This finishes the presentation of the scheme however

This finishes the presentation of the scheme. However, we have one outstanding item (i) is the scheme
stable?
5.1.2 von Neumann Stability Analysis
von Neumann was a pioneer of mathematics, physics and computing. He developed a method of stability
analysis which may be applied easily to the discretised form of the governing equation/s. This requires the
following steps:
1. Assume a solution to the governing equation which can represent a range of the possible physical
behaviour
2. Substitute this into the discretised governing equations, and simplify

f ( x )=
∞
∑
k = − ∞
c
k
e
Ik
n
x
(62)
where it should be noted that from Euler’s Formula:
e
Ik
n
x
= cos ( k
n
x )+ I sin ( k
n
x ) , e
− Ik
n
x
= cos ( k
n
x ) − I sin ( k
n
x ) (63)
where c
k
is a complex wave amplitude and k =
π n
L
is the wave number. By letting n be the temporal
index with increment Δ t and j be the spatial index with increment Δ x , the discrete version of the complex
exponential Fourier Series can be written as:
u ( x )=
∞
∑
k = − ∞
c
k
e
Ik
n
i Δ x
(64)
Using this, the solution to the PDE is now written as:
u ( x , t )=
∞
∑
k = − ∞
G
n
e
Ik
n
i Δ x
(65)
where G
n
( k , t ) is the time dependent amplitude of the mode. This will capture the potential oscillatory,
exponentially decaying, or exponentially growing behaviour of the PDE solution in time. If | G | > 1 then
the amplitude will grow exponentially in time, if | G | = 1 the amplitude will not grow, and if | G | ≤ 1 the
amplitude will decrease. Thus a stable scheme has | G | ≤ 1, and the boundary of stability is | G | = 1.
For linear PDEs in this course, all modes demonstrate the same physical behaviour. This means that by
understanding the behaviour of one single mode, you can understand the behaviour of all the modes in the
solution. Thus in von Neumann analysis we assume that the solution consists of just one single mode, thus:
u
n
i
= G
n
e
Ik
n
i Δ x
(66)
Step 2: Substitution into the Discretised Equation
Starting from Eqn. (61), substitute in Eqn. (66), giving:
G
n + 1
e
Ik
n
i Δ x
= G
n
e
Ik
n
i Δ x
+
c
2
Δ t
Δ x 2
parenleftBig
G
n
e
Ik
n
i + 1 Δ x
− 2 G
n
e
Ik
n
i Δ x
+ G
n
e
Ik
n
i − 1 Δ x
parenrightBig
(67)
Substantial simplification can be made by dividing through both sides by G
n
e
Ik
n
i Δ x
.