Solving Boundary Value Problems Using Numerical Analysis

Solving Boundary Value Problems using Numerical Analysis

December 4, 2003

Potential Theory

Solution of a Two dimensional Boundary Value Problem using Numerical Analysis

Numerical 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 2

Writing 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 3

Figure 4

Numbering 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: