Subsection 9.1.2 Definition
In the previous section, we mentioned the importance of integration by parts and \(e^{-st}\) to the Laplace transform process. To perform a Laplace transform is to multiply both sides of an equation by \(e^{-st}\) and integrate with respect to \(t\text{.}\) For instance, applying this process to the equation
\begin{equation*}
y'' + 3y' + 2y = -40e^{3t}
\end{equation*}
would look like this:
\begin{align*}
e^{-st} \left( y'' + 3y' + 2y \right)
=\amp\ e^{-st} \left( -40e^{3t} \right)\\
\int e^{-st} \left( y'' + 3y' + 2y \right) \, dt
=\amp\ \int e^{-st} \left( -40e^{3t} \right) \, dt\\
\ub{\int e^{-st} {\color{blue} y''} \, dt}_{\text{transform of } {\color{blue} y''}} +
3\ub{\int e^{-st} {\color{blue} y'} \, dt}_{\text{transform of } {\color{blue} y'}} +
2\ub{\int e^{-st} {\color{blue} y} \, dt}_{\text{transform of } {\color{blue} y}}
=\amp\ \ub{\int e^{-st} \left( {\color{blue} -40e^{3t}} \right) \, dt}_{\text{transform of } {\color{blue} -40e^{3t}}}\text{.}
\end{align*}
As this shows, applying the transform to the equation amounts to applying it to each term of our original equation (in blue). The following definition of the Laplace transform makes this more precise.
Definition 2. Laplace Transform.
The Laplace Transform of a given function \(f(t)\) is defined as
\begin{equation*}
\lap{ {\color{blue} f(t)} } = \int_0^{\infty} e^{-st} {\color{blue} f(t)}\ dt,
\end{equation*}
provided that the integral exists (i.e., the improper integral converges).This definition may appear daunting at first glance, but let’s break it down. The \(\infty\) in the upper limit signals that we are dealing with an improper integral. In calculus, we learned that improper integrals can be understood as limits. Specifically:
\begin{equation*}
\int_0^{\infty} e^{-st} f(t)\, dt = \ub{\lim_{b \to \infty} \ob{\int_0^b e^{-st} f(t)\, dt}^{\text{step 1: compute the integral}}}_{\text{step 2: compute the limit}}
\end{equation*}
Notice that while the integral involves both \(s\) and \(t\text{,}\) the integration is performed with respect to \(t\text{.}\) If you’re unfamiliar with handling multiple variables like this, simply treat \(s\) as a constant during the integration process. Now, let’s walk through an example to see this concept in action.
Reading Questions Check-Point Questions
1. The Laplace transform only exists if the improper integral converges.
The Laplace transform only exists if the improper integral converges
True.
This statement is true. The Laplace transform is defined by an improper integral, and it only exists when this integral converges. If the integral diverges, the Laplace transform is not defined.
False.
This statement is true. The Laplace transform is defined by an improper integral, and it only exists when this integral converges. If the integral diverges, the Laplace transform is not defined.
2. The Laplace transform is defined as \(\ds\lap{ f(t) } = \int_0^{\infty} e^{-st} f(t)\, dt\text{,}\) provided the exists.
The Laplace transform is defined as \(\ds\lap{ f(t) } = \int_0^{\infty} e^{-st} f(t)\, dt\text{,}\) provided the exists
- integral
Correct! The Laplace transform is defined only if the integral converges.
- function
No, it is the convergence of the integral that determines whether the Laplace transform is defined.
-
\(s\)-value
No, the Laplace transform is defined by an integral, not by a specific \(s\)-value.
- derivative
No, the Laplace transform is defined by an integral, not by a derivative.
3. Fill-in-the-Blank. The process of finding a Laplace transform involves multiplying the original function by \(\ul{\hspace{5em}}\) and \(\ul{\hspace{5em}}\) with respect to \(t\).
Fill-in-the-Blank. The process of finding a Laplace transform involves multiplying the original function by \(\ul{\hspace{5em}}\) and \(\ul{\hspace{5em}}\) with respect to \(t\)
-
\(\ds \ e^{-t}, \ \) integrating
No, the correct expression involves the exponential function \(\ds e^{-st}\text{.}\)
-
\(\ds \ e^{-st}, \ \) integrating
Correct! The Laplace transform involves multiplying the original function by \(\ds e^{-st}\) and integrating with respect to \(t\text{.}\)
-
\(\ds \ e^{st}, \ \) integrating
No, the correct expression involves the exponential function \(\ds e^{-st}\text{.}\)
-
\(\ds \ e^{-st}, \ \) differentiating
No, the Laplace transform involves integrating with respect to \(t\text{.}\)
-
\(\ds \ s, \ \) differentiating
No, the correct form involves the exponential function \(\ds e^{-st}\) and integrating with respect to \(t\text{.}\)
