---
title: "Step Functions, Discontinuous Forcing, Impulses, and Convolution"
module: The Laplace Transform
moduleNumber: 5
lessonNumber: 2
order: 502
summary: >
  The Heaviside step function and the second shifting theorem transform switches
  and discontinuous forcing into exponential factors on the transform. The Dirac
  delta idealizes an instantaneous impulse and transforms to a pure exponential.
  The convolution theorem inverts a product of transforms, writes the forced
  response as the impulse response convolved with the input, and solves Abel's
  tautochrone by transform.
topics: [The Laplace Transform]
sources:
  - book: Boyce
    ref: "Ch. 6 — §6.3 Step Functions; §6.4 Differential Equations with Discontinuous Forcing Functions; §6.5 Impulse Functions; §6.6 The Convolution Integral"
  - book: Simmons
    ref: "Ch. 9 — §52 Convolutions and Abel's Mechanical Problem; §53 More about Convolutions; The Unit Step and Impulse Functions"
---

The [transform-and-invert method](/differential-equations/laplace/laplace-definition-ivps)
handles smooth forcing directly. Harder inputs — a voltage switched on at a fixed
time, a force that jumps, a hammer blow delivered in an instant — get stitched
together interval by interval under classical methods, but transform to single
algebraic expressions. Three tools carry this: the step function for switching,
the delta function for impulses, and convolution for products of transforms.

## The unit step function

> **Definition (Unit step / Heaviside function).** For $c \ge 0$,
> $$
> u_c(t) = \begin{cases} 0, & t < c, \\ 1, & t \ge c. \end{cases}
> $$
> It models a switch turned **on** at time $c$; $1 - u_c(t)$ is a switch turned
> **off** at $c$. Its value at the jump is irrelevant to every integral.

Steps add up to any piecewise-constant function. A jump of height $h$ at time
$c$ contributes $h\, u_c(t)$, so reading the jumps off a graph reconstructs the
function. For instance, the function equal to $2$ on $[0,4)$, $5$ on $[4,7)$,
$-1$ on $[7,9)$, and $1$ thereafter has jumps $+3, -6, +2$ at $t = 4, 7, 9$:

$$
f(t) = 2 + 3\,u_4(t) - 6\,u_7(t) + 2\,u_9(t).
$$

The transform of a step is a decaying exponential over $s$:

$$
\mathcal{L}\{u_c(t)\} = \int_c^\infty e^{-st}\, \d t = \frac{e^{-cs}}{s},
  \qquad s > 0,
$$

which reduces to $\mathcal{L}\{1\} = 1/s$ when $c = 0$.

$$
% caption: The step $u_c(t)$ switches on at $c$; the difference $u_a(t)-u_b(t)$
% is a unit window, on over $[a,b)$ and off elsewhere.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % left panel: single step
  \begin{scope}
    \draw[->, black] (-0.2,0) -- (3.4,0) node[right, black!70] {$t$};
    \draw[->, black] (0,-0.2) -- (0,2.1) node[above, black!70] {$y$};
    \draw[acc, thick] (0,0) -- (1.5,0);
    \draw[acc, thick] (1.5,1.3) -- (3.1,1.3);
    \draw[dashed] (1.5,0) -- (1.5,1.3);
    \fill (1.5,1.3) circle (1.6pt);
    \draw[black] (0,1.3) -- (0.1,1.3);
    \node[black!70, anchor=east] at (0,1.3) {$1$};
    \node[black!70, anchor=north] at (1.5,-0.05) {$c$};
    \node[acc, anchor=south] at (2.3,1.35) {$u_c(t)$};
  \end{scope}
  % right panel: window
  \begin{scope}[xshift=4.6cm]
    \draw[->, black] (-0.2,0) -- (3.8,0) node[right, black!70] {$t$};
    \draw[->, black] (0,-0.2) -- (0,2.1) node[above, black!70] {$y$};
    \draw[acc, thick] (0,0) -- (1.0,0);
    \draw[acc, thick] (1.0,1.3) -- (2.6,1.3);
    \draw[acc, thick] (2.6,0) -- (3.5,0);
    \draw[dashed] (1.0,0) -- (1.0,1.3);
    \draw[dashed] (2.6,0) -- (2.6,1.3);
    \fill (1.0,1.3) circle (1.6pt);
    \node[black!70, anchor=north] at (1.0,-0.05) {$a$};
    \node[black!70, anchor=north] at (2.6,-0.05) {$b$};
    \node[acc, anchor=south] at (1.8,1.35) {$u_a(t)$-$u_b(t)$};
  \end{scope}
\end{tikzpicture}
$$

## The second shifting theorem

Translating a function to the right and switching it on at the translation point
multiplies its transform by an exponential.

> **Theorem (Second shifting / $t$-shift).** If $F(s) = \mathcal{L}\{f(t)\}$
> exists for $s > a \ge 0$ and $c > 0$, then
> $$
> \mathcal{L}\{u_c(t)\, f(t - c)\} = e^{-cs} F(s),
> $$
> and conversely $u_c(t)\, f(t-c) = \mathcal{L}^{-1}\{e^{-cs} F(s)\}$.

The delayed-and-switched function $u_c(t) f(t-c)$ is $f$ shifted a distance $c$
in the positive $t$ direction and held at zero before $c$.[^boyce-shift]
Substituting $\sigma = t - c$ into the defining integral produces the factor
$e^{-cs}$. Whereas the [first shifting rule](/differential-equations/laplace/laplace-definition-ivps)
translated the transform in $s$, this one translates the function in $t$ and
attaches $e^{-cs}$ to the transform.

$$
% caption: The second shifting theorem: $u_c(t)\,f(t-c)$ is the graph of $f$
% slid right by $c$ and zero before $c$; its transform is $e^{-cs}F(s)$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \begin{scope}
    \draw[->, black] (-0.2,0) -- (3.4,0) node[right, black!70] {$t$};
    \draw[->, black] (0,-0.2) -- (0,2.1) node[above, black!70] {$y$};
    \draw[thick] plot[smooth, domain=0:3.0, samples=40] (\x, {1.1*sin(60*\x) + 0.3*\x});
    \node[anchor=south] at (1.8,1.8) {$y = f(t)$};
  \end{scope}
  \begin{scope}[xshift=4.6cm]
    \draw[->, black] (-0.2,0) -- (4.4,0) node[right, black!70] {$t$};
    \draw[->, black] (0,-0.2) -- (0,2.1) node[above, black!70] {$y$};
    \draw[acc, thick] (0,0) -- (1.3,0);
    \draw[acc, thick] plot[smooth, domain=0:3.0, samples=40] ({\x+1.3}, {1.1*sin(60*\x) + 0.3*\x});
    \draw[black, dashed] (1.3,0) -- (1.3,1.7);
    \node[black!70, anchor=north] at (1.3,-0.05) {$c$};
    \node[acc, anchor=south] at (2.9,1.8) {$y = u_c(t) f(t$-$c)$};
  \end{scope}
\end{tikzpicture}
$$

Read backward, the rule inverts transforms carrying an $e^{-cs}$ factor. From
$\mathcal{L}^{-1}\{1/s^2\} = t$,

$$
\mathcal{L}^{-1}\!\left\{\frac{1 - e^{-2s}}{s^2}\right\}
  = t - u_2(t)\,(t - 2)
  = \begin{cases} t, & 0 \le t < 2, \\ 2, & t \ge 2, \end{cases}
$$

a ramp that saturates once the delayed term switches on.

## Discontinuous forcing

With the two shifting rules, a jump in the forcing function transforms as
cleanly as a smooth term.

> **Worked example.** A damped oscillator receives a unit pulse on for
> $5 \le t < 20$:
> $$
> 2y'' + y' + 2y = u_5(t) - u_{20}(t), \qquad y(0) = 0,\ y'(0) = 0.
> $$
> This is the charge on a capacitor in an [electric circuit](/differential-equations/first-order/modeling-first-order)
> given a voltage pulse, or a spring–mass system pushed by a finite-duration
> force. Transform, using $\mathcal{L}\{u_c\} = e^{-cs}/s$ and zero initial data:
> $$
> Y(s) = \frac{e^{-5s} - e^{-20s}}{s(2s^2 + s + 2)}
>   = \bigl(e^{-5s} - e^{-20s}\bigr) H(s),
>   \qquad H(s) = \frac{1}{s(2s^2 + s + 2)}.
> $$
> With $h(t) = \mathcal{L}^{-1}\{H(s)\}$, the second shifting theorem inverts each
> exponential factor as a delayed copy of $h$:
> $$
> y(t) = u_5(t)\, h(t - 5) - u_{20}(t)\, h(t - 20).
> $$
> Expand $H$ in partial fractions, $H(s) = a/s + (bs+c)/(2s^2+s+2)$; clearing
> denominators and matching gives $a = \tfrac12$, $b = -1$, $c = -\tfrac12$.
> Complete the square, $2s^2 + s + 2 = 2\bigl((s+\tfrac14)^2 + \tfrac{15}{16}\bigr)$,
> so the quadratic term inverts through the $e^{at}\sin/\cos$ rows:
> $$
> h(t) = \frac{1}{2} - \frac{1}{2}e^{-t/4}\cos\!\left(\frac{\sqrt{15}}{4}t\right)
>   - \frac{1}{2\sqrt{15}}\,e^{-t/4}\sin\!\left(\frac{\sqrt{15}}{4}t\right).
> $$

The solution runs in three regimes: at rest until $t = 5$; oscillating about the
forced level while decaying toward it for $5 \le t < 20$; and relaxing back to
$0$ after the forcing vanishes at $t = 20$.[^boyce-disc]

$$
% caption: Response of the pulse-forced oscillator from rest: flat until $t=5$,
% a decaying oscillation about the forced level while the pulse is on, then
% relaxation toward zero after the pulse switches off at $t=20$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.2,0) -- (7.4,0) node[right, black!70] {$t$};
  \draw[->, black] (0,-0.2) -- (0,2.0) node[above, black!70] {$y$};
  \draw[black, dashed] (1.1,0.9) -- (4.4,0.9);
  \node[black, anchor=east] at (-0.05,0.9) {$\tfrac12$};
  \draw[acc, thick] (0,0) -- (1.1,0);
  \draw[acc, thick] plot[smooth, domain=5:20, samples=140]
    ({\x*0.22}, {1.8*(0.5 - 0.5*exp(-(\x-5)/4)*cos(55.47*(\x-5)) - 0.1291*exp(-(\x-5)/4)*sin(55.47*(\x-5)))});
  \draw[acc, thick] plot[smooth, domain=20:32, samples=110]
    ({\x*0.22}, {1.8*((0.5 - 0.5*exp(-(\x-5)/4)*cos(55.47*(\x-5)) - 0.1291*exp(-(\x-5)/4)*sin(55.47*(\x-5))) - (0.5 - 0.5*exp(-(\x-20)/4)*cos(55.47*(\x-20)) - 0.1291*exp(-(\x-20)/4)*sin(55.47*(\x-20))))});
  \draw[black, dotted] (1.1,0) -- (1.1,0.9);
  \draw[black, dotted] (4.4,0) -- (4.4,0.95);
  \node[black!70, anchor=north] at (1.1,-0.05) {$5$};
  \node[black!70, anchor=north] at (4.4,-0.05) {$20$};
\end{tikzpicture}
$$

The regularity of $y$ across the switching points is the general lesson. For any
equation $y'' + p(t)y' + q(t)y = g(t)$ with $p, q$ continuous and $g$ only
piecewise continuous, the solution $y$ and its first derivative $y'$ stay
continuous, while $y''$ inherits the jumps of $g$: the jump in the highest-order
term $y''$ exactly balances the jump in $g$. Integrating a jump twice smooths it,
so the discontinuity climbs to the top derivative and no higher.

## The impulse function

An impulse is a force of large magnitude acting over a vanishingly short time,
with a finite total effect. Model it by a pulse of width $2\tau$ and height
$1/(2\tau)$, centered at $t_0$, so that its integral — the **impulse** — equals
$1$ for every $\tau$. Narrowing the pulse while preserving unit area defines the
idealized limit.

$$
% caption: Narrowing pulses of unit area (each rectangle integrates to $1$)
% converge to the delta function, drawn as a spike carrying unit impulse at $t_0$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.2,0) -- (5.6,0) node[right, black!70] {$t$};
  \draw[->, black] (0,-0.2) -- (0,3.1) node[above, black!70] {$y$};
  % three nested pulses of area 1, centered at t0 = 2.4
  \draw[black] (1.6,0) -- (1.6,0.55) -- (3.2,0.55) -- (3.2,0);
  \draw[black] (1.95,0) -- (1.95,1.1) -- (2.85,1.1) -- (2.85,0);
  \draw[thick] (2.2,0) -- (2.2,2.2) -- (2.6,2.2) -- (2.6,0);
  % idealized delta spike
  \draw[->, acc, very thick] (4.4,0) -- (4.4,2.7);
  \node[acc, anchor=south] at (4.4,2.7) {unit impulse};
  \node[black!70, anchor=north] at (2.4,-0.05) {$t_0$};
  \node[black!70, anchor=north] at (4.4,-0.05) {$t_0$};
\end{tikzpicture}
$$

> **Definition (Dirac delta function).** The unit impulse at $t_0$ is the
> generalized function $\delta(t - t_0)$ with
> $$
> \delta(t - t_0) = 0 \text{ for } t \ne t_0,
> \qquad \int_{-\infty}^{\infty} \delta(t - t_0)\, \d t = 1.
> $$
> No ordinary function meets both conditions; $\delta$ is a **generalized
> function**, defined by how it acts inside an integral.

The **sifting property** follows from the mean
value theorem applied to the narrowing pulse: against any continuous $f$,

$$
\int_{-\infty}^{\infty} \delta(t - t_0)\, f(t)\, \d t = f(t_0),
$$

so integrating against $\delta$ samples $f$ at the impulse location. Taking
$f(t) = e^{-st}$ gives the transform,

$$
\mathcal{L}\{\delta(t - t_0)\} = e^{-st_0}, \qquad t_0 > 0,
$$

which becomes $\mathcal{L}\{\delta(t)\} = 1$ as $t_0 \to 0^+$. The delta function
violates the exponential-order hypothesis of the existence theorem, so this
transform is defined by the limiting process rather than the raw integral.[^boyce-imp]

> **Worked example.** Drive the same oscillator with a unit impulse at $t = 5$:
> $$
> 2y'' + y' + 2y = \delta(t - 5), \qquad y(0) = 0,\ y'(0) = 0.
> $$
> Since $\mathcal{L}\{\delta(t-5)\} = e^{-5s}$, transforming gives
> $(2s^2 + s + 2)Y(s) = e^{-5s}$, so $Y(s) = e^{-5s}/(2s^2+s+2)$. The quadratic is
> the one from the pulse problem; inverting through the shifted $e^{at}\sin$ row,
> $$
> y(t) = \frac{2}{\sqrt{15}}\, u_5(t)\, e^{-(t-5)/4}
>   \sin\!\left(\frac{\sqrt{15}}{4}(t - 5)\right).
> $$

The response is continuous at $t = 5$, but $y'$ jumps there and $y''$ has an
infinite spike — the singularity in the forcing must appear somewhere on the
left side, and it lands in the top derivative.

## The convolution integral

A product $F(s)G(s)$ of transforms is **not** the transform of the product
$f(t)g(t)$.

> **Theorem (Convolution).** If $F(s) = \mathcal{L}\{f(t)\}$ and
> $G(s) = \mathcal{L}\{g(t)\}$ exist for $s > a$, then $F(s)G(s) = \mathcal{L}\{h(t)\}$
> where
> $$
> h(t) = (f \ast g)(t) = \int_0^t f(t - \tau)\, g(\tau)\, \d\tau
>   = \int_0^t f(\tau)\, g(t - \tau)\, \d\tau.
> $$
> The two integrals are equal; $h$ is the **convolution** of $f$ and $g$.

The proof writes $F(s)G(s)$ as a double integral over the first quadrant of the
$\xi\tau$-plane, substitutes $\xi = t - \tau$, and reverses the order of
integration over the resulting triangular region $0 \le \tau \le t$.[^boyce-conv]
The convolution behaves like a product: it is commutative, associative, and
distributive over addition, with $f \ast 0 = 0$. It is not, however, an
identity-preserving product — $f \ast 1 \ne f$ in general. For $f = \cos t$,
$(\cos \ast\, 1)(t) = \int_0^t \cos(t - \tau)\, \d\tau = \sin t \ne \cos t$.

$$
% caption: Convolution slides a reflected copy of $f$ across $g$; the value
% $(f \ast g)(t)$ is the area of their overlap (shaded) as the copy moves with $t$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.2,0) -- (6.0,0) node[right, black!70] {integration variable};
  \draw[->, black] (0,-0.2) -- (0,2.2) node[above, black!70] {};
  % g: fixed hump (triangle)
  \fill[black!10] (0.6,0) -- (1.9,1.5) -- (3.2,0) -- cycle;
  \draw[thick] (0.6,0) -- (1.9,1.5) -- (3.2,0);
  \node[anchor=south] at (1.9,1.5) {$g$};
  % f reflected/shifted hump
  \fill[black] (2.2,0) -- (3.6,1.2) -- (5.0,0) -- cycle;
  \draw[black, thick] (2.2,0) -- (3.6,1.2) -- (5.0,0);
  \node[black!70, anchor=south] at (3.9,1.2) {$f$};
  % overlap shaded darker
  \fill[acc!30] (2.2,0) -- (2.65,0.52) -- (3.2,0) -- cycle;
  \draw[->, black] (4.2,1.5) -- (5.1,1.5) node[right, black!70] {slides with $t$};
\end{tikzpicture}
$$

Convolution inverts a transform that factors into known pieces.

> **Worked example.** Invert $H(s) = a/\bigl(s^2(s^2+a^2)\bigr)$ by convolution.
>
> Factor $H$ into two known transforms, $H(s) = s^{-2}\cdot a/(s^2+a^2)$, the
> transforms of $t$ and $\sin(at)$. The convolution theorem inverts the product
> as the convolution of the pieces:
> $$
> \mathcal{L}^{-1}\{H(s)\} = (t \ast \sin(at))(t) = \int_0^t (t - \tau)\sin(a\tau)\, \d\tau.
> $$
> Split off the constant $t$ and integrate the two pieces, the second by parts:
> $$
> \int_0^t (t-\tau)\sin(a\tau)\,\d\tau
>   = t\,\frac{1 - \cos(at)}{a}
>   - \left(\frac{\sin(at)}{a^2} - \frac{t\cos(at)}{a}\right)
>   = \frac{at - \sin(at)}{a^2}.
> $$

## Transfer function and impulse response

Convolution gives the forced response of a linear system a clean structure. For
$a y'' + b y' + c y = g(t)$ with $y(0) = y_0$, $y'(0) = y_0'$, transforming and
solving splits $Y(s)$ into two parts,

$$
Y(s) = \underbrace{\frac{(as + b)y_0 + a y_0'}{as^2 + bs + c}}_{\Phi(s)}
  + \underbrace{H(s)\, G(s)}_{\Psi(s)},
  \qquad H(s) = \frac{1}{as^2 + bs + c}.
$$

The term $\Phi(s)$ carries the initial conditions and inverts to the solution of
the homogeneous equation; $\Psi(s)$ carries the forcing.

> **Definition (Transfer function and impulse response).** $H(s) = (as^2+bs+c)^{-1}$
> is the **transfer function**, fixed by the system coefficients alone. Its
> inverse $h(t) = \mathcal{L}^{-1}\{H(s)\}$ is the **impulse response**: the
> solution of $a y'' + b y' + c y = \delta(t)$ from rest, since $G(s) = 1$ when
> $g = \delta$.

By the convolution theorem, the forced response is the impulse response
convolved with the input,

$$
\psi(t) = \mathcal{L}^{-1}\{H(s)G(s)\} = \int_0^t h(t - \tau)\, g(\tau)\, \d\tau
  = (h \ast g)(t).
$$

$$
% caption: A linear system as an input-output map: the input $g$ produces the
% output $y = (h \ast g) + \phi$, where $h$ is the impulse response fixed by the
% transfer function $H(s)$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  box/.style={draw, minimum width=34mm, minimum height=13mm, align=center}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node (in) at (0,0) {$g(t)$};
  \node[box, draw=acc, fill=acc!10] (sys) at (3.4,0) {system\\ $H(s) = \dfrac{1}{as^2+bs+c}$};
  \node (out) at (7.3,0) {$y(t)$};
  \draw[->, thick] (in) -- (sys) node[midway, above] {input};
  \draw[->, thick] (sys) -- (out) node[midway, above] {output};
  \node[black!70, anchor=north, align=center] at (3.4,-1.15)
    {impulse response $h$};
\end{tikzpicture}
$$

This is the input–output picture: the coefficients $a, b, c$ describe the
system, $g$ is the input, $y$ is the output, and $H$ encodes the system's
response independent of any particular input. It generalizes
[variation of parameters](/differential-equations/second-order-linear/variation-of-parameters):
both express the forced solution as an integral against the forcing, but here the
kernel is the impulse response and the initial conditions are already absorbed.

## Abel's tautochrone

A convolution can also appear as the unknown side of an equation. The problem is
to find a frictionless wire down which a bead slides to the bottom in a time
independent of its starting height — the **tautochrone**. With $f(y) = \d s/\d y$
the unknown arc-length rate and $T(y)$ the descent time from height $y$,
conservation of energy gives a **Volterra integral equation**,

$$
T(y) = \frac{1}{\sqrt{2g}} \int_0^y \frac{f(v)}{\sqrt{y - v}}\, \d v,
$$

in which the integral is the convolution of $y^{-1/2}$ and $f(y)$.[^sim-abel]
Transforming and using $\mathcal{L}\{y^{-1/2}\} = \sqrt{\pi/s}$,

$$
\mathcal{L}\{T\} = \frac{1}{\sqrt{2g}}\, \mathcal{L}\{y^{-1/2}\}\, \mathcal{L}\{f\}
  \quad\Longrightarrow\quad
  \mathcal{L}\{f\} = \sqrt{\frac{2g}{\pi}}\, \sqrt{s}\, \mathcal{L}\{T\}.
$$

Demanding a constant descent time $T(y) = T_0$ makes the right side known; its
inverse transform is $f(y) = b/\sqrt{y}$ with $b = \sqrt{2g}\,T_0/\pi$. Solving
$1 + (\d x/\d y)^2 = b^2/y$ yields the parametric curve

$$
x = a(\theta + \sin\theta), \qquad y = a(1 - \cos\theta),
$$

the [cycloid](/differential-equations/history-variations/calculus-of-variations).
The tautochrone is an arc of a cycloid, the same curve that solves the
brachistochrone problem, recovered here without guessing the answer in advance.

## Transform rules for forcing

| Tool | Transform rule | Use |
| --- | --- | --- |
| Step $u_c(t)$ | $\mathcal{L}\{u_c\} = e^{-cs}/s$ | build piecewise-constant functions |
| Second shift | $\mathcal{L}\{u_c(t)f(t-c)\} = e^{-cs}F(s)$ | delayed / switched forcing; invert $e^{-cs}$ factors |
| Delta $\delta(t-t_0)$ | $\mathcal{L}\{\delta(t-t_0)\} = e^{-st_0}$ | instantaneous impulses; impulse response |
| Convolution | $\mathcal{L}\{f \ast g\} = F(s)G(s)$ | invert products; forced response $h \ast g$; integral equations |

[^boyce-shift]: **Boyce**, _Elementary Differential Equations and Boundary Value Problems_, §6.3, Theorem 6.3.1 — $\mathcal{L}\{u_c(t)f(t-c)\} = e^{-cs}F(s)$, proved by the substitution $\sigma = t - c$; contrasted with the first shifting theorem 6.3.2 in $s$.
[^boyce-disc]: **Boyce**, §6.4, Example 1 — the pulse-forced damped oscillator $2y'' + y' + 2y = u_5(t) - u_{20}(t)$, with the observation that $y$ and $y'$ stay continuous while $y''$ inherits the jumps of the forcing.
[^boyce-imp]: **Boyce**, §6.5 — the delta function as the limit of unit-area pulses $d_\tau$, its sifting property $\int \delta(t-t_0)f(t)\,\d t = f(t_0)$, and the transform $\mathcal{L}\{\delta(t-t_0)\} = e^{-st_0}$.
[^boyce-conv]: **Boyce**, §6.6, Theorem 6.6.1 — the convolution theorem and its proof by reversing the order of integration; the transfer-function and impulse-response decomposition $Y = \Phi + H G$. See also **Simmons**, §53 — the indicial and impulse response of a system at rest.
[^sim-abel]: **Simmons**, §52 — Convolutions and Abel's Mechanical Problem: the descent-time equation as a convolution of $y^{-1/2}$ and $f(y)$, solved by transform to show the tautochrone is a cycloid.