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midterm306

# midterm306 - CME 306 Numerical Solutions to Partial...

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CME 306 Numerical Solutions to Partial Differential Equations http://stanford.edu/ ~ pgarapon/cme306.html Part I - Finite difference techniques 1. Prove that the scheme above is consistent with equation ( ?? ) Using Taylor expansions, one shows: u n +1 j - 2 u n j + u n - 1 j t ) 2 = 2 u ∂t 2 + O t 2 ) δ 2 ( δ 2 ( u )) = h 4 4 u ∂x 4 + O ( h 6 ) Which gives the consistency. 2. Rewrite the scheme without the term σ . One substitutes and gets: u n +1 j - 2 u n j + u n - 1 j t ) 2 + ν u n j - 2 - 4 u n j - 1 + 6 u n j - 4 u n j +1 + u n j +2 ( h ) 4 = 0 3. Write the matrix D in M N ( R ) (matrix of size N , with real entries) corresponding to δ 2 . Write the scheme in a matrix form and explicitly define the matrices involved. Denoting ( U n ) j = u n j , one gets the vector equality: U n +1 = (2 I + cD 2 ) U n - U n - 1 where c = ν Δ t 2 h 4 and D is the matrix: D = 2 - 1 0 ... ... 1 - 1 2 - 1 ... ... .. 0 - 1 2 - 1 ... .. ... ... ... ... ... ... .. ... ... - 1 2 - 1 1 .. ... .. - 1 2 4. Following the Von Neumann approach, write a relation between the in- tensities of the Fourier mode k at times n - 1, n and n +1: ˆ u n +1 ( k ) , ˆ u n ( k ) , ˆ u n - 1 ( k ). Classic Fourier transform approach yields the recurrence: (after a little trigonometry) ˆ u n +1 ( k ) = - ˆ u n - 1 ( k ) + (2 - 16 c sin 4 ( kπh ))ˆ u n ( k ) 1

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which can be written in matrix form: ˆ u n +1 ( k ) ˆ u n ( k ) = 2 - 16 c sin 4 ( kπh ) - 1 1 0
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midterm306 - CME 306 Numerical Solutions to Partial...

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