test_that("Scaling model is set up correctly") not yet working

[gustavdelius]

The comparison of the trait-based model with perfect_scaling = TRUE against analytic solution is not yet working.

[gustavdelius]

This comment documents the analytic results that the test is checking against, to make it easier to eventually tighten the tolerances.

Setup

The test creates a trait-based model with perfect_scaling = TRUE, n = 2/3, λ = 2 + 3/4 - 2/3, and σ = 1. The underlying assumption is that the total community size spectrum is a perfect power law:

$$N_\text{total}(w) = \kappa, w^{-\lambda}$$

The predation kernel in mizer is the unnormalized log-normal (a function of the predator size $w$ and prey size $w_p$):

$$\phi_i(w,, w_p) = \exp!\left(-\frac{\bigl(\log(w / (w_p,\beta_i))\bigr)^2}{2\sigma_i^2}\right)$$

which peaks when $w/w_p = \beta_i$ (predator is $\beta_i$ times heavier than prey).

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1. Encounter rate

The volumetric encounter rate for predator species $i$ of size $w$ (units: g/year) is

$$E_i(w) = \gamma_i(w) \int_0^\infty \phi_i(w,, w_p); N_R(w_p); w_p; dw_p,$$

where $\gamma_i(w) = \gamma_i w^{q_i}$ is the search volume. Substituting $N_R(w_p) = \kappa, w_p^{-\lambda}$ and making the change of variables $u = \log!\bigl(w/(w_p,\beta_i)\bigr)/\sigma_i$, the Gaussian integral identity

$$\int_{-\infty}^{\infty} e^{-u^2/2;+; au},du = \sqrt{2\pi}; e^{a^2/2}$$

gives (setting $\ell \equiv \lambda - 2$):

$$\boxed{E_i(w) = \gamma_i,\kappa,\sqrt{2\pi},\sigma_i,\beta_i^{,\ell};\exp!\left(\frac{\ell^2\sigma_i^2}{2}\right) w^{q_i - \ell}.}$$

The test checks that getEncounter(p)[sp, ] * w^(λ−2−q) is constant and equal to this prefactor. Because the kernel is in practice truncated at 3σ, the actual computed value also includes the correction factor pnorm(3 − ℓσ) + pnorm(log(β)/σ + ℓσ) − 1.

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2. Feeding level

The Holling type-II feeding level is

$$f_i(w) = \frac{E_i(w)}{E_i(w) + h_i,w^{n_i}}.$$

For $f_i$ to be constant ($= f_0$) across all sizes, $E_i(w)$ must scale as $w^{n_i}$, i.e.

$$q_i - \ell = n_i \quad\Longrightarrow\quad q_i = n_i + \lambda - 2.$$

With this exponent choice the constant feeding level is

$$f_0 = \frac{\gamma_i,\kappa,\sqrt{2\pi},\sigma_i,\beta_i^{,\ell},e^{\ell^2\sigma_i^2/2}}{\gamma_i,\kappa,\sqrt{2\pi},\sigma_i,\beta_i^{,\ell},e^{\ell^2\sigma_i^2/2} + h_i},$$

which the test hard-codes as $f_0 = 0.6$. The test verifies that the computed feeding level is flat to tolerance 1e-14.

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3. Predation mortality

The predation rate (mizer docs, mizerPredRate) experienced by prey of size $w_p$ from species $j$ predators is

$$\text{pred_rate}_j(w_p) = \int_0^\infty \phi_j(w,,w_p);(1-f_j(w));\gamma_j(w);N_j(w);dw.$$

The factor $(1-f_j)$ arises from the Holling type-II response: the actual feeding rate equals $f h w^n$, and using the identity $f h w^n = (1-f),E(w)$ the clearance rate per unit prey density becomes $(1-f),\gamma,w^q,\phi$.

Substituting $N(w) = \kappa,w^{-\lambda}$ and $(1-f_j) = 1-f_0$ (constant), the same Gaussian integral with $a = \sigma(n-1)$ (because $q - \lambda + 1 = n - 1$) yields

$$\mu_p(w_p) = \mu_0; w_p^{,n-1},$$

where

$$\boxed{\mu_0 = (1-f_0),\sqrt{2\pi};\kappa,\gamma,\sigma;\beta^{n-1};\exp!\left(\frac{(n-1)^2\sigma^2}{2}\right).}$$

The test currently uses tolerance 0.2, marked as a TODO.

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4. Growth rate

The net somatic growth rate is

$$g(w) = \bigl(\alpha,h,f_0 - k_s\bigr),w^n,(1-\psi(w)) = \bar{h},w^n,(1-\psi(w)),$$

with $\bar{h} = \alpha h f_0 - k_s$. This follows directly from the constant feeding level: gross intake is $f_0 h w^n$, standard metabolism costs $k_s w^n$, and the fraction $(1-\psi)$ goes to somatic growth. The test checks this to tolerance 1e-4 (also a TODO).

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5. Perfect power-law community

The perfect_scaling = TRUE flag initialises the fish spectra so that

$$\sum_i N_i(w) = \kappa,w^{-\lambda} - N_R(w) \quad \text{for all } w \text{ covered by fish,}$$

i.e. fish exactly fill the gap between $\kappa w^{-\lambda}$ and the truncated resource spectrum. The test verifies this algebraic identity to tolerance 1e-15.

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6. Equal reproductive efficiencies

Self-similarity of the scaling solution requires all species to have the same reproductive efficiency $e_{\rm repro}$; the test checks this with expect_equal.

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The current loose tolerances (1e-1 for encounter, 0.2 for mortality, 1e-4 for growth) reflect the fact that the numerical solution is not yet close enough to the analytic predictions.