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### Laplace Transforms

#### Discussion

The next topic we will cover this semester is Laplace transforms. This is a method of solving differential equations that is very different from any of the methods we have used previously, but has a great many important applications. It is also the simplest of a whole family of techniques called transform techniques. The Laplace transform is the only one we will use in this class. Let $f(t)$ be a function. The**Laplace transform**${\mathcal L}\{f(t)\}$ of the function $f(t)$ is defined to be $${\mathcal L}\{f(t)\}=\int_0^{\infty}f(t)e^{-st}\,dt$$ We should worry about the convergence of the improper integral, but we won't in this course, as long as you promise me to remember that convergence is a possible problem that should be considered in later courses where you have more time to spend on this subject. If you want to discuss convergence now, you are welcome to stop by my office.

Note that the Laplace transform takes a function of a variable $t$ to another function of a variable $s$. (This is in fact what is meant by the term transform). Also note that the Laplace transform of a function only depends on the values of the function for $0\le t$ and not on the values of the function for $t<0$. This will be important when we apply Laplace transforms to initial value problems.

There are two ways to compute a Laplace transform. One way is to compute the improper integral.

EXAMPLE: What is ${\mathcal L}\{e^{rt}\}$? $$ \begin{align} {\mathcal L}\{e^{rt}\}&=\int_0^{\infty}e^{rt}e^{-st}\,dt \\ &=\int_0^{\infty}e^{(r-s)t}\,dt \\ &=\left.\frac{1}{r-s}e^{(r-s)t}\right|_0^{\infty} \\ &=\frac{1}{s-r} \end{align} $$ provided $s>r$ (otherwise the integral diverges and the Laplace transform is undefined).

This is the long way. The other way is to look up the answer in the Table of Laplace Transforms I've provided you with. This is the short way. Of course looking things up suffers from the disadvantage that the table can't list every function, but it lists enough for our purposes in this class. Using the Laplace transform table is similar to using a table of integrals; you are responsible for manipulating the expressions so they fit the forms in the table. To manipulate expressions so they fit the forms in the table, it is useful to note that the Laplace transform is a linear operator.

*Theorem* Let $f(t)$ be a function and $c$ a constant. Then
$$\begin{align}
{\mathcal L}\{f(t)+g(t)\} &= {\mathcal L}\{f(t)\} + {\mathcal L}\{g(t)\}
\\
{\mathcal L}\{cf(t)\} &= c{\mathcal L}\{f(t)\}
\end{align} $$
This is important
because it allows us to reduce the problem of finding the Laplace transform
of a complicated function to a collection of simpler problems by rewriting
the complicated function as a sum of simpler pieces. The simpler pieces can
then be dealt with by consulting our table.

EXAMPLE: Compute ${\mathcal L}\{3\sin(2t)\}$.

First we note that ${\mathcal L}\{3\sin(2t)\}=3{\mathcal L}\{\sin(2t)\}$. Now we look for ${\mathcal L}\{\sin(2t)\}$ in the table. We find that ${\mathcal L}\{\sin(at)\}=a/(s^2+a^2)$. So using this formula with $a=2$ we get ${\mathcal L}\{\sin(2t)\}=2/(s^2+4)$. Finally ${\mathcal L}\{3\sin(2t)\}=6/(s^2+4)$.

EXAMPLE: Compute ${\mathcal L}\{e^t-e^{2t}\}$.

We write ${\mathcal L}\{e^t-e^{2t}\}={\mathcal L}\{e^t\}-{\mathcal L}\{e^{2t}\}$. From the table we find ${\mathcal L}\{e^{at}\}=1/(s-a)$. So applying this with $a=1$ yields ${\mathcal L}\{e^t\}=1/(s-1)$ and with $a=2$ yields ${\mathcal L}\{e^{2t}\}=1/(s-2)$. Finally ${\mathcal L}\{e^t-e^{2t}\}=1/(s-1)-1/(s-2)=-1/(s-1)(s-2)$.

Of course, being able to take the Laplace transform of a function isn't very useful if you aren't going to be able to undo the transform later to recover the original function. We denote the inverse Laplace transform by ${\mathcal L}^{-1}\{F(s)\}$ and then $f(t)={\mathcal L}^{-1}\{{\mathcal L}\{f(t)\}\}$. It is possible to give a definition of the inverse Laplace transform similar to the definition of the Laplace transform, but it isn't very useful. $$ f(t)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}e^{st}{\mathcal L}\{f(t)\}(s)\,ds $$ for all sufficiently large $a$. This is a complex contour integral which you can learn about in Introduction to Complex Analysis (or stop by my office). We aren't going to use this formula in this class, instead we will just reverse the process of finding a Laplace transform in the table. It should be noted that the inverse of any invertible linear function is also linear. In particular $$ \begin{align} {\mathcal L}^{-1}\{F(s)+G(s)\}&={\mathcal L}^{-1}\{F(s)\}+{\mathcal L}^{-1}\{G(s)\} \\ {\mathcal L}^{-1}\{cF(s)\}&=c{\mathcal L}^{-1}\{F(s)\} \end{align} $$ Finally, since evaluating an inverse Laplace transform is evaluating a complicated integral, it should not be surprising that if you have an expression you don't know how to deal with, it is usually best to try the same manipulations you would use as if you were trying to integrate the expression (see the second example below).

EXAMPLE: Compute ${\mathcal L}^{-1}\left\{\displaystyle\frac{1}{(s-1)^2+4}\right\}$

In the table we find ${\mathcal L}\{e^{at}\sin(bt)\}=\displaystyle\frac{b}{(s-a)^2+b^2}$. Applying this with $a=1$ and $b=2$ we get ${\mathcal L}^{-1}\left\{\displaystyle\frac{2}{(s-1)^2+4}\right\} = e^t\sin(2t)$. So ${\mathcal L}^{-1}\left\{\displaystyle\frac{1}{(s-1)^2+4}\right\}= \displaystyle\frac 12e^t\sin(2t)$. Note that it is best to get the denominator right first and then multiply by a constant to get the numerator right.

EXAMPLE: Compute ${\mathcal L}^{-1}\left\{\displaystyle\frac{2}{s^2+3s+2}\right\}$

There isn't anything like this in the table at first glance. We have fractions with linear terms $(s-a)$ in the denominator and with sums of squares and the like but no general quadratic terms in the denominator. Now the general rule is to try to treat the expression as if we were going to integrate it. This is a case for partial fractions. $$ \begin{align} \frac{2}{s^2+3s+2}=\frac{2}{(s+2)(s+1)}&=\frac{A}{s+2}+\frac{B}{s+1} \\ 2&=A(s+1)+B(s+2) \\ 2&=(A+B)s+(A+2B) \end{align} $$ Equating the $s$ terms and the constant terms we get two equations in two unknowns $$ \begin{align} 0&=A+B \tag{$s$ terms} \\ 2&=A+2B \tag{constant terms} \end{align} $$ Solving we get $B=2$ and $A=-2$ so $$ \frac{2}{s^2+3s+2}=\frac{-2}{s+2}+\frac{2}{s+1} $$ Then $$ \begin{align} {\mathcal L}^{-1}\left\{\frac{2}{s^2+3s+2}\right\}&= {\mathcal L}^{-1}\left\{\frac{-2}{s+2}+\frac{2}{s+1}\right\} \\ &={\mathcal L}^{-1}\left\{\frac{-2}{s+2}\right\} +{\mathcal L}^{-1}\left\{\frac{2}{s+1}\right\} \\ &=-2e^{-2t}+2e^{-t} \end{align} $$

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©2010, 2014 Andrew G. Bennett