The Numerical Recipes codes have been rewritten in Fortran 90/95 and C/C++ by us. The original source codes are taken from the widely used software package LAPACK, which follows two other popular packages developed in the 1970s, namely EISPACK and LINPACK.
$$
\mathbf{A} =
\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44}
\end{bmatrix}\qquad
\mathbf{I} =
\begin{bmatrix} 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
The inverse of a matrix is defined by
$$
\mathbf{A}^{-1} \cdot \mathbf{A} = I
$$
| Relations | Name | matrix elements |
|---|---|---|
| \( A = A^{T} \) | symmetric | \( a_{ij} = a_{ji} \) |
| \( A = \left (A^{T} \right )^{-1} \) | real orthogonal | \( \sum_k a_{ik} a_{jk} = \sum_k a_{ki} a_{kj} = \delta_{ij} \) |
| \( A = A^{ * } \) | real matrix | \( a_{ij} = a_{ij}^{ * } \) |
| \( A = A^{\dagger} \) | hermitian | \( a_{ij} = a_{ji}^{ * } \) |
| \( A = \left (A^{\dagger} \right )^{-1} \) | unitary | \( \sum_k a_{ik} a_{jk}^{ * } = \sum_k a_{ki}^{ * } a_{kj} = \delta_{ij} \) |
For an \( N\times N \) matrix \( \mathbf{A} \) the following properties are all equivalent
The basic matrix operations that we will deal with are addition and subtraction
$$
\begin{equation}
\mathbf{A}= \mathbf{B}\pm\mathbf{C} \Longrightarrow a_{ij} = b_{ij}\pm c_{ij},
\tag{1}
\end{equation}
$$
scalar-matrix multiplication
$$
\begin{equation}
\mathbf{A}= \gamma\mathbf{B} \Longrightarrow a_{ij} = \gamma b_{ij},
\tag{2}
\end{equation}
$$
vector-matrix multiplication
$$
\begin{equation}
\mathbf{y}=\mathbf{Ax} \Longrightarrow y_{i} = \sum_{j=1}^{n} a_{ij}x_j,
\tag{3}
\end{equation}
$$
matrix-matrix multiplication
$$
\begin{equation}
\mathbf{A}=\mathbf{BC} \Longrightarrow a_{ij} = \sum_{k=1}^{n} b_{ik}c_{kj},
\tag{4}
\end{equation}
$$
and transposition
$$
\begin{equation}
\mathbf{A}=\mathbf{B}^T \Longrightarrow a_{ij} = b_{ji}
\tag{5}
\end{equation}
$$
Similarly, important vector operations that we will deal with are addition and subtraction
$$
\begin{equation}
\mathbf{x}= \mathbf{y}\pm\mathbf{z} \Longrightarrow x_{i} = y_{i}\pm z_{i},
\tag{6}
\end{equation}
$$
scalar-vector multiplication
$$
\begin{equation}
\mathbf{x}= \gamma\mathbf{y} \Longrightarrow x_{i} = \gamma y_{i},
\tag{7}
\end{equation}
$$
vector-vector multiplication (called Hadamard multiplication)
$$
\begin{equation}
\mathbf{x}=\mathbf{yz} \Longrightarrow x_{i} = y_{i}z_i,
\tag{8}
\end{equation}
$$
the inner or so-called dot product resulting in a constant
$$
\begin{equation}
x=\mathbf{y}^T\mathbf{z} \Longrightarrow x = \sum_{j=1}^{n} y_{j}z_{j},
\tag{9}
\end{equation}
$$
and the outer product, which yields a matrix,
$$
\begin{equation}
\mathbf{A}= \mathbf{yz}^T \Longrightarrow a_{ij} = y_{i}z_{j},
\tag{10}
\end{equation}
$$
We have an \( N\times N \) matrix A with \( N=100 \) In C/C++ this would be defined as
int N = 100;
double A[100][100];
// initialize all elements to zero
for(i=0 ; i < N ; i++) {
for(j=0 ; j < N ; j++) {
A[i][j] = 0.0;
Note the way the matrix is organized, row-major order.
We have \( N\times N \) matrices A, B and C and we wish to evaluate \( A=B+C \).
$$
\mathbf{A}= \mathbf{B}\pm\mathbf{C} \Longrightarrow a_{ij} = b_{ij}\pm c_{ij},
$$
In C/C++ this would be coded like
for(i=0 ; i < N ; i++) {
for(j=0 ; j < N ; j++) {
a[i][j] = b[i][j]+c[i][j]
We have \( N\times N \) matrices A, B and C and we wish to evaluate \( A=BC \).
$$
\mathbf{A}=\mathbf{BC} \Longrightarrow a_{ij} = \sum_{k=1}^{n} b_{ik}c_{kj},
$$
In C/C++ this would be coded like
for(i=0 ; i < N ; i++) {
for(j=0 ; j < N ; j++) {
for(k=0 ; k < N ; k++) {
a[i][j]+=b[i][k]*c[k][j];
ALLOCATE (a(N,N), b(N,N), c(N,N))
DO j=1, N
DO i=1, N
a(i,j)=b(i,j)+c(i,j)
ENDDO
ENDDO
...
DEALLOCATE(a,b,c)
Fortran 90 writes the above statements in a much simpler way
a=b+c
Multiplication
a=MATMUL(b,c)
Fortran contains also the intrinsic functions TRANSPOSE and CONJUGATE.
At least three possibilities in this course
int N;
double ** A;
A = new double*[N]
for ( i = 0; i < N; i++)
A[i] = new double[N];
Always free space when you don't need an array anymore.
for ( i = 0; i < N; i++)
delete[] A[i];
delete[] A;
#include <iostream>
#include <armadillo>
using namespace std;
using namespace arma;
int main(int argc, char** argv)
{
mat A = randu<mat>(5,5);
mat B = randu<mat>(5,5);
cout << A*B << endl;
return 0;
For people using Ubuntu, Debian, Linux Mint, simply go to the synaptic package manager and install armadillo from there. You may have to install Lapack as well. For Mac and Windows users, follow the instructions from the webpage http://arma.sourceforge.net. To compile, use for example (linux/ubuntu)
c++ -O2 -o program.x program.cpp -larmadillo -llapack -lblas
where the -l option indicates the library you wish to link to.
For OS X users you may have to declare the paths to the include files and the libraries as
c++ -O2 -o program.x program.cpp -L/usr/local/lib -I/usr/local/include -larmadillo -llapack -lblas
#include <iostream>
#include "armadillo"
using namespace arma;
using namespace std;
int main(int argc, char** argv)
{
// directly specify the matrix size (elements are uninitialised)
mat A(2,3);
// .n_rows = number of rows (read only)
// .n_cols = number of columns (read only)
cout << "A.n_rows = " << A.n_rows << endl;
cout << "A.n_cols = " << A.n_cols << endl;
// directly access an element (indexing starts at 0)
A(1,2) = 456.0;
A.print("A:");
// scalars are treated as a 1x1 matrix,
// hence the code below will set A to have a size of 1x1
A = 5.0;
A.print("A:");
// if you want a matrix with all elements set to a particular value
// the .fill() member function can be used
A.set_size(3,3);
A.fill(5.0); A.print("A:");
mat B;
// endr indicates "end of row"
B << 0.555950 << 0.274690 << 0.540605 << 0.798938 << endr
<< 0.108929 << 0.830123 << 0.891726 << 0.895283 << endr
<< 0.948014 << 0.973234 << 0.216504 << 0.883152 << endr
<< 0.023787 << 0.675382 << 0.231751 << 0.450332 << endr;
// print to the cout stream
// with an optional string before the contents of the matrix
B.print("B:");
// the << operator can also be used to print the matrix
// to an arbitrary stream (cout in this case)
cout << "B:" << endl << B << endl;
// save to disk
B.save("B.txt", raw_ascii);
// load from disk
mat C;
C.load("B.txt");
C += 2.0 * B;
C.print("C:");
// submatrix types:
//
// .submat(first_row, first_column, last_row, last_column)
// .row(row_number)
// .col(column_number)
// .cols(first_column, last_column)
// .rows(first_row, last_row)
cout << "C.submat(0,0,3,1) =" << endl;
cout << C.submat(0,0,3,1) << endl;
// generate the identity matrix
mat D = eye<mat>(4,4);
D.submat(0,0,3,1) = C.cols(1,2);
D.print("D:");
// transpose
cout << "trans(B) =" << endl;
cout << trans(B) << endl;
// maximum from each column (traverse along rows)
cout << "max(B) =" << endl;
cout << max(B) << endl;
// maximum from each row (traverse along columns)
cout << "max(B,1) =" << endl;
cout << max(B,1) << endl;
// maximum value in B
cout << "max(max(B)) = " << max(max(B)) << endl;
// sum of each column (traverse along rows)
cout << "sum(B) =" << endl;
cout << sum(B) << endl;
// sum of each row (traverse along columns)
cout << "sum(B,1) =" << endl;
cout << sum(B,1) << endl;
// sum of all elements
cout << "sum(sum(B)) = " << sum(sum(B)) << endl;
cout << "accu(B) = " << accu(B) << endl;
// trace = sum along diagonal
cout << "trace(B) = " << trace(B) << endl;
// random matrix -- values are uniformly distributed in the [0,1] interval
mat E = randu<mat>(4,4);
E.print("E:");
// row vectors are treated like a matrix with one row
rowvec r;
r << 0.59499 << 0.88807 << 0.88532 << 0.19968;
r.print("r:");
// column vectors are treated like a matrix with one column
colvec q;
q << 0.81114 << 0.06256 << 0.95989 << 0.73628;
q.print("q:");
// dot or inner product
cout << "as_scalar(r*q) = " << as_scalar(r*q) << endl;
// outer product
cout << "q*r =" << endl;
cout << q*r << endl;
// sum of three matrices (no temporary matrices are created)
mat F = B + C + D;
F.print("F:");
return 0;
#include <iostream>
#include "armadillo"
using namespace arma;
using namespace std;
int main(int argc, char** argv)
{
cout << "Armadillo version: " << arma_version::as_string() << endl;
mat A;
A << 0.165300 << 0.454037 << 0.995795 << 0.124098 << 0.047084 << endr
<< 0.688782 << 0.036549 << 0.552848 << 0.937664 << 0.866401 << endr
<< 0.348740 << 0.479388 << 0.506228 << 0.145673 << 0.491547 << endr
<< 0.148678 << 0.682258 << 0.571154 << 0.874724 << 0.444632 << endr
<< 0.245726 << 0.595218 << 0.409327 << 0.367827 << 0.385736 << endr;
A.print("A =");
// determinant
cout << "det(A) = " << det(A) << endl;
// inverse
cout << "inv(A) = " << endl << inv(A) << endl;
double k = 1.23;
mat B = randu<mat>(5,5);
mat C = randu<mat>(5,5);
rowvec r = randu<rowvec>(5);
colvec q = randu<colvec>(5);
// examples of some expressions
// for which optimised implementations exist
// optimised implementation of a trinary expression
// that results in a scalar
cout << "as_scalar( r*inv(diagmat(B))*q ) = ";
cout << as_scalar( r*inv(diagmat(B))*q ) << endl;
// example of an expression which is optimised
// as a call to the dgemm() function in BLAS:
cout << "k*trans(B)*C = " << endl << k*trans(B)*C;
return 0;
We start with the linear set of equations
$$
\mathbf{A}\mathbf{x} = \mathbf{w}.
$$
We assume also that the matrix \( \mathbf{A} \) is non-singular and that the
matrix elements along the diagonal satisfy \( a_{ii} \ne 0 \). Simple \( 4\times 4 \) example
$$
\begin{bmatrix}
a_{11}& a_{12} &a_{13}& a_{14}\\
a_{21}& a_{22} &a_{23}& a_{24}\\
a_{31}& a_{32} &a_{33}& a_{34}\\
a_{41}& a_{42} &a_{43}& a_{44}\\
\end{bmatrix} \begin{bmatrix}
x_1\\
x_2\\
x_3 \\
x_4 \\
\end{bmatrix}
=\begin{bmatrix}
w_1\\
w_2\\
w_3 \\
w_4\\
\end{bmatrix}.
$$
$$
\begin{align}
a_{11}x_1 +a_{12}x_2 +a_{13}x_3 + a_{14}x_4=&w_1 \nonumber \\
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 + a_{24}x_4=&w_2 \nonumber \\
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 + a_{34}x_4=&w_3 \nonumber \\
a_{41}x_1 + a_{42}x_2 + a_{43}x_3 + a_{44}x_4=&w_4. \nonumber
\end{align}
$$
The basic idea of Gaussian elimination is to use the first equation to eliminate the first unknown \( x_1 \) from the remaining \( n-1 \) equations. Then we use the new second equation to eliminate the second unknown \( x_2 \) from the remaining \( n-2 \) equations. With \( n-1 \) such eliminations we obtain a so-called upper triangular set of equations of the form
$$
\begin{align}
b_{11}x_1 +b_{12}x_2 +b_{13}x_3 + b_{14}x_4=&y_1 \nonumber \\
b_{22}x_2 + b_{23}x_3 + b_{24}x_4=&y_2 \nonumber \\
b_{33}x_3 + b_{34}x_4=&y_3 \nonumber \\
b_{44}x_4=&y_4. \nonumber
\tag{11}
\end{align}
$$
We can solve this system of equations recursively starting from \( x_n \) (in our case \( x_4 \)) and proceed with
what is called a backward substitution.
$$
\begin{equation}
x_m = \frac{1}{b_{mm}}\left(y_m-\sum_{k=m+1}^nb_{mk}x_k\right)\quad m=n-1,n-2,\dots,1.
\tag{12}
\end{equation}
$$
To arrive at such an upper triangular system of equations, we start by eliminating
the unknown \( x_1 \) for \( j=2,n \). We achieve this by multiplying the first equation by \( a_{j1}/a_{11} \) and then subtract
the result from the $j$th equation. We assume obviously that \( a_{11}\ne 0 \) and that
\( \mathbf{A} \) is not singular.
Our actual \( 4\times 4 \) example reads after the first operation
$$
\begin{bmatrix}
a_{11}& a_{12} &a_{13}& a_{14}\\
0& (a_{22}-\frac{a_{21}a_{12}}{a_{11}}) &(a_{23}-\frac{a_{21}a_{13}}{a_{11}}) & (a_{24}-\frac{a_{21}a_{14}}{a_{11}})\\
0& (a_{32}-\frac{a_{31}a_{12}}{a_{11}})& (a_{33}-\frac{a_{31}a_{13}}{a_{11}})& (a_{34}-\frac{a_{31}a_{14}}{a_{11}})\\
0&(a_{42}-\frac{a_{41}a_{12}}{a_{11}}) &(a_{43}-\frac{a_{41}a_{13}}{a_{11}}) & (a_{44}-\frac{a_{41}a_{14}}{a_{11}}) \\
\end{bmatrix} \begin{bmatrix}
x_1\\
x_2\\
x_3 \\
x_4 \\
\end{bmatrix}
=\begin{bmatrix}
y_1\\
w_2^{(2)}\\
w_3^{(2)} \\
w_4^{(2)}\\
\end{bmatrix},
$$
or
$$
\begin{align}
b_{11}x_1 +b_{12}x_2 +b_{13}x_3 + b_{14}x_4=&y_1 \nonumber \\
a^{(2)}_{22}x_2 + a^{(2)}_{23}x_3 + a^{(2)}_{24}x_4=&w^{(2)}_2 \nonumber \\
a^{(2)}_{32}x_2 + a^{(2)}_{33}x_3 + a^{(2)}_{34}x_4=&w^{(2)}_3 \nonumber \\
a^{(2)}_{42}x_2 + a^{(2)}_{43}x_3 + a^{(2)}_{44}x_4=&w^{(2)}_4, \nonumber \\
\tag{13}
\end{align}
$$
The new coefficients are
$$
\begin{equation}
b_{1k} = a_{1k}^{(1)} \quad k=1,\dots,n,
\tag{14}
\end{equation}
$$
where each \( a_{1k}^{(1)} \) is equal to the original \( a_{1k} \) element. The other coefficients are
$$
\begin{equation}
a_{jk}^{(2)} = a_{jk}^{(1)}-\frac{a_{j1}^{(1)}a_{1k}^{(1)}}{a_{11}^{(1)}} \quad j,k=2,\dots,n,
\tag{15}
\end{equation}
$$
with a new right-hand side given by
$$
\begin{equation}
y_{1}=w_1^{(1)}, \quad w_j^{(2)} =w_j^{(1)}-\frac{a_{j1}^{(1)}w_1^{(1)}}{a_{11}^{(1)}} \quad j=2,\dots,n.
\tag{16}
\end{equation}
$$
We have also set \( w_1^{(1)}=w_1 \), the original vector element.
We see that the system of unknowns \( x_1,\dots,x_n \) is transformed into an \( (n-1)\times (n-1) \) problem.
This step is called forward substitution. Proceeding with these substitutions, we obtain the general expressions for the new coefficients
$$
\begin{equation}
a_{jk}^{(m+1)} = a_{jk}^{(m)}-\frac{a_{jm}^{(m)}a_{mk}^{(m)}}{a_{mm}^{(m)}} \quad j,k=m+1,\dots,n,
\tag{17}
\end{equation}
$$
with \( m=1,\dots,n-1 \) and a
right-hand side given by
$$
\begin{equation}
w_j^{(m+1)} =w_j^{(m)}-\frac{a_{jm}^{(m)}w_m^{(m)}}{a_{mm}^{(m)}}\quad j=m+1,\dots,n.
\tag{18}
\end{equation}
$$
This set of \( n-1 \) elimations leads us to an equations which is solved by back substitution.
If the arithmetics is exact and the matrix \( \mathbf{A} \) is not singular, then the computed answer will be exact.
Even though the matrix elements along the diagonal are not zero, numerically small numbers may appear and subsequent divisions may lead to large numbers, which, if added to a small number may yield losses of precision. Suppose for example that our first division in \( (a_{22}-a_{21}a_{12}/a_{11}) \) results in \( -10^{-7} \) and that \( a_{22} \) is one. one. We are then adding \( 10^7+1 \). With single precision this results in \( 10^7 \).
Suppose we want to solve the following boundary value equation
$$
-\frac{d^2u(x)}{dx^2} = f(x,u(x)),
$$
with \( x\in (a,b) \) and with boundary conditions \( u(a)=u(b) = 0 \).
We assume that \( f \) is a continuous function in the domain \( x\in (a,b) \).
Since, except the few cases where it is possible to find analytic solutions, we
will seek after approximate solutions, we choose to represent the approximation to the second derivative
from the previous chapter
$$
f''=\frac{f_h -2f_0 +f_{-h}}{h^2} +O(h^2).
$$
We subdivide our interval \( x\in (a,b) \) into \( n \) subintervals by setting \( x_i = ih \), with \( i=0,1,\dots,n+1 \).
The step size is then given by \( h=(b-a)/(n+1) \) with \( n\in {\mathbb{N}} \).
For the internal grid points \( i=1,2,\dots n \) we replace the differential operator with the above formula
resulting in
$$
u''(x_i) \approx \frac{u(x_i+h) -2u(x_i) +u(x_i-h)}{h^2},
$$
which we rewrite as
$$
u^{''}_i \approx \frac{u_{i+1} -2u_i +u_{i-i}}{h^2}.
$$
We can rewrite our original differential equation in terms of a discretized equation with approximations to the derivatives as
$$
-\frac{u_{i+1} -2u_i +u_{i-i}}{h^2}=f(x_i,u(x_i)),
$$
with \( i=1,2,\dots, n \). We need to add to this system the two boundary conditions \( u(a) =u_0 \) and \( u(b) = u_{n+1} \).
If we define a matrix
$$
\mathbf{A} = \frac{1}{h^2}\begin{bmatrix}
2 & -1 & & & & \\
-1 & 2 & -1 & & & \\
& -1 & 2 & -1 & & \\
& \dots & \dots &\dots &\dots & \dots \\
& & &-1 &2& -1 \\
& & & &-1 & 2 \\
\end{bmatrix}
$$
and the corresponding vectors \( \mathbf{u} = (u_1, u_2, \dots,u_n)^T \) and
\( \mathbf{f}(\mathbf{u}) = f(x_1,x_2,\dots, x_n,u_1, u_2, \dots,u_n)^T \) we can rewrite the differential equation
including the boundary conditions as a system of linear equations with a large number of unknowns
$$
\mathbf{A}\mathbf{u} = \mathbf{f}(\mathbf{u}).
$$
We start with the linear set of equations
$$
\mathbf{A}\mathbf{u} = \mathbf{f},
$$
where \( \mathbf{A} \) is a tridiagonal matrix which we rewrite as
$$
\mathbf{A} = \begin{bmatrix}
b_1& c_1 & 0 &\dots & \dots &\dots \\
a_2 & b_2 & c_2 &\dots &\dots &\dots \\
& a_3 & b_3 & c_3 & \dots & \dots \\
& \dots & \dots &\dots &\dots & \dots \\
& & &a_{n-1} &b_{n-1}& c_{n-1} \\
& & & &a_n & b_n \\
\end{bmatrix}
$$
where \( a,b,c \) are one-dimensional arrays of length \( 1:n \).
In project 1 the arrays \( a \) and \( c \) are equal, namely \( a_i=c_i=-1/h^2 \).
The matrix is also positive definite.
We can rewrite as
$$
\mathbf{A} = \begin{bmatrix}
b_1& c_1 & 0 &\dots & \dots &\dots \\
a_2 & b_2 & c_2 &\dots &\dots &\dots \\
& a_3 & b_3 & c_3 & \dots & \dots \\
& \dots & \dots &\dots &\dots & \dots \\
& & &a_{n-1} &b_{n-1}& c_{n-1} \\
& & & &a_n & b_n \\
\end{bmatrix} \begin{bmatrix}
u_1\\
u_2\\
\dots \\
\dots \\
\dots \\
u_n\\
\end{bmatrix}
=\begin{bmatrix}
f_1\\
f_2\\
\dots \\
\dots \\
\dots \\
f_n\\
\end{bmatrix}.
$$
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. The above tridiagonal system can be written as
$$
a_iu_{i-1}+b_iu_i+c_iu_{i+1} = f_i,
$$
for \( i=1,2,\dots,n \). We see that \( u_{-1} \) and \( u_{n+1} \) are not required and we can set \( a_1=c_n=0 \).
In many applications the matrix is symmetric and we have \( a_i=c_i \).
The algorithm for solving this set of equations is rather simple and requires two steps only,
a forward substitution and a backward substitution. These steps are also
common to the algorithms based on
Gaussian elimination that
we discussed previously. However, due to its simplicity, the number of floating point operations
is in this
case proportional with \( O(n) \) while Gaussian elimination requires \( 2n^3/3+O(n^2) \) floating point operations.
In case your system of equations leads to a tridiagonal matrix, it is clearly an overkill to employ Gaussian elimination or the standard LU decomposition.
Our algorithm starts with forward substitution with a loop over of the elements \( i \) and gives an update of the diagonal elements \( b_i \) given by the new diagonals \( \tilde{b}_i \)
$$
\tilde{b}_i= b_i - \frac{a_ic_{i-1}}{\tilde{b}_{i-1}},
$$
and the new righthand side \( \tilde{f}_i \) given by
$$
\tilde{f}_i= f_i - \frac{a_i\tilde{f}_{i-1}}{\tilde{b}_{i-1}}.
$$
Recall that \( \tilde{b}_1=b_1 \) and \( \tilde{f}_1=f_1 \) always.
The backward substitution gives then the final solution
$$
u_{i-1}= \frac{\tilde{f}_{i-1}-c_{i-1}u_i}{\tilde{b}_{i-1}},
$$
with \( u_n=\tilde{f}_{n}/\tilde{b}_{n} \) when \( i=n \), the last point.
The matrix \( \mathbf{A} \) which rephrases a second derivative in a discretized form is much simpler than the general matrix
$$
\mathbf{A} = \begin{bmatrix}
2 & -1 & 0 & 0 &0 & 0\\
-1 & 2 & -1 &0 &0 &0 \\
0 & -1 & 2 & -1 & 0& 0 \\
0 & \dots & \dots & \dots &\dots & \dots \\
0 &0 &0 &-1 &2& -1 \\
0 & 0 &0 &0 &-1 & 2 \\
\end{bmatrix}.
$$
This matrix fulfills the condition of a weak dominance of the diagonal, with
\( |b_1| > |c_1| \), \( |b_n| > |a_n| \) and \( |b_k| \ge |a_k|+|c_k| \) for \( k=2,3,\dots,n-1 \).
This is a relevant but not sufficient condition to guarantee that the matrix \( \mathbf{A} \) yields a solution to a linear
equation problem. The matrix needs also to be irreducible. A tridiagonal irreducible matrix means that all the elements \( a_i \) and
\( c_i \) are non-zero. If these two conditions are present, then \( \mathbf{A} \) is nonsingular and has a unique LU decomposition.
When setting up the algo it is useful to note that the different operations on the matrix (here as a \( 4\times 4 \) case with diagonals \( d_i \) and off-diagonals \( e_i \)) give is an extremely simple algorithm, namely
$$
\begin{bmatrix}
d_1 & e_1 & 0 & 0 \\
e_2 & d_2 & e_2 & 0 \\
0 & e_3 & d_3 & e_3 \\
0 & 0 & e_4 & d_4
\end{bmatrix}\rightarrow
\begin{bmatrix}
d_1 & e_1 & 0 & 0 \\
0 & \tilde{d}_2 & e_2 & 0 \\
0 & e_3 & d_3 & e_3 \\
0 & 0 & e_4 & d_4
\end{bmatrix}\rightarrow
\begin{bmatrix}
d_1 & e_1 & 0 & 0 \\
0 & \tilde{d}_2 & e_2 & 0 \\
0 & 0 & \tilde{d}_3 & e_3 \\
0 & 0 & e_4 & d_4
\end{bmatrix}
$$
and finally
$$
\begin{bmatrix}
d_1 & e_1 & 0 & 0 \\
0 & \tilde{d}_2 & e_2 & 0 \\
0 & 0 & \tilde{d}_3 & e_3 \\
0 & 0 & 0 & \tilde{d}_4
\end{bmatrix}
$$
We notice the sub-blocks which get repeated
$$
\begin{bmatrix}
d_1 & e_1 & 0 & 0 \\
0 & \tilde{d}_2 & e_2 & 0 \\
0 & 0 & \tilde{d}_3 & e_3 \\
0 & 0 & 0 & \tilde{d}_4
\end{bmatrix}
$$
The matrices we often end up with in rewriting for for example partial differential equations,
have the feature that all leading principal submatrices are non-singular.
$$
\tilde{d}_i= 2 - \frac{1}{\tilde{d}_{i-1}}=\frac{i+1}{i},
$$
and the new righthand side \( \tilde{f}_i \) given by
$$
\tilde{f}_i= f_i + \frac{(i-1)\tilde{f}_{i-1}}{i}.
$$
Recall that \( \tilde{d}_1=2 \) and \( \tilde{f}_1=f_1 \). These arrays can be set up before computing \( u \).
The backward substitution gives then the final solution
$$
u_{i-1}= \frac{i-1}{i}\left(\tilde{f}_{i-1}+u_i\right),
$$
with \( u_n=\tilde{f}_{n}/\tilde{b}_{n} \).
#include <iostream>
#include <fstream>
#include <iomanip>
#include <cmath>
#include <string>
// use namespace for output and input
using namespace std;
// object for output files
ofstream ofile;
// Functions used
inline double f(double x){return 100.0*exp(-10.0*x);
}
inline double exact(double x) {return 1.0-(1-exp(-10))*x-exp(-10*x);}
// Begin main program
int main(int argc, char *argv[]){
int exponent;
string filename;
// We read also the basic name for the output file and the highest power of 10^n we want
if( argc <= 1 ){
cout << "Bad Usage: " << argv[0] <<
" read also file name on same line and max power 10^n" << endl;
exit(1);
}
else{
filename = argv[1]; // first command line argument after name of program
exponent = atoi(argv[2]);
}
// Loop over powers of 10
for (int i = 1; i <= exponent; i++){
int n = (int) pow(10.0,i);
// Declare new file name
string fileout = filename;
// Convert the power 10^i to a string
string argument = to_string(i);
// Final filename as filename-i-
fileout.append(argument);
double h = 1.0/(n);
double hh = h*h;
// Set up arrays for the simple case
double *d = new double [n+1]; double *b = new double [n+1]; double *solution = new double [n+1];
double *x = new double[n+1];
// Quick setup of updated diagonal elements and value of
d[0] = d[n] = 2; solution[0] = solution[n] = 0.0;
for (int i = 1; i < n; i++) d[i] = (i+1.0)/( (double) i);
for (int i = 0; i <= n; i++){
x[i]= i*h;
b[i] = hh*f(i*h);
}
// Forward substitution
for (int i = 2; i < n; i++) b[i] = b[i] + b[i-1]/d[i-1];
// Backward substitution
solution[n-1] = b[n-1]/d[n-1];
for (int i = n-2; i > 0; i--) solution[i] = (b[i]+solution[i+1])/d[i];
ofile.open(fileout);
ofile << setiosflags(ios::showpoint | ios::uppercase);
// ofile << " x: approx: exact: relative error" << endl;
for (int i = 1; i < n;i++) {
double xval = x[i];
double RelativeError = fabs((exact(xval)-solution[i])/exact(xval));
ofile << setw(15) << setprecision(8) << xval;
ofile << setw(15) << setprecision(8) << solution[i];
ofile << setw(15) << setprecision(8) << exact(xval);
ofile << setw(15) << setprecision(8) << log10(RelativeError) << endl;
}
ofile.close();
delete [] x; delete [] d; delete [] b; delete [] solution;
}
return 0;
}
The LU decomposition method means that we can rewrite this matrix as the product of two matrices \( \mathbf{L} \) and \( \mathbf{U} \) where
$$
\begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34} \\
a_{41} & a_{42} & a_{43} & a_{44}
\end{bmatrix}
= \begin{bmatrix}
1 & 0 & 0 & 0 \\
l_{21} & 1 & 0 & 0 \\
l_{31} & l_{32} & 1 & 0 \\
l_{41} & l_{42} & l_{43} & 1
\end{bmatrix}
\begin{bmatrix}
u_{11} & u_{12} & u_{13} & u_{14} \\
0 & u_{22} & u_{23} & u_{24} \\
0 & 0 & u_{33} & u_{34} \\
0 & 0 & 0 & u_{44}
\end{bmatrix}.
$$
LU decomposition forms the backbone of other algorithms in linear algebra, such as the solution of linear equations given by
$$
\begin{align}
a_{11}x_1 +a_{12}x_2 +a_{13}x_3 + a_{14}x_4=&w_1 \nonumber \\
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 + a_{24}x_4=&w_2 \nonumber \\
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 + a_{34}x_4=&w_3 \nonumber \\
a_{41}x_1 + a_{42}x_2 + a_{43}x_3 + a_{44}x_4=&w_4. \nonumber
\end{align}
$$
The above set of equations is conveniently solved by using LU decomposition as an intermediate step.
The matrix \( \mathbf{A}\in \mathbb{R}^{n\times n} \) has an LU factorization if the determinant is different from zero. If the LU factorization exists and \( \mathbf{A} \) is non-singular, then the LU factorization is unique and the determinant is given by
$$
det\{\mathbf{A}\}=det\{\mathbf{LU}\}= det\{\mathbf{L}\}det\{\mathbf{U}\}=u_{11}u_{22}\dots u_{nn}.
$$
There are at least three main advantages with LU decomposition compared with standard Gaussian elimination:
With the LU decomposition it is rather simple to solve a system of linear equations
$$
\begin{align}
a_{11}x_1 +a_{12}x_2 +a_{13}x_3 + a_{14}x_4=&w_1 \nonumber \\
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 + a_{24}x_4=&w_2 \nonumber \\
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 + a_{34}x_4=&w_3 \nonumber \\
a_{41}x_1 + a_{42}x_2 + a_{43}x_3 + a_{44}x_4=&w_4. \nonumber
\end{align}
$$
This can be written in matrix form as
$$ \mathbf{Ax}=\mathbf{w}. $$
where \( \mathbf{A} \) and \( \mathbf{w} \) are known and we have to solve for \( \mathbf{x} \). Using the LU dcomposition we write
$$ \mathbf{A} \mathbf{x} \equiv \mathbf{L} \mathbf{U} \mathbf{x} =\mathbf{w}. $$
The previous equation can be calculated in two steps
$$ \mathbf{L} \mathbf{y} = \mathbf{w};\qquad \mathbf{Ux}=\mathbf{y}. $$
To show that this is correct we use to the LU decomposition to rewrite our system of linear equations as
$$ \mathbf{LUx}=\mathbf{w}, $$
and since the determinat of \( \mathbf{L} \) is equal to 1 (by construction
since the diagonals of \( \mathbf{L} \) equal 1) we can use the inverse of
\( \mathbf{L} \) to obtain
$$
\mathbf{Ux}=\mathbf{L^{-1}w}=\mathbf{y},
$$
which yields the intermediate step
$$
\mathbf{L^{-1}w}=\mathbf{y}
$$
and as soon as we have \( \mathbf{y} \) we can obtain \( \mathbf{x} \)
through \( \mathbf{Ux}=\mathbf{y} \).
For our four-dimentional example this takes the form
$$
\begin{align}
y_1=&w_1 \nonumber\\
l_{21}y_1 + y_2=&w_2\nonumber \\
l_{31}y_1 + l_{32}y_2 + y_3 =&w_3\nonumber \\
l_{41}y_1 + l_{42}y_2 + l_{43}y_3 + y_4=&w_4. \nonumber
\end{align}
$$
and
$$
\begin{align}
u_{11}x_1 +u_{12}x_2 +u_{13}x_3 + u_{14}x_4=&y_1 \nonumber\\
u_{22}x_2 + u_{23}x_3 + u_{24}x_4=&y_2\nonumber \\
u_{33}x_3 + u_{34}x_4=&y_3\nonumber \\
u_{44}x_4=&y_4 \nonumber
\end{align}
$$
This example shows the basis for the algorithm needed to solve the set of \( n \) linear equations.
The algorithm goes as follows
ludcmp(double a, int n, int indx, double &d). This functions returns the LU decomposed matrix \( \bf A \), its determinant and the vector indx which keeps track of the number of interchanges of rows. If the determinant is zero, the solution is malconditioned.lubksb(double a, int n, int indx, double w) which uses the LU decomposed matrix \( \bf A \) and the vector \( \bf w \) and returns \( \bf x \) in the same place as \( \bf w \). Upon exit the original content in \( \bf w \) is destroyed. If you wish to keep this information, you should make a backup of it in your calling function.If the inverse exists then
$$
\mathbf{A}^{-1}\mathbf{A}=\mathbf{I},
$$
the identity matrix. With an LU decomposed matrix we can rewrite the last equation as
$$
\mathbf{LU}\mathbf{A}^{-1}=\mathbf{I}.
$$
If we assume that the first column (that is column 1) of the inverse matrix can be written as a vector with unknown entries
$$
\mathbf{A}_1^{-1}= \begin{bmatrix}
a_{11}^{-1} \\
a_{21}^{-1} \\
\dots \\
a_{n1}^{-1} \\
\end{bmatrix},
$$
then we have a linear set of equations
$$
\mathbf{LU}\begin{bmatrix}
a_{11}^{-1} \\
a_{21}^{-1} \\
\dots \\
a_{n1}^{-1} \\
\end{bmatrix} =\begin{bmatrix}
1 \\
0 \\
\dots \\
0 \\
\end{bmatrix}.
$$
In a similar way we can compute the unknow entries of the second column,
$$
\mathbf{LU}\begin{bmatrix}
a_{12}^{-1} \\
a_{22}^{-1} \\
\dots \\
a_{n2}^{-1} \\
\end{bmatrix}=\begin{bmatrix}
0 \\
1 \\
\dots \\
0 \\
\end{bmatrix},
$$
and continue till we have solved all \( n \) sets of linear equations.
Standard C/C++: fetch the files lib.cpp and lib.h. You can make a directory where you store
these files, and eventually its compiled version lib.o. The example here is program1.cpp from chapter 6 and performs the matrix inversion.
// Simple matrix inversion example
#include <iostream>
#include <new>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include "lib.h"
using namespace std;
/* function declarations */
void inverse(double **, int);
void inverse(double **a, int n)
{
int i,j, *indx;
double d, *col, **y;
// allocate space in memory
indx = new int[n];
col = new double[n];
y = (double **) matrix(n, n, sizeof(double));
ludcmp(a, n, indx, &d); // LU decompose a[][]
printf("\n\nLU form of matrix of a[][]:\n");
for(i = 0; i < n; i++) {
printf("\n");
for(j = 0; j < n; j++) {
printf(" a[%2d][%2d] = %12.4E",i, j, a[i][j]);
// find inverse of a[][] by columns
for(j = 0; j < n; j++) {
// initialize right-side of linear equations
for(i = 0; i < n; i++) col[i] = 0.0;
col[j] = 1.0;
lubksb(a, n, indx, col);
// save result in y[][]
for(i = 0; i < n; i++) y[i][j] = col[i];
} //j-loop over columns
// return the inverse matrix in a[][]
for(i = 0; i < n; i++) {
for(j = 0; j < n; j++) a[i][j] = y[i][j];
free_matrix((void **) y); // release local memory
delete [] col;
delete []indx;
} // End: function inverse()
For Fortran users:
PROGRAM matrix
USE constants
USE F90library
IMPLICIT NONE
! The definition of the matrix, using dynamic allocation
REAL(DP), ALLOCATABLE, DIMENSION(:,:) :: a, ainv, unity
! the determinant
REAL(DP) :: d
! The size of the matrix
INTEGER :: n
....
! Allocate now place in heap for a
ALLOCATE ( a(n,n), ainv(n,n), unity(n,n) )
For Fortran users:
WRITE(6,*) ' The matrix before inversion'
WRITE(6,'(3F12.6)') a
ainv=a
CALL matinv (ainv, n, d)
....
! get the unity matrix
unity=MATMUL(ainv,a)
WRITE(6,*) ' The unity matrix'
WRITE(6,'(3F12.6)') unity
! deallocate all arrays
DEALLOCATE (a, ainv, unity)
END PROGRAM matrix
#include <iostream>
#include "armadillo"
using namespace arma;
using namespace std;
int main()
{
mat A = randu<mat>(5,5);
vec b = randu<vec>(5);
A.print("A =");
b.print("b=");
// solve Ax = b
vec x = solve(A,b);
// print x
x.print("x=");
// find LU decomp of A, if needed, P is the permutation matrix
mat L, U;
lu(L,U,A);
// print l
L.print(" L= ");
// print U
U.print(" U= ");
//Check that A = LU
(A-L*U).print("Test of LU decomposition");
return 0;
}
// This program sets up a simple matrix with random values for the matrix elements
// Matrices are always given by upper case variables
#include <iostream>
#include <new>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
using namespace std;
#define ZERO 1.0E-15
/* function declarations */
double ** AllocateMatrix(int, int);
void DeallocateMatrix(double **, int, int);
void MatrixInverse(double **, int);
void WriteMatrix(double **, int);
void MatrixMultiplication(double **, double **, int);
void LUDecomposition(double **, int, int *);
void LUBackwardSubstitution(double **, int, int *, double *);
// Begin main function, reads from terminal mode the dimension
int main(int argc, char *argv[])
{
// Read from terminal the size of the matrix
int n = atoi(argv[1]);
// Memory for matrix to invert and copy of it
double **A = AllocateMatrix(n, n);
double **B = AllocateMatrix(n, n);
// Define period and seed for standard random number generator
double invers_period = 1./RAND_MAX; // initialise the random number generator
srand(time(NULL)); // This produces the so-called seed in MC jargon
// Setting general square matrices with random matrix elements
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++){
double x = double(rand())*invers_period;
A[i][j] = x;
B[i][j] = A[i][j];
}
}
// Write out original matrix
cout << " Initial matrix A:" << endl;
WriteMatrix(A, n);
// calculate and return inverse matrix
MatrixInverse(B, n);
// Write out inverse matrix
cout << "Inverse matrix of A:" << endl;
WriteMatrix(B, n);
// Check that A^-1A = identity matrix
cout << "Check that we get an identity matrix " << endl;
MatrixMultiplication(A,B,n);
return 0;
} // End: function main()
/*
The function MatrixInverse() performs a matrix inversion
of a square matrix a[][] of dimension n.
*/
void MatrixInverse(double **A, int n)
{
// allocate space in memory
int *indx;
double *column;
indx = new int[n];
column = new double[n];
double **Y = AllocateMatrix(n,n);
// Perform the LU decomposition
LUDecomposition(A, n, indx); // LU decompose a[][]
cout << "LU decomposed matrix A:" << endl;
WriteMatrix(A,n);
// find inverse of a[][] by columns
for(int j = 0; j < n; j++) {
// initialize right-side of linear equations
for(int i = 0; i < n; i++) column[i] = 0.0;
column[j] = 1.0;
LUBackwardSubstitution(A, n, indx, column);
// save result in y[][]
for(int i = 0; i < n; i++) Y[i][j] = column[i];
}
// return the inverse matrix in A[][]
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) A[i][j] = Y[i][j];
}
DeallocateMatrix(Y, n, n); // release local memory
delete [] column;
delete []indx;
} // End: function MatrixInverse()
// Allocate memory for a matrix and initialize the elements to zero
double ** AllocateMatrix(int m, int n){
double ** Matrix;
Matrix = new double*[m];
for(int i=0;i<m;i++){
Matrix[i] = new double[n];
for(int j=0;j<m;j++)
Matrix[i][j] = 0.0;
}
return Matrix;
}
// Free memory
void DeallocateMatrix(double ** Matrix, int m, int n){
for(int i=0;i<m;i++)
delete[] Matrix[i];
delete[] Matrix;
}
// Write out a given matrix
void WriteMatrix(double ** Matrix, int n){
for(int i=0;i < n;i++){
cout << endl;
for (int j=0 ; j < n;j++){
printf(" A[%2d][%2d] = %12.4E",i, j, Matrix[i][j]);
}
}
cout << endl;
}
// Straightforward matrix-matrix multiplication
void MatrixMultiplication(double ** a, double **b, int n){
double **c = AllocateMatrix(n, n);
for(int i=0;i < n;i++){
for (int j=0 ; j < n;j++){
double sum = 0.0;
for (int k = 0; k < n; k++) sum += a[i][k]*b[k][j];
c[i][j] = sum;
}
}
WriteMatrix(c,n);
}
/*
The function
void LUDecomposition(double **a, int n, int *indx)
takes as input a two-dimensional matrix a[][] of dimension n and
replaces it by the LU decomposition of a rowwise permutation of
itself. The results is stored in a[][]
The vector
indx[] records the row permutation effected by the partial pivoting.
*/
void LUDecomposition(double **a, int n, int *indx)
{
int i, imax, j, k;
double big, dum, sum, temp, *vv;
vv = new double [n];
for(i = 0; i < n; i++) { // loop over rows to get scaling information
big = ZERO;
for(j = 0; j < n; j++) {
if((temp = fabs(a[i][j])) > big) big = temp;
}
if(big == ZERO) {
printf("\n\nSingular matrix in routine ludcmp()\n");
exit(1);
}
vv[i] = 1.0/big;
}
for(j = 0; j < n; j++) { // loop over columns of Crout's method
for(i = 0; i< j; i++) { // not i = j
sum = a[i][j];
for(k = 0; k < i; k++) sum -= a[i][k]*a[k][j];
a[i][j] = sum;
}
big = ZERO; // initialization for search for largest pivot element
for(i = j; i< n; i++) {
sum = a[i][j];
for(k = 0; k < j; k++) sum -= a[i][k]*a[k][j];
a[i][j] = sum;
if((dum = vv[i]*fabs(sum)) >= big) {
big = dum;
imax = i;
}
} // end i-loop
if(j != imax) { // do we need to interchange rows ?
for(k = 0;k< n; k++) { // yes
dum = a[imax][k];
a[imax][k] = a[j][k];
a[j][k] = dum;
}
vv[imax] = vv[j]; // also interchange scaling factor
}
indx[j] = imax;
if(fabs(a[j][j]) < ZERO) a[j][j] = ZERO;
if(j < (n - 1)) { // divide by pivot element
dum = 1.0/a[j][j];
for(i=j+1;i < n; i++) a[i][j] *= dum;
}
} // end j-loop over columns
delete [] vv; // release local memory
}
/*
The function
void LUBackwardSubstitution(double **a, int n, int *indx, double *b)
solves the set of linear equations A X = B of dimension n.
a[][] is input, not as the matrix A[][] but rather as
its LU decomposition, indx[] is input as the permutation vector returned by
ludcmp(). b[] is input as the right-hand side vector B,
The solution X is returned in B. The input data a[][],
n and indx[] are not modified. This routine takes into
account the possibility that b[] will begin with many
zero elements, so it is efficient for use in matrix
inversion.
*/
void LUBackwardSubstitution(double **a, int n, int *indx, double *b)
{
int i, ii = -1, ip, j;
double sum;
for(i = 0; i< n; i++) {
ip = indx[i];
sum = b[ip];
b[ip] = b[i];
if(ii > -1) for(j = ii; j < i; j++) sum -= a[i][j] * b[j];
else if(sum) ii = i;
b[i] = sum;
}
for(i = n - 1; i >= 0; i--) {
sum = b[i];
for(j = i+1; j < n; j++) sum -= a[i][j] * b[j];
b[i] = sum/a[i][i];
}
}
It is a simple method for solving
$$
\mathbf{A}\mathbf{x}=\mathbf{b},
$$
where \( \mathbf{A} \) is a matrix and \( \mathbf{x} \) and \( \mathbf{b} \) are vectors. The vector \( \mathbf{x} \) is
the unknown.
It is an iterative scheme where we start with a guess for the unknown, and after \( k+1 \) iterations we have
$$
\mathbf{x}^{(k+1)}= \mathbf{D}^{-1}(\mathbf{b}-(\mathbf{L}+\mathbf{U})\mathbf{x}^{(k)}),
$$
with \( \mathbf{A}=\mathbf{D}+\mathbf{U}+\mathbf{L} \) and
\( \mathbf{D} \) being a diagonal matrix, \( \mathbf{U} \) an upper triangular matrix and \( \mathbf{L} \) a lower triangular
matrix.
If the matrix \( \mathbf{A} \) is positive definite or diagonally dominant, one can show that this method will always converge to the exact solution.
We can demonstrate Jacobi's method by this \( 4\times 4 \) matrix problem. We assume a guess for the vector elements \( x_i^{(0)} \), a guess which represents our first iteration. The new values are obtained by substitution
$$
\begin{align}
x_1^{(1)} =&(b_1-a_{12}x_2^{(0)} -a_{13}x_3^{(0)} - a_{14}x_4^{(0)})/a_{11} \nonumber \\
x_2^{(1)} =&(b_2-a_{21}x_1^{(0)} - a_{23}x_3^{(0)} - a_{24}x_4^{(0)})/a_{22} \nonumber \\
x_3^{(1)} =&(b_3- a_{31}x_1^{(0)} -a_{32}x_2^{(0)} -a_{34}x_4^{(0)})/a_{33} \nonumber \\
x_4^{(1)}=&(b_4-a_{41}x_1^{(0)} -a_{42}x_2^{(0)} - a_{43}x_3^{(0)})/a_{44}, \nonumber
\end{align}
$$
which after \( k+1 \) iterations reads
$$
\begin{align}
x_1^{(k+1)} =&(b_1-a_{12}x_2^{(k)} -a_{13}x_3^{(k)} - a_{14}x_4^{(k)})/a_{11} \nonumber \\
x_2^{(k+1)} =&(b_2-a_{21}x_1^{(k)} - a_{23}x_3^{(k)} - a_{24}x_4^{(k)})/a_{22} \nonumber \\
x_3^{(k+1)} =&(b_3- a_{31}x_1^{(k)} -a_{32}x_2^{(k)} -a_{34}x_4^{(k)})/a_{33} \nonumber \\
x_4^{(k+1)}=&(b_4-a_{41}x_1^{(k)} -a_{42}x_2^{(k)} - a_{43}x_3^{(k)})/a_{44}, \nonumber
\end{align}
$$
We can generalize the above equations to
$$
x_i^{(k+1)}=(b_i-\sum_{j=1, j\ne i}^{n}a_{ij}x_j^{(k)})/a_{ii}
$$
or in an even more compact form as
$$ \mathbf{x}^{(k+1)}= \mathbf{D}^{-1}(\mathbf{b}-(\mathbf{L}+\mathbf{U})\mathbf{x}^{(k)}),
$$
with \( \mathbf{A}=\mathbf{D}+\mathbf{U}+\mathbf{L} \) and
\( \mathbf{D} \) being a diagonal matrix, \( \mathbf{U} \) an upper triangular matrix and \( \mathbf{L} \) a lower triangular
matrix.
Our \( 4\times 4 \) matrix problem
$$
\begin{align}
x_1^{(k+1)} =&(b_1-a_{12}x_2^{(k)} -a_{13}x_3^{(k)} - a_{14}x_4^{(k)})/a_{11} \nonumber \\
x_2^{(k+1)} =&(b_2-a_{21}x_1^{(k)} - a_{23}x_3^{(k)} - a_{24}x_4^{(k)})/a_{22} \nonumber \\
x_3^{(k+1)} =&(b_3- a_{31}x_1^{(k)} -a_{32}x_2^{(k)} -a_{34}x_4^{(k)})/a_{33} \nonumber \\
x_4^{(k+1)}=&(b_4-a_{41}x_1^{(k)} -a_{42}x_2^{(k)} - a_{43}x_3^{(k)})/a_{44}, \nonumber
\end{align}
$$
can be rewritten as
$$
\begin{align}
x_1^{(k+1)} =&(b_1-a_{12}x_2^{(k)} -a_{13}x_3^{(k)} - a_{14}x_4^{(k)})/a_{11} \nonumber \\
x_2^{(k+1)} =&(b_2-a_{21}x_1^{(k+1)} - a_{23}x_3^{(k)} - a_{24}x_4^{(k)})/a_{22} \nonumber \\
x_3^{(k+1)} =&(b_3- a_{31}x_1^{(k+1)} -a_{32}x_2^{(k+1)} -a_{34}x_4^{(k)})/a_{33} \nonumber \\
x_4^{(k+1)}=&(b_4-a_{41}x_1^{(k+1)} -a_{42}x_2^{(k+1)} - a_{43}x_3^{(k+1)})/a_{44}, \nonumber
\end{align}
$$
which allows us to utilize the preceding solution (forward substitution). This improves normally the convergence
behavior and leads to the Gauss-Seidel method!
We can generalize
$$
\begin{align}
x_1^{(k+1)} =&(b_1-a_{12}x_2^{(k)} -a_{13}x_3^{(k)} - a_{14}x_4^{(k)})/a_{11} \nonumber \\
x_2^{(k+1)} =&(b_2-a_{21}x_1^{(k+1)} - a_{23}x_3^{(k)} - a_{24}x_4^{(k)})/a_{22} \nonumber \\
x_3^{(k+1)} =&(b_3- a_{31}x_1^{(k+1)} -a_{32}x_2^{(k+1)} -a_{34}x_4^{(k)})/a_{33} \nonumber \\
x_4^{(k+1)}=&(b_4-a_{41}x_1^{(k+1)} -a_{42}x_2^{(k+1)} - a_{43}x_3^{(k+1)})/a_{44}, \nonumber
\end{align}
$$
to the following form
$$
x^{(k+1)}_i = \frac{1}{a_{ii}} \left(b_i - \sum_{j > i}a_{ij}x^{(k)}_j - \sum_{j < i}a_{ij}x^{(k+1)}_j \right),\quad i=1,2,\ldots,n.
$$
The procedure is generally continued until the changes made by an iteration are below some tolerance.
The convergence properties of the Jacobi method and the Gauss-Seidel method are dependent on the matrix \( \mathbf{A} \). These methods converge when the matrix is symmetric positive-definite, or is strictly or irreducibly diagonally dominant. Both methods sometimes converge even if these conditions are not satisfied.
Given a square system of n linear equations with unknown \( \mathbf x \):
$$
\mathbf{A}\mathbf x = \mathbf b
$$
where
$$
\mathbf{A}=\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}, \qquad \mathbf{x} = \begin{bmatrix} x_{1} \\ x_2 \\ \vdots \\ x_n \end{bmatrix} , \qquad \mathbf{b} = \begin{bmatrix} b_{1} \\ b_2 \\ \vdots \\ b_n \end{bmatrix}.
$$
Then A can be decomposed into a diagonal component D, and strictly lower and upper triangular components L and U:
$$
\mathbf{A} =\mathbf{D} + \mathbf{L} + \mathbf{U},
$$
where
$$
D = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & a_{nn} \end{bmatrix}, \quad L = \begin{bmatrix} 0 & 0 & \cdots & 0 \\ a_{21} & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\a_{n1} & a_{n2} & \cdots & 0 \end{bmatrix}, \quad U = \begin{bmatrix} 0 & a_{12} & \cdots & a_{1n} \\ 0 & 0 & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & 0 \end{bmatrix}.
$$
The system of linear equations may be rewritten as:
$$
(D+\omega L) \mathbf{x} = \omega \mathbf{b} - [\omega U + (\omega-1) D ] \mathbf{x}
$$
for a constant \( \omega > 1 \).
The method of successive over-relaxation is an iterative technique that solves the left hand side of this expression for \( x \), using previous value for \( x \) on the right hand side. Analytically, this may be written as:
$$
\mathbf{x}^{(k+1)} = (D+\omega L)^{-1} \big(\omega \mathbf{b} - [\omega U + (\omega-1) D ] \mathbf{x}^{(k)}\big).
$$
However, by taking advantage of the triangular form of \( (D+\omega L) \), the elements of \( x^{(k+1)} \) can be computed sequentially using forward substitution:
$$
x^{(k+1)}_i = (1-\omega)x^{(k)}_i + \frac{\omega}{a_{ii}} \left(b_i - \sum_{j > i} a_{ij}x^{(k)}_j - \sum_{j < i} a_{ij}x^{(k+1)}_j \right),\quad i=1,2,\ldots,n.
$$
The choice of relaxation factor is not necessarily easy, and depends upon the properties of the coefficient matrix. For symmetric, positive-definite matrices it can be proven that \( 0 < \omega < 2 \) will lead to convergence, but we are generally interested in faster convergence rather than just convergence.
Cubic spline interpolation is among one of the most used methods for interpolating between data points where the arguments are organized as ascending series. In the library program we supply such a function, based on the so-called cubic spline method to be described below.
A spline function consists of polynomial pieces defined on subintervals. The different subintervals are connected via various continuity relations.
Assume we have at our disposal \( n+1 \) points \( x_0, x_1, \dots x_n \) arranged so that \( x_0 < x_1 < x_2 < \dots x_{n-1} < x_n \) (such points are called knots). A spline function \( s \) of degree \( k \) with \( n+1 \) knots is defined as follows
As an example, consider a spline function of degree \( k=1 \) defined as follows
$$
s(x)=\begin{bmatrix} s_0(x)=a_0x+b_0 & x\in [x_0, x_1) \\
s_1(x)=a_1x+b_1 & x\in [x_1, x_2) \\
\dots & \dots \\
s_{n-1}(x)=a_{n-1}x+b_{n-1} & x\in
[x_{n-1}, x_n] \end{bmatrix}.
$$
In this case the polynomial consists of series of straight lines
connected to each other at every endpoint. The number of continuous
derivatives is then \( k-1=0 \), as expected when we deal with straight lines.
Such a polynomial is quite easy to construct given
\( n+1 \) points \( x_0, x_1, \dots x_n \) and their corresponding
function values.
The most commonly used spline function is the one with \( k=3 \), the so-called cubic spline function. Assume that we have in adddition to the \( n+1 \) knots a series of functions values \( y_0=f(x_0), y_1=f(x_1), \dots y_n=f(x_n) \). By definition, the polynomials \( s_{i-1} \) and \( s_i \) are thence supposed to interpolate the same point \( i \), that is
$$
s_{i-1}(x_i)= y_i = s_i(x_i),
$$
with \( 1 \le i \le n-1 \). In total we have \( n \) polynomials of the
type
$$
s_i(x)=a_{i0}+a_{i1}x+a_{i2}x^2+a_{i2}x^3,
$$
yielding \( 4n \) coefficients to determine.
Every subinterval provides in addition the \( 2n \) conditions
$$
y_i = s(x_i),
$$
and
$$
s(x_{i+1})= y_{i+1},
$$
to be fulfilled. If we also assume that \( s' \) and \( s'' \) are continuous,
then
$$
s'_{i-1}(x_i)= s'_i(x_i),
$$
yields \( n-1 \) conditions. Similarly,
$$
s''_{i-1}(x_i)= s''_i(x_i),
$$
results in additional \( n-1 \) conditions. In total we have \( 4n \) coefficients
and \( 4n-2 \) equations to determine them, leaving us with \( 2 \) degrees of
freedom to be determined.
Using the last equation we define two values for the second derivative, namely
$$
s''_{i}(x_i)= f_i,
$$
and
$$
s''_{i}(x_{i+1})= f_{i+1},
$$
and setting up a straight line between \( f_i \) and \( f_{i+1} \) we have
$$
s_i''(x) = \frac{f_i}{x_{i+1}-x_i}(x_{i+1}-x)+
\frac{f_{i+1}}{x_{i+1}-x_i}(x-x_i),
$$
and integrating twice one obtains
$$
s_i(x) = \frac{f_i}{6(x_{i+1}-x_i)}(x_{i+1}-x)^3+
\frac{f_{i+1}}{6(x_{i+1}-x_i)}(x-x_i)^3
+c(x-x_i)+d(x_{i+1}-x).
$$
Using the conditions \( s_i(x_i)=y_i \) and \( s_i(x_{i+1})=y_{i+1} \) we can in turn determine the constants \( c \) and \( d \) resulting in
$$
\begin{align}
s_i(x) =&\frac{f_i}{6(x_{i+1}-x_i)}(x_{i+1}-x)^3+
\frac{f_{i+1}}{6(x_{i+1}-x_i)}(x-x_i)^3 \nonumber \\
+&(\frac{y_{i+1}}{x_{i+1}-x_i}-\frac{f_{i+1}(x_{i+1}-x_i)}{6})
(x-x_i)+
(\frac{y_{i}}{x_{i+1}-x_i}-\frac{f_{i}(x_{i+1}-x_i)}{6})
(x_{i+1}-x).
\tag{19}
\end{align}
$$
How to determine the values of the second derivatives \( f_{i} \) and \( f_{i+1} \)? We use the continuity assumption of the first derivatives
$$
s'_{i-1}(x_i)= s'_i(x_i),
$$
and set \( x=x_i \). Defining \( h_i=x_{i+1}-x_i \) we obtain finally
the following expression
$$
h_{i-1}f_{i-1}+2(h_{i}+h_{i-1})f_i+h_if_{i+1}=
\frac{6}{h_i}(y_{i+1}-y_i)-\frac{6}{h_{i-1}}(y_{i}-y_{i-1}),
$$
and introducing the shorthands \( u_i=2(h_{i}+h_{i-1}) \),
\( v_i=\frac{6}{h_i}(y_{i+1}-y_i)-\frac{6}{h_{i-1}}(y_{i}-y_{i-1}) \),
we can reformulate the problem as a set of linear equations to be
solved through e.g., Gaussian elemination
Gaussian elimination
$$
\begin{bmatrix} u_1 & h_1 &0 &\dots & & & & \\
h_1 & u_2 & h_2 &0 &\dots & & & \\
0 & h_2 & u_3 & h_3 &0 &\dots & & \\
\dots& & \dots &\dots &\dots &\dots &\dots & \\
&\dots & & &0 &h_{n-3} &u_{n-2} &h_{n-2} \\
& && & &0 &h_{n-2} &u_{n-1} \end{bmatrix}
\begin{bmatrix} f_1 \\
f_2 \\
f_3\\
\dots \\
f_{n-2} \\
f_{n-1} \end{bmatrix} =
\begin{bmatrix} v_1 \\
v_2 \\
v_3\\
\dots \\
v_{n-2}\\
v_{n-1} \end{bmatrix}.
$$
Note that this is a set of tridiagonal equations and can be solved
through only \( O(n) \) operations.
The functions supplied in the program library are spline and splint. In order to use cubic spline interpolation you need first to call
spline(double x[], double y[], int n, double yp1, double yp2, double y2[])
This function takes as input \( x[0,..,n - 1] \) and \( y[0,..,n - 1] \) containing a tabulation \( y_i = f(x_i) \) with \( x_0 < x_1 < .. < x_{n - 1} \) together with the first derivatives of \( f(x) \) at \( x_0 \) and \( x_{n-1} \), respectively. Then the function returns \( y2[0,..,n-1] \) which contains the second derivatives of \( f(x_i) \) at each point \( x_i \). \( n \) is the number of points. This function provides the cubic spline interpolation for all subintervals and is called only once.
Thereafter, if you wish to make various interpolations, you need to call the function
splint(double x[], double y[], double y2a[], int n, double x, double *y)
which takes as input the tabulated values \( x[0,..,n - 1] \) and \( y[0,..,n - 1] \) and the output y2a[0,..,n - 1] from spline. It returns the value \( y \) corresponding to the point \( x \).