Provides a mathematically rigorous, yet readily accessible treatment of applications of Laplace transforms. Includes the Dirac delta function given a completely justifiable treatment in the context of the Riemann-Stieltjes integral. DLC: Laplace transformation.
whose behaviour depended on properties of the Lebesgue integral.
established to justify this operation.
Another delicate problem in the literature has been the application of the Laplace transform to the so-called Dirac delta function.
entirely suited to this level of exposition.
variables is given in Chapter 3.
equation where y is the unknown function of the variable x.
conducted by Abel and many others.
calling it the Laplace transformation, in his work on theta functions.
1912), and bears many similarities to the Laplace transform method.
for γ lying to the right of all the singularities of the function x.
and to Betty Fong for turning my scribbling into a text.
1.1 The Laplace Transform . . . . . . . . . . .
1.2 Convergence . . . . . . . . . . . . . . . . .
1.3 Continuity Requirements . . . . . . . . . .
1.4 Exponential Order . . . . . . . . . . . . . .
1.5 The Class L . . . . . . . . . . . . . . . . . .
1.7 Inverse of the Laplace Transform . . . . .
1.8 Translation Theorems . . . . . . . . . . . .
Laplace Transform . . . . . . . . . . . . . .
1.10 Partial Fractions . . . . . . . . . . . . . . .
2.1 Gamma Function . . . . . . . .
2.2 Periodic Functions . . . . . . . .
2.3 Derivatives . . . . . . . . . . . .
2.5 Dirac Operator . . . . . . . . . .
Convolution . . . . . .
3.1 Complex Numbers . . . . . . . .
3.2 Functions . . . . . . . . . . . . .
3.3 Integration . . . . . . . . . . . .
3.4 Power Series . . . . . . . . . . .
initial conditions that describe the state of the system at time t  0.
properties of the Laplace transform.
notation s  x + iy.
functions f  f (t) and generates a new function, F(s)  L f (t) .
the same calculation, with Re(s) > 0, would have given L(1)  1/s.
(x cos yt + y sin yt) + i(x sin yt − y cos yt) .
which equals the left-hand side, and (1.5) follows.
killing off the limit term in (1.3).
fashion is readily handled as follows.
is given in the ﬁgures below.
bound as τ → ∞.
M < ∞ for 0 ≤ t < ∞.
uniformly if Re(s) ≥ x0 > 0.
(c) Show that F(s)  L f (t) → 0 as Re(s) → ∞.
2. Let f (t)  et on [0, ∞).
(a) Show that F(s)  L(et ) converges for Re(s) > 1.
(b) Show that L(et ) converges uniformly if Re(s) ≥ x0 > 1.
τ → ∞, τ  > τ.
(c) Show that F(s)  L(et ) → 0 as Re(s) → ∞.
does not exist for any value of s.
mean that t → t0 from the left and right, respectively (Figure 1.2).
neither limt→3− f (t) nor limt→3+ f (t) exists.
has a jump discontinuity at t  0 and is continuous elsewhere.
have a jump discontinuity at t  0.
transform deﬁned will have the following property.
in Example 1.7 is piecewise continuous on [0, ∞).
for ﬁnite constants Mi .
a > 0, n  0, 1, 2, . . .
+ 1, for t ≥ 0, a > 0, where [x]  greatest integer ≤ x.
of growth can be made explicit.
2. Show that f (t)  t n has exponential order α for any α > 0, n ∈ N.
3. Prove that the function g(t)  et does not have exponential order.
0 < t < t0 .
hence converges) for Re(s) > α.
The same calculation holds for a complex and Re(s) > Re(a).
is, the Laplace transform exists for all s in some right half-plane (cf.
f (t)  2t et cos(et ).
Then f (t) is continuous on [0, ∞) but not of exponential order.
nevertheless possesses a Laplace transform. See also Remark 2.8.
f (t)  √ .
context of the gamma function.
since f (t) → ∞ as t → 0+ , that is, t  0 is not a jump discontinuity.
1. Consider the function g(t)  t et sin(et ).
(a) Is g continuous on [0, ∞)? Does g have exponential order?
(b) Show that the Laplace transform F(s) exists for Re(s) > 0.
2. Without actually determining it, show that the following functions possess a Laplace transform.
for arbitrary constants c1 , c2 .
(f1 , f2 ∈ L).
−∞ < t < ∞.
and so the series diverges for all values of s.
inﬁnite series by term-by-term computation?
satisfy the hypothesis of the theorem.
overemphasized, as it is instrumental in the proofs of many results.
of this book (pp. 210–218) is the following.
and letting x → ∞ gives the result.
transform of any function f .
1. Find L(2t + 3e2t + 4 sin 3t).
4. Find L(3 cosh 2t − 2 sinh 2t).
9. Can F(s)  s/log s be the Laplace transform of some function f ?
need to know when the inverse transform is unique.
of points) does not alter the value of the Laplace (Riemann) integral.
is uniquely deﬁned and we can speak about the inverse, L−1 F(s) .
L and holds in the domain common to F and G.
there are other functions with the same L−1 F(s) .
of view are indistinguishable. When a  0, we will write ua (t)  u(t).
Another common notation for the unit step function ua (t) is u(t − a).
which is another variant of the unit step function.
Another interesting function along these lines is the following.
as shown in Figure 1.8.
1. Prove that L−1 is a linear operator.
for any null function N(t).
(d) How can part (c) be reconciled with Theorem 1.23?
Re(s) > 0, a > 0 .
(b) Write f (t) as an inﬁnite series of unit step functions.
series in (b), show that the same result as in (a) is attained.
s-domain and the second to a translation in the t-domain.
a real, Re(s) > a .
 L eat f (t) .
n  0, 1, 2, . . .
 e−2t cosh 3t − √ e−2t sinh 3 t.
In the ﬁrst step we have used the procedure of completing the square.
for F(s)  L f (t) , a ≥ 0.
This is just the function sin t, which gets “turned on” at time t  2.
(c) f (t)  uπ (t) cos(t − π).
n  1, 2, 3, . . . (s > α).
 L − tf (t) .
being uniformly convergent for s ≥ x0 > α.
for f (t)  L−1 F(s) . This formulation is also useful.
be integrated as well. Again the result is another Laplace transform.
existence of L f (t)/t is ensured by the hypotheses.
(d) L(t 2 sin ωt).
ﬁnd L−1 (e−a s ).
1  A(s − 3) + B(s − 2).
than n points are identically equal (Corollary A.8).
 −e2t + e3t .
ﬁnite sum of partial fractions.
A1 , A2 , . . . , An constants.
A1 , . . . , An , B1 , . . . , Bn constants.
P(s)/Q (s) has been represented by a partial fraction decomposition.
This can be achieved by several different methods.
A  −1, B  1 as before.
 −2 − t + 2et .
2s2  (As + B)(s − 1)2 + C(s2 + 1)(s − 1) + D(s2 + 1).
−1  B − C.
0  A − 2B + C.
 − cos t + et + tet .
Ai  lim (s − αi )F(s).
lim (s − αi ) F(s) eαi t .
purposes of solving all but PDEs (which we discuss in Chapter 5).
n  1, 2, 3, . . . .
bounded for all large values of t, the Laplace transform, L(t ν ), exists.
any elementary function (Figure 2.1).
ν > −1, s > 0.
(n + 1)  n!.
ν  0, −1, −2, . . .
(see Exercises 2.1, Question 1).
e−x log x dx  0.577215 . . .
is Euler’s constant. See also Exercises 2.1, Question 4.
|an | ≤ K r n .
 (n + ν + 1)  t ≤ n!
This also shows that f has exponential order.
translation theorem (1.27) and (2.5).
(d)  − 32 .
 t ν−1 log t.
s > 0, ν > 0.
is the Euler constant as in (2.6).
condition for periodicity as above for these functions as well.
multiple of this ﬁrst one.
x  Re(s) > 0 .
f (t)  ua (t) − u2a (t) + u3a (t) − u4a (t) + · · · .
that is, we can take the Laplace transform of f term by term.
(1 + e− ω ).
term by term and compare with Question 1(a).
 its Laplace transform term by term.
verify that indeed L f (t)  F(s).
L(f  ) is that it can be written in terms of L(f ).
L f  (t)  sL f (t) − f (0+ ).
 M e−(x−α)τ → 0 as τ → ∞.
L f  (t)  s L f (t) − f (0).
Example 2.9. Let us compute L(sin2 ωt) and L(cos2 ωt) from (2.18).
origin. This can be treated in the following way.
e−st f (t) dt .
L f  (t)  s L f (t) − f (0) − e−st1 f (t1+ ) − f (t1− ) .
e−stk f (tk+ ) − f (tk− ) .
exponential order, then it follows that the same is true of f itself .
t ≥ t0 , α  0.
is subsumed under this one.
 s2 L f (t) − s f (0) − f  (0).
In the general case we have the following result.
L f (n) (t)  sn L f (t) − sn−1 f (0+ ) − sn−2 f  (0+ ) − · · · − f (n−1) (0+ ).
n  0, 1, 2, . . . .
again by the ﬁrst translation theorem.
1. In Theorem 2.7, prove that f (0+ ) exists?
Re(s) > 0 , n  1, 2, 3, . . . .
by letting f (t)  sinh ωt and applying formula (2.20).
6. Write out the details of the proof of Theorem 2.12.
continuity cannot be replaced by piecewise continuity.
transform to ODEs will be found in ensuing sections.
y(0)  y (0)  0.
L(y ) + L(y)  L(1).
y  1 − cos t.
One may readily check that this is indeed the solution to the initialvalue problem.
the methods of variation of parameters or undetermined coefﬁcients).
results in what is called the transformed equation.
expression in the variable s.
conjunction with standard tables of Laplace transforms.
y(0)  y (0)  y (0)  0.
t − e + e.
line, −∞ < t < ∞, if f (t) has this domain.
that the input function f (t) can be discontinuous.
y(0)  0, y (0)  1.
sin(t − π) − (t − π) cos(t − π) .
again by the second translation theorem.
valid for −∞ < t < ∞.
t sin λt + cos λt + 1 − 2 sin λt.
(sin λt + λt cos λt) − λ sin λt + y (0) cos λt.
Systems of Differential Equations. Systems of differential equations can also be readily handled by the Laplace transform method.
We illustrate with a few examples.
 y, y(0)  1, z(0)  0.
i.e., s L(z)  L(y).
z  −y  sin t.
y(0)  −1, z(0)  2.
z  2et − t et .
compute the Laplace transform of an integral.
e f (t) dt .
x  Re(s) > α > 0.
by Example 1.34 (i). The function, Si(t), is called the sine integral.
sinh au du  2 (cosh at − 1).
Differential equations that involve integrals (known as integrodifferential equations) commonly arise in problems involving electrical circuits.
circuit problems throughout the sequel.
where L, R, E0 , and ω are constants. Find I  I(t) for t > 0 if I(0)  0.
for y(t) piecewise continuous on [0, ∞) and of exponential order α.
L ty(t)  −F  (s).
Suppose further that y (t) satisﬁes the hypotheses of the theorem.
 −sF  (s) − F(s).
 −s2 F  (s) − 2sF(s) + y(0).
− s F(s)  − 2 + 1.
µ(s)  e ( s −s)ds  s3 e−s /2 .
− + 3 es /2 .
which can be veriﬁed to be the solution.
There are pitfalls, however, of which the reader should be aware.
only the solution that is bounded at the origin.
ty + y + 2y  0.
1. Solve the following initial-value problems by the Laplace transform method.
(h) y − y  0, y(0)  1, y (0)  y (0)  y (0)  0.
2. Solve the boundary value problems.
y(0)  1, y λ  −1.
where L, R, E0 are positive constants.
(a) Find I(t) for t > 0 if I(0)  I0 > 0.
(b) Sketch a graph of the solution in (a) for the case I0 > E0 /R.
(c) Show that I(t) tends to E0 /R as t → ∞.
and L  1, R  3, C  12 , I(0)  1.
x(0)  0, y(0)  1, y (0)  −2.
7. Solve the following differential equations by the Laplace transform method.
required is an entirely new entity that is not really a function at all.
is just a natural extension of the conventional Riemann integral.
each subinterval [ti−1 , ti ] an arbitrary point xi with ti−1 ≤ xi ≤ ti .
f (xi )[ϕ(ti ) − ϕ(ti−1 )].
sums and we obtain the ordinary Riemann integral.
(see Protter and Morrey , Theorem 12.16).
since f (xj ) → f (a).
for constants c1 , c2 .
a contradiction for n sufﬁciently large (Figure 2.9).
However, there is an important relationship between the Riemann–Stieltjes and Riemann integrals under suitable conditions.
a function but a linear operator.
so that x(t) ≡ 1, t ≥ 0.
the unit step function compare with (2.32) .
uab (t) dt  1.
a (t)  lim uab (t).
f (t) uab (t) dt  f (a).
f (t) a (t) dt  f (a).
in the Riemann–Stieltjes sense of (2.33).
also achieved rigorously in (2.35).
L a (t)  e−as .
This is just the expression obtained in (2.37).
terms of the Riemann–Stieltjes integral.
L(δa )  e−as .
where x(t) is the displacement at time t ≥ 0, and k > 0 is a constant.
rest for t < 0.
x(0)  0, x (0)  v0 .
Solving this version yields the same results as above.
cos bt − sin bt .
cosh bt − sinh bt .
by x > 0, and upward displacement is shown by x < 0.
iii. F(t) represents all external impressed forces on m.
This equation is called the equation of motion.
the motion is called free; otherwise it is forced.
+ λ2 x  0.
x(t)  c e−bt cos( λ2 − b2 t − ϕ).
This represents the behavior of damped oscillation (Figure 2.15).
Let us apply a unit impulse force to the above situation.
which again is a case of damped oscillation.
x(0)  0, x (0)  0.
where L, and R are constants, and E(t) is the impressed voltage.
for m  1, a  2, k  1, x(0)  x (0)  0.
L−1 sF(s)  f  (t) + f (0) δ(t).
 δ(t) − 2ae−at .
by examining the behavior of L f (t) .
the conditions as in the derivative theorem (2.7), and F(s)  L f (t) .
tions), we know that L f  (t)  G(s) → 0 as s → ∞ (Theorem 1.20).
0  lim G(s)  lim s F(s) − f (0+ ) .
f (0+ )  lim s F(s).
where F(s)  L f (t) .
lim G(s)  lim s F(s) − f (0+ ).
 lim [f (τ) − f (0+ )].
lim f (t)  lim s F(s).
which of course exists if f and g are, say, piecewise continuous.
that is, the convolution is commutative.
(iii) f ∗ (g + h)  (f ∗ g) + (f ∗ h) (distributive property).
having reversed the order of integration.
 e − t − 1.
f (τ) g(u) du dτ.
e−st f (τ) g(t − τ) dt dτ.
e−st f (τ)g(t − τ) dt dτ.
(sin ωt − ωt cos ωt).
in evaluating inverse transforms that are products.
τ a−1 (t − τ)b−1 dτ.
ua−1 (1 − u)b−1 du.
since (1/2) > 0. See also Example 2.1.
The latter inequality for n  0, 1, 2, · · · can be veriﬁed by induction.
transform of J0 (at) term by term.