2802 Definition of Laplace Transform

Definition of Laplace Transform

The Laplace transform of a function $f(t)$ of a real variable $t$ is another function depending on a new variable $s$, which is in general complex. We will denote the Laplace transform of $f$ by $\mathcal{L}f$. It is defined by the integral $$(\mathcal{L}f)(s)=\int_{0^-}^\infty f(t)e^{-st}dt\tag{1}$$ for all values of $s$ for which the integral converges.

There are a few things to note. * $\mathcal{L}f$ is only defined for those values of $s$ for which the improper integral on the right-hand side of $(1)$ converges. * We will allow $s$ to be complex. * As with convolution the use of $0^-$, in the definition $(1)$ is necessary to accomodate generalized functions containing $\delta(t)$. Many textbooks do not do this carefully, and hence their definition of the Laplace transform is not consistent with the properties they assert. In those cases where $0^-$ isn't needed we will use the less precise form $$(\mathcal{L}f)(s)=\int_{0}^\infty f(t)e^{-st}dt\tag{1'}$$ * Also, as with convolution, the limits of integration mean that the Laplace transform is only concerned with functions on $(0^-, \infty)$.

Notation, $F(s)$

We will adopt the following conventions: * Writing $(\mathcal{L}f)(s)$ can be cumbersome so we will often use an uppercase letter to indicate the Laplace transform of the corresponding lowercase function: $$(\mathcal{L}f)(s)=F(s),(\mathcal{L}g)(s)=G(s),\text{ etc.}$$ For example, in the formula $$\mathcal{L}(f')=sF(s)-f(0^-)$$ it is understood that we mean $F(s) = \mathcal{L}(f)$. * If our function doesn't have a name we will use the formula instead. For example, the Laplace transform of the function $t^2$ is written $\mathcal{L}(t^2)(s)$ or more simply $\mathcal{L}(t^2)$. * If in some context we need to modify $f(t)$, e.g. by applying a translation by a number $a$, we can write $\mathcal{L}(f(t - a))$ for the Laplace transform of this translation of $f$. * You've already seen several different ways to use parentheses. Sometimes we will even drop them altogether. So, if $f(t) = t^2$ then the following all mean the same thing $$(\mathcal{L}f)(s) = F(s) = \mathcal{L}f(s) = \mathcal{L}(f(t))(s) = \mathcal{L}(t^2)(s); \mathcal{L}f = F = \mathcal{L}(t^2)$$

Examples

For the first few examples we will explicitly use a limit for the improper integral. Soon we will do this implicitly without comment.

Example 1. Let $f(t) = 1$, find $F(s) = \mathcal{L}f(s)$.
Solution. Using the definition $(1')$ we have $$\mathcal{L}(1)=F(s)=\int_0^\infty e^{-st}dt=\lim_{T \to \infty} \frac{e^{-st}}{-s}\bigg|_0^T=\lim_{T \to \infty} \frac{e^{-sT}-1}{-s}$$ The limit depends on whether s is positive or negative. $$\lim_{T \to \infty} e^{-sT}=\begin{cases} 0&\text{if }s>0\\ \infty&\text{if }s<0 \end{cases}$$ Therefore, $$\mathcal{L}(1)=F(s)=\begin{cases} \frac{1}{s}&\text{if }s>0\\ \text{diverges}&\text{if }s\leq 0 \end{cases}$$ (We didn't actually compute the case $s = 0$, but it is easy to see it diverges.)

Example 2. Compute $\mathcal{L}e^{at}$
Solution. Using the definition $(1')$ we have $$\mathcal{L}(e^{at})=F(s)=\int_0^\infty e^{at}e^{-st}dt=\lim_{T \to \infty} \frac{e^{(a-s)t}}{a-s}\bigg|_0^T=\lim_{T \to \infty} \frac{e^{(a-s)T}-1}{a-s}$$ The limit depends on whether $s > a$ or $s < a$. $$\lim_{T \to \infty} e^{(a-s)T}=\begin{cases} 0&\text{if }s>a\\ \infty&\text{if }s<a \end{cases}$$ Therefore, $$\mathcal{L}(e^{at})=\begin{cases} \frac{1}{s-a}&\text{if }s>a\\ \text{diverges}&\text{if }s\leq a \end{cases}$$ (We didn't actually compute the case $s = a$, but it is easy to see it diverges.)

We have the first two entries in our table of Laplace transforms: $$\begin{aligned} &f(t)=1 &\rArr &F(s)=1/s, &s>0\\ &f(t)=e^{at} &\rArr &F(s)=1/(s-a), &s>a \end{aligned}$$

Linearity

You will not be surprised to learn that the Laplace transform is linear. For functions $f,g$ and constants $c_1, c_2$ $$\mathcal{L}(c_1f+c_2g)=c_1\mathcal{L}(f)+c_2\mathcal{L}(g)$$ This is clear from the definition $(1)$ of $\mathcal{L}$ and the linearity of integration.