Theorem 11.7.1. Shifting the Variable \(\boldsymbol s\).
If \(F(s)\) is the Laplace transform of \(f(t)\text{,}\) then
\begin{equation*}
\mathcal{L}\big(e^{at}f(t)\big) = F(s-a).
\end{equation*}
Section 11.7 Shifting Theorems and the Step Function
We have seen how the Laplace transform can be used to solve linear differential equations. Familiar functions that arise in solutions to differential equations are \(e^{at}\cos bt\) and \(e^{at}\sin bt\text{.}\) The first shifting theorem will show how their transforms are related to those of \(\cos bt\) and \(\sin bt\) by shifting the variable \(s\) in \(F(s)\text{.}\) A companion result, called the second shifting theorem, will show how the transform of \(f(t-a)\) can be obtained by multiplying \(F(s)\) by \(e^{-as}\text{.}\) Loosely speaking, these results show that multiplication of \(f(t)\) by \(e^{at}\) corresponds to shifting \(F(s-a)\text{,}\) and shifting \(f(t-a)\) corresponds to multiplication of the transform \(F(s)\) by \(e^{as}\text{.}\)
Proof.
Using the integral definition \(\mathcal{L}\big(f(t)\big) = F(s) = \int_0^{\infty}f(t)e^{-st}\,dt\text{,}\) we see that
\begin{equation*}
\mathcal{L}\big(e^{at}f(t)\big) = \int_0^{\infty}e^{at}f(t)e^{-st}dt = \int_0^{\infty}f(t)e^{-(s-a)t}\,dt = F(s-a).
\end{equation*}
Definition 11.7.2. Unit Step Function.
For \(a \ge 0\) the unit step function \(U_a(t)\) is
\begin{equation*}
U_a(t) = \begin{cases}0 \amp \text{ for } t\lt a. \\ 1 \amp \text{ for } t>a. \end{cases}
\end{equation*}
Theorem 11.7.4. Shifting the Variable \(\boldsymbol t\).
If \(F(s)\) is the Laplace transform of \(f(t)\) and \(a \ge 0\text{,}\) then
\begin{equation*}
\mathcal{L}\big(U_a(t)f(t-a)\big) = e^{-as}F(s)\text{,}
\end{equation*}
Proof.
Using the definition of Laplace transform, we write
\begin{equation*}
e^{-as}F(s) = e^{-as}\int_0^{\infty}f(\tau)e^{-s\tau}\,d\tau = \int_0^{\infty}f(\tau)e^{-s(a+\tau)}\,d\tau\text{.}
\end{equation*}
Using the change of variable \(t=a+\tau\) and \(dt=d\tau\text{,}\) we obtain
\begin{equation*}
e^{-as}F(s) = \int_a^{\infty}f(t-a)e^{-st}\,dt\text{.}
\end{equation*}
Since \(U_a(t) \,f(t-a) =0\) for \(t\lt a\text{,}\) and \(U_a(t) \,f(t-a) =f(t-a)\) for \(t>a\text{,}\) we can write the last equation as
\begin{equation*}
e^{-as}F(s) = \int_0^{\infty}U_a(t)f(t-a)e^{-st}\,dt = \mathcal{L}\big(U_a(t)f(t-a)\big).
\end{equation*}
Example 11.7.6.
Show that \(\mathcal{L}(t^ne^{at}) =\frac{n!}{(s-a)^{n+1}}\text{.}\)
Solution.
Let \(f(t)=t^n\text{,}\) then \(F(s) = \mathcal{L}(t^n) = \frac{n!}{s^{n+1}}\text{.}\) Applying Theorem 11.7.1, we obtain the desired result:
\begin{equation*}
\mathcal{L}(t^ne^{at}) = F(s-a) = \frac{n!}{(s-a)^{n+1}}.
\end{equation*}
Example 11.7.7.
Show that \(\mathcal{L}(U_c(t)) = \frac{e^{-cs}}{s}\text{.}\)
Solution.
Set \(f(t) = 1\text{,}\) then \(F(s) = \mathcal{L}(1) = \frac{1}{s}\text{.}\) Now apply Theorem 11.7.4 and get
\begin{equation*}
\mathcal{L}\big(U_c(t)\big) = \mathcal{L}\big(U_c(t)f(t)\big) = \mathcal{L}\big(U_c(t)\cdot 1\big) = e^{-cs}\mathcal{L}(1) = \frac{e^{-cs}}{s}\text{.}
\end{equation*}
Example 11.7.8.
Solution.
We can represent \(f(t)\) in terms of step functions:
\begin{equation*}
f(t) = 1-U_1(t)+U_2(t)-U_3(t)+U_4(t)-U_5(t)\text{.}
\end{equation*}
Using the result of Example 10.6.1 and linearity, we obtain
\begin{equation*}
\mathcal{L}\big(f(t)\big) = \frac{1}{s} - \frac{e^{-s}}{s} + \frac{e^{-2s}}{s} - \frac{e^{-3s}}{s} + \frac{e^{-4s}}{s} -
\frac{e^{-5s}}{s}\text{.}
\end{equation*}
Example 11.7.10.
Use Laplace transforms to solve the initial value problem
\begin{equation*}
y\,''(t) + y(t) = U_{\pi}(t) \text{ with } y(0)=0 \text{ and } y\,'(0) =0\text{.}
\end{equation*}
Solution.
As usual, let \(Y(s)\) denote the Laplace transform of \(y(t)\text{.}\) Then we get
\begin{equation*}
s^2Y(s) + Y(s) = \frac{e^{-\pi s}}{s}\text{.}
\end{equation*}
Solving for \(Y(s)\text{,}\) we obtain
\begin{equation*}
Y(s) = e^{-\pi s}\frac{1}{s(s^2+1)} = \frac{e^{-\pi s}}{s} - \frac{e^{-\pi s}s}{s^2+1}\text{.}
\end{equation*}
We now use Theorem 11.7.4 and the facts that \(\frac{1}{s}\) and \(\frac{s}{s^2+1}\) are the transforms of \(1\) and \(\cos t\text{,}\) respectively. The solution \(y(t)\) computes as follows:
\begin{equation*}
y(t) = \mathcal{L}^{-1}\left(\frac{e^{-\pi s}}{s}\right) - \mathcal{L}^{-1}\left(\frac{e^{-\pi s}s}{s^2+1}\right) = U_{\pi}(t) - U_{\pi}(t) \cos(t-\pi)\text{,}
\end{equation*}
which can be written in the more familiar form:
\begin{equation*}
y(t) = \begin{cases} \; 0 \amp \text{ for } t \lt \pi, \\ 1-\cos t \amp \text{ for } t>\pi. \end{cases}
\end{equation*}
Exercises Exercises
1.
Find \(\mathcal{L}(e^t-te^t)\text{.}\)
2.
Find \(\mathcal{L}(e^{-4t}\sin 3t)\text{.}\)
3.
Show that \(\mathcal{L}(e^{at}\cos bt) =\frac{s-a}{(s-a)^2+b^2}\text{.}\)
4.
Show that \(\mathcal{L}(e^{at}\sin bt) =\frac{b}{(s-a)^2+b^2}\text{.}\)
5.
Find \(\mathcal{L}^{-1}(F(s))\) for the following:
(a)
\(F(s) =\frac{s+2}{s^2+4s+5}\text{.}\)
(b)
\(F(s) =\frac{8}{s^2-2s+5}\text{.}\)
(c)
\(F(s) =\frac{s+3}{(s+2)^2+1}\text{.}\)
Solution.
(d)
\(F(s) =\frac{2s+10}{s^2+6s+25}\text{.}\)
6.
Find \(\mathcal{L}^{-1}(\frac{1-e^{-s}+e^{-2s}}{s^2})\text{.}\)
7.
Find \(\mathcal{L}\big(f(t)\big)\) for the following:
(a)
\(f(t) = U_2(t)(t-2)^2\text{.}\)
(b)
\(f(t) = U_1(t)e^{1-t}\text{.}\)
(c)
\(f(t) = U_{3\pi}(t) \sin(t-3\pi)\text{.}\)
(d)
\(f(t) = 2U_1(t) - U_2(t) - U_3(t)\)
(e)
(f)
\hint{The function is the integral of the one in Exercise e.}
8.
Find \(\mathcal{L}^{-1}(\frac{e^{-s}+e^{-2s}}{s})\text{.}\)
9.
Solve the initial value problem for each of the following:
(a)
(b)
(c)
(d)
(e)
(f)
\(y\,''(t) + 2y\,'(t) + y(t) = 2U_1(t)e^{1-t}\text{,}\) with \(y(0) =0\) and \(y\,'(0) = 0\text{.}\)
