Powers D.L. Boundary Value Problems and Partial Differential Equations
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388 Chapter 6 Laplace Transform
EXERCISES
1. Find the persistent part of the solution of the heat problem
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
∂u
∂x
(0, t) = 0,
∂u
∂x
(1, t) = 1, 0 < t,
u(x, 0) = 0, 0 < x < 1.
2. Verify that the persistent part of the solution to Example 2 actually satisfies
the heat equation. What boundary condition does it satisfy?
3. Find the function v(x, t) whose transform is
cosh(
1
2
s) −cosh
s(
1
2
−x )
s
2
cosh(
1
2
s)
.
What boundary value–initial value problem does v(x, t) satisfy?
4. Solve
∂
2
u
∂x
2
=
∂
2
u
∂t
2
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 0, 0 < t,
u(x, 0) = 0,
∂u
∂t
(x, 0) = 1, 0 < x < 1.
5. a. Solve for ω =π :
∂
2
u
∂x
2
=
∂
2
u
∂t
2
−sin(π x) sin(ωt), 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 0, 0 < t,
u(x, 0) = 0,
∂u
∂t
(x, 0) = 0, 0 < x < 1.
b. Examine the special case ω = π .
6. Obtain the complete solution of Example 1 and verify that it satisfies the
boundary conditions and the heat equation.
6.5 Comments and References 389
7. a. Solve
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) =1 −e
−at
, 0 < t,
u(x, 0) = 0, 0 < x < 1.
b. Examine the special case where a = n
2
π
2
for some integer n.
6.5 Comments and References
The real development of the Laplace transform began in the late nineteenth
century, when engineer Oliver Heaviside invented a powerful, but unjustified,
symbolic method for studying the ordinary and partial differential equations
of mathematical physics. By the 1920s, Heaviside’s method had been legitima-
tized and recast as the Laplace transform that we now use. Later generalizations
are Schwartz’s theory of distributions (1940s) and Mikusinski’s operational
calculus (1950s). The former seems to be the more general. Both theories give
an interpretation of F(s) = 1, which is not the Laplace transform of any func-
tion, in the sense we use.
There are a number of other transforms, under the names of Fourier, Mellin,
Hänkel, and others, similar in intent to the Laplace transform, in which some
other function replaces e
−st
in the defining integral. Operational Mathematics,
by Churchill, has more information about the applications of transforms. Ex-
tensive tables of transforms will be found in Tables of Integral Transforms by
Erdelyi et al. (See the Bibliography.)
Miscellaneous Exercises
1. Solve the heat conduction problem
∂
2
u
∂x
2
−γ
2
(u −T) =
∂u
∂t
, 0 < x < 1, 0 < t,
∂u
∂x
(0, t) = 0,
∂u
∂x
(1, t) = 0, 0 < t,
u(x, 0) = T
0
, 0 < x < 1.
390 Chapter 6 Laplace Transform
2. Find the “persistent part” of the solution of
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
∂u
∂x
(0, t) = 0, u(1, t) =t, 0 < t,
u(x, 0) = 0, 0 < x < 1.
3. Find the complete solution of the problem in Exercise 2.
4. A solid object and a surrounding fluid exchange heat by convection. The
temperatures u
1
and u
2
are governed by the following equations. Solve
them by means of Laplace transforms.
du
1
dt
=−β
1
(u
1
−u
2
),
du
2
dt
=−β
2
(u
2
−u
1
),
u
1
(0) =1, u
2
(0) = 0.
5. Solve the following nonhomogeneous problem with transforms.
∂
2
u
∂x
2
=
∂u
∂t
−1, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 0, 0 < t,
u(x, 0) = 0, 0 < x < 1.
6. Find the transform of the solution of the problem
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = 1, 0 < t,
u(x, 0) = 0, 0 < x < 1.
7. Find the solution of the problem in Exercise 6 by using the extended
Heaviside formula.
8. Solve the heat problem
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x, 0 < t,
u(0, t) = 0, 0 < t,
u(x, 0) = sin(x), 0 < x.
Miscellaneous Exercises 391
9. Find the transform of the solution of
∂
2
u
∂x
2
=
∂u
∂t
, 0 < x, 0 < t,
u(0, t) = 0, 0 < t,
u(x, 0) = 1, 0 < x,
u(x, t) bounded as x →∞.
10. At the end of Section 2.12, the problem in Exercise 9 was solved by other
means. Use this fact to identify
1
s
1 −e
−
√
sx
=
L
erf
x
√
4t
and
1
s
e
−
√
sx
=L
1 −erf
x
√
4t
.
(The latter function is called the complementary error function,defined
by erfc(q) ≡ 1 −erf(q).)
11. Find the function of t whose Laplace transform is
F(s) =e
−x
√
s
.
12. Using the definition of sinh in terms of exponentials and a geometric
series, show that
sinh(
√
sx)
sinh(
√
s)
=
∞
n=0
e
−
√
s(2n+1−x)
−e
−
√
s(2n+1+x)
.
13. Use the series in Exercise 12 to find a solution of the problem in Exer-
cise 6 in terms of complementary error functions.
14. Show the following relation by using Exercise 11 and differentiating with
respect to s.
L
1
√
πt
exp
−k
2
4t
=
1
√
s
e
−k
√
s
.
15. Find the Laplace transform of the odd periodic extension of the function
f (t) = π − t, 0 < t <π,
by transforming its Fourier series term by term.
392 Chapter 6 Laplace Transform
16. Suppose that the function f (t) is periodic with period 2a. Show that the
Laplace transform of f is given by the formula
F(s) =
G(s)
1 −e
−2as
,
where
G(s) =
2a
0
f (t)e
−st
dt.
(Hint: See Section 6.1, Exercise 6.)
17. Apply the extended Heaviside method to the inversion of a transform
with the form
F(s) =
G(s)
1 −e
−2as
,
where G(s) does not become infinite for any value of s.
18. Show that for a periodic function f (t) the quantities
c
n
=
1
2a
G
inπ
a
(G(s) is defined in Exercise 16) are the complex Fourier coefficients.
19. How is it possible to determine that a Laplace transform F(s) corre-
sponds to a periodic f (t)?
20. Is this function the transform of a periodic function?
F(s) =
1
s
2
+a
2
.
21. Use the method of Exercise 16 to find the transform of the periodic ex-
tension of
f (t) =
1, 0 < t <π,
−1,π<t < 2π .
22. Same as Exercise 21, but use the function of Exercise 15.
23. Use the method of Exercise 16 to find the transform of
f (t) =
sin(t)
.
24. Find the transform of the solution of the problem
Miscellaneous Exercises 393
∂
2
u
∂x
2
=
∂
2
u
∂t
2
, 0 < x, 0 < t,
u(0, t) = h(t), 0 < t,
u(x, 0) = 0,
∂u
∂t
(x, 0) = 0, 0 < x,
u(x, t) bounded as x →∞.
Use the solution of the same problem as found in Section 3.6, to verify
the rule
L
−1
e
−sx
H(s)
=
h(t −x), t > x,
0, t < x.
25. Solve this wave problem with time-varying boundary condition, assum-
ing ω =nπ , n = 1, 2,....
∂
2
u
∂x
2
=
∂
2
u
∂t
2
, 0 < x < 1, 0 < t,
u(0, t) = 0, u(1, t) = sin(ωt), 0 < t,
u(x, 0) = 0,
∂u
∂t
(x, 0) = 0, 0 < x < 1.
26. SolvetheprobleminExercise25inthespecialcaseω = π .
27. Certain techniques for growing a crystal from a solution or a melt may
cause striations — variations in the concentration of impurities. Authors
R.T. Gray, M.F. Larrousse, and W.R. Wilcox [Diffusional decay of stria-
tions, Journal of Crystal Growth, 92 (1988): 530–542] use a material bal-
ance on a slice of a cylindrical ingot to derive this boundary value prob-
lem for the impurity concentration, C:
∂
∂x
D(x)
∂C
∂x
−V
∂C
∂x
=
∂C
∂t
, 0 < x, 0 < t,
C(0, t) = C
a
+A sin
2πt
t
C
, 0 < t,
C(x, 0) = C
a
, 0 < x.
Here, V is the crystal growth rate, t
C
is the striation period, C
a
is the
average concentration in the solid, and D(x) is the diffusivity of the im-
purity at distance x from the growth face (which is located at x = 0). Of
course, C(x, t) is bounded as x →∞.
394 Chapter 6 Laplace Transform
Next, the equations are made dimensionless by introducing new vari-
ables:
¯
C =
C −C
a
A
, ¯x =
V
x
D(0)
,
¯
t =
V
2
t
D(0)
.
The new problem is
∂
∂ ¯x
D(¯x)
D(0)
∂
¯
C
∂ ¯x
−
∂
¯
C
∂ ¯x
=
∂
¯
C
∂
¯
t
, 0 < ¯x, 0 <
¯
t,
¯
C
0,
¯
t
=sin
ω
¯
t
, 0 <
¯
t,
¯
C (¯x, 0) = 0, 0 < ¯x,
where ω =2πD(0)/V
2
t
C
.
Because D(¯x) depends in a complicated way on ¯x,anumericalsolution
was used. To check the numerical solution, the authors wished to find an
analytical solution of the problem corresponding to constant diffusivity,
D(¯x) = D(0).Letu be the solution of
∂
2
u
∂ ¯x
2
=
∂u
∂ ¯x
+
∂u
∂
¯
t
, 0 < ¯x, 0 <
¯
t,
u
0,
¯
t
=sin
ω
¯
t
, 0 <
¯
t,
u(¯x, 0) = 0, 0 < ¯x,
u bounded as x →∞.
Find the Laplace transform of the solution of this problem.
28. The authors of the paper mentioned in Exercise 27 were particularly in-
terested in the persistent part of the solution. Use the methods of Sec-
tion 6.4 to show that the persistent part of the solution is
u
1
=
1
2i
f (iω) − f (−iω)
,
where
f (iω) = exp
1
2
−
1
4
+iω
¯x +i ω
¯
t
.
29. Find the square root required in the foregoing expression by setting
1
4
+iω =α +iβ
Miscellaneous Exercises 395
so that
α
2
−β
2
+2iαβ =
1
4
+iω,
or
α
2
−β
2
=
1
4
,
2αβ =ω.
(To solve these equations: (i) solve the second for β; (ii) substitute the
expression found into the first; (iii) solve the resulting biquadratic for α.)
30. Noting that the persistent part is u
1
= Im(f (iω)) (see Exercise 28), de-
termine
u
1
¯x,
¯
t
=e
(
1
2
−α)¯x
sin
ω
¯
t −β ¯x
,
where α and β areasinExercise29.
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Numerical Methods
CHAPTER
7
7.1 Boundary Value Problems
More often than not, significant practical problems in partial — and even or-
dinary — differential equations cannot be solved by analytical methods. Dif-
ficulties may arise from variable coefficients, irregular regions, unsuitable
boundary conditions, interfaces, or just overwhelming detail. Now that ma-
chine computation is cheap and easily accessible, numerical methods provide
reliable answers to formerly difficult problems. In this chapter we examine
a few methods that are simple and equally adaptable to machine or manual
computation. Implementation of some of these methods with a spreadsheet
program is explained and carried out on the CD.
If we cannot find a simple analytic formula for the solution of a boundary
value problem, we may be satisfied with a table of (approximate) values of the
solution. For instance, the solution of the problem
d
2
u
dx
2
−12xu =−1, 0 < x < 1, (1)
u(0) = 1, u(1) =−1, (2)
may be written out in terms of Airy functions, but the values of u shown in
Table 1 are more informative for most of us. One way to obtain such a table
is to replace the original analytical problem by an arithmetical problem, as
described in what follows.
397