Section 5.1 Definitions
In functional analysis, there are a group of operations on functions called transforms. These operations act on sets of functions and transform them into new functions: through the transforms, behaviour that was previously hidden can be made clear.
The changes due to transforms are often radical: the resulting functions do not look anything like the originals. I can think of transforms as radically altering the environment of the function, so that everything changes into a surprisingly different form.
For those who like to think in linear algebra terms, think of the space of functions defined on some interval in \(\RR\) as a vector space. (The specific vector space depends on the interval and the class of function: continuous, piecewise-continuous, differentiable, etc). In this language, transforms are nothing but interesting linear transformations between vector spaces of functions.
In this section, I introduce the Laplace transform. It applies to a functions with certain controls on their asymptotic growth. I need a technical definition.
Definition 5.1.1.
A function \(f(t)\) defined on \([0, \infty)\) is of exponential order c if \(\exists M > 0 \ \
\exists T > 0\) such that \(\forall t > T \ \ \ |f(t)| \lt
Me^{ct}\text{.}\) Asymptotically, this is equivalent to \(f \in
\calO(e^{ct})\) for some real positive \(c\text{.}\)
Definition 5.1.2.
If \(f(t)\) is a piecewise-continuous function on \([0,
\infty)\) of exponential order \(c\text{,}\) then its Laplace transform is the new function in the variable \(s\) defined by the following integral.
\begin{equation*}
\calL \{ f(t) \} (s) = \int_0^\infty e^{-st} f(t) dt\text{.}
\end{equation*}
The Laplace transform is defined on the domain \((c,
\infty)\text{.}\) (You could check that this improper integral converges for all \(s \in (c, \infty)\text{.}\))
The restriction of exponential order is a fairly reasonable one: differential equations in applied mathematics are rarely concerned with functions which grow faster than the exponential.
The choice of variables is standard for Lapalce transforms. I will often refer to the original functions as functions in the \(t\)-domain, and to their transforms as functions in the \(s\)-domain. In addition, if \(f(t)\) and \(g(t)\) are function in the \(t\)-domain, I will also often write \(F(s)\) and \(G(s)\) for their Laplace transforms. By convention, I use lower case for the \(t\)-domain and uppercase for the matching function in the \(s\)-domain.
Most of the following examples are sub-exponential, so I expect a transform on \((0, \infty)\text{.}\) Note that I am never guaranteed convergence at \(0\text{.}\)
Example 5.1.3.
\begin{align*}
\calL \{ 1 \} \amp = \int_0^\infty e^{-st} dt = \left.
\frac{-e^{-st}}{s} \right|_0^\infty = \lim_{a \rightarrow
\infty} \frac{-e^{-sa} + 1}{s} = \frac{1}{s} \amp \amp\\
\calL \{ t \} \amp = \int_0^\infty te^{-st} dt = \lim_{a
\rightarrow \infty} \left[ \left. \frac{-te^{-st}}{s}
\right|_0^a + \int_0^a \frac{e^{-st}}{s} dt \right] \amp
\amp\\
\amp = \lim_{a \rightarrow \infty} \left[ \frac{0 -
ae^{-sa}}{s} + \left. \frac{-e^{-st}}{s^2} \right|_0^a
\right] = \frac{1}{s^2} \amp \amp\\
\calL \{ t^n \} \amp = \int_0^\infty t^ne^{-st} dt =
\frac{n!}{s^{n+1}} \amp \amp\\
\calL \{ e^{at} \} \amp = \int_0^\infty e^{-st}e^{at} dt =
\int_0^\infty e^{at-st} dt = \left. \frac{e^{at-st}}{a-s}
\right|_0^\infty = \frac{-1}{a-s} = \frac{1}{s-a}
\amp \amp s \in (a, \infty)
\end{align*}
My first observation about Laplace transforms is simply their strangeness. Powers of \(t\) turn into inverse powers of \(s\text{,}\) but exponentials (which are very different in the \(t\) domain) turn into very similar reciprocals.
Example 5.1.4.
Lapalce transforms exist for piecewise-continuous functions with a certain exponential order. Here is a piecewise continuous function and its transform.
\begin{align*}
f(t) \amp = \left\{ \begin{matrix} 0 \amp t \in [0,1]\\
t \amp t \in [1, \infty) \end{matrix} \right.\\
\calL \{ f(t) \} \amp = \int_0^\infty f(t) e^{-st} dt\\
\amp = \int_1^\infty te{-st} dt = \left.
\frac{-te^{-st}}{s} \right|_1^\infty + \int_1^\infty
\frac{e^{-st}}{s} dt\\
\amp = \frac{-e^{-s}}{s} + 0 + \left. \frac{-e^{-st}}{s^2}
\right|_1^\infty = \frac{-e^{-s}}{s} - \frac{e^{-s}}{s^2}
= \frac{-(s+1)e^{-s}}{s^2}
\end{align*}
The Laplace transform is not a piecewise function, even thought the original was.
Example 5.1.5.
The Laplace transform of \(t^n\) for \(n \in \ZZ\) involved factorials, so it is not surprising that \(t^\alpha\) for non-integer \(\alpha\) involves the extension of the factorial: the \(\Gamma\) function.
\begin{align*}
\calL \{ t^\alpha \} \amp = \int_0^\infty e^{-st}
t^{\alpha} dt\\
\amp = \int_0^\infty t^{(\alpha + a) - 1} e^{-st}\\
u \amp = st\\
\amp = \int_0^\infty \left( \frac{u}{s}
\right)^{(\alpha+1)-1} e^{-u} \frac{du}{s}\\
\amp = \frac{1}{s^{\alpha +1}} \int_0^\infty u^{(\alpha+
1)-1} e^{-u} du = \frac{\Gamma(\alpha +1)}{s^{\alpha +1}}
\end{align*}
Example 5.1.6.
Finally, Laplace transforms are defined for some important oscillating functions, such as the trigonometric functions and the Bessel functions.
\begin{align*}
\calL \{ \sin t \} \amp = \int_0^\infty e^{-st} \sin t dt
= \frac{1}{s^2+1}\\
\calL \{ \cos t \} \amp = \frac{s}{s^2+1}\\
\calL \{ J_0(kt) \} \amp = \frac{1}{\sqrt{s^2 + k^2}}
\end{align*}
Again, note the strangeness: even starting with transcendental and non-elementary functions like these, the Lapalce transforms are rational and algebraic functions.
As an aside, the Laplace transform is not the only transform in mathematics. The most well-known and well-used transform is the Fourier transform. It uses the same conventions about the \(s\) and the \(t\) domains, but it uses complex valued functions and integrals. In place of the \(e^{-st}\) term in the Laplace transform, the Fourier transform uses the complex \(e^{-2\pi \imath st}\text{.}\)
Definition 5.1.7.
The Fourier transform of a function \(f(s)\) is given by this integral.
\begin{equation*}
\hat{f}(s) = \int_{-\infty}^\infty f(t) e^{-2\pi \imath t s}
dt
\end{equation*}
