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Modified Patankar Euler preserves positivity and total mass

Statement

For a finite conservative PDS Finite conservative production destruction systems with domain $Ω$ , an admissible input $y^{n} \in Ω_{+}$ , step $h > 0$ , and any positive denominators $σ_{i} > 0$ , the step of Modified Patankar Euler for conservative transfer systems is uniquely defined, satisfies $y^{n + 1} \geq 0$ , and conserves total mass:

$\sum_{i = 1}^{m} y_{i}^{n + 1} = \sum_{i = 1}^{m} y_{i}^{n} .$

If $y^{n} > 0$ coordinatewise, then $y^{n + 1} > 0$ coordinatewise. These conclusions do not assert order for arbitrary denominators; the first-order choice and its accuracy hypotheses are treated separately.

Facts & Assumptions

Given: The finite conservative PDS identities, $y^{n} \in Ω_{+}$ , $h > 0$ , and $σ_{i} > 0$ for every compartment.

[F1]

Finite conservative production destruction systems: $p_{i j} = d_{j i} \geq 0$ , $p_{i i} = d_{i i} = 0$ , and $y_{i}^{'} = \sum_{j} (p_{i j} - d_{i j})$ .

[F2]

Modified Patankar Euler for conservative transfer systems: $M y^{n + 1} = y^{n}$ , with $M_{i i} = 1 + h \sum_{k \neq i} d_{i k} (y^{n}) / σ_{i}$ and $M_{i j} = - h p_{i j} (y^{n}) / σ_{j}$ for $i \neq j$ .

[F3]

Strict column dominance gives an entrywise nonnegative inverse: a finite matrix with positive diagonal, nonpositive off-diagonal entries, and strict column diagonal dominance has a nonnegative inverse; the inverse maps positive vectors to positive vectors.

Proof

1.1F1F2F3algebra

For every $j$ , $M_{j j} > 0$ and $M_{i j} \leq 0$ when $i \neq j$ by [F1]–[F2]. Moreover, $M_{j j} - \sum_{i \neq j} ∣ M_{i j} ∣ = 1 + \frac{h}{σ_{j}} \sum_{k \neq j} d_{j k} (y^{n}) - \frac{h}{σ_{j}} \sum_{i \neq j} p_{i j} (y^{n}) = 1$ , because $p_{i j} = d_{j i}$ makes the two finite sums equal. Thus [F3] applies: $M$ is invertible and $y^{n + 1} = M^{- 1} y^{n} \geq 0$ , with strict positivity when $y^{n} > 0$ .

2.1F1F2step 1.1algebra∎

The same identity gives each column sum $\sum_{i} M_{i j} = 1$ . Therefore $\sum_{i} y_{i}^{n} = \sum_{i} (M y^{n + 1})_{i} = \sum_{j} (\sum_{i} M_{i j}) y_{j}^{n + 1} = \sum_{j} y_{j}^{n + 1}$ . This proves exact summed conservation.

Depends on

Used by

A stiff reversible two-compartment transfer Example

Dependency tree · one level

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Sources

Stefan Kopecz and Andreas Meister, On order conditions for modified Patankar-Runge-Kutta schemes, §2 (standard reference, not scraped)
Sergio Blanes, Arieh Iserles, and Shev Macnamara, Positivity-preserving methods for ordinary differential equations, §3.4 (standard reference, not scraped)