The Laplace Transform - Armstrong

1 185 Chapter 7 The Laplace TransformIn this chapter we will explore a method for solving linear differential equationswith constant coefficients that is widely used in electrical engineering. It involvesthe transformation of an initial-value problem into an algebraic equation, whichis easily solved, and then the inverse transformation back to the solution of theoriginal problem, thereby bypassing the need to solve for arbitrary constants inthe general solution. The technique is especially well suited for finding gener-alized solutions to systems driven by impulses or by discontinuous or periodicforcing Definition and Basic PropertiesGiven a functionf(t) defined fort 0, itsLaplace transformF(s) is definedasF(s) = 0e stf(t)dt.(1)Notice that the variablesappears as a parameter in an improper integral.

2 Wesay that the Laplace Transform exists if this improper integral converges forall sufficiently larges. The notational convention used here is common, thoughnot universal: The uppercase version of the function s name denotes its Laplacetransform, andsis used for the Transform s independent variable. Before wego further, let s illustrate this definition and notation with a couple of simpleexamples. Example 1 Consider a constant functionf(t) =c, t 0. Its Laplace transformisF(s) = 0e stc dt=c se st t=0= cs(limt e st e0)= cs(0 1) =cs,provided thats >0. Note that the improper integral diverges whens 0. Example 2 Letf(t) =eat, t 0. Its Laplace Transform isF(s) = 0e steatdt= 0e (s a)tdt= 1s a(limt e (s a)t e0)= 1s a(0 1) =1s a,provided thats > a. Note that the improper integral diverges whens a.

3 186 Chapter 7. The Laplace TransformThe Laplace transformF(s) of a functionf(t) is the result of applying alinear operator tof. Denoting this linear operator byL, we can writeLf=F,orL[f](s) =F(s).whereFis given by (1). Using this notation, the result of Example 2, for in-stance, is thatL[eat](s) =1s a, s > is easy to check that the operatorLislinear:L[cf](s) = 0e stcf(t)dt=c 0e stf(t)dt=cF(s)L[f+g](s) = 0e st(f(t) +g(t))dt= 0e stf(t)dt+ 0e stg(t)dt=F(s) +G(s).Indeed, the linearity ofLis a simple consequence of the linearity properties study of Laplace transforms is primarily for the purpose of solving initial-value problemsy +py +qy=f, y(0) =y0, y (0) =v0,(2)wherepandqare constants. Later (starting in Section 3) we will considergen-eralized solutionsthat can arise when the nonhomogeneous termfispiecewisecontinuouson [0, ).]

4 A functionfispiecewise continuouson [0, ) if(i)fhas at most finitely many discontinuities in any bounded interval [0,T],(ii) limt 0+f(t) exists, and(iii) at every numbera >0 both limt a f(t) and limt a+f(t) that condition (iii) is automatically true at any point wherefis continuous,and it says that all off s discontinuities in (0, ) are simple jump discontinu-ities. Also, piecewise continuity guarantees integrability on any bounded interval[0,T].We also need a condition that will guarantee the existence of a function sLaplace Transform . We say that a functionfisexponentially boundedon[0, ) if there are constantsMandrsuch that|f(t)| Mertfor allt Definition and Basic Properties187 Iffis piecewise continuous, then T0e stf(t)dtwill exist for allT >0 and alls. Iffis also exponentially bounded, then T0e st|f(t)|dt T0Me(r s)tdt=Mr s(e(r s)T 1);thus T0e st|f(t)|dtconverges asT , provided thats > r.]]

5 This means thatthe Laplace Transform of|f|exists, which implies that the Laplace Transform the rest of this chapter we will be concerned almost exclusivelywith functions defined on [0, ). The phrases piecewise continuous and expo-nentially bounded should always be understood to mean piecewise continuouson [0, ) and exponentially bounded on [0, ). We also emphasize that throughout this chapter we will deal only with ex-ponentially bounded functions, since only those have Laplace transforms. Also,in this section and the next, all of our examples will involve ordinary elemen-tary functions polynomials, sinusoidal functions, exponential functions, etc. that are clearly continuous everywhere. Not until Section 3 will we begin toconsider differential equations with nonhomogeneous terms that are First Differentiation TheoremThe reason why Laplace transforms are useful in solving differential equationsis embodied in the following theorem, which (together with the corollary thatfollows) we will refer to as thefirst differentiation thatyis continuous and exponentially bounded andthaty is piecewise continuous.]]]

6 ThenL[y ]exists, andL[y ](s) =sY(s) y(0). , forT >0 we use integration by parts to compute T0e sty (t)dt=e sty(t) Tt=0 T0( s)e sty(t)dt=e sTy(T) y(0) +s T0e sty(t) computation is valid because of the piecewise continuity ofy and the con-tinuity ofy(at 0 andT). Now, sinceyis assumed to be exponentially bounded,it follows that, forssufficiently large,limT e sTy(T) = 0,andY(s) = limT T0e sty(t) 7. The Laplace TransformTherefore,L[y ] exists, andL[y ](s) = y(0) +sY(s).By repeated application of Theorem 1, we arrive at the following thatyandy are continuous and exponentiallybounded and thaty is piecewise continuous. ThenL[y ]exists, andL[y ](s) =s2Y(s) sy(0) y (0).Moreover, ify,y ,..,y(n 1)are continuous and exponentially bounded, and ify(n)is piecewise continuous, thenL[y(n)]exists, andL[y(n)](s) =snY(s) sn 1y(0) sn 2y (0) y(n 1)(0).

7 JBecause of the properties stated in Theorem 1 and Corollary 1, the Laplacetransform is particularly well suited for solving linear initial-value problemswith constant coefficients. The following are two examples that indicate thebasic idea. Example 3 Consider the first-order initial-value problemy + 2y= 4, y(0) = 1,and letYdenote the Transform of the solutiony. By applying the operatorLto both sides of the differential equation and using the result of Example 1 onthe right side, we find thatsY(s) y(0) + 2Y(s) = we use the given initial value and solve forY(s):(s+ 2)Y(s) =4s+ 1; soY(s) =s+ 4s(s+ 2).Now our job is to find the functiony(t) that has this Transform . A partialfraction expansion (consult your calculus book) reveals thatY(s) =2s 1s+ results of Examples 1 and 2 now tell us thatY(s) is the Transform ofy(t) = 2 e 2t.

8 Example 4 Consider the initial-value problemy + 3y + 2y= 0, y(0) = 1, y (0) = 0, Definition and Basic Properties189and letYdenote the Transform of the solutiony. By applying the operatorLto both sides of the differential equation, we find thats2Y(s) sy(0) y (0) + 3(sY(s) y(0)) + 2Y(s) = the given initial values and rearranging, this becomes(s2+ 3s+ 2)Y(s) s 3 = we solve forY(s), findingY(s) =s+ 3s2+ 3s+ 2=s+ 3(s+ 1)(s+ 2).Now we look for the functiony(t) that has this Transform . A partial fractionexpansion reveals thatY(s) =2s+ 1 1s+ the result of Example 2 above, we see that this is the Transform ofy(t) = 2e t e 2t. The General ProcedureExamples 3 and 4 each illustrate a general procedure for solving initial valueproblems with the help of Laplace transforms:(1) Transform each side of the differential equation, using the given initialvalues.

9 (2) Solve the resultingalgebraicequation for the transformY(s) of the solu-tiony(t).(3) Find the solutionyby identifying the transformY(s) with known final step amounts to finding theinverse transformofY(s). An importantquestion to ask is whether the operatorLis actually invertible. In other words,given a transformY(s), is there a unique functiony(t), t 0, such thatLy=Y?The answer to this is no,unless we require the functionyto be , this is not a difficulty in the context of solving differential equations,since solutions will be s now consider what happens in general when we apply the Laplacetransform technique to a linear second-order initial-value problem with constantcoefficients. Suppose that the differential equation isP(D)y=f,withP(D) =D2+pD+qI,and we have initial conditionsy(0) =y0, y (0) = 7.

10 The Laplace TransformThe Laplace Transform of the nonhomogeneous term is simplyLf=F. Trans-forming the left side producesLP(D)y=LD2y+pLDy+qLy=s2Y(s) sy0 v0+p(sY(s) y0) +q Y(s)=P(s)Y(s) (sy0+v0+py0).The transformed initial-value problem therefore becomesP(s)Y(s) (sy0+v0+py0) =F(s),and solving forY(s) gives usY(s) =F(s)P(s)+ (s)P(s),(3)where (s) =sy0+v0+ termF(s)/P(s) in (3) is the Transform of therest solution ofP(D)y=f, and the term (s)/P(s) in (3) is the Transform of thesolution of the homogeneous equationP(D)y= 0 that satisfies the given initialconditions. The same statements are true for all linear differential equationswith constant coefficients, regardless of Fraction ExpansionsSteps 1 and 2 of the general procedure just described are easy, provided that thetransform of the nonhomogeneous term is known.