Laplace Transform - Math

1 Chapter 7 Laplace TransformThe Laplace Transform can be used to solve differential equations. Be-sides being a different and efficient alternative to variation of parame-ters and undetermined coefficients, theLaplace methodis particularlyadvantageous for input terms that are piecewise-defined, periodic or Laplace transformor theLaplace integralof a functionf(t) defined for 0 t < is the ordinary calculus integration problem 0f(t)e stdt,succinctly denotedL(f(t)) in science and engineering literature. TheL notation recognizes that integration always proceeds overt= 0 tot= and that the integral involves anintegratore stdtinstead of theusualdt.

2 These minor differences distinguishLaplace integralsfromthe ordinary integrals found on the inside covers of calculus Introduction to the Laplace MethodThe foundation of Laplace theory isLerch s cancellation law 0y(t)e stdt= 0f(t)e stdtimpliesy(t) =f(t),orL(y(t) =L(f(t))impliesy(t) =f(t).(1)In differential equation applications,y(t) is the sought-after unknownwhilef(t) is an explicit expression taken from integral , we illustrate Laplace s method by solving the initial value prob-lemy = 1, y(0) = method obtains a relationL(y(t)) =L( t), whence Lerch s cancel-lation law implies the solution isy(t) = methodis advertised as atable lookup method, in whichthe solutiony(t))

3 To a differential equation is found by looking up theanswer in a special integral Introduction to the Laplace Method249 Laplace integral 0g(t)e stdtis called theLaplaceintegralof the functiong(t). It is defined by limN N0g(t)e stdtanddepends on variables. The ideas will be illustrated forg(t) = 1,g(t) =tandg(t) =t2, producing the integral formulas in Table 1. 0(1)e stdt= (1/s)e st t= t=0 Laplace integral ofg(t) = 1/sAssumeds >0. 0(t)e stdt= 0 dds(e st)dtLaplace integral ofg(t) = dds 0(1)e stdtUse ddsF(t, s)dt=dds F(t, s) dds(1/s)UseL(1) = 1 1/s2 Differentiate.

4 0(t2)e stdt= 0 dds(te st)dtLaplace integral ofg(t) = dds 0(t)e stdt= dds(1/s2)UseL(t) = 1 2/s3 Table 1. The Laplace integral 0g(t)e stdtforg(t) = 1,tandt2. 0(1)e stdt=1s 0(t)e stdt=1s2 0(t2)e stdt=2s3In summary,L(tn) =n!s1+nAn ideas of theLaplace methodwill be illus-trated for the solutiony(t) = tof the problemy = 1,y(0) = 0. Themethod, entirely different from variation of parameters or undeterminedcoefficients, uses basic calculus and college algebra; see Table 2. Laplace method details for the illustrationy = 1,y(0) = (t)e st= e stMultiplyy = 1bye st.

5 0y (t)e stdt= 0 e stdtIntegratet= 0tot= . 0y (t)e stdt= 1/sUse Table 0y(t)e stdt y(0) = 1/sIntegrate by parts on the left. 0y(t)e stdt= 1/s2 Usey(0) = 0and divide. 0y(t)e stdt= 0( t)e stdtUse Table (t) = tApply Lerch s cancellation TransformIn Lerch s law, the formal rule of erasing the integral signs is validpro-videdthe integrals are equal for largesand certain conditions hold onyandf see Theorem 2. The illustration in Table 2 shows that Laplacetheory requires an in-depth study of aspecial integral table, a tablewhich is a true extension of the usual table found on the inside coversof calculus books.

6 Some entries for the special integral table appear inTable 1 and also in section , Table for the direct Laplace Transform produces briefer details,as witnessed by the translation of Table 2 into Table 3 below. The readeris advised to move from Laplace integral notation to theL notation assoon as possible, in order to clarify the ideas of the Transform 3. Laplace methodL-notation details fory = 1,y(0) = 0translated from Table (y (t)) =L( 1)ApplyLacrossy = 1, or multiplyy = 1bye st, integratet= 0tot=.

7 L(y (t)) = 1/sUse Table (y(t)) y(0) = 1/sIntegrate by parts on the (y(t)) = 1/s2 Usey(0) = 0and (y(t)) =L( t)Apply Table (t) = tInvoke Lerch s cancellation Transform formal properties of calculus integralsplus the integration by parts formula used in Tables 2 and 3 leads to theserulesfor the Laplace Transform :L(f(t) +g(t)) =L(f(t)) +L(g(t))The integral of a sum is thesum of the (cf(t)) =cL(f(t))Constantscpass through theintegral (y (t)) =sL(y(t)) y(0)Thet-derivative rule, or inte-gration by parts.

8 See Theo-rem (y(t)) =L(f(t))impliesy(t) =f(t)Lerch s cancellation law. SeeTheorem Example ( Laplace method)Solve by Laplace s method the initial valueproblemy = 5 2t,y(0) = : Laplace s method is outlined in Tables 2 and 3. TheL-notation ofTable 3 will be used to find the solutiony(t) = 1 + 5t Introduction to the Laplace Method251L(y (t)) =L(5 2t)ApplyLacrossy = 5 (y (t)) =5s 2s2 Use Table (y(t)) y(0) =5s 2s2 Apply thet-derivative rule, page (y(t)) =1s+5s2 2s3 Usey(0) = 1and (y(t)) =L(1) + 5L(t) L(t2)Apply Table 1, (1 + 5t t2)Linearity, page (t) = 1 + 5t t2 Invoke Lerch s cancellation Example ( Laplace method)Solve by Laplace s method the initial valueproblemy = 10,y(0) =y (0) =.

9 TheL-notation of Table 3 will be used to find the solutiony(t) = (y (t)) =L(10)ApplyLacrossy = (y (t)) y (0) =L(10)Apply thet-derivative rule toy , that is,replaceybyy on page [sL(y(t)) y(0)] y (0) =L(10)Repeat thet-derivative rule, (y(t)) =L(10)Usey(0) =y (0) = (y(t)) =10s3 Use Table 1. Then (y(t)) =L(5t2)Apply Table 1, (t) = 5t2 Invoke Lerch s cancellation of the Laplace integral 0e stf(t)dtis known to exist in the sense of the improper integral definition1 0g(t)dt= limN N0g(t)dtprovidedf(t) belongs to a class of functions known in the literature asfunctions ofexponential order.

10 For this class of functions the relationlimt f(t)eat= 0(2)is required to hold for some real numbera, or equivalently, for someconstantsMand ,|f(t)| Me t.(3)In addition,f(t) is required to bepiecewise continuouson each finitesubinterval of 0 t < , a term defined as advanced calculus background is assumed for the Laplace Transform existenceproof. Applications of Laplace theory require only a calculus TransformDefinition 1 (piecewise continuous)A functionf(t) ispiecewise continuouson a finite interval [a, b] pro-vided there exists a partitiona=t0< < tn=bof the interval [a, b]and functionsf1,f2.