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ME 747 Introduction to computational
fluid dynamics
By Chainarong Chaktranond
Lecture 5Introduction to solve engineering
problems with finite-difference method
2Chainarong Chaktranond, Thammasat University
Lecture schedule
5. Introduction to solve engineering problems with finite-difference method
- Taylor series expansion, Approximation of the second derivative, Initial condition and
Boundary conditions
6 – 7
4. Introduction to numerical methods
- Finite different method, Finite volume method, Finite element method, etc.
5
3. Overviews of governing equations for flow and heat transfer
-Elliptic, Parabolic and Hyperbolic equations
4
2. Introduction to Fortran programming
- Basic commands in Fortran programming
2 - 3
1.Overviews of computational fluid dynamics
- Overviews and importance of heat transfer in real applications1
TopicsSession
3Chainarong Chaktranond, Thammasat University
Contents
Finite difference method
Numerical discretization schemes
4Chainarong Chaktranond, Thammasat University
Solving the fluid dynamic equations
2
2
u ut x
ν∂ ∂=
∂ ∂
Transient –diffusion term
กําหนดให 1ν =
11 1
2
2n n n n ni i i i iu u u u u
t x
+− +− − +
=Δ Δ
จัดรูปสมการไดเปน
11 12 2 21 2n n n n
i i i it t tu u u u
x x x+
− +
Δ Δ Δ⎡ ⎤= + − +⎢ ⎥Δ Δ Δ⎣ ⎦
(1)
(2)
(3)
Explicit Euler method
5Chainarong Chaktranond, Thammasat University
Explicit Euler method
11 12 2 21 2n n n n
i i i it t tu u u u
x x x+
− +
Δ Δ Δ⎡ ⎤= + − +⎢ ⎥Δ Δ Δ⎣ ⎦
กําหนดให ( )0, 5u x =
( ),0 1u t =
( ), 1u t L =0 1 2 m
11 1
n n n ni i i iu au bu cu+
− += + +
1niu −
niu 1
niu +
0 1 2 m
2
taxΔ
=Δ 21 2 tb
xΔ⎡ ⎤= −⎢ ⎥Δ⎣ ⎦ 2
tcxΔ
=Δ
โดยที่
6Chainarong Chaktranond, Thammasat University
11 1
n n n ni i i iu au bu cu+
− += + +1
niu −
niu 1
niu +
0 1 2 m
( )0, 5u x =Initial condition
กําหนดให Δt = 0.5 และ Δx = 1 ดังนั้น a = 0.5, b = 0, c = 0.5
Explicit Euler method
Boundary condition ( ),0 1u t = ( ), 1u t L =
[ ]{ } { }1n na u u +=
10 0 0 0 0
21 1 1 1 1
32 2 2 2 2
43 3 3 3
54 4 4
0
0
n n
n n
n n
n n
n n
a b c u ua b c u u
a b c u ua b u u
a u u
+
+
+
+
+
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭ ⎩ ⎭
7Chainarong Chaktranond, Thammasat University
Explicit Euler method
00 1u =01 5u =02 5u =
At t = 0
At t = t + Δt
( )( ) ( )( ) ( )( )11 0.5 1 0 5 0.5 5 3u = + + =
( )( ) ( )( ) ( )( )0.5 5 0 5 0.5 5 5+ + =12u =
13u = ( )( ) ( )( ) ( )( )0.5 5 0 5 0.5 5 5+ + =
( )( ) ( )( ) ( )( )0.5 5 0 5 0.5 5 5+ + =14u =
11 1
n n n ni i i iu au bu cu+
− += + +
10 1u =
8Chainarong Chaktranond, Thammasat University
At t = t + 2Δt
( )( ) ( )( ) ( )( )21 0.5 1 0 3 0.5 5 3u = + + =
( )( ) ( )( ) ( )( )0.5 3 0 5 0.5 5 4+ + =22u =
23u = ( )( ) ( )( ) ( )( )0.5 5 0 5 0.5 5 5+ + =
( )( ) ( )( ) ( )( )31 0.5 1 0 3 0.5 4 2.5u = + + =
( )( ) ( )( ) ( )( )0.5 3 0 4 0.5 5 4+ + =12u =
13u = ( )( ) ( )( ) ( )( )0.5 4 0 5 0.5 5 4.5+ + =
( )( ) ( )( ) ( )( )0.5 5 0 5 0.5 5 5+ + =14u =
At t = t + 3Δt
20 1u =
30 1u =
Explicit Euler method
9Chainarong Chaktranond, Thammasat University
.000E+00 .100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01
.100E+01 .500E+01.300E+01.300E+01.250E+01.250E+01.225E+01.225E+01.209E+01.209E+01.198E+01.198E+01
.200E+01 .500E+01.500E+01.400E+01.400E+01.350E+01.350E+01.319E+01.319E+01.295E+01.295E+01.276E+01
.300E+01 .500E+01.500E+01.500E+01.450E+01.450E+01.413E+01.413E+01.381E+01.381E+01.354E+01.354E+01
.400E+01 .500E+01.500E+01.500E+01.500E+01.475E+01.475E+01.444E+01.444E+01.413E+01.413E+01.383E+01
.500E+01 .500E+01.500E+01.500E+01.500E+01.500E+01.475E+01.475E+01.444E+01.444E+01.413E+01.413E+01
.600E+01 .500E+01.500E+01.500E+01.500E+01.475E+01.475E+01.444E+01.444E+01.413E+01.413E+01.383E+01
.700E+01 .500E+01.500E+01.500E+01.450E+01.450E+01.413E+01.413E+01.381E+01.381E+01.354E+01.354E+01
.800E+01 .500E+01.500E+01.400E+01.400E+01.350E+01.350E+01.319E+01.319E+01.295E+01.295E+01.276E+01
.900E+01 .500E+01.300E+01.300E+01.250E+01.250E+01.225E+01.225E+01.209E+01.209E+01.198E+01.198E+01
.100E+02 .100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01.100E+01
x t = 0 1 2 3 4 5 6 7 8 9 10
Explicit Euler method
6
5
4
3
2
1
0
u
1086420x
t = 0 t = 1 t = 2 t = 3 t = 4 t = 5
( )0, 5u x =
( ),0 1u t =
( ), 1u t L =
Initial condition
Boundary condition( )2 2
0.5 0.51
txΔ
= =Δ
program diffusion
parameter (n= 10,m=10)
real u(0:n+1,0:m+1)
open(1000,file='OUT-diffusion',status='unknown')
dt = 0.5
RX = 10.0
dx = RX/real(m)
dtdx = dt/dx/dx
do 10 i = 0,m
u(0,i) = 5.0
u(0,0) = 1.0
u(0,m) = 1.0
10 continue
do 110 j = 1,n
do 100 i = 1,m-1
u(j,0) = 1.0
u(j,m) = 1.0
u(j,i) = dtdx*u(j-1,i-1)+(1.0-2.0*dtdx)*u(j-1,i)+dtdx*u(j-1,i+1)
100 continue
110 continue
do 210 i = 0,m
write(1000,1001) real(i)*dx,(u(j,i),j = 0,n)
1001 format(e8.3,x,11e8.3)
210 continue
stop
end
12Chainarong Chaktranond, Thammasat University
Assignment 2
จงใชวิธี Explicit Euler method คํานวณและวาดกราฟการเปลี่ยนแปลง คา
ความเรว็ u ณ ตําแหนงและเวลาตางๆ เมื่อ Δt = 0.05 และ Δx = 0.1, 0.2, 0.4, 1, 2
2
2
u ut x
ν∂ ∂=
∂ ∂
กําหนดให 1ν =
( )0, 5u x =
( ),0 0u t =
( ), 0u t L =
Initial condition
Boundary condition
13Chainarong Chaktranond, Thammasat University
Implicit Euler method
1 1 1 11 1
2
2n n n n ni i i i iu u u u u
t x
+ + + +− +− − +
=Δ Δ
2
2
u ut x
ν∂ ∂=
∂ ∂กําหนดให 1ν = ( )0, 5u x =
( ),0 1u t =
( ), 1u t L =
1 1 1 11 12 2 22n n n n n
i i i i it t tu u u u u
x x x+ + + +− +
Δ Δ Δ= − + + −
Δ Δ Δ
1 1 1 11 12 2 22n n n n n
i i i i it t tu u u u u
x x x+ + + +
− +
Δ Δ Δ− = − +
Δ Δ Δ
1 1 11 12 2 21 2n n n n
i i i it t tu u u u
x x x+ + +− +
Δ Δ Δ⎛ ⎞= − + + −⎜ ⎟Δ Δ Δ⎝ ⎠
14Chainarong Chaktranond, Thammasat University
Implicit Euler method
1 1 11 1
n n n ni i i iu au bu cu+ + +
− += + +
2
taxΔ
= −Δ 21 2 tb
xΔ⎛ ⎞= +⎜ ⎟Δ⎝ ⎠
1 1 11 12 2 21 2n n n n
i i i it t tu u u u
x x x+ + +− +
Δ Δ Δ⎛ ⎞= − + + −⎜ ⎟Δ Δ Δ⎝ ⎠
2
tcxΔ
= −Δ
To solve this problem, need initial and boundary conditions
15Chainarong Chaktranond, Thammasat University
0 1 1 11 1 1 1 2 1 3u a u b u c u= + + +
0 15 5 5u a u=
1 2 3 54
1 01 1 1 1 1
1 02 2 2 2 2
1 03 3 3 3 3
1 04 4 4 4
1 05 5 5
a b c u ua b c u u
a b c u ua b u u
a u u
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬
⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭ ⎩ ⎭
[ ]{ } { }1n ni iA u u+ =
1 02 2 2 2 2
1 03 3 3 3
1 04 4 4
a b c u ua b u u
a u u
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎩ ⎭ ⎩ ⎭
0 1 1 12 2 2 2 3 2 4u a u b u c u= + +
0 1 1 13 3 3 3 4 3 5u a u b u c u= + +
0 1 14 4 4 4 5u a u b u= +
Implicit Euler method
เมื่อตัดคาที่ boundary conditions
สมมติวาพิจารณา 5 grid
16Chainarong Chaktranond, Thammasat University
11 1 1 1 1
12 2 2 2 2
13 3 3 3 3
14 4 4 4 4
15 5 5 5 5
16 6 6 6 6
17 7 7 7 7
18 8 8 8 8
19 9 9
110 10
n n
n n
n n
n n
n n
n n
n n
n n
n
n
a b c u ua b c u u
a b c u ua b c u u
a b c u ua b c u u
a b c u ua b c u u
a b u ua u
+
+
+
+
+
+
+
+
+
+
⎧ ⎫⎡ ⎤⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ =⎨ ⎬
⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎩ ⎭
9
10
n
nu
⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
Implicit Euler method
17Chainarong Chaktranond, Thammasat University
Implicit Euler method
Know initial and boundary conditions
12 2 2 2 2
13 3 3 3 3
14 4 4 4 4
15 5 5 5 5
16 6 6 6 6
17 7 7 7 7
18 8 8 8
19 9 9
n n
n n
n n
n n
n n
n n
n n
n n
a b c u ua b c u u
a b c u ua b c u u
a b c u ua b c u u
a b u ua u u
+
+
+
+
+
+
+
+
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎩ ⎭ ⎩ ⎭
10 1u =( )0, 5u x = 1 1u =
18Chainarong Chaktranond, Thammasat University
tridiagonal matrix algorithm (TDMA)
1 1i i i i i i ia x b x c x d− ++ + =
เมื่อ a1 = cn = 0
1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8 8
0
0
b c x da b c x d
a b c x da b c
a b ca b c
a b ca b x d
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
[ ]{ } { }i iA x d=
Concept
ใช Matrix operation เพื่อทํา
ใหสัมประสิทธิ์แถวทแยงมุม มี
คาเปน 1
19Chainarong Chaktranond, Thammasat University
ขั้นที่ 1 หารเมตริกซแถวแรกดวย b1
tridiagonal matrix algorithm (TDMA)
1 11
1 12
2 2 2 23
3 3 3 3
4 4 4
5 5 5
6 6 6
7 7 78
8 8 8
1 0
0
c dx
b bx
a b c dx
a b c da b c
a b ca b c
a b cx
a b d
⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
20Chainarong Chaktranond, Thammasat University
ขั้นที่ 2 คูณแถวที่หนึ่งดวย a2 แลวนําไปลบออกจากแถวที่สองและกําหนดให
1 1 1
2 1 2 2 2 2 1 2
3 3 3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8 8
1 00
0
c x db c a c x d d a
a b c x da b c
a b ca b c
a b ca b x d
′ ′⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥′ ′− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
11
1
ccb
′ =
tridiagonal matrix algorithm (TDMA)
11
1
ddb
′ =และ
ขั้นที่ 3 หารแถวที่สองดวย
1 11
2 2 1 22
2 1 2 2 1 23
3 3 3 3
4 4 4
5 5 5
6 6 6
7 7 78
8 8 8
1 0
0 1
0
c dx
c d d ax
b c a b c ax
a b c da b c
a b ca b c
a b cx
a b d
′ ′⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥′−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′ ′⎢ ⎥ ⎢ ⎥− −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
2 1 2b c a′−
และเริ่มตนทําซ้าํขั้นที่สอง กับแถวอื่นๆ จนถงึแถวที่ n หรือเขยีนไดเปน
1
1
1
; 1
; 2,3, ,i
i
i i i
c ib
cc
i nb c a−
⎧ =⎪⎪′ = ⎨⎪ =⎪ ′−⎩
…
1
1
1
1
; 1
; 2,3, ,i
i i i
i i i
d ib
dd d a i nb c a
−
−
⎧ =⎪⎪′ = ⎨ ′−⎪ =⎪ ′−⎩
…
2d ′
n nx d ′=
1 ; 1, 2, ,1i i i xx d c x i n n+′ ′= − = − − …
เมื่อคาสัมประสิทธิ์ bn = 1 แลว จะได
ขั้นสุดทายทําการแทนคา xn เพื่อหาคา xn-1, xn-2, …., x1
1
21
2 1 22 1 2
132 1 2
3 2 3 2
34
4 3 4
5
5 4 57 6 7
67 6 7
6 5 6 88
7
7 6 7
1 0
0 1
0 1
0 1
0 1
0 1
0 1
0 1
cc
db c a
d d axc
b c ab c a x
xcb c a
cb c a
d d ac
b c ab c a x
d dc
b c a
′⎡ ⎤⎢ ⎥⎢ ⎥ ′
′⎢ ⎥−′−⎢ ⎥
⎡ ⎤⎢ ⎥ ′−⎢ ⎥⎢ ⎥′− ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥′− ⎢ ⎥⎢ ⎥ =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥′− ⎢ ⎥⎢ ⎥ ′−⎢ ⎥⎢ ⎥⎢ ⎥ ′−⎢ ⎥′− ⎢ ⎥⎣ ⎦⎢ ⎥ ′−⎢ ⎥
⎢ ⎥′−⎢ ⎥
⎢ ⎥⎢ ⎥⎣ ⎦
7 8
8 7 8
ab c a
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥′−⎢ ⎥⎣ ⎦
nd ′=
12 2 2 2 2
13 3 3 3 3
14 4 4 4 4
15 5 5 5 5
16 6 6 6 6
17 7 7 7 7
18 8 8 8
19 9 9
n n
n n
n n
n n
n n
n n
n n
n n
a b c u ua b c u u
a b c u ua b c u u
a b c u ua b c u u
a b u ua u u
+
+
+
+
+
+
+
+
⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎩ ⎭ ⎩ ⎭
จากตัวอยาง Implicit Euler method
2 2 212
2 2 213
3 3 3 314
4 4 4 415
5 5 5 516
6 6 6 617
7 7 7 718
8 8 819
9 9
1n
n
nn
nn
nn
nn
nn
nn
nn
b c uua a au
a b c uu
a b c uu
a b c uu
a b c uu
a b c uu
a b uu
a u
+
+
+
+
+
+
+
+
⎧ ⎫⎡ ⎤⎧ ⎫ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎩ ⎭
หารแถวแรกดวย a2
22
2
b ba
′= 22
2
c ca
′= 22
2
nnu u
a′=
แลวกําหนดให
2 2 212
3 2 2 3 1 3 23
3 2 3 2 3 214
4 4 4 1 45
5 5 5 1 56
6 6 6 1 67
7 7 7 1 78
8 8 1 89
9 9
1
1
nn
n nn
nn
nn
nn
nn
nn
nn
b c uu
b c b c u uua b a b a bu
a b c uu
a b c uu
a b c uu
a b c uu
a b uu
a u
′+
′+
+
+
+
+
+
+
′ ′ ⎧⎡ ⎤⎧ ⎫ ⎪⎢ ⎥′ ′− −⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪′ ′⎢ ⎥− − ′−⎪ ⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ =⎨ ⎬ ⎨⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎢ ⎥⎣ ⎦
⎫⎪⎪⎪⎪
⎪ ⎪⎪ ⎪
⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
นําแถวที่สองลบออกดวยแถวที่หนึ่งและหารดวย 3 2a b′−
3nu ′
2 2
3 2 2 3
3 2 3 21
4 3 3 4 21
4 3 4 3 3
5 4 4 5
5 4 5 4
6 5 5 6
6 5 6 5
7 6 6 7
7 6 7 6
8 7 7
8 7
1
1
1
1
1
1
1
1
n
n
b cb c b ca b a b
b c b c ua b a b u
b c b ca b a b
b c b ca b a b
b c b ca b a b
b c ba b
+
+
′ ′⎡ ⎤⎢ ⎥′ ′−⎢ ⎥
′ ′⎢ ⎥− −⎢ ⎥′ ′−⎢ ⎥⎢ ⎥′ ′− −⎢ ⎥
′ ′−⎢ ⎥⎢ ⎥′ ′− −⎢ ⎥
′ ′−⎢ ⎥⎢ ⎥′ ′− −⎢ ⎥⎢ ′ ′ ⎥−⎢ ⎥′ ′− −⎢ ⎥⎢ ⎥′ ′−⎢ ⎥
′−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
2
3 2
3 2
4 3
4 3
5 414
5 415
1 6 56
1 6 57
1 7 68
1 7 69
8 7
8 7
9 8
9 8
n
n n
n n
n nn
nn n
n
nn n
n
n
n n
n n
uu ua bu ua bu u
ua b
uu uua b
uu uua buu ua bu ua b
′
′
′
′+
+′
+
+
′+
+
′
′
⎧ ⎫⎪ ⎪
−⎪ ⎪⎪ ⎪′−⎪ ⎪⎪ ⎪−
⎧ ⎫ ⎪ ⎪′−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪
′−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪= −⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ′−⎪ ⎪ ⎪⎪ ⎪ ⎪ −⎪ ⎪ ⎪
′−⎪ ⎪ ⎪⎩ ⎭ ⎪ −⎪
′⎪ −⎪
−⎪⎪ ′−⎪⎩
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭
ทําดงัขัน้ตอนทีผ่านมาจนกระทั่งคาสัมประสิทธิ์แถวทแยงมมุทุกตัวมีคาเปน 1
1 8 79 9
8 7
n nn nu u
u ua b
′′+ −
= =′−
จะไดวา 1 8 7 7
8 9 88 7
n n nb c bu u u
a b′ ′+ ′ ′−
= −′−
1 ; 1, 2, ,1i i i xx d c x i n n+′ ′= − = − − …
8 7 7
8 7
b c ba b
′ ′−′−
8 7
8 7
n nu ua b
′−′−
นําไปแทนคา
29Chainarong Chaktranond, Thammasat University
Other methods
1 11 1
2
22
n n n n ni i i i iu u u u u
t x
+ −− +⎡ ⎤− − +
= ⎢ ⎥Δ Δ⎣ ⎦
Richardson method
DuFort-Frankel method
1 1 1 11 1
22
n n n n n ni i i i i iu u u u u u
t x
+ − + −− +⎡ ⎤− − − +
= ⎢ ⎥Δ Δ⎣ ⎦
แทนเทอม diffusion ดวย 1 1
2
n ni iu u+ −+
Truncation error ( ) ( ) ( )2 2 2, , /O t x t x⎡ ⎤Δ Δ Δ Δ⎣ ⎦
30Chainarong Chaktranond, Thammasat University
Other methods
1 1 1 11 1
2
2n n n n ni i i i iu u u u u
t x
+ + + +− +⎡ ⎤− − +
= ⎢ ⎥Δ Δ⎣ ⎦
Lassonen method (Euler implicit method)
Crank-Nicolson method
1 1 1 11 1 1 1
2 2
2 21 12 2
n n n n n n n ni i i i i i i iu u u u u u u u
t x x
+ + + +− + − +⎡ ⎤ ⎡ ⎤− − + − +
= +⎢ ⎥ ⎢ ⎥Δ Δ Δ⎣ ⎦ ⎣ ⎦
Truncation error ( ) ( ),O t xΔ Δ⎡ ⎤⎣ ⎦
31Chainarong Chaktranond, Thammasat University
Assignment 3
จงใชวิธี Implicit Euler method คํานวณและวาดกราฟการเปลี่ยนแปลง คา
ความเรว็ u ณ ตําแหนงและเวลาตางๆ เมื่อ Δt = 0.05 และ Δx = 0.1, 0.2, 0.4, 1, 2
2
2
u ut x
ν∂ ∂=
∂ ∂
กําหนดให 1ν =
( )0, 5u x =
( ),0 0u t =
( ), 0u t L =
Initial condition
Boundary condition
