Step Functions and Discontinuous Forcing

The Laplace transform handles discontinuous forcing functions naturally through the Heaviside step function and the second shifting theorem.

The Heaviside Step Function

Definition4.3Heaviside step function

The Heaviside step function (or unit step function) is

$u_c(t) = u(t - c) = \begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases}$

for $c \geq 0$ . Its Laplace transform is $\mathcal{L}\{u_c(t)\} = e^{-cs}/s$ .

Theorem4.4Second shifting theorem

If $\mathcal{L}\{f(t)\} = F(s)$ , then

$\mathcal{L}\{u_c(t)f(t-c)\}(s) = e^{-cs}F(s).$

Conversely, $\mathcal{L}^{-1}\{e^{-cs}F(s)\} = u_c(t)f(t-c)$ .

ExamplePiecewise constant forcing

Solve $y' + 2y = g(t)$ , $y(0) = 0$ , where $g(t) = \begin{cases} 1 & 0 \leq t < 3 \\ 0 & t \geq 3\end{cases}$ .

Write $g(t) = 1 - u_3(t)$ . Taking transforms: $(s+2)Y = \frac{1}{s} - \frac{e^{-3s}}{s}$ .

$Y = \frac{1}{s(s+2)} - \frac{e^{-3s}}{s(s+2)} = \frac{1}{2}\left(\frac{1}{s} - \frac{1}{s+2}\right)(1 - e^{-3s}).$

Inverting: $y(t) = \frac{1}{2}(1 - e^{-2t}) - \frac{1}{2}u_3(t)(1 - e^{-2(t-3)})$ .

The Dirac Delta Function

Definition4.4Dirac delta function

The Dirac delta function $\delta(t - c)$ is the "generalized function" satisfying

$\int_{-\infty}^{\infty}\delta(t-c)\phi(t)\,dt = \phi(c)$

for any continuous $\phi$ . Its Laplace transform is $\mathcal{L}\{\delta(t-c)\} = e^{-cs}$ .

ExampleImpulse response

Solve $y'' + y = \delta(t - \pi)$ , $y(0) = 0$ , $y'(0) = 1$ .

Taking transforms: $s^2Y - 1 + Y = e^{-\pi s}$ , so $Y = \frac{1}{s^2+1} + \frac{e^{-\pi s}}{s^2+1}$ .

Inverting: $y(t) = \sin t + u_\pi(t)\sin(t - \pi) = \sin t - u_\pi(t)\sin t$ .

So $y(t) = \begin{cases}\sin t & t < \pi \\ 0 & t \geq \pi\end{cases}$ . The impulse at $t = \pi$ exactly cancels the oscillation.

Transfer Functions

RemarkTransfer function

For a linear constant-coefficient ODE $ay'' + by' + cy = g(t)$ with zero initial conditions, $Y(s) = H(s)G(s)$ where $H(s) = 1/(as^2 + bs + c)$ is the transfer function. The poles of $H$ (roots of the characteristic polynomial) determine stability:

All poles in $\text{Re}(s) < 0$ : stable (transients decay).
Any pole with $\text{Re}(s) > 0$ : unstable (solutions grow).
Poles on $\text{Re}(s) = 0$ : marginally stable.

ExampleResonance detection via transfer function

For $y'' + \omega_0^2 y = A\cos\omega t$ , $H(s) = 1/(s^2 + \omega_0^2)$ . The response amplitude is $|H(i\omega)| = 1/|\omega_0^2 - \omega^2|$ , which diverges as $\omega \to \omega_0$ (resonance). With damping ( $y'' + 2\zeta\omega_0 y' + \omega_0^2 y = \ldots$ ), the peak amplitude is finite: $1/(2\zeta\omega_0^2)$ .