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Unit 1

Numerical methods


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.

 

Taylor’s Series Method: 

The general first order differential equation

….(1)

With the initial condition …(2)

Let be the exact solution of equation (1), then the Taylor’s series for around is given by

   (3)

If the values of  are known, then equation (3) gives apowwer series   for y. By total derivatives   we have

   ,

  

                  

And other higher derivatives of y.  The method can easily  be extended to   simultaneous   and higher –order  differential  equations. In   general,

 

Putting   in these above results, we can obtain the values of finally, we substitute these values of  in equation (2) and obtain the approximate value of y; i.e., the solutions of (1).

 

Example1: Solve

,

using Taylor’s series method and compute .

 

Here This implies that .

Differentiating, we get

     .

     .

   

    .

The   Taylor’s series at ,

 

 

  (1)

At in equation (1) we get

 

 

At in equation (1) we get

 

 

 

Example2: Using Taylor’s series method, find the solution of

           

      At ?

Here

At implies that or or

Differentiating, we get

  implies that or .

  implies that or

  implies that or

   implies that or

The   Taylor’s series at ,

 

     

      (1)

At   in equation (1) we get

 

          

         

At in equation (1) we get

 

           

          

 

Example3: Solve

numerically, start from and carry to using Taylor’s series method.

 

Here .

 

We have

Differentiating, we get

  implies that or

  implies that or .

   implies that

  implies that

The   Taylor’s series at ,

 

 Or

              

            

Here

 

 

 

 

 

The   Taylor’s series

 

        

       .

 

Key takeaways

  1. The Taylor’s series for around is given by

 


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  

 

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

  1. Euler’s method:

, n=0,1,2,….

  1.   

 


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 RungeKutta 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 RungeKutta Method we have

A fourth order Runge Kutta formula:

 

Example2: Using RungeKutta 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 RungeKutta method with initial conditions

Given differential equation are

Let

And

Also

 By RungeKutta Method we have

A fourth order Runge Kutta formula:

And  

.

Key takeaways

  1. A fourth order Runge Kutta formula:

     

Where 

             

             

             

 


Milne’s predictor corrector formula-

For given dy/dx = f(x,y) and y = and x = , to find the value of y for x = , by using Milne’s method,

We follow the steps given below-

The value being given, here we calculate-

By Taylor’s series or Picard’s method.

Now we calculate-

Then to find
 

We substitute Newton’s forward interpolation formula-

In the relation-

By putting x = ,  dx = h dn

Neglecting fourth and higher order differences and expressing in terms of the function values, we get-

This is called a predictor.

Now having found we obtain a first approaximation to

Then the better value of is found by simpson’s rule as-

Which is called corrector.

Then an improved value of is computed and again corrector is applied to find a better value of .

We continues this step until remains unchanged.

Once and are obtained to desired degree of accuracy,

is found from the predictor as-

 

And

is calculated.

Then the better approximation to the value of we get from the corrector as-

We repeat until becomes stationary and we proceed to calculate .

This is called the Milne’s predictor-corrector method.

 

Adams - Bashforth predictor and corrector formula-

This is called Adams - Bashforth predictor formula.

And

This is called Adams - Bashforth corrector formula.

 

Example: Find the solution of the differential equation   in the range for the boundary conditions y = 0 and x = 0 by using Milne’s method.

Sol.

By using Picards method-

Where

To get the first approximation-

We put y = 0 in f(x, y),

Giving-

In order to find the second approximation, we put y  = in f(x,y)

Giving-

And the third approximation-

Now determine the starting  values of the Milne’s method from equation  (1), by choosing h = 0.2

Now using the predictor-

X = 0.8

,      

 

And the corrector-

,                 ................(2)

 

Now again using corrector-

Using predictor-

 

X = 1.0,  

,      

And the corrector-

 

,      

Again using corrector-

,       which is same as before

Hence

 

Example: Solve the intital value problem

y(0) = 1 to find y(0.4) by using Adams-Bashforth method.

Starting solutions required are to be obtained using Runge-Kutta method of order 4 using step value h = 0.1

Sol.

Here we have-

Here

So that-

Thus

To find y(0.2)-

Here

Thus,

Y(0.2) = 

To find y(0.3)-

Here

Thus,

Y(0.3) = 

Now the starting values of Adam’s method with h = 0.1-

Using predictor-

 

Using corrector-

Hence

 

 

Key takeaways

  1. Predictor-
     
  2. Corrector-
     
  3. Adams - Bashforth predictor formula-
     
  4. Adams - Bashforth corrector formula.
     

 

 

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

 

 

 

 

 

 

 

 


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