laplaceTransform
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A chapter (Chapter 7) on the Laplace transform, apparently from a differential equations textbook rather than Phil's own work. Section 7.1 introduces the Laplace integral, Lerch's cancellation law, the t-derivative rule, exponential order and piecewise continuity, with worked initial value problems and exercises. Section 7.2 begins tables of Laplace integrals, including Heaviside, Dirac delta and periodic functions. Only the first part of the text was seen.
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Chapter 7
Laplace Transform
The Laplace transform can be used to solve di erential equati ons. Be-
sides being a di erent and ecient alternative to variation o f parame-
ters and undetermined coecients, the Laplace method is particularly
advantageous for input terms that are piecewise-de ned, pe riodic or im-
pulsive.
Thedirect Laplace transform or the Laplace integral of a function
f(t) de ned for 0 t <1is the ordinary calculus integration problem
Z1
0f(t)e stdt;
succinctly denoted L(f(t)) in science and engineering literature. The
L{notation recognizes that integration always proceeds ove rt= 0 to
t=1and that the integral involves an integrator e stdtinstead of the
usual dt. These minor di erences distinguish Laplace integrals from
the ordinary integrals found on the inside covers of calculu s texts.
7.1 Introduction to the Laplace Method
The foundation of Laplace theory is Lerch's cancellation law
R1
0y(t)e stdt=R1
0f(t)e stdt implies y(t) =f(t);
or
L(y(t) =L(f(t)) implies y(t) =f(t):(1)
In di erential equation applications, y(t) is the sought-after unknown
while f(t) is an explicit expression taken from integral tables.
Below, we illustrate Laplace's method by solving the initia l value prob-
lem
y0= 1; y(0) = 0 :
The method obtains a relation L(y(t)) =L( t), whence Lerch's cancel-
lation law implies the solution is y(t) = t.
TheLaplace method is advertised as a table lookup method , in which
the solution y(t) to a di erential 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)e stdtis called the Laplace
integral of the function g(t). It is de ned by lim N!1RN
0g(t)e stdtand
depends on variable s. The ideas will be illustrated for g(t) = 1, g(t) =t
andg(t) =t2, producing the integral formulas in Table 1.
R1
0(1)e stdt= (1=s)e st t=1
t=0Laplace integral of g(t) = 1.
= 1=s Assumed s >0.
R1
0(t)e stdt=R1
0 d
ds(e st)dt Laplace integral of g(t) =t.
= d
dsR1
0(1)e stdtUseRd
dsF(t;s)dt=d
dsRF(t;s)dt.
= d
ds(1=s) UseL(1) = 1 =s.
= 1=s2Di erentiate.
R1
0(t2)e stdt=R1
0 d
ds(te st)dtLaplace integral of g(t) =t2.
= d
dsR1
0(t)e stdt
= d
ds(1=s2) UseL(t) = 1=s2.
= 2=s3
Table 1. The Laplace integralR1
0g(t)e stdtforg(t) = 1,tandt2.
R1
0(1)e stdt=1
sR1
0(t)e stdt=1
s2R1
0(t2)e stdt=2
s3
In summary, L(tn) =n!
s1+n
An Illustration. The ideas of the Laplace method will be illus-
trated for the solution y(t) = tof the problem y0= 1,y(0) = 0. The
method, entirely di erent from variation of parameters or un determined
coecients, uses basic calculus and college algebra; see Ta ble 2.
Table 2. Laplace method details for the illustration y0= 1,y(0) = 0 .
y0(t)e st= e stMultiply y0= 1bye st.
R1
0y0(t)e stdt=R1
0 e stdt Integrate t= 0tot=1.
R1
0y0(t)e stdt= 1=s Use Table 1.
sR1
0y(t)e stdt y(0) = 1=s Integrate by parts on the left.
R1
0y(t)e stdt= 1=s2Usey(0) = 0 and divide.
R1
0y(t)e stdt=R1
0( t)e stdt Use Table 1.
y(t) = t Apply Lerch's cancellation law.
248 Laplace Transform
In Lerch's law, the formal rule of erasing the integral signs is valid pro-
vided the integrals are equal for large sand certain conditions hold on y
andf{ see Theorem 2. The illustration in Table 2 shows that Laplac e
theory requires an in-depth study of a special integral table, a table
which is a true extension of the usual table found on the insid e covers
of calculus books. Some entries for the special integral tab le appear in
Table 1 and also in section 7.2, Table 4.
TheL-notation for the direct Laplace transform produces briefe r details,
as witnessed by the translation of Table 2 into Table 3 below. The reader
is advised to move from Laplace integral notation to the L{notation as
soon as possible, in order to clarify the ideas of the transfo rm method.
Table 3. Laplace method L-notation details for y0= 1,y(0) = 0
translated from Table 2.
L(y0(t)) =L( 1) Apply Lacross y0= 1, or multiply y0=
1bye st, integrate t= 0tot=1.
L(y0(t)) = 1=s Use Table 1.
sL(y(t)) y(0) = 1=s Integrate by parts on the left.
L(y(t)) = 1=s2Usey(0) = 0 and divide.
L(y(t)) =L( t) Apply Table 1.
y(t) = t Invoke 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 le ads to these
rules for 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)) Constants cpass 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))implies y(t) =f(t)Lerch's cancellation law. See
Theorem 2.
1 Example (Laplace method) Solve by Laplace's method the initial value
problem y0= 5 2t,y(0) = 1 .
Solution :Laplace's method is outlined in Tables 2 and 3. The L-notation of
Table 3 will be used to nd the solution y(t) = 1 + 5 t t2.
7.1 Introduction to the Laplace Method 249
L(y0(t)) =L(5 2t) Apply Lacross y0= 5 2t.
L(y0(t)) =5
s 2
s2Use Table 1.
sL(y(t)) y(0) =5
s 2
s2Apply the t-derivative rule, page 248.
L(y(t)) =1
s+5
s2 2
s3Usey(0) = 1 and divide.
L(y(t)) =L(1) + 5L (t) L(t2) Apply Table 1, backwards.
=L(1 + 5t t2) Linearity, page 248.
y(t) = 1 + 5 t t2Invoke Lerch's cancellation law.
2 Example (Laplace method) Solve by Laplace's method the initial value
problem y00= 10,y(0) = y0(0) = 0 .
Solution :TheL-notation of Table 3 will be used to nd the solution y(t) = 5 t2.
L(y00(t)) =L(10) Apply Lacross y00= 10.
sL(y0(t)) y0(0) = L(10) Apply the t-derivative rule to y0, that is,
replace ybyy0on page 248.
s[sL(y(t)) y(0)] y0(0) = L(10) Repeat the t-derivative rule, on y.
s2L(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) = 5 t2Invoke Lerch's cancellation law.
Existence of the Transform. The Laplace integralR1
0e stf(t)dt
is known to exist in the sense of the improper integral de nit ion1
Z1
0g(t)dt= lim
N!1ZN
0g(t)dt
provided f(t) belongs to a class of functions known in the literature as
functions of exponential order . For this class of functions the relation
lim
t!1f(t)
eat= 0 (2)
is required to hold for some real number a, or equivalently, for some
constants Mand ,
jf(t)j Me t: (3)
In addition, f(t) is required to be piecewise continuous on each nite
subinterval of 0 t <1, a term de ned as follows.
1An advanced calculus background is assumed for the Laplace tra nsform existence
proof. Applications of Laplace theory require only a calculu s background.
250 Laplace Transform
De nition 1 (piecewise continuous)
A function f(t) ispiecewise continuous on a nite interval [a;b ] pro-
vided there exists a partition a=t0<< tn=bof the interval [a;b ]
and functions f1,f2, .. ., fncontinuous on ( 1;1) such that for tnot
a partition point
f(t) =8
><
>:f1(t)t0< t < t 1;
......
fn(t)tn 1< t < t n:(4)
The values of fat partition points are undecided by equation (4). In
particular, equation (4) implies that f(t) has one-sided limits at each
point of a < t < b and appropriate one-sided limits at the endpoints.
Therefore, fhas at worst a jump discontinuity at each partition point.
3 Example (Exponential order) Show that f(t) =etcost+tis of expo-
nential order, that is, show that f(t)is piecewise continuous and nd >0
such that limt!1f(t)=e t= 0.
Solution :Already, f(t) is continuous, hence piecewise continuous. From
L'Hospital's rule in calculus, lim t!1p(t)=e t= 0 for any polynomial pand
any >0. Choose = 2, then
lim
t!1f(t)
e2t= lim
t!1cost
et+ lim
t!1t
e2t= 0:
Theorem 1 (Existence of L(f))
Letf(t)be piecewise continuous on every nite interval in t0and satisfy
jf(t)j Me tfor some constants Mand . Then L(f(t))exists for s >
andlims!1L(f(t)) = 0 .
Proof: It has to be shown that the Laplace integral of fis nite for s > .
Advanced calculus implies that it is sucient to show that the integran d is ab-
solutely bounded above by an integrable function g(t). Take g(t) =Me (s )t.
Then g(t)0. Furthermore, gis integrable, because
Z1
0g(t)dt=M
s :
Inequality jf(t)j Me timplies the absolute value of the Laplace transform
integrand f(t)e stis estimated by
f(t)e st Me te st=g(t):
The limit statement follows from jL(f(t))j R1
0g(t)dt=M
s , because the
right side of this inequality has limit zero at s=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)e stdt=R1
0f2(t)e stdtfor all s > s 0, then f1(t) =f2(t)fort0.
Proof : See Widder [ ?].
Theorem 3 ( t-Derivative Rule)
Iff(t)is continuous, lim
t!1f(t)e st= 0for all large values of sandf0(t)
is piecewise continuous, then L(f0(t))exists for all large sandL(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= 1 t,y(0) = 0.
6.y0= 1 + t,y(0) = 0.
7.y0= 3 2t,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= 1 t,y(0) = y0(0) = 0.
12.y00= 1 + t,y(0) = y0(0) = 0.
13.y00= 3 2t,y(0) = y0(0) = 0.
14.y00= 3 + 2t ,y(0) = y0(0) = 0.
Exponential order . Show that f(t)
is of exponential order, by nding a
constant 0 in each case such that
lim
t!1f(t)
e t= 0.
15.f(t) = 1 + t
16.f(t) =etsin(t)
17.f(t) =PN
n=0cnxn, for any choice
of the constants c0, . . . , cN.18.f(t) =PN
n=1cnsin(nt ), for any
choice of the constants c1, . . . ,cN.
Existence of transforms . Let f(t) =
tet2sin(et2). Establish these results.
19.The function f(t) is not of expo-
nential order.
20.The Laplace integral of f(t),R1
0f(t)e stdt, converges for all
s >0.
Jump Magnitude. For fpiecewise
continuous, de ne the jump attby
J(t) = lim
h!0+f(t+h) lim
h!0+f(t h):
Compute J(t) for the following f.
21.f(t) = 1 for t0, else f(t) = 0
22.f(t) = 1 for t1=2, else f(t) = 0
23.f(t) =t=jtjfort6= 0,f(0) = 0
24.f(t) = sin t=jsintjfort6=n,
f(n) = ( 1)n
Taylor series. The series relation
L(P1
n=0cntn) =P1
n=0cnL(tn) often
holds, in which case the result L(tn) =
n!s 1 ncan 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=P1
n=0(2t)n=n!
26.f(t) =e t=P1
n=0( t)n=n!
252 Laplace 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 rul es appear-
ing in Table 6, page 257, e ectively increase the table size ma nyfold,
making it possible to solve typical di erential equations fr om electrical
and mechanical problems. The combination of Laplace tables plus the
table manipulation rules is called the Laplace transform calculus.
Table 4 is considered to be a table of minimum size to be memori zed.
Table 5 adds a number of special-use entries. For instance, t he Heaviside
entry in Table 5 is memorized, but usually not the others.
Derivations are postponed to page 270. The theory of the gamma func-
tion (x) appears below on page 255.
Table 4. A minimal Laplace integral table with L-notation
R1
0(tn)e stdt=n!
s1+nL(tn) =n!
s1+n
R1
0(eat)e stdt=1
s aL(eat) =1
s a
R1
0(cosbt)e stdt=s
s2+b2L(cosbt) =s
s2+b2
R1
0(sinbt)e stdt=b
s2+b2L(sinbt) =b
s2+b2
Table 5. Laplace integral table extension
L(H(t a)) =e as
s(a0) Heaviside unit step, de ned by
H(t) =1fort0;
0otherwise.
L((t a)) =e asDirac delta, (t) =dH(t).
Special usage rules apply.
L(
oor(t=a)) =e as
s(1 e as)Staircase function,
oor(x) =greatest integer x.
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
0e xx dx.
L(t 1=2) =r
sBecause (1=2) =p.
7.2 Laplace Integral Table 253
4 Example (Laplace transform) Letf(t) =t(t 1) sin2t+e3t. Compute
L(f(t))using the basic Laplace table and transform linearity properties .
Solution :
L(f(t)) =L(t2 5t sin 2t+e3t) Expand t(t 5).
=L(t2) 5L(t) L(sin 2t ) +L(e3t) Linearity applied.
=2
s3 5
s2 2
s2+ 4+1
s 3Table lookup.
5 Example (Inverse Laplace transform) Use the basic Laplace table back-
wards plus transform linearity properties to solve for f(t)in the equation
L(f(t)) =s
s2+ 16+2
s 3+s+ 1
s3:
Solution :
L(f(t)) =s
s2+ 16+ 21
s 3+1
s2+1
22
s3Convert 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) = cos4 t+ 2e3t+t+1
2t2Lerch's cancellation law.
6 Example (Heaviside) Find the Laplace transform of f(t)in Figure 1.
1
3 1 55
Figure 1. A piecewise de ned
function f(t)on0t <1:f(t) = 0
except for 1t <2and3t <4.
Solution :The details require the use of the Heaviside function formula
H(t a) H(t b) =1at < b;
0 otherwise :
The formula for f(t):
f(t) =8
<
:1 1t <2;
5 3t <4;
0 otherwise=1 1t <2;
0 otherwise+ 51 3t <4;
0 otherwise
Then f(t) =f1(t) + 5 f2(t) where f1(t) =H(t 1) H(t 2) and f2(t) =
H(t 3) H(t 4). The extended table gives
L(f(t)) =L(f1(t)) + 5 L(f2(t)) Linearity.
=L(H(t 1)) L(H(t 2)) + 5L (f2(t)) Substitute for f1.
254 Laplace Transform
=e s e 2s
s+ 5L(f2(t)) Extended table used.
=e s e 2s+ 5e 3s 5e 4s
sSimilarly for f2.
7 Example (Dirac delta) A machine shop tool that repeatedly hammers a
die is modeled by the Dirac impulse model f(t) =PN
n=1(t n). Show
thatL(f(t)) =PN
n=1e ns.
Solution :
L(f(t)) =LPN
n=1(t n)
=PN
n=1L((t n)) Linearity.
=PN
n=1e nsExtended Laplace table.
8 Example (Square wave) A periodic camshaft force f(t)applied to a me-
chanical system has the idealized graph shown in Figure 2. Show that
f(t) = 1 + sqw(t)andL(f(t)) =1
s(1 + tanh(s= 2)).
02
1 3Figure 2. A periodic force f(t)applied
to a mechanical system.
Solution :
1 +sqw(t) =1 + 1 2n t <2n+ 1,n= 0;1; : : :,
1 1 2n + 1t <2n+ 2,n= 0;1; : : :,
=
2 2n t <2n+ 1,n= 0;1; : : :,
0 otherwise,
=f(t):
By the extended Laplace table, L(f(t)) =L(1) +L(sqw(t)) =1
s+tanh(s= 2)
s.
9 Example (Sawtooth wave) Express the P-periodic sawtooth wave repre-
sented in Figure 3 as f(t) =ct=P c
oor(t=P)and obtain the formula
L(f(t)) =c
Ps2 ce Ps
s se Ps:
0c
P 4PFigure 3. A P-periodic sawtooth
wave f(t)of height c >0.
7.2 Laplace Integral Table 255
Solution :The representation originates from geometry, because the period ic
function fcan be viewed as derived from ct=Pby subtracting the correct con-
stant from each of intervals [P; 2P], [2P;3P], etc.
The technique used to verify the identity is to de ne g(t) =ct=P c
oor(t=P)
and then show that gisP-periodic and f(t) =g(t) on 0 t < P . Two P-
periodic functions equal on the base interval 0 t < P have to be identical,
hence the representation follows.
The ne details: for 0 t < P ,
oor(t=P) = 0 and
oor(t=P+k) =k. Hence
g(t+kP) =ct=P+ck c
oor(k) =ct=P =g(t), which implies that gis
P-periodic and g(t) =f(t) for 0 t < P.
L(f(t)) =c
PL(t) cL(
oor(t=P)) Linearity.
=c
Ps2 ce Ps
s se PsBasic and extended table applied.
10 Example (Triangular wave) Express the triangular wave fof Figure 4 in
terms of the square wave sqwand obtain L(f(t)) =5
s2tanh(s= 2).
05
2Figure 4. A 2-periodic triangular
wave f(t)of height 5.
Solution :The representation of fin terms of sqwisf(t) = 5Rt=
0sqw(x)dx.
Details: A 2-periodic triangular wave of height 1 is obtained by integrating
the square wave of period 2. A wave of height cand period 2 is given by
ctrw(t) =cRt
0sqw(x)dx. Then f(t) =ctrw(2t=P ) =cR2t=P
0sqw(x)dxwhere
c= 5 and P= 2.
Laplace transform details: Use the extended Laplace table as follows.
L(f(t)) =5
L(trw(t=)) =5
s2tanh(s= 2):
Gamma Function. In mathematical physics, the Gamma func-
tionor the generalized factorial function is given by the identity
(x) =Z1
0e ttx 1dt; x > 0: (1)
This function is tabulated and available in computer langua ges like For-
tran, C and C++. It is also available in computer algebra syst ems and
numerical laboratories. Some useful properties of (x ):
(1 + x) = x (x) (2)
(1 + n) = n! for integers n1: (3)
256 Laplace Transform
Details for relations (2) and (3): Start withR1
0e tdt= 1, which gives
(1) = 1. Use this identity and successively relation (2) to obtain rela tion (3).
To prove identity (2), integration by parts is applied, as follows:
(1 + x) =R1
0e ttxdt De nition.
= txe tjt=1
t=0+R1
0e txtx 1dt Useu=tx,dv=e tdt.
=xR1
0e ttx 1dt Boundary terms are zero
forx >0.
=x (x).
Exercises 7.2
Laplace transform. Compute
L(f(t)) using the basic Laplace table
and the linearity properties of the
transform. Do not use the direct
Laplace transform!
1.L(2t)
2.L(4t)
3.L(1 + 2t +t2)
4.L(t2 3t+ 10)
5.L(sin 2t )
6.L(cos 2t )
7.L(e2t)
8.L(e 2t)
9.L(t+ sin 2t )
10.L(t cos2t)
11.L(t+e2t)
12.L(t 3e 2t)
13.L((t+ 1)2)
14.L((t+ 2)2)
15.L(t(t+ 1))
16.L((t+ 1)(t+ 2))
17.L(P10
n=0tn=n!)
18.L(P10
n=0tn+1=n!)
19.L(P10
n=1sinnt)
20.L(P10
n=0cosnt)Inverse Laplace transform. Solve
the given equation for the function
f(t). Use the basic table and linearity
properties of the Laplace transform.
21.L(f(t)) =s 2
22.L(f(t)) = 4 s 2
23.L(f(t)) = 1 =s+ 2=s2+ 3=s3
24.L(f(t)) = 1 =s3+ 1=s
25.L(f(t)) = 2 =(s2+ 4)
26.L(f(t)) =s=(s2+ 4)
27.L(f(t)) = 1 =(s 3)
28.L(f(t)) = 1 =(s+ 3)
29.L(f(t)) = 1 =s+s=(s2+ 4)
30.L(f(t)) = 2 =s 2=(s2+ 4)
31.L(f(t)) = 1 =s+ 1=(s 3)
32.L(f(t)) = 1 =s 3=(s 2)
33.L(f(t)) = (2 + s)2=s3
34.L(f(t)) = ( s+ 1)=s2
35.L(f(t)) =s(1=s2+ 2=s3)
36.L(f(t)) = ( s+ 1)(s 1)=s3
37.L(f(t)) =P10
n=0n!=s1+n
38.L(f(t)) =P10
n=0n!=s2+n
39.L(f(t)) =P10
n=1n
s2+n2
40.L(f(t)) =P10
n=0s
s2+n2
7.3 Laplace Transform Rules 257
7.3 Laplace Transform Rules
In Table 6, the basic table manipulation rules are summarize d. Full
statements and proofs of the rules appear in section 7.7, pag e 275.
The rules are applied here to several key examples. Partial f raction
expansions do not appear here, but in section 7.4, in connect ion with
Heaviside's coverup method.
Table 6. Laplace transform rules
L(f(t) +g(t)) =L(f(t)) +L(g(t)) Linearity.
The Laplace of a sum is the sum of the Laplaces.
L(cf(t)) =cL(f(t)) Linearity.
Constants move through the L-symbol.
L(y0(t)) =sL(y(t)) y(0) Thet-derivative rule.
Derivatives L(y0)are replaced in transformed equations.
LRt
0g(x)dx
=1
sL(g(t)) Thet-integral rule.
L(tf(t)) = d
dsL(f(t)) Thes-di erentiation rule.
Multiplying fbytapplies d=ds to the transform of f.
L(eatf(t)) =L(f(t))js!(s a)First shifting rule.
Multiplying fbyeatreplaces sbys a.
L(f(t a)H(t a)) =e asL(f(t)),
L(g(t)H(t a)) =e asL(g(t+a))Second shifting rule.
First and second forms.
L(f(t)) =RP
0f(t)e stdt
1 e PsRule for P-periodic functions.
Assumed here is f(t+P) =f(t).
L(f(t))L(g(t)) =L((fg)(t)) Convolution rule.
De ne (fg)(t) =Rt
0f(x)g(t x)dx.
11 Example (Harmonic oscillator) Solve by Laplace's method the initial value
problem x00+x= 0,x(0) = 0 ,x0(0) = 1 .
Solution :The solution is x(t) = sin t. The details:
L(x00) +L(x) =L(0) Apply Lacross the equation.
sL(x0) x0(0) +L(x) = 0 Use the t-derivative rule.
s[sL(x) x(0)] x0(0) +L(x) = 0 Use again the t-derivative rule.
(s2+ 1)L (x) = 1 Usex(0) = 0 ,x0(0) = 1.
L(x) =1
s2+ 1Divide.
=L(sint) Basic Laplace table.
x(t) = sin t Invoke Lerch's cancellation law.
258 Laplace Transform
12 Example ( s-di erentiation rule) Show the steps for L(t2e5t) =2
(s 5)3.
Solution :
L(t2e5t) =
d
ds
d
ds
L(e5t) Apply s-di erentiation.
= ( 1)2d
dsd
ds1
s 5
Basic Laplace table.
=d
ds 1
(s 5)2
Calculus power rule.
=2
(s 5)3Identity veri ed.
13 Example (First shifting rule) Show the steps for L(t2e 3t) =2
(s+ 3)3.
Solution :
L(t2e 3t) =L(t2)
s!s ( 3)First shifting rule.
=2
s2+1
s!s ( 3)Basic Laplace table.
=2
(s+ 3)3Identity veri ed.
14 Example (Second shifting rule) Show the steps for
L(sint H(t )) =e s
s2+ 1:
Solution :The second shifting rule is applied as follows.
L(sint H(t )) =L(g(t)H(t a)Choose g(t) = sin t,a=.
=e asL(g(t+a)Second form, second shifting theorem.
=e sL(sin(t +))Substitute a=.
=e sL( sint) Sum rule sin(a+b) = sin acosb+
sinbcosaplussin= 0,cos= 1.
=e s 1
s2+ 1Basic Laplace table. Identity veri ed.
15 Example (Trigonometric formulas) Show the steps used to obtain these
Laplace identities:
(a)L(tcosat) =s2 a2
(s2+a2)2(c)L(t2cosat) =2(s3 3sa2)
(s2+a2)3
(b)L(tsinat) =2sa
(s2+a2)2(d)L(t2sinat) =6s2a a3
(s2+a2)3
7.3 Laplace Transform Rules 259
Solution :The details for (a):
L(tcosat) = (d=ds)L(cosat) Uses-di erentiation.
= d
dss
s2+a2
Basic Laplace table.
=s2 a2
(s2+a2)2Calculus quotient rule.
The details for (c):
L(t2cosat) = (d=ds)L(( t)cosat) Uses-di erentiation.
=d
ds
s2 a2
(s2+a2)2
Result of (a).
=2s3 6sa2)
(s2+a2)3Calculus quotient rule.
The similar details for (b)and(d)are left as exercises.
16 Example (Exponentials) Show the steps used to obtain these Laplace
identities:
(a)L(eatcosbt) =s a
(s a)2+b2(c)L(teatcosbt) =(s a)2 b2
((s a)2+b2)2
(b)L(eatsinbt) =b
(s a)2+b2(d)L(teatsinbt) =2b(s a)
((s a)2+b2)2
Solution :Details for (a):
L(eatcosbt) = L(cosbt)js!s aFirst shifting rule.
=s
s2+b2
s!s aBasic Laplace table.
=s a
(s a)2+b2Veri ed (a).
Details for (c):
L(teatcosbt) = L(tcosbt)js!s aFirst shifting rule.
=
d
dsL(cosbt)
s!s aApply s-di erentiation.
=
d
dss
s2+b2
s!s aBasic Laplace table.
=s2 b2
(s2+b2)2
s!s aCalculus quotient rule.
=(s a)2 b2
((s a)2+b2)2Veri ed (c).
Left as exercises are (b)and(d).
260 Laplace Transform
17 Example (Hyperbolic functions) Establish these Laplace transform facts
about coshu= (eu+e u)=2andsinhu= (eu e u)=2.
(a)L(coshat) =s
s2 a2(c)L(tcoshat) =s2+a2
(s2 a2)2
(b)L(sinhat) =a
s2 a2(d)L(tsinhat) =2as
(s2 a2)2
Solution :The details for (a):
L(coshat) =1
2(L(eat) +L(e at)) De nition plus linearity of L.
=1
21
s a+1
s+a
Basic Laplace table.
=s
s2 a2Identity (a)veri ed.
The details for (d):
L(tsinhat) = d
dsa
s2 a2
Apply the s-di erentiation rule.
=a(2s)
(s2 a2)2Calculus power rule; (d)veri ed.
Left as exercises are (b)and(c).
18 Example ( s-di erentiation) Solve L(f(t)) =2s
(s2+ 1)2forf(t).
Solution :The solution is f(t) =tsint. The details:
L(f(t)) =2s
(s2+ 1)2
= d
ds1
s2+ 1
Calculus power rule (un)0=nun 1u0.
= d
ds(L(sint)) Basic Laplace table.
=L(tsint) Apply the s-di erentiation rule.
f(t) =tsint Lerch's cancellation law.
19 Example (First shift rule) Solve L(f(t)) =s+ 2
22+ 2s+ 2forf(t).
Solution :The answer is f(t) =e tcost+e tsint. The details:
L(f(t)) =s+ 2
s2+ 2s+ 2Signal for this method: the denom-
inator has complex roots.
=s+ 2
(s+ 1)2+ 1Complete the square, denominator.
7.3 Laplace Transform Rules 261
=S+ 1
S2+ 1Substitute Sfors+ 1.
=S
S2+ 1+1
S2+ 1Split into Laplace table entries.
=L(cost) +L(sint)js!S=s+1Basic Laplace table.
=L(e tcost) +L(e tsint) First shift rule.
f(t) =e tcost+e tsint Invoke Lerch's cancellation law.
20 Example (Damped oscillator) Solve by Laplace's method the initial value
problem x00+ 2x0+ 2x= 0,x(0) = 1 ,x0(0) = 1.
Solution :The solution is x(t) =e tcost. The details:
L(x00) + 2L (x0) + 2L (x) =L(0) Apply Lacross the equation.
sL(x0) x0(0) + 2L (x0) + 2L (x) = 0 Thet-derivative rule on x0.
s[sL(x) x(0)] x0(0)
+2[L(x) x(0)] + 2L (x) = 0Thet-derivative rule on x.
(s2+ 2s+ 2)L (x) = 1 + s Usex(0) = 1 ,x0(0) = 1.
L(x) =s+ 1
s2+ 2s+ 2Divide.
=s+ 1
(s+ 1)2+ 1Complete the square in the de-
nominator.
=L(cost)js!s+1Basic Laplace table.
=L(e tcost) First shifting rule.
x(t) =e tcost Invoke Lerch's cancellation law.
21 Example (Recti ed sine wave) Compute the Laplace transform of the
recti ed sine wave f(t) =jsin!tjand show it can be expressed in the
form
L(jsin!tj) =!coth s
2!
s2+!2:
Solution :The periodic function formula will be applied with period P=
2=!. The calculation reduces to the evaluation of J=RP
0f(t)e stdt. Because
sin!t0 on=!t2=!, integral Jcan be written as J=J1+J2, where
J1=Z=!
0sin!t e stdt; J 2=Z2=!
=! sin!t e stdt:
Integral tables give the result
Z
sin!t e stdt= !e stcos(!t)
s2+!2 se stsin(!t)
s2+!2:
Then
J1=!(e s=!+ 1)
s2+!2; J2=!(e 2s=!+e s=!)
s2+!2;
262 Laplace Transform
J=!(e s=!+ 1)2
s2+!2:
The remaining challenge is to write the answer for L(f(t)) in terms of coth.
The details:
L(f(t)) =J
1 e PsPeriodic function formula.
=J
(1 e Ps= 2)(1 +e Ps= 2)Apply 1 x2= (1 x)(1 + x),
x=e Ps= 2.
=!(1 +e Ps= 2)
(1 e Ps= 2)(s2+!2)Cancel factor 1 +e Ps= 2.
=ePs=4+e Ps= 4
ePs=4 e Ps= 4!
s2+!2Factor out e Ps= 4, then cancel.
=2 cosh(Ps= 4)
2 sinh(Ps= 4)!
s2+!2Apply cosh,sinhidentities.
=!coth(Ps=4)
s2+!2Usecothu= cosh u=sinhu.
=!coth s
2!
s2+!2Identity veri ed.
22 Example (Half{wave recti cation) Compute the Laplace transform of the
half{wave recti cation of sin!t, denoted g(t), in which the negative cycles
ofsin!thave been canceled to create g(t). Show in particular that
L(g(t)) =1
2!
s2+!2
1 + coths
2!
Solution :The half{wave recti cation of sin !tisg(t) = (sin !t+jsin!tj)=2.
Therefore, the basic Laplace table plus the result of Example 21 give
L(2g(t)) =L(sin!t) + L(jsin!tj)
=!
s2+!2+!cosh(s=(2!))
s2+!2
=!
s2+!2(1 + cosh(s= (2!))
Dividing by 2 produces the identity.
23 Example (Shifting rules) Solve L(f(t)) =e 3ss+ 1
s2+ 2s+ 2forf(t).
Solution :The answer is f(t) =e3 tcos(t 3)H(t 3). The details:
L(f(t)) =e 3ss+ 1
(s+ 1)2+ 1Complete the square.
=e 3sS
S2+ 1Replace s+ 1byS.
=e 3S+3(L(cost))js!S=s+1Basic Laplace table.
7.3 Laplace Transform Rules 263
=e3
e 3sL(cost)
s!S=s+1Regroup factor e 3S.
=e3(L(cos(t 3)H(t 3)))js!S=s+1Second shifting rule.
=e3L(e tcos(t 3)H(t 3)) First shifting rule.
f(t) =e3 tcos(t 3)H(t 3) Lerch's cancellation law.
24 Example () Solve L(f(t) =s+ 7
s2+ 4s+ 8forf(t).
Solution :The answer is f(t) =e 2t(cos2t +5
2sin 2t). The details:
L(f(t)) =s+ 7
(s+ 2)2+ 4Complete the square.
=S+ 5
S2+ 4Replace s+ 2byS.
=S
S2+ 4+5
22
S2+ 4Split into table entries.
=s
s2+ 4+5
22
s2+ 4
s!S=s+2Prepare for shifting rule.
=L(cos2t ) +5
2L(sin 2t )
s!S=s+2Basic Laplace table.
=L(e 2t(cos 2t +5
2sin 2t)) First shifting rule.
f(t) =e 2t(cos2t +5
2sin 2t) Lerch's cancellation law.
264 Laplace Transform
7.4 Heaviside's Method
This practical method was popularized by the English electr ical engineer
Oliver Heaviside (1850{1925). A typical application of the method is to
solve
2s
(s+ 1)(s2+ 1)=L(f(t))
for the t-expression f(t) = e t+cost+sint. The details in Heaviside's
method involve a sequence of easy-to-learn college algebra steps.
More precisely, Heaviside's method systematically converts a polyno-
mial quotient
a0+a1s++ansn
b0+b1s++bmsm(1)
into the form L(f(t)) for some expression f(t). It is assumed that
a0;::;a n;b0;::: ;b mare constants and the polynomial quotient (1) has
limit zero at s=1.
Partial Fraction Theory
In college algebra, it is shown that a rational function (1) c an be ex-
pressed as the sum of terms of the form
A
(s s0)k(2)
where Ais a real or complex constant and ( s s0)kdivides the denomi-
nator in (1). In particular, s0is arootof the denominator in (1).
Assume fraction (1) has real coecients. If s0in (2) is real, then Ais
real. Ifs0= +i in (2) is complex, then ( s s0)kalso appears, where
s0= i is the complex conjugate of s0. The corresponding terms
in (2) turn out to be complex conjugates of one another, which can be
combined in terms of realnumbers BandCas
A
(s s0)k+A
(s s0)k=B+C s
((s )2+ 2)k: (3)
Simple Roots. Assume that (1) has real coecients and the denomi-
nator of the fraction (1) has distinct real roots s1, .. .,sNanddistinct
complex roots 1+i 1, .. ., M+i M. The partial fraction expansion
of (1) is a sum given in terms of realconstants Ap,Bq,Cqby
a0+a1s++ansn
b0+b1s++bmsm=NX
p=1Ap
s sp+MX
q=1Bq+Cq(s q)
(s q)2+ 2q: (4)
7.4 Heaviside's Method 265
Multiple Roots. Assume (1) has real coecients and the denomi-
nator of the fraction (1) has possibly multiple roots . Let Npbe the
multiplicity of real root spand let Mqbe the multiplicity of complex root
q+i q, 1pN, 1qM. The partial fraction expansion of (1)
is given in terms of realconstants Ap;k,Bq;k,Cq;kby
NX
p=1X
1kNpAp;k
(s sp)k+MX
q=1X
1kMqBq;k+Cq;k(s q)
((s q)2+ 2q)k: (5)
Heaviside's Coverup Method
The method applies only to the case of distinct roots of the de nominator
in (1). Extensions to multiple-root cases can be made; see pa ge 266.
To illustrate Oliver Heaviside's ideas, consider the probl em details
2s+ 1
s(s 1)(s+ 1)=A
s+B
s 1+C
s+ 1(6)
=L(A) +L(Bet) +L(Ce t)
=L(A+Bet+Ce t)
The rst line (6) uses college algebra partial fractions. Th e second and
third lines use the Laplace integral table and properties of L.
Heaviside's mysterious method. Oliver Heaviside proposed to
nd in (6) the constant C=1
2by acover{up method :
2s+ 1
s(s 1)
s+1=0=C:
Theinstructions are to cover{up the matching factors ( s+ 1) on the left
and right with box , then evaluate on the left at the rootswhich
makes the contents of the box zero. The other terms on the righ t are
replaced by zero.
To justify Heaviside's cover{up method, multiply (6) by the denominator
s+ 1 of partial fraction C=(s+ 1):
(2s+ 1)(s+ 1)
s(s 1)(s+ 1)=A(s+ 1)
s+B(s+ 1)
s 1+C(s+ 1)
(s+ 1):
Set(s+ 1)= 0 in the display. Cancellations left and right plus annihi-
lation of two terms on the right gives Heaviside's prescript ion
2s+ 1
s(s 1)
s+1=0=C:
266 Laplace Transform
The factor ( s+ 1) in (6) is by no means special: the same procedure
applies to nd AandB. The method works for denominators with
simple roots, that is, no repeated roots are allowed.
Extension to Multiple Roots. An extension of Heaviside's method
is possible for the case of repeated roots. The basic idea is t ofactor{out
the repeats . To illustrate, consider the partial fraction expansion de tails
R=1
(s+ 1)2(s+ 2)A sample rational function having
repeated roots.
=1
s+ 11
(s+ 1)(s+ 2)
Factor{out the repeats.
=1
s+ 11
s+ 1+ 1
s+ 2
Apply the cover{up method to the
simple root fraction.
=1
(s+ 1)2+ 1
(s+ 1)(s+ 2)Multiply.
=1
(s+ 1)2+ 1
s+ 1+1
s+ 2Apply the cover{up method to the
last fraction on the right.
Terms with only one root in the denominator are already parti al frac-
tions. Thus the work centers on expansion of quotients in whi ch the
denominator has two or more roots.
Special Methods. Heaviside's method has a useful extension for the
case of roots of multiplicity two. To illustrate, consider t hese details:
R=1
(s+ 1)2(s+ 2)A fraction with multiple roots.
=A
s+ 1+B
(s+ 1)2+C
s+ 2See equation (5).
=A
s+ 1+1
(s+ 1)2+1
s+ 2FindBandCby Heaviside's cover{
up method.
= 1
s+ 1+1
(s+ 1)2+1
s+ 2Multiply by s+1. Sets=1. Then
0 =A+ 1.
The illustration works for one root of multiplicity two, bec auses=1
will resolve the coecient not found by the cover{up method.
In general, if the denominator in (1) has a root s0of multiplicity k, then
the partial fraction expansion contains terms
A1
s s0+A2
(s s0)2++Ak
(s s0)k:
Heaviside's cover{up method directly nds Ak, but not A1toAk 1.
7.5 Heaviside Step and Dirac Delta 267
7.5 Heaviside Step and Dirac Delta
Heaviside Function. Theunit step function orHeaviside func-
tionis de ned by
H(x) =(
1 for x0;
0 for x <0:
The most often{used formula involving the Heaviside functi on is the
characteristic function of the interval at < b, given by
H(t a) H(t b) =(
1at < b;
0t < a; t b:(1)
To illustrate, a square wave sqw(t) = ( 1)
oor (t)can be written in the
series form
1X
n=0( 1)n(H(t n) H(t n 1)):
Dirac Delta. A precise mathematical de nition of the Dirac delta,
denoted , is not possible to give here. Following its inventor P. Dira c,
the de nition should be
(t) =dH(t):
The latter is nonsensical, because the unit step does not hav e a cal-
culus derivative at t= 0. However, dH(t) could have the meaning of
a Riemann-Stieltjes integrator, which restrains dH(t) to have meaning
only under an integral sign. It is in this sense that the Dirac delta is
de ned.
What do we mean by the di erential equation
x00+ 16x= 5(t t0)?
The equation x00+ 16x=f(t) represents a spring-mass system without
damping having Hooke's constant 16, subject to external for cef(t). In
a mechanical context, the Dirac delta term 5 (t t0) is an idealization
of a hammer-hit at time t=t0>0 with impulse 5.
More precisely, the forcing term f(t) can be formally written as a Riemann-
Stieltjes integrator 5 dH(t t0) where His Heaviside's unit step function.
The Dirac delta or \derivative of the Heaviside unit step," n onsensical
as it may appear, is realized in applications via the two-sid ed or central
di erence quotient
H(t+h) H(t h)
2hdH(t):
268 Laplace Transform
Therefore, the force f(t) in the idealization 5 (t t0) is given for h >0
very small by the approximation
f(t)5H(t t0+h) H(t t0 h)
2h:
Theimpulse2of the approximated force over a large interval [a;b ] is
computed from
Zb
af(t)dt5Zh
hH(t t0+h) H(t t0 h)
2hdt= 5;
due to the integrand being 1 =(2h) on jt t0j< hand otherwise 0.
Modeling Impulses. One argument for the Dirac delta idealization
is that an in nity of choices exist for modeling an impulse. T here are in
addition to the central di erence quotient two other popular di erence
quotients, the forward quotient ( H(t+h) H(t))=hand the backward
quotient ( H(t) H(t h))=h(h >0 assumed). In reality, his unknown
in any application, and the impulsive force of a hammer hit is hardly
constant, as is supposed by this naive modeling.
The modeling logic often applied for the Dirac delta is that t he external
forcef(t) is used in the model in a limited manner, in which only the
momentum p=mvis important. More precisely, only the change in
momentum or impulse is important,Rb
af(t)dt= p =mv(b) mv(a).
The precise force f(t) is replaced during the modeling by a simplistic
piecewise-de ned force that has exactly the same impulse p . The re-
placement is justi ed by arguing that if only the impulse is i mportant,
and not the actual details of the force, then both models shou ld give
similar results.
Function or Operator ? The work of physics Nobel prize winner P.
Dirac (1902{1984) proceeded for about 20 years before the ma thematical
community developed a sound mathematical theory for his imp ulsive
force representations. A systematic theory was developed i n 1936 by
the soviet mathematician S. Sobolev. The French mathematic ian L.
Schwartz further developed the theory in 1945. He observed t hat the
idealization is not a function but an operator or linear functional, in
particular, maps or associates to each function (t) its value at t= 0, in
short, ( ) = (0). This fact was observed early on by Dirac and others,
during the replacement of simplistic forces by . In Laplace theory, there
is a natural encounter with the ideas, because L(f(t)) routinely appears
on the right of the equation after transformation. This term , in the case
2Momentum is de ned to be mass times velocity. If the force fis given by Newton's
law as f(t) =d
dt(mv(t)) and v(t) is velocity, thenRb
af(t)dt=mv(b) mv(a) is the
net momentum or impulse.
7.5 Heaviside Step and Dirac Delta 269
of an impulsive force f(t) =c(H(t t0 h) H(t t0+h))=(2h), evaluates
fort0>0 and t0 h >0 as follows:
L(f(t)) =Z1
0c
2h(H(t t0 h) H(t t0+h))e stdt
=Zt0+h
t0 hc
2he stdt
=ce st0
esh e sh
2sh!
The factoresh e sh
2shis approximately 1 for h >0 small, because of
L'Hospital's rule. The immediate conclusion is that we shou ld replace
the impulsive force fby an equivalent one fsuch that
L(f(t)) =ce st0:
Well,there is no such function f!
The apparent mathematical
aw in this idea was resolved by th e work
of L. Schwartz on distributions . In short, there is a solid foundation
for introducing f, but unfortunately the mathematics involved is not
elementary nor especially accessible to those readers whos e background
is just calculus.
Practising engineers and scientists might be able to ignore the vast lit-
erature on distributions, citing the example of physicist P . Dirac, who
succeeded in applying impulsive force ideas without the dis tribution the-
ory developed by S. Sobolev and L. Schwartz. This will not be t he case
for those who wish to read current literature on partial di er ential equa-
tions, because the work on distributions has forever change d the required
background for that topic.
270 Laplace Transform
7.6 Laplace Table Derivations
Veri ed here are two Laplace tables, the minimal Laplace Tab le 7.2-4
and its extension Table 7.2-5. Largely, this section is for r eading, as it is
designed to enrich lectures and to aid readers who study alon e.
Derivation of Laplace integral formulas in Table 7.2-4, page 252.
Proof of L(tn) =n!=s1+n:
The rst step is to evaluate L(tn) forn= 0.
L(1) =R1
0(1)e stdt Laplace integral of f(t) = 1 .
= (1=s)e stjt=1
t=0Evaluate the integral.
= 1=s Assumed s >0to evaluate limt!1e st.
The value of L(tn) forn= 1 can be obtained by s-di erentiation of the relation
L(1) = 1=s , as follows.
d
dsL(1) =d
dsR1
0(1)e stdt Laplace integral for f(t) = 1 .
=R1
0d
ds(e st)dt Usedd
dsRb
aFdt=Rb
adF
dsdt.
=R1
0( t)e stdt Calculus rule (eu)0=u0eu.
= L(t) De nition of L(t).
Then
L(t) = d
dsL(1) Rewrite last display.
= d
ds(1=s) UseL(1) = 1=s .
= 1=s2Di erentiate.
This idea can be repeated to give L(t2) = d
dsL(t) and hence L(t2) = 2=s3.
The pattern is L(tn) = d
dsL(tn 1) which gives L(tn) =n!=s1+n.
Proof of L(eat) = 1= (s a):
The result follows from L(1) = 1=s , as follows.
L(eat) =R1
0eate stdt Direct Laplace transform.
=R1
0e (s a)tdt UseeAeB=eA+B.
=R1
0e Stdt Substitute S=s a.
= 1=S Apply L(1) = 1=s .
= 1=(s a) Back-substitute S=s a.
Proof of L(cosbt) = s=(ss+b2)andL(sinbt) = b=(ss+b2):
Use will be made of Euler's formula ei= cos +isin, usually rst introduced
in trigonometry. In this formula, is a real number (in radians) and i=p 1
is the complex unit.
7.6 Laplace Table Derivations 271
eibte st= (cos bt)e st+i(sinbt)e stSubstitute =btinto Euler's
formula and multiply by e st.R1
0e ibte stdt=R1
0(cosbt)e stdt
+iR1
0(sinbt)e stdtIntegrate t= 0tot=1. Use
properties of integrals.
1
s ib=R1
0(cosbt)e stdt
+iR1
0(sinbt)e stdtEvaluate the left side using
L(eat) = 1= (s a),a=ib.
1
s ib=L(cosbt) +iL(sinbt) Direct Laplace transform de ni-
tion.
s+ib
s2+b2=L(cosbt) +iL(sinbt) Use complex rule 1=z=z=jzj2,
z=A+iB,z=A iB,jzj=p
A2+B2.
s
s2+b2=L(cosbt) Extract the real part.
b
s2+b2=L(sinbt) Extract the imaginary part.
Derivation of Laplace integral formulas in Table 7.2-5, page 252.
Proof of the Heaviside formula L(H(t a)) =e as=s.
L(H(t a)) =R1
0H(t a)e stdtDirect Laplace transform. Assume a0.
=R1
a(1)e stdt Because H(t a) = 0 for0t < a.
=R1
0(1)e s(x+a)dxChange variables t=x+a.
=e asR1
0(1)e sxdxConstant e asmoves outside integral.
=e as(1=s) Apply L(1) = 1=s .
Proof of the Dirac delta formula L((t a)) =e as.
Thede nition of the delta function is a formal one, in which every occurrence of
(t a)dtunder an integrand is replaced by dH(t a). The di erential symbol
dH(t a) is taken in the sense of the Riemann-Stieltjes integral. This integral
is de ned in [ ?] for monotonic integrators (x) as the limit
Zb
af(x)d (x) = lim
N!1NX
n=1f(xn)( (xn) (xn 1))
where x0=a,xN=bandx0< x1<< xNforms a partition of [a; b ] whose
mesh approaches zero as N! 1.
The steps in computing the Laplace integral of the delta function app ear below.
Admittedly, the proof requires advanced calculus skills and a certain level of
mathematical maturity. The reward is a fuller understanding of the D irac
symbol (x).
L((t a)) =R1
0e st(t a)dt Laplace integral, a >0assumed.
=R1
0e stdH(t a) Replace (t a)dtbydH(t a).
= lim M!1RM
0e stdH(t a)De nition of improper integral.
272 Laplace Transform
=e saExplained below.
To explain the last step, apply the de nition of the Riemann-Stieltjes integral:
ZM
0e stdH(t a) = lim
N!1N 1X
n=0e st n(H(tn a) H(tn 1 a))
where 0 = t0< t1<< tN=Mis a partition of [0 ; M] whose mesh
max1nN(tn tn 1) approaches zero as N! 1. Given a partition, if tn 1<
atn, then H(tn a) H(tn 1 a) = 1, otherwise this factor is zero. Therefore,
the sum reduces to a single term e st n. This term approaches e saasN! 1,
because tnmust approach a.
Proof of L(
oor(t=a)) =e as
s(1 e as):
The library function
oor present in computer languages C and Fortran is
de ned by
oor(x) = greatest whole integer x, e.g.,
oor(5:2) = 5 and
oor( 1:9) = 2. The computation of the Laplace integral of
oor(t) requires
ideas from in nite series, as follows.
F(s) =R1
0
oor(t)e stdt Laplace integral de nition.
=P1
n=0Rn+1
n(n)e stdt Onnt < n + 1,
oor(t) =n.
=P1
n=0n
s(e ns e ns s) Evaluate each integral.
=1 e s
sP1
n=0ne snCommon factor removed.
=x(1 x)
sP1
n=0nxn 1De ne x=e s.
=x(1 x)
sd
dxP1
n=0xnTerm-by-term di erentiation.
=x(1 x)
sd
dx1
1 xGeometric series sum.
=x
s(1 x)Compute the derivative, simplify.
=e s
s(1 e s)Substitute x=e s.
To evaluate the Laplace integral of
oor(t=a), a change of variables is made.
L(
oor(t=a)) =R1
0
oor(t=a)e stdt Laplace integral de nition.
=aR1
0
oor(r)e asrdr Change variables t=ar.
=aF(as) Apply the formula for F(s).
=e as
s(1 e as)Simplify.
Proof of L(sqw(t=a)) =1
stanh(as= 2):
The square wave de ned by sqw(x) = ( 1)
oor (x)is periodic of period 2 and
piecewise-de ned. Let P=R2
0sqw(t)e stdt.
7.6 Laplace Table Derivations 273
P=R1
0sqw(t)e stdt+R2
1sqw(t)e stdtApplyRb
a=Rc
a+Rb
c.
=R1
0e stdt R2
1e stdt Usesqw(x) = 1 on0x <1and
sqw(x) = 1on1x <2.
=1
s(1 e s) +1
s(e 2s e s) Evaluate each integral.
=1
s(1 e s)2Collect terms.
An intermediate step is to compute the Laplace integral of sqw(t):
L(sqw(t)) =R2
0sqw(t)e stdt
1 e 2sPeriodic function formula, page 275.
=1
s(1 e s)21
1 e 2s. Use the computation of Pabove.
=1
s1 e s
1 +e s. Factor 1 e 2s= (1 e s)(1 +e s).
=1
ses=2 e s=2
es=2+e s=2. Multiply the fraction by es=2=es=2.
=1
ssinh(s= 2)
cosh(s=2). Use sinhu= (eu e u)=2,
coshu= (eu+e u)=2.
=1
stanh(s= 2). Use tanhu= sinh u=coshu.
To complete the computation of L(sqw(t=a)), a change of variables is made:
L(sqw(t=a)) =R1
0sqw(t=a)e stdt Direct transform.
=R1
0sqw(r)e asr(a)dr Change variables r=t=a.
=a
astanh(as= 2) SeeL(sqw(t))above.
=1
stanh(as= 2)
Proof of L(atrw(t=a)) =1
s2tanh(as= 2):
The triangular wave is de ned by trw(t) =Rt
0sqw(x)dx.
L(atrw(t=a)) =1
s(f(0) +L(f0(t))Letf(t) =atrw(t=a). Use L(f0(t)) =
sL(f(t)) f(0), page 251.
=1
sL(sqw(t=a)) Usef(0) = 0, (aRt=a
0sqw(x)dx)0=
sqw(t=a).
=1
s2tanh(as= 2) Table entry for sqw.
Proof of L(t ) = (1 + )
s1+ :
L(t ) =R1
0t e stdt Direct Laplace transform.
=R1
0(u=s) e udu=s Change variables u=st,du=sdt.
274 Laplace Transform
=1
s1+ R1
0u e udu
=1
s1+ (1 + ). Where (x) =R1
0ux 1e udu, by
de nition.
Thegeneralized factorial function (x) is de ned for x >0 and it agrees with
the classical factorial n! = (1)(2) (n) in case x=n+ 1 is an integer. In
literature, ! means (1+ ). For more details about the Gamma function, see
Abramowitz and Stegun [? ], ormaple documentation.
Proof of L(t 1=2) =r
s:
L(t 1=2) = (1 + ( 1=2))
s1 1=2Apply the previous formula.
=ppsUse (1=2) =p.
7.7 Transform Properties 275
7.7 Transform Properties
Collected here are the major theorems and their proofs for th e manipu-
lation of Laplace transform tables.
Theorem 4 (Linearity)
The Laplace transform has these inherited integral properties:
(a) L(f(t) +g(t)) =L(f(t)) +L(g(t));
(b) L(cf(t)) =cL(f(t)):
Theorem 5 (The t-Derivative Rule)
Lety(t)be continuous, of exponential order and let f0(t)be piecewise
continuous on t0. Then L(y0(t))exists and
L(y0(t)) =sL(y(t)) y(0):
Theorem 6 (The t-Integral Rule)
Letg(t)be of exponential order and continuous for t0. Then
LRt
0g(x)dx
=1
sL(g(t)):
Theorem 7 (The s-Di erentiation Rule)
Letf(t)be of exponential order. Then
L(tf(t)) = d
dsL(f(t)):
Theorem 8 (First Shifting Rule)
Letf(t)be of exponential order and 1< a < 1. Then
L(eatf(t)) =L(f(t))js!(s a):
Theorem 9 (Second Shifting Rule)
Letf(t)andg(t)be of exponential order and assume a0. Then
(a) L(f(t a)H(t a)) =e asL(f(t));
(b) L(g(t)H(t a)) =e asL(g(t+a)):
Theorem 10 (Periodic Function Rule)
Letf(t)be of exponential order and satisfy f(t+P) =f(t). Then
L(f(t)) =RP
0f(t)e stdt
1 e Ps:
Theorem 11 (Convolution Rule)
Letf(t)andg(t)be of exponential order. Then
L(f(t))L(g(t)) =LZt
0f(x)g(t x)dx
:
276 Laplace Transform
Proof of Theorem 4 (linearity):
LHS=L(f(t) +g(t)) Left side of the identity in (a).
=R1
0(f(t) +g(t))e stdt Direct transform.
=R1
0f(t)e stdt+R1
0g(t)e stdtCalculus integral rule.
=L(f(t)) +L(g(t)) Equals RHS; identity (a) veri ed.
LHS=L(cf(t)) Left side of the identity in (b).
=R1
0cf(t)e stdt Direct transform.
=cR1
0f(t)e stdt Calculus integral rule.
=cL(f(t)) Equals RHS; identity (b) veri ed.
Proof of Theorem 5 ( t-derivative rule): Already L(f(t)) exists, because
fis of exponential order and continuous. On an interval [a; b ] where f0is
continuous, integration by parts using u=e st,dv=f0(t)dtgives
Rb
af0(t)e stdt=f(t)e stjt=b
t=a Rb
af(t)( s)e stdt
= f(a)e sa+f(b)e sb+sRb
af(t)e stdt:
On any interval [0 ; N], there are nitely many intervals [a; b ] on each of which
f0is continuous. Add the above equality across these nitely many inte rvals
[a; b]. The boundary values on adjacent intervals match and the integrals add
to give
ZN
0f0(t)e stdt= f(0)e0+f(N)e sN+sZN
0f(t)e stdt:
Take the limit across this equality as N! 1. Then the right side has limit
f(0) +sL(f(t)), because of the existence of L(f(t)) and lim t!1f(t)e st= 0
for large s. Therefore, the left side has a limit, and by de nition L(f0(t)) exists
andL(f0(t)) = f(0) +sL(f(t)).
Proof of Theorem 6 ( t-Integral rule): Letf(t) =Rt
0g(x)dx. Then fis of
exponential order and continuous. The details:
L(Rt
0g(x)dx) =L(f(t)) By de nition.
=1
sL(f0(t)) Because f(0) = 0 implies L(f0(t)) =sL(f(t)).
=1
sL(g(t)) Because f0=gby the Fundamental theorem of
calculus.
Proof of Theorem 7 ( s-di erentiation): We prove the equivalent relation
L(( t)f(t)) = ( d=ds)L(f(t)). If fis of exponential order, then so is ( t)f(t),
therefore L(( t)f(t)) exists. It remains to show the s-derivative exists and
satis es the given equality.
The proof below is based in part upon the calculus inequality
e x+x 1 x2; x0: (1)
7.7 Transform Properties 277
The inequality is obtained from two applications of the mean value theorem
g(b) g(a) =g0(x)(b a), which gives e x+x 1 =xxe x1with 0 x1xx.
In addition, the existence of L(t2jf(t)j) is used to de ne s0>0 such that
L(t2jf(t)j)1 fors > s 0. This follows from the transform existence theorem
for functions of exponential order, where it is shown that the trans form has
limit zero at s=1.
Consider h6= 0 and the Newton quotient Q(s; h) = ( F(s+h) F(s))=hfor the
s-derivative of the Laplace integral. We have to show that
lim
h!0jQ(s; h) L(( t)f(t))j= 0:
This will be accomplished by proving for s > s 0ands+h > s 0the inequality
jQ(s; h) L(( t)f(t))j jhj:
Forh6= 0,
Q(s; h) L(( t)f(t)) =Z1
0f(t)e st ht e st+the st
hdt:
Assume h >0. Due to the exponential rule eA+B=eAeB, the quotient in the
integrand simpli es to give
Q(s; h) L(( t)f(t)) =Z1
0f(t)e ste ht+th 1
h
dt:
Inequality (1) applies with x=ht0, giving
jQ(s; h) L(( t)f(t))j jhjZ1
0t2jf(t)je stdt:
The right side is jhjL(t2jf(t)j), which for s > s 0is bounded by jhj, completing
the proof for h >0. Ifh <0, then a similar calculation is made to obtain
jQ(s; h) L(( t)f(t))j jhjZ1
0t2jf(t)e st htdt:
The right side is jhjL(t2jf(t)j) evaluated at s+hinstead of s. Ifs+h > s 0,
then the right side is bounded by jhj, completing the proof for h <0.
Proof of Theorem 8 ( rst shifting rule): The left side LHS of the equality
can be written because of the exponential rule eAeB=eA+Bas
LHS =Z1
0f(t)e (s a)tdt:
This integral is L(f(t)) with sreplaced by s a, which is precisely the meaning
of the right side RHS of the equality. Therefore, LHS = RHS.
Proof of Theorem 9 (second shifting rule): The details for (a) are
LHS=L(H(t a)f(t a))
=R1
0H(t a)f(t a)e stdtDirect transform.
278 Laplace Transform
=R1
aH(t a)f(t a)e stdtBecause a0andH(x) = 0 forx <0.
=R1
0H(x)f(x)e s(x+a)dx Change variables x=t a,dx=dt.
=e saR1
0f(x)e sxdx UseH(x) = 1 forx0.
=e saL(f(t)) Direct transform.
=RHS Identity (a) veri ed.
In the details for (b), let f(t) =g(t+a), then
LHS=L(H(t a)g(t))
=L(H(t a)f(t a)) Usef(t a) =g(t a+a) =g(t).
=e saL(f(t)) Apply (a).
=e saL(g(t+a)) Because f(t) =g(t+a).
=RHS Identity (b) veri ed.
Proof of Theorem 10 (periodic function rule):
LHS=L(f(t))
=R1
0f(t)e stdt Direct transform.
=P1
n=0RnP+P
nPf(t)e stdt Additivity of the integral.
=P1
n=0RP
0f(x+nP)e sx nPsdx Change variables t=x+nP.
=P1
n=0e nPsRP
0f(x)e sxdx Because fisP-periodic and
eAeB=eA+B.
=RP
0f(x)e sxdxP1
n=0rnCommon factor in summation.
De ne r=e Ps.
=RP
0f(x)e sxdx1
1 rSum the geometric series.
=RP
0f(x)e sxdx
1 e PsSubstitute r=e Ps.
=RHS Periodic function identity veri ed.
Left unmentioned here is the convergence of the in nite series on lin e 3 of the
proof, which follows from fof exponential order.
Proof of Theorem 11 (convolution rule): The details use Fubini's in-
tegration interchange theorem for a planar unbounded region, an d therefore
this proof involves advanced calculus methods that may be outside t he back-
ground of the reader. Modern calculus texts contain a less genera l version of
Fubini's theorem for nite regions, usually referenced as iterated integrals. The
unbounded planar region is written in two ways:
D=f(r; t) :tr <1;0t <1g;
D=f(r; t) : 0 r <1;0rtg:
Readers should pause here and verify that D=D.
7.7 Transform Properties 279
The change of variable r=x+t,dr=dxis applied for xed t0 to obtain
the identity
e stR1
0g(x)e sxdx=R1
0g(x)e sx stdx
=R1
tg(r t)e rsdr:(2)
The left side of the convolution identity is expanded as follows:
LHS=L(f(t))L(g(t))
=R1
0f(t)e stdtR1
0g(x)e sxdxDirect transform.
=R1
0f(t)R1
tg(r t)e rsdrdt Apply identity (2).
=R
Df(t)g(r t)e rsdrdt Fubini's theorem applied.
=R
Df(t)g(r t)e rsdrdt Descriptions DandDare the same.
=R1
0Rr
0f(t)g(r t)dte rsdr Fubini's theorem applied.
Then
RHS=LRt
0f(u)g(t u)du
=R1
0Rt
0f(u)g(t u)due stdtDirect transform.
=R1
0Rr
0f(u)g(r u)due srdrChange variable names r$t.
=R1
0Rr
0f(t)g(r t)dt e srdr Change variable names u$t.
=LHS Convolution identity veri ed.
280 Laplace Transform
7.8 More on the Laplace Transform
Model conversion and engineering. Adi erential equation model
for a physical system can be subjected to the Laplace transfo rm in order
to produce an algebraic model in the transform variable s. Lerch's the-
orem says that both models are equivalent, that is, the solut ion of one
model gives the solution to the other model.
In electrical and computer engineering it is commonplace to dealonly
with the Laplace algebraic model. Engineers are in fact capa ble of hav-
ing hour-long modeling conversations, during which di eren tial equations
arenever referenced! Terminology for such modeling is necessarily spe-
cialized, which gives rise to new contextual meanings to the terms input
andoutput . For example, an RLC-circuit would be discussed with input
F(s) =!
s2+!2;
and the listener must know that this expression is the Laplac e transform
of the t-expression sin !t. Hence the RLC-circuit is driven by a sinu-
soindal input of natural frequency !. During the modeling discourse, it
could be that the output is
X(s) =1
s+ 1+10!
s2+!2:
Lerch's equivalence says that X(s) is the Laplace transform of e t+
10sin !t, but that is extra work, if all that is needed from the model is a
statement about the transient and steady-state responses t o the input.