# LT1. LAPLACE TRANSFORM by zuu19905

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```									18.034 - Honors Differential Equations, Spring 2007                                              Dr. Vera Mikyoung Hur

LT1. LAPLACE TRANSFORM

The Laplace transform of the function f , deﬁned for all real numbers t � 0, is the function F (s)
deﬁned as
� ∞                  � b
st
(1)                    F (s) = L[f (t)](s) =     e f (t)dt = lim      est f (t)dt
0−                   a→0+ −a
b→∞

provided that the limit exists for all sufﬁciently large s. The lower limit 0− ensures the inclusion
of the unit impulse function (or the Dirac function)∗. The parameter s is in general considered to
be complex.
As an integral operator, the Laplace transform forms an important part of functional analysis.
It is named in honor of Pierre-Simon Laplace, who used the transfrom in his work on probability
theory.

Functions of exponential type. A real- or complex-valued function f (t) is said to be of exponential
type, and we write f ∈ E if there are positive constants A and B, depending on f such that:
(a) f (t) is piecewise continuous on [0, ∞) and f (t) is bounded on the set of jump discontinuities
where it is deﬁned,
(b) |f (t)| � AeBt at all points t ∈ [0, ∞) where it is deﬁned.

If f ∈ E and s > B then
�   ∞                       �   ∞
−st                                           A
e     |f (t)|dt �           e−st AeBt dt =       .
0−                        0−                       s−B
This suggests that the integral in (1) exists for s > B, and thus, if f ∈ E then its Laplace transform
2
exists. (Note that et does not possess a Laplace transform.) Moreover, if f is continuous on [0, ∞)
and f � ∈ E then f ∈ E and
(2)                                              L[f ](s) = sL[f ](s) − f (0−).
This relation enables us to transform an initial value problem of a differential equation for y into
an algebraic equation for Ly, which is much easier to solve. Once Lf is found, the inverse Laplace
transform reverts back to the solution f to the differential equation. The theory takes its form
from a symbolic method developed by the English engineer Oliver Heaviside. It necessitates the
following justiﬁcations.
Theorem 1. If y is a solution of p(D)y = f on [0, ∞) with f ∈ E then y and its derivative up to the order
of p(D) are also in E.
Its proof is left as an exercise.
Theorem 2 (Uniqueness Theorem). If f and g are functions of class E and their Laplace transforms
agree for all large s, then f (t) = g(t) at all points t � 0 where both functions are continuous.
The proof depends on another theorem which is of independent interest.

∗
Kent H. Lundberg, Haynes R. Miller, and David L. Trumper, Initial conditions, generalized functions, and the
Laplace transform, IEEE Control Systems magazine, 27 (2007) pp. 22-35.
1
Theorem 3. If q is continuous for 0 � x � 1, then
� 1
xn q(x)dx = 0, for n = 0, 1, 2, . . .               implies          q(x) = 0    for 0 � x � 1.
0
Proof of Theorem 2. Given f and g as in the statement of Theorem 1, let u, v, w with their Laplace
transforms U, V, W , respectively, be deﬁned by
� t
u(t) = f (t) − g(t),   v(t) =     u(τ )dτ,  w(t) = e−ct v(t).
0
By assumption, then, U (s) = 0 for all large s, and our goal is to prove that u(t) = 0 at all points
of continuity. Note that V (s) = U (s)/s = 0 for large s. Accordingly, if v(t) = 0 then by the
fundamental theorem of calculus follows that u(t) = 0 at all points of continuity. Hence, we will
work with the continuous function v rather than u.
For the constant c large enough, by the shift theorem of the Laplace transform, W (s) = V (s +
c) = 0 for all s > 0 (not only for large s). Since u ∈ E follows v ∈ E, and for large c, it yields
(3)                                                lim w(t) = 0.
t→∞

By the change of vaiable x = e−t we obtain that
� ∞              �                   1
−st
W (s) =      e w(t)dt =                           xs−1 w(− ln x)dx.
0                        0
Let                                      �
w(− ln x)               for 0 < x � 1
q(x) =                                        .
0                       for x = 0
Equation (3) guarantees that q is continuous for 0 � x � 1. Since W (s) = 0 for all s � 0, we
certainly have W (s) = 0 for s = 1, 2, 3, . . . . Then the assertion follows from Theorem 2. �
The proof of Theorem 3 is found at the end of this note.
Example 4. Here we consider the initial value problem
y �� + 4y = 0,      y(0) = 5, y � (0) = 6.
Taking the transform gives
s2 Ly − 5s − 6 + 4Ly = 0.
Hence,
5s + 6      s      2
Ly =            =5 2   +3 2   ,
s2 + 4   s +4   s +4
and y(t) = 5 cos 2t + 3 sin 2t.
Further application of the Laplace transform to solve initial value problems of differential equa­
tions and useful properties of the Lapace transform, the Laplace transforms of frequently used
functions are found in many standard textbooks on differential equations, for instance, [2].

The Laplace transform is particularly useful in dealing with discontinuous inputs (closing of a
switch) and with periodic functions (rectiﬁed sine waves).
A simple discontinuity at c has the left and right limits
lim f (t) = f (c−),            lim f (t) = f (c+)
t→c−                           t→c+

but they are not equal. A function f is piecewise continuous if it is deﬁned and continuous except at
ﬁnitely many points and all of its discontinuities are simple.
2
Theorem 5. If f � ∈ E and f is continuous except a simple discontinuity at a single value t = c > 0, then
L[f � ](s) = sLf [s] − f (0+) − (f (c+) − f (c−))e−sc .
Sketch of the proof. The assertion follows by adding
� c                           � c
−st �        −st    c
e f (t)dt = e f (t)|0 + s     f (t)e−st dt,
0                             0
� ∞                           � ∞
−st �      −st    ∞
e f (t) = e f (t)|c + s        f (t)e−st dt.
c                                     c
�
The unit step function. The unit step function or the Heaviside function is deﬁned by
�
0,       t<0
(4)                                 u(t) =
1,       t � 0.
One of the main uses of the unit step function is in connection with the shift theorem
L[f (t − c)] = e−sc L[f (t)],             c�0
with an additional condition that f (t) = 0. Note that
e−cs
(5)                                  L[u(t − c)] =     ,     c � 0.
s
Example 6. A rectiﬁed sine wave is f (t) = | sin t|. The function
�
sin t,         0�t<π
f0 (t) =
0,             π � t < 2π.
agrees with f (t) on 0 � t < π and is zero elsewhere. It is easy to see that
f0 (t) = u(t)f (t) − u(t − π)f (t).
Taking the Laplace transform gives L[f0 ] = L[f ] − e−πs L[f ]. On the other hand,
1 + e−sπ
L[f0 ](s) =               .
1 + s2
Therefore,
1 1 + e−πs
L[f ] =                   .
1 + s2 1 − e−πs
We end this chapter by the proof of Theorem 3.
�
Proof of Theorem 3. Suppose q(x0 ) = 0 at some point in (0, 1). We may assume that q(x0 ) > 0.
Continuity then gives positive constants �, δ such that
|x − x0 | � 2δ    implies      q(x) � �.
Let m be an arbitrary positive integer and deﬁne
�   1
p(x) = 1 + 4δ 2 − (x − x0 )2 ,          Im =             pm (x)q(x)dx.
0
By the binomial theorem pm is a polynomial in x, and hence Im = 0 under the hypothesis of
Theorem 3.
On the other hand Im = J1 + J2 + J3 , where J1 , J2 , J3 are integrals over the part of [0, 1] in which
|x − x0 | < δ,     δ � |x − x0 | � 2δ,              |x − x0 | > 2δ,
3
respectively. On these three intervals we have
p(x) � 1 + 3δ 2 ,        p(x) � 1,          |p(x)| � 1
and also, if M is a sufﬁciently large constant,
q(x) � �,        q(x) � �,            |q(x)| � M.
It is easy to check that
J1 � (1 + 3δ 2 )m δ�,           J2 � 0,       J3 � −M.
The ﬁrst expression tends to inﬁnity as m → ∞, and hence IM → ∞. This contradicts that

Im = 0.                                                                             �

R EFERENCES
[1] Garret Birkhoff & Gian-Carlo Rota, Ordinary Differential Equations, 4th ed., New York: Wiley, 1978.
[2] William Boyce & Richard DiPrima, Elementary differential equations and boundary value problems, 7th ed., New York;
John Wiley, 2001.
[3] Ray Redheffer & Dan Port Differential Equations, theory and applications, Jones and Barlett publishers, Boston, 1991.

Problems.
1. If f (t) ∈ E, show that L[f (t)](s) → 0 as s → ∞.
2   √ 2
2. Showt that the Laplace transform of e−t /4 is πes erts.
3. (a) Let F (s) = L[f (t)](s). Suppose that f (t)/t has a limit as t → 0+. Prove that
� ∞
L[f (t)/t](s) =       F (u)du.
s
sin t
(b) Use part (a) to ﬁnd the Laplace transform of f (t) =                 t .
� �       ��
4. Find L−1 ln s+a [t]. (Hint. F � (s) = −L[tf (t)].)
s−a

5. Let f (t) = t2 /2 for 0 < t � 1 and f (t) = (t − 2)2 /2 for 1 � t � 2, and t(t) = 0 elsewhere. Sketch
the graph of f (t). Find the Laplace transforms of f , f � and f �� . (Answers.
1 − e−2s 2e−s              1 − e−2s 2e−s                 1 − e−2s
L[f ] =      − 2 , L[f � ] =           −       , L[f �� ] =           .)
s3       s                 s2        s                   s
6. (This problem requires knowledge of the geometric series for 1/(1 − r). ) Reason that
L[f ]
= L[u(t)f (t) + u(t − c)f (t − c) + u(t − 2c)f (t − 2c) + · · · ].
1 − e−sc

4

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