By Gordon Reece (auth.)
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Extra resources for Microcomputer Modelling by Finite Differences
Example text
All of which gives us the next program. Program 7 (Suitable as listed for the IBM PC or the Apple II) 100 200 300 400 500 999 GOSUB GOSUB GOSUB GOSUB GOSUB STOP 1000: 2000: 3000: 4000: 5000: REM REM REM REM REM INITIALISE SET UP PROBLEM BOUNDARY CONDITIONS SOLVE THE PROBLEM PRINT THE SOLUTION 1000 REM SUBROUTINE FOR INITIALISATION 1010 DIM X(20),Y(20) 1015 DIM A(20),B(20),C(20),D(20) 1020 INPUT "NUMBER OF NODES "iN 1030 INPUT "LENGTH OF DOMAIN "iL 1040 DX = L / (N - 1): REM INTERVAL BETWEEN NODES 1050 REM CALCULATE THE GRID 1060 X(l) = 0 1070 FOR I = 2 TO N 38 Microcomputer Modelling by Finite Differences 1080 XII) = XII - 1) + OX 1090 NEXT I 1100 C3 = 0: REM INITIALISE THE COUNTER C3 1999 RETURN 2000 REM SUBROUTINE FOR THE PHYSICAL PROPERTIES 2010 DEF FN F(X) = 2 2020 INPUT "ENTER A fIlA 2030 INPUT "ENTER B "lB 2040 INPUT "ENTER C "lC 2100 REM CALCULATE G1,G2 AND G4 2110 G1 = - B / 2 / OX - A / OX / OX 2120 G2 = B / 2 / OX - A / OX / DX 2130 G4 = C - 2 * A / OX / OX 2999 RETURN 3000 3010 3020 3030 3999 REM SUBROUTINE FOR SETTING BOUNDARY CONDITIONS INPUT "Y(l) "lY(l) PRINT "VALUE OF Y AT X="lLl INPUT YIN) RETURN 4000 REM SUBROUTINE FOR CALCULATING THE SOLUTION 4010 REM CALCULATE THE Y'S 4015 D3 = 0: REM SET RESIDUAL TO ZERO 4020 FOR I = 2 TO N - 1: REM ONLY TO N-1 NOW!!
Before doing this, fmd out where the RESET or BREAK button is on your computer. Why? 1 should suggest to you that we ought really to be looking at fractional rather than absolute changes. In line 4035 of Program 4 we simply add up the absolute values of the differences between the old and the new values of the dependent variables at each node. This could be improved by replacing line 4035 by a line such as 4035 03=03 + AB5(O-Y(I»/AB5(O + 1E-6) This change will ensure that the values of the residual are 'normalised' before we use them.
But we shall not be using vectors any more, so it does not matter if you have never used the shorthand of vector operators. 1) written out in full for cartesian coordinates in two dimensions is then ~ ax (k axaT) + ~(k ay aT) ay You will often see this simplified to k(aaxT + a T) 2 ay2 2 2 but this form is true only if k is constant in space - that is, if it does not change its value from point to point. 1) is to allow for varying values of k, so we shall use the slightly more complicated form for the left-hand side of the equation.