1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

7.5 Two-Dimensional Problems 423

Figure 7 Mesh numbering for numerical solution of Eqs. (9)–(11).

Because x= y= 1 /5, the parameterrin Eq. (8) is

r=^ t

x^2

= 25     t.

Clearly, the longest stable time step is t= 1 /100, corresponding tor= 1 /4.
Using this value ofrsimplifies the typical replacement equation to

ui(m+ 1 )=

1

4

(

uN(m)+uS(m)+uE(m)+uW(m)

)

. (12)

Specifically, we have

u 1 (m+ 1 )=^1

4

(

u 2 (m)+u 5 (m)+ 2 f(tm)

)

,

u 2 (m+ 1 )=

1

4

(

u 1 (m)+u 3 (m)+u 6 (m)+f(tm)

)

,

and so on. Thef(tm)terms enter because point 1 is adjacent to two boundary
points and point 2 to one boundary point. Note that symmetry about the line
through points 4 and 7 makes it unnecessary to computeu 8 (m),...,u 12 (m).
Table 10 contains calculated values ofufor the first four time levels.

Wave Problems

In solving two-dimensional wave problems, we replace the Laplacian, as in the
foregoing, and use a central difference for the time derivative, as we did in
Section 7.3:
∂^2 u
∂t^2 →

ui(m+ 1 )− 2 ui(m)+ui(m− 1 )

t^2. (13)