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10 Laplace Transform Methods

10.1 Laplace Transforms and Inverse Transforms

In Chapter 5 we saw that linear differential equations with constant coefficients have numerous applications and can be solved systematically. There are common situations, however, in which the alternative methods of this chapter are preferable. For example, recall the differential equations

m x^{″} + c x^{′} + k x = F (t) and L I^{″} + R I^{′} + \frac{1}{C} I = E^{′} (t)

$m x^{″} + c x^{′} + k x = F (t) and L I^{″} + R I^{′} + \frac{1}{C} I = E^{′} (t)$

corresponding to a mass–spring–dashpot system and a series RLC circuit, respectively. It often happens in practice that the forcing term, F(t) or $E^{′} (t),$ $E^{′} (t),$ has discontinuities—for example, when the voltage supplied to an electrical circuit is turned off and on periodically. In this case the methods of Chapter 5 can be quite awkward, and the Laplace transform method is more convenient.

The differentiation operator D can be viewed as a transformation which, when applied to the function f(t), yields the new function $D {f (t)} = f^{′} (t) .$ $D {f (t)} = f^{′} (t) .$ The Laplace transformation $L$ $L$ involves the operation of integration and yields the new function $L {f (t)} = F (s)$ $L {f (t)} = F (s)$ of a new independent variable s. The situation is diagrammed in Fig. 10.1.1. After learning in this section how to compute the Laplace transform F(s) of a function f(t), we will see in Section 10.2 that the Laplace transform converts a differential equation in the unknown function f(t) into an algebraic equation in F(s). Because algebraic equations are generally easier to solve than differential equations, this is one method that simplifies the problem of finding the solution f(t).

FIGURE 10.1.1.

Transformation of a function: $L$ $L$ in analogy with `D`.

Recall that an improper integral over an infinite interval is defined as a limit of integrals over bounded intervals; that is,

\int_{a}^{\infty} g (t) d t = \lim_{b \to \infty} \int_{a}^{b} g (t) d t .

$\int_{a}^{\infty} g (t) d t = \lim_{b \to \infty} \int_{a}^{b} g (t) d t .$ (2)

If the limit in (2) exists, then we say that the improper integral converges; otherwise, it diverges or fails to exist. Note that the integrand of the improper integral in (1) contains the parameter s in addition to the variable of integration t. Therefore, when the integral in (1) converges, it converges not merely to a number, but to a function F of s. As in the following examples, it is typical for the improper integral in the definition of $L {f (t)}$ $L {f (t)}$ to converge for some values of s and diverge for others.

Example 1

With $f (t) \equiv 1$ $f (t) \equiv 1$ for $t ≧ 0,$ $t ≧ 0,$ the definition of the Laplace transform in (1) gives

L {1} = \int_{0}^{\infty} e^{- s t} d t = {[- \frac{1}{s} e^{- s t}]}_{0}^{\infty} = \lim_{b \to \infty} [- \frac{1}{s} e^{- b s} + \frac{1}{s}],

$L {1} = \int_{0}^{\infty} e^{- s t} d t = {[- \frac{1}{s} e^{- s t}]}_{0}^{\infty} = \lim_{b \to \infty} [- \frac{1}{s} e^{- b s} + \frac{1}{s}],$

and therefore

L {1} = \frac{1}{s} for s > 0.

$L {1} = \frac{1}{s} for s > 0.$ (3)

As in (3), it’s good practice to specify the domain of the Laplace transform—in problems as well as in examples. Also, in this computation we have used the common abbreviation

{[g (t)]}_{a}^{\infty} = \lim_{b \to \infty} {[g (t)]}_{a}^{b} .

${[g (t)]}_{a}^{\infty} = \lim_{b \to \infty} {[g (t)]}_{a}^{b} .$ (4)

Remark

The limit we computed in Example 1 would not exist if $s < 0,$ $s < 0,$ for then $(1 / s) e^{- b s}$ $(1 / s) e^{- b s}$ would become unbounded as $b \to + \infty .$ $b \to + \infty .$ Hence $L {1}$ $L {1}$ is defined only for $s > 0 .$ $s > 0 .$ This is typical of Laplace transforms; the domain of a transform is normally of the form $s > a$ $s > a$ for some number a.

Example 2

With $f (t) = e^{a t}$ $f (t) = e^{a t}$ for $t ≧ 0,$ $t ≧ 0,$ we obtain

L {e^{a t}} = \int_{0}^{\infty} e^{- s t} e^{a t} d t = \int_{0}^{\infty} e^{- (s - a) t} d t = {[- \frac{e^{- (s - a) t}}{s - a}]}_{t = 0}^{\infty} .

$L {e^{a t}} = \int_{0}^{\infty} e^{- s t} e^{a t} d t = \int_{0}^{\infty} e^{- (s - a) t} d t = {[- \frac{e^{- (s - a) t}}{s - a}]}_{t = 0}^{\infty} .$

If $s - a > 0,$ $s - a > 0,$ then $e^{- (s - a) t} \to 0$ $e^{- (s - a) t} \to 0$ as $t \to + \infty,$ $t \to + \infty,$ so it follows that

L {e^{a t}} = \frac{1}{s - a} for s > a .

$L {e^{a t}} = \frac{1}{s - a} for s > a .$ (5)

Note here that the improper integral giving $L {e^{a t}}$ $L {e^{a t}}$ diverges if $s ≦ a .$ $s ≦ a .$ It is worth noting also that the formula in (5) holds if a is a complex number. For then, with $a = α + i β$ $a = α + i β$ ,

e^{- (s - a) t} = e^{i β t} e^{- (s - α) t} \to 0

$e^{- (s - a) t} = e^{i β t} e^{- (s - α) t} \to 0$

as $t \to + \infty,$ $t \to + \infty,$ provided that $s > α = Re [a];$ $s > α = Re [a];$ recall that $e^{i β t} = \cos β t + i \sin β t$ $e^{i β t} = \cos β t + i \sin β t$ .

The Laplace transform $L {t^{a}}$ $L {t^{a}}$ of a power function is most conveniently expressed in terms of the gamma function $Γ (x),$ $Γ (x),$ which is defined for $x > 0$ $x > 0$ by the formula

Γ (x) = \int_{0}^{\infty} e^{- t} t^{x - 1} d t .

$Γ (x) = \int_{0}^{\infty} e^{- t} t^{x - 1} d t .$ (6)

For an elementary discussion of $Γ (x),$ $Γ (x),$ see the subsection on the gamma function in Section 11.4, where it is shown that

Γ (1) = 1

$Γ (1) = 1$ (7)

and that

Γ (x + 1) = x Γ (x)

$Γ (x + 1) = x Γ (x)$ (8)

for $x > 0 .$ $x > 0 .$ It then follows that if n is a positive integer, then

\begin{array}{rcl} Γ (n + 1) & = & n Γ (n) \\ = & n \cdot (n - 1) Γ (n - 1) \\ = & n \cdot (n - 1) \cdot (n - 2) Γ (n - 2) \\ ⋮ \\ = & n (n - 1) (n - 2) \dots 2 \cdot Γ (2) \\ = & n (n - 1) (n - 2) \dots 2 \cdot 1 \cdot Γ (1); \end{array}

$\begin{array}{rcl} Γ (n + 1) & = & n Γ (n) \\ = & n \cdot (n - 1) Γ (n - 1) \\ = & n \cdot (n - 1) \cdot (n - 2) Γ (n - 2) \\ ⋮ \\ = & n (n - 1) (n - 2) \dots 2 \cdot Γ (2) \\ = & n (n - 1) (n - 2) \dots 2 \cdot 1 \cdot Γ (1); \end{array}$

thus

Γ (n + 1) = n!

$Γ (n + 1) = n!$ (9)

if n is a positive integer. Therefore, the function $Γ (x + 1),$ $Γ (x + 1),$ which is defined and continuous for all $x > - 1,$ $x > - 1,$ agrees with the factorial function for $x = n,$ $x = n,$ a positive integer.

Example 3

Suppose that $f (t) = t^{a}$ $f (t) = t^{a}$ where a is real and $a > - 1 .$ $a > - 1 .$ Then

L {t^{a}} = \int_{0}^{\infty} e^{- s t} t^{a} d t .

$L {t^{a}} = \int_{0}^{\infty} e^{- s t} t^{a} d t .$

If we substitute $u = s t, t = u / s,$ $u = s t, t = u / s,$ and $d t = d u / s$ $d t = d u / s$ in this integral, we get

L {t^{a}} = \frac{1}{s^{a + 1}} \int_{0}^{\infty} e^{- u} u^{a} d u = \frac{Γ (a + 1)}{s^{a + 1}}

$L {t^{a}} = \frac{1}{s^{a + 1}} \int_{0}^{\infty} e^{- u} u^{a} d u = \frac{Γ (a + 1)}{s^{a + 1}}$ (10)

for all $s > 0$ $s > 0$ (so that $u = s t > 0$ $u = s t > 0$ ). Because $Γ (n + 1) = n!$ $Γ (n + 1) = n!$ if n is a nonnegative integer, we see that

L {t^{n}} = \frac{n!}{s^{n + 1}} for s > 0.

$L {t^{n}} = \frac{n!}{s^{n + 1}} for s > 0.$ (11)

For instance,

L {t} = \frac{1}{s^{2}}, L {t^{2}} = \frac{2}{s^{3}}, and L {t^{3}} = \frac{6}{s^{4}} .

$L {t} = \frac{1}{s^{2}}, L {t^{2}} = \frac{2}{s^{3}}, and L {t^{3}} = \frac{6}{s^{4}} .$

As in Problems 1 and 2, these formulas can be derived immediately from the definition, without the use of the gamma function.

Linearity of Transforms

It is not necessary for us to proceed much further in the computation of Laplace transforms directly from the definition. Once we know the Laplace transforms of several functions, we can combine them to obtain transforms of other functions. The reason is that the Laplace transformation is a linear operation.

The proof of Theorem 1 follows immediately from the linearity of the operations of taking limits and of integration:

\begin{array}{rcl} L {a f (t) + b g (t)} & = & \int_{0}^{\infty} e^{- s t} [a f (t) + b g (t)] d t \\ = & \lim_{c \to \infty} \int_{0}^{c} e^{- s t} [a f (t) + b g (t)] d t \\ = & a (\lim_{c \to \infty} \int_{0}^{c} e^{- s t} f (t) d t) + b (\lim_{c \to \infty} \int_{0}^{c} e^{- s t} g (t) d t) \\ = & a L {f (t)} + b L {g (t)} . \end{array}

$\begin{array}{rcl} L {a f (t) + b g (t)} & = & \int_{0}^{\infty} e^{- s t} [a f (t) + b g (t)] d t \\ = & \lim_{c \to \infty} \int_{0}^{c} e^{- s t} [a f (t) + b g (t)] d t \\ = & a (\lim_{c \to \infty} \int_{0}^{c} e^{- s t} f (t) d t) + b (\lim_{c \to \infty} \int_{0}^{c} e^{- s t} g (t) d t) \\ = & a L {f (t)} + b L {g (t)} . \end{array}$

Example 4

The computation of $L {t^{n / 2}}$ $L {t^{n / 2}}$ is based on the known special value

Γ (\frac{1}{2}) = \sqrt{π}

$Γ (\frac{1}{2}) = \sqrt{π}$ (13)

of the gamma function. For instance, it follows that

Γ (\frac{5}{2}) = \frac{3}{2} Γ (\frac{3}{2}) = \frac{3}{2} \cdot \frac{1}{2} Γ (\frac{1}{2}) = \frac{3}{4} \sqrt{π},

$Γ (\frac{5}{2}) = \frac{3}{2} Γ (\frac{3}{2}) = \frac{3}{2} \cdot \frac{1}{2} Γ (\frac{1}{2}) = \frac{3}{4} \sqrt{π},$

using the formula $Γ (x + 1) = x Γ (x)$ $Γ (x + 1) = x Γ (x)$ in (9), first with $x = \frac{3}{2}$ $x = \frac{3}{2}$ and then with $x = \frac{1}{2} .$ $x = \frac{1}{2} .$ Now the formulas in (10) through (12) yield

L {3 t^{2} + 4 t^{3/2}} = 3 \cdot \frac{2!}{s^{3}} + \frac{4 Γ (\frac{5}{2})}{s^{5/2}} = \frac{6}{s^{3}} + 3 \sqrt{\frac{π}{s^{5}}} .

$L {3 t^{2} + 4 t^{3/2}} = 3 \cdot \frac{2!}{s^{3}} + \frac{4 Γ (\frac{5}{2})}{s^{5/2}} = \frac{6}{s^{3}} + 3 \sqrt{\frac{π}{s^{5}}} .$

Example 5

Recall that $\cosh k t = (e^{k t} + e^{- k t}) / 2 .$ $\cosh k t = (e^{k t} + e^{- k t}) / 2 .$ If $k > 0,$ $k > 0,$ then Theorem 1 and Example 2 together give

L {\cosh k t} = \frac{1}{2} L {e^{k t}} + \frac{1}{2} L {e^{- k t}} = \frac{1}{2} (\frac{1}{s - k} + \frac{1}{s + k});

$L {\cosh k t} = \frac{1}{2} L {e^{k t}} + \frac{1}{2} L {e^{- k t}} = \frac{1}{2} (\frac{1}{s - k} + \frac{1}{s + k});$

that is,

L {\cosh k t} = \frac{s}{s^{2} - k^{2}} for s > k > 0.

$L {\cosh k t} = \frac{s}{s^{2} - k^{2}} for s > k > 0.$ (14)

Similarly,

L {\sinh k t} = \frac{k}{s^{2} - k^{2}} for s > k > 0.

$L {\sinh k t} = \frac{k}{s^{2} - k^{2}} for s > k > 0.$ (15)

Because $\cos k t = (e^{ikt} + e^{- ikt}) / 2,$ $\cos k t = (e^{ikt} + e^{- ikt}) / 2,$ the formula in (5) (with $a = i k$ $a = i k$ ) yields

L {\cos k t} = \frac{1}{2} (\frac{1}{s - i k} + \frac{1}{s + i k}) = \frac{1}{2} \cdot \frac{2 s}{s^{2} - {(i k)}^{2}},

$L {\cos k t} = \frac{1}{2} (\frac{1}{s - i k} + \frac{1}{s + i k}) = \frac{1}{2} \cdot \frac{2 s}{s^{2} - {(i k)}^{2}},$

and thus

L {\cos k t} = \frac{s}{s^{2} + k^{2}} for s > 0.

$L {\cos k t} = \frac{s}{s^{2} + k^{2}} for s > 0.$ (16)

(The domain follows from $s > Re [i k] = 0$ $s > Re [i k] = 0$ .) Similarly,

L {\sin k t} = \frac{k}{s^{2} + k^{2}} for s > 0.

$L {\sin k t} = \frac{k}{s^{2} + k^{2}} for s > 0.$ (17)

Example 6

Applying linearity, the formula in (16), and a familiar trigonometric identity, we get

\begin{array}{rcl} L {3 e^{2 t} + 2 \sin^{2} 3 t} & = & L {3 e^{2 t} + 1 - \cos 6 t} \\ = & \frac{3}{s - 2} + \frac{1}{s} - \frac{s}{s^{2} + 36} \\ = & \frac{3 s^{3} + 144 s - 72}{s (s - 2) (s^{2} + 36)} for s > 0. \end{array}

$\begin{array}{rcl} L {3 e^{2 t} + 2 \sin^{2} 3 t} & = & L {3 e^{2 t} + 1 - \cos 6 t} \\ = & \frac{3}{s - 2} + \frac{1}{s} - \frac{s}{s^{2} + 36} \\ = & \frac{3 s^{3} + 144 s - 72}{s (s - 2) (s^{2} + 36)} for s > 0. \end{array}$

Inverse Transforms

According to Theorem 3 of this section, no two different functions that are both continuous for all $t ≧ 0$ $t ≧ 0$ can have the same Laplace transform. Thus if F(s) is the transform of some continuous function f(t), then f(t) is uniquely determined. This observation allows us to make the following definition: If $F (s) = L {f (t)},$ $F (s) = L {f (t)},$ then we call f(t) the inverse Laplace transform of F(s) and write

f (t) = L^{- 1} {F (s)} .

$f (t) = L^{- 1} {F (s)} .$ (18)

Example 7

Using the Laplace transforms derived in Examples 2, 3, and 5 we see that

L^{- 1} {\frac{1}{s^{3}}} = \frac{1}{2} t^{2}, L^{- 1} {\frac{1}{s + 2}} = e^{- 2 t}, L^{- 1} {\frac{2}{s^{2} + 9}} = \frac{2}{3} \sin 3 t

$L^{- 1} {\frac{1}{s^{3}}} = \frac{1}{2} t^{2}, L^{- 1} {\frac{1}{s + 2}} = e^{- 2 t}, L^{- 1} {\frac{2}{s^{2} + 9}} = \frac{2}{3} \sin 3 t$

and so on.

Notation: Functions and Their Transforms. Throughout this chapter we denote functions of t by lowercase letters. The transform of a function will always be denoted by that same letter capitalized. Thus F(s) is the Laplace transform of f(t) and x(t) is the inverse Laplace transform of X(s).

A table of Laplace transforms serves a purpose similar to that of a table of integrals. The table in Fig. 10.1.2 lists the transforms derived in this section; many additional transforms can be derived from these few, using various general properties of the Laplace transformation (which we will discuss in subsequent sections).

Piecewise Continuous Functions

As we remarked at the beginning of this section, we need to be able to handle certain types of discontinuous functions. The function f(t) is said to be piecewise continuous on the bounded interval $a ≦ t ≦ b$ $a ≦ t ≦ b$ provided that [a, b] can be subdivided into finitely many abutting subintervals in such a way that

f is continuous in the interior of each of these subintervals; and
f(t) has a finite limit as t approaches each endpoint of each subinterval from its interior.

FIGURE 10.1.2.

A short table of Laplace transforms.

`f`(`t`)	`F`(`s`)
1	$\frac{1}{s}$ $\frac{1}{s}$	$(s > 0)$ $(s > 0)$
t	$\frac{1}{s^{2}}$ $\frac{1}{s^{2}}$	$(s > 0)$ $(s > 0)$
$t^{n} (n ≧ 0)$ $t^{n} (n ≧ 0)$	$\frac{n!}{s^{n + 1}}$ $\frac{n!}{s^{n + 1}}$	$(s > 0)$ $(s > 0)$
$t^{a} (a > - 1)$ $t^{a} (a > - 1)$	$\frac{Γ (a + 1)}{s^{a + 1}}$ $\frac{Γ (a + 1)}{s^{a + 1}}$	$(s > 0)$ $(s > 0)$
$e^{a t}$ $e^{a t}$	$\frac{1}{s - a}$ $\frac{1}{s - a}$	$(s > a)$ $(s > a)$
cos `k` `t`	$\frac{s}{s^{2} + k^{2}}$ $\frac{s}{s^{2} + k^{2}}$	$(s > 0)$ $(s > 0)$
sin `k` `t`	$\frac{k}{s^{2} + k^{2}}$ $\frac{k}{s^{2} + k^{2}}$	$(s > 0)$ $(s > 0)$
cosh `k` `t`	$\frac{s}{s^{2} - k^{2}}$ $\frac{s}{s^{2} - k^{2}}$	$(s > \| k \|)$ $(s > \| k \|)$
sinh `k` `t`	$\frac{k}{s^{2} - k^{2}}$ $\frac{k}{s^{2} - k^{2}}$	$(s > \| k \|)$ $(s > \| k \|)$
$u (t - a)$ $u (t - a)$	$\frac{e^{- a s}}{s}$ $\frac{e^{- a s}}{s}$	$(s > 0)$ $(s > 0)$

We say that f is piecewise continuous for $t ≧ 0$ $t ≧ 0$ if it is piecewise continuous on every bounded subinterval of $[0, + \infty) .$ $[0, + \infty) .$ Thus a piecewise continuous function has only simple discontinuities (if any) and only at isolated points. At such points the value of the function experiences a finite jump, as indicated in Fig. 10.1.3. The jump in f(t) at the point c is defined to be $f (c +) - f (c -),$ $f (c +) - f (c -),$ where

f (c +) = \lim_{ϵ \to 0 +} f (c + ϵ) and f (c -) = \lim_{ϵ \to 0 +} f (c - ϵ) .

$f (c +) = \lim_{ϵ \to 0 +} f (c + ϵ) and f (c -) = \lim_{ϵ \to 0 +} f (c - ϵ) .$

Perhaps the simplest piecewise continuous (but discontinuous) function is the unit step function, whose graph appears in Fig. 10.1.4. It is defined as follows:

u (t) = {\begin{array}{l} 0 & for t < 0, \\ 1 & for t ≧ 0. \end{array}

$u (t) = {\begin{array}{l} 0 & for t < 0, \\ 1 & for t ≧ 0. \end{array}$ (19)

Because $u (t) = 1$ $u (t) = 1$ for $t ≧ 0$ $t ≧ 0$ and because the Laplace transform involves only the values of a function for $t ≧ 0,$ $t ≧ 0,$ we see immediately that

L {u (t)} = \frac{1}{s} (s > 0) .

$L {u (t)} = \frac{1}{s} (s > 0) .$ (20)

The graph of the unit step function $u_{a} (t) = u (t - a)$ $u_{a} (t) = u (t - a)$ appears in Fig. 10.1.5. Its jump occurs at $t = a$ $t = a$ rather than at $t = 0;$ $t = 0;$ equivalently,

u_{a} (t) = u (t - a) = {\begin{array}{l} 0 & for t < a, \\ 1 & for t ≧ a . \end{array}

$u_{a} (t) = u (t - a) = {\begin{array}{l} 0 & for t < a, \\ 1 & for t ≧ a . \end{array}$ (21)

Example 8

Find $L {u_{a} (t)}$ $L {u_{a} (t)}$ if $a > 0$ $a > 0$ .

Solution

We begin with the definition of the Laplace transform. We obtain

L {u_{a} (t)} = \int_{0}^{\infty} e^{- s t} u_{a} (t) d t = \int_{a}^{\infty} e^{- s t} d t = \lim_{b \to \infty} {[- \frac{e^{- s t}}{s}]}_{t = a}^{b};

$L {u_{a} (t)} = \int_{0}^{\infty} e^{- s t} u_{a} (t) d t = \int_{a}^{\infty} e^{- s t} d t = \lim_{b \to \infty} {[- \frac{e^{- s t}}{s}]}_{t = a}^{b};$

consequently,

L {u_{a} (t)} = \frac{e^{- a s}}{s} (s > 0, a > 0) .

$L {u_{a} (t)} = \frac{e^{- a s}}{s} (s > 0, a > 0) .$ (22)

FIGURE 10.1.3.

The graph of a piecewise continuous function; the solid dots indicate values of the function at discontinuities.

FIGURE 10.1.4.

The graph of the unit step function.

FIGURE 10.1.5.

The unit step function $u_{a} (t)$ $u_{a} (t)$ has a jump at $t = a .$ $t = a .$

General Properties of Transforms

It is a familiar fact from calculus that the integral

\int_{a}^{b} g (t) d t

$\int_{a}^{b} g (t) d t$

exists if g is piecewise continuous on the bounded interval [a, b]. Hence if f is piecewise continuous for $t ≧ 0,$ $t ≧ 0,$ it follows that the integral

\int_{0}^{b} e^{- s t} f (t) d t

$\int_{0}^{b} e^{- s t} f (t) d t$

exists for all $b < + \infty .$ $b < + \infty .$ But in order for F(s)—the limit of this last integral as $b \to + \infty$ $b \to + \infty$ —to exist, we need some condition to limit the rate of growth of f(t) as $t \to + \infty .$ $t \to + \infty .$ The function f is said to be of exponential order as $t \to + \infty$ $t \to + \infty$ if there exist nonnegative constants M, c, and T such that

| f (t) | ≦ M e^{c t} for t ≧ T .

$| f (t) | ≦ M e^{c t} for t ≧ T .$ (23)

Thus a function is of exponential order provided that it grows no more rapidly (as $t \to + \infty$ $t \to + \infty$ ) than a constant multiple of some exponential function with a linear exponent. The particular values of M, c, and T are not so important. What is important is that some such values exist so that the condition in (23) is satisfied.

The condition in (23) merely says that $f (t) / e^{c t}$ $f (t) / e^{c t}$ lies between $- M$ $- M$ and M and is therefore bounded in value for t sufficiently large. In particular, this is true (with $c = 0$ $c = 0$ ) if f(t) itself is bounded. Thus every bounded function—such as cos kt or sin kt—is of exponential order.

If p(t) is a polynomial, then the familiar fact that $p (t) e^{- t} \to 0$ $p (t) e^{- t} \to 0$ as $t \to + \infty$ $t \to + \infty$ implies that (23) holds (for T sufficiently large) with $M = c = 1 .$ $M = c = 1 .$ Thus every polynomial function is of exponential order.

For an example of an elementary function that is continuous and therefore bounded on every (finite) interval, but nevertheless is not of exponential order, consider the function $f (t) = e^{t^{2}} = \exp (t^{2}) .$ $f (t) = e^{t^{2}} = \exp (t^{2}) .$ Whatever the value of c, we see that

\lim_{t \to \infty} \frac{f (t)}{e^{c t}} = \lim_{t \to \infty} \frac{e^{t^{2}}}{e^{c t}} = \lim_{t \to \infty} e^{t^{2} - c t} = + \infty

$\lim_{t \to \infty} \frac{f (t)}{e^{c t}} = \lim_{t \to \infty} \frac{e^{t^{2}}}{e^{c t}} = \lim_{t \to \infty} e^{t^{2} - c t} = + \infty$

because $t^{2} - c t \to + \infty$ $t^{2} - c t \to + \infty$ as $t \to + \infty .$ $t \to + \infty .$ Hence the condition in (23) cannot hold for any (finite) value M, so we conclude that the function $f (t) = e^{t^{2}}$ $f (t) = e^{t^{2}}$ is not of exponential order.

Similarly, because $e^{- s t} e^{t^{2}} \to + \infty$ $e^{- s t} e^{t^{2}} \to + \infty$ as $t \to + \infty,$ $t \to + \infty,$ we see that the improper integral $\int_{0}^{\infty} e^{- s t} e^{t^{2}} d t$ $\int_{0}^{\infty} e^{- s t} e^{t^{2}} d t$ that would define $L {e^{t^{2}}}$ $L {e^{t^{2}}}$ does not exist (for any s), and therefore that the function $e^{t^{2}}$ $e^{t^{2}}$ does not have a Laplace transform. The following theorem guarantees that piecewise functions of exponential order do have Laplace transforms.

Proof:

First we note that we can take $T = 0$ $T = 0$ in (23). For by piecewise continuity, $| f (t) |$ $| f (t) |$ is bounded on [0, T]. Increasing M in (23) if necessary, we can therefore assume that $| f (t) | ≦ M$ $| f (t) | ≦ M$ if $0 ≦ t ≦ T .$ $0 ≦ t ≦ T .$ Because $e^{c t} ≧ 1$ $e^{c t} ≧ 1$ for $t ≧ 0,$ $t ≧ 0,$ it then follows that $| f (t) | ≦ M e^{c t}$ $| f (t) | ≦ M e^{c t}$ for all $t ≧ 0$ $t ≧ 0$ .

By a standard theorem on convergence of improper integrals—the fact that absolute convergence implies convergence—it suffices for us to prove that the integral

\int_{0}^{\infty} | e^{- s t} f (t) | d t

$\int_{0}^{\infty} | e^{- s t} f (t) | d t$

exists for $s > c .$ $s > c .$ To do this, it suffices in turn to show that the value of the integral

\int_{0}^{b} | e^{- s t} f (t) | d t

$\int_{0}^{b} | e^{- s t} f (t) | d t$

remains bounded as $b \to + \infty .$ $b \to + \infty .$ But the fact that $| f (t) | ≦ M e^{c t}$ $| f (t) | ≦ M e^{c t}$ for all $t ≧ 0$ $t ≧ 0$ implies that

\begin{array}{rcl} \int_{0}^{b} | e^{- s t} f (t) | d t & ≦ & \int_{0}^{b} | e^{- s t} M e^{c t} | d t = M \int_{0}^{b} e^{- (s - c) t} d t \\ ≦ & M \int_{0}^{\infty} e^{- (s - c) t} d t = \frac{M}{s - c} \end{array}

$\begin{array}{rcl} \int_{0}^{b} | e^{- s t} f (t) | d t & ≦ & \int_{0}^{b} | e^{- s t} M e^{c t} | d t = M \int_{0}^{b} e^{- (s - c) t} d t \\ ≦ & M \int_{0}^{\infty} e^{- (s - c) t} d t = \frac{M}{s - c} \end{array}$

if $s > c .$ $s > c .$ This proves Theorem 2.

We have shown, moreover, that

| F (s) | ≦ \int_{0}^{\infty} | e^{- s t} f (t) | d t ≦ \frac{M}{s - c}

$| F (s) | ≦ \int_{0}^{\infty} | e^{- s t} f (t) | d t ≦ \frac{M}{s - c}$ (24)

if $s > c .$ $s > c .$ When we take limits as $s \to + \infty,$ $s \to + \infty,$ we get the following result.

The condition in (25) severely limits the functions that can be Laplace transforms. For instance, the function $G (s) = s / (s + 1)$ $G (s) = s / (s + 1)$ cannot be the Laplace transform of any “reasonable” function because its limit as $s \to + \infty$ $s \to + \infty$ is 1, not 0. More generally, a rational function—a quotient of two polynomials—can be (and is, as we shall see) a Laplace transform only if the degree of its numerator is less than that of its denominator.

On the other hand, the hypotheses of Theorem 2 are sufficient, but not necessary, conditions for existence of the Laplace transform of f(t). For example, the function $f (t) = 1 / \sqrt{t}$ $f (t) = 1 / \sqrt{t}$ fails to be piecewise continuous (at $t = 0$ $t = 0$ ), but nevertheless (Example 3 with $a = - \frac{1}{2} > - 1$ $a = - \frac{1}{2} > - 1$ ) its Laplace transform

L {t^{- 1/2}} = \frac{Γ (\frac{1}{2})}{s^{1/2}} = \sqrt{\frac{π}{s}}

$L {t^{- 1/2}} = \frac{Γ (\frac{1}{2})}{s^{1/2}} = \sqrt{\frac{π}{s}}$

both exists and violates the condition in (24), which would imply that $s F (s)$ $s F (s)$ remains bounded as $s \to + \infty$ $s \to + \infty$ .

The remainder of this chapter is devoted largely to techniques for solving a differential equation by first finding the Laplace transform of its solution. It is then vital for us to know that this uniquely determines the solution of the differential equation; that is, that the function of s we have found has only one inverse Laplace transform that could be the desired solution. The following theorem is proved in Chapter 6 of Churchill’s Operational Mathematics, 3rd ed. (New York: McGraw-Hill, 1972).

Thus two piecewise continuous functions of exponential order with the same Laplace transform can differ only at their isolated points of discontinuity. This is of no importance in most practical applications, so we may regard inverse Laplace transforms as being essentially unique. In particular, two solutions of a differential equation must both be continuous, and hence must be the same solution if they have the same Laplace transform.

Historical Remark

Laplace transforms have an interesting history. The integral in the definition of the Laplace transform probably appeared first in the work of Euler. It is customary in mathematics to name a technique or theorem for the next person after Euler to discover it (else there would be several hundred different examples of “Euler’s theorem”). In this case, the next person was the French mathematician Pierre Simon de Laplace (1749–1827), who employed such integrals in his work on probability theory. The so-called operational methods for solving differential equations, which are based on Laplace transforms, were not exploited by Laplace. Indeed, they were discovered and popularized by practicing engineers—notably the English electrical engineer Oliver Heaviside (1850–1925). These techniques were successfully and widely applied before they had been rigorously justified, and around the beginning of the twentieth century their validity was the subject of considerable controversy. One reason is that Heaviside blithely assumed the existence of functions whose Laplace transforms contradict the condition that $F (s) \to 0$ $F (s) \to 0$ as $s \to 0,$ $s \to 0,$ thereby raising questions as to the meaning and nature of functions in mathematics. (This is reminiscent of the way Leibniz two centuries earlier had obtained correct results in calculus using “infinitely small” real numbers, thereby raising questions as to the nature and role of numbers in mathematics.)

10.1 Problems

Apply the definition in (1) to find directly the Laplace transforms of the functions described (by formula or graph) in Problems 1 through 10.

$f (t) = t$ $f (t) = t$
$f (t) = t^{2}$ $f (t) = t^{2}$
$f (t) = e^{3 t + 1}$ $f (t) = e^{3 t + 1}$
$f (t) = \cos t$ $f (t) = \cos t$
$f (t) = \sinh t$ $f (t) = \sinh t$
$f (t) = \sin^{2} t$ $f (t) = \sin^{2} t$
FIGURE 10.1.6.
FIGURE 10.1.7.
FIGURE 10.1.8.
FIGURE 10.1.9.

Use the transforms in Fig. 10.1.2 to find the Laplace transforms of the functions in Problems 11 through 22. A preliminary integration by parts may be necessary.

$f (t) = \sqrt{t} + 3 t$ $f (t) = \sqrt{t} + 3 t$
$f (t) = 3 t^{5/2} - 4 t^{3}$ $f (t) = 3 t^{5/2} - 4 t^{3}$
$f (t) = t - 2 e^{3 t}$ $f (t) = t - 2 e^{3 t}$
$f (t) = t^{3/2} - e^{- 10 t}$ $f (t) = t^{3/2} - e^{- 10 t}$
$f (t) = 1 + \cosh 5 t$ $f (t) = 1 + \cosh 5 t$
$f (t) = \sin 2 t + \cos 2 t$ $f (t) = \sin 2 t + \cos 2 t$
$f (t) = \cos^{2} 2 t$ $f (t) = \cos^{2} 2 t$
$f (t) = \sin 3 t \cos 3 t$ $f (t) = \sin 3 t \cos 3 t$
$f (t) = {(1 + t)}^{3}$ $f (t) = {(1 + t)}^{3}$
$f (t) = t e^{t}$ $f (t) = t e^{t}$
$f (t) = t \cos 2 t$ $f (t) = t \cos 2 t$
$f (t) = \sinh^{2} 3 t$ $f (t) = \sinh^{2} 3 t$

Use the transforms in Fig. 10.1.2 to find the inverse Laplace transforms of the functions in Problems 23 through 32.

$F (s) = \frac{3}{s^{4}}$ $F (s) = \frac{3}{s^{4}}$
$F (s) = s^{- 3/2}$ $F (s) = s^{- 3/2}$
$F (s) = \frac{1}{s} - \frac{2}{s^{5/2}}$ $F (s) = \frac{1}{s} - \frac{2}{s^{5/2}}$
$F (s) = \frac{1}{s + 5}$ $F (s) = \frac{1}{s + 5}$
$F (s) = \frac{3}{s - 4}$ $F (s) = \frac{3}{s - 4}$
$F (s) = \frac{3 s + 1}{s^{2} + 4}$ $F (s) = \frac{3 s + 1}{s^{2} + 4}$
$F (s) = \frac{5 - 3 s}{s^{2} + 9}$ $F (s) = \frac{5 - 3 s}{s^{2} + 9}$
$F (s) = \frac{9 + s}{4 - s^{2}}$ $F (s) = \frac{9 + s}{4 - s^{2}}$
$F (s) = \frac{10 s - 3}{25 - s^{2}}$ $F (s) = \frac{10 s - 3}{25 - s^{2}}$
$F (s) = 2 s^{- 1} e^{- 3 s}$ $F (s) = 2 s^{- 1} e^{- 3 s}$
Derive the transform of $f (t) = \sin k t$ $f (t) = \sin k t$ by the method used in the text to derive the formula in (16).
Derive the transform of $f (t) = \sinh k t$ $f (t) = \sinh k t$ by the method used in the text to derive the formula in (14).
Use the tabulated integral

$\int e^{a x} \cos b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \cos b x + b \sin b x) + C$ $\int e^{a x} \cos b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \cos b x + b \sin b x) + C$

to obtain $L {\cos k t}$ $L {\cos k t}$ directly from the definition of the Laplace transform.
Show that the function $f (t) = \sin (e^{t^{2}})$ $f (t) = \sin (e^{t^{2}})$ is of exponential order as $t \to + \infty$ $t \to + \infty$ but that its derivative is not.
Given $a > 0,$ $a > 0,$ let $f (t) = 1$ $f (t) = 1$ if $0 ≦ t < a, f (t) = 0$ $0 ≦ t < a, f (t) = 0$ if $t ≧ a .$ $t ≧ a .$ First, sketch the graph of the function f, making clear its value at $t = a .$ $t = a .$ Then express f in terms of unit step functions to show that $L {f (t)} = s^{- 1} (1 - e^{- a s})$ $L {f (t)} = s^{- 1} (1 - e^{- a s})$ .
Given that $0 < a < b,$ $0 < a < b,$ let $f (t) = 1$ $f (t) = 1$ if $a ≦ t < b, f (t) = 0$ $a ≦ t < b, f (t) = 0$ if either $t < a$ $t < a$ or $t ≧ b .$ $t ≧ b .$ First, sketch the graph of the function f, making clear its values at $t = a$ $t = a$ and $t = b .$ $t = b .$ Then express f in terms of unit step functions to show that $L {f (t)} = s^{- 1} (e^{- a s} - e^{- b s})$ $L {f (t)} = s^{- 1} (e^{- a s} - e^{- b s})$ .
The unit staircase function is defined as follows:

$f (t) = n if n - 1 ≦ t < n, n = 1, 2, 3, \dots$ $f (t) = n if n - 1 ≦ t < n, n = 1, 2, 3, \dots$

(a) Sketch the graph of f to see why its name is appropriate. (b) Show that

$f (t) = \sum_{n = 0}^{\infty} u (t - n)$ $f (t) = \sum_{n = 0}^{\infty} u (t - n)$

for all $t ≧ 0 .$ $t ≧ 0 .$ (c) Assume that the Laplace transform of the infinite series in part (b) can be taken termwise (it can). Apply the geometric series to obtain the result

$L {f (t)} = \frac{1}{s (1 - e^{- s})} .$ $L {f (t)} = \frac{1}{s (1 - e^{- s})} .$
(a) The graph of the function f is shown in Fig. 10.1.10. Show that f can be written in the form

$f (t) = \sum_{n = 0}^{\infty} {(- 1)}^{n} u (t - n) .$ $f (t) = \sum_{n = 0}^{\infty} {(- 1)}^{n} u (t - n) .$

(b) Use the method of Problem 39 to show that

$L {f (t)} = \frac{1}{s (1 + e^{- s})} .$ $L {f (t)} = \frac{1}{s (1 + e^{- s})} .$

FIGURE 10.1.10.

The graph of the function of Problem 40.
The graph of the square-wave function g(t) is shown in Fig. 10.1.11. Express g in terms of the function f of Problem 40 and hence deduce that

$L {g (t)} = \frac{1 - e^{- s}}{s (1 + e^{- s})} = \frac{1}{s} \tanh \frac{s}{2} .$ $L {g (t)} = \frac{1 - e^{- s}}{s (1 + e^{- s})} = \frac{1}{s} \tanh \frac{s}{2} .$

FIGURE 10.1.11.

The graph of the function of Problem 41.
Given constants a and b, define h(t) for $t ≧ 0$ $t ≧ 0$ by

$h (t) = {\begin{array}{l} a & if n - 1 ≦ t < n and n is odd; \\ b & if n - 1 ≦ t < n and n is even . \end{array}$ $h (t) = {\begin{array}{l} a & if n - 1 ≦ t < n and n is odd; \\ b & if n - 1 ≦ t < n and n is even . \end{array}$

Sketch the graph of h and apply one of the preceding problems to show that

$L {h (t)} = \frac{a + b e^{- s}}{s (1 + e^{- s})} .$ $L {h (t)} = \frac{a + b e^{- s}}{s (1 + e^{- s})} .$

10.1 Application Computer Algebra Transforms and Inverse Transforms

If $f (t) = t \cos 3 t,$ $f (t) = t \cos 3 t,$ then the definition of the Laplace transform gives the improper integral

F (s) = L {f (t)} = \int_{0}^{\infty} t e^{- s t} \cos 3 t d t,

$F (s) = L {f (t)} = \int_{0}^{\infty} t e^{- s t} \cos 3 t d t,$

whose evaluation would appear to require a tedious integration by parts. Consequently a computer algebra Laplace transforms package is useful for the quick calculation of transforms. Maple contains the integral transforms package inttrans, and the commands

with(inttrans):
f := t*cos(3*t):
F := laplace(f, t, s);

yield immediately the Laplace transform $F (s) = (s^{2} - 9) / {(s^{2} + 9)}^{2},$ $F (s) = (s^{2} - 9) / {(s^{2} + 9)}^{2},$ as do the Mathematica commands

f = t*Cos[3*t];
F = LaplaceTransform[f, t, s]

and the Wolfram|Alpha query

laplace transform t*cos(3t)

We can recover the original function $f (t) = t \cos 3 t$ $f (t) = t \cos 3 t$ with the Maple command

invlaplace(F, s, t);

or the Mathematica command

InverseLaplaceTransform[F, s, t]

or the Wolfram|Alpha query

inverse laplace transform (s^2 - 9)/(s^2 + 9)^2

Remark

Note carefully the order of s and t in the preceding Maple and Mathematica commands—first t, then s when transforming; first s, then t when inverse transforming.

You can use these computer algebra commands to check the answers to Problems 11 through 32 in this section, as well as a few interesting problems of your own selection.

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Table of Contents for 10 Laplace Transform Methods

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Table of Contents for
10 Laplace Transform Methods