Using Mathematica to evaluate Laplace Transforms and

their inverses, plus Cauchy’s PV prescription

Laplace Transforms (straightforward in Mathematica)

Here are the examples discussed in class

LaplaceTransform@1, t, pD1

p

LaplaceTransform@t^n, t, pDp

-1-nGamma@1 + nD

LaplaceTransform@Exp@a tD, t, pD1

-a + p

LaplaceTransform@Sin@b tD, t, pDb

b2 + p2

The results for L.T. of derivatives also come out simply:

LaplaceTransform@y'@tD, t, pDp LaplaceTransform@y@tD, t, pD - y@0DLaplaceTransform@y''@tD, t, pDp2LaplaceTransform@y@tD, t, pD - p y@0D - y

¢@0D

Checking the solution to Sec. 8.9 #4 (solved in class using L.T.)

InverseLaplaceTransform@1 � Hp^2 + 1L^2 + p � Hp^2 + 1L, p, tD

Cos@tD +1

2

H-t Cos@tD + Sin@tDL

Alternatively, solve directly:

TrigExpand@DSolve@8y''@tD + y@tD � Sin@tD, y@0D == 1, y'@0D � 0<, y@tD, tDD

::y@tD ® Cos@tD -1

2

t Cos@tD +Sin@tD

2

>>

Convolutions

Convolutions are defined for functions with infinite range in Mathematica

To use for our functions (which vanish for t<0) need to append a unit step.

Plot@UnitStep@tD, 8t, -1, 1<, PlotStyle ® 8Red, Thick<, Exclusions ® NoneD

-1.0 -0.5 0.5 1.0

0.2

0.4

0.6

0.8

1.0

Convolve[ ] has the form Convolve[ f[x], g[x], x, y (new variable)]

ConvolveB1

x4 + 1

, x, x, yF

Π y

2

Which we see is equivalent to,

IntegrateB1

Hy - xL4 + 1

* x, 8x, -¥, ¥<, Assumptions ® Im@yD � 0F

Π y

2

The order of the functions in a convolution doesn’t matter:

IntegrateB1

x4 + 1

* Hy - xL, 8x, -¥, ¥<, Assumptions ® Im@yD � 0F

Π y

2

Convolution needed for Sec. 8.10 #6

conv1 = Convolve@Exp@-a tauD UnitStep@tauD, Sinh@b tauD UnitStep@tauD � b, tau, tDHb ã-a t - b Cosh@b tD + a Sinh@b tDL UnitStep@tD

b Ha2 - b2LHere is the result we found in lecture notes:

conv2 = UnitStep@tD HExp@-a tD � H2 bLLHHExp@Ha + bL tD - 1L � Ha + bL + H1 - Exp@Ha - bL tDL � Ha - bLL

ã-a t J 1-ãHa-bL t

a-b+

-1+ãHa+bL t

a+bN UnitStep@tD

2 b

2 RevisedLT.nb

conv2 �� FullSimplify

Hb ã-a t - b Cosh@b tD + a Sinh@b tDL UnitStep@tDb Ha2 - b2L

They agree!

FullSimplify@conv1 - conv2D0

Inverse Laplace Transforms (with no pain)

Examples discussed in class, done using Bromwich integral:

InverseLaplaceTransform@1 � Hp - aL, p, tDãa t

InverseLaplaceTransform@1 � Hp^2 Hp + 1LL, p, tD-1 + ã

-t+ t

Sometimes you’ll need to FullSimplify a result from an InverseLaplaceTransform[ ] to get back what you

started with:

LaplaceTransform@Sin@tD Exp@tD, t, sD1

1 + H-1 + sL2

InverseLaplaceTransform@%, s, tD

-1

2

ä ãH1-äL t I-1 + ã

2 ä tM

% �� FullSimplify

1

1 + H-1 + sL2

(Cauchy) Principle Value Integral

Typically Mathematica will not let you integrate over a pole as the integral diverges at that point

RevisedLT.nb 3

PlotB1

x - 2, 8x, 0, 3<F

0.5 1.0 1.5 2.0 2.5 3.0

-5

5

IntegrateB1

x - 2, 8x, 0, 3<F

Integrate::idiv : Integral of

1

-2 + x

does not converge on 80, 3<. �

à0

3 1

-2 + x

âx

If we wanted to make sense of this integral somehow, we can use the Cauchy principle value of the

integral where you integrate excluding the pole and add the parts together.

IntegrateB1

x - 2, 8x, 0, 2 - a<, Assumptions ® a > 0F

LogBa2

F

IntegrateB1

x - 2, 8x, 2 + a, 3<, Assumptions ® 1 > a > 0F

-Log@aD

LimitBLogBa

2

F - Log@aD, a ® 0F

-Log@2DMathematica has this as the built in option “PrincipleValue”

IntegrateB1

x - 2, 8x, 0, 3<, PrincipalValue ® TrueF

-Log@2DThis also works for other integrals which are improper.

Consider a very simple example:

4 RevisedLT.nb

Integrate@x, 8x, -¥, ¥<D

Integrate::idiv : Integral of x does not converge on 8-¥, ¥<. �

à-¥

¥

x âx

This divergese because improper integrals are defined as:

à-¥

¥

f HxL â x = lima®¥

à-a

0

f HxL â x +limb®¥

à0

b

f HxL â x

and,

Integrate@x, 8x, 0, b<Db2

2

LimitBb2

2

, b ® ¥F

¥

The principle value in this case is defined by:

lima®¥

à-a

a

f HxL â x

Integrate@x, 8x, -¥, ¥<, PrincipalValue ® TrueD0

The result vanishes because the integrand is odd.

Recap of PV derivation done in class:

Often to do a principle value integral by hand you would use a contour integral, for example

IntegrateBExp@I * xD

x

, 8x, -¥, ¥<, PrincipalValue ® TrueF

ä Π

The contour we would use is:

RevisedLT.nb 5

ShowB:ListPlot@880, 0<<, PlotStyle ® PointSize@.03D,PlotRange ® 88-2, 2<, 8-2, 2<<, AspectRatio ® 1D,

GraphicsB:Text@C, 81.5, 1.5<D, Text@S, 8.25, .25<D, Text@r, 8.09, .13<D,

Text@R, 8-.7, .9<D, LineB:80, 0<, :.25

2

,

.25

2

>>F, LineB:80, 0<, :-2

2

,

2

2

>>F,

Line@88-2, 0<, 8-.25, 0<<D, Line@88.25, 0<, 82, 0<<D,Arrow@881.5, 1<, 81.3, 1.3<<D, Arrow@88-1.25, .1<, 8-.75, .1<<D,Circle@80, 0<, .25, 80, Π<D, Circle@80, 0<, 2, 80, Π<D>F>F

C

S

r

R

-2 -1 1 2

-2

-1

1

2

Where we care about the integral of the part over the real line in the limit as as r ® 0 and R® ¥

No poles are contained in the contour so we know:

limR®¥

r® 0

àC

+ àS

+à-R

-r

+àr

R

= 0

The C integral vanishes by Jordan’s lemma so:

limR®¥

r® 0

à-R

-r

+àr

R

= PV à-¥

¥

= -limr® 0

àS

The integral S is simply half what you would get for a full circle, integrated clockwise (so you get a -

sign) around the pole:

- -äΠ * ResidueBExp@I * zD

z

, 8z, 0<F

äΠ

A semicircle being half the value of the full circle is a lemma to the residue theorem which can be

proven by looking at the Laurent series about the pole and bounding non-pole term.

6 RevisedLT.nb

A semicircle being half the value of the full circle is a lemma to the residue theorem which can be

proven by looking at the Laurent series about the pole and bounding non-pole term.

To show it explicitly in this case:

z=r eiΘ, dz=i z dΘ

àΠ

0

i ei r e

iΘ

â Θ

As r ® 0:

àΠ

0

i â Θ = -iΠ

RevisedLT.nb 7