Vel Tech Dr.RR & Dr.SR Technical University
Avadi
Department of Mathematics
NUMERICAL METHODS
(Common to EEE, Civil, Aero and CSE)
1
UNIT - I
SOLUTION OF EQUATION AND EIGEN VALUE PROBLEMS
Properties of equation:
(i)
If f(a) and f(b) have opposite signs then one root of f(x) = 0 lies between a and b.
(ii)
Every equation of an odd degree has at least one real root whose sign is opposite to
that of its last term.
(iii)
Every equation of an even degree with last term negative has at least a pair of real
roots one positive and other negative.
Formula for false position (or) Regula falsi method:
x
a f(b)  bf(a)
f(b)  f(a)
Order of convergence:
Let x0, x1… Xn…. Be a sequence which converge to a number, and set en =  - xn. If
there exists a number p and a constant C  0 such that lim
n 
en1
 C , then P is called the
en p
order of convergence and C is known as the asymptotic error constant of the sequence.
Linear algebraic equations:
A system of m linear equation in n unknowns
x1, x2, x3 , …..xn is a set of equations of the form.
a11 x1 + a12 x2 +……….. + a1n xn = b1
a21 x1 + a22 x2 + ………. + a2n xn = b2
.
.
am1 x1 +am2 x2 +..……….+ amn xn = bm
Where the co efficient of x1 x2 ... xn and b1 b2 …. bm are constants. Then the above system can
be written as Ax = B
Where
 a11

A   a21
a
 m1
a12
a13
a22
a23
a m2
am 3
... a1n 
 x1 
 b1 

 
 
... a2 n  , X   x2  , B   b2 
x 
b 
... amn 
 n
 n
Example: 01:
Consider the following linear system 2x1 + 3x2 – x3 = 5; - 2x2 – x3 = -7; 5x3 = -15
find x1,x2 ,x3 ?
2
solution:
From the last equation x 3 

b3
a33
15
3
5
from the sec ond equation
x2 
b2  a23 x 3 7  3

2
a22
2
Hence the first equation
x1 
b1  a12 x 2 5  3.2  3

1
a11
2
Procedure to find Eigen value:
Let A be any square matrix of order n. Then for any Scalar  we can form the matrix
A - I. where I is the nth order unit matrix. The determinant of the matrix equal to zero is
called characteristic equation. This polynomial of degree n in x having n root for say 1 2
…. n These values are called eigen values of the given matrix A.
Latent or eigen vector:
First find out eigen values. For these eigen values, the system of equation (A - I) X = 0 has a
 x1 
 
x
non – trivial solution for the vector X   2  is called eigen vector of the corresponding
. 
 
 xn 
eigen values.
Gauss – seidal method is better than Gauss Jacobi method why?
In Gauss seidal method the latest values of unknowns at each stage of iteration are
used in proceeding to the next stage of iteration. Hence the convergence in Gauss – seidal
method is more rapid than Gauss – Jacobi method.
Regula falsi method:
Let f(x) = 0 be the given equation.
(i)
Find two number a and b such that f(a) and f(b) are of different signs. Then the root
lies between a and b.
(ii)
The first approximation to the root is given by
3
x1 
(iii)
af(b)  bf(a)
ba
If f(x1) and f(a) are of opposite signs, then the actual root lies between x 1 and a. Now
replacing b by x1 and keeping a as it is we get the next closer approximation x2 to the
actual root.
(iv)
This procedure is repeated till the root is found to be desired degree of accuracy.
Gauss Seidal method:
Let the rearranged form of a given set of equation be
x
1
 d1  b1y  c1z 
a1
y
1
 d2  a2 x  c 2 z 
b2
z
1
 d3  a3 x  b3 y 
c3
we start with the initial values y(0), z(0) for y and z get x(1) from x(1) 
Condition for Gauss – Jacobi method of converges:
Let the given equation be
a1x  b1y  c1z  d1
a2 x  b2 y  c 2 z  d2
a3 x  b3 y  c 3 z  d3
a1  b1  c1
The sufficient condition is
b2  a2  c 2
c 3  a3  b3
4
1
 d1  b1y(0)  c1z(0) 
a1
Newton’s algorithm for finding the Pth root of a Number N:
Pth
The
root
of
the
a
positive
number
N
is
the
root
of
the
equation.
xp  N  0
f(x)  x p  N
f 1(x)  pxp 1
By Newton' salgorithm
f(x)
xk  1 
f '(x)
 xP  N 
xk   k p1   1
 Pxk 
Pxk p  xPk  N

Pxkp 1

(p  1)xPk  N
Pxpk 1
Newton Raphson formula for cube root of a positive number k.
x 3k
f(x)  x 3  k  0
f 1(x)  3x 2
xn1  xn 

xn3  k
3xn2
1
k 
2xn  2 
3
xn 
Gauss – elimination method to solve Ax = B:
In this method the given system is transformed into an equivalent system with upper –
triangular co efficient matrix i.e. a matrix in which all elements below the diagonal elements
are zero which can be solved by back substitution.
Newtons – Raphson – formula
xn+1 =
a or
N Or
1
a
 xn   n = 0, 1, 2, …..
2
xn 
5
Let
x 
a,
x
2
a  0
f (x)  x
2
a
1
f (x)  2x
f (x n )
1
f (x n )
2
x a
 xn  n
2x n
2

x a
a  1
 n
  xn 
 2
2x n
xn 

x n1  x n 
Example:02
Evaluate 12 applying Newton formula.
Solution:
Let x = 12
x2= 12
x2 – 12 = 0
f(x) = x2 – 12
f(3) = -ve,
f(4) = +ve
take x0 = 3
 x1  x 0 
x 2  x1 
f(x 0 )
f(3)
3 1
 3.5
1
f (x 0 )
f (3)
f(x1 )
(3.5)2  12

3.5

 3.464
f 1(x1 )
2(3.5)
The root is 3.464
Iteration formula to find the reciprocal of a positive number N by Newton Raphson
method:
1
N
1
1
N   N  0
x
x
1
Let  N  0
x
1
f(x)   N
x
1
f 1(x) 
x2
x
6
by Newton' s formula
f(x )
x n1  x n  1 n
f (x n )
 1

N

x

x n1  x n   n
 1 
 x2 
 n 
 1

 xn    N  x 2
 xn

 x n  x n  x 2nN
 xn  2  N xn 
The order of converges in Newton Raphsons method:
Newton – Raphson process has a second – order convergence.
Comparison of Gauss elimination and Gauss seidel methods:
(i)
Gauss elimination method has only a finite number of computation, since it is a
direct method.
(ii)
Gauss seidel iteration method convergers only when the system is diagonally
dominant.
(iii)
Iteration method is a self-correcting method. Errors made at any step in the
computation are corrected in the subsequent iteration.
When should we not use Newton – Raphson method:
If x1 is the exact root and x0 is its approximate value of the equation f(x) =0. We know
that x1 = x0 -
f ( x0 )
f (x )
.If f1(x0) is small, the error 1 0 will be large and the computation of
1
f ( x0 )
f ( x0 )
the root by this method will be a slow process.
Hence the method should not be used in
case where the graph of the functions when it crosses the x axis is nearly horizontal.
Merits of Newton’s method of iteration:
Newton’s method is successfully used to improve the result obtained by other
methods. It is applicable to the solution of equations involving algebraical functions as well
as transcendental functions.
7
Compression of the Gaussian elimination method and Gauss-Jordan method:
Gauss elimination
1.
2.
Gauss Jordan
Coefficient matrix is transformed into
Coefficient matrix is transformed
3.
into diagonal matrix
upper triangular matrix.
4.
Direct method
Direct method
5.
We obtain the solutions by bank
No need of back substitution
7.
method.
6.
substitution method.
Comparession of Gauss Jacobi and Gauss seidel methods:
Gauss Jacobi
1.
Gauss seidel
Convergence rate is slow
The rate of convergence of Gauss.
seidel method roughly twice that of
Gauss-Jacobi
2
Indirect method
Indirect method
3.
Condition for con vergence is the
Condition for convergence is the
coefficient matrix is diagonally
coefficient
dominant.
dominant.
matrix
is
diagonally
Distinguish between Direct and iterative methods of solving simultaneous equations.
Direct methods involve a certain amount of fixed computation and they are exact
solutions. Iterative or indirect methods are those in which the solution is got by successive
approximations. But the method of iteration is not applicable to all systems of equations.
8
Numerical Examples:
01. Find the square root of 8. (by Newton – Raphson).
Solution:
Given N = 8 Clearly 2  8  3 taking x0  2.5 we get
x1 
1
N  1
8 
 2.85
 x 0    2.5 
2
x0  2 
2.5 
1
N 1
8 
 2.8285
 x1    2.85 
2
x1  2 
2.85 
1
N  1
8 
x 3   x 2    2.828 
 2.8284
2
x2  2 
2.828 
x2 
x4 
1
N  1
8 
 2.8284
 x 3    2.8284 
2
x3  2 
2.8284 
 8  2.8284
02. By applying Newton’s method twice, find the real root near 2 of the equation
x4 – 12x + 7 = 0
Solution:
Let f(x) = x4 – 12x + 7
f(x) = 4x3 – 12
Put x0 = 2, f(x0) = -1
f(x0) = 20
x1  x 0 
f(x 0 )
( 1) 41
 2

 2.05
1
f (x 0 )
20 20
f(x1 )
(2.05)4  12(2.05)  1
x 2  x1  1
 2.05 
f (x1 )
4(2.05)3  12
 2.6706
The root of the equation is 2.6706.
9
03. Find the approximately value of the root of equation x3 + x - 1 = 0 near x = 1, by the
method of false using the formula twice.
Solution:
f(x)  x 3  x  1
f(0.5)  0.675, f(1)  1
Hence theroot liesbetween 0.5 & 1
a  0.5, b  1
a f(b)  b f(a)
 0.64
f(b)  f(a)
f(0.64)  0.097a  0
Theroot lies between0.64 & 1
x0 
x1 
(0.64)  (1)( 0.0979)
 0.672
1  ( 0.0979)
04.Use the iteration method to find a root of the equation x = ½ + sin x?
Solution:
Let f(x) = sin x – x + ½
f(1) = sin 1 – 1 + ½ = 0.84 – 0.5 = +ve
f(2) = sin 2 – 2 + ½ = 0.9.9 – 1.5 = -ve.
A root lies between 1 and 2. The given equation can be written as
x = sin x + ½ =
(x)
1(x)  cos x  1 in (1,2) .
Hence the iteration method can be applied. Le the approximation be x0 = 1.
The successive approximation are as follows:
x1=
(x0) = sin 1 + ½ = 0.8414 + 0.5 = 1.3414
x2=
(x1) = sin (1.3414) + ½ = 0.9738 + 0.5 = 1.4738
x3=
(x2) = sin (1.4738) + ½ = 0.9952 + 0.5 = 1.4952
x4=
(x3) = sin (1.4952) + ½ = 0.9971 + 0.5 = 1.4971
x5=
(x4) = sin (1.4971) + ½ = 0.9972 + 0.5 = 1.4972
Since x4 and x5 are almost equal the required root is 1.497.
05. If an approximate root of the equation x ( l- logx) = 0.5 lies between 0.1 and 0.2 find
the value of the root correct to three decimal places.
Solution:
Given f(x) = x( 1- log x) -0.5
10
f1(x) = (1- logx) + x
 1
 
 x
= - log x
f(0.1) = 0.1 [1-log (0.1)] – 0.5 = 0.1697 (-ve)
f(0.2) = 0.2 [1-log (0.2)] – 0.5 = 0.02188 (+ve)
x0 = 0.9
x1  x 0 
f(x 0 )
f 1(x 0 )
 0.2 
0.2(1  log(0.2)  0.5
 log(0.2)
 0.2 
0.02188
 0.1864
1.6094
x 2  x1 
f(x1 )
f 1(x1 )
 0.1864 
0.1864(1  log(0.1864)  0.5
 log(0.1864)
 0.1864 
0.0004666
 0.1866
1.6799
x 3  x1 
f(x 2 )
f 1(x 2 )
 0.1866 
0.1866(1  log (0.1866)  0.5
 log(0.1866)
 0.1866
Hence the approximate root is 0.1866.
06. Find the root between (2, 3) of x3 – 2x – 5 = 0 by regula falsi method
Solution:
Given f(x) = x3 – 2x – 5
f(2) = 8 – 4 -5 = -1
f(3) = 27 – 6 -5 = 16
Let us take a = 2, b = 3
The first approximation to the root is x1 and is given by
x1 
=
af  b   bf  a 
f b   f  a 
2  16-3  -1
16   1

2f  3   3f  2 
f 3   f 2
 2.058
f  2.058    2.058   2  2.058   5
3
=-0.4
11
The root lies between 2.058 and 3
Taking a = 2.058 and b = 3 we have the second approximation to the root given by
x2 

a f(b)  b f(a)
f(b)  f(a)
 2.058   16  3(0.4)
16  ( 0.4)
 2.081
f (2.081)  (2.081)3  2  2.081  5
 0.15
The root lies between 2.081 and 3
Take a = 2.081, b = 3
The third approximation to the root is given by
a f(b)  b f(a)
f(b)  f(a)
2.081 16  3( 0.15)

16  ( 0.15)
 2.089
x3 
Now
f (2.089)  (2.089)3  2  2.089   5
 0.062
The root lies between 2.089 and 3
Take a = 2.089, b = 3
2.089  16  3( 0.062)
16  ( 0.062)
 2.093
x1 
The required root is 2.09.
07. Find the approximate root of xex = 3 by Newton’s Raphson method correct to three
decimal places.
Solution:
Given f(x) = xex – 3
f1(x) = xex + ex
f(1) = 1e-1 – 3 = 2.7182 – 3 = - 0.2817 (-ve)
12
f(1.5) = 1.5e1.5 – 3 = 3.7223 (+ve)
Here f(1) is –ve (Negative) and f(1.5) is +ve (positive). Therefore the root lies
between 1 ad 1.5. Since the magnitude of f(1) < f(1.5) we can take the initial approximate
x0 = 1. The first approximation is
x1  x0
 1
f(x 0 )
f 1(x 0 )
0.2817
 1.0518
5.4363
The second approximation
x 2  x1 
f(x1 )
f 1(x1 )
 1.0518 
0.0111
5.8739
 1.0499
The third approximation is
x3  x 2 
f(x 2 )
f 1(x 2 )
1.0499 e1.0499  3
 1.0499 
1.0499 e1.0499  e1.04999
 1.0499
Hence the root of xex is 1.0499
08. Solve the following system by Gaussian elimination method
x1 - x 2 + x 3 = 1
-3x1 + 2x 2 - 3x 3 = -6
2x1 - 5x 2 + 4x 3 = 5
Solution:
Write the given system as in the matrix form
 1 1 1 1 


2 3 6 
 3
 2 5 4 5 


From the first column with non – zero component select the component with the large
absolute value this component is called the pivot.
 1 1 1 1 


  3
2 3 6 
 2 5 4 5 


13
Rearrange the row to move the pivot to the top eg.
First column. Here we interchange the first and second row.
 3

 1
 2

2
1
5
 3  6

1 1
4 5 
Make the pivot as 1, by dividing the first row by the pivot.
 1

= 1
 2

 2 3 1 2 

1
1 1
5
4 5 
 1

 0
 0

2 3 1 2 
 R2  R2  R1
 1 3 0 1
R3  R3  2R1
 11 3 2 1 
Delete the first row and first column and perform turn the above procedure.
 1

 0
 0

 1

 0
 0

 1

 0
 0

2 3 1
2

1  New pivot
1 
1 3 0
 11 3 2
2 3 1
11 3
1 3
2 3
1
0
2

1  R 2  R3
1
2
0

3

 6 11  3 11  R 2  R 2 
11
 2 11  12 11
1
R3  R3  R 2
11
1
2
Delete first two row’s and first two columns
 1

 0
 0

2 3
1
0


 6 11  3 11 
 2 11  12 11
1
2
14
 1

 0
 0

2 3
1
2 

1
 6 11  3 11
0
1
6 
x 3  6 x 2  6 11 x 3  3 11
x2  3
x1  2 3 x 2  x 3  2  x1  2
09. Using the Gauss – Jordan method solve the following equation.
10x + y + z = 12
2x + 10y +z = 13
x + y + 5z = 7
Solution:
 10 1 1 12 


Step 1   2 10 1 13 
 1 1 5 7


1

 1 10

Step 2   2 10
1 1


1

12 10 
10

R
1
13  R1  1
10
5
7 


1

 1 10

49
Step 3   0

5

0 9

10

1

12 10 
10

4
106  R2  R2  2R1
5
10  R3  R3  R1

49
58 
10
10 
1

 1 10

Step 4   0 1


0 9

10

1

12 10 
10

4
53 
49
R2  R2 
49
49 
5

49
58 
10
10 
15
 1 0 0.0918 1.0918 


4
53  R3  R3  9 10R 2
Step 5   0 1

49
49  R1  R1  1 10R 2
 0 0 4.8265 4.8265 


 1 0 0.0918 1.0918 


4
53 

Step 6  0 1
R 2  R 3  4.8265

49
49 
0 0
1
1 

 1 0 0 1

 R  R2  0.0918R3
Step 7   0 1 0 1 2
 0 0 1 1 R2  R2  4 49R3


The matrix finally reduced to the form
 1 0 0  x   1

   
 0 1 0  y    1
 0 0 1  z   1

   
x  y  z 1
10. Solve the following equation using Jacobi iteration method:20x + y – 2z = 17
3x + 20y – z = - 18
2x – 3y + 20z = 25
Solution:
The above equation can be written as
16
1
17  y  2z 
20
1
y
 18  3x  z 
20
1
z
 25  2x  3y 
20
x
First approximation =
1
17  y 0  2z0 
20
1
y1 
 18  3x 0  z0 
20
1
z1 
 25  2x 0  3y 0 
20
x1 
Put x0 = y0 = z0 = 0
x1 = 0.85, y1 = - 0.9, z1 = 1.25
Second approximation
1
17  y1  2z1 
20
1
y2 
 18  3x1  z1 
20
1
z2 
 25  2x1  3y1 
20
x2 
x1 = 0.85, y1 = - 0.9, z1 = 1.25 we get
x2 = 1.02 y2 = - 0.965 z2 = 1.1515
Third approximation
1
17  y 2  2z2 
20
1
y3 
 18  3x 2  z2 
20
1
z3 
 25  2x 2  3y 2 
20
x 2  1.02, y 2  0.965, z 2  1.1515 we get
x3 
x 3  1.0134, y 3  0.9954, z3  1.0032
17
1
17  y 3  2z3 
20
1
y4 
 18  3x3  z3 
20
1
z4 
 25  2x 3  3y 3 
20
x 3  1.0134, y 3  0.9954, z 3  1.0032 we get
x4 
x 4  1.009, y 4  1.0018, z 4  0.9993
Fifth approximation
1
17  y 4  2z 4 
20
1
y5 
 18  3x 4  z 4 
20
1
z5 
 25  2x 4  3y 4 
20
x 4  1.009, y 4  1.0018, z 4  0.994 we get
x5 
x 5  1, y 5  1.0002, z 5  0.9996
 x  1,
y  1, z  1
11. Solve by Gauss – seidal method of iteration the equation.
10x1 + x2 + x3 = 12
2x1 + 10x2 + x3 = 13
2x1 +2x2 + 10x3 = 14
Solution:
From the above equation
1
12  x 2  x 3 
10
1
x2 
13  2x1  x 3 
10
1
x3 
14  2x1  2x 2 
10
Put x 2  x 3  0 we get x1  1.2, i.e x1(1)  1.2
x1 
1
10.6
13  2.4  0  
 1.06

10
10
 1.2 x (1)
2  1.06
Put x 2  
1
x1(1)
x (1)
3 
1
14  2.4  2.12  0.948
10
18
1
12  1.06  0.948   0.992
10
1

13  2  0.9992   0.948   1.00536
10
1

14  2  0.9992   2 1.00536    0.999098
10
x1(2) 
x(2)
2
x(2)
3
Thus the iteration process is continued.
i
x 1(i)
x (i)
2
x (i)
3
0
1.2000
0.000
0.000
1
1.2000
1.0600
0.9480
2
0.9992
1.0054
0.9991
3
0.9996
1.001
1.001
4
1.0000
1.0000
1.00
5
1.000
1.000
1.000
The exact values of the roots are
X1 = 1, x2 = 1, x3 = 1.
 2 1 1


12. Find the inverse of the matrix  3 2 3  using Gauss Jordan method.
 1 4 9


Solution:
 2 1 1


Let A   3 2 3 
 1 4 9


 x11 x12

X   x 21 222
x
 31 x32
x13 

x33 
x33 
Ax = I
19
2 1 1 1 0 0


Step 1   3 2 3 0 1 0 
 1 4 9 0 0 1



2 1 1

1 3
Step 2   0

2 2

7 17
0

2 2

0 0
3
 R2  R2  R1
3
2
1 0

2
1
R3  R3  R1

1
2
0 1
2

1
1
2 1 1

1 3 3
Step 3   0

2 2
2
 0 0 2 10

0

1 0  R3  R3  7R 2

7 1 
 2 0 0 6

1
Step 4   0
0 6

2
 0 0 2 10

5
17
4
7
1

3
1
R1  R1  R3
4
2

1
 2 0 0 6

1
Step 5   0
0 6

2
 0 0 2 10

5
17
4
7
1

3
R1  R1  2R 2
4
1 

 1 0 0 3

Step 6   0 1 0 12


 0 0 1 5


0
5
2
17
2
7
2
1 
2  R1  R 1

2
3 
R2  R2 x 2
2
 R  R3   Y2 
1 3
2 
Hence the inverse of the given matrix

 3

 12


 5


5
2
17
2
7
2
1 
2

3 
2

1 
2 
20
13. Using power method to find the dominant eigen values and eigen vector of
 4 5 
A 
.
 1 2
Solution:
 1
Let the initial even vector x 0   
 1
 1
 4 5   1  9 
 
Now x1  Ax 0  
       9  1   9x i
1
2
1
3
 

   
 3 
5   7 

 1   4    
 4 5   
3
3
x 2  Axi  
 
  1   
 1 2     1 2   1 
  
 3 
3   3 

 2.333 


 0.333 
 1 
 2.333  0.333 


 2.333 
 2.333 x12
 4 5  1   3.285 
x 3  Ax12  

 

 1 2  0.143   0.314 
 1 
 3.285  0.714 


 3.285 
 3.285 x13
 4 5  1   2.915 
x 4  Ax13  

 

 1 2  0.217   0.566 
 1 
 2.915 

 0.194 
 2.915 x14
 4 5  1   3.030 
x5  Ax14  

 

 1 2  0.194   0.612 
21
 1 
 3.03 

 0.202 
 3.03 x15
 4 5  1   2.99 
x 6  Ax15  

 

 1 2  0.202   0.596 
 1 
 2.99 

 0.199 
 2.99 x16
 4 5  1   3.005 
x 7  Ax16  

 

 1 2  0.199   0.602 
 1 
 3.005 

 0.200 
 3.005 U
 1 
U

 0.200 
The determine eigen vector (1, -0.2).
To find dominantevenvalue we have to solve
Au = , u for 1 where U = (1, -0.2)
 4 5   1 
 1 


  1 

 1 2   0.2 
 0.2 
 3   1 



 1.6   0.21 
 1   3
14. Using Newton raphson method to find correct to four decimals the root between 0
and 1 of the equation x3 – 6x + 4 = 0.
Solution:
f (x) = x3 – 6x + 4
Given
f (0) = 4, f (1) = -1
f (0) f (1) = 4 (-1) < 0
the root of f (x) = 0 lies between 0 and 1 the value of the root is near to 1. Let x 0 = 0.7
an approximate
22
f(x)  x 3  6x  4, f 1(x)  3x 2  6
f(0)  x 30  6x 0  4, f 1(x 0 )  3x 02  6
f(0  7)   0.7   6  0.7   4  0.143
3
f 1(0.7)  3  0.7   6  4.53
2
Then by Newton’s iteration formula we get
x1  x0 
f(x)
f  x0 
1
 0.7316
f  x1   0.0019805
f 1  x1   3   0.7316   6  4.394
2
The second approximate
x 2  x1 
f(x1 )
f 1  x2 
0.0019805
4.39428
 0.73250699
 0.7321
 0.7316 
the root of the equation = 0.7321
23
UNIT – II
INTERPOLATION AND APPROXIMATION
First differences of the function.
Let y = f(x) be any function given by the values y0, y1, ……yn. Which it takes for the
equidistant values x0, x1, ……xn of the independent variable x then y1 - y0, y2 – y1, …….yn –
yn-1 are called the first differences of the function y. Denoted by y0, y1, …..
Shift operator:
Let y = f(x) be function of x and x, x+h1 x + 2h … etc. be the consecutive values of x,
then the operator E is defined as E[f(x)] = f(x+h)
E is called Shift Operator.
Formula for Newton forward and Newton Backward differences:
The Newton’s forward interpolation formula is
y ( x0  nh)  y0  ny0 
n(n  1) 2
n(n  1)(n  2) 3
 y0 
 y0  .....
2!
3!
The Newton’s backward interpolation formula is
y ( x0  nh)  y0  ny0 
n(n  1) 2
 y0  .....
2!
Formula for Lagrange’s interpolation .
Let Y = f(x) be a function which assumes the values f(x0), f(x1) ….. f(xn)
corresponding to the values x: x1, x1 …..xn.
Y  f ( x) 
( x  x1 )( x  x2 )...( x  xn )
f ( x0 ) 
( x0  x1 )( x0  x2 )...( x0  xn )
( x  x0 )( x  x2 )...( x  xn )
f ( x1 )  .........
( x1  x0 )( x1  x2 )...( x1  xn )
Formula for inverse Lagrange’s interpolation:
x
( y  y1 )( y  y2 )...( y  yn )
x0 
( y0  y1 )( y0  y2 )...( y0  yn )
( y  y0 )( y  y2 )...( y  yn )
x1  .........
( y1  y0 )( y1  y2 )...( y1  yn )
24
Operator, , E, , .
(i)
yx = yx + h - yx
(ii)
yx = yx – yx – h
(iii)
yx = yx  1 h  yx 1
2

 = Y2 yx  1 h  yx  1
2
2
h
2
h

Relation between the operator  and E:
f(x)


f x  1 h  f ( x  1 h)
2
2
=
E
1

 E
1
  E E
2
1
2
E
E
1
1
f ( x)  E
2
1
2
E
1
2
( E  1)
2

2
1
2
f ( x)
 f ( x)
Relation between the operator  and E:
 [f(x)]= 1  f ( x  1 h)  f ( x  1 h) 
2
2
2 
1
1
 1  E 2  E 2  f ( x)
2 

1
1 1
   E 2  E 2 
2

Error formula in interpolation:
f (t )  f (t )   (t )   f ( x)   ( x) 
(t  x0 )(t  x1 )....(t  xn )
( x  x0 )( x  x1 )...( x  xn )
The above function f(t) is continuous in [x0, xn].
Numerical Examples:
01.Prove that if m and n are positive integers then
m n f(x) = m + n f(x).
Proof:
m n f(x) = (  …. m times) (   …..  n times) f(x)
= . …… (m + n) times f(x)
= m+n f(x).
25
02. Find  log x.
Solution:
log x =
log (x + h) – log x
 xh
 log 

 x 
log x = log (1 + h/x)
03. Find  tan-1 x
Solution: tan-1 x
=
tan-1 (x + h) – tan-1 (x)
 x h x 
 tan 1 

1  ( x  h) x 
h


 tan 1  2
.
 x  hx  1 
04. It n is a positive integer then
yn = y0 + nc1 y0 + nc2 2y0 + ……. + n y0.
Proof:
From the let
y1 = Ey0 = (1 + ) y0
= y0 + y0
y2 = E2 y0 = (1 + )2 y0
= (1 + 2c1  + 2) y0
= y0 + 2c1 y0 + 2y0
Similarly yn = Eny0 = (1 + )n y0
= (1 + nc1 + …….n) y0
= y0 + nc1 y0 + …..ny0
Hence proved.
26
05. Prove that E    E
Proof:= E [f(x + h) – f(x)]
E f(x)
= E [f(x + h) – E [f(x)]
= f(x + 2h) – f(x + h)
06. Prove that f(4) = f(3) +  f(2) + 2 f(1) + 3 f(1)
Solution:f(4) – f(3)
f(3)
=
=
 [f(2) + f(2)]
=
 f(2) + 2 f(2)
=
f(2) + 2 [f(1) + f(1)]
=
f(2) + 2 f(1) + 3 f(1).
f(3) – f(2) =  f(2)
f(4) = f(3) + f(2) + 2f(1) + 3f(1).
07. Express f(x) = x4 – 5x3 + 3x + 4 in terms of factorial polynomials.
Solution:
Synthetic division method:Given f(x) = x4 – 5x3 + 3x + 4
By Synthetic division
1
1 5 0 3
0 1 4 4
4(  E )
2
1 4 4 1  ( D)
0 2 4
3
1 2 8 ( C )
0 3
1(=A) 1(=B)
f(x) = x(4) + x(3) – 8x(2) – x(1) + 4.
08. Estimate y2 from the following table.
X
1
2
3
4
5
Yx
7
?
13
21
37
27
Solution:-
Here we are given four entries
Viz. y1, y3, y4 and y5. Therefore the function yx can be represented by a third degree
polynomial 3 yx = Constant and 4 yx = 0 in particular
4y1 = 0  (E – 1)4 y1 = 0;(E4 – 4E3 + 6E2 – 4E + 1) y1 = 0
y5 – 4y4 + 6y3 – 4y2 + 7 = 0; 38 – 4y2 = 0; y2 = 9.5
09. Obtain a function whose first differences is
6x2 + 16x + 11
Solution:Expressing the function in factorial notation, we get
6x2 + 16x + 11 = 6x(2) + 16x(1) + 11
 f(x) = 6x(2) + 16x(1) + 11
Integrating we get
6 x (3) 16 x (2) 11x(1)
f ( x) 


K
3
2
1
 2 x (3)  8 x (2)  11x (1)  K
which is the required function.
10. Find 10 (1 – ax) (1 – bx2) (1 – cx3) (1 – dx4).
Solution:Let f(x) = (1 – ax) (1 – bx2) (1 – cx3) (1 – dx4).
f(x) is a polynomial of degree 10 and the coefficient of x10 is a, b, c, d.
10 f(x)
=
10 (abcd x10)
=
abcd 10 x10 = 10! abcd.
11. Find n (1/x).
Solution:Now  (
1
)
x
1
2( )
x
=
1
1

x 1 x

1
x( x  1)
=
(1)2
x( x  1) ( x  2)
and so on
28
Preceding like this
n ( 1 ) =
x
(1)n
x( x  1)( x  2)....( x  n)
 2  3
12. Evaluate 
x
 E 
Solution:Let h be the interval of differencing
 2  3
2 1
3
  x  ( E ) x
 E 
 ( E  1)2 E 1 x3
 ( E 2  2 E  1) E 1 x3
 ( E  2  E 1 ) x3  Ex 3  2 x 3  E 1 x 3
 ( x  h )3  2 x 3  ( x  h )3
 6 xh.
13. Given u0 = 1, u1 = 11, u2 = 21, u3 = 28 and u4 = 29 find 4y0.
Solution:4y0
=
(E4 – 4C1 E3 + 4C2 E2 – 4C3 E + 1) y0
=
E4y0 – 4E3y0 + 6E2y0 – 4Ey0 + y0
=
y4 – 4y3 + 6y2 – 4y1 + y0
=
29 – 112 + 126 – 44 + 1
=
0.
14. Write the relation between E and .
Solution:
 f(x) =
=
f(x) – f(x – h)
=
f(x) – E-1 f(x)
=
(1 – E-1) f(x)
E 1
E
15. Prove that (1+ ) (1 - ) = 1
Solution:(1 + ) (1 - ) f(x) = E E-1 f(x)
29
= E f(x – h)
= f(x)
(1+ ) (1 - ) = 1
16. Prove that  =  - 
Proof:f(x) =
(E – 1) (1 – E-1) f(x)
=
(E – 1) [f(x) – f(x-h)]
=
E[f(x)] f(x) – E[f(x – h)] + f(x – h)
=
f(x+h) – f(x) – f(x) + f(x – h)
=
[Ef(x) – f(x)] – [f(x) – f(x – h)]
=
(E – 1) f(x) – (1 – E-1) f(x)
=
[(E – 1) – (1 – E-1)] f(x)
=
( -  ) f(x).
17. Prove that f(x) = (1 – E-1) f(x).
Proof: f(x)
=
E-1 [f(x)]
= E-1 [f(x + h) – f(x)]
f(x) – f(x – h)
= f(x) – f(x – h) 
 = E-1 
18. Prove that 023  6, 303  6
Solution:-
 0nr  0
n>r
 n 0n  n !
0r  1r  1
023  23  2.13  6
303  33  3.23  3.i 3  6
30
19. Calculate (i) 306 (ii) 506
Solution:-
306 =36 – 3.26 + 3.16
(i)
= 729 – 192 + 3 = 540
(ii) 506 =56 – 5.46 + 10.36 – 10.26 + 5.16
=15625 – 20480 + 7290 – 640 + 5
=1800.
20. Prove that E
1
2
-1
2
= μ+ 1 δ, E
2
= μ- 1 δ .
2
Solution:-

1
  12   12 E 2  E
1
  12  E
2
1
2
E
2
E
1
2

2

1
 1
 1 2 E 2   E 2
2



1
  12   12 E 2  E
E
1
1
  12  E
2
1
1
2
21. Prove that Δ = 1 δ2 +δ 1+ δ
2
4
2
Proof:-

 1 E
2
1      
1 E
2
4
2
2
1
1
2
2
E
E

1
1
2
2
2


2
1
 
  E
 1  E  E 1  2   E 2  E
2
1
1
2
E


E


1
E
2
 E 12  E 12
1
1
  E  E  2  
2

2

 E  1 = .
2

1
31
2
E

2
4  E  E 1 2
4
1
4
1
2
4
1
E 2 E
2
1
1
2

2

2
22. Prove that μ2 = 1+ 1 δ2
4
Proof:-
 2  E
2
  1
2
1
E
2
1
2

2
E  E 1  2 
4
1 2
 1

 1  E 2  E 2  4
4

2
 1   4 
4
1


 2  1  4
2
23. Prove that (E + 1)  = 2 (E – 1) .
Solution:-

1
( E  1)     2  E

1
 E 2 .E
E
1
E
1
2
2
E
E
1
1
1
1
1
2
2
 E 2 .E
2
E
2
E
1
1
2
2

1
2
 E
 E
 E
1
1
1
2
2
E
2
E
 E 1 2  E 1 2
 ( E  1)  2. 

2

=2( E  1)  
E
1
1
2
2
1
2







24. Find the missing yx values from the first differences provided.
yx
0
a
b
c
d
e
yx
0
1
2
4
7
11
Solution:-
By def:
a–0=1
a=1
b – a = 2, b – 1 = 2
c – b = 4, c – 3 = 4
b=3
c=7
32
d – c = 7, d – 7 = 7
d = 14
e – d = 11, e – 14 = 11
e = 25
25. Evaluate  n (ax n  bx n1 )
Solution:(n (axn + bxn-1)
=
(n (axn) + (n (bxn-1)
=
a n! + b(0)
=
an!
26. Find f(x) if x2 + 2x + 2 = f(x) and the interval of differencing as unity.
Solution:f(x) =
=
f (x + 1) – f (x)
( x + 1)2 + 2 (x + 1) + 2 – [x2 + 2x + 2]
 x 2  2x  1  2 x  2  2  x 2  2x  2
 2x  3
27. Find the second degree polynomial fitting the following data.
x 1 2 4
y 4 5 13
Solution:-
x0 = 1, x1 = 2, x2 = 4, y0 = 4, y1 =5, y2 = 13
By Lagrange’s formula
f ( x) 
( x  x1 )( x  x2 )
( x0  x1 )( x0  x2 )
y0 
( x  x0 )( x  x2 )
( x1  x0 )( x1  x2 )
y1 
( x  x0 )( x  x1 )
( x2  x0 )( x2  x0 )
( x  2)( x  4)
( x  1)( x  4)
( x  1)( x  2)
4
5
13
3
2
6
1
 8 x 2  48 x  64  15 x 2  75 x  60  13 x 2  39 x  26 
6

1
6 x 2  12 x  30 
6
 x2  2 x  5

33
y2
 Δ 2  x Ee x
x
28. Prove that 
e . 2 x = e
E
Δ
e


Solution:-
(  E 1 )e x .
x
Ee x
 x  h Ee


e
 e x
 e x
x
 x  h Ee
  e .e
 e h Ee x
 x
e
h
xh
 e  x  ex
3
1
1
29. Show that     
bcd a
abcd
 
Solution:
If f(x) = 1/x, f(a) = 1/a
1 1

1
1 b a
f(a, b) =    

b
ab
a ba
f(a, b, c) 
f  b, c   f  a , b 
ca
=
f (a, b, c, d ) 


1
1

bc ab
ca
1
abc
f  b, c, d   f  a, b, c 
d a
1
1

1
= bcd abc  
d-a
abcd
30. A function f(x) is given by the following table. Find f(0.2) by a suitable formula.
x
0
1
2
3
4
5
6
F(x)
176
185
194
203
212
220
229
Solution:
34
The difference table is follows:-
y = f(x)

0
176
(y0
1
185
9
(2y0
2
194
9
0
(3 y0
3
203
9
0
0
 4 y0
4
212
9
0
0
0
 5 y0
5
220
8
-1
-1
-1
-1
 6 y0
6
229
9
1
2
3
4
5
x
2
(3
(4
(5
(6
Here x0 = 0, h = 1, y0 = 176 = f(x)
We have to find the value of f (0.2). By Newton’s forward interpolation formula we
have
f ( x0  nh)  y0  ny0 
n(n  1) 
 y0  ........
2!
f (0.2)  ?
x0 + nh = 0.2
0 + n = 0.2  n = 0.2
f (0.2)  176  (0.2) 9 
(0.2)(0.2  1)
0
2
 176  1.8
 177.8
31. From the given table compute the value of sin 38.
x
0
10
20
30
40
Sin x
0
0.17365
0.34202
0.5
0.64276
Solution:As we have to determine the value of y = sin x near the lower end, we apply Newton’s
backward interpolation formula.
The difference table is as given below.
35
x0
y
y(x) = Sin x0
2y
3y
4y
0
0
10
0.17365
0.17365
20
0.34202
0.16837
- 0.00528
30
0.5000
0.15798
- 0.01039
- 0.00511
40
0.64279
0.14279
- 0.01519
- 0.0048
0.00031
X0
Y0
y0
2y0
3y0
4y0
Here x0 = 40, h = 0.64279, h = 10
Newton’s backward differences formula
y ( x0  nh)  y0  ny0 
n(n  1) 2
 y0  ....
2!
y (38) 
x0  nh  38
40  n(10)  38
y (38)  0.64279  ( 0.2) (0.14279) 
n  0.2
( 0.2) ( 0.2  1)
( 0.01519)
2!
(0.2) (0.21) (0.22)
(0.0048)  neglible term
3!
 0.64279  0.028558  0.0012152  0.0002304
 0.61566

32. If Ux = ax2 + bx + C, then show that
U2n - nC12U2n-1 +nC2 2U2n-2 ........ + (-2)n un = (-1)n (1- 2an)
Solution:Given that
Ux=ax2 + bx + C
Un =an2 + bn + C, Uni > a polynomial of
degree 2 in n
3un = 4 un = 0
Let the interval of differencing be equal to 1. Now
Un
= an2 + bn + C
Un
=
a (n + 1)2 + b (n + 1) + C – an2 – bn – C
36
=
2Un =
2 an + a + b
[Un)
=
2a (n + 1) + a + b – 2an – a – b
=
2a
LHS :U2n – nC1 2U2n-1 + nC2 U2n-2 - …….
=
[En Un – nC1 2En-1 Un + nC2 En-2 …..] Un
=
[En – nC1 2En-1 Un + nC2 En-2 ……..] Un
=
(E – 2) n Un
=
(E – 1 -1 )n un
=
( - 1)n Un
=
(-1)n (1 - )n Un
=
(-1)n [1 – nC1  + nC2 2 + …….]Un
=
(-1)n [Un – n  Un +
=

n2  n 
(-1)n  an2  bn  C  n(2an  a  b) 
2a 
2


E–1=
n( n  1) 2
 Un + ….]
2!
= (-1)n (C – 2an)
= RHS.
33. In an examination the number of candidates who obtained marks between certain
limits were as follows:Marks
No. of
Students
30 – 40
40 – 50
50 – 60
60 – 70
70 – 80
31
42
51
35
31
Find the number at candidate whose scores lie between 45 and 50.
Solution:-
First we construct a cumulative frequency table for the given table.
Upper limits 40 50 60
C.F .
31 73 124
70
159
80
190
37
The difference table is
x
y
Marks
C.F.
40
31
50
73
42
60
124
51
9
70
159
35
-16
- 25
80
190
31
-4
12
y
2y
3y
4y
37
We have x0 = 40, x = 45, h = 10
U
x  x0 45  40

 0.5
h
10
y0 = 73, y0 = 42, 2y0 = 9, 3y0 = - 25, 4y0 = 37
From Newton’s forward interpolation formula
U (U  1) 2
U (U  1) (U  2) 3
 y 
 y
0
0
2!
3!
U (U  1) (U  2) (U  3) 4

 y  ......
0
4!
f ( x)  y  U y 
0
0
(0.5) (0.5)
(0.5) (0.5  1) (0.5  2)
(9) 
( 25) 
2!
6
(0.5) (0.5) (1.5) ( 2.5)
(37)
24
f (45)  31  (0.5) 42 
=
47.8673 = 48 approximately.
The number of students who obtained marks less than 45 = 48, and the number of students
who scored marks between 45 and 50 = 73 – 48 = 25.
38
34. Find the form of the function f(x) under suitable assumption from the following
Solution:-
The divided differences table is given below.
X
f(x)
0
2
1
3
2
12
5
147
f(x)
32
1 0
2f(x)
3f(x)
1
12  3
2 1
9
147  12
52
 45
9 1
20
4
45  9
5 1
9
94
1
50
We have x0 = 0, f(x0 = 2, f(x0, x1) = 1, f(x0, x1, x2) = 4
f(x0, x1, x2, x3) = 1
The Newton’s divided difference interpolation formula is
X
0
1
2
5
f(x)
2
3
12
147
f ( x)  ( fx0 )  ( x  x0 ) f ( x0 , x1 )  ( x  x0 ) ( x  x1 ) f ( x0 , x1 , x2 )
 ( x  x0 ) ( x  x1 ) ( x  x2 ) f ( x0 , x1, x2 , x3 )
 2  ( x  0) 1  ( x  0) ( x  1) 4  ( x  0) ( x  1) ( x  2) 1
 x3  x 2  x  2
35. Using Lagrange’s interpolation formula, find the value of y corresponding to x = 10
from the following table.
x
5
6
9
11
f(x)
12
13
14
16
Solution:We have x0 = 5, x1 = 6, x2 = 9, x3 = 11
Y0 = 12, y1 = 13, y2 = 14, y3 = 16
Using Lagrange’s interpolation formula, we have
y  f ( x) 
( x  x1 ) ( x  x2 ) ( x  x3 )
( x0  x1 ) ( x0  x2 ) ( x0  x3 )
y0 
( x  x0 ) ( x  x2 ) ( x  x3 )
( x1  x0 ) ( x1  x2 ) ( x1  x3 )
39
y1

( x  x0 )( x  x1 )( x  x3 )
( x  x0 )( x  x1 )( x  x2 )
y2 
( x2  x0 )( x2  x1 )( x2  x3 )
( x3  x0 )( x3  x1 )( x3  x2 )
y3
Substitute
f (10) 
(10  6) (10  9) (10  11)
(5  6) (5  9) (5  11)
(10  5) (10  6) (10  11)

(9  6) (9  6) (9  11)
 2
(10  5) (10  9) (10  11)
(12) 
(6  5) (6  9) (6  11)
14 
(10  5) (10  6) (10  9)
(11  5) (11  6) (11  9)
(13)
 16
13 35 16 42
  
3 3 3
3
36. Find the value of x when y = 85, using Lagrange’s formula from the following table.
x
2
5
8
14
y
94.8
87.9
81.3
68.7
Solution:-
x0 = 2, x1 = 5, x2 = 8, x3 = 14
y0 = 94.8, y1 = 87.9, y2 = 81.3, y3 = 68.7
y = 85
We know that the Lagrange’s inverse formula is
x
( y  y1 ) ( y  y 2 ) ( y  y3 )
( y0  y1 ) ( y0  y 2 ) ( y0  y3 )

x0 
( y  y0 )( y  y1 )( y  y3 )
( y 2  y0 )( y 2  y1 )( y 2  y3 )
( y  y 0 ) ( y  y 2 ) ( y  y3 )
( y1  y0 ) ( y1  y 2 ) ( y1  y3 )
x2 
( y  y0 )( y  y1 )( y  y 2 )
( y3  y0 )( y3  y1 )( y3  y 2 )
Substituting the above values we get,
x
(85  87.9) (85  81.3) (85  68.7)
2
(94.8  87.9) (94.8  81.3) (94.8  68.7)
(85  94.8) (85  81.3) (85  68.7)

.5
(87.9  94.8) (87.9  81.3) (87.9  68.7)
(85  94.8) (85  87.9) (85  68.7)
(8)
(81.3  94.8) (81.3  87.9) (81.3  68.7)
(85  94.8) (85  87.9) (85  68.7)

14
(68.7  94.8) (68.7  87.9) (68.7  81.3)

x = 6.5928.
40
x1
x3
37. Find the first term of the series whose second and subsequent terms are 8, 3, 0, -1, 0
……..
Solution:-
Given f(2) = 8, f(3) = 3, f(4) = 0, f(4) = -1, f(5) = 0
We are to find f(1)
We construct the difference table with the given values.
f(x)
2 f(x)
3 f(x)
X
f(x)
2
8
3
3
-5
4
0
-3
2
5
-1
-1
2
0
6
0
1
2
0
4 f(x)
0
We have 3 f(x) = 4 f(x) = 0
Using the displacement operator
f(1)
= E-1 (f(2))
=
(1 + )-1 f(2)
=
(1 -  + 2 - 3 + ….) f(2)
=
f(2) - f(2) + 2 f(2) - 3 f(2) + ……..
=
8 – (- 5) + 2
=
15
f(1) = 15
41
38. Find f(x) as a polynomial in x for the following data by Newton’s divided difference
formula
X
:
F(x) :
-4
-1
0
2
5
1245
33
5
9
1335
Solution:
X
F(x)
-4
1245
f ( x )
2 f  x 
3 f  x 
4 f  x 
-404
-1
33
94
-28
0
5
-14
10
3
2
2
9
13
88
442
5
1335
By Newton’s divided difference interpolation formula
f(x)= 1245 + (x+4) (-404) + (x+4) (x+1) 94 + (x+4)
(x+1) x(-14) + (x+4) (x+1) x(x-2)3
= 3x4 + x3 – 14x+5
39. The following table gives same relation between steam pressure and temperature.
find the pressure at temperature 372.10
T
3610
3670
3780
3870
3990
P
154.9 167.9 191.0 212.5 244.2
Solution:
T
P
361
154.9
367
167.0
378
191.0
p
2 p
3 P
4 p
2.016666
0.0097147
2.18181818
0.000024
0.0103535
2.388889
387
212.5
0.00000073
0.000052
0.01203703
2.641667
399
244.2
42
By Newton’s divided difference formula
P(T=372.10) = 154.9 + (11.1)(2.016666) + (11.1)
(5.1) (0.009914) + (11.1) (5.1) (-5.9) (0.000024)
+ (11.1) (5.1) (-5.9) (-14.9) (0.00000073)
= 177.8394819
40. From the data given below, find the number of students whose weight is between 60
to 70
Weight
No
of
0-40
40-60
60-80
80-100
100-120
250
120
100
70
50
Students
Solution:
x
Y
Weight
No of Students
Below 40
250
Below 60
370
Below 80
470
y
2 y
3 y
4 y
120
-20
100
-10
-30
70
Below 100
540
20
10
-20
50
Below 120
590
x  x0 70  40

 1.5
h
20
u  u  1 2
y  70   y0  u y0 
 y0  .......
2!
1.51.5
 0.5 0.5 0.5
=250+ 1.5120  
 20  
 10 
2
6
1.5 0.5 0.5 1.5

24
=424
u
 Number of students whose weight is between
60 and 70
= y(70) – y(60) = 424 – 370 = 54
43
41. Explain Cubic Sp line.
The concept of the sp line originated from the mechanical drafting tool called spline
used by designers for drawing smooth curves. It is a splender flexible bar made of wood or
some other elastic material this curves resemble cubic curves and hence the name cubic
spline has been given to the piecewise cubic interpolating polynornials. Cubic splines are
popular because of their ability to interpolate data with smooth curves.
we consider here the construction of cubic spline function which would interpolate the points
(x0, f0) (x1, f1) ……(xn, fn). The cubic sp line s(x) consists of (n-1)3 corresponding to (n-1)
sub intervals. If we denote such cubic by si(x)
then s(x)= s;(x)
i = 1,2…….n
As pointed out earlier, there cubic must satisfied the following conditions
(i)
S(x) must interpolate f at all points x0, x1, ……xn i.e. for each i, S(xi) = fi
(ii)
The function values must be equal at all the interior knots i.e. Si(xi) = Si+1
(xi)
The first derivative at the interior knots must be equal.(i.e) si’(xi)
(iii)
= Si+1’(xi)
(iv)
The second derivatives at the
interior knots must be equal i.e. Si”(xi)=
Si+1”(xi)
The second derivatives at the end joint are 2000. i.e. S”(x0)
(v)
=S”(xn) = 0
A Cubic splines with zero second derivatives at the end joints are called the natural cubic
splines.
42.. Estimate the function value f at x=7 using cubic splines
Xi
4
9
16
Fi
2
3
4
Solution:
Here
h1 = x1 – x0 = 9 – 4 = 5
h2 = x2 – x1 = 16 – 9 = 7
f0 = 2; f1 = 3, f2 =4
44
ai 1
a
 hi 2ui  ui 3   i ui31  hi2ui 1 
6hi
6hi
1
  fi ui 1  f i 1ui   (1)
hi
 f  fi fi  f i 1 
hi a i-1  2  hi  hi 1  ai  hi 1ai 1  6  i 1

  (2)
hi 
 hi 1
Si ( x) 
.
put i = 1in (2)
f f
f f 
h1 a 0  2  h1  h2  a1  h2 a2  6  2 1  1 0   0
h1 
 h2
43 3 2
0  2  5  7  a1  0  6 

5 
 7
1 1
24a 1 = 6   
 7 5
a 1  0.0142
Since n  3 there are two cubic splines namely,
S1 ( x), x 0  x  x, and S2  x  , x1  x  x2 .
The target point x = 7 is in the dornain of S1(x) and therefore we need to use only S1(x)
for estimation put i=1 in (1)
S1  x  
a0 2
a
1
 h1 a1  u13   1 u03  h12u0    f1u0  f 0u1  
6
6h1
h1
u0  x  x0  7  4  3;
S1  x   0 
u1  x  x1  7  9  2
 0.0142  33  52 3  1 3 3  2 2
      

6  5
5
S1  x   2.6229
43. From the following table
X
1
2
3
Y
-8
-1
18
compute y(1.5) and y’(1) using cubic splines
Solution:
here x0=1,
t0=-8,
x1=2, x2 =3
t1=-1, t2= 18
45
h1=h2=1
t t t t 
W.K.T. h1a0 +2(h1+h2 )a1+h2 a2 = 6  2 1  1 0 
h1 
 h2
4a1=6(12)
a1= 18
u0 = x - x0 = 0.5;
S ( x) 
u1=x –x1 = -0.5
a0 2
a
1
 h1 u1  u13   1 u03  h12 u1   t1u0  t0u1 
6h1
6h1
h1
S 1.5  0 
18 
1
3
 0.5  12  0.5   1 0.5   8 0.5

6
1
S 1.5   5.625
S  x 
a0 2
a
1
h1 u1  u13   1 u03  h12 u1   t1u0  t0u1 

6h1
6h1
h1
18 
3
 x  1   x  1   1 x  1  8  x  2 
6 
 3 x3  9 x 2  13 x  15

S11 ( x)  9 x 2  18 x  13
S11 1  4
46
UNIT – III
NUMERICAL DIFFERENTIATION AND INTEGRATION
Numerical differentiation:
Numerical differentiation is the process of calculating the derivatives of a given
function by means of a table given values of that function. i.e., if y (xi, yi) are the given set of
values, then the process of computing the values of
dy d2 y
,
......... is called numerical
dx dx 2
differentiation.
Formula of forward difference formula to compute the derivatives:
f 1  x 0   1  y 0  1  2 y 0  1  3 y 0  1  4 y 0 ........
h
2
3
4

11
2
3
4
f  x 0   1 2   y 0   y 0  11  y 0 ..........
12

h 
f 111  x 0   1 3   3 y 0  3  4 y 0  ............
2

h 
Formula of backward difference formula to computer the derivatives:
f 1  x 0   1 y 0  1  2 y 0  1  4 y 0 ........
h
2
3

f 11  x 0   1 2  2 y 0  3 y 0  11  4 y 0  ..........
12

h 
f 111  x 0   1 3 3 y 0  3  4 y 0  ............
2

h 
Numerical Integration:
The term numerical integration is the numerical evaluation of a definite integral.
b
A   f  x dx Where a and b are given constants and f(x) is a function.
a
Formula for Trapezoidal Rule:
x0 nh
 y  x dx  h 2  y
0
 y n   2  y1  y 2  .......  y n1  
x0
47
Formula for Simpson’s 1/3 rule:
x0 nh
 f  x dx  h 3  y
0
 yn  4  sum of odd ordinates 
x0
+2  sum of even ordinates  
Formula for Simpson’s 3/8 rule:
x0 nh
 f  x dx  3h 8  y
0
 yn   3  y1  y 2  y 4  y 5  ......yn6 
x0
+2  y3  y 6  .........yn3  
Truncation error in the Trapezoidal rule:
The total error
h3 11 11
 y1  y 2  ..........  y11

E
n 
12
nh3 11
E<
y 
12
Error in the trapezoidal rule is of the order h2.
Truncation error in Simpson rules:
The error in the interval (r1, r3)
5 
 4

  h5 y1iv
 15 18 
 24  25  5 iv

h y
 90 

h5 iv
y1
90
The total error: 
nh5 iv
y 
90
Error in the Simpson 1/3 rule is of the order h4.
48
Formula for Romberg method:
I=I2 
I2  I1 
3
 2
I1  dividing h into two parts h
I2  dividing h into four parts
I3  dividing h into eight parts
Formula for Gauss Quadrature 2 point:
b
Gauss two point formula is  f  x  dx
a

1
b  a  b  a   a  b  
f
z
dz
2 1  2   2  
 1   1 
  f 

3 
3
1
 f  x  dx  f 
-1
Formula for Gaussian Quadrature 3 point:
b
1
a
1
 f  x  dx   f  t dt Where the interval (a, b) is changed in fo(-1, 1) by the transformation.
x
ba ba

t
2
 2 
1
Then
 f  t  dt  0.5555 f  0.77459   f 0.77459   0.8888 f 0 
1
Formula for Evaluation of Double integration using Trapezoidal:
Let the given double integral be of the form
b d
I    f  x,y  dxdy
a c
I
hK
4
[sum of values of f at the four corners +2 (sum of the values f at the remaining nodes
on the boundary) +4 (sum of the values of f at the interior nodes)]
49
Numerical Examples:
5.2
01. Compute the value of the definite integral
 logxdx using trapezoidal rule?
4
Solution:
Here f(x) = log x, a = x0=4, b=xn=5.2
Divide the internal of interaction into six equal parts each of width 0.2.
h
5.2  4
 0.2
6
X
4.0
4.2
4.4
4.6
4.8
5.0
5.2
F(x)
1.3863
1.4351
1.4816
1.5261
1.5686
1.6094
1.6457
We know that Trapezoidal rule
b
h
 f  x  dx  2  y
0
 y n   2  y1  y 2  .........y n1  
a
52
h
 log x dx = 2  y
0
 y 6   2  y1  y 2  y 3  y 4  y 5  
4
=
0.2
1.386294  1.6486   2 1.435084
2 
+1.481604  1.526056  1.568615  1.609237  
=  0.1 3.034952  15.241562
=1.876544
2 2
dxdy
using Trapezoidal rule with h =0.5.
x
+
y
1 1
02. Evaluate the integral I =  
Solution:
Using Trapezoidal rule
2 2
dxdy
xy
1 1
I 

1
 f 1,1  f  2,1  f 1,2   f  2,2 
16 

 
 
 



+  f 3 ,1  f 1, 3  f 2, 3  f 3 ,2   4f 3 , 3 
2
2
2
2 
2 2 

 1 0.5  1  1  0.25  2 0.4  0.4  2  2   4 
16 
3
3

7
7
3
 0.323304
50
2
03. Evaluate
dx
using Gaussian 2 point formula.
x
1

Solution:
Transform the variable x to t by the transformation
ab ba
x

t
 2   2 
 1+2   2  1 
=

t
 2   2 
=3  t
2
2
dx  dt
2
2
1
1
dx
2 dt
dt
I 


x 1 3  t 2 1 3  t
1
Here
1
 1 
f

 3  3 1
 0.2795
3
1
f   1  
 0.41288
3



1
3 

3

I  f  1   f   1 
2 
3

=0.6923
2
04. Evaluate
dx
using Gaussian 3 – point formula.
x
1

Solution:
Transform the variable from x to t by the transformation.
ba ba
x

t
 2   2 
3 t
 
2 2
dt
dx 
2
1
3t
dt
x

2
3t
1
51
2
1
dx
I 
 f  t  dt  A1f  x1   A 2 f  t 2   A 3 f  A 3 
x 1
1
A1  A 2  0.555
A 3  0.888
f  t1   f  0.7745  
1
 0.4493
3  0.7745
f  t2   f 0  1
3
1
f  t3  
 0.2649
3  0.7745
I= 0.555(0.4493) +0.555(0.2649) +0.333(0.888)
= 0.6931
4
05. Using Simpson rule find  ex dx given that e0 = 1, e1 = 2.72, e2= 7.39, e3=20.09,
0
e4 = 54.6.
Solution:
By Simpson rule we have
4
h
 e dx  u  y
x
0
 y 4   2y 2  4  y1  y 3  
0
= 1  54.6  1  2  7.39   4  2.72  20.04  
3
=53.8733
06. Write the polynomial to calculate the value of x when?
X
3
5
7
9
Y
6
24
58
108
Solution:
X
Y
3
6
y
2y
3y
18
5
24
16
34
3
58
0
16
50
9
108
x0  nx  x
3  n2  x; 2n  x  3;
n
x 3
2
52
y  x 0  nx   y 0  ny 0 
n  n  1
 2 y 0  .....
2!
 x  3  x  3 
 2   2  1
x

3



 16
y x  6  
18  
 

2
 2 
y  x   2x 2  3x  9
1
07. Evaluating
dx
 1+ x
2
by a numerical iteration method, Find this value?
0
Solution:
1
1
dx
1



tan
x


0 1  x2 
0
=tan-1 1  tan1  0 
=tan-1 1
=
2
1
09. Find the value of log
21/3
x2
from 
dx using Simpson’s 1/3 rule with h = 0.25.
1+ x 3
0
Solution:
Given h = 0.25
X
0
0.25
0.5
0.75
1
Y
0
0.06154
0.222
0.395
0.5
By Simpson’s 1/3 rule
1
x2
h
0 1 x3 dx  3  y0  y 4   2y2  4  y1  y3 
=0.23log3
We know that
1
1
x2
 1 log 1  x 3 
dx

0 1  x3
 3
0

 
log 2
1
3

= 1 loge2
3
1
x2

dx  0.231083
1  x3
0
53
10. Find the first second and 3rd derivatives of the function tabulated below at the point
x=1.5?
X
1.5
2.0
2.5
3.0
3.5
4.0
F(x)
3.325
7.0
13.625
24.0
38.87
59.0
Solution:
The table of difference is as follows:
X
F(x)
1.5
3.375
y
2y
3y
4y
3.625
2.0
7.0
3.0
6.625
2.5
13.625
0.75
3.25
0
10.375
3.0
24.0
0.75
4.50
0
14.875
3.5
38.875
0.75
5.25
20.125
4.0
59.0
Here we have to find the derivatives at the point x = 1.5 which is the starting value of the
table. Therefore newton’s forward differences formula for derivatives at x = x0, we have
f 1  x 0   1  y 0  1  2 y 0  1  3 y 0 .......
h
2
3

f0  1.5, h = 0.5
3.625  1  3   1  0.75  
f 1 1.5   1
0.5 
2
3

=4.75
f 11  x 0   1 2   2 y 0   3 y 0  11  4 y 0 .......
12

h 
1
=
3.0  0.75 
2 
 0.5 
=9.0
f 111  x 0   1 3   3 y 0  3  4 y 0 
2

h 
1
=
0.75
3 
 0.5 
=6.0
54
11. The population of certain town is shown in the following table.
Year
1931
1941
1951
1961
1971
Population
40.6
60.8
79.9
103.6
132.7
(in thousands)
Find the rate of growth of the population in the year 1961.
Solution:
The table of difference is as follows:
X
Y
1931
40.6
y
2y
3y
4y
20.2
1941
60.8
-1.1
19.1
1951
79.9
5.7
4.6
23.7
1961
103.6
-4.9
0.8
5.4
29.7
1971
132.7
Here h = 10, x0 = 1971
We know that
2
3
2

2n  1 2
3r  6r  2 3
2r  9r  11r  3 4 
1
y x0  nh  1 y0 
 y0 
 y0 
 y0 
h
2
6
12




 
1
y 1961  ?
x0  nh  1961
1971  n10  1961
n10  10  n 1
55






y1 1961  1 29.1   1 5.4   1  0.8    1  4.9  
10 
2
6
12

= 1  29.1  2.7  0.1334  0.4083 
10
=2.6775
= 1  4  0.7899  0.7943   0.7884
3
we get


I h1, h , h  I  0.6,0.3,0.15 
2 4
= 1  4  I  0.15  0.3  I  0.3, 0.6  
3
= 1  4  0.7884  0.7886 
3
=0.7883
The table of these values
0.8113
0.7886
0.7948
0.7883
0.7884
0.7899
2
dx
 0.7883
1 x
0
I 
12
12. Using Romberg’s Method compute I =
dx
 1+ x correct to 4 decimal places.
0
Solution:
Here f(x) = 1
1 x
Take h = 0.6, 0.3, 0.15
H = 0.6, h/2=0.3, h/4=0.15
X
0
0.15
0.30
0.45
0.60
0.75
0.9
1.05
1.20
F(x)
1
0.8695
0.7692
0.6896
0.6250
0.5014
0.5263
487
434
Using Trapezoidal method,
56
0.6
1  0.4545   2  0.6256  
2 
=0.8113
I  h   I  0.6   I1 
with h=
0.6
 0.3
2
we get
 2  I0.3   I
I h
2
0.3
1  0.4545   2  0.7692  0.625  0.5263  
2 
=0.7943
=
with
h
0.6
 0.15we get
4
 4   I0.15   I
I h

3
0.15
1  0.4545    0.8695  0.7692  0.6896 
2 
+  0.6250+0.5714+0.5263+0.4878  
=0.7899
Now,

I h, h
2
  I0.6, 0.3 
= I  4  I  0.3   I  0.6  
3
= I  4  0.7943   0.8113 
3
=0.7886


I h , h  I0.3,0.15 
2 4
= I 4  I  0.15   I  0.3  
3
1
13. Evaluate
e
-x2
dx by dividing the range of initiation in to 4 equal parts using (i)
0
Trapezoidal rule (ii) Simpson’s rule.
Solution:
Here the length of the interval is h =
1 0
= 0.25. The values of the function y = e-x2
u
for each point of sub divisions are given below.
57
X
0
0.25
0.5
0.75
1
-x2
1
0.9394
0.7784
0.5694
0.3628
Y0
Y1
Y2
Y3
Y4
e
(i) By Trapezoidal rule
1
x
 e dx 
2
0
h
 y 0  y 4   2  y1  y 2  y 3  
2
0.25
1.3678  2  2.8761 
2 
=  0.125  5.943 
=
=0.7428
(ii) By Simpson’s rule:
1
e
 x2
dx 
0
h
 y 0  y 4   2y 2  4  y1  y 3  
3
0.25
1.3678  1.5576  6.0352
2
=0.7467
=
14. The velocity v of a partial ad distances from a point on its path is given by the table.
F
0
10
20
30
40
50
60
Feet
V
47
58
64
65
61
52
38
Feet/fer
Estimate the time taken to travel 60feet by using Simpson’s one third value. Compare
the result with Simpson’s 3/8 rule.
Solution:
We know that the rate of charge of displacement is velocity
ds
V
dt
ds  vdt
1
dt  ds
v
 1
Here we find to find the time taken to travel 60 feet. Therefore interstate (1) from 0 to 60
We get
60
60
0
0
1
 dt   v ds
The time taken to travel 60 feet is
58
60

t
0
1
ds 
v
60
 ydx
0
The given table can write as given below:
X(5)
0
10
20
30
40
50
60
1
0.2127
0.017
0.0156
0.01538
0.0164
0.01923
0.0263
Y0
Y1
Y2
Y3
Y4
Y5
Y6
Y=
v
Simpson’s one third we have
60
h
 ydx  3  y
0
 y 6   2  y 2  y 4   4  y1  y 3  y 5  
0
=
10
 0.02127  0.0263   2  0.01563  0.0164 
3 
+4  0.01724+0.01538+0.01923  
10
0.04758  0.020740  0.06406 
3
 1.06 sec

Hence time taken to travel 60 feet is 1.063 feets
By Simpson’s 3/8 rule
60
 ydx  3h 8  y
0

0
 y 6   3  y1  y 2  y 4  y 5   2  y 3  
3 10 
 0.02127  0.02630   3  0.01723  0.0156 
8 
+  0.01640+0.01923   2  0.01538  
 3.75 0.04757  0.20547  0.03076
 1.064 sec
1.4
15. Evaluate
  sinx - ln x + e dx by Simpson’s 1/3 rule.
x
0.2
Solution:
Let us divide the interval at integration in to twelve equal parts by taking h = 0.1.
Now the table of values of the given function y = sinx – lnx + ex at each point of subdivision
is as given below.
X
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Y
3.0291
2.8493
2.7975
2.8213
2.8915
3.014
3.348
3.559
3.559
X
1.1
1.2
1.3
1.4
Y
3.8007
4.0698
4.3705
4.7041
59
Simpson’s rule
1.4
h
 ydx  3  y
0
 y12   2  y 2  y 4  y 6  y 8  y10 
0.2
+4  y1  y 3  y 5  y 7  y 9  y11  
0.1
7.73369  2 16.49077   4  20.20418  
3 
=4.05106
=
2
16. Evaluate
dx
 1+ x
3
by Gauss 3 point formula:-
1
Solution:
Transform the variable from x to t by the transformation.
ba ha

t
2
 2 
3 t 3t
=  
2 2
2
2
1
dx
1
dt


3
3

1 x
3t 2
1
1
1 

 2 
x
1
=4 
-1
1
=4 
1
8  3  t 
3
dt
3
  A1f  t1   A 2 f  t 2   A 3 f  t 3  
dt
-1 8   3  t 
where
60
f t 
1
3  t 
3
8
A1  A 2  0.55
A 3  0.888
f  t1   f  0.7745  
f  t2   f 0 
1
 3  0.7745 
3
8
 0.0525
1
 0.0285
35
f  t 3   f  0.7745  
1
 3  0.7745 
3
8
 0.0162
I  4 0.5555  0.0525  0.555  0.0285  0.888  0.0162 
=4 0.0292+0.0158+0.0144 
=0.0594  4
=0.2376
2 2
dxdy
Using the trapezoidal rule with h = k = 0.5 and h =
x+y
1 1
17. Evaluate to integral I =  
k = 0.25
Solution:
When h = k = 0.5
Y0=1
y
Y1=1.5
Y2=2
x
X0=1
f01 = 0.4
f00=0.5
X1=1.5
f10=0.4
X2 = 2
f02=0.33
f11=0.33
f12=0.285
f21=0.285
f20=0.33
f22=0.25
hk
f00  f02  f21  f22  2  f01  f10  f21   4  f11 
u 
1
=  0.5  0.33  0.25  0.33   2  0.4  0.4  0.285   0.285  4  0.33 
16
=0.3418
I
When
61
Y0=1
y
Y1=1.25
Y2=1.5
Y3=1.25
Y4=2.0
x
X0=1
f(1,1)
f(1,1.25)
f(1.25,1.25)
f(1,1.5)
f(1.25,1.5)
f(1,1.75)
f(1,2.0)
f(1.25,1.75)
f(1.25,2)
f(1.25,1)
X1=1.25
f(1.5,1.25)
f(1.5,1.5)
f(1.5,1.75)
f(1.5,2)
f(1.5,1)
X2=1.5
f(1.75,1.25)
f(1.75,1.5)
f(1.75,1.75)
f(1.75,2)
f(1.75,1)
X3=1.75
f(2,1)
f(2,1.25)
f(2,1.5)
f(2,1.75)
f(2,2)
X4=2.0
I
1
f 1,1  f 1,2  f  2,1  f  2,2   2f 1,1.25   f 1,1.5 
65
+f 1,1.75   f 1.25,1  f 1.5,1  f 1.75,1  f  2,1.25 
+f  2,1.5   f  2,1.75   f 1.25,2   f 1.5,2   f 1.75,2 
4 f 1.25,1.25   f 1.25,1.5   f 1.25,1.25   f 1.5,1.25 
+f 1.5,1.25   f 1.5,1.5   f 1.5,1.75  
f 1.75,1.25   f 1.25,1.5   f 1.25,1.25 
 0.3401
4.4 2.6
18. Using Trapezoidal and Simpson’s rule evaluate I =

4
2
dydx
xy
Solution:
Taking h = 0.3 along x direction and k =0.2 along y direction we can construct the following
table where
f  x,y  
1
xy
62
2.0
2.3
2.6
X0=4.0
f(x0,y0)=0.125
f(x0,y1)=0.108
f(x0,y2)=0.0961
X1=4.2
f(x1,y0)=0.119
f(x1,y1)=0.1035
f(x1,y1)=0.0916
X2=4.4
f(x2,y0)=0.1236
f(x2,y1)=0.0988
f(x2,y2)=0.0874
y
x
I
0.2  0.3
0.1250  0.0961  0.1136  0.0874
4
+2  0.1190+0.1087+0.0916+0.0988   4  0.1035  
 0.0250
using Simpson’s rule
hk
 f00  f20  f22  4  f10  f01  f11  f21   16f11 
9 
0.2  0.3

0.1250  0.0961  0.1136  0.0874  4(0.1087
9
+0.1190+0.0916+0.988)+16 0.1035  
I
=0.0250
19. If D, E,  and  be the operators with usual meaning and if hD = u where h13 the
interval at differencing. Prove that the following relations between the operator’s.
(i) E= eu
 2
(iii) μ = cosh u 
2
(ii) δ = 2sinh u
(iv) (E+1) =2(E-1)
Solution:
(i) E = ehD
=Eu (hD=u)
(ii) 2 sin h u
2
 eu 2  eu 2 

 2
2


 
 Eu
1
2
 
 Eu
1
2
   by dt 
63
(iii) cos u
2

1 E
2
E 

u

E
1
2
u
1
2
E
2
u
 
 Eu

2
1
2
2
E
2
1
2

(iv)
E  1   E  1
=E
1
2

E
E
1
2
1
2
E
E
1
2
1
2
E

1
2
E
1
2

 E 12  E 12 

= E-1 2 


2


=2 E-1 
20.
12500 = 111.8034,
Given that
125110 = 111.8481
12516 .
find the value of
The difference table
X
12500
y
x
y
2y
111.8034
0.0471
12510
111.8481
0
0.0447
12520
111.8928
0
0.0447
12530
111.9375
We have x0 = 12500, h = 10, x = 12516
u
x  x0 12516  12510

 1.6
h
10
64
u  u  1
f  x   y 0  uy 0 
 2 y 0  .....
21
f 12516   111.8034  1.6  0.0447
=111.8034+0.07152
=111.87492
12516  111.87492
21.Use Newton’s forward interpolation and find value of sin 52 from the following data.
Estimate the error.
X
45
50
55
60
Y=sinx
0.7071
0.7660
0.8192
0.8660
2y
 3y
Solution:The difference table
X
Sin x
45
0.7071
y
0.0589
50
0.7660
-0.0057
0.0532
55
0.8192
-0.0007
-0.0064
0.0462
60
0.8660
We have x0 = 45, x1 = 52, y0= 0.7071, y0 = 0.0589,
2y0 = -0.0057, 3y0 = -0.0007
u
x  x0 52  45

 1.4
h
5
Newton’s formula
y  u0  uy 0 
u  u  1
2!
2 y0 
u  u  1 u  2 
f  52   0.7071  1.4  0.0589  
+
3!
1.4 1.4  1
sin52  0.7880032
u u  1u  2 .... u  n 
n  1!
 0.0057 
2
1.4 1.4  11.4  2 
3!
=0.7071+0.8246-0.001596+0.000392
Error 
 3 y 0  .....
n1y0
65
 0.0007 
using n = 2 we get


u u  1u  2 
3 y0
3!
1.4 1.4  11.4  2
6
22.The
y=
following
 0.0007   0.0000392
table
gives
the
values
of
the
e
π
-x2
dx corresponding to certain values of x. For what value of x is this integral
equation of to ½?
X
2
π
n
-x
 e dx
2
0.46
0.47
0.48
0.49
0.4846
0.4937
0.5027
0.5116
0
Solution:
Here x0=0.46, x1=0.47, y0 = 0.487 y = ½
From Laurens’s inverse interpolation formula
 y  y1  y  y 2  y  y3  x   y  y0  y  y 2  y  y 3  x
 y0  y1  y0  y 2  y0  y3  0  y1  y0  y1  y 2  y1  y3  1
 y  y0  y  y1  y  y3  x   y  y0  y  y1  y  y 2  x

 y 2  y0  y 2  y1  y 2  y3  2  y3  y0  y3  y1  y3  y 2  3
x

integral
n
2
0
y=
probability
 0.5  0.49  0.5  0.5274  0.5  0.51 0.46 


 0.48  0.49  0.48  0.52  0.48  0.51
 0.5  0.48  0.5  0.502  0.5  0.511 0.47 


 0.49  0.48  0.49  0.502  0.49  0.511
 0.5  0.48  0.5  0.49  0.5  0.511 0.48 


 0.52  0.48  0.52  0.49  0.52  0.511
 0.5  0.48   0.5  0.49  0.5  0.502  0.49


 0.51  0.48  0.51  0.49  0.51  0.502 
 0.0207787  0.157737  0.369928  0.0299495
 0.476937
66
UNIT - IV
INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL
EQUATIONS
Initial value problem:
A general solution of a differential equation of nth order has n arbitrary constants. It
will be of the form f(x,y,c1,c2,…cn) =0. if n conditions are given we can obtain the values of
the constants c1,c2,…cn. If all the n conditions are specified at the initial point only, then the
problem is called an initial value problem
Boundary value problem:
A general solution of a differential equation of nth order has n arbitrary constants. It
will be of the form f(x1, y1, c1, c2, ….cn) = 0. If n conditions are given we can obtain the
values of the constants c1, c2….. cn If n conditions are specified at more than one point, then
the problem is called a boundary value problem
Particular solution:
A most general from of an ordinary differential equation is given by Q (x, y,
dy d 2 y
dny
,
,.... n ) = 0 we know that the general solution of a differential equation of nth order
dx dx 2
dx
has n arbitrary constants. If we give particular values to the constants, the solution is said to
be a particular solution.
Formula for Taylor series:
Yn+1  yn 
h 1 h 2 II h3 III
yn  yn  yn  ......
1!
2!
3!
Formula for Euler’s method or Euler’s algorithm:
Yn+1= yn +h f (xn,yn), n = 0,1,2,…..
Formula for improved Euler’s method?:
Yn+1 =yn + ½ h [f (xn, yn)+ f (xn+h, yn +h f (xn, yn))]
Formula for modified Eulers method:
Yn+1 =yn +h[f(xn+h/2, yn +h/2 f (xn, yn))
67
Formula for fourth order Runge-kutta method:
K1= h f(x,y)
K2 = h f (x+ h/2, y+k1/2)
K3 = h f(x+h/2, y+k2/ 2)
K4 = h f (x+h, y+k3)
1
y  ( K1  2k2  2 K3  k4 )
6
y ( x  h)  y ( x)  y
Runge-kutta method for simultaneous first order differential equations:
To solve numerically the simultaneous equations
dy
dx
 f1( x, y , z ), and
dz
dx
 f2 ( x, y , z ) given the initial conditions y(x 0 )  y0 ,
z(r0 )  Z o
we starting from (x0, Y0, z0) the increments y and Z in y and z respectively are given by
formulae
K1=hf, (x0,y0,z0)
l1 = hf2 (xo, yo, zo)
k
l
k
l
h
h
K 2 = hf1 (x 0 + ,y 0 + 1 +Z0 + 1 ) l 2  hf 2 ( x0  , y0  1 , z0  1 )
2
2
2
2
2
2
k
l
k
l
h
h
K 3 = hf1 (x 0 + ,y 0 + 2 +Z0 + 2 ) l3  hf 2 ( x0  , y0  2 , z0  2 )
2
2
2
2
2
2 where h =  x
K 4 = hf1 (x 0 +h,y 0 +k 3 ,Z0 +l3 ) l 4  hf 2 ( x0  h, y0  k3 , z0  l3 )
1
1
y  (k1  2k2  2k3  k4 ) z= (l1  2l2  2l3  l4 )
6
6
y1=y0+y and z1=z0+z
having got (x1,y1,z1 ) we get (x2,y2,z2) by repeating the above algorithm once again starting
from (x1,y1,z1)
Runge-kutta method for second order differential equation (or R-K-method of order
from to solve y|| = f(x,y, y1), given y(x0)= y0 and y1(x0) =y01?
To solve yII =f(x,y,y1), given y(x0) =y0 y1(x0) = y1 0
Now, set y1=Z and y” = z1
Hence, differential equation reduce to
dy
 y1  z and
dx
68
dz
 z1  y " = f (x, y, y”) = f (x, y, z)
dx

dy
dz
 z and
 f (x, y, z) are simultaneous equation Where f1 (x, y, z) = z, f2 (x, y, z) = f
dx
dy
(x, y, z) given
Also y (0) and z (0) are given
Starting from these equations, we can use the R – K method for simultaneous equation and
solve the problem.
Milne’s predictor formula:
Yn+1, P = yn-3 +
4h
(2yn-21 – y1n-1 +2y1n)
3
Milne’s corrector formula:
Yn+1, C = y n -1 +
h 1
(y n-1 + 4y1n +y1n+1)
3
Adam – Bashforth predictor formula:
Yn+1, P = yn +
h
[55y1n - 59y1n + 37y1n-2 - 9 y1n-3]
24
Adam – Bashforth corrector formula:
Yn+1, C = yn +
h
[9y1n+1 + 19y1n – 5y1n-1 + y1n-2]
24
Relation between Runge – kutta method of second order and modified Euler’s method:
In second order Runge – kutta method,
k
h
y0 = k2 = hf  x 0  , y 0 + 1 

2
2
y0 = hf  x 0  h , y 0 + 1 h f (x 0 , y 0 ) 

2
2
y1 = y0 +y0 + y0 + hf

h
h


f (x 0 , y 0 ) 
 x 0  , y0 +
2
2


This is exactly the modified Euler method
So, the Runge – kutta method of second order is nothing but the modified Euler method.
69
Numerical Examples:
01. Using Taylor series method, find correct to four decimal places, the values of y (0.1),
given
dy
= x2 +y2 and y (0) = 1
dx
Solution:
We have y1 = x2 + y2
Yii = 2x + 2yy’
Yiii = 2 +2yy” +2’2
Yiv = 2yyiii + 2yiyii + 4yiyii
= 2yyiii + 6yiyii
x0 = 0, y0 =1, h = 0.1
x1 = 0.1, y1 = y (0.1) =?
Y01= x02 + y02 = 0 +1 = 1
Y0ii = 2x0 + 2y0y01 =2
Y0iii = 2 + 2(1) (2) + 2 (1)2 = 8
Y0iv = 2  1  8 + 6 (1) (2) = 28
By Taylor series method
Y1 =y0 +
h 1
h 2 2 h3 3
y0 
y0 
y 0  ......
1!
2!
3!
Y (0.1) = y1 = 1+
0.1
(0.1) 2
(0.1)3
(0.1) 4
(1) 
(2) 
(8) 
(28)  ....
1
2
6
24
= 1 + 0.1 +0.01 +0.0013333 + 0.000116666
= 1.11144999
= 1.11145
02. Using Taylor series method, find y (1.1) correct to four decimal places given
dy
=xy1/3 and y (1) = 1
dx
Solution:
Take x0 =1, y0 = 1, h =0.1
Y1 = xy1/3
Yii =
1 -2/3 1
xy y + y2/3
3
70
=
1 2 -1/3
x y + y1/3
3
yiii =
x 2  1   43 1 2 x  13 1  23 1
y  y y

y y 
3  3 
3
3
y01 = 1 (1)1/3 =1
By Taylor series Y1 = y (1.1) = 1+0.1 +
(0.2)2  4  (0.1)3  8 
 
   ......
2 3
6 9
= 1+0.1 + 0.00666 + 0.000148 + …….
= 1.10681
03. Using Taylor series method, find y (0.1) given
dy
= x2 – y, y (0) = 1 (correct to 4
dx
decimal places)
Solution:
X0 = 0, y0 = 1, h =0.1, x1 =0.1
Y1 = x2 – y
Yii = 2x – y1
Yiii = 2 - yii
Yiv = - yiii
Y01 = x02 – y0 = 0 -1 = -1
Y011 = 2x0 – y01 = 0 – (-1) =1
Y0iii = 2 -1 = 1
Y0iv = - 1
y (o.1) = 1+0.1 (-1)+
0.01
(0.001)
(0.0001)
(1) 
(1) 
(1, .....)
2
6
24
=0.905125
04. Given y1 = - y and y (0) = 1, determine the value of y at x = (0.01) (0.01) (0.04) by
Euler method
Solution:
Y1 = - y, x0 =0, y0 =1, x1 = 0.01, x2 = 0.02, x3 = 0.03, x4 = 0.04
We have to find y1, y2, y3, y4 takes h = 0.01
By Euler algorithm, yn+1 = yn + hyn1 = yn + hf (xn, yn)
Y1 = y0 + h f (x0, y0) = 1 + (0.01) (-1) = 0.99
Y2 = y1 + hy11 = 0.99 + (0.01) (-y1)
71
= 0.99 + (0.01) (-0.99)
=0.9801
y3 = y2 +hf (x2, y2) = 0.9801 + (0.01) (-0.9801)
= 0.9703
y4 = y3 +h f (x3, y3) = 0.9703 + (0.01) (-0.9703) = 0.9606
05. Compute y at x = 0.25 by modified Euler method given y1 = 2xy, y (0) = 1
Solution:
Here f (x, y) = 2xy, x0 = 0, y0 = 1
Take h = 0.25, x1 = 0.25
By modified Euler method
h
h


Y1 = y0 + h [f  x 0  , y 0 + f (x 0 , y 0 ) 
2
2


f (x0, y0) = f (0, 1) = 2 (0) (1) = 0
y1 = 1 +0.25 [6 (0.125, 1)]
= 1+0.25 [2  0.125, 1]
=1 = 0.25 [2  0.125 1]
= 1.0625
06. Solve
dy
=-2x – y, y (0) = -1 by Taylor series method to find y (0.1) compare it with
dx
exact solution?
Solution:
Here x0 = 0, y0 = -1, h = 0.1
Y1 = -2x –y
Yii = -2 –y1
Yiii = - yii
Yiv = - yiii
Y01 = -2x1 –y0 = 1
Y011 = -2 – 1=-3
Y0iii = 3
Y0iv = -3
y1 =1+
0.1
(0.1) 2
(0.1)3
(0.1) 4
1 
 (3) 
3
 (3)  ....
1!
2!
3!
4!
= 1+ 0.1 -0.015 +0.0005 – 0.0000125
= -0.91451
72
07. Solve
dy
=x (1+x3y), y (0) = 3 by Euler’s method for y (0.1)
dx
Solution:
X0 = 0, y0 = 3, h = 0.1, x1 = 0.1
By Euler’s algorithm is y1 = y0 + hf (x0, y0)
= 3 +0.1 f (0, 3) = 3 + 0.1(0)
=3
08. Solve
dy
= 2x +3y, y(0) = 1 by Euler’s method for y (0.1), y (o.2)
dx
Solution:
X0 =0, y0 = 1, x1 = 0.1
By Euler algorithm, y1 = 1+0.1 [2  0 + 3 1] = 1.3
Y2 =y1 + hf (x1, y1)
= 1.3 + 0.1f [0.1, 1.3]
= 1.3 + 0.1 [2  0.1 + 3  1.3] = 1.71
09. Obtain the values of y at x = 0.1 using Runge – kutta method of fourth order for the
differential equation y1 = - y, given y (0) = 1
Solution:
Here f (x, y) = - y, x0 = 0, y0 = 1, x1 = 0.1
K1 = hf (x0, y0) = 0.1 f (0, 1) = -0.1
k
h
K2 = hf (x0 + , y0 + 1 ) = (0.1) f (0.05, 0.95) = - 0.095
2
2
K 
h

K3 = hf  x 0 + , y 0 2  = (0.1) f (0.05, 0.9525) = -0.09525
2
2 

K4 = hf (x0+h, y0 +K3) = (0.1) f (0.1, 0.90475) = - 0.090475
y =
1
(k1+2k2 + 2k3 + k4)
6
y1 = y0 + y = 0.9048375
10. Compute y (0.3) given
dy
+y+xy2 = 0, y (0) = 1 by taking h = 0.1 using R.K method of
dx
fourth order?
Solution:
Y1= - (xy2 +y) = f (x, y), x0 = 0, y0 =1, h =0.1 x1 = 0.1
K1 =h f (x0, y0) = 0.1 [- (x0y02 + y0)] = -0.1
K2 = hf
k 
h

, y0  1 
 x0 
2
2 

= -0.1 [(0.05) (0.95)2 + 0.95] = -0.0995
73
K3 = hf
k2  =
h

 x 0  , y0 

2
2 

 y1 =1 +
(0.1) f (o.1, 0.9005) = - 0.0982
1
[-0.1 +2 (-0.0995) + 2 ( -0.0995) – 0.0982]
6
= 0.9006
11. What are the values of k1 and l1 to solve y11 +xy1 + y = 0; y (0) =1, y1 (0) =0 by Runge
kutta method of fourth order
y11 = -xy1 – y, x0 = 0, y0 =1
Setting y1 = z, the equation becomes y11 =z1 = -xz – y

dy
dz
= z = 6, (x, y, z),
=-xz – y = f2 (x, y, z)
dx
dx
given y0 =1, z0 = y01 = 0
By algorithm, k1= hf1 (x0, y0, z0) = 0.1 f1 (0, 1, 0) = 0
L1 = hf2 (x0, y0, z0) = 0.1 f2 (0, 1, 0) = -1 (0.1) = -0.1
12. What are the values of k1 and l1 solve y11 +2xy1-4y=0 ,y(0)=0.2,y1(0)=0.5.
Solution:
dy
Let
 z then
dx
dz
d
2
dx
2
 2 xz  4 y now
dx

dy
the given differential equation becomes
dx
dy
 z and
dx
dz
 2 xz  4 y
dx
x0  0, y0  0.2 h  0.2 f1(x,y,z)=z, f2(x1,x2,x3)=-2x2+4yK1=hf1(x0,y0,z0)=0.10.5=0.05,
l1 =ht2(x0,y0,z0)=0.1[-200.5+4.5]=0.8
13. What are the values of k1 and l1 to solve y11 – x2y1 – 2xy = 1 y (0) = 1, y1 (0) = 0
Solution:
Let
dy
=z
dx
The given differential equation becomes
d2y
= x2y1 +2xy+1
2
dx
dz 2
=x z +2xy +1, x0 =0, y0= 1, z0 =0, f1 (x, y, z) = z
dx
f2 (x, y, z) = x2z +2xy +1, h = 0.1
x1 =hf1(x0 , y0, z0)= 0.1 f (0, 1, 0) = 0.1  0 = 0
l1 =hf2 (x0, y0, z0) = 0.1
74
14. What are the values of k1, k2, l1 and l2 from the system of equations,
dy
dz
= x +z,
=
dx
dx
x – y given y (0) =2, z (0) = 1 using Runge – Kutta method of fourth order.
Solution:
f1 (x, y, z) = x + z; f2 (x, y, z) = x –y
X0 = 0, y0 =2, z0 =1, h = 0.1
Now
K1 = hf1 (x0, y0, z0)
= (0.2) f1 (0, 2, 1)
= (0.1) (0+1)
= 0.1
K2 = hf1  x 0  h , y 0 + k1 , z 0  l1 

2
2
2
= 0.1 f1 (0.05, 2.05, 0.8)
= 0.085
l1 = (0.1) f2 (0, 2, 1)
= (0.1) (0 -22)
= - 0.4
l2 =hf2
k1
l 
h

, z0  1 
 x 0  , y0 +
2
2
2

= (0.1) f2 (0.05, 2.05, 0.8)
= - 0.41525
15. Solve by Euler’s method
dy
= x2 + y, y (0) = 1 of x = 0.02, 0.04
dx
Solution:
Here x0 = 0, y0 =1, f (x, y) = x2 + y, h = 0.2
By Euler’s algorithm, y1 = y0 + h f (x0, y0)
i.e. y1 =1 + 0.02 (x02 + y0) = 1.02
y2 = y1 + hf (x1, y1)
= 1.02 +0.02 [(0.02)2 + 1.02]
=1.04041
16. Solve
dy
=x + y, given y (1) = 0 and get y (1.1) by Taylor series method?
dx
Solution:
Here x0 = 1, y0 =0, h = 0.1
Y1 =x + y
75
Yii = 1+y1
Yiii = yii
Yiv = yiii
Y01 = x0 + y0 = 1 + 0 =1
Y011 = 1 +y01 = 2
Y0iii = 2
Y0iv =
By Taylor series, we have
Y1 =y0 +
h 1 h 2 11 h 3 111
y0  y0  y0  ....
1!
2!
3!
Y1 = y (1.1) = 0 +
0.1
(0.1) 2
(0.1)3
(0.1) 4
(1) 
2
2
 2  ......
1
2
6
24
= 0.11033847
17. Using Taylor method, compute y (0.2) correct to 4 decimal places given
and y (0) = 0
Solution:
Here x0 = 0, y0 = 0, h =0.2
Y1 = 1 – 2xy
Y11 = -2 (xy1 + y)
Yiii = -2 [xy11 + 2y1]
Yiv = -2 [xyiii + 3y11]
Yv = -2 (xyiv + 4yiii]
Y01 = 1 – 2.0.0 = 1
Y011 = 0
Y0111 = -4
Y0iv = 0
Y0v = 32
By Taylor series,
Y1 =y (0.2) = 0 +
0.2
(0.2) 2
(0.2)3
(0.2)5
(1) 
(0)
( 4)  0 
(32)  ....
1
2
6
120
= 0.1948
76
dy
= 1 – 2xy
dx
18. Solve dy/dx = x+y, given y(1) =0, and get y(1.1), y(1.2) by Taylor series method.
Compare your result with the analysis.
Solution:
Here x0= 1,
yo =0x =0.1
Y1=x + y
y0I= x0+y0=1+0 = 1
yII = 1+y1
y0II=1xy01 = 2
yIII = yII
y0III = y0II = 2
yIV = yIII
y0iv = 2 etc
By Taylor series, are have
y1  y0 
h 1 h 2 II h3 III
y0 
y0  y0  ......
1!
Z!
3!
 0.1 (2)  (0.1)3  (0.1)4 (2)
0.1
 y1  y (1.1)  0 
(1) 
1
2
6
24
2

0.15
.2  ....    (2)
120
= 0.1+0.01+0.00033+0.00000833+0.000000166+….
Y(1.1) = 0.11033847
Now, take x0 = 0.1103847
Now, take x0 = 1.1 h = 0.1,
y2  y1 
h 1 h 2 II h3 III h 4 IV
y1  yI  yI  y1  ....    (3)
1!
2!
3!
4!
we calculate y1I , y1II , y1III ,.....,
x1  1.1, y1  0.11033847
y1I  x1  y1  1.1  0.11033847  1.21033847
y1II  1  y1I  2.21033847
y1III  y1II  y1IV  y1V  ....  2.21033847
using in (3),
y2 = y(1.2) = 0.11033847 + 0.1 / 1 (1.21033847)
(0.1)2
(0.1)3
(0.1) h

(2.21033847) 
(2.21033847) 
(2.21033847)  ....
2
6
2h
= 0.11033847+2.21033847(0.005+0.0016666+….)
= 0.2461077
77
The exact solution
dy
 x  y is y = -x-1+2e x-1
dx
Y (1.1) = -1 1 -1 +2e 0.1
= 0.11034
y (1.2) = -1.2 – 1+ze0.2 = 0.2428
y (1.1) = 0.11033847
y (1.2) = 0.2461077
Exact values: y(1.1) = 0.110341876
Y(1.2) = 0.24280552
19.Using Taylor method compute y (0.2) and y(0.4) correct to 4 decimal places given
dy
 1  2 xy and y(0) =0
dx
Soln
We know y1=1-2xy
Here x0=0 y0 =0, h=0.2
yII = -2(xy1+y)
y01=1-zx0y0 = 1
yIII =-2(xyII+2yI)
y0II = 0
yIV = -2(xyIII +JyII)
yoIII =-4
yV = -2(xyiv+4yIII)
yivo =0
y0II =J 32
by Taylor series
h 1 h 2 II h3 III
yI  y0  y0  yo  y0  ....      (!)
I!
2!
3!
(0.2)  0.2 
(0.2)3 (0.2) 4 0 (0.2)5
y1  y (0.2)  0 
1
(0) 


(32)  ....
1
2
6
24
120
2
yI=Z-x
Z1 = x+y
yII =Z1-1
ZII = 1+y1
yIII = ZIIetc
ZIII = yII etc
By Taylor series, for y1and z we have
Y1 = y (0.1) = y0+hy01+
h 2 II h3 III
y0  y0  ......    (1)
2!
3!
78
And Z1=Z (0.1) = Z0 +hZ01+
h 2 II h3 III
Z 0  Z 0  .....(2)
2
6
Y0 = 1
z0 =1
Y01=Z0-x0= 1-0 = 1
z01 = x0 + y0 + 0 =1=1
Y0II =Z01-1=1-1 = 0
z0II = 1 + yo1 =1 + 1 =2
Y0III =z0II = 2
z0III = y0II = 0
Substituting in (1) and (2), we get z0IV = y0III = 2
Y1 =y (0.1) = 1 + (0.1) +
(0.01)
(0.001)
(0) 
2  .....
2
6
= 1 + 0.1 + 0.000333+…. =1.1007 (correct to 4 decimals)
z1 = z (0.1) = 1 + (0.1) +
(0.01)
(0.001)
0.0001
2
(0) 
 2  ....
2
6
24
=1+0.1+0.01 +0.0000083+….
=1.1100 (correct to 4 decimal places)
 y (0.1) = 1.1003 and z (0.1) = 1.1100
20. Solve
dy
dz
 z - x,
 y + x with y (0) = 1, z (0) = 1, by taking h = 0.1, to get y (0.1)
dx
dx
and z (0.1). Here y and z are dependent variables and x is independent.
Solution:
Y1 = z –x
and z1 = x + y
Take x0 = 0, y0 = 1
take x0 = 0, z0 = 1 and h = 0.1
Y1= y (0.1) =?
Z1 = z (0.1) =?
Using in (6)
Y1 =y (0.1) = 0 +
Y2 = y1 +
0.1
0.19
[1  0.9] 
 0.095
2
2
1
h [f(x1, y1) +f (x2, y1 +hf (x1, y1)] (7)
2
F (x1, y1) = 1 – y1= 1 – 0.095 = 0.905
F(x2, y1 + h f (x1, y1) = f (0.2, 0.095 + (0.1) (0.905)) = 0.8145
Using in (7) we get y2 = y (0.2) = 0.095 +
0.1
[0.905 +0.8145]
2
Y (0.2) = 0.18098
79
Y3 = y2 +
1
h [f (x2, y2) +6 (x3, x2 +h f (x2, y2))] (8)
2
Using in (8)
Y3 =y (0.3) = 0.18098 +
0.1
(0.81902 + 1 – 0.26288)
2
Y (0.3) = 0.258787
The values are tabulated
X
Modified
Improved
Exact solution
Euler
Euler
0.1
0.095
0.095
0.09516
0.2
0.18098
0.18098
0.18127
0.3
0.258787
0.258787
0.25918
Modified Euler and improved Euler methods give the same values come A to sin decimal
places.
21. Evaluate the values of y (0.1) and y (0.2) given yII – x (y1)2 +y2 = 0; y (0) =1, y1 (0) =0
by using Taylor series method?
Solution:
YII – x (y1)2 + y2 = 0
Put y1 =z  (1)
Hence the eqn reduces to z1 – xz2 +y2 = 0
 z1 = xz2 – y2  (2)
By initial condition, y0 =y (0) = 1, z0 = y01 = 0  (3)
Y1 = 0.2 – 0.00533333 + 0. 000085333
= 0.194752003
Now again starting with x = 0.2 as the starting value so, use again eqn (1)
Now y0 = 0.2, y0 = 0.194752003, h = 0.2
Y01 = 1 – 2x0y0 = 1 – 2(0.2) (0.194752003) = 0.9220992
Y0II = -2 (x0y01 + y0) = -2 [(0.2) (0.9220992) + 0.194752003]
= -0.758343686
y0III = -2 [x0y0II + 2y01]
= -2 [(0.2) (-0.758343686) +2 (0.9220992)]
= -3.38505933
y0IV = -2 [(0.2) (-3.38505933) + 3 (-0.758343686)]
= 5.90408585
80
Using eqn (1), again
Y2 = y (0.4) = 0.194752003 + (0.2) (0.9220992)
(0.2)2
(0.2)3
(0.2)4
(0.758343686) 
(3.38505933) 
(5.90408585) = 0.359883723
2
6
24
22. Using improved Euler method find y at x =0.1 and y at x =0.2 give
dy
2x
=y,
dx
y
y (0) = 1
Solution:
By improved Euler method,
1
Yn+1 = yn + h [f (xn, yn) +f (xn + h, yn + hf (xn, yn))]  (1)
2
1
y1 = y0 + h [f (x0, y0 + h f(x0, y0))] (2)
2
2x 0
=1–0=1
y0
f (x0, y0) = y0 -
f (x1, y0 + h f (x0, y0)) = f (0.1, 1.1) = 1.1 -
2  (0.1)
 0.91818
1.1
y (0.1) = y1 = 1 +
0.1
[1+0.91818] = 1.095909
2
y2 =y (0.2) =y1 +
1
h [f (x1, y1) + f (x2, x1 +hf (x1, y1))] (3)
2
f (x1, y1) = y1 -
2x1
2  0.1
=1.095909 1.095909
y1
= 0.913412
f (x2, y1 + h f (x1, y1) = f (0.2, 1.095909 + (0.1) (0.9134121)
= f (0.2, 1.18732) = 1.18732 -
2  0.2
=0.8594268
1.18732
Using in (3), y2 = 1.095909 +
0.1
[0.913412 + 0.850427]
2
= 1. 1841009
X
0
0.1
Y
1 1.095907
0.2
1.1841009
23. Apply the fourth order Runge – kutta method, to find y (0.2) given that y1 = x + y,
y (0) = 1
Solution:
81
Since h is not mentioned in the question we take h = 0.1
Y1 = x + y; y (0) = 1
f (x, y) = x+y, x0 = 0, y0 =1
x1 = 0.1, x2 = 0.2
By fourth order Runge – kutta method, for the first interative
K1 = h f (x0, y0) = (0.1) (x0 + y0) = (10.1) (0+1) = 0.1
K2 =h f (x0 +
1
1
h, y0 + k1)
2
2
= (0.1) f (0.05, 1.05) = (1.0) (0.05 +1.05) = 0.11
k3 = h f (x0 +
1
1
h, y0 + k2) = (0.1) f (0.05, 1.055)
2
2
= (0.1) (0.05 + 1.055) = 0.1105
k4 = hf (x0 +h, y0 + k3)
= (0.1) f (0.1, 11 05) = (0.1) (0.1 + 1.1105)
= 0.12105
y =
=
1
(k1 +2k2 + 2k3 + k4)
6
1
[0.1 + 0.22 + 0.2210 + 0.12105) = 0.110341667
6
y (0.1) = y1 = y0 +y = 1.110341667
1.110342
Now starting from (x1, y1) we get (x2, y2) again
Apply Runge kutta algorithm replacing (x0, y0) by (x1, y1)
K1 = h f (x1, y1) = (0.1) (x1+y1) = (0.1) (0.1 + .110342) = 0.1210342
h
1
K2 = h f (x1+ , y1 + k1) = (0.1) f (0.15, 1.170859)
2
2
= (0.1) (0.15 + 1.170859) = 0.1320859
h
1
k3 = hf (x1 + , y1 + k2) = (0.1) f (0.15, 1.1763848
2
2
= (0.1) (0.15 +1.1763848) = 0.13262848
k4 =hf (x1 + h, y1 +k3) = (0.1) f (0.2, 1.24298048)
= 0.144298048
Y (0.2) = y (0.1) +
= 1.110342 +
1
[k1 +2k2 + 2k3 + k4]
6
1
(0.794781008
6
Y (0.2) = 1.2428055.Correct to four decimals places, y (0.2) = 1.2428
82
24. Using the Runge – kutta method, tabulate the solution of the system
dz
= x-y, y = 0, z =1 when x = 0 at intervals of h = 0.1 from x = 0.0 to x =0.2.
dx
Solution:
Given f (x, y, z) = x + z, g (x, y, z) = x – y, x0 =0, y0 =0, z0 =1 and h = 0.1
To compute y (0.1) and z (0.1)
K1 = hf (x0, y0, z0)
L1 = hg (x0, y0, z0)
= h (x0 + z0)
= h (x0 – y0)
= (0.1) (0 +1) = 0.1
= (0.1) (0 – 0) = 0
k
l
h
K2 = hf (x0 + , y0 + 1 , z0 + 1 )
2
2
2
k
l 
h

L2 = hg  x0  , y0  1 , z0  1 
2
2
2


l 
h 
=h  x 0     z 0  1  
2 
2 


k 
h 
=h  x0     y0  2  
2 
2 


0.1   0  
= (0.1)  0 
  1  
2   2  


0.1  
0.1  
= (0.1)  0 
 0

2  
2  

=0.105
k
l 
h

K3 = hf  x0  , y0  2 , z0  2 
2
2
2

k
l 
h

L3 =hg  x0  , y0  2 , z0  2 
2
2
2


l 
h 
= h  x0     z0  2  
2 
2 


h 
kl 2  
= 4  x0     y0 

2 
2  


0.1   0  
= (0.1)  0 
  1  
2   2  


0.1  
0.105  
= (0.1)  0 
 0

2  
2  

= 0.105
= - 0.00026
K4 = hf (x0 + h, y0 + k3, z0 +l3)
L4 = hg (x0 +h, y0 +k3, z0 +l3)
= h [x0 +h) + (z0 + l3)
= h [x0 +h) – (y0 +k3)
= (0.1) [ (0+0.1) + (1 – 0.00026)
= (0.1) [(0 +0.1) – (0 +0.105)]
= 0.1099
= - 0.0005
83
dy
= x +z,
dx
y =
=
1
[k1 + 2k2 + 2k3 + k4]
6
1
z = [l1 +2l2+ 2l3 +l4]
6
1
[0.1+2 (0.105) + 2 (0.105) +0.1099)]
6
=
1
[0+0+2 (- 0.00026) -0.0005]
6
=0.1050
= 0.00017
Y1 = y0 + y
Z1 =z0 +z
= 0 + 0 .1050
= 1 – 0.00017
y (0.1) = 0.1050
z (0.1) = 0.9998
To compute y (0.2) and z (0.2)
Here x1 = 0.1, y1= 0.1050, z1 =0.9998
K1 = hf (x1, y1, z1)
L1 = hg (x1, y1, z1)
= h (x1 +z1)
= h (x1 – y1)
= (0.1) (0.1 + 0.9998)
= (0.1) (0.1 – 0.1050)
= 0.1099
= - 0.0005


K2 = hf  x1 

= h  x1 


k
l 
h
, y1  1 , z1  1 
2
2
2
l1  
h 
   z1   
2 
2 
= (0.1)  0.1 



L2=hg  x1 

=h  x1 

0.1  
0.0005  
   0.9998 

2  
2  
=h


0.1 
0.1099  
 0.105 

2 
2  
= -0.0099


k2
l2 
h

 x1  , y1  , z1  
2
2
2

L3 = hg  x1 

l2  
h 
  x1  2    z1  2  
 


= (0.1)
k1  
h 
   y1   
2 
2 
=(0.1)  0.1 
= 0.1149
K3 = hf
k
l 
h
, y1  1 , z1  1 
2
2
2

= h  x1 


0.1  
0.00099  
 0.1  2    0.9998  2  
 



k2  
h 
   y1   
2 
2 
=(0.1)  0.1 

k
l 
h
, y1  2 , z1  2 
2
2
2
0.1  
0.1149  
   0.1050 

2  
2  
= 0.1149
= -0.00125
K4 = hf (x1 + h, y1+k3, z1+l3)
L4 = hg [x1+h, y1+k3, z1 + l3]
= h [(x1 +h) + z1+l3)]
= h [(x1+h) – (y1 + k3)]
= (0.1) [(0.1+0.1) + (0.9998 – 0.00125)]
= (0.1) [(0.1 +0.1) – (0.1050 + 0.1149)]
= 0.1198
= -0.00199
84
y =
1
[k1 +2k2 + 2k3 +k4]
6
z =
=
1
[0.1099 + 2 (0.1149) + 2 (0.1149) + 0.1198)]
6
=
1
[0.1099+0.2298 + 0.2298 + 0.1198]
6
=
1
[l1 + 2l2 + 2l3 +l4]
6
1
[-0.0005 +2(-0.00049)+2 (-0.00125) –
6
0.001199]
=
1
[-0.0005 -0.00198 -0.00199]
6
=
1
[-0.0005 – 0.00198 – 0.00199
6
= 0.1149
= -0.00116
Y2 = y1 + y
Z2 = z1 + z
= 0.1050 +0.1149
= 0.9998 – 0.00116
= 0.2199
= 0.9986
y (0.2) = 0.2199
z (0.1) = 0.9986
X=0 X = 0.1
X = 0.2
Y
0
0.1050
0.2199
X
1
0.9998
0.9986
2
d2 y
 dy 
 x    y 2  0 using Runge – kutta method for x = 0.2 correct to 4
25. Solve
2
dx
 dx 
decimal places. Initial condition are x = 0, y =1, y1 = 0
Solution:
Given:
2
d2y
 dy 
 x   +y2 = 0  (1)
2
dx
 dx 
Put

dy
=z  (2)
dx
d2y d2
 (3)

dx dx
Substituting (2) and (3) in (1), we get
dz
= xz2 – y2
dx
Let
dz
=xz2 – y2 = g (x, y, z)
dx
85
dz
=xz2 – y2 = g (x, y, z)
dx
Also we are give that y0 = 0, y0 = 1, y01 = 0 (or) z0 = 0, h = 0.2
Now
K1 = hf (x0, y0, z0)
= hz0 = 0
l1 = hg (x0, y0, z0)
= h (x0z02 – y02)
= (0.2) (0 -1) = - 0.2
k
l 
h

k2 = hf  x0  , y0  1 , z0  1 
2
2
2

l 

= h  z0  1  = (0.2)
2

0.2 

0
 = - 0.02
2 

k
l 
h

l2 = hg  x0  , y0  1 , z0  1 
2
2
2

2
2

l  
k  
h 
= h  x0   z0  1    y0  1  
2 
2 
2  

2
2

0.2 
0.2   0  
l2 = (0.2)  0 
 0 
  1   
2 
2   2  

= -0.1998
k
l 
h

k3 = hf  x0  , y0  2 , z0  2 
2
2
2

l 

= h  z0  1  = (0.2)
2

0.1998 

0

2 

= - 0.01998
k
l 
h

l3 = hg  x0  , y0  2 , z0  2 
2
2
2

2
2

l2  
k2  
h 
= h  x0   z0     y0   
2 
2 
2  

2
2

0.2 
0.01998   0.02  
= (0.2)  0 
 0 
  1 
 
2 
2
2  
 

= - 0.1958
k4 = hf (x0 +h, y0 + k3, z0 + h2)
= h (z0 +l3) = (0.2) (0 -0.1958)
= - 0.0392
l4 = hg (x0 +h, y0 + k3, z0 +l3)
86
= h [(x0 +h) (z0 + l3)2 – (y0 + k3)2]
= (0.2) [(0.2) (0 – 0.1958)2 – (1 – 0.01998)2]
= - 0.1906
y =
=
1
[k1 + 2k2 +2k3 +k4]
6
1
[0 +2 (-0.02) + 2 (-0.01998) – 0.0392]
6
= - 0.0199
y (0.2) = y1 = y0 +y
=1 – 0.0199
= 0.9801
y (0.2) = 0.9801
26. The differential equation
dy
= y –x2 is satisfied by y (0) = 1, y (0.2) = 1.12186, y (0.4)
dx
= 1.46820, y (0.6) = 1.7379 compute the value of y (0.8) by Milne’s predictor corrector
formula?
Solution:
Given
dy 1
=y = y – x2 and h = 0.2
dx
X0 = 0
y0 =1
X1 =0.2
y1 = 1.12186
X2 = 0.4
y2 = 1.46820
X3 = 0.6
y3 = 1.7379
X4 = 0.8
y4 =?
By Milne’s predictor formula, we have
Yn+1, P = y n-3 +
4h
[zyn-21 – y1n-1 + 2y1n]  (1)
3
To get yn, put n = 3 in (1) we get
Yn, P =y0 +
4h
[2y11 – y12 +2y13]  (2)
3
Now y11 = (y – x) 12 = y1 – x12
= 1.12186 – (0.2)2 = 1.08186  (3)
87
y21 = (y – x2)2 = y2 – x12
= 1. 46820 – (0.4)2 = 1.3082  (4)
y31 = (y –x2)3 = y3 – x32
=1.7379 – (0.6)2 = 1.3779  (5)
Substituting (3), (4) and (5) and (2), we get
Yh, g = 1+
4(0.2)
[2(1.08186) – 1.3082 + 2 (1.3779)]
3
= 1+0.266 [2.1637 – 1.3082 + 2.7558]
= 1.9630187
y (0.8) = 1.9630187 (by predictor formula)
By Milne’s corrector formula we have
Yn+1, C = yn-1 +
h 1
(y n-1 +4y1n + y1n+1)
3
To get yh, put n = 3, we get
Yh, C = y2 +
h 1
(y2 +hy31+yn1)  (6)
3
Now yn1 = (y –x2) h = yh – xh2
= 1.96277 – (0.8)2
= 1.3230187  (7)
Substituting (4), (5), (7) in (6) we get
Y4, C = 1.46820 +
0.2
[1.3082 +4 (1.3779) + 1.3230187]
3
= 2.0110546
i.e. y (0.8) = 2.0110546
27. Using Taylor’s series method, solve
dy
=xy+y2, y (0) = 1 at x = 0.1, 0.2 and 0.3
dx
continue the solve at x = 0.4 by Milne’s predictor corrector method?
Solution:
Given y1 = xy +y2, and x0 = 0, y0 = 1 and h = 0.1
Now y1 = xy + y2
Y11= xy1 + y + 2yy1
YIII = xyII + 2y1 + 2yyII + 2y12
To find y (0.1)
By Taylor series we have
88
y (0.1) = y1 + hy01 +
h 2 II h3 III
y0 +
yo + …… (1)
2!
3!
y0II = (xy + y2)0 = (x0y0 + y02) = 1 ……. (2)
y0II = (xy1 + y + 2yy1)
y0II = (x0y01 + y0 + 2y0y01) = 3 ……. (3)
y0III = (xy0II + 2y1 + 2yyII + 2y12)0 = 10 …… (4)
Substituting (2). (3) and (4) in (1) we get
(0.1) 2
(0.1)3
Y (0.1) = 1 + 0.1 +
3+
 10
2
6
= 1 + 0.1 + 0.016 + 0.001666
y (0.1) = 1.11666
To find y (0.2)
By Taylor series we have
Y2 = y1 + hy11 +
h 2 II h3 III
y1 +
y1 + ….. (5)
2!
3!
Now y11 = (xy + y2) = x1y1 + y12
= (0.1) (1.11666) + (1.11666)2
= 0.111666 + 1.2469
= 1.3585 …… (6)
y1II = (xy1 + y + 2yy1)
= x1y11 + y1 + 2y1 y11
= (0.1) (1.3585) + 1.11666 + 2 (1.11666) (1.3585)
= 0.13585 + 1.11666 + 3.0339
= 4.2865 …… (6)
y1III = (xyII + 2yI + 2yyII + 2y12)
= (x1 y1II + 2y1I + 2yIyIII + 2y112)
= (0.1) (4.2865) + 2 (1.3585) + 2 (1.1167) (4.2865) + 2 (1.3585)2
= 0.4287 + 2.717 + 9.5735 + 3.6916
= 16. 4102 …… (8)
Substituting (6), (7) and (8) in (5) we get
Y (0.2) = 1.1167 + (0.1) (1.3585) +
(0.1) 2
(0.1)3
(4.2865) +
(16. 4102)
2
6
Y (0.2) = 1.1167 + 0. 13585 + 0. 0214 + 0.002735
Y (0.2) = 1.27668
To find y (0.3)
89
By Taylor series we have
Y3 = y2 + hy21 +
h 2 II h3 III
y2 + y2 + …. (9)
2!
3!
Now y21 = (xy +y2)2 = (x2y2 + y22)
= (0.2) (1.2767) + (1.2767)2
= 0.2553 + 1.6299
= 1. 8852 …… (10)
y2II = (xy1 + y + 2yy1)2
= x2y21 + y2 +2y2y21
= (0.2) (1.8852) + 1.2767 + 2 (1.2767) (1.8852)
= 0.33770 + 1.2767 + 4.8136
= 6.4674 …… (11)
y2III = (xyIII + 2y1 + 2yyII 2y12)2
= (x2y2II + 2y21 + 2y2y2II + 2y212)
= (0.2) (6.4674) + 2 (1.8852) + 2 (1.2767) (6.4774) + 2 (1.8852)2
= 1.2974 + 3.7704 + 16.5138 + 7.1079
= 28.6855
Substituting (10), (11) and (12) in (9), we get
0.12
(0.1)3
Y (0.3) = 1.2767 + (0.1) (1.8852) +
(6.4674) +
(28.6855)
2
6
= 1.2767 + 0.18852 + 0.0323 + 0.004780
= 1.5023
y (0.3) = 1.5023
We have the following values
X0= 0
y0 =1
X1 = 0.1
y1 = 1.11666
X2 = 0.2
y2 = 1.27668
X3 = 0.3
y3 = 1.50233
To find y (0.4) by Milne’s predictor formula
Yn+1, P = yn+3 +
4h
[2y1n-2 – y1n-2 + 2y1n] ….. (1)
3
Y31 = (xy +y2)3
= (x3y3 + y32)
= [(0.3) (1.5023) + (1.5023)2]
= 0.45069 + 2.2569
90
= 2. 7076
Putting n =3, we get
Y4, P = y0 +
=1+
4h
[2y11 – y21 + 2y31]
3
4(0.1)
[2 (1.3585) – 1.8852 + 2 (2.7076)]
3
= 1 + 0.1333 [2.717 – 1.0852 + 5.4152]
y4, P = 1.8329
To find y (.04) by Milne’s corrector formula
By Milne’s corrector formula we have
yn+1, C = yn-1 +
h 1
[y n-1 + 4yn1 + y1n+1] …… (14)
3
Now y41 = (x2 + y2)4 = (x4y4 + y42)
= [(0.4) (1.8327) + (1.8327)2]
= 0.7330 + 3.3588
= 4.0918
Putting n = 3 in (14) we get
Y4, C = y2 +
h 1
[y2 + 4y31 + y41]
2
Y4, C =1.27668 +
(0.1)
[1.8852  4(2.7076)  4.0918]
3
= 1.27668 + 0.0333 [1.8852 + 10.8304 + 4.0918]
= 1.8369
28. Solve and get y (2) given
dy 1
= (x + y), y (0) = 2
dx 2
y (0.5) = 2.636, y (1) = 3.595; y (1.5) = 4.968 by Adam’s
method?
Solution:
By Milne’s method, we have y01 =
1
(0 +2) =1
2
Y11 = 1.5680, y21 = 2.2975, y31 = 3.2340
By Adam’s predictor formula
Yn+1, P = yn +
h
[55yn1 – 59y1n-1 + 37y1n-2 – 9y1n-3]
24
91
y4, P = y3 +
=4.968 +
h
[55y1n – 59y21 + 37y11 – 9y10] ….. (1)
24
0.5
[55 (3.2340) – 59 (2.2975) + 37 (1.5680) – 9 (1)]
24
= 68708
y41 =
1
1
(x4 + y4) =
(2+6.8708) = 4.4354
2
2
By corrector, y4, C = y3 +
= 4.968 +
h
[9yn1 + 19y31 – 5y21 + y11].. (2)
24
0.5
[9 (4.4354) + 19 (3.234) – (2.2975) + 1.5680]
24
= 6.8731
29. Find y (0.1), y (0.2), y (0.3) from
dy
= xy + y2, y (0) = 1 by using Runge – kutta
dx
method and hence obtain y (0.4) using Adam’s method?
Solution:
f (x, y) = xy + y2, x0 = 0, x1 = 0.1, x2 = 0.2,
xy = 0.4, x4 = 0.4, y0 =1
k1 = hf (x0, y0) = (0.1) f (0, 1) = (0.1) 1 = 0.1
k2 = hf (0.05, y0 +
k1
) = (0.1) f (0.05, 1.05)
2
= (0.1) [(0.05) (1.05) + (1.05)2] = 0.1155
k3 = hf (0.05, y0 +
k2
) = (0.1) f (0.05, 1.0578)
2
= (0.1) [(0.5) (1.0578) + (1.0578)2]
= 0.1172
k4 = hf (x0 + h, y0 + k3)
= (0.1) f (0.1, 1.1172)
= (0.1) [(0.10 (1.1172) + (1.1172)2] = 0.13598
y1 = y0 +
1
[k1 + 2k2 + 2k3 + k4]
6
= 1.1169
y (0.1) = 1.1169
Again, start from y1
k1 = hf (x1, y1) = (0.1) f (0.1, 1.1169)
= 0.1359
92
k2 = hf (x1+
k
h
, y1 + 1 ) = (0.1) f (0.15, 1.1849)
2
2
= 0.1582
k3 = hf (0.15, y1+
k3
) = (0.1) f (0.15, 1.196)
2
= 0.16098
k4 = (0.1) f (0.2, 1.2779) = 0.1889
y2 = 1.1169 +
1
[0.1359 + 2(0.1582 + 0.16098) + 0.1889]
6
y (0.2) = 1.2774
Start from (x2, y2) to get y3
K1 = hf (x2, y2) = (0.1) f (0.2, 1.2774) = 0.1887
k
h
K2 = hf (x2 + , y2 + 1 ) = (0.1) f (0.25, 1.3718) = 0.2225
2
2
K3 = hf (x2 +
k
h
, y2 + 2 )
2
2
= (0.1) f (0.25, 1.3887) = 0.2274
k4 = hf (x3, y2 +
k3
) = (0.1) f (0.3, 1.5048)
2
=0.2716
y3 = 1.2774 +
1
[0.1887 + 2 (0.2225) + 2 (0.2274) + 0.2716] = 1.5041
6
Now we use Adam’s predictor formula
Y4, P = y3 +
h
[55y31 – 59y21 + 37y11 – 9y01] …. (2)
24
Y01 = x0y0 + y02 =1
Y11= x1y1 +y12 = 1.3592
Y21 = x2y2 + y22 = 1.8872
Y31 = x3y3 + y32 = 2.7135
Using (2)
Y4, P = 1.5041 +
0.1
[55 (2.7135) – 59 (1.8872) + 37 (1.3592) – 9 (1)]
2
= 1.8341
y14, P = x4y4 + y24 = (0.4) (1.8341) + (1.8341)2 = 4.0976
93
y4, P = y3 +
h
[9y41 + 19y31 – 5y21 + y11]
24
= 1.5041 +
0.1
[9 (4.0976) +19 (2.7135) – 5 (1.8872) + 1.3592)
24
= 1.8389
y (0.4) = 1.8389
30. Solve y1 =
y 2  x2
; y (0) = 1 by Runge – kutta method of fourth order to find y (0.2)
y 2  x2
Solution:
Y1 =f (x, y) =
y 2  x2
, x0 = 0, h = 0.2, x1 = 0.2
y 2  x2
f (x0, y0) = f (0, 1) =
1 0
=1
1 0
k1 = hf (x0, y0) = 0.2  1 = 0.2
k
h
k2 = hf (x0 + , y0 + 1 ) = (0.2) f (0.1, 1.1)
2
2
1.21  0.01 
= 0.2 
= 0.9167213
1.21  0.01 
k
h
k3 = hf (x0 + , y0 + 2 ) = (0.2) f (0.1, 1.0983606)
2
2
= 0.1967
k4 =hf (x0 +4, y0 +k3) = 0.2 f (0.2, 1.1967)
= 0.1891
y =
=
1
[k1 + 2k2 + 2k3 +k4]
6
1
[0.2 + 2 (0.19672) + 2 (1.1967) + 0.1891]
6
= 0.19598
y (0.2) = y1 = y0 + y = 1.19598
94
UNIT – V
BOUNDARY VALUE PROBLEMS IN ORDINARY AND PARTIAL DIFFERENTIAL
EQUATIONS
Classification of a partial differential equations:
A general two dimensional second order partial diff. equation is of the form.
 2u
 2u
 2u
u
u
A 2 B
C 2  D  E
 Fu  G
x
xy
y
x
y
where A, B, C, D, E, F, G are functions of x and y only.
This equation is said to be
(i)
elliptic if
(ii)
Parabolic if
(iii)
Hyperbolic
B 2  4 AC  0
B 2  4 AC  0
B 2  4 AC  0
Standard five point formula (SFPF)
ui , j 
1
ui 1, j  ui 1, j  ui , j 1  ui , j 1 
4
Diagonal five point formula (DFPF)
ui , j 
1
ui 1, j 1  ui 1, j 1  ui 1, j 1  ui 1, j 1 
4
Explicit formula for Bender – Schmidt method:
ui , j 1  ui 1, j  (1  2 )ui , j  ui 1, j , where 
it is valid if 0< 
1
2
Bender-Schmidt recurrence equation :
ui , j 1 
1
1
K
ui 1, j  ui 1, j  , where    2
2
2 ah
a
is K= h 2
2
Formula for Poisson equation:
ui 1, j  ui 1, j  ui , j 1  ui , j 1  4ui , j  h2 f (ih, jh)
95
k
ah 2
Crank-Nicholson difference scheme or methods:
 ui 1, j  ui 1, j  ui 1, j 1  ui 1, j 1   2(  1)ui , j  2(  1)ui , j 1
Explicit scheme or Explicit formula for hyperbolic equations:
ui , j 1  2(1   2 a2 )ui , j   2 a2 (ui 1, j  ui 1, j )  ui , j 1 I
Standard five point formula is showing laplace equation over a region.
ui , j 
1
ui 1, j  ui 1, j  ui , j 1  ui , j 1 
4
Numerical Examples:
01. Classify the partial differential equation xfxx+yfyy=0, x>0,y>0
Here A=x, B=0, C=y
 B 2  4 AC  0  4 xy  0
The given pde. Is elliptic
02.Classify the pde fxx+2fxy+4fyy=0
Here A=1,B=2, C=4
B 2  4 AC  4  16  12  0
 The given pde is elliptic
03. Classify the pde fxx-2fxy+fyy=0
Here A=1, B=-2, C=1
B 2  4 AC  4  4  0
 The given pde is parabolic.
04. Classify one dimensional wave equation.
The one dimensional wave equation is
2
 2u
2  u

a
t 2
x 2
here A=a2, B=0, C=-1
Now B2 -4Ac=0-4a2(-1) =4a2>0
The dimensional wave equation is hyperbolic.
96
05. Classify one dimensional heat equation.
The one dimensional heat equation is
u
 2u
 a2 2
t
x
here A=a2, B=0, C=0
Now B2-4Ac=0-0=0
The dimensional wave equation is parabolic.
06. Classify Laplace equation is two dimensions Laplace equation is
Here A=1, B=0, C=1
Now B 2  4 AC  0  4.1.1  4  0
Laplace equation is elliptic.
07. Classify Poisson’s equation.
The Poisson’s equation is
 2u  2u

 f ( x, y )
x 2 y 2
Here A=1, B=0, C=1
Now B 2  4 AC  0  4.1.1  4  0
 Poisson equation is elliptic.
08. Classify
 2u
 2u  2u

2

0
x 2
xy y 2
Here A=1, B=2, C=1
Now B 2  4 AC  4  4  0, x, y
Hence, the given equation is parabolic.
09. Classify x2fxx+(1-y2)fyy=0
Here A=x2, B=0, C=1-y2
Now,
B 2  4 AC  0  4 x 2 (1  y 2 )
=4 x 2 ( y 2  1)
For all x except x=0,x2 is +ve
If -1<y<1, y2-1 is negative
 B 2  4 AC is negative if -1<y<1, x 0
For   x  ( x  0), 1  y  1, the given equation is elliptic.
For   x  , x  0, y  1or  y  1, the equation is hyperbolic
For x=0 for all y or for all x, y=I1 the equation is parabolic.
97
 2u  2u

0
x 2 y 2
10. Classify Uxx+4uxy+(x2+4y2)uyy=sin(x+y)
Here A=1, B=4, C=x2+4y2
Now, B 2  4 AC =16-4.1.(x2+4y2)
= 4(4- x2+4y2)
The given equation is elliptic if 4- x2+4y2 <0
Is x2+4y2>4
x2 y 2

1
4 1
x2 y 2
If is elliptic is the region outside the ellipse

1
4 1
x2 y 2
It is hyperbolic inside the ellipse

1
4 1
x2 y 2
It is parabolic on the ellipse

1
4 1
11. Classify (x+1) Uxx – 2(x+2)uxy +(x+3)uyy=0
Here A= x+1, B=-2(x+2), C=x+3
Now, B 2  4 AC = 4(x+2)2-4(x+1)(x+3)
 x 2  16  16 x  x 2  12 x  4 x  12
40
The given equation is hyperbolic at all points of the region.
12. Classify fxx-2fxy=0,x>0, y>0
here A=1. B=-2, C=0
Now, B 2  4 AC =4-0=4>0
Given pde is hyperbolic
13. Classify xfxx +fyy=0
Here A=x, B=0, C=1
Now, B 2  4 AC =0-4x=-4x
Given pde is (i) elliptic if x>0
(ii) Parabolic if x=0
(iii) hyperbolic if x<0
98
14. Given
 2u u
 , f (0, t )  f (5, t )  0 f ( x, 0)  x 2 (25  x 2 )
2
x
t
find f in the range taking h=1 and up to 2 seconds
Solution:
To use bender- Schmidt method, K 
Have a=1, h=1 K 
Step size of time =
ah 2
2
1
2
1
2
Step size of x=1
F(0,0)=0,f(1,0)=2h,f(2,0)=84,f(3,0)=144 f(4,0)=144, d(5,0)=0
We have ui , j 1 
1
ui 1, j  ui 1, j 
2
15.Using crank-Nicholson method, solve
u  2u

subject to u( x, 0)  0, u (0, t )  0 and u(1, t )  t
t x 2
1
taking h=0.5 and k=
8
1
k
1
Solution:Here a=1   2  8
ah 1 1 2
4
Since  
1
, we can not use simplified formula
2
we use formula
ui 1, j 1  ui 1, j 1  6ui , j 1  2ui , j ui 1, j  ui 1, j 
1
 0  6u,  2(0)  (0  0)
8
1
 u1 
 0.020833
48
99
16. Solve the differential equation
d 2x
t
 (1  ) x  t , x(1)=2, x(3)=-1.
2
dt
5
Solution:
Replace
x  2 xi  xi 1
d 2x
by i+1
2
dt
h2
the differential equation becomes
t
x  2 xi  xi 1
d 2x
- (1  i ) xi  ti
by i+1
2
2
5
dt
h
t 

xi 1   2  h 2 (1  i )  xi  xi 1  h 2ti ..........(1)
5 

Let us subdivide the interval (1,3) into 4 equal parts Note that the interval given in the
problem is (1,3).
T1=1, t2=1.5, t,=2,tn=2.5,t5=3
We write the difference equation (1) for the intermediates joints of ti At t=t2=1.5,i=2
2
  1 2  1.5  
1
X1 -  2+   1-   x2  x,    1.5 
2
  2   5  
At t  t ,  2.0, i  3
yi 1  2 yi  yi  1
2
 y i ...............(1)
2
  1 2  2.0  
1
X 2 -  2+   1  x3  xn     2.0 
2
  2   5  
At t  tn  2.5, i  4
2
  1 2  2.5  
1
X3 -  2+   1  x4  x5     2.5 
2
  2   5  
But the from the boundaries condition, we have x1=x(t1)=x(1)=2 and x5=x(+5)=x(3)=-1
putting there values of x1 and x5 in the above equation.
1


2   2   0.7  x2  x3  0.375
4


We get
1


x2   2   0.6  x3  x4  0.5
4


100
1


x3   2   0.5 x4  1  0.625
4


is  2.175 x2  x3  0.x4  1.625
x2  2.15 x3  x4  0.5
0 x2  x3  2.125 x4  1.625
solving the above there equation we get x2=0.552,
x3  0.424, x4  0.964
d2y
17. Solve
 y given y1(0)=0 and y(1)-1.
dx 2
Solution:
Replacing
d2y
by the different formula,
dx 2
yi 1  2 yi  yi 1
 yi ........(1)
h2
Let us divide the interval (0,1) into two parts so that
X1=0,x2=0.5,x3=1. Note that y1 is not given y3=1 using (1) at x1, (i=1)
Y0-2y1+y2=1/4 y1…….(2)
(Have y0 is not mentioned in the problem)
using (1) at x2,(i=2)
yi-2y2+y3=1/4y2
is y1-2y2+1=1/4y2……(3)
This we have a system of two equation is three unknown y0,y1 and y2.
We need one more equation to solve for y0,y1 and wy2 for this, we have to use the
other boundary condition is derivative, y1 (0)=0
Here x1=0
 y1(x1)=0
is 0=y1(x1) =
y ( x2 )  y ( x0 )
 1y ( x2 )  y ( x0 )or y 2  y0
2h
using this in (2), we have 2y2-2y1=1/4 y1
the system to be solved in 8 ½ -9y1=0 and turn (3), 4y1-9 ½ =-4.
Is -9y1+8 ½ =0
Hy1 -9 ½ =-4
Showing this, we get
y2 
36
32
32
36
and y1  is y(o)= and y (1) 
49
49
49
49
101
18. Solve the Laplace equation 2u=0 at the interior points of the square region given.
Solution:
Let u1,u2……u9 be the values of f at the interior lattice points.
1
u5  [17.0  12.1  0  21.0]  12.5[ SFPF ]
4
1
u1  [17.0  0  0  12.5]  7.4( DFPF )
4
1
1
u3  [18.6  12.5  17  21]  17.3[ DFPF ] u9  [21.0  12.1  12.5  9.0]  13.7[ DFPF ]
4
4
1
1
u7  [12.5  0  0  12.1]  6.2( DFPF ) u2  [17.0  12.5  7.4  17.3]  13.6( SFPF )
4
4
1
u4  [0  12.5  7.4  6.2]  6.5[ SFPF ]
4
1
u6  [17.3  17.7  12.5  12]  16.1( SFPF )
4
1
u8  [12.5  12.1  6.2  13.7]  11.1( SFPF )
4
We have got the rough values at all interior grid points. Now we can start Liebmanns
iteration process. We use only standard two point formula to evaluate the in.
First iteration:
1
1
u1(1)  [o  11.1  u2  un ]  [0  11.1  13.6  6.5]  7.8
4
4
1
u2(1)  [17.0  12.5  7.8  17.3]  13.7
4
1
u3(1)  [13.7  21.9  19.7  16.1]  17.9
4
1
u4(1)  [0  12.5  7.8  6.2]  6.6
4
1
u5(1)  [13.7  11.1  6.6  16.1]  11.9
4
1
u6(1)  [17.9  13.7  11.9  21.0]  16.1
4
1
u7(1)  [6.6  8.7  0  11.1]  6.6
4
1
u8(1)  [11.9  12.1  6.6  13.7]  11.1
4
1
u9(1)  [16.1  12.8  17.0  11.1]  14.3
4
102
Second iteration:
1
u1(2)  [o  11.1  13.7  6.6]  7.9
4
1
u2(2)  [17.0  17.9  7.9  17.9]  13.7
4
1
u3(2)  [13.7  19.7  21.9  16.1]  17.9
4
1
u4(2)  [7.9  0  11.9  6.6]  6.6
4
1
u5(2)  [13.7  11.1  6.6  16.1]  11.9
4
1
u6(2)  [11.9  17.9  21.0  14.3]  16.3
4
1
u7(2)  [0  6.6  11.1  8.7]  6.6
4
1
u8(2)  [6.6  11.9  14.3  12.1]  11.2
4
1
u9(2)  [11.2  16.3  17.0  12.8]  14.3
4
Third iteration:
1
u1(3)  [o  11.1  13.7  6.6]  7.9
4
1
u2(3)  [7.9  17.9  17.0  11.9]  13.7
4
1
u3(3)  [13.7  21.9  19.7  16.3]  17.9
4
1
u4(3)  [6.6  0  7.9  11.9]  6.6
4
1
u5(3)  [13.7  11.2  6.6  16.3]  11.95
4
1
u6(3)  [11.9  21.0  17.9  14.3]  16.3
4
1
u7(3)  [6.6  8.7  0  11.2]  6.6
4
1
u8(3)  [11.9  12.1  6.6  14.3]  11.2
4
1
u9(3)  [11.2  16.3  17.0  12.8]  14.3
4
103
The second and third iterated values are same. Thus we can stop the iteration. We can write
the iterated values of is in the table given below.
u1  7.9, u2  13.7, u3  17.9, u4  6.6u5  11.95, u6  16.3
u7  6.7, u8  11.2, u9  14.3
19.Solve 2u=8x2y2 in the square mesh given u=0 on the 4 boundaries dividing the
square into 16 sub – squares of length. 1 unit.
Solution:
Take the co-ordinate system with origin at the centre of the square.
Since the P.D.E. and boundary conditions are symmetrical about x, y axes and y=x
we have, u1= U3=u7=u9
U2=u4=u6=u8
we need to find u1= U2= U3=u5 only
(have h=1)
As per the five point formula, (have f(x.y)=8x2y2)
ui 1, j  ui 1, j  ui 1, j  ui , j 1  4ui , j  h2 f (ih, jh)  f (i, j )......(1)
At (i=-I,j=-1), is have,
u2  u4  4u1  8(1) 2 (1) 2  8 is u 2  2u1  4......(2)
At (i=0,j=1), is have,
u1  u3  u5  4u2  0
2u1  4u2  u5  0......(3)
At (i=0,j=0),
u2  u4  u6  u8  4u5  0
4u5  4u5  0
u2  u5  0......(4)
from (2), u1 
1
 u2  4 
2
from (4) u5=u2
using in (3),
u2  4  4u2  u2  0.
 u2  2
 u5  2, u1  3,
 u1  3, u2  2  u5 ,
104
20.Solve ut=uxx subject to u(0,t)=0,u(l,t)=0 and u(x,0)= sin nx, 0<x<1.
Solution:
Since h, and K are not given in will select then properly and use bender –schmidt
method. k 
a 2 1 2
h  h
2
2
a 1
since range of x is (0,1), takes h=0.2
hence K 
(0.2)
 0.02
2
the formula is ui , j 1 
u(o,o)=0,u(0.2,0)=sin
1
 ui1, j 1  ui1, j 
2

5
 0.5878
2
3
 0.9511;sin(0.6, 0,  sin
 0.9511
5
5
4
sin(0.8, 0)  sin
0.5878
5
sin(1, 0)  0
we form the table
u (0.4, o)  sin
21. Solve numerically, 4uxx=utt with the boundary conditions u(0,t)=0, u(4,t)=0 and the
initial conditions ut (x,0) and u(x,0)=x(4-x), taking h=1 (for 4 time steps)
Solution:
Since a 2  4, h  1, k 
h 1
 ........(1)
a 2
Taking k= ½ , we use the formula,
ui , j 1  ui 1, j  ui 1, j  ui , j 1......(2)
From u(0,t)=0  u along x= 0 are all zero
From u (e,t)=0  u along x=4 are all zero
U(x,0)=x(4-x) implies that u(0,0)=0, u(1,0)=3, u(2,0)=4, u(3,0)=3.
Now, we fill up the row t=0 using the above values,
Ut(x,o)=o implies ui ,1 
ui 1,0  ui 1,0
2
..........(3)
Now we draw the table ; for that we require
105
u1,1 
u2,1 
u3,1 
u2,0  u0,0
2
u3,0  u1,0
2
u4,0  u2,0
2
40
2
2
33

3
2

2
u4,1  0
1
Period is h seconds N 8  h   8    4sec.
2
 2u  2u
22. Solve 2  2 , 0  x  1, t  o, givenu ( x, 0)  ut  x, 0 
t
x
 u (0, t )  0andu (1, t )  100 sin t
compute u for 4 times steps with h=0.252
Solution:
Here, the two ends are not fixed
x=1 is a free end with displacement not equal to zero.
Since the differential equation is same, we have
ui , j 1  ui 1, j  ui 1, j  ui , j 1......(1)
if k 
h 0.25

 0.25sin ce a =1
a
1
step-size is 0.25 in both variable
range of x is (0,1)
At t=0, all u’s are zero
At x=0, all u’s are zero; since u(1,t)=100 sint, u(1,0)=0.
u (1, 0.25)  100sin
u (1, 0.5)  100sin


4
 70.7106
 100
2
3
u (1, 0.75)  100sin
 70.7106
4
u (1,1)  0
106
ut ( x, 0)  0 )ui ,1 
u1,1 
u2,0  u0,0
ui 1,0  ui 1,0
2
0
2
u2,1  0
u2,1  0
u4,1  70.7106  u (4, 0.25)
Please note u(x,t)  u(i,j)
since x=ih, t= jk
23. Solve, uxx +uyy=0 over the square mesh of 4 units, satisfying the following boundary
conditions.
(i ) u (o, y )  0, 0  y  4
(ii ) u (4, y )  12  y, 0  y  4
(iii) u ( x, 0)  3x,
0 x4
(iv) u ( x, 4)  x ,
0 x4
3
Solution:
Let the internal grid points be u1,u2, ……u9
Initial values
1
 4  6  0  14  6
4
1
u1   0  6  4  0  2.5
4
u5 
1
16  6  14  4  10
4
1
u7   0  6  0  6   3
4
u3 
u9 
1
6  14  6  12  9.5
4
1
 4  6  2.5  10  5.625
4
1
u4   0  6  2.5  3  3.125
4
Now, u2 
1
 6  14  10  9.5  9.875
4
1
u8   6  6  3  9.5  6.125
4
u6 
Let us start the iteration now.
107
After three iterations, we can see that u1=2.37,
u2  5.59, u3  9.87,
u4  2.88, u5  6.13, u6  9.88,
u7  3.01, u8  6.16, u9  9.51,
24. Solve the Poisson equation 2u=-(x+y)2 over the square region bounded by the lines
x=0,y=0,x=0,y=3 given that u=0 throughout the boundaries taking h=1.
Solution:
Here f(x,y)=-(x+y)2=-(i+j)2
The standard five point formula for solving Poisson equation.
ui 1, j  ui 1, j  ui , j 1  ui , j 1 4ui , j  h2 f (ih, jh)
Using the formula at A, (i=1,j=2)
0+0+u2+u3 -4u1=-(1+2)2=-9
u2+u3-4u1=-9-----------(1)
Using the formula is, (i=2,J=2)
u1+0+0+u4 -4u2=-(2+2)2=-16
u1+u4-4u2=16 -----------(2)
Using the formula at C, (i=1,j=1)
0+u1+u4 +0-4u3=-(1+1)2=-4
u1+u4-4u3=-4 -----------(3)
Using the formula at u1(i=2,j=1)
U3+u2 0+0 -4u4=-(2+1)2=-9
U3+u2-4u4=-9-----------(4)
From (1) and (4), u1=u4 from(2), u2 
from(3), u3 
1
u1  u4  16............(6)
4
1
u1  u4  4............(7)
4
(6) can be written as
u2 
1
1
 2u1  16 or u2  u1  2 .......(8)
4
2
(7) can be written as
u3 
1
1
 2u1  4 = u1  2 .......(9)
4
2
Now we have to solve u1 
1
1
1
u2  u3  9 , u2  u1  8 , u3  u1  2
4
2
2
108
Taking the initial values, for u1,u2,u3, as (0,0,0) and applying gauss seidel method, we get the
following table of approximation.u1=4.66,u2=6.33,u3 = 3.33, u4 = 4.66
u 1  2u
25.Solve

, 0  x,  12, 0  t  12 with boundary conditions and initial conditions.
t 2 x 2
1
x(15  x), 0  x  12
4
=u (o, t )  o, u (12, t )  9, u (o, t )  o, o  x  12
u ( x, 0) 
using Bender Schmidt relation using h=k=3
Solution:

k
3
1


2
ah
29 6
we use the explicit formula
ui , j 1  ui 1, j  (1  2 )ui , j  ui 1, j
1
1
1

= ui 1, j  1  2   ui , j  ui 1, j
6
6
6

1
2
= ui 1, j  ui 1, j   ui , j
6
3
12
0
6.604
10.639
11.104
9
9
0
7.083
11.292
11.583
9
6
0
7.625
12
12.125
9
3
0
8.25
12.75
12.75
9
0
0
9
13.5
13.5
9
t/x
0
3
6
9
12
1
u (3, 0)   3 15  3  9
4
1
u (6, 0)   6 15  6   13.5
4
1
u (9, 0)   9 15  9   13.5
4
109
26.Use crank-Nicholson method, solve ut=uxx, subject to u(x,o)=0,u(o,t)=o and u(1,t)=t,
1
1
taking h  , K  for one time step.
4
8
1
1
1
k
Solution:Here a  1, h  , K  ,   2  8  2
4
8
ah 1 1
6
we can not use the simplified formulas.
Using the explicit formula
  ui 1, j 1  ui 1, j 1   2( x  1)ui , j 1  2    1 ui , j   ui 1, j  ui 1, j 
  ui 1, j 1  ui 1, j 1   6ui , j 1  2ui , j  2  ui 1, j  ui 1, j 
ui 1, j 1  ui 1, j 1  3ui , j 1  ui , j  ui 1, j  ui 1, j
1/8
0
U1
U2
1/3
1/8
0
0
0
0
0
0
t/x
0
¼
½
¾
1
Ui-1,j+1
Ui-1,j+1
Ui-1,j+1
0  u2  u1  0
u1  u3  3u2  0
1
u2   3u3  0
8
Solving these equation, we get
U1=0.00595
U2=0.001786
U3=0.04762
27. Solve
t=
u  2u

with the conditions U(x,0)=0, u(o,t)=o and u(1,t)=t compute is for
t x 2
1
in two steps using crank –Nicholson formula.
8
Solution:
Uxx=ut
Here a=1; k=
1
choose h such that =1 Now
16
110
k
1
h2a
 h 2 a  k or h 2  k

the simplified crank-Nicholson scheme can be used.
ui , j 1 
1
ui 1, j 1  ui 1, j 1  ui 1, j  ui 1, j 
4
1/8
0
U4
U5
U6
1/8
1/16
0
U1
U2
U3
1/16
0
0
0
0
0
0
t/x
0
¼
½
¾
1
1
step, weget
16
1
At u1  (0  u2  0  0)
4
4u1  u2  0.........(1)
t
1
u2  (u1  u3  0  0)
4
4u2  u1  u3  .........(2)
1
1
(u2   0  0)
4
16
1
4u3  u2  ........(3)
16
u3 
solving (1), (2) and (3) we get u1=0.0011116,
u2=0.004464,u3=0.01674
At t=1/8 step, we have u4 = ¼[0+u5+0+u2]
4u4=u5+0.004464……….(4)
u5=1/4[u4+u6+u1+u3]
4u5=u4+u6+0.0011116+0.004464…….(5)
1
1
1
u4   u2  

4
8
16 
1
1
4u6  u4   0.004464  .........(6)
8
16
u6 
solving (4) ,(5) and (6) we get
U4=0.005899
U5=0.019132
U6= 0.052771
111
28. Solving 25 uxx-utt=0 for u at this pivotal points given
u(0,t)=u(5,t)=0, ut(x,0)=0 and
u(x,0)=2x for 0 x  2.5
=10-2x for 2.5  x  5 for one half period of vibration.
Solution:
2l 2  5

 2sec.
a
5
Here a2=25, a=5 period of vibration =
Half period =1 second.
Thus we have to find the values of u up to t= 1

k
1
h 1
   , K   for min gh  1
h
a
a 5
5/5
0
-2
-4
-4
-2
0
4/5
0
-2
-3
-3
-2
0
3/5
0
-1
-1
-1
-1
0
2/5
0
1
1
1
0
(0+3-2)
(2+3-4)
3+0-2
1/5
0
2
3
3
2
0
0
0
2
4
4
2
0
t/x
0
1
3
2
4
5
U(1) = 2x1=2
U(2)=2x2=4
U(3)=10-2x3=4
U(4)=10-2x4=2
Now we get the values of u at t= 1/5 using the relation
1
ui ,1  (ui 1,0  ui 1,0 )
2
1
40
ui ,1  (u2,0  u0,0 ) 
2
2
1
u2,1  (u3,0  u1,0 )  3
2
1
u3,1  (u4,0  u2,0 )  3
2
1
u4,1  (u5,0  u3,0 )  3
2
2
3
we have to get the values of u at t = , t  etc.
5
5
112
for this we use ui,j+1= ui+1,j+ ui-1,j+ ui,j
29. Solve
u  2u

in0  x,  15; t  o given that u(x,o)=20, u(0,t)=0, u(5,t)=100. compute
t x 2
u for one time step with h=1 by crank – Nicholson method.
Solution:
Here a=1, h=1
K=ah2 to use simple form k=1.12=1
ui , j 1 
1
ui 1, j 1  ui 1, j 1  ui 1, j  ui 1, j 
4
1
1
u1  [0  20  20  u2 ]  u1  (40  u2 ).....(1)
4
4
1
1
u2  [u1  20  20  u3 ]  u 2  (40  u1  u3 ).....(2)
4
4
1
1
u3  [u2  20  20  u4 ]  u 3  (40  u2  u4 ).....(3)
4
4
1
1
u4  [u3  20  20  100]  u 4  (140  u3 ).....(4)
4
4
form(1) 4u1-u2=40……..(5)
form(2) u1-4u2+u3=-40……..(6)
form(3) u2-4u2+u4=-40……..(7)
4  (6)  4u1  16u2  4u3  160.....(8)
(5)  (6)  15u2  4u3  200.....(9)
1
from(7)u2  4u3  (140  u3 )  40
4
4u 2  16u3  140  u3  160
4u 2  15u3  300......(10)
4  (9)  60u2  16u3  800.....(11)
15  (10)  60u2  225u3  4500.....(12)
(11)  (12)  20943  5300
 u 3  25.358852
From (10) 4u2=-300+380.38278
80.38278
u2 = 20.095695
From (5) 4u1=40+20.095695
=60.095695
113
u1 = 15.023924
1
(140  25.358852)
4
=41.339712
From(4)u4 
is the values are u1 = 15.023924,u2=20.095695
u3=25.358852, u4 = 41.339712
30. Solve
 2u  2u

 10( x 2  y 2  10) over the square mesh with sides x=0,y=0,x=3,y=3
x 2 y 2
with u=0 on the boundary and mesh length 1 unit
solution:
The P.D.E. in 2u=-101x2+y2+10) here h=1 ……..(1)
ui 1, j  ui 1, j  ui , j 1  ui , j  4ui , j  10(i 2  j 2  10)......(2)
Applying formula (2) at 1) (i=1,j=z)
0+0+u2+u3-4u1=-10(15)=-150
u2+u3-4u1=-150…..(3)
Applying (2) at E (i=2,j=2)
U1+u4-4u2=-180…..(4)
Applying (2) at F, (i=1, j=1)
U1+u4-4u3=-120…..(5)
Applying (2) at G, (i=2, j=1)
U2+u3-4u4=-10(22+12+10)=-150…..(6)
We can solve the equation (3), (4), (5), (6) either by direct
elimination or by Gauss-seidel method.
Method (1)
(5)-(4) gives, (Eliminate u1)
4(u2-u3)=60
u2-u3=15------------(7)
Eliminate u1 from (3) and (4);
(3)+4(4) gives, -15u2+u3+4u4=-870…….(8)
Adding (6) and (8) -7u2+u3=-510…….(9)
From (7),(9) adding, u2 = 82.5
Using (7), u3=u2-15=82.5-15=67.5
Put in (3), 4u1=300
u1=754u4=150+150; u4=75
u1=u4=75, u2=82.5,u3=67.5
114