Unbounded-grid fixed-pivot dynamics are globally well posed and stable

The grid $G$ has no finite accumulation point, so it is closed in $E$. A weighted-total-variation limit of measures supported on $G$ assigns zero variation to $E\setminus G$, hence the grid-supported subspace is closed and complete by [F5]. For $z\ge a$, [F4] gives $\int w(u)P_{\mathrm{\Delta }}(z)(du)\le w(z)+{\mathrm{\Delta }}^2\mathrm{/}4\le (5\mathrm{/}4)w(z)$. Thus projection maps weighted measures boundedly into this subspace. The positive cone is closed: if ${\nu }_n\ge 0$ and ${\nu }_n\to \nu$ in weighted total variation, then ${\nu }^{-}(E)\le \mathrm{\parallel }\nu -{\nu }_n{\mathrm{\parallel }}_{TV}\to 0$.

Testing the equation with $1$ gives $M_0^{\mathrm{\prime }}\le {\beta }_{\ast }D{M}_0$, since each aggregation event changes number by $-1$ and each fragmentation increment is at most $D$. Testing with $x$ gives zero coagulation increment by [F4], and zero fragmentation increment because $\int z{B}_x(dz)=x$ by [F1] and projection preserves each daughter's mass by [F4]. Thus $M_1^{\mathrm{\prime }}=0$. Testing with $x^2$, each coagulation event contributes at most $2xy+{\mathrm{\Delta }}^2\mathrm{/}4$, and each fragmentation event contributes at most $D{\mathrm{\Delta }}^2\mathrm{/}4$ because daughter second moment before projection is at most $x^2$. Integrating the resulting inequalities proves all three displayed moment bounds.

For an admissible test $\phi$ in [F3], let $\bar{\phi }(z)=\int \phi (u)P_{\mathrm{\Delta }}(z)(du)$. On each cell this is the linear interpolant of the endpoint values; its slope has absolute value at most $1$. Continuity at the pivots gives $\mathrm{\parallel }\bar{\phi }{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and $\mathrm{Lip}(\bar{\phi })\le 1$ on all of $E$. The coagulation test integrand $\frac{1}{2}K(x,y)[\bar{\phi }(x+y)-\phi (x)-\phi (y)]$ is bounded and Lipschitz in each variable with constants depending only on $K_{\ast },L_K$.

Write $F_{\mathrm{\Delta }}(\nu )=G_{\mathrm{\Delta }}(\nu )-q_{\nu }\nu$, where $q_{\nu }(x)=\int K(x,y)\nu (dy)+\beta (x)$ and $G_{\mathrm{\Delta }}$ is the sum of the positive aggregation and fragmentation gains in [F1]. On a signed-measure ball $\mathrm{\parallel }\nu {\mathrm{\parallel }}_w\le R$, $\mathrm{\parallel }{q}_{\nu }{\mathrm{\parallel }}_{\mathrm{\infty }}\le K_{\ast }R+{\beta }_{\ast }$, and $q_{\nu }\ge 0$ when $\nu \ge 0$. The bound in step 1.1 and $w(x+y)+w(x)+w(y)\le 3w(x)w(y)$ bound the aggregation gain by a constant times $K_{\ast }{R}^2$; for fragmentation, $\int z^2{B}_x(dz)\le x^2$ and $B_x(E)\le D$ bound the gain by a constant times ${\beta }_{\ast }(D+1)R$. Bilinearity makes the aggregation gain locally Lipschitz, and the fragmentation gain is bounded linear.

Decompose $\nu \otimes \nu -\eta \otimes \eta =(\nu -\eta )\otimes \nu +\eta \otimes (\nu -\eta )$. Each section of the step 1.3 integrand, scaled by its bounded-Lipschitz norm, is an admissible test in [F3]. The resulting coagulation difference is bounded by $C_K(M_0(\nu )+M_0(\eta ))d_{BL}(\nu ,\eta )$. For fragmentation the integrand $\beta (x)[\int \bar{\phi }d{B}_x-\phi (x)]$ has supremum at most ${\beta }_{\ast }(D+1)$ and Lipschitz constant at most $L_{\beta }(D+1)+{\beta }_{\ast }(L_B+1)$, so its difference is bounded by a constant times $d_{BL}(\nu ,\eta )$.

On the closed positive path ball of radius $R>2\mathrm{\parallel }{\nu }_0{\mathrm{\parallel }}_w$, let $C_R$ bound $\mathrm{\parallel }{G}_{\mathrm{\Delta }}(\xi ){\mathrm{\parallel }}_w$ and $L_R$ be its Lipschitz constant, as obtained in step 2.1. For positive paths in that ball, set $\mathrm{\Phi }(\xi {)}_t=e^{-{\int }_0^t{q}_{{\xi }_s}(\cdot )ds}{\nu }_0+{\int }_0^t{e}^{-{\int }_s^t{q}_{{\xi }_r}(\cdot )dr}{G}_{\mathrm{\Delta }}({\xi }_s)ds\mathrm{.}$ Since $\mathrm{\parallel }{q}_{{\xi }_t}-q_{{\xi }_s}{\mathrm{\parallel }}_{\mathrm{\infty }}\le K_{\ast }\mathrm{\parallel }{\xi }_t-{\xi }_s{\mathrm{\parallel }}_w$, $q_{\xi }$ is continuous in time in $L^{\mathrm{\infty }}(E)$; the gain path is continuous in ${\mathcal{M}}_w(E)$ by step 2.1. For bounded $b$, multiplication $M_b\sigma =b\sigma$ has $\mathrm{\parallel }{M}_b\mathrm{\parallel }\le \mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }}$, so the survival multipliers are continuous in operator norm and the formula defines a continuous path. They lie in $[0,1]$, so $\mathrm{\Phi }$ preserves positivity. Its norm is at most $\mathrm{\parallel }{\nu }_0{\mathrm{\parallel }}_w+T{C}_R$, hence it maps the ball into itself when $T{C}_R\le R-\mathrm{\parallel }{\nu }_0{\mathrm{\parallel }}_w$. Since $\mathrm{\mid }{q}_{\xi }-q_{\zeta }{\mathrm{\mid }}_{\mathrm{\infty }}\le K_{\ast }\mathrm{\parallel }\xi -\zeta {\mathrm{\parallel }}_w$, the inequalities $\mathrm{\mid }{e}^{-u}-e^{-v}\mathrm{\mid }\le \mathrm{\mid }u-v\mathrm{\mid }$ for $u,v\ge 0$ and the triangle inequality give $\mathrm{\parallel }\mathrm{\Phi }(\xi )-\mathrm{\Phi }(\zeta ){\mathrm{\parallel }}_{C{\mathcal{M}}_w}\le (T{L}_R+T{K}_{\ast }\mathrm{\parallel }{\nu }_0{\mathrm{\parallel }}_w+\frac{1}{2}{T}^2{K}_{\ast }{C}_R)\mathrm{\parallel }\xi -\zeta {\mathrm{\parallel }}_{C{\mathcal{M}}_w}\mathrm{.}$ Choose $T$ so this coefficient is $q<1$. The positive path ball is nonempty because it contains ${\xi }_t^0={\nu }_0$. Iteration ${\xi }^{j+1}=\mathrm{\Phi }({\xi }^j)$ from ${\xi }^0$ satisfies $\mathrm{\parallel }{\xi }^{j+1}-{\xi }^j\mathrm{\parallel }\le q^j\mathrm{\parallel }{\xi }^1-{\xi }^0\mathrm{\parallel }$; for $m>n$, $\mathrm{\parallel }{\xi }^m-{\xi }^n\mathrm{\parallel }\le q^n\mathrm{\parallel }{\xi }^1-{\xi }^0\mathrm{\parallel }\mathrm{/}(1-q)\to 0$. Completeness from step 1.1 gives a limit, and the contraction bound makes $\mathrm{\Phi }$ continuous, so the limit is fixed. Any two fixed points have distance at most $q$ times itself, so they coincide. For a fixed point, $S_{\xi }(t,s):=M_{e^{-{\int }_s^t{q}_{{\xi }_r}(\cdot )dr}}$ has operator-norm derivative ${\mathrm{\partial }}_t{S}_{\xi }(t,s)=-M_{q_{{\xi }_t}}{S}_{\xi }(t,s)$, since the scalar exponential difference quotients converge uniformly in size. Splitting the difference quotient of the source integral at $s=t$ gives endpoint term $G_{\mathrm{\Delta }}({\xi }_t)$ and integral term $-M_{q_{{\xi }_t}}{\int }_0^t{S}_{\xi }(t,s)G_{\mathrm{\Delta }}({\xi }_s)ds$; hence the fixed point solves the equation in [F1]. Conversely, for any positive solution $\zeta$, the inverse multiplier $A_{\zeta }(t):=M_{e^{{\int }_0^t{q}_{{\zeta }_s}(\cdot )ds}}$ is operator-norm differentiable with $A_{\zeta }^{\mathrm{\prime }}=M_{q_{{\zeta }_t}}{A}_{\zeta }$. The product difference quotient gives $(A_{\zeta }\zeta {)}^{\mathrm{\prime }}=A_{\zeta }{G}_{\mathrm{\Delta }}(\zeta )$; integrating this continuous derivative by telescoping partitions and multiplying by $A_{\zeta }(t{)}^{-1}$ recovers the same mild formula. Continuity at $0$ puts any competing solution in this ball on a sufficiently short interval, where the fixed-point uniqueness just proved applies.

On $[0,T]$ the number and second-moment estimates bound $\mathrm{\parallel }{\nu }_t{\mathrm{\parallel }}_w=M_0({\nu }_t)+M_2({\nu }_t)$. The local Lipschitz estimates in step 2.1 bound $\mathrm{\parallel }{F}_{\mathrm{\Delta }}({\nu }_t){\mathrm{\parallel }}_w$ there, so ${\nu }_t$ is Cauchy in weighted total variation at any finite endpoint. Completeness gives a positive limit; the local construction restarts there and uniqueness glues the extension. Therefore no finite maximal endpoint exists.

Integrating the two weak equations and using the number bound in the Statement gives $d(t)\le d(0)+C_T{\int }_0^{t}d(s)ds$, where $d(t)=d_{BL}({\nu }_t,{\eta }_t)$. With $g(t)=d(0)+C_T{\int }_0^{t}d(s)ds$, one has $d\le g$ and $g^{\mathrm{\prime }}\le C_{T}g$ almost everywhere. Thus $(e^{-C_{T}t}g(t){)}^{\mathrm{\prime }}\le 0$ and $d(t)\le e^{C_{T}t}d(0)$.