Laplace Transform

The Laplace transform can be used to solve diﬀerential equations. Besides being a diﬀerent and eﬃcient alternative to variation of parameters and undetermined coeﬃcients, the Laplace method is particularly advantageous for input terms that are piecewise-deﬁned, periodic or impulsive. The direct Laplace transform or the Laplace integral of a function f (t) deﬁned for 0 ≤ t < ∞ is the ordinary calculus integration problem ∞ 0

f (t)e−st dt,

succinctly denoted L(f (t)) in science and engineering literature. The L–notation recognizes that integration always proceeds over t = 0 to t = ∞ and that the integral involves an integrator e−st dt instead of the usual dt. These minor diﬀerences distinguish Laplace integrals from the ordinary integrals found on the inside covers of calculus texts.

7.1 Introduction to the Laplace Method

The foundation of Laplace theory is Lerch’s cancellation law ∞ −st dt 0 y(t)e

=

∞ −st dt 0 f (t)e

(1) L(y(t) = L(f (t))

implies or implies

y(t) = f (t), y(t) = f (t).

In diﬀerential equation applications, y(t) is the sought-after unknown while f (t) is an explicit expression taken from integral tables. Below, we illustrate Laplace’s method by solving the initial value problem y = −1, y(0) = 0. The method obtains a relation L(y(t)) = L(−t), whence Lerch’s cancellation law implies the solution is y(t) = −t. The Laplace method is advertised as a table lookup method, in which the solution y(t) to a diﬀerential equation is found by looking up the answer in a special integral table.

7.1 Introduction to the Laplace Method

247

Laplace Integral. The integral

is called the Laplace N integral of the function g(t). It is deﬁned by limN →∞ 0 g(t)e−st dt and depends on variable s. The ideas will be illustrated for g(t) = 1, g(t) = t and g(t) = t2 , producing the integral formulas in Table 1. ∞ −st dt 0 (1)e

∞ −st dt 0 g(t)e

= −(1/s)e−st = 1/s = = = =

t=∞ t=0

Laplace integral of g(t) = 1. Assumed s > 0. Laplace integral of g(t) = t. Use d ds F (t, s)dt

∞ −st dt 0 (t)e

∞ d −st )dt 0 − ds (e d ∞ − ds 0 (1)e−st dt d − ds (1/s) 1/s2 ∞ d −st )dt 0 − ds (te d ∞ − ds 0 (t)e−st dt d − ds (1/s2 ) 2/s3

=

d ds

F (t, s)dt.

Use L(1) = 1/s. Diﬀerentiate. Laplace integral of g(t) = t2 . Use L(t) = 1/s2 .

∞ 2 −st dt 0 (t )e

= = = =

Table 1. The Laplace integral

∞ 0

g(t)e−st dt for g(t) = 1, t and t2 .

∞ −st 0 (1)e

dt =

1 s

∞ −st 0 (t)e

dt =

1 s2 n! s1+n

∞ 2 −st dt 0 (t )e

=

2 s3

In summary, L(tn ) =

An Illustration. The ideas of the Laplace method will be illustrated for the solution y(t) = −t of the problem y = −1, y(0) = 0. The method, entirely diﬀerent from variation of parameters or undetermined coeﬃcients, uses basic calculus and college algebra; see Table 2. Table 2. Laplace method details for the illustration y = −1, y(0) = 0.

y (t)e−st = −e−st

∞ −st dt = ∞ −e−st dt 0 y (t)e 0 ∞ y (t)e−st dt = −1/s 0 ∞ s 0 y(t)e−st dt − y(0) = −1/s ∞ −st dt = −1/s2 0 y(t)e ∞ −st dt = ∞ (−t)e−st dt 0 y(t)e 0

Multiply y = −1 by e−st . Integrate t = 0 to t = ∞. Use Table 1. Integrate by parts on the left. Use y(0) = 0 and divide. Use Table 1. Apply Lerch’s cancellation law.

y(t) = −t

248

Laplace Transform

In Lerch’s law, the formal rule of erasing the integral signs is valid provided the integrals are equal for large s and certain conditions hold on y and f – see Theorem 2. The illustration in Table 2 shows that Laplace theory requires an in-depth study of a special integral table, a table which is a true extension of the usual table found on the inside covers of calculus books. Some entries for the special integral table appear in Table 1 and also in section 7.2, Table 4. The L-notation for the direct Laplace transform produces briefer details, as witnessed by the translation of Table 2 into Table 3 below. The reader is advised to move from Laplace integral notation to the...