Properties of the Laplace Transform

Section 5.2 Properties of the Laplace Transform

Subsection 5.2.1 First Properties

Proposition 5.2.1.

The Laplace transform (like all reasonable operations in analysis) is linear.

\begin{align*} \calL \{ f+ g \} \amp = \calL \{ f \} + \calL \{ g \}\\ \ \calL \{ c f(t)\} \amp = c \calL \{ f \} \end{align*}

As I will demonstrate, the Laplace transform is not multiplicative. There is no good, general way to deal with (\(\calL \{ fg \}\)), though particular product have identities which I will show later.

Second, the asymptotic order of \(f\) was important to allowing the definition of the Laplace transform. However, all the transforms I’ve shown so far produce decaying functions. This is a general statment: for any function \(f(t)\) of exponential order \(c > 0\text{,}\) it is true that

\begin{equation*} \lim_{s \rightarrow \infty} \calL \{ f(t) \} (s) = 0\text{.} \end{equation*}

This property is a useful check: if you ever get a Laplace transform which doesn’t have this limit, you’ve made a mistake.

For same students who like linear algebra, I can use linear algebra language to describe the Laplace transform. Let \(EOF(0,\infty)\) be the space of piecewise-continuous functions of exponential order defined on \((0, \infty)\) (\(EOF\) for exponential order functions). Let \(DF^+\) be the set of continuous functions which are defined on \((a, \infty)\) for some \(a > 0\) and have limit \(0\) as \(s \rightarrow \infty\) (DF for decay functions). These are both linear spaces, since addition, subtraction and multiplication by constants preserve the properties of definition. The the Laplace transform is a linear transformation between these vector spaces.

\begin{equation*} \calL : EOF(0,\infty) \rightarrow DF^+ \end{equation*}

Subsection 5.2.2 Shifts

One of the product that does work well for Laplace transforms is multiplication by exponential functions.

Proposition 5.2.2.

\begin{align*} \calL \{ e^{at} f(t) \} (s) \amp = \int_0^\infty e^{at} f(t) e^{-st} dt\\ \amp = \int_0^\infty f(t) e^{at-st} dt = \int_0^\infty f(t) e^{(a-s)t} dt\\ \amp = \calL \{ f(t) \} (s-a) \end{align*}

Multiplication by \(e^{at}\) in the \(t\) domain results in a shift by \(a\) in the \(s\) domain. This also adjust the domain of \(s\text{,}\) reflected the fact that multiplication by \(e^{at}\) adjusts the exponential order.

Shifts are quite important in the whole theory. What happens when the shift is in the \(t\) domain?

Definition 5.2.3.

The unit step function \(u_a(t)\) is a piecewise function defined as follows.

\begin{equation*} u_a(t) = \left\{ \begin{matrix} 0 \amp t \lt a \\ 1 \amp t \geq a. \end{matrix} \right. \end{equation*}

The unit step function multiplied by some other function (\(u_a(t) f(t)\)) cancels any part of \(f\) for \(t\lt a\text{,}\) but preserves \(f\) for \(t \geq a\text{.}\) To define a shift of \(f\text{,}\) I can write \(u_a(t) f(t-a)\) to remove anything before \(t=0\) from \(f\text{.}\) (Remember that piecewise continuous functions work with Laplace transforms.) I calculate to find the result of this shift in the \(t\)-domain.

\begin{align*} \calL \{ u_a(t) \} (s) \amp = \int_0^\infty u_a(t) e^{-st} dt = \int_a^\infty e^{-st} dt\\ \amp = \left. \frac{-e^{-st}}{s} \right|_a^\infty = \frac{e^{-as}}{s}\\ \calL \{ u_a (t) f(t-a)\} (s) \amp = \int_a^\infty e^{-st} f(t-a) dt\\ v \amp = t-a\\ \amp = \int_0^\infty e^{-s(v+a)} f(v) dv = e^{-sa} \int_0^\infty e^{-sv} f(v) dv\\ \amp = e^{-sa} \calL \{ f(t) \} (s) = e^{-sa} F(s) \end{align*}

There is a nice parallel here, though it’s not a perfect symmetry. Shifts in one domain correspond to multiplication by exponentials in the other domain.

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