414 Chapter 7 Numerical Methods
7.4 Potential Equation
In this section, we will be concerned with approximate solutions of the po-
tential equation and related equations in a regionRof thexy-plane. For the
sake of simplicity, we will limit ourselves to regions whose boundaries can be
made to coincide with the lines on a sheet of graph paper with square divi-
sions. Thus, we admit such shapes as rectangles,L’s andT’s, but not circles or
triangles. The graph paper provides us with a ready-made mesh of points in
the regionRand on its boundary, at which we wish to know the solution of
ourproblem.Thesepointsaretobenumberedinsomefashion—usuallyleft
to right and bottom to top.
On such a mesh, the replacement for the Laplacian operator is the following:
∂^2 u
∂x^2
+∂
(^2) u
∂y^2
→uW−^2 ui+uE
( x)^2
+uN−^2 ui+uS
( y)^2
, (1)
where the subscriptsE,Wstand for the indices of the mesh points to the left
and right of pointiand the subscriptsN,Sstand for those above and below
(see Fig. 1). The result is sometimes called thefive-point approximation to the
Laplacian. Because we are assuming that x= y,weobtainafurthersim-
plification in the replacement:
∂^2 u
∂x^2 +∂^2 u
∂y^2 →uN+uS+uE+uW− 4 ui
( x)^2. (2)Example.
Solve this problem numerically (see Chapter 4 for the analytical solution):
∂^2 u
∂x^2+∂
(^2) u
∂y^2
= 0 , 0 <x< 1 , 0 <y< 1 , (3)
u( 0 ,y)= 0 , u( 1 ,y)= 0 , 0 <y< 1 , (4)
u(x, 0 )=f(x), u(x, 1 )=f(x), 0 <x< 1 , (5)
f(x)=
{ 2 x, 0 <x< 1
2 ,
2 ( 1 −x),^12 ≤x<1. (6)
Let us take x= y= 1 /4 and number the mesh points inside the 1× 1
square as shown in Fig. 2.
At each of the nine mesh points, we will have the replacement equation
uN+uS+uE+uW− 4 ui= 0. (7)
Together, these make up a system of nine equations in the nine unknowns
u 1 ,u 2 ,...,u 9. Referring to Fig. 2, where the values ofuat boundary points are