using the Gronwall’s lemma, we obtain that, for all $t\in[\overline{\tau},\overline{\tau}+\underline{\tau}]$ and $m\in[0,b]$, $$N_{1}(t,m)=N_{2}(t,m).$$ Using the method of steps, we obtain this result for $t\geq\overline{\tau}$.

We define the mapping $\alpha:[0,1]\to[0,1]$ by $$\alpha(m)=\Delta(\underline{\tau},m).$$ Notice that $\alpha$ is an increasing function such that $\alpha(m) < m$ for all $m \in (0,1]$ and $\alpha(g(1))=h^{-1}(e^{-\underline{\tau}})$. From (\[delta\]) and since $\underline{\tau}>\tau_0$, $h^{-1}(e^{-\underline{\tau}})<g(1)$. Hence, $\alpha$ is invertible and we denote by $\Lambda$ its inverse function on $[0,h^{-1}(e^{-\underline{\tau}})]$. We set $\Lambda(m)=g(1)$ for $m \in [h^{-1}(e^{-\underline{\tau}}),g(1)]$. Since $\alpha$ is non-negative and increasing on the interval $[0,1]$, $\Lambda$ is increasing on $[0,h^{-1}(e^{-\underline{\tau}})]$.

We first suppose that $b\in(0,h^{-1}(e^{-\underline{\tau}})]$.

We consider the increasing sequences $(b_{n})_{n \in \mathbb{N}}$ and $(t_{n})_{n \in \mathbb{N}}$ defined by $$\left\{ \begin{array}{rcll}
b_{n+1} &=& \Lambda(b_{n}),& \textrm{ for } n \in \mathbb{N},\\
b_{0} &=& b,&
\end{array} \right.$$ and $$\left\{ \begin{array}{rcll}
t_{n+1} &=& t_{n}+\overline{\tau}+\displaystyle\ln\Big(\frac{h(b_{n+1})}{h(b_{n})}\Big),& \textrm{ for } n \in \mathbb{N},\\
t_{0} &=& 0.&
\end{array} \right.$$ We show by induction that, if $m\in[0,b_{n}]$ and $t\geq t_{n}$, then $N_{1}(t,m)=N_{2}(t,m)$.\
Suppose the equality is satisfied for $n\geq 0$. Let $m \in [0,b_{n+1}]$ and $t\geq t_{n+1}$. Then $t\geq t_{n}+\overline{\tau}$. The solutions $N_{i}$, $i=1,2$, of Problem (\[mildformulation\]) can be rewritten as $$\begin{array}{rcl}
N_{i}(t,m) & = & N_{i}(t_{n}+\overline{\tau},\pi_{-(t-t_{n}-\overline{\tau})}(m)) K(t-t_{n}-\overline{\tau},m)-J(N_{i})(t,m) \\
       & + & 2\displaystyle\int_{t_{n}+\overline{\tau}}^{t}\Big(\int_{\underline{\tau}}^{\overline{\tau}} \zeta(\pi_{-(t-s)}(m),a)N_{i}(s-a,\Delta(a,\pi_{-(t-s)}(m))) \times   \\
       