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A short note on Numerical Analysis II meant for self-taught
students until face to face class will resume
Complied by
Hailu Muleta (Assistant Professor of Mathematics)
April, 2020
Jimma, Ethiopia
2
Chapter One
Revision on Numerical Integration
This chapter is highly devoted to find the numerical integration for a
given set of data points. The topics covered here are
- Numerical Integration
- Newton-Cote’s quadrature formula
- Trapezoidal Rule
- Simpson’s one-third Rule
- Simpson’s three-eighths Rule
- Weddle’s rule
Consider the definite integral .
This integral represents the area between , the –axis and the lines & .
This integration is possible as far as is explicitly given and the function is integrable.
Now suppose set of (n+ 1) paired values are given. First as we did in the case of numerical
differentiation, we find by an interpolating polynomial and obtain which
can approximate the value for .
A General Quadrature Formula for Equidistant Spacing (Newton-Cote’s Formula)
For equally spaced intervals, we have Newton’s forward difference formula as
dxxfb
a
)(xfy x ax bx
)(xf
)(xf )(xPn dxxp
b
an
dxxpb
an
1...!3
21
!2
10
3
0
2
00
yuuu
yuu
yuyxy
3
Here, is the interval of differencing.
Now, instead of , we will replace it by this interpolating polynomial of Newton.
Since
Thus,
. Where is interpolating polynomial of degree n.
This equation is called Newton-cote’s quadrature formula and is a general quadrature
formula.
By taking different values for n we get a number of special formulas. Here we try to look for
some values of n for which their practical application is very important in different disciplines of
science. Detail information and results are explained below.
hh
xxu ,0
)(xf )(xy
h
xxnu
h
xxuandnhxxn
000
dxxfdxxfnhx
x
x
x
n
0
00
dxxPnhx
xn
0
0
)(xPn
hduy
uuuy
uuyuy
n
00
3
0
2
00 ...!3
21
!2
1
duy
uuuy
uuyuyh
n
00
323
0
2
00 ...!3
23
!2
1
n
y
uuu
y
uu
yu
uyh
0
0
3
234
0
2
23
0
2
0 ...!3
4
!2
23
2
)2(...46
1
232
1
20
3234
0
223
0
2
0
ynn
ny
nny
nnyh
4
Trapezoidal Rule
Put n = 1, in the quadrature formula
Since other differences do not exist if n = 1.
This is known as a trapezoidal rule.
Even though this method is very simple for calculation, the error in this case is significant.
Truncation Error in Trapezoidal Rule
In the neighborhood of , we can expand by Taylor series in powers of .
That is,
002
1.1
0
0
0
0
yyhdxxfdxxfhx
x
nhx
x
010
2
1yyyh
102
yyh
dxxfdxxfnx
x
nhx
x
0
0
0
dxxfdxxfdxxfnhx
hnx
hx
hx
nhx
x
0
0
0
0
0
0 1
2
...
nn yyh
yyh
yyh
121102
...22
1210 ...22
nn yyyyyh
0xx )(xfy 0xx
1...!2
''
0
2
0'
000
yxx
yxxyxy
5
(2)
If is the equidistant length, then also
Area of the first trapezium = A0 say
Putting in (1), we get
Subtracting A0 from (2), we obtain
dxyxx
yxxyydxx
x
x
x
1
0
1
0
...!2
''
0
2
0'
000
1
0
...!3!2
''
0
3
0'
0
2
00
x
x
yxx
yxx
xy
...!3!2
''
0
3
01'
0
2
01010
y
xxy
xxxxy
...!3!2
''
0
3'
0
2
0 yh
yh
hy
h
10
1
0 2yy
hydx
x
x
1xx
...!2
''
0
2
01'
001011
yxx
yxxyyxy
...!2
''
0
2'
00 yh
hyy
...
!222
''
0
2'
000100 yh
hyyyh
yyh
A
...!2.22
''
0
3'
0
2
0 yh
yh
hy
...!2.2
1
!3
1''
0
3
0
1
0
yhAydx
x
x
6
The error made in the first interval is
Similarly the error in the ith
interval
Hence, the total cumulative error E is
if the interval is and
The error in the trapezoidal rule is of order .
The accuracy of the result can be improved by increasing the number of intervals and
decreasing the value of .
Simpson’s One-Third Rule
Setting n = 2 in Newton-cotes quadrature formula, we have
since the other terms vanish (become zero).
...2
1 ''
0
3 yh
),( 10 xx ...2
1 ''
0
3 yh
''
1
3
2
1 iyh
''
1
''
2
''
1
''
0
3 ...2
1 nyyyyhE
,...,,max12
''2
''1
''0
3
yyyMwhereMnh
E
M
hab
12
2 ),( ba
n
abh
2h
h
0
2
002
4
3
8
2
142
2
yyyhdxxfx
x
0
2
010 13
122 yEyyyh
012010 2
3
1222 yyyyyyh
7
Similarly, and
If n is an even integer, then the last integral will be
Adding all these integrals, if n is even positive integer, then are odd in
number; we have
Simpson’s Three-Eighths Rule
Putting n = 3 in Newton-cotes formula, we get
012
3
1
3
4
3
1yyyh
012 43
yyyh
432 43
4
2
yyyh
dxxfx
x 214
3
2
iii
x
xyyy
hdxxf
i
i
nnn
x
xyyy
hdxxf
n
n
12 432
nyyyyy ,...,,,, 3210
nx
nx
x
x
x
x
nx
xdxxfdxxfdxxfdxxf
20
4
20...)()(
2
nnn yyyyyyyyyh
12432210 4...443
1312420 ...4...23
nnn yyyyyyyyh
0
3
0
2
00 9274
81
6
1
2
9
2
1
2
93)(3
0yyyyhdxxf
x
x
0
3
0
2
010 18
31
4
9
2
93 yEyEyyyh
0123012010 33
8
32
4
9
2
9
2
93 yyyyyyyyyyh
0123 338
3yyyy
h
8
If n is a multiple of 3,
This is Simpson’s three-eighths rule and is applicable only when n is a multiple of 3.
Weddle’s Rule
Putting n = 6 in Newton-cotes formula
Now replace the term by doing this, the error introduced is only
which is negligible when and are small.
Using and replacing all differences in terms of ’s,
we get
Similarly,
and
Adding all these integrals, we get
nhx
hnx
hx
x
hx
hx
nhx
xdxxfdxxfdxxfdxxf 0
)3(0
30
0
60
30
0
0
...)()(
nnnn yyyyyyyyyyyyh
12365433210 33...33338
3
nnnn yyyyyyyyyyyh
631254210 2...38
3
...36216324
6
11872
2
1186)( 0
3
0
2
00
60
0yyyyhdxxf
hx
x
0
6
0
5
0
4
0
3
0
2
00140
41
10
33
10
1232427186 yyyyyyyh
,140
42
140
410
6
0
6 ybyy 0
6
140y
h
h0
6 y
1 E y
6543210 56510
360
0
yyyyyyyh
dxxfhx
x
1211109876 56510
3120
60
yyyyyyyh
dxxfhx
hx
nnnnnnn yyyyyyyh
dxxfnhx
hnx
123456 56510
30)6(
0
9
This equation is called Weddle’s rule.
Truncation Error in Simpson’s Formula
By Taylor expansion of in the neighborhood of , we obtain
Now let , by Simpson’s rule
Putting in (1), to get
Putting in (1), we have
nnnnnnn yyyyyyy
yyyyyyyyyyyyh
dxxfnhx
x
123456
1109876543210
5652...
565256510
30
0
)(xfy 0xx
1......!3!2
'''
0
3
0''
0
2
0'
000
yxx
yxx
yxxyy
dxy
xxy
xxy
xxyydx
x
x
x
x
2
0
2
0...
!3!2!1
'''
0
3
0''
0
2
0'
00
0
2
0
..!4!3!21
'''
0
4
0''
0
3
0'
0
2
00
x
x
yxx
yxx
yxx
xy
...!4!3!2
'''
0
4
02''
0
3
02'
0
2
02020
y
xxy
xxy
xxxxy
...!4
16
!3
8
!2
42 '''
0
2''
0
3'
0
2
0 yhyh
yh
hy
...15
4
3
2
3
422 4
0
5'''
0
4''
0
3'
0
2
0 yh
yh
yhyhhy
2101 43
area2
0
yyyh
ydxAx
x
1xx
...!2
''
021'
00101
yxx
yxxyy
...!3!2
'''
0
3''
0
2'
00 yh
yh
hyy
2xx
10
Omitting the remaining terms involving and higher powers of .
This means that the error made in is
Similarly, the error made in and so on.
Hence the total error E is
, where M is the maximum value of
Since is the last paired value because we require odd number of ordinates to apply
Simpson’s one-third rule.
If the interval is , then , using this
Hence, the error in Simpson’s one-third rule is of the order .
Examples
...!3
8
!2
42 '''
0
3''
0
2'
002 yh
yh
hyyy
...22...
!3!24
3
''
0
2'
00
'''
0
3''
0
2'
0001 yhhyyyh
yh
hyyyh
A
...18
5
3
2
3
422 4
0
5'''
0
4''
0
3'
0
2
0 yhyhyhyhhy
...18
5
15
4 4
0
5
1
2
0
yhAydx
x
x
...90
4
0
5
yh 6h h
20 , xx ...90
4
0
5
yh
4
2
5
4290
, yh
isxx
...9
, 4
2
4
0
5
0 yyh
isxx n
Mnh
E90
5
4
22
4
2
4
0 ...,, nyyy
nn yx 22 ,
),( ba )2( nhab
M
habE
180
4
4h
11
1. Evaluate by using
a) Trapezoidal rule
b) Simpson’s rule and verify your results by actual integration
Solution
Here and the interval length is 7-1 = 6 so we divide this interval into 6 equal parts
with
: 1 2 3 4 5 6 7
: 1 4 9 16 25 36 49
a) By Trapezoidal rule
b) By Simpson’s one-third rule
=
Since n = 6, we can also use Simpson’s three-eighth rule.
So
dxx7
1
2
2)( xxy
16
6h
x
y
6543271
7
1
2 22
yyyyyyyh
dxx
1153625169424912
1
6425371
7
1
2 (4)23
1yyyyyyydxx
36164425924913
1
114
4653271
7
1
2 238
3yyyyyyydxx
16236259434918
3
12
=
2. Evaluate using Trapezoidal rule with , and hence
obtain an approximate value of .
Solution
: 0 0.2 0.4 0.6 0.8 1
: 1 0.961538461 0.862068965 0.735294117 0.609756097 0.5
=
But by actual integration
To compare this approximated value of with its actual value using calculator ,
the error is which is .
114
1143
342
3
17
3
1
3
3
7
1
37
1
2 x
dxx
,1
1
0 2 x
dx2.0h
21
1
xxyLet
x
y
432150
1
0 22
21yyyyyy
h
x
dx
6097566097.0735294117.0862068965.0961538461.025.012
2.0
80.78373152
4tan
1
1
0
11
0 2
xx
dx
783731528.04
134926112.3
43.14159265
1590.00666654 %6.0
13
3. From the table below, find the area bounded by the curve and
the -axis from to
:
:
Solution
Let us compare the results obtained by different methods
i) By Trapezoidal rule
=
ii) By Simpson’s one-third rule
iii) By Simpson’s three-eighths rule
iv) By Weddle’s rule
As we can see from these rules the area is (correct to four decimal places)
x 47.7x 53.7x
x 7.53 7.52 7.51 7.50 7.49 7.48 7.47
y 2.08 2.06 2.03 2.01 1.98 1.95 1.93
06.203.201.298.195.1208.293.12
01.053.7
47.7 dxxf
0.12035
06.201.295.14403.298.1208.293.13
01.053.7
47.7 dxxf
70.12036666
01.2206.203.298.195.1308.293.18
)01.0(353.7
47.7 dxxf
1203375.0
08.2)06.2(503.201.2698.195.1593.110
)01.0(353.7
47.7 dxxf
12039.0
0.1203
14
4. Evaluate the integral using the rules so far developed.
Solution
Since let us divide the interval into 6 equal parts, i.e.;
i) By Trapezoidal rule
=
ii) By Simpson’s one-third rule
iii) By Simpson’s three-eighths rule
iv) By Weddle’s Rule
2.5
4nxdxI
2.142.5 ab 2.06
2.1h
x xln x xln
61.64865862 5.2
21.60943791 5.0 11.48160454 4.4
81.56861591 4.8 51.43508452 4.2
31.52605630 4.6 11.38629436 4
)]609437912.1568615918.1526056303.1
481604541.1435084525.1(2648658626.1386294361.12
2.02.5
4
nxdx
91.82765513
)]609437912.1526056303.1435084525.1(
4)568615918.1481604541.1(2648658626.1386294361.13
2.02.5
4
nxdx
827847258.1
)526056303.1(2)609437912.1568615918.1(
3)481604541.1435084525.1(3648658626.1386294361.13
2.02.5
4
dxnx
827847258.1
15
By actual integration,
Here Weddle’s rule best approximates the exact value.
5. Evaluate by Simpson’s one-third rule correct to five decimal places.
Solution
The interval
Since error where M = Max in the range
Now we require
Hence we take
648658626.1609437912.15568615918.1.1)526056303.1(6
526056303.16481604541.1435084525.15386294361.110
)2.0(32.5
4
dxnx
827847407.1
827847409.112.5
4
2.5
4 nxxnxdx
dxex
1
0
1 ab
,
180
4Mhab
E
)( xe
eh4
180
1
610E
64
10180
eh
1.0090207886.010180 4
16
eh
1.0h
16
By the actual integration,
Correct to five decimal places which is the same as the exact value.
6. A curve passes through the points
and . Obtain the area bounded by the curve, the -axis, and .
Area =
sq. units
Volume =
= 64.1041981 cubic units.
7. A river is 80 meters wide. The depth ‘d’ in meters at a distance
meters from one bank is given by the following table.
: 0 10 20 30 40 50 60 70 80
: 0 4 7 9 12 25 14 8 3
Calculate the area of cross-section of the river using Simpson’s rule.
718282782.1
4213
1.0 9.07.05.03.01.08.06.04.02.01
0
eeeeeeeeeedxex
718281828.111
0
1
0 eedxe xx
71828.11
0 dxex
(2.5,2.8), (2.0,2.7), (1.5,2.4), (1,2), (3.5,2.6) (3,3),
(4,2.1) x 1x 4x
b
aydxydx
4
1
)6.28.24.2(437.221.223
5.0
7833.72.314.111.46
1
dxydxyb
a 4
1
22
2222222 6.28.24.2437.221.223
5.0
405.2044.8158.3241.86
x
x
d
17
Solution
Area of cross-section =
= sq. meters
8. The table below gives the velocity of a moving particle at time second. Find the
distance (S) covered by the particle in 12 seconds and also the acceleration at
seconds.
: 0 2 4 6 8 10 12
: 4 6 16 34 60 94 136
Solution
We know that and
To get S
= meters
To find first form the difference table
t v
0 4
2 6
4 16
6 34
8 60
80
0ydx
)81594(4)14127(2303
10 710
v t
2t
t
v
dt
dsv
dt
dva
9434646016213643
212
0 vdtS
552
2
,
tdt
dvaa
v v2 v3
2
10
18
26
34
8
8
8
8
0
0
0
0
18
10 94
12 136
Taking = 6
Exercise
1. Evaluate taking , using Trapezoidal rule. Can you use
Simpson’s rule? Justify your reasons.
2. Compute the value of correct to four decimal places
taking .
3. Find the value of from using Simpson’s one-third rule
with .
4. When a train is moving at steam is shut off and brakes are applied. The speed of the
train per second after seconds is given by
Time :
Speed :
Using Simpson’s rule, determine the distance moved by the train in seconds.
5. Evaluate a) dividing the range into four equal parts
0
3
0
2
0
2 3
1
2
11vvv
hdt
dv
t
0v
2sec/382
110
2
1m
2
1
21 x
dx2.0h
1
0
cossin dxxx
8.0h
3
1
2log
1
0
3
2
1dx
x
x
25.0h
.sec/30m
t
t 0 5 10 15 20 25 30 35 40
v 30 24 5.19 16 6.13 7.11 10 5.8 0.7
40
1
0
2
dxe x
42 8
19
b) dividing the range into ten equal parts by
i) Trapezoidal rule and
ii) Simpson’s one-third rule
6. Evaluate taking .
7. Calculate taking .
8. Evaluate taking four strips.
9. Calculate taking 5 ordinates by Simpson’s rule.
10. Evaluate by Weddle’s rule, dividing the range into six
parts.
11. Evaluate dividing into six equal parts using Simpson’s rule,
Weddle’s rule and Trapezoidal rule.
2
0
sin
dxe x
6
h
0
3sin xdx6
h
7
3
2 log xdxx
7.0
5.0
dxxe x
5.0
021
dxx
dx
0
sindx
x
x
20
Chapter Two
Curve Fitting
Fitting of curves to a set of numerical data is of considerable importance- theoretical as well as
practical. Theoretically it is useful in the study of correlation and regression. In practice it
enables us to represent the relationship between two variables by simple algebraic expressions
(polynomials, exponential or logarithmic functions or any). Besides, it may be used to estimate
the values of one variable which would correspond to the specified values of the other
variable(s).
This chapter covers how to fit a curve for a given set of data points using different methods and
it focuses on the following points:
Regression
- Linear regression
- quadratic regression
- polynomial regression
- multiple regression
- fitting an exponential curve
- curve fitting with Sinusoidal Functions
In most of the fields of engineering and science, we come across experiments which
involve many variables, and most of the time data is collected or given for discrete values
along a continuum; the relation between these variables can be discussed so easily and for
many of these variables it is very difficult to identify the relation unless we can model the
system mathematically. When the system is explained in terms of mathematical models we
have the following relationships about the variables:
1. The relationship between these variables is given in terms of
21
mathematical rules, formulae if any, to determine the quantities of these variables. Actually
it is simple to use these rules for application.
2. The quantities/ variables are given so that we will be interested in finding the
relationships between these variables. This process is a little bit difficult because to write
one variable in terms of the other variables (called empirical equation). Most of the time
we may not be able to get an exact relation between these variables and we may get only
an approximate relation or curve.
This approximating curve is an empirical equation and the method of finding such an
approximating curve is called curve fitting.
Suppose , be sets of observations and the law relating and
can be determined by different mathematical systems that clearly explains the relationship
between these sets of observations . Actually, here we may have different approaches to
fit the given data, and one system may approximate better than the other system on the same given
set of data points.
Now we will see some of these different approaches:
REGRESSION
1. LINEAR REGRESSION
Suppose that the relationship is given
, . (1)
Equation (1) represents a family of straight lines for different values of the arbitrary constants ' '
and ' '. The problem now is to determine ' ' and ' ' so that the line (1) is the line of "best fit".
The term best fit is interpreted in accordance with the Legendre's principle of least squares which
consists of the deviations of the actual values as given by the line of best fit. As a matter of
ii yx , ni ,...,3,2,1 n x
y
n ii yx ,
ii bxay ni ,...,3,2,1
a
b a b
22
chance all the points may lie on a straight line and in this case the line is a 'perfect fit' and the
sum of the squares of the deviations is zero.
Let be any point in the scatter diagram. Draw perpendicular to the x-axis meeting
the line in . The coordinates of are .
PiHi is called the error of estimates or the residual for .
According to the principle of least squares, we have to determine a and b so that
is the minimum.
Using the principle of maxima and minima what we have studied in calculus, the partial
derivatives of E with respect to and should vanish separately.
That is,
),( iii yxP MPi
bxay iH iH ii bxax ,
)( ii
iiii
bxay
MHMPHP
iy
n
i
n
i
iiii bxayHPE1 1
22)(
a b
n
i
ii bxaya
E
1
0)(2
ixbna
xbay
n
i
i
n
i
i
n
i
n
i
i
1
111
23
.
Equations (i) and (ii) are called normal equations.
Solving for and from (i) and (ii), we get the values of and , and with these values of
and so obtained, equation (1) is the line of best fit to the given set of points ,
.
Now let us see some examples to illustrate the above discussion.
Example 1. By the method of least squares find the best fitting
straight line to the data given below:
: 5 10 15 20 25
: 15 19 23 26 30
Solution
Let the line of best be
The normal equations are
We calculate and form the table below
iixbxayx
bxayxb
E
n
i
i
n
i
i
n
i
ii
n
i
iii
1
2
11
1
0)(2
a b a b a
b ii yx ,
ni ,...,3,2,1
x
y
bxay
n
i
i
n
i
i xbay11
5
n
i
i
n
i
i
n
i
ii xbxayx1
2
11
xyxyx ,,, 2
24
5 15 25 75
10 19 100 190
15 23 225 345
20 26 400 520
25 30 625 750
75 114 1375 1885
Using these values in the normal equations, we get
Solving for and we get a=12.3 and b=0.7 and thus the line of best fit is
Example 2. Find the best fitting straight line to the data given below
by the method of least squares and also estimate when
is 70.
: 71 68 73 69 67 65 66 67
: 69 72 70 70 68 67 68 64
Solution
First transform the values of and to
and the normal equations are
Calculations:
71 69 3 -1 9 -3
68 72 0 2 0 0
xyxyx 2
1885137575
114755
ba
ba
a b xy 7.03.12
y
x
x
y
x y 70 and 68 yYxX
XYXaXb
YaXb
2
8
YXYXx X y 2
25
73 70 5 0 25 0
69 70 1 0 1 0
67 68 -1 -2 1 2
65 67 -3 -3 9 9
66 68 -2 -2 4 4
67 64 -1 -6 1 6
2 -12 50 18
Substituting these values in the normal equations, we get
Solving for and , we get
Thus the line of best fit is of the form
This implies
When
1. Quadratic Regression (Fitting Of Second Degree Parabola)
Let be the second degree parabola of best fit to set of points ,
.
Using the principle of least squares, we have to determine and so that
is minimum.
18250
1282
ab
ab
a b99
16a and
99
35b
99
16
99
35 XY
99
16)68(
99
3570 xy
99
4566
99
35 xy
87.7070 yx
2cxbxay n ii yx ,
ni ,...,3,2,1
,,ba c
2
1
2
n
i
iii cxbxayE
26
Equating to zero the partial derivatives of E with respect to and separately, we get the
normal equations for estimating and as
…………………. (1)
Solving for and from (1), (2) and (3), we get with these values of and the
parabola of best fit.
Example 1 Fit a parabola of second degree to the following data
X: 0 1 2 3 4
Y: 1 1.8 1.3 2.5 6.3
Solution
X Y XY Y
0 1 0 0 0 0 0
1 1.8 1 1 1 1.8 1.8
2 1.3 4 8 16 2.6 5.2
,,ba c
,,ba c
021
2
n
i
iii cxbxaya
E
n
i
i
n
i
i
n
i
i xcxbnay1
2
11
021
2
i
n
i
iii xcxbxayb
E
)2....(....................1
3
1
2
11
n
i
i
n
i
i
n
i
i
n
i
ii xcxbxayx
02 2
1
2
i
n
i
iii xcxbxayc
E
)3...(....................1
4
1
3
1
2
1
2
n
i
i
n
i
i
n
i
ii
n
i
i xcxbxayx
,,ba c ,,ba c
2X 3X 4X 2X
27
3 2.5 9 27 81 7.5 22.5
4 6.3 16 64 256 25.2 100.8
10 12.9 30 100 354 37.1 130.3
Substituting these values in the normal equations, we get
Solving for and , we get
, and
is the best fit.
Exercise
1. Find the best fitting parabola to the data given below by the
method of least squares and also estimate when is 70.
: 71 68 73 69 67 65 66 67
: 69 72 70 70 68 67 68 64
2. Polynomial Regression Fitting Of A Polynomial Of kth
Degree
If is the kth
degree polynomial of best fit to the set of points
; the constants are to be obtained so that
is minimum.
Thus the normal equations for estimating are obtained on equating to zero the
partial derivatives of E with respect to separately.
cba
cba
cba
354100303.130
10030101.37
301059.12
,,ba c
07.1,42.1 ba 55.0c
255.007142.1 xxy
y x
x
y
k
k xaxaxaay ...2
210
ii yx , ni ,...,3,2,1naaaa ,...,,, 210
n
i
k
ikiii xaxaxaayE1
22
210 ...
naaaa ,...,,, 210
naaaa ,...,,, 210
28
…
Exercise
1. Find the polynomial of degree three that best fits the data given below by the method of
least squares and also estimate when is 70.
: 71 68 73 69 67 65 66 67
: 69 72 70 70 68 67 68 64
4. Multiple Regressions
There are different multiple regression forms. For the sake of discussion let us see the following
regression type.
Suppose we want to fit the set of data points by the relation
, here we need the points to be of the form .
We determine the values of and , from
k
ikiii xaxaxanaya
E...0 2
210
0
12
10
11
...0 k
kii
n
i
ii xaxaxayxa
E
k
ik
k
i
k
i
n
i
i
k
i
k
xaxaxayxa
E 21
10
1
...0
y x
x
y
cybxyaxZ iii zyx ,,
,,ba c
n
i
iiiii cyybxaxzE1
2
021
i
n
i
iiiii xcyybxaxza
E
)1(..............................2
1
2
1
iiii
n
i
i
n
i
ii yxcyxbxazx
29
By solving (1), (2) and (3) for and , we get the best approximation.
5. Fitting an Exponential Curve
Let , be the sets of observations of related data and let be the
best fit for the data.
Then taking logarithm on both sides,
(*)
Let and , then (*) reduces to
which is linear in and , we can find since and are known, and from
we can get and hence is found out.
Fitting a curve of the form
Using this linear fit, we find .
ii
n
i
iiiii yxcyybxaxzb
E
1
2
)2(..............................1
22
1
2
1
2
1
n
i
iii
n
i
ii
n
i
i
n
i
iii yxcyxbyxayxz
i
n
i
iiiii ycyybxaxzb
E
1
2
)3(..............................1
22
111
n
i
ii
n
i
ii
n
i
i
n
i
ii ycyxbyxayz
,,ba c
ii yx , ni ,...,3,2,1 nxaby
logloglog101010
bay
x
,, loglog1010
ay
AY log10
b
B
BxAY x Y BA, x Y
BA, ba,xaby
baxy
baxy
logloglog101010
xay
b
logloglog101010
,,,xay
xAylettingbxAY
bA,
30
are known and thus is found out.
Example 1. From the table given below, find the best values of
a and b in the law by the method of least squares.
: 0 5 8 12 20
: 3.0 1.5 1.0 0.55 0.18
Solution
Let be the approximating curve.
where B = b
So the normal equations are
y Y Y
0 3.0 0.4771 0 0
5 1.5 0.1761 25 0.8805
8 1.0 0 64 0
12 0.55 -0.2596 144 -3.1152
20 0.18 -0.7447 400 -14.894
45 -0.3511 633 -17.1287
Substituting these values, we get
Solving for and , we get
ba,baxy
bxaey
x
y
bxaey
BxAYbxAYe
log10
log10
e
xYxAxB
YAxB
2
5
x 2x x
1287.1763345
3511.0455
BA
BA
A B
31
So
Hence the curve is
Exercise
1. Fit a straight line to the following data and hence find
: 0 5 10 15 20
: 7 11 16 20 26
2. Fit a straight line to the data
: 0.5 1.0 1.5 2.0 2.5 3.0
: 0.31 0.82 1.29 1.85 2.51 3.02
3. Fit a parabola to the data
: 1 2 3 4 5
: 2 3 5 8 10
4. Fit a curve of the form to the data given below:
: 1 2 3 4 5 6 7 8
: 15.3 20.5 27.4 36.6 49.1 65.6 87.8 117.6
5. Fit a curve of the form to the data given below
4815.0A
0613.0B
0304.31010 4815.0 Aa
0613.0log10
e
bB
1411.0)0613.0( log10
e
b
xey 1411.00304.3
)25( xy
x
y
x
y
x
y
bxaey
x
y
cxybyaxz
32
(0,0,1), (0,1,2), (1,0,4), (1,1,1), (2,0,4), (1,2,5)
6. It is given that , and are related by to the data below and obtain the best
values of a and b.
: 1 2 4 6 8
: 5.43 6.28 10.32 14.86 19.5
x y bxx
ay
x
y
33
Chapter Three
Numerical Solution of Ordinary Differential Equations
In this chapter we are highly interested in finding the solution of ordinary differential
equation numerically using different methods.
The topics covered here is
- Taylor series method
- Taylor series method for simultaneous first order DE
- Taylor series method for second order differential equation
- Picard’s method of successive approximations
- Euler’s Method
- Runge-Kutta method
- Predictor-corrector method
In the fields of Engineering and Science, we come across through the natural phenomena that can
be represented by mathematical models which happen to be in the form of differential equations;
for instance, the equation of motion, the equation of deflection of a beam, etc. The solution of
these differential equations is very essential in the studies of such phenomena.
While finding the solution of these differential equations there are number of differential
equations that we cannot solve analytically; however, in such situations, depending on the nature
of the model, we go for numerical solutions of these differential equations. In many researches,
especially after the advent of modern computers, the numerical solutions of the differential
equations have become very easy for manipulation.
34
Thus, in this part we try to look some of the methods of numerical solutions that are
approximate solutions and in many cases these solutions are in the required (desired) degree of
accuracy and are quite sufficient.
Suppose we want to solve with the initial condition .
Let , , , … be the solution of at
Let be the exact solution. If we plot and draw the graph of
, (the exact curve) and also draw the approximate curve by plotting
(the approximated solution graph) we get two curves.
Suppose
The equation subject to the initial condition
is called an initial-value problem.
Using Taylor series we can expand in the neighborhood of as a power
series of . That is, if is close to , then by Taylor’s series, we have
),( yxfdx
dy 00 )( yxy
00 )( yxy 11)( yxy 22 )( yxy y ,...,, 210 xxxx
)(xyy
)(xyy
),...,(),,(),,( 221100 yxyxyx
)1........(,' yxfdx
dyy
),( yxfy
00)( yxy
)(xy 0x
0xx x0x
M
Q
P (x0,y0)
Approximate solution
Exact solution
QM = approximate value
PM = exact value at x = xi
Then QP = QM-PM
= yi – y(xi) = is called
35
where etc
If is close to , substitute in (*) and get
Again starting from , express in a power series of and then substitute to
get . In this way we can get the sequence of values
If , we get the Maclaurin’s series expansion,
Example 1. Evaluate the solution of the differential equation
by taking four terms of its Maclaurin’s series for , ,
given and compare this result with its exact
solution.
Solution
By Maclaurin’s series, we have
....!3
'''
!2
'''
3
00
2
00000
xxxyxxxyxxxyxyxy
,'','
00
2
2
00
xxxx dx
ydxy
dx
dyxy
1xx 0x1xx )( 11 xyy
1x )(xy 1xx 2xx
)( 22 xyy y ,...,, 210 yyy
00 xx
...0''!2
0'02
yx
xyyxy
12 ydx
dy
0x 2.0x
6.0x 0)0( y
1' 2 yy 10' y
'2'' yyy 00'020'' yyy
''2'2''' 2 yyyy 20''' y
'''6'''24 yyyyy 004 y
...0'''!3
0''!2
0'032
yx
yx
xyyxy
36
Exact solution
Let us compare the actual value with the approximate value
Values of x 0 0.2 0.4 0.6
Actual value of y 0 0.2027 0.4228 0.6841
Approximate value of y 0 0.2027 0.4213 0.6720
Error 0 0 0.0015 0.0121
Percentage of error 0 0 0.35 1.77
This table shows that when the distance of from increases the error also increases.
In this example, we have expanded in the neighborhood of and used the same
result to find when , and .
Now instead of doing this, after getting , expand again in the
neighborhood of and use this result to get . In doing so, we can minimize the
error.
Thus in the neighborhood of
...!5
16
!320
53
xx
x
0)0( y
00032.015
2
3
008.02.02.0
15
2
3
2.02.02.0
53
y 4213.0
6720.06.015
2
3
6.06.06.0
53
y
xycxydxy
dytantan
1
1
2
6942.06.0tan,2927.02.tan,00tan o
x0x
)(xy 0x
)(xy 2.0x 6.0x
)2.0()( 1 yxy )(xy
2.0x )6.0(y
2.0x
37
= = (0.2027)2 + 1 = 1.0411
…, etc.
Putting these values and using in (1), we get
When we compare this value with the actual one i.e. 0.4228, we see that the error is only 0.0003,
nearly 0.07%
The error has decreased from 0.35% to 0.07%
Therefore, to reduce the error, each time obtain the power series of at and use this
to get and so on.
This method is called the method of starting the solution.
Point Wise Methods
Consider the previous example
First we got in terms of x and then we substituted
...!3
2.02.0'''
!2
2.02.0''2.02.0'2.0
32
xyxy
xyyxy
2027.0)2.0( y
)2.0(y 1))2.0(( 2 y
4221.00411.12027.022.0'2.022.0'' yyy
2.0''2.022.0'22.0'''2
yyyy
,3389.24221.02027.020411.122
4.0x
...2.04.06
3389.22.04.0
2
4221.02.04.00411.12027.04.0
32y
...)389871.0)(008.0()21105.0)(04.0()0411.1(2.02027.0
4225.0422480536.0
)(xy 1 ixx
)( 2ixy
0)0(,1' 2 yyy
...3
)(3
x
xxy
38
. Instead, without getting as a function of we can directly get =
as
That is we get directly. So, a point wise solution is a series of points
which satisfy approximately a pre-assigned but not known particular solution.
Solution Using Taylor Series
AIM- To find the numerical solution of the equation
given the initial condition ……….. (1)
Now, we expand about the point using Taylor’s series in powers of .
That is,
Where
, where or
To find we use (1) and its derivatives at .
Even though the series in (2) is an infinite series, we can truncate it at any convenient term, if h
is small, and the accuracy can be obtained.
2.01 xx )(xy x )( 1xy )2.0(y
...!3
2.00'''2.0
!2
)0(''2.0)0(')0(2.0
32
yy
yyy
),(),,( 2211 yxyx
),...,(),,( 2211 yxyx
yxfdx
dy, 00 yxy
)(xy 0xx 0xx
...''!2
' 0
2
0000
xy
xxxyxxxyxy
0
0
xxdx
ydxy
r
rr
)2(...!3!2
'''
0
3''
0
21
0011 yh
yh
hyyxyy
01 xxh hxx 01
,..., ''
0
'
0 yy 0xx
39
Now, once if we get , we can calculate etc by using
Again, expanding , in a Taylor’s series about the point
, we get
Proceeding in the same way, we get
Since this is an infinite series, to get an approximate value we have to truncate it at some term to
have a calculated numerical value.
Now, let us consider the terms up to and including and neglect terms involving and
higher powers of . The Taylor algorithm used this way is said to be of nth
order.
Thus, the truncation error is . If h is small enough we can neglect terms after the nth
term and get the error as
where
Example 1. Solve given , and find by using
Taylor’s method.
Solution
Here and
y ,,..., ''
1
'
1 yy
),( yxfy
)(xy
1xx
...!3!2
'''
1
3''
1
2'
112 yh
yh
hyyy
...!3!2
'''3
''2
'
1 nnnnn yh
yh
hyyy
nh 1nh
h
)( 1nhO
nn
fn
h
!hxxifxx 0110
,yxdx
dy 0)1( y )2.1(),1.1( yy
10 x 0.1 ,00 hy
yxydx
dy ' 010 xyy
40
Thus by Taylor’s method, we have
Now again take and
Let us check for its exact solution
1''' yy 10100
'
0 yxy
''''' yy 2110'' y
etcy ,20'''
...!3!2!1
'''
0
3''
0
2'
001 yh
yh
yh
yy
...2!5
1.02
!4
1.02
!3
1.02
!2
1.011.001.1
5432
y
...000000166.000000833.000033.001.01.0
11033847.0
1.10 x 1.0h
21033847.111033847.01.111
'
1 yxy
21033847.21 '
1
''
1 yy
...
2103384.2
5
1
4
1
'''
1
1
'''
1
yyy
yy
...21033847.2
!4
1.0
21033847.2!3
1.021033847.2
2
1.021033847.11.011033847.0
4
32
2
y
...)21033847.2()0016667.0()005.0(21033847.2121033847.011033847.0
2461077.0
41
Let
and
Integrating both sides,
Since
Hence
Example 2. Using Taylor series method, compute correct to
four decimal places, given and .
Solution
here and
yxdx
dy
dx
dy
dx
dvvyx 1
1 vdx
dv
1
v
dvdx
1 vncx
1 yxncx
cxeyx 1
0)1( y
102 ce
1221 ncnc
121 xexy
60.11034183 211.1)1.1( 11.1 ey
60.24280551 212.1)2.1( 12.1 ey
)1.0(y
22 yxdx
dy 10 y
22' yxydx
dy 00 x 1.0,10 hy
1100' 22 y
222'' '
000 yyxy
42
Example 3. Using Taylor method, compute and
correct to four decimal places given
and
Solution
etc
By Taylor’s series, we have
8222 ''
00
'
0
'''
0 yyyy
2826224 ''
0
'
0
''
0
'
0
'''
00
''
0
'
0
''
0
'
0
4
0 yyyyyyyyyyy
...28!4
1.08
!3
1.02
2
1.01
1
1.011.0
432
y
70.00011666330.001333330.010.11 1.11144999
11145.1
)2.0(y )4.0(y
xydx
dy 1 00 y
xyy 21' 121 00
'
0 yxy
'2'' xyyy 01002''
0 y
''''2''' xyyyy 4022'''
0 y
'''''324 xyyy 04
0 y
45 '''42 xyyy ,325
0 y
...!4!3!2
4
0
4'''
0
3''
0
2'
001 yh
yh
yh
hyyy
...33!5
2.00
!4
2.04
!3
2.00
!2
2.012.002.0
5432
y
000085333.000533333.02.0
43
Now again starting with , we have
,
.
Example 4. Using Taylor series method, find at
and given (correct to five decimal places)
Solution
Here
194752003.0
2.0x
2.0x 2.0,194752003.00 hy
9220992.0194752003.02.02121 00
'
0 yxy
758343806.0194752063.09220992.02.022 0
'
00
''
0 yyxy
9220992.027583436806.02.0222 ''
00
'
0
'''
0 yxyy
38505933.3
90408585.5758343686.0()38505933.3)(2.0(24
0 y
...90408585.5
!4
2.038505933.3
!3
2.0
)758343686.0(2
2.0)9220992.0)(2.0(194752003.04.0
43
2
y
359883723.0
y 2.0,1.0 xx
4.0x 10,2 yyxdx
dy
,...2.0,1.0,1.0,1,0 2100 xxhyx
yxy 2' 1'
0 y
'2'' yxy 1''
0 y
''2''' yy 112'''
0 y
44
= 1+ (-0.1) + 0.005 + 0.0001666 + (-0.0000416) +…
= 0.905125
Now again using and , we have
etc.
= 0.821235167
Similarly, and
Taylor series method for simultaneous first order differential equations
The equation of the type
and
with initial conditions
'''4 yy 14
0 y
45 yy etcy ,10 5
0
...1!5
1.01
!4
1.01
!3
1.01
2
1.011.01
2
1.0)1)(1.0(1)1.0(
5432
2
y
1.01 x 905125.01 y
895125.0090512501.01
2
1
'
1 yxy
095125.1)895125.0(2.02 '
11
''
1 yxy
904875.0090512522 ''
1
'''
1 yy
,904875.0'''
1
4
1 yy
095125.12
1.0895125.01.0905125.02.0
2
y
...904875.0
!4
1.0904875.0
!3
1.043
7492.0)3.0( y 6897.0)4.0( y
zyxfdx
dy,, zyxg
dx
dz,,
0000 ,)( zxzyxy
45
can be solved by Taylor series method as given below:
Example 5. Solve with by taking
, to get and .
Solution
and
and and
Using Taylor series, for and , we have
…… (1)
and
Substituting these in (1) and (2), we get
xydx
dzxz
dx
dy , 1)0(,10 zy
1.0h )1.0(y )1.0(z
xzy ' yxz '
00 x 10 y 1,0 00 zx 1.0h
?)1.0(1 yy ?1.01 zz
xzy ' '1'' yz
1''' zy etcyz '''''
''''' zy
1y 1z
...!3!2
1.0 0
3''
0
2'
001 yh
yh
hyyyy
)2(........!3!2
)1.0( '''
0
3''
0
2'
001 zh
zh
hyyzz
10 y 10 z
100
1
0 xzy 11000
'
0 yxz
011'
0
''
0 zy 2111 '
0
''
0 yz
2''
0
'''
0 zy 0''
0
'''
0 yz
2'''
0
4
0 yz
46
= 1+0.1+0.000333+…
= 1.1003
=1+ 0.1 + 0.01+0.0000083+… = 1.1100
Example 6. Find given
and
Solution
Here and
etc
...024
0001.02
6
001.00
2
01.01.011 y
...224
0001.00
6
001.02
2
01.01.011 z
2.0,2.0,1.0 zyy
2, yxdx
dzzx
dx
dy 10,20 zy
2,0 00 yx 10 z
zxy '2' yxz
'1'' zy '21'' yyz
''''' zy ''2'2'''2
yyyz
etczy '''4 ''2'''6'''2'''2'''44 yyyyyyyyyyz
10 00
''
0 zxyy 420 2'
0 z
341'
0 y 31221''
0 z
3''
0 y 10322122'''
0 z
104
0 y 303223164
0 z
305
0 y 745
0 z
...30
120
1.010
24
1.03
6
1.03
2
1.0)1)(1.0(2)1.0(
5432
y
47
= 2.084544167
= 2.084544167+0.267132966-0.016226621-0.0003105449027
- (0.000003091501458) +…
=0.152817221
Taylor series method for second order differential equation
Any differential equation of the second order or higher can be solved by reducing it to
a lower order differential equation. A second order differential equation can be reduced to a first
order differential equation by transforming and then the given equation can be solved so
easily.
Suppose …… (1)
is the given differential equation together with the given initial
conditions
and where , are known values at .
Setting , we get and (1) becomes
with initial condition and =
Using Taylor series method, we get
where …… (*)
...74
120
1.030
24
1.010
6
1.03
2
1.0)4)(1.0(1)1.0(
5432
z
...
24
1.074196035.0
6
1.0863269416.1
2
1.0)245324384.4)(1.0(5867855.0)2.0(
432
y
zy
dx
dyyxf
dx
yd,,
2
2
',,'' yyxfy
00)( yxy 00)( yxy 0y 0y 0x
py py
),,(' pyxfp
00)( yxy 00)( pxp 0y
...!32
'''
0
3''
0
2'
001 yh
ph
hppp hxxxxpp 0111 &
48
becomes
Since and taking the derivative of again and again with respect to x, we get
, etc.
Hence can be solved using (**) and (*), so that we can get
& .
Once knowing and we can get
Again using
we get some value for and using
we get still some value for .
Example 7. Evaluate the values of and given
by using Taylor series method.
Solution
First put and hence the equation reduces to
Using the initial condition
Now can be solved given that
Here,
...!32
'''
0
3''
0
2'
001 yh
yh
hyyy
...(**)...!3!2
0
3
0
2
001
ph
ph
hpyy
),,(' pyxfp p
pp ,
,...,, '''
0
''
0
'
0 ppp
1y 1p
1y 1p 11
'''
1
''
1
'
1 ,,...,, yxatppp
...!3!2
'''
1
3''
1
2'
112 ph
ph
hppp2p
...!3!2
'''
1
3''
1
2'
112 yh
yh
hyyy2y
)1.0(y )2.0(y
00',10,0'''' 22 yyyyxy
zy 0' 22 yxzz
22' yxzz
0,10 '
00 yzy
22' yxzz 0,00 00 xandzz
1......2
''
0
2'
001 zh
hzzz
49
and
Substituting, these values in (1), we get
= -0.0997
By Taylor series for , we get
= 0.995
Similarly,
22' yxzz zy '
'2'22'' yyxzzzz ,....''''',''' zyzy
22'''2''''2'2''' yyyzxzzxzzzzz
12
0
2
00
'
0 yzxz
0''
0 z 2'''
0 z
...26
001.00
2
01.011.001 z
1y
...2
)1.0( ''
0
2'
001 yh
hyyyy
...6
1.0
2
1.0))(1.0(1 ''
0
3
'
0
2
0 zzz
...)0(6
001.0)1(
2
01.001.01
...2
''
1
2'
1122 yh
hyyxyy
2......6
001.0
2
01.01.0995.0 ''
1
'
11 zzz
9890.0995.00997.01.0222
1
2
11
'
1 yzxz
1687.0''
1 z
50
Using (2),
= 0.9801
Example 8. Solve and calculate .
Solution
Here and
Differentiating with respect to x,
Here
Exercise
Using Taylor series method, find the values required in each problem.
1. Find given
....1687.06
001.09890.0
2
01.00997.01.0995.02 y
00''' ygivenxyyy )1.0(y
0,1,0 '
000 yyx ''' xyyy
'''2''''''' xyyxyyyy 1'
000
''
0 yxyy
'''''3'''''''24 xyyxyyyy 02 ''
00
'
0
'''
0 yxyy
45 '''4 xyyy 34
0 y
546 5 xyyy 05
0 y ,...156
0 y
...!32
'''
0
3''
0
2'
00 yx
yx
xyyxy
...3!4
01!2
0142
xx
...4882
1642
xxx
00501252.1....
48
1.0
8
1.0
2
1.01
642
)1.0(y 1)0( , yyxdx
dy
51
2. Find given
3. Obtain and given taking .
4. Find given .
5. Find given
and
6. Evaluate given
given at
7. Solve for x and y given at .
8. Find at given
9. Express as a power series given,
Picard’s Method of Successive Approximations
AIM: To solve subject to
Now
Integrating,
Setting , we have
)1.0(y 1)0( ,12 yyxy
)2.4(y )4.4(y 4)4( ,1
2
y
yxdx
dy2.0h
)3.0(),2.0(),1.0( yyy 1)0( ,23
ye
xyxy
x
)2.0(),1.0(),2.0(),1.0( zzyy
2, yxdx
dzzx
dx
dy 1)0(,2)0( zy
)2.0(),2.0(),1.0(),1.0( yxyx
txdt
dyty
dt
dx ,1 1,0 yx 0t
txdt
dytyx
dt
x 2 , 1,0 yx 1t
y 1.3 x,2.1 ,1.1 xx 1)1( ,1)1( ,32 yyxyyy
y 1)0( ),)(1.0( 22 yyxy
yxfdx
dy, 00)( yxy
dxyxfdyyxfdx
dy,,
cdxyxfyx
,
0xx
cdxyxfyx
0
,0
dxyxfyyx
x0
,0
52
this type of equation is called an integral equation.
As the integration is not possible as it is, we will solve it numerically by successive
approximation. Now substitute the initial values of namely in the integrand in
place of and then integrate it to get an approximate value of .
i.e.
After getting the first approximation for , use this value of in place of in
and then again integrate to get the second approximation of namely .
Thus
Repeating this procedure again and again, we eventually get
This equation is called Picard’s iteration formula. This formula gives the general iterative
formula for y.
The sequence , ,…, should converge to ; otherwise the process is not valid.
The condition for the convergence of the sequence is both and are continuous.
i.e. in a region containing the point where , are
constants.
Example 1. Solve , , by Picard’s method up to the
third approximation. Hence find the value of .
dxyxfyyx
x0
,0
y0y ),( yxf
y y
dxyxfyyx
x0
00
1 ,
)1(y y )1(y y
),( yxf y )2(y
dxyxfyyx
x0
1
0
2 ,
dxyxfyyx
x
nn
0
1
0 ,
)1(y )2(y )(ny )(xy
),( yxfy
f
21, ky
fdnakyxf
),( 00 yx
1k 2k
2xyy 1)0( y
)2.0(),1.0( yy
53
Solution
Hence, x0 = 0, y0 = 1 and
Using y(1)
again in (1), we get
Using again this result, we get
Now put
2' xyy
....0
2
0 dxxyyyx
x
2),( xyyxf
1.....10
2 dxxyyx
3
1113
0
21 xxdxxy
x
dxxyyx
0
212 1
dxxx
xx
0
23
311
12321
432 xxxx
dxxyyx
0
223 1
dxxxxx
xx
0
2432
123211
6012621
5432 xxxxx
1.0x
60
1.0
12
1.0
6
1.0
2
1.01.011.0
5432
y
54
=
Note- In getting the value we could have started with = 0.1 and
= 1.104824833 to get a closer value of
i.e.
Example 2. Solve given . Obtain the values of
and using Picard’s method and check your
104824833.1
60
2.0
12
2.0
6
2.0
2
2.02.012.0
5432
y
1.218528
)2.0(y 0x
0y )2.0(y
dxxyyx
1.0
2
0104824833.1
x
xxyy
1.0
3
0
1
3104824833.1
3
1.0)1.0(104824833.1
3104824833.1104824833.1
33
x
x
3104824833.199467574.0
3xx
dxxyyx
1.0
212 104824833.1
dxxx
xx
0
23
3104824833.199467574.0104824833.1
22 1.0104824833.1)1.0(99467574.0104824833.1 xx
3344 1.03
11.0
12
1 xx
2184066.12.02 y
,yxdx
dy 1)0( y
)1.0(y )2.0(y
55
result with the exact solution.
Solution
Here and
=
Now put
0 ,),( 0 xyxyxf 10 y
dxyxfyyx
x0
,0
dxyxfx
0
,1
dxyxfyx
0
1 ,1
dxxx
0
1,1
2
1112
0
xxdxx
x
dxxx
xyx
0
22
211
61
32 x
xx
dxx
xxxyx
0
323
611
2431
432 xx
xx
...243
143
2 xx
xxxy
1.0x
...24
1.0
3
1.01.01.011.0
432
y
56
=
To find we use = and =
=
=
...24
0001.0
3
001.001.01.01
1.1103374
)2.0(y 0x 0.10y 1.1103374
x
dxyxfyy1.0
0 ,
x
dxyx1.0
1103374.1
x
dxxy1.0
1 1103374.11103374.1
))1.0(()1.0(1103374.11103374.1 22 xx
21103374.198930366.0 xx
x
dxxxxy1.0
22 1103374.198930366.01103374.1
3322 1.03
1))1.0((1103374.2)1.0(98930366.01103374.1 xxx
32
3
11103374.298930366.0989970326.0 xxx
x
dxxxxy1.0
323
3
11103374.298930366.1989970326.01103374.1
))1.0((98930366.1)1.0(989970326.01103374.1 22 xx
4433 1.012
11.01103374.2 xx
0015.012
1)007.0(1103374.2
)03.0(98930366.1)1.0(989970326.01103374.12.03
y
41.28391090
57
Solving analytically, we get
= and
=
Example 3. Solve by Picard’s method
Solution
Here
=
Now
Example 4. Solve , by Picard’s method
Solution
By Picard’s method
yxdx
dy 12 xey x
)1.0(y 61.11034183
)2.0(y 61.24280551
10,22 yyxdx
dy
1,0 00 yx
dxyxfyyx
x0
,0 dxyx
x
0
221
3
1113
0
21 xxdxxy
x
dxx
xxyx
0
23
22
311
5743
2
15
2
63
1
6
1
3
21 xxx
xxx
...63
1
15
2
6
1
3
21 75432 xxxxxx
0)0(, yeyy x
x
xdxyxfyy
0
),(0 dxye
xx
0
100
1 x
xx edxey
58
Exercise
1. Using Picard’s iterative formula,
a) Solve , given .
b) Obtain given and
c) Solve given
2. Find the values of for and given and which passes
through .
3. Given and , find
xdxeeyx
xx 02 1
12
2
0
3 x
edxxey xx
x
dxx
eeyx
xx
0
24 1
2
xx
6
3
dxxx
eyx
x
0
35
6
1242
42
xx
ex
1242
42
xx
exy x
12 yxdx
dy0)0( y
)1.0(yxy
xyy
1)0( y
xyy 21 0)0( y
y 1.0,0 xx 5.0x 22 yxy
)1,0(
2
2
1 y
xy
0)0( y )5.0(),25.0( yy
59
Euler’s Method
In solving a first order differential equation by numerical methods, we have two types of
solutions:
i) Values of at specified values of
ii) A series solution of in terms of , which will yield the value of at a particular
value of by direct substitution in the series solution.
Taylor and Picard’s method studied so far belong to the first category in finding the numerical
solution of differential equations; the methods due to Euler, Runge-kutta, Adam-Bash Forth and
Milne come under the second category.
The methods of second category are called step-by-step methods because the values of are
calculated by short steps ahead of equal interval of the independent variable .
Euler’s method
Suppose we want to solve with initial conditions
Let us take the points
i.e. ,
Let the actual solution of the differential equation be denoted by the graph lies on
the curve. We require the value of the curve at .
The equation of the tangent line at to the curve is
y x
y x y
x
y
h x
yxfdx
dy, 00)( yxy
hxxxxxx ii 1210 where,...,,
ihxxi 0 ,...2,1,0i
),( iiii yxPP
yixx
),( 00 yx
0
P0
P1
P2
Q2
(x1,y1) Q1
M0 M1 M2
y
x0 x1 x2 x
60
=
In the interval , the curve is approximated by the tangent.
The value of on the curve is approximately equal to the value of on the tangent
at corresponding to .
i.e.
Again, we approximate the curve by the line through and whose slope is f , we
get
Thus,
This formula is called Euler’s algorithm.
In other words,
In this method, the actual curve is approximated by a sequence of short straight lines.
As the intervals increase the straight line deviates much from the actual curve. Hence the
accuracy cannot be obtained as the number of intervals increase.
Referring to the above graph
error at
= it is of order h2.
0
'
,0 00xxyyy yx
000 , xxyxf
0000 , xxyxfyy
),( 10 xx
y y
),( 00 yx ixx
010001 , xxyxfyy
'
001 hyyy
),( 11 yx ),( 11 yx
hyxfyy 1112 ,
'
11 hyy
,...2,1,0,,1 nyxhfyy nnnn
),()()( yxhfxyhxy
11PQ 1xx
11
2
11
2
01 ,''2
,''!2
yxyh
yxyxx
61
Example 1. Given and , determine the values of at
and by Euler method.
Solution
and ;
Here
We have to find . Take
By Euler algorithm,
Let us compare the results
Example 2. Using Euler’s method, solve numerically the equation,
for and .
Check your answer with the exact solution
yy 1)0( y y
02.0,01.0 xx 04.0x
yy 1)0( y yyxf ),(
04.0,03.0,02.0,01.0,1,0 432100 xxxxyx
4321 ,,, yyyy 01.0h
nnnnnn yxhfyhyyy ,'
1
99.001.01101.01, 0001 yxhfyy
))(01.0(99.0 1
'
112 yhyyy
9801.099.001.099.0
9801.001.09801.0, 2223 yxhfyy 0.9606
9703.001.09703.0, 3334 yxhfyy 0.9606
x 0.04 0.03 0.02 0.01 0
y 0.9606 0.9703 0.9801 0.9900 1
xey 0.9608 0.9704 0.9802 0.9900 1
1)0(, yyxy 2.0,0 xx 0.1x
62
Solution
Here ,
= , = , = , = , =
By Euler algorithm,
=
=
=
=
=
Exact solution is
Euler
Exact
As you can see from the table, the values of deviate from the exact values as increases. To
avoid this discrepancy we need to improve Euler’s method.
Improved Euler Method
Let the tangent at to the curve be P0A. In the interval by the previous
Euler’s method, we approximate the curve by the tangent P0A.
2.0h 1,0 ,),( 00 yxyxyxf
1x 2.02x 4.0
3x 6.04x 8.0
5x 0.1
0000001 , yxhyyxhfyy
1.2 1)0.2(01
)2.12.0)(2.0(2.1)( 1112 yxhyy
48.1
)48.14.0)(2.0(48.12223 yxhyy
856.1
)856.16.0)(2.0(856.1)( 3334 yxhyy
3472.2
)3472.28.0(2.03472.25 y 94664.2
12 xey x
x 1.0 0.8 0.6 0.4 0.2 0
y 2.94664 2.3472 1.856 1.48 1.2 1
y 3.4366 2.6511 2.0442 1.5836 1.2428 1
y x
),( 00 yx ),( 10 xx
63
Let Q1C be the line at Q1 whose slope is
Now take the average of the slopes at P0 and Q1
i.e.
Draw a line P0D through P0 with this as the slope.
That is,
and this line intersects at
In general,
This is what is known as improved Euler’s method.
11
1
1000
1
1 where, QMyyxhfyy
1
11, yxf
1
1100 ,,2
1yxfyxf
),( 00 yx
0
1
11000 ,,2
1xxyxfyxfyy
1xx
1
110001 ,,2
1yxfyxfhyy
0001000 ,,,2
1yxhfyxfyxfhy
nnnnnnnn yxhfyhxfyxfhyy ,,,2
11
(x0,y0)
P1
P0
x0
M0
Q1
x1
M1
(x1y )
A C
B
D
x
y
64
Notice that the difference between Euler’s method and the improved Euler’s method
is that in the improved one we take the average of the slopes at and instead of
the slope at in the former method.
Modified Euler’s Method
In the improved Euler method, we arranged the slopes, whereas in modified Euler method, we
will average the points.
Let P0 be the point on the solution curve
Let P0A be the tangent at to the curve. Now let this tangent meet the ordinate at
and coordinate of . Calculate the slope at N1. i.e.
.
Let this line meet at .
This is taken as the approximate value of at
),( 00 yx 1
11, yx
),( 00 yx
),( 00 yx
),( 00 yx
102
1Nathxx y 0001 ,
2
1yxhfyN
0000 ,
2
1,
2
1yxhfyhxf
1xx 1
111 , yxk
1
1y y1xx
00000
1
1 ,2
1,
2
1yxhfyhxfhyy
P0 (x0,y0)
k1(x1,y1)
M0
N1
B
A
y
0 x0 x0+ h x1=x0+h
x
65
In general,
or
This is called modified Euler’s formula.
Note: The Euler predictor is and the corrector is in the
improved Euler method.
When you read some literature there is some confusion among the authors. Some take the
improved Euler method as the modified Euler method and the modified Euler method is not
mentioned at all.
Example 3 Solve numerically for by improved
Euler’s method.
Solution
and h = 0.2.
By improved Euler method,
nnnnnn yxhfyhxfhyy ,
2
1,
2
11
yxhfyhxfhxyhxy ,
2
1,
2
1
'
1 nnn hyyy '
1
'
12
nnnn yyh
yy
00,' yeyy x4.0 ,2.0 xx
4.0,2.0,0,000,' 2100 xxyxyeyy x
00010001 ,,,2
1yxhfyxfyxfhyy
hxxxeeyhyeyy
000
00012
2.00
2.0102.00101.0 e
24214.02.0 1 yy
66
Here
Example 4. Compute at by Modified Euler’s method
, .
Solution
Take
By modified Euler method,
=
1111111
2 ,,,2
1yxhfyhxfyxfhyy
46354.124214.0, 2.0
1111 eeyyxf
x
53485.046354.12.024214.0, 111 yxhfy
02667.253485.0,, 4.0
1111 eyxhfyhxf
02667.246354.11.024214.02 y
59116.0)4.0( 2 yy
y 25.0x
xyy 2 1)0( y
1 ,0 ;2),( 00 yxxyyxf
25.0 ,25.0 1 xh
nnnnnn yxhfy
hxfhyy ,
2
1,
21
000001 ,
2
1,
2yxhfy
hxfhyy
0)1,0(),( 00 fyxf
)1,125.0()25.0(11 fy
)1)(125.0)(2)(25.0(1
67
Using
The error is only .
Example 5. Solve the equation given using modified
and Euler’s method and tabulate the solutions at ,
and . Compare these results with the exact solutions.
Solution
Here
i) Modified Euler method
0625.1)25.0( y
xdxy
dyxyy 22'
1)0( y
1)0( cy
2xey
064494459.125.02
25.0 ey
58910.00199944
,1 ydx
dy 0)0( y
2.0,1.0 xx
3.0x
1.0,3.0,2.0,1.0,0,0 32100 hxxxyx
yy 1
11,,1, 000 yyxfyyxf
000001 ,
2
1,
2yxfhy
hxfhyy
05.0,2
1and05.01.0
2
1
2
10000 yxhfyhx
095.005..011.005.0,05.01.001 fy
68
Now again = 1- 0.95 = 0.905
=
=
=
ii) Improved Euler method
=
111 1),( yyxf
111112 ,
2
1,
2
1yxfhyhxfhyy
14025.015.01.0095.0 f
14025.011.0095.0
0.180975
222223 ,
2
1,
2
1yxfhyhxfhyy
180975.0,2.0
2
1.0180975.0,
2
1.02.01.0180975.0 ff
)0.22192625 f(0.25, (0.1) 0.180975
50.25878237
nnnnnnnn yxfhyhxfyxfhyy ,,,2
11
00000001 ,,,2
1yxfhyhxfyxfhyy
11, 00 yyxf
1.0,1.011.00,1.0,, 0000 ffyxfhyhxf
0.9 0.1-1
095.09.012
1.001.01 yy
11111112 ,,,2
1yxfhyhxfyxfhyy
69
= = 0.8145
Exact solution
At
Modified improved exact
905.0095.011, 111 yyxf
095.011.0905,0,2.0,, 1111 fyxhfyhxf
)1855.0,2.0(f
180975.0905.08145.02
1.0095.02 y
22222223 ,,,2
1yxfhyhxfyxfhyy
180975.011.0180975.0,3.0180975.012
1.0180975.0 f
)2628775.01819025.0(05.0180975.0
258782375.0
dxy
dyy
dx
dy
11
xceycxyn 11 1
101 ,0 0 ccex
xey 1
095162581.01.0 y
181269246.0)2.0( y
259181779.0)3.0( y
x
10.09516258 0.095 0.095 0.1
60.18126924 0.180975 0.180975 0.2
70
Example 6. Find, correct to four decimal places, the value of
Given , by using improved Euler method.
Solution
Here, , , and
By the improved Euler method,
Example 7. Using improved Euler method find y at and at
given
Solution
By improved Euler theorem,
Here
90.25918177 50.25878237 50.25878237 0.3
)1.0(y
yxy 21)0( y
1.0h 00 x 1.01 x 10 y
nnnnnnnn yxhfyhxfyxfhyy ,,,2
11
00000001 ,,,2
1yxhfyhxfyxfhyy
0
2
000
2
0 1.0,1.02
1.01 yxyfyx
11.011.0105.012
0.9055
1.0x
2.0x 10,2
yy
xy
dx
dy
nnnnnnnn yxfhyhxfyxfhyy ,,,2
11
1.0,1,0 00 hyx
00000001 ,,,2
1yxhfyhxfyxfhyy
0
1.1
211.01
1
021
2
1.01
71
=
Thus
Example 8. Using modified Euler method, find
given
Solution
Here, ,
)1.1,1.0(11.01,1.0,, 00 ffyxhfyhxf nn
...918181818.0
1.1
1.021.1
0.918182
095909091.1918182.012
1.011 y
11111112 ,,,2
yxhfyhxhfyxfh
yy
913412201.0
09590909.1
1.02095909091.1, 11
yxf
913412201.01.0095909091.1,2.0,, 1111 fyxhfyhxf
187250311.1,2.0f
850337365.0
187250311.1
2.02187250311.1
850337365.0913412201.02
1.0095909091.12 y
184096569.1
)2.0(),1.0( yy
10,22 yyxdx
dy
1.0,1.0,1,0 100 xhyx 22),( yxyxf
72
By modified Euler method,
=
Exercise
1. Use Euler’s method to find
a) given
b) taking given
2. Compute taking given using improved Euler
method.
3. Find and given taking by improved
Euler method.
4. Use improved Euler method to find given .
5. Using improved Euler method find given .
6. Use improved Euler method to calculate ,taking and
000001 ,
2,
2
1yxf
hyhxhfyy
22 105.0105.01.01
)1.0(y 1.1105
111112 ,
2,
2
1yxf
hyhxhfyy
22 172660513.115.01.01105.1
250263268.1
)4.0(y 1)0(, yxyy
)5.1(y 5.0h 1.1)0(,1 yyy
)3.0(y 1.0h 1)0(,2
yy
xy
dx
dy
)8.0(),6.0( yy )1(y 0)0(, yyxy 2.0h
)1.0(y 1)0(,
y
xy
xyy
)4.0(),2.0( yy 1)0(, yyxdx
dy
)5.0(y 1.0h
2)0(,sin yxyy
73
7. Use modified Euler method and obtain given .
8. Use improved and modified Euler method, to get if , if .
9. Solve given if to obtain .
10. Given find if .
11. Find by improved Euler method, given
if .
Runge-Kutta Method
i) Second order Runge-kutta method
Suppose
By Taylor series, we have
Differentiating (1) with respect to x,
Using the values of and derived from (1) and (3), in (2) we get
Let
)2.0(y 2.0,1)0(),log( hyyxy
)6.1(yx
yy
dx
dy 2 1)1( y
yxy 23 4)0( y 25.0h )5.0(),25.0( yy
2
1)1(,
2
5 32 yyxx
yy )2(y 125.0h
)2.0(y 2)0(,2 yxyy
1.0h
1..........given, 00 yxyyxfdx
dy
)2(..........0''!2
' 32
hxyh
xhyxyhxy
3........''' yxyx ffffyfdx
dy
y
f
x
fy
y y
32
2
1hOfffhhfxyhxy yx
4........2
1 32 hOfffhhfy yx
11 ,, kxyxfyxhfy
212 , kmkymhxhfy
74
and let where and are constants to be determined to get the better
accuracy of
Now expand and in powers of h, by Taylor series for two variables, we have
= higher powers of .
Substituting , in we get
Equating (4) and (5), we have
21 bkaky ba, m
.y
2k y
12 , mkymhxfhk
...
!2,
2
1
1
fy
mkx
mh
fy
mkx
mhyxfh
...!2
2
1 fy
mkx
mh
mhffmhffh yx
...22 yx ffmhfhmfh h
1k 2k ,y
32 hOfffmhhfbahfy yx
5.....)( 21
32 bkakhOfffbmhhfba yx
2
1,1 mbba
bmandba
2
11
211 kbkby
75
Where
Now
i.e.
From this general second order Runge-kutta formula, setting , , , we get the
second order Runge-kutta algorithm.
and where
ii) Third order Runge-kutta Method
For n = 3, a similar derivation to the one as the second-order method can be performed. Since
the derivation is tedious we state simply the formula.
, where
iii) Fourth Order Runge-kutta method
),(1 yxhfk
b
fhy
b
hxfhk
2,
22
xyhxyy
b
hfy
b
hxbhfhfbxyhxy
2,
21
3
1 ,2
,2
,1 hOyxfb
hy
b
hxfhbyxfhbyy nnnnnnnn
0a 1b2
1m
12
1
2
1,
2
1
),(
kyhxhfk
yxhfk
2ky xh
3211 46
1kkkyy nn
),(1 yxfhk
hkyhxhfk 12
2
1,
2
1
)2,( 23 kkyhxhfk
76
The most popular and commonly used form is the classical fourth-order Runge-kutta method.
, where
Note 1. The second order Runge-kutta method,
this is exactly the Modified Euler method.
Thus, the second order Runge-kutta method is simply the modified Euler method.
2. If , a function of alone, then the fourth order Runge-kutta method
reduces to
43211 226
1kkkkyy nn
yxhfk ,1
hkyhxhfk 12
2
1,
2
1
hky
hxhfk 23
2
1,
2
hkyhxhfk 34 ,
10020
2
1,
2ky
hxhfky
0000 ,
2
1,
2yxhfy
hxhf
000001 ,
2
1,
2yxhfy
hxhfyy
)(),( xfyxf x
01 xhfk
hxf
hxfxfhy 000
24
6
1
77
= the area of between and with two equal intervals of
length by Simpson’s one-third rule. i.e. reduces to the area by Simpson’s one-third rule.
Example 1. Apply the fourth order Runge-kutta method to find
given that
Solution
Since h is not mentioned, we can take
By fourth-order Runge-kutta method, for the first interval
=
hxf
hxfxf
h
0002
43
2
)(xfy 0xx hxx 0
2
hy
)2.0(y 1)0(, yyxy
1.0h
2.0,1.0,1,0,, 2100 xxyxyxyxf
1.0101.0, 001 yxhfk
11.005.0105.01.02
1,
2
11002
kyhxhfk
1105.0055.0105.01.02
1,
2
12003
hkyhxhfk
12105.01105.011.01.0, 3004 kyhxhfk
432101 226
11.0 kkkkyyy
12105.01105.0211.021.06
11
110341667.1
78
Now starting from we get . Again apply Runge-kutta algorithm replacing
by .
=
Remember that the exact solution is
The difference between the exact solution and the fourth order Runge-kutta method is
.
As compared to other methods, this method is the best one.
Example 2. Find the values of at using Runge-kutta
method of i) second order ii) third order and
iii) fourth order for the given that and .
Solution
i) Second order
),( 11 yx ),( 22 yx
),( 00 yx ),( 11 yx
121034166.0110341667.11.01.0, 111 yxhfk
176384605.115.01.02
,2
2112
ky
hxhfk
13263846.0
144298012.013263846.0132085875.02121034166.06
1110341667.12.0 y
21.24280514
12 xey x
61.24280551 )2.0( y
4320.00000037
y 2.0,1.0x
yy 1)0( y
1.0,2.0,1.0,1,0,),( 2100 hxxyxyyxf
1.011.0, 001 yxhfk
79
Again let
=
ii) Third order
=
Now take and repeat the process
095.005.011.02
1,
2
11002
kyhxhfk
201 kyy
905.0095.01
),( 11 yx 0.905) (0.1,
1k 0.0905- (-0.905) (0.1)
085975.00905.02
1905.01.02
k
)2.0(y 0.819025 )(-0.0859750.905
1.0),( 001 yxhfk
2k 0.095-
)2,( 12003 kkyhxhfk
0.091- (-0.1)) 2(-0.095)(-(1 (0.1)
321 46
111.0 kkky
091.0()095.0(41.06
11
30.90483333
),( 11 yx 833333)(0.1,0.904
0904833333.0904833333.0,1.0, 111 yxhfk
80
iii) Fourth order
Again taking and repeating the process, we get
1112
2
1,
2)1.0( ky
hxk
0904833333.0
2
1904833333.01.0
60.08595916-
0904833333.0085959166.02904833333.01.03 k
30.08233983-
32112 46
1kkkyy
082339833.0082339833.040904833333.06
1904833333.0
90.82113624
1.01 k
095.02 k
09525.00475.011.02
1,
2
12003
kyhxhfk
090475.009525.011.0, 3004 kyhxhfk
43211 226
11 kkkky
9048375 0.
),( 11 yx 8375)(0.1,0.904
1k 0.09048375- 5)(-0.904837 (0.1)
81
2nd
order 3rd
order 4th
order exact
As we can see from the table fourth order values are closer to the exact values.
Example 3. Compute given by taking
using Runge-kutta method of fourth order.
Solution
1002
2
1,
2
11.0 kyhxfk
20.08595956- 5))0.04524187-75(-(0.90483 (0.1)
2003
2
1,
2
1)1.0( kyhxfk
086185771.0085959562.02
19048375.01.0
3114 , kyhxhfk
720.08186551- 10.08618577 048375(0.1)(-0.9
432112 226
1kkkkyy
818730902.0081865172.0
086185771.0085959562.0209048375.06
19048375.0
x xey
80.90483741 0.9048375 30.90483333 0.905 0.1
30.81873075 20.81873090 90.82113624 0.819025 0.2
)3.0(y 10,2 yxyydx
dy1.0h
)(),( 22 xyyyxfxyyy
82
Now = (0.1) (-1) = -0.1
Now again taking in place of , we get
3.0,2.0,1.0,1.0,1,0 32100 xxxhyx
1k
1
2
22
1.0
2
1.01
2
)1.0(11.0k
295.005.095.01.0
0.0995125 -
2003
2
1,
2ky
hxhfk
95024375.0,05.01.0 f
295024375.005.095024375.01.0
0.09953919-
3004 , kyhxhfk
900460809.0,1.01.0 f
098154377.0
098154377.009953919.020995125.021.06
111 y
900623707.01.0 y
),( 11 yx ),( 00 yx
900623707.0,1.01.0, 111 fyxhfk
098173601.0
83
Again, using = we find
=
=
2,
2
1112
ky
hxhfk
851536906.0,15.01.0 f
096030417.0
2,
2
2113
ky
hxhfk
2852608498.015.0852608498.01.0
096164968.0
3114 , kyhxhfk
804458738.0,2.01.0 f
10.09338895-
093388951.0096164968.0
2096030417.02098173601.06
1900623707.02
y
60.80463148
),( 22 yx 6)0.80463148 (0.2, )3.0(3 yy
221 , yxhfk
2)804631486.0)(2.0(804631486.01.0
50.09341178-
757925593.0,25.0)1.0(2
,2
2222 f
ky
hxhfk
90.09015383-
84
=
Example 4. Using Runge-kutta method of fourth order, find
if
Solution
Here and
759554566.0,25.01.02
, 2223 f
kyhxhfk
35-0.0903785
714252951.0,25.01.0, 3224 fkyhxhfk
70.08417922-
084179227.0090378535.0
2090153839.02093411785.06
1804631486.03
y
714855526.0
)8.0(y
7379.1)6.0(,2 yxyy
8.0,7.0,1.0,7379.1,6.0 2100 xxhyx 2),( xyyxf
13779.06.07379.11.0, 2
001 yxhfk
2,
2
1002
ky
hxhfk
1384295.0806795.1,65.01.0
2,
2
2003
ky
hxhfk
38461475.065.080711475.11.0 2
3004 , kyhxhfk
85
=
Now using = we continue to find
Example 5. Using Runge-kutta method of fourth order, solve
given at
Solution
and
138636147.0138461475.021384295.0213779.06
17379.17.0 y
61.87626801
),( 11 yx 6)1.87626801 (0.7, )8.0(y
138626801.07.0876268016.11.0, 2
111 yxhfk
2,
2
1112
ky
hxhfk
2,
2
2113
ky
hxhfk
138292208.0945422087.1,75.01.0 f
3114 , kyhxhfk
137456022.0014560224.2,8.01.0 f
13746022.0138292208.0
2138308141.02138626801.06
1876268016.18.0
y
62.01448193
22
22
xy
xy
dx
dy
1)0( y 4.0,2.0x
4.0,2.0,2.0,1,0 2100 xxhyx 22
22
,xy
xyyxf
86
y
=
To find we use =
=
2.001
012.0, 201
yxhfk
2,
2
1002
ky
hxhfk
196721311.01.01.1
1.01.12.0
22
22
2,
2
2003
ky
hxhfk
196711597.01.0098360656.1
1.0098360656.12.0
22
22
3004 , kyhxhfk
18913131.02.0196711598.1
2.0196711598.12.0
22
22
18913131.0196711597.02196721311.022.06
112.0
11.19599952
)4.0(y ),( 11 yx 999521)(0.2,1.195
22
22
1112.0195999521.1
2.0195999521.12.0,2.0 yxfk
70.18911871
29055888.1,3.02.02
,2
2.0 1112 f
ky
hxfk
179493515.0
87
=
Runge-kutta method for simultaneous first order differential equations
Aim- To solve numerically the simultaneous equations
given the initial conditions
Now starting from the increments in and respectively are given
by formulae
and
285746279.1,3.02.02
,2
2113 f
ky
hxhfk
179347655.0
375347176.1,4.02.0, 3114 fkyhxhfk
168804528.0
168804528.0179347655.0
2179493515.02189118717.06
1195999521.14.0
y
91.37526711
zyxfdx
dzandzyxf
dx
dy,,,, 21
0000 )(,)( zxzyxy
),,( 000 zyx zandy y z
00011 ,, zyxhfk 00021 ,, zyxhfm
2,
2,
2
10
10012
mz
ky
hxhfk
2,
2,
2
10
10022
mz
ky
hxhfm
2,
2,
2
20
20013
mz
ky
hxhfk
2,
2,
2
20
20023
mz
ky
hxhfm
3030014 ,, mzkyhxhfk 3030024 ,, mzkyhxhfm
4321 226
1kkkky 4321 22
6
1mmmmz
88
By repeating this algorithm once again we can find starting from .
Example 6. Find , from the system of equations,
given using Runge-kutta
method of fourth order.
Solution
Here
So
zzzandyyy 0101
),,( 222 zyx ),,( 111 zyx
)1.0(y )1.0(z
2, yxdx
dzzx
dx
dy 1)0(,2)0( zy
2
21000 ),,(,),,(,1.0,1,2,0 yxzyxfzxzyxfhzyx
00011 ,, zyxhfk 00021 ,, zyxhfm
0.4- )2-(0 (0.1) 0.1 1)(0 (0.1) 2
2,
2,
2
10
10012
mz
ky
hxhfk
2,
2,
2
10
10022
mz
ky
hxhfm
4.2025)-(0.05 (0.1) (-0.2))1(0.1)(0.05
0.41525- 0.085
2,
2,
2
20
20013
mz
ky
hxhfk
2,
2,
2
20
20023
mz
ky
hxhfm
)(2.0425)-(0.05 (0.1) 0.792375)(0.05 (0.1) 2
50.41218062- 0.0842375
3030014 ,, mzkyhxhfk 3030024 ,, mzkyhxhfm
))(2.0842375 -(0.1 (0.1) 5)0.58781937 (0.1 (0.1) 2
50.42440459- 70.06878193
)068781937.0)0842375.0(2085.0(21.06
121 y
89
=
=
Runge-kutta method for second order differential equation
Aim – To solve given
Now, let and so
and are the two simultaneous equations with
and itself.
Once again by applying Runge-kutta method for solving simultaneous first order differential
equations we can get the solution of the given problem.
Example 7. Given find by using
Runge-kutta method of fourth order.
Solution
Let
given
32.08454282
424404595.0412180625.0241525.024.06
111 z
50.58678902
',, yyxfy 0000 )(,)( yxyyxy
zy ' zy
',,' yyxfzy zy zzyxf ),,(1
),,(),,(2 zyxfzyxf
,0)0(,1)0(,0 yyyyxy )1.0(y
'''0''' xyyyyxyy
xzyzzy ''
zyxfzdx
dy,,1
zyxfxzydx
dz,,2
0,1.0,,0,1 0
'
000 xhyzy
001.0,, 00011 zyxhfk 1.0,, 0001 zyxhfm
90
=
Exercise
In the exercise below, unless specified use fourth order Runge-Kutta method.
1. Find given taking .
2. Obtain the value of at if satisfies
taking .
3. Solve for , taking .
2,
2,
2
10
10012
mz
ky
hxhfk
2,
2,
2
10
10022
mz
ky
hxhfm
0.05) (0.05,1,- (0.1) = 0.005- = (0.05) (0.1) =
049875.0,
2
005.01,05.013 hfk 05)](0.05)(-0.-[-1 (0.1)
0.09975- )(-0.049875 (0.1)
0.0049875 - 049875.0,9975.0,05.023 hfm
3030114 ,, mzkyxhfk 975)049875)0.9((0.05)(0. (0.1)
50.09950062- 625)(-0.099500 (0.1)
250.00995006- 303024 ,, mzkyxhfm i
)099500625.0,09950125.0,1.0)(1.0(
)9950125.0)09950125.0)(1.0)((1.0(
30.09850624-
)0099500625.0()0049875.0(2)005.0(206
111 y
90.99501248
)2.0(y 2)0(, yxyy 1.0h
y 2.0x y 1)0(,2 yxyxdx
dy
1.0h
xydx
dy 4.1x 2.0,2)1( hy
91
4. Solve given , to obtain .
5. Solve the initial-value problem with on the interval .
6. Evaluate given .
7. Solve for taking .
8. Find given taking by
i) second order ii) third order
iii) fourth order Runge-Kutta methods.
9. Solve as given taking .
10. Solve the system: for taking
.
11. Solve , given for to taking
12. Evaluate given .
13. Using Runge-Kutta method determine
given .
14. Solve using Runge-Kutta method for given
taking .
15. Find given by Runge-Kutta method.
16. Find given .
xy
xy
dx
dy
1)0( y )2.0(y
1)0(,2 2 utudt
du2.0h )6.0,0(
)3.0(),2.0(),1.0( yyy 1)0(,)1(2
1 2 yyxy
1)1(,1
2 y
xx
y
dx
dy)1.1(y 05.0h
)2.0(),1.0( yy 1)0(,2 yyxy 1.0h
1 xyy 6.0,4.0,2.0 xxx 2)0( y 2.0h
xydx
dzxz
dx
dy ,1 9.0,6.0,3.0 xxx
1,0,0 zyx
yxdx
dzzx
dx
dy , 1)0(,0)0( zy 0.0x 2.0 .1.0h
)1.1(),1.1( zy 1)1(,5.0)1(,, zyz
xy
dx
dzxyz
dx
dy
)1.0(),1.0( yx
1)0(,,1)0(, yxtydt
dyxtxy
dt
dx
0)( 22 yyxy 2.0x
0)0(,1)0( yy 2.0h
)1.0(y 5)0(,10)0(,2 yyyy
)2.0(,1.0( yy 0)0(,1)0(,122 yyxyyxy
92
17. Obtain the value of given .
18. Compute the value of given .
Predictor-Corrector Method
The methods so far discussed are called single step methods because they use only the
information from the last step computed. But now we try to discuss multi-step methods.
Consider
We have used Euler’s formula to solve differential equation of the form
, and we have improved the Euler method by
In (2), to get the value of we require on the right hand side.
To overcome this difficulty, we calculate using Euler’s formula (1) and then we use it on the
right hand side of (2), to get the left hand side of (2). This can be used further to get refined
on the left hand side. Here, we predict the value of from the rough formula (1) and use
in (2) to correct the value. Every time, we improve using (2). Hence equation (1) Euler’s
formula is a predictor and (2) is a corrector.
A predictor formula is used to predict the value of at and a corrector formula is used
to correct the error and to improve that value of .
Milne’s Predictor- Corrector Formulae
Suppose our aim is to solve numerically.
)1.0(x 0)0(,3)0(,42
2
xxxdt
tdx
dt
xd
)2.0(y 0)0(,1)0(, yyyy
00,, yxyyxfdx
dy
1... ,2,1,0,,'1 iyxhfyy iiii
2... ,,2
1111 iiiiii yxfyxfhyy
1iy 1iy
1iy
1iy
1iy 1iy
y1iy
1iy
00,, yxyyxfdx
dy
93
, where is a suitable
accepted spacing, which is very small.
By Newton’s forward interpolation formula, we have
, where
Changing to
Integrating both sides from to ,
Since
Now
nhxyyhxyyxyxyhxyy n 00211001 ...,,2, h
...
!3
21
!2
10
3
0
2
00
yuuu
yuu
yuyy
uhxxh
xxu
0
0
y y
)2(...
!3
21
!2
1' '
0
3'
0
2'
0
'
0
yuuu
yuu
yuyy
0x4x
hx
x
hx
x
x
xydxy
uuyuydxy
44
'
0
2'
0
'
00
0
0
0
4
0
|...!2
1'
duy
uuyuyh
4
0
'
0
2'
0
'
0 ...2
1
hdudxuhxx 0
...
46
1
232
1
2
234
'
0
323
'
0
22
'
0
'
004 uuu
yuu
yu
yuyhyy
...166464
6
18
3
64
2
184 '
0
3'
0
2'
0
'
0 yyyyh
...
3
8
3
2084 '
0
3'
0
2'
0
'
0 yyyyh
...
45
141
3
81
3
20)1(84 '
0
4'
0
3'
0
2'
0
'
0 yyEyEyEyh
...
45
1433
3
82
3
20)(84 '
0
4'
0
'
1
'
2
'
3
'
0
'
1
'
0
'
0
'
1
'
0 yyyyyyyyyyyh
94
Taking into account only up to the third order equation, (3) gives
…………… (4)
The error happened in (4) is and this can be proved to be , where
.
Since for small values of .
The error is and hence (3) becomes
……..(5)
In general,
, where ……(6)
This equation is called Milne’s predictor formula.
To get Milne’s corrector formula, integrate equation (2) between the limits to .
..45
14
3
88
3
208
3
408
3
8
3
2084 '
0
4'
32
'
1
'
0
yhyyyyh
...45
14
3
8
3
4
3
8 '
0
4'
3
'
2
'
1
yhyyyh
3.......45
1422
3
4 '
0
4'
3
'
2
'
1 yh
yyyh
3210
'
3
'
2
'
104
223
4
223
4
fffh
y
yyyh
yy
...45
14 '
0
4 yh vy
h
45
14
40 , xx
hDeE hD 11 h
vy
h
45
14 5
vyh
yyyh
yy45
1422
3
4 '
3
'
2
'
104
1
5''
1
'
23145
1422
3
4v
nnnnn yh
yyyh
yy 131 , nn xx
0x hx 20
dxy
uuyuydxy
hx
x
hx
x
2 2'
0
2'
0
'
0
0
0
0
0
...2
1'
95
…….…(7)
Again here taking into account only up to third order, we get
Here the error is and this can be proved to be
where and thus (7) becomes
In general,
where .
This equation is called Milne’s corrector formula.
Example 1. Find if is the solution of given
and
duy
uuyuyhyy
2
0
'
0
2'
0
'
002 ...2
1
2
0
2
0
'
0
223
'
0
2'
0 ...232
1
2
y
uuy
uuyh
...
24
1
15
412
3
8
2
1122 '
0
4'
0
2'
0
'
0 yyEyEyh
...
90
12
3
122 '
0
4'
0
'
1
'
2
'
0
'
1
'
0 yyyyyyyh
...90
43
'
0
4'
2
'
1
'
0 yh
yyyh
)8...(43
'
2
'
1
'
002 yyyh
yy
...90
'
0
4 yh vy
h
90
5
;20 xx
vyh
yyyh
yy90
43
5'
2
'
1
'
002
2
5
1
''
11190
'43
v
nnnnn yh
yyyh
yy 121 nn xx
)2(y )(xy yxdx
dy
2
1
595.3)1(,636.2)5.0(,0)0( yyy 968.4)5.1( y
96
Solution
Here
By Milne’s predictor formula,
Now,
=
Using Milne’s corrector formula, we get,
=
5.0,2,5.1,1,5.0,0 43210 hxxxxx
968.4,595.3,636.2,2 3210 yyyy
'2
1, yyxyxf
''
1
'
231 223
4nnnnn yyy
hyy
3
'
2
'
104 223
4yyy
hyy
568.1636.25.02
1
2
111
'
1 yxy
2975.2595.312
1
2
122
'
2 yxy
234.3968.45.12
1
2
133
'
3 yxy
234.322975.2568.12
3
5.0424 y
6.871
'
4
'
3
'
224 43
yyyh
yy
4355.4871.622
1
2
144
'
4 yxy
4355.4234.342975.23
5.0595.34 y
76.87316666
97
Example 2. Using Milne’s method find given
given and .
Solution
By Milne’s predictor formula,
=
Using the corrector formula, we get
)4.4(y 025 2 yyx
0097.1)2.4(,0049.1)1.4(,1)4( yyy 0143.1)3.4( y
1.0,4.4,3.4,2.4,1.4,4,5
2' 43210
2
hxxxxxx
yy
0143.1,0097.1,0049.1,1 3210 yyyy
048301267.0
1.45
0049.12
5
2 2
1
2
1'
1
x
yy
046690757.0
2.45
0097.12
5
2 2
2
2
1'
2
x
yy
045171884.0
3.45
0143.12
5
2 2
2
2
3'
3
x
yy
'
3
'
2
'
104 223
4yyy
hyy
00451718842046690757.0048301267.02
3
1.041
91.01870073
043738581.0
4.45
018700739.12
5
2 2
4
2
4'
4
x
yy
'
4
'
3
'
224 43
yyyh
yy
98
=
Example 3. Determine the value of using Milne’s method given
use Taylor series to get the values of and
Solution
Here
Now
=
Once again
043738581.0045171884.04046690757.03
1.00097.1
91.01873722
)4.0(y ;10,' 2 yyxyy
),2.0(),1.0( yy )3.0(y
5.0,1,4.0,3.0,2.0,1.0,0 043210 hyxxxxx
1' 2
000
'
0
2 yyxyyxyy
32'2''' '
000
'
00
''
0 yyyyxyyyyxyy
10)'(2''2'2''''' '''
0
2 yyyyyxyy
...!32
'''
0
3''
0
2'
001 yh
yh
hyyy
116666667.1...106
001.03
2
01.011.01
358611111.12
111
'
1 yyxy
28675926.42 1
111
'
11
''
1 yyyyxy
4113088.162222'
1
'1
111
''
11
'''
1 yyyyyxy
...4113088.166
001.028675926.4
2
01.0358611111.11.011666667.12 y
31.27669679
885294059.12
222
'
2 yyxy
467653362.62 '
222
'
22
''
2 yyyyxy
99
=
By Milne’s predictor formula
Since
=
Now using Milne’s corrector formula,
Example 4. Given and ,
evaluate by Milne’s predictor-corrector method.
68725078.282222'
2
''
22
'
2
''
22
'''
2 yyyyyxy
68725078.286
001.0467653362.6
2
01.0885294059.11.0276696793.13 y
41.50234567
'
3
'
2
'
104 223
4yyy
hyy
358611111.1'
1 y
885294059.1'
2 y
707746226.22
333
'
3 yyxy
70774626.22885294059.1358611111.12
3
1.0414 y
41.83298942
093046.42
444
1
4 yyxy
'
4
'
3
'
224 43
yyyh
yy
093046.470774626.24885294059.13
1.0276696793.1
83700763.1
2212
1yx
dx
dy 21.1)3.0(,12.1)2.0(,06.11.0,10 yyyy
)4.0(y
100
Solution
By Milne’s predictor formula
=
By Milne’s corrector formula
=
Once again if we use this value of , we get
4.0,3.0,2.0,1.0,0 43210 xxxxx
1.0,21.1,12.1,06.1,1 3210 hyyyy
2212
1' yxy
2
1101
2
11
2
1 2
0
2
0
'
0 yxy
567418.006.11.012
11
2
1 222
1
22
1
'
1 yxy
7979345.0)21.1)(3.01(2
1)1(
2
1 222
3
2
33
yxy
'
3
'
2
'
104 223
4yyy
hyy
7979345.02652288.0367418.02
3
1.041
71.27712226
'
4
'
3
'
224 43
yyyh
yy 946003944.0'
4 y
946003944.0)7979345.0(4652288.03
1.012.1
51.27966766
4y
949778612.0279667665.14.012
1 22'
4 y
101
So
=
If we repeat this procedure once again using this ,
By Milne’s corrector formula
=
Continuing this process again and again, after some steps we get
Example 5. Given , and , and
i) by Euler method; using that value obtained
ii) by Modified Euler method
iii) Obtain by Improved Euler method and find
iv) by Milne’s method.
Solution
By Euler method
By Modified Euler
949778612.07979345.04652288.03
1.012.14 y
71.27979348
4y
949965394.0)279793487.1(4.012
1 22'
4 y
949965394.07979345.04652288.03
1.012.14 y
31.27979971
71.279800034 y
yy 1 0)0( y
)1.0(y
)2.0(y
)3.0(y
)4.0(y
1.0011.00, 0001 yxhfyy
111112 ,
2
1,
2yxhfy
hxhfyy
1855.01.011.02
11.011.01.0
102
By improved Euler method
=
Now using Milne’s predictor formula
=
By Milne’s corrector formula
=
Exercise
1. Using Milne’s method, find given
22232223 ,,,2
1yxhfyxfyxfhyy
1855.011.01855.011855.012
1.01855.0
0.2628775
'
3
'
2
'
104 223
4yyy
hyy
3210 121123
4yyy
hy
2628775.01(21855.011.012
3
1.040
0.327966
672034.0327966.011 4
'
4 yy
'
4
'
3
'
224 43
yyyh
yy
'
4322 1413
yyyh
y
672034.0)2628775.0(4)1855.01((3
1.01855.0
30.33333413
)2.0(y
0323.2)15.0(,0211.2)1.0(,0103.2)05.0(,2)0(,1.02.0 yyyyyxdx
dy
103
2. Find given .
3. Using Runge-Kutta method of fourth order, find at given
. Continue your work to get by Milne’s method.
4. Solve by Taylor series method at and
hence find and by Milne’s method.
5. If , find and
by Milne’s method.
6. Estimate and using Milne’s method correct to three decimal places, given
.
7. Solve to obtain by Milne’s method. Obtain the data you
require by any method you like.
8. Using both predictor-corrector methods, estimate if satisfies , and
.
9. Given , find
by Milne’s predictor-corrector method.
10. Compute by Milne’s method given with . Obtain the
required data by Taylor series method.
11. Given , find by Euler method; by Taylor series
method; by Runge-Kutta method and by Milne’s method.
)8.0(y 7379.1)6.0(,46820.1)4.0(,12186.1)2.0(,1)0(,2 yyyyxyy
y 3.0,2.0,1.0 xxx
1)0(,2 yyxyy )4.0(y
1)0(,)1(2
1 2 yyxy 6.0,4.0,2.0 xxx
)8.0(y )1(y
090.2)3.0(,040.2)2.0(,010.2)1.0(,2)0(,2 yyyyyedx
dy x )4.0(y
)5.0(y
)8.0(y )1(y
6841.0)6.0(,4228.0)4.0(,2027.0)2.0(,0)0(,1 2 yyyyyy
1)0(,2 yyxy )4.0(y
)4.1(y y2
1
xx
y
dx
dy
972.0)3.1(,986.0)2.1(,996.0)1.1(,1)1( yyyy
2493.2)6.0(,1755.2)4.0(,0933.2)2.0(,2)0(,1
yyyyyx
y
)8.0(y
)6.0(y 1)0(, yyxy 2.0h
0)0(,23 yyey x)1.0(y )2.0(y
)3.0(y )4.0(y
104
12. Find by Taylor series method; by modified Euler method; by
Runge-Kutta method and by Milne’s method, given .
13. Given obtain by Picard’s method; by modified Euler
method; by Runge-Kutta method; by Milne’s predictor-corrector method.
14. Given obtain power series by Picard’s method; using Milne’s
method, estimate and show that .
Estimate given using .
Boundary-Value Problem (BVP)
BVPs can be solved numerically by using either the Shooting Method or the Finite-
Difference Method (FDM). Here we consider the numerical solution of PVP using FDM. The
former method is left for the students as a reading assignment.
Some simple examples of two-point linear BVPs are:
'' 'y x f x y x g x y x r x (1)
with boundary conditions
0y x a and ny x b (2)
'' 21 1
2
2i i ii
y y yy O h
h
ivy x p x y x q x
(3)
with boundary conditions
'
0 0y x y x A and '
n ny x y x B . (4)
Problems of the type in Eq. (3) and Eq. (4), which involve the fourth-order differential
equation, are much involved and will not be discussed here. There exist many methods for
)2.0(y )4.0(y )6.0(y
)8.0(y 0)0(,1 2 yyy
1)0(,1 yxyy )1.0(y )2.0(y
)3.0(y )4.0(y
,10)0(,2 2 yxyy
2.0,6505.1)1( hy
)4.0(),5.0( yy 1)0(,2 yyxy 1.0h
105
solving second-order BVPs of the type in Eq. (1) and Eq. (2). Of these, the Finite-Difference
Method is popular one and will be discussed.
Finite-Difference Method (FDM)
The FDM for the solution for a two-point BVP consists in replacing the derivatives occurring
in the differential equation (and the boundary conditions as well) by means of their finite-
difference approximations and then solving the resulting linear system of equations by a
standard procedure.
To obtain the approximate finite-difference approximations to the derivatives, we proceed as
follows:
Expanding y x h in Taylor’s series, we have
2 3' '' '''
2 6...h hy x h y x hy x y x y x (5)
from which we obtain
' '' ...2
y x h y x hy x y x
h
Thus we have
'y x h y x
y x O hh
(6)
Which is forward difference approximation for 'y x .
Similarly, expansion of y x h in Taylor’s series gives
2 3' '' '''
2 6...h hy x h y x hy x y x y x (7)
from which we obtain
'y x y x h
y x O hh
(8)
106
Which is the backward difference approximation for 'y x .
A central difference approximation for 'y x can be obtained by subtracting Eq. (7) from
Eq. (5). We thus have
' 2
2
y x h y x hy x O h
h
(9)
It is clear that Eq. (9) is better approximation to 'y x than either Eq. (6) or Eq. (8). Again,
adding Eq. (5) and Eq. (7) we get an approximation for ''y x as
'' 2
2
2y x h y x y x hy x O h
h
(10)
In a similar manner, it is possible to derive finite-difference approximations to higher
derivatives.
To solve the BVP defined by Eq. (1) and Eq. (2), we divide the range 0 , nx x into n equal
subintervals of width h so that
0 , 0,1,2,..., .ix x ih i n
The corresponding values of y at these points are denoted by
0 , 0,1,2,..., .i iy x y y x ih i n
From Eq. (9) and Eq. (10), values of 'y x & ''y x at the point ix x can now be written as
' 21
2
i ii
y yy O h
h
and '' 21 1
2
2i i ii
y y yy O h
h
107
In many applied problems; however, derivative boundary conditions may be prescribed, and
this requires a modification of the procedures described above. The following examples
illustrate the application of the FDM.
Example 1: A boundary-value problem is defined by
1 0, 0 1iiy y x ,
where
0 1 0, 0.5y y and h
Use the FDM to determine the value of y(0.5). Its exact solution is given by
1 cos1cos sin 1
sin1y x x x
,
from which, we obtain
0.5 0.139493927y .
Here 1nh . The difference equation is approximated as
1 1
2
21 0i i i
i
y y yy
h
, and this gives after simplification
2 2
1 12 , 1,2,..., 1i i iy h y y h i n ,
which together with the boundary conditions 0 0ny y , comprises a system of 1n
equations for 1n unknowns 0 1, ,...,
ny y y .
Choosing 1
2h (i.e. 2n ), the above system becomes
108
0 1 2
1 12
4 4y y y
.
With 0 2 0y y , this gives
1 0.5 0.142857142y y .
Comparing with exact solution given above shows that the error in the computed solution is
0.00336 .
On the other hand, if we choose 1
4h (i.e. n=4), we obtain the three equations:
0 1 2
1 2 3
2 3 4
31 1
16 16
31 1
16 16
31 1
16 16
y y y
y y y
y y y
Where 0 4 0y y . Solving the system we obtain 2 0.5 0.140311804y y , the error in
which 0.00082 . Since the ration of the error is about 4 , it follows that the order of
convergence is 2h .
These results show that the accuracy obtained by the finite difference method depends upon the
width of the subinterval chosen and also on the order of approximations. As h is reduced, the
accuracy increases but the number equations to be solved also increases.
Example 2: Solve the boundary-value problem
2
20, 0 2
d yy x
dx ,
with 0 0 & 2 3,62686y y .
109
The exact solution of this problem is sinhy x . The finite-difference approximation is given by
1 12
12i i i iy y y y i
h
We subdivide the subinterval 0,2 the four equal parts so that 0.5h . Let the values of y at the
five points be 0 1 2 3 4, , , ,&y y y y y . We are given that 0 0y , and 4 3.62686y .
Writing the difference equations at the three interval points (which are the unknowns), we obtain
0 1 2 1
1 2 3 2
2 3 3 3
4 2
4 2
4 2
y y y y
y y y y ii
y y y y
, respectively. Substituting for 0 4&y y and rearranging, we get the system
1 2
1 2 3
2 3
9 4 0
4 9 4 0
4 9 14.50744
y y
y y y iii
y y
The solution of (iii) is given in the table below.
x Computed solution of
y
Exact value
sinhy x
Error
0.5 0.52635 0.52110 0.00525
1.0 1.18428 1.17520 0.00908
1.5 2.13829 2.12928 0.00901
Exercise
1. Solve the boundary value defined by
'' 0, 0 0, 1 1y y y y ,
110
by finite-difference method. Compare the computed solution at 0.5y with the exact
value. Take 0.25h .
2. Project work. Shooting Method. This is a popular method for the solution of two-point
boundary-value problems. If the problem is defined by
''
0 1, 0y x f x y x and y x A ,
Then it is first transformed into the initial value problem
' ',y x z z x f x with 0 0y x and 0 0z x m , where 0m is a guess
for the value '
0y x . Let the solution corresponding to 1x x be 0Y . If 1Y is the value
obtained by another guess 1m for '
0y x , then 0 1Y and Y and are related linearly. Thus,
linear interpolation can be carried out between the values 0 0,x y and 1 1,m y .
Obviously, the process can be repeated till we obtain the value for 1y x is close to A .
Apply the Shooting method to solve the BVP
'' , 0 0, 1 1y x y x y y
