ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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Half-size division preserves the discrete biomass budget

Example

Use the disjoint partition $[0, 0.5), [0.5, 1.5), [1.5, 2.5), [2.5, 3.5]$ , whose last three cell centers are $1, 2, 3$ . Take $g = 0.1$ , $μ = 0$ , $γ = 1$ , time step $Δ t = 0.1$ , and the half-size daughter kernel $K_{y} = 2 γ (y) δ_{y / 2}$ . Put mass $3$ in the cell centered at size $2$ and zero mass in the other cells. Verify one finite-volume division step and its number and weighted-size budget.

Verification

Given: The size-structured model, finite-volume update and exact discrete moment identities.

[F1] Size transport with a nonlocal division kernel: $K_{y}$ records daughter production and $γ$ records removal of the dividing parent.

[F2] Finite-volume size transport with positive division quadrature: the mass update uses upwind flux, loss $μ + γ$ and nonnegative kernel coefficients.

[F3] Size characteristics, positivity, and discrete number and size-moment balances: the number and center-weighted-size balances include flux, total loss and kernel zeroth/first moments.

1.1F1F2constructalgebra

The parent center $2$ lies in the third cell and its daughters have size $1$ , exactly the second cell center. Thus the positive midpoint kernel quadrature gives $A_{2, 3} = 2$ and all other daughter coefficients for the occupied parent column vanish. The CFL number is $0.1 (0.1 / 0.5 + 1) = 0.12$ in cell 1 and $0.1 (0.1 / 1 + 1) = 0.11$ in each of cells 2–4, so the global maximum is $0.12 \leq 1$ .

2.1F2step 1.1algebra

The rightward internal flux from cell $C_{3}$ into $C_{4}$ is $F_{3} = g M_{3} / h_{3} = 0.3$ ; the absorbing boundary flux is $F_{4} = 0$ . The daughter source is $Δ t A_{2, 3} M_{3} = 0.6$ , and parent removal is $Δ t γ M_{3} = 0.3$ . Therefore the updated cell masses are $M_{2}^{'} = 0.6$ , $M_{3}^{'} = 3 - 0.03 - 0.3 = 2.67$ , and $M_{4}^{'} = 0.03$ , with $M_{1}^{'} = 0$ ; they are all nonnegative.

3.1F1F3step 2.1algebra∎

The discrete number changes from $3$ to $0.6 + 2.67 + 0.03 = 3.3$ , an increase $0.3$ equal to $Δ t (2 - 1) γ M_{3}$ . The weighted-size total changes from $2 \cdot 3 = 6$ to $1 \cdot 0.6 + 2 \cdot 2.67 + 3 \cdot 0.03 = 6.03$ . Formula [F3] gives the same $0.03$ increase from transport, while division contributes $Δ t (2 \cdot 1 - 2 \cdot 1) M_{3} = 0$ . Thus the first moment is conserved by division and changes only by the size-growth flux in this step.

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Sources

Benoît Perthame, Introduction to Structured Equations in Biology, full lecture notes (standard reference, not scraped)
MATH 373 lecture notes, §8 Approximation of Integrals (standard reference, not scraped)