308 10 Partial Differential Equations
bi j= 2 δi j−δi+ 1 j−δi− 1 j,meaning that we have the following set of eigenequations forcomponenti
(Bˆxˆ)i=μixi,resulting in
(Bˆxˆ)i=n
∑
j= 1(
2 δi j−δi+ 1 j−δi− 1 j)
xj= 2 xi−xi+ 1 −xi− 1 =μixi.If we assume thatxcan be expanded in a basis ofx= (sin(θ),sin( 2 θ),...,sin(nθ))withθ=
lπ/n+ 1 , where we have the endpoints given byx 0 = 0 andxn+ 1 = 0 , we can rewrite the last
equation as
2 sin(iθ)−sin((i+ 1 )θ)−sin((i− 1 )θ) =μisin(iθ),
or
2 ( 1 −cos(θ))sin(iθ) =μisin(iθ),
which is nothing but
2 ( 1 −cos(θ))xi=μixi,
with eigenvaluesμi= 2 −2 cos(θ).
Our requirement in Eq. (10.8) results in
− 1 < 1 −α 2 ( 1 −cos(θ))< 1 ,which is satisfied only ifα<( 1 −cos(θ))−^1 resulting inα≤ 1 / 2 or∆t/∆x^2 ≤ 1 / 2.
A more general tridiagonal matrix
Aˆ=
a b 0 0 ...
c a b 0 ...
... ... ... ... b
0 0... c a
,
has eigenvaluesμi=a+s
√
bccos(iπ/n+ 1 )withi=1 :n, see for example Ref. [56] for a deriva-
tion using a finite difference scheme.
10.2.2Implicit Scheme.
In deriving the equations for the explicit scheme we startedwith the so-called forward for-
mula for the first derivative, i.e., we used the discrete approximation
ut≈
u(xi,tj+∆t)−u(xi,tj)
∆t.
However, there is nothing which hinders us from using the backward formula
ut≈
u(xi,tj)−u(xi,tj−∆t)
∆t,
still with a truncation error which goes likeO(∆t). We could also have used a midpoint ap-
proximation for the first derivative, resulting in
ut≈
u(xi,tj+∆t)−u(xi,tj−∆t)
2 ∆t