VONNEUMANN STABILITY ANALYSIS


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1 universitylogo VONNEUMANN STABILITY ANALYSIS School of Mathematics Semester
2 OUTLINE 1 REVIEW 2 EXPONENTIAL GROWTH Stability under exponential growth 3 VONNEUMANN STABILITY ANALYSIS 4 SUMMARY
3 Errors will propogate through the solution; e k = A k e 0. We can bound the errors by the equivalent conditions A 1, or max λ s 1 s Using the norm condition, we find the explicit scheme requires β 1 2 and from eigenvalue analysis the CrankNicolson scheme has no restrictions. universitylogo
4 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.
5 WHY VONNEUMANN? 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.
6 FOURIER SERIES A finite sine or cosine series expansion in the interval a x b takes the form a s sin s where L = b a. ( sπx ) L, or b s cos s ( sπx ) L Now consider an individual component written in complex exponential form at a mesh point x = x j = a + j x A s e isxπ L = A s e isaπ isj xπ L e L = Ā s e iα sj x where α s = sπ/l. universitylogo
7 EXPRESS THE INITIAL CONDITION AS A FOURIER SERIES Given initial data we can express the initial values as wj 0 n = Ā s e iα sj x s=0 j = 0, 1,...,n, We use the n + 1 equations above to determine the n + 1 unknowns Ā.
8 ERRORS GROWING WITH TIME We want to find out how each Fourier mode develops in time. Assume a simple separable solution of the form wj k n = s=0 n = s=0 where ξ = e Ω t. Ā s e iα sj x e Ωt k = Ā s e iα sj x ξ k, Here ξ is called the amplification factor. n Ā s e iαsj x e Ωk t s=0
9 ERRORS GROWING WITH TIME 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.
10 FULLY IMPLICIT SCHEME Consider the fully implicit scheme w k+1 j w k j t = κ [ w k+1 j+1 2wk+1 j + wj 1 k+1 ] x 2 Let us look at just one of the components of the Fourier series = ξ k e iα sj x s w k j Then substituting into the above gives 1 t ξk (ξ 1)e iα sj x = κξk+1 x 2 (e iα s x 2 + e iα s x )e iα sj x. universitylogo
11 universitylogo Thus with β = tκ/ x 2 ( ) ξ 1 = βξ(2 cos(α s x) 2) = 4βξ sin 2 αs x. 2 This gives ξ = β sin 2 ( α s x 2 ), and clearly 0 < ξ 1 for all β > 0 and for all α s. Thus the fully implicit scheme is unconditionally stable.
12 VONNEUMANN ANALYSIS The Richardson scheme is given by w k+1 j w k 1 j t = κ Use vonneumann analysis and write [ ] w k j+1 2wj k + wj 1 k x 2 w k j = ξ k e iα sj x,
13 VONNEUMANN ANALYSIS After substitution we have ξ k 1 (ξ 2 1)e iα sj x = βξ k (e iα s x 2 + e iα s x )e iα sj x. Which gives ( ) ξ 2 1 = 4βξ sin 2 αs x. 2 where β = tκ/ x 2.
14 VONNEUMANN ANALYSIS Thus ( ) ξ 2 + 4βξ sin 2 αs x 1 = 0. 2 This quadratic has two roots ξ 1, ξ 2. The sum and product of the roots is given by ( ) ξ 1 + ξ 2 = 4β sin 2 αs x, ξ 1 ξ 2 = 1. 2
15 universitylogo VONNEUMANN ANALYSIS 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 viceversa. Also if ξ 1 = 1 and ξ 2 = 1 then we must have β = 0. Thus the Richardson scheme is unconditionally unstable.
16 universitylogo VonNeumann 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
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