[PDF] [PDF] The Inverse Laplace Transform - UAH

[PDF] Chapter 7 Laplace Transforms Section 74 Inverse Laplace

Example 1 Determine the inverse Laplace transform of the given function (a) F(s ) = 2 s3 SOLUTION L−1 { 2 s3 } = L−1 {2 s3 } = t2 (b) F(s) = 2 s2+4

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[PDF] The Inverse Laplace Transform - UAH

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[PDF] Section 74: Inverse Laplace Transform A natural question to ask

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7 1 30 - Find the inverse Laplace transform of the function F(s) = 9 + s 4 − s2 Solution - If we break up this fraction we get: L−1 ( 9 + s 4 − s2 ) = −92L−1 ( 2

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26

The Inverse Laplace Transform

We now know how to find Laplace transforms of "unknown" functions satisfying various initial- value problems. Of course, it's not the transforms of those unknown function which are usually

of interest. It's the functions, themselves, that are of interest. So let us turn to the general issue

of finding a functiony(t)when all we know is its Laplace transformY(s).

26.1 Basic Notions

On Recovering a Function from Its Transform

In attempting to solve the differential equation in example25.1, we got

Y(s)=4

s-3, which, since

Y(s)=L[y(t)]|sand4

s-3=L?4e3t???s, we rewrote as L [y(t)]=L?4e3t?.

From this, seemed reasonable to conclude that

y(t)=4e3t. But, what if there were another functionf(t)with the same transform as 4e3t? Then we could not be sure whether the abovey(t)should be 4e3tor that other functionf(t). Fortunately, someone has managed to prove the following:

Theorem 26.1 (uniqueness of the transforms)

Suppose

fandgare any two piecewise continuous functions on[0,∞)of exponential order and having the same Laplace transforms,

L[f]=L[g].

Then, as piecewise continuous functions,

f(t)=g(t)on[0,∞). 521

522The Inverse Laplace Transform

(You may want to quickly review the discussion of "equality of piecewise continuous func- tions" on page 493.) The proof of this theorem goes beyond our abilities at this time.1What is important is that it assures us that, if L [y(t)]|s=L?4e3t???s, then y(t)=4e3t, at least fort≥0. What about fort<0? Well, keep in mind that the Laplace transform of any functionf,

F(s)=L[f]|s=?

0 f(t)e-stdt, involves integration only over the positiveT-axis. The behavior offon the negativeT-axis has no effect on the formula forF(s). In fact,f(t)need not even be defined fort<0. So, even if theyexist, therecanbeno way torecoverthevaluesoff(t)onthenegativeT-axisfrom F(s). But that is not a real concern because we will just use the Laplace transform for problems over the positiveT-axis - problems in which we have initialvalues att=0 and want to know what happenslater. What all this means is that we are only interested in functions oftwitht≥0. That was hinted at when we began our discussions of the Laplace transform (see note 3 on page 473), but we did not make an issue of it to avoid getting too distracted by technical details. Now, with the inverse transform, requiringt≥0 becomes more of an issue. Still, there is no need to obsess about this any more than necessary, or to suddenly start including "fort≥0" with every formula oft. Let us just agree that the negativeT-axis is irrelevant to our discussions, and that in all formulas involvingt, it is assumed thatt≥0.

Example 26.1:Somewhere above, we have

y(t)=4e3t.

What we really mean is that

y(t)=4e3tfort≥0.

Wehavenoideawhat

y(t)isfort<0. Wedon'tevenknowwhether,inwhateverapplication this may have arisen, it makes sense to talk about y(t)fort<0, nor do we care.2

The Inverse Laplace Transform Defined

We can now officially define the inverse Laplace transform: Given a functionF(s), theinverse Laplace transform of F, denoted byL-1[F], is that functionfwhose Laplace transform isF.

1It is proven inOperational Mathematicsby Ruel Churchill, which was mentioned in an earlier footnote.

2For example: Whatify(t)denoted thetemperatureinacup of coffeetminutesafter beingpoured? Doesitmake

sense to consider the temperature of the coffee before it exists? (Answer this assuming you are not a Zen master.)

Basic Notions523

More succinctly:

f(t)=L-1[F(s)]|t??L[f(t)]|s=F(s). Our theoremonuniqueness(theorem26.1) (alongwithour understandingabout "alwaysassum- ingt≥0") assures us that the above definition forL-1[F]is unambiguous. In this definition, of course, we assumeF(s)can be given asL[f(t)]for some functionf.

Example 26.2:We have

L-1?4s-3?

?t=4e3t because4 s-3=L?4e3t???s.

Likewise, since

L?t3???s=6s4,

we have t3=L-1?6s4? ?t.

The fact that

f(t)=L-1[F(s)]|t??L[f(t)]|s=F(s) means that any table of Laplace transforms (such as table 24.1 on page 480) is also a table of inverse Laplace transforms. Instead of reading off theF(s)for eachf(t)found, read off the f(t)for eachF(s). As you may have already noticed, we take inverse transforms of "functions ofsthat are denoted by upper case Roman letters" and obtain "functions oftthat are denoted by the cor- responding lower case Roman letter". These notational conventions are consistent with the notational conventions laid down earlier for the Laplace transform. We should also note that the phrase "inverse Laplace transform" can refer to either the `inverse transformed function'for to the process of computingffromF. By the way, there is a formula for computing inverse Laplace transforms. If you must know, it is L -1[F(s)]|t=1

2π?

et(ξ+iy)F(ξ+iy)dy. Here, the integral is over a line in the complex plane, andξis a suitably chosen positive value.3 Don't pretend to understand it, and don't try to use it until you've had a course in complex variables. Besides, it is not nearly as useful as a good tableof transforms.

3again, see the text by Churchill

524The Inverse Laplace Transform

26.2 Linearity and Using Partial Fractions

Linearity of the Inverse Transform

The fact that the inverse Laplace transform is linear follows immediately from the linearity of the Laplace transform. To see that, let us considerL-1[αF(s)+βG(s)]whereαandβare any two constants andFandGare any two functions for which inverse Laplace transforms exist. Following our conventions, we'll denote those inverse transforms byfandg. That is, f(t)=L-1[F(s)]|tandg(t)=L-1[G(s)]|t. Remember, this is completely the same as stating that L [f(t)]|s=F(s)andL[g(t)]|s=G(s). Because we already know the Laplace transform is linear, we know L This, along the definition of the inverse transform and the above definitions offandg, yields L Redoing these little computations with as many functions and constants as desired then gives us the next theorem: Theorem 26.2 (linearity of the inverse Laplace transform) The inverse Laplace transform transform is linear. That is,

L-1[c1F1(s)+c2F2(s)+ ··· +cnFn(s)]

=c1L-1[F1(s)]+c2L[F2(s)]+ ··· +cnL[Fn(s)] when eachckis a constant and eachFkis a function having an inverse Laplace transform. Let's now use the linearity to compute a few inverse transforms.

Example 26.3:Let's find

L-1?1s2+9?

?t.

We know (or found in table 24.1 on page 480) that

L-1?3s2+9?

?t=sin(3t), which is almost what we want. To use this in computing our desired inverse transform, we will combine linearity with one of mathematics' oldest tricks (multiplying by

1with, in this

case,

1=3/3):

L-1?1s2+9?

?t=L-1?13·3s2+9? ?t=13L-1?3s2+9? ?t=13sin(3t).

Linearity and Using Partial Fractions525

The use of linearity along with `multiplying by 1' will be used again and again. Get used to it. Example 26.4:Let's find the inverse Laplace transform of 30
s7+8s-4.

We know

L-1?6!s7?

?t=t6andL-1?1s-4? ?t=e4t. So,

L-1?30s7+8s-4?

?t=30L-1?1s7? ?t+8L-1?1s-4? ?t =30L-1?1

6!·6!s7?

?t+8e4t 30

6!L-1?6!s7?

?t+8e4t=306·5·4·3·2t6+8e4t, which, after a little arithmetic, reduces to

L-1?30s7+8s-4?

?t=124t6+8e4t.

Partial Fractions

When using the Laplace transform with differential equations, we often get transforms that can be converted via `partial fractions' to forms that are easily inverse transformed using the tables andlinearity,asabove. Thismeansthatthegeneralmethod(s)of partialfractionsareparticularly important. By now, you should well-acquainted with using partial fractions - remember, the basic idea is that, if we have a fraction of two polynomials Q(s) P(s) andP(s)can be factored into two smaller polynomials

P(s)=P1(s)P2(s),

then two other polynomialsQ1(s)andQ2(s)can be found so that Q(s) Moreover, if (as will usually be the case for us) the degree ofQ(s)is less than the degree of P(s), then the degree of eachQk(s)will be less than the degree of the correspondingPk(s). You probably used partial fractions to compute some of the integrals in the earlier chapters of this text. We'll go through a few examples to both refresh our memories of this technique, and to see how it naturally arises in using the Laplace transform to solve differential equations.

526The Inverse Laplace Transform

!?Example 26.5:In exercise 25.1e on page 517, you found that the Laplace transform of the solution to y??+4y=20e4twithy(0)=3andy?(0)=12 is

Y(s)=3s2-28(s-4)?s2+4?.

The partial fraction expansion of this is

Y(s)=3s2-28(s-4)?s2+4?=As-4+Bs+Cs2+4

for some constantsA,BandC. There are many ways to find these constants. The basic method is to "undo" the partial fraction expansion by getting a common denominator and adding up the fractions on the right:

3s2-28

(s-4)?s2+4?=As-4+Bs+Cs2+4

A?s2+4?

(s-4)?s2+4?+(s-4)(Bs+C)(s-4)?s2+4? (A+B)s2+(C-4B)s+4(A-C) (s-4)?s2+4?. Cutting out the middle and canceling out the common denominator leads to the equation which, in turn, means that our constants satisfy the three bythree system 3=A+B

0=C-4B

-28=4A-4C This is relatively simple system. Solving it however you wish, you obtain

A=1andB=2andC=8.

Hence

Y(s)=As-4+Bs+Cs2+4=1s-4+2s+8s2+4,

and y(t)=L-1[Y(s)]|t=L-1?1s-4+2s+8s2+4? ?t =L-1?1 s-4? ?t+2L-1?ss2+4? t+8L-1?1s2+4? ?t =e4t+2L-1?s s2+22? t+8·12L-1?2s2+22? ?t =e4t+2cos(2t)+4sin(2t)

Linearity and Using Partial Fractions527

!?Example 26.6:In example 25.2 on page 505 we obtained

Y(s)=16(s-2)(s2-7s+12)+6s-38s2-7s+12

and, equivalently,

Y(s)=6s2-50s+92(s-2)?s2-7s+12?

as the Laplace transform of the solution to some initial-value problem. While we could find partialfractionexpansionsfor eachtermofthefirstexpressionabove, it willcertainlybemore convenient to simply find the single partial fraction expansion for the second expression for Y(s). But before attempting that, we should note that one factor in the denominator can be further factored, s2-7s+12=(s-3)(s-4), giving us

Y(s)=6s2-50s+92(s-2)(s-3)(s-4).

Now we can seek the partial fraction expansion of

Y(s):

6s2-50s+92

(s-2)(s-3)(s-4)=As-2+Bs-3+Cs-4 =A(s-3)(s-4)+B(s-2)(s-4)+C(s-2)(s-3) (s-2)(s-3)(s-4). Cutting out the middle and canceling out the common denominator leaves

6s2-50s+92

=A(s-3)(s-4)+B(s-2)(s-4)+C(s-2)(s-3) .(26.1) Rather than multiplying out the right side of this equation and setting up the system thatA, BandCmust satisfy for this equation to hold (as we did in the previous example), let find these constants after making clever choices for the value of sin this last equation.

Letting

s=2in equation (26.1):

6?22?-50·2+92

=A(2-3)(2-4)+B(2-2)(2-4)+C(2-2)(2-3) ??16=2A+0B+0C??A=8

Letting

s=3in equation (26.1):

6?32?-50·3+92

=A(3-3)(3-4)+B(3-2)(3-4)+C(3-2)(3-3) ?? -4=0A-B+0C??B=4

528The Inverse Laplace Transform

Lettings=4in equation (26.1):

6?42?-50·4+92

=A(4-3)(4-4)+B(4-2)(4-4)+C(4-2)(4-3) ?? -12=0A+0B+2C??C= -6

Combining the above results, we have

Y(s)=6s2-50s+76(s-2)(s-3)(s-4)

A s-2+Bs-3+Cs-4=8s-2+4s-3-6s-4.

Hence,

y(t)=L-1[Y(s)]|t=L-1?8s-2+4s-3-6s-4? ?t =8L-1?1 s-2? ?t+4L-1?1s-3? ?t-6L-1?1s-4? ?t =8e2t+4e3t-6e4t Do recallhow todeal withrepeatedfactorsin thedenominator. In particular,if your denom- inator has factors of the form (s+c)nor?s2+bs+c?n for some positive integernand constantsbandc, then the corresponding partial fraction expansions are A 1 (s+c)n+A2(s+c)n-1+A3(s+c)n-2+ ··· +Ans+c and A 1s+B1 ?s2+bs+c?n+A2s+B2?s2+bs+c?n-1+A3s+B3?s2+bs+c?n-2+ ··· +Ans+Bns2+bs+c, respectively.

Example 26.7:The partial fraction expansion of

Y(s)=2s2(s-6)3

is of the formA (s-6)3+B(s-6)2+Cs-6.

To find the constants

A,BandC, we proceed as in the previous examples:

2s2 (s-6)3=A(s-6)3+B(s-6)2+Cs-6 A

A+B(s-6)+C(s-6)2

(s-6)3.

Inverse Transforms of Shifted Functions529

So we must have

2s2=A+B(s-6)+C(s-6)2.

The value of

Acan be easily found by lettings=6in this equation, and the values ofB andCcan be found by lettings=6after taking derivatives of both sides of this equation. Or we can multiply out the right side and rewrite the left sidemore explicitly, obtaining

2s2+0s+0=Cs2+(B-12C)s+(A-6B+36C).

This tells us that the constants can be obtained by solving the system C=2

B-12C=0

A-6B+36C=0

In either case, you will discover that

A=72,B=24andC=2.

Thus,

Y(s)=2s2(s-6)3

A (s-6)3+B(s-6)2+Cs-6 72
(s-6)3+24(s-6)2+2s-6. In the next section, we will discuss an easy way to find the inverse transform of each of the terms in this partial fraction expansion.

26.3 Inverse Transforms of Shifted Functions

transform. Of particular value to us is the first shifting identity L ?eatf(t)???s=F(s-a) whereF=L[f(t)]andais any fixed real number. In terms of the inverse transform, this is L -1[F(s-a)]|t=eatf(t). wheref=L-1[F(s)]andais any fixed real number. Viewed this way, we have a nice way to find inverse transforms of functions that can be written as "shifts" of functions in our tables.

Example 26.8:Consider

L-1?1(s-6)3?

530The Inverse Laplace Transform

Here, the `shift' is clearly bya=6, and we have, by the above identity,

L-1?1(s-6)3?

?t=F-1[F(s-6)]|t=e6tf(t).(26.2) We now need to figure out thef(t)from the fact that

F(s-6)=1(s-6)3.

Letting

X=s-6in this equation, we have

F(X)=1X3.

Thus,

F(s)=1s3,

and f(t)=L-1[F(s)]|t=L-1?1s3? ?t =L-1?1

2!·2!s2+1?

?t=12!L-1?

·2!s2+1?

?t=12t2. Plugging this back into equation (26.2), we obtain

L-1?1(s-6)3?

?t= ··· =e6tf(t)=e6t12t2=12t2e6t. In many cases, determining the shift is part of the problem. Example 26.9:Consider finding the inverse Laplace transform of 1 s2-8s+25. If the denominator could be factorednicely, we would use partialfractions. This denominator does not factor nicely (unlesswe use complex numbers). Whenthat happens, try "completing the square" to rewrite the denominator in terms of " s-a" for some constanta. Here, s2-8s+25=s2-2·4s+?42-42?+25 =s2-2·4s+42? (s-4)2-42+25=(s-4)2+9.

Hence,

L-1?1s2-8s+25?

?t=L-1?1(s-4)2+9? ?t =L-1[F(s-4)]|t=e4tf(t) (26.3) Again, we need to findf(t)from a shifted version of its transform. Here,

F(s-4)=1(s-4)2+9.

Additional Exercises531

LettingX=s-4in this equation, we have

F(X)=1X2+9,

which means the formula for

F(s)is

F(s)=1s2+9.

Thus, f(t)=L-1[F(s)]|t=L-1?1s2+9? ?t =L-1?1

3·3s2+9?

?t=13L-1?3s2+32? ?t=13sin(3t).

Plugging this back into equation (26.3), we get

L-1?1s2-8s+25?

?t= ··· =e4tf(t)=e4t13sin(3t)=13e4tsin(3t).

Additional Exercises

26.1.Using the tables (mainly, table 24.1 on page 480) or your own memory, find the inverse

Laplace transform for each of the following:

a. 1 s-6b.1 s+2c.1 s2 d.6 s4e.5 s2+25f.s s2+3π2

26.2.Using the tables and `linearity', find the inverse Laplace transform for each of the

following: a. 6 s+2b.1 s4c.3⎷s-8s-4 d.4s2-4 s5e.3s+1 s2+25f.1-e-4s s

26.3.Verifythe following inverseLaplacetransformsassumingωis any real constant:

a. L-1? s?s2+ω2?2? ?t=t2ωsin(ωt) b.L-1?

1?s2+ω2?2?

?t=12ω3[sin(ωt)-ωtsin(ωt)]

26.4.Solve each of the following initial-value problems using the Laplace transform:

a. y?+9y=0withy(0)=4 b.y??+9y=0withy(0)=4andy?(0)=6

532The Inverse Laplace Transform

26.5.Using the tables and partial fractions, find the inverse Laplace transform for each of the

following: a. 7s+5 (s+2)(s-1)b.s-1 s2-7s+12c.1 s2-4 d.3s2+6s+27 s3+9se.1 s3-4s2f.8s3 s4-81 g.5s2+6s-40 (s+1)?s2+10s+21?

26.6.Solve each of the following initial-value problems using the Laplace transform (and

partial fractions): a. y??-9y=0withy(0)=4andy?(0)=9 b.y??+9y=27t3withy(0)=0andy?(0)=0 c.y??+8y?+7y=165e4twithy(0)=8andy?(0)=1

26.7.Using the translation identity (and the tables), find the inverse Laplace transform for

each of the following: a. 1 (s-7)5b.1 s2-6s+45c.s s2-6s+45d.1⎷s+2

26.8.Using the Laplace transform with the translation identity,solve the following initial-

value problems: a. y??-8y?+17y=0withy(0)=3andy?(0)=12 b.y??-6y?+9y=e3tt2withy(0)=0andy?(0)=0 c.y??+6y?+13y=0withy(0)=2andy?(0)=8 d.y??+8y?+17y=0withy(0)=3andy?(0)= -12

26.9.Using the Laplace transform, solve the following initial-value problems:

a. y??=etsin(t)withy(0)=0andy?(0)=0 c.y??-9y=24e-3twithy(0)=6andy?(0)=2

26.10.The inverse transforms of the following could be computed using partial fractions.

Instead, find the inverse transform of each using the appropriate integration identity from section 25.3.quotesdbs_dbs6.pdfusesText_12