Name: FY4150-Project 1 Solved
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The aim of this project is to get familiar with various vector and matrix operations, from dynamic memory allocation to the usage of programs in the library package of the course. For Fortran users memory handling and most matrix and vector operations are included in the ANSI standard of Fortran 90/95. Array handling in Python is also rather trivial. For C++ user however, there are several possible options. Two are listed here.

For this exercise we recommend that you make your own functions for dynamic memory allocation of a vector and a matrix. You don’t need to write a class for this operations. Use then the library package lib.cpp with its header file lib.hpp for obtaining LU-decomposed matrices, solve linear equations etc.
A very good and often recommended library for C++ handling of arrays is the library Armadillo, to be found at \url{arma.sourceforge.net}. We will discuss the usage of this library during the lab sessions and lectures. Armadillo has also an interface to Lapack functions for solving systems of linear equations. Alternatively you can use the Eigen library, which has much of the same functionality as Armadillo.

Your program, whether it is written in C++, Python, Fortran or other languages, should include dynamic memory handling of matrices and vectors.

The material needed for this project is covered by chapter 6 of the lecture notes, in particular section 6.4 and subsequent sections.

Many important differential equations in Science can be written as linear second-order differential equations

d 2 y d x 2 + k 2 (x) y = f (x),

where $f$ is normally called the inhomogeneous term and $k^{2}$ is a real function.

A classical equation from electromagnetism is Poisson’s equation. The electrostatic potential $Φ$ is generated by a localized charge distribution $ρ (r)$ . In three dimensions it reads

\nabla 2 Φ = - 4 π ρ (r) .

With a spherically symmetric $Φ$ and $ρ (r)$ the equations simplifies to a one-dimensional equation in $r$ , namely

1 r 2 d d r (r 2 d Φ d r) = - 4 π ρ (r),

which can be rewritten via a substitution $Φ (r) = ϕ (r) / r$ as

d 2 ϕ d r 2 = - 4 π r ρ (r) .

The inhomogeneous term $f$ or source term is given by the charge distribution $ρ$ multiplied by $r$ and the constant $- 4 π$ .

We will rewrite this equation by letting $ϕ \to u$ and $r \to x$ . The general one-dimensional Poisson equation reads then

- u'' (x) = f (x) .

Project 1 a):

In this project we will solve the one-dimensional Poisson equation with Dirichlet boundary conditions by rewriting it as a set of linear equations.

To be more explicit we will solve the equation

- u'' (x) = f (x), x \in (0, 1), u (0) = u (1) = 0.

and we define the discretized approximation to $u$ as $v_{i}$ with grid points $x_{i} = i h$ in the interval from $x_{0} = 0$ to $x_{n + 1} = 1$ . The step length or spacing is defined as $h = 1 / (n + 1)$ . We have then the boundary conditions $v_{0} = v_{n + 1} = 0$ . We approximate the second derivative of $u$ with

- v i + 1 + v i - 1 - 2 v i h 2 = f i f o r i = 1, \dots, n,

where $f_{i} = f (x_{i})$ . Show that you can rewrite this equation as a linear set of equations of the form

A v = b ~,

where $A$ is an $n \times n$ tridiagonal matrix which we rewrite as

A = ⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢ 2 - 1 000 - 1 2 - 1 \dots \dots \dots 0 - 1 2 \dots \dots 0 - 1 \dots - 1 0 \dots \dots 0 \dots 2 - 1 0 \dots \dots \dots - 1 2 ⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥,

and ${\tilde{b}}_{i} = h^{2} f_{i}$ .

In our case we will assume that the source term is $f (x) = 100 e^{- 10 x}$ , and keep the same interval and boundary conditions. Then the above differential equation has a closed-form solution given by $u (x) = 1 - (1 - e^{- 10}) x - e^{- 10 x}$ (convince yourself that this is correct by inserting the solution in the Poisson equation). We will compare our numerical solution with this result in the next exercise.

Project 1 b):

We can rewrite our matrix $A$ in terms of one-dimensional vectors $a, b, c$ of length $1 : n$ . Our linear equation reads

A = ⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢ b 1 a 1 c 1 b 2 a 2 \dots 0 c 2 b 3 \dots \dots \dots c 3 \dots a n - 2 \dots \dots \dots \dots b n - 1 a n - 1 \dots \dots \dots \dots c n - 1 b n ⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥ ⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢ v 1 v 2 \dots \dots \dots v n ⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥ = ⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢ b ~ 1 b ~ 2 \dots \dots \dots b ~ n ⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥ .

Note well that we do not include the endpoints since the boundary conditions are used resulting in a fixed value for $v_{i}$ . A tridiagonal matrix is a special form of banded matrix where all the elements are zero except for those on and immediately above and below the leading diagonal. Develop a general algorithm first which does not assume that we have a matrix with the same elements along the diagonal and the non-diagonal elements. The algorithm for solving this set of equations is rather simple and requires two steps only, a decomposition and forward substitution and finally a backward substitution.

Before we proceed with the solution of the differential equation, you should now plan the organization of your data flow. Here you will find it convenient to define vectors that will contain the matrix elements, the solution to the problem and the function $f (x)$ when discretized. You should also plan on to read input data, whether you do this from the command line or from a selected file. We recommend strongly that you use dynamical memory allocation. An example of a C++ program which reads from the command line various input parameters can be found here

Your first task is to set up the general algorithm (assuming different values for the matrix elements) for solving this set of linear equations. Find also the precise number of floating point operations needed to solve the above equations. For the general algorithm you need to specify the values of the array elements $a$ , $b$ and $c$ by inserting their explicit values.

Then you should code the above algorithm and solve the problem for matrices of the size $10 \times 10$ , $100 \times 100$ and $1000 \times 1000$ . That means that you select $n = 10$ , $n = 100$ and $n = 1000$ grid points.

Compare your results (make plots) with the closed-form solution for the different number of grid points in the interval $x \in (0, 1)$ . The different number of grid points corresponds to different step lengths $h$ .

Project 1 c):

Use thereafter the fact that the matrix has identical matrix elements along the diagonal and identical (but different) values for the non-diagonal elements. Specialize your algorithm to the special case and find the number of floating point operations for this specific tri-diagonal matrix. Compare the CPU time with the general algorithm from the previous point for matrices up to $n = 10^{6}$ grid points.

Project 1 d):

Compute the relative error in the data set $i = 1, \dots, n$ ,by setting up

ϵ i = l o g 10 (∣∣∣ v i - u i u i ∣∣∣),

as function of $l o g_{10} (h)$ for the function values $u_{i}$ and $v_{i}$ . For each step length extract the max value of the relative error. Try to increase $n$ to $n = 10^{7}$ . Make a table of the results and comment your results. You can use either the algorithm from b) or c).

Project 1 e):

Compare your results with those from the LU decomposition codes for the matrix of sizes $10 \times 10$ , $100 \times 100$ and $1000 \times 1000$ . Here you should use the library functions provided on the webpage of the course. Alternatively, if you use armadillo as a library, you can use the similar function for LU decomposition. The armadillo function for the LU decomposition is called $L U$ while the function for solving linear sets of equations is called $s o l v e$ . Use for example the unix function time when you run your codes and compare the time usage between LU decomposition and your tridiagonal solver. Alternatively, you can use the functions in C++, Fortran or Python that measure the time used.

Make a table of the results and comment the differences in execution time How many floating point operations does the LU decomposition use to solve the set of linear equations? Can you run the standard LU decomposition for a matrix of the size $10^{5} \times 10^{5}$ ? Comment your results.

To compute the elapsed time in c++ you can use the following statements

...
#include "time.h"   //  you have to include the time.h header
int main()
{
    // declarations of variables 
    ...
    clock_t start, finish;  //  declare start and final time
    start = clock();
    // your code is here, do something and then get final time
    finish = clock();
    ( (finish - start)/CLOCKS_PER_SEC );
...

Similarly, in Fortran, this simple example shows how to compute the elapsed time.

PROGRAM time
 REAL :: etime          ! Declare the type of etime()
 REAL :: elapsed(2)     ! For receiving user and system time
 REAL :: total          ! For receiving total time
 INTEGER :: i, j

 WRITE(*,*) 'Start'

 DO i = 1, 5000000  
      j = j + 1
 ENDDO

 total = ETIME(elapsed)
 WRITE(*,*) 'End: total=', total, ' user=', elapsed(1), &
              ' system=', elapsed(2)

END PROGRAM time

Your results may depend on the granularity of the clock.

[SOLVED] FY4150-Project 1

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