VON-NEUMANN STABILITY ANALYSIS

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1 university-logo VON-NEUMANN STABILITY ANALYSIS School of Mathematics Semester

2 OUTLINE 1 REVIEW 2 EXPONENTIAL GROWTH Stability under exponential growth 3 VON-NEUMANN 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 Crank-Nicolson scheme has no restrictions. university-logo

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 VON-NEUMANN? 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. university-logo

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. university-logo

11 university-logo 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 VON-NEUMANN ANALYSIS The Richardson scheme is given by w k+1 j w k 1 j t = κ Use von-neumann analysis and write [ ] w k j+1 2wj k + wj 1 k x 2 w k j = ξ k e iα sj x,

13 VON-NEUMANN 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 VON-NEUMANN 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 university-logo VON-NEUMANN 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 vice-versa. Also if ξ 1 = 1 and ξ 2 = 1 then we must have β = 0. Thus the Richardson scheme is unconditionally unstable.

16 university-logo 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

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