11—Numerical Analysis 346
Then-fold iteration of this, therefore involves only thenthpower of the bracketed expression; that’s why the
exponential form is easier to use in this case. Ifk∆xis small, the first term in the expansion of the sine says that
this is approximately
eikx
[
1 −ikc∆t
]n
,
and with small∆tandn=t/∆ta large number, this is
eikx
[
1 −
ikct
n
]n
≈eik(x−ct).
Looking more closely though, the object in brackets in Eq. ( 65 ) has magnitude
r=
[
1 +
c^2 (∆t)^2
(∆x)^2
sin^2 k∆x
] 1 / 2
> 1. (66)
so the magnitude of the solution grows exponentially. This instability can be pictured as a kind of negative
dissipation. This growth is reduced by requiringkc∆t 1.
Given a finite fixed time interval, is it possible to get there with arbitrary accuracy by making∆tsmall
enough? Withnsteps=t/∆t,rnis
r=
[
1 +
c^2 (∆t)^2
(∆x)^2
sin^2 k∆x
]t/2∆t
= [1 +α]β
=
[
[1 +α]^1 /α
]αβ
≈eαβ
= exp
[
c^2 t∆t
2(∆x)^2
sin^2 k∆x
]
,
so by shrinking∆tsufficiently, this is arbitrarily close to one.
There are several methods to avoid some of these difficulties. One is the Lax-Friedrichs method:
u(t+ ∆t,x) =
1
2
[
u(t,x+ ∆x) +u(t,x−∆x)
]
−
c∆t
2∆x
[
u(t,x+ ∆x)−u(t,x−∆x)