Weak consistency of age-cohort characteristic quadrature

Let $\widehat{n}$ be the exact solution with initial density ${\widehat{n}}_0$. Applying the characteristic formula [F2] and changing variables from age $a$ to birth time $s=t-a$ gives, for every bounded Lipschitz $\varphi$, $\int \varphi (a)\widehat{n}(t,a)da={\int }_0^{A-T}\varphi (x+t)S(x,t)n_0(x)dx+{\int }_0^t\varphi (t-s)S(0,t-s)\widehat{B}(s)ds$. The birth kernel $K(s)=\beta (s)S(0,s)$ is bounded and Lipschitz, and $\widehat{G}(t)={\int }_0^{A-T}\beta (x+t)S(x,t)n_0(x)dx$ is bounded and Lipschitz because $\beta ,\mu$ have bounded first derivatives and $n_0$ is integrable. The mass estimate [F5] bounds $\widehat{B}$ by ${\beta }_{\mathrm{\infty }}{e}^{{\beta }_{\mathrm{\infty }}T}\mathrm{\parallel }{n}_0{\mathrm{\parallel }}_1$. Subtracting its Volterra equation at $t$ and $t+h$, the old-history kernel difference is at most $\mathrm{Lip}(K)h{\int }_0^T\mathrm{\mid }\widehat{B}\mathrm{\mid }$ and the new interval at most $h\mathrm{\parallel }K{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }\widehat{B}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus $\widehat{B}$ is Lipschitz on $[0,T]$.

If $N=0$, there are no time steps and the survival-error claim is zero. Suppose $N\ge 1$, and put $T_h=N\mathrm{\Delta }t>0$. A cohort born with mass $m_0$ and initial age $a_0$ has exact survival mass solving $m^{\mathrm{\prime }}(s)=-\mu (a_0+s)m(s)$; let $q=m_0$ be its reference mass. To apply [F4] at $a_0=0$, define the extension $\bar{\mu }(r)=\mu (r)$ for $r\ge 0$ and $\bar{\mu }(r)=\mu (0)+{\mu }_{+}^{\mathrm{\prime }}(0)r{e}^r$ for $r<0$. The one-sided derivative ${\mu }_{+}^{\mathrm{\prime }}(0)$ exists; the two formulas and their first derivatives match at zero, and $\bar{\mu }$ and ${\bar{\mu }}^{\mathrm{\prime }}$ are bounded on $\mathbb{R}$. For a cohort whose reference mass is $q>0$, the vector field $f(s,m)=-\bar{\mu }(a_0+s)m$ is $C^1$ on the open tube $V=(-1,T_h+1)\times (-q,2q)$, where $T_h=N\mathrm{\Delta }t$, with all bounds required by [F4] uniform in $a_0$ and proportional to $q$ except for the state-Lipschitz constant, which is bounded by $\mathrm{\parallel }\bar{\mu }{\mathrm{\parallel }}_{\mathrm{\infty }}$. The exact continuation $m(s)=m_{\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{t}}\exp (-{\int }_0^s\bar{\mu }(a_0+r)dr)$ exists on an open interval containing $[0,T_h]$; after shrinking that interval at the endpoints its graph lies in $V$. Its exact forward solution from any $m\in [0,q]$ and its Euler input/output states lie in $K=[0,T_h]\times [0,q]$: mortality is nonnegative for ages $a_0+s\ge 0$, and $\mathrm{\Delta }t\mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$. Thus [F4] gives one-step defect at most $C\mathrm{\Delta }{t}^{2}q$ and a stability factor $1+L\mathrm{\Delta }t$, with constants independent of $a_0$ and proportional to $q$. Applying [L1] on the aligned horizon $T_h=N\mathrm{\Delta }t$ gives survival-weight error at most $C_T\mathrm{\Delta }tq$ over the full run; the zero-mass case is exact. For the discrete total cohort mass $N_h^k$, positivity and $B_h^k\le {\beta }_{\mathrm{\infty }}{N}_h^k$ give $N_h^{k+1}\le (1+\mathrm{\Delta }t{\beta }_{\mathrm{\infty }})N_h^k$, hence $N_h^k\le e^{{\beta }_{\mathrm{\infty }}T}{N}_h^0$. Since $N_h^0\le \mathrm{\parallel }{n}_0{\mathrm{\parallel }}_1+{ϵ}_0$ (test the initial BL defect with $\varphi \equiv 1$), the sum of the initial cohort masses and all inserted newborn masses ${\sum }_{j<k}\mathrm{\Delta }t{B}_h^j$ is at most $C_T(\mathrm{\parallel }{n}_0{\mathrm{\parallel }}_1+{ϵ}_0)$. Thus the sum of all cohort survival-weight errors is at most $C_T\mathrm{\Delta }t(\mathrm{\parallel }{n}_0{\mathrm{\parallel }}_1+{ϵ}_0)$.

Write $e_k=\mathrm{\mid }{B}_h^k-\widehat{B}(t_k)\mathrm{\mid }$. The initial-cohort contribution is at most $C_T({ϵ}_0+\mathrm{\Delta }t)$: the test $x\mapsto \beta (x+t_k)S(x,t_k)$ has bounded-Lipschitz norm at most $C_T$, and step 2.1 bounds the accumulated Euler survival error. Each cohort born during $[t_j,t_{j+1}]$ is inserted at $t_{j+1}$ with mass $\mathrm{\Delta }t{B}_h^j$, so at $t_k$ its age is $(k-j-1)\mathrm{\Delta }t$. The exact renewal convolution is the integral of $K(t_k-s)\widehat{B}(s)$ over $s\in [0,t_k]$. Since $K$ and $\widehat{B}$ are Lipschitz, replacing this integral by the left-endpoint sum with age $(k-j-1)\mathrm{\Delta }t$ costs at most $C_T\mathrm{\Delta }t$ (the age sample differs from $t_k-t_j$ by one step). Replacing $\widehat{B}(t_j)$ by $B_h^j$ contributes at most $\mathrm{\parallel }K{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\Delta }t{\sum }_{j<k}{e}_j$, and replacing exact newborn survival by its Euler product contributes at most $C_T\mathrm{\Delta }t$ by step 2.1. Therefore $e_k\le C_T({ϵ}_0+\mathrm{\Delta }t)+\mathrm{\parallel }K{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\Delta }t{\sum }_{j<k}{e}_j$. Set $r_k=C_T({ϵ}_0+\mathrm{\Delta }t)+\mathrm{\parallel }K{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\Delta }t{\sum }_{j<k}{e}_j$. Then $e_k\le r_k$ and $r_{k+1}=r_k+\mathrm{\parallel }K{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\Delta }t{e}_k\le (1+\mathrm{\parallel }K{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\Delta }t)r_k$, so $e_k\le e^{\mathrm{\parallel }K{\mathrm{\parallel }}_{\mathrm{\infty }}T}{C}_T({ϵ}_0+\mathrm{\Delta }t)$.

For any test function with $\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{Lip}(\varphi )\le 1$, the initial cohort pushforward differs from the first integral for $\widehat{n}$ in step 1.1 by at most $C_T({ϵ}_0+\mathrm{\Delta }t)$, using step 2.1 and the bounded-Lipschitz norm of $x\mapsto \varphi (x+t)S(x,t)$. The newborn cohort sum differs from the second integral by its left-Riemann error $C_T\mathrm{\Delta }t$ plus $\mathrm{\Delta }t{\sum }_{j<k}{e}_j\le C_T({ϵ}_0+\mathrm{\Delta }t)$ by step 3.1. Thus $\mathrm{\parallel }{\nu }_h^k-\widehat{n}(t_k,a)da{\mathrm{\parallel }}_{B{L}^{\ast }}\le C_T({ϵ}_0+\mathrm{\Delta }t)$. The initial data difference is $\mathrm{\parallel }{n}_0-{\widehat{n}}_0{\mathrm{\parallel }}_1={\tau }_A$, so [F5] gives $\mathrm{\parallel }n(t_k)-\widehat{n}(t_k){\mathrm{\parallel }}_1\le e^{{\beta }_{\mathrm{\infty }}T}{\tau }_A$. The inequality $\mathrm{\parallel }f(a)da{\mathrm{\parallel }}_{B{L}^{\ast }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$ and the triangle inequality prove the stated bound.