7.1: Introduction to the Laplace Transform (2024)

Let \(f\) be defined for \(t\ge0\) and let \(s\) be a real number. Then the Laplace transform of \(f\) is the function \(F\) defined by

\[\label{eq:8.1.2} F(s)=\int_0^\infty e^{-st} f(t)\,dt, \]

for those values of \(s\) for which the improper integral converges.

Find the Laplace transform of \(f(t)=1\).

It is convenient to combine the steps of integrating from \(0\) to \(T\) and letting \(T → ∞\). Therefore, instead of writing Equation \ref{eq:8.1.3} and \ref{eq:8.1.4} as separate steps we write

\[\int_{0}^{\infty} e^{-s t} d t=-\left.\frac{1}{s} e^{-s t}\right|_{0} ^{\infty}=\left\{\begin{array}{cl}{\frac{1}{s},} & {s>0} \\[4pt] {\infty,} & {s<0}\end{array}\right. \nonumber \]

We'll follow this practice throughout this chapter.

Find the Laplace transform of \(f(t)=t\).

Find the Laplace transform of \(f(t)=e^{at}\), where \(a\) is a constant.

[Find the Laplace transforms of \(f(t)=\sin\omega t\) and \(g(t)=\cos\omega t\), where \(\omega\) is a constant.

Use the table of Laplace transforms to find \({\cal L}(t^3e^{4t})\).

Suppose \({\cal L}(f_i)\) is defined for \(s>s_i,\) \(1\le i\le n).\) Let \(s_0\) be the largest of the numbers \(s_1\), \(s_{2},\) …,\(s_n,\) and let \(c_1\), \(c_2\),…, \(c_n\) be constants. Then

\[{\cal L}(c_1f_1+c_2f_2+\cdots+c_nf_n)=c_1{\cal L}(f_1)+c_2{\cal L}(f_2) +\cdots+c_n{\cal L}(f_n)\mbox{ for } s>s_0.\nonumber \]

Proof

We give the proof for the case where \(n=2\). If \(s>s_0\) then

\[\begin{aligned} {\cal L}(c_1f_1+c_2f_2)&= \int_0^\infty e^{-st}\left(c_1f_1(t)+c_2f_2(t))\right)\,dt\\[4pt] &= c_1\int_0^\infty e^{-st}f_1(t)\,dt+c_2\int_0^\infty e^{-st}f_2(t)\,dt\\[4pt] &= c_1{\cal L}(f_1)+c_2{\cal L}(f_2).\end{aligned}\nonumber \]

Use Theorem 8.1.2 and the known Laplace transform

\[{\cal L}(e^{at})={1\over s-a} \nonumber \]

to find \({\cal L}(\cosh bt)\,(b\ne0)\).

If

\[\label{eq:8.1.10} F(s)=\int_0^\infty e^{-st} f(t)\,dt \]

is the Laplace transform of \(f(t)\) for \(s>s_0\), then \(F(s-a)\) is the Laplace transform of \(e^{at}f(t)\) for \(s >s_0+a\).

Proof

Replacing \(s\) by \(s-a\) in Equation \ref{eq:8.1.10} yields

\[\label{eq:8.1.11} F(s-a)=\int_0^\infty e^{-(s-a)t}f(t)\,dt \]

if \(s-a>s_0\); that is, if \(s>s_0+a\). However, Equation \ref{eq:8.1.11} can be rewritten as

\[F(s-a)=\int_0^\infty e^{-st}\left(e^{at}f(t)\right)\,dt,\nonumber \]

which implies the conclusion.

Use Theorem 8.1.3 and the known Laplace transforms of \(1\), \(t\), \(\cos\omega t\), and \(\sin\omega t\) to find

\[{\cal L}(e^{at}),\quad {\cal L}(te^{at}),\quad {\cal L}(e^{\lambda t}\sin \omega t),\mbox{and } {\cal L}(e^{\lambda t}\cos\omega t).\nonumber \]

We know from calculus that a definite integral is not affected by changing the values of its integrand at isolated points. Therefore, redefining a function f to make it continuous at removable discontinuities does not change \(\cal{L}(f)\).

• A function \(f\) is said to be piecewise continuous on a finite closed interval \([0,T]\) if \(f(0+)\) and \(f(T-)\) are finite and \(f\) is continuous on the open interval \((0,T)\) except possibly at finitely many points, where \(f\) may have jump discontinuities or removable discontinuities.
• A function \(f\) is said to be piecewise continuous on the infinite interval \([0,\infty)\) if it is piecewise continuous on \([0,T]\) for every \(T>0\).

A function \(f\) is said to be of exponential order \(s_0\) if there are constants \(M\) and \(t_0\) such that

\[\label{eq:8.1.14} |f(t)|\le Me^{s_0t},\quad t\ge t_0. \]

In situations where the specific value of \(s_0\) is irrelevant we say simply that \(f\) is of exponential order.

If \(f\) is piecewise continuous on \([0,\infty)\) and of exponential order \(s_0,\) then \({\cal L}(f)\) is defined for \(s>s_0\).

We emphasize that the conditions of Theorem 8.1.6 are sufficient, but not necessary, for \(f\) to have a Laplace transform. For example, Exercise 8.1.14(c) shows that \(f\) may have a Laplace transform even though \(f\) isn’t of exponential order

If \(f\) is bounded on some interval \([t_0,\infty)\), say

\[|f(t)|\le M,\quad t\ge t_0,\nonumber \]

then Equation \ref{eq:8.1.14} holds with \(s_0=0\), so \(f\) is of exponential order zero. Thus, for example, \(\sin\omega t\) and \(\cos \omega t\) are of exponential order zero, and Theorem 8.1.6 implies that \({\cal L}(\sin\omega t)\) and \({\cal L}(\cos \omega t)\) exist for \(s>0\). This is consistent with the conclusion of Example 8.1.4.

It can be shown that if \(\lim_{t\to\infty}e^{-s_0t}f(t)\) exists and is finite then \(f\) is of exponential order \(s_0\) (Exercise 8.1.9). If \(\alpha\) is any real number and \(s_0>0\) then \(f(t)=t^\alpha\) is of exponential order \(s_0\), since

\[\lim_{t\to\infty}e^{-s_0t}t^\alpha=0,\nonumber \]

by L’Hôpital’s rule. If \(\alpha\ge 0\), \(f\) is also continuous on \([0,\infty)\). Therefore Exercise 8.1.9 and Theorem 8.1.6 imply that \({\cal L}(t^\alpha)\) exists for \(s\ge s_0\). However, since \(s_0\) is an arbitrary positive number, this really implies that \({\cal L}(t^\alpha)\) exists for all \(s>0\). This is consistent with the results of Example 8.1.2 and Exercises 8.1.6 and 8.1.8.

Find the Laplace transform of the piecewise continuous function

\[f(t)=\left\{\begin{array}{cl} 1,&0\le t<1,\\[4pt] -3e^{-t},&t\ge 1.\end{array}\right.\nonumber \]

In Section 8.4 we’ll develop a more efficient method for finding Laplace transforms of piecewise continuous functions.

We stated earlier that

\[\int_0^\infty e^{-st} e^{t^2} dt=\infty \nonumber \]

for all \(s\), so Theorem 8.1.6 implies that \(f(t)=e^{t^2}\) is not of exponential order, since

\[\lim_{t\to\infty} {e^{t^2}\over Me^{s_0t}}=\lim_{t\to\infty} {1\over M} e^{t^2-s_0t}=\infty, \nonumber \]

so

\[e^{t^2}>Me^{s_0t} \nonumber \]

for sufficiently large values of \(t\), for any choice of \(M\) and \(s_{0}\) (Exercise 8.1.3).