The Laplace Transform

Abstract

A function f(t) is called original function if:

• 1. f(t) ≡ 0 for t < 0,

• 2. $|f(t)| < Me^{s_0t}$ for t > 0 with $M > 0, s_0\in \mathbb {R}$.

• 3. For every closed interval [a, b], the function satisfies the Dirichlet conditions:

  • (a) is bounded,

  • (b) or is continuous, or has a finite number of discontinuities of first kind,

  • (c) has a finite number of extremes.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

A function f(t) is called original function if [8, 9]:

• 1. f(t) ≡ 0 for t < 0,

• 2. $|f(t)| < Me^{s_0t}$ for t > 0 with $M > 0, s_0\in \mathbb {R}$.

• 3. For every closed interval [a, b], the function satisfies the Dirichlet conditions:

  • (a) is bounded,

  • (b) or is continuous, or has a finite number of discontinuities of first kind,

  • (c) has a finite number of extremes.

We consider the complex variable s = α + iβ, where Re(s) = α ≥ s₁ ≥ s₀. Then

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

(3.1)

is called the Laplace¹ integral, or Laplace transform (LT), or image of the original function f(t). In the follow-up we denote by L[f(t)] = F(s) or simply by Laplace transform (L) the Laplace transform. In Table 3.1 the LT of some elementary usual functions are listed.

Table 3.1 Images of basic elementary functions

The corresponding inverse Laplace transform is [1, 4]:

$$\displaystyle \begin{aligned} \begin{array}{rcl} f(t) = \frac{1}{2 \pi i} \lim_{t\to \infty} \int_{\gamma - i t}^{\gamma + i t} F(s) e^{s t} dt = L^{-1}[F(s)], {} \end{array} \end{aligned} $$

(3.2)

where $i=\sqrt {-1}$ and $\gamma \in \mathbb {R}$, so that the contour path of integration is contained in the convergence region.

3.1 Calculus of the Images

Example 1

Establish the image of f(t) = t^λ:

$$\displaystyle \begin{aligned}F(s) = \displaystyle \int_0^{\infty}e^{-t s}t^{\lambda} \, dt.\end{aligned} $$

We introduce the change of variable x = ts. We have also dx = s dt. It results:

$$\displaystyle \begin{aligned} F(s) &= \displaystyle \int_0^{\infty}e^{- x}\frac{x^{\lambda}}{s^{\lambda}}\frac{dx}{s},\\ F(s) &= \displaystyle \frac{1}{s^{\lambda + 1}}\int_0^{\infty}e^{- x}x^{\lambda}dx = \displaystyle \frac{\Gamma (\lambda + 1)}{s^{\lambda + 1}}.\end{aligned} $$

The direct and inverse LT are:

$$\displaystyle \begin{aligned}L( t^{\lambda}) = \displaystyle \frac{\Gamma (\lambda + 1)}{s^{\lambda + 1}},\hspace {1 cm} L^{-1}\left(\displaystyle \frac{1}{s^{\lambda + 1}}\right) = \frac{t^{\lambda}}{\Gamma (\lambda + 1)}.\end{aligned} $$

Example 2

Find the image of: $f(t) = \sin ^2 (t)$.

Using the identity $\sin ^2 t =\displaystyle\frac {1 - \cos (2t)}{2}$, it results:

$$\displaystyle \begin{aligned}F(s) = \displaystyle\frac{1}{2s} - \displaystyle\frac{2s}{2(s^2 + 4)} = \frac{2}{s(s^2 + 4)}.\end{aligned} $$

Example 3

Find the image of $f(t) = \displaystyle\frac {1}{2}te^{bt} + \displaystyle\frac {1}{2}te^{-bt}$.

$$\displaystyle \begin{aligned}F(s) = \displaystyle\frac{1}{2(s - b)^2} + \displaystyle\frac{1}{2(s + b)^2} = \displaystyle\frac{s^2 + b^2}{(s^2 - b^2)^2}. \end{aligned}$$

3.2 Calculus of the Original Function

3.2.1 Calculus of Original Using Residues

If we denote by L[f(t)] = F(s), then if we consider all residues of the function F(s)e^st, denoted by r₁, r₂, …r_n we can use the theorem:

$$\displaystyle \begin{aligned}f(t) = r_1 + r_2 + \ldots + r_n. \end{aligned}$$

The residues , denoted by Residues (Res), can be calculated using the following procedure (theorem):

If a is a simple pole of the function, then:

$$\displaystyle \begin{aligned}\mathop{\mbox{Res}}\limits_a[ e^{st}F(s)] = \lim_{s \to a}[(s - a) e^{st}F(s)]. \end{aligned}$$

If a is a simple pole of order n of the function, then:

$$\displaystyle \begin{aligned}\mathop{\mbox{Res}}\limits_a[ e^{s t}F(s)] = \displaystyle \frac{1}{(n - 1)!}\lim_{s \to a} [(s - a)^ne^{st}F(s)]^{(n - 1)}. \end{aligned}$$

Example 1

Find the original function of the image:

$$\displaystyle \begin{aligned}F(s) = \displaystyle \frac{1}{(s - 3)^2(s + 1)}. \end{aligned}$$

Solution

The residue of the function F(s)e^st is

$$\displaystyle \begin{aligned} r_1 &= \displaystyle \lim_{t \to - 1} (s + 1)\displaystyle \frac{e^{st}}{(s - 3)^2(s + 1)} = \displaystyle \frac{e^{- t}}{16},\\ r_2 &= \displaystyle \lim_{t \to 3} \frac{d}{ds}\displaystyle \left(\displaystyle \frac{e^{st}}{s + 1}\right) \\ &= \displaystyle \frac{t e^{3t} - e^{3t}}{16}, \end{aligned} $$

resulting:

$$\displaystyle \begin{aligned}f(t) = r_1 + r_2 = \displaystyle \frac{e^{- t}}{16} + \displaystyle \frac{te^{3t} -e^{3t}}{16}. \end{aligned}$$

Example 2

Find the original function of the following image:

$$\displaystyle \begin{aligned}F(s) = \displaystyle\frac{s^2}{s^4 - 1}. \end{aligned}$$

Solution

The function F has singularities: 1, − 1, − i, i. The residues will be:

$$\displaystyle \begin{aligned}r_1 = \mathop{\mbox{Res}}\limits_1 F(s)e^{st} = \displaystyle\lim_{s\to 1}(s - 1) \displaystyle\frac{s^2e^{st}}{(s - 1)(s + 1)(s^2 + 1)} = \displaystyle\frac{e^t}{4}, \end{aligned}$$

$$\displaystyle \begin{aligned}\qquad r_2 = \mathop{\mbox{Res}}\limits_{-1}F(s)e^{st} = \displaystyle\lim_{s\to -1}(s + 1) \displaystyle\frac{s^2e^{st}}{(s - 1)(s + 1)(s^2 + 1)} = - \displaystyle\frac{e^{-t}}{4}, \end{aligned}$$

$$\displaystyle \begin{aligned}\qquad r_3 = \mathop{\mbox{Res}}\limits_{-i}F(s)e^{st} = \displaystyle\lim_{s\to -i}(s + i)\displaystyle\frac{s^2e^{st}}{(s^2 - 1)(s - i)(s + i)} = \displaystyle\frac{-e^{-it}}{4i}, \end{aligned}$$

$$\displaystyle \begin{aligned}r_4 = \mathop{\mbox{Res}}\limits_{i}F(s)e^{st} = \displaystyle\lim_{s\to i}(s + i) \displaystyle\frac{s^2e^{st}}{(s^2 - 1)(s - i)(s + i)} = \displaystyle\frac{e^{it}}{4i}. \end{aligned}$$

It results finally

$$\displaystyle \begin{aligned}f(t) = \displaystyle\frac{1}{2}\left(\displaystyle\frac{e^t - e^{-t}}{2}\right) + \displaystyle\frac{1}{2}\left(\displaystyle\frac{e^{it} - e^{-it}}{2i}\right), \end{aligned}$$

or:

$$\displaystyle \begin{aligned}f(t) =\displaystyle\frac{1}{2}(\sinh t +\sin t). \end{aligned}$$

3.2.2 Calculus of Original with Post’s Inversion Formula

E. Post² obtained the formula [7, 10]:

$$\displaystyle \begin{aligned}f(t) = \displaystyle \lim_{k \to \infty} \displaystyle \frac{(- 1)^k}{k!}\displaystyle \left(\displaystyle \frac {k}{t}\right)^{k + 1} F^{(k)}\left( \displaystyle \frac{k}{t}\right),\quad t > 0. \end{aligned}$$

Example 3

Find the original function of the image:

$$\displaystyle \begin{aligned}F(s) = \displaystyle \frac{n!}{{s}^{n + 1}} n!s^{-1 - n}. \end{aligned}$$

Solution

With the aid of Post formula we have

$$\displaystyle \begin{aligned}f(t) = t^n\displaystyle \lim_{k \to \infty}\displaystyle \frac{k^{k + 1}(n + k)!}{k!t^{k + 1}}\left(\displaystyle \frac{k}{t}\right)^{-n - k - 1}. \end{aligned}$$

Using the Stirling³ formula:

$$\displaystyle \begin{aligned}\displaystyle \lim_{k \to \infty}\displaystyle \frac{k!}{\sqrt{2\pi k}}k^ke^{- k} = 1, \end{aligned}$$

it results:

$$\displaystyle \begin{aligned}f(t) = t^ne^{- n}\displaystyle \lim_{k \to \infty}\displaystyle \sqrt{1 + \displaystyle \frac{n}{k}}\left( 1 + \displaystyle \frac {n}{k}\right)^k\left( 1 + \displaystyle \frac{n}{k} \right)^n = t^n. \end{aligned}$$

3.3 The Properties of the Laplace Transform

In this section, we will use the notations F(s) = L[f(t)] and L(s) = G[g(t)]. In the follow-up are discussed properties of the LT.

3.3.1 The Property of Linearity

$$\displaystyle \begin{aligned} L[a f(t) + b g(t)] = a F(s) + b G(s), \hspace{0.2 cm} a, b \in \mathbb{R}. \end{aligned} $$

(3.3)

3.3.2 Similarity Theorem

$$\displaystyle \begin{aligned} L[f(\alpha t)] = \displaystyle \frac{1}{\alpha}F\left(\displaystyle \frac{s}{\alpha}\right),\qquad\alpha > 0. \end{aligned} $$

(3.4)

3.3.3 The Differentiation and Integration Theorems

Theorem (Differentiation of an Original)

The LT of the derivative of order k from f(t) gives:

$$\displaystyle \begin{aligned} L [f^{(k)}(t)] = s^k F(s) - \left[s^{k - 1}f(0) + s^{k - 2}f'(0) + \ldots + f^{(k-1)}(0)\right]. \end{aligned} $$

(3.5)

Proof

For k = 1, using the definition and integrating by parts, we have:

$$\displaystyle \begin{aligned} L[f'(t)] &= \displaystyle \int_0^\infty e^{- st}f'(t)dt = e^{- st}f(t)\Big |{}_0^\infty + s\displaystyle \int_0^\infty e^{- st}\, f(t)\,dt \\ &= -f(0) + s L[f(t)],\\ L [f'(t)] &= s F(s) - f(0), \end{aligned} $$

and for k = 2, using L[f″(t)] = L[(f′(t))′] we obtain:

$$\displaystyle \begin{aligned}L [f''(t)] = s^2 F(s) - s f(0) - f'(0). \end{aligned}$$

Using mathematical induction method, we have: L[f^(k)(t)].

Example 1

Find the LT for original f(t) = t².

Solution

• f′(t) = 2t, f^′′(t) = 2,

• f(0) = 0, f′(0) = 0, f^′′(0) = 2,

• L[f″(0)] = s²F(s) − sf(0) − f^′(0),

• $L[2] = \displaystyle\frac {2}{s} = s^2F(s)\hspace {0.2 cm} \Rightarrow \hspace {0.2 cm} F(s) = \displaystyle\frac {2}{s^3}$.

Example 2

Find the LT of following original $f(t) =\cos 2t$.

Solution

$f'(t) = - 2\sin (2t),\hspace {0.2 cm} f^{\prime \prime }(t) = - 4\cos (2t)$,

f(0) = 1, f′(0) = 0, f^′′(0) = −4,

L[f″(t)] = s²F(s) − sf′(0) − f(0),  ⇒ − 4F(s) = s²F(s) − s,

$$\displaystyle \begin{aligned}F(s) = \displaystyle \frac{s}{s^2 + 4}. \end{aligned}$$

Theorem (Integration of an Original)

It can be obtained:

$$\displaystyle \begin{aligned}L \left[\displaystyle\int_{0}^{t}f(\tau )d\tau \right] = \displaystyle\frac{F(s)}{s}, \end{aligned}$$

Proof

Let be $g(t) = \displaystyle\int _{0}^{t}f(\tau )d\tau$. Then:

$$\displaystyle \begin{aligned}L[g'(t)] = sL[g] - g(0). \end{aligned}$$

Also, we have g′(t) = f(t) and g(0) = 0. Then:

$$\displaystyle \begin{aligned}L[g] = \displaystyle \frac{F(s)}{s}. \end{aligned}$$

Theorem (Differentiation of a Transform)

We have:

$$\displaystyle \begin{aligned}F^{(n)}(s) = L [(-t)^nf(t)]. \end{aligned}$$

Proof

The theorem can be proved by induction. For n = 1, we have, successively:

$$\displaystyle \begin{aligned}F'(s) = L [-t f(t)], \end{aligned}$$

$$\displaystyle \begin{aligned}\displaystyle \frac{d F(s)}{ds} = \displaystyle \frac{d}{ds} \int_0^\infty e^{- st}f(t)dt = - \displaystyle \int_0^\infty e^{- s t}t f(t)dt = - L[t f(t)], \end{aligned}$$

and, finally:

$$\displaystyle \begin{aligned}F^{(n)}(s) = \displaystyle \frac{d}{ds}[F^{(n - 1)}(s)]. \end{aligned}$$

3.3.4 Delay Theorem

For a positive number a we have:

$$\displaystyle \begin{aligned}L[f(t - a)] = e^{- a s}F(s)\end{aligned}$$

3.3.5 Displacement Theorem

It is valid the formula:

$$\displaystyle \begin{aligned}L[e^{\lambda t}f(t)] = F(s - \lambda). \end{aligned}$$

3.3.6 Multiplication Theorem

The convolution product of two functions f(t) and g(t) is designated by the symbol ∗. We have:

$$\displaystyle \begin{aligned}(f * g)(t) = \displaystyle \int_0^t f(\tau)g(t - \tau)\, d\tau, \end{aligned}$$

$$\displaystyle \begin{aligned}L[(f * g)(t)] = F(s) G(s),\hspace{0.2 cm} F(s) = L[f(t)],\hspace{0.2 cm} G(s) = L[g(t)]. \end{aligned}$$

3.3.7 Properties of the Inverse Laplace Transform

The following formula is valid:

The inverse LT is not unique. We have:

$$\displaystyle \begin{aligned} L^{- 1}\left[\displaystyle \frac{s^{-(\alpha - \beta)}}{s^\beta - a}\right] = t^{\alpha - 1}E_{\beta,\alpha}(at^\beta), \, \alpha,\beta > 0,s^\alpha > |a|, {} \end{aligned} $$

(3.6)

$$\displaystyle \begin{aligned} L^{- 1}\left[\displaystyle \frac{s^{-(\alpha - 1)}}{s - a}\right] = t^{\alpha - 1}E_{1,\alpha}(at) = E(t,\alpha - 1,a), {} \end{aligned} $$

(3.7)

$$\displaystyle \begin{aligned} L^{- 1}\left[\displaystyle \frac{s^{-\alpha }}{(s - a)^2}\right]= tE(t,\alpha,a) - \alpha E(t,\alpha + 1,a), {} \end{aligned} $$

(3.8)

$$\displaystyle \begin{aligned} L^{- 1}\left[\displaystyle \frac{s^{-\alpha }}{(s - a)^3}\right] = \displaystyle \frac{1}{2}t^2E(t,\alpha,a) - \alpha tE(t,\alpha + 1,a) + \frac {\alpha (\alpha + 1)}{2}E(t,\alpha + 2,a) {}, \end{aligned} $$

(3.9)

$$\displaystyle \begin{aligned} L^{- 1}\left[\displaystyle \frac{1}{(s^\alpha + as^\beta)^{n + 1}}\right] = t^{\alpha(n + 1) - 1}\displaystyle \sum_{k = 0}^{\infty} \displaystyle \frac{(- a)^k\Big ({}_{\;\; k}^{n + k}\Big )}{\Gamma[k(\alpha - \beta) + ( n + 1)\alpha]}t^{k(\alpha - \beta)}, {} \end{aligned} $$

(3.10)

where 0 < β ≤ α.

$$\displaystyle \begin{aligned} \begin{array}{rcl}L^{- 1}\left[\displaystyle \frac{s^\gamma}{s^\alpha + as^\beta +b}\right] = t^{\alpha - \gamma - 1}\displaystyle \sum_{n = 0}^{\infty} \displaystyle \sum_{k = 0}^{\infty} \displaystyle \frac{(-b)^n(-a)^k\Big({}_{\;\; k}^{n + k}\Big)} {\Gamma[k(\alpha - \beta) + ( n + 1)\alpha - \gamma]}t^{k(\alpha - \beta) + n\alpha},\\ {} \end{array} \end{aligned} $$

(3.11)

where $\beta \le \alpha ,\hspace {0.2 cm}\gamma < \alpha ,\, a \in \mathbb {R}$ or: |a| < s^α−β, |b| < |s^α + as^β|.

Proof

Proof of the identity (3.6):

$$\displaystyle \begin{aligned} \begin{array}{rcl}L\left[t^{\alpha - 1}E_{\beta,\alpha}(at^\beta)\right] &\displaystyle =&\displaystyle \displaystyle \int_0^{\infty}e^{-st}t^{\alpha - 1}E_{\beta,\alpha}(at^\beta)dt \\ &\displaystyle =&\displaystyle \displaystyle \sum_{k = 0}^{\infty}\frac{a^k}{\Gamma(k\beta + \alpha)}\int_0^{\infty}e^{-st}t^{k\beta + \alpha - 1}dt \\ &\displaystyle =&\displaystyle \sum_{k = 0}^{\infty}\frac{a^k}{\Gamma(k\beta + \alpha)}L[t^{k\beta + \alpha - 1}] \\ &\displaystyle =&\displaystyle \displaystyle \sum_{k = 0}^{\infty}\frac{a^k}{\Gamma(k\beta + \alpha)}\frac{\Gamma(k\beta + \alpha)}{s^{k\beta + \alpha}} = \frac{1}{s^\alpha}\sum_{k = 0}^{\infty}\Big (\frac{a}{s^\beta}\Big )^k = \frac{s^{-(\alpha - \beta)}}{s^\beta - a}. \end{array} \end{aligned} $$

Proof of the identity (3.8):

$$\displaystyle \begin{aligned} \begin{array}{rcl}L[t \, E(t,\alpha,a) - \alpha E(t,\alpha + 1,a)] &\displaystyle =&\displaystyle - \frac{d}{ds}L[E(t,\alpha,a)] - \alpha L[E(t,\alpha + 1,a)] \\ &\displaystyle =&\displaystyle -\displaystyle \frac{d}{ds}\left[\frac{s^{- \alpha}}{s - a}\right] - \alpha \left[\frac{s^{-(\alpha + 1)}}{s - a}\right] = \frac{1}{s^\alpha(s - a)^2}. \end{array} \end{aligned} $$

Proof of the identity (3.9):

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle &\displaystyle L\left[\displaystyle \frac{1}{2}t^2E(t,\alpha,a) - \alpha tE(t,\alpha + 1,a) + \frac {1}{2}\alpha (\alpha + 1)E(t,\alpha + 2,a)\right] \\ {}\quad &\displaystyle =&\displaystyle \displaystyle \frac{1}{2}\frac{d^2}{ds^2}L\left[E(t,\alpha,a)\right] + \alpha \frac{d}{ds}L\left[E(t,\alpha +1,a)\right] + \frac{\alpha (\alpha + 1)}{2}L\left[E(t,\alpha + 2,a)\right]\\ {}\quad &\displaystyle =&\displaystyle \displaystyle \frac{1}{2}\frac{d^2}{ds^2}\left(\frac{s^{- \alpha}}{s-a}\right) + \alpha \frac{d}{ds}\left(\frac{s^{-(\alpha + 1)}}{s - a}\right) + \frac {\alpha(\alpha + 1)}{2}\left(\frac{s^{-(\alpha + 2)}}{s - a}\right) = \frac {1}{s^\alpha (s - a)^3}. \end{array} \end{aligned} $$

For the identities (3.10) and (3.11) the reader can use the reference [5].

Proof of the identity (3.10). We will apply the well-known identity [4]:

$$\displaystyle \begin{aligned}\displaystyle \frac{1}{(1 + x)^n} = \displaystyle \sum_{k = 0}^{\infty}\Big({}_{\;\; k}^{n + k}\Big)(-x)^k. \end{aligned}$$

It follows:

$$\displaystyle \begin{aligned} \begin{array}{rcl}\displaystyle \frac{1}{(s^\alpha + as^\beta)^{n + 1}} &\displaystyle =&\displaystyle \displaystyle \frac{1}{(s^\alpha )^{n + 1}}\displaystyle \frac{1}{\left( 1 + \displaystyle \frac{a}{s^{\alpha \beta}}\right) ^{n + 1}} \\ &\displaystyle =&\displaystyle \displaystyle \frac{1}{(s^\alpha )^{n + 1}}\displaystyle \sum_{k = 0}^{\infty} \Big ({}_{\;\; k}^{n + k}\Big )\left(\displaystyle \frac{-a}{s^{\alpha - \beta}}\right)^k. \end{array} \end{aligned} $$

Proof of the identity (3.11):

$$\displaystyle \begin{aligned}\displaystyle\frac{s^{\gamma}}{s^\alpha + as^\beta + b} = \displaystyle \frac{s^\gamma}{s^\alpha + as^\beta}\displaystyle \frac{1}{1 + \displaystyle \frac{b}{s^\alpha + as^\beta}} = \displaystyle \sum_{n = 0}^{\infty}\displaystyle \frac{s^\gamma(-b)^n}{(s^\alpha + as^\beta)^{n + 1}}, \end{aligned}$$

and for the case of (3.10) we obtain:

$$\displaystyle \begin{aligned} \begin{array}{rcl}\displaystyle \frac{s^\gamma}{(s^\alpha +s^\beta a)^{n + 1}} &\displaystyle =&\displaystyle \displaystyle \frac{s^\gamma}{s^{\alpha(n + 1)}}\displaystyle \frac{1}{\left( 1 + \displaystyle \frac{a}{s^{\alpha - \beta}}\right)^{n + 1}} \\ &\displaystyle =&\displaystyle \displaystyle \frac{1}{s^{\alpha (n + 1) - \gamma}}\displaystyle \sum_{k = 0}^{\infty}\Big ({}_{\;\; k}^{n + k}\Big )\left( \displaystyle \frac{- a}{s^{\alpha - \beta}}\right)^k = \displaystyle \sum_{k = 0}^\infty \Big ({}_{\;\; k}^{n + k}\Big ) \displaystyle \frac{(- a)^k}{s^{\alpha (n+ 1) + k(\alpha - \beta)- \gamma}}, \end{array} \end{aligned} $$

$$\displaystyle \begin{aligned}\displaystyle\frac{s^{\gamma}}{s^\alpha + as^\beta + b} = \displaystyle \sum_{n = 0}^\infty (-b)^n \displaystyle\sum_{k = 0}^\infty \Big ({}_{\;\; k}^{n + k}\Big ) \displaystyle \frac{(- a)^k}{s^{\alpha (n+ 1) + k(\alpha - \beta)- \gamma}}, \end{aligned}$$

$$\displaystyle \begin{aligned}L^{- 1}\left[\displaystyle\frac{s^{\gamma}}{s^\alpha + as^\beta + b}\right] = t^{\alpha - \gamma - 1}\displaystyle \sum_{n = 0}^{\infty} \displaystyle \sum_{k = 0}^{\infty} \displaystyle \frac{(-b)^n(-a)^k\Big ({}_{\;\; k}^{n + k}\Big )} {\Gamma[k(\alpha - \beta) + ( n + 1)\alpha - \gamma]}t^{k(\alpha - \beta) + n \alpha}. \end{aligned}$$

Lemma

The following identities are valid:

$$\displaystyle \begin{aligned} L^{- 1}\left[\displaystyle \frac{1}{s^\alpha + as +b} \right] = t^{\alpha - 1}\displaystyle \sum_{n = 0}^{\infty} \displaystyle \sum_{k = 0}^{\infty} \displaystyle \frac{(-b)^n(-a)^k\Big ({}_{\;\; k}^{n + k}\Big )} {\Gamma[k(\alpha - 1) + ( n + 1)\alpha ]}t^{k(\alpha - 1) + n\alpha}, {} \end{aligned} $$

(3.12)

for 1 ≤ α, 0 < α, $a \in \mathbb {R}$ , and |a| < s^α−1, |b| < |s^α + as|, respectively:

$$\displaystyle \begin{aligned} L^{- 1}\left[\displaystyle \frac{s^{\alpha - 1}}{s^\alpha + as + b} \right] = \displaystyle \sum_{n = 0}^{\infty} \displaystyle \sum_{k = 0}^{\infty} \displaystyle \frac{(-b)^n(-a)^k\Big ({}_{\;\; k}^{n + k}\Big )} {\Gamma[k(\alpha - 1) + n \alpha + 1 ]}t^{k(\alpha - 1) + n\alpha}, {} \end{aligned} $$

(3.13)

for 1 ≤ α, $a \in \mathbb {R}$ and for |a| < s^α−1, |b| < |s^α + as|.

Proof

Proof of the identity (3.12). In (3.11) we take γ = 0, β = 1.

Proof of the identity (3.13). In (3.11) we take γ = α − 1, and β = 1.

Example 1

Establish the LT of:

$$\displaystyle \begin{aligned}f(t) = y''(t) - 2y'(t) - 3y(t);\qquad\mbox{where: } y(0) = y'(0)= 0. \end{aligned}$$

Solution

$$\displaystyle \begin{aligned}F(s) = s^2 Y(s) - s y(0) -y'(0) - 2[s Y(s) -y(0)] - 3 Y(s), \end{aligned}$$

and finally:

$$\displaystyle \begin{aligned}F(s) = (s^2 - 2 s - 3) Y(s). \end{aligned}$$

Example 2

Establish the LT of:

$$\displaystyle \begin{aligned}y = \displaystyle\int_0^tydt + 1. \end{aligned}$$

Solution

$$\displaystyle \begin{aligned}Y(s) = \frac{Y(s)}{s} + \displaystyle\frac{1}{s}\hspace {0.2 cm}\Rightarrow Y(s) = \displaystyle\frac{1}{s - 1}. \end{aligned}$$

Example 3

Establish the LT of:

$$\displaystyle \begin{aligned}\displaystyle\int_0^ty(\tau )\sin (t - \tau )d\tau = 1 -\cos t. \end{aligned}$$

Solution

$$\displaystyle \begin{aligned}Y(s)\displaystyle\frac{1}{s^2 + 1} = \displaystyle\frac{1}{s} - \displaystyle\frac{s}{s^2 + 1} = \displaystyle\frac{1}{s(s^2 + 1)},\qquad\Rightarrow Y(s) = \displaystyle\frac{1}{s}. \end{aligned}$$

Example 4

Establish the LT of:

$$\displaystyle \begin{aligned}\displaystyle\int_0^ty(\tau )e^{t-\tau }d\tau = y(t) - e^t. \end{aligned}$$

Solution

$$\displaystyle \begin{aligned}Y(s)\displaystyle\frac{1}{s - 1} = Y(s) - \displaystyle\frac{1}{s - 1};\qquad\Rightarrow \qquad Y(s) = \displaystyle\frac{1}{s - 2}. \end{aligned}$$

3.4 Laplace Transform of the Fractional Integrals and Derivatives

3.4.1 Fractional Integrals

If α > 0, the Riemann–Liouville and Caputo FI are the same for both cases:

$$\displaystyle \begin{aligned}I = I^{\alpha}f(t) = \displaystyle \frac{1}{\Gamma(\alpha)}\int_0^t(t - y)^{\alpha - 1}f(y)dy. \end{aligned}$$

Using the LT of the convolution product formula, we have:

$$\displaystyle \begin{aligned}L[I] = \displaystyle \frac{1}{\Gamma(\alpha)} L[t^{\alpha - 1}]L[f(t)] = \frac{F(s)}{s^\alpha}. \end{aligned}$$

3.4.2 Fractional Derivatives

• The Riemann–Liouville FD is

  $$\displaystyle \begin{aligned} \begin{array}{rcl}L\left[D_t^\alpha f(t)\right] &\displaystyle =&\displaystyle L\left[\displaystyle \frac{1}{\Gamma(n - \alpha)}\displaystyle \left(\displaystyle \frac {dt^n}{d^nt}\right)\int_0^t(t - u)^{n - \alpha - 1} f(u) du\right] \\ &\displaystyle =&\displaystyle L\left[\displaystyle \left(\displaystyle \frac{dt^n}{d^nt}\displaystyle \right)I^{n - \alpha}f(t)\right], \end{array} \end{aligned} $$

  where we can apply the classical formula:

  $$\displaystyle \begin{aligned}L[f^{(n)}(t)] = s^n F(s) - s^{n - 1} f'(0) - {\ldots} - f^{(n - 1)}(0), \end{aligned}$$
  $$\displaystyle \begin{aligned}L[D_t^\alpha f(t)] = s^n\displaystyle \frac{F(s)}{s^{n - \alpha}} - \displaystyle \sum_{k = 0}^{n - 1}s^{n - \alpha - 1}[D^kI^{n - \alpha}f(t)]_{t = 0}, \end{aligned}$$
  $$\displaystyle \begin{aligned}L[D_t^\alpha f(t)] = s^\alpha \displaystyle F(s) - \displaystyle \sum_{k = 0}^{n - 1}s^{n - \alpha - 1}[D^k I^{n - \alpha}f(t)]_{t = 0}. \end{aligned}$$

• The Caputo [2, 3] FD is

  $$\displaystyle \begin{aligned}L[D^\alpha f(t)] = L\left[\displaystyle \frac{1}{\Gamma(n - \alpha)}\int_0^t(t - u)^{n - \alpha - 1} f^{(n)}(u) du\right] = L[I^{n - \alpha}f^{(n)}(t)], \end{aligned}$$

  where we can apply the classical formula:

  $$\displaystyle \begin{aligned}L[f^{(n)}(t)] = s^n F(s) - s^{n - 1}f'(0) - \ldots - f^{(n - 1)}(0), \end{aligned}$$

  and

  $$\displaystyle \begin{aligned}L[D^\alpha f(t)] = s^n\displaystyle \frac{F(s)}{s^{n - \alpha}} - \displaystyle \sum_{k = 0}^{n - 1}s^{n - \alpha - 1}[I^{n - \alpha}f^{(k)}(t)]_{t = 0}. \end{aligned}$$
  $$\displaystyle \begin{aligned}L[D^\alpha f(t)] = s^\alpha F(s) - \displaystyle \sum_{k = 0}^{n - 1}s^{n - \alpha - 1}f^{(k)}(0). \end{aligned}$$

Exercise 5

For the function f(t) = t², calculate the Caputo L[D^α]. It results:

• 1. $\alpha = \displaystyle\frac {1}{2},$

• 2. $\alpha = \displaystyle- \frac {1}{2}.$

Solution

• 1.
  $$\displaystyle \begin{aligned} L[D^{1/2}t^2] &= \displaystyle \frac{1}{s^{1 - \frac{1}{2}}}L[2t],\\ L[D^{1/2}t^2] &= \displaystyle \frac{2}{s^{\frac{5}{2}}},\\ D^{1/2}t^2 &= L^{- 1}\left[\displaystyle \frac{2}{s^{\frac{5}{2}}}\right] = \frac{2t^{\frac {3}{2}}}{\Gamma\left(\frac{5}{2}\right)}= \frac{8t^{\frac{3}{2}}}{3\sqrt{\pi}}. \end{aligned} $$

In MAPLE, the FD of order 1∕2 of t from t² can be evaluated using the command: fracdiff(t ̂2,t,1/2).

For 0 < α < 1, f^α(0) = 0, and

$$\displaystyle \begin{aligned}F(t) = \displaystyle \int_0^tf(u)(du)^\alpha = \alpha \int_0^t(t - u)^{\alpha - 1} f(u)\,du \end{aligned}$$

we define:

$$\displaystyle \begin{aligned}L_{\alpha}[f(t)] = F_{\alpha}(s) = \displaystyle \int_0^\infty E_{\alpha} (-s^\alpha t^\alpha)f(t)\,(dt)^\alpha. \end{aligned}$$

The following formulae can be obtained without difficulty [6]:

• 1. L_α[t^αf(t)] = −D^αL_α[f(t)].

• 2. $L_\alpha [f(at)] = \displaystyle\frac {1}{a^\alpha }L_\alpha [f(t)].$

• 3. L_α[f(t − b)] = E_α(−s^αb^α)L_α[f(t)].

• 4. $L_\alpha \left [\displaystyle\int _0^t f(u)\,(du)^\alpha \right ] = \displaystyle\frac {1}{a^\alpha \Gamma (\alpha + 1) }L_\alpha [f(t)].$

• 5. L_α[g^(α)(t)] = s^αL_α[g(t)] − Γ(α + 1)g(0).

For

$$\displaystyle \begin{aligned}\left(f(t)*g(t)\right)_\alpha = \displaystyle \int_0^tf(t - u)g(u)(du)^\alpha, \end{aligned}$$

we have:

$$\displaystyle \begin{aligned}L_\alpha \left[\left(f(t)*g(t)\right)_\alpha\right] = L_\alpha [f(t)] \, L_\alpha [g(t)]. \end{aligned}$$