PIECE-WISE OR SECTIONAL CONTINUITY
A function $f(x)$ is called sectionally continuous or piece-wise continuous in any interval $[a, b]$ if it is continuous and has finite left and right hand limits in every subinterval $\left[a_{1}, b_{1}\right]$.
FUNCTION OF EXPONENTIAL ORDER
A function $f(x)$ is said to be of exponential order $a$ as $x \rightarrow \infty$
if
$$\operatorname{Lim}_{x \rightarrow \infty} e^{-a x} f(x)=\text { finite quantity }$$
i.e., if given a positive integer $n_{0}$, there exists a real number $M>0$ s.t.
$$\begin{aligned}\left|e^{-a x} f(x)\right|<M & \quad \forall x \geq n_{0} \\or \quad|f(x)|<M e^{a x} & \quad \forall x \geq n_{0}\end{aligned}$$
Briefly we say that $f(x)$ is of exponential order.
Sometimes we write
$$f(x)=O\left(e^{a x}\right), x \rightarrow \infty$$
Example 1. Show that $x^{n}$ is of exponential order as $x \rightarrow \infty, n$ being any positive integer.
Solution.
$$\begin{aligned}\operatorname{Lim}_{x \rightarrow \infty} e^{-a x} x^{n}, a>0 &=\operatorname{Lim}_{x \rightarrow \infty} \frac{x^{n}}{e^{a x}}\\& =\operatorname{Lim}_{x \rightarrow \infty} \frac{n!}{a^{n} e^{a x}} \quad \text{(by L. Hospital’s rule)}\\& =\frac{n!}{\infty}=0\end{aligned}$$
$\therefore \quad \operatorname{Lim}_{x \rightarrow \infty} e^{a x} x^{n}=0=$ finite number.
Consequently $x^{n}$ is of exponential order as $x \rightarrow \infty$.
Example 2. Show that $F(t)=e^{t^{2}}$ is not of exponential order as $t \rightarrow \infty$.
Solution. $\lim_{t \rightarrow \infty} e^{-a t} F(t), a>0=\lim_{t \rightarrow \infty} e^{t(t-a)}=\infty$.
This $\Rightarrow \quad \lim _{t \rightarrow \infty} e^{-a t} F(t)=\infty$ if $a>0$
$\therefore \quad F(t)$ is not of exponential order.
FUNCTION OF CLASS $A$
A function $f(t)$ is said to be of class $A$ if $(i)$ it is sectionally continuous over every finite interval in the range $t \geq 0$.
(ii) $f(t)$ is of exponential order as $t \rightarrow \infty$.
Example 3. Show that the function $F(t)=t^{n}$, $-1<n<0$ is not a function of class A.
Solution. Given $-1<n<0, \quad \therefore \quad n+1>0$.
$F(t)=t^{n}=\frac{1}{t^{-n}}$ as $-n>0$
which is not sectionally continuous in every finite interval of range for $t \geq 0$.
Here $F(t)$ is not continuous at $t=0$.
Consequently $F(t)$ is not a function of class $A$.
THE TRANSFORM CONCEPT
An improper integral of the form
$$\int_{-\infty}^{\infty} K(s, t) F(t) d t$$
is called integral transform of $F(t)$ if it is convergent. Sometimes it is denoted by $f(s)$ or $T\{F(t)\}$. Thus
$$f(s)=T\{F(t)\}=\int_{-\infty}^{\infty} K(s, t) F(t) d t\quad \quad \ldots (1)$$
The function $K(s, t)$ appearing in the integrand is called Kernel of the transform. Here $s$ is a parameter and is independent of $t, s$ may be real or complex number.
If we take
$$K(s, t)= \begin{cases}e^{-s t} & , t \geq 0 \\ 0 & , t<0\end{cases}$$
Then (1) becomes
$$f(s)=T\{F(t)\}=\int_{0}^{\infty} F(t) e^{-s t} d t$$
This transform is known as Laplace transform.
Hankel transform of $F(t)$ is
$$f(s)=\int_{0}^{\infty} F(t) t J_{n}(s t) d t$$
Mellin transform of $F(t)$ is
$$f(s)=\int_{0}^{\infty} F(t) t^{s-1} J_{n}(s t) d t$$
Fourier transform of $F(t)$ is
$$f(s)=\int_{-\infty}^{\infty} F(t) e^{-i s t} d t$$
LAPLACE TRANSFORM
Suppose $F(t)$ is a real valued function defined over the interval $(-\infty, \infty)$ s.t. $F(t)=0, \forall t<0$.
The Laplace transform of $F(t)$, denoted by $L\{F(t)\}$, is defined as
$$L\{F(t)\}=\int_{0}^{\infty} e^{-s t} F(t) d t\quad \ldots (1)$$
We also write
$$L\{F(t)\}=f(s)=\int_{0}^{\infty} e^{-s t} F(t) d t$$
Here $L$ is called Laplace transformation operator. The parameter $s$ is a real or complex number. In general, the parameter $s$ is taken to be a real positive number. Sometimes we use symbol $p$ for parameter $s$.
Thus
$$L[F(t)]=f(p)=\int_{0}^{\infty} e^{-p t} F(t) d t$$
The Laplace transform is said to exist if the integral (1) is convergent for some value of $s$.
The operation of multiplying $F(t)$ by $e^{-s t}$ and integrating from $0$ to $\infty$ is called Laplace transformation.
NOTATION
We follow two types of notations :
(i) Functions are denoted by capital letters
$$F(t), G(t), H(t), \ldots$$
and their Laplace transforms are denoted by corresponding lower case letters
$$f(s), g(s), h(s), \ldots \text { or by } f(p), g(p) h(p), \ldots$$
(ii) Functions are denoted by lower case letters
$$f(t), g(t), h(t), \ldots$$
and their Laplace transforms are denoted by
$\bar{f}(s), \bar{g}(s), \bar{h}(s), \ldots$ respectively or $\bar{f}(p), \bar{g}(p), \bar{h}(p)$.
Theorem 1. Existence of Laplace Transform.
If $F(t)$ is a function of class $A$, then Laplace transform of $F(t)$ exists.
Or, Suppose $F(t)$ is piece-wise continous in every finite interval and is of exponential order $a$ as $t \rightarrow \infty$. Then $f(s)$ exists $\forall s>a$, i.e., Laplace transform exists.
Proof. Let $F(t)$ be piece-wise continuous in every finite interval and of exponential order $a$ as $t \rightarrow \infty$.
To show that $f(s)=L[F(t)]$ exists $\forall s>a$.
Let $t_{0}>0$. Then
$$f(s)=\int_{0}^{\infty} e^{-s t} F(t) d t=\int_{0}^{t_{0}} e^{-s t} F(t) d t+\int_{t_{0}}^{\infty} e^{-s t} F(t) d t$$
Continuity of $F(t)$ in the finite interval $\left(0, t_{0}\right)$ implies that
$$\int_{0}^{t_{0}} e^{-s t} F(t) d t$$
exists. It remains to show that
$$\int_{t_{0}}^{\infty} e^{-s t} F(t) d t$$
exists $\forall s>a$. $F(t)$ is of exponential order $a$ as $t \rightarrow \infty$ implies
$$\operatorname{Lim}_{t \rightarrow \infty} e^{-a t} F(t)$$
is finite, i.e., given a number $t_{0}$, there exists a real number $M>0$ s.t.
$$\begin{aligned}\left|e^{-a t} F(t)\right|<M,\; & \forall t \geq t_{0} \\\text{i.e.,}\quad |F(t)|<M e^{a t},\; & \forall t \geq t_{0}\end{aligned}$$
Now $\left|\int_{t_{0}}^{\infty} e^{-s t} F(t) d t\right| \leq \int_{t_{0}}^{\infty}\left|e^{-s t} F(t)\right| d t$
$$\begin{aligned}& =\int_{t_{0}}^{\infty} e^{-s t}|F(t)| d t \leq \int_{t_{0}}^{\infty} e^{-s t} M e^{a t} d t=M \int_{t_{0}}^{\infty} e^{-(s-a) t} d t \\& =\frac{M e^{-(s-a) t_{0}}}{s-a} \text { if } s>a\end{aligned}$$
Finally, $\quad\left|\int_{t_{0}}^{\infty} e^{-s t} F(t) d t\right| \leq \frac{M e^{-(s-a) t_{0}}}{s-a}$ if $s>a$.
$\frac{M e^{-(s-a) t_{0}}}{s-a}$ can be made as small as we please choosing $t_{0}$ sufficiently large. Hence $\int_{t_{0}}^{\infty} e^{-s t} F(t) d t$ exists $\forall s>a$.
REMARK : The conditions given in the last theorem are sufficient for the existence of $L\{ F(t)\}$, but are not the necessary conditions.
LAPLACE TRANSFORMS OF STANDARD FUNCTIONS
Result 1. $L(1)=\frac{1}{p}, p>0$.
Proof. $L(1)=\int_{0}^{\infty} e^{-p t} \cdot 1 \cdot d t=\left\{-\frac{e^{-p t}}{p}\right\}_{t=0}^{\infty}=\frac{1}{p}$.
Result 2. $L\left(t^{n}\right)=\frac{\Gamma(n+1)}{p^{n+1}}$ if $n+1>0$.
Proof. $L\left(t^{n}\right)=\int_{0}^{\infty} e^{-p t} \cdot t^{n} d t, p t=x, p d t=d x$
$$\begin{aligned}& =\int_{0}^{\infty} e^{-x}\left(\frac{x}{p}\right)^{n} \frac{d x}{p}=\frac{1}{p^{n+1}} \int_{0}^{\infty} e^{-x} x^{n+1-1} d x \\& =\frac{\Gamma(n+1)}{p^{n+1}}\end{aligned}$$
Also $L\left(t^{n}\right)=\frac{n!}{p^{n+1}}$ if $n$ is a positive integer.
Result 3. $L\left\{e^{a t}\right\}=\frac{1}{s-a}$, if $s>a$.
For $L\left\{e^{a t}\right\}=\int_{0}^{\infty} e^{-s t} e^{a t} d t=\int_{0}^{\infty} e^{-(s-a) t} d t$
$$=\left[-\frac{e^{-(s-a)}}{s-a}\right]_{t=0}^{\infty}=\frac{1}{s-a}, \text { if } s>a$$
Result 4. $L(\cos a t)=\frac{s}{s^{2}+a^{2}}, \quad L(\sin a t)=\frac{a}{s^{2}+a^{2}}$.
Proof. $L\left(e^{-i a t}\right)=\int_{0}^{\infty} e^{-p t} \cdot e^{-i a t} d t=\int_{0}^{\infty} e^{-t(p+i a)} d t$
$$\begin{gathered}=\int_{0}^{\infty} e^{-t s} d t=\left\{\frac{e^{-t s}}{-s}\right\}_{t=0}^{\infty}=\frac{1}{s}, \text { where } s=p+i a \\=\frac{1}{p+i a}=\frac{p-i a}{p^{2}+a^{2}} \\\Rightarrow \quad L(\cos a t-i \sin a t)=\frac{p-i a}{p^{2}+a^{2}} \\\Rightarrow \quad L(\cos a t)=\frac{p}{p^{2}+a^{2}}, \quad L(\sin a t)=\frac{a}{p^{2}+a^{2}} \\\Rightarrow \quad L(\cos a t)=\frac{s}{s^{2}+a^{2}}, \quad L(\sin a t)=\frac{a}{s^{2}+a^{2}} .\end{gathered}$$
Result 5. Find $L\{\sinh (a t)\}, L\{\cosh (a t)\}$.
Proof. (i) $L\{\sinh (a t)\}=L\left\{\frac{1}{2}\left(e^{a t}-e^{-a t}\right)\right\}=\frac{1}{2}\left\{\frac{1}{p-a}-\frac{1}{p+a}\right\}=\frac{a}{p^{2}-a^{2}}$.
(ii) $L\{\cosh (a t)\}=L\left\{\frac{1}{2}\left(e^{a t}+e^{-a t}\right)\right\}=\frac{1}{2}\left\{\frac{1}{p-a}+\frac{1}{p+a}\right\}=\frac{p}{p^{2}-a^{2}}$.
Theorem 2. Linear Property.
Suppose $f_{1}(s)$ and $f_{2}(s)$ are Laplace transforms of $F_{1}(t)$ and $F_{2}(t)$ respectively:
Then
$$L\left\{c_{1} F_{1}(t)+c_{2} F_{2}(t)\right\}=c_{1} L\left\{F_{1}(t)\right\}+c_{2} L\left\{F_{2}(t)\right\}$$
where $c_{1}$ and $c_{2}$ are any constants.
Proof. Let $$\begin{aligned}L\left\{F_{1}(t)\right\}&=f_{1}(s)=\int_{0}^{\infty} e^{-s t} F_{1}(t) d t\\\text{and}\quad L\left\{F_{2}(t)\right\}&=f_{2}(s)=\int_{0}^{\infty}e^{-s t} F_{2}(t) d t\end{aligned}$$
Also let $c_{1}$ and $c_{2}$ be arbitrary constants.
We claim
$L\left\{c_{1} F_{1}(t)+c_{2} F_{2}(t)\right\}=c_{1} L\left\{F_{1}(t)\right\}+c_{2} L\left\{F_{2}(t)\right\}$
Now,
$L\left\{c_{1} F_{1}(t)+c_{2} F_{2}(t)\right\}=\int_{0}^{\infty} e^{-s t}\left[c_{1} F_{1}(t)+c_{2} F_{2}(t)\right] d t$
$$\begin{aligned}& =c_{1} \int_{0}^{\infty} e^{-s t} F_{1}(t) d t+c_{2} \int_{0}^{\infty} e^{-s t} F_{2}(t) d t \\& =c_{1} L\left\{F_{1}(t)\right\}+c_{2} L\left\{F_{2}(t)\right\}\end{aligned}$$
REMARK : Generalizing this result, we obtain
$$L\left\{\sum_{r=1}^{n} c_{r} F_{r}(t)\right\}=\sum_{r=1}^{n} c_{r} L\left\{F_{r}(t)\right\}$$
Theorem 3. First Shifting Theorem (First Translation)
If $L\{F(t)\}=f(s)$, then $L\left\{e^{a t} F(t)\right\}=f(s-a)$.
Proof. Let $L\{F(t)\}=f(s)=\int_{0}^{\infty} e^{-s t} F(t) d t$.
Then, $\quad L\left\{e^{a t} F(t)\right\}=\int_{0}^{\infty} e^{-s t} e^{a t} F(t) d t=\int_{0}^{\infty} e^{-(s-a) t} F(t) d t$
$$\begin{aligned}& =\int_{0}^{\infty} e^{-u t} F(t) d t, \text { where } u=s-a>0 \\& =f(u)=f(s-a)\end{aligned}$$
REMARK : This theorem can also be restated as :
If $f(s)$ is the Laplace transform of $F(t)$ and $a$ is any real, or complex number, then $f(s+a)$ is the Laplace transform of $e^{-a t} F(t)$. That is to say,
$$f(s)=L\{F(t)\} \Rightarrow f(s+a)=L\left\{e^{-a t} F(t)\right\}$$
Example 4. Find $L\left(t^{n} e^{a t}\right), L\left(e^{b t} \cos a t\right), L\left(e^{b t} \sin a t\right)$.
Solution. By first shifting theorem.
$$\begin{aligned}L\left\{e^{a t} F(t)\right\} & =f(p-a) \text { if } L\{F(t)\}=f(p) \\\text{Also}\quad L\left(t^{n}\right) & =\frac{\Gamma(n+1)}{p^{n+1}}\end{aligned}$$
These results $\Rightarrow \quad L\left(t^{n} e^{a t}\right)=\frac{\Gamma(n+1)}{(p-a)^{n+1}}$,
$$L\left(e^{b t} \cos a t\right)=\frac{p-b}{(p-b)^{2}+a^{2}}, \quad L\left(e^{b t} \sin a t\right)=\frac{a}{(p-b)^{2}+a^{2}}$$
Example 5. Find $L\left(t^{5} e^{3 t}\right)$.
Solution. $\quad L\left(t^{5}\right)=\frac{\Gamma(n+1)}{s^{n+1}}$ if $n=5$
$$\begin{array}{ll}\Rightarrow & L\left(t^{5}\right)=\frac{\Gamma(6)}{s^{6}}=\frac{5!}{s^{6}} \\\Rightarrow & L\left(t^{5} e^{3 t}\right)=\frac{120}{(s-3)^{6}} \text { as } L\left\{e^{a t} F(t)\right\}=f(s-a)\end{array}$$
Example 6. Find $L\left\{e^{-t}(3 \sinh 2 t-5 \cosh 2 t)\right\}$.
Solution. We know that
$$\text{if}\quad L\{F(t)\}=f(s) \text {, then } L\left\{e^{a t} F(t)\right\}=f(s-a)$$
Also
$$L\{\sinh 2 t\}=\frac{2}{s^{2}-2^{2}}, \quad L\{\cosh 2 t\}=\frac{s}{s^{2}-2^{2}}$$
Therefore $L\left\{e^{-t}(3 \sinh .2 t-5 \cosh 2 t)\right\}$
$$\begin{aligned}& =3 \cdot \frac{2}{(s+1)^{2}-2^{2}}-5 \cdot \frac{(s+1)}{(s+1)^{2}-2^{2}} \\& =\frac{1-5 s}{s^{2}+2 s-3}\end{aligned}$$
Example 6(a). Prove that
$$L\left\{e^{2 t}(\cos 4 t+3 \sin 4 t)\right\}=\frac{p+10}{p^{2}-4 p+20}$$
Solution. We know that $L\{\cos (a t)\}=\frac{p}{p^{2}+a^{2}}$,
$$L\{\sin (a t)\}=\frac{a}{p^{2}+a^{2}}$$
This $\Rightarrow L(\cos 4 t+3 \sin 4 t)=\frac{p}{p^{2}+4^{2}}+\frac{3(4)}{p^{2}+4^{2}}=\frac{p+12}{p^{2}+16}$
Using the result $L\left\{e^{a t} F(t)\right\}=f(p-a)$, we get
$$L\left\{e^{2 t}(\cos 4t+3\sin 4 t)\right\}=\frac{(p-2)+12}{(p-2)^{2}+16}=\frac{p+10}{p^{2}-4 p+20}$$
Example 6 (b). Prove that $L\left\{t e^{a t} \cdot \sin (a t)\right\}=\frac{2 a(p-a)}{\left(p^{2}-2 a p+2 a^{2}\right)^{2}}$
Solution. $\because L\{\sin (a t)\}=\frac{a}{p^{2}+a^{2}}$
$$\begin{gathered}\therefore L\left\{e^{a t} \sin (a t)\right\}=\frac{a}{(p-a)^{2}+a^{2}}=\frac{a}{p^{2}-2 a p+2 a^{2}} \\\Rightarrow \quad L\left\{t \cdot e^{a t} \sin (a t)\right\}=(-1)^{1} \frac{d}{d p}\left\{\frac{a}{p^{2}-2 a p+2 a^{2}}\right\} \\=\frac{2 a(p-a)}{\left(p^{2}-2 a p+2 a^{2}\right)^{2}}\end{gathered}$$
Theorem 4. Second Shifting Theorem (Second Translation).
If $L\{F(t)\}=f(s)$ and $G(t)=\left\{\begin{array}{cc}F(t-a), & \text{if}\;t>a \\ 0, & \text{if}\;t<a\end{array}\right.$
then $L\{G(t)\}=e^{-as}f(s)$
Or $L\{F(t-a) H(t-a)\}=e^{-a s} f(s) .$
Proof. Let $L\{F(t)\}=f(s)$
$$\begin{aligned}\text{and}\quad G(t) & =\left\{\begin{array}{cc}F(t-a), & \text { if } t>a \\0, & \text { if } t<a\end{array}\right. \\L\{G(t)\} & =\int_{0}^{\infty} e^{-s t} G(t) d t \\& =\int_{0}^{a} e^{-s t} G(t) d t+\int_{a}^{\infty} e^{-s t} G(t) d t \\& =\int_{0}^{a} e^{-s t} \cdot 0 d t+\int_{a}^{\infty} e^{-s t} F(t-a) d t \\& =0+\int_{a}^{\infty} e^{-s t} \cdot F(t-a) d t\end{aligned}$$
Put $t-a=p$ so that $d t=d p$.
If $t=a$, then $p=t-a=a-a=0$.
If $t=\infty$, then $p=\infty-a=\infty$.
$$\begin{aligned}\therefore \quad L\{G(t)\} & =\int_{0}^{\infty} e^{-s(p+a)} F(p) d p \\& =e^{-s a} \int_{0}^{\infty} e^{-s p} F(p) d p=e^{-s a} f(s)\end{aligned}$$
Example 7. Find the Laplace transform of $F(t)$, where
$$F(t)=\left\{\begin{array}{cl}\cos \left(t-\frac{2 \pi}{3}\right), & \text { if } t>\frac{2 \pi}{3} \\0, &\text { if } t<\frac{2 \pi}{3}\end{array}\right.$$
Solution. Let $a=\frac{2 \pi}{3}, G(t)=\cos t$, then
$$\begin{aligned}& L\{G(t)\}=\frac{p}{p^{2}+1}=g(p), \text { as } L(\cos a t)=\frac{p}{p^{2}+a^{2}} \\\text{Also}\quad & F(t)=\left\{\begin{array}{cc}G(t-a), & \text { if } t>a \\0, & \text { if } t<a\end{array}\right.\end{aligned}$$
By second shifting theorem, $L\{F(t)\}=e^{-a p} g(p)$
$$\Rightarrow \quad L\{F(t)\}=e^{-\frac{2 \pi p}{3}} \cdot \frac{p}{p^{2}+1}$$
Example 8. Evaluate $L\{F(t)\}$, if
$$F(t)=\left\{\begin{array}{cl}(t-1)^{2} &\text{if } t>1 \\0, & \text { if } 0<t<1\end{array}\right.$$
Solution. Let $a=1, G(t)=t^{2}, g(s)=L\{G(t)\}=\frac{2!}{s^{2+1}}$
$$\Rightarrow \quad g(s)=\frac{2}{s^{3}}$$
Also
$$F(t)=\left\{\begin{array}{cc}(t-a)^{2} & , t>a \\0 & , t<a\end{array}\right.$$
By second shifting theorem,
$$L\{F(t)\}=e^{-a s} g(s)=e^{-1 s} \cdot \frac{2}{s^{3}}=\frac{2 e^{-s}}{s^{3}}$$
Example 9. Prove that $L\{H(t)\}=\frac{2\left(1-e^{-\pi s}\right)}{s^{2}+4}$
where
$$H(T)=\left\{\begin{array}{cl}\sin 2 t & , 0<t<\pi \\0 & , t>\pi\end{array}\right.$$
Solution. $L\{H(t)\}=\int_{0}^{\infty} e^{-s t} H(t) d t$
$$\begin{aligned}& =\int_{0}^{\pi} e^{-s t} H(t) d t+\int_{\pi}^{\infty} e^{-s t} H(t) d t \\& =\int_{0}^{\pi} e^{-s t} \sin 2 t d t+\int_{\pi}^{\infty} e^{-s t} \cdot 0 \cdot d t \\& =\left[e^{-s t} \frac{(-s \sin 2 t-2 \cos 2 t)}{s^{2}+4}\right]_{t=0}^{\pi}=\frac{2\left(1-e^{-\pi s}\right)}{s^{2}+4}\end{aligned}$$