Step Functions, Discontinuous Forcing, Impulses, and Convolution

The transform-and-invert method 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

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 $t=4,7,9$:

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

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

The second shifting theorem

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

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 $c$.¹ Substituting $\sigma =t-c$ into the defining integral produces the factor $e^{-cs}$. Whereas the first shifting rule translated the transform in $s$, this one translates the function in $t$ and attaches $e^{-cs}$ to the transform.

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

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.

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 $t=20$.²

The regularity of $y$ across the switching points is the general lesson. For any equation $y^{\prime \prime }+p(t)y^{\prime }+q(t)y=g(t)$ with $p,q$ continuous and $g$ only piecewise continuous, the solution $y$ and its first derivative $y^{\prime }$ stay continuous, while $y^{\prime \prime }$ inherits the jumps of $g$: the jump in the highest-order term $y^{\prime \prime }$ 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.

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

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

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.³

The response is continuous at $t=5$, but $y^{\prime }$ jumps there and $y^{\prime \prime }$ 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 $f(t)g(t)$.

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 $0\le \tau \le t$.⁴ 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 $f=\cos t$, $(\cos \ast 1)(t)={\int }_0^t\cos (t-\tau )\mathbf{d}\tau =\sin t\ne \cos t$.

Convolution inverts a transform that factors into known pieces.

Transfer function and impulse response

Convolution gives the forced response of a linear system a clean structure. For $a{y}^{\prime \prime }+b{y}^{\prime }+cy=g(t)$ with $y(0)=y_0$, $y^{\prime }(0)=y_0^{\prime }$, transforming and solving splits $Y(s)$ into two parts,

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

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

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: 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 $f(y)=\mathbf{d}s/\mathbf{d}y$ the unknown arc-length rate and $T(y)$ the descent time from height $y$, conservation of energy gives a Volterra integral equation,

in which the integral is the convolution of $y^{-1/2}$ and $f(y)$.⁵ Transforming and using $\mathcal{L}\{y^{-1/2}\}=\sqrt{\pi /s}$,

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 $1+(\mathbf{d}x/\mathbf{d}y{)}^2=b^2/y$ yields the parametric curve

the cycloid. 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