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Gauss-Seidel method

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Introduction

We seek the solution to set of linear equations:

 A  \phi = b

In matrix terms, the the Gauss-Seidel iteration can be expressed as

 
\phi^{(k+1)}  = \left( {D - L} \right)^{ - 1} \left( {U\phi^{(k)}  + b} \right),

where D, L, and U represent the diagonal, lower triangular, and upper triangular parts of the coefficient matrix A and k is the iteration count. This matrix expression is mainly of academic interest, and is not used to program the method. Rather, an element-based approach is used:

 
\phi^{(k+1)}_i  = \frac{1}{a_{ii}} \left(b_i - \sum_{j<i}a_{ij}\phi^{(k+1)}_j-\sum_{j>i}a_{ij}\phi^{(k)}_j\right),\, i=1,2,\ldots,n.

Note that the computation of \phi^{(k+1)}_i uses only those elements of \phi^{(k+1)} that have already been computed and only those elements of \phi^{(k)} that have yet to be advanced to iteration k+1. This means that no additional storage is required, and the computation can be done in place (\phi^{(k+1)} replaces \phi^{(k)}). While this might seem like a rather minor concern, for large systems it is unlikely that every iteration can be stored. Thus, unlike the Jacobi method, we do not have to do any vector copying should we wish to use only one storage vector. The iteration is generally continued until the changes made by an iteration are below some tolerance.

Algorithm

Chose an initial guess \phi^{0}
for k := 1 step 1 until convergence do
for i := 1 step until n do
 \sigma = 0
for j := 1 step until i-1 do
 \sigma  = \sigma  + a_{ij} \phi_j^{(k)}
end (j-loop)
for j := i+1 step until n do
 \sigma  = \sigma  + a_{ij} \phi_j^{(k-1)}
end (j-loop)
  \phi_i^{(k)}  = {{\left( {b_i  - \sigma } \right)} \over {a_{ii} }}
end (i-loop)
check if convergence is reached
end (k-loop)

Example Calculation

In some cases, we need not even explicitly represent the matrix we are solving. Consider the simple heat equation problem

\nabla^2 T = 0

on the unit interval subject to the boundary conditions T(0)=0 and T(1)=1. The standard second-order finite difference discretization is

 T_{i-1}-2T_i+T_{i+1} = 0,

where T_i is the (discrete) solution available at uniformly spaced nodes. In matrix terms, this can be written as

 
\left[ 
\begin{matrix}
   {1  } & {0  } & {0  } & \cdot & \cdot & { 0 } \\ 
   {1  } & {-2 } & {1  } & {   } & {   } & \cdot \\ 
   { 0 } & {1  } & {-2 } & { 1 } & {   } & \cdot \\ 
   {   } & {   } & \cdot & \cdot & \cdot & \cdot \\
   {   } & {   } & {   } & {1  } & {-2 } & {1  }\\
   { 0 } & \cdot & \cdot & { 0 } & { 0 } & {1  }\\ 
\end{matrix}
\right]
\left[ 
\begin{matrix}
   {T_1 }  \\ 
   {T_2 }  \\ 
   {T_3 }  \\
   \cdot   \\
   {T_{n-1} }  \\
   {T_n }  \\
\end{matrix}
\right]
=
\left[ 
\begin{matrix}
   {0}  \\ 
   {0}  \\ 
   {0}   \\
   \cdot   \\
   {0}  \\ 
   {1}  \\
\end{matrix}
\right].

However, for any given T_i for 1 < i < n, we can write

 T_i = \frac{1}{2}(T_{i-1}+T_{i+1}).

Then, stepping through the solution vector from i=2 to i=n-1, we can compute the next iterate from the two surrounding values. Note that (in this scheme), T_{i+1} is from the previous iteration, while T_{i-1} is from the current iteration. The following table gives the results of 10 iterations with 5 nodes (3 interior and 2 boundary) as well as L2 norm error.

Iteration T(1) T(2) T(3) T(4) T(5) L2 error
0 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 1.0000E+00 1.0000E+00
1 0.0000E+00 0.0000E+00 0.0000E+00 5.0000E-01 1.0000E+00 6.1237E-01
2 0.0000E+00 0.0000E+00 2.5000E-01 6.2500E-01 1.0000E+00 3.7500E-01
3 0.0000E+00 1.2500E-01 3.7500E-01 6.8750E-01 1.0000E+00 1.8750E-01
4 0.0000E+00 1.8750E-01 4.3750E-01 7.1875E-01 1.0000E+00 9.3750E-02
5 0.0000E+00 2.1875E-01 4.6875E-01 7.3438E-01 1.0000E+00 4.6875E-02
6 0.0000E+00 2.3438E-01 4.8438E-01 7.4219E-01 1.0000E+00 2.3438E-02
7 0.0000E+00 2.4219E-01 4.9219E-01 7.4609E-01 1.0000E+00 1.1719E-02
8 0.0000E+00 2.4609E-01 4.9609E-01 7.4805E-01 1.0000E+00 5.8594E-03
9 0.0000E+00 2.4805E-01 4.9805E-01 7.4902E-01 1.0000E+00 2.9297E-03
10 0.0000E+00 2.4902E-01 4.9902E-01 7.4951E-01 1.0000E+00 1.4648E-03
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