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Unformatted text preview: F1.4ZH2 Numerical Solution of PDEs Page 23 2.6.1 LTE analysis of the θmethod The PDE u t = u xx is approximated on a uniform grid of size Δ x = 1 /N in space and Δ t in time, with approximate solution w n j ≈ u ( x j , t n ) , x j = x + j Δ x , t n = n Δ t, where u ( x, t ) is an exact solution of the PDE. The θmethod scheme is w n +1 j w n j Δ t = (1 θ ) δ 2 x Δ x 2 w n j + θ δ 2 x Δ x 2 w n +1 j where δ 2 x w n j = w n j 1 2 w n j + w n j +1 . The scheme can be written as L Δ w n j ≡ F t Δ t w n j (1 θ ) δ 2 x Δ x 2 w n j θ δ 2 x Δ x 2 w n +1 j = 0 . (2.9) The LTE is found in the usual way by plugging in an exact solution u ( x j , t n ) in place of the approximate solution w n j (for all j , n ) into L Δ w n j , Taylor expanding about ( x, t ) = ( x j , t n ) and eliminating terms using the PDE. Remember, do not multiply or divide (2.9) by Δ t or Δ x when working out the LTE. Recall that when applied to a smooth enough function u ( x, t ) we can write F t u ( x j , t n ) = u ( x j , t n +Δ t ) u ( x j , t n ) = Δ t ∂ ∂t + Δ t 2 2! ∂ 2 ∂t 2 + Δ t 3 3! ∂ 3 ∂t 3 + O (Δ t 4 ) u ( x j , t n ) = Δ t u t + Δ t 2 2! u tt + Δ t 3 3! u ttt ( x j ,t n ) + O (Δ t 4 ) . At line 2 above we have expanded in Taylor series and cancelled terms. Expanding the 2nd central space difference term δ 2 x u ( x j , t n ) gives δ 2 x u ( x j , t n ) = u ( x j +Δ x, t n ) 2 u ( x j , t n ) + u ( x j Δ x, t n ) = Δ x 2 ∂ 2 ∂x 2 + 2Δ x 4 4! ∂ 4 ∂x 4 + O (Δ x 6 ) u ( x j , t n ) = Δ x 2 u xx + Δ x 4 12 u xxxx ( x j ,t n ) + O (Δ x 6 ) . (We have again missed out the details of the Taylor series expansion). The similar term δ 2 x u ( x j , t n +1 ) (time level n + 1) becomes δ 2 x u ( x j , t n +Δ t ) = u ( x j +Δ x, t n +Δ t ) 2 u ( x j , t n +Δ t ) + u ( x j Δ x, t n +Δ t ) = Δ x 2 u xx + Δ x 4 12 u xxxx + O (Δ x 6 ) ( x j ,t n +Δ t ) . F1.4ZH2 Numerical Solution of PDEs Page 24 Note that this term is evaluated at time t = t n + Δ t and so it must also be expanded in Δ t . That is δ 2 x u ( x j , t n +Δ t ) = Δ x 2 u xx + Δ x 4 12 u xxxx + O (Δ x 6 ) ( x j ,t n +Δ t ) = 1 + Δ t ∂ ∂t + Δ t 2 2! ∂ 2 ∂t 2 + . . . Δ x 2 u xx + Δ x 4 12 u xxxx + O (Δ x 6 ) ( x j ,t n ) = Δ x 2 u xx + Δ t Δ x 2 u xxt + Δ x 4 12 u xxxx ( x j ,t n ) + O (Δ t Δ x 4 , Δ t 2 Δ x 2 , Δ x 6 ) We now substitute each of the three terms into (2.9) and collect terms together to get LTE = ( u t u xx ) + Δ t 1 2 u tt θu txx Δ x 2 12 u xxxx + O (Δ t 2 , Δ t Δ x 2 , Δ x 4 ) . Since u is a solution of the PDE, this eliminates u t u xx . Also, differentiating the PDE once with respect to t gives u tt = u txx = u xxxx and hence LTE = Δ t 1 2 θ Δ x 2 12 u xxxx + O (Δ t 2 , Δ t Δ x 2 , Δ x 4 ) ....
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This note was uploaded on 11/14/2011 for the course MATH 480 taught by Professor Sd during the Spring '11 term at Middle East Technical University.
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