The unit step function \(\displaystyle \mathcal {U}(t)=\left \{\begin{array}{lr} 0 & \mbox { if }t<0\\ \\ 1 & \mbox { if }t\geq 0 \end{array}\right .\) is a basic piecewise continuous function.

Consider a piecewise continuous function \(\displaystyle f(t)=\left \{\begin{array}{lr} f_{1}(t) & \mbox { if }0\leq t<t_{1}\\ \\ f_{2}(t) & \mbox { if }t\geq t_{1} \end{array}\right .\).

We rewrite \(f(t)\) using the unit step function: \(f(t)=f_{1}(t)+\mathcal {U}(t-t_{1})\;[f_{2}(t)-f_{1}(t)]\).

Here is the graph of \(\displaystyle f(t)=\left \{\begin{array}{lr} \frac {\pi }{2} -t& \mbox { if }0\leq t<\frac {\pi }{2}\\ \\ 0 & \mbox { if }t\geq \frac {\pi }{2} \end{array}\right .\).

Using the unit step function, \(f(t)=\frac {\pi }{2}-t+\mathcal {U}\left (t-\frac {\pi }{2}\right )\;\left [0-\left (\frac {\pi }{2}-t\right )\right ]\) \(=\frac {\pi }{2}-t+\mathcal {U}\left (t-\frac {\pi }{2}\right )\;\left [t-\frac {\pi }{2}\right ]\).

Consider a piecewise continuous function \(\displaystyle f(t)=\left \{\begin{array}{lr} f_{1}(t) & \mbox { if }0\leq t<t_{1}\\ \\ f_{2}(t) & \mbox { if } t_{1}\leq t<t_{2} \\ \\ f_{3}(t) & \mbox { if }t\geq t_{2} \end{array}\right .\).

Using the unit step function:
\(f(t)=f_{1}(t)+\mathcal {U}(t-t_{1})[f_{2}(t)-f_{1}(t)]+\mathcal {U}(t-t_{2})[f_{3}(t)-f_{2}(t)]\).

2nd Shifting Property: Consider that
\(\displaystyle \mathcal {L}(\mathcal {U}(t-a)g(t))=\displaystyle \int _{0}^{\infty }e^{-st}\mathcal {U}(t-a)g(t)\;dt=\displaystyle \int _{a}^{\infty }e^{-st}g(t)\;dt\).

Using the substitution \(u=t-a\), \(du=dt\) we get

\(\displaystyle \int _{a}^{\infty }e^{-st}g(t)\;dt=\int _{0}^{\infty }e^{-s(u+a)}g(u+a)\;du=\) \(\displaystyle e^{-sa}\int _{0}^{\infty }e^{-su}g(u+a)\;du=e^{-sa}\int _{0}^{\infty }e^{-st}g(t+a)\;dt\).

Hence we have justified that: \(\mathcal {L}(\mathcal {U}(t-a)g(t))=e^{-sa}\mathcal {L}(g(t+a))\).

An equivalent reformulation is \(\mathcal {L}(\mathcal {U}(t-a)g(t-a))=e^{-sa}\mathcal {L}(g(t))\).

This is the \(``\)2nd Shifting Property" of the Laplace transform. In particular, this means \(\mathcal {L}^{-1}(e^{-as}G(s))=\mathcal {U}(t-a)g(t-a)\) where \(g=\mathcal {L}^{-1}(G)\).

Let’s find the Laplace transform of \(\displaystyle f(t)=\left \{\begin{array}{lr} 2-t & \mbox { if }0\leq t<1\\ \\ t & \mbox { if }t\geq 1 \end{array}\right .\).

First we rewrite it as \(f(t)=2-t+\mathcal {U}(t-1) \;[t-(2-t)]=2-t+2\,\mathcal {U}(t-1)\;[t-1]\). Then
\(\displaystyle \mathcal {L}(f(t))=\dfrac {2}{s}-\dfrac {1}{s^{2}}+2\mathcal {L}(\mathcal {U}(t-1)\;[t-1])\) \(\displaystyle =\dfrac {2}{s}-\dfrac {1}{s^{2}}+2e^{-s}\mathcal {L}(t)\) \(\displaystyle =\dfrac {2}{s}-\dfrac {1}{s^{2}}+\dfrac {2}{s^{2}}e^{-s}\).

Let’s use linearity and the 2nd Shifting Property to find \(\mathcal {L}^{-1}\left (\dfrac {1-e^{-s}}{s^{3}}\right )\).

First we use linearity and then apply the second shifting property:

\(\mathcal {L}^{-1}\left (\dfrac {1-e^{-s}}{s^{3}}\right )=\frac {1}{2}\mathcal {L}^{-1}\left (\dfrac {2}{s^{3}}\right )-\frac {1}{2}\mathcal {L}^{-1}\left (e^{-s}\dfrac {2}{s^{3}}\right )\) \(\displaystyle =\frac {1}{2}t^2-\frac {1}{2}\mathcal {L}^{-1}\left (e^{-s}\mathcal {L}(t^2)\right )=\frac {1}{2}\left (t^2-\mathcal {U}(t-1)[(t-1)^2]\right )\).

We rewrite this in the more familiar form \(\displaystyle \mathcal {L}^{-1}\left (\dfrac {1-e^{-s}}{s^{3}}\right )=\left \{\begin{array}{lr} \frac {1}{2}t^2 & \mbox { if }0\leq t<1\\ \\ t-\frac {1}{2} & \mbox { if }t\geq 1 \end{array}\right .\).

Let’s solve the IVP \(\displaystyle y''+y=\left \{\begin{array}{lr} t & \mbox { if }0\leq t<\pi \\ \\ \pi & \mbox { if }t\geq \pi \end{array}\right .\),   \(y(0)=0\), \(y'(0)=1\).

Let’s first find \(\mathcal {L}(f)\) where \(\displaystyle f(t)=\left \{\begin{array}{lr} t & \mbox { if }0\leq t<\pi \\ \\ \pi & \mbox { if }t\geq \pi \end{array}\right .\).

We can rewrite this function as \(f(t)=t+\mathcal {U}(t-\pi )\;[\pi -t]\) \(\displaystyle =t-\mathcal {U}(t-\pi )\;[t-\pi ]\).

Then \(\displaystyle \mathcal {L}(f)=\mathcal {L}(t)-\mathcal {L}\left (\mathcal {U}(t-\pi )[t-\pi ]\right )\) \(\displaystyle =\frac {1}{s^{2}}-e^{-\pi s}\mathcal {L}(t)\) \(\displaystyle =\frac {1}{s^{2}}-e^{-\pi s}\,\frac {1}{s^{2}}\).

Let \(Y=\mathcal {L}(y)\) and let’s take the Laplace transform of both sides of the differential equation:

\((s^2+1)Y=1+\frac {1}{s^{2}}-e^{-\pi s}\,\frac {1}{s^{2}}\rightarrow \) \(\displaystyle Y=\frac {1+\frac {1}{s^{2}}}{s^2+1}-e^{-\pi s}\,\frac {1}{s^{2}(s^2+1)}\rightarrow \) \(\displaystyle Y=\frac {1}{s^2}-e^{-\pi s}\,\frac {1}{s^{2}(s^2+1)}\rightarrow \)

\(\displaystyle Y=\frac {1}{s^2}-e^{-\pi s}\,\left (\frac {1}{s^{2}}-\frac {1}{s^2+1}\right )\) using a PFD.

Therefore \(y=\mathcal {L}^{-1} (Y)=t-\mathcal {U}(t-\pi )\;g(t-\pi )\) where \(g=\mathcal {L}^{-1}\left (\frac {1}{s^{2}}-\frac {1}{s^2+1}\right )=t-\sin {(t)}\).

So \(y=t-\mathcal {U}(t-\pi )\;[(t-\pi )-\sin {(t-\pi )}]\). Since \(\sin {(t-\pi )}=-\sin {(t)}\), \(y=t-\mathcal {U}(t-\pi )\;[(t-\pi )+\sin {(t)}]\).

In a more familiar form, \(\displaystyle y(t)=\left \{\begin{array}{lr} t & \mbox { if }0\leq t<\pi \\ \\ \pi -\sin {(t)} & \mbox { if }t\geq \pi \end{array}\right .\).