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Unformatted text preview: Chapter 7 Laplace Transform The Laplace transform can be used to solve differential equations. Be sides being a different and efficient alternative to variation of parame ters and undetermined coefficients, the Laplace method is particularly advantageous for input terms that are piecewisedefined, periodic or im pulsive. The direct Laplace transform or the Laplace integral of a function f ( t ) defined for 0 ≤ t < ∞ is the ordinary calculus integration problem Z ∞ 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 R ∞ y ( t ) e st dt = R ∞ f ( t ) e st dt implies y ( t ) = f ( t ) , or L ( y ( t ) = L ( f ( t )) implies y ( t ) = f ( t ) . (1) In differential equation applications, y ( t ) is the soughtafter unknown while f ( t ) is an explicit expression taken from integral tables. Below, we illustrate Laplace’s method by solving the initial value prob lem y = 1 , y (0) = 0 . The method obtains a relation L ( y ( t )) = L ( t ), whence Lerch’s cancel lation 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 249 Laplace Integral. The integral R ∞ g ( t ) e st dt is called the Laplace integral of the function g ( t ). It is defined by lim N →∞ R N 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 ) = t 2 , producing the integral formulas in Table 1. R ∞ (1) e st dt = (1 /s ) e st fl fl t = ∞ t =0 Laplace integral of g ( t ) = 1 . = 1 /s Assumed s > . R ∞ ( t ) e st dt = R ∞ d ds ( e st ) dt Laplace integral of g ( t ) = t . = d ds R ∞ (1) e st dt Use R d ds F ( t,s ) dt = d ds R F ( t,s ) dt . = d ds (1 /s ) Use L (1) = 1 /s . = 1 /s 2 Differentiate. R ∞ ( t 2 ) e st dt = R ∞ d ds ( te st ) dt Laplace integral of g ( t ) = t 2 . = d ds R ∞ ( t ) e st dt = d ds (1 /s 2 ) Use L ( t ) = 1 /s 2 . = 2 /s 3 Table 1. The Laplace integral R ∞ g ( t ) e st dt for g ( t ) = 1 , t and t 2 . R ∞ (1) e st dt = 1 s R ∞ ( t ) e st dt = 1 s 2 R ∞ ( t 2 ) e st dt = 2 s 3 In summary, L ( t n ) = n ! s 1+ n An Illustration. The ideas of the Laplace method will be illus trated 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....
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 Winter '08
 JaberAbdualrahman
 Differential Equations, Equations, est dt

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