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A stiff reversible two-compartment transfer

Example

Take state domain $Ω = R^{2}$ and two compartments with transfers $p_{21} = d_{12} = 1000 y_{1}$ and $p_{12} = d_{21} = y_{2}$ , all other transfers zero. This is a stiff reversible conservative system: its fast difference mode has decay rate $1001$ . Starting from $y^{n} = (1, 1)$ and taking $h = 1$ , verify one modified Patankar Euler step and the finite column-generator convention.

Verification

Given: The transfer rates in the Example, the positive input $(1, 1)$ , and step $h = 1$ .

[F1] Finite conservative production destruction systems: the PDS equations are $y_{i}^{'} = \sum_{j} (p_{i j} - d_{i j})$ with $p_{i j} = d_{j i}$ .

[F2] Modified Patankar Euler for conservative transfer systems: for positive input and denominators $σ_{i} = y_{i}^{n}$ , the step solves the transfer-weighted linear system.

[F3] Modified Patankar Euler preserves positivity and total mass: that solve is positive and preserves the sum.

[F4] Finite column generator: a finite column generator has nonnegative off-diagonal entries and each column sums to zero; column $j$ is the source state.

1.1F1F4algebra

The rates satisfy $p_{21} = d_{12} = 1000 y_{1}$ and $p_{12} = d_{21} = y_{2}$ , so [F1] gives $y_{1}^{'} = - 1000 y_{1} + y_{2}$ and $y_{2}^{'} = 1000 y_{1} - y_{2}$ . In the state order $(1, 2)$ the coefficient matrix is $Q = (\begin{array}{cc} - 1000 & 1 \\ 1000 & - 1 \end{array})$ ; its off-diagonal entries are nonnegative and both column sums are zero, so it is a finite column generator under [F4].

1.2F2F3algebra

With $h = 1$ and $y^{n} = (1, 1)$ , [F2] gives $(\begin{array}{cc} 1001 & - 1 \\ - 1000 & 2 \end{array}) y^{n + 1} = (\begin{array}{c} 1 \\ 1 \end{array})$ . The determinant is $1001 \cdot 2 - 1000 = 1002$ , and direct elimination gives $y_{1}^{n + 1} = 3 / 1002 = 1 / 334$ and $y_{2}^{n + 1} = 2001 / 1002 = 667 / 334$ . Both coordinates are positive and their sum is $2$ , as also predicted by [F3].

2.1F1step 1.2algebra∎

For total mass $M = y_{1} + y_{2}$ and $s = 1001$ , the continuous first coordinate obeys $y_{1}^{'} = - s (y_{1} - y_{1}^{*})$ with equilibrium $y_{1}^{*} = M / 1001$ . The one-step formula in step 1.2 satisfies $y_{1}^{n + 1} - y_{1}^{*} = (y_{1}^{n} - y_{1}^{*}) / (1 + h s)$ ; for $h = 1$ this positive damping factor is $1 / 1002$ although $h s = 1001$ . This verifies the large-rate damping for the specified example without inferring a general stability theorem.

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