Unit – 2

Numerical Methods

2.1 Numerical solutions of system of linear equations: Gauss elimination method

In this method we eliminate successively the unknown so that the equation (1) remain with the single unknown and reduce to upper triangular system. At last with help of back substitution we calculate the values of the remaining unknowns.

Consider a system of n linear equation in n unknown

…          ……                …..         … (1)

…          ……                 ….         …

To convert the above system into upper triangular matrix we eliminate from the second, third, fourth …., n equations above by multiplying the first equation by added them to the corresponding equations second, third, fourth,…., n equation. We get

                                         …          ……                …..         … 

                                         …          ……                 ….         …

Repeating the above method for we get finally the upper triangular form.

Upper Triangular form of above

                                                      ………………………..                        (i)

Thus .

Then we calculate the values of .

Example1 Apply Gauss Elimination method to solve the equations:

Given                                                    Check Sum (sum of coefficient and constant)

                        -1                            …. (I)

                      -16                          …. (ii)

                          5 …. (iii)

(I)We eliminate x from (ii) and (iii)

Apply eq(ii)-eq(i) and eq(iii)-3eq(i) we get

                        -1                            ….(i)

                      -15                          ….(iv)

                          8            ….(v)

(II) We eliminate  y from eq(v)

Apply

                        -1                            ….(i)

                      -15                          ….(iv)

                      73            ….(vi)

(III) Back Substitution we get

From (vi) we get

From (iv) we get

From (i) we get 

Hence the solution of the given equation is

Example2 Solve the equation by Gauss Elimination Method:

Given

Rewrite the given equation as

                 … (i)

                             ….(ii)

                             ….(iii)

                                 …(iv)

(I)                We eliminate x from (ii),(iii) and (iv) we get

Apply eq(ii) + 6eq(i), eq(iii) -3eq(i), eq(iv)-5eq(i) we  get

                 …(i)

                             ….(v)

                       ….(vi)

                             …(vii)

(II)             We eliminate y   from (vi) and (vii) we get

Apply 3.8 eq(vi)-3.1eq(v) and 3.8eq(vii)+5.5eq(v) we get

                 …(i)

                             ….(v)

                         …(viii)

                             …(ix)

(III)           We eliminate z from eq (ix) we get

Apply 9.3eq (ix) + 8.3eq (viii), we get

                 … (i)

                             ….(v)

                         …(viii)

350.74u=350.74

Or u = 1

(IV)          Back Substitution

From eq(viii)

Form eq(v), we get

From eq(i) ,

Hence the solution of the given equation is x=5, y=4, z=-7 and u=1.

Example3 Apply Gauss Elimination Method to solve the following system of equation:

Given                    … (i)

              … (ii)

              … (iii)

(I)                We eliminate x from (ii) and (iii)

Apply we get

                … (i)

              … (iv)

             … (v)

(II)             We eliminate y from  (v)

Apply we get

                … (i)

              … (vi)

                      … (vii)

(III)           Back substitution 

From (vii)  

From (vi)

From (i)

Hence the solution of the equation is

2.2 Cholesky

The Cholesky decomposition of a Matrix A is a decomposition of the form

Where, L = Lower Triangular Matrix

conjugate transpose of L

Working Rule to solve prroblems:

Step 1:

Step 2:

Step 3:

Step 4:

Consider

Que. Solve

Solution. Step I. Consider AX= B

Step 2: Consider

By solving

,     

Put all the values

Step 3: LY = B, Let

By multiplying and equating

g=3

and,

and,

Therefore,

Step 4:

Now we are just cross check our Ans. by putting given equations of question

Given equation:

Our answer are correct

Que. Solve

Solution. Step I. consider AX =B

Step 2. Consider

BY solving,

Put all the values we get

Step 3: LY = B, Let

By multiplying and equating

Step 4:

BY MULTIPLYING WE GET

Now we are just cross check our ans by putting values in given equations of question,

Given equation

Our answer is correct.

Que. Solve

Solution. Step I Consider AX = B

Step 2: Consider

Put all the values we get

Step 3:

By multiplying and equating

Step 4:

By multiplying we get

Now we are just cross check our answer by putting values of in equation

Given equation:

And we have

Our correct answer is

Que 4. Solve

Solution. Step 1. Consider AX = B

Step 2: Consider,

Let,

Put all the values we get

Step 3: LY = B, Let

By multiplying and equating

Step 4:

By Multiplying we get

Now we just cross check our answer by putting values in given equation

Given equation:

Final answer,

2.3 Jacobi and Gauss-Seidel methods

Jacobi’s Iteration method :

Let us consider the system of simultaneous linear equation

        (1)

The coefficients of the diagonal elements are larger than the all other coefficients and are non-zero. Rewrite the above equation we get

          (2)

Take the initial approximation we get the values of the first approximation of .

By the successive iteration we will get the desired the result.

Example1Use Jacobi’s method to solve the system of equations:

Since

So, we express the unknown with large coefficient in terms of other coefficients.

              (1)

Let the initial approximation be

2.35606

0.91666

1.932936

0.831912

3.016873

1.969654

3.010217

1.986010

1.988631

0.915055

1.986532

0.911609

1.985792

0.911547

1.98576

0.911698

Since the approximation in ninth and tenth iteration is same up to three decimal places, hence the solution of the given equations is

Example2 Solve by Jacobi’s Method, the equations

Given equation can be rewrite in the form

     … (i)

    ..(ii)

    ..(iii)

Let the initial approximation be

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

0.90025

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Hence   solution approximately is

Example3Use Jacobi’s method to solve the system of the equations

Rewrite the given equations

        (1)

Let the initial approximation be

1.2

1.3

0.9

1.03

0.9946

0.9934

1.0015

Hence the solution of the above equation correct to two decimal places is

Gauss Seidel method:   

This is the modification of the Jacobi’s Iteration.  As above in Jacobi’s Iteration, we take first approximation as and put in the right hand side of the first equation of (2) and let the result be  . Now we put right hand side of second equation of (2) and suppose the result is now put in the RHS of third equation of (2) and suppose the result be the above method is repeated till the values of all the unknown are found up to desired accuracy.

Example1 Use Gauss –Seidel Iteration method to solve the system of equations

Since

So, we express the unknown of larger coefficient in terms of the unknowns with smaller coefficients.

Rewrite the above system of equations

      (1)

Let the initial approximation be

3.14814

2.43217

2.42571

2.4260

Hence the solution correct to three decimal places is

Example2 Solve the following system of equations

  By Gauss-Seidel method.

Rewrite the given system of equations as

      (1)

Le t the initial approximation be

Thus the required solution is

Example3 Solve the following equations by Gauss-Seidel Method

Rewrite the above system of equations

          (1)

Let the initial approximation be

Hence the required solution is

2.4 Numerical solutions of ordinary differential equations: Euler’s

The general first order differential equation

        With the initial condition

 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  

2.5 Modified Euler’s

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

2.6 Runge- Kutta 4th order and Predictor-Corrector methods

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

Reference Books