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MA 1506 Mathematics II

Tutorial 7The Laplace transformation

Groups: B03 & B08March 14, 2012

Ngo Quoc AnhDepartment of Mathematics

National University of Singapore

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Reviews of the Laplace transformation

Definition

The Laplace transformation

The Laplace transform of a function f(t), defined for all realnumbers t > 0, is the function F (s), defined by

F (s) = L(f(t)

)=

∫ +∞

0e−stf(t)dt.

Properties

Linearity;Transform of derivatives and integrals;s-Shifting and t-Shifting.

Some known transforms

Standard functions;The Heaviside function u(t− a);The Dirac delta function δ(t− a).

Application to the theory of ODEs.

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Question 1: Finding the Laplace transform of givenfunctions

Generally, the standard approach is to use definition.However, in order to simplify calculation (otherwise stated),we can make use of known properties for the Laplacetransform.(a) Since the given function t2e−3t is of the form ectf(t), wecan make use of the s-Shifting property, that is,

L(ectf(t)

)= F (s− c), where F (s) = L

(f(t)

).

Precisely,L(e−3t t2︸︷︷︸

f(t)

) = F (s+ 3).

It suffices to find L(t2). Since

L(tn) = n!

sn+1,

we find that F (t2) = 2s3

. Thus, L(e−3tt2) = 2

(s+ 3)2.

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Question 1: Finding the Laplace transform of givenfunctions

(b) In this question, we encounter L(f(t− a)u(t− a)) wheref is given and u is the Heaviside function. Remember thet-Shifting property saying that

L(f(t− a)u(t− a)

)= e−asF (s), where F (s) = L

(f(t)

).

In our context, clearly a = 2 which forces f(t− 2) = t.Hence f(t) = t+ 2. Therefore,

L(f(t)

)= L(t+ 2) = L(t) + L(2) = 1

s2+

2

s.

In conclusion,

L(tu(t−2)

)= L

((t+ 2︸ ︷︷ ︸f(t)

− 2︸︷︷︸a

)u(t−2))= e−2s

(1

s2+

2

s

).

NB: The Heaviside function is a cut-off function, see also .

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Question 2: Finding the inverse Laplace transform ofgiven functions

The best way to find the inverse Laplace transformL−1

(F (s)

)is to simplify F (s) as much as possible and is to

use some known Laplace transforms, check this for a listof selected Laplace transforms.(a) For the function F (s) = s

s2+10s+26, it is routine to write

F (s) =s− 5 + 5

(s+ 5)2 + 1.

Therefore, by denoting F1(s) =s−5s2+1

, we get that

L−1(F (s)) = L−1(F1(s− (−5))) = e−5tL−1(F1(s)).

To simplify F1(s), we first write F1(s) =s

s2+1− 5

s2+1. Then

L−1(F1(s)) = L−1(

s

s2 + 1

)−5L−1

(1

s2 + 1

)= cos t−5 sin t.

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Question 2: Finding the inverse Laplace transform ofgiven functions

(b) In this case, we aim to find L−1(e−2s 1+2s

s3

)which is of

the form L−1(e−asF (s)). In order to do this, we make useof the t-Shifting property, that is,

L−1(e−asF (s)) = f(t− a)u(t− a).

Before we apply that formula, it is clear to see thatF (s) = 1+2s

s3. Therefore, we can write

F (s) =1

s3+

2

s2.

Hence,

f(t) = L−1(F (s)) = L−1(

1

s3

)+ 2L−1

(1

s2

)=t2

2+ 2t.

Since a = 2, we get L−1(e−2s 1+2s

s3

)= ( t

2

2 − 2)u(t− 2).

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Question 3: Solving initial value problems for ODEs

To solve IVPs for ODEs,

we simply take the Laplace transform of both sides toget an algebraic equation for the Laplace transform ofthe solution;

once this algebraic equation is solved, taking the inverseLaplace transform leads us to the solution we need.

(a) For the equation y′ = tu(t− 2), we get by taking theLaplace transform both sides that

sY (s)− y(0)︸ ︷︷ ︸L(y′(t))

= e−2s(

1

s2+

2

s

)︸ ︷︷ ︸L(tu(t−2))

.

Therefore, thanks to y(0) = 4,

Y (s) = e−2s1 + 2s

s3+

4

s.

By taking the inverse transform, y(t) = ( t2

2 − 2)u(t− 2) + 4.

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Question 3: Solving initial value problems for ODEs

(b) For the equation y′′ − 2y′ = 4, we get after taking theLaplace transform both sides that

s2Y (s)− sy(0)− sy′(0)︸ ︷︷ ︸L(y′′(t))

−2(sY (s)− y(0)︸ ︷︷ ︸L(y′(t))

) =4

s︸︷︷︸L(4)

.

Thanks to y(0) = 1 and y′(0) = 0, we find that

Y (s) =s2 − 2s+ 4

s(s2 − 2s).

To find L−1(Y (s)

), we should write

Y (s) =1

s− 2− 2

s2.

Therefore,

y(t) = L−1(Y (s)) = L−1(

1

s− 2

)− 2L−1

(1

s2

)= e2t− 2t.

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Question 4: The Laplace transform for the function tf(t)

By definition, we need to prove∫ +∞

0e−sttf(t)dt = −

(∫ +∞

0e−stf(t)dt

)′.

Notice that the integral on the right in the above equation isthe so-called parameter-dependent integral. By the chainrule (keep in mind that ′ = d

ds), we get(∫ +∞

0e−stf(t)dt

)′=

∫ +∞

0(e−stf(t))′dt

= −∫ +∞

0te−stf(t)dt,

which concludes the proof.

Since L(sin t

)= 1

1+s2, we obtain

L(t sin t) = −(

1

1 + s2

)′=

2s

(1 + s2)2.

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Question 4: The Laplace transform for the function tf(t)

By taking the Laplace transform of both sides ofy′′ + y = cos t, we get

s2Y (s)− sy(0)− sy′(0)︸ ︷︷ ︸L(y′′(t))

+ sY (s)− y(0)︸ ︷︷ ︸L(y′(t))

=s

1 + s2︸ ︷︷ ︸L(cos t)

.

Thanks to the initial conditions, we obtain

Y (s) =s

(1 + s2)2.

Thus, from the previous calculation, there holds

y(t) =1

2t sin t.

NB: This is a particular case of the general solution

x(t) =F0

2mωt sin(ωt)

found in Chapter 2 when we solve mx+ kx = F0 cos(αt).

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Question 5: The effect of rogue waveIn view of the Newton second law, weget

Mx =Mg − ρA(d+ x)g,

representing the situation in the ab-sence of friction. We need an ex-tra term to account for the suddenforce exerted by the rogue wave, sayF. Since the force is exerted sud-dently, this suggests that we need aDirac delta function, so the force willbe proportional to δ(t− T ).

Gravity F=Mg

M(mass of the tanker)

Buoyancy F=ρA(d+x)g

ρ(density of seawater)

object

(horizontal cross-section area)

Tanker

A

dx(t)

Since P = mv, v = a, and F = ma, we find that

P =

∫ +∞

0Fdt.

Since∫ +∞0 δ(t− T )dt = 1, one can conclude that the force

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Question 5: The effect of rogue wave

F is nothing but F = −Pδ(t− T ). Why? By taking intoaccount, one gets that

Mx =Mg − ρA(d+ x)g − Pδ(t− T ),

which is equivalent to, thanks to Mg = ρAdg,

x = −ρAgM

x− P

Mδ(t− T ).

To solve the above ODE, we take the Laplace transform ofboth sides,

s2X(s)− sX(0)−X ′(0) = −ρAgM

X(s)− P

Me−Ts

which yields, since X(0) = 0 and X ′(0) = 0,

X(s) = − PM

e−Ts

s2 +

(√ρAgM

)2 = − P

ωM

ωe−Ts

s2 + ω2, ω =

√ρAg

M.

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Question 5: The effect of rogue wave

In order to find x(t) = L−1(X(s)), we can make use of thet-Shifting property as follows. First we have

x(t) = L−1(− P

ωM

ωe−Ts

s2 + ω2

)= − P

ωML−1

(ωe−Ts

s2 + ω2

).

By using f(t− a)u(t− a) = L−1(e−asF (s)) we get that

L−1(ωe−Ts

s2 + ω2

)= L−1

(ω

s2 + ω2

)︸ ︷︷ ︸

sin(ω ·)

(t− T )u(t− T ).

Thus,

x(t) = − P

ωMsin(ω(t− T ))u(t− T ).

As can be seen, before t = T , the ship is at rest. At t = T ,the ship is hit by rogue wave; then simple harmonic motionoccurs with the amplitude P

ωM .

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Question 6: The effect of rogue wave

Following the previous argument, we can derive the formulafor the force F, formed by the rogue wave, as follows

F = −Pτ

(u(t− T )− u(t− (T + τ))

).

Note that we need the scaling factor 1τ since

∫ +∞0 Fdt = P .

Our corresponding ODE reads as follows

x = −ρAgM

x− P

Mτ

(u(t− T )− u(t− (T + τ))

).

To solve this ODE, we take the Laplace transform of bothsides as usual, X(0) = X ′(0) = 0,

s2X(s) = −ρAgM

X(s)− P

Msτ

(e−Ts − e−(T+τ)s

)which yields

X(s) = − P

Msτ

e−Ts − e−(T+τ)s

s2 + ω2, ω =

√ρAg

M.

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Question 6: The effect of rogue wave

By using the L’Hopital rule, one sees that

limτ→0+

X(s) = − P

Ms(s2 + ω2)limτ→0+

e−Ts − e−(T+τ)s

τ

= − P

Ms(s2 + ω2)se−Ts

= − PM

e−Ts

s2 + ω2.

Hence, our expression for X(s) does indeed tend to thesame expression for X(s) as in Q5 if we let τ tend to 0.That’s as expected because the impulse should be like adelta function if τ is small. In this case,

x(t) = − P

Mτω

[L−1

(e−Ts

ω

s2 + ω2

)−

L−1(e−(T+τ)s

ω

s2 + ω2

)].

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Other transformation: The Fourier transformation

In physics and engineering, the Fourier transform plays animportant role.

The Fourier transformation

If the function f : R→ C is integrable, we can define theFourier transform by

f(ξ) = F(f) =∫ +∞

−∞f(x) e−2πixξdx, i =

√−1

and the inverse Fourier transform by

f(x) = F−1(f) =∫ +∞

−∞f(ξ) e2πiξxdξ, i =

√−1.

The motivation for the Fourier transform comes from thestudy of Fourier series where complicated functions arewritten as the sum of simple waves mathematicallyrepresented by sines and cosines.

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Other transformation: The Mellin transformation

The Mellin transform is an integral transform that may beregarded as the multiplicative version of the two-sidedLaplace transform.

The Mellin transformation

For a function f(t), defined for all real numbers t > 0, wecan define the Mellin transform by

M(f) =

∫ +∞

0xs−1f(x)dx.

By letting x = e−t, then dx = −e−tdt, the Mellin transformbecomes

M(f) =

∫ −∞+∞

e−(s−1)tf(e−t)(−e−tdt)

=

∫ +∞

−∞e−stf(e−t)dt = L(f(e−t)).

See: The Handbook of Formulas and Tables for Signal....