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Laplace Transforms

  • Martin Hermann
  • Masoud Saravi
Chapter

Abstract

The main focus of Chap. 4 is the Laplace transform. At first, the Laplace transform operator is introduced and applied to some nonperiodic and periodic functions. Then, the Laplace transforms of derivatives and integrals, the derivatives and integrals of the Laplace transforms, as well the convolution theorem are presented. It is shown how the Laplace transform can be used to solve scalar differential equations.

Keywords

Integral Equation Periodic Function Laplace Transform Piecewise Continuous Function Convolution Theorem 
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.

4.1 Introduction

One of the very useful tools in solving linear ODEs and the corresponding initial and boundary values problems is the method of Laplace transforms. Laplace transforms are well adapted for the treatment of inhomogeneous equations that result when systems involving discontinuous or periodic input functions are modeled. They also can be used to solve certain types of improper integrals and integral equations.

We are familiar with the operator D defined in Sect.  3.3.3, and we observed that using this operator many functions can be transformed into other functions. In this chapter, we introduce another class of transformations that can be defined by
$$\displaystyle{ T\{f(x)\} =\int _{ -\infty }^{\infty }k(s,t)f(t)\,dt = F(s), }$$
(4.1)
where the function k(s, t) is called the kernel of the transformation. Here T{f} represents an entire class of transformations, since if k is changed, a new transform is defined. If we let
$$\displaystyle{ k(s,t) = \left \{\begin{array}{ll} 0, &\quad t < 0\\ {e}^{-st }, &\quad t \geq 0. \end{array} \right. }$$
(4.2)
then we have a member of this class that is called the Laplace transform. This chapter is devoted to this transform, but before dealing with it, we give two definitions which prepare the formulation of a sufficient condition for the existence of a Laplace transform.

Definition 4.1.

A function f(x) is said to be a piecewise continuous function on a closed interval \([a,b] \in \mathbb{R}\) if there exist a finite number of points \(a = x_{0} < x_{1} < x_{2} < \cdots < x_{N} = b\) such that f(x) is continuous in each of the intervals (x i−1, x i ) for 1 ≤ i ≤ N and has finite limits as x approaches the endpoints. □ 

An example of a piecewise continuous function is the greatest integer function f(x) = [x].

Definition 4.2.

A function f(x) is said to be of exponential order if there exist a constant α and positive constants x 0 and M such that
$$\displaystyle{\left \vert {e}^{-\alpha x}f(x)\right \vert < M}$$
for all x > x 0 at which f(x) is defined. □ 
An example of an exponential order function is f(x) = x n where n > 0. This can be shown as follows. Let α > 0 be given. Using L’Hopital’s rule n times, we obtain
$$\displaystyle{\lim _{x\rightarrow \infty }{e}^{-\alpha x}{x}^{n} = 0.}$$
Thus, there exist constants M > 0 and x 0 such that
$$\displaystyle{\vert {e}^{-\alpha x}{x}^{n}\vert \leq M\quad \mbox{ for all}\;\,x \geq x_{ 0}.}$$
This shows that f(x) = x n is of exponential order with the constant α any positive number.
In contrast, the function \(f(x) = {e}^{{x}^{2} }\) is not of exponential order, since
$$\displaystyle{\left \vert {e}^{-\alpha x}{e}^{{x}^{2} }\right \vert = {e}^{{x}^{2}-\alpha x }}$$
can be made larger than any given constant by increasing x.

4.2 Laplace Transform

The Laplace transform of the function f(x) denoted by \(\mathcal{L}\{f(x)\}\) is defined by
$$\displaystyle{ \mathcal{L}\{f(x)\} \equiv \int _{0}^{\infty }{f(t)e}^{-st}dt \equiv F(s),\quad s > 0, }$$
(4.3)
where \(\mathcal{L}\) denotes the Laplace transform operator. The variable s may be complex valued, but in this chapter we restrict s to real numbers.
The Laplace transform operator has a number of properties. For example, \(\mathcal{L}\) is a linear operator, i.e.,
$$\displaystyle{\mathcal{L}\left \{af(x) \pm bg(x)\right \} = a\mathcal{L}\left \{f(x)\right \} \pm b\mathcal{L}\left \{g(x)\right \},}$$
where a and b are constants and f(x) and g(x) are functions whose Laplace transforms exist. The second property of the Laplace transform is called the first shifting property. It is expressed as
$$\displaystyle{ \mathcal{L}\left \{{e}^{ax}f(x)\right \} = F(s - a),\quad s > 0. }$$
(4.4)
The third characteristic is the second shifting property which can be expressed as
$$\displaystyle{ \mathcal{L}\left \{f(x)\right \} = {e}^{-as}F(s). }$$
(4.5)

Before considering the existence of the Laplace transform, we present some examples.

Example 4.3.

Find the Laplace transforms of the following functions:
  1. 1.

    f(x) = k,

     
  2. 2.

    f(x) = e ax ,

     

where k and a are constants.

Solution.

  1. 1.
    Here, we have
    $$\displaystyle{\mathcal{L}\left \{k\right \} =\int _{ 0}^{\infty }k{e}^{-st}dt =\lim _{ c\rightarrow \infty }\int _{0}^{c}k{e}^{-st}dt =\lim _{ c\rightarrow \infty }\left [\left.\left (-\frac{k} {s}{e}^{-st}\right )\right \vert _{ 0}^{c}\right ] = \frac{k} {s}.}$$
     
  2. 2.
    For the second function f(x) = e ax we compute
    $$\displaystyle{\begin{array}{ll} \mathcal{L}\left \{{e}^{ax}\right \}& =\int _{ 0}^{\infty }{e}^{at}{e}^{-st}dt =\lim _{ c\rightarrow \infty }\int _{0}^{c}{e}^{-(s-a)t}dt \\ & =\lim _{c\rightarrow \infty }\left.\left [\left (- \frac{1} {s - a}{e}^{-(s-a)t}\right )\right \vert _{ 0}^{c}\right ] = \frac{1} {s - a},\quad \mbox{ where}\;\,s > a. \end{array} }$$
     

 □ 

Remark 4.4.

If we replace a by − a in 2, we obtain \(\mathcal{L}\left \{{e}^{-ax}\right \} = \frac{1} {s + a}\). □ 

Remark 4.5.

Since \(\cosh (ax) = \frac{{e}^{ax} + {e}^{-ax}} {2}\) and \(\sinh (ax) = \frac{{e}^{ax} - {e}^{-ax}} {2}\), it is easy to show that
  • \(\mathcal{L}\left \{\cosh (ax)\right \} = \frac{s} {{s}^{2} - {a}^{2}}\),

  • \(\mathcal{L}\left \{\sinh (ax)\right \} = \frac{a} {{s}^{2} - {a}^{2}}\).

Furthermore, since \(\cos (ax) = \frac{{e}^{iax} + {e}^{-iax}} {2}\) and \(\sin (ax) = \frac{{e}^{iax} - {e}^{-iax}} {2i}\), one can show that
  • \(\mathcal{L}\left \{\cos (ax)\right \} = \frac{s} {{s}^{2} + {a}^{2}}\),

  • \(\mathcal{L}\left \{\sin (ax)\right \} = \frac{a} {{s}^{2} + {a}^{2}}\). □ 

Remark 4.6.

Using the first shifting formula (4.4), we obtain
  • \(\mathcal{L}\left \{{e}^{ax}\cos (bx)\right \} = \frac{s - a} {{(s - a)}^{2} + {b}^{2}}\),

  • \(\mathcal{L}\left \{{e}^{ax}\sin (bx)\right \} = \frac{b} {{(s - a)}^{2} + {b}^{2}}\). □ 

Example 4.7.

Show that \(\mathcal{L}\left \{{x}^{n}\right \} = \frac{n!} {{s}^{n+1}}\), where n is a positive integer.

Solution.

We have
$$\displaystyle{\begin{array}{ll} \mathcal{L}\left \{{x}^{n}\right \}& =\int _{ 0}^{\infty }{t}^{n}{e}^{-st}dt =\lim _{ c\rightarrow \infty }\int _{0}^{c}{t}^{n}{e}^{-st}dt \\ & =\lim _{c\rightarrow \infty }\left (\left.-\frac{{t}^{n}{e}^{-st}} {s} \right \vert _{0}^{c} + \frac{n} {s}\int _{0}^{c}{t}^{n-1}{e}^{-st}dt\right ) \\ & = \cdots = \frac{n(n - 1)\cdots 3 \cdot 2 \cdot 1} {{s}^{n}} \lim _{c\rightarrow \infty }\int _{0}^{c}{e}^{-st}dt = \frac{n!} {{s}^{n}} \frac{1} {s} = \frac{n!} {{s}^{n+1}}.\end{array} }$$

 □ 

Remark 4.8.

By Example 4.7, we conclude that if
$$\displaystyle{f(x) = a_{0} + a_{1}x + a_{2}{x}^{2} + \cdots \,,}$$
then
$$\displaystyle{\mathcal{L}\left \{f(x)\right \} = \frac{a_{0}} {s} + \frac{a_{1}} {{s}^{2}} + 2\frac{a_{2}} {{s}^{3}} + \cdots =\sum _{n=0} \frac{n!\,a_{n}} {{s}^{n+1}}.}$$
 □ 

Example 4.9.

Find \(\mathcal{L}\left \{\frac{\sin (x)} {x} \right \}\)

Solution.

The series expansion of sin(x) is given by
$$\displaystyle{\sin (x) = \frac{x} {1!} -\frac{{x}^{3}} {3!} + \frac{{x}^{5}} {5!} -\frac{{x}^{7}} {7!} + \cdots }$$
Therefore,
$$\displaystyle{\begin{array}{ll} \mathcal{L}\left \{\frac{\sin (x)} {x} \right \}& = \mathcal{L}\left \{\frac{1} {1!} -\frac{{x}^{2}} {3!} + \frac{{x}^{4}} {5!} -\frac{{x}^{6}} {7!} + \cdots \,\right \} = \frac{\frac{1} {s}} {1!} -\frac{\frac{2!} {{s}^{3}} } {3!} + \frac{ \frac{4!} {{s}^{5}} } {5!} -\frac{\frac{6!} {{s}^{7}} } {7!} \cdots \\ & = \frac{\frac{1} {s}} {1} -\frac{\frac{1} {{s}^{3}} } {3} + \frac{ \frac{1} {{s}^{5}} } {5} -\frac{\frac{1} {{s}^{7}} } {7} \cdots {=\tan }^{-1}\left (\frac{1} {s}\right ).\end{array} }$$
Now, one can write
$$\displaystyle{\mathcal{L}\left \{\frac{\sin (x)} {x} \right \} {=\tan }^{-1}\left (\frac{1} {s}\right ) = \frac{\pi } {2} {-\tan }^{-1}(s).}$$

 □ 

Example 4.10.

A function that plays an important role in statistics is the so-called error function
$$\displaystyle{\mathrm{erf}(x) = \frac{2} {\sqrt{\pi }}\int _{0}^{x}{e}^{-{t}^{2} }dt.}$$
Show that \(\mathcal{L}\left \{\mathrm{erf}(\sqrt{x})\right \} = \frac{1} {s\,\sqrt{s + 1}}\)

Solution.

We have
$$\displaystyle{\begin{array}{ll} \mathcal{L}\left \{\mathrm{erf}\left (\sqrt{x}\,\right )\big)\right \}& = \mathcal{L}\left \{\frac{2} {\sqrt{\pi }}\int _{0}^{\sqrt{x}}{e}^{-{t}^{2} }dt\right \} \\ & = \frac{2} {\sqrt{\pi }}\mathcal{L}\int _{0}^{\sqrt{x}}\left (1 -\frac{{t}^{2}} {1!} + \frac{{t}^{4}} {2!} -\frac{{t}^{6}} {3!} + \cdots \,\right )dt \\ & = \frac{2} {\sqrt{\pi }}\mathcal{L}\left \{{x}^{\frac{1} {2} } - \frac{{x}^{\frac{3} {2} }} {3 \cdot 1!} + \frac{{x}^{\frac{5} {2} }} {5 \cdot 2!} - \frac{{x}^{\frac{7} {2} }} {7 \cdot 3!} + \cdots \,\right \} \\ & = \frac{2} {\sqrt{\pi }}\left (\frac{\varGamma \left (\frac{3} {2}\right )} {{s}^{\frac{3} {2} }} - \frac{\varGamma \left (\frac{5} {2}\right )} {3 \cdot 1!\,{s}^{\frac{5} {2} }} + \frac{\varGamma \left (\frac{7} {2}\right )} {5 \cdot 3!\,{s}^{\frac{7} {2} }} - \frac{\varGamma \left (\frac{9} {2}\right )} {7 \cdot 5!\,{s}^{\frac{9} {2} }} + \cdots \,\right ) \\ & = \frac{1} {{s}^{\frac{3} {2} }} - \frac{1} {2 \cdot {s}^{\frac{5} {2} }} + \frac{1 \cdot 3} {2 \cdot 4 \cdot {s}^{\frac{7} {2} }} - \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 \cdot {s}^{\frac{9} {2} }} + \cdots \\ & = \frac{1} {{s}^{\frac{3} {2} }} \left (1 -\frac{1} {2} \cdot \frac{1} {s} + \frac{1 \cdot 3} {2 \cdot 4} \cdot \frac{1} {{s}^{2}} -\frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \cdot \frac{1} {{s}^{3}} + \cdots \,\right ) \\ & = \frac{1} {{s}^{\frac{3} {2} }}{ \left (1 + \frac{1} {s}\right )}^{-1/2} = \frac{1} {s\sqrt{1 + s}}. \end{array} }$$
Here we used the identities \(\varGamma (x + 1) = x\varGamma (x)\) and \(\varGamma \left (\frac{1} {2}\right ) = \sqrt{\pi }\), where x is a positive real number.

 □ 

We come now to the following:

Theorem 4.11.

If the function f(x) is piecewise continuous in any finite interval 0 ≤ x ≤ N and \(\left \vert {e}^{-\alpha x}f(x)\right \vert < M\) as x →∞, then the Laplace transform \(\mathcal{L}\{f(x)\} = F(s)\) defined by Eq. (4.3) exists for s > α.

Proof.

Since f(x) is piecewise continuous, the function e α x f(x) is integrable on any finite interval of the x-axis. Moreover, f(x) is an exponential order function, i.e., | f(x) |  < e α x M.

Hence, for s > α it holds
$$\displaystyle{\begin{array}{ll} \vert \mathcal{L}\{f(x)\}\vert & = \left \vert \int _{0}^{\infty }f(t){e}^{-st}dt\right \vert \leq \int _{ 0}^{\infty }\left \vert f(t)\right \vert {e}^{-st}dt \\ & \leq M\int _{0}^{\infty }{e}^{\alpha t}{e}^{-st}dt = \frac{M} {s-\alpha }.\end{array} }$$
This proves the theorem. ■ 
It must be noted that the stated conditions are sufficient to guarantee the existence of the Laplace transform. For example, consider \(f(x) = \frac{1} {\sqrt{x}}\). This function is not finite in x = 0, but its Laplace transform exists, because
$$\displaystyle{\left \vert \mathcal{L}\left \{ \frac{1} {\sqrt{x}}\right \}\right \vert =\int _{ 0}^{\infty } \frac{1} {\sqrt{t}}{e}^{-st}dt = \frac{1} {\sqrt{s}}\int _{0}^{\infty }{x}^{-\frac{1} {2} }{e}^{-x}dx = \frac{1} {\sqrt{s}}\varGamma \left (\frac{1} {2}\right ) = \sqrt{ \frac{\pi } {s}}.}$$
Here we used the relation \(\varGamma \left (\frac{1} {2}\right ) = \sqrt{\pi }\) and in the first integral the change of variables x ≡ st.

Example 4.12.

Find \(\mathcal{L}\{u_{c}(t)\}\) and \(\mathcal{L}\{P_{c}(t)\}\), where u c (t) and P c (t) are Heaviside’s unit function and unit impulse (Dirac delta) function, respectively, given by
$$\displaystyle{u_{c}(t) = \left \{\begin{array}{l@{}l} 1,&t \geq c\\ 0, &t < c \end{array} \right.,\qquad P_{c}(t) = \left \{\begin{array}{l@{}l} \frac{1} {c},&0 \leq t \leq c \\ 0, &t > c \end{array} \right.}$$

Solution.

We have
$$\displaystyle{\begin{array}{ll} \mathcal{L}\left \{u_{c}(t)\right \} & =\int _{ 0}^{\infty }u_{ c}(t){e}^{-st}dt =\int _{ c}^{\infty }{e}^{-st}dt = \frac{{e}^{-cs}} {s}, \\ \mathcal{L}\left \{P_{c}(t)\right \}& =\int _{ 0}^{\infty }P_{ c}(t){e}^{-st}dt =\int _{ 0}^{c}\frac{{e}^{-st}} {c} dt = \frac{1 - {e}^{-cs}} {cs}. \end{array} }$$
One can prove that
$$\displaystyle{ \mathcal{L}\left \{u_{c}(t)f(t - c)\right \} = {e}^{-cs}F(s),\quad c \geq 0. }$$
(4.6)
Relation (4.6) can be useful when we deal with a several terms function, in particular if the forcing function in the ODE is a several terms function or a periodic function.

 □ 

Example 4.13.

Find the Laplace transform of the function
$$\displaystyle{f(t) = \left \{\begin{array}{l@{}l} 1,&0 < t < 1\\ t, &t > 1 \end{array} \right.}.$$

Solution.

Since for 0 < t < 1 we have f(t) = 1, we denote the function by f 1(t). For t > 1 we must have f(t) = t. Thus, we choose f 2(t) such that for t ∈ (0, 1) it has the value zero. Clearly, f 2(t) = tu 1(t). But if we write \(f(t) = f_{1}(t) + f_{2}(t)\), then for t > 1, we obtain \(f(t) = 1 + t\). Hence, we choose \(f_{2}(t) = (t - 1)u_{1}(t)\) and obtain
$$\displaystyle{f(t) = f_{1}(t) + f_{2}(t) = 1 + (t - 1)u_{1}(t).}$$
Now, everything is ready to use Eq. (4.6). We have
$$\displaystyle{\mathcal{L}\{f(t)\} = \mathcal{L}\{1\} + \mathcal{L}\{(t - 1)u_{1}(t)\} = \frac{1} {s} + \frac{{e}^{-s}} {{s}^{2}}.}$$

 □ 

Example 4.14.

Find the Laplace transform of the function
$$\displaystyle{f(t) = \left \{\begin{array}{l@{}l} 1, &0 < t < 1 \\ t, &1 < t < 2 \\ {t}^{2},&t > 2 \end{array} \right.}.$$

Solution.

Again for 0 < t < 1 we let f 1(t) = 1. For 1 < t < 2, we must have f(t) = t. Thus, we choose f 2(t) such that for t ∈ (1, 2) it has the value zero. Clearly, \(f_{2}(t) = (t - 1)u_{1}(t)\). Therefore, \(f_{3}(t) = ({t}^{2} - t)u_{2}(t)\). Then
$$\displaystyle{f(t) = f_{1}(t) + f_{2}(t) + f_{3}(t) = 1 + (t - 1)u_{1}(t) + ({t}^{2} - t)u_{ 2}(t).}$$
But for using Eq. (4.6), we must have for the third term a function of (t − 2). Hence, we write
$$\displaystyle{f(t) = 1 + (t - 1)u_{1}(t) + [({t - 2)}^{2} + 3(t - 2) + 2]u_{ 2}(t).}$$
Now if we apply (4.6), we obtain
$$\displaystyle{\begin{array}{ll} \mathcal{L}\{f(t)\}& = \mathcal{L}\{1\} + \mathcal{L}\{(t - 1)u_{1}(t)\} + \mathcal{L}\{{(t - 2)}^{2}u_{2}(t)\} \\ & + 3\mathcal{L}\{(t - 2)u_{2}\left (t\right )\} + 2\mathcal{L}\{u_{2}(t)\} \\ & = \frac{1} {s} + \frac{{e}^{-s}} {{s}^{2}} + {e}^{-2s}\left ( \frac{2} {{s}^{3}} + \frac{3} {{s}^{2}} + \frac{2} {s}\right ). \end{array} }$$

 □ 

4.3 The Laplace Transform of Periodic Functions

Many problems in applied mathematics deal with periodic functions. Most inhomogeneous ODEs in engineering involve forcing functions that are periodic. We wish to find the Laplace transform of such functions.

A periodic function f(x) is one that is piecewise continuous and for some p > 0 satisfies
$$\displaystyle{f(x) = f(x + p) = f(x + 2p) = \cdots = f(x + np) = \cdots }$$
Using this relation, we can write the transform of f(x) as a series of integrals
$$\displaystyle{\mathcal{L}\{f(x)\} =\int _{ 0}^{\infty }f(t){e}^{-st}dt =\int _{ 0}^{p}f(t){e}^{-st}dt +\int _{ p}^{2p}f(t){e}^{-st}dt + \cdots }$$
Now if in the first integral we let t = u and in second integral we let \(t = u + p\), and in third one \(t = u + 3p\) etc., we obtain
$$\displaystyle{\begin{array}{ll} &\mathcal{L}\{f(x)\} \\ & =\int _{ 0}^{p}f(u){e}^{-su}du +\int _{ 0}^{p}f(u){e}^{-s(u+p)}du +\int _{ 0}^{p}f(u){e}^{-s(u+2p)}du + \cdots \\ & =\int _{ 0}^{p}f(u){e}^{-su}du + {e}^{-sp}\int _{ 0}^{p}f(u){e}^{-su}du + {e}^{-2sp}\int _{ 0}^{p}f(u){e}^{-su}du + \cdots \\ & = \left (1 + {e}^{-sp} + {e}^{-2sp} + \cdots \,\right )\int _{0}^{p}f(u){e}^{-su}du \\ & ={ \left (1 - {e}^{-sp}\right )}^{-1}\int _{0}^{p}f(u){e}^{-su}du. \end{array} }$$
Therefore, we have proved

Theorem 4.15.

If f(x) is piecewise continuous with period p > 0, then
$$\displaystyle{ \mathcal{L}\{f(x)\} ={ \left (1 - {e}^{-sp}\right )}^{-1}\int _{ 0}^{p}f(u){e}^{-su}du. }$$
(4.7)

Example 4.16.

Find the Laplace transform of the half-way rectification given by
$$\displaystyle{f(x) = \left \{\begin{array}{c@{}l} \sin (x),&0 < x <\pi \\ 0, &\pi < x < 2\pi \end{array} \right.\,\quad p = 2\pi.}$$

Solution.

Using Eq. (4.7), we obtain
$$\displaystyle{\begin{array}{ll} &\mathcal{L}\{f(x)\} \\ & ={ \left (1 - {e}^{-2\pi s}\right )}^{-1}\int _{0}^{2\pi }f(u){e}^{-su}du ={ \left (1 - {e}^{-2\pi s}\right )}^{-1}\int _{ 0}^{\pi }\sin (u){e}^{-su}du \\ & = \frac{1 - {e}^{-\pi s}} {\left (1 - {e}^{-2\pi s}\right )(1 + {s}^{2})} = \frac{1} {\left (1 + {e}^{-\pi s}\right )(1 + {s}^{2})}.\end{array} }$$

 □ 

Example 4.17.

Find the Laplace transform of the wave function with period 2p.

Solution.

We have
$$\displaystyle{f(x) = \left \{\begin{array}{r@{}l} k,&0 < x < p\\ - k, &p < x < 2p. \end{array} \right.}$$
Hence,
$$\displaystyle{\begin{array}{ll} \mathcal{L}\{f(x)\}& ={ \left (1 - {e}^{-2ps}\right )}^{-1}\int _{0}^{2p}f(u){e}^{-su}du \\ & ={ \left (1 - {e}^{-2ps}\right )}^{-1}\left [\int _{0}^{p}k{e}^{-su}du -\int _{ p}^{2p}k{e}^{-su}du\right ] \\ & = \frac{k(1 - {e}^{-ps})} {s(1 + {e}^{-ps})}.\end{array} }$$

 □ 

Example 4.18.

Find the Laplace transform of the saw-tooth wave function defined by
$$\displaystyle{f(x) = \frac{x} {p}\quad \mbox{ when}\;\,0 < x < p\quad \mbox{ and}\;\,f(x + p) = f(x).}$$

Solution.

We have
$$\displaystyle{\begin{array}{ll} \mathcal{L}\{f(x)\}& = \frac{1} {p(1 - {e}^{-ps})}\int _{0}^{p}u{e}^{-su}du \\ & = \frac{1} {p(1 - {e}^{-ps})}\left [\frac{{ue}^{-su}} {s} \right ]_{0}^{p} -\frac{1} {s}\int _{0}^{p}{e}^{-su}du \\ & = \frac{1 - (1 + sp){e}^{-ps}} {p{s}^{2}(1 - {e}^{-ps})}. \end{array} }$$

 □ 

Definition 4.19.

Let us suppose that there exists a function f(x) such that \(\mathcal{L}\{f(x)\} = F(s)\). Then f(x) is called the inverse Laplace transform of F(s) and will be denoted by \({\mathcal{L}}^{-1}\), i.e. \(f(x) = {\mathcal{L}}^{-1}\{F(s)\}\). □ 

It must be noted that for finding the inverse Laplace transform, in most problems, we may factoring the denominator and expressing the function in terms of its partial fraction expansion.

Example 4.20.

Find the inverse of the transformed function
$$\displaystyle{F(s) = \frac{2s - 3} {{s}^{2} - 3s - 4}.}$$

Solution.

We have
$$\displaystyle{\begin{array}{ll} {\mathcal{L}}^{-1}\{F(s)\}& = {\mathcal{L}}^{-1}\left \{ \frac{2s - 3} {{s}^{2} - 3s - 4}\right \} = {\mathcal{L}}^{-1}\left \{ \frac{1} {s+1} + \frac{1} {s-4}\right \} \\ & = {\mathcal{L}}^{-1}\left \{ \frac{1} {s + 1}\right \} + {\mathcal{L}}^{-1}\left \{ \frac{1} {s - 4}\right \} = {e}^{-x} + {e}^{4x}. \end{array} }$$

 □ 

Example 4.21.

Find \({\mathcal{L}}^{-1}\left \{ \frac{2s} {({s}^{2} + 1)({s}^{2} + 4)}\right \}\)

Solution.

Employing partial fractions, we get
$$\displaystyle{\begin{array}{ll} \frac{2s} {({s}^{2} + 1)({s}^{2} + 4)} & = \frac{As + B} {{s}^{2} + 1} + \frac{Cs + D} {{s}^{2} + 4} \\ & = \frac{(As + B)({s}^{2} + 4) + (Cs + D)({s}^{2} + 1)} {({s}^{2} + 1)({s}^{2} + 4)}, \end{array} }$$
Thus we obtain
$$\displaystyle{2s = (As + B)({s}^{2} + 4) + (Cs + D)({s}^{2} + 1).}$$
Ordering the terms in powers of s and comparing both sides it is an easy but tedious task to show that
$$\displaystyle{A = -B = \frac{2} {3},\quad C = D = 0.}$$
Hence
$$\displaystyle{\begin{array}{ll} {\mathcal{L}}^{-1}\left \{ \frac{2s} {({s}^{2} + 1)({s}^{2} + 4)}\right \}& = \frac{2} {3}{\mathcal{L}}^{-1}\left \{ \frac{s} {{s}^{2} + 1}\right \} -\frac{2} {3}{\mathcal{L}}^{-1}\left \{ \frac{s} {{s}^{2} + 4}\right \} \\ & = \frac{2} {3}(\cos (x) -\cos (2x)).\end{array} }$$
In general this process of finding the constants takes time. Later, by using the convolution, we try to reduce this task.

 □ 

4.4 The Laplace Transforms of Derivatives and Integrals

For x ≥ 0, let the function f(x) be piecewise continuous and of exponential order. However, the exponential order of the function f(x) does not imply that its derivatives are also of exponential order. For example, \(f(x) =\sin ({e}^{{x}^{2} })\) is of exponential order but its derivative has not this property. By imposing stronger conditions on the function as in Theorem 4.11, one can prove the following result.

Theorem 4.22.

Assume that for x ≥ 0 the function f(x) is continuous and f (x) is piecewise continuous. Furthermore, let f and f be of exponential order when x →∞. Then
$$\displaystyle{ \mathcal{L}\left \{{f}^{{\prime}}(x)\right \} = s\mathcal{L}\{f(x)\} - f(0). }$$
(4.8)
In a similar way, if f and f are continuous functions and in Theorem 4.22 the conditions posed on f are applied to f ′ ′ , it holds
$$\displaystyle{ \mathcal{L}\{{f}^{{\prime\prime}}(x)\} = {s}^{2}\mathcal{L}\{f(x)\} - sf(0) - {f}^{{\prime}}(0). }$$
(4.9)
These results can be extended to higher-order derivatives and give us
$$\displaystyle{ \mathcal{L}\left \{{f}^{(n)}(x)\right \} = {s}^{n}\mathcal{L}\{f(x)\} - {s}^{n-1}f(0) - {s}^{n-2}{f}^{{\prime}}(0) -\cdots - {f}^{(n-1)}(0). }$$
(4.10)

Equation (4.10) represents a fundamental relation for the application of the Laplace transform to solve an ODE with constant coefficients. The differential equation is transformed into an algebraic expression in terms of s, and the solution of the ODE itself will result from the determination of the inverse Laplace transform which we discussed before.

Example 4.23.

Solve the initial value problem given by
$$\displaystyle{{x}^{{\prime\prime}}(t) {+\alpha }^{2}x(t) = A\sin (\omega t),\quad x(0) = {x}^{{\prime}}(0) = 0.}$$

Solution.

This initial value problem describes the forced, undamped, and resonant motion of a mass on a spring such that \({\alpha }^{2} = k/m\), where k is the spring constant. The mass starts from rest. We want to find the displacement x in terms of t. Under the given initial conditions, we apply the Laplace transform to both sides of this equation. This yields
$$\displaystyle{X(t) = \frac{A\omega } {({s}^{2} {+\omega }^{2})({s}^{2} {+\alpha }^{2})}.}$$
Now, since \(X(t) = \mathcal{L}\{x(t)\}\), the inverse Laplace transform gives
$$\displaystyle{\begin{array}{ll} x(t)& = {\mathcal{L}}^{-1}\left \{ \frac{A\omega } {({s}^{2} {+\omega }^{2})({s}^{2} {+\alpha }^{2})}\right \} ={ \frac{A} {\alpha }^{2} {-\omega }^{2}}{\mathcal{L}}^{-1}\left \{ \frac{\omega } {{s}^{2} {+\omega }^{2}} - \frac{\omega } {{s}^{2} {+\alpha }^{2}}\right \} \\ & ={ \frac{A} {\alpha }^{2} {-\omega }^{2}}\left (\sin (\omega t) -\frac{\omega } {\alpha }\sin (\alpha t)\right ). \end{array} }$$

 □ 

Example 4.24.

Solve the initial value problem
$$\displaystyle{{x}^{{\prime\prime}}(t) + 4x(t) = f(t),\quad x(0) = {x}^{{\prime}}(0) = 0,}$$
where \(f(t) = \left \{\begin{array}{l@{}l} \sin (2t),&0 \leq t <\pi \\ 0, &t \geq \pi \end{array} \right.\).

Solution.

Here the forcing function can be interpreted as an external force which is acting on a mechanical system only for a short time and is then removed. We write
$$\displaystyle{{x}^{{\prime\prime}}(t) + 4x(t) =\sin (2t) - u_{\pi }\sin (2(t-\pi )).}$$
Taking the Laplace transform and implying the initial conditions, we obtain
$$\displaystyle{\mathcal{L}\left \{x(t)\right \} = \frac{2} {{({s}^{2} + 4)}^{2}} - \frac{2{e}^{-\pi s}} {{({s}^{2} + 4)}^{2}}.}$$
Hence
$$\displaystyle{\begin{array}{ll} x(t)& = {\mathcal{L}}^{-1}\left \{ \frac{2} {{({s}^{2} + 4)}^{2}}\right \} -{\mathcal{L}}^{-1}\left \{ \frac{2{e}^{-\pi s}} {{({s}^{2} + 4)}^{2}}\right \} \\ & = \frac{\sin (2t) - 2t\cos (2t)} {8} -\frac{(t-\pi )\sin (2(t-\pi ))} {4} u_{\pi }.\end{array} }$$

 □ 

Example 4.25.

Find the response of an RL circuit to a unit pulse function u a u b . Assume that the initial current is zero.

Solution.

We know that the initial value problem describing this circuit is
$$\displaystyle{L{i}^{{\prime}} + Ri = u_{ a} - u_{b},\quad i(0) = 0.}$$
Taking the Laplace transform, we obtain
$$\displaystyle{LsI(s) + RI(s) = \frac{{e}^{-as}} {s} -\frac{{e}^{-bs}} {s},}$$
where \(I(s) = \mathcal{L}\left \{i(t)\right \}\). Writing
$$\displaystyle{I(s) = \frac{{e}^{-as} - {e}^{-bs}} {s(Ls + R)} }$$
and factoring the denominator to express the right-hand side in terms of its partial fraction expansion, we get
$$\displaystyle{I(s) = \frac{{e}^{-as}} {R} \left (\frac{1} {s} - \frac{1} {s + R/L}\right ) -\frac{{e}^{-bs}} {R} \left (\frac{1} {s} - \frac{1} {s + R/L}\right ).}$$
Taking the inverse Laplace transform, we obtain
$$\displaystyle{i(t) = \frac{1} {R}\left [\left (1 - {e}^{-\frac{R(t-a)} {L} }\right )u_{a} -\left (1 - {e}^{-\frac{R(t-b)} {L} }\right )u_{b}\right ].}$$

 □ 

Example 4.26.

Let a beam whose ends are at x = 0 and x = l is coincident with the x-axis and a vertical load, given by W(x) per unit length, acts transversely on it. Then the axis of the beam has a transverse deflection y(x) at the point x which satisfies the ODE
$$\displaystyle{EI{y}^{(4)}(x) = W(x),\quad 0 < x < l,}$$
where EI is assumed to be constant. The boundary conditions associated with this equation depend on the manner in which the beam is supported. For example, for a free end, we have \({y}^{{\prime\prime}} = {y}^{{\prime\prime\prime}} = 0\); for a hinged or supported end, \(y = {y}^{{\prime\prime}} = 0\); and for a clamped, built-in, or fixed end, \(y = {y}^{{\prime}} = 0\). Let l = 2. If the beam is clamped at x = 0 and is free at the other end, find the deflection if it carries a load given by
$$\displaystyle{W(x) = \left \{\begin{array}{l@{}l} 8, &0 < x < 1\\ 0, &1 < x < 2 \end{array}.\right.}$$

Solution.

We have
$$\displaystyle{EI{y}^{(4)}(x) = \left \{\begin{array}{l@{}l} 8, &0 < x < 1 \\ 0, &1 < x < 2 \end{array},\right.}$$
with the initial conditions \(y(0) = {y}^{{\prime}}(0) = 0\), \({y}^{{\prime\prime}}(2) = {y}^{{\prime\prime\prime}}(2) = 0\). We write the right-hand side in terms of the unit function. Thus, we get
$$\displaystyle{EI{y}^{(4)}(x) = 8(u_{ 0} - u_{1}).}$$
Taking the Laplace transform and implying the initial conditions, we obtain
$$\displaystyle{{s}^{4}\mathcal{L}\left \{y(x)\right \} - s{y}^{{\prime\prime}}(0) - {y}^{{\prime\prime\prime}}(0) = \frac{8} {EI}\left (\frac{1 - {e}^{-s}} {s} \right ).}$$
Replacing the two unknown initial conditions by y ′ ′ (0) = A, y ′ ′ ′ (0) = B, we get
$$\displaystyle{\mathcal{L}\left \{y(x)\right \} = \frac{A} {{s}^{3}} + \frac{B} {{s}^{4}} + \frac{8} {EI}\left (\frac{1 - {e}^{-s}} {{s}^{5}} \right ).}$$
Thus,
$$\displaystyle{\begin{array}{ll} y(x)& = {\mathcal{L}}^{-1}\left \{ \frac{A} {{s}^{3}}\right \} + {\mathcal{L}}^{-1}\left \{ \frac{B} {{s}^{4}}\right \} + \frac{8} {EI}{\mathcal{L}}^{-1}\left \{\frac{1 - {e}^{-s}} {{s}^{5}} \right \} \\ & = \frac{A{x}^{2}} {2} + \frac{B{x}^{3}} {6} + \frac{u_{1}} {3EI}{(x - 1)}^{4}. \end{array} }$$
Using \({y}^{{\prime\prime}}(2) = {y}^{{\prime\prime\prime}}(2) = 0\), one can show that \(A = \frac{1} {EI}\) and \(B = \frac{-4} {EI}\).

 □ 

We may now turn our attention to the integration process, and we expect to have a division of the transform by s since the integration of f(x) is the inverse operation of the differentiation. We continue the present section with the following theorem.

Theorem 4.27.

If the function f(x) satisfies the assumptions of Theorem 4.11, then
$$\displaystyle{ \mathcal{L}\left \{\int _{0}^{x}f(t)dt\right \} = \frac{F(s)} {s}. }$$
(4.11)

Proof.

Let \(g(x) \equiv \int _{0}^{x}f(t)dt\). Thus, g (x) = f(x). Since g(0) = 0, we have
$$\displaystyle{\mathcal{L}\left \{{g}^{{\prime}}(x)\right \} = s\mathcal{L}\left \{g(x)\right \} - g(0) = s\mathcal{L}\left \{g(x)\right \}.}$$
Hence
$$\displaystyle{\mathcal{L}\left \{g(x)\right \} = \frac{1} {s}\mathcal{L}\left \{{g}^{{\prime}}(x)\right \} = \frac{1} {s}\mathcal{L}\left \{f(x)\right \} = \frac{F(s)} {s}.}$$
 ■ 

Example 4.28.

Find \(\mathcal{L}\left \{\int _{0}^{x}\frac{\sin (t)} {t} dt\right \}\).

Solution.

Since \(\mathcal{L}\left \{\frac{\sin (x)} {x} \right \} {=\tan }^{-1}\left (\frac{1} {s}\right )\), we have
$$\displaystyle{\mathcal{L}\left \{\int _{0}^{x}\frac{\sin (t)} {t} dt\right \} ={ \frac{1} {s}\tan }^{-1}\left (\frac{1} {s}\right ).}$$

 □ 

Remark 4.29.

Looking at (4.11), one may write
$$\displaystyle{ {\mathcal{L}}^{-1}\left \{\frac{F(s)} {s} \right \} =\int _{ 0}^{x}f(t)dt. }$$
(4.12)
Equation (4.12) can be used to find f(x). □ 

Example 4.30.

If \(f(x) = {\mathcal{L}}^{-1}\left \{ \frac{1} {s({s}^{2} + {a}^{2})}\right \}\), find f(x). 

Solution.

Since
$$\displaystyle{{\mathcal{L}}^{-1}\left \{ \frac{1} {{s}^{2} + {a}^{2}}\right \} = \frac{\sin (ax)} {a} }$$
it holds
$$\displaystyle{f(x) = {\mathcal{L}}^{-1}\left \{ \frac{1} {s({s}^{2} + {a}^{2})}\right \} =\int _{ 0}^{x}\frac{\sin (at)} {a} dt = \frac{1 -\cos (ax)} {{a}^{2}}.}$$

 □ 

4.5 Derivatives and Integrals of the Laplace Transforms

We described the Laplace transform of derivatives and integrals. In advanced calculus, it can be shown that differentiation and integration of the Laplace transform are possible. For example, if \(F(s) =\int _{ 0}^{\infty }f(t){e}^{-st}dt\), then using Leibniz’s rule of differentiation and integration we have
$$\displaystyle{{F}^{{\prime}}(s) = -\int _{ 0}^{\infty }tf(t){e}^{-st}dt =\int _{ 0}^{\infty }(-tf(t)){e}^{-st}dt = \mathcal{L}\left \{-xf(x)\right \}.}$$
That is, a Laplace transform corresponds to f(x) times (−x). By the same manner, the second derivative is
$$\displaystyle{{F}^{{\prime\prime}}(s) =\int _{ 0}^{\infty }{t}^{2}f(t){e}^{-st}dt =\int _{ 0}^{\infty }({t}^{2}f(t)){e}^{-st}dt = \mathcal{L}\left \{{x}^{2}f(x)\right \}.}$$
These operations enable us to formulate the following theorem.

Theorem 4.31.

If f(x) and its derivatives satisfy the conditions of Theorem 4.11, then
$$\displaystyle{{F}^{(n)}(s) = \mathcal{L}\left \{{(-x)}^{n}f(x)\right \}.}$$
This equation can be rewritten as
$$\displaystyle{ \mathcal{L}\left \{{x}^{n}f(x)\right \} = {(-1)}^{n}{F}^{(n)}(s). }$$
(4.13)

Example 4.32.

Find \(\mathcal{L}\left \{{x}^{2}\sin (3x)\right \}\)

Solution.

We have
$$\displaystyle{\mathcal{L}\left \{{x}^{2}\sin (3x)\right \} = \frac{{d}^{2}} {d{s}^{2}}\left ( \frac{3} {{s}^{2} + 9}\right ) = \frac{6s(3{s}^{2} - 9)} {{({s}^{2} + 9)}^{3}}.}$$

 □ 

The relation (4.13) can also be used to evaluate some improper integrals.

Example 4.33.

Show that \(\int _{0}^{\infty }{x}^{2}\cos (x){e}^{-x}dx = -\frac{1} {2}\).

Solution.

By relation (4.13), we have
$$\displaystyle{\begin{array}{ll} \int _{0}^{\infty }{x}^{2}\cos (x){e}^{-sx}dx& = \mathcal{L}\left \{{x}^{2}\cos (x)\right \} = \frac{{d}^{2}} {d{s}^{2}}\mathcal{L}\left \{\cos (x)\right \} \\ & = \frac{{d}^{2}} {d{s}^{2}}\left ( \frac{s} {{s}^{2} + 1}\right ) = \frac{2s({s}^{2} - 3)} {{({s}^{2} + 1)}^{3}}. \end{array} }$$
Now set s = 1 to get the desired result.

 □ 

Example 4.34.

Let \(\mathcal{L}\left \{f(x)\right \} =\ln \left (\frac{s + 1} {s + 2}\right ) = F(s)\). Find f(x).

Solution.

We write \(\ln \left (\frac{s + 1} {s + 2}\right ) =\ln \left (s + 1\right ) -\ln \left (s + 2\right )\). Thus,
$$\displaystyle{{F}^{{\prime}}(s) = \frac{1} {s + 1} - \frac{1} {s + 2} = \mathcal{L}\left \{{e}^{-x}\right \} -\mathcal{L}\left \{{e}^{-2x}\right \} = \mathcal{L}\left \{{e}^{-x} - {e}^{-2x}\right \}.}$$
Now, if we use the relation (4.13) with n = 1, we obtain
$$\displaystyle{xf(x) = {e}^{-2x} - {e}^{-x},\quad \mbox{ i.e.}\;\,f(x) = \frac{{e}^{-2x} - {e}^{-x}} {x}.}$$
Suppose g(x) ≡ f(x)∕x, then f(x) = xg(x). But we have
$$\displaystyle{\mathcal{L}\left \{xg(x)\right \} = -{G}^{{\prime}}(s).}$$
Thus, \(F(s) = -{G}^{{\prime}}(s)\). That is, \(-dG = F(s)ds\). Integrating both sides, we get
$$\displaystyle{-G(s) =\int _{ \infty }^{s}F(t)dt,}$$
where G(s) → 0 as s → . Hence \(G(s) =\int _{ s}^{\infty }F(t)dt\). Thus,
$$\displaystyle{ \mathcal{L}\left \{\frac{f(x)} {x} \right \} =\int _{ s}^{\infty }F(t)dt, }$$
(4.14)
provided \(\lim _{x\rightarrow {0}^{+}} \frac{f(x)} {x}\) exists.

 □ 

Example 4.35.

Find \(\mathcal{L}\left \{\frac{{e}^{-4x} - 1} {x} \right \}\).

Solution.

Since \(\lim _{x\rightarrow {0}^{+}}\left (\frac{{e}^{-4x} - 1} {x} \right ) = 1\) and \(\mathcal{L}\left \{{e}^{-4x} - 1\right \} = \frac{1} {s + 4} -\frac{1} {s}\), we have
$$\displaystyle{\mathcal{L}\left \{\frac{{e}^{-4x} - 1} {x} \right \} =\int _{ s}^{\infty }\left ( \frac{1} {t + 4} -\frac{1} {t}\right )dt =\ln \left ( \frac{s} {s + 1}\right ).}$$

 □ 

Example 4.36.

Evaluate \(\int _{0}^{\infty }\frac{{e}^{-2x}\sin (x)\cosh (x)} {x} dx\).

Solution.

We have \(\lim _{x\rightarrow {0}^{+}} \frac{\sin (x)\cosh (x)} {x} = 1\), hence
$$\displaystyle{\mathcal{L}\left \{\frac{\sin (x)\cosh (x)} {x} \right \} =\int _{ 0}^{\infty }\frac{{e}^{-st}\sin (t)\cosh (t)} {t} dt = F(s).}$$
But
$$\displaystyle{\mathcal{L}\left \{\frac{\sin (x)\cosh (x)} {x} \right \} =\int _{ s}^{\infty }\mathcal{L}\left \{\sin (t)\cosh (t)\right \}dt = F(s).}$$
Comparing these results, we obtain
$$\displaystyle{\begin{array}{ll} \int _{0}^{\infty }\frac{{e}^{-st}\sin (t)\cosh (t)} {t} dt& =\int _{ s}^{\infty }\mathcal{L}\left \{\sin (t)\cosh (t)\right \}dt \\ & = \frac{1} {2}\int _{s}^{\infty }\left [\mathcal{L}\left \{{e}^{t}\sin (t)\right \} + \mathcal{L}\left \{{e}^{-t}\sin (t)\right \}\right ]dt \\ & = \frac{1} {2}\int _{s}^{\infty }\left [ \frac{1} {{(t - 1)}^{2} + 1} + \frac{1} {{(t + 1)}^{2} + 1}\right ]dt \\ & = \frac{1} {2}\left[{\tan }^{-1}(s - 1) {+\tan }^{-1}(s + 1)\right ]. \end{array} }$$
Now, if we set s = 2, we get
$$\displaystyle{\int _{0}^{\infty }\frac{{e}^{-2x}\sin (x)\cosh (x)} {x} dx = \frac{1} {2}\left ( \frac{\pi } {4} {+\tan }^{-1}(3)\right ).}$$

 □ 

Example 4.37.

Solve the initial value problem given by
$$\displaystyle{x{y}^{{\prime\prime}} + 2{y}^{{\prime}} + xy = 0,\quad y({0}^{+}) = 1,\;\,y(\pi ) = 0.}$$

Solution.

By taking the Laplace transform of the given equation, we obtain
$$\displaystyle{\begin{array}{ll} &\mathcal{L}\left \{x{y}^{{\prime\prime}}\right \} + 2\mathcal{L}\left \{{y}^{{\prime}}\right \} + \mathcal{L}\left \{xy\right \} \\ & = -{\left [{s}^{2}Y - sy({0}^{+}) - {y}^{{\prime}}({0}^{+})\right ]}^{{\prime}} + 2\left [sY - y({0}^{+})\right ] - {Y }^{{\prime}} = 0, \end{array} }$$
where \(Y \equiv \mathcal{L}\left \{y\right \}\). Since \(y({0}^{+}) = 1\), we have
$$\displaystyle{{Y }^{{\prime}} = - \frac{1} {{s}^{2} + 1},\;\mbox{ i.e.}\quad Y (s) = c {-\tan }^{-1}(s).}$$
But, Y → 0 as s → , hence \(c = \frac{\pi } {2}\). Thus,
$$\displaystyle{Y (s) = \frac{\pi } {2} {-\tan }^{-1}(s) {=\tan }^{-1}\left (\frac{1} {s}\right ).}$$
Since \(\mathcal{L}\left \{\frac{\sin (x)} {x} \right \} {=\tan }^{-1}\left (\frac{1} {s}\right )\), we obtain
$$\displaystyle{{\mathcal{L}}^{-1}\left \{Y \right \} = {\mathcal{L}}^{-1}\left \{{\tan }^{-1}\left (\frac{1} {s}\right )\right \} \rightarrow y(x) = \frac{\sin (x)} {x}.}$$
This solution satisfies y(π) = 0.

 □ 

4.6 The Convolution Theorem and Integral Equations

As we observed, for obtaining the inverse Laplace transform of F(s), the partial fraction expansion can be applied. In general, such an expansion is not difficult, but it may be tedious to develop it. In this section, we introduce the concept of the convolution which can be used not only for finding the inverse of the product of two transform functions but also to solve some integral equations that are of the form of a convolution integral.

Definition 4.38.

Suppose f(x) and g(x) are piecewise continuous functions. The convolution fg of the functions f and g is defined by
$$\displaystyle{f {\ast} g \equiv \int _{0}^{x}f(t)g(x - t)dt.}$$
 □ 
The convolution fg has the following properties:
  • fg = gf (commutative law),

  • f ∗ (gh) = (fg) ∗ h (associative law),

  • \(f {\ast} (g + h) = f {\ast} g + f {\ast} h\) (distributive law).

The next theorem contains a fundamental statement which is concerned with the transform of a convolution function.

Theorem 4.39.

If \(\mathcal{L}\left \{f(x)\right \} = F(s)\) and \(\mathcal{L}\left \{g(x)\right \} = G(s)\) , then
$$\displaystyle{ {\mathcal{L}}^{-1}\left \{F(s)G(s)\right \} =\int _{ 0}^{x}f(t)g(x - t)dt. }$$
(4.15)

Proof.

We have
$$\displaystyle{\mathcal{L}\left \{\int _{0}^{x}f(t)g(x - t)dt\right \} =\int _{ 0}^{\infty }{e}^{-sx}\left \{\int _{ 0}^{x}f(t)g(x - t)dt\right \}dx.}$$
It follows
$$\displaystyle{ \mathcal{L}\left \{\int _{0}^{x}f(t)g(x - t)dt\right \} =\int _{ 0}^{\infty }\int _{ 0}^{x}{e}^{-sx}f(t)g(x - t)dt\,dx. }$$
(4.16)
Now, if we assume that \(x = u + t\), then (4.16) becomes
$$\displaystyle{\mathcal{L}\left \{\int _{0}^{x}f(t)g(x - t)dt\right \} =\int _{ 0}^{\infty }\int _{ 0}^{\infty }{e}^{-s(u+t)}f(t)g(u)dt\,du.}$$
The double integral of the right-hand side can be written as the product of two integrals. Then
$$\displaystyle{\begin{array}{ll} \mathcal{L}\left \{\int _{0}^{x}f(t)g(x - t)dt\right \}& =\int _{ 0}^{\infty }{e}^{-st}f(t)dt\int _{ 0}^{\infty }{e}^{-su}g(u)du \\ & = \mathcal{L}\left \{f(x)\right \}\mathcal{L}\left \{g(x)\right \}. \end{array} }$$
But \(\mathcal{L}\left \{f(x)\right \}\mathcal{L}\left \{g(x)\right \} = F(s)G(s)\). Thus,
$$\displaystyle{\mathcal{L}\left \{\int _{0}^{x}f(t)g(x - t)dt\right \} = F(s)G(s),}$$
and the proof is completed. ■ 
Let us return to Example 4.21. Our aim was to find
$$\displaystyle{{\mathcal{L}}^{-1}\left \{ \frac{2s} {({s}^{2} + 1)({s}^{2} + 4)}\right \}.}$$
Hence we write
$$\displaystyle{{\mathcal{L}}^{-1}\left \{ \frac{2s} {({s}^{2} + 1)({s}^{2} + 4)}\right \} = {\mathcal{L}}^{-1}\left \{ \frac{s} {{s}^{2} + 1} \times \frac{2} {{s}^{2} + 4}\right \}}$$
and use Theorem 4.39. The result is
$$\displaystyle{{\mathcal{L}}^{-1}\left \{ \frac{s} {{s}^{2} + 1} \times \frac{2} {{s}^{2} + 4}\right \} =\int _{ 0}^{x}\cos (x - t)\sin (2t)dt = \frac{2} {3}(\cos (x) -\cos (2x)).}$$

Example 4.40.

Find \({\mathcal{L}}^{-1}\left \{ \frac{1} {({s}^{2} + 9){s}^{5}}\right \}\).

Solution.

We can write
$$\displaystyle{{\mathcal{L}}^{-1}\left \{ \frac{1} {({s}^{2} + 9){s}^{3}}\right \} = \frac{1} {6}{\mathcal{L}}^{-1}\left \{ \frac{3} {{s}^{2} + 9} \times \frac{2} {{s}^{3}}\right \} = \frac{1} {6}\int _{0}^{x}{(x - t)}^{2}\sin (3t)dt.}$$
Integrating by parts leads to
$$\displaystyle{{\mathcal{L}}^{-1}\left \{ \frac{1} {({s}^{2} + 9){s}^{3}}\right \} = \frac{{x}^{2}} {18} + \frac{1} {81}(\cos (3x) - 1).}$$

 □ 

As we mentioned before, Theorem 4.39 may be applied to solve some integral equations that are of the form of a convolution integral.

Consider the integral equation given by
$$\displaystyle{ f(x) = h(x) +\lambda \int _{ 0}^{x}f(t)g(x - t)dt. }$$
(4.17)
Taking the Laplace transform, we obtain
$$\displaystyle{F(s) = H(s) +\lambda F(s)G(s),}$$
where \(F(s) = \mathcal{L}\left \{f(x)\right \}\), \(G(s) = \mathcal{L}\left \{g(x)\right \}\), \(H(s) = \mathcal{L}\left \{h(x)\right \}\), and λ is a constant. We can write
$$\displaystyle{F(s) = \frac{H(s)} {1 +\lambda G(s)}.}$$
Now, take the inverse Laplace transform to find f(x).

Example 4.41.

Solve the integral equation given by
$$\displaystyle{f(x) = x +\int _{ 0}^{x}f(t){e}^{x-t}dt.}$$

Solution.

We have \(G(s) = \mathcal{L}\left \{{e}^{x}\right \} = \frac{1} {s - 1}\), \(H(s) = \mathcal{L}\left \{x\right \} = \frac{1} {{s}^{2}}\) and λ = 1. Thus,
$$\displaystyle{F(s) = \frac{H(s)} {1 +\lambda G(s)} = \frac{s - 1} {{s}^{3}} = \frac{1} {{s}^{2}} - \frac{1} {{s}^{3}}.}$$
It follows
$$\displaystyle{f(x) = {\mathcal{L}}^{-1}\left \{ \frac{1} {{s}^{2}}\right \} -{\mathcal{L}}^{-1}\left \{ \frac{1} {{s}^{3}}\right \} = x -\frac{{x}^{2}} {2}.}$$

 □ 

Example 4.42.

Solve the integral equation given by
$$\displaystyle{{f}^{{\prime}}(x) =\sin (x) +\int _{ 0}^{x}f(x - t)\cos (t)dt,\quad {f}^{{\prime}}(0) = 0.}$$

Solution.

Taking the Laplace transform and using the initial condition, we get
$$\displaystyle{sF(s) = \frac{1} {{s}^{2} + 1} + \frac{sF(s)} {{s}^{2} + 1},\;\mbox{ i.e.}\quad F(s) = \frac{1} {{s}^{3}}.}$$
From this we obtain
$$\displaystyle{f(x) = \frac{{x}^{2}} {2}.}$$

 □ 

The equation in Example 4.42 is a so-called integro-differential equation. That is, the Laplace transform can also be used to solve such equations. In the next chapter, we use the Laplace transform to solve two or more simultaneous ODEs.

4.7 Exercises

Exercise 4.1.

Show that:
  1. (1)

    \(\mathcal{L}\left \{f\left (\frac{a} {x}\right ){e}^{-\frac{bx} {a} }\right \} = F(as + b)\),

     
  2. (2)

    \(\mathcal{L}\left \{{x}^{n}{e}^{ax}\right \} = \frac{n!} {{(s - a)}^{n+1}}\),

     
  3. (3)

    \(\mathcal{L}\left \{{e}^{-ax}\sin (bx)\right \} = \frac{n!} {{(s - a)}^{n+1}}\),

     
  4. (4)

    \(\mathcal{L}\left \{\frac{1 -\cos (ax)} {x} \right \} = \frac{1} {2}\ln \left (1 + \frac{{a}^{2}} {{s}^{2}} \right )\),

     
  5. (5)

    \(\mathcal{L}\left \{\sqrt{x}\right \} = \frac{1} {2s}\sqrt{ \frac{\pi } {s}}\),

     
  6. (6)

    \(\mathcal{L}\left \{\sin (\sqrt{x})\right \} = \frac{{e}^{-1/4s}} {2s} \sqrt{ \frac{\pi } {s}}\),

     
  7. (7)

    \(\mathcal{L}\left \{\frac{\cos (\sqrt{x})} {\sqrt{x}} \right \} = {e}^{-1/4s}\sqrt{ \frac{\pi } {s}}\),

     
  8. (8)

    \(\mathcal{L}\left \{\int _{0}^{\infty }\frac{f(t){x}^{t}} {t!} dt\right \} = \frac{1} {2s}\sqrt{ \frac{\pi } {\mathrm{s}}}\),

     
  9. (9)

    \(\mathcal{L}\left \{f({x}^{2})\right \} = \frac{1} {2\sqrt{\pi }}\int _{0}^{\infty }\frac{f(t){e}^{-{s}^{2}/4t }} {t\sqrt{t}} dt\),

     
  10. (10)

    \(\mathcal{L}\left \{ \frac{1} {\sqrt{\pi x}}\int _{0}^{\infty }f(t){e}^{-{t}^{2}/4x }dt\right \} = \frac{F(\ln (s))} {\sin (s)}\).

     

Exercise 4.2.

Determine the solution of the following initial value problems:

$$\displaystyle{\begin{array}{r@{\quad }l@{\qquad }l} (1)\quad &{y}^{{\prime\prime}}- 3{y}^{{\prime}} + 2y = 4\cos (x), \qquad &y(0) = 0,\;\,{y}^{{\prime}}(0) = 1, \\ (2)\quad &{y}^{{\prime\prime}} + 4{y}^{{\prime}} + 3y = x + 4, \qquad &y(0) = 1,\;\,{y}^{{\prime}}(0) = 0, \\ (3)\quad &{y}^{{\prime\prime\prime}}- 3{y}^{{\prime\prime}}- 4{y}^{{\prime}} = 2{e}^{x} -\sin (x), \qquad &y(0) = {y}^{{\prime}}(0) = {y}^{{\prime\prime\prime}}(0) = 0, \\ (4)\quad &{y}^{{\prime\prime}} + 4y =\sin (x) -\sin (x - 2\pi )u_{2\pi }, \qquad &y(0) = {y}^{{\prime}}(0) = 0, \\ (5)\quad &{y}^{{\prime\prime}} + y = (x - 2)u_{2}, \qquad &y(0) = {y}^{{\prime}}(0) = 0, \\ (6)\quad &x{y}^{{\prime\prime}} + (3x - 1){y}^{{\prime}}- (4x - 9)y = 0, \qquad &y(0) = {y}^{{\prime}}(0) = 0, \\ (7)\quad &x{y}^{{\prime\prime}} + (x - 1){y}^{{\prime}}- y = 0, \qquad &y(0) = 5,\;\,{y}^{{\prime}}(\infty ) = 0, \\ (8)\quad &{y}^{{\prime\prime}}- x{y}^{{\prime}} + y = 1, \qquad &y(0) = 1,\;\,{y}^{{\prime}}(0) = -2, \\ (9)\quad &{y}^{{\prime\prime}} + y = \left \{\begin{array}{lc} 4, &0 \leq x \leq 2\\ x + 2, & x > 2 \end{array} \right., \qquad &y(0) = 1,\;\,{y}^{{\prime}}(0) = 0, \\ (10)\quad &{y}^{{\prime\prime}} + 4y = \left \{\begin{array}{lc} x,&0 \leq x \leq 1\\ 1, & x > 1 \end{array} \right.,\qquad &y(0) = 1,\;\,{y}^{{\prime}}(0) = 0.\\ \quad \end{array} }$$

Exercise 4.3.

Find the inverse Laplace transform of following expressions:

$$\displaystyle{\begin{array}{l@{\quad }l@{\qquad }l@{\quad }l} (1)\quad & \frac{{s}^{2}} {({s}^{2} + 1)({s}^{2} + {4}^{2})}, \qquad &(2) \quad & \frac{{s}^{2}} {{({s}^{2} + 9)}^{4}}, \\ (3)\quad &{e}^{-2s}\ln \left ( \frac{s} {s - 1}\right ), \qquad &(4) \quad &\frac{1} {s}\ln \left (\frac{{s}^{2} + {a}^{2}} {{s}^{2} + {b}^{2}} \right ), \\ (5)\quad &\frac{1 - s{e}^{-\pi s}} {{s}^{2} + 1}, \qquad &(6) \quad & \frac{1} {{s}^{3} + 1}, \\ (7)\quad &\frac{3({s}^{2} + 1){e}^{-4s}} {{s}^{6}}, \qquad &(8) \quad &\frac{(1 - {e}^{-s}){e}^{-s}} {s({s}^{2} + 1)}, \\ (9)\quad &{e{}^{-as}\tan }^{-1}\left (s - 1\right ),\;a > 0,\qquad &(10)\quad &{ \frac{s} {{s}^{2} + 1}\cot }^{-1}\left (s + 1\right ). \\ \quad \end{array} }$$

Exercise 4.4.

Verify that:

$$\displaystyle{\begin{array}{l@{\quad }l@{\qquad }l@{\quad }l} (1)\quad &\int _{0}^{\infty }\frac{\cos (ax)\! -\! cos(bx)} {x} dx\! =\!\ln \left (\frac{a} {b}\right ),\qquad &(2) \quad &\int _{0}^{\infty } \frac{\cos (ax)} {1\! +\! {x}^{2}}dx\! =\! \frac{\pi } {2}{e}^{\!-\!a}\!. \\ (3)\quad &\int _{0}^{\infty } \frac{x\sin (ax)} {1 + {x}^{2}}dx = \frac{\pi } {2}{e}^{-a},\;a > 0,\qquad &(4) \quad &\int _{ 0}^{\infty }\frac{\sin (ax)} {x} dx = \frac{\pi } {2}, \\ (5)\quad &\int _{0}^{\infty }\frac{{e}^{\!-\!ax}\sin (bx)} {x} dx\! {=\!\cot }^{\!-\!1}\left ( \frac{b} {a}\right )\!, \qquad &(6) \quad &\int _{0}^{\infty }\sin ({x}^{2})dx\! =\! \frac{1} {2}\sqrt{ \frac{\pi } {2}}, \\ (7)\quad &\!\int _{0}^{\infty }{x}^{3}{e}^{\!-\!x}\sin (x)d\!x\! =\! 0, \qquad &(8) \quad &\!\int _{ 0}^{\infty }x{e}^{\pm \!2x}\cos (x)d\!x\! =\! 0, \\ (9)\quad &\int _{0}^{\infty }\frac{{e}^{-ax} - {e}^{-bx}} {x} dx =\ln \left ( \frac{b} {a}\right ), \qquad &(10)\quad &\int _{0}^{\infty }{x}^{2}\cos (x)dx = 0, \\ \quad \end{array} }$$

Exercise 4.5.

Evaluate the following integrals:

$$\displaystyle{\begin{array}{l@{\quad }l@{\qquad }l@{\quad }l} (1)\quad &\int _{0}^{\infty }{x}^{3}{e}^{-x}\cos (x)dx,\qquad &(2)\quad &\int _{ 0}^{\infty }{(-\ln (x))}^{\frac{1} {2} }dx, \\ (3)\quad &\mathcal{L}\left \{\frac{1} {\pi } \int _{0}^{\infty }\cos (x\cos (x))dx\right \},\qquad &(4)\quad &\int _{ 0}^{\infty }x{e}^{-x}\mathrm{erf}(x)dx, \\ (5)\quad &\multicolumn{3}{l}{\int _{0}^{\infty }x{e}^{-3x}\left({\sin }^{2}(x)\right )\left [{\mathcal{L}}^{-1}\left \{\ln \left ( \frac{s} {s - 1}\right )\right \}\right ]dx.} \\ \quad \end{array} }$$

Exercise 4.6.

Solve the following integral and integro-differential equations:

  1. (1)

    \(f(x) = {x}^{2} +\int _{ 0}^{x}f(x - t)\sin (t)dt\),

     
  2. (2)

    \(f(x) =\cos (x) +\int _{ 0}^{x}f(t){e}^{x-t}dt\),

     
  3. (3)

    \(f(x) =\cos (x) + {e}^{2x}\int _{0}^{x}f(t){e}^{-2t}dt +\int _{ 0}^{x}f(x - t){e}^{-2t}dt\),

     
  4. (4)

    \({y}^{{\prime}}(x) = x +\int _{ 0}^{x}y(x - t)\cos (t)dt,\quad {y}^{{\prime}}(0) = 0\),

     
  5. (5)

    \({y}^{{\prime}}(x) + 2y - 3\int _{0}^{x}y(t)dt = 5(1 + x),\quad {y}^{{\prime}}(0) = 2\),

     
  6. (6)

    \({y}^{{\prime\prime}}(x) =\cos (3x) +\int _{ 0}^{x}(x - t){y}^{{\prime}}(t)dt,\quad y(0) = {y}^{{\prime}}(0) = 0\),

     
  7. (7)

    \({y}^{{\prime\prime}}(x) + {y}^{{\prime}}(x) =\int _{ 0}^{x}\sin (x - t){y}^{{\prime}}(t)dt,\quad y(0) = {y}^{{\prime}}(0) = 1\),

     
  8. (8)

    \({y}^{{\prime\prime}}(x) + {y}^{{\prime}}(x) =\cos (x) +\int _{ 0}^{x}\sin (t){y}^{{\prime}}(x - t)dt,\quad y(0) = {y}^{{\prime}}(0) = 0\),

     
  9. (9)

    \(y(x) =\sin (x) -\int _{0}^{x}ty(x - t){\mathcal{L}}^{-1}\left \{{\cot }^{-1}(s)\right \}dt,\quad y(0) = {y}^{{\prime}}(0) = 1\),

     
  10. (10)

    \({y}^{{\prime\prime}}(x) - {y}^{{\prime}}(x) =\int _{ 0}^{x}{y}^{{\prime\prime}}(t){y}^{{\prime}}(x - t)dt,\quad y(0) = {y}^{{\prime}}(0) = 0\).

     

Exercise 4.7.

The differential equation
$$\displaystyle{{y}^{{\prime\prime}}(t) + 5y(t) = \frac{1} {2}\delta (t - 2)}$$
describes the equation of motion for a mechanical system where δ(t) is the unit impulse function. Show that
$$\displaystyle{y(t) =\cos (\sqrt{5}t) + \frac{u_{2}} {2\sqrt{5}}\sin (\sqrt{5}(t - 2)),}$$
if \(y(0) = 1,\quad {y}^{{\prime}}(0) = 0\).

Exercise 4.8.

Assume that the deflection of a beam is governed by the following equation:
$$\displaystyle{EI{y}^{(4)}(x) = \frac{W_{0}} {c} \left [c - x + (x - c)u_{c}\right ],\quad 0 < x < 2c.}$$
Solve this equation under the boundary conditions
$$\displaystyle{y(0) = {y}^{{\prime}}(0) = {y}^{{\prime\prime}}(2c) = {y}^{{\prime\prime\prime}}(2c) = 0.}$$

Exercise 4.9.

The differential equation
$$\displaystyle{L{i}^{{\prime\prime}}(t) + \frac{1} {C}i = {v}^{{\prime}}(t)}$$
arises from Kirchhoff’s second law for an electrical circuit. Solve this equation under the initial conditions
$$\displaystyle{i(0) = {i}^{{\prime}}(0) = 0,}$$
where \(v(t) = \left \{\begin{array}{l @{}l} - 4t, &0 \leq t \leq 3\\ - 12, &t > 3 \end{array} \right.\).

Exercise 4.10.

Suppose that the driving voltage for an RC-circuit is sin(t) for t ≥ 0. Find the current of the circuit if the initial current is 1∕R and the equation governing the circuit is \(R{i}^{{\prime}}(t) + \frac{1} {C}\int _{0}^{t}i(x)dx =\sin (t)\).

Exercise 4.11.

A simple satellite-tracking system is modeled by the differential equation
$$\displaystyle{{I\theta }^{{\prime\prime}}(t) = -k[\theta (t) -\varPsi (t)].}$$
In this expression Ψ(t) is an observed input angle, θ(t) is the system response output, I is an appropriate moment of inertia, and k denotes a positive constant that relates the system’s induced compensating torque to the instantaneous pointing θ(t) −Ψ(t). If Ψ(t) is assumed to be known, determine the solution of the initial value problem with
$$\displaystyle{\theta (0) =\theta _{0},{\quad \theta }^{{\prime}}(0) =\theta _{ 0}^{{\prime}}}$$
by the Laplace transform method and show that it can be expressed in convolution form as
$$\displaystyle{\theta (t) =\theta _{0}\cos (\omega t) + \left (\frac{\theta _{0}^{{\prime}}} {\omega } \right )\sin (\omega t) +\omega \int _{ 0}^{t}\varPsi (u)\sin (t - u)du,}$$
where \(\omega \equiv \sqrt{k/I}\).

Copyright information

© Springer India 2014

Authors and Affiliations

  • Martin Hermann
    • 1
  • Masoud Saravi
    • 2
  1. 1.Institute of Applied MathematicsFriedrich Schiller UniversityJenaGermany
  2. 2.Department of MathematicsIslamic Azad University Nour BranchNourIran

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