AMME2960 Assignment 2.docx - AMME2960 Assignment 2 Part1Q1 2 U tt =c U xx 1 The following numerical scheme will be able to approximate the above

AMME2960 Assignment 2.docx - AMME2960 Assignment 2 Part1Q1...

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AMME2960 Assignment 2 Part1Q1 U tt = c 2 U xx ( 1 ) The following numerical scheme will be able to approximate the above equation: U i n + 1 = 2 U i n U i n 1 + ∆t 2 c 2 ∆x 2 ( U i + 1 n 2 U i n + U i 1 n ) This scheme is appropriate because it is first derived using Talyor Series then fitted into the finite difference approximation both of which are excellent approximation tools hence making this scheme appropriate. Hence as dt or dx gets smaller the model becomes more accurate. From Taylor Series: U i + 1 n = U i n + ∆ xU x n 1 ! + ∆ x 2 U xx n 2 ! + ∆ x 3 U xxx n 3 ! + ∆ x 4 U xxxx n 4 ! + ( a ) ∆ x ( ¿¿ 4 ) U xxxx n 4 ! + ( b ) U i 1 n = U i n + (− ∆ x ) U x n 1 ! + (− ∆ x 2 ) U xx n 2 ! + (− ∆x 3 ) U xxx n 3 ! + ¿ (a)+(b)= ∆ x ( ¿¿ 4 ) U xxxx n 4 ! + ∆x U x n 1 ! + ∆ x 2 U xx n 2 ! + ∆ x 3 U xxx n 3 ! + ∆ x 4 U xxxx n 4 ! + U i + 1 n + U i 1 n = U i n + U i n + (− ∆x ) U x n 1 ! + (− ∆x 2 ) U xx n 2 ! + (− ∆x 3 ) U xxx n 3 ! + ¿ All odd powers will cancel out leaving only the even powers remaining U i + 1 n + U i 1 n = 2 U i n + ∆ x 2 U xx n 2 ! + ∆x 4 U xxxx n 4 ! + U i + 1 n 2 U i n + U i 1 n = 2 U i n + 2 ∆ x 2 U xx n 2 ! + 2 ∆ x 4 U xxxx n 4 ! + U i + 1 n 2 U i n + U i 1 n ∆ x 2 = 2 U i n + U xx n 1 + 2 ∆ x 2 U xxxx n 4 ! +
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U xx n = U i + 1 n 2 U i n + U i 1 n ∆ x 2 ∆ x 4 U xxxx n 12 ( c ) For t U i n + 1 = U i n + ∆t U t n 1 ! + ∆t 2 U tt n 2 ! + ∆t 3 U ttt n 3 ! + ∆t 4 U tttt n 4 ! + ( d ) ∆t ¿ U t n ¿ ∆t ¿ ¿ ¿ 2 U tt n ¿ ∆t ¿ ¿ ¿ 4 U tttt n ¿ ¿ U i n 1 = U i n + ¿ (d)+(f)= ∆t ¿ U t n ¿ ∆t ¿ ¿ ¿ 2 U tt n ¿ ∆t ¿ ¿ ¿ 4 U tttt n ¿ ¿ U i n + 1 + U i n 1 = U i n + U i n + ¿ U i n + 1 + U i n 1 = 2 U i n + 2 ∆t 2 U tt n 2 ! + 2 ∆t 4 U tttt n 4 ! + U i n + 1 2 U i n + U i n 1 = ∆t 2 U tt n + 2 ∆t 4 U tttt n 4 ! +
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U i n + 1 2 U i n + U i n 1 ∆t 2 = U tt n + ∆t 2 U tttt n 12 + (This is essential adding a forward in time model with a backward in time model) U tt n = U i n + 1 2 U i n + U i n 1 ∆t 2 ∆t 2 U tttt n 12 ( g ) (g) & (c) (1) , assuming error terms are not included U i n + 1 2 U i n + U i n 1 ∆t 2 c 2 ( U i + 1 n 2 U i n + U i 1 n ∆x 2 ) = 0 U ∆ x 2 ( ¿¿ i n + 1 2 U i n + U i n 1 )− c 2 ∆t 2 ( U i + 1 n 2 U i n + U i 1 n )= 0 ¿ ∆ x 2 U i n + 1 = c 2 ∆t 2 ( U i + 1 n 2 U i n + U i 1 n ) ∆ x 2 ( 2 U i n + U i n 1 ) U i n + 1 = c 2 ∆t 2 ∆ x 2 ( U i + 1 n 2 U i n + U i 1 n ) + 2 U i n U i n 1 ( h ) Part1Q2 BC’s: U ( 0, t ) = U ( L,t ) = 0 Von Neuman Analysis U i n = G n e Iik ∆ x ( i ) Let α 2 = c 2 ∆t 2 ∆ x 2 (i) (h) G n + 1 e Iik ∆ x = 2 G n e Iik ∆ x G n 1 e Iik∆ x + α 2 ( G n e I ( i + 1 ) k ∆ x 2 G n e Iik ∆x + G n e I ( i 1 ) k ∆ x ) G 2 = 2 G 1 + α 2 ( G e Ik ∆x 2 G + G e Ik ∆ x ) ¿ 2 G ( 1 α 2 ) 1 + 2 α 2 Gcos ( k ∆x )
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1 cos ( k ∆ x ) 1 α 2 ( ¿ )− 1 ¿ 2 G ¿ 1 cos ( k ∆ x ) = 2sin 2 ( k ∆ x 2 ) Identity Cos2A =1–2 sin 2 ( A ) G 2 = 2 G ( 1 2 α 2 sin 2 ( k ∆x 2 ) ) 1 Let 1 2 α 2 sin 2 ( k ∆ x 2 ) = β G 2 2 βG + 1 = 0 Quadratic Equation G = 2 β ± 4 β 2 4 2 = β± β 2 1 If β > 1
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