This finishes the presentation of the scheme However we have one outstanding

This finishes the presentation of the scheme however

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This finishes the presentation of the scheme. However, we have one outstanding item (i) is the scheme stable? 5.1.2von Neumann Stability Analysisvon Neumann was a pioneer of mathematics, physics and computing. He developed a method of stabilityanalysis which may be applied easily to the discretised form of the governing equation/s. This requires thefollowing steps:1. Assume a solution to the governing equation which can represent a range of the possible physicalbehaviour2. Substitute this into the discretised governing equations, and simplify
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f ( x )= k = c k e Ik n x (62) where it should be noted that from Euler’s Formula: e Ik n x = cos ( k n x )+ I sin ( k n x ) , e Ik n x = cos ( k n x ) I sin ( k n x ) (63) where c k is a complex wave amplitude and k = π n L is the wave number. By letting n be the temporal index with increment Δ t and j be the spatial index with increment Δ x , the discrete version of the complex exponential Fourier Series can be written as: u ( x )= k = c k e Ik n i Δ x (64) Using this, the solution to the PDE is now written as: u ( x , t )= k = G n e Ik n i Δ x (65) where G n ( k , t ) is the time dependent amplitude of the mode. This will capture the potential oscillatory, exponentially decaying, or exponentially growing behaviour of the PDE solution in time. If | G | > 1 then the amplitude will grow exponentially in time, if | G | = 1 the amplitude will not grow, and if | G | ≤ 1 the amplitude will decrease. Thus a stable scheme has | G | ≤ 1, and the boundary of stability is | G | = 1. For linear PDEs in this course, all modes demonstrate the same physical behaviour. This means that by understanding the behaviour of one single mode, you can understand the behaviour of all the modes in the solution. Thus in von Neumann analysis we assume that the solution consists of just one single mode, thus: u n i = G n e Ik n i Δ x (66) Step 2: Substitution into the Discretised Equation Starting from Eqn. (61), substitute in Eqn. (66), giving: G n + 1 e Ik n i Δ x = G n e Ik n i Δ x + c 2 Δ t Δ x 2 parenleftBig G n e Ik n i + 1 Δ x 2 G n e Ik n i Δ x + G n e Ik n i 1 Δ x parenrightBig (67) Substantial simplification can be made by dividing through both sides by G n e Ik n i Δ x .
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