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\fancyhead[CO]{Notes on the Laplace transform}
\fancyhead[CE]{David Pierce}
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\title{Notes on the Laplace transform}
\author{David Pierce}
\date{\today}

\begin{document}
\maketitle

%These notes are prepared by a fallible human; the reader should check
%all computations.

Our aim is to develop a method for solving initial-value problems
\begin{equation}\label{eqn:ivp}
\begin{gathered}
  y^{(n)}+a_1y^{(n-1)}+\dotsb+a_{n-1}y'+a_ny=g,\\
y(t_0)=y_0,\quad
y'(t_0)=y'_0,\quad
\dots,\quad
y^{(n-1)}(t_0)=y^{(n-1)}_0,
\end{gathered}
\end{equation}
where the coefficients $a_i$ are constant, although the
\textbf{forcing function} $g$ need not be constant.  To give the
conditions that the forcing function will have to satisfy, we need
some definitions:

A function is \textbf{piecewise continuous} on an interval $[a,b]$ if
the interval has a partition
\begin{equation}
  a=t_0<t_1<\dotsb<t_{n-1}<t_n=b
\end{equation}
such that the function is continuous on each sub-interval
$(t_k,t_{k+1})$ and has finite limits at the endpoints of these
intervals.  The value of the function \emph{at} the points $t_k$ is
irrelevant.  Indeed, we may say that $f$ is piecewise continuous on
$[a,b]$ even if it is undefined at the points $t_k$.  

\begin{thm-defn*}
  Suppose $f$ is piecewise continuous on $[0,b]$ for each positive
  $b$, and
  \begin{equation}\label{eqn:exp-ord}
    \lim_{t\to\infty}\me^{-at}f(t)=0
  \end{equation}
for some $a$.  Then $f$ has a \textbf{Laplace transform}, namely the
function $F$ on $(a,\infty)$ given by
\begin{equation}\label{eqn:F}
  F(s)=\int_0^{\infty}\me^{-st}f(t)\dee t.
\end{equation}
\end{thm-defn*}

\begin{proof}
  By~\eqref{eqn:exp-ord}, there is some (positive) $R$ such that
  $\size{\me^{-at}f(t)}<1$ when $R<t$.  
But then $f$ is piecewise continuous on $[0,R]$, so
  $\int_0^R\me^{-st}f(t)\dee t$
  exists.  Also, when $R<t$, we have
  \begin{equation}
    \size{\me^{-st}f(t)}\leq\me^{(a-s)t};
  \end{equation}
and when $a<s$, the integral $\int_R^{\infty}\me^{(a-s)t}\dee t$ converges;
so $\int_R^{\infty}\me^{-st}f(t)\dee t$ converges.  Therefore the
integral in~\eqref{eqn:F} converges when $a<s$.
\end{proof}

The property expressed by equation \eqref{eqn:exp-ord} is that
$f$ has \textbf{exponential order}.
If $f$ has the Laplace transform $F$, then we may write 
\begin{equation}
F=\Lpt f.
\end{equation}
Often $F$ is expressed in terms of its values $F(s)$, and $f$ in terms
of $f(t)$; then we may write
\begin{equation}\label{eqn:notation}
  F(s)=\Lpt{f(t)}.
\end{equation}

The solution of problems~\eqref{eqn:ivp} by means of Laplace transforms
involves three essential lemmas:

\begin{lemma}\label{fact:linear}
The operation $f\mapsto\Lpt{f}$ is linear, that is, 
\begin{equation}
\Lpt{c_0f_0+c_1f_1}=c_0\cdot\Lpt{f_0}+
  c_1\cdot\Lpt{f_1}.
\end{equation} 
\end{lemma}

\begin{proof}
  Multiplication by $\me^{-st}$ and integration are linear.
\end{proof}

Remembering the proof of the following may be an aid in remembering
the lemma itself:

\begin{lemma}\label{fact:der}
If the derivatives $f^{(k)}$ are of
  exponential order and are continuous on $[0,\infty)$ when $k<n$,
  and if $f^{(n)}$ is
  piecewise continuous on $[0,\infty)$, then $\Lpt{f^{(n)}}$ exists
  and is given by 
  \begin{equation}\label{eqn:fact2}
    \Lpt{f^{(n)}}= s^n\cdot\Lpt{f}-(s^{n-1}\cdot
    f(0)+s^{n-2}\cdot f'(0)+\dotsb+ f^{(n-1)}(0)).
  \end{equation}
In particular,
\begin{align}\label{eqn:der}
  \Lpt{f'}&=s\cdot\Lpt{f}-f(0),\\
  \Lpt{f''}&=s^2\cdot\Lpt{f}-s\cdot f(0)-f'(0).
\end{align}
\end{lemma}

\begin{proof}
Use integration by parts and induction.
Equation~\eqref{eqn:fact2} holds trivially when $n=0$.  By integration
by parts, 
  \begin{align}
    \Lpt{f^{(m+1)}}&=\int_0^{\infty}\me^{-st}f^{(m)}(t)\dee t\\
&=\bndry{\me^{-st}f^{(m)}(t)}
    0{\infty}+s\cdot\int_0^{\infty}\me^{-st}f^{(m)}(t)\dee t\\ 
&=-f^{(m)}(0)+s\cdot\Lpt{f^{(m)}},
  \end{align}
assuming $f^{(m)}$ has exponential order.
Hence, if~\eqref{eqn:fact2} holds when $n=m$, then it holds when
$n=m+1$ (under the stated conditions).
\end{proof}

\begin{lemma}\label{fact:inv}
  Suppose $f$ and $g$ are piecewise continuous on each interval
  $[0,a]$.  Then $\Lpt f=\Lpt g$ if and only if $f(t)=g(t)$ for all
  but finitely many $t$ in each interval $[0,a]$.
\end{lemma}

\begin{proof}
  Part of this is easy, since the integral of a function is not
  changed if the value of the function is changed at finitely many
  points.  We shall not try to prove the hard part.
\end{proof}

Lemma~\ref{fact:inv} means that, if $\Lpt f=F$, then we can write
\begin{equation}
  \Lptinv F=f:
\end{equation}
in a word, the operation $f\mapsto\Lpt f$ is \textbf{invertible.}

To simplify matters, let us say we want to solve the
intermediate-value problem 
\begin{equation}\label{eqn:simple}
  y''+ay'+by=g,\quad y(0)=y_0,\quad y'(0)=y'_0.
\end{equation}
(What is important here is that the differential equation is linear with
constant coefficients.)  By Lemma~\ref{fact:linear} we get
\begin{equation}
  \Lpt{y''}+a\cdot\Lpt{y'}+ b\cdot\Lpt{y}=\Lpt{g}.
\end{equation}
Writing $Y$ for $\Lpt y$, by Lemma~\ref{fact:der} we get
\begin{equation}
  s^2\cdot Y-s\cdot y_0-y'_0+as\cdot Y-ay_0+b\cdot Y=\Lpt{g},
\end{equation}
which we can solve for $Y$:
\begin{equation}\label{eqn:sol1}
  Y=\frac{y_0s+ay_0+y'_0}{s^2+as+b} +\frac{\Lpt{g}}{s^2+as+b}.
\end{equation}
So, to be able to solve~\eqref{eqn:simple} by means of Laplace
transforms, we need:
\begin{enumerate}
  \item
to be able to find $\Lpt g$ easily;
\item
to be able to recover $y$ as $\Lptinv Y$.
\end{enumerate}
If $g$ is a reasonable function, then (because of
 Lemma~\ref{fact:inv}) we can accomplish these tasks
 by knowing some standard Laplace transforms.

First, we find the transforms of polynomials:

 \begin{theorem}
For all non-negative integers $n$,
\begin{equation}
  \Lpt{t^n}=\frac{n!}{s^{n+1}}.
\end{equation}
 \end{theorem}

 \begin{proof}
Just compute:
   \begin{align}\label{eqn:1}
  \Lpt 1
&=\int_0^{\infty}\me^{-st}\dee
  t=\bndry{\frac{\me^{-st}}{-s}}0{\infty}=\frac 1s;\\
  \Lpt{t^{k+1}}
&= \int_0^{\infty}\me^{-st}t^{k+1}\dee t\\
&= \bndry{\frac{\me^{-st}t^{k+1}}{-s}}0{\infty}+ \frac{k+1}{s}
  \int_0^{\infty} \me^{-st}t^k\dee t\\
&= \frac{k+1}s\cdot\Lpt{t^k};
   \end{align}
Now the claim follows by induction.
 \end{proof}

If we know that $\Lpt f=F$, then we can recover the function whose
Laplace transform is $F(s-c)$:

\begin{theorem}\label{thm:shift}
\begin{equation}
  \Lpt{\me^{ct}f(t)}=F(s-c).
\end{equation}
\end{theorem}

\begin{proof}
Apply definition~\eqref{eqn:F} twice:
  \begin{equation}
  \Lpt{\me^{ct}f(t)}=\int_0^{\infty}\me^{-(s-c)t}f(t)\dee t,
  \end{equation}
which is $F(s-c)$.
\end{proof}

\begin{theorem}\label{thm:th}
  \begin{align*}
\Lpt{\cosh ct}&=\frac{s}{s^2-c^2}, & \Lpt{\cos ct}&=\frac s{s^2+c^2},\\
\Lpt{\sinh ct}&=\frac c{s^2-c^2}, &\Lpt{\sin ct}&=\frac c{s^2+c^2}.
\end{align*}
\end{theorem}

\begin{proof}
By Theorem~\ref{thm:shift} and~\eqref{eqn:1},
\begin{equation}\label{eqn:ct}
  \Lpt{\me^{ct}}=\frac 1{s-c}.
\end{equation}
Hence, by using the definitions
\begin{equation}\label{eqn:th}
\begin{aligned}
  \cosh t&= \frac{\me^t+\me^{-t}}2, & \cos t&=\cosh it,\\
  \sinh t&= \frac{\me^t-\me^{-t}}2, & \sin t&=\frac{\sinh it}{i},
\end{aligned}
\end{equation}
we can compute the desired transforms.
\end{proof}

An alternative proof of Theorem~\ref{thm:th} uses the following rule.

\begin{theorem}
\begin{equation}
  \Lpt{f(ct)}= \frac 1c\cdot F\left(\frac
  sc\right). 
\end{equation}
\end{theorem}

\begin{proof}
The computation is
\begin{equation}
  \Lpt{f(ct)}=\int_0^{\infty}\me^{-st}f(ct)\dee t= \frac
  1c\int_0^{\infty} \me^{-su/c}f(u)\dee u,
\end{equation}  
which is $(1/c)\cdot F(s/c)$.
\end{proof}

A special case of~\eqref{eqn:ct} is
\begin{equation}
   \Lpt{\exp}=\displaystyle\frac1{s-1}.
\end{equation}
Then, from~\eqref{eqn:th}, we get
\begin{align*}
\Lpt{\cosh}&=\frac{s}{s^2-1}, & \Lpt{\cos}&=\frac s{s^2+1},\\
\Lpt{\sinh}&=\frac 1{s^2-1}, &\Lpt{\sin}&=\frac 1{s^2+1}.
\end{align*}
Now Theorem~\eqref{thm:th} follows.

\begin{theorem}
  If $F$ is a rational function, then $\Lptinv{F}$ exists and
 is a linear
  combination of
$\me^{ct}$ and $\me^{ct}\cos bt$ and $\me^{ct}\sin bt$ (for various
  $b$ and $c$).
\end{theorem}

\begin{proof}
If $F$ is a rational function, then we can write $F(s)$ as a sum of
`partial' fractions of the forms
\begin{equation}
  \frac A{s-a}\quad\text{ and }\quad \frac{As+B}{s^2+as+b},
\end{equation}
where $a^2<4b$ in the latter.  Hence we can write $F$ as a linear
combination of fractions of the forms
\begin{equation}
  \frac 1{s-c}\quad\text{ and }\quad \frac{s-c}{(s-c)^2+b^2}\quad
  \text{ and }\quad \frac b{(s-c)^2+b^2}.
\end{equation}
These are respectively the transforms of
\begin{equation}
  \me^{ct}\quad\text{ and }\quad \me^{ct}\cos bt\quad\text{ and }\quad
  \me^{ct}\sin bt;
\end{equation}
so, by Lemma~\ref{fact:linear}, we can compute $\Lptinv F$.
\end{proof}

Now we can use Laplace transforms to solve~\eqref{eqn:simple} (and
higher-order problems) in case $g$ is a linear combination of
polynomial and exponential and trigonometric functions.  The solution
does not require solving a homogeneous equation first, or finding a
general solution before using the initial conditions.

\subsection*{Piecewise-continuous forcing functions}

If $g$ is merely piecewise continuous,
will still may be able to compute $\Lpt g$ by expressing $g$ in terms
of \textbf{step-functions}.  Let $u_c$ be the function on
$(-\infty,\infty)$ given by
\begin{equation}
  u_c(t)=
  \begin{cases}
    0,&\text{ if }t<c;\\
1,&\text{ if }c\leq t.
  \end{cases}
\end{equation}
If, for example, $g$ is given by
\begin{equation}
  g(t)=
  \begin{cases}
    g_0(t),&\text{ if }\phantom{t_1< {}}t\leq t_1;\\
g_1(t),&\text{ if }t_1<t<t_2;\\
g_2(t),&\text{ if }t_2\leq t;
  \end{cases}
\end{equation}
then
\begin{equation}
  g(t)=g_0(t)+u_{t_1}(t)\cdot(g_1(t)-g_0(t)) +
  u_{t_2}(t)\cdot(g_2(t)-g_1(t)), 
\end{equation}
unless $t=t_1$; this one difference won't affect computation of the
Laplace transform, by Lemma~\ref{fact:inv}.  We can perform this
computation by means of the following.

\begin{theorem}\label{thm:piece}
  \begin{equation}
  \Lpt{u_c(t)\cdot f(t-c)}=\me^{-cs}F(s).
  \end{equation}
\end{theorem}

\begin{proof}
Compute:
\begin{equation}
  \Lpt{u_c(t)\cdot f(t-c)}=\int_c^{\infty}\me^{-st}f(t-c)\dee t
= \int_0^{\infty} \me^{-s(u+c)}f(u)\dee u,
\end{equation}
which is $\me^{-cs}F(s)$.
\end{proof}

\subsection*{Impulse functions}

We have encountered no function whose Laplace transform is a non-zero
constant.
In calculus, when we note that $1/x$ is not the derivative of a
rational function, we \emph{define} $\log x=\int_1^x\dee t/t$.
Similarly, we look for a function $\delta$ (the
\textbf{Dirac delta}) satisfying
\begin{equation}
  \Lpt{\delta}=1
\end{equation}
and---so that Theorem~\ref{thm:piece} is satisfied---
\begin{equation}
  \Lpt{\delta(t-c)}=\me^{-cs}.
\end{equation}
These properties are achieved if $\delta(t)=0$ when $t\neq0$, but
$\int_{-\infty}^{\infty}\delta =1$.
There is no such function, literally; but the equation
\eqref{eqn:sol1} still makes sense if $g$ is a
multiple of $\delta(t-c)$, and then we can find $y$ by means of
Theorem~\ref{thm:piece}. 

\subsection*{Convolution}

\begin{thm-defn*}
  If $\Lpt f=F$ and $\Lpt g=G$,
then
\begin{equation}
  F\cdot G=\Lpt{f*g},
\end{equation}
where the \textbf{convolution} $f*g$ is defined by
\begin{equation}
  f*g(t)=\int_0^tf(t-x)\cdot g(x)\dee x.
\end{equation}
\end{thm-defn*}
In particular, Equation \eqref{eqn:sol1} has the solution
$y=\phi+g*h$, where
\begin{equation}
  \Lpt{\phi}=\frac{y_0s+ay_0+y'_0}{s^2+as+b}\quad \text{ and }
  \Lpt{h}=\frac1{s^2+as+b}. 
\end{equation}
Note that $\phi$ and $h$ are independent of $g$; also $\phi=0$ if
$y_0=0$ and $y'_0=0$.  Once $\phi$ and $h$ are found, then solutions
to \eqref{eqn:sol1} can be obtained by integration.

\end{document}
