Table of Contents

Example of Inverse Power Iteration

Author: Philip Nelson

Language: C++

Description

This demonstrates the condition number method on the tri-diagonal matrix used to solve the Elliptic ODE.

Output

The condition number of the NxN matrix with increasing N. Observe that as the size of the matrix increases, the condition number increases unbounded

Example

int main()
{
  std::cout << "A\n" << SecOrdFinDifMethEllipticMat<double, 5>() << std::endl << std::endl;
  std::cout << "3x3" << std::endl;
  std::cout << conditionNumber(SecOrdFinDifMethEllipticMat<double, 3>()) << std::endl << std::endl;
  std::cout << "5x5" << std::endl;
  std::cout << conditionNumber(SecOrdFinDifMethEllipticMat<double, 5>()) << std::endl << std::endl;
  std::cout << "10x10" << std::endl;
  std::cout << conditionNumber(SecOrdFinDifMethEllipticMat<double, 10>()) << std::endl << std::endl;
  std::cout << "15x15" << std::endl;
  std::cout << conditionNumber(SecOrdFinDifMethEllipticMat<double, 15>()) << std::endl << std::endl;
  std::cout << "25x25" << std::endl;
  std::cout << conditionNumber(SecOrdFinDifMethEllipticMat<double, 25>()) << std::endl << std::endl;
  std::cout << "75x75" << std::endl;
  std::cout << conditionNumber(SecOrdFinDifMethEllipticMat<double, 75>()) << std::endl << std::endl;
  std::cout << "100x100" << std::endl;
  std::cout << conditionNumber(SecOrdFinDifMethEllipticMat<double, 100>()) << std::endl << std::endl;
}

Result

A
| -2     1     0  |
|  1    -2     1  |
|  0     1    -2  |

3x3
-5.82843

5x5
-13.9282

10x10
-48.3742

15x15
-103.087

25x25
-273.306

75x75
-2340.09

100x100
-4133.2

Last Modification date: 27 February 2018