© 2008 Zachary S Tseng C-1 - 1 The Laplace Transform Definition and properties of Laplace Transform, piecewise continuous functions, the Laplace Transform method of solving initial value problems The method of Laplace transforms is a system that relies on algebra (rather than calculus-based methods) to solve linear differential equations. While it might seem to be a somewhat cumbersome method at times, it is a very powerful tool that enables us to readily deal with linear differential equations with discontinuous forcing functions. Definition: Let f (t) be defined for t ≥ 0. The Laplace transform of f (t), denoted by F(s) or L{f (t)}, is an integral transform given by the Laplace integral: L {f (t)} =∫∞−=0)()( dttfesFst. Provided that this (improper) integral exists, i.e. that the integral is convergent. The Laplace transform is an operation that transforms a function of t (i.e., a function of time domain), defined on [0, ∞), to a function of s (i.e., of frequency domain)*. F(s) is the Laplace transform, or simply transform, of f (t). Together the two functions f (t) and F(s) are called a Laplace transform pair. For functions of t continuous on [0, ∞), the above transformation to the frequency domain is one-to-one. That is, different continuous functions will have different transforms. * The kernel of the Laplace transform, e−st in the integrand, is unit-less. Therefore, the unit of s is the reciprocal of that of t. Hence s is a variable denoting (complex) frequency.© 2008 Zachary S Tseng C-1 - 2 Example: Let f (t) = 1, then ssF1)( =, s > 0. L{f (t)} = ∞−∞−∞−−==∫∫0001)(stststesdtedttfe The integral is divergent whenever s ≤ 0. However, when s > 0, it converges to ())(1)1(1010sFsses==−−=−−. Example: Let f (t) = t, then 21)(ssF =, s > 0. [This is left to you as an exercise.] Example: Let f (t) = e at, then assF−=1)(, s > a. L{f (t)} = ∞−∞−∞−−==∫∫0)(0)(01tsatsaatstesadtedtee The integral is divergent whenever s ≤ a. However, when s > a, it converges to ())(1)1(1010sFassaesa=−=−−=−−.© 2008 Zachary S Tseng C-1 - 3 Definition: A function f (t) is called piecewise continuous if it only has finitely many (or none whatsoever – a continuous function is considered to be “piecewise continuous”!) discontinuities on any interval [a, b], and that both one-sided limits exist as t approaches each of those discontinuity from within the interval. The last part of the definition means that f could have removable and/or jump discontinuities only; it cannot have any infinity discontinuity. Theorem: Suppose that 1. f is piecewise continuous on the interval 0 ≤ t ≤ A for any A > 0. 2. │f (t)│ ≤ K e at when t ≥ M, for any real constant a, and some positive constants K and M. (This means that f is “of exponential order”, i.e. its rate of growth is no faster than that of exponential functions.) Then the Laplace transform, F(s) = L{f (t)}, exists for s > a. Note: The above theorem gives a sufficient condition for the existence of Laplace transforms. It is not a necessary condition. A function does not need to satisfy the two conditions in order to have a Laplace transform. Examples of such functions that nevertheless have Laplace transforms are logarithmic functions and the unit impulse function.© 2008 Zachary S Tseng C-1 - 4 Some properties of the Laplace Transform 1. L {0} = 0 2. L {f (t) ± g(t)} = L {f (t)} ± L {g(t)} 3. L {c f (t)} = c L {f (t)}, for any constant c. Properties 2 and 3 together means that the Laplace transform is linear. 4. [The derivative of Laplace transforms] L {(−t)f (t)} = F ′(s) or, equivalently L {t f (t)} = −F ′(s) Example: L {t2} = − (L {t})′ = 332221sssdsd=−−=− In general, the derivatives of Laplace transforms satisfy L {(−t)nf (t)} = F (n)(s) or, equivalently L {t n f (t)} = (−1)n F (n)(s) Warning: The Laplace transform, while a linear operation, is not multiplicative. That is, in general L {f (t) g(t)} ≠ L {f (t)} L {g(t)}. Exercise: (a) Use property 4 above, and the fact that L { e at} = as−1, to deduce that L {t e at} = 2)(1as −. (b) What will L {t 2 e at} be?© 2008 Zachary S Tseng C-1 - 5 Exercises C-1.1: 1 – 5 Use the (integral transformation) definition of the Laplace transform to find the Laplace transform of each function below. 1. t 2 2. t e 6t 3. cos 3t 4. e −t sin 2t 5.* e iαt, where i and α are constants, 1−=i. 6 – 8 Each function F(s) below is defined by a definite integral. Without integrating, find an explicit expression for each F(s). [Hint: each expression is the Laplace transform of a certain function. Use your knowledge of Laplace Transformation, or with the help of a table of common Laplace transforms to find the answer.] 6. ∫∞+−0)7(dtets 7. ∫∞−−0)3(2dtetts 8. ∫∞−06sin4 dttest Answers C-1.1: 1. 32s 2. 2)6(1−s 3. 92+ss 4. 5222++ss 5. 2222ααα+++siss Note: Since the Euler’s formula says that e iαt = cos αt + i sin αt, therefore, L{e iαt} = L{cos αt + i sin αt}. That is, the real part of its Laplace transform corresponds to that of cos αt, the imaginary part corresponds to that of sin αt. (Check it for yourself!) 6. 71+s 7. 3)3(2−s 8. 36242+s© 2008 Zachary S Tseng C-1 - 6 Solution of Initial Value Problems We now shall meet “the new System”: how the Laplace transforms can be used to solve linear differential equations algebraically. Theorem: [Laplace transform of derivatives] Suppose f is of exponential order, and that f is continuous and f ′ is piecewise continuous on any interval 0 ≤ t ≤ A. Then L {f ′(t)} = s L {f (t)} − f (0) Applying the theorem multiple times yields: L {f ″(t)} = s2 L {f (t)} − s f (0) − f ′(0), L {f ″′(t)} = s3 L {f (t)} − s2 f (0) − s f ′(0) − f ″(0), : : L {f (n)(t)} = s n L {f (t)} − s n − 1 f (0) − s n − 2 f ′(0) − … − s 2 f (n −3)(0) − s f (n −2)(0) − f (n −1)(0). This is an extremely useful aspect of the Laplace transform: that it changes differentiation with respect to t into multiplication by s (and, as seen a little earlier, differentiation with respect to s into multiplication by −t, on the other hand). Equally importantly, it
View Full Document
Unlocking...