Unit-5

Boundary Value problems

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

Solving Eigen value problems (Power method)-

To find the approximate values of all the Eigen values and Eigen vectors, iteration method or power method is used.

Power method is used when only the largest and/or the smallest eigen values of a matrix are desired.

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

How do we determine the smallest Eigen value?

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.

Example: 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.

Sol.

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-

Key takeaways-

4. 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.

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

Then there are three conditions-

Example: Classify the equation-

Sol.

Here A =

Now

That means,

The equation is hyperbolic.

Example: Classify the equation

Sol.

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.

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 .

PDEs – Elliptical explicit method:

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 Seidal 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 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

.

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

.

PDEs- parabolic explicit method:

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 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

Key takeaways-

5. The Poisson’s equation

References