Unit - 2

Numerical Solution of Differential Equation

2.1 Ordinary Differential Equation: Taylor’s Series 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.

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 ofare 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

.

2.2 Euler’s Method

The general first order differential equation

With the initial condition

4.2.1 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  

4.2.2 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

2.3 Runge Kutta Method fourth order:

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

2.4 Simultaneous equation using Runge Kutta 2nd order 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  

.

2.5 Partial Differential Equation: The finite Difference Method

The general linear partial differential equation of the second order in two independent variables is of the form.

Such a PDE is said to be

1. Elliptic: if
2. Parabolic: if
3. Hyperbolic: if

Example: Classify the equation

Here

Hence the equation is parabolic.

Finite Difference Approximation

We construct a rectangular region R in the xy- plane and divide into network of sides and . The intersection points of the dividing lines are called the mesh point, nodal point or grid points.

Then the finite difference approximation for the partial derivative in x-direction is

And

For the above approximation is

   …. (1)

     … (2)

   … (3)

And    … (4)

Similarly, we have the approximations for the derivatives with respect to y

    …. (5)

  …. (6)

   … (7)

And   …. (8)

Replacing the derivatives in any partial differential equation by their corresponding difference approximation (1) to (8), we obtain the finite difference similar to the given equations.

2.6 Simple Laplace’s Method

The Laplace’s equation

Consider the Laplace’s equation in a region R with boundary C. Let R be a square region so that it can be divided into network of small squares of side h. Let the values of on the boundary C be given by and let the interior mesh points be as in figure

The approximate function values at the interior point of the mesh can be calculated by the diagonal five-point formula of   in this order

   (bigger +)

     (X form)

        (X form)

   (X form)

  (X form)

Similarly, the remaining quantities are calculated by using standard five-point diagonal formulas.

  (+ form)

   (+ form)

    (+ form)

  (+ form)

In this way all are computed.

Example1: Solve the Laplace’s equation  in the domain

The initial values using five diagonal formula we have

Here , 

The remaining quantities are calculated by using standard five-point diagonal formulas.

Hence and .

Example2: Solve the Laplace’s equation for

The initial values using five diagonal formula we have

Here , 

The remaining quantities are calculated by using standard five-point diagonal formulas.

Hence and .

2.7 Elliptic explicit solution

The Laplace’s equation

And the Poisson’s equation

Are the example of elliptic partial differential equation.

The Poisson’s equation

This can be solved by interior mesh points of a square network when the boundary values are known. The standard five-point formula for Poisson’s equation is

After using the above formula, we get the linear equations in the pivotal values

Then these are solved by Gauss Seidel Iteration formula

.

Example1: Solve the elliptical equation for

The initial values using five diagonal formula we have

Here , 

The remaining quantities are calculated by using standard five-point diagonal formulas.

The Above is symmetric about PQ, so that .

We will have iteration process using the Gauss Seidel Formula

First iteration: Putting we get

Second Iteration: Putting , we get

Third Iteration: Putting , we get

Fourth Iteration: Putting , we get

Fifth iteration: Putting n=4 we get

.

Example2: Solve the Poisson equation

Let the point be defined by At the point A, . The standard five-point formula at point A is

Or  

Or       ….(i)

Again, the standard five-point formula at the point B is

Or 

Or   ...(ii)

Similarly, the standard five-point formula at the point C

Or

Or     …...(iii)

Similarly, the standard five-point formula at the point D

Or

Or      …. (iv)

From (ii) and (iii) we get =. Hence the iteration formula we have

.

First iteration: Putting . Hence, we obtain

Second iteration: Putting n=1, we get

Third iteration: Putting n=2, we get

Fourth iteration: Putting n=3, we get

Fifth iteration: Putting n=4, we get

Sixth iteration: Putting n=5, we get

Since last two iteration are approximately equal, hence

.

2.8 PDEs- parabolic explicit solution

The example of parabolic is one dimensional heat conduction equation

   ... (1)

Where is the diffusivity of the substance.

Consider a rectangular mesh in plane.

The spacing in x-direction is h and in t direction is k. Let the mesh point or simply we get

And

Substituting these in equation (1) we get

Or      … (2)

Where   is the mesh ratio parameter.

This formula gives the value of u at position mesh points I nterms of known values of and at the instant . It gives the relation between two-time level therefore known as two level formula.

Also named as Schmidt explicit formula and is true for .

At the above formula reduces to

..(3)

Is known as Bendre-Schmidt recurrence relation which gives the value of u at internal mesh points with the help of boundary conditions.

Example1:  Solve the equation with the conditions . Assume. Tabulate u for choosing appropriate value of k?

Here and let   ,

Since

The Bendre-Schmidt recurrence formula we have

  ….(i)

Also given .

for all values of j, i.e., the entries in the first and the last columns are zero.

Since

  (Using

For .

Putting

Putting successively we get

These will give the entries in the second row.

Putting in equation (i), we will get the entries of the third row.

Similarly, successively in (i), the entries of the fourth rows are

Obtained.

Hence the values of  are as given in the below the table:

	0	1	2	3	4	5	6	7	8	9	10
0	0	0.09	0.16	0.21	0.24	0.25	0.24	0.21	0.16	0.09	0
1	0	0.08	0.15	0.20	0.23	0.24	0.23	0.20	0.15	0.08	0
2	0	0.075	0.14	0.19	0.22	0.23	0.22	0.19	0.14	0.075	0
3	0	0.07	0.133	0.18	0.21	0.22	0.21	0.18	0.133	0.07	0

Example2: Solve the heat equation

Subject to the conditions and

.

Take and k according to Bendre-Schmidt equation.

Here and let   ,

Since

The Bendre-Schmidt recurrence formula we have

  …. (i)

Also given .

for all values of j, i.e., the entries in the first and the last columns are zero.

Since  

.

.

For  

Putting

Putting successively we get

These will give the entries in the second row.

Putting in equation (i), we will get the entries of the third row.

Similarly, successively in (i), the entries of the fourth rows are

Obtained.

Hence the values of  are as given in the below the table:

	0	1	2	3	4
0	0	0.5	1	0.5	0
1	0	0.5	0.5	0.5	0
2	0	0.25	0.5	0.25	0
3	0	0.25	0.25	0.25	0

Example3: Use the Bendre-Schmidt formula to solve the heat conduction problem

With the condition and .

Let we see when .

The initial condition are .

Also .

The iteration formula is

=

First iteration: Putting n=0, we get

Second iteration: Putting n=1, we get

Third Iteration: putting n=3, we get

Fourth Iteration: putting n=3, we get

Fifth Iteration: putting n=4, we get

Hence the approximate solution is

References:

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