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

$\nabla^2 T(x) = 0,\ x\in [0,1]$

subject to the boundary conditions $T(0)=0$ and $T(1)=1$ . The exact solution to this problem is $T(x)=x$ . 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:

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

The following table gives the results of 10 iterations with 5 nodes (3 interior and 2 boundary) as well as $L_2$ norm error.

Gauss-Seidel Solution (i=2,3,4)
Iteration	$T_1$	$T_2$	$T_3$	$T_4$	$T_5$	$L_2$ 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

In this situation, the direction that we "sweep" is important - if we step though the solution vector in the opposite direction, the solution moves away from the chosen initial condition (zero everywhere in the interior) more quickly. The iteration is defined by

$T_i^{k+1} = \frac{1}{2}(T_{i-1}^{k}+T_{i+1}^{k+1}),$

and this gives us (slightly) faster convergence, as shown in the table below.

Gauss-Seidel Solution (i=4,3,2)
Iteration	$T_1$	$T_2$	$T_3$	$T_4$	$T_5$	$L_2$ error
0	0.0000E+00	0.0000E+00	0.0000E+00	0.0000E+00	1.0000E+00	1.0000E+00
1	0.0000E+00	1.2500E-01	2.5000E-01	5.0000E-01	1.0000E+00	3.7500E-01
2	0.0000E+00	1.8750E-01	3.7500E-01	6.2500E-01	1.0000E+00	1.8750E-01
3	0.0000E+00	2.1875E-01	4.3750E-01	6.8750E-01	1.0000E+00	9.3750E-02
4	0.0000E+00	2.3438E-01	4.6875E-01	7.1875E-01	1.0000E+00	4.6875E-02
5	0.0000E+00	2.4219E-01	4.8438E-01	7.3438E-01	1.0000E+00	2.3438E-02
6	0.0000E+00	2.4609E-01	4.9219E-01	7.4219E-01	1.0000E+00	1.1719E-02
7	0.0000E+00	2.4805E-01	4.9609E-01	7.4609E-01	1.0000E+00	5.8594E-03
8	0.0000E+00	2.4902E-01	4.9805E-01	7.4805E-01	1.0000E+00	2.9297E-03
9	0.0000E+00	2.4951E-01	4.9902E-01	7.4902E-01	1.0000E+00	1.4648E-03
10	0.0000E+00	2.4976E-01	4.9951E-01	7.4951E-01	1.0000E+00	7.3242E-04

For this toy example, there is not large penalty for choosing the wrong sweep direction. For some of the more complicated variants of Gauss-Seidel, there is a substantial penalty - the sweep direction determines (in a vague sense) the direction in which information travels.

Gauss-Seidel method

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Introduction

Algorithm

Example Calculation

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