Preview: Preliminary and Engineering Problems
Professor K.T. Chau
•Binomial theorem
•Euler’s formula
•Differentiation
•Integration
•Complex variables
•Tensor and vector calculus
•Hyperbolic functions
•Simplest Differential Equations
P-1
Review on fundamentals (something you MUST know)
Binomial theorem
n(n − 1) n − 2 2
( x + h) = x + nx h +
x h + ⋅⋅⋅ + nxh n −1 + h n
2
n
n
n −1
( a + b)0 = 1
P-2
(1261)
賈憲三角
杨辉三角
Apianus (1527)
Pascal’s triangle (1654)
(a + b)1 = a + b
(a + b) 2 = a 2 + 2ab + b 2
(a + b)3 = a 3 + 3a 2 b + 3ab 2 + b3
(a + b) 4 = a 4 + 4a 3b + 6a 2b 2 + 4ab3 + b 4
...
Proof of power series of ex
x
n x n(n − 1) x 2 n(n − 1)(n − 2) x 3
e x = lim (1 + ) n = lim [1 + ( ) +
( ) +
( ) + ...]
n →∞
n
→∞
n
1! n
2!
n
3!
n
x (1 − 1/ n) x 2 (1 − 1 / n)(1 − 2 / n) x3
= lim [1 + +
+
+ ...]
n →∞
1!
2!
3!
x x 2 x3
= 1+ +
+
+ ...
1! 2! 3!
Yang Hui
Pascal
Jia Xian
n(n − 1) n − 2 2
( x + h) = x + nx h +
x h + ⋅⋅⋅ + nxh n −1 + h n
2
= x n + Cn1 x n −1h + Cn2 x n − 2 h 2 + ⋅⋅⋅ + Cnr x n − r h r + ... + Cnn −1 xh n −1 + h n
n
n −1
n
Related to probability
Cnr =
n(n − 1)...(n − r + 1)
n!
=
r!
r !(n − r )!
Example P-1
C
r −1
n −1
+C
r
n −1
Basic identity
(n − 1)!
(n − 1)!
=
+
(r − 1)!(n − r )! r !(n − 1 − r )!
(n − 1)!
=
[r + (n − r )]
(r − 1)!(n − r )!
n!
=
= Cnr
r !(n − r )!
Problem P-1
Find S?
S = 1 + Cn1 + Cn2 + ⋅⋅⋅ + Cnr + ... + Cnn −1 + 1
Hua Luogeng
P-3
Problem P-2
Find S1?
S1 = 1 − Cn1 + Cn2 + ... + (−1) n −1 Cnn −1 + (−1) n
P-4
Review on compound interest and the birth of e
Growth of money in bank
Euler
r
=
annual
interest
rate
S = P (1 + r )
Pr = interest at the end of 1 year
After ½ year, take out money and redeposit (a bigger P)
r
r
r 2
S = [ P (1 + )](1 + ) = P(1 + )
Compound interest
2
2
2
After 1/n year, take out money and redeposit (a bigger P)
r
r
r n
S = [ P (1 + )...](1 + ) = P (1 + )
n
n
n
Euler’s number
The best you can earn in compound interest (in 1 year)
r n
S = P lim (1 + ) = Pe r
1 n
e = lim (1 + ) = 2.71828...
n →∞
n
n →∞
n
P-5
History of Complex Numbers
Italian mathematician Gerolamo Cardano is the first
known to have introduced complex numbers in 1545.
He called them "fictitious", during his attempts to find
solutions to cubic equations in the 16th century
ax3 + bx 2 + cx + d = 0
General formula of roots for cubic equation
Gerolamo Cardano
(1501-1576)
i = −1
Review on Euler’s formula
P-6
x + iy = reiθ = r (cos θ + i sin θ )
Demoivre’s Formula
(cos θ + i sin θ ) n = cos n θ + i sin nθ
Example P-2
Problem P-3
Special case of r =1 and
eiπ = −1
Find
i = −1
θ =π
eiπ + 1 = 0
Euler’s formula
ii = ?
(Answer: Infinite answers and the smallest one is 0.207879576…)
Problem P-4
Find
i
i =?
(Answer: Infinite answers and the smallest one is 4.810477381…)
{
}
d
U ( k −1)V − U ( k − 2)V (1) + U ( k −3)V (2) − ... + (−1)( k −1) UV ( k −1) + (−1) k UV ( k )
dx
L’Hopital’s Rule (actually by Johann Bernoulli)
0 ∞
,
0 ∞
f ( x), g ( x) → ∞
f ( x), g ( x)
VU ( k ) =
x→a
f ( x), g ( x) → 0
are differentiable lim
x→a
f ( x)
f ′( x)
= lim
g ( x) x → a g ′( x)
Bernoulli
L’Hopital
Swiss mathematician and scientist
P-7
Teacher & inspirer
Saturday afternoon lessons from
Johann Bernoulli
1727
Russia
Euler (1707-1783)
Catherine I
Johann Bernoulli
(1667 – 1748)
Father-son
1741
Basel University
(1459)
Euler cannot get a job
at Basel University
the oldest university
in Switzerland
Prussia
Frederick II
friends
Nicolaus II Bernoulli Daniel Bernoulli
P-8
Swiss franc
P-9
Seven Bridges of
Königsberg
Graph theory
Letters of Euler on different Subjects in Natural Philosophy
Addressed to a German Princess
German Princess
886 publications of Euler available here
http://www.math.dartmouth.edu/~euler/tour/tour_00.html
Review on differentiation
Power rule
Definition df
f ( x + h) − f ( x)
= f ′ ( x ) = lim
h →0
dy
h
d
(c) = 0
dx
Product rule
Sum rule
du dv
d
dv
du d
(u + v ) = +
( uv ) = u + v
dx dx
dx
dx
dx dx
Circular functions (Trigonometric Functions )
d
du
sin ( u ) = cos u
dx
dx
Chain Rule:
Constant multiple
dy dy du
=
⋅
dx du dx
d n
d n
du
x ) = nx n −1
u ) = nu n −1
(
(
dx
dx
dx
Quotient rule
d
du
( cu ) = c
dx
dx
d
du
cos ( u ) = − sin u
dx
dx
d
1 du
cos −1 u = −
dx
1 − u 2 dx
d
1 du
sin −1 u =
dx
1 − u 2 dx
P-10
du
dv
−u
d ⎛u⎞
dx
dx
⎜ ⎟=
2
dx ⎝ v ⎠
v
v
d
du
tan ( u ) = sec 2 u
dx
dx
d
1 du
tan −1 u =
dx
1 + u 2 dx
y = y (u ), u = u ( x)
Hyperbolic functions
d
du
(sinh u ) = cosh u
dx
dx
d
du
(cosh u ) = sinh u
dx
dx
d
du
(tanh u ) = sech 2u
dx
dx
d
1
du
(sinh −1 u ) =
dx
1 + u 2 dx
d
(cosh −1 u ) =
dx
d
1 du
(tanh −1 u ) =
dx
1 − u 2 dx
du
u 2 − 1 dx
1
Derivatives of Exponential and Logarithmic Functions
d
1 du
ln u =
dx
u dx
d
1 du
log a u =
dx
u ln a dx
d u
du
a = a u ln a
dx
dx
( )
d u
du
e = eu
dx
dx
P-11
Leibniz's rule of differentiation under integral sign
h ( x ) df ( x, ξ )
d h( x)
dh( x)
dg ( x)
ξ
ξ
=
ξ
+
−
f
(
x
,
)
d
d
f
[
x
,
h
(
x
)]
f
[
x
,
g
(
x
)]
∫g ( x ) dx
dx ∫g ( x )
dx
dx
General Leibniz rule
Example P-3
k
n−k
n
dn
g
k d f d
(
fg
)
=
C
∑
n
dx n
dx k dx n − k
k =0
Leibniz
0
d9
d 9 sin x
d
d 8 sin x 9 ⋅ 8 d 2
d 7 sin x
+ 9 ( x)
+
+ ...
( x sin x) = x
( x)
9
9
8
2
7
2! dx
dx
dx
dx
dx
dx
= x cos x + 9sin x
Partial differentiation
Total differential
Proof
∂f
f ( x + Δ x, y ) − f ( x, y )
∂f
= lim
= ( )y = fx
∂x Δ x→0
∂x
Δx
∂f
∂f
f ( x, y + Δ y ) − f ( x, y )
= lim
= ( )x = f y
Δy
∂y Δ y →0
∂y
x → x + Δ x and y → y + Δ y ⇒ f → f + Δ f
∂f
∂f
Δ f = f ( x + Δ x, y + Δ y ) − f ( x , y )
df =
dx + dy
∂x
∂y
= f ( x + Δ x, y + Δ y ) − f ( x, y + Δ y ) + f ( x, y + Δ y ) − f ( x , y )
∂f
∂f
∂f
f ( x + Δ x, y + Δ y ) − f ( x, y + Δ y )
f ( x, y + Δ y ) − f ( x , y )
df =
dx1 +
dx2 + ... +
dxn
]Δ x + [
]Δ y
=[
∂x1
∂x2
∂xn
Δx
Δy
as Δ x → 0 and Δ y → 0, the total differential df is
∂f
∂f
df =
dx + dy
∂x
∂y
Chain rule
f = f ( x, y ) and x = x(u ), y = y (u )
df =
∂f
∂f
df ∂f dx ∂f dy
dx + dy ⇒
=
+
du ∂x du ∂y du
∂x
∂y
Problem P-5 Show that
∂2
∂ 2v
∂u ∂v
∂u ∂v
∂ 2u
+2
+2
+v
(uv) = u
∂x1∂x2
∂x1∂x2
∂x1 ∂x2
∂x2 ∂x1
∂x1∂x2
Example P-4
Polar coordinates ρ and φ, Cartesian coordinates x and y, x=ρcos φ ,
2
2
∂
f
∂
f
y=ρsinφ, f ( x, y ) → f ( ρ , ϕ ) transform
into one in ρ and φ
+
2
2
∂x
∂y
ρ 2 = x2 + y2 ,
x
∂ρ
∂ρ
cos
,
= 2
=
ϕ
= sin ϕ
∂x ( x + y 2 )1/2
∂y
∂φ
− y / x2
−y
− ρ sin ϕ − sin ϕ ∂φ cos φ
,
ϕ = tan (y / x ),
=
=
=
=
=
2
2
2
2
x +y
ρ
∂x 1 + ( y / x)
ρ
ρ
∂y
∂ ∂ρ ∂ ∂ϕ ∂
∂ sin ϕ ∂
∂
∂ cos ϕ ∂
,
=
+
= cos ϕ
−
= sin ϕ
+
∂x ∂x ∂ρ ∂x ∂ϕ
∂ρ
ρ ∂ϕ ∂y
∂ρ
ρ ∂ϕ
−1
∂ ∂f
∂ 2 f ∂ 2 f ∂ 2 f 1 ∂f
1 ∂2 f
∂2 f
∂ ∂f
∂2
2
= ( ) ⇒ ∇ f ( x, y ) = 2 + 2 = 2 +
+ 2
= ( ) and
2
2
∂y
∂y ∂y
∂x
∂y
∂ρ
ρ ∂ρ ρ ∂ϕ 2
∂x
∂x ∂x
P-12
∞
Taylor series expansion
MaClaurin series
Examples
f ( x) = ∑
n =1
∞
f ( x) = ∑
n =1
P-13
f ( n ) ( x0 )
(x − x0 )n ,
n!
f ( n ) (0)
n
( x)
n!
x2 x3
x2 x4 x6
x3 x5 x7
e = 1+ x +
+
+ " cos( x) = 1 −
+
− + " sin( x) = x −
+ − +"
2! 3!
2! 4! 6!
3! 5! 7!
x
Example P-5
Expand f ( x) = sin x
f ( x) = sin x
f '( x) = cos x
f (0) = 0
f '(0) = 1
f ''( x) = − sin x
f ''(0) = 0
f '''( x) = − cos x
f '''(0) = −1
f (4) ( x) = sin x
f (4) (0) = 0
Example P-6
Second approach
f ( x) = f ′( x) = f ′′( x) = f ′′′( x) = ... = e x
f (0) = f ′(0) = f ′′(0) = f ′′′(0) = ... = 1
f '(0)
f ''(0) 2 f '''(0) 3
x+
x +
x + ⋅⋅⋅
1!
2!
3!
x3 x5 x 7
= x − + − + ⋅⋅⋅
3! 5! 7!
∞
x 2 n +1
x3 x5 x 7
n
sin x = ∑ (−1)
= x − + − + ⋅⋅⋅
(2n + 1)!
3! 5! 7!
n=0
f (0) +
f ( x) = e x
f '(0)
f ''(0) 2 f '''(0) 3
x+
x +
x + ⋅⋅⋅
1!
2!
3!
x 2 x3
= 1 + x + + +"
2! 3!
e x = f (0) +
All 3 formulas are derived
i 2 x 2 i 3 x3 i 4 x 4 i 5 x5
e = 1 + ix +
+
+
+
+ " Euler formula
x2 x4 x6
2!
3!
4!
5!
ix
e = cos x + i sin x cos( x) = 1 − 2! + 4! − 6! + "
2
4
3
5
x
x
x
x
3
5
7
= 1 − + + " + i ( x − + + ")
x
x
x
2! 4!
3! 5!
sin( x) = x − + −
+"
ix
3!
5!
7!
Review on integration
Definition
P-14
Summing area under a function
Multiple integral
Finding volume
Integration by parts
∫ u dv = uv − ∫ v du
Generalized integration by parts
b
V = ∫ π y 2dx
a
∫
d
V = ∫ π x 2dy
f ( n ) g dx = f ( n −1) g − f ( n − 2) g ′ + f ( n −3) g ′′ − ...(−1) n ∫ fg ( n ) dx
c
P-15
General rule of integration
∫
f (ax)dx =
∫
F { f ( x)}dx = F (u )
1
a
∫ f (u)du
∫
∫
∫
a u du = eu ln a du =
dx
F (u )
du =
du
du
f ′( x)
∫
u ln a
u
e
a
=
ln a ln a
∫
u = f ( x)
a > 0, a ≠ 1
∫
Transformation rule
∫
∫
∫
∫
1
u = ax + b
F (u )du
a
2
F ( ax + b )dx =
uF (u )du
u = ax + b
a
n n −1
u = n ax + b
F ( n ax + b )dx =
u F (u )du
a
F (ax + b)dx =
∫
∫
∫
∫ F(
∫ F(
∫
+ x )dx = a ∫ F (a sec u ) sec udu
− a )dx = a ∫ F (a tan u ) sec u tan udu
F ( a 2 − x 2 )dx = a F (a cos u ) cos udu
a2
x2
2
2
2
∫
x = a sin u
x = a tan u
x = a sec u
u n +1
u dx =
n +1
1
du = ln u
u
n
eu du = eu
n ≠ −1
∫
F (u )
du
u
x = eax
F (ln x)dx = F (u )eu du
u = ln x
1
F (e )dx =
a
ax
∫
∫ F (sin
∫
P-16
∫
∫
−1
x
)dx = a F (u ) cos udu
a
∫
2u
u = sin −1
1− u2
du
,
)
F (sin x, cos x)dx = 2 F (
1+ u2 1+ u2 1+ u2
∫
x
a
u = tan
x
2
Definite integrals
b
∫
∫
f ( x)dx = (b − a) f (c),
a
b
a
f ( x) g ( x)dx = f (c)
∫
b
a
g ( x)dx,
a<c<b
Mean value theorem
a<c<b
Generalized mean value theorem
Improper integrals
∫
∫
b
a
∞
a
∫
∞
0
b
∫ ε
ε
f ( x)dx = lim ∫ f ( x)dx
f ( x)dx = lim
→0 a +
b
f ( x)dx
f (a)
singular
b →∞ a
f (ax) − f (bx)
b
dx = { f (0) − f (∞)}ln
x
a
Frullani’s integral
f ( x ) − f (∞ )
dx
1
x
converges
∫
∞
Review on complex variables
P-17
z = x + iy
f ( z ) = u ( x, y ) + iv( x, y )
Single valued function if there is one-to-one
coorespondence between z and f(z), otherwise multivalued
df ( z )
f ( z + Δz ) − f ( z )
f ′( z ) =
= lim
Δz →0
Δz
dz
If derivative exists, it is called analytic. (just another term for differentiable for
real variables)
Cauchy-Riemann equation (necessary condition for analytic)
∂u ∂v
= ,
∂x ∂y
Example P-7
∂ 2u
∂x 2
+
∂ 2u
∂y 2
Problem P-5
∂u
∂v
=−
∂y
∂x
(just another term for left-hand limit equals to = right
hand limit for real variables)
∂ ∂u
∂2v
( )=
,
∂x ∂x
∂x∂y
∂ ∂u
∂2v
( )=−
∂y ∂y
∂y∂x
since
= ∇ 2 u = 0 This is called Laplace equation and
will be discussed in later chapter
∂2v
2
Show that ∂x
+
∂2v
∂y 2
= ∇2v = 0
∂2v
∂ 2v
=
∂y∂x ∂x∂y
(order of differentiation is
not important)
This indicates that Laplace equation
can be solved by analytic functions!!
Cauchy’s Theorem
v∫
C
Closed curve = C
f ( z ) is analytic within C
f ( z )dz = 0
Cauchy’s integral formulas
f ( z) =
1
2π i
f (n) ( z ) =
>∫
n!
2π i
C
C
f ( z)
dz
z−a
>∫
P-18
f ( z)
C
( z − a)
n +1
dz
Remarkable results: Value of
f(z) and its higher derivatives
only depends on boundary
values on C
Singular points
f (a ) is not analytic then a = isolated singular point
Poles
f ( z) =
Example P- 8
φ ( z)
( z − a)n
f ( z) =
φ (a ) ≠ 0
(or f(z) has singularity at a)
z = a is a pole of order n
z = a is a simple pole if n =1
z
( z − 3) 2 ( z + 1)
has two singularities, a pole of order 2 at z=2
and a simple pole at z = −1.
P-19
Laurent’s series
f ( z) =
a− n
( z − a)
n
a− n +1
+
( z − a ) n −1
∞
a
+ ... + −1 + a0 + a1 ( z − a ) + a2 ( z − a )2 + ... =
ak ( z − a ) k
( z − a)
k =−∞
∑
Principal part
Analytic part
This the principal part has infinite terms, it is an essential singularity
Example P-9
1
1
has is an essential singularity at z = 0
e1/ z = 1 + +
+
...
z 2! z 2
Residue
a−1
1
d n −1
n
= lim
−
{(
z
a
)
f ( z )}
z → a ( n − 1)! dz n −1
a−1 = residue
a−1 = lim( z − a ) f ( z )
For simple pole
Residue Theorem
>∫
b
z →a
C
c
f ( z )dz = 2π i (a−1 + b−1 + c−1 + ...)
a
C
P-20
Removable singularity
z −π = u
Example P-10
u3 u5
u2 u4
sin z sin(u + π )
sin u
1
=
=−
= − (u − + − ...) = −1 +
−
+ ...
z −π
u
u
u
3! 5!
3! 5!
( z − π )2 ( z − π )4
= −1 + +
−
+ ...
(no singularity)
3!
5!
Branch point & Branch cut
w= n z
z = ρ eiθ
w = ρ 1/ n eiθ / n
0-x is a branch cut (restrict θ to ensure single-valuedness)
n branches of single-valued functions
n
n
n
{ z }0 ,{ z }1 ,...,{ z }n −1
2π
n
ρ
θ /n
θ
Branch cut
ρ 1/n
w − plane
2π
z − plane
n different points in w-plane corresponds
to the same point in z-plane
Example P-11
Consider
>∫
∫
r
x p −1
π
dx =
, 0 < p <1
1+ x
sin pπ
Contour C = ABEGDJA
R
J
z p −1
lim (1 + z )
= (eπ i ) p −1 = e( p −1)π i
z →−1
1+ z
z p −1
dz =
C 1+ z
x p −1
dx +
1+ x
∫
2π
0
∫
AB
... +
∫
... +
BDEFG
GH
(Reiθ ) p −1 iR iθ dθ
1 + Reiθ
∫
... +
+
∫
0
x p −1
dx +
1+ x
[1 − e
2π i ( p −1)
∫
0 e 2π i ( p −1) x p −1dx
∞
∫
∞
]
0
1+ x
H G
... = 2π ie( p −1)π i
r
( xe2π i ) p −1 dx
R
1 + xe 2π i
∫
= 2π ie( p −1)π i
x p −1
dx = 2π ie( p −1)π i
1+ x
B
D
+
0
(reiθ ) p −1 ir iθ dθ
2
1 + reiθ
∫π
= 2π ie( p −1)π i
dz = ir iθ dθ r → 0, p > 0
... → 0
∫ ... → 0
HJA
BDEFG
∞
A
HJA
∫
dz = iRiθ dθ R → ∞, p < 1
∫
r
−1
E
P-21
y
C
Branch point: z=0
Residue:
R
0
z p −1
dz , 0 < p < 1
C 1+ z
Pole: z=−1
>∫
∫
Show that
∞
∫
∞
0
x p −1
2π ie( p −1)π i
2π i
dx =
=
1+ x
1 − e 2π i ( p −1) e pπ i − e− pπ i
=
π
sin pπ
QED
x
Elementary Tensor Analysis
P-22
Scalar (independent of direction)
1st order (vector) (e.g. displacement)
zeroth rank or order (e.g. temperature, pressure)
3
u = u1e1 + u2e2 + u3e3 =
First rank or order (e.g. displacement, velocity)
2nd order (e.g. stress, strain)
Fourth rank or order (e.g. elastic tensor )
σ=
Normal vector on plane 3 is e3
3
3
∑∑ σ
ij eie j
= σ ij eie j
i =1 j =1
Plane 3
There are two direction senses
Plane 1
σ ij
Plane 2
Plane i
i = 1, 2,3; j = 1, 2,3
3× 3 = 9
Direction j
4th order tensor (stiffness tensor)
e3
e2
Einstein notation
(drop summation)
We need 9 components to fully
describe stress at a point!!
2nd order tensor (e.g. stress)
x3
= ui ei
3 components to fully describe a vector
Third rank or order (e.g. permutation tensor )
x1
i i
i =1
Second rank or order (e.g. stress, strain)
e1
∑u e
3
x2
C=
3
3
3
∑∑ ∑ ∑ C
ijkl eie j e k el
i =1 j =1
k =1
l =1
= Cijkl eie j ek el
3 × 3 × 3 × 3 = 81
Woldemar Voigt
1850 –1919
P-23
Vector analysis
3
u = u1e1 + u2 e2 + u3e3 =
∑u e
i i
= ui ei
i =1
1st PhD in USA in
engineering in 1863
u(u1,u2,u3)
e3
e1
Gibbs, J.W.
(1839-1903)
Riemann, G.F.B.
Heaviside O.
(1850-1925)
(1826 –1866) Age =39
The Clay Mathematics Institute (Cambridge, Massachusetts)
e2
Riemann Hypothesis
$1 million award
Story
This formula says that the zeros of the Riemann zeta function control
the oscillations of primes around their "expected" positions.
Permutation tensor
eijk
1
= (i − j )( j − k )(k − i )
2
e133 = e221 = e131 = 0
odd
This is not a tensor equation!
e123 = e231 = e312 = 1
Even permutation
3
1
even
2
e132 = e213 = e321 = −1
Odd permutation
A Brief Review of Vector Analysis
dot product
| u ⋅ v | = | u | | v|cosθ
u = u1e1 +u2e 2 +u3e3 ,
P-24
(0 ≤ θ ≤ π )
v = v1e1 +v2 e2 +v3e3
3
u ⋅ v = u1v1 + u2 v2 + u3v3 =
∑
uk vk = uk vk
k =1
Repeated indices in component
form means summation
cross product
w = u × v =| u || v | sin θ
Polyadic form
(0 ≤ θ ≤ π / 2)
German
w = wi ei = eijk u j vk ei
component form
wi = eijk u j v k
same
wi = eimn um vn
Rule in tensor notation
j is a dummy index (repeated)
k is a dummy index (repeated)
i is a free index (not repeated)
Balance in free index on both size of “=“.
Leopold Kronecker
(1823 –1891)
Kronecker delta
ei ⋅ e j = δ ij
δ ij = 0, i ≠ j
= 1, i = j
Problem P-6
P-25
Show that these are the same
w = u × v = (u2 v3 − u2 v3 ) e1 + (u3v1 − u1v3 ) e 2 + (u1v2 − u2 v1 ) e3
wi e i = eijk u j v k e i
e1 e 2
u × v = u1 u2
v1
v2
e3
u3
v3
= (u2 v3 − u3v2 )e1 + (u3v1 − u1v3 )e 2 + (u1v2 − u2 v1 )e3
u × v = −(v × u)
u × (v + w) = u × v + u × w
u× u = 0
ku × v = u × kv = k (u × v )
determinant
det( Aij ) = e ijk Ai1 A j 2 A k 3
e-δ identity
e ijk e irs = δ jrδ ks − δ jsδ kr
e3
e1 × e2 = e3
e2 × e3 = e1
e1
e3 × e1 = e2
e2
Vector calculus
Derivatives of tensors
Vector differential operator
comma-subscript convention
∇ = ei
∂
∂xi
σ=σ(x,t)
∂ϕ
ei = ϕ,i ei
∂xi
grad ϕ = ∇ϕ =
Curl
Divergence
curl v = ∇ × v = eijk vk , j ei
Useful formulas
div v = ∇ • v = vi ,i
∇ 2ϕ = ∇ • ∇ϕ = ϕ,ii
Identities exist for the differential operator ∇
∇ ( fg ) = f ∇g + g ∇f
2
∇ • ( fv ) = (∇f ) • v + f ∇ • v
∇ • ( f ∇g ) = f ∇ 2 g + ∇f • ∇ g
∇ × (∇f ) = 0
∇ • (∇ × v ) = 0
∇ • (a × b) = (∇ × a) • b − a • ∇ × b
∇ ( fg ) = f ∇ g + 2(∇f ) • (∇g ) + g ∇ f
2
P-26
∂σ ij
∂ε ij
∂v i
= v i, j ,
= σ ij, k ,
= ε ij, j
∂x j
∂xk
∂x j
Gradient
Laplacian
∂
∂
∂
+ e2
+ e3
∇ = e1
∂x1
∂x2
∂x3
2
∇ × ( fv ) = ∇ f × v + f ∇ × v
∇ × (∇ × v ) = ∇ (∇ • v) − ∇ 2 v
∇ • ∇ 2 a = ∇ 2 (∇ • a )
∇ 2 (∇φ ) = ∇ (∇ 2φ )
∇ 2 (a • r ) = 2∇ • a + r ∇ 2 a
∇ 2 (φ r ) = 2∇φ + r ∇ 2φ
∇ • σ = (e k
∇ × σ = (e i
∂
∂ xk
) • (σ ij e ie j) = σ ij,i e j
∂
) × (σ jk e je k )= σ
∂xi
The Divergence Theorem
n
V
dS
jk ,i (e i × e j)e k
= elijσ jk ,i e l e k
∫ n ⋅ TdS = ∫ ∇ ⋅ TdV
S
Stokes Theorem.
P-27
S
S
V
n
dS
C
∫ CT ⋅ ds = ∫ S (∇ ×T ) ⋅ n dS
ds
Some Formulae in Cylindrical Coordinate
r = x1 e1 + x2 e 2 + x3 e 3 = r cos ϕ e1 + r sin φ e 2 + z e 3
1 ∂r
1 ∂r
= cos ϕ e1 + sin ϕ e 2 , e ϕ =
= − sin ϕ e 1 + cos ϕ e 2
h r ∂r
hϕ ∂ϕ
1 ∂r
= e3
ez =
h z ∂z
er =
∂ eϕ
∂e r
= eϕ ,
= − er
∂ϕ
∂ϕ
eα =
1 ∂r
hα ∂ xα
hα =
ez
φ
r
er
er
∂r
∂ xα
Some formulas in cylindrical coordinate
∂
∂
∂
+ eϕ
+ e z )(u r e r + u ϕeϕ + u ze z )
∂r
∂z
r ∂ϕ
∂u ϕ
∂u ϕ
∂u
∂u
∂u r
1
1 ∂u z
)eϕ eϕ + z e ze z +
= r e re r + (u r +
+
+
eϕ e z
e z eϕ
e ze r
∂r
∂ϕ
∂z
∂z
∂z
r
r ∂ϕ
∂u ϕ
∂u z
1 ∂u r
+
− u ϕ )eϕ e r
e re z +
e r eϕ + (
∂r
∂r
r ∂ϕ
∇u = (e r
ε=
1
(∇u+ u∇ )
2
ε zz =
∂u z
∂u
1 ∂u θ
u
, ε rr = r , ε θθ = r +
r r ∂θ
∂z
∂r
1 ∂u ϕ 1 ∂u r u ϕ
1 1 ∂u z ∂u ϕ
1 ∂u r ∂u z
+
+
+
) , ε rz = (
)
− ), εϕ z = (
ε rϕ = (
∂z
∂r
r
2 ∂r r ∂ϕ
2 r ∂ϕ
2 ∂z
∂ur 1
1 ∂uϕ ∂u z
+ ur +
+
∇•u =
r ∂ϕ
∂r r
∂z
∂u ϕ u ϕ 1 ∂u r
1 ∂u z ∂u ϕ
∂u r ∂u z
) + eϕ (
) + e z(
)
∇ × u = er(
−
−
−
+
∂z
∂z
∂r
∂r
r ∂ϕ
r r ∂ϕ
∂ 2 f 1 ∂f 1 ∂ 2 f ∂ 2 f
+ 2
+ 2
∇ f = ∇ • ∇f = 2 +
2
r ∂r r ∂ϕ
∂r
∂z
2
P-28
Example P-12 Laplace equation in 2-D polar form
∂
∂
∂
1 ∂
+ ey
= er
+ eθ
∇ = ex
r ∂θ
∂x
∂y
∂r
P-29
∂eθ
= − er
∂θ
eθ
∂
∂
∂
∂
∂ 2u ∂ 2u
∇ u = ∇ ⋅ ∇ u = (e x
+ e y ) ⋅ (e x
+ e y )u = ( 2 + 2 )
∂x
∂y
∂x
∂y
∂x
∂y
2
∇ 2 u = ∇ ⋅ ∇ u = (er
∂
er ⋅ (er
∂r
1 ∂
⋅ (er
eθ
r ∂θ
∂u
+ eθ
∂r
∂u
+ eθ
∂r
∂
1 ∂
∂u
1 ∂u
+ eθ
) ⋅ (er
+ eθ
)
r ∂θ
r ∂θ
∂r
∂r
∂er
= eθ
∂θ
dθ
er
θ
1 ∂u
∂ 2u
)= 2
eθ ⋅ eθ = 1
er ⋅ er = 1 eθ ⋅ er = 0
∂r
r ∂θ
1 ∂u
1 ∂er ∂u
1 ∂ 2u 1 ∂eθ 1 ∂u
1 ∂ 2u
) = eθ ⋅ [ (
) + er
)
]
+ (
+ eθ 2
2
r ∂θ
r ∂θ ∂r
r ∂θ∂r r ∂θ r ∂θ
r ∂θ
1 ∂u
1 ∂ 2u
1 ∂u
1 ∂ 2u
]
= eθ ⋅ [eθ
+ er
− er 2
+ eθ 2
2
r ∂r
r ∂θ∂r
r ∂θ
r ∂θ
1 ∂u 1 ∂ 2u
=
+ 2
r ∂r r ∂θ 2
1
1
urr + u r + 2 uθθ = 0
r
r
Will be useful later!
Review on Hyperbolic functions
sinh x =
csch x =
e −e
2
x
1
sinh x
−x
x
cosh x =
e +e
2
sech x =
1
cosh x
−x
tanh x =
coth x =
P-30
sinh x
cosh x
cosh x
cosh x tanh x
sinh x
sinh x
sinh( − x) = − sinh x
cosh( − x ) = cosh x
cosh 2 x − sinh 2 x = 1
sinh 2 x = cosh 2 x − 1
cos 2 x + sin 2 x = 1
sinhx, coshx versus sinx, cosx
P-31
Lambert (1728–1777)
V Riccati (1707–1775)
P-32
sinhx, coshx versus sinx, cosx
Hyperbolic
Circular
functions
functions
Resemblance
hyperbola
e −e
sinh x =
2
x
−x
circle
e x + e− x
cosh x =
2
1 − tanh 2 x = sech 2 x
sinh( x + y ) = sinh x cosh y + cosh x sinh y
cosh( x + y ) = cosh x cosh y + sinh x sinh y
d
(sinh x) = cosh x
dx
d
(cosh x) = sinh x
dx
d
(tanh x) = sech 2 x
dx
d
(sechx) = −sechx tanh x
dx
d
(csch x) = −csch x coth x
dx
d
(coth x) = −csch 2 x
dx
d
1
(sinh −1 x ) =
dx
1 + x2
d
(cosh −1 x ) =
dx
1
x2 − 1
d
1
(tanh −1 x ) =
dx
1 − x2
d
1
(sech −1 x) = −
dx
x 1 − x2
d
1
(csch −1 x) = −
dx
x 1 + x2
d
1
(coth −1 x) =
dx
1 − x2
P-33
∫ sinh udu = cosh u + C
∫ cosh udu = sinh u + C
∫ sech udu = tanh u + C
2
∫
∫
∫
du
a +u
du
2
2
= sinh
−1
∫ csch u cot udu = −csch u + C
∫ sech u tanh udu = −sech u + C
∫ csch udu = − coth u + C
2
∫u
u
+C
a
u
= cosh −1 + C
a
u 2 − a2
⎧1
2⎫
2
−1 u
C
a
tanh
+
<
if
u
⎪⎪ a
⎪⎪
du
a
= ⎨
⎬
u
1
a2 − u 2
⎪ coth −1 + C if u 2 > a 2 ⎪
⎪⎩ a
⎪⎭
a
Problem P-7 Prove
∫u
du
a −u
du
2
u +a
2
2
2
=
−1
u
sec h −1 + C
a
a
0<u<a
=
u
−1
csc h −1 + C
a
a
u≠0
x3
sinh x = x +
+
3!
x2
cosh x = 1 +
+
2!
x5
+ ...
5!
x4
+ ...
4!
sinh x = −i sin ix, cosh x = cos ix
Most fundamental differential equations
Differential Equation for circular
functions
d2y
2
+
λ
y=0
2
dx
Try
y ( x) = eα x
Differential Equation for hyperbolic
functions
d2y
2
−
λ
y=0
2
dx
Try
y ( x) = eα x
(α 2 − λ 2 )eα x = 0
(α 2 + λ 2 )eα x = 0
α = ±iλ
y ( x) = C1eiλ x + C1e −iλ x
P-34
α = ±λ
Left-hand
side is real
y ( x) = C1eλ x + C2 e − λ x
eix = cos x + i sin x
eλ x = cosh λ x + sinh λ x
e −ix = cos x − i sin x
e − λ x = cosh λ x − sinh λ x
y ( x) = D1 sin λ x + D2 cos λ x
d 2 sin λ x
dx 2
d 2 cos λ x
dx 2
= −λ 2 sin λ x,
= −λ cos λ x,
2
y ( x) = D1 sinh λ x + D2 cosh λ x
d 2 sinh λ x
dx 2
d 2 cosh λ x
dx 2
= λ 2 sinh λ x,
= λ 2 cosh λ x,
P-35
Morris Kline
(1908-1992)
Milton Abramowitz Irene Stegun
1915-1958
1919-
The best Tables of Special
functions
The best Tables of integrals
Gradshteyn, I.S. and Ryzhik, I.M.
The best history book
on differential
equations
Bateman manuscript project
Bateman
P-36
Arthur Erdélyi
Higher Transcendental Functions, 3 Volumes
Thank you
P-37
謝謝各位 !!