Solving Boundary Value Problems Using Numerical Analysis
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Solving Boundary Value Problems using Numerical AnalysisDecember 4, 2003Potential TheorySolution of a Two dimensional Boundary Value Problem using Numerical AnalysisNumerical analysis will be used to solve a two dimensional boundary-valued problem.  We will solve for steady-state temperatures of a slab.  All edges are kept at 0 degrees, except one side which is 100 degrees as shown in the figure. The Laplacian equation governs this situation and is given by [pic 1].  We will use the central-difference approximation for the second derivative to solve for the Laplace Equation.  The central-difference approximation is  with an error approximation depending on the value of h.  The approximations for [pic 3]becomes [pic 4]and [pic 5].  Add the two equations and set equal to zero.  The equations lead to the difference equation [pic 6] ,which requires each [pic 7]value to be the average of its four nearest neighbors.  We focus our attention on a square lattice of points with horizontal and vertical separation h.  Our difference equation can be abbreviated to [pic 8] with points labeled as in figure 2.[pic 2]                                [pic 9]                                                                                              [pic 10][pic 11][pic 12][pic 13]                             [pic 14]                   [pic 15]            L          [pic 16][pic 17]                                  [pic 18][pic 19]                                      L    Figure 1                                    [pic 20][pic 21][pic 22][pic 23][pic 24][pic 25][pic 26][pic 27][pic 28][pic 29][pic 30][pic 31][pic 32][pic 33][pic 34]

Figure 2Writing such an equation for each interior point E, we have a linear system in which each equation involves five unknowns, except when a known boundary value reduces the number.  We choose h so that there are only nine interior points, as in Figure 2.  Numbering these points from left to right, top row first, our nine equations for the boundary conditions become:[pic 35][pic 36][pic 37][pic 38][pic 39][pic 40][pic 41][pic 42][pic 43]Solving the nine equations leads to the following solution:[pic 44]  7.1429,    [pic 45]9.8214,   [pic 46]  7.1429[pic 47]18.7500,    [pic 48]25.000,   [pic 49]18.7500[pic 50]42.8571,    [pic 51]52.6786,  [pic 52]42.8571 This problem is homework #1 problem #3 and the analytical solution is [pic 53], with [pic 54]25 which is equivalent to [pic 55].We will now solve for a steady state temperature of a rectangle slab.  We will assign dimensions to the variable L.  Assume that the slab is 20 cm wide and 10 cm high with all edges at 0 degrees except the right edge, which is at 100 degrees.  The boundary condition is shown in Figure 3.  [pic 56]Figure 3ADEBCFigure 4Numbering these points from left to right, top row first, our twenty-one equations for the boundary conditions become:[pic 57][pic 58][pic 59][pic 60][pic 61][pic 62][pic 63][pic 64][pic 65][pic 66][pic 67][pic 68][pic 69][pic 70][pic 71][pic 72][pic 73][pic 74][pic 75][pic 76][pic 77]Solving the twenty-one solutions leads to the following solution:ColumnRow 1Row 2Row 310.3500.49890.35020.91321.28940.913232.01032.83242.010344.29576.01944.295759.153212.65389.1532619.663226.289419.6632743.210153.177443.2101

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