Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
V ON -N EUMANN S TABILITY A NALYSIS
Dr. Johnson
School of Mathematics
Semester 1 2008
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Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
O UTLINE
1
R EVIEW
2
E XPONENTIAL G ROWTH
Stability under exponential growth
3
V ON -N EUMANN S TABILITY A NALYSIS
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
4
S UMMARY
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Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Errors will propogate through the solution;
ek = Ak e0 .
We can bound the errors by the equivalent conditions
||A|| ≤ 1,
or
max |λs | ≤ 1
s
Using the norm condition, we find the explicit scheme requires
β≤
1
2
and from eigenvalue analysis the Crank-Nicolson scheme has
no restrictions.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Stability under exponential growth
The condition ||A|| ≤ 1 does not make allowance for solutions
of the pde which may be growing exponentially in time.
A necessary and sufficient condition for stability when the
solution of the pde is increasing exponentially in time is:
||A|| ≤ 1 + M∆t = 1 + O (∆t )
where M is a constant independent of ∆x and ∆t.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
W HY V ON -N EUMANN ?
A very versatile tool for analysing stability is the Fourier
method developed by von Neumann.
Here initial values at mesh points are expressed in terms of a
finite Fourier series, and we consider the growth of individual
Fourier components.
We do not need to find eigenvalues, or matrix norms.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
F OURIER S ERIES
A finite sine or cosine series expansion in the interval
a ≤ x ≤ b takes the form
∑ as sin
s
³ sπx ´
L
or
,
∑ bs cos
s
³ sπx ´
L
where L = b − a.
Now consider an individual component written in complex
exponential form at a mesh point x = xj = a + j ∆x
As e
isx π
L
= As e
isaπ
L
e
isj ∆x π
L
= Ās e iαs j ∆x
where αs = sπ/L.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
E XPRESS THE INITIAL CONDITION AS A F OURIER
SERIES
Given initial data we can express the initial values as
wj0 =
n
∑ Ās e iα j ∆x
s
j = 0, 1, . . . , n,
s =0
We use the n + 1 equations above to determine the n + 1
unknowns Ā.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
E RRORS G ROWING WITH T IME
We want to find out how each Fourier mode develops in time.
Assume a simple separable solution of the form
n
wjk =
∑
Ās e iαs j ∆x e Ωtk =
s =0
n
=
n
∑ Ās e iα j ∆x e Ωk∆t
s
s =0
∑ Ās e iα j ∆x ξ k ,
s
s =0
where ξ = e Ω∆t .
Here ξ is called the amplification factor.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
E RRORS G ROWING WITH T IME
For stability we thus require:
|ξ | ≤ 1.
If the exact solution of the pde grows exponentially, then the
difference scheme will allow such solutions if
|ξ | ≤ 1 + M∆t
where M does not depend on ∆x or ∆t.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
F ULLY I MPLICIT S CHEME
Consider the fully implicit scheme
" k +1
#
wj +1 − 2wjk +1 + wjk−+11
wjk +1 − wjk
=κ
∆t
∆x 2
Let us look at just one of the components of the Fourier series
¯
¯
wjk ¯ = ξ k e iαs j ∆x
s
Then substituting into the above gives
1 k
κξ k +1 −iαs ∆x
ξ (ξ − 1)e iαs j ∆x =
(e
− 2 + e iαs ∆x )e iαs j ∆x .
∆t
∆x 2
Dr. Johnson
MATH65241
university-logo
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
Thus with β = ∆tκ/∆x 2
2
ξ − 1 = βξ (2 cos(αs ∆x ) − 2) = −4βξ sin
µ
αs ∆x
2
¶
.
This gives
ξ=
1
1 + 4β sin2
³
αs ∆x
2
´,
and clearly 0 < ξ ≤ 1 for all β > 0 and for all αs .
Thus the fully implicit scheme is unconditionally stable.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
V ON -N EUMANN A NALYSIS
The Richardson scheme is given by
wjk +1 − wjk −1
∆t
=κ
"
wjk+1 − 2wjk + wjk−1
∆x 2
#
Use von-Neumann analysis and write
wjk = ξ k e iαs j ∆x ,
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
V ON -N EUMANN A NALYSIS
After substitution we have
ξ k −1 (ξ 2 − 1)e iαs j ∆x = βξ k (e −iαs ∆x − 2 + e iαs ∆x )e iαs j ∆x .
Which gives
2
2
ξ − 1 = −4βξ sin
µ
αs ∆x
2
¶
.
where β = ∆tκ/∆x 2 .
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
V ON -N EUMANN A NALYSIS
Thus
2
2
ξ + 4βξ sin
µ
αs ∆x
2
¶
− 1 = 0.
This quadratic has two roots ξ 1 , ξ 2 . The sum and product of
the roots is given by
ξ 1 + ξ 2 = −4β sin2
µ
αs ∆x
2
¶
,
ξ 1 ξ 2 = −1.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Fourier Series
Growth of Errors
The Fully Implicit Scheme
Richardson’s Scheme
V ON -N EUMANN A NALYSIS
For stability we require |ξ 1 | ≤ 1 and |ξ 2 | ≤ 1
The product of the roots show that if |ξ 1 | < 1 then |ξ 2 | > 1,
and vice-versa.
Also if ξ 1 = 1 and ξ 2 = −1 then we must have β = 0.
Thus the Richardson scheme is unconditionally unstable.
university-logo
Dr. Johnson
MATH65241
Review
Exponential Growth
Von-Neumann Stability Analysis
Summary
Von-Neumann is one of the simplest ways to evaluate stability.
We look at the growth rate of the Fourier components.
The stability conditions on the amplification factor ξ are:
|ξ | ≤ 1
and if the exact solution of the pde grows exponentially:
|ξ | ≤ 1 + M∆t
university-logo
Dr. Johnson
MATH65241