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Chapter 7
Laplace Transform
The Laplace transform can be used to solve dierential equations. Be- sides being a dierent and ecient alternative to variation of parame- ters and undetermined coecients, theLaplace methodis particularly advantageous for input terms that are piecewise-dened, periodic or im- pulsive. Thedirect Laplace transformor theLaplace integralof a function f(t) dened for 0t <1is the ordinary calculus integration problemZ1
0f(t)estdt;
succinctly denotedL(f(t)) in science and engineering literature. The L{notation recognizes that integration always proceeds overt= 0 to t=1and that the integral involves anintegratorestdtinstead of the usualdt. These minor dierences distinguishLaplace integralsfrom the ordinary integrals found on the inside covers of calculus texts.
7.1 Introduction to the Laplace Method
The foundation of Laplace theory isLerch's cancellation law R1
0y(t)estdt=R1
0f(t)estdtimpliesy(t) =f(t);
or
L(y(t) =L(f(t)) impliesy(t) =f(t):(1)In dierential equation applications,y(t) is the sought-after unknown
whilef(t) is an explicit expression taken from integral tables. Below, we illustrate Laplace's method by solving the initial value prob- lem y
0=1; y(0) = 0:
The method obtains a relationL(y(t)) =L(t), whence Lerch's cancel- lation law implies the solution isy(t) =t. TheLaplace methodis advertised as atable lookup method, in which the solutiony(t) to a dierential equation is found by looking up the answer in a special integral table.
7.1 Introduction to the Laplace Method 247
Laplace Integral.The integralR1
0g(t)estdtis called theLaplace
integralof the functiong(t). It is dened by limN!1RN
0g(t)estdtand
depends on variables. The ideas will be illustrated forg(t) = 1,g(t) =t andg(t) =t2, producing the integral formulas in Table 1. R 1
0(1)estdt=(1=s)estt=1
t =0Laplace integral ofg(t) = 1. = 1=sAssumeds >0.R1
0(t)estdt=R1
0d ds(est)dtLaplace integral ofg(t) =t. =d dsR 1
0(1)estdtUseRddsF(t;s)dt=ddsRF(t;s)dt.
=d ds(1=s)UseL(1) = 1=s. = 1=s2Dierentiate.R1
0(t2)estdt=R1
0d ds(test)dtLaplace integral ofg(t) =t2. =d dsR 1
0(t)estdt
=d ds(1=s2)UseL(t) = 1=s2. = 2=s3
Table 1. The Laplace integral
R1
0g(t)estdtforg(t) = 1,tandt2.
R1
0(1)estdt=1
sR 1
0(t)estdt=1s2R
1
0(t2)estdt=2s3
In summary,L(tn) =n!
s1+n An Illustration.The ideas of theLaplace methodwill be illus- trated for the solutiony(t) =tof the problemy0=1,y(0) = 0. The method, entirely dierent from variation of parameters or undetermined coecients, uses basic calculus and college algebra; see Table 2. Table 2. Laplace method details for the illustrationy0=1,y(0) = 0. y0(t)est=estMultiplyy0=1byest. R1
0y0(t)estdt=R1
0estdtIntegratet= 0tot=1.R1
0y0(t)estdt=1=sUse Table 1.
sR1
0y(t)estdty(0) =1=sIntegrate by parts on the left.R1
0y(t)estdt=1=s2Usey(0) = 0and divide.R1
0y(t)estdt=R1
0(t)estdtUse Table 1.
y(t) =tApply Lerch's cancellation law.
248Laplace Transform
In Lerch's law, the formal rule of erasing the integral signsis validpro- videdthe integrals are equal for largesand certain conditions hold ony andf{ see Theorem 2. The illustration in Table 2 shows that Laplace theory requires an in-depth study of aspecial integral table, a table which is a true extension of the usual table found on the inside covers of calculus books. Some entries for the special integral table appear in
Table 1 and also in section 7.2, Table 4.
TheL-notation for the direct Laplace transform produces briefer details, as witnessed by the translation of Table 2 into Table 3 below.The reader is advised to move from Laplace integral notation to theL{notation as soon as possible, in order to clarify the ideas of the transform method. Table 3. Laplace methodL-notation details fory0=1,y(0) = 0 translated from Table 2.
L(y0(t)) =L(1)ApplyLacrossy0=1, or multiplyy0=
1byest, integratet= 0tot=1.
L(y0(t)) =1=sUse Table 1.
sL(y(t))y(0) =1=sIntegrate by parts on the left.
L(y(t)) =1=s2Usey(0) = 0and divide.
L(y(t)) =L(t)Apply Table 1.
y(t) =tInvoke Lerch's cancellation law. Some Transform Rules.The formal properties of calculus integrals plus the integration by parts formula used in Tables 2 and 3 leads to these rulesfor the Laplace transform: L(f(t) +g(t)) =L(f(t)) +L(g(t))The integral of a sum is the sum of the integrals.
L(cf(t)) =cL(f(t))Constantscpass through the
integral sign. L(y0(t)) =sL(y(t))y(0)Thet-derivative rule, or inte- gration by parts. See Theo- rem 3. L(y(t)) =L(f(t))impliesy(t) =f(t)Lerch's cancellation law. See
Theorem 2.
1 Example (Laplace method)Solve by Laplace's method the initial value
problemy0= 52t,y(0) = 1. Solution:Laplace's method is outlined in Tables 2 and 3. TheL-notation of Table 3 will be used to nd the solutiony(t) = 1 + 5tt2.
7.1 Introduction to the Laplace Method 249
L(y0(t)) =L(52t)ApplyLacrossy0= 52t.
L(y0(t)) =5
s2s2Use Table 1. sL(y(t))y(0) =5 s2s2Apply thet-derivative rule, page 248.
L(y(t)) =1
s+5s22s3Usey(0) = 1and divide. L(y(t)) =L(1) + 5L(t) L(t2)Apply Table 1, backwards. =L(1 + 5tt2)Linearity, page 248. y(t) = 1 + 5tt2Invoke Lerch's cancellation law.
2 Example (Laplace method)Solve by Laplace's method the initial value
problemy00= 10,y(0) =y0(0) = 0. Solution:TheL-notation of Table 3 will be used to nd the solutiony(t) = 5t2.
L(y00(t)) =L(10)ApplyLacrossy00= 10.
sL(y0(t))y0(0) =L(10)Apply thet-derivative rule toy0, that is, replaceybyy0on page 248. s[sL(y(t))y(0)]y0(0) =L(10)Repeat thet-derivative rule, ony. s
2L(y(t)) =L(10)Usey(0) =y0(0) = 0.
L(y(t)) =10
s3Use Table 1. Then divide.
L(y(t)) =L(5t2)Apply Table 1, backwards.
y(t) = 5t2Invoke Lerch's cancellation law.
Existence of the Transform.The Laplace integralR1
0estf(t)dt
is known to exist in the sense of the improper integral denition1 Z 1
0g(t)dt= limN!1Z
N
0g(t)dt
providedf(t) belongs to a class of functions known in the literature as functions ofexponential order. For this class of functions the relation lim t !1f(t) eat= 0(2)is required to hold for some real numbera, or equivalently, for some constantsMand, jf(t)j Met:(3)In addition,f(t) is required to bepiecewise continuouson each nite subinterval of 0t <1, a term dened as follows.
1An advanced calculus background is assumed for the Laplace transform existence
proof. Applications of Laplace theory require only a calculus background.
250Laplace Transform
Denition 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 fortnot a partition point f(t) =8><>:f
1(t)t0< t < t1;
f n(t)tn1< t < tn:(4)The values offat partition points are undecided by equation (4). In particular, equation (4) implies thatf(t) has one-sided limits at each point ofa < t < band appropriate one-sided limits at the endpoints. Therefore,fhas at worst ajump discontinuityat each partition point.
3 Example (Exponential order)Show thatf(t) =etcost+tis of expo-
nential order, that is, show thatf(t)is piecewise continuous and nd >0 such thatlimt!1f(t)=et= 0. Solution:Already,f(t) is continuous, hence piecewise continuous. From
L'Hospital's rule in calculus, lim
t!1p(t)=et= 0 for any polynomialpand any >0. Choose= 2, then lim t!1f(t) e2t= limt!1costet+ limt!1te2t= 0:
Theorem 1 (Existence ofL(f))
Letf(t)be piecewise continuous on every nite interval int0and satisfy jf(t)j Metfor some constantsMand. ThenL(f(t))exists fors > andlims!1L(f(t)) = 0. Proof:It has to be shown that the Laplace integral offis nite fors > . Advanced calculus implies that it is sucient to show that the integrand is ab- solutely bounded above by an integrable functiong(t). Takeg(t) =Me(s)t.
Theng(t)0. Furthermore,gis integrable, because
Z 1 0 g(t)dt=M s: Inequalityjf(t)j Metimplies the absolute value of the Laplace transform integrandf(t)estis estimated by f(t)estMetest=g(t):
The limit statement follows fromjL(f(t))j R1
0g(t)dt=M
s, because the right side of this inequality has limit zero ats=1. The proof is complete.
7.1 Introduction to the Laplace Method 251
Theorem 2 (Lerch)
Iff1(t)andf2(t)are continuous, of exponential order andR1
0f1(t)estdt=R1
0f2(t)estdtfor alls > s0, thenf1(t) =f2(t)fort0.
Proof: See Widder [?].
Theorem 3 (t-Derivative Rule)
Iff(t)is continuous,limt!1f(t)est= 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 276.
Exercises 7.1
Laplace method. Solve the given
initial value problem using Laplace's method.
1.y0=2,y(0) = 0.
2.y0= 1,y(0) = 0.
3.y0=t,y(0) = 0.
4.y0=t,y(0) = 0.
5.y0= 1t,y(0) = 0.
6.y0= 1 +t,y(0) = 0.
7.y0= 32t,y(0) = 0.
8.y0= 3 + 2t,y(0) = 0.
9.y00=2,y(0) =y0(0) = 0.
10.y00= 1,y(0) =y0(0) = 0.
11.y00= 1t,y(0) =y0(0) = 0.
12.y00= 1 +t,y(0) =y0(0) = 0.
13.y00= 32t,y(0) =y0(0) = 0.
14.y00= 3 + 2t,y(0) =y0(0) = 0.
Exponential order. Show thatf(t)
is of exponential order, by nding a constant0 in each case such that lim t!1f(t) et= 0.
15.f(t) = 1 +t
16.f(t) =etsin(t)
17.f(t) =PNn=0cnxn, for any choice
of the constantsc0, ...,cN.
18.f(t) =PNn=1cnsin(nt), for any
choice of the constantsc1, ...,cN.
Existence of transforms. Letf(t) =
te t2sin(et2). Establish these results.
19.The functionf(t) is not of expo-
nential order.
20.The Laplace integral off(t),R1
0f(t)estdt, converges for all
s >0.
Jump Magnitude
. Forfpiecewise continuous, dene thejumpattby
J(t) = limh!0+f(t+h)limh!0+f(th):
ComputeJ(t) for the followingf.
21.f(t) = 1 fort0, elsef(t) = 0
22.f(t) = 1 fort1=2, elsef(t) = 0
23.f(t) =t=jtjfort6= 0,f(0) = 0
24.f(t) = sint=jsintjfort6=n,
f(n) = (1)n
Taylor series
. The series relation
L(P1n=0cntn) =P1n=0cnL(tn) often
holds, in which case the resultL(tn) = n!s1ncan be employed to nd a series representation of the Laplace transform. Use this idea on the fol- lowing to nd a series formula for
L(f(t)).
25.f(t) =e2t=P1n=0(2t)n=n!
26.f(t) =et=P1n=0(t)n=n!
252Laplace Transform
7.2 Laplace Integral Table
The objective in developing a table of Laplace integrals, e.g., Tables 4 and 5, is to keep the table size small. Table manipulation rules appear- ing in Table 6, page 257, eectively increase the table size manyfold, making it possible to solve typical dierential equations from electrical and mechanical problems. The combination of Laplace tablesplus the table manipulation rules is called theLaplace transform calculus. Table 4 is considered to be a table of minimum size to be memorized. Table 5 adds a number of special-use entries. For instance, the Heaviside entry in Table 5 is memorized, but usually not the others. Derivationsare postponed to page 270. The theory of thegamma func- tion(x) appears below on page 255. Table 4. A minimal Laplace integral table withL-notation R1
0(tn)estdt=n!
s1+nL(tn) =n!s1+n R 1
0(eat)estdt=1
saL(eat) =1saR1
0(cosbt)estdt=s
s2+b2L(cosbt) =ss2+b2 R 1
0(sinbt)estdt=b
s2+b2L(sinbt) =bs2+b2
Table 5. Laplace integral table extension
L(H(ta)) =eass(a0) Heaviside unit step, dened by
H(t) =1fort0;
0otherwise.
L((ta)) =easDirac delta,(t) =dH(t).
Special usage rules apply.
L( oor(t=a)) =eas s(1eas)Staircase function, oor(x) =greatest integerx.
L(sqw(t=a)) =1
stanh(as=2)Square wave, sqw(x) = (1) oor(x).
L(atrw(t=a)) =1
s2tanh(as=2)Triangular wave, trw(x) =Rx
0sqw(r)dr.
L(t) =(1 +)
s1+Generalized power function,(1 +) =R1
0exxdx.
L(t1=2) =r
sBecause(1=2) =p.
7.2 Laplace Integral Table253
4 Example (Laplace transform)Letf(t) =t(t1)sin2t+e3t. Compute
L(f(t))using the basic Laplace table and transform linearity properties.
Solution:
L(f(t)) =L(t25tsin2t+e3t)Expandt(t5).
=L(t2)5L(t) L(sin2t) +L(e3t)Linearity applied. 2 s35s22s2+ 4+1s3Table lookup.
5 Example (Inverse Laplace transform)Use the basic Laplace table back-
wards plus transform linearity properties to solve forf(t)in the equation
L(f(t)) =s
s2+ 16+2s3+s+ 1s3:
Solution:
L(f(t)) =s
s2+ 16+ 21s3+1s2+122s3Convert to table entries. =L(cos4t) + 2L(e3t) +L(t) +1
2L(t2)Laplace table (backwards).
=L(cos4t+ 2e3t+t+1
2t2)Linearity applied.
f(t) = cos4t+ 2e3t+t+1
2t2Lerch's cancellation law.
6 Example (Heaviside)Find the Laplace transform off(t)in Figure 1.
1 3 155
Figure 1. A piecewise dened
functionf(t)on0t <1:f(t) = 0 except for1t <2and3t <4. Solution:The details require the use of the Heaviside function formula
H(ta)H(tb) =1at < b;
0 otherwise:
The formula forf(t):
f(t) =8<:1 1t <2;
5 3t <4;
0 otherwise=1 1t <2;
0 otherwise+ 51 3t <4;
0 otherwise
Thenf(t) =f1(t) + 5f2(t) wheref1(t) =H(t1)H(t2) andf2(t) =
H(t3)H(t4). The extended table gives
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