Laplace Transform - Math

1 Chapter 7 Laplace TransformThe Laplace Transform can be used to solve di erential equations. Be-sides being a di erent and e cient alternative to variation of parame-ters and undetermined coe cients, theLaplace methodis particularlyadvantageous for input terms that are piecewise-de ned, periodic or Laplace transformor theLaplace integralof a functionf(t) de ned for 0 t <1is the ordinary calculus integration problemZ10f(t)e stdt;succinctly denotedL(f(t)) in science and engineering literature. TheL{notation recognizes that integration always proceeds overt= 0 tot=1and that the integral involves anintegratore stdtinstead of theusualdt. These minor di erences distinguishLaplace integralsfromthe ordinary integrals found on the inside covers of calculus Introduction to the Laplace MethodThe foundation of Laplace theory isLerch's cancellation lawR10y(t)e stdt=R10f(t)e stdtimpliesy(t) =f(t);orL(y(t) =L(f(t))impliesy(t) =f(t):(1)In di erential 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-lemy0= 1.)}

2 Y(0) = 0:The 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) to a di erential equation is found by looking up theanswer in a special integral Introduction to the Laplace Method247 Laplace integralR10g(t)e stdtis called theLaplaceintegralof the functiong(t). It is de ned by limN!1RN0g(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)e stdt= (1=s)e st t=1t=0 Laplace integral ofg(t) = 1=sAssumeds > (t)e stdt=R10 dds(e st)dtLaplace integral ofg(t) = ddsR10(1)e stdtUseRddsF(t; s)dt=ddsRF(t; s) dds(1=s)UseL(1) = 1= 1=s2Di (t2)e stdt=R10 dds(te st)dtLaplace integral ofg(t) = ddsR10(t)e stdt= dds(1=s2)UseL(t) = 1= 2=s3 Table 1.

3 The Laplace integralR10g(t)e stdtforg(t) = 1, (1)e stdt=1sR10(t)e stdt=1s2R10(t2)e stdt=2s3In summary,L(tn) =n!s1+nAn ideas of theLaplace methodwill be illus-trated for the solutiony(t) = tof the problemy0= 1,y(0) = 0. Themethod, entirely di erent from variation of parameters or undeterminedcoe cients, uses basic calculus and college algebra; see Table 2. Laplace method details for the illustrationy0= 1,y(0) = (t)e st= e stMultiplyy0= 1bye (t)e stdt=R10 e stdtIntegratet= 0tot= (t)e stdt= 1=sUse Table (t)e stdt y(0) = 1=sIntegrate by parts on the (t)e stdt= 1=s2 Usey(0) = 0and (t)e stdt=R10( t)e stdtUse Table (t) = tApply Lerch's cancellation TransformIn Lerch's law, the formal rule of erasing the integral signsis validpro-videdthe integrals are equal for largesand certain conditions hold onyandf{ see Theorem 2.}

4 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. 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 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 fory0= 1,y(0) = 0translated from Table (y0(t)) =L( 1)ApplyLacrossy0= 1, or multiplyy0= 1bye st, integratet= 0tot= (y0(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 .}

5 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 (y0(t)) =sL(y(t)) y(0)Thet-derivative rule, or inte-gration by parts. 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 valueproblemy0= 5 2t,y(0) = : Laplace 's method is outlined in Tables 2 and 3. TheL-notation ofTable 3 will be used to nd the solutiony(t) = 1 + 5t Introduction to the Laplace Method249L(y0(t)) =L(5 2t)ApplyLacrossy0= 5 (y0(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 valueproblemy00= 10,y(0) =y0(0) =.

6 TheL-notation of Table 3 will be used to nd the solutiony(t) = (y00(t)) =L(10)ApplyLacrossy00= (y0(t)) y0(0) =L(10)Apply thet-derivative rule toy0, that is,replaceybyy0on page [sL(y(t)) y(0)] y0(0) =L(10)Repeat thet-derivative rule, (y(t)) =L(10)Usey(0) =y0(0) = (y(t)) =10s3 Use Table 1. Then (y(t)) =L(5t2)Apply Table 1, (t) = 5t2 Invoke Lerch's cancellation of the Laplace integralR10e stf(t)dtis known to exist in the sense of the improper integral de nition1Z10g(t)dt= limN!1ZN0g(t)dtprovidedf(t) belongs to a class of functions known in the literature asfunctions ofexponential order. For this class of functions the relationlimt!1f(t)eat= 0(2)is required to hold for some real numbera, or equivalently, for someconstantsMand ,jf(t)j M e t:(3)In addition,f(t) is required to bepiecewise continuouson each nitesubinterval of 0 t <1, a term de ned as advanced calculus background is assumed for the Laplace Transform existenceproof.

7 Applications of Laplace theory require only a calculus TransformDe nition 1 (piecewise continuous)A functionf(t) ispiecewise continuouson a nite interval [a; b] pro-vided there exists a partitiona=t0< < tn=bof the interval [a; b]and functionsf1,f2, .. ,fncontinuous on ( 1;1) such that fortnota partition pointf(t) =8> <>:f1(t)t0< t < t1;..fn(t)tn 1< t < tn:(4)The values offat partition points are undecided by equation (4). Inparticular, equation (4) implies thatf(t) has one-sided limits at eachpoint ofa < t < band appropriate one-sided limits at the ,fhas at worst ajump discontinuityat each partition Example (Exponential order)Show thatf(t) =etcost+tis of expo-nential order, that is, show thatf(t)is piecewise continuous and nd >0such thatlimt!

8 1f(t)=e t= :Already,f(t) is continuous, hence piecewise continuous. FromL'Hospital's rule in calculus, limt!1p(t)=e t= 0 for any polynomialpandany >0. Choose = 2, thenlimt!1f(t)e2t= limt!1costet+ limt!1te2t= 0:Theorem 1 (Existence ofL(f))Letf(t)be piecewise continuous on every nite interval int 0and satisfyjf(t)j M e tfor some constantsMand . ThenL(f(t))exists fors > andlims!1L(f(t)) = :It has to be shown that the Laplace integral offis nite fors > .Advanced calculus implies that it is su cient to show that the integrand is ab-solutely bounded above by an integrable functiong(t). Takeg(t) =M e (s ) (t) 0. Furthermore,gis integrable, becauseZ10g(t)dt=Ms :Inequalityjf(t)j M e timplies the absolute value of the Laplace transformintegrandf(t)e stis estimated by f(t)e st M e te st=g(t):The limit statement follows fromjL(f(t))j R10g(t)dt=Ms , because theright side of this inequality has limit zero ats=1.

9 The proof is Introduction to the Laplace Method251 Theorem 2 (Lerch)Iff1(t)andf2(t)are continuous, of exponential order andR10f1(t)e stdt=R10f2(t)e stdtfor alls > s0, thenf1(t) =f2(t)fort : See Widder [?].Theorem 3 (t-Derivative Rule)Iff(t)is continuous,limt!1f(t)e st= 0for all large values ofsandf0(t)is piecewise continuous, thenL(f0(t))exists for all largesandL(f0(t)) =sL(f(t)) f(0).Proof: See page method. Solve the giveninitial value problem using Laplace ' 2,y(0) = 1,y(0) = t,y(0) = ,y(0) = 1 t,y(0) = 1 +t,y(0) = 3 2t,y(0) = 3 + 2t,y(0) = 2,y(0) =y0(0) = 1,y(0) =y0(0) = 1 t,y(0) =y0(0) = 1 +t,y(0) =y0(0) = 3 2t,y(0) =y0(0) = 3 + 2t,y(0) =y0(0) = order.

10 Show thatf(t)is of exponential order, by nding aconstant 0 in each case such thatlimt!1f(t)e t= (t) = 1 + (t) =etsin(t) (t) =PNn=0cnxn, for any choiceof the constantsc0, .. , (t) =PNn=1cnsin(nt), for anychoice of the constantsc1, .. , of transforms. Letf(t) =tet2sin(et2). Establish these functionf(t) is not of expo-nential Laplace integral off(t),R10f(t)e stdt, converges for alls > Magnitude. Forfpiecewisecontinuous, de ne thejumpattbyJ(t) = limh!0+f(t+h) limh!0+f(t h):ComputeJ(t) for the (t) = 1 fort 0, elsef(t) = (t) = 1 fort 1=2, elsef(t) = (t) =t=jtjfort6= 0,f(0) = (t) = sint=jsintjfort6=n ,f(n ) = ( 1)nTaylor series. The series relationL(P1n=0cntn) =P1n=0cnL(tn) oftenholds, in which case the resultL(tn) =n!