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Given an FTBS scheme

$\phi_j^{(n+1)} = \phi_j^{(n)} - c \left(\phi_j^{(n)} - \phi_{j-1}^{(n)} \right)$

where $c$ is the courant number, $n$ is the timestep, and $x$ is the spatial index, how can I prove that no new extrema are created at timestep $n+1$? I have only done this informally assuming $0 < c \leq 1$ by looking at the minimum and maximum values of $\phi$:

$ \phi_j^{(n+1)} = 1 - c(1 - 0) = 0 \\ \phi_j^{(n+1)} = 0 - c(0 - 1) = 1 $

hertzsprung
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1 Answers1

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I think you have already proved this (pretty much). FTBS is only stable (and bounded) for 0<=c<=1. FTBS can be re-written:

$\phi^{(n+1)}_j = (1-c)\phi^{(n)}_j + c \phi^{(n)}_{j−1}$

So $ϕ^{(n+1)}_j$ cannot lie outside $ϕ^{(n)}_j$ and $ϕ^{(n)}_{j−1}$. (It is a linear combination of $ϕ^{(n)}_j$ and $ϕ^{(n)}_{j−1}$). So new extrema cannot be generated.

Hilary
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