ): *the graph of a [maximal solution](#thm-b3-ode-maximal) eventually leaves every [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) subset of $U$* . In particular, for $U = \R\times\R^d$ and $T_+ < +\infty$ : $\norm{x(t)} \to +\infty$ as $t \to T_+^-$ ( *blow-up* ).

**Proof.** (1) Let $\mathcal S$ be the set of all solutions through $(t_0, x_0)$; by [Corollary 19.4](#cor-b3-ode-uniqueness) any two agree on the intersection of their intervals, so they glue: on $J =
\bigcup_{y \in \mathcal S}I_y$, define $x(t) = y(t)$ for any $y$ defined at $t$: a well-defined solution, evidently maximal and unique. $J$ is open: a solution defined at an endpoint could be prolonged by [Theorem 19.2](#thm-b3-ode-cauchylipschitz) at that endpoint.

(2) Suppose the claim fails at $T_+$: there are $t_n \to
T_+$ with $(t_n, x(t_n)) \in K$; note this forces $T_+ <
\infty$ or, if $T_+ = \infty$, there is nothing to prove ($K$ is bounded in time). So let $T_+ < \infty$. Compactness: uniform constants $M, L, a,
b$ work for all Cauchy data in a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $K$ — concretely, cover $K$ by finitely many boxes $Q_i$ as in the local theorem’s proof and let $T^* > 0$ be the minimum of the corresponding existence times: any Cauchy datum in $K$ launches a solution living at least $T^*$ beyond its initial time. Applying this at $(t_n, x(t_n))$ with $t_n > T_+ -
T^*/2$ extends $x$ past $T_+$ (the extension agrees with $x$ by uniqueness, then prolongs it): contradiction with maximality. So the graph leaves $K$ definitively before $T_+$. For $U = \R\times\R^d$: if $\norm{x(t)}\not\to\infty$, a sequence $t_n \to T_+$ keeps $(t_n, x(t_n))$ in the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $[t_0, T_+]\times\bar B(0, R)$: excluded. ∎

**Corollary 19.6 (Global existence under linear growth).**

If $U = I\times\R^d$ ($I$ open interval) and $\norm{f(t, x)}
\leq \alpha(t)\norm x + \beta(t)$ with $\alpha, \beta$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), then every [maximal solution](#thm-b3-ode-maximal) is defined on all of $I$.

**Proof.** On a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $[t_0, T] \subseteq I$: $\norm{x(t)} \leq
\norm{x_0} + \int_{t_0}^t(\alpha\norm x + \beta)$, so by Grönwall (with $\alpha, \beta$ bounded by $A, B$ there) $\norm{x(t)} \leq (\norm{x_0} + B(T - t_0))\eu^{A(T - t_0)}$: bounded. If $T_+ < \sup I$, the graph stays in a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) of $I\times\R^d$ near $T_+$: contradicts escape ([Theorem 19.5](#thm-b3-ode-maximal)). ∎

## 19.3 Linear systems

Throughout this section $A \colon I \to M_d(\R)$ and $b
\colon I \to \R^d$ are [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity); the system is $x' = A(t)x
+ b(t)$ — linear growth: all [maximal solutions](#thm-b3-ode-maximal) live on all of $I$ ([Corollary 19.6](#cor-b3-ode-globallinear)).

**Theorem 19.7 (Structure).**

The solutions of the homogeneous system $x' = A(t)x$ form a $d$-dimensional vector space $S_H$; for each $t_0$, the evaluation $x \mapsto x(t_0)$ is an isomorphism $S_H \to
\R^d$. The *resolvent* $R(t, s) \in GL_d(\R)$, defined by: $t \mapsto R(t,
s)v$ is the solution with value $v$ at $s$, satisfies

$$
R(s,s) = I,\quad R(t, u)R(u, s) = R(t, s),\quad
\partial_tR(t,s) = A(t)R(t,s),
$$

and the inhomogeneous problem is solved by *Duhamel’s formula*:

$$
x(t) = R(t, t_0)\,x_0 + \int_{t_0}^{t}R(t, s)\,b(s)\,\dd s .
$$

Finally the *Wronskian* $w(t) = \det
R(t, s)$ obeys Liouville’s formula $w' =
\operatorname{tr}A(t)\,w$, so $w(t) =
\exp\bigl(\int_s^t\operatorname{tr}A\bigr) > 0$.

**Proof.** Linearity of the equation makes solutions a vector space; evaluation is linear, injective (uniqueness: a solution vanishing at $t_0$ is $\equiv 0$) and surjective (existence): dimension $d$. The [resolvent](#thm-b3-ode-linearstructure) properties restate uniqueness (both sides of each identity solve the same Cauchy problem); invertibility from $R(s,t)R(t,s) = I$. Duhamel: differentiate the formula — $x'(t) = A(t)R(t,t_0)x_0 + R(t,t)b(t) +
\int_{t_0}^tA(t)R(t,s)b(s)\dd s = A(t)x(t) + b(t)$ (differentiation under the integral is legitimate: the integrand is $\mathcal C^1$ in $t$ with [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) derivative in $(t,s)$; or verify by the integral equation). Liouville: $w(t + h) = \det\bigl(R(t+h, t)\bigr)w(t)$ and $R(t + h, t) =
I + hA(t) + o(h)$ (from the integral equation), so $\det = 1
+ h\operatorname{tr}A(t) + o(h)$ (expansion of $\det$ at $I$): $w'(t) = \operatorname{tr}A(t)\,w(t)$; integrate the scalar linear ODE. ∎

**Theorem 19.8 (Matrix exponential).**

For $A \in M_d(\C)$, the series $\eu^{A} =
\sum_{n\geq0}\frac{A^n}{n!}$ converges (absolutely, in any submultiplicative norm), $\eu^{A+B} = \eu^A\eu^B$ whenever $AB = BA$, and $t \mapsto \eu^{tA}$ is the [resolvent](#thm-b3-ode-linearstructure) of the constant system: $R(t,s) = \eu^{(t-s)A}$; it is $\mathcal
C^\infty$ with $\frac{\dd}{\dd t}\eu^{tA} = A\eu^{tA}$. Moreover: if $\operatorname{Re}\lambda < -\alpha < 0$ for every eigenvalue $\lambda$ of $A$, then $\vertiii{\eu^{tA}}
\leq C\eu^{-\alpha t}$ for $t \geq 0$.

**Proof.** Convergence: $\vertiii{A^n/n!} \leq \vertiii A^n/n!$, summable ([Exercise 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#exo-b3-complete-1)(b) in the Banach algebra $M_d$). For commuting $A, B$: the Cauchy product of the two absolutely convergent series rearranges, via the binomial theorem (valid when $AB = BA$), into $\sum_n\frac{(A+B)^n}{n!}$. Differentiability, directly: $\frac{\eu^{(t+h)A} - \eu^{tA}}h = \eu^{tA}\frac{\eu^{hA} -
I}{h} \to \eu^{tA}A$ since $\norm{\frac{\eu^{hA} - I}h - A}
\leq \sum_{n\geq2}\frac{\abs h^{n-1}\vertiii A^n}{n!} =
O(h)$. Hence $t \mapsto \eu^{(t - s)A}v$ solves the Cauchy problem defining $R(t,s)v$. Spectral bound: by the [Jordan form](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#thm-b3-modules-jordan) ([Theorem 3.18](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#thm-b3-modules-jordan)), $A = P(D + N)P^{-1}$ with $D$ diagonal carrying the eigenvalues, $N$ nilpotent, and $DN =
ND$. Then $\eu^{tA} = P\,\eu^{tD}\eu^{tN}P^{-1}$ with $\vertiii{\eu^{tD}} \leq \eu^{-(\alpha + \delta)t}$ for some $\delta > 0$ ($t \geq 0$) and $\eu^{tN}$ polynomial in $t$ (nilpotence): the product is $\leq C\eu^{-\alpha t}$ (polynomial beaten by $\eu^{-\delta t}$). ∎

**Example 19.9 (The plane, classified).**

For $x' = Ax$ with $A \in M_2(\R)$ invertible, the phase portrait near $0$ is decided by $\tau = \operatorname{tr}A$ and $\delta = \det A$, through the eigenvalues $\lambda_\pm
= \frac{\tau \pm \sqrt{\tau^2 - 4\delta}}2$:

- $\delta < 0$ : real eigenvalues of opposite signs — a *saddle* ; two trajectories enter, two leave, all others fly by. Always unstable.
- $\delta > 0$ , $\tau^2 \geq 4\delta$ : real eigenvalues of the same sign ( $= \operatorname{sign}\tau$ ) — a *node* , [stable](#def-b3-ode-stability) iff $\tau < 0$ ; trajectories are tangent to the slow eigendirection.
- $\delta > 0$ , $\tau^2 < 4\delta$ , $\tau \neq 0$ : complex conjugate eigenvalues $\frac\tau2 \pm  \iu\omega$ — a *spiral* (focus), [stable](#def-b3-ode-stability) iff $\tau < 0$ ; the solutions are $\eu^{\tau t/2}\times$ rotations of period $\frac{2\pi}\omega$ .
- $\tau = 0$ , $\delta > 0$ : purely imaginary eigenvalues — a *[center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions)* : closed [orbits](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) (ellipses), stability without asymptotic stability, exactly the borderline that [Theorem 19.12](#thm-b3-ode-linearization) cannot decide for nonlinear systems (the pendulum’s bottom equilibrium, [Problem 19.1](#pb-b3-ode-1) , sits here).

The boundary parabola $\tau^2 = 4\delta$ carries the degenerate nodes (Jordan blocks: trajectories with a single tangent direction). Everything is read off two numbers — which is why the first reflex before any planar phase portrait is to compute $\operatorname{tr}$ and $\det$; e.g. $A = \bigl(\begin{smallmatrix}0 & 1\\ -1 &
-c\end{smallmatrix}\bigr)$ (damped oscillator): $\delta = 1 >
0$, $\tau = -c$: [stable](#def-b3-ode-stability) spiral for $0 < c < 2$, [stable](#def-b3-ode-stability) node for $c \geq 2$ — underdamping versus overdamping, in one glance.

## 19.4 Flows, equilibria, stability

Consider now the *autonomous* equation $x' = F(x)$, $F
\colon \Omega \to \R^d$ [locally Lipschitz](#def-b3-ode-lipschitz) on the open $\Omega
\subseteq \R^d$. Write $\varphi_t(x_0) = x(t)$ for the [maximal solution](#thm-b3-ode-maximal) with $x(0) = x_0$ (the *flow*); autonomy gives the group property $\varphi_{t+s} =
\varphi_t\circ\varphi_s$ where defined (both sides solve the same problem at time $s$).

**Definition 19.10.**

An *equilibrium* is a point $\bar x$ with $F(\bar x) =
0$ (so $\varphi_t(\bar x) = \bar x$). It is *stable* if for every $\varepsilon > 0$ there is $\delta > 0$ such that $\norm{x_0 - \bar x} < \delta$ implies that the solution exists for all $t \geq 0$ with $\norm{\varphi_t(x_0) - \bar
x} < \varepsilon$; *asymptotically stable* if moreover $\varphi_t(x_0) \to \bar x$ for all $x_0$ near $\bar x$.

**Theorem 19.11 (Lyapunov functions).**

Let $\bar x$ be an equilibrium and $V \colon \mathcal V \to
\R$ a $\mathcal C^1$ function on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $\bar x$ with:

$$
V(\bar x) = 0,\qquad V(x) > 0 \text{ for } x \neq \bar x,
\qquad \dot V(x) = \nabla V(x)\cdot F(x) \leq 0 .
$$

Then $\bar x$ is [stable](#def-b3-ode-stability). If moreover $\dot V < 0$ off $\bar
x$, then $\bar x$ is asymptotically [stable](#def-b3-ode-stability).

**Proof.** Along a solution, $\frac{\dd}{\dd t}V(x(t)) = \dot V(x(t))
\leq 0$: $V$ decreases. Given $\varepsilon$ (small enough that $\bar B(\bar x, \varepsilon) \subseteq \mathcal V$), let $m = \min\{V(x) : \norm{x - \bar x} = \varepsilon\} > 0$ (compactness, positivity) and pick $\delta < \varepsilon$ with $V < m$ on $B(\bar x, \delta)$ ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)). A solution starting in $B(\bar x, \delta)$ has $V(x(t)) < m$ for all later times, so it can never reach the sphere $\norm{x -
\bar x} = \varepsilon$ (where $V \geq m$): it stays in the ball — and then exists for all $t \geq 0$: the solution remains in the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\bar B$, so [Theorem 19.5](#thm-b3-ode-maximal)(2) ([escape from compacts](#thm-b3-ode-maximal)) forces $T_+ = +\infty$. Stability.

Asymptotic case: let $x(t)$ start in $B(\bar x, \delta)$; $V(x(t))$ decreases to some $c \geq 0$. If $c > 0$: the trajectory stays in $K = \{x \in \bar B(\bar x,
\varepsilon): V(x) \geq c\}$, a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) set that excludes a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $\bar x$ ($V$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with $V(\bar x)
= 0 < c$). On $K$, the function $\dot V$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), strictly negative, hence $\mu = \max_K\dot V < 0$ (compactness); then $V(x(t)) \leq V(x(0)) + \mu t \to
-\infty$: absurd, $V \geq 0$. So $c = 0$, and $x(t) \to \bar x$ (points at distance $\geq
\rho$ from $\bar x$ within the ball have $V \geq m_\rho >
0$). ∎

**Theorem 19.12 (Stability by linearization).**

Let $F$ be $\mathcal C^1$, $F(\bar x) = 0$, $A = DF(\bar x)$. If every eigenvalue of $A$ has $\operatorname{Re}\lambda <
0$, then $\bar x$ is asymptotically [stable](#def-b3-ode-stability).

**Proof.** Translate $\bar x$ to $0$ and write $F(x) = Ax + g(x)$ with $g(x) = o(\norm x)$ ($\mathcal C^1$ differentiability). Pick $\alpha > 0$ with $\vertiii{\eu^{tA}} \leq C\eu^{-\alpha t}$ ($t \geq 0$; [Theorem 19.8](#thm-b3-ode-matrixexp)) and $r > 0$ with $\norm{g(x)} \leq \frac{\alpha}{2C}\norm x$ for $\norm x \leq
r$. Duhamel with $b(s) = g(x(s))$:

$$
x(t) = \eu^{tA}x_0 + \int_0^t\eu^{(t-s)A}g(x(s))\,\dd s,
$$

valid as long as $\norm{x(s)} \leq r$. Then $u(t) =
\eu^{\alpha t}\norm{x(t)}$ satisfies

$$
u(t) \leq C\norm{x_0} +
\int_0^{t}C\,\frac{\alpha}{2C}\,u(s)\,\dd s ,
$$

so Grönwall gives $u(t) \leq C\norm{x_0}\eu^{\alpha t/2}$, i.e. $\norm{x(t)} \leq C\norm{x_0}\eu^{-\alpha t/2}$. If $\norm{x_0} < r/C$, the a priori bound keeps $\norm{x(t)} <
r$ for all $t$ (a [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)/bootstrap argument: the set of times where $\norm x \leq r$ is open and closed in $\intco0{T_+}$ given the strict estimate), the solution is global, and it converges to $0$ exponentially: asymptotic stability. ∎

**Method 19.13.**

Facing an ODE: (1) existence/uniqueness — check local Lipschitz (usually $\mathcal C^1$); (2) globality — linear growth, boundedness, or an invariant [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) via a [Lyapunov function](#thm-b3-ode-lyapunov) or first integral; failing that, suspect blow-up and test on the scalar caricature $x' = x^2$; (3) linear systems — [resolvent](#thm-b3-ode-linearstructure), Duhamel, and for constant coefficients the eigenstructure of $A$; (4) qualitative questions — equilibria, linearize, and hunt for a [Lyapunov function](#thm-b3-ode-lyapunov) (energy, when the system is mechanical) or a first integral whose level sets trap trajectories. The weekend problem walks the whole method through the pendulum.

![Phase portrait of the pendulum x'' = - x in the (x, x')-plane: level curves of the energy E = x'22 - x. Closed curves (blue): oscillations, E < 1; running curves (orange): full rotations, E > 1; between them the separatrix (red), E = 1, connecting the unstable equilibria (±π, 0). The weekend problem proves everything this picture suggests.](https://one-course.com/images/onecourse/chapters/math-5/b3-ode/fig-dd339003c539.svg)

*Phase portrait of the pendulum $x'' = -\sin x$ in the $(x, x')$-plane: level curves of the energy $E = \frac{x'^2}2
- \cos x$. Closed curves (blue): oscillations, $E < 1$; running curves (orange): full rotations, $E > 1$; between them the *separatrix* (red), $E = 1$, connecting the unstable equilibria $(\pm\pi, 0)$. The weekend problem proves everything this picture suggests.*

## 19.5 Exercises

**Exercise 19.1 ★.**

Solve explicitly and determine the maximal interval: (a) $x' = x^2$, $x(0) = 1$; (b) $x' = 1 + x^2$, $x(0) = 0$; (c) $x' = x(1-x)$, $x(0) = \frac12$. Reconcile each answer with [Theorem 19.5](#thm-b3-ode-maximal)(2) and [Corollary 19.6](#cor-b3-ode-globallinear).

**Solution of Exercise 19.1.**

(a) Separating variables: $x(t) = \frac1{1 - t}$ on $\intoo{-\infty}1$: blow-up at $T_+ = 1$, with $x(t) \to
+\infty$ — exactly [Theorem 19.5](#thm-b3-ode-maximal)(2). (b) $x(t) = \tan t$ on $\intoo{-\pi/2}{\pi/2}$: blow-up at both ends. (c) $x(t) = \frac{1}{1 + \eu^{-t}}$, global: the solution stays in $\intoo01$, a bounded set, so the graph cannot escape every [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) of $\R\times\R$ in finite time — $T_\pm
= \pm\infty$. Note (a), (b) do not contradict [Corollary 19.6](#cor-b3-ode-globallinear): $x^2$ and $1 + x^2$ have superlinear growth.

**Exercise 19.2 ★.**

Let $x, y$ solve $x' = f(t,x)$ with $f$ globally $L$-Lipschitz in $x$ on $\R\times\R^d$. (a) Prove $\norm{x(t) - y(t)} \leq \norm{x(0) -
y(0)}\eu^{L\abs t}$, and show by example (linear!) that the factor $\eu^{L\abs t}$ is attained. (b) Deduce that the [flow](#ex-b3-ode-planeclassification) map $x_0 \mapsto x(t; x_0)$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), in fact Lipschitz on bounded sets.

**Solution of Exercise 19.2.**

(a) This is [Corollary 19.4](#cor-b3-ode-uniqueness)’s estimate with $t_0
= 0$. Sharpness: for $x' = Lx$ (globally $L$-Lipschitz), two solutions differ by exactly $(x_0 - y_0)\eu^{Lt}$. (b) The estimate reads: the time-$t$ [flow](#ex-b3-ode-planeclassification) map is $\eu^{L\abs t}$-Lipschitz in the initial condition — [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), uniformly for $t$ in [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact); on bounded sets of non-globally-Lipschitz $f$, run the same on a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) tube around the trajectories with the local constant.

**Exercise 19.3 ★★.**

Show that each of the following has all [maximal solutions](#thm-b3-ode-maximal) global on $\R$, quoting the right theorem: (a) $x' = \sin(tx)$; (b) $x' = \frac{t\,x}{1 + x^2}$; (c) $x'' + q(t)x = 0$ with $q$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (convert to a first-order system); (d) $x' = A(t)x$ with $A$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and bounded — and give the Grönwall bound on $\norm{x(t)}$.

**Solution of Exercise 19.3.**

(a) $\abs{\sin(tx)} \leq 1$: bounded, i.e. linear growth with $\alpha = 0$, $\beta = 1$: [Corollary 19.6](#cor-b3-ode-globallinear) on $U = \R\times\R$. (b) $\abs{tx/(1 + x^2)} \leq \abs t\cdot\frac12$: again sublinear (in fact bounded on [time-compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)): global. (c) $X = (x, x')$: $X' = \bigl(\begin{smallmatrix}0 & 1\\ -q(t)
& 0\end{smallmatrix}\bigr)X$: linear with [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) coefficients: global ([Theorem 19.7](#thm-b3-ode-linearstructure)’s setting). (d) Global; Grönwall as in [Corollary 19.6](#cor-b3-ode-globallinear): $\norm{x(t)} \leq \norm{x(t_0)}\,\eu^{M\abs{t - t_0}}$ with $M = \sup\vertiii{A}$.

**Exercise 19.4 ★★.**

(a) Compute $\eu^{tA}$ for $A = \bigl(\begin{smallmatrix}0 & -1\\ 1 &
0\end{smallmatrix}\bigr)$, $\bigl(\begin{smallmatrix}\lambda & 1\\ 0 &
\lambda\end{smallmatrix}\bigr)$, and $\bigl(\begin{smallmatrix}0 & 1\\ 1 &
0\end{smallmatrix}\bigr)$. (b) Solve the forced oscillator $x'' + x = \cos(\omega t)$ by Duhamel (system form), for $\omega \neq 1$ and $\omega =
1$: resonance appears as the secular term $t\sin t$.

**Solution of Exercise 19.4.**

(a) $A^2 = -I$ for the first: $\eu^{tA} = \cos t\,I + \sin
t\,A = \bigl(\begin{smallmatrix}\cos t & -\sin t\\ \sin t &
\cos t\end{smallmatrix}\bigr)$. Jordan block: $\lambda I$ and $N$ commute: $\eu^{tA} =
\eu^{\lambda t}\bigl(\begin{smallmatrix}1 & t\\ 0 &
1\end{smallmatrix}\bigr)$. Third: $A^2 = I$: $\eu^{tA} =
\cosh t\,I + \sinh t\,A$.

(b) System $X' = \bigl(\begin{smallmatrix}0&1\\-1&0
\end{smallmatrix}\bigr)X + \bigl(\begin{smallmatrix}0\\
\cos\omega t\end{smallmatrix}\bigr)$; Duhamel with the rotation [resolvent](#thm-b3-ode-linearstructure) gives the particular solutions: for $\omega \neq 1$, $x_p(t) = \frac{\cos(\omega t)}{1 -
\omega^2}$ (verify directly); for $\omega = 1$ the integral $\int_0^t\sin(t - s)\cos s\,\dd s = \frac t2\sin t$ produces the *secular* growth $x_p = \frac t2\sin t$: resonance — the forcing pumps energy at the natural frequency and the amplitude grows linearly.

**Exercise 19.5 ★★.**

For the scalar equation $x'' + p(t)x' + q(t)x = 0$: (a) Show that the [Wronskian](#thm-b3-ode-linearstructure) $w = x_1x_2' - x_1'x_2$ of two solutions satisfies $w' = -p\,w$ (Abel), and deduce that two solutions with $w \neq 0$ somewhere form a basis. (b) Given one nonvanishing solution $x_1$, find the general solution by *reduction of order*: set $x_2 = x_1\int
\frac{\eu^{-\int p}}{x_1^2}$ and verify. Apply to $t^2x'' -
2x = 0$ on $\intoo0{+\infty}$ with $x_1(t) = t^2$.

**Solution of Exercise 19.5.**

(a) $w' = x_1x_2'' - x_1''x_2 = x_1(-px_2' - qx_2) - (-px_1'
- qx_1)x_2 = -p\,w$: $w(t) = w(t_0)\eu^{-\int_{t_0}^tp}$, never zero or identically zero. If $w \neq 0$, the vectors $(x_i, x_i')(t_0)$ are independent in $\R^2$, and since the solution space has dimension $2$ ([Theorem 19.7](#thm-b3-ode-linearstructure) for the system), $(x_1,
x_2)$ is a basis.

(b) With $u = \int\frac{\eu^{-\int p}}{x_1^2}$: $x_2 = x_1u$, $x_2' = x_1'u + \frac{\eu^{-\int p}}{x_1}$, and

$$
x_2'' + px_2' + qx_2
= u\,(x_1'' + px_1' + qx_1) +
\Bigl(-\,p\frac{\eu^{-\int p}}{x_1} +
p\frac{\eu^{-\int p}}{x_1}\Bigr) = 0
$$

(the cross terms cancel exactly; expand carefully). For $t^2x'' - 2x = 0$, i.e. $x'' - \frac2{t^2}x = 0$ ($p = 0$) with $x_1 = t^2$: $u = \int t^{-4} = -\frac1{3t^3}$, so $x_2
= -\frac1{3t}$: general solution $at^2 + \frac bt$.

**Exercise 19.6 ★★.**

(Logistic) For $x' = x(1 - x)$: determine all equilibria and their stability (by [Theorem 19.12](#thm-b3-ode-linearization) and directly); show every solution with $x(0) \in \intoo01$ is increasing, global, with limits $0$ and $1$ at $\mp\infty$; and solve explicitly to confirm. Show more generally that scalar autonomous solutions are monotone, and conclude: no nonconstant periodic solutions in dimension $1$.

**Solution of Exercise 19.6.**

Equilibria $0, 1$; $f'(x) = 1 - 2x$: $f'(0) = 1 > 0$ (unstable — nearby solutions move away, as the explicit form shows), $f'(1) = -1 < 0$: asymptotically [stable](#def-b3-ode-stability) ([Theorem 19.12](#thm-b3-ode-linearization) in dimension $1$). For $x(0) \in \intoo01$: $f > 0$ there, so as long as the solution stays in $\intoo01$ it increases; it can never reach $0$ or $1$ (uniqueness: those are trajectories), so it stays, is bounded — hence global — and increases to a limit $L
\in \intoc{x(0)}1$. If $f(L) \neq 0$, then $x' \geq c > 0$ near the limit, forcing $x$ past $L$: so $f(L) = 0$, $L = 1$; symmetrically $x \to 0$ at $-\infty$. Explicitly $x(t) =
\frac1{1 + C\eu^{-t}}$ confirms everything. Generally: if a scalar autonomous solution had $x'(t_0) = 0$, then $x(t_0)$ is an equilibrium and uniqueness makes $x$ constant; otherwise $f(x(t))$ keeps a fixed sign (it never vanishes, and $t \mapsto f(x(t))$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)): $x$ is strictly monotone — so a nonconstant periodic solution is impossible.

**Exercise 19.7 ★★.**

(First integrals) Let $H \colon \Omega \to \R$ be $\mathcal
C^1$ and consider the planar Hamiltonian system $x' =
\partial_yH$, $y' = -\partial_xH$. (a) Show that $H$ is constant along solutions. (b) For $H = \frac{y^2}2 + \frac{x^4}4$: show all solutions are global and bounded, and that the origin is [stable](#def-b3-ode-stability) (Lyapunov: $H$) though the linearization ($\bigl(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\bigr)$) is *not* asymptotically [stable](#def-b3-ode-stability): linearization can be inconclusive.

**Solution of Exercise 19.7.**

(a) $\frac{\dd}{\dd t}H(x,y) = H_xx' + H_yy' = H_xH_y +
H_y(-H_x) = 0$. (b) The level sets of $H = \frac{y^2}2 + \frac{x^4}4$ are [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) ($H$ coercive), so solutions are trapped in [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): global and bounded ([Theorem 19.5](#thm-b3-ode-maximal)). Stability of $(0,0)$: $V = H$ is positive definite ($H = 0$ only at the origin) with $\dot V = 0$: [Theorem 19.11](#thm-b3-ode-lyapunov). The linearization $x' = y$, $y' = 0$ has the non-diagonalizable nilpotent matrix with eigenvalue $0$: [Theorem 19.12](#thm-b3-ode-linearization) is silent (its hypothesis $\operatorname{Re}\lambda < 0$ fails), and indeed the linearized system is unstable ($y_0 \ne 0$ drifts) while the nonlinear one is [stable](#def-b3-ode-stability): linearization at a non-hyperbolic equilibrium proves nothing.

**Exercise 19.8 ★★★.**

(Damped pendulum) $x'' + cx' + \sin x = 0$, $c > 0$; system: $x' = y$, $y' = -\sin x - cy$. (a) Show $V(x, y) = \frac{y^2}2 + 1 - \cos x$ satisfies $\dot
V = -cy^2 \leq 0$: the origin is [stable](#def-b3-ode-stability). (b) $\dot V$ vanishes on the whole axis $y = 0$: Lyapunov’s strict criterion fails. Prove asymptotic stability anyway, by linearization ([Theorem 19.12](#thm-b3-ode-linearization)): compute the eigenvalues of the linearized matrix at $(0,0)$ and check $\operatorname{Re} < 0$ for every $c > 0$. (c) What happens at the equilibrium $(\pi, 0)$? Compute the linearization and conclude (one eigenvalue positive: instability — you may use the instability statement informally or produce an explicit escaping solution of the linear system).

**Solution of Exercise 19.8.**

(a) $\dot V = y\,y' + \sin x\cdot x' = y(-\sin x - cy) +
y\sin x = -cy^2 \leq 0$, and $V = \frac{y^2}2 + (1 - \cos x)$ is positive definite on $\{\abs x < 2\pi\}$ around the origin: [stable](#def-b3-ode-stability) ([Theorem 19.11](#thm-b3-ode-lyapunov)). (b) The linearized matrix at $(0,0)$ is $\bigl(\begin{smallmatrix}0 & 1\\ -1 &
-c\end{smallmatrix}\bigr)$, with characteristic polynomial $\lambda^2 + c\lambda + 1$: roots $\frac{-c \pm \sqrt{c^2 -
4}}2$ — both real negative if $c \geq 2$, complex with real part $-\frac c2 < 0$ if $0 < c < 2$. In all cases $\operatorname{Re}\lambda < 0$: [Theorem 19.12](#thm-b3-ode-linearization) gives asymptotic stability (despite the degenerate $\dot V$). (c) At $(\pi, 0)$: $\sin(\pi + u) = -\sin u$, linearization $\bigl(\begin{smallmatrix}0&1\\1&-c\end{smallmatrix}\bigr)$, characteristic $\lambda^2 + c\lambda - 1$: roots of opposite signs ($\lambda_+\lambda_- = -1$). Along the unstable eigenvector, the linear system has the explicitly escaping solution $\eu^{\lambda_+t}v_+$ with $\lambda_+ > 0$: the inverted pendulum is unstable for every damping.

**Exercise 19.9 ★★.**

(Uniqueness frontier) For $\alpha \in \intoo01$, show that the problem $x' = \abs x^{\alpha}$, $x(0) = 0$ has infinitely many solutions (adapt [Problem 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#pb-b3-complete-1), Part III). Show on the contrary that for $\alpha = 1$ (i.e. $x' = \abs
x$) the solution through $0$ is unique, and identify precisely which hypothesis of [Theorem 19.2](#thm-b3-ode-cauchylipschitz) distinguishes the two cases.

**Solution of Exercise 19.9.**

For $\alpha \in \intoo01$: besides $x \equiv 0$, each

$$
x_c(t) = \begin{cases}0 & t \leq c,\\
\bigl((1-\alpha)(t - c)\bigr)^{1/(1-\alpha)} & t \geq c,
\end{cases}
$$

is $\mathcal C^1$ and solves the equation (the exponent $\frac1{1-\alpha} > 1$ makes the derivative vanish at $c$): a continuum of solutions through $(0,0)$. For $\alpha = 1$: $x
\mapsto \abs x$ is globally $1$-Lipschitz ($\abs{\abs a - \abs b} \leq \abs{a - b}$), so [Theorem 19.2](#thm-b3-ode-cauchylipschitz) applies and the only solution through $0$ is $x \equiv 0$. The frontier is exactly the local Lipschitz condition at $0$: $\abs x^\alpha$ has unbounded difference quotients there for $\alpha < 1$.

**Exercise 19.10 ★★★.**

(A priori bounds trap solutions) Let $F \colon \R^d \to \R^d$ be [locally Lipschitz](#def-b3-ode-lipschitz) with $\langle F(x), x\rangle \leq 0$ whenever $\norm x \geq R$. (a) Show that the closed ball $\bar B(0, R)$ is positively invariant: solutions starting inside stay inside for $t \geq
0$. *(If $\norm{x(t_2)} > R$, consider the last time $t_1 < t_2$ with $\norm{x(t_1)} = R$ and study $\frac{\dd}{\dd t}\norm{x(t)}^2$ on $\intcc{t_1}{t_2}$.)* (b) Deduce global forward existence for data in the ball. Then treat the gradient system $x' = -\nabla G(x)$, $G \in
\mathcal C^2$ with $G(x) \to +\infty$ as $\norm x \to
\infty$: show $G$ decreases along solutions, that each solution stays in the (bounded) sublevel set $\{G \leq
G(x_0)\}$, and conclude global forward existence.

**Solution of Exercise 19.10.**

(a) Suppose $\norm{x(t_2)} > R$ for some $t_2 > 0$ with $\norm{x(0)} \leq R$, and let $t_1 = \sup\{t \leq t_2 :
\norm{x(t)} \leq R\}$: then $\norm{x(t_1)} = R$ and $\norm{x(t)} > R$ on $\intoc{t_1}{t_2}$. On that interval $g(t) = \norm{x(t)}^2$ has $g'(t) = 2\langle x(t),
F(x(t))\rangle \leq 0$ (the hypothesis applies: $\norm{x(t)}
\geq R$), so $g(t_2) \leq g(t_1) = R^2$: contradiction. The ball is positively invariant. (b) A solution trapped in the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) ball cannot have $T_+ < \infty$ ([Theorem 19.5](#thm-b3-ode-maximal)(2)): global forward. Gradient system: $\frac{\dd}{\dd t}G(x(t)) =
\langle\nabla G, -\nabla G\rangle = -\norm{\nabla G(x(t))}^2
\leq 0$: $G$ decreases, so the solution stays in $\{G \leq
G(x_0)\}$, which is bounded (coercivity: outside a large ball, $G > G(x_0)$) and closed: [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact). Escape is impossible: every solution of a coercive gradient system is global forward, sliding downhill forever.

**Exercise 19.11 ★★.**

(Blow-up by comparison) Consider $x' = x^2 + t^2$, $x(0) =
1$. (a) Show the [maximal solution](#thm-b3-ode-maximal) exists on some $\intco0{T_+}$ with $T_+ < \infty$: compare with $y' = y^2$, $y(0) = 1$ *(prove the comparison lemma you need: if $x' \geq
F(x)$ and $y' = F(y)$ with $x(0) \geq y(0)$, then $x \geq
y$ where both live)*, and deduce $T_+ \leq 1$. (b) Bound $T_+$ from below: on $\intcc01$, $x' \leq x^2 +
1$; compare with the supersolution $z' = z^2 + 1$, $z(0) =
1$, solved by $z(t) = \tan\bigl(t + \frac\pi4\bigr)$, and conclude $T_+ \geq \frac\pi4$. (c) Assemble $\frac\pi4 \leq T_+ \leq 1$ and frame the moral: superlinear growth of the right-hand side is what kills global existence ([Exercise 19.3](#exo-b3-ode-3) being the counterpoint), the frontier being the convergence of $\int^{\infty}\frac{\dd s}{F(s)}$.

**Solution of Exercise 19.11.**

(a) Comparison lemma: let $w = x - y$ on the common interval; $w(0) \geq 0$ and $w' = x' - y' \geq F(x) - F(y) =
c(t)w$ with $c(t) = \frac{F(x) - F(y)}{x - y}$ bounded on [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) time intervals ($F$ [locally Lipschitz](#def-b3-ode-lipschitz)); then $(w\eu^{-\int c})' \geq 0$, so $w \geq 0$ throughout. With $F(x) = x^2$: $y(t) = \frac1{1 - t}$ blows up at $1$, and $x
\geq y$ as long as $x$ lives; if $T_+ > 1$, then $x$ would be finite at $t = 1$ while dominating $y \to \infty$: absurd. $T_+ \leq 1$.

(b) The reversed comparison (same lemma, roles swapped): on $\intcc01\cap\intco0{T_+}$, $t^2 \leq 1$ gives $x' \leq
x^2 + 1$, while $z(t) = \tan(t + \frac\pi4)$ satisfies $z' =
z^2 + 1$, $z(0) = 1 = x(0)$: hence $x \leq z$ as long as both are defined. Since $z$ is finite on $\intco0{\frac\pi4}$, $x$ cannot blow up before $\frac\pi4$: $T_+ \geq \frac\pi4$.

(c) Together: $\frac\pi4 \leq T_+ \leq 1$ (numerically $T_+
\approx 0.96$). Moral: for $x' = F(t, x)$ with $F$ superlinear in $x$, solutions explode in finite time whenever $\int^\infty\frac{\dd s}{F(s)} < \infty$ (the comparison solution reaches infinity in that finite time); linear growth, where the integral diverges, forces global existence ([Exercise 19.3](#exo-b3-ode-3)). It is the same Osgood integral as in [Problem 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#pb-b3-complete-1), now governing the escape to infinity rather than the escape from zero.

**Exercise 19.12 ★★★.**

(Sturm’s comparison theorem) Let $q_1 \leq q_2$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on an interval $I$, and let $u \neq 0$ solve $u''
+ q_1u = 0$, $v \neq 0$ solve $v'' + q_2v = 0$. (a) Establish the [Wronskian](#thm-b3-ode-linearstructure) identity: with $W = uv' - u'v$, $W' = (q_1 - q_2)\,uv$. (b) (Sturm) Show that between two consecutive zeros $a < b$ of $u$, either $v$ vanishes somewhere in $\intoo ab$, or $q_1 = q_2$ and $v \propto u$ there *(assume $u > 0$ on $\intoo ab$ and $v > 0$ too; integrate (a) from $a$ to $b$ and inspect the signs of the boundary terms $W(a),
W(b)$)*. (c) Deduce: solutions of $u'' + qu = 0$ with $q \geq m > 0$ vanish at least once in every interval of length $\pi/\sqrt m$ (compare with $v'' + mv = 0$); solutions with $q \leq 0$ vanish at most once on $\R$. Test both on $q \equiv \pm1$.

**Solution of Exercise 19.12.**

(a) $W' = uv'' - u''v = u(-q_2v) - (-q_1u)v = (q_1 -
q_2)\,uv$.

(b) Let $a < b$ be consecutive zeros of $u$; normalize $u >
0$ on $\intoo ab$ (so $u'(a) > 0$, $u'(b) < 0$ — nonzero by uniqueness, since $u(a) = u'(a) = 0$ would force $u
\equiv 0$). Suppose $v$ has no zero in $\intoo ab$; normalize $v > 0$ there (hence $v(a), v(b) \geq 0$ by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)). Integrate (a):

$$
W(b) - W(a) = \int_a^b(q_1 - q_2)\,uv \;\leq\; 0 .
$$

But $W(a) = u(a)v'(a) - u'(a)v(a) = -u'(a)v(a) \leq 0$ and $W(b) = -u'(b)v(b) \geq 0$: so $W(b) - W(a) \geq 0$. Equality throughout: $\int(q_1 - q_2)uv = 0$ with $uv > 0$ on the open interval forces $q_1 = q_2$ there; and $W(a) =
W(b) = 0$ forces $v(a) = v(b) = 0$; then $W \equiv 0$ on $\intcc ab$ (its derivative vanishes), i.e. $(v/u)' = -W/u^2 = 0$ on $\intoo ab$: $v \propto u$.

(c) Take $q_1 = m$ and $u = \sin(\sqrt m(t - t_0))$, whose consecutive zeros are $\pi/\sqrt m$ apart, and $q_2 = q \geq
m$: by (b), every solution $v$ of $v'' + qv = 0$ vanishes in each open interval of length $\pi/\sqrt m$ (in the degenerate alternative $q \equiv m$, $v \propto u$ vanishes too). If instead $q \leq 0$: apply (b) with $q_1 = q$, $u =
v$, and $q_2 = 0$ with the zero-free solution $\mathbf 1$ of $v'' = 0$. If $v$ had two consecutive zeros, (b) would force either a zero of $\mathbf 1$ between them or the degenerate case $\mathbf 1 \propto v$ — both absurd: $v$ vanishes at most once. Tests: for $q = 1$, $\sin t$ vanishes every $\pi
= \pi/\sqrt1$, as predicted; for $q = -1$, $\sinh t$ vanishes exactly once and $\eu^t$ never — at most one zero, as predicted.

## 19.6 Problem: the pendulum, completely solved

**Problem 19.1.**

Weekend problem — oscillations, rotations, separatrix, and the period

The pendulum equation $x'' = -\sin x$ — as a system: $x' =
y$, $y' = -\sin x$ on $\R^2$ — is the drosophila of dynamics: simple to write, impossible to solve by elementary formulas, yet completely understandable by the qualitative method. Let $E(x, y) = \frac{y^2}2 - \cos x$ (the *energy*).

**Part I — Global structure.**

1. Show that all [maximal solutions](#thm-b3-ode-maximal) are global ( $\R$ -defined): use $\dot E = 0$ and [Theorem 19.5](#thm-b3-ode-maximal) . Equilibria: $(k\pi, 0)$ ; classify their linearizations (center-type for even $k$ , saddle for odd $k$ ).
2. Show that the trajectories are contained in the level sets $\{E = E_0\}$ , and sketch/describe them by the value of $E_0 \in \intco{-1}{+\infty}$ : $E_0 = -1$ (equilibria), $-1 < E_0 < 1$ (closed curves around $(2k\pi, 0)$ ), $E_0 = 1$ (the separatrix through $(\pm\pi, 0)$ ), $E_0 > 1$ (graphs over $x$ : rotations).
3. Prove that the bottom equilibrium $(0,0)$ is [stable](#def-b3-ode-stability) but *not* asymptotically [stable](#def-b3-ode-stability) . *(Lyapunov with $V = E + 1$; non-asymptotic: energy conservation traps [orbits](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) on level curves away from the origin.)*

**Part II — Oscillations and their period.** Fix $-1 < E_0 < 1$ and write $E_0 = -\cos a$ with $a \in
\intoo0\pi$ (the amplitude).

4. Show that the solution with $x(0) = a$ , $y(0) = 0$ oscillates: $x(t) \in \intcc{-a}a$ , and the [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) is the closed curve $y^2 = 2(\cos x - \cos a)$ . Justify that the solution is periodic: the [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) is a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) curve without equilibria, traversed at speed bounded below — make this an argument (the solution returns to its initial point in finite time, then uniqueness forces periodicity).
5. Establish the period formula $$T(a) = 4\int_0^{a}\frac{\dd x}{\sqrt{2(\cos x -  \cos a)}}$$ *(on a [quarter-orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action), $y = \frac{\dd x}{\dd t} >  0$ and separate variables; justify the improper convergence at $x = a$)*.
6. (Small oscillations) Substitute $\sin\frac x2 =  \sin\frac a2\,\sin\varphi$ and show $$T(a) = 4\int_0^{\pi/2}\frac{\dd\varphi}{\sqrt{1 -  k^2\sin^2\varphi}},  \qquad k = \sin\frac a2$$ (a [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) elliptic integral). Deduce by dominated convergence that $T(a) \to 2\pi$ as $a \to 0^+$: the harmonic limit, independent of amplitude — Galileo’s approximate isochronism, with its exact correction $T(a) = 2\pi\bigl(1 + \frac{k^2}4 +  O(k^4)\bigr)$ (expand the integrand and integrate term by term, justifying by normal convergence).
7. Show that $T(a) \to +\infty$ as $a \to \pi^-$ *(bound the integrand below near $\varphi =  \frac\pi2$ when $k \to 1$, or apply monotone convergence)* : approaching the separatrix, the pendulum slows without bound.

**Part III — The separatrix.**

8. For $E_0 = 1$, $y = 2\cos\frac x2$ on the upper branch: separate variables and integrate to find the explicit solution $$x(t) = 4\arctan\bigl(\eu^{t}\bigr) - \pi$$ (with $x(0) = 0$, $y(0) = 2$). Verify directly that it solves the pendulum equation, and compute its limits and the limits of $y(t)$ as $t \to  \pm\infty$.
9. Conclude: the separatrix [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) connects the saddle $(-\pi, 0)$ (as $t \to -\infty$ ) to the saddle $(\pi, 0)$ (as $t \to +\infty$ ) but reaches neither in finite time — consistent with uniqueness (why would reaching a saddle in finite time contradict [Corollary 19.4](#cor-b3-ode-uniqueness) ?).

**Part IV — Rotations, and the full picture.**

10. For $E_0 > 1$: show $y$ never vanishes, $x$ is strictly monotone and global with $x(t) \to  \pm\infty$, and $t \mapsto y(t)$ is periodic with period $$\tau(E_0) = \int_{-\pi}^{\pi}  \frac{\dd x}{\sqrt{2(E_0 + \cos x)}} .$$
11. Assemble the [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) phase portrait (the chapter’s figure) with full justification of each feature, and write a ten-line summary of the method: energy, level sets, compactness, uniqueness — how each theorem of the chapter entered. Where did we ever need a formula for the general solution?

**Part V — The period function under the microscope.**

12. Prove the Wallis moments $$W_n = \int_0^{\pi/2}\sin^{2n}\varphi\,\dd\varphi  = \frac\pi2\cdot\frac{(2n)!}{4^n\,(n!)^2}$$ by induction (integrate by parts), expand the integrand of question 6 by the binomial series, and justify term-by-term integration to obtain the full series $$T(a) = 2\pi\sum_{n\geq0}  \Bigl(\frac{(2n)!}{4^n(n!)^2}\Bigr)^{\!2}k^{2n}  = 2\pi\Bigl(1 + \frac{k^2}4 + \frac{9k^4}{64}  + O(k^6)\Bigr),  \qquad k = \sin\frac a2 .$$
13. Convert to the amplitude: $$T(a) = 2\pi\Bigl(1 + \frac{a^2}{16} +  \frac{11\,a^4}{3072} + O(a^6)\Bigr)$$ *(substitute the expansion of $\sin\frac a2$ and collect)*. Isochronism fails at order $a^2$, and the failure is now quantified to order $a^4$.
14. Show that $a \mapsto T(a)$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and strictly increasing on $\intoo0\pi$ , and conclude with questions 6–7 that $T$ is a bijection from $\intoo0\pi$ onto $\intoo{2\pi}{+\infty}$ : every supercritical period is realized by exactly one amplitude.
15. (Clockmaker’s arithmetic) A pendulum regulated at vanishing amplitude keeps [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal) time; show that run at amplitude $a$ it lags by the fraction $\frac{a^2}{16} + O(a^4)$ of [ideal](https://one-course.com/books/math/5/en/chapter/2-rings-and-arithmetic#def-b3-rings-ideal) time, and compute the drift for $a = 0.2$ rad: about $216$ seconds per day. (Huygens’ cycloidal cheeks and the small constant amplitudes of escapements are both answers to this number.)
16. Return to the rotation period $\tau$ of question 10: show that $\tau$ is strictly decreasing on $\intoo1{+\infty}$ , that $\tau(E_0) \to +\infty$ as $E_0 \to 1^+$ (monotone convergence), and that $\sqrt{2E_0}\,\tau(E_0) \to 2\pi$ as $E_0 \to  +\infty$ (dominated convergence): fast whirling is asymptotically free rotation at angular speed $\sqrt{2E_0}$ .

**Part VI — The method exported: Lotka–Volterra.** The pendulum’s recipe — first integral, [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) level curves, uniqueness — solves an ecosystem. Fix $\alpha, \beta, \gamma, \delta > 0$ and consider, on the open quadrant $Q = \intoo0{+\infty}\times\intoo0{+\infty}$,

$$
x' = x\,(\alpha - \beta y), \qquad
y' = y\,(\delta x - \gamma)
$$

($x$ prey, $y$ predators).

17. Show that $Q$ is invariant — the axes are unions of [orbits](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) , explicitly computable, that no solution may cross ( [Corollary 19.4](#cor-b3-ode-uniqueness) ) — and that the unique equilibrium in $Q$ is $(x_*, y_*) =  \bigl(\frac\gamma\delta, \frac\alpha\beta\bigr)$ .
18. Show that $$H(x, y) = \delta x - \gamma\ln x + \beta y -  \alpha\ln y$$ is a first integral, that $H = f(x) + g(y)$ with $f, g$ strictly convex and proper on $\intoo0{+\infty}$ with minima at $x_*$, $y_*$, and deduce that all [maximal solutions](#thm-b3-ode-maximal) in $Q$ are global.
19. Show that for $h > h_* = H(x_*, y_*)$ the level set $\{H = h\}\cap Q$ is a closed curve around the equilibrium: two [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) branches $y_\pm(x)$ over a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) interval $\intcc{x_-}{x_+} \ni x_*$ , glued at the endpoints — the analogue of the pendulum’s ovals.
20. Prove that every nonequilibrium [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) in $Q$ is periodic: establish the counterclockwise circulation through the four regions cut by the lines $x = x_*$ and $y = y_*$ , bound the crossing time of each arc by an integral with a convergent square-root singularity (as in question 5), and close with uniqueness (as in question 4).
21. (Volterra’s law of averages) If $T$ is the period of such an [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action), show that $$\frac1T\int_0^Tx(t)\,\dd t = \frac\gamma\delta,  \qquad  \frac1T\int_0^Ty(t)\,\dd t = \frac\alpha\beta :$$ the time averages equal the equilibrium values, whatever the amplitude *(integrate $(\ln x)' =  \alpha - \beta y$ over one period)*.
22. (The fishing paradox) Harvest both species at rate $\varepsilon \in \intoo0\alpha$ : the system keeps its form with $\alpha - \varepsilon$ and $\gamma +  \varepsilon$ in place of $\alpha$ and $\gamma$ . What happens to the average populations? Explain d’Ancona’s observation (1914–1918): when Adriatic fishing decreased during the war, the proportion of predators (sharks) in the catch *increased* — and why moderate fishing favors the prey.
23. Write the ten-line moral: which theorems of the chapter power each step, what replaces the pendulum’s energy, and why neither system needed — or admits — an elementary closed-form solution.
24. (Speed modulation) In the rotation regime $E_0 > 1$ , show that $y = x'$ oscillates between $\sqrt{2(E_0 - 1)}$ (at $x \equiv \pi \bmod 2\pi$ ) and $\sqrt{2(E_0 + 1)}$ (at $x \equiv 0$ ), that the time average of $y$ over one period is exactly $\frac{2\pi}{\tau(E_0)}$ , and that the modulation ratio $\sqrt{\frac{E_0 + 1}{E_0 - 1}} \to 1$ as $E_0  \to \infty$ : fast rotation is asymptotically uniform.
25. (Monotonicity of the rotation period) Show that $\tau(E_0)$ is $\mathcal C^1$ and strictly decreasing on $\intoo1{+\infty}$ *(differentiate under the integral sign, with domination on every $\intco{1 + \delta}\infty$)* , with $\tau \to \infty$ as $E_0 \to 1^+$ and $\tau  \to 0$ as $E_0 \to \infty$ . Assemble the full bifurcation picture of the pendulum along the energy axis: equilibria at $E_0 = -1$ , librations with period increasing from $2\pi$ to $\infty$ on $-1 < E_0 < 1$ , the separatrix at $E_0 = 1$ , and rotations with period decreasing from $\infty$ to $0$ beyond.

**Solution of Problem 19.1.**

**1.** $\dot E = yy' + \sin x\cdot x' = -y\sin x +
y\sin x = 0$: energy is a first integral. On a [maximal solution](#thm-b3-ode-maximal), $y^2 = 2(E_0 + \cos x) \leq 2(E_0 + 1)$: $y$ is bounded; then $\abs{x(t)} \leq \abs{x(0)} +
t\sup\abs y$ grows at most linearly: on any finite time interval the trajectory stays in a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) of $\R^2$, so [Theorem 19.5](#thm-b3-ode-maximal)(2) forces $T_\pm = \pm\infty$. Equilibria $(k\pi, 0)$; linearization $\bigl(\begin{smallmatrix}0&1\\
\mp1&0\end{smallmatrix}\bigr)$ with $-\cos(k\pi) = \mp1$: eigenvalues $\pm\iu$ for even $k$ ([center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) type, inconclusive by itself) and $\pm1$ for odd $k$ (saddle).

**2.** $E$ constant along solutions confines each trajectory to a level set $\{y^2 = 2(E_0 + \cos x)\}$. For $E_0 = -1$: only the points $(2k\pi, 0)$. For $-1 < E_0 <
1$: writing $E_0 = -\cos a$, the set is a disjoint union of closed curves $y = \pm\sqrt{2(\cos x - \cos a)}$ over $x \in
[2k\pi - a, 2k\pi + a]$, one around each [stable](#def-b3-ode-stability) equilibrium. For $E_0 = 1$: the curves $y = \pm2\cos\frac x2$ joining consecutive saddles — the separatrix — together with the saddles themselves. For $E_0 > 1$: two graphs $y =
\pm\sqrt{2(E_0 + \cos x)}$, defined for all $x$, never touching $y = 0$.

**3.** $V = E + 1 = \frac{y^2}2 + (1 - \cos x)$ vanishes at $(0,0)$, is positive on a punctured [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) ($\abs x < 2\pi$), and $\dot V = 0 \leq 0$: [Theorem 19.11](#thm-b3-ode-lyapunov) gives stability. Not asymptotic: the solution through $(a, 0)$ ($0 < a$ small) remains on the level curve $E = -\cos a$, whose distance to the origin is positive (the curve meets the $x$-axis only at $\pm a$): $\varphi_t(a, 0) \not\to (0,0)$.

**4.** On the level curve $C_a$: no equilibria ($y = 0$ forces $x = \pm a$ with $\sin(\pm a) \neq 0$ for $0 < a <
\pi$), so the speed $\norm{(y, -\sin x)}$ has a positive minimum $m$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $C_a$. Follow the solution from $(a, 0)$: in the lower half plane $x' = y < 0$, so $x$ decreases from $a$ to $-a$ in finite time (the quarter/half period integrals converge: question 5’s analysis), reaching $(-a, 0)$; by the symmetry $(x, y) \mapsto (x, -y)$, $t
\mapsto -t$ of the equation, the upper half is traversed back in the same time $\frac T2$: the solution returns to $(a, 0)$ at time $T$. Uniqueness ([Corollary 19.4](#cor-b3-ode-uniqueness)) then propagates: $x(t + T) =
x(t)$ for all $t$: periodic.

**5.** On the branch where $y > 0$: $y = \sqrt{2(\cos x
- \cos a)}$ and $\dd t = \frac{\dd x}{y}$; integrating $x$ from $-a$ to $a$ gives the half period, and the symmetry $x
\mapsto -x$ halves the integral again:

$$
T(a) = 4\int_0^a\frac{\dd x}{\sqrt{2(\cos x - \cos a)}} .
$$

Convergence at $x = a^-$: $\cos x - \cos a = \sin(a)(a - x) +
O((a-x)^2)$ with $\sin a > 0$: the integrand behaves like $\bigl(2\sin a\,(a - x)\bigr)^{-1/2}$, [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1).

**6.** With $\sin\frac x2 = k\sin\varphi$, $k =
\sin\frac a2$: $\cos x - \cos a = 2(k^2 - \sin^2\frac x2)
= 2k^2\cos^2\varphi$ and $\frac12\cos\frac x2\,\dd x =
k\cos\varphi\,\dd\varphi$, so

$$
T(a) = 4\int_0^{\pi/2}
\frac{\dd\varphi}{\sqrt{1 - k^2\sin^2\varphi}} .
$$

As $a \to 0^+$, $k \to 0$: for $k \leq k_0 < 1$ the integrand is dominated by $(1 - k_0^2\sin^2\varphi)^{-1/2}$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $\intcc0{\pi/2}$: DCT gives $T \to
4\cdot\frac\pi2 = 2\pi$. Expanding $(1 - u)^{-1/2} = 1 +
\frac u2 + \frac{3u^2}8 + \cdots$ with $u =
k^2\sin^2\varphi$ (normal convergence for $k < 1$) and $\int_0^{\pi/2}\sin^2 = \frac\pi4$:

$$
T(a) = 2\pi\Bigl(1 + \frac{k^2}{4} + O(k^4)\Bigr) :
$$

isochronism holds only to first order; the period grows with amplitude.

**7.** As $k \uparrow 1$ the integrands increase to $(1
- \sin^2\varphi)^{-1/2} = \frac1{\cos\varphi}$, whose integral diverges: by monotone convergence, $T(a) \to
+\infty$ as $a \to \pi^-$.

**8.** On the branch $y = 2\cos\frac x2$ ($\abs x <
\pi$): $\frac{\dd x}{2\cos(x/2)} = \dd t$; with $u = \frac
x2$, $\int\frac{\dd u}{\cos u} =
\ln\tan\bigl(\frac u2 + \frac\pi4\bigr)$, so $t =
\ln\tan\bigl(\frac x4 + \frac\pi4\bigr)$, i.e.

$$
x(t) = 4\arctan(\eu^t) - \pi,
\qquad y(t) = x'(t) = \frac{4\eu^t}{1 + \eu^{2t}} =
\frac{2}{\cosh t} .
$$

Verification via the energy: with $\theta = \arctan\eu^t$, $\sin2\theta = \frac1{\cosh t}$, so $\cos x = -\cos4\theta =
-1 + \frac{2}{\cosh^2t}$ and $E = \frac{y^2}2 - \cos x =
\frac2{\cosh^2t} + 1 - \frac2{\cosh^2t} = 1$: the trajectory lies on the separatrix, and differentiating $y^2 = 2(1 +
\cos x)$ where $y > 0$ reproduces $y' = -\sin x$. Limits: $x
\to \pm\pi$ and $y \to 0$ as $t \to \pm\infty$.

**9.** The [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) tends to the saddle $(\pi, 0)$ forward and $(-\pi, 0)$ backward but never arrives: if it reached $(\pi, 0)$ at a finite time $t^*$, two distinct [maximal solutions](#thm-b3-ode-maximal) — the separatrix solution and the constant solution at the saddle — would pass through the same point $(t^*, (\pi, 0))$, contradicting [Corollary 19.4](#cor-b3-ode-uniqueness). Saddles are approached only asymptotically.

**10.** For $E_0 > 1$: $y^2 = 2(E_0 + \cos x) \geq
2(E_0 - 1) > 0$: $y$ keeps its sign, and $\abs{x'} = \abs y
\geq \sqrt{2(E_0 - 1)}$: $x$ is strictly monotone, global (question 1), with $x(t) \to \pm\infty$. Since $y(t) =
\pm\sqrt{2(E_0 + \cos x(t))}$ and $\cos$ is $2\pi$-periodic, $y$ returns to its value each time $x$ advances by $2\pi$; the time needed is

$$
\tau(E_0) = \int_{x_0}^{x_0 + 2\pi}\frac{\dd
x}{\sqrt{2(E_0 + \cos x)}} =
\int_{-\pi}^{\pi}\frac{\dd x}{\sqrt{2(E_0 + \cos x)}}
$$

(substitution; periodicity): $y$ is $\tau$-periodic — the pendulum whirls with asymptotically constant rotation rate $2\pi/\tau \approx \sqrt{2E_0}$ for large energies.

**11.** The method, in order: *energy* ($\dot E =
0$) reduces the two-dimensional [flow](#ex-b3-ode-planeclassification) to one-dimensional level curves; *boundedness* of $y$ on each level plus escape-from-compacts gives global existence; *compactness* of the closed levels gives speed bounds and hence periodicity; *uniqueness* converts the first-return into exact periodicity, forbids finite-time arrival at saddles, and separates the [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) types; *linearization and Lyapunov* classify the equilibria; the *period integral* is analyzed with the convergence theorems of [Chapter 10](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ch-b3-lebesgue). At no point did we possess — or need — a closed-form general solution: the qualitative theory extracted every feature of the motion from the equation itself.

**12.** Parts: $W_n = \int_0^{\pi/2}\sin^{2n-1}\varphi
\cdot\sin\varphi\,\dd\varphi = (2n-1)\int_0^{\pi/2}
\sin^{2n-2}\varphi\cos^2\varphi\,\dd\varphi = (2n-1)(W_{n-1}
- W_n)$, so $W_n = \frac{2n-1}{2n}W_{n-1}$; with $W_0 =
\frac\pi2$ and $\prod_{j=1}^n\frac{2j-1}{2j} =
\frac{(2n)!}{4^n(n!)^2}$ (split $(2n)! = 2^nn!\prod(2j-1)$), induction gives the displayed value. Binomial series: $(1 -
u)^{-1/2} = \sum_{n\geq0}c_nu^n$ with $c_n =
\frac{(2n)!}{4^n(n!)^2} \in \intoc01$, radius $1$. For $k
\leq k_0 < 1$ and $u = k^2\sin^2\varphi$, the series $\sum c_nk^{2n}\sin^{2n}\varphi$ converges normally in $\varphi$ ($c_nk^{2n} \leq k_0^{2n}$), so term-by-term integration in question 6’s formula is legitimate:

$$
T(a) = 4\sum_{n\geq0}c_nk^{2n}W_n
= 4\cdot\frac\pi2\sum_{n\geq0}c_n^2\,k^{2n}
= 2\pi\sum_{n\geq0}
\Bigl(\frac{(2n)!}{4^n(n!)^2}\Bigr)^{\!2}k^{2n} .
$$

With $c_0 = 1$, $c_1 = \frac12$, $c_2 = \frac38$: $T(a) =
2\pi\bigl(1 + \frac{k^2}4 + \frac{9k^4}{64} + O(k^6)\bigr)$, the remainder uniform for $k \leq k_0$ (tail dominated by a geometric series).

**13.** $k = \sin\frac a2 = \frac a2 - \frac{a^3}{48} +
O(a^5)$, so

$$
k^2 = \frac{a^2}4 - \frac{a^4}{48} + O(a^6),
\qquad
k^4 = \frac{a^4}{16} + O(a^6) ,
$$

and

$$
\frac{T(a)}{2\pi} = 1 + \frac14\Bigl(\frac{a^2}4 -
\frac{a^4}{48}\Bigr) + \frac9{64}\cdot\frac{a^4}{16} +
O(a^6)
= 1 + \frac{a^2}{16} + \frac{11\,a^4}{3072} + O(a^6) ,
$$

since $-\frac1{192} + \frac9{1024} = \frac{-16 + 27}{3072}
= \frac{11}{3072}$.

**14.** In the elliptic form of question 6, $a \mapsto
k = \sin\frac a2$ is a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) strictly increasing bijection from $\intoo0\pi$ onto $\intoo01$, and for each $\varphi \in \intoc0{\pi/2}$ the integrand $(1 -
k^2\sin^2\varphi)^{-1/2}$ is strictly increasing in $k$: $T$ is strictly increasing. [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): on $k \leq k_0 < 1$ the integrand is dominated by the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $(1 -
k_0^2\sin^2\varphi)^{-1/2}$, so dominated convergence applies along $k' \to k$. With the limits $T \to 2\pi$ as $a
\to 0^+$ (question 6) and $T \to +\infty$ as $a \to \pi^-$ (question 7), strict monotonicity and the intermediate value theorem make $T$ a bijection from $\intoo0\pi$ onto $\intoo{2\pi}{+\infty}$.

**15.** A clock counts swings; regulated at vanishing amplitude, it books the harmonic period $2\pi$ per swing (in the pendulum’s time unit). Run at amplitude $a$, the true period is $T(a) = 2\pi\bigl(1 + \frac{a^2}{16} +
O(a^4)\bigr)$: the clock books $2\pi$ while $T(a)$ really elapses, so it lags by the fraction

$$
\frac{T(a) - 2\pi}{T(a)} = \frac{a^2}{16} + O(a^4) .
$$

For $a = 0.2$ rad (about $11.5$ degrees): $\frac{a^2}{16} =
\frac{0.04}{16} = \frac1{400}$, and a day has $86400$ s: the clock loses $86400/400 = 216$ seconds — some three and a half minutes — per day. Hence the two historical remedies: enforce a tiny constant amplitude (the escapement), or bend the constraint so the period is exactly amplitude-free (Huygens’ cycloidal cheeks, 1657).

**16.** In $\tau(E_0) = \int_{-\pi}^{\pi}\frac{\dd
x}{\sqrt{2(E_0 + \cos x)}}$ the integrand is, for each fixed $x$, strictly decreasing in $E_0$: $\tau$ is strictly decreasing. As $E_0 \downarrow 1$ the integrands increase pointwise to $\bigl(2(1 + \cos x)\bigr)^{-1/2} =
\frac1{2\abs{\cos\frac x2}}$, whose integral over $\intoo{-\pi}\pi$ diverges ($\cos\frac x2$ vanishes to first order at $\pm\pi$): monotone convergence gives $\tau(E_0)
\to +\infty$. As $E_0 \to +\infty$:

$$
\sqrt{2E_0}\,\tau(E_0) = \int_{-\pi}^{\pi}
\frac{\dd x}{\sqrt{1 + \cos x/E_0}} \longrightarrow 2\pi
$$

by dominated convergence (for $E_0 \geq 2$ the integrand is at most $\sqrt2$). So $\tau \approx 2\pi/\sqrt{2E_0}$: one turn takes the time of free rotation at speed $\sqrt{2E_0}$, the potential reduced to a ripple — matching question 10’s rotation rate.

**17.** The axes carry the explicit solutions $t
\mapsto (x_0\eu^{\alpha t}, 0)$ and $t \mapsto (0,
y_0\eu^{-\gamma t})$, together with the equilibrium $(0, 0)$: they are unions of [orbits](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action). The field is $\mathcal
C^1$, hence [locally Lipschitz](#def-b3-ode-lipschitz); a solution starting in $Q$ that touched an axis would pass through a point of one of those [orbits](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) and, by [Corollary 19.4](#cor-b3-ode-uniqueness), coincide with it — impossible, one living on the axis and the other not. So $Q$ is invariant in both time directions. Equilibria in $Q$: $x > 0$ forces $\alpha - \beta y = 0$ and $y > 0$ forces $\delta x - \gamma = 0$: the single point $(x_*, y_*) = \bigl(\frac\gamma\delta,
\frac\alpha\beta\bigr)$.

**18.** Along a solution,

$$
\dot H = \Bigl(\delta - \frac\gamma x\Bigr)x' +
\Bigl(\beta - \frac\alpha y\Bigr)y'
= (\delta x - \gamma)(\alpha - \beta y) +
(\beta y - \alpha)(\delta x - \gamma) = 0 .
$$

$f(x) = \delta x - \gamma\ln x$ has $f'' = \gamma/x^2 > 0$, $f'$ vanishing only at $x_*$, and $f \to +\infty$ both at $0^+$ and at $+\infty$: strictly convex and proper, minimum $f(x_*)$; likewise $g(y) = \beta y - \alpha\ln y$, minimum $g(y_*)$. So $H \geq h_*$ with equality only at $(x_*,
y_*)$, and each sublevel $\{H \leq h\}\cap Q$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): $f(x) \leq h - g(y_*)$ confines $x$ to a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) interval of $\intoo0{+\infty}$ by properness, likewise $y$, and the set is closed in $\R^2$ since $H \to +\infty$ at the boundary of $Q$. A [maximal solution](#thm-b3-ode-maximal) stays on its [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) level set, so it cannot leave every [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) in finite time: [Theorem 19.5](#thm-b3-ode-maximal) makes it global.

**19.** Fix $h > h_*$ and set $c = h - g(y_*) >
f(x_*)$. Since $f$ decreases strictly from $+\infty$ to $f(x_*)$ on $\intoc0{x_*}$ and increases strictly back to $+\infty$ on $\intco{x_*}{+\infty}$, the equation $f(x) = c$ has exactly two roots $x_- < x_* < x_+$, and $\{f \leq c\} =
\intcc{x_-}{x_+}$. For $x \in \intoo{x_-}{x_+}$: $g(y) = h -
f(x) > g(y_*)$ has exactly two roots $y_-(x) < y_* <
y_+(x)$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) in $x$ (inverses of the strictly monotone [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) restrictions of $g$ on either side of $y_*$), with $y_\pm(x) \to y_*$ as $x \to x_\pm$; at $x =
x_\pm$ the unique solution is $y = y_*$. So $\{H = h\}\cap
Q$ is the union of the graphs of $y_+$ and $y_-$ over $\intcc{x_-}{x_+}$, glued at $(x_\pm, y_*)$: a closed curve around $(x_*, y_*)$ — the analogue of the pendulum’s ovals.

**20.** Let $C_h = \{H = h\}\cap Q$ with $h > h_*$: the only equilibrium of $Q$ is off $C_h$, so the field never vanishes on it. Signs: $x' = \beta x(y_* - y)$, $y' =
\delta y(x - x_*)$: the motion goes right below the line $y
= y_*$, up to the right of $x = x_*$, left above, down on the left — counterclockwise circulation. Follow the solution from a point $(x_0, y_-(x_0))$ of the open lower branch, where $x' > 0$: the time to reach the right corner $B = (x_+, y_*)$ is

$$
\int_{x_0}^{x_+}\frac{\dd x}{\beta x\,\bigl(y_* -
y_-(x)\bigr)} .
$$

Near $x_+$, choose $x' \in \intoo{x_*}{x_+}$; for $x \in
\intcc{x'}{x_+}$, $f(x_+) - f(x) \geq f'(x')\,(x_+ - x)$ ($f'$ is increasing and positive past $x_*$), while the level relation and Taylor’s inequality give $g(y_-(x)) -
g(y_*) \leq \frac12\,\bigl(\max g''\bigr)\,(y_* -
y_-(x))^2$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $y$-range of $C_h$: hence $y_* -
y_-(x) \geq c\,\sqrt{x_+ - x}$ with $c > 0$, and the integrand is $O\bigl((x_+ - x)^{-1/2}\bigr)$: [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) — question 5’s convergence, transposed. Elsewhere on the branch the integrand is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). So $B$ is reached in finite time; there $y' = \delta y_*(x_+ - x_*) > 0$, the [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) enters the region $x > x_*$, $y > y_*$, climbs to the top corner $(x_*, y_+)$ by the symmetric estimate (roles of $f$ and $g$ exchanged), and so on around the four arcs: after a finite time $T > 0$ the solution returns to its starting point. By [Corollary 19.4](#cor-b3-ode-uniqueness) it is $T$-periodic — question 4’s argument, verbatim.

**21.** On a $T$-periodic [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) in $Q$, $t \mapsto \ln
x(t)$ is $\mathcal C^1$ and $T$-periodic, so

$$
0 = \int_0^T(\ln x)'\,\dd t
= \int_0^T(\alpha - \beta y)\,\dd t
= \alpha T - \beta\int_0^Ty\,\dd t ,
$$

giving $\frac1T\int_0^Ty = \frac\alpha\beta$; likewise $0 =
\int_0^T(\ln y)' = \delta\int_0^Tx\,\dd t - \gamma T$ gives $\frac1T\int_0^Tx = \frac\gamma\delta$. The time averages are the equilibrium values, for every [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) regardless of amplitude — a conservation law nobody put in by hand.

**22.** With harvesting the system is again of Lotka–Volterra form, with parameters $\alpha -
\varepsilon$, $\beta$, $\gamma + \varepsilon$, $\delta$ (the [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) equilibrium persists since $\varepsilon <
\alpha$). Question 21 applied to the new system:

$$
\bar x = \frac{\gamma + \varepsilon}{\delta}
\ \ (\text{prey average rises}),
\qquad
\bar y = \frac{\alpha - \varepsilon}{\beta}
\ \ (\text{predator average falls}) :
$$

indiscriminate harvesting shifts the balance toward the prey. D’Ancona’s data read this backwards: the war cut fishing, $\varepsilon$ dropped, so the predator average $(\alpha - \varepsilon)/\beta$ rose and the prey average fell — a larger shark fraction in the catch, exactly what the Adriatic fish markets recorded. This is Volterra’s principle, the same mechanism behind pesticide paradoxes: culling both trophic levels benefits the level being eaten.

**23.** The recipe, both times: (i) a first integral — $E$ for the pendulum, $H$ here, found by separating $\dd y/\dd x$ — collapses the plane onto curves; (ii) properness and compactness of the level sets give global existence through [Theorem 19.5](#thm-b3-ode-maximal); (iii) the geometry of the levels — ovals, from the shape of $\cos$ there and the strict convexity of $f$ and $g$ here — is read off the integral, not the [flow](#ex-b3-ode-planeclassification); (iv) a nonvanishing field on a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) oval plus [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) corner singularities forces a finite return time; (v) uniqueness ([Corollary 19.4](#cor-b3-ode-uniqueness)) converts return into periodicity and forbids finite-time arrival at equilibria; (vi) the dividends — period expansions, laws of averages — come from the convergence theorems applied to the resulting integrals. Neither $x'' = -\sin x$ nor Lotka–Volterra admits an elementary closed-form solution (elliptic integrals in one case, transcendental level curves in the other), and at no point was one needed: the equation itself, interrogated qualitatively, surrendered the entire motion.

**24.** On an energy level $E_0 > 1$, $y^2 = 2(E_0 +
\cos x) > 0$: extremes of $y^2$ at $\cos x = \pm1$, giving the stated bounds, attained at $x \equiv 0, \pi$. Time average over one period $\tau = \tau(E_0)$: $x$ advances by exactly $2\pi$, so

$$
\frac1\tau\int_0^\tau y\,\dd t = \frac{x(\tau) -
x(0)}{\tau} = \frac{2\pi}\tau .
$$

The ratio of extreme speeds is $\sqrt{\frac{E_0 + 1}{E_0 -
1}} = 1 + O(E_0^{-1}) \to 1$: at high energy the potential’s $\pm1$ ripple is negligible against $E_0$, and the pendulum spins almost uniformly — the washboard flattens.

**25.** On $E_0 \geq 1 + \delta$, the integrand of $\tau(E_0) = \int_{-\pi}^{\pi}\frac{\dd x}{\sqrt{2(E_0 +
\cos x)}}$ is dominated by $(2\delta)^{-1/2}$ and its $E_0$-derivative

$$
\partial_{E_0}\frac1{\sqrt{2(E_0 + \cos x)}}
= -\frac{1}{\bigl(2(E_0 + \cos x)\bigr)^{3/2}}
$$

by $(2\delta)^{-3/2}$, both [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\intcc{-\pi}\pi$: differentiation under the integral ([Theorem 10.15](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#thm-b3-lebesgue-paramdiff)) applies and gives $\tau'(E_0) < 0$ (the integrand is strictly negative): strictly decreasing, $\mathcal C^1$. Limits: as $E_0 \to
\infty$, $\tau \leq \frac{2\pi}{\sqrt{2(E_0 - 1)}} \to 0$; as $E_0 \to 1^+$: write $E_0 + \cos x = (E_0 - 1) +
2\cos^2\frac x2$; the integrand increases as $E_0$ decreases, so by monotone convergence

$$
\tau(E_0) \;\nearrow\;
\int_{-\pi}^{\pi}\frac{\dd x}{2\,\abs{\cos\frac x2}}
= +\infty,
$$

the limit integral diverging at $x = \pm\pi$ (there $\abs{\cos\frac x2} \sim \frac{\abs{x \mp \pi}}2$, a [non-integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) $\frac1{\abs\cdot}$): the period blows up approaching the separatrix, matching Part II’s $T(a) \to
\infty$ from the libration side. The energy axis reads: rest at $E_0 = -1$; librations, $2\pi \nearrow
\infty$, on $\intoo{-1}1$; the infinitely slow separatrix at $E_0 = 1$; rotations, $\infty \searrow 0$, beyond. One integral, the entire life of the pendulum.