6. Laplace Transforms - NCU

Advanced Engineering Mathematics6. Laplace transforms16. Laplace Transforms6.1 Laplace transform, inverse transform, linearity6.2 Transforms of derivatives and integralsAdvanced Engineering Mathematics6. Laplace transforms6.1 Laplace transform, inverse transform, linearity Laplace transformof the function f(t), and be denoted by L (f). Inverse Laplace transformf(t) L-1(F). EX.1.f(t) 1, when t 0.2

Advanced Engineering Mathematics6. Laplace transforms3 The notation EX.2. Sometimes we may obtain the Laplace transform of a functionindirectly from the definition. Theorem 1 (Linearity of the Laplace transform)The Laplace transform is a linear operation; that is, forany functions f(t) and g(t) whose Laplace transformsexist and any constant a and b,L {af(t) bg(t)} a L {f(t)} b L {g(t)}L -1{aF(s) bG(s)} a L -1{F(s)} b L -1{G(s)}.Advanced Engineering Mathematics EX.3. EX. EX.6. Laplace transforms4

Advanced Engineering Mathematics6. Laplace transforms5 Ex.4.Prove thatsinceBy Euler formula: e i t cos t i sin t , we haveAdvanced Engineering Mathematics6. Laplace transforms6First shifting theorem Theorem 2 (First shifting theorem)If f(t) has the transform F(s) (where s k), then eat f(t) has thetransform F(s - a) (where s - a k). In formulas,L {eat f(t)} F(s - a) andeat f(t) L -1{F(s - a)}.Proof EX.5.f(t)F(s)

Advanced Engineering Mathematics6. Laplace transforms7n Using induction method to prove that the Laplace transform of t is.By integration by part [d(uv) udv vdu udv uv - vdu]dv Prove thatvuuwhere gamma functionTaking st xAdvanced Engineering Mathematics6. Laplace transforms The Laplace transforms of some important elementary functionsare listed in Table 6.1 (on page 224 of the textbook), and a moreextensive list is given in Section 6.8 (on pages 264 267 of thetextbook). EX.6.From Table 6.1 and the first shifting theorem, weimmediately obtain another useful formula,Sincen(i) L {t } (ii) L {eat f(t)} F(s - a)8

Advanced Engineering Mathematics6. Laplace transforms9Existence of Laplace transform Piecewise continuousA function f(t) is piecewise continuous on a finite interval a t b,if f is continuous on [a,b], except possibly at finitely many pointsc1, ,cn, at each of which f has a finite left and right limit.Advanced Engineering Mathematics6. Laplace transforms10 Theorem 3 (Existence theorem for Laplace transforms)(i) f(t) is piecewise continuous on [0, )(ii) There exist some constants k and M such that f(t) Me kt L [f](s) exists for s k .boundedProoff(t) piecewise continuous on [0, ) e -stf(t) is integrable on [0, ).Assume s k ,

Advanced Engineering Mathematics6. Laplace transforms11 Note 1Not all function f(t) satisfy f(t) Me kt.Some functions satisfy the boundary condition; for example,cosh t e t , t n n! e t (n 0,1, ) , 2but we can not find a M and k such that e t Me kt Note 2The conditions in Theorem 3 are sufficient rather than necessary( i.e., 充分非必要 or say ”satisfy conditionse.g.,L[f](s) exits” .doesn’t satisfy the conditionTaking st xAdvanced Engineering Mathematics6. Laplace transforms Note 3If the Laplace transform of a given function exists, it is uniquelydetermined. If L [f1] L [f2], then f1 f2 on a piecewise interval.f1(t)F1(s)f2(t)F2 (s)If F1(s) F2(s), then f1(t) f2(t).It means that Laplace transform is a one-to-one transformation. Problems of Section 6.1.12

Advanced Engineering Mathematics6. Laplace transforms136.2 Transforms of derivatives and integrals Theorem 1 [L (f ’(t))]1. f(t) continuous on t 0,2. f(t) Me kt for some constants k and M3. f ’(t) exists and is piecewise continuous on t 0 L (f ’(t)) exits, when s k, andL (f ’(t)) s L (f) - f(0).Proof (integrating by parts)Advanced Engineering Mathematics6. Laplace transforms14 Remark 1.and so on. Theorem 2 [L (f (n)(t))]1. f (t), f ‘(t) , . , f (n-1)(t) continuous on t 0.2. (f (m)(t) Me kt for m 0, 1, 2, ., n -1.3. f (n)(t) exists and piecewise continuous on t 0. L ((f (n)(t)) exists, when s k, andL ((f (n)(t)) s n L (f) – s n-1 f (0) – s n-2 f ‘(0) - - f (n-1)(0)

Advanced Engineering Mathematics6. Laplace transforms15 Ex.4. f t sin t L (t sin t) ?f (0) 0f ‘(t) sin t t cos t, f ‘(0) 02f “(t) cos t cos t t (-sin t)2 2 cos t - t sin t2 2 cos t - f (t)2L (f ’’) 2 L (cos t) - L (f (t))According to Theorem 2, we haveL (f ’’) s 2 L (f(t)) – s 2 f(0) – f ‘(0) s 2 L (f(t))(1)(2)From Eqs.(1) and (2), we haveAdvanced Engineering Mathematics6. Laplace transformsDifferential equations, initial value problems Purpose: using Laplace transform to solve DE( initial value problem)y ’’ ay’ by r (t), y(0) k0 , y‘(0) k1step 1. L [y“ ay‘ by] L [r(t)]L (y) Y, L (r) R[s 2Y - sy(0) – y’(0)] a [sY - y(0)] bY R(call the subsidiary equation) (s 2 as b)Y (s) (s a) y(0) y’(0) R(s)16

Advanced Engineering Mathematics6. Laplace transforms17step 2. dividing by (s 2 as b) and partial fractionthat is, multiplyY(s) [(s a) y(0) y’(0)] Q(s) R(s) Q(s) [(s a) k 0 k1] Q(s) R(s) Q(s) . . (7)(partial fraction) reduce Eq.(7) to a sum of terms whoseinverses can be found from the table.step 3. (inverse Laplace transform)y (t) L -1(Y(s)).Advanced Engineering Mathematics6. Laplace transforms Ex.5.y ’’ - y t, y(0) 1, y‘(0) 1.step 1. (Laplace transform)step 2. (dividing by (s2 - 1) and partial fraction)step 3. (inverse Laplace transform)18

Advanced Engineering Mathematics6. Laplace transforms19Laplace transform of the integral of a functionTheorem 3 [L ( f(t)dt)]1. f(t) is piecewise continuous on t 02. f(t) Me kt for some M and kproofLet, g(t) is continuous.That is g (t) also satisfies the condition g(t) N e kt for someN and k. Thus L {g(t)} exists. g‘(t) f(t), except for points atwhich f(t) is discontinuous. Hence g‘(t) is piecewise continuous.By theorem 1, L {f (t)} L {g‘(t)} s L {g(t)} - g(0),clearly g(0) 0. so that L (f) s L {g(t)}Advanced Engineering Mathematics Remark Ex.7.6. Laplace transforms20

Advanced Engineering Mathematics6. Laplace transforms216. Laplace transforms22 Ex.8.Advanced Engineering Mathematics Shifted data problem an initial value problem with initial conditions refer to some laterconstant instead of t 0.For example, y” ay‘ by r(t), y(t1) k1, y‘(t1) k2. Ex.9.step 1.step 2.

Advanced Engineering Mathematicsstep 3. y L-1(Y)6. Laplace transforms23 2t - 2 sin t y(0) cos t y‘(0) sin t 2t y(0) cos t (y‘(0) - 2) sin tLet A y(0), B y‘(0) – 2 2 t A cos t B sin tsince ProblemsofSection 6.2.Advanced Engineering Mathematics The Hermann grid illusion6. Laplace transforms24

Advanced Engineering Mathematics 6. Laplace transforms 21 Ex.8. Advanced Engineering Mathematics 6. Laplace transforms 22 Shifted data problem an initial value problem with initial conditions refer to some later constant instead of t 0. For example, y” ay‘ by r(t), y(t1) k1, y‘(t1) k2. Ex.9. step 1.