Unit - 2

Numerical differentiation, integration and solution of differential equations

2.1 Numerical differentiation

Formulae for derivatives:

Newton’s forward Difference formula:

This method is useful for interpolation near the beginning of a set of tabular values.

    Where

Differentiating both side with respect to p, we get

h

This formula is applicable to compute the value of for non tabular values of x.

For tabular values of x , we can get formula by putting 

    Therefore

In similar manner we can get the formula for higher order by differentiating the previous order formulas

   Again, differentiating with respect to p, we get

Hence

Also

And so on.

Example1: Given that

Find at .

Here the first derivative is to be calculated at the beginning of the table, therefore forward difference formula will be used

Forward difference table is given below:

X	Y
1.0   1.1   1.2   1.3	0.841   0.891   0.932   0.962	0.050   0.041   0.031	-0.009   -0.010	-0.001

By Newton’s forward differentiation formula for differentiation

Here 

Example2: Find the first and second derivatives of the function given below at the point :

Here the point of the calculation is at the beginning of the table,

Forward difference table is given by:

X	Y
1   2   3   4   5	0   1   5   6   8	1   4   1   2	3   -3   1	-6   4	-10

By Newton’s forward differentiation formula for differentiation

Here  , 0.

Again 

At

Example3: From the following table of values of x and y find   for 

Here the value of the derivative is to be calculated at the beginning of the table.

Forward difference table is given by

X	Y
1.00   1.05   1.10   1.15   1.20   1.25   1.30	1.0000   1.02470   1.04881   1.07238   1.09544   1.11803   1.14017	0.02470   0.02411   0.02357   0.02306   0.02259   0.02214	-0.00059   -0.00054   -0.00051   -0.00047   -0.00045	0.00005   0.00003   0.00004   0.00002	-0.00002   0.00001   -0.00002	0.00003   -0.00003	-0.00006

From Newton’s forward difference formula for differentiation, we get

Here

     =0.48763

Newton Backward Difference Method:

This method is useful for interpolation near the ending of a set of tabular values.

Where 

Differentiating both side with respect to p, we get

This formula is applicable to compute the value of for non tabular values of x.

For tabular values of x, we can get formula by putting 

    Therefore

In similar manner we can get the formula for higher order by differentiating the previous order formulas

Differentiating both side with respect to p, we get

Also

Example1: Given that

Find ?

Backward difference table:

X	Y
0.1   0.2   0.3   0.4	1.10517   1.22140   1.34986   1.49182	0.11623   0.12846   0.14196	0.01223   0.01350	0.00127

Newton’s Backward formula for differentiation

Here

Example2: Given that

Find the derivative of y at ?

The difference table is given below:

X	Y
1.0   1.2   1.4   1.6   1.8   2.0	0   0.128   0.544   1.296   2.432   4.0	0.128   0.416   0.752   0.136   1.568	0.288   0.336   0.384   0.432	0.048   0.048   0.048	0   0

Since the point  is at the beginning of the table therefore

From Newton’s forward difference formula for differentiation, we get

Here

Since the point is at the end of the table therefore

Backward difference table is:

X	Y
1.0   1.2   1.4   1.6   1.8   2.0	0   0.128   0.544   1.296   2.432   4.000	0.128   0.416   0.752   0.136   1.568	0.288   0.336   0.384   0.432	0.048   0.048   0.048	0   0

Newton’s Backward formula for differentiation

Key takeaways-

Newton’s forward Difference formula:

2.2 The trapezoidal rule

Numerical Integration

Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x). In case of function of single variable, the process is called quadrature.

Trapezoidal Method:

Let the interval [a, b] be divided into n equal intervals such that <<…. <=b.

Here .

To find the value of .

Setting n=1, we get

Or I =

The above is known as Trapezoidal method.

Note: In this method second and higher difference are neglected and so f(x) is a polynomial of degree 1.

Geometrical Significance: The curve y=f(x), is replaced by n straight lines with the points ();() and ();…….;() and ().

The area bounded by the curve y=f(x), the ordinates ,and the x axis is approximately equivalent to the sum of the area of the n trapeziums obtained.

Example1: State the trapezoidal rule for finding an approximate area under the given curve. A curve is given by the points (x, y) given below:

(0, 23), (0.5, 19), (1.0, 14), (1.5, 11), (2.0, 12.5), (2.5, 16), (3.0, 19), (3.5, 20), (4.0, 20).

Estimate the area bounded by the curve, the x axis and the extreme ordinates.

We construct the data table:

Here length of interval h =0.5, initial value a = 0 and final value b = 4

By Trapezoidal method

Area of curve bounded on x axis =

Example2: Compute the value of  

Using the trapezoidal rule with h=0.5, 0.25 and 0.125.

Here

For h=0.5, we construct the data table:

By Trapezoidal rule

For h=0.25, we construct the data table:

By Trapezoidal rule

For h = 0.125, we construct the data table:

By Trapezoidal rule

[(1+0.5) +2(0.98461+0.94117+0.87671+0.8+0.71910+0.64+0.56637)]

Example3: Evaluate using trapezoidal rule with five ordinates

     Here

We construct the data table:

X	0
Y	0	0.3693161	1.195328	1.7926992	1.477265	0

Key takeaways

1. Numerical integration is a process of evaluating or obtaining a definite integral from a set of numerical values of the integrand f(x).
2. 

2.3 The Simpson’s rule

Overview-

Generally fundamental theorem of calculus is used find the solution for definite integrals, but sometime integration becomes too hard to evaluate, numerical methods are used to find the approximated value of the integral.

Simpson’s rules are very useful in numerical integration to evaluate such integrals.

Here we will understand the concept of Simpson’s rule and evaluate integrals by using numerical technique of integration.

We find more accurate value of the integration by using Simpson’s rule than other methods

Simpson’s rule

We will study about Simpson’s one-third rule and Simpson’s three-eight rules.

But in order to get these two formulas, we should have to know about the general quadrature formula-

General quadrature formula-

The general quadrature formula is gives as-

Simpson’s one-third and three-eighth formulas are derived by putting n = 2 and n = 3 respectively in general quadrature formula.

Simpson’s one-third and three-eighth formulas are derived by putting n = 2 and n = 3 respectively in general quadrature formula.

Simpson’s one-third rule-

 Put n = 2 in general quadrature formula-

We get-

Note- the given interval of integration has to be divided into an even number of sub-intervals.

Simpson’s three-eighth rule-

 Put n = 3 in general quadrature formula-

We get-

Note- the given interval of integration has to be divided into sub-intervals whose number n is a multiple of 3.

Example: Evaluate the following integral by using Simpson’s 1/3rd and 3/8th rule.

Solution-

First, we will divide the interval into six parts, where width (h) = 1, the value of f(x) is given in the table below-

x	0	1	2	3	4	5	6
f(x)	1	0.5	0.2	0.1	1/17 = 0.05884	1/26 = 0.0385	1/37 = 0.027

Now using Simpson’s 1/3rd rule-

We get-

And now

Now using Simpson’s 3/8th rule-

Example: Find the approximated value of the following integral by using Simpson’1/3rd rule.

Solution-

The table of the values-

x	1	1.25	1.5	1.75	2
f(x)	0.60653	0.53526	0.47237	0.41686	0.36788

Now using Simpson’s 1/3rd rule-

We get-

Example1: Estimate the value of the integral

By Simpson’s rule with 4 strips and 8 strips respectively.

For n=4, we have

E construct the data table:

By Simpson’s Rule

For n = 8, we have

By Simpson’s Rule

Example2: Evaluate Using Simpson’s 1/3 rule with .

For , we construct the data table:

X	0
	0	0.50874	0.707106	0.840896	0.930604	0.98281	1

By Simpson’s Rule

Example3: Using Simpson’s 1/3 rule with h = 1, evaluate

For h = 1, we construct the data table:

X	3	4	5	6	7
	9.88751	22.108709	40.23594	64.503340	95.34959

By Simpson’s Rule

            = 177.3853

Example1: Evaluate

By Simpson’s 3/8 rule.

Let us divide the range of the interval [4, 5.2] into six equal parts.

For h=0.2, we construct the data table:

By Simpson’s 3/8 rule

            = 1.8278475

Example2: Evaluate

Let us divide the range of the interval [0,6] into six equal parts.

For h=1, we construct the data table:

X	0	1	2	3	4	5	6
	1	0.5	0.2	0.1	0.0588	0.0385	0.027

By Simpson’s 3/8 rule

+3(0.0385) +0.027]

              =1.3571

Error in Integration

Error in Trapezoidal method

The total error in trapezoidal method is given by

Let is the largest value of the n quantities on the right-hand side of the above equation then

Error in Simpson’s Rule

The error in the Simpson’s rule is given by

Where is the largest value of the fourth derivative of y(x).

Error in Simpson’s 3/8 Rule

The error in this rule is given by

Where is the largest value of the derivative of y(x).

Key takeaways

1. The general quadrature formula is gives as-

2.     Simpson’s one-third rule-

3.     Simpson’s three-eighth rule-

2.4 Gauss integration formulas

Quadrature: It is the process to evaluate the value of the functions at the chosen point, to its exact value for polynomial up to higher degree as possible.

The general form Gaussian quadrature is given by

Where depends on the choice of n (number of points).

Also note that the possible polynomial of degree up to .

  . …... (1)

For point

Which gives exact value of the polynomial up to degree degree

i.e.,

For point and using (1)

Which gives exact value of the polynomial up to degree degree

i.e.,

On solving we get

 .

Gauss Quadrature 3-point method:

The general form Gaussian quadrature is given by

Where depends on the choice of n (number of points).

Also note that the possible polynomial of degree up to .

  . …... (1)

For point

Which gives exact value of the polynomial up to degree degree

i.e.,

On solving we get

 .

 For n=3,

Note:  To evaluate 

The above integral can be converted into Gauss quadrature by substituting

Hence  .

Example: Evaluate

Here

Using =

Also

For

For

Hence

Here

By Gauss quadrature 3-point rule

Example: Evaluate  by 2-point Gaussian rule.

Here

Using =

Also

For

For

Hence

Here

By Gauss quadrature 2-point rule

                      =0.99847

Example: Solve by Gauss quadrature 3-point method

Given

Here

Using =

Also

For

For

Hence

Here

By Gauss quadrature 3-point rule

Hence

2.5 Solution of ODE: Euler’s method

The general first order differential equation

…. (1)

With the initial condition … (2)

In general, the solution of first order differential equation in one of the two forms:

a)     A series for y in terms of power of x, from which the value of y can be obtained by direct solution.

b)    A set of tabulated values of x and y.

The case (a) is solved by Taylor’s Series or Picard method whereas case (b) is solved by Euler’s, Runge Kutta Methods etc.

Euler’s method:

In this method the solution is in the form of a tabulated values

Integrating both side of the equation (i) we get

Assuming that  in  this gives Euler’s formula

In general formula

                 , n=0,1, 2...

Error estimate for the Euler’s method

Example1: Use Euler’s method to find y (0.4) from the differential equation

                                            with h=0.1

Given equation  

Here

We break the interval in four steps.

So that

By Euler’s formula

          , n=0,1,2,3   ……(i)

For n=0 in equation (i) we get

For n=1 in equation (i) we get

                                                               .01

For n=2 in equation (i) we get

For n=3 in equation (i) we get

Hence y (0.4) =1.061106.

Example2: Using Euler’s method solve the differential equation for y at x=1 in five steps

Given equation  

Here       

No. Of steps n=5 and so that

So that

Also

By Euler’s formula

          , n=0,1,2,3,4   ……(i)

For n=0 in equation (i) we get

For n=1 in equation (i) we get

For n=2 in equation (i) we get

For n=3 in equation (i) we get

For n=4 in equation (i) we get

Hence 

Example3: Given  with the initial condition y=1 at   x=0. Find y for x=0.1 by Euler’s   method (five steps).

Given equation is  

Here   

No. Of steps n=5 and so that

So that

Also

By Euler’s formula

          , n=0,1,2,3,4   ……(i)

For n=0 in equation (i) we get

For n=1 in equation (i) we get

For n=2 in equation (i) we get

For n=3 in equation (i) we get

For n=4 in equation (i) we get

Hence  

Key takeaways

Euler’s method:                 , n=0,1, 2…...

2.6 Improvement of Euler’s method

Modified Euler’s Method:

Instead of approximating as in Euler’s method.  In the modified Euler’s method, we have the iteration formula

Where is the nth approximation to .The iteration started with the Euler’s formula

Example1: Use modified Euler’s method to compute y for x=0.05.  Given that

Result correct to three decimal places.

Given equation  

Here

Take h =  = 0.05

By modified Euler’s formula the initial iteration is

                              )

The iteration formula by modified Euler’s method is

          -----(i)

For n=0 in equation (i) we get

Where   and  as above

For n=1 in equation (i) we get

For n=3 in equation (i) we get

Since third and fourth approximation are equal.

Hence y=1.0526 at x = 0.05 correct to three decimal places.

Example2: Using modified   Euler’s method, obtain a solution of the equation

Given equation 

Here

By modified Euler’s formula the initial iteration is

The iteration formula by modified Euler’s method is

          -----(i)

For n=0 in equation (i) we get

Where   and  as above

For n=1 in equation (i) we get

For n=2 in equation (i) we get

For n=3 in equation (i) we get

Since third and fourth approximation are equal.

Hence y=0.0952 at x=0.1

To calculate the value of  at x=0.2

By modified Euler’s formula the initial iteration is

The iteration formula by modified Euler’s method is

          -----(ii)

For n=0 in equation (ii) we get

                                                                  1814

For n=1 in equation (ii) we get

                                                                  1814

Since first and second approximation   are equal.

Hence y = 0.1814 at x=0.2

To calculate the value of  at x=0.3

By modified Euler’s formula the initial iteration is

The iteration formula by modified Euler’s method is

          -----(iii)

          For n=0 in equation (iii) we get

For n=1 in equation (iii) we get

For n=2 in equation (iii) we get

For n=3 in equation (iii) we get

Since third and fourth approximation are same.

Hence y = 0.25936 at x = 0.3

Key takeaways

Modified Euler’s Method:

2.7 Runge-kutta method, multi step method

Solution of Second order ODE using 4th order Runge-Kutta method (Numerical), Runge-Kutta fourth order method (Numerical)

This method is more accurate than Euler’s method.

Consider the differential equation of first order

Let  be the first interval.

A second order Runge Kutta formula  

Where

Rewrite as 

A fourth order Runge Kutta formula:

Where 

Example1: Use Runge Kutta method to find y when x=1.2 in step of h=0.1 given that

Given equation

Here

Also

By Runge Kutta formula for first interval

Again

A fourth order Runge Kutta formula:

To find y at 

A fourth order Runge Kutta formula:

Example2: Apply Runge Kutta fourth order method to find an approximate value of y for x=0.2   in step of 0.1, if

Given equation 

Here 

Also

By Runge Kutta formula for first interval

A fourth order Runge Kutta formula:

Again

A fourth order Runge Kutta formula:

Example3: Using Runge Kutta method of fourth order, solve

Given equation 

Here 

Also

By Runge Kutta formula for first interval

         )

A fourth order Runge Kutta formula:

Hence at x = 0.2 then y = 1.196

To find the value of y at x=0.4. In this case

A fourth order Runge Kutta formula:

Hence at x = 0.4 then y=1.37527

Solution of Second order ODE using 4th order Runge-Kutta method:

The second order differential equation

Let then the above equation reduces to first order simultaneous differential equation

Then

This can be solved as we discuss above by Runge- Kutta Method. Here for and for .               

A fourth order Runge- Kutta formula:

Where 

Example1: Using Runge- Kutta method of order four, solve   to find

Given second order differential equation is

Let   then above equation reduces to

     Or

(say)

Or .

By Runge- Kutta Method we have

A fourth order Runge Kutta formula:

Example2: Using Runge Kutta method, solve

for correct to four decimal places with initial condition .

Given second order differential equation is

Let   then above equation reduces to

     Or

(say)

Or .

By Runge-Kutta Method we have

A fourth order Runge Kutta formula:

And  

.

Example3: Solve the differential equations

for

Using four order Runge Kutta method with initial conditions

Given differential equation are

Let

And

Also

By Runge Kutta Method we have

A fourth order Runge Kutta formula:

And  

.

Key takeaways

A fourth order Runge Kutta formula:

Where 

2.8 Method for system and higher order ordinary differential equations

Problems involving second and higher order differential equations, in which the conditions at two or more points are specified, are called boundary-value problems.

Finite difference method

This method is used to solve bounded value problem.

The general two-point linear boundary value problem

With boundary condition

To solve this problem by using finite difference, we divide the interval [a, b] into n sub interval so that .

The approximated function f(x) at the points

………

We use the following central difference formula:

We substitute these value in the given equation and apply the boundary condition to calculate the value of the   unknown.

Example1: Solve the boundary value problem defined by

By finite difference method. Compare the solution at y (0.5) by taking h=0.5 and h=0.25.

Given equation

With boundary   condition   

By finite difference method

       …. (iii)

Putting(iii) in (i) we get

       …. (iv)

For h=0.5, here for which corresponds to

For i=1 in equation (iv) we get

For h=0.25, here

Which corresponds to 

For i=1 in equation (iv) we get

For i=2 in equation (iv) we get

For i=3 in equation (iv) we get

From   equation (v), (vi) and (vii) we get

On solving above triangular   equation we get

Hence for h=0.5 we   get    y (0.5) =0.44444

And for h=0.25 we get y (0.5) =0.443674

Example2:  Solve the bounded value problem

With boundary condition 

Given equation

With 

By finite difference method

       …. (3)

Putting (3) in equation (1) we have

By finite difference method

…… (4)

Let h=1, we have

Corresponds to 

For i=1 in equation (4) we get

For i=2 in equation (4) we get

For i=3 in equation (4) we get

From equation (5), (6) and (7) we get

On solving we get

Example3: Solve the boundary value problem

With y (0) =0 and y (2) =3.62686

Given equation   …... (1)

With boundary condition y (0) =0 and y (2) =3.62686…. (2)

By finite difference method

       …. (3)

Substituting (3) in equation (1) we get

…. (3)

Let h=0.5 then for 

Which corresponds   to 

For   i=1 in equation (3) we get

For   i=2 in equation (3) we get

For   i=3 in equation (3) we get

From equation (4), (5) and (6) we get

On solving we get 

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port and C. J. Stone, “Introduction to Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction to Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
7. T. Veerarajan, “Engineering Mathematics”, Tata McGraw-Hill, New Delhi, 2010
8. Higher engineering mathematics, HK Dass
9. Higher engineering mathematics, BV Ramana.
10. Computer based numerical & statistical techniques, M goyal