Transform of discontinuous functions

The Laplace transform handles discontinuous functions gracefully — provided the discontinuities are jumps (the function and its limits are finite). This is one of Laplace’s biggest practical advantages: discontinuous forcing in ODEs is no harder than continuous forcing.

The trick: rewrite piecewise functions in terms of Heaviside step functions and rectangular windows, then transform each piece using the second shifting theorem.

What kinds of discontinuities are OK

A function $f(t)$ has a jump discontinuity at $t_0$ if $f$ is discontinuous at $t_0$ but the one-sided limits

${\lim }_{t\to t_0^{-}}f(t),{\lim }_{t\to t_0^{+}}f(t)$

both exist and are finite. Functions with finitely many jump discontinuities on any bounded interval are piecewise continuous — and the Laplace transform exists for them under the usual exponential-order condition.

Worse discontinuities (infinite jumps, oscillation singularities) usually break the Laplace transform.

Useful properties for discontinuous functions

Heaviside transform: $\mathcal{L}\{u(t-a)\}(s)=\frac{e^{-as}}{s}$ for $a\ge 0$.

Rectangular window: $\mathcal{L}\{{\Pi }_{a,b}\}(s)=\frac{e^{-as}-e^{-bs}}{s}$ for $0<a<b$.

Second shifting theorem: $\mathcal{L}\{u(t-a)f(t-a)\}(s)=e^{-as}F(s)$.

The shifting theorem is the workhorse: when a function is “turned on” at $t=a$ with its argument shifted, the Laplace transform picks up an $e^{-as}$ factor.

Example 1: piecewise function

Find $\mathcal{L}\{f\}$ for

$f(t)=\{\begin{array}{ll}\sin t& 0\le t\le \pi /4\\ \sin t+\cos (t-\pi /4)& t>\pi /4\end{array}$

Express using Heaviside:

$f(t)=\sin t+u(t-\pi /4)\cos (t-\pi /4)$

(The $\sin t$ is the same in both branches; only the cosine “turns on” at $\pi /4$.)

Take Laplace:

$F(s)=\mathcal{L}\{\sin t\}+\mathcal{L}\{u(t-\pi /4)\cos (t-\pi /4)\}$

Use the standard table and the second shifting theorem:

$F(s)=\frac{1}{s^2+1}+e^{-\pi s/4}\cdot \frac{s}{s^2+1}=\frac{1+s{e}^{-\pi s/4}}{s^2+1}$

Example 2: full IVP with discontinuous forcing

Solve $y^{\prime \prime }+4y=g(t)$, $y(0)=y^{\prime }(0)=0$, where

$g(t)=\{\begin{array}{ll}0& t<5\\ (t-5)/5& 5<t<10\\ 1& t\ge 10\end{array}$

Express $g$ using Heavisides:

$g(t)=\frac{t-5}{5}{\Pi }_{5,10}(t)+u(t-10)$

Expanding the window:

$g(t)=\frac{t-5}{5}u(t-5)-\frac{t-10}{5}u(t-10)$

(The $u(t-10)$ pieces from the window and the $+u(t-10)$ term combine in a way that gives the right values; the second piece is essentially the slope-1 ramp from time 10 onward, which together with the linear ramp from 5–10 gives the desired piecewise structure.)

Take Laplace of both sides of the ODE. Using $\mathcal{L}\{(t-a)u(t-a)\}(s)=e^{-as}\mathcal{L}\{t\}(s)=e^{-as}/s^2$:

$s^{2}Y(s)+4Y(s)=\frac{1}{5}\cdot \frac{e^{-5s}}{s^2}-\frac{1}{5}\cdot \frac{e^{-10s}}{s^2}$

(Initial conditions are zero, so they don’t contribute.)

Solve for $Y$:

$Y(s)=\frac{e^{-5s}-e^{-10s}}{5s^2(s^2+4)}$

Define $H(s)=\frac{1}{s^2(s^2+4)}$. Then $Y(s)=\frac{1}{5}[e^{-5s}-e^{-10s}]H(s)$.

Invert via the second shifting theorem:

$y(t)=\frac{1}{5}[h(t-5)u(t-5)-h(t-10)u(t-10)]$

where $h(\tau )={\mathcal{L}}^{-1}\{H(s)\}$.

Compute $h$: partial fractions on $H(s)=\frac{1}{s^2(s^2+4)}$:

$H(s)=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+4}$

Solving: $A=0$, $B=1/4$, $C=0$, $D=-1/4$. So:

$H(s)=\frac{1/4}{s^2}-\frac{1/4}{s^2+4}$

Inverse: $h(\tau )=\tau /4-(1/8)\sin (2\tau )$.

Final:

$y(t)=\frac{1}{5}\left[\left(\frac{t-5}{4}-\frac{1}{8}\sin (2(t-5))\right)u(t-5)-\left(\frac{t-10}{4}-\frac{1}{8}\sin (2(t-10))\right)u(t-10)\right]$

Looks complicated, but each piece “turns on” at the appropriate moment, and the system smoothly responds to the changing forcing.

Why this is the right tool

For ODEs with discontinuous forcing — switches turning on/off, impulse-like inputs, multi-stage processes — the Laplace transform produces one unified algebraic expression. Compare with the alternative: solve the ODE separately on each interval, then patch solutions together at boundaries to maintain continuity of $y$ and $y^{\prime }$. Tedious and error-prone.

Laplace handles all the bookkeeping automatically through Heavisides and the shifting theorem.

For the impulse limit (a “function” that’s zero everywhere except at one instant where it has unit area), see Dirac delta function. For the convolution-based approach when forcing is arbitrary but unknown, see Convolution integral.