Unit-5

Boundary value problems

Q1) 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.

A1) 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

Q2) Solve the bounded value problem

With boundary condition

A2) 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

Q3) Solve the boundary value problem

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

A3) 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

Q4) Explain how do we find the smallest Eigen value.

A4) If is the Eigen value of A, then the reciprocal is the Eigen value of , then the reciprocal of the largest Eigen value of will be the smallest Eigen value of A.

Q5) Explain power method.

A5) Procedure for Power method-

Step-1: First we choose an arbitrary real vector , basically is chosen as-

Step-2: Compute , , , , ………… Put

Step-3: compute , ,

Step-4: The largest Eigen value is

The error in can be find as-

The Eigen vector corresponding to is

Q6) Find the largest Eigen value and the corresponding Eigen vector of the matrix

Also find the error in the value of the largest Eigen value.

A6)

Let us choose the initial vector

Then

Now put , then-

Hence the largest Eigen value is-

And the corresponding Eigen vector is-

The error can be calculated as-

Q7) Give the classification of PDE’s.

A7) The general linear PDE of the second order in two independent variables is of the form-

Then there are three conditions-

Q8) Classify the equation-

A8) Here A =

Now

That means,

The equation is hyperbolic.

Q9) Classify the equation

A9) Here

Hence the equation is parabolic

Q10) Solve the Laplace’s equation in the domain

A10) The initial values using five diagonal formula we have

Here ,

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

Hence and .

Q11) Solve the Laplace’s equation for

A11) The initial values using five diagonal formula we have

Here ,

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

Hence and .

Q12) Solve the elliptical equation for

A12) 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 Seidal 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

.

Q13) Solve the Poisson equation

A13) 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

.

Q14) Solve the equation with the conditions . Assume. Tabulate u for choosing appropriate value of k?

A14) 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 |

Q15) Solve the heat equation

Subject to the conditions and

.

A15) 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 |

Q16) Use the Bendre-Schmidt formula to solve the heat conduction problem

With the condition and .

A16) Let we see when .

The initial condition is .

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