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Laplace - Chapter 6 The Laplace Hansform The main...

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Unformatted text preview: Chapter 6. The Laplace Hansform The main difficulty that occurs in solving initial value problems by the transform technique lies in the problem of determining the function y = qb (t) corresponding to the transform Y(s). This problem is known as the inversion problem for the Laplace transform; (1) (t) is called the inverse transform corresponding to Y(s), and the process of finding d) (r) from Y(s) is known as inverting the transform. We also use the notation £‘1{Y(s)} to denote the inverse transform of Y(s). There is a general formula for the inverse Laplace transform, but its use requires a knowledge of the theory of functions of a complex variable, and we do not consider it in this book. However, it is still possible to develop many important properties of the Laplace transform, and to solve many interesting problems, without the use of complex variables. In solving the initial value problem (4), (5), we did not consider the question of whether there may be functions other than the one given by Eq. (8) that also have the transform (13). In fact, it can be shown that if f is a continuous function with the Laplace transform F, then there is no other continuous function having the same transform. In other words, there is essentially a one—to-one correspondence between functions and their Laplace transforms. This fact suggests the compilation of a table, such as Table 6.2.1, giving the transforms of functions frequently encountered, and vice versa. The entries in the second column of Table 62.1 are the transforms of those in the first column. Perhaps more important, the functions in the first column are the inverse transforms of those in the second column. Thus, for example, if the transform of the solution of a differential equation is known, the solution itself can often be found merely by looking it up in the table. Some of the entries in Table 6.2.1 have been used as examples, or appear as problems in Section 6.1, while others will be developed later in the chapter. The third column of the table indicates where the derivation of the given transforms may be found. Although Table 6.2.1 is sufficient for the examples and problems in this book, much larger tables are also available (see the list of references at the end of the chapter). Transforms and inverse transforms can also be readily obtained electronically by using a computer algebra system. Frequently, a Laplace transform F (s) is expressible as a sum of several terms, F(S) = F1 (S) + F2(S) + ‘ ' ' + F/1(S)- (17) Suppose that f1(t) : £_1{F1(S)}, . . . ,f,,(t) = £4 {F,,(s)}. Then the function f(r)=f1(t)+-~+fn(t) ' has the Laplace transform F (s). By the uniqueness property stated previously, no other continuous function f has the same transform. Thus Elms» = £"1{F1(s)} + - - - + film. (Sn; (18) that is, the inverse Laplace transform is also a linear operator. In many problems it is convenient to make use of this property by decomposing a given transform into a sum of functions whose inverse transforms are already known or can be found in the table. Partial fraction expansions are particularly useful in this connection, and a general result covering many cases is given in Problem 38. Other useful properties of Laplace transforms are derived later in this chapter. As further illustrations of the technique of solving initial value problems by means of the Laplace transform and partial fraction expansions, consider the following examples. 52 Solution of Initial Value Problems 319 TABLE 6.2.1 Elementary Laplace Transforms f0) = L‘1 {Fm} F(s) = cum} Notes 1 1. 1 E, s > 0 Sec. 6.1;Ex. 4 (II 1 2. e , s > (1 Sec. 6.1;Ex. 5 s — a I 3. t”, n 2 positive integer %, s > 0 Sec. 6.1;Prob. 27 F 1 4. :11, p > —1 (1” ), s > 0 Sec. 6.1;Prob. 27 Sp+1 5. sinat 7 a 7, s > 0 Sec. 6.1;Ex. 6 s— —— a- 6. cos at 7 S 7, s > 0 Sec. 6.1;Prob. 6 s— —— a- 7. smhm 7 a 7, s > In] Sec. 6.1;Prob. s s- — a- 8. coshat 7,S 7, s > [at Sec. 6.1;Prob. 7 s- — a- 9. 6‘” sin bt b s > :1 Sec. 6.1'Prob. 13 (s — a)2 —— b2 ’ ’ 10. 6‘” cos bt S _ a s > (1 Sec. 6.1‘Prob. 14 (s — a)2 —— by j r 11. t"e”,’, n = positive integer —n'-—, s,> :1 Sec. 6.1;Prob. 18 (S _ [DIM-1 e—CS 12. 1150‘) S , s > 0 Sec. 6.3 13. u,(t)f(t — c) e‘”F(s) Sec. 6.3 14. ec’f(t) F(s — c) Sec. 6.3 1 15. f(ct) —F (5), c > 0 Sec. 6.3;Prob. 19 c c f 16. / f(t — r)g(r)dr F(S)G(s) Sec. 6.6 o 17. 6(t — c) e‘” Sec. 6.5 18. f‘")(t) s”F(s) — s”‘]f(0) — ~ . ~ —f‘”‘1)(0) Sec. 6.2 19. (—t)“f(t) F(")(s) Sec. 6.2; Prob. 28 ...
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