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Modified Patankar Euler converges globally with first-order error on a positive tube

Statement

Let $T > 0$ , $N \geq 1$ , $h = T / N$ , and $t_{n} = n h$ . Use sets $U, V$ satisfying the domain and regularity hypotheses of Modified Patankar Euler has a uniform first-order defect and one-step bound on positive tubes, and let $K \subset U$ be a compact positive tube. Let $y : [0, T] \to U$ solve the ODE of a finite conservative PDS Finite conservative production destruction systems, let $y_{0} = y (0)$ , and obtain $y_{n + 1}$ by Modified Patankar Euler for conservative transfer systems with $σ_{i} = (y_{n})_{i}$ . Suppose the exact path and every numerical state lie in the common compact tube $K$ , and $h \leq h_{0}$ for the local constants in that lemma. Then there is a constant $C_{T}$ , independent of $N$ , such that

$\max_{0 \leq n \leq N} ∥ y_{n} - y (t_{n}) ∥_{2} \leq C_{T} h .$

The conclusion is conditional on the stated positive compact tube; positivity and mass conservation alone do not establish that a trajectory remains in such a tube for a prescribed interval.

Facts & Assumptions

Given: A solution of the finite PDS on $[0, T]$ , exact initial data, a fixed uniform grid, and the assumption that exact and numerical states stay in the common compact positive tube $K$ .

[F1]

Finite conservative production destruction systems: the ODE has finite paired internal transfer rates.

[F2]

First-order systems, initial value problems, and solutions on intervals: the exact trajectory is a solution of the stated first-order IVP on its interval.

[F3]

Modified Patankar Euler for conservative transfer systems: the standard first-order step uses $σ_{i} = (y_{n})_{i}$ and solves its frozen-rate linear system.

[F4]

Modified Patankar Euler has a uniform first-order defect and one-step bound on positive tubes: on $K$ and for $h \leq h_{0}$ , the exact-start defect is at most $C h^{2}$ and the step map is Lipschitz with factor $1 + L h$ .

[F5]

Global error from uniform one-step defect and stability: on a fixed grid and a common compact tube containing the exact path and numerical states, initial error $C_{0} h^{p}$ , exact-start defect $C h^{p + 1}$ , and one-step factor $1 + L h$ imply a uniform global $O (h^{p})$ error with the stated exponential constant.

Proof

1.1givenF1F2F3F4F5

The assumption $y_{n} \in K \subset (0, \infty)^{m}$ makes every standard denominator $σ_{i} = (y_{n})_{i}$ positive, so [F3] is admissible at every step. By [F4] the local defect is bounded by $C h^{2}$ and the one-step map has factor $1 + L h$ , with constants independent of $n$ and $N$ . The initial error is zero, and the tube hypothesis supplies the common-tube condition of [F5].

2.1F5step 1.1algebra∎

Apply [F5] with $p = 1$ and $C_{0} = 0$ . If $L > 0$ , its bound is $C (\exp (L T) - 1) h / L$ ; if $L = 0$ , it is $C T h$ . Both are $C_{T} h$ for a constant independent of $N$ .

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Sources

Ernst Hairer, course notes, Chapter III: Équations Différentielles Ordinaires (standard reference, not scraped)
Mark Embree, Lecture Notes on Numerical Analysis, global error analysis for one-step methods (standard reference, not scraped)