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MATH 2270 (28)
Lecture

# Week 7.pdf

15 Pages
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School
University of Guelph
Department
Mathematics
Course
MATH 2270
Professor
Matthew Demers
Semester
Fall

Description
1 4.1 Laplace Transforms Now, we turn to a totally di▯erent method of solving ODEs: One that involves transforming the di▯erential equation into an algebraic equation, whereupon we can solve it and transform back. De▯nition: Consider a function F(t) for t ▯ 0. Then the Laplace transform of F(t), denoted by LfF(t)g, is given by Z 1 f(s) = LfF(t)g = e▯stF(t)dt; (4.1) 0 provided that the improper integral converges. The following criteria guarantee the convergence of this integral: ▯ F(t) must be piecewise-continuous; ▯ F(t) is exponentially bounded; that is, jF(t)j ▯ Ke at (4.2) for t ▯ M, where K, a, and M are constants. This condition simply means that the function F(t) can grow no more quickly than an exponential function. If these two hypotheses hold, then the Laplace Transform of F(t) exists for s > a (we will not prove this). Note that as it is de▯ned by an (improper) integral, the Laplace Transform is a linear operator; that is, Lfc 1 1 c f2 2= c L1f g 1 c Lf2 g: 2 (4.3) 2 Laplace Transforms simply take one function F(t) as input, and transform it into another function f(s) as output. We may take the Laplace Transform of many di▯erent functions. Here are some examples: Example: Laplace transform of a constant Let F(t) = 1. Then, Z 1 ▯st Lf1g = (1)e dt (4.4) 0 1 ▯N = lim ▯ e▯s▯ (4.5) N!1 s ▯t=0 ▯ ▯ ▯ ▯ = lim ▯ 1 e▯s(N) ▯ ▯ 1e▯s(0) (4.6) N!1 s s = 0 ▯ (▯ )1 (4.7) s 1 = : (4.8) s Example: Find the Laplace transform of ▯3. Note that since the Laplace transform is linear, multiplicative constants can easily move inside or outside the transform. Hence, ▯3 Lf▯3g = ▯3Lf1g = : (4.9) s Laplace transform of an exponential function 3 Let F(t) = e , where a is a constant. Then, Z 1 Lfe g = (e )e▯stdt (4.10) 0 Z 1 = e(a▯sdt (4.11) 0 ▯N 1 (a▯s▯t = lim e ▯ (4.12) ▯!1 a ▯ s t▯0 ▯ ▯ 1 1 = lim e(a▯s)N ▯ e(a▯s)(0) (4.13) N!1 a ▯ s a ▯ s (4.14) The ▯rst term goes to zero, since s > a and thus (a ▯ s) < 0. We obtain at 1 Lfe g = 0 ▯ (4.15) a ▯ s 1 = ;s > a: (4.16) s ▯ a Example: ▯t Find the Laplace transform of e . Applying the rule, we have a = ▯1, giving ▯ ▯t 1 L e = s + 1: (4.17) Laplace transform of trig functions Let F(t) = cos(!t), where ! is a constant. Then, Z 1 Lfe g = (cos(!t))e▯sdt: (4.18) 0 This integral is not pretty, and requires using integration by parts twice. It is doable, but instead, let’s take a di▯erent approach. Recall that from Euler’s formula, we have that ej!x= cos(!x) + j sin(!x): (4.19) 4 Thus, Lfe j!xg = Lfcos(!x) + j sin(!x)g (4.20) = Lfcos(!x)g + jLfsin(!x)g; (4.21) since L is a linear operator. But, we can use the Laplace transform of an exponential, given in Equation (4.16), to give us Lfe j!xg = 1 (4.22) s ▯ j! = s + j! ; (4.23) s + ! 2 multiplying both the numerator and denominator by s + j!. Thus, we have s + j! 2 2 = Lfcos(!x)g + jLfsin(!x)g; or (4.24) s + ! s + j ! = Lfcos(!x)g + jLfsin(!x)g: (4.25) s + ! 2 s + ! 2 From this equation, it is easy to deduce that s Lfcos(!x)g = s + ! 2;s > j!j (4.26) and ! Lfsin(!x)g = s + ! 2;s > j!j: (4.27) Example: Find the Laplace transform of sin(5t). Applying the rule, we have 5 Lfsin(5t)g = 2 : (4.28) s + 25 Example: Find the Laplace transform of cos(2t). 5 Applying the rule, we have s Lfcos(2t)g = s + 4 : (4.29) Laplace transform of derivatives Given a function F(t), consider its derivative F (t). Then, Z 1 LfF (t)g = (F (t))e▯stdt: (4.30) 0 Using integration by parts, we have that Z 0 ▯ ▯st▯N 1 ▯st LfF (t)g = lim F(t)e ▯0 + s(F(t))e dt: (4.31) N!1 0 Z ▯ ▯ 1 = lim F(N)e ▯s(N)▯ F(0)e ▯s(0) + s (F(t))e▯stdt: (4.32) N!1 0 = ▯F(0) + sLfF(t)g; or (4.33) = sLfF(t)g ▯ F(0): (4.34) Similarly, it is not hard to prove that 00 2 0 LfF (t)g = s LfF(t)g ▯ sF(0) ▯ F (0); (4.35) we need only to integrate by parts twice instead. In general, we have that (n) n n▯1 n▯2 0 (n▯2) (n▯1) LfF (t)g = s LfF(t)g▯s F(0)▯s F (0)▯:::▯sF (0)▯F (0): (4.36) Example: d f Find the Laplace transform of , where f is some function. dt3 There isn’t much we can do except apply our new rule. Keeping with the notation, we have ▯ ▯ d f df d f L = s Lff(t)g ▯ s f(0) ▯ s (0) ▯ s (0): (4.37) dt 3 dt dt2 6 Laplace transform of powers Let F(t) = t . While the Laplace transform could be found using integration by parts, we take a di▯erent approach to make things simpler. From the last section, we have LfF (t)g = sLfF(t)g ▯ F(0): (4.38) If we substitute F(t) = t into this equation, we obtain Lf1g = sLftg ▯ (0); or (4.39) 1 Lftg = sLf1g (4.40) ▯ ▯▯ ▯ = 1 1 (4.41) s s 1 = s2: (4.42) If we substitute F(t) = t into equation (4.38), we obtain 2 Lf2tg = sLft g ▯ (0) (4.43) 2Lftg = sLft g; 2 or (4.44) ▯ ▯ Lft g = 2 (Lftg) (4.45) s ▯ ▯▯ ▯ 2 2 1 Lft g = s s2 (4.46) 2 = 3: (4.47) s 3 6 In a similar way, Lft g = 4 (do it yourself), and in general, we ▯nd that s n n! Lft g = n+1;s > 0: (4.48) s Example: 5 Find the Laplace transform of the function F(t) = 3t . 5 Remember that the Laplace Transform is linear, so that we know Lf3t g is the 5 same as 3Lft g. So, don’t worry about that 3. Now, we have 7 5! 120 360 3Lft g = 3 = 3 = : (4.49) s 6 s6 s6 at Laplace transform of e F(t) If F(t) is a function, then we know that the Laplace Transform of F is Z 1 ▯st LfF(t)g = e F(t)dt; (4.50) 0 which de▯nes a function of s. So, let LfF(t)g = f(s). Then, we have that Z 1 ▯(s▯a)t f(s ▯ a) = e F(t)dt: (4.51) 0 But, this can be rewritten as Z 1 ▯st at f(s ▯ a) = e e F(t)dt; (4.52) 0 at which is the Laplace transform for e F(t). Thus, we have the property that at Lfe F(t)g = f(s ▯ a); (4.53) where f(s) is the Laplace transform for F(t). This may sound a little complicated, but an example will clear things up: Example: 4t Find the Laplace transform of the function F(t) = te . 4t Seeing the e present as part of this function, we know to use the property we have just detailed. Hence, the Laplace transform for this function is the same as the 1 Laplace transform for t (that is, s2), except with an \s ▯ 4" subbed in for the usual \s". Hence,
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