Approximating the dynamics of communicating cells in a diffusive medium by ODEs—homogenization with localization

Abstract

Bacteria may change their behavior depending on the population density. Here we study a dynamical model in which cells of radius $R$ within a diffusive medium communicate with each other via diffusion of a signalling substance produced by the cells. The model consists of an initial boundary value problem for a parabolic PDE describing the exterior concentration $u$ of the signalling substance, coupled with $N$ ODEs for the masses $a_i$ of the substance within each cell. We show that for small $R$ the model can be approximated by a hierarchy of models, namely first a system of $N$ coupled delay ODEs, and in a second step by $N$ coupled ODEs. We give some illustrations of the dynamics of the approximate model.

Appendices

Appendix. Proof of approximation results: one cell

1.1 The idea

We want to approximate a solution $a$ of (1) with $N=1$, i.e.,

$$\begin{aligned} u_t&= D\Delta u \text{ in} \Omega ,\quad u(\cdot ,0)=u_0=\frac{d_2}{D+d_1} a_0\psi _c|_{\Omega }, \quad B u = \frac{d_2}{R^2}a \text{ on} \partial \Omega ,\qquad \end{aligned}$$

(39a)

$$\begin{aligned} a^{\prime }&= f(a)-4\pi d_2 a+ \int _{\partial \Omega }\frac{d_1u}{R} do,\qquad a(0)=a_0, \end{aligned}$$

(39b)

by a solution $b$ of (4), i.e., $b^{\prime }=f(b)-Mb, b(0)=a_0$. For $u_0$ we assume the compatibility conditions (12). The main idea is to consider the auxiliary problem

$$\begin{aligned} p_t&= D\Delta p +\tilde{b}(t)\delta _0(x),\qquad \qquad p(\cdot ,0)=\frac{d_2}{D+d_1} a_0\psi _c(\cdot ), \end{aligned}$$

(40a)

$$\begin{aligned} b^{\prime }&= f(b)-4\pi d_2 b+ \int _{\partial \Omega }\frac{d_1 p}{R} do,\qquad b(0)=a_0. \end{aligned}$$

(40b)

and to calculate a suitable $\tilde{b}$. Since (40a) can be explicitly solved to leading order, the ODE (40b) for $b$ can then be written as

$$\begin{aligned} b^{\prime }(t)=f(b(t))-4\pi d_2 b(t)+d_1(\widehat{T}_R\tilde{b})(t)+r(t) \end{aligned}$$

(41)

with $\widehat{T}_R$ defined in (16), and small $r$ due to the initial conditions in (40a). The main step is to derive a priori estimates for the difference between $T_Ra$ and $\widehat{T}_R\tilde{b}$, which, to lowest order, lead to the optimal choice $\tilde{b}=Mb$. With the notation $b_R$ for the solution of (41) we then show that $b_R\rightarrow b$ where $b$ solves (4) and thus prove Corollary 5.

1.2 The auxiliary problem

We compare solutions of (39) with solutions of (40). As already said, it turns out that in lowest order $\tilde{b}(t)=Mb(t)$ is optimal, but first we keep $\tilde{b}$ free. To simplify notation we introduce

$$\begin{aligned} \rho = R^2/(4D) \end{aligned}$$

and define

$$\begin{aligned} (F_{\rho ,k}f)(t)&:= \frac{1}{\rho }\int _0^t f(t-\tau ) (\rho /\tau )^{k/2}e^{-\rho /\tau }\mathrm{d}\tau = \int _{\rho /t}^\infty f(t-\rho /\zeta ) \zeta ^{k/2-2}e^{-\zeta }d\zeta ,\\ (K_\rho f)(t)&:= d_1\pi ^{-3/2} (F_{\rho , 3} f)(t) + 2 D \pi ^{-3/2}(F_{\rho , 5} f)(t). \end{aligned}$$

Lemma 13

Let $w(x,t) = u(x,t)-p(x,t)|_{\Omega }$. Then $w(x,t)$ satisfies

$$\begin{aligned} w_t=D\Delta w \text{ in} \Omega , \quad w(x,0)=0,\quad \quad Bw= -g(x,t) \text{ on} \partial \Omega , \end{aligned}$$

(42)

where

$$\begin{aligned} g(x,t) = \frac{1}{16D^2} \frac{1}{\rho }\left\{ (K_\rho [\tilde{b}-4\pi D\alpha _0])(t)- 4 D\,d_2 (a(t)-a_0)\right\} +\frac{1}{R} g_2(x,t), \end{aligned}$$

and $\Vert g_2\Vert _{\infty }={\mathcal O}(1)$.

Proof

We have $Bw=Bu-Bp=\frac{d_2}{R^2}a-Bp=:-g(x,t)$. The leading order terms of $p$ can be calculated explicitly. For $\tilde{p}=p-\alpha _0\psi _c$, where we recall $\alpha (t)=\frac{d_2}{D+d_1}a(t)=\frac{M}{4\pi D}a(t)$, we find

$$\begin{aligned} \tilde{p}_t=D\Delta \tilde{p}+\hat{c}\delta _0+\alpha _0 h, \quad \tilde{p}_0=0. \end{aligned}$$

(43)

Here $\hat{c}=\tilde{b}-4\pi D\alpha _0$, using $D\Delta \psi _c=-4\pi D\delta _0+h$ with $\Vert h\Vert _{H^2}\le C$, cf. (7). Letting

$$\begin{aligned} k(x,t) = \frac{1}{(4\pi Dt)^{3/2}} e^{-x^2 /(4Dt)}, \end{aligned}$$

we obtain, since $\tilde{p}(0)=0$,

$$\begin{aligned} \tilde{p}(x,t)&= \int _0^t\int _{\mathbb R ^3}k(x{-}y,t{-}\tau ) [\hat{c}(\tau )\delta _0(y)+\alpha _0 h(y,\tau )] dy \mathrm{d}\tau \\&= \int _0^tk(x,t{-}\tau ) \hat{c}(\tau ) \mathrm{d}\tau +p_2(x,t) \end{aligned}$$

with $\Vert p_2(\cdot ,t)\Vert _{H^2}\le C\alpha _0$, and in particular $\Vert p_2\Vert _{L^\infty (\partial \Omega )}\le C\alpha _0$. Thus

$$\begin{aligned} p(x,t)=p_1(x,t)+p_2(x,t)= \int _0^tk(x,t-\tau ) \hat{c}(\tau ) \mathrm{d}\tau +\alpha _0\psi _c+p_2(x,t). \end{aligned}$$

(44)

Therefore,

$$\begin{aligned} g(x,t)&= \frac{1}{R} \int _0^t (\tilde{b}(\tau )-4\pi D\alpha _0) \left[d_1 k(R, t-\tau )- R D k_x(R, t-\tau ) \right] d\tau \\&-\frac{d_2(a(t)-a_0)}{R^2}+\frac{1}{R} g_2 \end{aligned}$$

with $\Vert g_2(\cdot ,t)\Vert _{L^\infty (\partial \Omega )}\le C\alpha _0$. The integral kernels read

$$\begin{aligned} \frac{d_1}{R} k(R, t)&= \frac{d_1}{R\,(4\pi Dt)^{3/2}} e^{-R^2/(4Dt)} = \frac{d_1}{16D^2} \,\,\frac{1}{R^4/(4D)^2 (\pi t\, 4D/R)^{3/2}} e^{-\rho /t}\\&= \frac{d_1}{16D^2} \,\,\frac{1}{\rho ^2}\,\, \pi ^{-3/2} (\rho /t)^{3/2} e^{-\rho /t}\\ Dk_x(R, t)&= \frac{1}{(4\pi Dt)^{3/2}} e^{-R^2/(4Dt)}\,\, \frac{2DR}{4 Dt} = \frac{2D}{R(4\pi Dt)^{3/2}} e^{-R^2/(4Dt)}\,\, \frac{R^2}{4 Dt}\\&= \frac{2D}{16D^2} \,\,\frac{1}{\rho ^2}\,\, \pi ^{-3/2} (\rho /t)^{5/2} e^{-\rho /t},\\ \end{aligned}$$

and thus

$$\begin{aligned} g(x,t)&= \frac{1}{16D^2} \frac{1}{\rho }(K_\rho [\tilde{b}-4\pi D\alpha _0])(t)- \frac{4 D}{R^2}\,d_2 (a(t)-a_0) +\frac{1}{R} g_2(x,t)\\&= \frac{1}{16D^2} \frac{1}{\rho }\left\{ (K_\rho [\tilde{b}-4\pi D\alpha _0])(t)- 4 D\,d_2 (a(t)-a_0)\right\} +\frac{1}{R} g_2(x,t). \end{aligned}$$

$\square$

Lemma 14

For solutions $w$ of (42) we have

$$\begin{aligned} \Vert w\Vert ^2_{L^2(\Omega )}&\le \frac{2R}{d_1} \int _0^t \int _{\partial \Omega } g^2(x,\tau ) do d\tau ,\end{aligned}$$

(45)

$$\begin{aligned} \Vert w\Vert ^2_{L^2([0, t],H^1(\Omega ))}&\le \frac{2R}{d_1 D}\left( \int _0^t \int _{\partial \Omega } g^2(x,\tau ) do d\tau + D \int _0^t \int _0^\tau \int _{\partial \Omega } g^2(x,\sigma ) do d\sigma d\tau \right) .\nonumber \\ \end{aligned}$$

(46)

Proof

We have the a priori estimate

$$\begin{aligned}&D\int _\Omega |\nabla w|^2 dx+\frac{d}{dt}\frac{1}{2}\int _{\Omega }w^2(x,t) dx = D\int _\Omega |\nabla w|^2 dx + D\int _{\Omega }w\Delta w dx\\&\quad = D\left[ \int _\Omega |\nabla w|^2 dx - \int _\Omega |\nabla w|^2 dx + \int _{\partial \Omega } w\partial _\nu w do \right] =-\int _{\partial \Omega } w\left[ \frac{d_1}{R} w+ g(x,t)\right] do\\&\quad \le -\frac{d_1}{R}\int _{\partial \Omega } w^2\mathrm{d}o + \frac{d_1}{R}\int _{\partial \Omega } w^2\mathrm{d}o + \frac{R}{d_1} \int _{\partial \Omega } g^2(x,t) \mathrm{d}o = \frac{R}{d_1} \int _{\partial \Omega } g^2(x,t) do. \end{aligned}$$

Integrating over time we find (45), and integrating a second time w.r.t. time yields (46). $\square$

By (43), the key step to minimize the right hand sides in (45), (46) for given $a$, is to approximately solve for $f$ the integral equation

$$\begin{aligned} (K_\rho f)(t) := d_1\pi ^{-3/2} (F_{\rho , 3} f)(t) + 2 D \pi ^{-3/2}(F_{\rho , 5} f)(t) \stackrel{!}{=}4 D\,d_2\hat{a}(t) \end{aligned}$$

(47)

with $\hat{a}(t) = a(t)-a_0$, and $f = \tilde{b} +4\pi D \alpha _0$. Formally, for $\tau >0$

$$\begin{aligned} \lim _{\rho \rightarrow 0} (F_{\rho ,k}f)(\tau ) = \Gamma (k/2-1)f(\tau )=:(F_{0,k}f)(\tau ). \end{aligned}$$

With $\Gamma (1/2)=\sqrt{\pi }$ and $\Gamma (3/2) = \sqrt{\pi }/2$, we may write

$$\begin{aligned} \lim _{\rho \rightarrow 0} (K_\rho f) = \pi ^{-1}(d_1+D)f(\tau ) =: (K_0f)(\tau ). \end{aligned}$$

In order to establish this equation not only formally we define the residual

$$\begin{aligned} ({\mathcal R}_{\rho }f)(\tau ):=(K_{\rho }f)(\tau )-(K_0f)(\tau ). \end{aligned}$$

Lemma 15

Let $f\in C^{0,1/2}(\mathbb R _+), f(0)=0$. Then for all $t_1>0$ there exists a $C>0$ such that

$$\begin{aligned} \Vert {\mathcal R}_{\rho }f\Vert _{C^0[0,t_1]} \le C \Vert f\Vert _{C^{0,1/2}[0,t_1]} \rho ^{1/2}. \end{aligned}$$

Proof

We have

$$\begin{aligned}&(F_{\rho ,k}f)(\tau )- (F_{0,k}f)(\tau )\\&\quad =\frac{1}{\rho }\int _0^\tau (f(\tau -x) -f(\tau ))(\rho /x)^{k/2}e^{-\rho /x}dx + \frac{f(\tau )}{\rho } \int _\tau ^\infty (\rho /x)^{k/2}e^{-\rho /x}dx\\&\quad = \int _{\rho /\tau }^\infty [f(\tau -\rho /\zeta )-f(\tau )] \zeta ^{k/2-2}e^{-\zeta }dy+\,f(\tau ) \int _0^{\rho /\tau } \zeta ^{k/2-2} e^{-\zeta }d\zeta \\&\quad = \rho ^\alpha \, \int _{\rho /\tau }^\infty \frac{f(\tau -\rho /\zeta )-f(\tau )}{|(\tau -\rho /\zeta )-\tau |^\alpha } \zeta ^{k/2-2-\alpha }e^{-\zeta }dy + f(\tau ) \int _0^{\rho /\tau } \zeta ^{k/2-2} e^{-\zeta }d\zeta \end{aligned}$$

for $\alpha \in (0,1)$. Thus, with $A = \pi ^{-3/2} d_1$ and $B = 2 \pi ^{-3/2} D$,

$$\begin{aligned} |(K_\rho f)(\tau )-(K_0f)(\tau )|&\le \left| \rho ^{\alpha _1}\, A \int _{\rho /\tau }^\infty \frac{f(\tau -\rho /\zeta )-f(\tau )}{|(\tau -\rho /\zeta )-\tau |^{\alpha _1}} \zeta ^{3/2-2-{\alpha _1}} e^{-\zeta }dy\right| \\&+\left| \rho ^{\alpha _2}\, B \int _{\rho /\tau }^\infty \frac{f(\tau -\rho /\zeta )-f(\tau )}{|(\tau -\rho /\zeta )-\tau |^{\alpha _2}} \zeta ^{5/2-2-\alpha _2} e^{-\zeta }dy\right| \\&+ \left| \,f(\tau ) \int _0^{\rho /\tau } (A \zeta ^{3/2-2}+ B \zeta ^{5/2-2}) e^{-\zeta }d\zeta \right|\\&=:I_1(\tau )+I_2(\tau )+I_3(\tau ). \end{aligned}$$

Select $\alpha _1\in (0,1/2)$. We find by partial integration

$$\begin{aligned} I_1(\tau )&\le \rho ^{\alpha _1}\, A \int _{\rho /\tau }^\infty \frac{|f(\tau -\rho /\zeta )-f(\tau )|}{|(\tau -\rho /\zeta ) -\tau |^{\alpha _1}} \zeta ^{3/2-2-{\alpha _1}}e^{-\zeta }d\zeta \\&\le \rho ^{\alpha _1}A|f|_{C^{0,{\alpha _1}}[0,\tau ]} \left(\, \int _{\rho /\tau }^\infty \zeta ^{-1/2-{\alpha _1}}e^{-\zeta }d\zeta \right)\\&= \rho ^{\alpha _1} A|f|_{C^{0,{\alpha _1}}[0,\tau ]} \left( \frac{-1}{(1/2-{\alpha _1})} (\rho /\tau )^{1/2-{\alpha _1}} e^{-\rho /\tau } \right.\\&\left.+ \frac{1}{(1/2-{\alpha _1})}\int _{\rho /\tau }^\infty \zeta ^{1/2-{\alpha _1}}e^{-\zeta }d\zeta \right)\\&\le \rho ^{{\alpha _1}} C\Vert f\Vert _{C^{0,{\alpha _1}}[0,\tau ]} (\rho /\tau )^{1/2-{\alpha _1}} = \rho ^{1/2}C\, \Vert f\Vert _{C^{0,{\alpha _1}}[0,\tau ]}/\tau ^{1/2-{\alpha _1}}\\&\le \rho ^{1/2}C\,\Vert f\Vert _{C^{0,1/2}[0,\tau ]}. \end{aligned}$$

In $I_2$ we may directly choose $\alpha _2=1/2$, and obtain $I_2(\tau ) \le \rho ^{1/2}\, B\,\Vert f\Vert _{C^{0,1/2}}\Gamma (3/2-1/2)$. Last,

$$\begin{aligned} I_3(\tau )&\le |f(\tau )| \int _0^{\rho /\tau } (A \zeta ^{-1/2}+ B\zeta ^{1/2}) e^{-\zeta }d\zeta \\&= |f(\tau )| \left(\int _0^{\rho /\tau } (A+B/2) \zeta ^{-1/2} e^{-\zeta }d\zeta - B (\rho /\tau )^{1/2} e^{-\rho /\tau }\right) \\&\le |f(\tau )| \int _0^{\rho /\tau } (A+B/2) \zeta ^{-1/2}d\zeta - B (\rho /\tau )^{1/2} e^{-\rho /\tau } \\&\le \rho ^{1/2}\,\frac{f(\tau )}{\tau ^{1/2}}\, C \le C\rho ^{1/2}\,\Vert f\Vert _{C^{0,1/2}} \end{aligned}$$

where we used in the last step that $f(0)=0$. $\square$

Lemma 16

For all $t_1>0$ there exists a $C>0$ such that if $\tilde{b}(t) = M\, a(t)$ then

$$\begin{aligned} \int _0^t \int _{\partial \Omega } g^2(x,\tau ) do d\tau \le C t\left(1+\Vert a\Vert _{C^{0,1/2}}^2\right) \quad \text{ for} t\le t_1. \end{aligned}$$

(48)

Proof

First note that without suitable choice of $\tilde{b}$, e.g., for $\tilde{b}\equiv 0$ and $a\not \equiv 0$, we obviously have $\int _0^t \int _{\partial \Omega } g^2(x,\tau )\mathrm{d}o\mathrm{d}\tau \sim R^{-2}$. Denote

$$\begin{aligned} \tilde{a}(t):=a(t)-a_0 \end{aligned}$$

such that $\hat{c}(t):=\tilde{b}-4\pi D\alpha _0=\hat{\tilde{a}}$. Then, since $\tilde{a}(0)=0$, we have, for $|x|=R$, and using $\rho = R^2/(4D)$,

$$\begin{aligned} \int _0^t g^2(x,\tau ) d\tau&= \int _0^t \left[ \frac{1}{R^4} \int _0^\tau \hat{\tilde{a}}(\tau -\sigma ) \phi \big (\frac{R^2}{4D\sigma }\big ) d\sigma -\frac{d_2}{R^2}\tilde{a}(\tau ) +\frac{1}{R} g_2(x,\tau )\right]^2 d\tau \\&= \int _0^t\left[\frac{1}{\rho } \frac{1}{4D} (K_\rho \hat{\tilde{a}}(\tau )-4Dd_2\tilde{a}(\tau ))+\frac{1}{R} g_2(x,t)\right]^2d\tau \\&\le \frac{1}{\rho ^2}\frac{1}{16D^2}\int _0^t |(K_\rho \hat{\tilde{a}})(\tau )-4Dd_2\tilde{a}(\tau )|^2\mathrm{d}\tau +\frac{1}{R^2}\int _0^t g_2^2(x,\tau )\mathrm{d}\tau . \end{aligned}$$

Thus, choosing $\hat{\tilde{a}}=M\tilde{a}(t)$ the first integrand can be estimated, for each $\tau$, as

$$\begin{aligned} |(K_\rho \hat{\tilde{a}})-4Dd_2\tilde{a}|&= |K_\rho \hat{\tilde{a}}-K_0\hat{\tilde{a}}|\le \Vert {\mathcal R}\hat{\tilde{a}}\Vert _{C^0} \le C\,\rho ^{1/2}\,\Vert \hat{\tilde{a}}\Vert _{C^{0,1/2}}\\&\le CR^2 (1+\Vert a\Vert _{C^{0,1/2}}). \end{aligned}$$

Moreover, $\frac{1}{R^2}\int _0^t g_2^2(x,\tau )\mathrm{d}\tau \le CtR^{-2}$, thus

$$\begin{aligned} \int _0^t g^2(x,\tau ) d\tau \le CtR^{-2}(1+\Vert a\Vert _{C^{0,1/2}}^2), \end{aligned}$$

and integrating over $\partial \Omega$ now yields the result. $\square$

To compare solutions of the full problem (39) and the auxiliary problem (40) we finally need to estimate the differences of the traces of $u$ and $p$ on $\partial \Omega$. For this we start with some explicit calculations involving $\widehat{T}_R$. For $b\in C^1(\mathbb R _+)$, let $p=p_1+p_2$ be the solution of

$$\begin{aligned} \partial _tp=D\Delta p+M\, b(t)\delta _0, \quad p_0=\alpha _0\psi , \end{aligned}$$

i.e., $p_1(x,t)=\int _0^t k(x,t-\tau )(Mb(\tau )-4\pi D\alpha _0)\mathrm{d}\tau +\alpha _0\psi _c$, cf. (44), such that

$$\begin{aligned} (\widehat{T}_RMb)(t) = \frac{1}{R} \int _{\partial \Omega }p_1(x,t) do. \end{aligned}$$

We find

$$\begin{aligned} (\widehat{T}_R Mb)(t)&= \frac{1}{R} \int _{\partial \Omega }\int _0^t \frac{1}{(4\pi D (t-\tau ))^{3/2}} e^{-x^2 /(4D(t-\tau ))}\\&\times (M b(\tau )-4\pi D\alpha _0) \mathrm{d}\tau \mathrm{d}o +4\pi \alpha _0\\&= 4\pi R\int _0^t \frac{1}{(4\pi D (t-\tau ))^{3/2}} e^{-R^2 /(4D(t-\tau ))} ( Mb(\tau )\!-\!4\pi D\alpha _0) d\tau \!+\!4\pi \alpha _0\\&= \frac{1}{\sqrt{\pi }D}\int _{R^2/4Dt}^\infty \xi ^{-1/2}e^{-\xi }(M b(t-R^2/(4D\xi )-4\pi D\alpha _0)\mathrm{d}\xi +4\pi \alpha _0. \end{aligned}$$

Moreover, using

$$\begin{aligned} \int _{R^2/4Dt}^\infty \xi ^{-1/2}e^{-\xi }\mathrm{d}\xi =\Gamma (1/2)- \int _0^{R^2/4Dt}\xi ^{-1/2}e^{-\xi }\mathrm{d}\xi \end{aligned}$$

and $M b(0)=4\pi D\alpha _0$ we obtain

$$\begin{aligned}&\biggl |(\widehat{T}_RMb)(t)-\frac{M}{D} b(t)\biggr |\nonumber \\&\quad =\biggl |\frac{M}{\sqrt{\pi }D}\biggl [( b(t){-}4\pi D\alpha _0)\int _0^{R^2/4Dt}\xi ^{-1/2}\mathrm{e}^{-\xi }\mathrm{d}\xi \nonumber \\&\qquad +\int _{R^2/4Dt}^\infty \xi ^{-1/2}\mathrm{e}^{-\xi } (b(t{-}\frac{R^2}{4D\xi }){-}b(t))\mathrm{d}\xi \biggr ]\biggr |\nonumber \\&\quad \le C\Vert b^{\prime }\Vert _{C^0}\int _0^{R^2/4Dt}\xi ^{-1/2}\mathrm{e}^{-\xi }\mathrm{d}\xi +CR^2\Vert b^{\prime }\Vert _{C^0}\int _{R^2/4Dt}^\infty \xi ^{-3/2}\mathrm{e}^{-\xi }\mathrm{d}\xi \nonumber \\&\quad \le CR\sqrt{t}\Vert b^{\prime }\Vert _{C^0}. \end{aligned}$$

(49)

Finally, for $a,b\in C^0$ we have

$$\begin{aligned} \Vert \widehat{T}_Ra-\widehat{T}_Rb\Vert _{L^\infty (0,t_1)} \le C \Vert a-b\Vert _{L^\infty (0,t_1)}. \end{aligned}$$

(50)

Setting $\zeta = a-b$ and assuming w.l.o.g. $a_0=b_0$ this follows from

$$\begin{aligned} \Vert \widehat{T}_R a-\widehat{T}_R b\Vert _{L^\infty (0,t_1)}&= \frac{1}{R}\sup _{t\in (0,t_1)} \int _{\partial \Omega }\int _0^t \frac{1}{(4\pi D (t-\tau ))^{3/2}} e^{-x^2 /(4D(t-\tau ))} \zeta (\tau ) d\tau do \\&\le 4\pi \Vert \zeta \Vert _{L^\infty (0,t_1)} R\sup _{t\in (0,t_1)} \int _0^t \frac{1}{(4\pi D(t\!-\!\tau ))^{3/2}} e^{-R^2 /(4D(t\!-\!\tau ))} d\tau \\&\le C\Vert \zeta \Vert _{L^\infty (0,t_1)} \sup _{t\in (0,t_1)} \int _{R^2/(4Dt)}^\infty \xi ^{-1/2} e^{-\xi } d\xi \\&\le C\Vert \zeta \Vert _{L^\infty (0,t_1)} \Gamma (1/2) = C\Vert \zeta \Vert _{L^\infty (0,T)}. \end{aligned}$$

Lemma 17

For $a,b$ as above we have $\Vert T_R a-\widehat{T}_RMa\Vert _{L^1(0,t_1)} \le CRt_1(1+\Vert a\Vert _{C^{0,1/2}}).$

Proof

We have

$$\begin{aligned} (T_Ra)(t)-(\widehat{T}_RMa)(t)= \frac{1}{R} \int _{\partial \Omega } u(x,t)-p_1(x,t) do = \frac{1}{R} \int _{\partial \Omega } w(x,t)+p_2(x,t) do, \end{aligned}$$

with $\Vert p_2\Vert _{L^\infty (\partial \Omega )}\le C$, cf. Lemma 13. Next, reasoning like in Lemma 14 we have

$$\begin{aligned} 0&\le D\int _0^t \int _\Omega |\nabla w|^2 dx d\tau + \frac{1}{2}\int _{\Omega }w^2(x,t) dx \\&= -\int _0^t \frac{d_1}{R} \int _{\partial \Omega } w^2 do - \int _{\partial \Omega } [R^{-1/2}d_1^{1/2}w][ g(x,t)R^{1/2}d_1^{-1/2}] do d\tau \\&\le -\frac{d_1}{2R} \int _0^t \int _{\partial \Omega } w^2 do d\tau + \frac{R}{2d_1}\int _0^t \int _{\partial \Omega } g^2(x,t) do d\tau \end{aligned}$$

and hence

$$\begin{aligned} \int _0^t \int _{\partial \Omega } w^2 do d\tau \le \frac{R^2}{d_1^2}\int _0^t\int _{\partial \Omega } g^2 do d\tau . \end{aligned}$$

Thus

$$\begin{aligned} \Vert T_Ra-\widehat{T}_RMa\Vert _{L^1(0,t_1)}&\le \frac{1}{R} \int _0^{t_1}\left|\,\int _{\partial \Omega }w(x,\tau ) do\right|\mathrm{d}\tau +CRt_1\\&\le \frac{1}{R}\int _0^{t_1} \left(\,\, \int _{\partial \Omega }1^2do\right)^{1/2} \left(\,\, \int _{\partial \Omega }w(x,\tau )^2 do\right)^{1/2} \mathrm{d}\tau +CRt_1\\&\le C \int _0^{t_1} \left(\,\,\int _{\partial \Omega } w^2(x,\tau ) do\right)^{1/2} d\tau +CRt_1\\&\le \sqrt{t_1}\,C \left(\,\, \int _0^{t_1} \int _{\partial \Omega } w^2(x,\tau ) do\right)^{1/2} d\tau +CRt_1\\&\le CR\sqrt{t_1} \left(\int _0^{t_1}\int _{\partial \Omega } g^2(x,\tau ) do d\tau \right)^{1/2}+CRt_1 \end{aligned}$$

and the result now follows from Lemma 16. $\square$

1.3 The full problem

Proof of Theorem 4

Let $a,b$ be the solutions of (14), (15), respectively. We write the ODEs for $a$ resp. $b$ as

$$\begin{aligned} a^{\prime }(t)&= f(a(t)) - 4\pi d_2 a(t) + d_1 (T_Ra)(t),\qquad a(0) = a_0,\\ b^{\prime }(t)&= f(b(t)) - 4\pi d_2 b(t) + d_1 (\widehat{T}_RM b)(t), \qquad b(0)=a_0, \end{aligned}$$

where we know that

$$\begin{aligned} \Vert T_Ra-\widehat{T}_R Ma\Vert _{L^1(0,t_1)}&\le CRt_1(1+\Vert a\Vert _{C^{0,1/2}}),\\ \Vert \widehat{T}_Ra -\widehat{T}_R b\Vert _{L^\infty (0,t_1)}&\le C_2 \Vert a-b\Vert _{L^\infty (0, t_1)}. \end{aligned}$$

For $\zeta =a-b$ we obtain

$$\begin{aligned} |\zeta (t)|&= \left|\zeta (0)+\int _0^t \zeta ^{\prime }\mathrm{d}\tau \right|\\&\le \int _0^t\Big [ |f(a){-}f(b)|+4\pi d_2 |a-b| +d_1|T_R a-\widehat{T}_R Ma|\\&+d_1|\widehat{T}_R Ma -\widehat{T}_R Mb|\Big ]\mathrm{d}\tau \\&\le d_1\Vert T_R Ma -\widehat{T}_R Ma \Vert _{L^1}+C\int _0^t \eta \mathrm{d}\tau \le CRt(1+\Vert a\Vert _{C^{0,1/2}})+C\int _0^t \eta \mathrm{d}\tau \end{aligned}$$

where $\eta (\tau )=\sup _{0\le \sigma \le \tau }\zeta (\sigma )$. In particular, $\eta (t)\le CRt(1+\Vert a\Vert _{C^{0,1/2}})+C\int _0^t\eta (\tau )\mathrm{d}\tau$, and Gronwall’s inequality yields the result. $\square$

Proof of Corollary 5

The solutions $a,b$ of (14), (15) depend on $R$ and thus henceforth in particular $b$ is denoted by $b_R$. From Theorem 4 we have $b_R(t)=a(t)+Rc_R(t)$ with $\Vert c_R(t)\Vert _{C^0}\le C$ and $a\in C^0([0,t_1])$ uniformly bounded by Theorem 1 and also equicontinuous since also $\Vert a\Vert _{C^1}\le C$ independent of $R$. Therefore, $(b_R)_{0<R<R_0}$ is also uniformly bounded and equicontinuous, and thus by Arzela–Ascoli we have $b_R\rightarrow b\in C^0([0,t_1])$ as $R\rightarrow 0$, at least for a subsequence. It remains to show that $b$ fulfills (4).

Equation (15) for $b_R$ is equivalent to

$$\begin{aligned} b_R(t)=b_R(0)+\int _0^t h_R(b_R)(\tau )\mathrm{d}\tau \end{aligned}$$

with $h_R(b_R)(\tau )=f(b_R(\tau ))-4\pi d_2 b_R(\tau )+ d_1(\widehat{T}_RMb_R)(\tau )$. We show that

$$\begin{aligned} \int _0^t h_R(b_R)(\tau )\mathrm{d}\tau \rightarrow \int _0^t h(b(\tau ))\mathrm{d}\tau \text{ for} R\rightarrow 0 \end{aligned}$$

with $h(b(\tau ))=f(b(\tau ))-Mb=f(b(\tau ))-4\pi d_2 b(\tau ) +d_1 \frac{M}{D} b(\tau )$. For the first two terms in $h_R(b_R)$ we have uniform convergence $|f(b_R(\tau ))-4\pi d_2 b_R(\tau )-f(b(\tau ))-4\pi d_2 b(\tau )| \le C\Vert b_R-b\Vert _{C^0}\le CR$. Finally, using (49), and setting $b_R=b+Re_R$ with $\Vert e_R\Vert _{C^0}\le C$ and $e_R(0)=0$ we also find

$$\begin{aligned}&\int _0^t \left|(\widehat{T}_RMb_R)(\tau ){-}\frac{M}{D} b(\tau )\right|\mathrm{d}\tau \le \int _0^t\left|(\widehat{T}_RMb)(\tau ){-}\frac{M}{D} b(\tau )\right|\nonumber \\&\quad +R\left|(\widehat{T}_R e_R)(\tau )\right|\mathrm{d}\tau \le CR\Vert b\Vert _{C^1} \end{aligned}$$

(51)

uniformly in $0\le t\le t_1$. $\square$

1.4 Improved approximation with delay

In Lemma 15 we showed that, for $f(0)=0$,

$$\begin{aligned} \Vert R_\rho f\Vert _{C^0}\le C\rho ^{1/2} \Vert f\Vert _{C^{0,1/2}[0,t_1]}, \end{aligned}$$

(52)

with the residual $R_\rho f$ defined by

$$\begin{aligned} R_\rho f=K_\rho f-K_0f \end{aligned}$$

where we recall $(K_\rho f)(t)= d_1\pi ^{-3/2} (F_{\rho , 3} f)(t) + 2 D \pi ^{-3/2}(F_{\rho , 5} f)(t), K_0f(t)=\pi ^{-1}(d_1+D)f(t)$. This was used in Lemma 16 to construct and estimate the approximation $\tilde{b}(t)=Ma(t)$ of the solution $b$ of the integral equation

$$\begin{aligned} K_\rho b=4Dd_2a. \end{aligned}$$

(53)

The purpose of this appendix is to find the improved approximation $b_{\mathrm{del}}$ of (53), i.e., to prove Theorem 6. From $K_\rho \rightarrow K_0$ we may use, for $\rho$ sufficiently small, a formal Neumann’s series

$$\begin{aligned} K_\rho ^{-1}&= K_0^{-1}(\mathrm{Id}\!-\!(\mathrm{Id}\!-\!K_0^{-1}K_\rho ))^{-1} \!=\!K_0^{-1} \left(2\mathrm{Id}-K_0^{-1}K_\rho \!+\!{\mathcal O}\bigl ((\mathrm{Id}-K_0^{-1}K_\rho )^2\bigr )\right)\nonumber \\&= 2K_0^{-1}\mathrm{Id}-K_0^{-2}K_\rho +{\mathcal O}\bigl (K_0^{-1}(\mathrm{Id}-K_0^{-1}K_\rho )^2\bigr ). \end{aligned}$$

(54)

The problem with this formula is the loss of regularity in (52). To iterate, i.e., to estimate the second order terms $K_0^{-2}(K_0-K_\rho )^2$ in (54), we need a $C^{0,1/2}$-bound for ${\mathcal R}_{\rho }f$.

Lemma 18

Let $f\in C^{1,1/2}(\mathbb R _+), f(0)=0$. Then

$$\begin{aligned} \Vert {\mathcal R}_{\rho }f\Vert _{C^{0,1/2}[0,t_1]} \le C \Vert f\Vert _{C^{1,1/2}[0,t_1]} \rho ^{1/2}. \end{aligned}$$

(55)

We postpone the proof to the end of the section and first show that we obtain an improved approximation of $a$.

Corollary 19

Let $a\in C^{1,1/2}$ with $a(0)=0$. Defining

$$\begin{aligned} b_{\mathrm{del}}(t)&= ((2K_0^{-1}{-}K_0^{-2}K_\rho )4 Dd_2a)(t) \nonumber \\&= 2Mb(t){-}\frac{ \pi M}{d_1+D} \, \int _0^t b(t{-}\sigma )\,\frac{4D}{R^2}\,\phi (R^2/(4D\sigma ))\,d\sigma \end{aligned}$$

(56)

we find

$$\begin{aligned} \Vert (K_{\rho }b_{\mathrm{del}})(\tau )-4Dd_2a(\tau )\Vert _{C^0}\le C\rho \Vert a\Vert _{C^{1,1/2}}. \end{aligned}$$

Proof

This follows from combining Lemma 15 and Lemma 18, i.e.,

$$\begin{aligned}&(\sqrt{\pi }(A+2B))^2|K_{\rho }b_{\mathrm{del}}-a| \\&\quad =(\sqrt{\pi }(A+B/2))^2\left|\frac{-K_\rho K_\rho a}{\pi (A+B/2)^2} + \frac{2 K_{\rho } a }{\sqrt{\pi }(A+B/2)} -a\right|\\&\quad =\left|-K_\rho (K_\rho a-\sqrt{a}(A+B/2)a)+\sqrt{\pi }(A+B/2) (K_\rho a-\sqrt{\pi }(A+B/2)a)\right|\\&\quad =|R_\rho (R_\rho a)|\le C\rho ^{1/2}\Vert R_\rho a\Vert _{C^{0,1/2}}\le C\rho \Vert a\Vert _{C^{1,1/2}}. \end{aligned}$$

$\square$

Proof of Theorem 6

We now compare the solution $a$ of (14) for $a_0=0$ and $u_0=0$ with the solution of the delayed ODE

$$\begin{aligned} b^{\prime }(t)=f(b(t))-4\pi d_2b+d_1(\widehat{T}_R b_{\mathrm{del}})(t) ,\qquad b(0)=0. \end{aligned}$$

(57)

Since $a\in C^{1,1/2}([0,t_1])$, see Remark 2, and as

$$\begin{aligned} g(x,t)&= \frac{1}{R^4} \int _0^t b(\tau ) \phi \big (R^2 /(4D(t-\tau ))\big ) d\tau -\frac{d_2a(t)}{R^2} \\&= \frac{1}{4D R^2}\left[K_{\frac{R^2}{4D}}b_{\mathrm{del}})(t)-4Dd_2a(t)\right], \end{aligned}$$

we find

$$\begin{aligned} \int _0^t \int _{\partial \Omega } g^2(x,\tau ) do&\le \frac{1}{16D^2 R^4} \int _0^t \int _{\partial \Omega }[K_{R^2/(4D)}b_{\mathrm{del}})(t) -4Dd_2a(t)]^2\, dt\\&\le \frac{1}{16D^2 R^4} \int _0^t \int _{\partial \Omega } [ C \Vert a\Vert _{C^{1,1/2}} R^2/(4D)]^2\, dt = C t R^2\, \Vert a\Vert _{C^{1,1/2}}^2, \end{aligned}$$

and the remainder of the proof works as the one of Theorem 4. $\square$

It remains to give the somewhat lengthy

Proof of Lemma 18

We claim that $\Vert {\mathcal R}_\rho f\Vert _{C^{0,1/2}}=\Vert K_\rho f-K_0f\Vert _{C^{0,1/2}} \le C\rho ^{1/2} \Vert f\Vert _{C^{1,1/2}}$ where

$$\begin{aligned} ({\mathcal R}_\rho f)(\tau )=(F_{\rho ,3}f)(\tau )+F_{\rho ,3}f)(\tau ) -(F_{0,3}f(\tau )+F_{0,5}f(\tau )). \end{aligned}$$

For $k=3,5$ we split

$$\begin{aligned}&|\tau _2-\tau _1|^{-1/2} \biggl [(F_{\rho ,k}f)(\tau _1)- (F_{\rho ,k}f)(\tau _2) - (F_{0,k}f(\tau _1)- F_{0,k}f(\tau _2))\bigg ]\\&\quad = |\tau _2-\tau _1|^{-1/2}\left(\,\,\int _{\rho /\tau _1}^\infty [f(\tau _1-\rho /\zeta )-f(\tau _1)] \zeta ^{k/2-2}e^{-\zeta }dy \right.\\&+ f(\tau _1) \times \left. \int _0^{\rho /\tau _1} \zeta ^{k/2-2} e^{-\zeta }d\zeta \right)\\&\qquad -|\tau _2-\tau _1|^{-1/2}\left(\,\,\int _{\rho /\tau _2}^\infty [f(\tau _2-\rho /\zeta )-f(\tau _2)] \zeta ^{k/2-2}e^{-\zeta }dy \right.\\&+ \,f(\tau _2)\times \left. \int _0^{\rho /\tau _2} \zeta ^{k/2-2} e^{-\zeta }d\zeta \right)\\&\quad =T_{1,k}+T_{2,k}+T_{3,k}+T_{4,k}, \end{aligned}$$

where $0<\tau _1<\tau _2<t_1$ and

$$\begin{aligned} T_{1,k}&= |\tau _2\!-\!\tau _1|^{-1/2} \int _{\rho /\tau _1}^\infty [f(\tau _1\!-\!\rho /\zeta )\!-\!f(\tau _2\!-\!\rho /\zeta )-f(\tau _1) \!+\!f(\tau _2)] \zeta ^{k/2-2}e^{-\zeta }\, d\zeta , \\ T_{2,k}&= |\tau _2-\tau _1|^{-1/2}\int _{\rho /\tau _2}^{\rho /\tau _1} [f(\tau _2-\rho /\zeta )+f(\tau _2)] \zeta ^{k/2-2}e^{-\zeta } d\zeta ,\\ T_{3,k}&= |\tau _2-\tau _1|^{-1/2} f(\tau _1) \int _{\rho /\tau _2}^{\rho /\tau _1} \zeta ^{k/2-2}e^{-\zeta } d\zeta ,\\ T_{4,k}&= |\tau _2-\tau _1|^{-1/2} [f(\tau _1)-f(\tau _2)] \int _0^{\rho /\tau _2} \zeta ^{k/2-2}e^{-\zeta } d\zeta . \end{aligned}$$

In the following we estimate term by term, always assuming $0<\tau _1<\tau _2<t_1$. The critical terms are $T_{1,k}$ which yield $\Vert f\Vert _{C^{1,1/2}}$ on the right hand side of (55), while the estimates for all other $T_{i,k}$ involve only $\Vert f\Vert _{C^{1}}$.

• a) ad $T_{1,k}$. We have

  $$\begin{aligned} T_{1,k}&= \int _{\rho /\tau _1}^\infty \left(\int _0^{\rho /\zeta } \frac{|f^{\prime }(\tau _2-x)-f^{\prime }(\tau _1-x)|}{\sqrt{|\tau _2-\tau _1|}}\mathrm{d}x\right)\zeta ^{k/2-2}\mathrm{e}^{-\zeta }\mathrm{d}\zeta \\&\le \rho \int _{\rho /\tau _1}^\infty \Vert f^{\prime }\Vert _{C^{0,1/2}}\zeta ^{k/2-3}\mathrm{e}^{-\zeta } \mathrm{d}\zeta \le C\rho ^{1/2}\Vert f\Vert _{C^{1,1/2}}, \end{aligned}$$

  where for $k=3$ we used

  $$\begin{aligned} \int _{\rho /\tau _1}^\infty \zeta ^{3/2-3}\mathrm{e}^{-\zeta }\mathrm{d}\zeta =\int _{\rho /\tau _1}^1 \zeta ^{3/2-3}\mathrm{e}^{-\zeta }\mathrm{d}\zeta +\int _{1}^\infty \zeta ^{3/2-3}\mathrm{e}^{-\zeta }\mathrm{d}\zeta \le 2\rho ^{-1/2}\tau _1^{1/2}+C. \end{aligned}$$

  while for $k=5$ we integrate by parts.

• b) ad $T_{3,k}$. For $k=3$ we have

  $$\begin{aligned} |T_{3,3}|&= \frac{|f(\tau _1)|}{\sqrt{\tau _2-\tau _1}} \int _{\rho /\tau _2}^{\rho /\tau _1} \zeta ^{-1/2}e^{-\zeta } d\zeta \le \frac{|f(\tau _1)|}{\sqrt{\tau _2-\tau _1}} \int _{\rho /\tau _2}^{\rho /\tau _1} \zeta ^{-1/2} d\zeta \\&= |f(\tau _1)| 2\,\rho ^{1/2} \,\left( \frac{1}{\sqrt{\tau _1}} - \frac{1}{\sqrt{\tau _2}} \right) \frac{1}{\sqrt{\tau _2-\tau _1}} = 2\rho ^{1/2} \frac{|f(\tau _1)|}{\sqrt{\tau _1\tau _2}} \frac{\sqrt{\tau _2} - \sqrt{\tau _1}}{\sqrt{\tau _2-\tau _1}}\\&\le 2\rho ^{1/2} \frac{|f(\tau _1)|}{\tau _1} \frac{(\sqrt{\tau _2} - \sqrt{\tau _1})(\sqrt{\tau _1} + \sqrt{\tau _2})}{\sqrt{\tau _2-\tau _1}(\sqrt{\tau _1} + \sqrt{\tau _2})} = 2\rho ^{1/2} \frac{|f(\tau _1)|}{\tau _1} \frac{\sqrt{\tau _2-\tau _1}}{\sqrt{\tau _1} + \sqrt{\tau _2}}. \end{aligned}$$

  Now, $f(\tau _1)/\tau _1 = f^{\prime }(\theta )$ for some $\theta \in (0,\tau _1)$ as $f(0)=0$, and thus $|f(\tau _1)/\tau _1|\le \Vert f\Vert _{C^1}$. The second factor $\sqrt{\tau _2-\tau _1}/(\sqrt{\tau _1} + \sqrt{\tau _2})=1$ for $\tau _1=0$ and is monotonously decreasing in $\tau _1$, thus bounded on $0<\tau _1<\tau _2$. Hence $|T_{3,3}|\le C\rho ^{1/2}\Vert f\Vert _{C^1}$ For $k=5$ we use integration by parts to find

  $$\begin{aligned} |T_{3,5}|&= \frac{|f(\tau _1)|}{\sqrt{\tau _2-\tau _1}} \int _{\rho /\tau _2}^{\rho /\tau _1} \zeta ^{1/2}e^{-\zeta } d\zeta \\&\le \frac{|f(\tau _1)|}{\sqrt{\tau _2-\tau _1}} \frac{1}{2} \left[\!-\!\left(\frac{\rho }{\tau _1}\right)^{1/2} e^{-\rho /\tau _1} \!+\! \left(\frac{\rho }{\tau _2}\right)^{1/2}e^{-\rho /\tau _2} \!+\! \int _{\rho /\tau _2}^{\rho /\tau _1} \zeta ^{-1/2}e^{-\zeta } d\zeta \right]\!. \end{aligned}$$

  From $k=3$ we already know that $|f(\tau _1)| \int _{\rho /\tau _2}^{\rho /\tau _1} \zeta ^{-1/2}e^{-\zeta } d\zeta /\sqrt{\tau _2-\tau _1} \le C \sqrt{\rho }\Vert f\Vert _{C^1}$, and it remains to estimate the remaining part

  $$\begin{aligned}&\frac{|f(\tau _1)|}{\sqrt{\tau _2-\tau _1}} \frac{1}{2} \left[-\left(\frac{\rho }{\tau _1}\right)^{1/2} e^{-\rho /\tau _1} + \left(\frac{\rho }{\tau _2}\right)^{1/2}e^{-\rho /\tau _2} \right]\\&\quad =\frac{\sqrt{\rho }}{2} \frac{|f(\tau _1)|}{\sqrt{\tau _1\tau _2}} \frac{-\sqrt{\tau _2}e^{-\rho /\tau _1} -\sqrt{\tau _1}e^{-\rho /\tau _2}}{\sqrt{\tau _2-\tau _1}} \\&\quad = \frac{\sqrt{\rho }}{2} \frac{|f(\tau _1)|}{\sqrt{\tau _1\tau _2}} \frac{\sqrt{\tau _1}-\sqrt{\tau _2}}{\sqrt{\tau _2-\tau _1}} e^{-\rho /\tau _1} + \frac{\sqrt{\rho }}{2} \frac{|f(\tau _1)|}{\sqrt{\tau _1\tau _2}} \sqrt{\tau _1}\frac{e^{-\rho /\tau _2} - e^{-\rho /\tau _1}}{\sqrt{\tau _2-\tau _1}}. \end{aligned}$$

  The first term in the last sum is again known from case $k=3$. Concerning the second we have $|f(\tau _1)|/\sqrt{\tau _1\tau _2}\le \Vert f\Vert _{C^1}$ as before, and it remains to show that $\sqrt{\tau _1}\frac{e^{-\rho /\tau _2} - e^{-\rho /\tau _1}}{\sqrt{\tau _2-\tau _1}}$ is bounded. With $x=\tau _1/\tau _2\in (0,1)$ and $z=\rho /\tau _2\in \mathbb R _+$ we have

  $$\begin{aligned} \sqrt{\tau _1}\frac{e^{-\rho /\tau _2} - e^{-\rho /\tau _1}}{\sqrt{\tau _2-\tau _1}}&= \sqrt{\tau _1/\tau _2} \frac{e^{-\rho /\tau _2} - e^{-(\rho /\tau _2)(\tau _2/\tau _1)}}{\sqrt{1-\tau _1/\tau _2}}\\&= \sqrt{x}\frac{e^{-z}-e^{-z/x}}{\sqrt{1-x}}=h_1(x,z). \end{aligned}$$

  We fix $x_0\in (0,1)$ and compute $z_0=z_0(x_0)= -x_0\ln (x_0)/(1-x_0)$ which maximizes $h_1(x_0,\cdot )$. This defines

  $$\begin{aligned} h_2(x)&= \sqrt{x}\frac{e^{\frac{x\ln (x)}{1-x}} - e^{\frac{x\ln (x)}{x(1-x)}}}{\sqrt{1-x}} = \sqrt{x} e^{\frac{\ln (x)}{1-x}} \frac{1/x-1}{\sqrt{1-x}}\\&= \frac{e^{\frac{\ln (x)}{1-x}} }{ \sqrt{x}}\sqrt{1-x} = x^{1/(1-x)-1/2} \sqrt{1-x}, \end{aligned}$$

  which on $[0,1]$ is bounded by $x^{1/2}(1-x)^{1/2}\le 1/2$ Alltogether, $|T_{3,k}|\le C\rho ^{1/2}\Vert f\Vert _{C^1}$.

• c) ad $T_{2,k}$. For $k=3$ we have

  $$\begin{aligned} |T_{2,3}|\le \frac{\rho \Vert f^{\prime }\Vert _{C^0}}{\sqrt{\tau _2-\tau _1}} \int _{\rho /\tau _2}^{\rho /\tau _1} \zeta ^{3/2-3} d\zeta \le \rho ^{1/2} \Vert f^{\prime }\Vert _{C^0} \frac{\sqrt{\tau _2}-\sqrt{\tau _1}}{\sqrt{\tau _2-\tau _1}}. \end{aligned}$$

  The last factor equals $\frac{\sqrt{\tau _2-\tau _1}}{\sqrt{\tau _1}+\sqrt{\tau _2}}$ and is therefore bounded as in b). Thus $|T_{2,3}|\le C\rho ^{1/2}\Vert f\Vert _{C^1}$. Also $|T_{2,5}|\le C\rho ^{1/2}\Vert f\Vert _{C^1}$ by a similar estimate.

• d) ad $T_{4,k}$. Here $|T_{4,3}|\le \Vert f^{\prime }\Vert _{C^0}\sqrt{\tau _2-\tau _1}\int _0^{\rho /\tau _2} \zeta ^{3/2-2}\mathrm{d}\zeta \le C\rho ^{1/2}\Vert f\Vert _{C^1}$, and integrating by parts for $k=5$ yields a similar result.$\square$

Appendix. Proof of approximation results: several cells

The basic idea for $N\ge 2$ cells is to introduce a delta source for each cell, i.e., to consider

$$\begin{aligned} p_t&= D\Delta p +\sum _{i=1}^N \tilde{b}_i(t)\delta _{x_i}(x),\quad p|_{t=0}= \psi _c|_{t=0}, \end{aligned}$$

(58a)

$$\begin{aligned} b_i^{\prime }&= f(b_i)-4\pi d_2 b_i+ \int _{\partial \Omega _i}\frac{d_1}{R}p\mathrm{d}o, \quad b_i(0)=a_{i0}, \quad i=1,\ldots ,N, \end{aligned}$$

(58b)

with $\psi _c(x,t)=\sum _{i=1}^N \frac{d_2a_i(t)}{d_1{+}D} \left.\frac{\chi (\Vert x-x_i\Vert )}{\Vert x-x_i\Vert }\right|_{\Omega }$, cf. (24). The ODEs (58b) can then be rewritten as

$$\begin{aligned} b_i^{\prime }&= f(b_i)-4\pi d_2 b_i+d_1(\widehat{T}_R^i\tilde{b})(t)+r, \end{aligned}$$

(59)

$$\begin{aligned} (\widehat{T}_R^i\tilde{b})(t)&= \frac{1}{R} \int _{\partial \Omega _i} p(x,t)\mathrm{d}o\nonumber \\&= \frac{1}{R} \int _{\partial \Omega _i} \left(\int _0^t \sum _j \frac{1}{(4\pi D(t{-}\tau ))^{3/2}} \mathrm{e}^{-\Vert x-x_j\Vert ^2/4D(t-\tau )}\tilde{b}_j(\tau )\mathrm{d}\tau \right)\mathrm{d}o, \qquad \end{aligned}$$

(60)

with $r$ due to the initial conditions, i.e., $r\equiv 0$ if $\psi _c|_{t=0}=0$. The proof Theorem 10 again consists of two parts: first, given $a := (a_1(t),..,a_N(t))$, we need a good choice of $\tilde{b} := (\tilde{b}_1,..,\tilde{b}_N)$ to control the difference between the solution $p$ of (58) and the outer field $u(x,t)$ on the boundary $\partial \Omega$; see Lemma 23. Second, the communication terms $\int _{\partial \Omega _i}\frac{d_1}{R} p\, do$ are to be replaced by functionals of $\tilde{b}$; see Lemma 25. The proofs parallel that for one cell; most computations are straight forward (though often tedious) generalization of the one-cell-case. We only sketch the differences, and start with the scaled case $\Vert x_i-x_j\Vert =\delta _{ij}=R^{2\eta }\tilde{\delta }_{ij}$.

Using explicit heat kernel calculations we first obtain the following generalization of Lemma 13.

Lemma 20

Let $w(x,t) = u(x,t)-p(x,t)|_{\Omega }$. Then $w(x,t)$ satisfies

$$\begin{aligned} w_t&= D\Delta w \text{ in} \Omega , \quad w(0,x)=0,\quad \quad B_iw= -g_i(x,t) \text{ on} \partial \Omega _i, \end{aligned}$$

(61)

where

$$\begin{aligned} g_i(x,t)&= \frac{1}{16D^2} \frac{1}{\rho }\left\{ (K_\rho [\tilde{b}_i-4\pi D\alpha _{i0}])(t)- 4 D\,d_2 (a_i(t)-a_{i0})\right\} \\&+ \sum _{j\not = i }\left\{ \!\frac{d_1}{16 D^2} \rho ^{-1/2-\eta } (H_{1,\rho }^j \tilde{b}_j)(x,t) \!- \!2 D(4D)^{4-2\eta }\rho ^{-\eta /2}(H_{2,\rho }^{i,j} \tilde{b}_j)(x,t)\!\right\} \\&+\rho ^{-1/2} g_2^i(x,t), \end{aligned}$$

with

$$\begin{aligned} (H_{1,\rho }^jf)(x,t)&= \rho ^{\eta } \pi ^{-3/2} \int _0^t f(t-\tau )\, {\tau }^{-3/2}e^{-\Vert x-x_j\Vert ^2/(4D\tau )} \, d\tau ,\\ (H_{2,\rho }^{i,j} f)(x,t)&= \rho ^{\eta -1/2} \pi ^{-3/2} \int _0^t f(t-\tau )\, \tau ^{-5/2} \, e^{-\Vert x-x_j\Vert ^2/(4D\tau )}d\tau \,\bigl \langle x\!-\!x_i, x\!-\!x_j\bigr \rangle , \end{aligned}$$

and $\Vert g_2^i\Vert _{\infty }={\mathcal O}(1)$.

Remark 21

The reason for splitting off $\rho ^{-1/2-\eta }$ respectively $\rho ^{-\eta /2}$ from $H_{1,\rho }^jf$ and $H_{2,\rho }^{i,j}f$ is that this way both terms are of order $\rho ^0$. For $H_{1,\rho }^jf$ we show this explicitly in Lemma 26 below, while for $H_{2,\rho }^{i,j}f$ we may estimate, using $\Vert x-x_i\Vert ={\mathcal O}(\rho ^{1/2})$ and $\Vert x_i-x_j\Vert ={\mathcal O}(\rho ^{\eta /2})$,

$$\begin{aligned} \Vert H_{2,\rho }^{i,j}f\Vert _{C^0}\le C\rho ^{3\eta /2} \Vert f\Vert _{C^0}\int _0^t \tau ^{-5/2}\mathrm{e}^{-\rho ^\eta /\tau } \le C\Vert f\Vert _{C^0}\Gamma (3/2). \end{aligned}$$

Thus, the second interaction term $H_{2,\rho }^{i,j} \tilde{b}_j$ can be neglected, while the first interaction term has to be taken into account, also if we scale the distances of cells $\Vert x_i-x_j\Vert$ by $\rho ^\eta , \eta \in (0,1/2)$.

1.1 Low order approximation

Let $\delta _{ij}=R^{2\eta }\tilde{\delta }_{ij}$. Similarly to $K_\rho$ in Lemma 15, the interaction delay terms $H^j_{1,\rho }f$ may be approximated by undelayed terms as $\rho \rightarrow 0$. Also the proof parallels that of Lemma 15.

Lemma 22

Assume $\eta \in (0,1/2)$ and $f\in C^{0,1/2}[0,t]$. There exists a $C>0$ such that for all $x\in \partial \Omega _i$ and $j\ne i$

$$\begin{aligned} \Vert H_{1,\rho }^jf(x,\cdot ) \!- \!H_{1,0}^{i,j}f(\cdot )\Vert _{C^0} \!\le \! C \rho ^{\eta }\Vert f\Vert _{C^{0,1/2}[0,t]}, \text{ where} H_{1,0}^{i,j}f(t) \!:= \! \frac{(4 D)^{1/2-\eta }}{\pi \tilde{\delta }_{i,j}}f(t). \end{aligned}$$

We now transfer Lemma 16 and explain the (to this order) optimal choice $\tilde{b}_i$.

Lemma 23

Let $\eta \in (0,1/2)$. For all $t_1>0$ there exists a $C>0$ such that if

$$\begin{aligned} \tilde{b}_i(t) = M a_i(t) - (4D\rho )^{1/2-\eta }\frac{d_1M}{d_1+D} \sum _{j\not = i}\frac{a_j(t)}{\tilde{\delta }_{i,j}}, \end{aligned}$$

(62)

then for $i=i,\ldots ,N$ and $t\le t_1$

$$\begin{aligned} \int _0^t \int _{\partial \Omega _i} g_i^2(x,\tau )do d\tau \le C t (1+\Vert a\Vert _{C^{0, 1/2}}^2). \end{aligned}$$

Proof

We know that

$$\begin{aligned} \int _0^t g_i(x,\tau )^2\, \mathrm{d}\tau&= \frac{1}{(16D^2)^2} \int _0^t \bigg \{ \frac{1}{\rho }\bigg [ (K_\rho [\tilde{b}_i-4\pi D\alpha _{i0}])(\tau ) - 4 D\,d_2(a_i(\tau )-a_{i0}) \\&\qquad + \sum _{j\not = i } d_1 \rho ^{1/2-\eta } (H_{1,\rho }^j \tilde{b}_j)(x,\tau ) \!\bigg ] \!+\! {\mathcal O}(\rho ^{-1/2})\bigg \}^2\, \mathrm{d}\tau \\&\le \left(\frac{1}{16D^2}\right)^2 \,\, \rho ^{-2}\,\,I_{i,\rho }(x,t)+ Ct\rho ^{-1}, \end{aligned}$$

with

$$\begin{aligned} I_{i,\rho }(x,t)&= \int _0^t \bigg [ (K_\rho [\tilde{b}_i{-}4\pi D\alpha _{i0}])(\tau ) {-} 4 D\,d_2(a_i(\tau ){-}a_{i0}) \\&+ \sum _{j\not =i} d_1\rho ^{1/2-\eta } (H_{1,\rho }^j \tilde{b}_j)(x,\tau ) \bigg ]^2 \, \mathrm{d}\tau \end{aligned}$$

The aim is to select $\tilde{b}_i$ in such a way that $I_{i,\rho } = {\mathcal O}(\rho )$. We know that, for $x\in \partial \Omega _i$,

$$\begin{aligned} | K_\rho f - K_0 f | \le C \rho ^{1/2}\Vert f\Vert _{C^{0,1/2}},\qquad | H_{1,\rho }^j f - H_{1,0}^j f | \le C \rho ^{\eta }\Vert f\Vert _{C^{0,1/2}} \end{aligned}$$

where $K_0 f = \frac{d_1+D}{\pi }\, f$ and $H_{1,0}^j f = \frac{(4 D)^{1/2-\eta }}{\pi \tilde{\delta }_{i,j}}\, f$. Thus, if we define $I_0$ by

$$\begin{aligned} I_{i,0}(t)&= \int _0^t \bigg [ (K_0 [\tilde{b}_i{-}4\pi D\alpha _{i0}])(\tau ) {-} 4 D\,d_2(a_i(\tau ){-}a_{i0}) \\&+ \sum _{j\not = i}d_1 \rho ^{1/2-\eta } (H_{1,0}^j \tilde{b}_j)(\tau ) \bigg ]^2 \, d\tau \\&= \int _0^t \bigg [\frac{d_1+D}{\pi } [\tilde{b}_i(t){-}4\pi D\alpha _{i0}]) {-} 4 D\,d_2(a_i(t){-}a_{i0}) \\&+ \sum _{j\not = i} (4D\rho )^{1/2-\eta } \frac{d_1}{\pi \tilde{\delta }_{i,j}} \tilde{b}_j(t) \bigg ]^2 \, dt, \end{aligned}$$

then $| I_\rho -I_0| \le C \rho \Vert \tilde{b}_i\Vert _{C^{0,1/2}}$, using $\alpha _{i0}=\frac{M}{4\pi D}a_{i0}$. It is sufficient to choose $\tilde{b}_i$, such that $I_0 = {\mathcal O}(\rho )$. Defining $\tilde{b}_i = Ma_i+\rho ^{1/2-\eta } B_i$ and solving for $B_i$ at ${\mathcal O}(\rho ^{1/2-\eta })$ yields (62). $\square$

Given $a_i(t)$ we have an approximation $p$ of $u$ such that $g_i=(Bu-Bp)|_{\Omega _i}$ is ${\mathcal O}(1)$. Next we control the inflow into the cell by a lemma paralleling Lemma 17; we skip the proof.

Lemma 24

Given $a=(a_1,\ldots ,a_N)$, let $\tilde{b}_i$ be defined by (62). Let $T_{i,R} a = \frac{1}{R}\int _{\partial \Omega _i} u(t,x)$ and $\widehat{T}_{i,R} \tilde{b} = \frac{1}{R}\int _{\partial \Omega _i} p_1(t,x)$ where $p_1$ is the solution of (58) with zero initial data, as in (44) for $N=1$. There exists a $C>0$ such that

$$\begin{aligned} \Vert T_{i,R} a - \widehat{T}_{R,i} \tilde{b}\Vert _{L^1(0,t_1)} \le C\rho ^{1/2} (1+\Vert a\Vert _{C^{0,1/2}}^2). \end{aligned}$$

The Lipschitz-continuity of $\widehat{T}_R^i$ can be shown similarly like that of $\widehat{T}_R$, and the proof of Theorem 10, i.e., the justification of the delayed ODE system (32), now follows from Gronwalls inequality, exactly as in the proof of Theorem 4, replacing $\rho$ by $R^2/(4D)$ to obtain $\tilde{b}_i(t) = M a_i(t) - R^{1-2\eta } \frac{d_1\, M}{d_1+D}\sum _{j\not = i}\frac{a_j(t)}{\tilde{\delta }_{i,j}}.$

In order to obtain an ODE from (32) we approximate the delays

$$\begin{aligned} C_{ij}(t):=\frac{1}{R} \int _{\partial \Omega _i} \int _0^t \frac{1}{(4\pi D(t-\tau ))^{3/2}} \mathrm{e}^{-\Vert x-x_j\Vert ^2/4D(t-\tau )}\tilde{b}_j(\tau )\mathrm{d}\tau \mathrm{d}o. \end{aligned}$$

Lemma 25

There exists a $C>0$ such that

$$\begin{aligned} \left|C_{ii}(t)-\frac{\tilde{b}_i(t)}{D}\right|&\le CRt^{1/2}\Vert \tilde{b}^{\prime }\Vert _{C^0},\quad \text{ and,} \text{ for} i\not = j\text{,} \\ \left|R^{-(1-2\eta )}C_{ij}-\frac{\tilde{b}_j(t)}{D\tilde{\delta }_{ij}}\right|&\le CRt^{1/2}\Vert \tilde{b}^{\prime }\Vert _{C^0}. \end{aligned}$$

Proof

The first estimate has been already derived in (49). Recalling $\Vert x_i-x_j\Vert =\delta _{ij}=\tilde{\delta }_{ij}R^{2\eta }$ we obtain for $j\not = i$

$$\begin{aligned}&R^{-(1-2\eta )}C_{ij}= R^{-(1-2\eta )} \frac{1}{R} \int _{\partial \Omega _i} \int _0^t \frac{1}{(4\pi D(t-\tau ))^{3/2}} \mathrm{e}^{-\Vert x-x_j\Vert ^2/4D(t-\tau )}\tilde{b}_j(\tau )\mathrm{d}\tau \mathrm{d}o\\&\quad =4\pi R^{2\eta }\int _0^t (4\pi D\tau )^{-3/2}\mathrm{e}^{-(\delta _{ij}+{\mathcal O}(R))^2/4D\tau } \tilde{b}_j(t-\tau )\mathrm{d}\tau \\&\quad =\frac{R^{2\eta } }{D\pi ^{1/2}(\delta _{ij}+{\mathcal O}(R))}\int _{(\delta _{ij}+{\mathcal O}(R))/4Dt}^\infty \xi ^{-1/2}e^{-\xi } \tilde{b}_j(t-(\delta _{ij}+{\mathcal O}(R))^2/4D\xi )\mathrm{d}\xi \\&\quad \rightarrow \frac{1}{D\pi ^{1/2}\tilde{\delta }_{ij}}\Gamma (1/2) \tilde{b}_j(t) =\frac{1}{D\tilde{\delta }_{ij}} \tilde{b}_j(t)\quad \text{ as} R\rightarrow 0, \end{aligned}$$

as in (49). $\square$

Similar to (51), this last estimate is used in the proof of Corollary 11, that is, the justification of the approximate system (31), i.e.,

$$\begin{aligned} b_i^{\prime }&= f(b_i)-4\pi d_2 b_i+d_1\sum _{i=1}^N C_{ij} =f(b_i)-Mb_i+R^{1-2\eta }\frac{d_1M}{d_1+D}\sum _{j=\ne i} \frac{1}{D\tilde{\delta }_{ij}}b_j, \nonumber \\ \end{aligned}$$

(63)

where in the second equality we dropped the ${\mathcal O}(R)$ resp.  ${\mathcal O}(R^{2-2\eta })$ terms from Lemma 25.

1.2 Improved approximation

We now do not scale the distances between cells ($\eta =0$), and aim at an error bound of order $R^2$. For simplicity we again assume zero initial conditions, s.t. $\alpha _{i0}=a_{i0}=0$ and consequently $g_{i,2}=0$ in Lemma 20. The analysis proceeds similar to that before; the pertinent approximation of $H_{1,\rho }^j$, however, is different.

Lemma 26

There exists a $C>0$ such that $\Vert H_{1,\rho }^jf -I_{1,0}^{i,j}f\Vert _{C^0} \le C \rho ^{1/2}\Vert f\Vert _{C^{0}[0,t]}$ for $x\in \partial \Omega _i$, where

$$\begin{aligned} I_{1,0}^{i,j} f := \int _0^t f(t-\tau )\, \pi ^{-3/2} \, \tau ^{-3/2} \, e^{-\frac{\Vert x_i-x_j\Vert ^2}{4D\tau }}\, d\tau . \end{aligned}$$

(64)

Proof

For $x\in \partial \Omega _i$,

$$\begin{aligned}&\Vert H_{1,\rho }^j f -I_{1,0}^{i,j} f\Vert _{C^0[0, t]}\\&= \sup _{t_1\in [0,t]} \left| \int _0^{t_1} f(t_1-\tau ) \pi ^{-3/2} \, \tau ^{-3/2} \, \left(e^{-\frac{\Vert x_i-x_j\Vert ^2}{4D\tau }}-e^{-\frac{\Vert x-x_j\Vert ^2}{4D\tau }} \right)\, d\tau \right|\\&\le C \Vert f\Vert _{C^0[0,t]} \int _0^{\infty } \tau ^{-3/2} \, \left|e^{-\frac{\Vert x_i-x_j\Vert ^2}{4D\tau }}-e^{-\frac{\Vert x-x_j\Vert ^2}{4D\tau }} \right|\, d\tau \\&\le C \Vert f\Vert _{C^0[0,t]} \left(\int _0^{1/\sqrt{\rho }} \zeta ^{-1/2} \, \left|e^{-\zeta }-e^{-\frac{\Vert x-x_j\Vert ^2}{\Vert x_i-x_j\Vert ^2}\zeta } \right|\, d\zeta \right.\\&\quad + \left. \int _{1/\sqrt{\rho }}^{\infty } \zeta ^{-1/2} \, \left|e^{-\zeta }-e^{-\frac{\Vert x-x_j\Vert ^2}{\Vert x_i-x_j\Vert ^2}\zeta } \right|\, d\zeta \right) \end{aligned}$$

Now, as $|\Vert x-x_j\Vert -\Vert x_i-x_j\Vert |\le C \rho$,

$$\begin{aligned} T_1&:= \int _{1/\sqrt{\rho }}^{\infty } \zeta ^{-1/2} \, \left|e^{-\zeta }-e^{-\frac{\Vert x-x_j\Vert ^2}{\Vert x_i-x_j\Vert ^2}\zeta } \right|\, d\zeta \\&= \int _{1/\sqrt{\rho }}^{\infty } \zeta ^{-1/2} \, e^{-\zeta /2} \left|e^{-\zeta /2}-e^{-\left(\frac{\Vert x-x_j\Vert ^2}{\Vert x_i-x_j\Vert ^2}-\frac{1}{2} \right)\zeta } \right|\, d\zeta \\&\le 2 \sqrt{\rho }\,\, \rho ^{-1/2} \int _{1/\sqrt{\rho }}^{\infty } \zeta ^{-1/2} \, e^{-\zeta /2} \, d\zeta \le C\sqrt{\rho } \end{aligned}$$

by l’Hospital’s rule. Similarly,

$$\begin{aligned} T_2&:= \int _0^{1/\sqrt{\rho }} \zeta ^{-1/2} \, \left|e^{-\zeta }-e^{-\frac{\Vert x-x_j\Vert ^2}{\Vert x_i-x_j\Vert ^2}\zeta } \right|\, d\zeta \le C \rho ^{1/2} \int _0^{1/\sqrt{\rho }} \zeta ^{-1/2} \, e^{-\zeta } \, d\tau , \end{aligned}$$

using $\left|\left( \frac{\Vert x-x_j\Vert ^2}{\Vert x_i-x_j\Vert ^2} -1 \right) \zeta \right| \le C \rho \rho ^{-1/2} = C \rho ^{1/2}$ for $0\le \zeta \le 1/\sqrt{\rho }$. $\square$

Lemma 23 is slightly modified in order to define the appropriate approximation.

Lemma 27

For all $t_1>0$ there exists a $C>0$ such that if

$$\begin{aligned} \tilde{c}_{i,del}^0(t)&= M ( 2I - K_0^{-1}K_\rho ) (a_i)(t)\\ \tilde{c}_{i,del}^1(t)&= \frac{ \pi d_1}{d_1+D} \sum _{j\not = i} ((2 I-K_0^{-1}K_\rho )\circ I_{1,0}^{i,j}) (\tilde{c}_{j,del}^0)(t)\\ \tilde{c}_{i,del}(t)&= c_{i,del}^0(t) - \rho ^{1/2}\quad \tilde{c}_{i,del}^1(t) \end{aligned}$$

then, replacing $\tilde{b}_j$ by $\tilde{c}_{j,del}$ in the definition of $g(x,t)$, for $i=i,\ldots ,N$ and $t\le t_1$

$$\begin{aligned} \int _0^t \int _{\partial \Omega _i} g_i^2(x,\tau )do d\tau \le C \rho ^{1/2} t (1+\Vert a\Vert _{C^{1, 1/2}}^2). \end{aligned}$$

Proof

We start off as in the proof of Lemma 23, and find that ($a_{i,0}=0, u_0=0$)

$$\begin{aligned} \int _0^t g_i(x,\tau )^2\, d\tau \le \left(\frac{1}{ (4D)^2}\right)^2 \quad \frac{1}{\rho ^2}\,\, I_{i,\rho } \end{aligned}$$

where $I_{i,\rho }$ assumes the form ($\eta =0, a_{i,0}=0, u_0=0$ and replacing $\tilde{b}_j$ by $\tilde{c}_j$)

$$\begin{aligned} I_{i,\rho }(x,t)&= \int _0^t \bigg [ (K_\rho \tilde{c}_i)(\tau ) {-} 4 D\,d_2 a_i(\tau ) + \sum _{j\not =i} d_1\rho ^{1/2} (H_{1,\rho }^j \tilde{c}_j)(x,\tau ) \bigg ]^2 \, \mathrm{d}\tau \end{aligned}$$

We aim at a choice of $\tilde{c}_i$ that leads to $I_{i,\rho } = {\mathcal O}(\rho ^{2})$. As we know from Corollary 19 that the integral equation $K_\rho \tilde{c}_i = 4 D\,d_2 a_i$ is solved up to order $\rho$ by the choice $\tilde{c}_{i,del}^0(t) = M ( 2I - K_0^{-1}K_\rho ) (a_i)(t)$, we plug in the ansatz $\tilde{c}_{i,del}(t) = c_{i,del}^0(t) - \rho ^{1/2} B_i(t)$ to obtain

$$\begin{aligned}&I_{i,\rho }(x,t)\\&=\int _0^t \bigg [(K_\rho \tilde{c}_i^0)(\tau ) {-} 4 D\,d_2 a_i(\tau ) {-} \rho ^{1/2} K_\rho B_i(\tau ) + \sum _{j\not =i} d_1\rho ^{1/2} (H_{1,\rho }^j \tilde{c}_j^0)(x,\tau )+ {\mathcal O}(\rho ) \bigg ]^2 \, \mathrm{d}\tau \\&= \int _0^t \bigg [{-} \rho ^{1/2}\left(K_\rho B_i(\tau ) + \sum _{j\not =i} d_1 (I_{1,0}^{i,j} \tilde{c}_j^0)(\tau ) \right)\\&\qquad + \sum _{j\not =i} d_1\rho ^{1/2} ((H_{1,\rho }^j{-}H_{1,0}^{i,j}) \tilde{c}_j^0)(x,\tau ) + {\mathcal O}(\rho ) \bigg ]^2 \, \mathrm{d}\tau \end{aligned}$$

The natural choice for $B_i$ that guarantees the necessary approximation order reads

$$\begin{aligned} B_i = \frac{\pi d_1}{d_1+D} ( 2I - K_0^{-1}K_\rho ) \sum _{j\not =i} (I_{1,0}^{i,j} \tilde{c}_j^0)(\tau ). \end{aligned}$$

As for $x\in \partial \Omega _i$ we know $\Vert (H_{1,\rho }^j-I_{1,0}^{i,j}) f \Vert _{C^0} \le C \rho ^{1/2}\Vert f\Vert _{C^{0}}$, we find $I_{i,\rho } = {\mathcal O}(\rho ^{2})$.

$\square$

This lemma implies

$$\begin{aligned} \Vert T_R^i a-\widehat{T}_R^i \tilde{c}_{del}\Vert _{L^1(0,t_1)} \le C \rho (1+\Vert a\Vert ^2_{C^{1+1/2}(0,t_1)}), \end{aligned}$$

and as $\widehat{T}_R^i$ is Lipschitz-continuous, Gronwall’s lemma yields the approximation theorem as before, i.e., the solution $\tilde{c}_{i,del}$ of

$$\begin{aligned} \tilde{c}_{i,del}^{\prime } = f(\tilde{c}_{i,del}) - 4\pi d_2 \tilde{c}_{i,del} + \widehat{T}_R^i \tilde{c}_{del} \end{aligned}$$

(65)

approximates $a(t)$ up to an error of ${\mathcal O}(\rho ) = {\mathcal O}(R^2)$. In order to finish the proof of Theorem 12 it only remains to computate explicitly an approximation of

$$\begin{aligned} \widehat{T}_R^i \tilde{c}_{i,del}&= R^{-1}(4 D \pi )^{-3/2} \int _{\partial \Omega _i} \int _0^t \tau ^{-3/2} \sum _{j=1}^N\mathrm{e}^{-\Vert x-x_j\Vert ^2/(4D\tau )}\tilde{c}_{j,del}(t-\tau ) \mathrm{d}\tau \mathrm{d}o\\&= \sum _{j\not = i} \rho ^{-1/2} (4 D)^{-2} \int _{\partial \Omega _i} (H_{1,\rho }^{j}\tilde{c}_{i,del})(t, x) \mathrm{d}o + \widehat{T}_R \tilde{c}_{i,del} (t) \end{aligned}$$

where, of course we take $\eta =0$ in the definition of $H_{1,\rho }^{j}$. Lemma 26 indicates that in case $j\not = i$ and $x\in \partial \Omega _i$ we have the estimate $\Vert (H_{1,\rho }^{j} f)(t,x)-(I_{1,0}^{i,j}f)(t)\Vert _{C^0} \le C \rho ^{1/2}\Vert f\Vert _{C^0}$. Hence,

$$\begin{aligned} \widehat{T}_R^i \tilde{c}_{i,del} = \widehat{T}_R \tilde{c}_{i,del} (t) + \frac{\pi }{D} \sum _{j\not = i} \rho ^{1/2} (I_{1,0}^{i,j}\tilde{c}_{j,del})(t)+{\mathcal O}(\rho ). \end{aligned}$$

If we take into account that $\tilde{c}_{i,del}(t) = \tilde{c}_{i,del}^0(t)-\rho ^{1/2}\tilde{c}_{i,del}^1(t)$, we find

$$\begin{aligned} \widehat{T}_R^i \tilde{c}_{i,del} = \widehat{T}_R \tilde{c}_{i,del}^0 (t) + \rho ^{1/2}\left(\frac{\pi }{D} \sum _{j\not = i} (I_{1,0}^{i,j}\tilde{c}_{j,del}^0)(t) -(\widehat{T}_R \tilde{c}_{i,del}^1)(t)\right) +{\mathcal O}(\rho ). \end{aligned}$$

Thus we may replace $\widehat{T}_R^i \tilde{c}_{i,del}$ by this expression in (65) without increasing the approximation error and this completes the proof of Theorem 12.

Remark 28

In order to find stationary solutions of (35) we consider a constant function $c(t) \equiv f_0$ as input into the pertinent delays and consider the limit $t\rightarrow \infty$. We either already proved, or it is possible to prove with similar methods, that

$$\begin{aligned} \tilde{c}_{i,del}^0(t) = M ( 2I - K_0^{-1}K_\rho ) f_0&\rightarrow M f_0\\ I_{1,0}^{i,j} f&\rightarrow f_0 \frac{\sqrt{4D}}{\pi \delta _{ij}}\\ \frac{ \pi d_1}{d_1+D} \sum _{j\not = i} ((2 I-K_0^{-1}K_\rho )\circ I_{1,0}^{i,j}) (c_{j,del}^0)(t)&\rightarrow M \frac{d_1 \sqrt{4D}}{(d_1+D)\delta _{ij}} f_0\\ \widehat{T}_R&\rightarrow f_0/D. \end{aligned}$$

Therefore, the stationary solutions of the DDE (35) and the ODE (31) agree (up to the scaling of the cell distances in the ODE case).