L1 convergence for BV data without renewal-boundary compatibility

Write $u_j^k=M_j^k\mathrm{/}\mathrm{\Delta }a$, $B_k=B_h^k$, $\lambda =\mathrm{\Delta }t\mathrm{/}\mathrm{\Delta }a$, and $V_k=\mathrm{\mid }{u}_1^k-B_k\mathrm{\mid }+{\sum }_{j=2}^J\mathrm{\mid }{u}_j^k-u_{j-1}^k\mathrm{\mid }$. The initial cell averages satisfy $V_0\le \mathrm{TV}(n_0)+\mathrm{\parallel }{n}_0{\mathrm{\parallel }}_{\mathrm{\infty }}+{\beta }_{\mathrm{\infty }}\mathrm{\parallel }{n}_0{\mathrm{\parallel }}_1$, which is finite because an integrable function of finite total variation on the half-line has finite essential supremum; the first term bounds differences of adjacent cell averages by averaging the variation measure.

Set $u_0^k=B_k$ and put $d_1^k=u_1^k-B_k$ and $d_j^k=u_j^k-u_{j-1}^k$ for $2\le j\le J$. For $j\ge 2$, subtract adjacent updates from [F3] to obtain $u_j^{k+1}-u_{j-1}^{k+1}=(1-\lambda -\mathrm{\Delta }t{\mu }_j)(u_j^k-u_{j-1}^k)+\lambda (u_{j-1}^k-u_{j-2}^k)+\mathrm{\Delta }t({\mu }_{j-1}-{\mu }_j)u_{j-1}^k$. The CFL makes the first two coefficients nonnegative, and $\mathrm{\mid }{\mu }_j-{\mu }_{j-1}\mathrm{\mid }\le \mathrm{Lip}(\mu )\mathrm{\Delta }a$. For $J\ge 2$, summing their absolute-value bounds for $j=2,\dots ,J$ gives at most ${\sum }_{j=2}^{J-1}\mathrm{\mid }{d}_j^k\mathrm{\mid }+(1-\lambda )\mathrm{\mid }{d}_J^k\mathrm{\mid }+\lambda \mathrm{\mid }{d}_1^k\mathrm{\mid }+\mathrm{\Delta }t\mathrm{Lip}(\mu )N_k$: for $2\le j<J$ the two transport coefficients add to at most $1$, the last-cell coefficient is at most $1-\lambda$, and the boundary difference contributes $\lambda \mathrm{\mid }{d}_1^k\mathrm{\mid }$. When $J=1$, omit this adjacent-difference sum and use only the following first-cell estimate. The first-cell jump is $d_1^{k+1}=(1-\lambda -\mathrm{\Delta }t{\mu }_1)d_1^k+(1-\mathrm{\Delta }t{\mu }_1)B_k-B_{k+1}$. The quadrature weights are fixed and nonnegative, so $\mathrm{\mid }{B}_{k+1}-B_k\mathrm{\mid }\le {\beta }_{\mathrm{\infty }}{\sum }_j\mathrm{\mid }{M}_j^{k+1}-M_j^k\mathrm{\mid }$. The update and the definition of $V_k$ give ${\sum }_j\mathrm{\mid }{M}_j^{k+1}-M_j^k\mathrm{\mid }\le \mathrm{\Delta }t(V_k+{\mu }_{\mathrm{\infty }}{N}_k)$. Also $\mathrm{\mid }(1-\mathrm{\Delta }t{\mu }_1)B_k-B_{k+1}\mathrm{\mid }\le \mathrm{\mid }{B}_k-B_{k+1}\mathrm{\mid }+\mathrm{\Delta }t{\mu }_{\mathrm{\infty }}{B}_k$ and $B_k\le {\beta }_{\mathrm{\infty }}{N}_k$. Adding the first-cell bound to the preceding sum combines the $d_1^k$ coefficients as $(1-\lambda -\mathrm{\Delta }t{\mu }_1)+\lambda \le 1$; all other variation coefficients are at most $1$. Therefore $V_{k+1}\le (1+C\mathrm{\Delta }t)V_k+C\mathrm{\Delta }t{N}_k$, with $C$ depending only on the rate bounds and Lipschitz constants.

Iterating [F4] gives $N_k\le (1+\mathrm{\Delta }t{\beta }_{\mathrm{\infty }}{)}^k{N}_0\le e^{k\mathrm{\Delta }t{\beta }_{\mathrm{\infty }}}{N}_0\le e^{{\beta }_{\mathrm{\infty }}T}{N}_0$. Iterating the recurrence in step 2.1 gives $V_k\le (1+C\mathrm{\Delta }t{)}^k{V}_0+C\mathrm{\Delta }t{\sum }_{r=0}^{k-1}(1+C\mathrm{\Delta }t{)}^{k-1-r}{N}_r\le e^{CT}(V_0+CT{e}^{{\beta }_{\mathrm{\infty }}T}{N}_0)$. The same update gives ${\sum }_j\mathrm{\Delta }a\mathrm{\mid }{u}_j^{k+1}-u_j^k\mathrm{\mid }\le \mathrm{\Delta }t(V_k+{\mu }_{\mathrm{\infty }}{N}_k)$; hence the linear time interpolation is uniformly Lipschitz into $L^1$.

On any fixed interval $[0,R]$, the uniform $L^1$ and variation bounds make the time slices precompact in $L^1$: replacing each slice by averages on blocks of width $\rho$ changes its $L^1$ norm by at most $\rho V_k$, while those block averages lie in a bounded finite-dimensional set. A diagonal extraction at rational times and the uniform time-Lipschitz bound from step 3.1 give a subsequence converging uniformly in time in $L^1$ on every bounded age interval. A newborn cohort has age at most $T$; under the upwind update it can move at most one cell per step, so all newborn mass lies below age $T\mathrm{/}{c}_{-}+\mathrm{\Delta }a$. For $R>T\mathrm{/}{c}_{-}+\mathrm{\Delta }a$, births therefore contribute nothing above $R$. An initial cell moves by at most $k$ cells, so any initial mass reconstructed above $R$ came from initial cells whose lower endpoint exceeds $R-k\mathrm{\Delta }a-\mathrm{\Delta }a\ge R-T\mathrm{/}{c}_{-}-\mathrm{\Delta }a$. Its mass is at most ${\int }_{R-T\mathrm{/}{c}_{-}-2\mathrm{\Delta }a}^{\mathrm{\infty }}{n}_0$ (extend $n_0$ by zero to negative ages); mortality and outflow only decrease that inherited mass. This tail tends to zero uniformly as $R\to \mathrm{\infty }$. For the exact solution, [F2] shows that newborns have age at most $T$, while mass above $R>T$ comes only from initial ages above $R-T$ and is at most ${\int }_{R-T}^{\mathrm{\infty }}{n}_0$. Hence the subsequence converges in $C([0,T];L^1(0,\mathrm{\infty }))$.

Let ${\varphi }_j^k=\varphi (t_k,(a_{j-1}+a_j)\mathrm{/}2)$ for a compactly supported $C^1$ test function with terminal value zero. Multiplying [F3] by ${\varphi }_j^{k+1}$ and summing by parts gives the exact discrete weak identity $-{\sum }_{k,j}{M}_j^k({\varphi }_j^{k+1}-{\varphi }_j^k)-\mathrm{\Delta }t{\sum }_{k,j<J}{F}_j^k({\varphi }_{j+1}^{k+1}-{\varphi }_j^{k+1})+\mathrm{\Delta }t{\sum }_{k,j}{\mu }_j{M}_j^k{\varphi }_j^{k+1}={\sum }_j{M}_j^0{\varphi }_j^0+\mathrm{\Delta }t{\sum }_k{B}_k{\varphi }_1^{k+1}$; the outflow term vanishes because the test is supported below the cutoff. Uniform $L^1$ convergence, Lipschitz continuity of $\mu ,\beta$, and mesh refinement make the sums converge to the weak transport/loss terms. Since each positive birth rule is supported in its cell and its weights sum to $\mathrm{\Delta }a$, $\mathrm{\mid }{B}_k-\int \beta (a)u_h(t_k,a)da\mathrm{\mid }\le 2\mathrm{Lip}(\beta )\mathrm{\Delta }a{N}_k$, so the birth sum converges to the boundary term. The limit therefore satisfies [F1] with the given initial trace.

In characteristic coordinates $x=a-t$ on $a>t$, the weak identity [F1] becomes ${\mathrm{\partial }}_{t}v(t,x)=-\mu (x+t)v(t,x)$ in distributions; multiplication by $\exp ({\int }_x^{x+t}\mu )$ makes its distributional $t$-derivative zero, and the initial trace gives $v(t,x)=n_0(x)S(x,t)$. On $0<a<t$, use $s=t-a$ and obtain ${\mathrm{\partial }}_{a}w(s,a)=-\mu (a)w(s,a)$ with trace $w(s,0)=B(s)$; the same integrating-factor argument gives $w(s,a)=B(s)S(0,a)$. Substitution into the renewal relation in [F1] yields the Volterra equation of [F2], whose solution is unique. Thus every subsequential limit equals the mild solution. Since every subsequence has a further subsequence with this same limit, the entire sequence converges.