If the mesh is connected, and if there is at least one fixed value, the methods will still converge to the correct answer, but I will not prove that here - although it is not much harder than the proof I give.Ĭonvergence of the Jacobi and Gauss-Seidel methods for strictly diagonally dominant matrices Thus, this matrix is strictly diagonally dominant and the methods will converge to the correct solution.įor the mesh on a lamina, the diagonal entries for each non-fixed point are 4, and the absolute values of the other elements of each such row sum to 4. Similarly, for row two: 11 > 1+5 and for row three: 13 > 2+1. In our 3 × 3 example, the diagonal entry in row one, 10, is strictly greater than the sum of the absolute values of the other two entries: 10 > 1+3. Strictly diagonally dominant if the absolute value of each diagonal element is strictly greater than the sum of the absolute values of the remaining entries in the same row. One should alos have hope that the method will converge if the matrix is diagonally dominant.Ī matrix is diagonally dominant if the absolute value of each diagonal element is at least as large as the sum of the absolute values of the remaining entries in the same row. This is the class of strictly diagonally dominant matrices. However, there is a class of square matrices for which we can prove they do work. What makes the Jacobi and Gauss-Seidel methods work? Keep in mind that our first example might involve 10,000 linear equations in 10,000 variables. This was a rather small system - small enough that one could find the exact solution Again, for this example, starting with zeroes: X 1, we use that in the calculations for x 2, rather than waiting for the next round. The only different in the implementation of Gauss-Seidel from that of Jacobi is that once we have the new value for Thus, this would seem to be very close to the correct answer. Now use the equations listed above to find new values for each variable.Īnd, we repeat it a few more times (without showing the intermediate steps):Īnd the only digits that have changed since the previous iteration are the last digits of x 1 and x 2. Pick an arbitrary set of starting values for each variable. Rather than subtracting equations, or substituting one equation into another, as is done in the methods you probably saw in previous courses, we will first solve each equation for one of the variables. I will discuss a method for solving these larger problems, if certain conditions hold. They would certainly be fairly bad for 10,000 equations in 10,000 unknowns. The traditional methods (Gaussian elimination, Gauss-Jordan elimination) for solving systems of linear equations given in linear algebra or matrix theory classes would not be of much use for this type of problem since the solutions generated by them tend to have poor control of errors, once the number of equations is more than 20.
There are slight modifications of these for the edge points which are not fixed, but I will ignore that for this example. Our basic assumption is that the temperature at each interior point is an average of the temperatures of its four neighbours on the mesh.
Thus, we have labelled hundreds or thousands of points in the lamina with co-ordinates. Determine what the temperature is at each point of the plate, assuming that we have waited long enough for the temperatures to stabilize.Ī typical way to solve this type of problem is the draw a copy of the lamina in the plane and then place a rectangular grid (or mesh) across the lamina. For example, the left edge might be 100 degrees Fahrenheit and the right edge 0 degrees. Suppose that you have a sheet of metal (a lamina) of uniform thickness but not necessarily rectangular, and that the temperature at some points of the sheet is fixed and known. Let us start with an example from Mechanical Engineering. MAD 2502 Introduction to Computational Mathematics Gauss-Seidel Jacobi and Gauss-Seidel Methodsįor solving large systems of linear equations
0 kommentar(er)
