Nair S. Advanced Topics in Applied Mathematics: For Engineering and the Physical Sciences
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Laplace Transforms 209
The difference form of Eq. (4.213)is
v
n+2
−2v
n+1
+(1 +h
2
ω
2
)v
n
=0. (4.217)
Comparing this with our second-order prototype, Eq. (4.204),
b +c =2, bc = 1 +h
2
ω
2
. (4.218)
Solving for b and c,
b =1 +ihω, c = 1 −ihω. (4.219)
Substituting in the general solution, Eq. (4.208), with v
0
=0,
v
n
=v
1
(1 +ihω)
n
−(1 −ihω)
n
2ihω
. (4.220)
To satisfy the condition v
N
=0, we require
(1 +ihω)
N
=(1 −ihω)
N
. (4.221)
Taking the Nth root,
1 +ihω =(1 −ihω)e
2iπ k/N
, k = 1,2,...,N −1. (4.222)
Solving for ω,
ω =
1
ih
e
iπ k/N
−e
−iπ k/N
e
iπ k/N
+e
−iπ k/N
=
1
h
tan(πk/N) =N tan(πk/N). (4.223)
For any finite values of nodes N, this equation gives approximate
values of the frequency. We may identify k with the mode shape. For
large values of N, we may expand this as
ω =N
πk
N
−
π
3
k
3
3N
3
+···
=πk
1 −
π
2
k
2
3N
2
+···
. (4.224)
210 Advanced Topics in Applied Mathematics
For the kth mode, the error is given by
ω −ω
exact
ω
exact
≈−
π
2
k
2
3N
2
. (4.225)
Of course, as N →∞, we recover the exact frequency.
4.10.4 Brilluoin Approximation for Crystal Acoustics
To study the propagation of acoustic waves (phonons) in crystals, the
physicist, Brilluoin, used an idealized crystal lattice where each atom is
considered as a particle of mass m, and the interaction between atoms
is approximated using a spring of stiffness k. This is shown in Fig. 4.6.
If the displacement of the nth particle is denoted by u
n
(t), its motion
is governed by
m¨u
n
=k(u
n+1
+u
n−1
−2u
n
). (4.226)
By changing n to n +1, we write this (in our standard form) as
u
n+2
−2u
n+1
+u
n
=
m
k
¨u
n+1
. (4.227)
We assume steady-state motion of the particles in the form
u
n
(t) =v
n
sint, (4.228)
to obtain the Helmholtz equation
v
n+2
−(2 −m
2
/k)v
n+1
+v
n
=0. (4.229)
Let
ω =
k/m
, (4.230)
u
0
u
1
u
2
u
3
mmmmkkkk
Figure 4.6. A spring-mass representation of a crystal lattice.
Laplace Transforms 211
where the denominator represents the angular frequency of a simple
spring-mass system.
Comparing this with Eq. (4.204),
b +c =2 −ω
2
, bc =1. (4.231)
From this we get
b,c = 1 −
ω
2
2
±
&
1 −
ω
2
2
2
−1. (4.232)
When ω<2, we may denote
cosθ =1 −
ω
2
2
, (4.233)
and write
b =e
iθ
, c =e
−iθ
. (4.234)
The solution of the Helmholtz equation is
v
n
=
v
1
−e
−iθ
v
0
e
iθ
−e
−iθ
e
inθ
−
v
1
−e
iθ
v
0
e
iθ
−e
−iθ
e
−inθ
=v
1
sinnθ
sinθ
−v
0
sin(n −1)θ
sinθ
. (4.235)
As long as θ is positive real, this solution is bounded for all values
of n. When ω>2, the angle θ is complex, that is the imaginary part
of θ is nonzero, and we get unbounded solutions, which cannot be
sustained with finite power input. This illustrates the concept of a cut-
off frequency,
cut−off
=2
k/m. (4.236)
When is less than the cut-off frequency, we see that our solu-
tion is periodic. The significance of a cut-off frequency is that periodic
vibration of the system is not possible for frequencies above it.
If n is increased by 2π/θ,wefind
v
n
=v
n+N
, N =2π/θ. (4.237)
212 Advanced Topics in Applied Mathematics
Here, N is a measure of the wave length, and the relation
θ =2π/N (4.238)
is known as the dispersion relation. From
cosθ =1 −ω
2
/2, (4.239)
we get, explicitly,
ω
2
=2
1 −cos
2π
N
ω =2sin π/N. (4.240)
Using the periodicity, we may choose v
0
=0 and obtain
v
n
=v
1
sinnθ
sinθ
. (4.241)
SUGGESTED READING
Andrews, L. C., and Shivamoggi, B. K. (1988). Integral Transforms for
Engineers and Applied Mathematicians, Macmillan.
Bender, C. M., and Orzag, S. A. (1978). Advanced Mathematical Methods for
Scientists and Engineers, McGraw-Hill.
Brillouin, L. (1946). Wave Propagation in Periodic Structures, McGraw-Hill.
Davies, B. (1985). Integral Transforms and Their Applications, 2nd ed.,
Springer-Verlag.
Ewing, W. M., Jardetzky, W. S., and Press, F. (1957). Elastic Waves in Layered
Media, McGraw-Hill.
Fung, Y. C. (1965). Foundations of Solid Mechanics, Prentice-Hall.
Miles, J. W. (1961). Integral Transforms in Applied Mathematics, Cambridge
University Press.
Sneddon, I. N. (1972). The Use of Integral Transforms, McGraw-Hill.
EXERCISES
4.1 Find the Laplace transforms of
f (t) =e
at
cosbt, g(t) =e
at
sinbt.
Laplace Transforms 213
4.2 Using the relation
∞
0
e
−(a
2
x
2
+b
2
/x
2
)
dx =
√
π
2a
e
−2ab
, a,b > 0,
obtain the Laplace transforms of
f (t) =
1
√
t
e
−x
2
/(4t)
, g(t) =erfc
x
2
√
t
.
4.3 Find the Laplace transforms of
f (t) =a
t
, g(t) =
cosat
√
t
,
4.4 Invert the transforms
¯
f (p) =
e
−2p
[(p +1)
2
+1]
2
, ¯g(p) =
√
p
p −a
.
4.5 Show that the Laplace transform of
f (t) =
∞
1
e
−tτ
dτ
τ
is
¯
f (p) =
log(p +1)
p
.
4.6 Invert the transforms
¯
f (p) =
coshap
p
3
coshp
, ¯g(p) =
sinhap
p
2
sinhp
,0< a < 1.
4.7 Show that the Laplace transform of the Theta function,
θ(t) =
∞
n=−∞
e
−n
2
π
2
t
,
can be expressed as
¯
θ(p) =
coth
√
p
√
p
.
214 Advanced Topics in Applied Mathematics
4.8 Using the expansion
1
1 −e
−2
√
p
=1 +e
−2
√
p
+e
−4
√
p
+···,
invert the transform
¯
f (p) =
coth
√
p
√
p
.
4.9 Assume p = a is a branch point of the transform
¯
f (p), and near
p =a the transform may be expanded as
¯
f (p) =c
0
(p −a)
ν
0
+c
1
(p −a)
ν
1
+c
2
(p −a)
ν
2
+···,
where ν
0
<ν
1
<ν
2
< ···. Show that, as t →∞, f (t) has the
expansion
f (t) ∼e
at
c
0
(−ν
0
)t
ν
0
+1
+
c
1
(−ν
1
)t
ν
1
+1
+
c
2
(−ν
2
)t
ν
2
+1
+···
.
Use the inversion formula
L
−1
[p
ν
]=
1
(−ν)t
ν+1
for this purpose.
4.10 Solve
d
2
u
dt
2
+3
du
dt
+2u = e
2t
cost.
with u(0) =du/dt(0) =0.
4.11 Solve
d
3
u
dt
3
−6
d
2
u
dt
2
+11
du
dt
−6u =t,
with
u(0) =
du
dt
(0) =
d
2
u
dt
2
(0) =0.
4.12 Solve
d
2
u
dt
2
+2
du
dt
+2u = e
−t
sint, u(0) =
du
dt
(0) =0.
Laplace Transforms 215
4.13 Solve the boundary value problem
d
2
u
dx
2
+
du
dx
−2u = e
−x
, u(0) =1,
du
dx
(1) =0.
4.14 Solve the system of differential equations
du
dt
−u +v =0,
dv
dt
+v −u =e
t
,
subject to the initial conditions
u(0) =1, v(0) =0.
4.15 To solve
∂
2
ψ
∂r
2
+
2
r
∂ψ
∂r
=
∂ψ
∂t
, ψ(a,t) =T
0
, ψ(r,0) =0,
we use a substitution
ψ = φ/r.
Obtain the equation for φ. Using the Laplace transform solve
for ψ.
4.16 Obtain the solution of
∂
2
u
∂x
2
=
∂
2
u
∂t
2
, u(0,t) = U
0
sinωt, u(,t) = 0,
u(x,0) =
∂u
∂t
(x,0) =0.
4.17 The propagation of elastic stress in a bar: 0 < x <, is governed by
∂
2
σ
∂x
2
=
1
c
2
∂
2
σ
∂t
2
,
where c is the wave speed. At time t =0, the end x =0 is subjected
to a stress σ
0
, while the end x = is subject to ∂σ/∂x = 0. The
initial conditions are σ(x,0) =0and∂σ/∂t(x,0) =0. Obtain σ(x,t)
by expanding (1 −e
−2pl/c
)
−1
in a binomial series.
4.18 The viscoelastic motion of a semi-infinite bar: 0 < x < ∞,is
governed by
c
2
∂
2
u
∂x
2
=
∂
2
u
∂t
2
+β
∂u
∂t
.
216 Advanced Topics in Applied Mathematics
If the boundary and initial conditions are
u(0,t) = U
0
sinωt, u(x →∞,t) =0, u(x,0) =
∂u
∂x
(x,0) =0,
show that the solution can be written as
u(x,t) =U
0
ω
t
0
k(x,τ)cos ω(t −τ)dτ ,
where, using a branch cut in the p-plane between p = 0and
p =−β, an expression for k(x,t) can be obtained as
k(x,t) =1 −
1
π
β
0
1
r
e
−rt
sin
r(β −r) dr.
4.19 Unsteady heat conduction in a semi-infinite bar is governed by
the equation
κ
∂
2
u
∂x
2
=
∂u
∂t
,
with the conditions
u(x,0) =0, u(0, t) = U
0
te
−at
, u(x →∞,t) =0.
Find u(x,t).
4.20 Two semi-infinite bars, A: −∞ < x < 0 and B: 0 < x < ∞ have
thermal diffusivities κ
A
and κ
B
, respectively. Their conductivities
are k
A
and k
B
and they are at uniform temperatures, T
0
A
and T
0
B
when t < 0. At time t =0, their ends are made to contact. Obtain
the transient temperatures T
A
(x,t) and T
B
(x,t) for t > 0.
4.21 A semi-infinite bar is made of two materials: A and B. Material
A occupies 0 < x < 1 and B occupies 1 < x < ∞. Heat conduction
in the two materials is governed by
κ
A
∂
2
T
A
∂x
2
=
∂T
A
∂t
, κ
B
∂
2
T
B
∂x
2
=
∂T
B
∂t
.
The boundary and initial conditions are:
T
A
(0,t) =T
0
h(t), T
A
(x,0) =T
B
(x,0) =0, T
B
(x →∞) =0,
Laplace Transforms 217
where h(t) is the Heaviside step function. At the interface x =1,
∂T
A
∂x
=α
∂T
B
∂x
, T
A
=T
B
,
where a is a constant. Obtain the temperatures in the two sections
of the bar.
4.22 If u is the solution of
∂
2
u
∂x
2
=
∂u
∂t
, u(0,t) =
∂u
∂x
(0,t) =0, u(x,0) =f (x),
and v is the solution of
∂
2
v
∂x
2
=
∂
2
v
∂t
2
, v(0,t) =
∂v
∂x
(0,t) =0, v(x,0) =0,
∂v
∂t
(x,0) =f (x),
show that
u(x,t) =
1
√
4πt
3
∞
0
v(x, τ)e
−τ
2
/4t
τ dτ
(fromSnedddon,1972).
4.23 If u is the solution of
∂
2
u
∂x
2
=
∂u
∂t
, u(0,t) =
∂u
∂x
(0,t) =0, u(x,0) =f (x),
and v is the solution of
∂
2
v
∂x
2
=
∂
2
v
∂t
2
, v(0,t) =
∂v
∂x
(0,t) =0, v(x,0) =f (x),
∂v
∂t
(x,0) =0,
show that
u(x,t) =
1
√
πt
∞
0
v(x, τ)e
−τ
2
/4t
τ dτ.
(fromSnedddon,1972).
4.24 Solve the integral equation
u(t) −a
t
0
e
−aτ
u(t −τ)dτ =e
−bt
,
where a and b are constants.
218 Advanced Topics in Applied Mathematics
4.25 Solve the integral equation
t
0
cos[k
√
t
2
−τ
2
]
√
t
2
−τ
2
f (τ )dτ =g(t).
4.26 Solve the integral equation
x
0
u(ξ )dξ
(x −ξ)
2/3
=
x
0
f (ξ )dξ
(x −ξ)
1/3
.
4.27 Solve the integral equation
x
0
u(ξ )dξ
x
2
−ξ
2
=
1
√
x
.
4.28 Find the solution of
u
n+1
−au
n
=nb
n
, u
0
=1.
4.29 Find the solution of the difference equation
u
n+2
−2bu
n+1
+b
2
u
n
=b
n
, u
0
=u
1
=0.
4.30 Obtain the frequencies ω corresponding to the periodic solu-
tions of
u
n+2
−(2 −ω
2
)u
n+1
+u
n
=0,
subject to the conditions
u
0
=u
4
=1, u
2
=u
5
.