7.5 Two-Dimensional Problems 425
The running equation is Eq. (20), which, withρ^2 = 1 /2, simplifies to
ui(m+ 1 )=1
2
[
uE(m)+uW(m)+uN(m)+uS(m)]
−ui(m− 1 ). (21)To find the starting equation we solve Eq. (21) withm=0 together with the re-
placement equation for the initial-velocity condition, Eq. (18). The equations
are
ui( 1 )+ui(− 1 )=^12[
uE( 0 )+uW( 0 )+uN( 0 )+uS( 0 )]
, (22)
ui( 1 )−ui(− 1 )= 2 tgi. (23)Becauseg(x,y)=0 in this instance, we find
ui( 1 )=^14[
uE( 0 )+uW( 0 )+uN( 0 )+uS( 0 )]
as the starting equation; the right-hand side contains known values ofuonly.
In Fig. 8 are representations of the numerical solution at various time levels.
Thesimplenumericaltechniquewehavedevelopedcanbeadaptedeasily
to treat inhomogeneities, boundary conditions involving derivatives ofu,or
time-varying boundary conditions. Even nonrectangular regions can be han-
dled, provided they fit neatly on a rectangular grid. Several exercises illustrate
these points.
EXERCISES
In Exercises 1–5, set up replacement equations using the given space mesh and
the numbering shown in the figure cited. Then find theui(m)for a few values
ofmusing the largest stable value ofr. Let boundary conditions override the
initial condition if there is a disagreement.
- ∇^2 u=∂u
∂t
,0<x<1, 0 <y< 0 .75, 0 <t,
u( 0 ,y,t)=0, u( 1 ,y,t)=0, 0 <y< 0 .75, 0 <t,
u(x, 0 ,t)=0, u(x, 0. 75 ,t)=1, 0 <x<1, 0 <t,
u(x,y, 0 )=0, 0 <x<1, 0 <y< 0 .75,
x= y= 1 /4. (See Fig. 9a.)- ∇^2 u=∂∂utinR,0<t
u=0onboundary,0<t
u=1inR,t=0.