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The discrete product-kernel solution has finite-time second-moment blow-up

Statement

Define concentrations $c_{k} (t) \geq 0$ by the discrete product-kernel equations ${\dot{c}}_{k} = \frac{1}{2} \sum_{i + j = k} i j c_{i} c_{j} - k c_{k} \sum_{j \geq 1} j c_{j}, k \geq 1,$ with monomer initial data $c_{1} (0) = 1$ and $c_{k} (0) = 0$ for $k \geq 2$ . For $r = 1, 2$ , define $M_{r} (t) = \sum_{k \geq 1} k^{r} c_{k} (t)$ . The pre-gelation solution is $c_{k} (t) = \frac{k^{k - 2}}{k!} t^{k - 1} e^{- k t}, 0 \leq t < 1,$ where the value at $t = 0$ is understood by continuity. Its first and second moments satisfy $M_{1} (t) = 1$ and $M_{2} (t) = 1 / (1 - t)$ . Thus the second moment becomes unbounded at the finite time $t = 1$ although the pre-gelation first moment is constant.

Facts & Assumptions

Given: The discrete product-kernel equations and the rooted-tree generating series.

[F1]

Rooted-tree series and the inverse of u exp(-u): $T (z) = \sum_{k \geq 1} k^{k - 1} z^{k} / k!$ , $U (z) = \sum_{k \geq 1} k^{k - 2} z^{k} / k!$ , $T = z e^{T}$ , $z T^{'} = T / (1 - T)$ , and $z U^{'} = T$ for $∣ z ∣ < e^{- 1}$ .

[F2]

Rooted-tree series and the inverse of u exp(-u): for $0 \leq z < e^{- 1}$ , $T (z)$ is the unique $u \in [0, 1)$ satisfying $u e^{- u} = z$ .

Proof

1.1F1F2algebra

Put $q (t) = t e^{- t}$ and $a_{k} = k^{k - 2} / k!$ . For $0 < t < 1$ , define $c_{k} (t) = a_{k} q (t)^{k} / t$ . Since $q (t) < e^{- 1}$ , [F1] gives the convergent generating function $C (z, t) = \sum_{k \geq 1} c_{k} (t) z^{k} = U (z q (t)) / t$ . Its mass generating function is $z \partial_{z} C = T (z q) / t$ . Because $q = t e^{- t}$ and $t \in [0, 1)$ , [F2] gives $T (q) = t$ , so $M_{1} = z \partial_{z} C (1, t) = 1$ .

2.1F1step 1.1algebra

For every fixed $k$ and $0 < t < 1$ , differentiation of $a_{k} q^{k} / t$ gives ${\dot{c}}_{k} = c_{k} ((k - 1) / t - k)$ . The gain coefficient in the $k$ th product-kernel equation is $\frac{1}{2} \sum_{i + j = k} i j c_{i} c_{j} = \frac{q^{k}}{t^{2}} [z^{k}] \frac{T (z)^{2}}{2} = \frac{k - 1}{t} c_{k},$ because $U = T - T^{2} / 2$ implies $[z^{k}] T^{2} / 2 = (k - 1) a_{k}$ . The loss is $k c_{k} M_{1} = k c_{k}$ by step 1.1. Their difference is exactly ${\dot{c}}_{k}$ , verifying every coefficient equation for $0 < t < 1$ . At $t = 0$ , only $c_{1}$ is nonzero: the right-hand side is $- 1$ for $k = 1$ , $1 / 2$ for $k = 2$ , and $0$ for $k \geq 3$ . The displayed formula has the same right derivatives, since $a_{2} = 1 / 2$ and $t^{k - 1}$ has zero derivative at $0$ for $k \geq 3$ . Thus the equations and initial data hold at the initial time as well.

3.1F1step 1.1algebra∎

Applying $z \partial_{z}$ once more to $C = U (z q) / t$ and using [F1] gives $M_{2} (t) = (z \partial_{z})^{2} C (1, t) = \frac{q T^{'} (q)}{t} = \frac{T (q)}{t (1 - T (q))} = \frac{1}{1 - t} .$ Therefore $M_{2} (t) \to \infty$ as $t ↑ 1$ , while step 1.1 gives $M_{1} (t) = 1$ throughout the pre-gelation interval.

Depends on

Rooted-tree series and the inverse of u exp(-u)

Used by

Finite-grid mass conservation does not bound the product-kernel continuum second moment Counterexample

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Philippe Laurençot, Weak compactness techniques and coagulation equations, §§1 and 3.3 (standard reference, not scraped)