HMS211 Laplace Transforms

# HMS211 Laplace Transforms - Chapter 2 Laplace Transforms...

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Chapter 2 Laplace Transforms 2.1 Preliminary Remarks Laplace transforms provide a very e ffi cient method for finding the solution of linear di ff erential equations, particularly those with constant coe ffi cients. 2.1.1 Linear Di ff erential Equations The following are examples of linear di ff erential equations: 1. dx dt + ax = 0 a const. 2. dx dt + ax = e - t 3. d 2 x dt 2 + a dx dt + bx = 0 a, b const. 4. d 2 x dt 2 + 3 dx dt - 12 x = e - t cos 2 t The first equation can be solved by separation of variables: The equation can be re-written 1 x dx dt = - a and integrating with respect to t gives ln | x | = - at + c so | x | = e - at + c = e c e - at where c is arbitrary. Hence the solution is x = Ae - at where | A | = e c . We can determine the constant A if we know the value of x at some particular point, usually the initial value x (0) of x. 2.1.2 Some Useful Limits We will sometimes need to consider the behaviour of functions such as f ( t ) = e - t t, as t → ∞ . More generally we consider limits such as lim t →∞ e - at t n . The general result that we need is the following. Suppose a > 0 and n is an integer, then lim t →∞ t n e a t = lim t →∞ e - a t t n = 0 . 53

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54 CHAPTER 2. LAPLACE TRANSFORMS Essentially, this is saying that the exponential e a t function eventually grows more rapidly than any power function t n . For example lim t →∞ e - 5 t t 2 = 0 , lim t →∞ e - 2 t t 7 = 0 , and so on. 2.1.3 Improper Integrals Integrals such as a f ( t ) dt where the region of integration is infinite, or 3 1 1 t - 2 dt where the integrand is infinite in the interval of integration are called improper integrals. Integrals of the form a f ( t ) dt are evaluated by defining a f ( t ) dt = lim R →∞ R a f ( t ) dt Example 1 Evaluate 1 1 t 2 dt Solution 1 1 t 2 dt = lim R →∞ R 1 1 t 2 dt = lim R →∞ R 1 t - 2 dt = lim R →∞ - t - 1 R 1 = lim R →∞ - R - 1 - ( - (1) - 1 ) = lim R →∞ 1 - 1 R = 1 since lim R →∞ 1 R = 0 . Example 2 Evaluate 0 e - 3 t dt Solution We have
2.1. PRELIMINARY REMARKS 55 0 e - 3 t dt = lim R →∞ R 0 e - 3 t dt = lim R →∞ e - 3 t - 3 R 0 = lim R →∞ e - 3 R - 3 - e 0 - 3 = lim R →∞ 1 3 - 1 3 e - 3 R = 1 3 since lim R →∞ e - 3 R = 0 . Example 3 Evaluate 0 1 1 + t 2 dt Solution We have 0 1 1 + t 2 dt = lim R →∞ R 0 1 1 + t 2 dt = lim R →∞ arctan( t ) R 0 = lim R →∞ arctan( R ) - arctan(0) = lim R →∞ arctan( R ) = π 2 2.1.4 Laplace Transforms If f ( t ) is a function then its Laplace transform is defined to be the function F ( s ) , s > 0, where F ( s ) = 0 e - st f ( t ) dt We write L{ f ( t ) } = F ( s ) to indicate F ( s ) is the Laplace transform of f ( t ) . The transform operation transforms a function f ( t ) of the independent variable t to a function F ( s ) of the variable s . The domain of f ( t ) is often called the time domain and the domain of F ( s ) is called the frequency domain. It is a useful convention that if a function is denoted by a lower case letter, then the Laplace transform of the function is denoted by the corresponding capital letter. Thus L{ g ( t ) } = G ( s ) , L{ x ( t ) } = X ( s ) , and so on.

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