advanced engineering mathematics presentation
DESCRIPTION
Advanced Engineering MathematicsTRANSCRIPT
Advanced Engineering Advanced Engineering MathematicsMathematics
A.EmamzadehA.Emamzadeh
Spring Spring 20092009
Syllabus# Topic Reference Duration (weeks)
1 Numerical solution of ODE [1] Chapter 5 & 11 2
2 Numerical solution of PDE [1] Chapter 12 2
3 Matrix Algebra [1] Chapters 6, 7 and 10 2
4 Integral equations [2] Chapter 16 1
5 Fourier series and Integrals [3] Chapter 1 1
6 Partial differential equations [3] Chapters 2 to 6 2
7 Special functions [4] Appendix B 1
8 Complex functions [5] Chapter 2 to 10 2
9 Monte Carlo method [2] Chapter 19 1
10 Integral transforms [4] Chapter 13 1
Total 15
Reference texts
Reference texts
[1] Faires & Burden Numerical Methods 3rd ed. ,Brooks/Cole 2003
[2] FrobergIntroductions to Numerical Analysis, Addison-Wesley
[3] Powers Boundary Value Problems
[4] Pipes & HarvillApplied Mathematics for Engineers and Physicists, McGraw-Hill
[5] Churchill Complex variable and applications
Ordinary Differential Equations
1) Introduction
2) Initial Value Problems (IVP)
3) Boundary Value Problems (BVP)
1) Introduction
1- Sources of errors1.1- Initial data error1.2- Round off error1.3- Truncation error
2- Type of problems2.1- well – behaved2.2- ill – conditioned
3- Stability
4- Big Oh
2) Methods of solving IVP
1- Introduction
2- Taylor series method
3- Single step methods
4- Multi step methods
5- Extrapolation methods
3) Methods of solving BVP
1- Shooting methods
2- Finite difference methods
3- Variational methods
4- Eigen Value Problems (EVP)
Types of problems in IVP
Single first order equation System of n first order equations Single n th order equation
Single first order equation
, with initial condition (IC),),( yxfy
00 )( yxy
To find y at x1 to a specified accuracy.
System of n first order equations
),,...,,(
),...,,(
1
111
nnn
n
yyxfy
yyxfy
With initial conditions
00
1001
)(
)(
nn yxy
yxy
To find y1,…yn at x1 to a specified accuracy.
Where tn1 y,,yY
and tn1 f,,fF
In matrix form
00 Y)Y(x
Y)F(x,Y
Single n th order equation
),,,,,,( )1()( nn yyyyxfy with initial conditions
00)1(
100 )(,,)( nn yxyyxy
This equation with the given initial conditions can be transformed into a system of n first order equations as follows,
Let nn yyyyyy )1(
21 then
nn yyyxfy ,,,, 21
and
001001 )(,)( nn yxyyxy
The whole system can be written in the matrix form
00 )(
),(
YxY
YxFY
Where tnt
n fyyyFyyY ,,,,,, 321 and
Finally we are dealing with the form
00 )(
),(
yxy
yxfy
as a single first order equation or
as a system of n first order equations
Taylor series method
00 )(
),(
yxy
yxfy
To find y at x1 correct to md
Let x1 – x0 = h then the Taylor expansion of y about
x = x0 is
yh
yh
yhyhxyxy!3!2
3
0
2
0001
0
0
0
0
000
0
),(
y
ff
x
f
yf
y
ff
x
f
xy
y
ff
x
fy
yxfy
y givenishere
20
10
3 2
y
y
yxyGiven
:Example
onsoand
Find y(h) correct to 2d where h=0.1, 0.5,1 and 2.
14)0(32
12)0(2
5)0(21
2)0(
)5()5(
)4(2)4(
yyyyyy
yyyyy
yyyy
yhavewe
5432
60
7
2
1
6
521 hhhhhhy
Taking number of terms n=6
When h=0.1 y(0.1)=.8107845000
h=0.5 y(0.5)=.3265625000
h=1 y(1)=.4500000000
h=2 y(2)=3.400000000
Taking number of terms n=7
y(0.1)= .8107846111
y(0.5)= .3282986111
y(1)= .5611111111
y(2)= 10.51111111
Taking n=10
y(0.1)= .8107846019
y(0.5)= .3276448999
y(1.0)= .4951609347
y(2.0)= 9.887477949
Compare the accuracy
Single step methodsDef.:
Local truncation error
Global truncation error
Order of a method
First order method (Euler's method)
Statement of the problem
00
,
yxy
yxfy
given
To find y at x=b correct to md
Let b-x0=h then
200000 , hyxfhyyhyby o
Here O(h2) is the local truncation error
In the standard form
,1,0,1
11
nyxfhk
kyy
nn
nn where
Second order methods
errortruncationglobaltheisand
where1.
2
12
1
221
2,
2
,
hO
ky
hxfhk
yxfhk
hOkyy
nn
nn
nn
g.t.e.beforeas
where2.
2
12
1
2211
,
,2
1
2
1
hO
kyhxfhk
yxfhk
hOkkyy
nn
nn
nn
A third order method
123
12
1
33211
2,
2,
2
,
,2,146
1
kkyhxfhk
ky
hxfhk
yxfhk
nhOkkkyy
nn
nn
nn
nn
where
A 4th order method (runge – kutta)
34
23
12
1
443211
,
2,
2
2,
2
,
226
1
kyhxfhk
ky
hxfhk
ky
hxfhk
yxfhk
hOkkkkyy
nn
nn
nn
nn
nn
where
Example:
20,10
2sin
yy
yyxy
Find y(0.1) and y’(0.1) using h=0.1 and a 2nd order method.
20
10
2sin
p
y
pyxp
py thenLet
Take k for the slope of y and l for the slope of p then
101002102
000101
2101
2101
2sin
2sin
2
12
1
lpkyhxhllphk
pyxhlphk
llpp
kkyy
where
Now calculate [Note the argument of sine must the in radiance].
Modified Euler's method
g.t.e.theiswhere
2
2111
1
,,2
,
hO
hOyxfyxfh
yy
yxfhyy
pnnnnn
cn
nnnpn
error.truncationlocaltheiswhere
2
21 ,2
hO
hOyxhfyy nnnn 1
Mid - Point rule
How to control the truncation error1) By lowering the step size and repeat the
whole calculations: Runge-Kutta method
2) By considering the 1st neglected term in the Taylor expansion:
Merson’s method
A 2nd and third order method
Fehlberge’s method
A single-step method with error estimator Merson’s method
5431
431
5
314
213
12
1
55411
89230
1
22
32
,
83
8,
2
66,
3
3,
3
,
46
1
kkkkE
kkk
yhxhfk
kky
hxhfk
kky
hxhfk
hy
hxhfk
yxhfk
hOkkkyy
nn
nn
nn
nn
nn
nn
where
Fehlberg method
543216
43215
3214
213
12
1
54311
654311
~
40
11
4104
1859
2565
35442
27
8,
2
4104
845
513
36808
216
439,
2197
7296
2197
7200
2197
1932,
13
12
32
9
32
3,
8
3
4
1,
4
,
5
1
4104
2197
2565
1408
216
2555
2
50
9
56430
28561
12825
6656
135
16
kkkkkyh
xhfk
kkkkyhxhfk
kkkyh
xhfk
kkyh
xhfk
kyh
xhfk
yxhfk
kkkkyy
kkkkkyy
nn
nn
nn
nn
nn
nn
nn
nn
where
Multi step methods Predictor formulas Corrector formulas
Predictor formulas
543211
43211
3211
211
11
1
1440
475
1440
2877
1440
7298
1440
9982
1140
7923
1140
4277
720
251
720
1274
720
2616
720
2774
720
1901
24
9
24
37
24
59
24
55
12
5
12
16
12
23
2
1
2
3
nnnnnnnn
nnnnnnn
nnnnnn
nnnnn
nnnn
nnn
ffffffhyy
fffffhyy
ffffhyy
fffhyy
ffhyy
fhyy
Corrector formulas
432111
32111
2111
111
11
1440
27
1440
173
1440
482
1440
798
1440
1427
1440
475
720
19
720
106
720
264
720
646
720
251
24
1
24
5
24
19
24
9
12
1
12
8
12
5
2
1
2
1
nnnnnnnn
nnnnnnn
nnnnnn
nnnnn
nnnn
ffffffhyy
fffffhyy
ffffhyy
fffhyy
ffhyy
Extrapolation methods1.Introduction:Consider calculating as the area of a unit circle
(Fig.1), by calculation of the area of n sided polygon inscribed in the circle.
n
1
Fig. 1nA0Let be the area of n sided polygon.
n
nnA n
n 2sin
2sin
20
Using Taylor expansion of Sin x
66
44
22
753
0 !7
2
!5
2
!3
22
2
n
a
n
a
n
a
nnnn
nAn
Where a2, a4, a6, … are constants, do not depend on n.
Now if we double n then
6
64
42
220 64164 n
a
n
a
n
aA n
To form a linear combination of
in such away that in the linear combination the first term be and the second term vanishes, that is, for some and
nn AA 200 and
88
66
44
1200 n
b
n
b
n
bAAA n
saynn
Where b4, b6, b8, … are constants, do not depend on n.
04
1 andThen
The implies that 34
31 and
To go further we can repeat the same procedure by doubling the sides again and hence.
8
86
64
421 2566416 n
b
n
b
n
bA n
Again to form a linear combination of
in such away that in the linear combination the first term be p and the second term vanishes, that is for some and .
nn AA 211 and
,88
66
22
11 n
c
n
cAAA n
saynn
Where c6, c8, … are constants, do not depend on n.
,88
66
22
11 n
c
n
cAAA n
saynn
1,2,nfor
andgeneralIn
andthatimpliesthis
andThen
14
4
14
1
15
16
15
1
,016
1
n
n
n
The result of above operations can be presented in an extrapolation table as follows.
nnnn
nnn
nn
n
AAAA
AAA
AA
A
322
41
80
22
140
120
0 31
34
151
1516
Note that truncation error of the columns respectively are
on.soand,1
,1
,1
642
nO
nO
nO
631
6364
Numerically1. Starting with a triangle (the least accuracy),
we have
Are these coefficients ( and ) general?
30A
60A
120A
240A
480A
1.299038
2.598077
3.000001
3.105829
3.132629
3.03109
3.133975
3.141105
3.141563
3.140834
3.141581
3.141593
3.141593
3.141593
3.141593
40A
80A
160A
320A
640A
2
2.828427
3.061468
3.121446
3.136549
3.10457
3.139148
3.141439
3.141584
3.141453
3.141591
3.141593
3.141593
3.141593
3.141593
50A
100A
200A
400A
800A
2. 377642
2.938927
3.09017
3.12869
3.138364
3.126022
3.140584
3.3014153
3.141489
3.141555
3.141593
3.141593
3.141593
3.141594
3.141594
1600A 3.14078
63.14159
33.14159
33.14159
33.14159
33.14159
3
100A
200A
400A
800A
1600A
2. 938926
3.09017
3.12869
3.138364
3.140786
3.140584
3.14153
3.141589
3.141593
3.141593
3.141593
3.141593
3.141593
3.141593
3.141593
3200A 3.14139
13.14159
33.14159
33.14159
33.14159
33.14159
3
1. yes, whenever the accuracy parameter is changed by factor of 2. and the terms of the Taylor expansion are alternatively zero.
2. No otherwise
Extrapolation in differentiation Starting with central difference formula for
the first and second derivative
)2(2
)1(2
22
2
hOh
hxfxfhxfxf
hOh
hxfhxfxf
Denote the RHS of (1) as and the RHS of (2) as hF0hS0
Similar extrapolation table looks like
hhh
hh
h
FFF
FF
F
22
14
0
12
0
0 31
34
151
1516
The truncation error of the columns in order are
.onsoandand642 , hOhOhO
Similarly for the second derivative
Extrapolation in integrationConsider the trapezoidal rule
1
100
20
22
n
iin
h
hb
a
fffh
T
hOTdxxf
where
hhh
hh
h
TTT
TT
T
22
14
0
12
0
0 31
34
151
1516
Extrapolation in IVPSuppose y'=f(x,y), y(x0)=y0 to find y(b)
correct to md, with extrapolation method.
Start with Mid-Point rule
20011 ,2 hOyxhfyy
Either y-1is known previously or it should be
calculated by, say, Euler's method and corrected by modified Euler's method to have the same accuracy as O(h2)
therefore with h<0
Ec
E
yxfyxfh
yy
yxhfyy
110001
0001
,,2
,
and
If we denote then in extrapolation notationhYy 01 by
hhh
hh
h
YYY
YY
Y
22
14
0
12
0
0 31
34
151
1516
And so on.
Compare this method with single step methods and multi step methods.
ExampleConsider y'=y , y(0)=1 . To find y(1)
correct to 3d .
start with h=1 , y0=1 , x0=0
21
021
11
10
11
8
21
8
131
8
131
8
5
8
5
2
11
4
11
2
1
2
11
2
1
5.222
12
101
2
11,0
Yyy
yy
h
Y
yy
cE
cE
with
To continue calculate and and extrapolate.
6666.26
16
2
7
6
5
8
21
3
4
2
5
3
13
4
3
1 21
01
01
1
YYY
41
0Y 81
0Y
Shooting methodsStatement of a boundary value problem
byay
yyxfy
,
,,Given
Find y for a<x<b correct to md.
Shooting method changes the BVP into an IVP by assuming a value for y'(a) , say S, then using a root finder to find the
root of y(b,S)-=0 up to a desired accuracy.
1. Secant method
2. Newton’s method
Let us start with secant method as the root finder
Step 1. Guess S0=y'(a)Step 2. Solve the IVP
,,,,, 0Sayayyyxfy
to find y(b, S0) .
Is y(b, S0)=? if yes then stop, otherwise continue
001
001
,
,
SbyifSayS
SbyifSayS otherwiseGuess
Step 4. Solve the IVP
,,,,, 1Sayayyyxfy
to find y(b, S1).
Step 3.
Is y(b, S1)=If yes then stop, otherwise continue.
Step 5. Find the next Sn+1 ,using secant method.
,2,1
,,
,
1
11
n
SbySby
SSSbySS
nn
nnnnn
Step 6. Solve the IVP
1
1
,
,,,,,
n
n
Sby
Sayayyyxfy
findto
Step 7. Test for convergence
stopthenyesifIs TolSby n 1,
Step 8. Go to step 5.
Example 1:Given
31,32
1
22
yy
yx
y [note that this is a linear differential equation]
To find y for 12
1 x correct to 3d.
0001.3,1
27500.36250.3
21
36250.3
2
1
6250.3,12
1
7500.3,11
2
2
01
00
Sy
S
SyS
SyS
x y
0.5 3
0.6 2.8667
0.7 2.8287
0.8 2.8501
0.9 2.9112
1 3.0001
Example 2:
Given
82,41
21
yy
y
y
xyy [this is a nonlinear differential equation]
To find y for 1<x<2 correct to 3d.
,2,1
,2,2
8,2
08,2
1
11
n
SySy
SSSySS
SySF
nn
nnnnn
000000.8,2333333.1
9902356.7,23331706.1
9688716.7,2328125.1
2580645.8,2375.1
111111.7,216667.1
4.6,21
16,22
66
55
44
33
22
11
00
SyS
SyS
SyS
SyS
SyS
SyS
SyS
x y
1 4
1.1 4.1450
1.2 4.3165
1.9 7.0793
2 7.9994
In case of Newton's method as the root finder the steps will be as follows
Step 1. Guess S0=y'(a)Step 2. Solve the system
1
0
,,
0
S
ayS
ay
Say
ay
S
y
y
f
S
y
y
f
S
y
yyxfy
(1)
By any IVP method find y(b, S0), y' (b, S0) and
00 ,,, SbS
ySb
S
y
Step 3. Find a new Sn using Newton’s method
,1,0,
,1
n
Sbsy
SbySS
n
nnn
Step 4. Solve the system(1) with new Sn=y'(a)
Step 5. Test for convergence, i.e.
stopthenyesifTolSby n ,
Step 6. Go to step 3.
Example: Find y for 1<x<2, given
thenanddenoteuslet
system(1)solvetonow
guess
US
yY
S
y
S
yy
y
y
xyy
2
82,41
21
0
if
11
01
1
41
412
21
0
2
2
U
Y
Sp
y
Uy
p
xY
y
pU
UY
y
p
xpp
py then
Solve by 4th order Runge-Kutta with h=0.01 (say) then
66666667.1
24,216,2
1
00
S
SS
ySy
impliesthis
and
Next five iterations yields
000000.6,2000000.8,233333333.1
00018279.6,200012208.8,2333353689.1
04715063.6,203137248.8,233854166.1
82666667.6,25333333.8,2416666667.1
6666667.10,26666667.10,266666667.1
555
444
333
222
111
SS
ySyS
SS
ySyS
SS
ySyS
SS
ySyS
SS
ySyS
Finite difference methodsFinite difference methodsStatement of the problem:Given the BVP
byay
yyxfy
,
,,
Find y for a<x<b correct to md.
There are two cases:
1. Linear equation, in general
byay
xryxqyxpy
,
2. Non linear equation
byay
yyxfy
,
,,
The above two cases are called BVPs with separated boundary conditions.
BVP with general linear boundary values are of the form
54321
54321
,,
BbyBbyBayBayB
AbyAbyAayAayA
yyxfy
BVP with general boundary conditions are of the form
0,,,,
0,,,,
,,
2
1
bybyayaybg
bybyayayag
yyxfy
The idea of finite difference methods is to discretise the equation by dividing the interval [a,b] into n equal divisions, i.e.
hn
ab say
bxniihxxax ni ,1,,1, 00 then
The BVP becomes
n
iiii
yyax
yyxfy
,,
,,
00
Now we replace the derivatives by an approximate value of finite difference such as:
42112
32112
211
11
4642
22
22
h
yyyyyy
h
yyyyy
h
yyyy
h
yyy
iiiiii
iiiii
iiii
iii
Central-difference expressions with error of order h2
Central-difference expressions with error of order h4
4321123
3321123
22112
2112
6
12395639128
81313812
16301612
88
h
yyyyyyyy
h
yyyyyyy
h
yyyyyy
h
yyyyy
iiiiiiii
iiiiiii
iiiiii
iiiii
Forward-difference expressions with error of order h
41234
3123
2112
1
464
33
2
h
yyyyyy
h
yyyyy
h
yyyy
h
yyy
iiiiii
iiiii
iiii
iii
4
12345
31234
2123
12
31426241122
51824143
2542
34
h
yyyyyyy
h
yyyyyy
h
yyyyy
h
yyyy
iiiiii
i
iiiiii
iiiii
iiii
Forward-difference expressions with error of order h2
Backward-difference expressions with error of order h
44321
3321
211
1
464
33
2
h
yyyyyy
h
yyyyy
h
yyyy
h
yyy
iiiiii
iiiii
iiii
iii
454321
34321
2321
21
21124261432
31424185
4522
43
h
yyyyyyy
h
yyyyyy
h
yyyyy
h
yyyy
iiiiiii
iiiiii
iiiii
iiii
Backward-difference expressions with error of order h2
In linear case we have, using the central difference forms with truncation error of O(h2),
1,,1,2
2
0
112
11
niyy
xryxqh
yyxp
h
yyy
n
iiiii
iiii
After some simplifications
1,,1,0
11
niyy
DyCyByA
n
iiiiiii
Where
ii
ii
ii
ii
rhD
hpC
qhB
hpA
2
2
2
2
24
2
In matrix form
11
2
11
1
2
1
11
222
11
nnnnn CD
D
AD
y
y
y
BA
CBA
CB
O
O
This linear system can be solved by
1.Direct methods (gauss elimination, …)
2. Iterative methods (Jacobi's method, Gauss Seidel method, Successive Over Relaxation method, SOR).
Example:
3d.tocorrectforFind 12
1
31,32
1
22
xy
yy
yx
y
Take n=5 h=0.1 then
96.2
04.0
04.0
46.1
8.18.000
9.06.17.00
08.04.16.0
007.02.1
4
3
2
1
y
y
y
y
Using Gauss elimination method
91111111.285000000.2
8285714.2866666667.2
43
21
yy
yy
In non linear case, using central difference formulas with truncation error O(h2)
1,,1,,
2/,,2
0
112
11
niyy
hyyyxfhyyy
n
iiiiiii
Using simple iteration method, with suitable initial guess
,1,0
1,,1,
2,,
22
1
0
)1(1
)(1)(
2)(1
)1(1
1
k
niyy
h
yyyxf
hyyy
n
ki
kik
iik
ik
iki
Example:
82,41
21
yy
y
y
xyy
Find y for 1<x<2 correct to 5d.
Upon discritisation we have,
,1,0,1,,18,4
1
222
1
0
)(
)1(1
)(1
)1(1
)(1
2)(
1)1(
1)1(
kniyy
yh
yy
xh
yyhyyy
n
ki
ki
ki
i
ki
kik
ik
ik
i
Now, takeN=5 h=0.2 After k= 50 iterations and Tol=
610
1.2 4.30868851
1.4 4.74347837
1.6 5.37398310
1.8 6.34221887
ix iy
Take N=10 h=0.1
After k= iterations
1.1
1.2 4.31464627
1.3
1.4 4.75747894
1.5
1.6 5.39795103
1.7
1.8 6.37355180
1.9
ix iy
To improve the accuracy1. Either increase n smaller hLarger system of equations2. Use extrapolation,In the above example
Improved 3
14.1 y less accurate
3
44.1 y more accurate y(1.4)
4762145797.4
75747894.43
474347837.4
3
1
hO
Continue to improve accuracy.
Iterative methods for system of linear equations.To solve
1) Ax=b
Where
nnnnn
n
b
b
b
x
x
aa
aa
A
11
1
111
,,x
2)
nnnnn
nn
bxaxa
bxaxa
11
11111
3)
nibxa
nibxaxan
jijij
inini
,,1
,,1
1
11
4)
Pivoting , ScalingJacobi’s method: guess nixi ,,1)0(
,1,0
,,11
1
)()1(
k
nixab
ax
n
ijj
kjiji
ii
ki
Test for convergence at each iteration
Tolxx kj
kj )()1( For all j=1,…,n
Guess – Seidel method
,1,0,,,1
1
,,1
1
)(1
1
)1()1(
)0(
kni
xaxaba
x
nix
n
ij
kjij
i
j
kjiji
ii
ki
iGuess
Test for convergence
Successive Over Relaxation (SOR) method
,1,0,,,1
,,1
)(1
1
)1()()1(
)0(
kni
xaxaba
xx
nix
n
ij
kjij
i
j
kjiji
ii
ki
ki
i
Guess
Test for convergence
required?isor bopt .
1. No general formula for
2. depends on the form of A.
3. 0< <2
When 1< <2 we have over relaxation and when 0< <1 we have under relaxation
4. to be calculated numerically.
bb
bb
b
b
Example: solve
96.2
04.0
46.1
8.18.000
9.06.17.00
08.04.16.0
007.02.1
4
3
2
1
x
x
x
x
w 1 1.1 1.2 1.3 1.4 1.5 1.6
k 34 27 21 15 18 24 30
bw
Minimum When k=1428.1
Variational methodsVariational methods IntroductionDistance between two points
2
1
2
1
222 1
x
x
x
xdx
dx
dydydxyI
To minimize I[y] set its derivative to zero.
There are certain restrictions on each y which must pass through ,etc. 2211 ,, yxyx &
2y
1y
1x 2x
4y
3y 2y1y
E-L eqn. yyxFy
yyxFydx
d
,,,,
If F is independent of 0
y
Fy
If F is independent of Cy
F
y
F
dx
dy
0
If F is independent of
0
y
FyF
dx
d
Cy
FyFx
or
Now to solve the BVP
010
yy
xfyxqyxp
The Rayleigh - Ritz method minimizes the E-L equation
1
0
22 2 dxxuxfxuxqxuxpuI
Choose u to be a linear combination of some basis function such as few first terms of Taylor expansion or in general iicu
Where are chosen in such away to satisfy the boundary conditions
Now for minimization find
and equate to zero
These are called the normal equations which can be solved by a method of linear systems.
i 010 ii ni
c
I
i
,,1
nic
I
i
,,10
Example:Let us choose the basis to be the linear functions
10
00
1
11
11
1
1
xx
xxxh
xx
xxxh
xx
xx
x
i
iii
ii
iii
i
i
i
xi
x0
1
1ix ix1ix 1
The normal equations yield a tridiagonal linear system which can be solved by previously introduced methods.
Example:
010
422 22
yy
xyyxyxGiven
Use h=0.1 and linear approximation.Now given
0
byay
xfQyy
Then E-L equation
b
a
dxfuQuuuI 222
to be minimized.
B.C.&If2
210 xcxccu
210210
,0,0,0 cccc
I
c
I
c
Igivesthen
B.C.&If3
32
210 xcxcxccu
on.so&
givesthen 3,2,1,03,2,1,00
icic
Ii
i
Eigen Value Problems(Homogeneous BVP)
Consider the problem:
010
2
yy
yy
To find the nontrivial solution of the above system
,2,1
,2,10sin0sin
01
sin
00
sincos
kk
kkB
y
xBy
y
xBxAy
k
B.C.&
B.C.&
The Eigen Values are
and are the Eigen functions.
Let us use finite difference method
222 kk xBy kkk sin
1,,10
2
0
2211
niyy
yhyyy
n
iiii
In matrix form, let th 222
0
0
1
1
1
11
1
1
1
ny
y
t
t
t
O
O
This homogeneous linear system has nontrivial solution when
0
1
1
11
1
det
t
t
t
A
O
O
Choosing n=2 then 21h
and an approximation to the analytical value
80 2 t8696.922
1
Choosing n=4 then 41h
6274.5422
32
3726.922
0
10
11
01
23
22
21
t
t
t
and
The analytical value of
8264.88
4784.39
8696.9
23
22
21
Again for improvement there are two methods
1) To increase n and have smaller h leading to a higher degree polynomial equation and hence more round off error.
2) To use extrapolation technique.
In this case
Improved
8301.9
3726.93
48
3
13
4
3
1 21
21
21
accuratemoreaccurateless
Numerical Solution of PDE(Finite difference method)
We are going to consider
1) Heat equation
2) Steady – State equation
3) Wave equation
Consider the heat equation in the form
xftxu
ttbu
ttau
tbxax
uD
t
u
,
,
,
0,2
2
To find u(x,t) using finite difference method.
1. Explicit method
As before let nixixxax i ,,1,0, 00
and
ijji
j
utxu
jtjttt
,
,1,0,0 00
denote
The partial derivatives will be as,
)difference(central
)difference(forward
2
2
1
2
2
1
2xO
x
uuu
x
u
tOt
uu
t
u
jiijiji
ij
ijij
ij
Finally, let 2.
x
Dtr
Upon substitution in the equation,
2111 2 xtOuuuruu jiijjiijij
Stability condition is
When 2
1r
2
1r
jijiij uuu 111 2
1
The explicit method looks like
i, ji-1, j i+1, j
i, j+1
In the whole problem
iju
a bx
t
Problem:Problem: show that when the local truncation error will reduce to
The restriction on r causes higher number of operations which leads to higher round off error.
Hence we move to implicit procedures.
6
1r
42 xtO
2.Implicit methods
Let us replace by a linear combination of the known time step and the unknown (next) time step as follows,
2
2
x
u
10
11
2
2
2
2
where
ijijij x
u
x
uD
t
u
When then we have the Explicit method.
When then we have the Implicit Crank - Nicolson method
2
1
When then we have the fully implicit method (backward difference method)
Implicit Crank – Nicolson method
11111111 222 jiijjijiijjiijij uuuuuur
uu
The diagram of this method looks like
i-1j+1 ij+1 i+1j+1
i-1j ij i+1j
Rearrangement of the equation yields
,1,0,1,,2,12
12
21
2
11
11111
jni
ur
urur
b
bur
urur
jiijjiij
ijjiijji
where
This system of (n-1) equation in (n-1) unknown must be solved for each time step j . Therefore j can be omitted at each time step, that is
For each j=0,1,…
1,,2,12
12 11
ni
bur
urur
iiii
Taking into consideration the boundary conditions then in matrix form,
nnn ur
b
b
ur
b
u
u
rr
rr
r
rr
O
O
2
2
12
21
2
21
1
2
01
1
1
This system can be solved by SOR method,
For each j=0,1,…
,1,0;1,,2,1
21
21)(1
)()1(1
)()1(
kni
ur
urur
br
uu ki
ki
kii
ki
ki
k is the number of iterations
In this special case is found to be.opt
nr
rb
cos1
,11
22
Example Given
201020
1000,
0100,20
16.0,00,0
0,2002
2
xx
xxxu
ttu
Dttu
txx
uD
t
u
Find at x=4,8,12,16
and at x=2,4,…,18
,xu
As you know
Take r=1 and then and for each j
216.0
x
tr
4x 100t
00
4
6
6
4
25.000
5.025.00
05.025.0
005.02
50
4
3
2
1
uu
u
u
u
u
&thatnote
100
27273.3
09091.5
09091.5
27273.3
4
3
2
1
t
u
u
u
u
at
For the next time step and00 u 1005 u
545455.52
18182.4
18182.4
545455.2
4
3
2
1
b
b
b
b
Now at t=200
7373.55
8582.17
33188.7
1057.3
50
4
3
2
1
4
3
2
1
4
3
2
1
u
u
u
u
b
b
b
b
u
u
u
u
A
& so on.
Elliptic ProblemsConsider the Poisson equation
mjni
jkcyihax
yxgyxu
dycbxa
yxfy
u
x
uyxu
ji
,,1,0,,1,0
,,
,
,,2
2
2
22
andfor
andlet
boundariestheonwith
on
ax 0 1x 2x 3x 4x nxb
cy 0
1y2y
dym
y
x
Let us use the central differential formula for both derivatives, then
1,,1;1,,1
2
22
222211
211
2
2
11
2
11
mjni
fkhukhuuhuuk
fk
uuu
h
uuu
ijijijijjiji
ijijijijjiijji
or
The diagram looks like
ij+1
i+1jiji-1j
ij-1
Five point formula.
This linear system of (n-1)(m-1) equations and unknowns can be solved by Gauss – Seidel method or by SOR method with
mnc
cb
coscos2
1
11
22
where
Example:Example:Find , given the Poisson equation ji yxu ,
80,6022
2
2
2
yxy
u
x
u
u=0 on the boundaries.
Take a) h=k=2 b) h=k=1
For the case a) there are 6 unknowns ijui=1,2 ; j=1,2,3
2111
2221
2331
uu
uu
uu
8
6 x
y
3,2,1,2,1
84
11111
ji
uuuuu ijijjijiij
Using the symmetric property (in this particular problem) we have
56.456.4
72.572.5
56.456.4
2313
2212
2111
uu
uu
uu
For the case b) there are 35 linear equations in 35 unknowns 7,,1;5,,1? jiuij
Using the symmetric property the system will be reduced to 12 equations in 12 unknowns.
x x x
x x x
x x x
x x x
8
6 x
y
Preparing the system to be solved by SOR it looks like
b
kij
kij
kij
kji
kji
bkij
kij
ji
uuuuuuu
7,,15,,1
484
)()(1
)1(1
)(1
)1(1
)()1(
The results are as follows
647.6960.5181.3
335.6686.5657.3
319.5794.4123.3
353.3047.3042.2
Hyperbolic Problem Hyperbolic Problem (Wave equation)(Wave equation)
Two dimensional heat equation in Cartesian coordinates
(AAlternative lternative DDirection irection IImplicit methodmplicit method) ADIADI
Fourier SeriesFourier SeriesPower series play an integral part in real(& complex)
analysis.
Taylor Series in ODE
Fourier Series in PDE
Definition:Definition: Two nonzero f(x) and g(x) are said to be orthogonal on the interval with respect to the weight function (x) if their scalar product vanishes:
bxa
b
adxxgxfx 0
ExampleExample
0sinsin
2,0
1sinsin
2
0
dxmxnx
xnmmxnx
onarthogonal
areand
2l – periodic function means xflxf 2
Problem:Problem: the set of function
l
xn
l
xn sin,cos,1
are orthogonal over the interval with respect to the weight function (x)=1 .
lxl
Furthermore,
l
l
l
l
l
lldx
l
xndx
xnldxl
222 sin
2cos;2
Let f(x) be sufficiently smooth function defined on and periodic with period 2l.
Then we seek to express f(x) as an infinite linear sum of sines and cosines having the same period as f(x), namely 2l.
We assume that
lxl
10 sincos
2
1
nnn l
xnb
l
xnaaxf
Where and the coefficients and are found to be
0a na nb
l
ln
l
ln
l
l
dxl
xnxf
lb
n
dxl
xnxf
la
dxxfl
a
sin1
,2,1
cos1
10
CorollaryCorollaryWhen f(x) is periodic, piecewise smooth function, its
Fourier series converges to [f(x+)+f(x-)]/2.
Example 1. Find the Fourier series of f(x)=x 0<x<2l and is 2l – periodic
02
sin1
00cos1
21
2
0
2
0
2
00
nn
ldx
l
xnx
lb
ndxl
xnx
la
lxdxl
a
l
n
l
n
l
Therefore
1 1
sin12
1sin2
n n l
xn
nl
l
xn
n
llxf
y
x
L
2L
2L 4L 6L
Example 2.Example 2.Find the Fourier series of
and is of period 2l lxlxxf 2
1
2
2
122
222
12
2
2
22
2
22
22
22
0
6
1
,14
3
cos14
3
00sin1
014
cos1
3
21
n
n
n
l
ln
nl
ln
l
l
n
n
llllx
l
xn
n
llxf
ndxl
xnx
lb
nn
ldx
l
xnx
la
ldxx
la
hence
At
Complex FormComplex Form
Knowing that
2cos
2sin
ii
ii
ee
i
ee
and
Then the Fourier series of f(x) takes the form
lxin
n
nnlxin
n
nn
n
lixn
lixn
n
lxin
lxin
n
eiba
eibaa
i
eeb
eea
axf
11
0
1
0
222
222
l
l
lxin
n
n
lxin
n
nnn
nnn
dxexfl
c
ecxf
niba
c
niba
cac
2
1
02
02
,20
0
where
then
andLet
Fourier Sine and Cosine SeriesFourier Sine and Cosine Series1. When f(x) is an even function then
1
0
0
cos2
0,cos2
nn
l
nn
l
xna
axf
bdxl
xnxf
la
The cosine expansion of f(x)
2. When f(x) is an odd function then
1
00
sin
sin2
,0
nn
l
nn
l
xnbxf
dxl
xnxf
lbaa
The sine expansion of f(x)
Fourier IntegralFourier IntegralThe Fourier series extension to none periodic
functions leads to Fourier Integral
Consider a periodic function of period 2l. Its Fourier expansion is
xfl
.,sincos2
0
l
nxbxa
axf nnnnnl
where
Now let and set
that is
llnn
1
l
1
1
sinsincoscos1
1
n
l
l nln
l
l nln
l
l ll
duuufxduuufx
duufl
xf
If and f is absolutely integrable
then using the idea of definite integrals,
xfxf ll
lim
0sinsincoscos
1
dduuufxduuufxxf
0
sincos dxBxAxfor
duuufBduuufA
sin1
,cos1
where
Example:Example:Find the Fourier integral representation of the function
0sin1
sin2cos
1cos
1
10
11
1
1
duuufB
duuduuufA
x
xxf
if
if
dx
xf
0
sincos2
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