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188 4 Initial value problems
and substitute these expansions into (4.159),
N N N N
[k+1] [ j] [k] [ j] [k] [ j] [k+1] [ j]
c w = c w + ( t) (1 − θ)A c w + θ A c w
j j j j
j=1 j=1 j=1 j=1
(4.161)
[ j]
Using Aw [ j] = λ j w , collecting terms yields
N
[k+1] [k] [k] [k+1] [ j]
0 = − c + c + c ( t)(1 − θ)λ j + c ( t)θλ j w (4.162)
j j j j
j=1
Setting to zero each component of (4.162) yields
[k+1]
c 1 + ( t)λ j (1 − θ)
j
= (4.163)
[k]
c 1 − ( t)λ j θ
j
For simplicity, let us assume that each λ j of interest for the determination of absolute stability
is real. Then, λ j < 0 and absolute stability requires
[k+1]
c
j 1 − ω j (1 − θ)
= ≤ 1 ω j =−( t)λ j > 0 (4.164)
[k]
c 1 + θω j
j
Let us consider the three cases of interest:
[k+1]
c
j 1 − ω j
explicit Euler [k] = ≤ 1 only when ω j ≤ 2 (4.165)
c 1
j θ=0
[k+1]
c
j 1
implicit Euler [k] = ≤ 1 ∀ω j > 0 (4.166)
c 1 + ω j
j θ=1
[k+1] 1
c
j 1 − ω j
2
Crank–Nicholson [k] = 1 ≤ 1 ∀ω j > 0 (4.167)
c j θ=1/2 1 + ω j
2
For both the implicit Euler and the Crank–Nicholson method, we find that for any ω j > 0,
(4.164) is satisfied and the methods are absolutely stable.
Definition A method is A-stable if it is absolutely stable for all positive time steps.
Thus, the implicit Euler and Crank–Nicholson methods, or more generally (4.150) when-
ever θ ≥ 1/2, are A-stable. Here, we have only shown this to be true when all stable eigen-
values are real. For four values of θ, Figure 4.11 plots the modulus of the growth coefficient
[k+1] [k]
µ j = c /c as a function of ω j in the complex plane. The region of absolute stability,
j j
|µ j |≤ 1, lies within the contour µ j = 1. When the rules are biased more towards the
future state than the old state, θ ≥ 1/2, the method is A-stable.
By contrast, the explicit Euler method loses absolute stability whenever ω j > 2 for
any stable eigenvalue. To see what happens in this case, Figure 4.12 plots the numerical
−t
trajectories for ˙ x =−x, x(0) = 1, for which the true solution is x(t) = e . The explicit