화공수학 및 전산응용
Mathematics for chemical engineers
Ch. 3 Series solution of ODEs
• Teacher: Young-il Lim – Phone: 031-670-5207 – Email: limyi@hknu.ac.kr
– Homepage: http://facs.maru.net
• Text book:
– V.G. Jenson and G.V. Jeffreys,
Mathematical methods in chemical engineering, 2
ndeds., Academic Press, 1994
Ch. 3 Solution by series
3.1 Intro
3.2 Infinite series (무한급수) 3.3 Power series (멱급수) 3.4 Method of Frobenius 3.5 Bessel’s equation
3.6 Properties of Bessel functions
Section 3.1 Intro
Linear steady-state
convection-diffusion-reaction ODE with constant coefficient (상수 계수를 갖는 선형 정상상태 대류-확산-반응 상미방)
2 0
2 + =
∂ + ∂
∂
− ∂
∂ =
∂ km
z D m z
u m t
m
Section 3.1 Intro
2
0
2
=
+ + Ry
dx Q dy dx
y
P d 1) Unequal real roots to Auxiliary eq.
x m x
m
A e
e A
y =
1 1+
2 23) Equal roots to Auxiliary eq.
e
mx) D Cx
(
y = +
See Ch. 2, p44-48
2) Unequal complex roots to Auxiliary eq.
)) x m sin(
iA )
x m cos(
A ( e
y =
m1x 1 2+
2 2What are the main features?
For variable coefficients, its solution
has the form of infinite series
The series has to be convergent
rapidly for practical solution of ODEs
Section 3.2 Infinite series
Convergent, divergent, oscillatory - Convergent:
- Divergent:
- Oscillatory:
n
n u u u ... u
S = 1 + 2 + 3 + +
a S
lim
nn
=
∞
→
= ±∞
∞
→ n
n
S
lim
b S
lim
a
nn
≤
≤
→∞Section 3.2 Infinite series
Properties of infinite series
- if a series has only positive real number or zero, it must be convergent or divergent (not oscillatory)
- if a series is convergent, then
- if a series is absolutely convergent, then it is also convergent.
= 0
∞
→ n
n
u
lim
Convergence test
Comparison test and ratio test
Section 3.2.2-4 Comparison and ratio test
Comparison test consists of two parts + see p78
+ see p78
+ see p79 for examples
Ratio test also consists of two parts
+ the series is absolutely convergent
+ the series is not convergent then
, N n all for u k
if u
n
n
≥ > >
+
1
1
then ,
N n all u for
if u
n
n
< >
+
1
1
Section 3.2.2-4 Comparison and ratio test
Alternating series
+
+ if the absolute value of successive terms decreases for all values of n and u
n 0 as n ∞, then the series is convergent
i n i
i
n
u u u u .... ( ) u
S
1
1 4
3 2
1
1
+
∑
=−
=
− +
−
=
Section 3.3 Power series
n n n
n
n
a z a a z a z ... a z
S = ∑
∞=
0+
1+
2 2+ +
0
∞
→
→
=
+ +
n as z ,
R z
a a u
u
n n n
n
1
1 1
then ,
n as a R
if a
n
n
→ → ∞
+1
Binomial series +
+ the binomial series is convergent for |z|<1.
n n
n
z
n ... p
! z
) p )(
p ( z p
! ) p ( pz p
S ∑
∞=
=
− + + −
+ − +
=
0 3
2
3
2 1
2
1 1
Section 3.3 Power series
∑
∞=
−
−= +
− +
−
= +
1
1 4
3
2
1
4 3 1 2
n
n n
n z ) .. (
z z z z
) z ln(
Logarithmic series
The series is always convergent.
Exponential series
The exponential series is always convergent.
∑
∞=
= + + +
−
=
0 3
2
3 1 2
n n z
! n ... z
! z
! z z e
Trigonometric series
∑
∞=
+
+
= −
− +
−
=
0
1 2 5
3
1 2
1 5
3
nn n
)!
n (
z ) ... (
! z
! z z z sin
∑
∞=
= −
− +
−
=
0
2 4
2
2 1 4
1 2
n
n n
)!
n (
z ) ... (
! z
! z z
cos
Hyperbolic series
∑
∞=
+
= +
− + +
=
0
1 2 5
3
1 2 5
3
nn
)!
n ( ... z
! z
! z z z sinh
∑
∞=
= + +
+
=
0 2 4
2
2 4
1 2
n
n
)!
n ( ... z
! z
! z z
cosh
Section 3.3.6 Taylor’s theorem
...
) a (
! f n ... x
) a ( f x )
a ( f x ) a ( f ) a x (
f
(n)n
+
+
′′ +
′ + +
=
+
22 1
...
) a (
! f n
) a x ... (
) a ( f ) a x ( )
a ( f ) a x ( ) a ( f ) x (
f
(n)n
+
+ −
′′ +
−
′ +
− +
=
22
1
Section 3.3.6 Taylor’s theorem
Proof of Taylor expansion let
...
x A x
A x
A A
) a x (
f + =
0+
1+
2 2+
3 3+
...
x A x
A A
) a x (
f ′ + =
1+ 2
2+ 2
3 2+
0
0 A )
a ( f
x put
=
=
1
0 A )
a ( f
x put
′ =
=
...
x A A
) a x (
f ′′ + = 2
2+ 4
3+
) a ( f A
, A )
a ( f
x put
= ′′
′′ =
=
2 2 1
0
2 2
Section 3.3.6 Taylor’s theorem
...
)
! ( ) ... (
) ( )
2 ( ) 1 ( ) (
) ( )
(
2−
( )+
+
′′ +
−
′ +
− +
= f a
n a a x
f a x a
f a x a
f x
f
nn
...
) a (
! f n ... x
) a ( f x )
a ( f x ) a ( f ) a x (
f
(n)n
+
+
′′ +
′ + +
=
+
22
1
Section 3.3.6 Taylor’s theorem
Team Working
+ express the exponential function (y=e
x) as a series.
+ express the sine function (y=sin x) as a series.
+ express the cosine function (y=cos x) as a series.
+ demonstrate Euler’ formula:
θ θ
θ θ
θ θ
sin cos
sin cos
j e
j e
j j
−
=
+
=
−
Section 3.3.6 Taylor’s theorem
...
) a (
! f n
) a x ... (
) a ( f ) a x ( )
a ( f ) a x ( ) a ( f ) x (
f
(n)n
+
+ −
′′ +
−
′ +
− +
=
22 1
...
) a (
! f n ... x
) a ( f x )
a ( f x ) a ( f )
a x
(
f
(n)n
+
+
′′ +
′ + +
=
+
22
1
Section 3.3.6 Taylor’s theorem
∑
∞=
=
+ +
+ +
+
=
′′′ + + ⋅
+ ′′
+ ′
=
=
0
3 2
3 2
6 1 2
3 0 0 2
0 2 0
n
n
n x
! n x
! ...
n ... x x
x x
....
) ( x f
) ( x f
) ( f x ) ( f e ) x ( f Ex 1. f(x)=e
xSee page 82
Section 3.3.6 Taylor’s theorem
...
) a (
! f n
) a x ... (
) a ( f ) a x ( )
a ( f ) a x ( ) a ( f ) x (
f
(n)n
+
+ −
′′ +
−
′ +
− +
=
22 1
...
) a (
! f n ... x
) a ( f x )
a ( f x ) a ( f ) a x (
f
(n)n
+
+
′′ +
′ + +
=
+
22
1
Section 3.3.6 Taylor’s theorem
∑
∞=
−
−
−
=
+
− +
− +
− +
=
′′′ + + ⋅
+ ′′
+ ′
=
+
=
1
1
1 4
3 2
3 2
1
4 1 3
0 2
3 0 0 2
0 2 0
1
n
n n
n n
n ) x
(
n ...
) x x ...(
x x x
....
) ( x f
) ( x f
) ( f x ) ( f
) x
ln(
) x ( f Ex 2. f(x)=ln(x+1)
See page 82
Section 3.3.6 Taylor’s theorem
...
) a (
! f n
) a x ... (
) a ( f ) a x ( )
a ( f ) a x ( ) a ( f ) x (
f
(n)n
+
+ −
′′ +
−
′ +
− +
=
22 1
...
) a (
! f n ... x
) a ( f x )
a ( f x ) a ( f ) a x (
f
(n)n
+
+
′′ +
′ + +
=
+
22
1
Section 3.3.6 Taylor’s theorem
∑
∞=
+
+
− +
=
+ +
− +
+
− +
=
′′′ + + ⋅
+ ′′
+ ′
=
=
0
1 2
1 2 5
3
3 2
1 1 2
1 1 2
5 0 3
3 0 0 2
0 2 0
n
n n
n n
)!
n ( ) x (
)! ...
n ( ) x
! ...(
x
! x x
....
) ( x f
) ( x f
) ( f x ) ( f
x sin )
x ( f Ex 3. f(x)=sin(x)
See page 82
Section 3.3.6 Taylor’s theorem
...
) a (
! f n
) a x ... (
) a ( f ) a x ( )
a ( f ) a x ( ) a ( f ) x (
f
(n)n
+
+ −
′′ +
−
′ +
− +
=
22 1
...
) a (
! f n ... x
) a ( f x )
a ( f x ) a ( f ) a x (
f
(n)n
+
+
′′ +
′ + +
=
+
22
1
Section 3.3.6 Taylor’s theorem
∑
∞=
−
=
+
− +
+
−
=
′′′ + + ⋅
+ ′′
+ ′
=
=
0
2
2 4
2
3 2
1 2
1 2 4
1 2
3 0 0 2
0 2 0
n
n n
n n
)!
n ( ) x (
)! ...
n ( ) x
! ...(
x
! x
....
) ( x f
) ( x f
) ( f x ) ( f
x cos )
x ( f Ex 4. f(x)=cos(x)
See page 82
Section 3.3.6 Taylor’s theorem
...
) a (
! f n
) a x ... (
) a ( f ) a x ( )
a ( f ) a x ( ) a ( f ) x (
f
(n)n
+
+ −
′′ +
−
′ +
− +
=
22 1
...
) a (
! f n ... x
) a ( f x )
a ( f x ) a ( f )
a x
(
f
(n)n
+
+
′′ +
′ + +
=
+
22
1
Section 3.3.6 Taylor’s theorem
− + +
+ +
− +
−
=
+ +
−
− +
=
′′′ + + ⋅
+ ′′
+ ′
=
=
! ...
x
! x x
i
! ...
x
! x x
! ...
x
! i x ix x
....
) ( x f
) ( x f
) ( f x ) ( f e ) x (
f
ix5 3
6 4
1 2
4 3
1 2
3 0 0 2
0 2 0
5 3
6 4
2
4 3
2
3 2
Ex 5. f(x)=e
ix) x
cos( sin( x )
x sin i
x cos
e ix ≡ +
Euler’s formula
Section 3.3.7 L’Hopital’s rule
) x ( g
) x ( lim f
) x ( g
) x ( lim f
) , a ( g
) a ( f )
x ( g
) x ( lim f
when
a x a
x
a x
′
≡ ′
=
=
→
→
→ 0
0
Section 3.3.7 L’Hopital’s rule
Ex 1. f(x)=sin(x)/x, f(0)=?
1 1
0
0 = =
→
→
x lim cos
x x lim sin
x
x
주별 강의 세부 내용 - 12-13 주차 Objectives
- Chapter 3 ODE solution by series - 3.3 Power series and Talyor's theorem - 3.4 Method of Frobenius
Course Introduction
- 1차 상미방: exact solution by inspection, variable separation, homogeneous equations, integrating factor
- 2차 상미방: 비선형 상미방, 선형 상미방
Contents
General solution of 2nd-order ODE.
1. 어떤 함수가 무한 미분이 가능하다면, 이 함수는 테일러 급수로 표현 가능함.
2. 미분방정식의 해를
∞
로 간주할 수 있음.
3. dy/dx, d2y/dx2 을 미분식에 대입하여 정리하면 (se example 1, p 87) 항등식이 나오고, 이 항등식의 최소차수에 대한 식을 indicial equation 이라 정의함.
4. Indicial equation에서 c 의 근에 따라 여러 형태의 해가 존재함.
5. see example 1. y=y(c1) + y(c2) -
- indicial equaiton, c=0, c=-1/2 (xc 항에 대하여) - recurrence relation, ... (xc+r 항에 대하여)
6. 어느 특별한 형태의 2차 상미방의 해를 구하면, 특별한 형태의 해가 구해지는데, 그 해의 이름을, Bessel
function (see p102) 이라고 부름.
Terminology
- indicial equation - recurrence relation
Exercises/Homework
- example 1 (page 87) 풀기 Erratum
Info
- exam of chapter 3
2
0
2 2
1
- r + hA ( T - T ) =
dx VC dT
dx u T
kLA d
p wair
T
in=20
oC Q=0.009m
3/s C
p=1kJ/kg/
oC
L=1.2m
d=0.1m T
w=300
oC
주별 강의 세부 내용 - Homework Ch. 3.4
Example 6 (page 100-101): Solution of Heat transfer in a tubular gas preheater
A supply of hot air is to be obtained by drawing cool air through a heated cylindrical pipe. The wall of pipe
(i.d.=0.1m and L=1.2m) is maintained at Tw=300 oC throughout its length. Assuming that heat transfer takes place by conduction within the gas in an axial direction, mass flow of the gas in an axial direction, and by the above variable heat transfer coefficient from the walls of the tube, find T(x).
dL A2 =p 4
2 1
A pd
= h 10= x
2 2 2
2
dx T d dx
T , d dx
T d dx
), dT T T (
T = w- =- =- b
kLA , A kLA a
VC
u p
= r =
1 2 1
10
where , k=0.035 W/m/K, ρ=0.8 kg/m3, u=Q/A1, and V=LA1.
Solve the above 2nd-order ODE using Frobenius method, draw a graph of T(x), and analyze this problem.
Hint: and
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