Feeding kernels from stomach distributions

Gustav Delius

05/11/2021

Feeding kernel

Now we will discuss how the stomach content is related to the feeding kernel.

The stomach content is the result of a balance between the rate at which prey items are ingested and the rate at which they are digested. So we start by looking at ingestion and digestion rates.

Ingestion

The rate density $I(w|w')$ at which a predator of size $w'$ swallows prey of size $w$ is proportional to the product of the feeding kernel $\phi(w'/w)$ and the prey number density $N_c(w)$: $$I(w|w') = i(w') \phi(w'/w)\ N_c(w)$$ This is a rate density, fin the sense that if one integrates this over a range of prey sizes one gets the rate at which the predator swallows individuals from this size range.

The proportionality constant $i(w')$ only depends on the predator size $w'$ and, as we will see, we will not need it in our analysis. The prey number density $N_c(w)$ is defined as usual such that $N_c(w)dw$ is the total number of potential prey items with sizes between $w$ and $w+dw$.

In a multi-species model in which the predator has different interaction strengths with different prey species, given by an interaction matrix $\theta_{ij}$, the prey number density for a predator of species $i$ would be obtained from the number densities $N_j(w)$ of individual prey species as $$N_c(w) = \sum_j \theta_{ij}N_j(w).$$ Of course, we do not know the prey density that the predators experienced that were sampled in the stomach observation. So the best we can do is to assume that it was reasonably close to the Sheldon spectrum $$N_c(w) = n_c\ w^{-\lambda}$$ with $\lambda \approx 2$.

Digestion

Prey items will stay identifiable in the predator stomach only for a certain period of time before they disintegrate. After that time they will not contribute to the observation of prey items in the stomach. Of course we do not know in detail the rate at which prey items are digested. However it is reasonable to make the approximation that the rate at which a prey item has its body mass digested away scales with body size as $w^{2/3}$ because digestion acts on the surface of the prey item and this surface scales approximately as $w^{2/3}$. Let us assume that a prey item becomes unidentifiable when a fixed percentage of its body mass is digested. The time until that has happened is proportional to the body mass divided by the rate at which mass is digested: $T\propto w/w^{2/3} = w^{1/3}$. The rate at which a prey item of size $w$ disintegrate is equal to the inverse of the disintegration time. Denoting that rate by $D(w)$ we have $$D(w) = d\ w^{-1/3}.$$ The proportionality constant $d$ will not be important in what follows.

Balance equation

We describe the size distribution of prey in stomachs of predators of size $w'$ by the density function $N_w(w|w')$. See the Density functions vignette for a discussion of this density function.

The observed stomach content is such that the rate at which prey items of a particular size are ingested is equal to the rate at which such items disintegrates due to digestion and thus $I(w|w') = D(w)N_w(w|w')$. Using our expressions for these rates we obtain $$i(w') \phi(w'/w)\ n_c\ w^{-\lambda} = d\ w^{-1/3} N_w(w|w').$$ This we can solve for the feeding kernel: $$\phi(w'/w) = \frac{d}{i(w')\,n_c}\ w^{\lambda - 1/3}N_w(w|w').$$ Using further that (see Density functions) $$N_w(w|w') \propto e^lf_l(l)$$ where $l=\log(w'/w)$ and $w\propto e^{- l}$, we find $$\phi(w'/w) \propto e^{(4/3 - \lambda)l}f_l(l)$$ This tells us that if the stomach distribution $f_l(l)$ is described by the truncated exponential distribution with rate $\alpha$, the feeding kernel as a function of $l$ is described by the truncated exponential distribution with rate $\alpha + 4/3 - \lambda$.

If the stomach distribution $f_l(l)$ is the normal distribution with mean $\mu$ and variance $\sigma^2$ then the feeding kernel as a function of $l$ is also a Gaussian with the same variance and with mean $\mu + (4/3 - \lambda)\sigma^2$.

We assume that on average the stomach observations reflect the steady state with $dN/dt=0$. We now substitute our expressions for the various rates into the resulting balance equation: $$0 = i(w') \phi(w'/w)\ w^{-\lambda} + \frac{\partial}{\partial_w}\left(d(w')w^{2/3}N_w(w|w')\right) - u(w')w^{-1/3}N_w(w|w')$$