By Hugo. Rossi

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We should be crazy to solve such a simple system by iteration when we know several direct and quick methods (such as substitution or elimination). But if we do solve them by iteration we fmd that, because we know what the result should be, we can see just what is happening when we use iteration and just where it is liable to fail. 3) Does it matter which of the schemes we choose? In fact it matters a great deal. We can write a very simple program to compute the solution of two simultaneous equations Al*X + Bl*Y = Cl A2*X + B2*Y = C2 by the two possible iterative methods.

11 ) 42 Microcomputer Modelling by Finite Differences which shows that we still need only the L(I - 1) and M(I - 1), so that there is in fact no need to carry round all four of the A(I), B(I), C(I) and D(I): we need only two items of information, not four. Putting L(I) as C(I) and M(I) as B(I), we can write 1015 DIM B(20),C(20) and 4025 4030 4035 4040 4045 4050 4052 4053 4055 4060 4065 4070 4075 4080 4082 4084 4086 4088 4090 4095 A1=G2 B(I)=-G3 C(I)=G1 D1=-G4 NEXT I B(2)=B(2)-A1*Y(1) B(2)=B(2)/D1 C(2)=C(2)/D1 B(N-1)=B(N-1)-C(N-1)*Y(N) FOR 1=3 TO N-1 D=D1-A1*C(I-1) B(I)=(B(I)-A1*B(I-1»/D C(I)=C(I)/D NEXT I Y(N-1)=B(N-1) FOR I=N-2 TO 2 STEP -1 O=Y(I) Y(I)=B(I)-C(I)*Y(I+1) D3=D3+ABS(O-Y(I» NEXT I Using this chunk of program instead of the corresponding one in Program 7 should lead to exactly the same answers as before, while using fewer arrays.

We shall need to store the coefficients A(I), B(I), e(I), D(I), replacing them as we modify them. These arrays are dimensioned in line 1015 and set in lines 4025-4040. 5): this happens in lines 4060-4080. 6) to give us the values ofY(I) for I = N - 1, N - 2, ... , 2. 7) are strictly tri-diagonal only if we move the terms involving Y(1) and Y(N) over to the right-hand side, which is where we keep our 'knowns'. 8). This is done in lines 4050 and 4055. 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!!