Powers D.L. Boundary Value Problems and Partial Differential Equations
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358 Chapter 5 Higher Dimensions and Other Coordinates
20. Observe that the function φ in Exercise 19 is the difference of two differ-
ent eigenfunctions of the 1×1 square (see Section 5.3) corresponding to
the same eigenvalue. Use this idea to construct other eigenfunctions for
the triangle
T of Exercise 19.
21. Let
T be the equilateral triangle in the xy-plane whose base is the interval
0 < x < 1ofthex-axis and whose sides are segments of the lines y =
√
3x
and y =
√
3(1 −x). Show that for n =1, 2, 3,...,thefunction
φ
n
(x, y) = sin
4nπy/
√
3
+sin
2nπ
x −y/
√
3
−sin
2nπ
x + y/
√
3
is a solution of the eigenvalue problem ∇
2
φ =−λ
2
φ in T , φ =0onthe
boundary of
T . What are the eigenvalues λ
2
n
corresponding to the func-
tion φ
n
that is given? [See “The eigenvalues of an equilateral triangle,”
SIAM Journal of Mathematical Analysis, 11 (1980): 819–827, by Mark A.
Pinsky.]
22. In Comments and References, Section 5.11, a theorem is quoted that re-
lates the least eigenvalue of a region to that of a smaller region. Confirm
the theorem by comparing the solution of Exercise 19 with the smallest
eigenvalue of one-eighth of a circular disk of radius 1:
1
r
∂
∂r
r
∂φ
∂r
+
1
r
2
∂
2
φ
∂θ
2
=−λ
2
φ, 0 <θ<
π
4
, 0 < r < 1,
φ(r, 0) = 0,φ
r,
π
4
=0, 0 < r < 1,
φ(1,θ)= 0, 0 <θ<
π
4
.
23. Same task as Exercise 22, but use the triangle of Exercise 21 and the
smallest eigenvalue of one-sixth of a circular disk of radius 1.
24. Show that u(ρ, t) = t
−3/2
e
−ρ
2
/4t
is a solution of the three-dimensional
heat equation ∇
2
u =
∂u
∂t
, in spherical coordinates.
25. For what exponent b is u(r, t) = t
b
e
−r
2
/4t
a solution of the two-
dimensional heat equation ∇
2
u =
∂u
∂t
? (Use polar coordinates.)
26. Suppose that an estuary extends from x = 0tox = a, where it meets the
open sea. If the floor of the estuary is level but its width is proportional
to x, then the water depth u(x, t) satisfies
1
x
∂
∂x
x
∂u
∂x
=
1
gU
∂
2
u
∂t
2
, 0 < x < a, 0 < t,
Miscellaneous Exercises 359
where g is the acceleration of gravity and U is mean depth. The tidal
motion of the sea is represented by the boundary condition
u(a, t) = U +h cos(ωt).
Find a bounded solution of the partial differential equation that satisfies
the boundary condition by setting
u(x, t) = U +y(x) cos(ωt).
(See Lamb, Hydrodynamics, pp. 275–276.)
27. Is there any combination of parameters for which the solution of Exer-
cise 26 does not exist in the form suggested?
28. If the estuary of Exercise 26 has uniform width but variable depth
h =Ux/a, then the equation for u is
∂
∂x
x
∂u
∂x
=
a
gU
∂
2
u
∂t
2
, 0 < x < a, 0 < t,
subject to the same boundary condition as in Exercise 26. Find a
bounded solution in the form suggested. (See Eq. (1) of Section 5.8.)
29. The equation for radially symmetric waves in n-dimensional space is
1
r
n−1
∂
∂r
r
n−1
∂u
∂r
=
1
c
2
∂
2
u
∂t
2
where r is distance to the origin. Find product solutions of this equation
that are bounded at the origin.
30. Show that the equation of Exercise 29 has solutions of the form
u(r, t) = α(r)φ(r − ct)
for n = 1andn = 3. [See “A simple proof that the world is three-
dimensional” by Tom Morley, SIAM Review, 27 (1985): 69–71.]
31. A certain kind of chemical reactor contains particles of a solid catalyst
and a liquid that reacts with a gas bubbled through it. M. Chidambaran
[“Catalyst mixing in bubble column slurry reactors,” Canadian Journal
of Chemical Engineering, 67 (1989): 503–506] uses the following problem
to model the catalyst concentration C in a cylindrical reactor:
D
r
1
r
∂
∂r
r
∂C
∂r
+D
z
∂
2
C
∂z
2
+U
∂C
∂z
= 0, 0 < r < R, 0 < z < L,
∂C
∂r
(R, z) = 0, 0 < z < L.
360 Chapter 5 Higher Dimensions and Other Coordinates
Here, D
r
and D
z
are the diffusion constants in the radial and axial di-
rections, respectively. The term containing ∂C/∂z represents physical
movement of particles at speed U.
Show that the change of variables ρ = r/R, ζ = z/L, u(ρ, ζ ) = C(r, z)
leads to the equivalent equations
b
1
ρ
∂
∂ρ
ρ
∂u
∂ρ
+
∂
2
u
∂ζ
2
+p
∂u
∂ζ
= 0, 0 <ρ<1, 0 <ζ <1,
∂u
∂ρ
(1,ζ)= 0, 0 <ζ <1,
and identify the parameters b and p.
32. When a boundedness condition at ρ = 0isadded,productsolutionsof
the foregoing equation are found to have the form u(ρ, ζ ) = R(ρ)Z(ζ ):
R
0
(ρ) = 1, Z
0
(ζ ) =
e
−pζ
1
,
R
n
(ρ) = J
0
(λ
n
ρ), Z
n
(ζ ) =
e
m
1
ζ
e
m
2
ζ
,
where m
1
< 0 < m
2
are the roots of the equation m
2
+pm −λ
2
n
b =0and
λ
n
is chosen to satisfy J
0
(λ
n
) =0.
a. Check the details of the solution.
b. Show that the λ’s also satisfy J
1
(λ
n
) =0.
33. The solution of the problem in Exercise 31 has the form
u(ρ, ζ ) =a
0
e
−pζ
+b
0
+
∞
n=1
a
n
e
m
1
ζ
+b
n
e
m
2
ζ
J
0
(λ
n
ρ).
The coefficients would normally be found by applying boundary condi-
tions
u(ρ, 0) = f (ρ), u(ρ, 1) = g(ρ), 0 <ρ<1.
In this case, however, information is scarce. The author suggests discard-
ing the solutions that do not approach 0 as ζ →∞. The justification is
that g(ρ) is approximately 0. The solution then becomes
u(ρ, ζ ) =a
0
e
−pζ
+
∞
n=1
a
n
e
m
1
ζ
J
0
(λ
n
ρ),
Miscellaneous Exercises 361
andthecoefficientsshouldbedeterminedby
a
0
=2
1
0
f (ρ)ρ dρ, a
n
=
1
0
f (ρ)J
0
(λ
n
ρ)ρ dρ
1
0
J
2
0
(λ
n
ρ)ρ dρ
.
The function f is known only roughly through experiment. Use the
numbers in the following table to find a
0
and a
1
by the trapezoidal rule
of numerical integration.
ρ 0 0.1 0.2 0.3 0.4
f (ρ) 8.8 8.9 9.2 9.8 10.3
J
0
(λ
1
ρ) 10 0.964 0.858 0.696 0.493
ρ 0.5 0.6 0.7 0.8 0.9 1.0
f (ρ) 11.2 12.0 13.1 14.1 14.8 15
J
0
(λ
1
ρ) 0.273 0.056 −0.135 −0.281 −0.373 −0.403
34. In the article “Asymptotic analysis of intraparticle diffusion in GAC
batch reactors” [D.A. Lyn, Journal of Environmental Engineering, 122
(1996): 1013–1022], the author analyzes chemical diffusing into a spher-
ical particle, with a view to determining some parameter. The concen-
tration q is modeled in dimensionless variables by
1
r
2
∂
∂r
r
2
∂q
∂r
=
∂q
∂t
, 0 < r < 1, 0 < t,
q(r, t) bounded as r →0,
∂q
∂t
(1, t) =−D
∂q
∂r
(1, t).
Separate the variables and find the eigenvalue problem, assuming that
q(r, t) = R(r)T(t).
35. The eigenvalue problem that comes from Exercise 34 has a peculiar
boundary condition that prevents the eigenfunctions from being orthog-
onal. However, the author needs only the first terms of a series solution.
Find an equation for the eigenvalues. Confirm that λ = 0isasolution.
Find the next one numerically for D = 0, 1, 10.
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Laplace Transform
CHAPTER
6
6.1 Definition and Elementary Properties
The Laplace transform serves as a device for simplifying or mechanizing the
solution of ordinary and partial differential equations. It associates a function
f (t) with a function of another variable F(s) from which the original function
can be recovered.
Let f (t) be sectionally continuous in every interval 0 ≤ t < T.TheLaplace
transform of f , written
L(f ) or F(s),isdefinedbytheintegral
L(f ) = F(s) =
∞
0
e
−st
f (t) dt. (1)
We use the convention that a function of t is represented by a lowercase letter
and its transform by the corresponding capital letter. The variable s may be
real or complex, but in the computation of transforms by the definition, s is
usually assumed to be real. Two simple examples are
L(1) =
∞
0
e
−st
·1 dt =
1
s
,
L
e
at
=
∞
0
e
−st
e
at
dt =
−e
−(s−a)t
s −a
∞
0
=
1
s −a
.
Not every sectionally continuous function of t has a Laplace transform, for
the defining integral may fail to converge. For instance, exp(t
2
) has no trans-
form. However, there is a simple sufficient condition, as expressed in the fol-
lowing theorem.
363
364 Chapter 6 Laplace Transform
Theorem. Let f (t) be sectionally continuous in every finite interval 0 ≤t < T. If,
for some constant k, it is true that
lim
t→∞
e
−kt
f (t) =0,
then the Laplace transform of f exists for Re(s)>k.
A function that satisfies the limit condition in the hypotheses of the theorem
is said to be of exponential order.
The Laplace transform inherits two important properties from the integral
used in its definition:
L
cf (t)
= c
L
f (t)
, c constant, (2)
L
f (t) +g(t)
=
L
f (t)
+
L
g(t)
. (3)
By exploiting these properties, we easily determine that
L
cosh(at)
=
L
1
2
e
at
+e
−at
=
1
2
1
s −a
+
1
s +a
=
s
s
2
−a
2
,
L
sin(ωt)
=
L
1
2i
e
iωt
−e
−iωt
=
1
2i
1
s −iω
−
1
s +iω
=
ω
s
2
+ω
2
.
Notice that the linearity properties work with complex constants and func-
tions.
Because of the factor e
−st
in the definition of the Laplace transform, expo-
nential multipliers are easily handled by the “shifting theorem”:
L
e
bt
f (t)
=
∞
0
e
−st
e
bt
f (t) dt
=
∞
0
e
−(s−b)t
f (t) dt = F(s −b),
where F(s) =
L(f (t)). For instance, since L(sin(ωt)) = ω/(s
2
+ω
2
),
L
e
bt
sin(ωt)
=
ω
(s −b)
2
+ω
2
=
ω
s
2
−2sb + b
2
+ω
2
.
The real virtue of the Laplace transform is seen in its effect on derivatives.
Suppose f (t) is continuous and has a sectionally continuous derivative f
(t).
Chapter 6 Laplace Transform 365
Then by definition
L
f
(t)
=
∞
0
e
−st
f
(t) dt.
Integrating by parts, we get
L
f
(t)
=e
−st
f (t)
∞
0
−
∞
0
(−s)e
−st
f (t) dt.
If f (t) is of exponential order, e
−st
f (t) must tend to 0 as t tends to infinity (for
large enough s), so the foregoing equation becomes
L
f
(t)
=−f (0) + s
∞
0
e
−st
f (t) dt
=−f (0) + s
L
f (t)
.
(If f (t) has a jump at t =0, f (0) is to be interpreted as f (0+).)
Similarly, if f and f
are continuous, f
is sectionally continuous; and if all
three functions are exponential order, then
L
f
(t)
=−f (0) + sL
f
(t)
=−f (0) − sf (0) + s
2
L
f (t)
.
An easy generalization extends this formula to the nth derivative,
L
f
(n)
(t)
=−f
(n−1)
(0) −sf
(n−2)
(0) −···−s
n−1
f (0) + s
n
L
f (t)
, (4)
on the assumption that f and its first n − 1 derivatives are continuous, f
(n)
is
sectionally continuous, and all are of exponential order.
We may apply Eq. (4) to the function f (t) = t
k
, k being a nonnegative inte-
ger. Here we have
f (0) = f
(0) =···=f
(k−1)
(0) = 0, f
(k)
(0) = k!, f
(k+1)
(t) = 0.
Thus, Eq. (4) with n = k +1yields
0 =−k ! +s
k+1
L
t
k
,
or
L
t
k
=
k!
s
k+1
.
A different application of the derivative rule is used to transform integrals. If
f (t) is sectionally continuous, then
t
0
f (t
) dt
is a continuous function, equal
to zero at t = 0, and has derivative f (t).Hence
L
f (t)
=s
L
t
0
f (t
) dt
,
366 Chapter 6 Laplace Transform
L(f ) = F(s) =
∞
0
e
−st
f (t) dt
L(cf (t)) = cL( f (t))
L( f (t) +g(t)) = L( f (t)) + L( g(t))
L( f
(t)) =−f (0) +sF(s)
L( f
(t)) =−f
(0) −sf (0) +s
2
F(s)
L( f
(n)
(t)) =−f
(n−1)
(0) −sf
(n−2)
(0) −···−s
n−1
f (0) +s
n
F(s)
L(e
bt
f (t)) = F(s −b)
L
t
0
f (t
) dt
=
1
s
F(s) L
1
t
f (t)
=
∞
s
F(s
) ds
L(tf (t)) =−
dF
ds
Tab le 1 Properties of the Laplace transform
or
L
t
0
f (t
) dt
=
1
s
L
f (t)
. (5)
Differentiation and integration with respect to s may produce transforma-
tions of previously inaccessible functions. We need the two formulas
−
de
−st
ds
=te
−st
,
∞
s
e
−s
t
ds
=
1
t
e
−st
to derive the results
L
tf (t)
=−
dF(s)
ds
,
L
1
t
f (t)
=
∞
s
F(s
) ds
. (6)
(Note that, unless f (0) = 0, the transform of f (t)/t will not exist.) Examples
of the use of these formulas are
L
t sin(ωt)
=−
d
ds
ω
s
2
+ω
2
=
2sω
(s
2
+ω
2
)
2
,
L
sin(t)
t
=
∞
s
ds
s
2
+1
=
π
2
−tan
−1
(s) =tan
−1
1
s
.
Significant properties of the Laplace transform are summarized in Table 1.
When a problem is solved by use of Laplace transforms, a prime difficulty
is computation of the corresponding function of t. Methods for computing
the “inverse transform” f (t) =
L
−1
(F(s)) include integration in the complex
plane, convolution, partial fractions (discussed in Section 6.2), and tables of
Chapter 6 Laplace Transform 367
f (t) F(s) f (t) F(s)
00 t
k
k!
s
k+1
1
1
s
e
bt
cos(ωt)
s −b
s
2
−2bs + b
2
+ω
2
e
at
1
s −a
e
bt
sin(ωt)
ω
s
2
−2bs + b
2
+ω
2
cosh(at)
s
s
2
−a
2
e
bt
t
k
k!
(s −b)
k+1
sinh(at)
a
s
2
−a
2
e
at
−1
a
s(s −a)
cos(ωt)
s
s
2
+ω
2
t cos(ωt)
s
2
−ω
2
(s
2
+ω
2
)
2
sin(ωt)
ω
s
2
+ω
2
t sin(ωt)
2sω
(s
2
+ω
2
)
2
t
1
s
2
Tab le 2 Laplace transforms
transforms. The last method, which involves the least work, is the most popu-
lar. The transforms in Table 2 were all calculated from the definition or by use
of formulas in this section.
EXERCISES
1. By using linearity and the transform of e
at
, compute the transform of each
of the following functions.
a. sinh(at);
d. sin(ωt −φ);
b. cos(ωt);
e. e
2(t+1)
;
c. cos
2
(ωt);
f. sin
2
(ωt).
2. Use differentiation with respect to t to find the transform of
a. te
at
from L(e
at
),
b. sin(ωt) from L(cos(ωt)),
c. cosh(at) from L(sinh(at)).
3. Compute the transform of each of the following directly from the defini-
tion.
a. f (t) =
0, 0 < t < a,
1, a < t;