Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 7 ppt

We set
α
2
=
U − 1
U
.
β is then
β =
α
1 − α
2
=

(U − 1)/U
1 − (U − 1)/U)
=

(U − 1)U
U − (U − 1)
=

(U − 1)U.
The solution for η becomes
αβ
2
sech
2

αx − βt

2

where
β =
α
1 − α
2
.
2.
u
tt
− u
xx
−

3
2
u
2

xx
− u
xxxx
= 0
We make the substitution
u(x, t) = z(X), X = x −Ut.
2214
(U
2
− 1)z


−

3
2
z
2


− z

= 0
(U
2
− 1)z

−

3
2
z
2


− z

= 0
(U
2
− 1)z −

3
2
z
2
− z

= 0
We multiply by z

and integrate.
1
2
(U
2
− 1)z
2
−
1
2
z
3
−
1
2
(z

)
2
= 0
(z


)
2
= (U
2
− 1)z
2
− z
3
z = (U
2
− 1) sech
2

1
2
√
U
2
− 1X

u(x, t) = (U
2
− 1) sech
2

1
2

√

U
2
− 1x − U
√
U
2
− 1t


The linearized equation is
u
tt
− u
xx
− u
xxxx
= 0.
Substituting u = e
−αx+βt
into this equation yields
β
2
− α
2
− α
4
= 0
β
2
= α

2
(α
2
+ 1).
We set
α =
√
U
2
− 1.
2215
β is then
β
2
= α
2
(α
2
+ 1)
= (U
2
− 1)U
2
β = U
√
U
2
− 1.
The solution for u becomes
u(x, t) = α

2
sech
2

αx − βt
2

where
β
2
= α
2
(α
2
+ 1).
3.
φ
tt
− φ
xx
+ 2φ
x
φ
xt
+ φ
xx
φ
t
− φ
xxxx

We make the substitution
φ(x, t) = z(X), X = x −Ut.
(U
2
− 1)z

− 2Uz

z

− Uz

z

− z

= 0
(U
2
− 1)z

− 3Uz

z

− z

= 0
(U
2

− 1)z

−
3
2
(z

)
2
− z

= 0
Multiply by z

and integrate.
1
2
(U
2
− 1)(z

)
2
−
1
2
(z

)
3

−
1
2
(z

)
2
= 0
(z

)
2
= (U
2
− 1)(z

)
2
− (z

)
3
z

= (U
2
− 1) sech
2

1

2
√
U
2
− 1X

φ
x
(x, t) = (U
2
− 1) sech
2

1
2

√
U
2
− 1x − U
√
U
2
− 1t


.
2216
The linearized equation is
φ

tt
− φ
xx
− φ
xxxx
Substituting φ = e
−αx+βt
into this equation yields
β
2
= α
2
(α
2
+ 1).
The solution for φ
x
becomes
φ
x
= α
2
sech
2

αx − βt
2

where
β

2
= α
2
(α
2
+ 1).
4.
u
t
+ 30u
2
u
1
+ 20u
1
u
2
+ 10uu
3
+ u
5
= 0
We make the substitution
u(x, t) = z(X), X = x −Ut.
−Uz

+ 30z
2
z


+ 20z

z

+ 10zz

+ z
(5)
= 0
Note that (zz

)

= z

z

+ zz

.
−Uz

+ 30z
2
z

+ 10z

z


+ 10(zz

)

+ z
(5)
= 0
−Uz + 10z
3
+ 5(z

)
2
+ 10zz

+ z
(4)
= 0
Multiply by z

and integrate.
−
1
2
Uz
2
+
5
2
z

4
+ 5z(z

)
2
−
1
2
(z

)
2
+ z

z

= 0
2217
Assume that
(z

)
2
= P (z).
Differentiating this relation,
2z

z

= P


(z)z

z

=
1
2
P

(z)
z

=
1
2
P

(z)z

z

z

=
1
2
P

(z)P (z).

Substituting this expressions into the differential equation for z,
−
1
2
Uz
2
+
5
2
z
4
+ 5zP (z) −
1
2
1
4
(P

(z))
2
+
1
2
P

(z)P (z) = 0
4Uz
2
+ 20z
4

+ 40zP (z) −(P

(z))
2
+ 4P

(z)P (z) = 0
Substituting P (z) = az
3
+ bz
2
yields
(20 + 40a + 15a
2
)z
4
+ (40b + 20ab)z
3
+ (4b
2
+ 4U)z
2
= 0
This equation is satisfied by b
2
= U, a = −2. Thus we have
(z

)
2

=
√
Uz
2
− 2z
3
z =
√
U
2
sech
2

1
2
U
1/4
X

u(x, t) =
√
U
2
sech
2

1
2
(U
1/4

x − U
5/4
t)

2218
The linearized equation is
u
t
+ u
5
= 0.
Substituting u = e
−αx+βt
into this equation yields
β − α
5
= 0.
We set
α = U
1/4
.
The solution for u(x, t) becomes
α
2
2
sech
2

αx − βt
2


where
β = α
5
.
2219
Part VIII
Appendices
2220
Appendix A
Greek Letters
The following table shows the greek letters, (some of them have two typeset variants), and thei r corresponding
Roman letters.
Name Roman Lower Upper
alpha a α
beta b β
chi c χ
delta d δ ∆
epsilon e 
epsilon (variant) e ε
phi f φ Φ
phi (variant) f ϕ
gamma g γ Γ
eta h η
iota i ι
kappa k κ
lambda l λ Λ
mu m µ
2221
nu n ν

omicron o o
pi p π Π
pi (variant) p 
theta q θ Θ
theta (variant) q ϑ
rho r ρ
rho (variant) r 
sigma s σ Σ
sigma (variant) s ς
tau t τ
upsilon u υ Υ
omega w ω Ω
xi x ξ Ξ
psi y ψ Ψ
zeta z ζ
2222
Appendix B
Notation
C class of continuous functions
C
n
class of n-times continuously differentiable functions
C set of complex numbers
δ(x) Dirac delta function
F[·] Fourier transform
F
c
[·] Fourier cosine transform
F
s

[·] Fourier sine transform
γ Euler’s constant, γ =

∞
0
e
−x
Log x dx
Γ(ν) Gamma function
H(x) Heaviside function
H
(1)
ν
(x) Hankel function of the first kind and order ν
H
(2)
ν
(x) Hankel function of the second kind and order ν
ı ı ≡
√
−1
J
ν
(x) Bessel function of the first kind and order ν
K
ν
(x) Modified Bessel function of the first kind and order ν
L[·] Laplace transform
2223
N set of natural numbers, (positive integers)

N
ν
(x) Modified Bessel function of the second kind and order ν
R set of real numbers
R
+
set of positive real numbers
R
−
set of negative real numbers
o(z) terms smaller than z
O(z) terms no bigger than z
−

principal value of the integral
ψ(ν) digamma function, ψ(ν) =
d
dν
log Γ(ν)
ψ
(n)
(ν) polygamma function, ψ
(n)
(ν) =
d
n
dν
n
ψ(ν)
u

(n)
(x)
∂
n
u
∂x
n
u
(n,m)
(x, y)
∂
n+m
u
∂x
n
∂y
m
Y
ν
(x) Bessel function of the second kind and order ν, Neumann function
Z set of integers
Z
+
set of positive integers
2224
Appendix C
Formulas from Complex Variables
Analytic Functions. A function f(z) is analytic in a domain if the derivative f

(z) exists in that domain.

If f(z) = u(x, y) + ıv(x, y) is defined in some neighborhood of z
0
= x
0
+ ıy
0
and the partial d erivatives of u and v
are continuous and satisfy the Cauchy-Riemann equations
u
x
= v
y
, u
y
= −v
x
,
then f

(z
0
) exists.
Residues. If f(z) has the Laurent expansion
f(z) =
∞

n=−∞
a
n
z

n
,
then the residue of f(z) at z = z
0
is
Res(f(z), z
0
) = a
−1
.
2225
Residue Theorem. Let C be a positively oriented, simple, closed contour. If f(z) is analytic in and on C except
for isolated singularities at z
1
, z
2
, . . . , z
N
inside C then

C
f(z) dz = ı2π
N

n=1
Res(f(z), z
n
).
If in addition f(z) is analytic outside C in the finite complex plane then


C
f(z) dz = ı2π Res

1
z
2
f

1
z

, 0

.
Residues of a pole of order n. If f(z) has a pole of order n at z = z
0
then
Res(f(z), z
0
) = lim
z→z
0

1
(n − 1)!
d
n−1
dz
n−1
[(z −z

0
)
n
f(z)]

.
Jordan’s Lemma.

π
0
e
−R sin θ
dθ <
π
R
.
Let a be a positive constant. If f(z) vanishes as |z| → ∞ then the integral

C
f(z)
e
ıaz
dz
along the semi-circle of radius R in the upper half plane vanishes as R → ∞.
Taylor Series. Let f(z) be a function that is analytic and single valued in the disk |z −z
0
| < R.
f(z) =
∞


n=0
f
(n)
(z
0
)
n!
(z −z
0
)
n
The series converges for |z −z
0
| < R.
2226
Laurent Series. Let f(z) be a function that is analytic and single valued in the annulus r < |z − z
0
| < R. In this
annulus f(z) has the convergent series,
f(z) =
∞

n=−∞
c
n
(z −z
0
)
n
,

where
c
n
=
1
ı2π

f(z)
(z −z
0
)
n+1
dz
and the path of integration is any simple, closed, positive contour around z
0
and lying in the annulus. The path of
integration is shown in Figure
C.1.
C
Im(z)
Re(z)
R
r
Figure C.1: The Path of Integration.
2227
Appendix D
Table of Derivatives
Note: c denotes a constant and

denotes differentiation.

d
dx
(fg) =
df
dx
g + f
dg
dx
d
dx
f
g
=
f

g −fg

g
2
d
dx
f
c
= cf
c−1
f

d
dx
f(g) = f


(g)g

d
2
dx
2
f(g) = f

(g)(g

)
2
+ f

g

d
n
dx
n
(fg) =

n
0

d
n
f
dx

n
g +

n
1

d
n−1
f
dx
n−1
dg
dx
+

n
2

d
n−2
f
dx
n−2
d
2
g
dx
2
+ ··· +


n
n

f
d
n
g
dx
n
2228
d
dx
ln x =
1
|x|
d
dx
c
x
= c
x
ln c
d
dx
f
g
= gf
g−1
df
dx

+ f
g
ln f
dg
dx
d
dx
sin x = cos x
d
dx
cos x = −sin x
d
dx
tan x = sec
2
x
d
dx
csc x = −csc x cot x
d
dx
sec x = sec x tan x
d
dx
cot x = −csc
2
x
d
dx
arcsin x =

1
√
1 − x
2
, −
π
2
≤ arcsin x ≤
π
2
2229
d
dx
arccos x = −
1
√
1 − x
2
, 0 ≤ arccos x ≤ π
d
dx
arctan x =
1
1 + x
2
, −
π
2
≤ arctan x ≤
π

2
d
dx
sinh x = cosh x
d
dx
cosh x = sinh x
d
dx
tanh x = sech
2
x
d
dx
csch x = −csch x coth x
d
dx
sech x = −sech x tanh x
d
dx
coth x = −csch
2
x
d
dx
arcsinh x =
1
√
x
2

+ 1
d
dx
arccosh x =
1
√
x
2
− 1
, x > 1, arccosh x > 0
d
dx
arctanh x =
1
1 − x
2
, x
2
< 1
2230
d
dx

x
c
f(ξ) dξ = f(x)
d
dx

c

x
f(ξ) dξ = −f(x)
d
dx

h
g
f(ξ, x) dξ =

h
g
∂f(ξ, x)
∂x
dξ + f(h, x)h

− f(g, x)g

2231
Appendix E
Table of Integrals

u
dv
dx
dx = uv −

v
du
dx
dx


f

(x)
f(x)
dx = log f(x)

f

(x)
2

f(x)
dx =

f(x)

x
α
dx =
x
α+1
α + 1
for α == −1

1
x
dx = log x

e

ax
dx =
e
ax
a
2232

a
bx
dx =
a
bx
b log a
for a > 0

log x dx = x log x −x

1
x
2
+ a
2
dx =
1
a
arctan
x
a

1

x
2
− a
2
dx =

1
2a
log
a−x
a+x
for x
2
< a
2
1
2a
log
x−a
x+a
for x
2
> a
2

1
√
a
2
− x

2
dx = arcsin
x
|a|
= −arccos
x
|a|
for x
2
< a
2

1
√
x
2
± a
2
dx = log(x +
√
x
2
± a
2
)

1
x
√
x

2
− a
2
dx =
1
|a|
sec
−1
x
a

1
x
√
a
2
± x
2
dx = −
1
a
log

a +
√
a
2
± x
2
x



sin(ax) dx = −
1
a
cos(ax)

cos(ax) dx =
1
a
sin(ax)
2233

tan(ax) dx = −
1
a
log cos(ax)

csc(ax) dx =
1
a
log tan
ax
2

sec(ax) dx =
1
a
log tan


π
4
+
ax
2


cot(ax) dx =
1
a
log sin(ax)

sinh(ax) dx =
1
a
cosh(ax)

cosh(ax) dx =
1
a
sinh(ax)

tanh(ax) dx =
1
a
log cosh(ax)

csch(ax) dx =
1
a

log tanh
ax
2

sech(ax) dx =
i
a
log tanh

iπ
4
+
ax
2


coth(ax) dx =
1
a
log sinh(ax)
2234

x sin ax dx =
1
a
2
sin ax −
x
a
cos ax


x
2
sin ax dx =
2x
a
2
sin ax −
a
2
x
2
− 2
a
3
cos ax

x cos ax dx =
1
a
2
cos ax +
x
a
sin ax

x
2
cos ax dx =
2x cos ax

a
2
+
a
2
x
2
− 2
a
3
sin ax
2235
Appendix F
Definite Integrals
Integrals from −∞ to ∞. Let f(z) be analytic except for isolated singularities, none of which lie on the real
axis. Let a
1
, . . . , a
m
be the singularities of f(z) in the upper half plane; and C
R
be the semi-circle from R to −R in
the upper half plane. If
lim
R→∞

R max
z∈C
R
|f(z)|


= 0
then

∞
−∞
f(x) dx = ı2π
m

j=1
Res (f(z), a
j
) .
Let b
1
, . . . , b
n
be the singularities of f(z) in the lower half plane. Let C
R
be the semi-circle from R to −R in the lower
half plane. If
lim
R→∞

R max
z∈C
R
|f(z)|

= 0

then

∞
−∞
f(x) dx = −ı2π
n

j=1
Res (f(z), b
j
) .
2236
Integrals from 0 to ∞. Let f (z) be analytic except for isolated singularities, none of which lie on the positive real
axis, [0, ∞). Let z
1
, . . . , z
n
be the singularities of f(z). If f(z)  z
α
as z → 0 for some α > −1 and f(z)  z
β
as
z → ∞ for some β < −1 then

∞
0
f(x) dx = −
n

k=1

Res (f(z) log z, z
k
) .

∞
0
f(x) log dx = −
1
2
n

k=1
Res

f(z) log
2
z, z
k

+ ıπ
n

k=1
Res (f(z) log z, z
k
)
Assume that a is not an integer. If z
a
f(z)  z
α

as z → 0 for some α > −1 and z
a
f(z)  z
β
as z → ∞ for some
β < −1 then

∞
0
x
a
f(x) dx =
ı2π
1 −
e
ı2πa
n

k=1
Res (z
a
f(z), z
k
) .

∞
0
x
a
f(x) log x dx =

ı2π
1 −
e
ı2πa
n

k=1
Res (z
a
f(z) log z, z
k
) , +
π
2
a
sin
2
(πa)
n

k=1
Res (z
a
f(z), z
k
)
Fourier Integrals. Let f(z) be analytic except for isolated singularities, none of which lie on the real axis. Suppose
that f(z) vanishes as |z| → ∞. If ω is a positive real number then

∞

−∞
f(x)
e
ıωx
dx = ı2π
n

k=1
Res(f(z)
e
ıωz
, z
k
),
where z
1
, . . . , z
n
are the singularities of f(z) in the upper half plane. If ω is a n egative real nu mber then

∞
−∞
f(x)
e
ıωx
dx = −ı2π
n

k=1
Res(f(z)

e
ıωz
, z
k
),
where z
1
, . . . , z
n
are the singularities of f(z) in the lower half plane.
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