1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

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416 Chapter 7 Numerical Methods


Figure 3 Numerical solution of Eqs. (3)–(6).

Example.
Set up the replacement equations for the problem


∂^2 u
∂x^2 +

∂^2 u
∂y^2 =^16 (u−^1 ),^0 <x<^1 ,^0 <y<^1 , (9)
u(x, 0 )= 0 , u(x, 1 )= 0 , 0 <x< 1 , (10)
u( 0 ,y)= 0 , u( 1 ,y)= 0 , 0 <y< 1. (11)

We may use the same numbering as in the first example (Fig. 2). At each mesh
point, the replacement is


uN+uS+uE+uW− 4 ui
( x)^2

= 16 (ui− 1 ). (12)

Because x= 1 /4,( 1 / x)^2 =16, and the typical replacement equation be-
comes


uN+uS+uE+uW− 4 ui=ui− 1 ,

or


uN+uS+uE+uW− 5 ui=− 1. (13)

Finally, we may write out the equations to be solved. The first four of the nine
equations, corresponding to Eq. (13) withi= 1 , 2 , 3 ,4, are


u 2 +u 4 − 5 u 1 =− 1
u 1 +u 3 +u 5 − 5 u 2 =− 1
u 2 +u 6 − 5 u 3 =− 1
u 1 +u 5 +u 7 − 5 u 4 =− 1.

(14)

The solution of this problem is left as an exercise. 


On more complicated regions, the replacement for the Laplacian operator
hasexactlythesameform,sincewestillusethe“graph-papermesh.”Thesys-

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