Renewal age and size structured numerical methods — Examples

Direct integration gives $\int {\rho }_{\epsilon }(a)da=1$ and symmetry about $a_{\ast }$ gives $\int a{\rho }_{\epsilon }(a)da=a_{\ast }$. Its second centered moment is $\int (a-a_{\ast }{)}^2{\rho }_{\epsilon }(a)da={\epsilon }^2\mathrm{/}6$.

Let $b$ be Lipschitz on $[0,T]$ and take $t\ge a_{\ast }+\epsilon$. The support of ${\rho }_{\epsilon }$ is contained in $[0,t]$, so the convolution in [F2] is $H_{\epsilon }(t)=R\int {\rho }_{\epsilon }(a)e^{-ma}b(t-a)da$. For each supported $a$, $\mathrm{\mid }{e}^{-ma}b(t-a)-e^{-m{a}_{\ast }}b(t-a_{\ast })\mathrm{\mid }\le (\mathrm{Lip}(b)+m\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }})\mathrm{\mid }a-a_{\ast }\mathrm{\mid }$. Since $\mathrm{\mid }a-a_{\ast }\mathrm{\mid }\le \epsilon$ and the kernel has unit mass, $\mathrm{\mid }{H}_{\epsilon }(t)-R{e}^{-m{a}_{\ast }}b(t-a_{\ast })\mathrm{\mid }\le R\epsilon (\mathrm{Lip}(b)+m\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }})$.

A finite-width fertility is not an exact delay: set $m=0$ and $b(s)=s^2$ on the relevant nonnegative interval. By step 1.1, $H_{\epsilon }(t)=R((t-a_{\ast }{)}^2+{\epsilon }^2\mathrm{/}6)$, while the point-delay term is $Rb(t-a_{\ast })=R(t-a_{\ast }{)}^2$; their difference is $R{\epsilon }^2\mathrm{/}6>0$.

The parent center $2$ lies in the third cell and its daughters have size $1$, exactly the second cell center. Thus the positive midpoint kernel quadrature gives $A_{2,3}=2$ and all other daughter coefficients for the occupied parent column vanish. The CFL number is $0.1(0.1\mathrm{/}0.5+1)=0.12$ in cell 1 and $0.1(0.1\mathrm{/}1+1)=0.11$ in each of cells 2–4, so the global maximum is $0.12\le 1$.

The rightward internal flux from cell $C_3$ into $C_4$ is $F_3=g{M}_3\mathrm{/}{h}_3=0.3$; the absorbing boundary flux is $F_4=0$. The daughter source is $\mathrm{\Delta }t{A}_{2,3}{M}_3=0.6$, and parent removal is $\mathrm{\Delta }t\gamma M_3=0.3$. Therefore the updated cell masses are $M_2^{\mathrm{\prime }}=0.6$, $M_3^{\mathrm{\prime }}=3-0.03-0.3=2.67$, and $M_4^{\mathrm{\prime }}=0.03$, with $M_1^{\mathrm{\prime }}=0$; they are all nonnegative.

The discrete number changes from $3$ to $0.6+2.67+0.03=3.3$, an increase $0.3$ equal to $\mathrm{\Delta }t(2-1)\gamma M_3$. The weighted-size total changes from $2\cdot 3=6$ to $1\cdot 0.6+2\cdot 2.67+3\cdot 0.03=6.03$. Formula [F3] gives the same $0.03$ increase from transport, while division contributes $\mathrm{\Delta }t(2\cdot 1-2\cdot 1)M_3=0$. Thus the first moment is conserved by division and changes only by the size-growth flux in this step.

The triangle has base length $1$ and height $1$, so $B(0)={\int }_1^2\beta (a)da=1\mathrm{/}2$. The chosen initial density has right-hand trace $n_0(0+)=0$, which differs from $B(0)$; hence smooth renewal-corner compatibility fails. The example uses the characteristic/weak formulation rather than assuming that compatibility.

Here $S(x,t)=e^{-t}$, so [F2] gives $G(t)=e^{-t}{\int }_1^2\beta (x+t)dx$ and $B(t)=G(t)+{\int }_0^t\beta (a)e^{-a}B(t-a)da$. The first formula records the initially present age cohort; the second is the renewal feedback term.

Use cells $[0,1],[1,2],[2,3],[3,4]$, each of width $1$, and initial masses $(0,1,0,0)$. Four-panel composite midpoint quadrature on the second cell samples fertility values $1\mathrm{/}4,3\mathrm{/}4,3\mathrm{/}4,1\mathrm{/}4$, so $B_h^0=(1\mathrm{/}4)(1\mathrm{/}4+3\mathrm{/}4+3\mathrm{/}4+1\mathrm{/}4)=1\mathrm{/}2$.

With $\lambda =\mathrm{\Delta }t\mathrm{/}\mathrm{\Delta }a=1\mathrm{/}4$ and ${\mu }_j\mathrm{\Delta }t=1\mathrm{/}4$, the CFL sum is $1\mathrm{/}2$. The fluxes are $F_0=1\mathrm{/}2,F_1=0,F_2=1,F_3=0,F_4=0$, so the first updated mass vector is $(1\mathrm{/}8,1\mathrm{/}2,1\mathrm{/}4,0)$. Every entry is nonnegative, in agreement with [F4].

The exact solution in [F1] has total mass ${\int }_t^{\mathrm{\infty }}{e}^{-(a-t)}da=1$ at every $t$. The exact initial cell masses are $M_j^0=e^{-(j-1)}-e^{-j}$ for $1\le j\le 4$.

Since $\lambda =1$ and $\mu =0$, [F2] gives $M_1^{k+1}=0$ and $M_j^{k+1}=M_{j-1}^k$ for $2\le j\le 4$. After $k$ steps, the retained total is ${\sum }_{j=1}^{4-k}{M}_j^0=1-e^{-(4-k)}$, so the error from total mass $1$ is $e^{-(4-k)}$.

For $T=k=1,2,3$, the errors are respectively $e^{-3},e^{-2},e^{-1}$. The cumulative outflow is ${\sum }_{r=0}^{k-1}{M}_{4-r}^0=e^{-(4-k)}-e^{-4}$, and the initial mass omitted above $A$ is $e^{-4}$. Their sum equals the exact tail ${\int }_{4-k}^{\mathrm{\infty }}{e}^{-a}da=e^{-(4-k)}$ in [F3]. The larger error at the longer horizon is the age-cutoff tail, not a time-discretization defect.