4.5 Laplace Transform of Piecewise Functions

A. Step function

In this section, we explore how to apply the Laplace Transform to piecewise continuous functions. In the next section, we will address solving initial value problems that involve second-order differential equations with constant coefficients where the forcing function ${\displaystyle f\left(t\right)}$ is a continuous piecewise function.

Jump discontinuities often occur in physical situations like switching mechanisms or abrupt changes in forces acting on the system. To handle such discontinuities in the Laplace domain, we utilize the unit step function to transform piecewise functions into a form amenable to Laplace transforms and subsequently find piecewise continuous inverses of Laplace transforms for the solution.

The unit step function (Heaviside function) ${\displaystyle u\left(t\right)}$ is defined as

 ${\displaystyle u\left(t\right)=\{\begin{array}{ll}0& t<0\\ 1& t\ge 0\end{array}}$

It steps (or jumps) from 0 to 1 at ${\displaystyle t=0}$. By shifting the input argument ${\displaystyle t}$, we can move the step to different locations.

 ${\displaystyle u(t-a)=\{\begin{array}{ll}0& t-a<0\\ 1& t-a\ge 0\end{array}}$     ${\displaystyle \to }$     ${\displaystyle u_a\left(t\right)=u(t-a)=\{\begin{array}{ll}0& t<a\\ 1& t\ge a\end{array}}$

The step function can also be transformed, e.g., shifted, stretched, or compressed. For example, by multiplying ${\displaystyle u\left(t\right)}$ by some constant ${\displaystyle M>1}$, we can stretch it vertically.

 ${\displaystyle Mu(t-a)=\{\begin{array}{ll}0& t<a\\ M& t\ge a\end{array}}$

Or by combined shifting and reflecting ${\displaystyle u\left(t\right)}$, we can opposite the way the function switches on and off.

 ${\displaystyle 1-u(t-a)=\{\begin{array}{ll}1& t<a\\ 0& t\ge a\end{array}}$

The step function enables us to represent any piecewise continuous function conveniently. For instance, consider the function

 ${\displaystyle f\left(t\right)=\{\begin{array}{ll}{f}_0\left(t\right)& 0\le t<a\\ f_1\left(t\right)& t\ge a\end{array}}$

 ${\displaystyle =f_0\left(t\right)\{\begin{array}{ll}1& 0\le t<a\\ 0& t\ge a\end{array}}$ ${\displaystyle +f_1\left(t\right)\{\begin{array}{ll}0& 0\le t<a\\ 1& t\ge a\end{array}}$

 ${\displaystyle =f_0\left(t\right)(1-u(t-a))+f_1\left(t\right)u(t-a)}$

It can be rewritten as

 ${\displaystyle f\left(t\right)=f_0\left(t\right)+u(t-a)(f_1\left(t\right)-f_0\left(t\right))}$ (4.5.1)

We can extend Equation 4.5.1 to more general continuous piecewise functions.

 ${\displaystyle f\left(t\right)=\{\begin{array}{ll}{f}_0\left(t\right)& 0\le t<a\\ f_1\left(t\right)& a\le t<b\\ f_2\left(t\right)& t\ge b\end{array}}$

 ${\displaystyle f\left(t\right)=f_0\left(t\right)+u(t-a)(f_1\left(t\right)-f_0\left(t\right))}$ ${\displaystyle +u(t-b)(f_2\left(t\right)-f_1\left(t\right))}$ (4.5.2)

B. Laplace Transform of Piecewise Functions

The Laplace Transform of the step-modulated function is key in solving differential equations with piecewise forcing functions.

Theorem: Laplace Transform of a Step-Modulated Function. Let ${\displaystyle g\left(t\right)}$  be defined on ${\displaystyle [0,\infty )}$, suppose ${\displaystyle a\ge 0}$, and assume ${\displaystyle L\left\{g(t+a)\right\}}$ exists for ${\displaystyle s>s_0}$. Then

 ${\displaystyle L\left\{u(t-a)g\left(t\right)\right\}=e^{-as}L\left\{g(t+a)\right\}}$  (4.5.3)

This theorem enables the transformation of step-modulated functions into the Laplace domain, which can then be manipulated algebraically.

C. Inverse Laplace Transform of Piecewise Functions

The previous theorem also allows us to determine the inverse Laplace Transform of functions that arise from piecewise functions. However, it will be more convenient to shift the argument of ${\displaystyle g\left(t\right)}$ and replace ${\displaystyle g\left(t\right)}$ with ${\displaystyle g(t-a)}$.

Translation in ${\displaystyle t}$ Theorem. If ${\displaystyle a\ge 0}$  and ${\displaystyle L\left(g\right)}$ exists for ${\displaystyle s>s_0}$, then

 ${\displaystyle L\left\{u(t-a)g(t-a)\right\}=e^{-as}L\left\{g\left(t\right)\right\}}$

Given ${\displaystyle G\left(s\right)=L\left\{g\left(t\right)\right\}}$, it is equivalent to

 ${\displaystyle u(t-a)g(t-a)}$ ${\displaystyle \leftrightarrow }$ ${\displaystyle e^{-as}G\left(s\right)}$ (4.5.4)

Section 4.5 Exercises