To define the Laplace transform, we first recall the definition of an improper integral. If \(g\) is integrable over the interval \([a,T]\) for every \(T>a\), then the improper integral of \(g\) over \([a,\infty)\) is defined as
We say that the improper integral converges if the limit in Equation \ref{eq:8.1.1} exists; otherwise, we say that the improper integral diverges or does not exist. Here’s the definition of the Laplace transform of a function \(f\).
Defintion 8.1.1
: Laplace Transform
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
for those values of \(s\) for which the improper integral converges.
It is important to keep in mind that the variable of integration in Equation \ref{eq:8.1.2} is \(t\), while \(s\) is a parameter independent of \(t\). We use \(t\) as the independent variable for \(f\) because in applications the Laplace transform is usually applied to functions of time.
The Laplace transform can be viewed as an operator \({\cal L}\) that transforms the function \(f=f(t)\) into the function \(F=F(s)\). Thus, Equation \ref{eq:8.1.2} can be expressed as
\[F={\cal L}(f).\nonumber \]
The functions \(f\) and \(F\) form a transform pair, which we’ll sometimes denote by
\[f(t)\leftrightarrow F(s).\nonumber \]
It can be shown that if \(F(s)\) is defined for \(s=s_0\) then it is defined for all \(s>s_0\).
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
Extensive tables of Laplace transforms have been compiled and are commonly used in applications. The brief table of Laplace transforms in the Appendix will be adequate for our purposes.
Example 8.1.5
Use the table of Laplace transforms to find \({\cal L}(t^3e^{4t})\).
\[\cal L (t^3e^{4t})={3!\over(s-4)^4}={6\over(s-4)^4}.\nonumber \]
We’ll sometimes write Laplace transforms of specific functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms.
Linearity of the Laplace Transform
The next theorem presents an important property of the Laplace transform.
Theorem 8.1.2
Linearity Property
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
where the first transform on the right is defined for \(s>b\) and the second for \(s>-b\); hence, both are defined for \(s>|b|\). Simplifying the last expression in Equation \ref{eq:8.1.9} yields
In the following table the known transform pairs are listed on the left and the required transform pairs listed on the right are obtained by applying Theorem 8.1.3
.
for every real number \(s\). Hence, the function \(f(t)=e^{t^2}\) does not have a Laplace transform.
Our next objective is to establish conditions that ensure the existence of the Laplace transform of a function. We first review some relevant definitions from calculus.
If \(f(t_0+)\) and \(f(t_0-)\) have finite but distinct values, we say that \(f\) has a jump discontinuity at \(t_0\), and
\[f(t_0+)-f(t_0-)\nonumber \]
is called the jump in \(f\) at \(t_0\) (Figure 8.1.1
).
Figure 8.1.1
: A jump discontinuity.
If \(f(t_0+)\) and \(f(t_0-)\) are finite and equal, but either \(f\) isn’t defined at \(t_0\) or it is defined but
\[f(t_0)\ne f(t_0+)=f(t_0-),\nonumber \]
we say that \(f\) has a removable discontinuity at \(t_0\) (Figure 8.1.2
). This terminology is appropriate since a function \(f\) with a removable discontinuity at \(t_0\) can be made continuous at \(t_0\) by defining (or redefining)
\[f(t_0)=f(t_0+)=f(t_0-).\nonumber \]
Figure 8.1.2
Note
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 \(\cal{L}(f)\).
Definition 8.1.4
: Piecewise Continuous
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 \(T>0\).
Figure 8.1.3
shows the graph of a typical piecewise continuous function.
It is shown in calculus that if a function is piecewise continuous on a finite closed interval then it is integrable on that interval. But if \(f\) is piecewise continuous on \([0,\infty)\), then so is \(e^{-st}f (t)\), and therefore\[\int_0^T e^{-st}f(t)\,dt \nonumber \]
Figure 8.1.3
exists for every \(T>0\). However, piecewise continuity alone does not guarantee that the improper integral
converges for \(s\) in some interval \((s_0,\infty)\). For example, we noted earlier that Equation \ref{eq:8.1.13} diverges for all \(s\) if \(f(t)=e^{t^2}\). Stated informally, this occurs because \(e^{t^2}\) increases too rapidly as \(t\to\infty\). The next definition provides a constraint on the growth of a function that guarantees convergence of its Laplace transform for \(s\) in some interval \((s_0,\infty)\).
Definition 8.1.5
: of exponential order
A function \(f\) is said to be of exponential order \(s_0\) if there are constants \(M\) and \(t_0\) such that
In situations where the specific value of \(s_0\) is irrelevant we say simply that \(f\) is of exponential order.
The next theorem gives useful sufficient conditions for a function \(f\) to have a Laplace transform. The proof is sketched in Exercise 8.1.10.
Theorem 8.1.6
If \(f\) is piecewise continuous on \([0,\infty)\) and of exponential order \(s_0,\) then \({\cal L}(f)\) is defined for \(s>s_0\).
Note
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
Example 8.1.8
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
.
Example 8.1.9
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
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.
Example 8.1.10
Find the Laplace transform of the piecewise continuous function