Definition of Laplace Transform

Let f(t) be a given function which is defined for t ≥ 0. If there exists a function F(s) so that
 

$\displaystyle F(s) = \int_0^\infty e^{-st} \, f(t) \, dt$,

 

then F(s) is called the Laplace Transform of f(t), and will be denoted by $\mathcal{L} \left\{f(t)\right\}$. Notice the integrator e-st dt where s is a parameter which may be real or complex.
 

Thus,

$\mathcal{L} \left\{f(t)\right\} = F(s)$

 

The symbol $\mathcal{L}$ which transform f(t) into F(s) is called the Laplace transform operator.
 

Laplace transformation is a powerful method of solving linear differential equations. It reduces the problem of solving differential equations into algebraic equations. For more information about the application of Laplace transform in engineering, see this Wikipedia article and this Wolfram article.
 

Laplace Transform by Direct Integration

To get the Laplace transform of the given function   $f(t)$,   multiply   $f(t)$   by   $e^{-st}$   and integrate with respect to   $t$   from zero to infinity. In symbol,
 

$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$.

 

Example
Find the Laplace transform of f(t) = 1 when t > 0.
 

Solution

 

Table of Laplace Transforms of Elementary Functions

Below are some functions f(t) and their Laplace transforms F(s).
 

$f(t)$ $F(s) = \mathcal{L} \left\{f(t)\right\}$
$1$ $\dfrac{1}{s}$
$t$ $\dfrac{1}{s^2}$
$t^2$ $\dfrac{2}{s^3}$
$t^n, \, (n = 1, \, 2, \, 3\, ...)$ $\dfrac{n!}{s^{n + 1}}$
$t^a, \, (a \, \text{ is positive})$ $\dfrac{\Gamma(a + 1)}{s^{a + 1}}$
$e^{at}$ $\dfrac{1}{s - a}$
$\sin bt$ $\dfrac{b}{s^2 + b^2}$
$\cos bt$ $\dfrac{s}{s^2 + b^2}$
$\sinh at$ $\dfrac{a}{s^2 - a^2}$
$\cosh at$ $\dfrac{s}{s^2 - a^2}$

 

Properties of Laplace Transform

Constant Multiple

If a is a constant and f(t) is a function of t$ then
 

$\mathcal{L} \left\{ a \, f(t) \right\} = a \, \mathcal{L} \left\{ f(t) \right\}$

 

Example: $\mathcal{L} (4 \cos t) = 4 \, \mathcal{L} (\cos t)$
 

Linearity Property

If a and b are constants while f(t) and g(t) are functions of t whose Laplace transform exists, then
 

$\mathcal{L} \left\{ a \, f(t) + b \, g(t) \right\} = a \, \mathcal{L} \left\{ f(t) \right\} + b \, \mathcal{L} \left\{ g(t) \right\}$

 

Example

$\mathcal{L} (3t^2 - 4t + 9) = 3 \, \mathcal{L} (t^2) - 4 \, \mathcal{L} (t) + 9 \, \mathcal{L} (1)$

 

Proof of Linearity Property
$\displaystyle \mathcal{L} \left\{ a \, f(t) + b \, g(t) \right\} = \int_0^\infty e^{-st}\left[ a \, f(t) + b \, g(t) \right] \, dt$

$\displaystyle \mathcal{L} \left\{ a \, f(t) + b \, g(t) \right\} = a\int_0^\infty e^{-st} f(t) \, dt + b\int_0^\infty e^{-st} g(t) \, dt$

$\mathcal{L} \left\{ a \, f(t) + b \, g(t) \right\} = a \, \mathcal{L} \left\{ f(t) \right\} + b \, \mathcal{L} \left\{ g(t) \right\}$       okay
 

This property can be easily extended to more than two functions as shown from the above proof. With the linearity property, Laplace transform can also be called the linear operator.
 

First Shifting Property

If $\mathcal{L} \left\{ f(t) \right\} = F(s)$, when s > a then,
 

$\mathcal{L} \left\{ e^{at} \, f(t) \right\} = F(s - a)$

 

In words, the substitution s - a for s in the transform corresponds to the multiplication of the original function by eat.
 

Proof of First Shifting Property
$\displaystyle F(s) = \int_0^\infty e^{-st} f(t) \, dt$

$\displaystyle F(s - a) = \int_0^\infty e^{-(s - a)t} f(t) \, dt$

$\displaystyle F(s - a) = \int_0^\infty e^{-st + at} f(t) \, dt$

$\displaystyle F(s - a) = \int_0^\infty e^{-st} e^{at} f(t) \, dt$

$F(s - a) = \mathcal{L} \left\{ e^{at} f(t) \right\}$   ←   okay
 

Second Shifting Property

If $\mathcal{L} \left\{ f(t) \right\} = F(s)$, and $g(t)
= \begin{cases} f(t - a) & t \gt a \\ 0 & t \lt a \end{cases}$
 

then,

$\mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$

 

Proof of Second Shifting Property
$g(t) = \begin{cases} f(t - a) & t \gt a \\ 0 & t \lt a \end{cases}$
 

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-st} g(t) \, dt$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^a e^{-st} (0) \, dt + \int_a^\infty e^{-st} f(t - a) \, dt$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_a^\infty e^{-st} f(t - a) \, dt$
 

Let
$z = t - a$

$t = z + a$

$dt = dz$
 

when $t = a, \, z = 0$

when $t = \infty, \, z = \infty$
 

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-s(z + a)} f(z) \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-sz - sa} f(z) \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-sz} e^{-sa} f(z) \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-sa} \int_0^\infty e^{-sz} f(z) \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-as} \mathcal{L} \left\{ f(z) \right\}$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-as} \mathcal{L} \left\{ f(t - a) \right\}$

$\mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$   ←   okay
 

Change of Scale Property

If $\mathcal{L} \left\{ f(t) \right\} = F(s)$, then,
 

$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$

 

Proof of Change of Scale Property
$\displaystyle \mathcal{L} \left\{ f(at) \right\} = \int_0^\infty e^{-st} f(at) \, dt$
 

Let
$z = at$

$t = z/a$

$dt = \dfrac{dz}{a}$
 

when $t = 0, \, z = 0$

when $t = \infty, \, z = \infty$
 

$\displaystyle \mathcal{L} \left\{ f(at) \right\} = \int_0^\infty e^{-s(z/a)} f(z) \, \dfrac{dz}{a}$

$\displaystyle \mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a}\int_0^\infty e^{-(s/a)z} f(z) \, dz$
 

Hence,
$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F\left( \dfrac{s}{a} \right)$   ←   okay
 

Multiplication by Power of t

If $\mathcal{L} \left\{ f(t) \right\} = F(s)$, then,
 

$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s) = (-1)^n F^{(n)}(s)$

where $n = 1, \, 2, \, 3, \, ...$
 

Proof of Multiplication by Power of t
$F(s) = \mathcal{L} \left\{ f(t) \right\}$

$\displaystyle F(s) = \int_0^\infty e^{-st} f(t) \, dt$
 

Differentiate both sides in s,
$\displaystyle \dfrac{dF}{ds} = \dfrac{d}{ds} \int_0^\infty e^{-st} f(t) \, dt$
 

From Leibniz rule of differentiation under integral sign,
$\displaystyle \dfrac{dF}{ds} = \int_0^\infty \dfrac{\partial}{\partial s} e^{-st} f(t) \, dt$

$\displaystyle \dfrac{dF}{ds} = \int_0^\infty (-te^{-st}) f(t) \, dt$

$\displaystyle \dfrac{dF}{ds} = -\int_0^\infty e^{-st} [ \, t \, f(t) \, ] \, dt$

$\dfrac{dF}{ds} = -\mathcal{L} \left\{ t \, f(t) \right\}$
 

Thus,
$\mathcal{L} \left\{ t \, f(t) \right\} = -\dfrac{dF}{ds} = -F'(s)$   ←   Equation (1)

which proves the theorem for n = 1.
 

Assuming the theorem is true for n = k then,
$\mathcal{L} \left\{ t^k f(t) \right\} = (-1)^k \dfrac{d^kF}{ds^k} = (-1)^k F^{(k)}(s)$

$\displaystyle \int_0^\infty e^{-st} [ \, t^k f(t) \, ] \, dt = (-1)^k \dfrac{d^kF}{ds^k} = (-1)^k F^{(k)}(s)$   ←   Equation (2)
 

Differentiate both sides in s,
$\dfrac{d}{ds} \displaystyle \int_0^\infty e^{-st} [ \, t^k f(t) \, ] \, dt = (-1)^k \dfrac{d}{ds} \left[ \dfrac{d^kF}{ds^k} \right] = (-1)^k \dfrac{d}{ds} \left[ F^{(k)}(s) \right]$

$\displaystyle \int_0^\infty \dfrac{\partial}{\partial s} e^{-st} [ \, t^k f(t) \, ] \, dt = (-1)^k \dfrac{d^{(k + 1)}F}{ds^{(k + 1)}} = (-1)^k F^{(k + 1)}(s)$

$\displaystyle \int_0^\infty (-te^{-st}) [ \, t^k f(t) \, ] \, dt = (-1)^k \dfrac{d^{(k + 1)}F}{ds^{(k + 1)}} = (-1)^k F^{(k + 1)}(s)$

$-\displaystyle \int_0^\infty e^{-st} [ \, t^{(k + 1)} f(t) \, ] \, dt = (-1)^k \dfrac{d^{(k + 1)}F}{ds^{(k + 1)}} = (-1)^k F^{(k + 1)}(s)$

$\displaystyle \int_0^\infty e^{-st} [ \, t^{(k + 1)} f(t) \, ] \, dt = (-1)^{(k + 1)} \dfrac{d^{(k + 1)}F}{ds^{(k + 1)}} = (-1)^k F^{(k + 1)}(s)$   ←   Equation (3)
 

This shows that the theorem is true for n = k from Equation (2) and for n = k + 1, from Equation (3). From Equation (1), the theorem is true for n = 1. Hence, it is true for n = 1 + 1 = 2 and n = 2 + 1 = 3, and so on, and thus, for all positive integer values of n.
 

Therefore,
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s) = (-1)^n F^{(n)}(s)$

where   $n = 1, \, 2, \, 3, \, ...$       okay
 

Division by t

If $\mathcal{L} \left\{ f(t) \right\} = F(s)$, then,
 

$\displaystyle \mathcal{L} \left\{ \dfrac{f(t)}{t} \right\} = \int_s^\infty F(u) \, du$

 

provided $\displaystyle \lim_{t \rightarrow 0} \left[ \dfrac{f(t)}{t} \right]$ exists.
 

Proof of Division by t
Let
$g(t) = \dfrac{f(t)}{t}$,   then

$f(t) = t \, g(t)$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left\{ t \, g(t) \right\}$
 

From Multiplication by Power of t
$\mathcal{L} \left\{ f(t) \right\} = (-1)^1 \dfrac{d}{ds}\mathcal{L} \left\{ g(t) \right\}$

$F(s) = - \dfrac{d}{ds}\mathcal{L} \left\{ g(t) \right\}$

$F(s) \, ds = -d\big[ \mathcal{L} \left\{ g(t) \right\} \big]$

$d\big[ \mathcal{L} \left\{ g(t) \right\} \big] = -F(s) \, ds$
 

Integrate both sides
$\displaystyle \int d\big[ \mathcal{L} \left\{ g(t) \right\} \big] = -\int F(s) \, ds$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = -\int_\infty^s F(u) \, du$

$\displaystyle \mathcal{L} \left\{ \dfrac{f(t)}{t} \right\} = \int_s^\infty F(u) \, du$
 

Laplace Transform of Derivatives

The Laplace transform of the derivative f'(t) exists when s > a, and
 

$\mathcal{L} \left\{ f'(t) \right\} = s\mathcal{L} \left\{ f(t) \right\} - f(0)$

 

In general, the Laplace transform of nth derivative is

$\mathcal{L} \left\{ f^n(t) \right\} = s^n\mathcal{L} \left\{ f(t) \right\} - s^{n - 1}f(0) - s^{n - 2}f'(0) - s^{n - 3}f''(0) - \, ... \, - f^{n - 1}(0)$

 

For first-order derivative:
$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$
 

For second-order derivative:
$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$
 

For third-order derivative:
$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$
 

Hence, for nth order derivative:
$\mathcal{L} \left\{ f^n(t) \right\} = s^n \mathcal{L} \left\{ f(t) \right\} - s^{n - 1} f(0) - s^{n - 2} \, f'(0) - \dots - f^{n - 1}(0)$
 

Proof of Laplace Transform of Derivatives
$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \int_0^\infty e^{-st} f'(t) \, dt$
 

Using integration by parts,
$u = e^{-st}$
$du = -se^{-st} \, dt$

$dv = f'(t) \, dt$
$v = f(t)$
 

Thus,
$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \Big[ e^{-st} f(t) \Big]_0^\infty - \int_0^\infty f(t) \, (-se^{-st} \, dt)$

$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \left[ \dfrac{f(t)}{e^{st}} \right]_0^\infty + s\int_0^\infty e^{-st} f(t) \, dt$

$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \left[ \dfrac{f(t)}{e^{st}} \right]_0^\infty + s \, \mathcal{L} \left\{ f(t) \right\}$
 

Apply the limits from 0 to :
$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = \left[ \dfrac{f(\infty)}{e^\infty} - \dfrac{f(0)}{e^0} \right] + s \, \mathcal{L} \left\{ f(t) \right\}$

$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = -f(0) + s \, \mathcal{L} \left\{ f(t) \right\}$

$\displaystyle \mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$           okay
 

Laplace Transform of Integrals

Theorem
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then
 

$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$

 

Proof
Let   $\displaystyle g(t) = \int_0^t f(u) \, du$

then,   $g'(t) = f(t)$   and   $g(0) = 0$
 

Taking the Laplace transform of both sides,
$\mathcal{L} \left\{ g'(t) \right\} = \mathcal{L} \left\{ f(t) \right\}$
 

From Laplace transform of derivative, $\mathcal{L} \left\{ g'(t) \right\} = s \, \mathcal{L} \left\{ g(t) \right\} - g(0)$ and from the Theorem above, $\mathcal{L} \left\{ f(t) \right\} = F(s)$
 

Thus,
$s \, \mathcal{L} \left\{ g(t) \right\} - g(0) = F(s)$

$s \, \mathcal{L} \left\{ g(t) \right\} - 0 = F(s)$

$s \, \mathcal{L} \left\{ g(t) \right\} = F(s)$

$\mathcal{L} \left\{ g(t) \right\} = \dfrac{F(s)}{s}$

$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$
 

Evaluation of Integrals
If $F(s) = \mathcal{L}\left\{ f(t) \right\}$, then $\displaystyle \int_0^\infty e^{-st} f(t) \, dt = F(s)$.
 

Taking the limit as $s \to 0$, then $\displaystyle \int_0^\infty f(t) \, dt = F(0)$ assuming the integral to be convergent.
 

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