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MATH 251

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Step Functions; and Laplace Transforms of Piecewise Continuous Functions The present objective is to use the Laplace transform to solve differential equations with piecewise continuous forcing functions (that is, forcing functions that contain discontinuities). Before that could be done, we need to learn how to find the Laplace transforms of piecewise continuous functions, and how to find their inverse transforms. Step Functions Definition: The unit step function (or Heaviside function), is defined by u t = 0, t < c c 1, t ≥ c , c ≥ 0 . Often the unit step function c (t) is also denoted as u(t − c)c H (t), or H(t − c). The step could also be negative (going down). The complement function is  1, t < c 1−u (tc =  , c ≥ 0 .  0, t ≥ c © 2008 Zachary S Tseng C▯2 ▯ 1 The Laplace transform of the unit step function is e −cs L{u (t)} = , s > 0, c≥ 0 c s Notice that when c = 0,0u) has the same Laplace transform as the constant function f(t) = 1. (Why?) Therefore, for our purpo0e, u (t) = 1. (Keep in mind that a Laplace transform is only defined for t ≥ 0.) Note: The calculation oL{u ct)} goes as follow (given thc≥ 0): ∞ ∞ u (t)e −stdt = 1⋅e −stdt = −1 e −st ∞ L{u ct)} = ∫0 c ∫c s c −1 −cs e−cs = 0 −e = ) , s > 0. s s © 2008 Zachary S Tseng C▯2 ▯ 2 The unit step function is much more useful than it first appears to be. When put in a product with a second function, the unit step function acts like a switch to turn the other function on or off: 0, t < c u ct) f (t) =  f (t), t ≥ c, (an “on” switch)   f (t), t < c (1−u (c)) f (t) =  0, t ≥ c, (an “off” switch). By combining two unit step functions, we can also selectively make a function appears only for a finite duration, then it disappears. That is, the function is switched “on” at a, then is switched “off” at a later time b.  0, t < a (u (t)−u (t)) f (t) = f (t), a ≤ t < b a b  ,  0, t ≥ b where 0 ≤ a < b. We could think this combination as an “on▯off” toggle switch that controls the appearance of the second function f(t). By cascading the above types of products, we can now write any piecewise▯ defined function in a succinct form in terms of unit step functions. © 2008 Zachary S Tseng C▯2 ▯ 3 Suppose  f1(t), t < a  f2(t), a ≤ t < b  F(t) =  f3(t), b ≤ t < c  .  : :  f (t), t ≥ d  n Then, we can rewrite F(t), succinctly, as F(t) = (1 − u at))f 1t) + (u at) − u bt))f 2t) + (u bt) − u (c))f 3t) + … + u (t)f (t). d n Example: 2  3t−2, t < 4  5t F t ) = e + t 4 ≤
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