The Laplace Transform - UH

1 IntroductionThe Laplace Transform provides an effective method of solving initial-value problems forlinear differential equations with constant coefficients. However, the usefulness of Laplacetransforms is by no means restricted to this class of problems. Some understanding of thebasic theory is an essential part of the mathematical background of engineers, scientists Laplace Transform is defined in terms of an integral over the interval [0, ). In-tegrals over an infinite interval are calledimproper integrals,a topic studied in a continuous function on [0, ).]]

2 The Laplace Transform off,denoted byL[(f(x)], or byF(s), is the function given byL[f(x)] =F(s)= 0e sxf(x)dx.(1)The domain ofFis the set of all real numberssfor which the improper integral more advanced treatments of the Laplace Transform the parametersassumes com-plex values, but the restriction to real values is sufficient for our purposes here. Note thatLtransforms a functionf=f(x) into a functionF=F(s) of the parameters. Thecontinuity assumption onfwill hold throughout the first three sections. It is made forconvenience in presenting the basic properties ofLand for applying the Laplace transformmethod to solving initial-value problems.)

3 In the last two sections of this chapter we extendthe definition ofLto a larger class of functions, the piecewise continuous functions on[0, ). There we will applyLto the problem of solving nonhomogeneous equations inwhich the nonhomogeneous term is piecewise continuous. This will involve some extensionof our concepts of differential equation and indicated above, the primary application of Laplace transforms of interest to us issolving linear differential equations with constant coefficients. Referring to our work inChapter 3, the functions which arise naturally in the treatment of these equations are:p(x)erx,p(x) cos x, p(x) sin x, p(x)erxcos x, p(x)erxsin xwherepis a begin by calculating the Laplace transforms of some simple cases of these (x)=1 e0x 1on[0, ).]]

4 By the Definition,L[1] = 0e sx 1dx= limb b0e sxdx= limb [e sx s b0]= limb [e sb s]+1s= limb [ 1sesb]+ , limb 1/sesbexists if and only ifs>0, and in this caselimb 1sesb= ,L[1] =1s,s>0. Example (x)=erxon [0, ). Then,L[erx]= 0e sx erxdx= limb b0e (s r)xdx= limb e (s r)x (s r) b0 = limb [e (s r)b (s r)]+1s limit exists (and has the value 0) if and only ifs r>0. ThereforeL[erx]=1s r, s> that ifr= 0, then we have the result in Example 1. Example (x) = cos xon [0, ). Then,L[cos x]= 0e sx cos x dx= limb b0e sxcos x dx= limb e sx[ scos x sin x]s2+ 2 b0.]]

5 (Note the integral was calculated using integration by parts; also, it is a standard entry ina table of integrals.)Now,L[cos x]= [limb 1e sb scos b+ sin bs2+ 2+ss2+ 2].Since [scos b+ sin b]/(s2+ 2) is bounded, the limit exists (and has the value 0)if and only ifs>0. Therefore,L[cos x]=ss2+ 2,s>0. 116 The following table gives a basic list of the Laplace transforms of functions that we willencounter in this chapter. While the entries in the table can be verified using the Definition,some of the integrations involved are complicated. The properties of the Laplace transformpresented in the next section provide a more efficient way to obtain many of the entries in thetable.

6 Handbooks of mathematical functions, for example the CRC Standard MathematicalTables, give extensive tables of Laplace of Laplace Transformsf(x)F(s)=L[f(x)]11s,s>0e x1s ,s> cos xss2+ 2,s>0sin x s2+ 2,s>0e xcos xs (s )2+ 2,s> e xsin x (s )2+ 2,s> xn,n=1,2,..n!sn+1,s>0xnerx,n=1,2,..n!(s r)n+1,s>rxcos xs2 2(s2+ 2)2,s>0xsin x2 s(s2+ 2)2,s>0 Exercises the definition of the Laplace Transform to find the Laplace Transform of the (x)= (x)= (x) = (x)= (x) = (x) = Use the fact that eaxcosbx dx=eaxa2+b2[acosbx+bsinbx] to findL[erxcos x]by the Use the fact that eaxsinbx dx=eaxa2+b2[asinbx bcosbx] to findL[erxsin x]by the Show thatL[sinx]= 2 0e sxsinxdx1 e 2 Letfbe a continuous function on [0, ) and suppose thatfis periodic withperiodp.]

7 That isfis continuous andf(x+p)=f(x),p>0, for allx. ShowthatL[f(x)] = p0e sxf(x)dx1 e Basic Properties of the Laplace TransformIn the preceding section we defined the Laplace Transform and calculated the Laplace trans-forms of some of the functions that occur in solving linear differential equations with con-stant coefficients. In this section we consider the basic question of the existence of theLaplace Transform of a functionf, and we develop the properties of the Laplace transformthat will be used in solving initial value motivate the material in this section, consider the differential equationy +ay +by=f(x)(2)whereaandbare constants andfis a continuous function on [0, ).]

8 If we assumethaty=y(x) is a solution of (1) and formally applyL, we obtainL[y (x)+ay (x)+by(x)]=L[f(x)].(3)The right-hand side of this equation suggests the basic question of the existence ofL[f(x)].That is, for what functionsfdoesL[f] exist?DEFINITIONA functionf, continuous on [0, ), is said to be ofexponential order , a real number, if there exists a positive numberMand a nonnegative numberAsuch that|f(x)| Me xon [A, ).Example 1.(a) Iffis a bounded function on [0, ) [for example,f(x) = cos xorf(x)=sin x], thenfis of exponential order implies that there exists a positive numberMsuch that|f(x)| Mfor allx [0, ).]]]]

9 Therefore,|f(x)| M=Me0xon [0, ).[Note: iff(x) = cos xorf(x) = sin x, then we could takeM= 1.](b) Letf(x)=xon [0, ). For any positive number ,limx xe x=0by L H opital s rule. Therefore, there exists a nonnegative numberAsuch thatxe x 1on[A, ).This implies thatx e x=1 e xon [A, )andf(x)=xis of exponential order . The same argument can be used to showthatf(x)=x , any real number, is of exponential order for any positivenumber . In general, ifp=p(x) is a polynomial, thenpis of exponential order for any positive number .119(c) Iff(x)=erx, thenfis of exponential order for any r.]]]]

10 (d) Consider the functionf(x)= of exponential order for some , thenthere exists a positive numberMand a nonnegative numberAsuch thatex2 Me xon [A, ) which impliese xex2 Mon [A, ).But,limx e xex2= limx ex(x )= ,a contradiction. Thusf(x)=ex2is not of exponential order for any positivenumber . Our first property ofLis a sufficient condition forL[f(x)] to exist. We shall omitthe a continuous function on [0, ). Iffis of exponential order , then the Laplace transformL[f(x)] =F(s) exists fors> .We now turn to the left-hand side of equation (2) where we haveLapplied to thelinear combinationy (x)+ay (x)+by(x).]]]