---
title: "Differential Equations"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 16
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/16-differential-equations
---

# Chapter 16 — Differential Equations

Year 1 solved the linear equations that admit formulas. This chapter supplies what formulas cannot: the *Cauchy–Lipschitz theorem* — existence and uniqueness for $y' = f(t, y)$ — proved with the Banach fixed point theorem, exactly as promised in [Chapter 4](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#ch-b2-metric); then the complete theory of *linear systems* $X' = A(t)X + B(t)$, with the [matrix exponential](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ex-b2-nvs-matrixexp) and the [wronskian](#def-b2-diffeq-wronskian) as computational engines.

## 16.1 The Cauchy–Lipschitz theorem

**Theorem 16.1 (Cauchy–Lipschitz, global Lipschitz version).**

Let $I$ be a segment, $f \colon I \times \R^n \to \R^n$ [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and *[Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in the second variable*, uniformly in the first: $\norm{f(t, y) - f(t, z)} \leq k\,\norm{y - z}$ for all $t
\in I$. Then for every $(t_0, y_0) \in I \times \R^n$, the Cauchy problem

$$
y' = f(t, y), \qquad y(t_0) = y_0
$$

has exactly one solution $y \colon I \to \R^n$ of class $C^1$.

**Proof.** *Reformulation.* A [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $y$ solves the problem iff it satisfies the integral equation

$$
y(t) = y_0 + \int_{t_0}^{t} f\bigl(s, y(s)\bigr)\,\dd s
=: T(y)(t)
$$

(fundamental theorem of calculus both ways; a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) solution of the integral equation is automatically $C^1$).

*A contraction, after re-norming.* On the [Banach space](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-banach) $E =
C(I, \R^n)$ with the *weighted* [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm)

$$
N(y) = \sup_{t \in I}\; \eu^{-2k\abs{t - t_0}}\,\norm{y(t)} ,
$$

(equivalent to the sup [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm): the weight is bounded above and below on the segment $I$, so $E$ stays [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete)), estimate for $y, z \in
E$ and, say, $t \geq t_0$:

$$
\norm{T(y)(t) - T(z)(t)}
\leq \int_{t_0}^{t} k\,\norm{y(s) - z(s)}\,\dd s
\leq k\,N(y - z)\int_{t_0}^{t} \eu^{2k(s - t_0)}\dd s
\leq \frac{N(y-z)}{2}\,\eu^{2k(t - t_0)} .
$$

Multiplying by $\eu^{-2k(t - t_0)}$ and taking the sup (the case $t
< t_0$ is [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint)): $N\bigl(T(y) - T(z)\bigr) \leq \frac12 N(y -
z)$: $T$ is a $\frac12$-contraction of the [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) $(E, N)$. The Banach fixed point theorem ([Theorem 4.12](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-banach)) yields a unique fixed point: the unique solution. ∎

**Remark 16.2.**

For $f$ merely $C^1$ (locally [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)), the theorem holds *locally*, with a maximal solution on a maximal [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) interval; solutions can explode in finite time ($y' = y^2$, $y(0) = 1$: $y(t)
= \frac{1}{1-t}$, gone at $t = 1$). The global [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) hypothesis is what buys the whole segment. Two consequences worth engraving: solution curves of an ODE with [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) field *never cross*; and the zero function is the only solution vanishing anywhere of a linear homogeneous equation.

**Example 16.3 (Uniqueness is a theorem: a field that leaks).**

Consider $y' = 2\sqrt{\abs y}$ with $y(0) = 0$. The zero function solves it; so does

$$
y(t) = \begin{cases} 0 & t \leq 0,\\ t^2 & t \geq 0,
\end{cases}
$$

which is $C^1$ (both pieces have derivative $0$ at the glue point) and satisfies $y'(t) = 2t = 2\sqrt{t^2}$ for $t > 0$ — indeed, delaying the take-off gives a solution for *every* release time $c \geq 0$: infinitely many solutions through the same initial data. No contradiction with [Theorem 16.1](#thm-b2-diffeq-cauchylipschitz): near $y = 0$,

$$
\frac{\abs{2\sqrt y - 2\sqrt z}}{\abs{y - z}}
= \frac{2}{\sqrt y + \sqrt z} \longrightarrow +\infty ,
$$

the field is not [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in $y$, and the theorem is silent. Closing insight: the physical reading is a bucket draining under gravity run backwards — from the empty state, one cannot tell when it started filling; determinism of ODEs is exactly the [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) condition, not a law of nature.

## 16.2 Linear systems

**Theorem 16.4 (Structure of linear systems).**

Let $A \colon I \to \mathcal{M}_n(\R)$ and $B \colon I \to \R^n$ be [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on an interval $I$. For every $(t_0, X_0)$ the problem

$$
X' = A(t)X + B(t), \qquad X(t_0) = X_0
$$

has exactly one solution on *all* of $I$. The solutions of the homogeneous system ($B = 0$) form a vector space $\mathcal{S}_H$ of dimension exactly $n$, and evaluation $X \mapsto X(t_0)$ is an isomorphism $\mathcal{S}_H \to \R^n$; general solution $=$ particular $+$ homogeneous.

**Proof.** On every segment $J \subseteq I$ containing $t_0$: $f(t, X) =
A(t)X + B(t)$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), and [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in $X$ with constant $k
= \sup_J \vertiii{A(t)}$ (finite: [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on a segment): [Theorem 16.1](#thm-b2-diffeq-cauchylipschitz) applies on $J$; letting $J$ exhaust $I$, uniqueness glues the solutions into one on $I$. Linearity of the solution set and of the evaluation map are clear; evaluation is bijective by existence (surjective) and uniqueness (injective): $\dim \mathcal{S}_H = n$. The affine structure is Year 1’s argument verbatim. ∎

**Example 16.5 (The evaluation isomorphism, concretely).**

For $y'' + y = 0$, viewed as the system $X' =
\begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix}X$ with $X = (y,
y')$: the theorem says the solution space is a plane, and that $X \mapsto X(0) = (y(0), y'(0))$ is an isomorphism onto $\R^2$. The solutions $\cos$ and $\sin$ evaluate to $(1, 0)$ and $(0, 1)$ — the canonical basis of $\R^2$ — so they form a basis of the solution space, and *every* solution is

$$
y(t) = y(0)\cos t + y'(0)\sin t ,
$$

with the coefficients read directly off the initial data, no linear system to solve. Closing insight: choosing the [fundamental system](#def-b2-diffeq-wronskian) whose initial values are the canonical basis (here $\cos, \sin$) is exactly choosing the columns of $\eu^{tA}$; the evaluation isomorphism is why initial conditions parametrize trajectories — the geometric content of “deterministic dynamics” for linear equations.

**Definition 16.6 (Wronskian).**

For solutions $X_1, \dots, X_n$ of the homogeneous system, the *wronskian* is $W(t) = \det\bigl(X_1(t),
\dots, X_n(t)\bigr)$. By the isomorphism above, either $W$ vanishes identically (the family is linked) or never (a *fundamental system*); quantitatively, $W' = \operatorname{tr}\bigl(A(t)\bigr)
W$, so

$$
W(t) = W(t_0)\,\exp\Bigl(\int_{t_0}^{t}
\operatorname{tr} A(s)\,\dd s\Bigr)
\quad \text{(Liouville's formula)}.
$$

**Example 16.7 (Liouville checked on an Euler equation).**

On $\intoo{0}{\infty}$, the equation $t^2y'' + ty' - y = 0$ has the solutions $y_1(t) = t$ and $y_2(t) = \frac1t$ (substitute). Their [wronskian](#def-b2-diffeq-wronskian):

$$
W(t) = \det\begin{pmatrix} t & \tfrac1t\\[2pt]
1 & -\tfrac{1}{t^2}\end{pmatrix}
= -\frac1t - \frac1t = -\frac2t ,
$$

never zero: a [fundamental system](#def-b2-diffeq-wronskian). Now check Liouville: in normalized form $y'' + \frac1t\,y' - \frac{1}{t^2}\,y = 0$, the companion matrix $A(t) = \begin{pmatrix} 0 & 1\\ \frac{1}{t^2}
& -\frac1t\end{pmatrix}$ has trace $-\frac1t$, so

$$
W(t) = W(1)\exp\Bigl(-\int_1^t\frac{\dd s}{s}\Bigr)
= -2\,\eu^{-\ln t} = -\frac2t . \checkmark
$$

Closing insight: Liouville predicts the *shape* of the [wronskian](#def-b2-diffeq-wronskian) before any solution is known — here, that $W$ must be $\frac{c}{t}$; this is what powers the reduction-of-order method ([Proposition 16.15](#prop-b2-diffeq-secondorder)), where knowing $y_1$ and the [wronskian](#def-b2-diffeq-wronskian)’s form determines $y_2$ by one quadrature.

**Proof of Liouville’s formula.** $W(t) = \det M(t)$ with $M' = AM$. Differentiating the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) as a multilinear function of the columns,

$$
W'(t) = \sum_j \det(X_1, \dots, X_j', \dots, X_n)
= \sum_j \det(X_1, \dots, AX_j, \dots, X_n) .
$$

Now the map $(C_1, \dots, C_n) \mapsto \sum_j \det(C_1, \dots,
AC_j, \dots, C_n)$ is $n$-linear and alternating (with two equal columns $C_i = C_k$, the terms $j \notin \{i, k\}$ vanish outright, and the terms $j = i$ and $j = k$ cancel in pairs after one column swap): by the uniqueness theorem ([Theorem 2.14](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#thm-b2-linalg-detspace)) it is $c \cdot \det$, with $c$ read on the canonical columns: $c = \sum_j \det(e_1, \dots, Ae_j, \dots,
e_n) = \sum_j a_{jj} = \operatorname{tr} A$. Hence $W' =
\operatorname{tr}\bigl(A(t)\bigr)W$: a scalar linear ODE, solved by the Year 1 formula. ∎

## 16.3 Constant coefficients: the matrix exponential

**Theorem 16.8.**

For $A \in \mathcal{M}_n(\R)$ (or $\C$), the exponential $\eu^{tA}
= \sum_k \frac{(tA)^k}{k!}$ ([Example 5.22](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#ex-b2-nvs-matrixexp)) satisfies: $t \mapsto \eu^{tA}$ is $C^1$ (indeed $C^\infty$) with

$$
\frac{\dd}{\dd t}\,\eu^{tA} = A\,\eu^{tA} = \eu^{tA}A ,
\qquad
\eu^{(s+t)A} = \eu^{sA}\,\eu^{tA},
\qquad
(\eu^{A})^{-1} = \eu^{-A} ;
$$

and $\eu^{A + B} = \eu^A\eu^B$ *when* $AB = BA$. The Cauchy problem $X' = AX$, $X(0) = X_0$ has the unique solution $X(t) =
\eu^{tA}X_0$; with a source, the *variation of constants* formula holds:

$$
X(t) = \eu^{(t - t_0)A}X_0 + \int_{t_0}^{t} \eu^{(t-s)A}B(s)\,\dd
s .
$$

**Proof.** *[Differentiability](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#def-b2-diffcalc-differential):* the series $\sum \frac{t^kA^k}{k!}$ and its term-derived series $\sum \frac{t^{k-1}A^k}{(k-1)!} = A\sum
\frac{(tA)^{k-1}}{(k-1)!}$ converge [normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) on every segment ([norms](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) $\leq \frac{(\abs t\,\vertiii A)^k}{k!}$): differentiate term by term ([Theorem 10.11](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#thm-b2-funcseq-seriestransfer), vector-valued). The two orderings $A\eu^{tA}$ and $\eu^{tA}A$ agree since every partial sum commutes with $A$.

*Group law:* for commuting $A, B$, the [Cauchy product](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-fubini) of the two exponential series reorganizes by the binomial theorem exactly as in [Example 7.15](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#ex-b2-series-exp) (absolute convergence in the Banach algebra justifies it): $\eu^{A+B} = \eu^A\eu^B$; with $B = sA$ this gives the one-parameter group law, and $B = -A$ the inverse.

*Cauchy problem:* $X(t) = \eu^{tA}X_0$ solves it (differentiate); uniqueness by [Theorem 16.4](#thm-b2-diffeq-linear). Variation of constants: set $Y(t) = \eu^{-tA}X(t)$; differentiating, $Y' = \eu^{-tA}(X' -
AX) = \eu^{-tA}B(t)$; integrate from $t_0$ to $t$ and multiply back by $\eu^{tA}$. ∎

**Method 16.9 (Computing etA\eu^{tA}etA).**

Reduce $A$ ([Chapter 3](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#ch-b2-reduction)): if $A = PDP^{-1}$ diagonal, $\eu^{tA} = P\,\eu^{tD}P^{-1}$ with $\eu^{tD}$ diagonal of $\eu^{t\lambda_i}$; in general use Dunford $A = D + N$ (commuting): $\eu^{tA} = \eu^{tD}\,\eu^{tN}$ with $\eu^{tN}$ a *polynomial* in $t$ (nilpotence truncates the series). Complex [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) pair into rotation-times-exponential blocks ([Exercise 16.5](#exo-b2-diffeq-5)).

**Remark 16.10 (Common pitfalls).**

*(i) $\eu^{A+B} \neq \eu^A\eu^B$ without commutation:* take $A = \begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}$, $B =
\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}$. Then $\eu^A = I +
A$, $\eu^B = I + B$ (nilpotence), so

$$
\eu^A\eu^B = \begin{pmatrix} 2 & 1\\ 1 & 1\end{pmatrix},
\qquad\text{while}\qquad
\eu^{A+B} = \cosh(1)\,I + \sinh(1)\,(A + B)
= \begin{pmatrix} \cosh 1 & \sinh 1\\ \sinh 1 & \cosh 1
\end{pmatrix},
$$

using $(A+B)^2 = I$; and $\cosh 1 \approx 1.54 \neq 2$. The group law of [Theorem 16.8](#thm-b2-diffeq-matrixexp) carries a genuine hypothesis. *(ii) Nonlinear intuition on linear turf:* solutions of a *linear* system with [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) coefficients live on the whole interval ([Theorem 16.4](#thm-b2-diffeq-linear)) — if a candidate solution explodes inside $I$, the equation was not linear or the computation is wrong; conversely, for nonlinear equations never promise globality without an argument ($y' = y^2$). *(iii) Dividing by the unknown:* separating variables in $y' = y(1-y)$ silently discards the constant solutions $0$ and $1$ — exactly the ones that organize the phase line ([Exercise 16.3](#exo-b2-diffeq-3)); list constant solutions first. *(iv) Initial data fix vectors, not scalars:* an $n$-th order scalar equation needs $n$ conditions ($y, y', \dots$ at $t_0$); matching only $y(t_0)$ leaves an $(n-1)$-parameter family, a classic source of “lost” constants.

**Example 16.11 (A 3×33\times33×3 exponential by Dunford).**

Solve $X' = AX$ for $A = \begin{pmatrix} 2 & 1 & 0\\ 0 & 2 &
0\\ 0 & 0 & 3\end{pmatrix}$. Dunford by blocks: $A = D + N$ with $D = \operatorname{diag}(2, 2, 3)$ and $N = E_{12}$, which commute ($N$ lives inside the eigenvalue-$2$ block), and $N^2 =
0$:

$$
\eu^{tA} = \eu^{tD}\,\eu^{tN}
= \begin{pmatrix}
\eu^{2t} & t\,\eu^{2t} & 0\\
0 & \eu^{2t} & 0\\
0 & 0 & \eu^{3t}
\end{pmatrix} .
$$

The general solution reads off column by column: $X(t) =
\bigl(\eu^{2t}(x_0 + ty_0),\ \eu^{2t}y_0,\
\eu^{3t}z_0\bigr)$. Sanity checks: at $t = 0$ the matrix is $I$; its [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) is $\eu^{7t} =
\eu^{t\operatorname{tr}A}$, as Liouville demands; and the factor $t$ appears exactly where the [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $2$ is defective. Closing insight: polynomials times exponentials are not a guess to be memorized — they are the truncated series $\eu^{tN}$, and their degree is bounded by the nilpotency index, never more.

**Method 16.12 (Solving X′=AX+B(t)X' = AX + B(t)X′=AX+B(t), start to finish).**

1. [Spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $A$ ; then $\eu^{tA}$ by [Method 16.9](#met-b2-diffeq-computeexp) (diagonalize; or Dunford as in [Example 16.11](#ex-b2-diffeq-dunford3) ; or a polynomial trick like $A^2 = -I$ ).
2. A particular solution: variation of constants $\int_{t_0}^t\eu^{(t-s)A}B(s)\dd s$ always works; for exponential-polynomial $B$ , an ansatz of the same shape (degree raised on [resonance](#pb-b2-diffeq-1) , [Exercise 16.10](#exo-b2-diffeq-10) ) is faster.
3. General solution $= \eu^{(t-t_0)A}X_0 +$ particular; fit the initial data *last* , on the complete formula.
4. Sanity checks: $X(t_0)$ correct; the homogeneous part’s growth matches the [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) real parts ( [Exercise 16.8](#exo-b2-diffeq-8) ); and $\det$ of a fundamental matrix obeys Liouville.

**Example 16.13 (A phase portrait).**

$X' = AX$ with $A = \begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix}$: $A^2 = -I$, so the series splits into

$$
\eu^{tA} = (\cos t)\,I + (\sin t)\,A
= \begin{pmatrix} \cos t & \sin t\\ -\sin t & \cos t
\end{pmatrix} :
$$

trajectories are circles run clockwise — the harmonic oscillator $x'' + x = 0$ in first-order clothes. [Eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\pm\iu$ on the imaginary axis: a *center*. More generally the real parts of the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $A$ decide growth or decay of $\norm{X(t)}$ ([Exercise 16.8](#exo-b2-diffeq-8)).

![Two linear phase portraits. Left: a center (eigenvalues ±) — closed circular orbits of the harmonic oscillator. Right: a stable node (eigenvalues -1, -2) — all trajectories fall into the origin tangentially to the slow eigendirection.](https://one-course.com/images/onecourse/chapters/math-4/b2-diffeq/fig-e6a5565f0d22.svg)

![Two linear phase portraits. Left: a center (eigenvalues ±) — closed circular orbits of the harmonic oscillator. Right: a stable node (eigenvalues -1, -2) — all trajectories fall into the origin tangentially to the slow eigendirection.](https://one-course.com/images/onecourse/chapters/math-4/b2-diffeq/fig-f41c08dcc309.svg)

*Two linear phase portraits. Left: a *center* ([eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\pm\iu$) — closed circular orbits of the harmonic oscillator. Right: a *stable node* ([eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $-1, -2$) — all trajectories fall into the origin tangentially to the slow eigendirection.*

![The trace–determinant plane for X' = AX in dimension 2: below the horizontal axis, saddles; between the axis and the parabola = 2/4, nodes; inside the parabola, spirals; on the positive -axis, centers. The weekend problem proves this classification and follows one vertical line of it — the damped oscillator — into resonance.](https://one-course.com/images/onecourse/chapters/math-4/b2-diffeq/fig-356f301d726d.svg)

*The trace–determinant plane for $X' = AX$ in dimension $2$: below the horizontal axis, saddles; between the axis and the parabola $\delta = \tau^2/4$, nodes; inside the parabola, spirals; on the positive $\delta$-axis, centers. The weekend problem proves this classification and follows one vertical line of it — the damped oscillator — into [resonance](#pb-b2-diffeq-1).*

**Remark 16.14 (Where this is used).**

Linear systems are the local model for everything nonlinear: near an equilibrium, a smooth vector field behaves (in the hyperbolic cases) like its linearization, whose portrait the trace–determinant plane classifies. The weekend problem works out the oscillator story in full — damping, forcing, [resonance](#pb-b2-diffeq-1), and Sturm’s comparison theorems for variable coefficients — the mathematics behind shock absorbers, AC circuits, and spectral gaps alike. The Year 3 volume returns with the qualitative theory (flows, stability, first integrals) on manifolds.

## 16.4 Second order with variable coefficients

**Proposition 16.15.**

The equation $y'' + a(t)y' + b(t)y = c(t)$ ($a, b, c$ [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $I$) is the system $X' = A(t)X + B(t)$ for $X = (y, y')$: solutions exist and are unique on all of $I$ for any initial data $(y(t_0),
y'(t_0))$; homogeneous solutions form a plane. If one nonvanishing homogeneous solution $y_1$ is known, a second independent one is found by *lowering the order*: setting $y = y_1 z$ turns the homogeneous equation into a first-order equation for $z'$, solved by quadratures.

**Proof.** The system form and [Theorem 16.4](#thm-b2-diffeq-linear) give everything structural. Lowering: substituting $y = y_1z$,

$$
y_1 z'' + (2y_1' + a y_1)z' + \underbrace{(y_1'' + ay_1' +
by_1)}_{=\,0}\,z = 0 :
$$

a first-order linear equation in $u = z'$, solvable by the Year 1 formula; integrating $u$ gives $z$, hence $y_2 = y_1 z$, independent of $y_1$ whenever $z$ is nonconstant. ∎

**Example 16.16.**

$t^2y'' - 2y = 0$ on $\intoo{0}{\infty}$: $y_1 = t^2$ is a solution. Substitute $y = t^2z$: from $y' = t^2z' + 2tz$ and $y''
= t^2z'' + 4tz' + 2z$,

$$
t^2y'' - 2y = t^4 z'' + 4t^3z' = 0,
\qquad\text{i.e.}\qquad \frac{z''}{z'} = -\frac4t :
$$

$z' = t^{-4}$ (up to a constant), $z = -\frac{1}{3t^3}$, and $y_2
= t^2z = -\frac{1}{3t}$. General solution: $y = \alpha t^2 +
\frac{\beta}{t}$.

## 16.5 Exercises

**Exercise 16.1 ★.**

Solve $X' = AX$, $X(0) = (1, 0)^{\mathsf T}$, for $A =
\begin{pmatrix} 1 & 1\\ 0 & 2\end{pmatrix}$ (diagonalize) and $A =
\begin{pmatrix} 2 & 1\\ 0 & 2 \end{pmatrix}$ (Dunford).

**Solution of Exercise 16.1.**

First matrix: [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $1, 2$, [eigenvectors](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $(1,0)$ and $(1,1)$. Decompose $X_0 = (1,0) = 1\cdot(1,0) + 0\cdot(1,1)$: the solution is

$$
X(t) = \eu^{t}\begin{pmatrix}1\\ 0\end{pmatrix} .
$$

(The initial vector is itself an [eigenvector](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen).)

Second: $A = 2I + N$, $N = E_{12}$, $N^2 = 0$: $\eu^{tA} =
\eu^{2t}(I + tN)$, so

$$
X(t) = \eu^{2t}\begin{pmatrix} 1 & t\\ 0 & 1\end{pmatrix}
\begin{pmatrix}1\\ 0\end{pmatrix}
= \eu^{2t}\begin{pmatrix}1\\ 0\end{pmatrix}.
$$

**Exercise 16.2 ★.**

Which Cauchy problems have unique global solutions on $\R$ by [Theorem 16.1](#thm-b2-diffeq-cauchylipschitz)? $y' = \sin(ty)$; $\;y' =
y^2$; $\;y' = \abs y$. For the last, solve explicitly with $y(0) =
0$ and $y(0) = 1$.

**Solution of Exercise 16.2.**

$y' = \sin(ty)$: $\abs{\sin(ty) - \sin(tz)} \leq \abs t\,\abs{y -
z}$ — [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in $y$ [uniformly](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) on every segment of times: unique global solutions on $\R$ (apply the theorem on every segment).

$y' = y^2$: locally [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) only; no global theorem, and indeed $y(0) = 1$ explodes at $t = 1$.

$y' = \abs y$: $\abs\cdot$ is $1$-Lipschitz: global existence and uniqueness. With $y(0) = 0$: $y \equiv 0$ (uniqueness!). With $y(0)
= 1$: $y$ stays positive (cannot cross the zero solution), so $y' =
y$: $y = \eu^t$.

**Exercise 16.3 ★.**

Prove that two distinct maximal solutions of $y' = f(t,y)$ ($f$ [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in $y$) never take the same value at the same time, and deduce that solutions of $y' = y(1 - y)$ starting in $\intoo{0}{1}$ remain in $\intoo{0}{1}$ forever.

**Solution of Exercise 16.3.**

If $y(t_1) = z(t_1)$ at some time, then $y$ and $z$ solve the same Cauchy problem at $t_1$: uniqueness forces $y = z$ on their common interval — distinct solutions never meet.

For $y' = y(1-y)$: the constants $0$ and $1$ are solutions. A solution starting in $\intoo{0}{1}$ can never reach $0$ or $1$ (it would collide with a constant solution): it stays in $\intoo{0}{1}$, hence is global (bounded: no explosion — e.g. by [Exercise 16.9](#exo-b2-diffeq-9)’s criterion, or because the vector field is bounded on the trapped strip).

**Exercise 16.4 ★★.**

Compute $\eu^{tA}$ for $A = \begin{pmatrix} 3 & 1\\ -1 &
1\end{pmatrix}$ *(Dunford: $(A - 2I)^2 = 0$)*, and solve $X' =
AX + \begin{pmatrix} \eu^{2t}\\ 0\end{pmatrix}$, $X(0) = 0$, by variation of constants.

**Solution of Exercise 16.4.**

$(A - 2I)^2 = \begin{pmatrix}1 & 1\\ -1 & -1\end{pmatrix}^2 = 0$: Dunford with $D = 2I$, $N = A - 2I$:

$$
\eu^{tA} = \eu^{2t}\,(I + tN)
= \eu^{2t}\begin{pmatrix} 1 + t & t\\ -t & 1 - t\end{pmatrix}.
$$

Variation of constants with $B(s) = (\eu^{2s}, 0)^{\mathsf T}$:

$$
X(t) = \int_0^t \eu^{(t-s)A}B(s)\,\dd s
= \eu^{2t}\int_0^t \begin{pmatrix} 1 + (t-s)\\ -(t-s)
\end{pmatrix}\dd s
= \eu^{2t}\begin{pmatrix} t + \frac{t^2}{2}\\[2pt] -\frac{t^2}{2}
\end{pmatrix},
$$

using $\eu^{(t-s)A}B(s) = \eu^{2(t-s)}(I + (t-s)N)\,(\eu^{2s},
0)^{\mathsf T} = \eu^{2t}\bigl(1 + (t-s),\, -(t-s)\bigr)^{\mathsf
T}$. (Check: $X(0) = 0$; $X' - AX = (\eu^{2t}, 0)^{\mathsf T}$ by differentiation.)

**Exercise 16.5 ★★.**

For $A = \begin{pmatrix} \alpha & -\beta\\ \beta &
\alpha\end{pmatrix}$, prove $\eu^{tA} =
\eu^{\alpha t}\begin{pmatrix} \cos\beta t & -\sin\beta t\\
\sin\beta t & \cos\beta t\end{pmatrix}$ — spiral trajectories — two ways: via the series (write $A = \alpha I + \beta J$, $J^2 =
-I$), and via the complex identification $z' = (\alpha +
\iu\beta)z$.

**Solution of Exercise 16.5.**

*Series:* $A = \alpha I + \beta J$ with $J =
\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$, $J^2 = -I$; the two summands commute, so $\eu^{tA} = \eu^{\alpha t}\,\eu^{\beta tJ}$, and the series of $\eu^{\beta t J}$ splits along even/odd powers into $\cos(\beta t)I + \sin(\beta t)J$: the stated rotation-scaling matrix.

*Complex:* identify $(x, y) \in \R^2$ with $z = x + \iu y$; the system $X' = AX$ reads $z' = (\alpha + \iu\beta)z$, whose solution $z(t) = \eu^{\alpha t}\eu^{\iu\beta t}z_0$ is exactly the spiral: modulus $\eu^{\alpha t}$, argument advancing at speed $\beta$.

**Exercise 16.6 ★★.**

(Gronwall’s lemma) Let $u$ be [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) nonnegative with $u(t)
\leq C + k\int_{t_0}^{t} u(s)\,\dd s$ on $\intco{t_0}{T}$. Prove $u(t) \leq C\,\eu^{k(t - t_0)}$ *(differentiate $v(t) =
\eu^{-kt}\int_{t_0}^t u$)*. Deduce again the uniqueness in Cauchy–Lipschitz and the [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) dependence $\norm{y(t) -
z(t)} \leq \norm{y_0 - z_0}\,\eu^{k\abs{t - t_0}}$ for two solutions with different initial data.

**Solution of Exercise 16.6.**

Let $v(t) = \eu^{-k(t - t_0)}\int_{t_0}^t u$. Then

$$
v'(t) = \eu^{-k(t-t_0)}\Bigl(u(t) - k\int_{t_0}^t u\Bigr)
\leq C\,\eu^{-k(t-t_0)},
$$

by the hypothesis. Integrating from $t_0$ to $t$ ($v(t_0) = 0$): $v(t) \leq \frac{C}{k}\bigl(1 - \eu^{-k(t - t_0)}\bigr)$, i.e. $k\int_{t_0}^t u \leq C\bigl(\eu^{k(t-t_0)} - 1\bigr)$; feeding this back into the hypothesis: $u(t) \leq C\eu^{k(t-t_0)}$.

Uniqueness/dependence: two solutions $y, z$ of the integral equation satisfy

$$
\norm{y(t) - z(t)} \leq \norm{y_0 - z_0} +
k\int_{t_0}^{t}\norm{y - z},
$$

and Gronwall with $C = \norm{y_0 - z_0}$ gives the exponential bound; $C = 0$ gives uniqueness.

**Exercise 16.7 ★★.**

Knowing that $y_1(t) = \frac{\sin t}{t}$ solves $ty'' + 2y' + ty =
0$ on $\intoo{0}{\pi}$, find a second independent solution by lowering the order, and give the general solution.

**Solution of Exercise 16.7.**

Substitute $y = y_1 z$ with $y_1 = \frac{\sin t}{t}$: the general lowering formula ([Proposition 16.15](#prop-b2-diffeq-secondorder)) gives, for $u
= z'$,

$$
y_1 u' + \Bigl(2y_1' + \frac{2}{t}\,y_1\Bigr)u = 0
$$

(the equation normalized as $y'' + \frac2t y' + y = 0$). Compute $2y_1' + \frac2t y_1 = 2\,\frac{t\cos t - \sin t}{t^2} +
\frac{2\sin t}{t^2} = \frac{2\cos t}{t}$: so

$$
\frac{u'}{u} = -\frac{2\cos t}{t}\cdot\frac{t}{\sin t}
= -2\cot t
\quad\Longrightarrow\quad
u = \frac{1}{\sin^2 t}
\quad (\text{up to a constant}),
$$

and $z = -\cot t$, giving $y_2 = y_1 z = -\frac{\cos t}{t}$. General solution on $\intoo{0}{\pi}$:

$$
y(t) = \alpha\,\frac{\sin t}{t} + \beta\,\frac{\cos t}{t} .
$$

(These are the spherical Bessel functions of order zero.)

**Exercise 16.8 ★★★.**

Let $A \in \mathcal{M}_n(\C)$ with all [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of (strictly) negative real part. Prove that every solution of $X' = AX$ tends to $0$ as $t \to +\infty$, with an exponential rate: $\norm{X(t)}
\leq C\,\eu^{-\alpha t}$ for some $\alpha > 0$. *(Trigonalize; treat the triangular system from the last row up, or use Dunford: $\eu^{tA} = \eu^{tD}\eu^{tN}$ with $\norm{\eu^{tD}} \leq \eu^{-\alpha' t}$ and $\eu^{tN}$ polynomial in $t$.)*

**Solution of Exercise 16.8.**

Dunford: $A = D + N$ commuting, $D$ [diagonalizable](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-diag) with the same [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen), $N$ nilpotent, so

$$
\eu^{tA} = \eu^{tD}\,\eu^{tN},
\qquad
\eu^{tN} = \sum_{k < n} \frac{t^kN^k}{k!}
\ \text{(a matrix polynomial in } t).
$$

Let $-2\alpha = \max_i \Re\lambda_i < 0$. In a basis diagonalizing $D$, $\vertiii{\eu^{tD}} \leq \eu^{-2\alpha t}$ (entries $\eu^{t\lambda_i}$ of modulus $\eu^{t\Re\lambda_i}$); [norms](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) in different bases differ by constants. Hence

$$
\norm{X(t)} \leq \vertiii{\eu^{tA}}\,\norm{X_0}
\leq C'\,\eu^{-2\alpha t}\,(1 + t)^{n-1}\,\norm{X_0}
\leq C\,\eu^{-\alpha t}\norm{X_0} ,
$$

absorbing the polynomial into one exponential factor ($\eu^{-\alpha t}(1+t)^{n-1} \to 0$, hence bounded).

**Exercise 16.9 ★★★.**

(No escape in finite time for linear growth) Suppose $f$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) with $\norm{f(t, y)} \leq a\norm y + b$ on $\intco{0}{\infty} \times \R^n$, locally [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in $y$. Using Gronwall ([Exercise 16.6](#exo-b2-diffeq-6)) on the integral form, prove that maximal solutions are global (defined on all of $\intco{0}{\infty}$).

**Solution of Exercise 16.9.**

Let $y$ be a maximal solution on $\intco{0}{T}$, $T \leq \infty$, and suppose $T < \infty$. The integral form gives, for $t < T$,

$$
\norm{y(t)} \leq \norm{y_0} + \int_0^t \bigl(a\norm{y(s)} +
b\bigr)\dd s
\leq \bigl(\norm{y_0} + bT\bigr) + a\int_0^t\norm{y(s)}\,\dd s ,
$$

and Gronwall bounds $\norm{y(t)} \leq (\norm{y_0} +
bT)\,\eu^{aT} =: M$ on $\intco{0}{T}$: the solution stays in a [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) ball. Then $y' = f(t, y)$ is bounded near $T$, so $y$ is [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) near $T$ and extends [continuously](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) to $T$ (Cauchy criterion); solving the Cauchy problem at $(T, y(T))$ prolongs $y$ beyond $T$, contradicting maximality. Hence $T = \infty$: no finite-time escape under linear growth.

**Exercise 16.10 ★.**

Solve $y'' - 3y' + 2y = \eu^{t}$: homogeneous solutions, then a particular solution of the form $\alpha t\,\eu^{t}$ *(why does the naive guess $\alpha\eu^t$ fail?)*; general solution and the solution with $y(0) = y'(0) = 0$.

**Solution of Exercise 16.10.**

Characteristic roots of $r^2 - 3r + 2$: $1$ and $2$, so the homogeneous solutions are $a\eu^t + b\eu^{2t}$. The guess $\alpha\eu^t$ fails because $\eu^t$ already solves the homogeneous equation (the root $r = 1$ “resonates” with the right-hand side). With $y = \alpha t\,\eu^t$: $y' = \alpha(1 +
t)\eu^t$, $y'' = \alpha(2 + t)\eu^t$, and

$$
y'' - 3y' + 2y = \alpha\eu^t\bigl(2 + t - 3 - 3t + 2t\bigr)
= -\alpha\,\eu^t :
$$

$\alpha = -1$, $y_p = -t\,\eu^t$. General solution: $y = a\eu^t
+ b\eu^{2t} - t\eu^t$. Initial data $y(0) = y'(0) = 0$: $a + b
= 0$ and $a + 2b - 1 = 0$: $b = 1$, $a = -1$:

$$
y(t) = \eu^{2t} - (1 + t)\,\eu^{t} .
$$

**Exercise 16.11 ★★.**

Compute $\eu^{tA}$ for the Jordan block

$$
A = \begin{pmatrix} \lambda & 1 & 0\\ 0 & \lambda & 1\\
0 & 0 & \lambda\end{pmatrix},
$$

and describe all solutions of $X' = AX$: exponentials times polynomial vectors, with degrees up to $2$. Where does the polynomial degree come from?

**Solution of Exercise 16.11.**

$A = \lambda I + N$ with $N = E_{12} + E_{23}$: $N^2 = E_{13}$, $N^3 = 0$, and $\lambda I$ commutes with $N$:

$$
\eu^{tA} = \eu^{\lambda t}\Bigl(I + tN +
\frac{t^2}{2}N^2\Bigr)
= \eu^{\lambda t}\begin{pmatrix}
1 & t & \frac{t^2}{2}\\
0 & 1 & t\\
0 & 0 & 1
\end{pmatrix}.
$$

Solutions: $X(t) = \eu^{\lambda t}\bigl(X_0 + tNX_0 +
\frac{t^2}2N^2X_0\bigr)$ — each component is $\eu^{\lambda
t}$ times a polynomial of degree $\leq 2$. The degree bound is the nilpotency index minus one: the series of $\eu^{tN}$ truncates at $N^2$.

**Exercise 16.12 ★★★.**

(Periodic forcing, periodic response) Let $A \in
\mathcal{M}_n(\R)$ and $B \colon \R \to \R^n$ [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and $T$-periodic.

1. Show that a solution of $X' = AX + B(t)$ is $T$ -periodic if and only if $X(T) = X(0)$ *(compare $X(\cdot + T)$ and $X$)* .
2. Show that [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $\eu^{TA}$ are the $\eu^{T\lambda}$ , $\lambda \in \operatorname{Sp}A$ *(trigonalize over $\C$)* . Deduce: if no [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $A$ lies in $\frac{2\iu\pi}{T}\Z$ , then $I - \eu^{TA}$ is invertible.
3. Under that hypothesis, prove that the system has exactly one $T$-periodic solution, with $$X(0) = \bigl(I - \eu^{TA}\bigr)^{-1}  \int_0^{T}\eu^{(T-s)A}B(s)\,\dd s .$$ What does the excluded case correspond to, for the harmonic oscillator? (The weekend problem answers: [resonance](#pb-b2-diffeq-1).)

**Solution of Exercise 16.12.**

1. If $X(T) = X(0)$ , then $Y(t) = X(t + T)$ solves $Y' =  AY + B(t + T) = AY + B(t)$ with $Y(0) = X(0)$ : uniqueness ( [Theorem 16.4](#thm-b2-diffeq-linear) ) gives $Y = X$ , i.e. $X$ is $T$ -periodic. The converse is trivial.
2. Trigonalize over $\C$ : $A = PT'P^{-1}$ with $T'$ upper triangular, diagonal $(\lambda_i)$ . Every power of a triangular matrix is triangular with diagonal $(\lambda_i^k)$ , so $\eu^{TA} = P\eu^{TT'}P^{-1}$ is triangular in the same basis with diagonal $(\eu^{T\lambda_i})$ : those are the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) . Then $I - \eu^{TA}$ is invertible iff $\eu^{T\lambda} \neq  1$ for every [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) , i.e. iff $T\lambda \notin  2\iu\pi\Z$ , which is the stated hypothesis.
3. Variation of constants: $X(T) = \eu^{TA}X(0) +  \int_0^T\eu^{(T-s)A}B(s)\dd s$, so $X(T) = X(0)$ reads $$\bigl(I - \eu^{TA}\bigr)X(0) =  \int_0^{T}\eu^{(T-s)A}B(s)\,\dd s ,$$ which has a unique solution $X(0)$ under the invertibility hypothesis: exactly one $T$-periodic solution. For the harmonic oscillator ($\lambda =  \pm\iu\omega$), the excluded case is $\omega T \in  2\pi\Z$: forcing whose period is a multiple of the natural period — [resonance](#pb-b2-diffeq-1), as the weekend problem quantifies.

## 16.6 Problem: Oscillations, resonance, and Sturm’s comparison theorems

**Problem 16.1.**

One equation rules the mechanical and electrical world:

$$
x'' + 2\zeta\omega\,x' + \omega^2 x = F(t),
\qquad \omega > 0,\ \zeta \geq 0 .
$$

This problem studies it completely — via the trace–determinant classification of planar linear systems, the three damping regimes, the steady-state response to periodic forcing with its *resonance* peak and the [resonance](#pb-b2-diffeq-1) catastrophe — then leaves constant coefficients for *Sturm’s separation and comparison theorems*, which control the zeros of solutions of $y'' + q(t)y = 0$ with no formula at all.

**Part I — The trace–determinant plane.** Let $A \in \mathcal M_2(\R)$, $\tau = \operatorname{tr}A$, $\delta
= \det A$, $\Delta = \tau^2 - 4\delta$.

1. Show that the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $A$ are $\frac{\tau\pm\sqrt\Delta}{2}$ and classify: two real [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of opposite signs iff $\delta < 0$ ; real [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of the same sign iff $\delta > 0$ , $\Delta  \geq 0$ (sign of $\tau$ ); nonreal conjugate pair iff $\Delta < 0$ (real part $\frac\tau2$ ).
2. (Saddle, $\delta < 0$ ) With [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\mu < 0 <  \lambda$ and [eigenvectors](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $v_\pm$ , write the general solution and describe the trajectories: two stable and two unstable rays, all other orbits asymptotic to both. Why can no solution but $0$ stay bounded on all of $\R$ ?
3. (Nodes, $\delta > 0$ , $\Delta > 0$ ) For $\mu < \lambda  < 0$ : show every nonzero solution tends to $0$ and that all orbits except those on the fast axis arrive *tangent to the slow eigendirection* *(compare $\eu^{\mu t}$ and $\eu^{\lambda t}$)* .
4. (Spirals and centers, $\Delta < 0$ ) Writing the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\alpha \pm \iu\beta$ , use [Exercise 16.5](#exo-b2-diffeq-5) (after a real change of basis, admitted in that generality or proved for the systems of Part II, which are the ones used below) to describe the orbits: spirals converging for $\alpha =  \frac\tau2 < 0$ , diverging for $\tau > 0$ , closed curves (center) for $\tau = 0$ .
5. (Boundary cases) For a double [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) ( $\Delta =  0$ ): show $\eu^{tA} = \eu^{\lambda t}(I + tN)$ with $N  = A - \lambda I$ nilpotent, and distinguish the star ( $N = 0$ ) from the improper node ( $N \neq 0$ ). Summarize Part I in the trace–determinant picture of this chapter’s figure.

**Part II — The damped oscillator.** Now $F =
0$: $x'' + 2\zeta\omega x' + \omega^2x = 0$, i.e. $X' = AX$ with $A = \begin{pmatrix} 0 & 1\\ -\omega^2 &
-2\zeta\omega\end{pmatrix}$.

6. Compute $\tau, \delta, \Delta$ and place the three regimes in the trace–determinant plane: *underdamped* $0 < \zeta < 1$ (stable spiral), *critically damped* $\zeta = 1$ (double [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) ), *overdamped* $\zeta > 1$ (stable node); $\zeta = 0$ is the center.
7. Solve the three regimes explicitly: $$\zeta < 1:\ \eu^{-\zeta\omega t}\bigl(a\cos\omega_d t  + b\sin\omega_dt\bigr),\ \omega_d =  \omega\sqrt{1-\zeta^2};  \qquad  \zeta = 1:\ (a + bt)\,\eu^{-\omega t};$$ $\zeta > 1$: two real exponentials. Define the pseudo-period $\frac{2\pi}{\omega_d}$ and show that the ratio of successive maxima of $\abs x$ is the constant $\eu^{-2\pi\zeta/\sqrt{1-\zeta^2}}$ (the logarithmic decrement).
8. (The door-closer principle) For $\zeta \geq 1$ the decay rate is governed by the slowest [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\lambda_{\mathrm{slow}} = -\omega\bigl(\zeta -  \sqrt{\zeta^2-1}\bigr)$ . Show that $\abs{\lambda_{\mathrm{slow}}} =  \frac{\omega}{\zeta + \sqrt{\zeta^2 - 1}}$ is a *decreasing* function of $\zeta \geq 1$ : critical damping $\zeta = 1$ gives the fastest non-oscillating return to rest.
9. (Energy) Let $E(t) = \frac12x'^2 +  \frac12\omega^2x^2$ . Prove $E' = -2\zeta\omega\,x'^2  \leq 0$ , and deduce that for $\zeta > 0$ the equation has no nonzero periodic solution *(a period would force $E$ constant, hence $x' \equiv 0$)* .
10. Explain in two sentences why the center $\zeta = 0$ is *structurally fragile* : any $\zeta > 0$ , however small, destroys periodicity — and where that shows in the trace–determinant plane (the center line has empty interior).

**Part III — Forced oscillations and [resonance](#pb-b2-diffeq-1).** Now $F(t) = F\cos(\gamma t)$ with $F, \gamma > 0$.

11. ($\zeta > 0$: the steady state) Seek $x_p =  \Re\bigl(z\,\eu^{\iu\gamma t}\bigr)$: show $$z = \frac{F}{\omega^2 - \gamma^2 +  2\iu\zeta\omega\gamma},  \qquad  A(\gamma) := \abs z = \frac{F}{\sqrt{(\omega^2 -  \gamma^2)^2 + 4\zeta^2\omega^2\gamma^2}} ,$$ and write $x_p = A(\gamma)\cos(\gamma t - \varphi)$ with $\tan\varphi =  \frac{2\zeta\omega\gamma}{\omega^2-\gamma^2}$.
12. Show that *every* solution is $x_p$ plus a transient from Part II, which tends to $0$ : whatever the initial data, the system locks onto the steady state — amplitude $A(\gamma)$ , phase lag $\varphi$ .
13. (The [resonance](#pb-b2-diffeq-1) curve) Maximize $A$: show that $A(\gamma)$ has an interior maximum iff $\zeta <  \frac{1}{\sqrt2}$, at $$\gamma_* = \omega\sqrt{1 - 2\zeta^2},  \qquad  A(\gamma_*) =  \frac{F}{2\zeta\omega^2\sqrt{1-\zeta^2}} ,$$ and that for small $\zeta$ the peak amplifies the static response $A(0) = \frac F{\omega^2}$ by the factor $\approx \frac{1}{2\zeta}$.
14. ($\zeta = 0$, off [resonance](#pb-b2-diffeq-1)) For $\gamma \neq \omega$, show that the solution with $x(0) = x'(0) = 0$ is $$x(t) = \frac{F}{\omega^2 -  \gamma^2}\bigl(\cos\gamma t - \cos\omega t\bigr)  = \frac{2F}{\omega^2-\gamma^2}  \sin\frac{(\omega-\gamma)t}{2}  \sin\frac{(\omega+\gamma)t}{2} :$$ bounded, with *beats* — a fast oscillation under a slow envelope — when $\gamma$ is close to $\omega$.
15. ( $\zeta = 0$ , [resonance](#pb-b2-diffeq-1) ) For $\gamma = \omega$ , show that $x_p(t) = \frac{F}{2\omega}\,t\sin(\omega t)$ is a solution, and recover it as the limit of question 14 as $\gamma \to \omega$ : the amplitude grows linearly forever — the [resonance](#pb-b2-diffeq-1) catastrophe.
16. (Fourier link) A general periodic forcing decomposes into harmonics (the Fourier chapter); by linearity the steady state is the sum of the harmonic responses. For an undamped oscillator of frequency $\omega = 3$ forced by the square wave of [Exercise 14.1](https://one-course.com/books/math/4/en/chapter/14-fourier-series#exo-b2-fourier-1) -type (harmonics at all odd integers), which harmonic resonates? One sentence on why engineers fear square waves.

**Part IV — Sturm’s theorems.** Consider $y''
+ q(t)\,y = 0$ on an interval $I$, $q$ [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity). (Any equation $y'' + ay' + by = 0$ reduces to this normal form by the substitution $y = u\exp\bigl(-\frac12\int a\bigr)$; question 21 shows a variant of the trick in action.)

17. For two solutions $y_1, y_2$ , show that the [wronskian](#def-b2-diffeq-wronskian) $W = y_1y_2' - y_1'y_2$ is *constant* , zero iff the solutions are proportional; and that a nonzero solution has only *simple, isolated* zeros.
18. (Sturm separation) Let $y_1, y_2$ be independent solutions and $a < b$ two consecutive zeros of $y_1$ . Prove that $y_2$ vanishes exactly once in $\intoo{a}{b}$ *(evaluate the constant $W$ at $a$ and $b$: $W = y_1'y_2$ there, and $y_1'(a)$, $y_1'(b)$ have opposite signs)* : zeros of independent solutions interlace.
19. (Sturm comparison) Let $q_1 \leq q_2$ on $I$ , $y \neq  0$ with $y'' + q_1y = 0$ , $z \neq 0$ with $z'' + q_2z  = 0$ , and $a < b$ consecutive zeros of $y$ . Show that $z$ vanishes in $\intcc{a}{b}$ — strictly inside if $q_1 < q_2$ somewhere on $\intoo ab$ *(if $z \neq  0$ on $\intoo ab$, study $(yz' - y'z)' = (q_1 -  q_2)yz$ with fixed signs for $y, z$ and compare the boundary values)* .
20. Deduce the *spacing bounds*: if $0 < m^2 \leq  q(t) \leq M^2$ on $I$, then any two consecutive zeros $a < b$ of a nonzero solution of $y'' + qy = 0$ satisfy $$\frac{\pi}{M} \;\leq\; b - a \;\leq\; \frac{\pi}{m}$$ *(compare with $u'' + M^2u = 0$ and $u'' + m^2u  = 0$, whose zeros are spaced $\frac\pi M$ and $\frac\pi m$)*. Check on the harmonic oscillator.
21. Transform $ty'' + 2y' + ty = 0$ ( [Exercise 16.7](#exo-b2-diffeq-7) ) by $u = ty$ into $u'' + u =  0$ , recover its solutions $\frac{\sin t}t$ , $\frac{\cos t}{t}$ instantly, and conclude that the zeros of every nonzero solution are spaced exactly $\pi$ : Sturm’s world view — zeros are controlled by the coefficient $q$ , formulas or not.

**Part V — Duhamel and the boundedness frontier.**

22. (Duhamel for the oscillator) Show that for [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $F$, the solution of $x'' + \omega^2x =  F(t)$ with $x(0) = x'(0) = 0$ is $$x(t) = \frac1\omega\int_0^t\sin\bigl(\omega(t -  s)\bigr)F(s)\,\dd s ,$$ and re-derive the resonant solution of question 15 from it with $F(s) = F\cos(\omega s)$ *(product-to-sum)*.
23. ( $\zeta > 0$ : bounded input, bounded output) Show that for $\zeta > 0$ and *any* bounded [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $F$ , every solution of the damped equation is bounded on $\intco{0}{\infty}$ *(variation of constants plus the exponential decay $\vertiii{\eu^{tA}} \leq  C\eu^{-\alpha t}$ of [Exercise 16.8](#exo-b2-diffeq-8))* .
24. ( $\zeta = 0$ ) Show that with no damping, bounded periodic forcing keeps all solutions bounded *except* exactly at [resonance](#pb-b2-diffeq-1) ( $\gamma =  \omega$ , question 15 versus question 14): damping is what turns the boundedness frontier into uniform stability.
25. Synthesis. In one sentence each: (i) how the trace–determinant plane organizes Parts I–II and where forcing (Part III) leaves it; (ii) the physical meaning of $\gamma_*$ , $A(\gamma_*)$ and the factor $\frac1{2\zeta}$ ; (iii) what Sturm’s theorems say that explicit formulas cannot; (iv) which two results of this problem the rest of the book will quietly reuse (Liouville-constant [wronskians](#def-b2-diffeq-wronskian) ; bounded-input stability).

**Solution of Problem 16.1.**

**1.** The [characteristic polynomial](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-charpoly) is $\lambda^2 -
\tau\lambda + \delta$, with roots $\frac{\tau\pm\sqrt\Delta}{2}$. If $\delta < 0$ then $\Delta =
\tau^2 - 4\delta > 0$ and the two real roots have product $\delta < 0$: opposite signs. If $\delta > 0$ and $\Delta \geq
0$: real roots of product $> 0$ and sum $\tau$: both of the sign of $\tau$. If $\Delta < 0$: conjugate pair $\alpha \pm
\iu\beta$ with $\alpha = \frac\tau2$, $\beta =
\frac{\sqrt{-\Delta}}2$.

**2.** $X(t) = a\,\eu^{\mu t}v_- + b\,\eu^{\lambda t}v_+$. Orbits with $a = 0$ (resp. $b = 0$) run along the unstable (resp. stable) eigenline; all others have $\norm X \to \infty$ in both time directions, asymptotic to $\R v_+$ as $t \to
+\infty$ and to $\R v_-$ as $t \to -\infty$: the saddle picture. Boundedness on all of $\R$ forces $b = 0$ (else explosion at $+\infty$) and $a = 0$ (at $-\infty$): only the origin.

**3.** With $\mu < \lambda < 0$, both exponentials decay: $X(t) \to 0$. If $b \neq 0$, factor $\eu^{\lambda t}$:

$$
X(t) = \eu^{\lambda t}\bigl(b\,v_\lambda + a\,\eu^{(\mu -
\lambda)t}v_\mu\bigr),
\qquad \eu^{(\mu-\lambda)t} \to 0 :
$$

the direction of $X(t)$ tends to $\R v_\lambda$, the *slow* eigendirection — all orbits but the fast axis arrive tangent to it (the right panel of the chapter’s phase portraits).

**4.** In the basis where $A = \begin{pmatrix} \alpha &
-\beta\\ \beta & \alpha\end{pmatrix}$ ([Exercise 16.5](#exo-b2-diffeq-5); for the oscillator systems of Part II this form is reached by an explicit real change of basis), the solution is $\eu^{\alpha t}$ times a rotation of angle $\beta
t$: logarithmic spirals, contracting when $\alpha = \frac\tau2
< 0$, expanding when $\tau > 0$, and closed curves (ellipses in the original coordinates) when $\tau = 0$: the center.

**5.** $\Delta = 0$ gives the double [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\lambda =
\frac\tau2$; by Cayley–Hamilton ([Theorem 3.21](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#thm-b2-reduction-cayleyhamilton)), $(A - \lambda I)^2 =
0$, so $N = A - \lambda I$ is nilpotent, commutes with $\lambda I$, and $\eu^{tA} = \eu^{\lambda t}(I + tN)$. If $N =
0$: $A = \lambda I$, all rays are orbits (star node). If $N
\neq 0$: $X(t) = \eu^{\lambda t}(X_0 + tNX_0)$, and for $NX_0
\neq 0$ the direction converges to the single eigendirection $\operatorname{im}N$: improper node. This completes the trace–determinant picture.

**6.** $\tau = -2\zeta\omega$, $\delta = \omega^2 > 0$, $\Delta = 4\omega^2(\zeta^2 - 1)$. So: $0 < \zeta < 1$ gives $\Delta < 0$, $\tau < 0$: stable spiral; $\zeta = 1$: $\Delta =
0$: degenerate stable node; $\zeta > 1$: $\Delta > 0$, $\tau <
0$, $\delta > 0$: stable node; $\zeta = 0$: $\tau = 0$, $\delta > 0$: center. A vertical journey in the plane at $\delta = \omega^2$.

**7.** Roots $r = -\zeta\omega \pm
\omega\sqrt{\zeta^2-1}$. For $\zeta < 1$: $r = -\zeta\omega
\pm \iu\omega_d$, $\omega_d = \omega\sqrt{1-\zeta^2}$:

$$
x(t) = \eu^{-\zeta\omega t}\bigl(a\cos\omega_dt +
b\sin\omega_dt\bigr)
= R\,\eu^{-\zeta\omega t}\cos(\omega_dt - \varphi) .
$$

For $\zeta = 1$: $x = (a + bt)\eu^{-\omega t}$. For $\zeta >
1$: $x = a\eu^{r_-t} + b\eu^{r_+t}$, both rates negative. Successive maxima of $\abs x$ in the underdamped case occur at times separated by the pseudo-period $\frac{2\pi}{\omega_d}$ (same phase of the cosine), and their ratio is $\eu^{-\zeta\omega\cdot2\pi/\omega_d} =
\eu^{-2\pi\zeta/\sqrt{1-\zeta^2}}$: the logarithmic decrement, a damping meter readable on an oscilloscope.

**8.** Rationalizing,

$$
\abs{\lambda_{\mathrm{slow}}} = \omega\bigl(\zeta -
\sqrt{\zeta^2-1}\bigr)
= \frac{\omega}{\zeta + \sqrt{\zeta^2 - 1}} ,
$$

whose denominator increases with $\zeta \geq 1$: the decay rate is largest at $\zeta = 1$, where it equals $\omega$. An overdamped door closes without slamming but *slowly*; critical damping is the engineer’s optimum.

**9.** $E' = x'x'' + \omega^2xx' =
x'\bigl(-2\zeta\omega x' - \omega^2x\bigr) + \omega^2xx' =
-2\zeta\omega\,x'^2 \leq 0$. If $x$ were periodic and nonconstant, $E$ would be periodic and nonincreasing, hence constant, forcing $x' \equiv 0$: $x$ constant, and then $\omega^2x = 0$: $x \equiv 0$. So for $\zeta > 0$ the only periodic solution is rest: damping kills every [cycle](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-sn).

**10.** The center $\zeta = 0$ lives on the line $\tau =
0$ of the trace–determinant plane — a set with empty interior: an arbitrarily small perturbation of the matrix (any physical damping) moves $\tau$ off zero and turns the closed orbits into spirals. Periodicity of the undamped oscillator is thus a razor’s-edge phenomenon, not a robust one.

**11.** Substituting $x_p = \Re(z\eu^{\iu\gamma t})$ into the equation:

$$
\bigl(-\gamma^2 + 2\iu\zeta\omega\gamma +
\omega^2\bigr)z = F
\quad\Longrightarrow\quad
z = \frac{F}{\omega^2 - \gamma^2 + 2\iu\zeta\omega\gamma},
$$

so $x_p = \abs z\cos(\gamma t - \varphi)$ with $\varphi =
\arg(\omega^2 - \gamma^2 + 2\iu\zeta\omega\gamma)$, i.e. $\tan\varphi = \frac{2\zeta\omega\gamma}{\omega^2 -
\gamma^2}$, and the stated $A(\gamma) = \abs z$.

**12.** The difference of two solutions solves the homogeneous equation, which for $\zeta > 0$ decays to $0$ (question 7): every solution equals $x_p$ plus a transient vanishing at infinity. The steady state is a global attractor: initial conditions are forgotten, only $A(\gamma)$ and the phase lag $\varphi$ remain.

**13.** Minimize $g(u) = (\omega^2 - u)^2 +
4\zeta^2\omega^2u$ over $u = \gamma^2 \geq 0$: $g'(u) =
-2(\omega^2 - u) + 4\zeta^2\omega^2 = 0$ at $u = \omega^2(1 -
2\zeta^2)$, interior iff $\zeta < \frac{1}{\sqrt2}$. There

$$
g(u_*) = 4\zeta^4\omega^4 + 4\zeta^2\omega^4(1 - 2\zeta^2)
= 4\zeta^2\omega^4(1 - \zeta^2),
\qquad
A(\gamma_*) = \frac{F}{2\zeta\omega^2\sqrt{1 - \zeta^2}} .
$$

Against the static response $A(0) = \frac{F}{\omega^2}$: amplification $\frac{1}{2\zeta\sqrt{1-\zeta^2}} \approx
\frac{1}{2\zeta}$ for small $\zeta$ — a lightly damped system near $\gamma_* \approx \omega$ multiplies the input a hundredfold when $\zeta = 0.005$.

**14.** The stated $x$ satisfies $x(0) = x'(0) = 0$ and

$$
x'' + \omega^2 x = \frac{F(\omega^2 -
\gamma^2)\cos\gamma t}{\omega^2 - \gamma^2} = F\cos\gamma t
$$

(the $\cos\omega t$ parts cancel). The product form follows from $\cos p - \cos q = 2\sin\frac{q+p}{2}\sin\frac{q-p}2$ with $p = \gamma t$, $q = \omega t$. For $\gamma$ close to $\omega$, the factor $\sin\frac{(\omega-\gamma)t}2$ is a slow envelope modulating the fast oscillation $\sin\frac{(\omega+\gamma)t}2$: beats, with amplitude $\frac{2F}{\abs{\omega^2-\gamma^2}}$ — large, but bounded.

**15.** For $x_p = \frac{F}{2\omega}t\sin\omega t$:

$$
x_p'' = \frac{F}{2\omega}\bigl(2\omega\cos\omega t -
\omega^2t\sin\omega t\bigr)
= F\cos\omega t - \omega^2x_p :
$$

a solution. And at fixed $t$, letting $\gamma \to \omega$ in question 14:

$$
\frac{2F\sin\frac{(\omega-\gamma)t}2
\sin\frac{(\omega+\gamma)t}{2}}
{(\omega-\gamma)(\omega+\gamma)}
\longrightarrow
\frac{2F\cdot\frac{(\omega-\gamma)t}2\big/(\omega-\gamma)
\cdot\sin\omega t}{2\omega}
= \frac{F\,t\sin\omega t}{2\omega} .
$$

The amplitude grows linearly without bound: the [resonance](#pb-b2-diffeq-1) catastrophe — the reason soldiers break step on bridges.

**16.** The square wave carries harmonics at every odd frequency $n = 1, 3, 5, \dots$; by linearity, each harmonic $n$ is amplified by the oscillator’s response at $\gamma = n$. For $\omega = 3$ the third harmonic hits [resonance](#pb-b2-diffeq-1) exactly. Engineers fear square (and sawtooth) inputs because they excite *all* odd harmonics at once: whatever the natural frequency of the structure, some harmonic is waiting for it.

**17.** $W' = y_1y_2'' - y_1''y_2 = -qy_1y_2 + qy_1y_2 =
0$: $W$ is constant (Liouville with a trace-zero companion matrix). $W = 0$ at one point makes the initial data of $y_2$ proportional to those of $y_1$, hence $y_2$ proportional to $y_1$ (uniqueness); $W \neq 0$ iff independent. If $y(t_0) =
y'(t_0) = 0$ then $y \equiv 0$ (uniqueness): a nonzero solution has simple zeros, and a simple zero is isolated ($y'$ of fixed sign nearby).

**18.** Between consecutive zeros $a < b$, $y_1$ keeps one sign, say $y_1 > 0$ on $\intoo ab$: then $y_1'(a) > 0$ and $y_1'(b) < 0$ (simple zeros). Evaluating the constant $W =
y_1y_2' - y_1'y_2$ at $a$ and $b$:

$$
W = -y_1'(a)\,y_2(a) = -y_1'(b)\,y_2(b) ,
$$

so $y_2(a)$ and $y_2(b)$ have opposite signs ($W \neq 0$ forbids either to vanish): $y_2$ vanishes in $\intoo ab$ (intermediate values). It cannot vanish twice there: two zeros of $y_2$ would flank a zero of $y_1$ by the same argument with the roles exchanged, contradicting consecutiveness: exactly one zero — interlacing.

**19.** Suppose $z$ has no zero in $\intoo ab$; replacing $y, z$ by their negatives, assume $y > 0$ and $z > 0$ on $\intoo ab$. Set $\varphi = yz' - y'z$: $\varphi' = yz'' -
y''z = (q_1 - q_2)\,yz \leq 0$ on $\intoo ab$: $\varphi$ is nonincreasing. But $\varphi(a) = -y'(a)z(a) \leq 0$ (as $y'(a) > 0$, $z(a) \geq 0$) and $\varphi(b) = -y'(b)z(b) \geq
0$ (as $y'(b) < 0$, $z(b) \geq 0$): a nonincreasing function running from $\leq 0$ to $\geq 0$ vanishes identically, so $(q_1 - q_2)yz \equiv 0$ on $\intoo ab$. If $q_1 < q_2$ somewhere in $\intoo ab$, this is absurd ($y, z > 0$ there): $z$ must vanish strictly inside. In general ($q_1 \leq q_2$), either $z$ vanishes in $\intoo ab$, or $\varphi \equiv 0$ forces $z$ proportional to $y$, which vanishes at $a$ and $b$: in all cases $z$ has a zero in $\intcc ab$.

**20.** Upper bound: compare $y$ (coefficient $q \geq
m^2$) with $u(t) = \sin(m(t - a))$ (coefficient $m^2 \leq q$, so $y$ plays the role of $z$ in question 19): if $y$ had no zero in $\intoc{a}{a + \pi/m}$, the zeros $a$ and $a +
\frac\pi m$ of $u$ would be consecutive with $y \neq 0$ between them, contradicting question 19: consecutive zeros of $y$ are at distance $\leq \frac\pi m$. Lower bound: if two consecutive zeros $a < b$ of $y$ had $b - a < \frac\pi M$, then $z(t) = \sin(M(t-a))$ (coefficient $M^2 \geq q$) would have to vanish in $\intcc ab \subset \intoo{a}{a + \pi/M}
\cup\{a\}$, where its only zero is $a$ itself — but question 19 applied on $\intoo{a}{b}$ with strictness at the endpoints gives a zero in $\intcc ab$, and $z > 0$ on $\intoc ab$: contradiction. Hence $\frac\pi M \leq b - a \leq
\frac\pi m$; for $q \equiv \omega^2$ both bounds collapse to the exact spacing $\frac\pi\omega$ of the harmonic oscillator.

**21.** With $u = ty$: $u'' = ty'' + 2y'$, so $ty'' + 2y'
+ ty = u'' + u = 0$: $u = A\sin t + B\cos t = R\sin(t +
\varphi)$, and $y = \frac{u}{t}$ recovers $\frac{\sin t}t$ and $\frac{\cos t}t$ ([Exercise 16.7](#exo-b2-diffeq-7)) with no lowering of order. The zeros of any nonzero solution are those of $R\sin(t + \varphi)$: spaced exactly $\pi$ — Sturm’s philosophy in action: the coefficient $q \equiv 1$ dictates the zeros, formula or no formula.

**22.** Set $x(t) =
\frac1\omega\int_0^t\sin(\omega(t-s))F(s)\dd s$. Then $x(0) =
0$;

$$
x'(t) = \frac1\omega\sin(0)F(t) +
\int_0^t\cos(\omega(t-s))F(s)\dd s
= \int_0^t\cos(\omega(t-s))F(s)\dd s ,
$$

so $x'(0) = 0$; and $x''(t) = F(t) -
\omega\int_0^t\sin(\omega(t-s))F(s)\dd s = F(t) - \omega^2x(t)$ (differentiation of a [parameter integral](https://one-course.com/books/math/4/en/chapter/9-integration#thm-b2-integration-continuity) with variable limit, as in the integration chapter). With $F(s) = F\cos\omega s$, product-to-sum gives

$$
\int_0^t\sin(\omega(t-s))\cos(\omega s)\dd s
= \frac12\int_0^t\bigl(\sin\omega t + \sin(\omega t -
2\omega s)\bigr)\dd s
= \frac{t}{2}\sin\omega t ,
$$

(the second piece integrates to zero), so $x =
\frac{F}{2\omega}t\sin\omega t$: question 15 again, from Duhamel.

**23.** In system form $X' = AX + (0, F(t))^{\mathsf T}$ with $\operatorname{Sp}A$ of negative real parts ($\zeta >
0$): variation of constants and [Exercise 16.8](#exo-b2-diffeq-8) ($\vertiii{\eu^{tA}} \leq C\eu^{-\alpha t}$) give

$$
\norm{X(t)} \leq C\eu^{-\alpha t}\norm{X_0}
+ \int_0^t C\eu^{-\alpha(t-s)}\norm{F}_\infty\dd s
\leq C\norm{X_0} + \frac{C\norm F_\infty}{\alpha} :
$$

bounded input, bounded output — uniformly in the initial data after the transient.

**24.** For $\zeta = 0$ and $\gamma \neq \omega$, question 14’s solution is bounded, and adding any homogeneous solution (bounded: the center’s orbits are circles) keeps it bounded; at $\gamma = \omega$, question 15 grows linearly. So for the undamped oscillator, boundedness under periodic forcing fails at exactly one frequency — [resonance](#pb-b2-diffeq-1) — while question 23 shows any positive damping restores boundedness for *all* bounded inputs.

**25.** (i) The trace–determinant plane classifies all autonomous planar linear dynamics, and Part II’s oscillator walks a vertical line of it; forcing leaves the plane (non-autonomous), and Duhamel takes over. (ii) $\gamma_*$ is the frequency the system prefers, $A(\gamma_*)$ the price of exciting it, and $\frac{1}{2\zeta}$ the amplification factor — the [resonance](#pb-b2-diffeq-1) sharpness engineers call the quality factor. (iii) Sturm’s theorems read oscillation off the sign and size of $q$ alone: they govern equations (Bessel, Schrödinger) whose solutions have no elementary formulas. (iv) Constant [wronskians](#def-b2-diffeq-wronskian) (question 17, via Liouville) and bounded-input stability (question 23) are reused silently whenever the book meets variable-coefficient equations or perturbed systems.