Mathematical Tools for Physics

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)

]

. (67)