Proof. Let us define the hitting time \[\begin{align} T := \inf \left\{t \geq 0 : \lVert a\rVert+\varepsilon^{-1/2}\lVert s\rVert \geq C\beta \quad or \quad \lVert a_{\perp}-b\rVert+\varepsilon^{-1/2}\lVert s_{\perp}-g\rVert \geq \varepsilon^{\eta}\beta\right\}, \end{align}\] where \(C\) and \(\eta\) are constants to be determined in the course of the proof. Throughout the argument, we write \(D = D(\mathrm{data}) < +\infty\) for a large constant whose value may change from place to place; we stress in particular that the constant \(D\) does not depend on the choice of the constant \(C\). Although we will not always state it explicitly, all the derivations in this proof are performed for times \(t \leq T \wedge t_{\varepsilon}\) only.

7.0.0.1 Controlling \(\overline{a},\overline{s}\)

Recall that \[\partial_{t} \overline{a} = \frac{1}{\varepsilon}\sum_{k=0}^{\infty}\sigma_{k}\overline{s^{k}}\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right).\] Then, note that, on the considered time interval \[\lVert a\rVert+\varepsilon^{-1/2}\lVert s\rVert \leq C\beta \leq CD\varepsilon^{-1/4},\] where we used the fact that \(\beta(t)\) is increasing and that \(\beta(t_{\varepsilon}) \leq D\varepsilon^{-1/4}\). Thus, \[\vert \partial_{t}\overline{a}-\left(\frac{\sigma_{0}}{\varepsilon}\left(\varphi_{0}-\sigma_{0}\overline{a}\right)+\frac{\sigma_{1}\overline{s}}{\varepsilon}\left(\varphi_{1}-\sigma_{1}\overline{as}\right)\right) \vert \leq \big \vert \frac{1}{\varepsilon}\sum_{k=2}^{\infty}\sigma_{k}\overline{s^{k}}\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right) \big \vert.\] But then, for any \(k \geq 2\), \(\overline{as^{k}}\leq \overline{as^{2}} \leq C^{3}D^{3}\varepsilon^{1/4}\), so that, for \(\varepsilon\) sufficiently small, \[\frac{1}{\varepsilon}\sum_{k=2}^{\infty}\sigma_{k}\overline{s^{k}}\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right) \leq C^{2}D\beta^{2}.\] Owing to the definition of \(\overline{b}\), we may write \[\vert \partial_{t}(\overline{a}-\overline{b})+\frac{\sigma_{0}}{\varepsilon}(\overline{a}-\overline{b}) \vert \leq \frac{\sigma_{1}\vert \overline{s}(t) \vert }{\varepsilon}\vert \varphi_{1}-\sigma_{1}\overline{as} \vert +C^{2}D\beta,\] which, upon noting that \(\overline{as} \leq C^{2}\sqrt{\varepsilon}\beta^{2} \leq C^{2}D\), leads to \[\vert \partial_{t}(\overline{a}-\overline{b})+\frac{\sigma_{0}}{\varepsilon}(\overline{a}-\overline{b}) \vert \leq \frac{C^{2}D \vert \overline{s} \vert}{\varepsilon}+C^{2}D\beta^{2}.\] Integrating the above and using the fact that \(\beta\) is increasing, we find \[\begin{align} \vert \overline{a}(t)-\overline{b}(t) \vert &\leq e^{-\frac{\sigma_{0}}{\varepsilon}t}\int_{0}^{t}e^{\frac{\sigma_{0}}{\varepsilon}u}\left(\frac{C^{2}D \vert \overline{s}(u) \vert}{\varepsilon}+C^{2}D\beta^{2}(u)\right)du, \\ & \leq e^{-\frac{\sigma_{0}}{\varepsilon}t}\left(\frac{C^{2}D \vert \overline{S}(t) \vert}{\varepsilon}+C^{2}D\beta^{2}(t)\right)\int_{0}^{t}e^{\frac{\sigma_{0}}{\varepsilon}u}du, \end{align}\] where we introduced \(\overline{S}(t) = \sup_{r \leq t} \vert \overline{s}(r) \vert\). Finally, we obtain \[\label{eq:bound95a95bar} \vert \overline{a}(t)-\overline{b}(t) \vert \leq C^{2}D\left(\overline{S}(t)+\varepsilon\beta^{2}(t)\right).\tag{18}\] Turning to \(\overline{s}\), we write \[\begin{align} \partial_{t}\overline{s} = \overline{a}\sigma_{1}(\varphi_{1}-\sigma_{1}\overline{as})+\sum_{k=2}^{\infty}k\sigma_{k}\overline{as^{k-1}}\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right)-\sum_{k=1}^{\infty}k\sigma_{k}\overline{as^{k+1}}\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right). \end{align}\] Recalling that \(\overline{as^{2}} \leq C^{3}D^{3}\varepsilon^{1/4}\), and that \(\sum_{k=1}^{\infty}k\sigma_{k} < \infty\), we find that, for \(\varepsilon\) sufficiently small \[\vert \partial_{t}\overline{s}-\overline{a}\sigma_{1}\left(\varphi_{1}-\sigma_{1}\overline{as}\right) \vert \leq C^{2}D\sqrt{\varepsilon}\beta^{2}+C^{3}D\varepsilon^{1/4},\] so that \[\vert \partial_{t} \overline{s} \vert \leq C^{2}D(|\overline{a}|+1).\] Using the decomposition \(\vert \overline{a} \vert \leq \vert \overline{b} \vert +\vert \overline{a}-\overline{b} \vert\) along with the bound Eq.@eq:eq:bound95a95bar , we find \[\vert \overline{a} \vert \leq C^{2}D\left(\overline{S}+\varepsilon\beta^{2}+1\right),\] which, recalling that \(\beta^{2} \leq \frac{D}{\sqrt{\varepsilon}}\), leads to \[\partial_{t}\overline{s} \leq C^{4}D\left(\overline{S}+1\right).\] This, along with Lemma 21, implies that \[\partial_{t}\overline{S} \leq C^{4}D\left(\overline{S}+1\right),\] and \[\label{eq:bound95s95bar} \overline{S}(t) \leq C^{4}Dt.\tag{19}\] Finally, using the above bound in Eq.@eq:eq:bound95a95bar , we find \[\vert \overline{a}(t)-\overline{b}(t) \vert \leq C^{4}D\left(t+\varepsilon\beta^{2}(t)\right).\]

7.0.0.2 Controlling \(a_{\perp},s_{\perp}\)

We now write \[\label{eq:dt95a95p} \partial_{t}a_{\perp} = \frac{1}{\varepsilon}\sigma_{1}s_{\perp}\left(\varphi_{1}-\sigma_{1}\overline{as}\right)+\frac{1}{\varepsilon}\sum_{k=2}^{\infty}\sigma_{k}\left(s^{k}-\overline{s^{k}}\right)\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right),\tag{20}\] and \[\begin{align} \label{eq:dt95s95p} \partial_{t}s_{\perp} = \sigma_{1}a_{\perp}(\varphi_{1}-\sigma_{1}\overline{as})&+\sum_{k=2}^{\infty}k\sigma_{k}(as^{k-1}-\overline{as^{k-1}})(\varphi_{k}-\sigma_{k}\overline{as^{k}}) \notag \\ &-\sum_{k=1}^{\infty}k\sigma_{k}(as^{k+1}-\overline{as^{k+1}})(\varphi_{k}-\sigma_{k}\overline{as^{k}}). \end{align}\tag{21}\] Now note that \(\overline{as} = \overline{a_{\perp}s_{\perp}}+\bar{a}\bar{s}\), so that \[\begin{align} \partial_{t}a_{\perp} &= \frac{1}{\varepsilon}\sigma_{1}s_{\perp}(\varphi_{1}-\sigma_{1}\overline{a_{\perp}s_{\perp}})+e_{1}(t), \\ \partial_{t}s_{\perp} &= \sigma_{1}a_{\perp}(\varphi_{1}-\sigma_{1}\overline{a_{\perp}s_{\perp}})+e_{2}(t), \end{align}\] where \[\begin{align} e_{1}(t) &= -\frac{1}{\varepsilon}\sigma^{2}_{1}s_{\perp}\bar{a}\bar{s}+\frac{1}{\varepsilon}\sum_{k=2}^{\infty}\sigma_{k}\left(s^{k}-\overline{s^{k}}\right)\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right), \\ e_{2}(t) &= -\sigma^{2}_{1}a_{\perp}\bar{a}\bar{s}+\sum_{k=2}^{\infty}k\sigma_{k}(as^{k-1}-\overline{as^{k-1}})(\varphi_{k}-\sigma_{k}\overline{as^{k}}) -\sum_{k=1}^{\infty}k\sigma_{k}(as^{k+1}-\overline{as^{k+1}})(\varphi_{k}-\sigma_{k}\overline{as^{k}}). \end{align}\] To establish upper bounds on \(e_{1},e_{2}\), we use 18 and 19 along with the definition of \(T\) to find \[\begin{align} s_{\perp}\bar{a}\bar{s} \leq C\sqrt{\varepsilon}\beta C^{4}D\left(t+\varepsilon\beta^{2}(t)\right)C^{4}Dt, \\ a_{\perp}\bar{a}\bar{s} \leq C\beta C^{4}D\left(t+\varepsilon\beta^{2}(t)\right)C^{4}Dt. \end{align}\] Similarly, for any \(t \leq T\) and \(\varepsilon\) sufficiently small, \[\begin{align} \tag{22} \vert \frac{1}{\varepsilon}\sum_{k=2}^{\infty}\sigma_{k}\left(s^{k}-\overline{s^{k}}\right)\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right) \vert &\leq C^{2}D\beta^{2}, \\ \vert \sum_{k=2}^{\infty}k\sigma_{k}(as^{k-1}-\overline{as^{k-1}})(\varphi_{k}-\sigma_{k}\overline{as^{k}}) \vert &\leq C^{2}D\sqrt{\varepsilon}\beta^{2}, \\ \vert \sum_{k=1}^{\infty}k\sigma_{k}(as^{k+1}-\overline{as^{k+1}})(\varphi_{k}-\sigma_{k}\overline{as^{k}}) \vert &\leq C^{3}D\varepsilon\beta^{3}. \tag{23} \end{align}\] Now, since \(t \leq D\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\) and \(\beta(t) \leq D\varepsilon^{-1/4}\), we reach that, for \(\varepsilon\) sufficiently small, \[\vert e_{1}(t) \vert+\varepsilon^{-1/2}\vert e_{2}(t) \vert \leq C^{2}D\beta^{2}(t).\] To linearize \(a_{\perp},s_{\perp}\) around \(b,g\), we introduce \(u:= a_{\perp}-b\) and \(v = s_{\perp}-g\), so that \[\overline{a_{\perp}s_{\perp}} = \overline{bg}+\overline{ug}+\overline{bv}+\overline{uv},\] and \[\begin{align} \partial_{t}a_{\perp} &= \frac{1}{\varepsilon}\sigma_{1}s_{\perp}(\varphi_{1}-\sigma_{1}\overline{bg})-\frac{1}{\varepsilon}\sigma_{1}^{2}s_{\perp}(\overline{ug}+\overline{bv}+\overline{uv})+e_{1}(t), \\ \partial_{t}s_{\perp} &= \sigma_{1}a_{\perp}(\varphi_{1}-\sigma_{1}\overline{bg})-\sigma_{1}^{2}a_{\perp}(\overline{ug}+\overline{bv}+\overline{uv})+e_{2}(t). \end{align}\] Owing to Proposition 17, we know that \(\lVert g\rVert \leq D\sqrt{\varepsilon}\beta\) and \(\lVert b\rVert \leq D\beta\), so that \[\vert \overline{ug}+\overline{bv} \vert \leq D\beta \left(\lVert u\rVert\sqrt{\varepsilon}+\lVert v\rVert\right),\] and, letting \(Z = \lVert u\rVert+\varepsilon^{-1/2}\lVert v\rVert\), \[\label{eq:bound95ug95bv95bar} \vert \overline{ug}+\overline{bv} \vert \leq D\beta \sqrt{\varepsilon}Z.\tag{24}\] Now, \[\begin{align} \partial_{t} (u+\varepsilon^{-1/2}v) &= \frac{1}{\varepsilon}\sigma_{1}v(\varphi_{1}-\sigma_{1}\overline{bg})-\frac{1}{\varepsilon}\sigma_{1}^{2}s_{\perp}(\overline{ug}+\overline{bv}+\overline{uv})+e_{1}(t) \\ &+\varepsilon^{-1/2}\left(\sigma_{1}u(\varphi_{1}-\sigma_{1}\overline{bg})-\sigma_{1}^{2}a_{\perp}(\overline{ug}+\overline{bv}+\overline{uv})+e_{2}(t)\right). \end{align}\] or equivalently \[\begin{align} \partial_{t}\left(u+\varepsilon^{-1/2}v\right) \leq \frac{\sigma_{1}}{\sqrt{\varepsilon}}(u+\varepsilon^{-1/2}v)(\varphi_{1}-\sigma_{1}\overline{bg})-\sigma_{1}^{2}\left(\frac{1}{\sqrt{\varepsilon}}a_{\perp}+\frac{1}{\varepsilon}s_{\perp}\right)(\overline{ug}+\overline{bv}+\overline{uv})+e_{1}(t)+\varepsilon^{-1/2}e_{2}(t). \end{align}\] Decomposing \(a_{\perp},s_{\perp}\) further (to eliminate the constant \(C\) in leading order terms), we note that \(\lVert a_{\perp}\rVert \leq \lVert b\rVert+\lVert u\rVert\) and \(\lVert s_{\perp}\rVert \leq \lVert g\rVert+\lVert v\rVert\). Using the bound Eq.@eq:eq:bound95ug95bv95bar , along with the fact that \(\frac{\lVert b\rVert}{\sqrt{\varepsilon}}+\frac{\lVert g\rVert}{\varepsilon} \leq \frac{D\beta}{\sqrt{\varepsilon}}\), we reach \[\begin{align} \partial_{t}Z \leq \frac{\sigma_{1}}{\sqrt{\varepsilon}}Z(\varphi_{1}-\sigma_{1}\overline{bg})&+\sigma_{1}^{2}D\beta^{2}Z+\sigma_{1}^{2}D\beta\sqrt{\varepsilon} Z\left(\frac{\lVert u\rVert}{\sqrt{\varepsilon}}+\frac{\lVert v\rVert}{\varepsilon}\right) \\ &+\sigma_{1}^{2}\varepsilon^{-1/2}C\beta(\lVert u\rVert\lVert v\rVert)+e_{1}(t)+\varepsilon^{-1/2}e_{2}(t). \end{align}\] Using the threshold on \(u,v\) defined in \(T\), and the bounds obtained on \(e_{1}(t),e_{2}(t)\), we find \[\begin{align} \partial_{t}Z &\leq \frac{\sigma_{1}}{\sqrt{\varepsilon}}(\varphi_{1}-\sigma_{1}\overline{bg})Z+\sigma_{1}^{2}D\beta^{2}Z+\sigma_{1}^{2}D\beta^{2}Z\varepsilon^{\eta}+\sigma_{1}^{2}\varepsilon^{2\eta}C\beta^{3}+C^{2}D\beta^{2}. \end{align}\] Thus, as long as \(\eta \geq \frac{1}{8}\), we find \[\label{eq:Z95IDE} \partial_{t} Z \leq \frac{\sigma_{1}}{\sqrt{\varepsilon}}(\varphi_{1}-\sigma_{1}\overline{bg})Z+D\beta^{2}Z+C^{2}D\beta^{2}.\tag{25}\] Letting \(t^{\circ}_{\varepsilon} = \frac{1}{4\sigma_{1}\varphi_{1}}\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\), we begin by considering the interval \([0,t^{\circ}_{\varepsilon}]\), for which it will be sufficient to study the differential inequality \[\label{eq:Z95IDE95cut} \partial_{t} Z \leq \frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}Z+D\beta^{2}Z+C^{2}D\beta^{2}.\tag{26}\] Defining \[\lambda_{\varepsilon}(t) = \int_{0}^{t}\left(\frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}+D\beta^{2}(u)\right)du,\] we thus have \[Z(t) \leq \exp(\lambda_{\varepsilon}(t))\int_{0}^{t}\exp(-\lambda_{\varepsilon}(u))C^{2}D\beta^{2}(u)du.\] Then, \[\begin{align} \lambda_{\varepsilon}(t)-\frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t &\leq D\int_{0}^{t}\exp\left(\frac{2\sigma_{1}\varphi_{1}u}{\sqrt{\varepsilon}}\right)du \\ & \leq D\sqrt{\varepsilon}\exp\left(\frac{2\sigma_{1}\varphi_{1}t}{\sqrt{\varepsilon}}\right). \end{align}\] Now, \[\begin{align} \int_{0}^{t}\exp\left(-\lambda_{\varepsilon}(u)\right)\beta^{2}(u)du &\leq \int_{0}^{t}\exp(-\lambda_{\varepsilon}(u))\exp\left(\frac{2\sigma_{1}\varphi_{1}u}{\sqrt{\varepsilon}}\right)du \\ &\leq \int_{0}^{t}\exp\left(\frac{\sigma_{1}\varphi_{1}u}{\sqrt{\varepsilon}}\right)du \\ &\leq D\sqrt{\varepsilon}\exp\left(\frac{\sigma_{1}\varphi_{1}t}{\sqrt{\varepsilon}}\right). \end{align}\] We thus reach \[Z(t) \leq C^{2}D\sqrt{\varepsilon}\exp\left(\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t+D\sqrt{\varepsilon}\exp\left(\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t\right)\right).\] Noting that, on the considered time interval, \(\frac{1}{D}\exp\left(\frac{\sigma_{1}\varphi_{1}t}{\sqrt{\varepsilon}}\right) \leq \beta(t) \leq D\exp\left(\frac{\sigma_{1}\varphi_{1}t}{\sqrt{\varepsilon}}\right)\), we finally reach \[\begin{align} Z(t) &\leq C^{2}D\beta(t)\sqrt{\varepsilon}\exp\left(\frac{\sigma_{1}\varphi_{1}t}{\sqrt{\varepsilon}}\right) \\ &\leq C^{2}D\varepsilon^{1/4}\beta(t). \end{align}\] Thus, as long as \(\eta \in [\frac{1}{8},\frac{1}{4}]\), the proof is complete on this time interval. We now turn to the time interval \([t^{\circ}_{\varepsilon},\tilde{t}^{\circ}_{\varepsilon}]\) for some \(\tilde{t}^{\circ}_{\varepsilon} \geq t^{\circ}_{\varepsilon}\) to be set later. Going back to the differential inequality Eq.@eq:eq:Z95IDE , we introduce \[\Gamma_{\varepsilon}(t) = \int_{0}^{t}\left(\frac{\sigma_{1}}{\sqrt{\varepsilon}}(\varphi_{1}-\sigma_{1}\overline{bg})(u)+D\beta^{2}(u)\right)du,\] so that \[\label{eq:Z95bound95G} Z(t) \leq \exp(\Gamma_{\varepsilon}(t))\int_{0}^{t}\exp(-\Gamma_{\varepsilon}(u))C^{2}D\beta^{2}(u)du.\tag{27}\] Beginning with an upper bound on \(\Gamma_{\varepsilon}(t)\), we write \[\begin{align} \Gamma_{\varepsilon}(t) &= \int_{0}^{t^{\circ}_{\varepsilon}}\left(\frac{\sigma_{1}}{\sqrt{\varepsilon}}(\varphi_{1}-\sigma_{1}\overline{bg})(u)+D\beta^{2}(u)\right)du+\int_{t^{\circ}_{\varepsilon}}^{t}\left(\frac{\sigma_{1}}{\sqrt{\varepsilon}}(\varphi_{1}-\sigma_{1}\overline{bg})(u)+D\beta^{2}(u)\right)du \\ &= \Gamma_{\varepsilon}(t^{\circ}_{\varepsilon})+\Gamma_{\varepsilon}(t)-\Gamma_{\varepsilon}(t^{\circ}_{\varepsilon}). \end{align}\] Using the results proven for the interval \([0,t^{\circ}_{\varepsilon}]\), we have \[\Gamma_{\varepsilon}(t^{\circ}_{\varepsilon}) \leq \frac{1}{4}\log(\frac{1}{\varepsilon})+D.\] Then, note that for any \(t \in [t^{\circ}_{\varepsilon},\infty)\), \(\beta(t) \leq D \varepsilon^{-1/4}\). Furthermore, Corollary 18 gives that \[0 \leq \varphi_{1}-\sigma_{1}\overline{bg} \leq \frac{D}{\sqrt{\varepsilon}}\exp\left(-\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t\right)+D\varepsilon.\] Thus, \[\begin{align} \Gamma_{\varepsilon}(t)-\Gamma_{\varepsilon}(t^{\circ}_{\varepsilon}) &\leq \frac{D}{\varepsilon}\int_{t^{\circ}_{\varepsilon}}^{t}\exp\left(-\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}u\right)du+\frac{D}{\sqrt{\varepsilon}}(t-t^{\circ}_{\varepsilon}) \\ &\leq D+\frac{D}{\sqrt{\varepsilon}}(t-t^{\circ}_{\varepsilon}), \end{align}\] and we reach the upper bound \[\label{eq:bound95G951} \exp(\Gamma_{\varepsilon}(t)) \leq D\varepsilon^{-1/4}\exp\left(\frac{D}{\sqrt{\varepsilon}}(t-t^{\circ}_{\varepsilon})\right).\tag{28}\] Turning to the integral term in Eq.@eq:eq:Z95bound95G , we begin with the term \[\begin{align} \int_{0}^{t^{\circ}_{\varepsilon}}\exp(-\Gamma_{\varepsilon}(u))C^{2}D\beta^{2}(u)du &\leq C^{2}D\int_{0}^{t^{\circ}_{\varepsilon}}\exp\left(\int_{0}^{u}\frac{\sigma_{1}}{\sqrt{\varepsilon}}\left(\varphi_{1}+\sigma_{1}\overline{bg}\right)\right)(u)du \\ &\leq C^{2}D\int_{0}^{t^{\circ}_{\varepsilon}}\exp\left(\frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}u+\frac{\sigma_{1}\sigma_{1}'}{\sqrt{\varepsilon}}\int_{0}^{u}\beta(r)\gamma(r)dr\right) \\ &\leq C^{2}D\int_{0}^{t^{\circ}_{\varepsilon}}\exp\left(\frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}u+\frac{\sigma_{1}\sigma_{1}'}{4}\int_{0}^{u}\exp\left(\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}r\right)dr\right), \end{align}\] where we used Corollary 18 in the third line. Since \(u \leq t^{\circ}_{\varepsilon}\), it holds that \(\int_{0}^{u}\exp\left(\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}r\right)dr \leq D\), and we reach \[\label{eq:bound95G952} \int_{0}^{t^{\circ}_{\varepsilon}}\exp(-\Gamma_{\varepsilon}(u))C^{2}D\beta^{2}(u)du \leq C^{2}D\varepsilon^{1/4}.\tag{29}\] Continuing with the term \(\int_{t^{\circ}_{\varepsilon}}^{t}\exp(-\Gamma_{\varepsilon}(u))C^{2}D\beta^{2}(u)du\), we require a lower bound on \(\Gamma_{\varepsilon}(u)\), for
\(t \in [t^{\circ}_{\varepsilon},\infty)\). We then write \[\begin{align} \Gamma_{\varepsilon}(t^{\circ}_{\varepsilon}) &\geq \frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t^{\circ}_{\varepsilon}-\frac{\sigma_{1}}{\sqrt{\varepsilon}}\sigma'_{1}\int_{0}^{t^{\circ}_{\varepsilon}}\beta(u)\gamma(u)du \\ &\geq \frac{1}{4}\log(\frac{1}{\varepsilon})-D\int_{0}^{t^{\circ}_{\varepsilon}}\exp(\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}u)du \\ &\geq \frac{1}{4}\log(\frac{1}{\varepsilon})-D, \end{align}\] where we used Corollary 18 again in the second line. Then, noting that, for any \(t \in [t^{\circ}_{\varepsilon},\infty)\), \(\beta(t) \geq \frac{1}{D}\varepsilon^{-1/4}\), we may write \[\begin{align} \Gamma_{\varepsilon}(t)-\Gamma_{\varepsilon}(t^{\circ}_{\varepsilon}) &\geq \frac{1}{D}\int_{t^{\circ}_{\varepsilon}}^{t}\beta^{2}(r)dr \\ &\geq \frac{1}{D}\varepsilon^{-1/2}(t-t^{\circ}_{\varepsilon}). \end{align}\] We thus reach \[\begin{align} \label{eq:bound95G953} \int_{t^{\circ}_{\varepsilon}}^{t}\exp(-\Gamma_{\varepsilon}(u))C^{2}D\beta^{2}(u)du &\leq C^{2}D\varepsilon^{-1/2}\exp\left(-\Gamma_{\varepsilon}(t_{\varepsilon}^{\circ})\right)\int_{t^{\circ}_{\varepsilon}}^{t}\exp\left(-\frac{1}{D\sqrt{\varepsilon}}(u-t^{\circ}_{\varepsilon})\right)du \\ &\leq C^{2}D\varepsilon^{1/4}. \end{align}\tag{30}\] Combining the bounds obtained at Eq.@eq:eq:bound95G951 ,29 ,30 , and recalling that, for \(t \in [t^{\circ}_{\varepsilon},\infty)\), \(\beta(t) \geq \frac{1}{D}\varepsilon^{-1/4}\), we obtain \[Z(t) \leq C^{2}D\beta(t)\varepsilon^{1/4}\exp\left(\frac{D}{\sqrt{\varepsilon}}(t-t^{\circ}_{\varepsilon})\right).\] We now set \(\tilde{t}^{\circ}_{\varepsilon} = t^{\circ}_{\varepsilon}+c\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\) for some small constant \(c\) to be determined in such a way that the above bound becomes \[Z(t) \leq C^{2}D\beta(t)\varepsilon^{1/4-cD}.\] Thus, as long as \(cD \leq \frac{1}{8}\), the proof is complete for the time interval \([0,\tilde{t}^{\circ}_{\varepsilon}]\). ◻

Proof. Using the notations of Theorem 12, we consider the hitting time \[\begin{align} T := \inf \bigg\{t \geq 0 : \lVert a\rVert+\varepsilon^{-1/2}\lVert s\rVert \geq C\beta \quad or \quad \vert \varphi_{1}-\sigma_{1}\overline{as} \vert \geq C\bigg\}, \end{align}\] for some constant \(C\) to be set later. First, note that the argument used to establish the inequalities \[\begin{align} \vert \overline{a}-\overline{b} \vert(t) &\leq Ct+C\varepsilon\beta^{2}(t), \\ \vert \overline{s}(t) \vert &\leq Ct, \end{align}\] in the proof of Theorem 12 is valid for any \(0 \leq t \leq T\). Thus, for any \(0 \leq t \leq t_{\varepsilon} \wedge T\), we find \[\label{eq:approx95phi0} \vert \frac{\varphi_{0}}{\sigma_{0}}- \overline{a}(t)\vert \leq KC\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})+C\sqrt{\varepsilon}.\tag{31}\] We then write \[\begin{align} a\partial_{t}s &= a^{2}(1-s^{2})\sum_{k=1}^{\infty}k\sigma_{k}s^{k-1}\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right) \\ s\partial_{t}a &= \frac{1}{\varepsilon}\sum_{k=0}^{\infty}\sigma_{k}s^{k+1}\left(\varphi_{k}-\sigma_{k}\overline{as^{k}}\right), \end{align}\] so that \[\begin{align} \big \vert \partial_{t} \overline{as}-\frac{1}{\varepsilon}\sigma_{0}\overline{s}\left(\varphi_{0}-\sigma_{0}\overline{a}\right)-\left(\overline{a^{2}}+\frac{1}{\varepsilon}\overline{s^{2}}\right)\sigma_{1}\left(\varphi_{1}-\sigma_{1}\overline{as}\right) \big \vert \leq C^{3}D\sqrt{\varepsilon}\beta^{3}, \end{align}\] and thus \[\begin{align} \big \vert \partial_{t} \overline{as}-\left(\overline{a^{2}}+\frac{1}{\varepsilon}\overline{s^{2}}\right)\sigma_{1}\left(\varphi_{1}-\sigma_{1}\overline{as}\right) \big \vert \leq C^{3}D\sqrt{\varepsilon}\beta^{3}+DKC^{2}\log^{2}(\frac{1}{\varepsilon}). \end{align}\] Recalling that there exists a constant \(D\) depending only on the problem data such that, for any \(t \geq 0\), \(\beta(t) \leq D\varepsilon^{-1/4}\), we find \[\big \vert \partial_{t}\left(\varphi_{1}-\sigma_{1}\overline{as}\right)^{2}+\sigma_{1}^{2}\left(\overline{a^{2}}+\frac{1}{\varepsilon}\overline{s^{2}}\right)\left(\varphi_{1}-\sigma_{1}\overline{as}\right)^{2} \big \vert \leq C^{3}D \varepsilon^{-1/4}\vert \varphi_{1}-\sigma_{1}\overline{as} \vert.\] Allowing the exact value of \(D\) to change from line to line in what follows, recall from Proposition 17 that \(\beta(t^{\circ}_{\varepsilon}) \geq \frac{1}{D}\varepsilon^{-1/4}\). This, along with Theorem 12, gives the existence of a (strictly positive) constant \(c\) such that, for any \(t\) in the interval \([t^{\circ}_{\varepsilon},t^{\circ}_{\varepsilon}+c\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})]\) and \(\varepsilon\) sufficiently small, \[\overline{a^{2}}+\frac{1}{\varepsilon}\overline{s^{2}} \geq \frac{1}{D}\varepsilon^{-1/2},\] giving, on the same time interval, \[\partial_{t}\left(\varphi_{1}-\sigma_{1}\overline{as}\right)^{2} \leq -\frac{1}{D\sqrt{\varepsilon}}\left(\varphi_{1}-\sigma_{1}\overline{as}\right)^{2}+C^{3}D\varepsilon^{-1/4}\big \vert \varphi_{1}-\sigma_{1}\overline{as} \big \vert.\] Lemma 22 started at \(t = t^{\circ}_{\varepsilon}\) then gives \[\label{eq:stable95phi1} \big \vert \varphi_{1}-\sigma_{1}\overline{as} \big \vert \leq \varphi_{1}\exp\left(-\frac{(t-t^{\circ}_{\varepsilon})}{D\sqrt{\varepsilon}}\right)+C^{3}D^{2}\varepsilon^{1/4}\left(1-\exp\left(-\frac{(t-t^{\circ}_{\varepsilon})}{D\sqrt{\varepsilon}}\right)\right).\tag{32}\] Now, note that Eq.@eq:eq:dt95a95p ,21 and the bounds at 22 23 are valid for any \(0 \leq t \leq T\), so that we may write \[\begin{align} \partial_{t}a_{\perp} &= \frac{1}{\varepsilon}\sigma_{1}s_{\perp}\left(\varphi_{1}-\sigma_{1}\overline{as}\right)+e_{1}(t) \\ \partial_{t}s_{\perp} &= \sigma_{1}a_{\perp}\left(\varphi_{1}-\sigma_{1}\overline{as}\right)+e_{2}(t), \end{align}\] for some continuous functions \(e_{1}(t),e_{2}(t)\) which verify \(\sup_{t \leq T\wedge t_{\varepsilon}} \vert e_{1}(t) \vert +\varepsilon^{-1/2} \vert e_{2}(t) \vert \leq C^{2}D\beta^{2}\). Then, adding and subtracting \(\overline{bg}\), \[\begin{align} \partial_{t}a_{\perp} &= \frac{\sigma_{1}}{\varepsilon}s_{\perp}\left(\varphi_{1}-\sigma_{1}\overline{bg}\right)-\frac{\sigma_{1}^{2}}{\varepsilon}s_{\perp}\left(\overline{as}-\overline{bg}\right)+e_{1}(t), \\ \partial_{t}s_{\perp} &= \sigma_{1}a_{\perp}\left(\varphi_{1}-\sigma_{1}\overline{bg}\right)-\sigma_{1}^{2}a_{\perp}\left(\overline{as}-\overline{bg}\right)+e_{2}(t). \end{align}\] Recalling the definition \(Z = \lVert u\rVert+\varepsilon^{-1/2}\lVert v\rVert\), with \(u = a_{\perp}-b\) and \(v = s_{\perp}-g\), we have, for any \(0 \leq t \leq T\), \[\label{eq:partial95Z95stable} \partial_{t}Z \leq \frac{\sigma_{1}}{\sqrt{\varepsilon}}\left(\left(\varphi_{1}-\sigma_{1}\overline{bg}\right)+\sigma_{1}\vert \overline{as}-\overline{bg}\vert\right)Z+\frac{D}{\sqrt{\varepsilon}}\beta\vert \overline{as}-\overline{bg}\vert+e_{1}+\varepsilon^{-1/2}e_{2}.\tag{33}\] At this point, we have established that the bounds 32 and 33 are valid on the time interval
\([t^{\circ}_{\varepsilon},t^{\circ}_{\varepsilon}+c\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})]\). Thus, at time \(\overline{t}^{\circ}_{\varepsilon} = t^{\circ}_{\varepsilon}+c\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\), the following estimate holds: \[\begin{align} \vert \varphi_{1}-\sigma_{1}\overline{as}\vert &\leq \varphi_{1}\varepsilon^{c/D}+C^{3}D\varepsilon^{1/4}, \end{align}\] and, recalling that Corollary 18 implies the bound \(0 \leq \varphi_{1}-\sigma_{1}\overline{bg} \leq \frac{D}{\sqrt{\varepsilon}}\exp(-\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t)+D\varepsilon\), we find that \[\vert \varphi_{1}-\sigma_{1}\overline{bg} \vert \leq D\varepsilon^{2\sigma_{1}\varphi_{1}c \wedge 1}.\] We conclude that there exists an exponent \(\delta>0\), depending only on the problem data, such that, at \(t=\overline{t}^{\circ}_{\varepsilon}\), \[\vert \overline{as}-\overline{bg} \vert \leq D\varepsilon^{\delta}.\] Let us now introduce the hitting time \[T' = \inf\left\{t \geq \overline{t}^{\circ}_{\varepsilon} : \overline{a^2}(t)+\frac{1}{\varepsilon}\overline{s^2}(t) \leq \frac{1}{D}\varepsilon^{-1/2}\right\}.\] Clearly \(T'>\overline{t}^{\circ}_{\varepsilon}\), and the inequality 32 remains valid on the time interval \([\overline{t}^{\circ}_{\varepsilon},T' \wedge T\wedge t_{\varepsilon}]\). Noting that the r.h.s. of the inequality 32 is decreasing for any \(t \geq \overline{t}^{\circ}_{\varepsilon}\), we find that, for any \(t\) in the time interval \([\overline{t}^{\circ}_{\varepsilon},T' \wedge T\wedge t_{\varepsilon}]\) and \(\varepsilon\) sufficiently small, \[\partial_{t}Z \leq D\varepsilon^{-1/2+\delta}Z+\frac{D}{\varepsilon^{3/4}}\left(\varphi_{1}\exp\left(-\frac{t}{D\sqrt{\varepsilon}}\right)+C^{3}D^{2}\varepsilon^{1/4}\right).\] Integrating the above, we find that, for any \(t \in [\overline{t}^{\circ}_{\varepsilon},T' \wedge T\wedge t_{\varepsilon}]\), \[\begin{align} Z(t) &\leq Z(\overline{t}^{\circ}_{\varepsilon})\exp\left(\varepsilon^{-1/2+\delta}(t-\overline{t}^{\circ}_{\varepsilon})\right)+\frac{D}{\varepsilon^{3/4}}\int_{\overline{t}^{\circ}_{\varepsilon}}^{t}\exp\left(\varepsilon^{-1/2+\delta}(t-s)\right)\left(D\exp\left(-\frac{s}{D\sqrt{\varepsilon}}\right)+C^{3}D^{2}\varepsilon^{1/4}\right)ds, \end{align}\] and, recalling from Theorem 12 that \(Z(\overline{t}^{\circ}_{\varepsilon}) \leq C\varepsilon^{1/8}\beta(\overline{t}^{\circ}_{\varepsilon})\), \[\begin{align} Z(t) &\leq C\varepsilon^{1/8}\beta(\overline{t}^{\circ}_{\varepsilon})\exp\left(\frac{\varepsilon^{\delta}}{\sqrt{\varepsilon}}t\right)+\frac{D}{\varepsilon^{3/4}}\exp\left(\frac{\varepsilon^{\delta}}{\sqrt{\varepsilon}}t\right)\left(D^{2}\sqrt{\varepsilon}\left[\exp\left(-\frac{\overline{t}_{\varepsilon}^{\circ}}{D\sqrt{\varepsilon}}\right)-\exp\left(-\frac{t}{D\sqrt{\varepsilon}}\right)\right]\right), \\ &+\frac{C^{3}D^{3}}{\varepsilon^{1/2}}\exp\left(\frac{\varepsilon^{\delta}}{\sqrt{\varepsilon}}t\right)\left(\frac{\sqrt{\varepsilon}}{\varepsilon^{\delta}}\left[\exp\left(-\frac{\varepsilon^{\delta}\overline{t}_{\varepsilon}^{\circ}}{\sqrt{\varepsilon}}\right)-\exp\left(-\frac{\varepsilon^{\delta}t}{\sqrt{\varepsilon}}\right)\right]\right). \end{align}\] Thus, for any \(K>0\) independent of \(\varepsilon\) and \(\varepsilon\) sufficiently small, we find \[\begin{align} Z(t_{\varepsilon}) &\leq C\varepsilon^{1/8}\beta(\overline{t}^{\circ}_{\varepsilon})+D\varepsilon^{-1/4+\delta}+C^{3}D^{3}\varepsilon^{-\delta}, \\ &\leq D\varepsilon^{\delta}\beta(t_{\varepsilon}), \end{align}\] where we have used that \(\frac{1}{D}\varepsilon^{-1/4} \leq \beta(t) \leq D\varepsilon^{-1/4}\) for any \(t \in [t^{\circ}_{\varepsilon},\infty)\). From this bound, we deduce that the desired bounds hold for any \(t\) in \([\overline{t}^{\circ}_{\varepsilon},T' \wedge T\wedge t_{\varepsilon}]\), and, by definition of \(Z\), that \(T \wedge T' > K\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\), concluding the proof. ◻

Proof. With \(T\) the same stopping time as the one introduced in the proof of Theorem 12, we have the upper bound 32 , valid for any \(t_{\varepsilon}^{\circ} \leq t \leq T \wedge t_{\varepsilon}\). Thus, the part of the statement regarding \(\vert \varphi_{1}-\sigma_{1}\overline{as} \vert\) is true for some time \(K_{1}\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\), with \(\frac{1}{4\sigma_{1}\varphi_{1}} < K_{1} < K\). Now, for any \(t \in [K_{1}\sqrt{\varepsilon}\log(\frac{1}{\varepsilon}),t_{\varepsilon}]\), \(\vert \varphi_{1}-\sigma_{1}\overline{as} \vert \leq C\varepsilon^{1/4}\) and \(\vert \overline{s}(t) \vert \leq CK_{1}\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\), we find \[\begin{align} \vert \partial_{t}\overline{a} - \frac{\sigma_{0}}{\varepsilon}(\varphi_{0}-\sigma_{0}\overline{a}(t))\vert \leq CDK_{1}\varepsilon^{-1/4}\log(\frac{1}{\varepsilon})+C^{2}D\varepsilon^{-1/2}, \end{align}\] so that, for \(\varepsilon\) sufficiently small, \[\begin{align} \partial_{t} \left(\varphi_{0}-\sigma_{0}\overline{a}\right)^{2} \leq -\frac{\sigma_{0}}{\varepsilon}\left(\varphi_{0}-\sigma_{0}\overline{a}\right)^{2}+\frac{C^{2}D}{\sqrt{\varepsilon}}\vert \varphi_{0}-\sigma_{0}\overline{a}\vert. \end{align}\] Lemma 22 started at \(t = K_{1}\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\) then gives \[\begin{align} \vert \varphi_{0}-\sigma_{0}\overline{a}(t) \vert &\leq C\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\exp\left(-\frac{\sigma_{0}}{2\varepsilon}\left(t-K_{1}\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\right)\right) \notag \\ &+C^{2}D\sqrt{\varepsilon}\left(1-\exp\left(-\frac{\sigma_{0}}{2\varepsilon}\left(t-K_{1}\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\right)\right)\right), \end{align}\] implying the result for \(\vert \varphi_{0}-\sigma_{0}\overline{a} \vert\), for any \(K^{\circ}>K_{1}\) independent of \(\varepsilon\). Finally, choosing any \(K^{\circ} \geq \frac{1}{2\sigma_{1}\varphi_{1}}\), Corollary 18 implies that there exists a constant \(D\) depending only on \(\varphi_{1},\sigma_{1},\sigma_{1}',M\) such that, for any \(t \geq K^{\circ}\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\), \[\lVert b(t,.)-\left(\frac{\varphi_{1}}{\sigma_{1}'}\right)^{1/2}\varepsilon^{-1/4}a_{init,\perp}(.)\rVert+\varepsilon^{-1/2}\lVert g(t,.)-\left(\frac{\varphi_{1}}{\sigma_{1}'}\right)^{1/2}\varepsilon^{1/4}a_{init,\perp}(.)\rVert \leq D\varepsilon^{1/4},\] implying the last part of the statement. ◻

Proof of Proposition 2. We begin by noting that, for any \(\eta \in (0,1)\), \[\vert \overline{b}(t)-\frac{\varphi_{0}}{\sigma_{0}} \vert \leq \varepsilon^{\eta} \iff \exp\left(\frac{\sigma_{0}^{2}}{\varepsilon}t\right) \geq \frac{\vert \overline{a}(0)-\frac{\varphi_{0}}{\sigma_{0}} \vert}{\varepsilon^{\eta}},\] so that \[\inf\left\{t \geq 0 : \vert \overline{b}(t)-\frac{\varphi_{0}}{\sigma_{0}} \vert \leq \varepsilon^{\eta} \right\} \leq \frac{\left(\eta\log(\frac{1}{\varepsilon})+\log (\vert \overline{a}(0)-\frac{\varphi_{0}}{\sigma_{0}} \vert )\right)_{+}}{\sigma_{0}^{2}}\varepsilon.\] Letting \(t_{0}\) denote the right-hand-side of the above inequality, this time horizon falls within the interval of validity of Theorem 12, so there exists a constant \(C\) depending only on the problem data such that \[\vert \overline{a}-\overline{b} \vert (t_{0}) \leq Ct_{0}+C\varepsilon\beta^{2}(t_{0}).\] Using the bounds on \(\beta(t)\) given at Proposition 17 and keeping the leading order terms in \(\varepsilon\) concludes the proof. ◻

Proof of Proposition 3. First note that, for all \(t \geq 0\), \[\begin{align} \vert \overline{as}(t)-\overline{bg}(t) \vert &\leq \vert \overline{a_{\perp}s_{\perp}}(t) -\overline{bg}(t) \vert +\vert \bar{a}(t)\bar{s}(t) \vert, \notag \\ &\leq \lVert a_{\perp}(t,.)\rVert\lVert s_{\perp}(t,.)-g(t,.)\rVert+\lVert g(t,.)\rVert\lVert a_{\perp}(t,.)-b(t,.)\rVert+\vert \overline{a}(t) \vert\vert \overline{s}(t) \vert. \end{align}\] Owing to Theorem 12 and 13, for any constant \(K\) and \(0 \leq t \leq K\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\), for \(\varepsilon\) sufficiently small, there exists a constant \(C\) depending only on the problem data and \(K\), and a constant \(\delta \in (0,\frac{1}{4})\) depending on the problem data, such that \[\begin{align} &\vert \overline{a}(t) \vert \leq C, \quad \vert \overline{s}(t) \vert \leq C\sqrt{\varepsilon}\log(\frac{1}{\varepsilon}), \\ &\lVert a_{\perp}(t,.)\rVert \leq C\beta(t), \quad \lVert g(t,.)\rVert \leq C\sqrt{\varepsilon}\beta(t), \\ &\lVert s_{\perp}(t,.)-g(t,.)\rVert \leq C\varepsilon{1/2+\delta}\beta(t), \quad \lVert a_{\perp}(t,.)-b(t,.)\rVert \leq C\varepsilon^{\delta}\beta(t). \end{align}\] In what follows, we will allow \(C\) to change from line to line (as long as it only depends on the problem data and \(K\)). Now, recall from Proposition 17 that there exists a constant \(D\) depending only on the problem data such that, for all \(t \geq 0\), \(\beta(t) \leq D\varepsilon^{-1/4}\), so that we reach, for any \(0 \leq t \leq K\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\), \[\vert \overline{as}(t)-\overline{bg}(t) \vert \leq C\varepsilon^{\delta}.\] Using Corollary 18 at time \(t_{1} = \left(\frac{1}{4\sigma_{1}\varphi_{1}}+c_{1}\right)\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})\), we find that \[\vert \frac{\varphi_{1}}{\sigma_{1}}-\overline{bg}(t_{1}) \vert \leq C\varepsilon^{2\sigma_{1}\varphi_{1}c_{1}}.\] We conclude that \[\vert \overline{as}(t_{1}) -\frac{\varphi_{1}}{\sigma_{1}}\vert \leq C\varepsilon^{\delta \wedge 2\sigma_{1}\varphi_{1}c_{1}},\] giving the first part of Theorem 4. The second part is a direct consequence of Corollary 14. ◻

Proof. For any \(t_{0} \leq t \leq T_{q}\), \[\begin{align} \partial_{t} a_{q}(t,\omega)^{2} = \frac{2\sigma_{2}\varphi_{2}}{\varepsilon}a_{q}(t,\omega)s_{q}^{2}(t,\omega), \\ \partial_{t} s_{q}(t,\omega)^{2} = 4\sigma_{2}\varphi_{2}a_{q}(t,\omega)s_{q}^{2}(t,\omega), \end{align}\] so that, \[\label{eq:quad95as95link} a_{q}(t,\omega)^{2} = \frac{1}{2\varepsilon}s_{q}(t,\omega)^{2}-\frac{1}{2\varepsilon}s_{q}(t_{0},\omega)^{2}+a_{q}(t_{0},\omega)^{2}.\tag{40}\] Now, recall that, owing to Theorem 13 and Corollary 14, there exist constants \(C,\delta\) independent of \(\varepsilon\) such that \[\begin{align} \vert \overline{a}_{q}(t_{0})-\frac{\varphi_{0}}{\sigma_{0}} \vert \leq C\sqrt{\varepsilon}, \quad \vert \overline{s}_{q}(t_{0}) \vert \leq CK^{\circ}\sqrt{\varepsilon}\log(\frac{1}{\varepsilon}), \\ \lVert(a_{q,\perp}-b)(t_{0},.)\rVert+\varepsilon^{-1/2}\lVert(s_{q,\perp}-g)(t_{0},.)\rVert \leq C\varepsilon^{\delta}\beta(t_{0}). \end{align}\] To compare \(s_{q}(t_{0},\omega)\) and \(a_{q}(t_{0},\omega)\), we write \[\begin{align} a_{q}(t_{0},\omega) = b(t_{0},\omega)+a_{q,\perp}(t_{0},\omega)-b(t_{0},\omega)+\overline{a_{q}}(t_{0}), \\ s_{q}(t_{0},\omega) = g(t_{0},\omega)+s_{q,\perp}(t_{0},\omega)-g(t_{0},\omega)+\overline{s_{q}}(t_{0}), \end{align}\] so that, allowing \(C\) to change from line to line, \[\begin{align} \tag{41} \lVert a_{q}(t_{0},.)-\frac{\varphi_{0}}{\sigma_{0}}-\beta(t_{0})a_{init,\perp}(.)\rVert &\leq C\varepsilon^{\delta}\beta(t_{0})+C\sqrt{\varepsilon}, \\ \lVert s_{q}(t_{0},.)-\gamma(t_{0})a_{init,\perp}(.)\rVert &\leq C\sqrt{\varepsilon}\log(\frac{1}{\varepsilon})+C\varepsilon^{1/2+\delta}\beta(t_{0}). \tag{42} \end{align}\] Choosing any \(K^{\circ} > \frac{1}{4\sigma_{1}\varphi_{1}}\), Proposition 17 then gives that \[\theta(t_{0}) = \frac{1}{2}\log(\frac{4\varphi_{1}}{\sigma_{1}'\sqrt{\varepsilon}}+o(\frac{1}{\sqrt{\varepsilon}})),\] so that there exists a constant \(D\) depending only on \(\sigma_{1},\varphi_{1},\sigma_{1}'\) such that \[\vert \beta(t_0) -\frac{1}{2}e^{\theta(t_{0})} \vert \leq D\varepsilon^{1/4}\quad and \quad \vert \gamma(t_0) - \frac{\sqrt{\varepsilon}}{2}e^{\theta(t_{0})} \vert \leq D\varepsilon^{3/4}.\] Combining the bounds 41 , 42 with the above, keeping the leading order in \(\varepsilon\) and absorbing the constant \(D\) in \(C\), we find that \[\lVert s_{q}(t_{0},.)-\sqrt{\varepsilon}a_{q}(t_{0},.)\rVert \leq C\varepsilon^{3/4},\] which, upon recalling Eq.@eq:eq:quad95as95link , implies in turn \[\begin{align} \lVert a_{q}(t,.)^{2}-\frac{1}{2\varepsilon}s_{q}(t,.)^{2}-\frac{1}{2}a_{q}(t_{0},.)^{2}\rVert &\leq C \lVert\frac{s_{q}(t_{0},.)}{\sqrt{\varepsilon}}-a_{q}(t_{0},.)\rVert\lVert\frac{s_{q}(t_{0},.)}{\sqrt{\varepsilon}}+a_{q}(t_{0},.)\rVert, \\ &\leq C. \end{align}\] Then, for any \(\omega \in \Omega\), we may bound the evolution of \(a_{q}(t,\omega)\) as follows \[2\sigma_{2}\varphi_{2}a_{q}^{2}(t,\omega)-\sigma_{2}\varphi_{2}a_{q}(t_{0},\omega)^{2}-C \leq \partial_{t}a_{q}(t,\omega) \leq 2\sigma_{2}\varphi_{2}a_{q}^{2}(t,\omega)-\sigma_{2}\varphi_{2}a_{q}(t_{0},\omega)^{2}+C,\] and, since \(a_{q}(t,\omega)\) is increasing for any \(\omega\), \(a_{q}(t,\omega)^{2}\) is increasing for any \(\omega\) in the set \[\mathcal{A} = \left\{\omega : a_{q}(t_{0},\omega)\geq 0\right\}.\] We may then write, for any \(\omega \in \mathcal{A}\), \[\sigma_{2}\varphi_{2}a_{q}^{2}(t,\omega)-C \leq \partial_{t}a_{q}(t,\omega) \leq 2\sigma_{2}\varphi_{2}a_{q}^{2}(t,\omega)+C.\] Thus, for \(\varepsilon\) sufficiently small, the upper and lower bounds at ?? hold for any \(\omega\) in the set \[\mathcal{B} = \left\{\omega : a_{q}(t_{0},\omega)^{2} \geq \frac{C}{c\sigma_{2}\varphi_{2}}\right\}.\] The bound 41 then ensures that \(\mathcal{U}(c) \subset \mathcal{A} \cap \mathcal{B}\), concluding the proof. ◻

Proof. We begin by constructing an appropriate initial condition \(z(t_{0},.) = (z_{1}(t_{0},.),z_{2}(t_{0},.))\). Recalling that \(\hat{a}(t_{0},.) = \varepsilon^{1/4}a(t_{0},.)\) and \(\hat{s}(t_{0},.) = \varepsilon^{-1/4}s(t_{0},.)\), we have \[\begin{align} \vert \overline{\hat{a}}(t_{0})-\varepsilon^{1/4}\frac{\varphi_{0}}{\sigma_{0}} \vert \leq C_{2}\varepsilon^{3/4}, \\ \vert \overline{\hat{a}\hat{s}}(t_{0})-\frac{\varphi_{1}}{\sigma_{1}} \vert \leq C_{2}\varepsilon^{1/4}. \end{align}\] Then, we choose \[\begin{align} z_{1}(t_{0},.) = \hat{a}_{\perp}(t_{0},.)+\varepsilon^{1/4}\frac{\varphi_{0}}{\sigma_{0}}, \\ z_{2}(t_{0},.) = \frac{\varphi_{1}}{\sigma_{1}\overline{\hat{a}_{\perp}\hat{s}_{\perp}}(t_{0})}\hat{s}_{\perp}(0,.), \end{align}\] so that \[\begin{align} \overline{z_{1}}(t_{0}) &= \varepsilon^{1/4}\frac{\varphi_{0}}{\sigma_{0}}, \\ \overline{z_{1}z_{2}}(t_{0}) &= \frac{\varphi_{1}}{\sigma_{1}}, \end{align}\] and \[\begin{align} \lVert z_{1}(t_{0},.)-\hat{a}(t_{0},.)\rVert &\leq C_{2}\varepsilon^{3/4}. \end{align}\] Also, \[\begin{align} \lVert z_{2}(t_{0},.)-\hat{s}(t_{0},.)\rVert \leq \frac{\vert \varphi_{1}-\sigma_{1}\overline{\hat{a}\hat{s}}(t_{0})\vert}{\sigma_{1}\overline{\hat{a}\hat{s}}(t_{0})}\lVert\hat{s}_{\perp}(t_{0},.)\rVert+ \vert \overline{\hat{s}}(t_{0}) \vert, \end{align}\] which, upon recalling that there exists \(\delta>0\) independent of \(\varepsilon\) such that, \[\begin{align} \lVert\hat{s}_{\perp}(t_{0},.)-\left(\frac{\varphi_{1}}{\sigma_{1}'}\right)^{1/2}a_{init,\perp}(.)\rVert \leq C\varepsilon^{\delta} \quad and \quad \vert \overline{\hat{s}}(t_{0}) \vert \leq C\varepsilon^{1/4}\log(\frac{1}{\varepsilon}), \end{align}\] implies that there exists a constant \(C'\) such that \[\lVert z_{2}(t_{0},.)-\hat{s}(t_{0},.)\rVert \leq C'\varepsilon^{1/4}\log(\frac{1}{\varepsilon}).\] Now, note that, owing to Theorem 13, there exists a constant \(D\) depending only on \(\sigma_{1},\varphi_{1},\sigma_{1}',M\) such that, for \(\varepsilon\) sufficiently small, \[\lVert a(t_{0},.)\rVert+\varepsilon^{-1/2}\lVert s(t_{0},.)\rVert \leq D\varepsilon^{-1/4}.\] Let \(T_{1}\) be the hitting time \[T_{1} = \inf\{t \geq t_{0} : \lVert a(t,.)\rVert+\varepsilon^{-1/2}\lVert s(t,.)\rVert \geq C_{1}\varepsilon^{-1/4}\},\] for any constant \(C_{1} >D\) (that does not depend on \(\varepsilon\)), and let \(T\) be a time horizon independent of \(\varepsilon\). Then, for any \(t_{0} \leq t \leq T\varepsilon^{1/4} \wedge T_{1}\) consider the following change in time and scale in the dynamical system 34 35 : \[\begin{align} \hat{a}(t,\omega) = \varepsilon^{1/4}a(\varepsilon^{1/4} t,\omega), \\ \hat{s}(t,\omega) = \varepsilon^{-1/4}s(\varepsilon^{1/4} t,\omega). \end{align}\] Thus, we are now interested in controlling the evolution of the pair \(\hat{a},\hat{s}\) on a timescale of order one, i.e. independent of \(\varepsilon\). The evolution 34 35 then becomes \[\begin{align} \partial_{t} \hat{a}(t,\omega) &= \frac{\sigma_{0}}{\sqrt{\varepsilon}}\left(\varphi_{0}-\sigma_{0}\overline{a}(t)\right)+\frac{\sigma_{1}}{\varepsilon^{1/4}}\hat{s}(t,\omega)\left(\varphi_{1}-\sigma_{1}\overline{as}(t)\right)+\sigma_{2}\varphi_{2}\hat{s}^{2}(t,\omega), \\ \partial_{t} \hat{s}(t,\omega) &= \hat{a}(t,\omega)\frac{\sigma_{1}}{\varepsilon^{1/4}}\left(\varphi_{1}-\sigma_{1}\overline{as}(t)\right)+2\sigma_{2}\varphi_{2}\hat{a}(t,\omega)\hat{s}(t,\omega), \end{align}\] and we may rewrite \(T_{1}\) as \[T_{1} = \inf\{t \geq t_{0} : \lVert\hat{a}(t,.)\rVert+\lVert\hat{s}(t,.)\rVert \geq C_{1}\}.\] Now, an argument similar to the one employed to prove the upper bound Eq.@eq:eq:stable95phi1 in the proof of Theorem 13 and Corollary 14 shows that there exist constants \(C_{3},C_{4}\) depending only on the problem data and \(C_{1}\) such that, for any \(t_{0} \leq t \leq T_{1}\), \[\begin{align} \vert \varphi_{0}-\sigma_{0}\overline{a}(t) \vert \leq C_{3}\sqrt{\varepsilon}, \quad and \quad \vert \varphi_{1}-\sigma_{1}\overline{as}(t) \vert \leq C_{4}\varepsilon^{1/4}. \end{align}\] We now note that, by construction, \[\lVert z_{1}(t_{0},.)\rVert+ \lVert z_{2}(t_{0},.)\rVert \leq D,\] and introduce the hitting time \[T_{2} = \inf\{t \geq t_{0} : \lVert z_{1}(t,.)\rVert+\lVert z_{2}(t,.)\rVert \geq C_{1}\}.\] Letting \[\begin{align} z(t,\omega) &= (z_{1}(t,\omega),z_{2}(t,\omega)), \\ z_{\varepsilon}(t,\omega) &= (\hat{a}(t,\omega),\hat{s}(t,\omega)), \end{align}\] we apply Corollary 20 to obtain that there exist constants \(C_{5},C_{6}\) independent of \(\varepsilon\) such that, for any \(t_{0} \leq t \leq T_{1} \wedge T_{2} \wedge T\), \[\lVert\lVert z_{\varepsilon}(t,.)-z(t,.)\rVert_{2}\rVert \leq \left(\lVert\lVert z_{\varepsilon}(t_{0},.)-z(t_{0},.)\rVert_{2}\rVert+C_{5}\varepsilon^{1/8}(t-t_{0})\right)\exp(C_{6}(t-t_{0})),\] concluding the proof. Proposition 7 is an immediate consequence of Lemma 16. ◻

Proof. A direct calculation using 47 48 shows that \[\begin{align} \partial_{t}\big(\beta^{2}(t)-\varepsilon^{-1} \gamma^{2}(t)\big) &= 2\beta(t)\partial_{t}\beta(t)-2\varepsilon^{-1}\gamma(t)\partial_{t}\gamma(t) \\ &=0, \end{align}\] and, since \(\beta(0) = 1\) and \(\gamma(0) = 0\), we find that \(\beta^{2}-\varepsilon^{-1}\gamma^{2} = 1\) for all times. This allows us to reparametrize the solution using a hyperbolic angle \(\theta(t)\) such that \(\beta(t) = \cosh \theta(t)\) and \(\gamma(t) = \sqrt{\varepsilon}\sinh \theta(t)\). Substituting this into 47 , we find \[\left(\sinh \theta(t)\right)\partial_{t}\theta = \varepsilon^{-1}\sigma_{1}(\sqrt{\varepsilon}\sinh \theta(t))\left(\varphi_{1}-\sigma'_{1} \cosh \theta(t)\sqrt{\varepsilon}\sinh \theta(t)\right).\] Dividing by \(\sinh(\theta)\) (for \(t>0\)) and using the identity \(2\sinh \theta \cosh \theta = \sinh 2\theta\), we obtain the scalar ODE ?? .
Now, the latter implies that, for all times, \[\frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}-\frac{\sigma_{1}\sigma'_{1}}{4}e^{2\theta(t)} \leq \partial_{t} \theta(t) \leq \frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}-\frac{\sigma_{1}\sigma'_{1}}{4}\left(e^{2\theta(t)}-1\right),\] so that \[l(t) \leq \theta(t) \leq u(t),\] where \(l,u\) are respectively the solutions of \[\begin{align} \partial_{t} l(t) &= \frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}-\frac{\sigma_{1}\sigma'_{1}}{4}e^{2l(t)}, \quad l(0) = 0, \\ \partial_{t} u(t) &= \frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}-\frac{\sigma_{1}\sigma'_{1}}{4}\left(e^{2u(t)}-1\right), \quad u(0) = 0. \end{align}\] Focusing on \(l(t)\), let us introduce \(\tilde{l}(t) = l(t)-\frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t\), so that \(\tilde{l}(t)\) solves \[\partial_{t}\tilde{l}(t) = -\frac{\sigma_{1}\sigma'_{1}}{4}e^{2\tilde{l}(t)}e^{\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t}, \quad \tilde{l}(0) = 0.\] To solve this ODE, we define \[\begin{align} F(x) &= -\int_{0}^{x}e^{-2y}dy \\ &= \frac{1}{2}\left(e^{-2x}-1\right), \\ G(t) &= \int_{0}^{t}\frac{\sigma_{1}\sigma'_{1}}{4}e^{\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}s} ds \\ &= \frac{\sigma'_{1}\sqrt{\varepsilon}}{8\varphi_{1}}\left(e^{\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t}-1\right), \end{align}\] so that \[\begin{align} \tilde{l}(t) = F^{-1}(G(t)) = -\frac{1}{2}\log\left(1+\frac{\sigma'_{1}\sqrt{\varepsilon}}{4\varphi_{1}}\left(e^{\frac{2\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}t}-1\right)\right), \end{align}\] yielding the expression for \(l(t)\). A similar procedure applied to \(\tilde{u}(t) = u(t)-\left(\frac{\sigma_{1}\varphi_{1}}{\sqrt{\varepsilon}}+\frac{\sigma_{1}\sigma'_{1}}{4}\right)t\) gives the expression for \(u(t)\). ◻

Proof. The first result is immediate from Eq.@eq:eq:theta95ode . For the second one, we write \[\begin{align} \varphi_{1}-\sigma'_{1}\beta(t)\gamma(t) &\leq \varphi_{1}-\sigma'_{1}\sqrt{\varepsilon}\cosh(l(t))\sinh(l(t)) \\ &= \varphi_{1}-\frac{\sigma'_{1}\sqrt{\varepsilon}}{2}\sinh(2l(t)) \\ &\leq \varphi_{1}-\frac{\sigma_{1}'\sqrt{\varepsilon}}{4}\left(\frac{\exp(\frac{2\sigma_{1}\varphi_{1}t}{\sqrt{\varepsilon}})}{1+\frac{\sigma_{1}'\sqrt{\varepsilon}}{4\varphi_{1}}\left(\exp(\frac{2\sigma_{1}\varphi_{1}t}{\sqrt{\varepsilon}})-1\right)}-\frac{1+\frac{\sigma_{1}'\sqrt{\varepsilon}}{4\varphi_{1}}\left(\exp(\frac{2\sigma_{1}\varphi_{1}t}{\sqrt{\varepsilon}})-1\right)}{\exp(\frac{2\sigma_{1}\varphi_{1}t}{\sqrt{\varepsilon}})}\right), \\ \end{align}\] implying the desired. ◻

Proof. We begin by writing, for any \(t <S\) and \(\delta >0\) such that \(t+\delta \leq S\), \[\begin{align} \tag{53} z_{\varepsilon}(t+\delta,\omega) &= z_{\varepsilon}(t,\omega)+\int_{0}^{\delta}f(z_{\varepsilon}(t+s,\omega))ds+\sum_{i=1}^{m}\int_{0}^{\delta}g_{i}(z_{\varepsilon}(t+s,\omega))\int_{\Omega}\frac{h_{i}(z_{\varepsilon}(t+s,\nu))}{v_{i}(\varepsilon)}d\rho(\nu)ds, \\ z(t+\delta,\omega) &= z(t,\omega)+\int_{0}^{\delta}f(z(t+s,\omega))ds+\sum_{i=1}^{m}\int_{0}^{\delta}\alpha_{i}(t+s)g_{i}(z(t+s,\omega))ds. \tag{54} \end{align}\] Let us introduce the vectors \(v_{z_{\varepsilon}},v_{z},\) defined as \[\begin{align} \tag{55} v_{z_{\varepsilon}}(t,\omega) &= z_{\varepsilon}(t+\delta,\omega)-z_{\varepsilon}(t,\omega)-\delta f(z_{\varepsilon}(t,\omega))-\sum_{i=1}^{m}g_{i}(z_{\varepsilon}(t,\omega))\int_{0}^{\delta}\int_{\Omega}\frac{h_{i}(z_{\varepsilon}(t+s,\nu))}{v_{i}(\varepsilon)}d\rho(\nu)ds, \\ v_{z}(t,\omega) &= z(t+\delta,\omega)-z(t,\omega)-\delta f(z(t,\omega))-\delta\sum_{i=1}^{m}\alpha_{i}(t)g_{i}(z(t,\omega)), \tag{56} \end{align}\] and turn to bounding their norms.

9.0.0.1 Bounding the norms of \(v_{z_{\varepsilon}},v_{z}\)

Combining Eq.@eq:eq:dev95z95e with Eq.@eq:eq:def95z95e , and Eq.@eq:eq:dev95z with Eq.@eq:eq:def95z , we find \[\begin{align} \label{eq:exp95vz95e} v_{z_{\varepsilon}}(t,\omega) &= \int_{0}^{\delta}f(z_{\varepsilon}(t+s,\omega))ds-\delta f(z_{\varepsilon}(t,\omega)) \notag \\ &+\sum_{i=1}^{m}\left(\int_{0}^{\delta}g_{i}(z_{\varepsilon}(t+s,\omega))\int_{\Omega}\frac{h_{i}(z_{\varepsilon}(t+s,\nu))}{v_{i}(\varepsilon)}d\rho(\nu)ds-g_{i}(z_{\varepsilon}(t,\omega))\int_{0}^{\delta}\int_{\Omega}\frac{h_{i}(z_{\varepsilon}(t+s,\nu))}{v_{i}(\varepsilon)}d\rho(\nu)ds\right) \end{align}\tag{57}\] and \[\begin{align} \label{eq:exp95vz} v_{z}(t,\omega) &= \int_{0}^{\delta}f(z(t+s),\omega)ds-\delta f(z(t,\omega))+\sum_{i=1}^{m}\left(\int_{0}^{\delta}\alpha_{i}(t+s)g_{i}(z(t+s,\omega))ds-\delta\alpha_{i}(t)g_{i}(z(t,\omega))\right). \end{align}\tag{58}\] Starting with \(v_{z_{\varepsilon}}\), a first order Taylor expansion then gives \[f(z_{\varepsilon}(t+s),\omega) = f(z_{\varepsilon}(t,\omega))+sDf[z(\tilde{t},\omega)]\partial_{t}z(\tilde{t},\omega),\] for some \(\tilde{t} \in [t,t+s]\). But then, owing to Eq.@eq:eq:model95dyn and the definition of \(S\), \[\label{eq:Lz95e} \sup_{0 \leq t \leq S, \thickspace \omega \in \Omega} \lVert\partial_{t}z_{\varepsilon}(t,\omega)\rVert_{2} \leq \sum_{i=1}^{m}\frac{u_{i}(\varepsilon)}{v_{i}(\varepsilon)}C_{g_{i}}(z_{0},R)+C_{f}(z_{0},R),\tag{59}\] so that, letting \(L_{z_{\varepsilon}}(z_{0},R)\) denote the right-hand-side of the above inequality, we find \[\begin{align} \lVert\int_{0}^{\delta}f(z_{\varepsilon}(t+s,\omega))ds-\delta f(z_{\varepsilon}(t,\omega))\rVert_{2} &\leq \sqrt{\delta}\left(\int_{0}^{\delta}s^{2}\lVert Df[z(\tilde{t}(s),\omega)]\rVert_{op}^{2}\lVert\partial_{t}z(\tilde{t}(s),\omega)\rVert_{2}^{2}ds\right)^{1/2} \\ &\leq \frac{\delta^{2}}{\sqrt{3}}C'_{f}(z_{0},R)L_{z_{\varepsilon}}(z_{0},R). \label{eq:b195vz95e} \end{align}\tag{60}\] Letting, for any \(1 \leq i \leq m\), \(\beta_{\varepsilon,i}(t) = \int_{\Omega}\frac{h_{i}(z_{\varepsilon}(t,\omega))}{v_{i}(\varepsilon)}d\rho(\omega)\), we turn to bounding the second term in Eq.@eq:eq:exp95vz95e . Starting again with a first order Taylor expansion, we find \[g_{i}(z_{\varepsilon}(t+s),\omega) = g_{i}(z_{\varepsilon}(t,\omega))+sDg_{i}[z_{\varepsilon}(\tilde{t},\omega)]\partial_{t}z_{\varepsilon}(\tilde{t},\omega),\] for some \(\tilde{t}(s) \in [t,t+\delta]\). Then, \[\begin{align} &\lVert\int_{0}^{\delta}g_{i}(z_{\varepsilon}(t+s,\omega))\beta_{\varepsilon,i}(t+s)ds-g_{i}(z_{\varepsilon}(t,\omega))\int_{0}^{\delta}\beta_{\varepsilon,i}(t+s)ds\rVert_{2}, \notag \\ &\leq \lVert\int_{0}^{\delta}sDg_{i}[z_{\varepsilon}(\tilde{t}(s),\omega)]\partial_{t}z_{\varepsilon}(\tilde{t}(s),\omega)\beta_{\varepsilon,i}(t+s)ds\rVert_{2}, \notag \\ &\leq \frac{\delta^{2}}{\sqrt{3}}C'_{g_{i}}(z_{0},R)L_{z_{\varepsilon}}(z_{0},R)\frac{u_{i}(\varepsilon)}{v_{i}(\varepsilon)}, \label{eq:b295vz95e} \end{align}\tag{61}\] where we have used the upper bound 59 and the fact that \(\sup_{0 \leq t \leq S}\vert \beta_{\varepsilon,i}(t) \vert \leq \frac{u_{i}(\varepsilon)}{v_{i}(\varepsilon)}\). Combining the bounds 60 and 61 , along with the triangle inequality, we reach \[\label{eq:final95vz95e} \lVert v_{z_{\varepsilon}}(t,\omega)\rVert_{2} \leq \frac{\delta^{2}}{\sqrt{3}}L_{z_{\varepsilon}}(z_{0},R)\left(\sum_{i=1}^{m}\frac{u_{i}(\varepsilon)}{v_{i}(\varepsilon)}C'_{g_{i}}(z_{0},R)+C'_{f}(z_{0},R)\right).\tag{62}\] Turning to \(v_{z}\), we begin by noting that, for any \(t \leq S\), \[\begin{align} \lVert\sum_{i=1}^{m}\alpha_{i}(t)g_{i}(z(t,\omega))\rVert_{2} &\leq \sum_{i=1}^{m}\vert \alpha_{i}(t) \vert C_{g_{i}}(z_{0},R) \\ &\leq \lVert\boldsymbol{\alpha}(t)\rVert_{2} \left(\sum_{i=1}^{m}C_{g_{i}}(z_{0},R)^{2}\right)^{1/2}, \end{align}\] owing to the Cauchy-Schwarz inequality. But then, recalling that \(\boldsymbol{\alpha}(t) = -\boldsymbol{G}(t)^{-1}\boldsymbol{F}(t)\), we find \[\label{eq:c95alpha} \lVert\boldsymbol{\alpha}(t)\rVert_{2} \leq \frac{\lVert\boldsymbol{F}(t)\rVert_{2}}{a},\tag{63}\] where, recalling the definition of \(\boldsymbol{F}(t) := \boldsymbol{F}(z(t,.))\), we have the upper bound \[\lVert\boldsymbol{F}(t)\rVert_{2} \leq C_{f}(z_{0},R)\left(\sum_{i=1}^{m}C'_{h_{i}}(z_{0},R)^{2}\right)^{1/2}\] Letting \(C_{F}\) denote the right-hand-side of the above inequality and \(C_{\alpha} = \frac{C_{F}}{a}\), we reach \[\begin{align} \label{eq:Lz} \sup_{0 \leq t \leq S, \omega \in \Omega} \lVert\partial_{t}z(t,\omega)\rVert_{2} &\leq C_{\alpha}\left(\sum_{i=1}^{m}C_{g_{i}}(z_{0},R)^{2}\right)^{1/2}+C_{f}(z_{0},R). \end{align}\tag{64}\] Letting \(L_{z}(z_{0},R)\) denote the right-hand-side of inequality 64 , an argument similar to the one used to obtain the bound 60 gives \[\lVert\int_{0}^{\delta}f(z(t+s),\omega)ds-\delta f(z(t,\omega))\rVert_{2} \leq \frac{\delta^{2}}{\sqrt{3}}C'_{f}(z_{0},R)L_{z}(z_{0},R).\] Now, for any \(0 < t,t' \leq S\) , the triangle inequality gives \[\begin{align} \lVert\boldsymbol{\alpha}(t)-\boldsymbol{\alpha}(t')\rVert_{2} \leq \lVert\boldsymbol{G}^{-1}(t)-\boldsymbol{G}^{-1}(t')\rVert_{op}\lVert\boldsymbol{F}(t)\rVert_{2}+\lVert\boldsymbol{G}^{-1}(t)\rVert_{op}\lVert\boldsymbol{F}(t)-\boldsymbol{F}(t')\rVert_{2}. \end{align}\] Then, for any \(1 \leq i \leq m\), the fact that \(f,\nabla h_{i}\) are bounded Lipschitz (on \(\mathcal{B}(z_{0},R)\)) gives, \[\begin{align} \big\vert \left[\boldsymbol{F}(t)-\boldsymbol{F}(t')\right]_{i} \big \vert &\leq \int_{\Omega}\big \vert \left(\langle f(z(t,\omega)),\nabla h_{i}(z(t,\omega)) \rangle-\langle f(z(t',\omega)),\nabla h_{i}(z(t',\omega)) \rangle\right)\big \vert d\rho(\omega) \\ &\leq \left(C'_{f}C'_{h_{i}}+C_{f}C''_{h_{i}}\right)L_{z}\vert t-t'\vert, \end{align}\] leading to \[\lVert\boldsymbol{F}(t)-\boldsymbol{F}(t')\rVert_{2} \leq \left(\sum_{i=1}^{m}\left(C'_{f}C'_{h_{i}}+C_{f}C''_{h_{i}}\right)^{2}\right)^{1/2}L_{z}\vert t-t' \vert.\] Then, noting that \(\boldsymbol{G}^{-1}(t)-\boldsymbol{G}^{-1}(t') = \boldsymbol{G}^{-1}(t')\left(\boldsymbol{G}(t')-\boldsymbol{G}(t)\right)\boldsymbol{G}^{-1}(t)\) \[\begin{align} \lVert\boldsymbol{G}^{-1}(t)-\boldsymbol{G}^{-1}(t')\rVert_{op} &\leq \frac{1}{a^{2}}\lVert\boldsymbol{G}(t)-\boldsymbol{G}(t')\rVert_{op} \\ &\leq \frac{m}{a^{2}}\sup_{1 \leq i,j \leq m}\vert\left[\boldsymbol{G}(t)-\boldsymbol{G}(t')\right]_{ij} \vert \\ &\leq \frac{m}{a^{2}}\sup_{1 \leq i,j \leq m}\left(C_{g_{i}}C''_{h_{j}}+C'_{g_{i}}C'_{h_{j}}\right)L_{z}\vert t-t' \vert. \end{align}\] Letting \[\label{eq:bound95dalpha} C'_{\alpha} = \left(\frac{1}{a}\left(\sum_{i=1}^{m}\left(C'_{f}C'_{h_{i}}+C_{f}C''_{h_{i}}\right)^{2}\right)^{1/2}+\frac{m}{a^{2}}\sup_{1 \leq i,j \leq n}\left(C_{g_{i}}C''_{h_{j}}+C'_{g_{i}}C'_{h_{j}}\right)C_{F}\right),\tag{65}\] we reach \[\begin{align} \label{eq:bound95dalpha95f} \lVert\boldsymbol{\alpha}(t)-\boldsymbol{\alpha}(t')\rVert_{2} \leq C'_{\alpha}L_{z}\vert t-t' \vert. \end{align}\tag{66}\] Then, for any \(1 \leq i \leq m\), we write \[\begin{align} \alpha_{i}(t+s)g_{i}(z(t+s),\omega)-\alpha_{i}(t)g_{i}(z(t,\omega)) &= \alpha_{i}(t+s)g_{i}(z(t+s),\omega)-\alpha_{i}(t+s)g_{i}(z(t),\omega), \\ &+\alpha_{i}(t+s)g_{i}(z(t,\omega))-\alpha_{i}(t)g_{i}(z(t,\omega)), \end{align}\] so that \[\lVert\alpha_{i}(t+s)g_{i}(z(t+s),\omega)-\alpha_{i}(t)g_{i}(z(t,\omega))\rVert_{2} \leq \vert \alpha_{i}(t+s) \vert C'_{g_{i}}L_{z}s+C_{g_{i}}\vert \alpha_{i}(t+s)-\alpha_{i}(t) \vert,\] and \[\begin{align} &\lVert\sum_{i=1}^{m}\left(\int_{0}^{\delta}\alpha_{i}(t+s)g_{i}(z(t+s,\omega))ds-\delta\alpha_{i}(t)g_{i}(z(t,\omega))\right)\rVert_{2} \leq \\ &\sqrt{2\delta}\sum_{i=1}^{m}\left(\int_{0}^{\delta}\left(\vert \alpha_{i}(t+s)|C'_{g_{i}}L_{z}s\right)^{2}ds +\int_{0}^{\delta}\left(C_{g_{i}}\vert \alpha_{i}(t+s)-\alpha_{i}(t) \vert\right)^{2}ds\right)^{1/2} \\ &\leq \sqrt{2\delta m}\left(\frac{\delta^{3}}{3}L_{z}^{2}\left(\sup_{1 \leq i \leq n}C'_{g_{i}}\right)^{2}\sum_{i=1}^{m}\alpha_{i}(t+s)^{2}+\left(\sup_{1 \leq i \leq n}C_{g_{i}}\right)^{2}\int_{0}^{\delta}\lVert\boldsymbol{\alpha}(t+s)-\boldsymbol{\alpha}(t)\rVert_{2}^{2}ds\right)^{1/2} \\ &\leq \frac{\sqrt{2m}}{\sqrt{3}}\delta^{2}\left(C_{\alpha}^{2}L_{z}^{2}\left(\sup_{1 \leq i \leq n}C'_{g_{i}}\right)^{2}+\left(\sup_{1 \leq i \leq n}C_{g_{i}}\right)^{2}(C'_{\alpha})^{2}\right)^{1/2} \end{align}\] We finally reach \[\label{eq:final95vz} \lVert v_{z}(t,\omega)\rVert_{2} \leq \frac{\delta^{2}}{\sqrt{3}}C'_{f}(z_{0},R)L_{z}(z_{0},R)+\frac{\delta^{2}\sqrt{2m}}{\sqrt{3}}\left(C_{\alpha}^{2}L_{z}^{2}\left(\sup_{1 \leq i \leq n}C'_{g_{i}}\right)^{2}+\left(\sup_{1 \leq i \leq n}C_{g_{i}}\right)^{2}(C'_{\alpha}L_{z})^{2}\right)^{1/2}.\tag{67}\] In what follows, we will respectively denote \(K_{v_{z_{\varepsilon}}}\) and \(K_{v_{z}}\) the bounds obtained at Eq.@eq:eq:final95vz95e and Eq.@eq:eq:final95vz . We now turn to showing that the vector \(\boldsymbol{\alpha}(t)\) is close to the vector \(\boldsymbol{\alpha}_{\varepsilon}(t)\) whose entries are given by \[\alpha_{\varepsilon,i}(t) = \frac{1}{\delta}\int_{0}^{\delta}\int_{\Omega}\frac{h_{i}(z_{\varepsilon}(t+s),\omega)}{v_{i}(\varepsilon)}d\rho(\omega)ds, \quad \forall \thickspace 1 \leq i \leq m.\]

9.0.0.2 Approximating the \(\alpha_{\varepsilon,i}(t)\) with the \(\boldsymbol{\alpha_{i}}(t)\)

Owing to the definitions of \(v_{z_{\varepsilon}},v_{z}\), we may write \[\begin{align} \tag{68} z_{\varepsilon}(t+\delta,\omega) &= z_{\varepsilon}(t,\omega)+\delta f(z_{\varepsilon}(t,\omega))+\delta\sum_{i=1}^{m} \alpha_{\varepsilon,i}(t)g_{i}(z_{\varepsilon}(t,\omega))+v_{z_{\varepsilon}}(t,\omega), \\ z(t+\delta,\omega) &= z(t,\omega)+\delta f(z(t,\omega))+\delta\sum_{i=1}^{m}\alpha_{i}(t)g_{i}(z(t,\omega))+v_{z}(t,\omega). \tag{69} \end{align}\] Then, for any \(1 \leq i \leq m\), an order two Taylor expansion gives that \[\begin{align} \label{eq:taylor95h95e} h_{i}(z_{\varepsilon}(t+\delta,\omega)) = h_{i}(z_{\varepsilon}(t,\omega))&+\langle \nabla h_{i}(z_{\varepsilon}(t,\omega)),z_{\varepsilon}(t+\delta,\omega)-z_{\varepsilon}(t,\omega) \rangle, \notag \\ &+\frac{1}{2}\nabla^{2}h_{i}(\tilde{z}_{\varepsilon})\left[(z_{\varepsilon}(t+\delta,\omega)-z_{\varepsilon}(t,\omega))^{\otimes 2}\right], \end{align}\tag{70}\] for some \(\tilde{z}_{\varepsilon} \in \mathcal{B}(z_{\varepsilon}(t,\omega),\lVert z_{\varepsilon}(t+\delta,\omega)-z_{\varepsilon}(t,\omega)\rVert_{2}) \subset \mathcal{B}(z_{\varepsilon,0}(\omega),3R)\). Letting \[\label{eq:s} s_{i}(t,\omega) = h_{i}(z_{\varepsilon}(t+\delta,\omega)) - h_{i}(z_{\varepsilon}(t,\omega)) - \frac{1}{2}\nabla^{2}h_{i}(\tilde{z}_{\varepsilon})\left[(z_{\varepsilon}(t+\delta,\omega)-z_{\varepsilon}(t,\omega))^{\otimes 2}\right]\tag{71}\] and replacing the increment \(z_{\varepsilon}(t+\delta,\omega)-z_{\varepsilon}(t,\omega)\) using Eq.@eq:eq:exp95z95e in Eq.@eq:eq:taylor95h95e , and integrating with respect to \(\omega\), we find, for any \(1 \leq j \leq m\), \[\begin{align} \label{eq:def95alpha95eps} \int_{\Omega}\langle \nabla h_{j}(z_{\varepsilon}(t,\omega)), \delta f(z_{\varepsilon}(t,\omega))+\delta \sum_{i=1}^{m}\alpha_{\varepsilon,i}(t)g_{i}(z_{\varepsilon}(t,\omega))+v_{z_{\varepsilon}}(t,\omega) \rangle d\rho(\omega) = \int_{\Omega} s_{j}(t,\omega)d\rho(\omega). \end{align}\tag{72}\] Now, defining the vectors \(\boldsymbol{s}(t)\) and \(\boldsymbol{V}_{z_{\varepsilon}}(t)\), whose entries are respectively given by \[\begin{align} &[\boldsymbol{s}(t)]_{1 \leq i \leq m} = \int_{\Omega}s_{i}(t,\omega)d\rho(\omega), \\ &[\boldsymbol{V}_{z_{\varepsilon}}(t)]_{1 \leq i \leq m} = \int_{\Omega}\langle \nabla h_{i}(z_{\varepsilon}(t,\omega)),v_{z_{\varepsilon}}(t,\omega) \rangle d\rho(\omega), \end{align}\] we may rewrite Eq.@eq:eq:def95alpha95eps as \[\delta \left(\boldsymbol{G}(z_{\varepsilon}(t,.))\boldsymbol{\alpha}_{\varepsilon}(t)+\boldsymbol{F}(z_{\varepsilon}(t,.))\right) = -\boldsymbol{V}_{z_{\varepsilon}}(t)+\boldsymbol{s}(t).\] Abbreviating \(\boldsymbol{G}(z_{\varepsilon}(t,.))\) with \(\boldsymbol{G}^{\varepsilon}(t)\) and \(\boldsymbol{F}(z_{\varepsilon}(t,.))\) with \(\boldsymbol{F}^{\varepsilon}(t)\), and letting \(\tilde{\boldsymbol{\alpha}}_{\varepsilon}(t) = - \boldsymbol{G}^{\varepsilon}(t)^{-1}\boldsymbol{F}^{\varepsilon}(t)\), we may write \[\begin{align} \lVert\boldsymbol{\alpha}(t)-\boldsymbol{\alpha}_{\varepsilon}(t)\rVert_{2} &\leq \lVert\boldsymbol{\alpha}(t)-\tilde{\boldsymbol{\alpha}}_{\varepsilon}(t)\rVert_{2}+\frac{1}{\delta}\lVert\boldsymbol{G}^{\varepsilon}(t)^{-1}\left(-\boldsymbol{V}_{z_{\varepsilon}}(t)+\boldsymbol{s}(t)\right)\rVert\notag \\ &\leq \lVert\boldsymbol{\alpha}(t)-\tilde{\boldsymbol{\alpha}}_{\varepsilon}(t)\rVert_{2}+\frac{1}{\delta a}\left(\lVert\boldsymbol{V}_{z_{\varepsilon}}(t)\rVert_{2}+\lVert\boldsymbol{s}(t)\rVert_{2}\right). \end{align}\] Now, using the definition of \(s_{i}(t)\) at Eq.@eq:eq:s , we find \[\sup_{0 \leq t \leq S} \vert \int_{\Omega}s_{i}(t,\omega)d\rho(\omega) \vert \leq 2u_{i}(\varepsilon)+C''_{h_{i}}L^{2}_{z}\delta^{2},\] and thus \[\lVert\boldsymbol{s}(t)\rVert_{2} \leq \left(\sum_{i=1}^{m}\left(2u_{i}(\varepsilon)+C''_{h_{i}}L^{2}_{z}\delta^{2}\right)^{2}\right)^{1/2}.\] Also, \[\lVert\boldsymbol{V}_{z_{\varepsilon}}(t)\rVert_{2} \leq K_{v_{z_{\varepsilon}}}\left(\sum_{i}^{m}(C'_{h_{i}}(z_{0},R))^{2}\right)^{1/2}.\] Furthermore, the argument employed to establish Eq.@eq:eq:bound95dalpha95f also implies that \[\lVert\boldsymbol{\alpha}(t)-\tilde{\boldsymbol{\alpha}}_{\varepsilon}(t)\rVert_{2} \leq C'_{\alpha}\lVert z(t)-z_{\varepsilon}(t)\rVert_{2}.\] In summary, we reach \[\begin{align} \lVert\boldsymbol{\alpha}(t)-\boldsymbol{\alpha}_{\varepsilon}(t)\rVert_{2} \leq &\frac{1}{\delta a}\left(\left(\sum_{i=1}^{m}\left(2u_{i}(\varepsilon)+C''_{h_{i}}L^{2}_{z}\delta^{2}\right)^{2}\right)^{1/2}+K_{v_{z_{\varepsilon}}}\left(\sum_{i}^{m}(C'_{h_{i}}(z_{0},R))^{2}\right)^{1/2}\right) \notag\\ &+C'_{\alpha}\lVert z(t,\omega)-z_{\varepsilon}(t,\omega)\rVert_{2}. \end{align}\]

9.0.0.3 Concluding the proof

To conclude the proof, we use Eq.@eq:eq:exp95z95e ,69 to obtain that, for all \(\omega \in \Omega\), \[\begin{align} \lVert z_{\varepsilon}(t+\delta,\omega)-z(t+\delta,\omega)\rVert_{2} &\leq \left(1+\delta C'_{f}\right)\lVert z_{\varepsilon}(t,\omega)-z(t,\omega)\rVert_{2}+K_{v_{z}}+K_{v_{z_{\varepsilon}}},\notag \\ &+\delta \lVert\sum_{i=1}^{m}\left(\alpha_{\varepsilon,i}(t)g_{i}(z_{\varepsilon}(t,\omega))-\alpha_{i}(t)g_{i}(z(t,\omega))\right)\rVert_{2}. \end{align}\] But then, \[\begin{align} &\lVert\sum_{i=1}^{m}\left(\alpha_{\varepsilon,i}(t)g_{i}(z_{\varepsilon}(t,\omega))-\alpha_{i}(t)g_{i}(z(t,\omega))\right)\rVert_{2} \leq \sum_{i=1}^{m} \lVert\vert \alpha_{\varepsilon,i}(t)-\alpha_{i}(t)\vert g_{i}(z_{\varepsilon}(t,\omega))\rVert_{2} \\ &+\sum_{i=1}^{m} \lVert\alpha_{i}(t)\left(g_{i}(z_{\varepsilon}(t,\omega))-g_{i}(z(t,\omega))\right)\rVert_{2} \\ &\leq \left(\sum_{i=1}^{m}C_{g_{i}}^{2}\right)^{1/2}\lVert\boldsymbol{\alpha}(t)-\boldsymbol{\alpha}_{\varepsilon}(t)\rVert_{2}+C_{\alpha}\left(\sum_{i=1}^{m}(C'_{g_{i}})^{2}\right)^{1/2}\lVert z(t,\omega)-z_{\varepsilon}(t,\omega)\rVert_{2}. \end{align}\] Gathering the above bounds, we introduce the quantities \[\begin{align} K &= \left(C'_{f}+C_{\alpha}\left(\sum_{i=1}^{m}(C'_{g_{i}})^{2}\right)^{1/2}+C'_{\alpha}\left(\sum_{i=1}^{m}C_{g_{i}}^{2}\right)^{1/2}\right), \\ \ell(\delta,\varepsilon) &= K_{v_{z}}+K_{v_{z_{\varepsilon}}}+\frac{\left(\sum_{i=1}^{m}C_{g_{i}}^{2}\right)^{1/2}}{a}\left(\left(\sum_{i=1}^{m}\left(2u_{i}(\varepsilon)+C''_{h_{i}}L^{2}_{z}\delta^{2}\right)^{2}\right)^{1/2}+K_{v_{z_{\varepsilon}}}\left(\sum_{i}^{m}(C'_{h_{i}}(z_{0},R))^{2}\right)^{1/2}\right), \end{align}\] so that we finally obtain the following control: \[\begin{align} &\lVert z_{\varepsilon}(t+\delta,\omega)-z(t+\delta,\omega)\rVert_{2} \leq \left(1+K\delta\right)\lVert z_{\varepsilon}(t,\omega)-z(t,\omega)\rVert_{2}+\ell(\delta,\varepsilon). \end{align}\] Applying the discrete Gronwall inequality then gives that, for any \(n \in \mathbb{N}\) such that \(n\delta \leq S\), \[\lVert z_{\varepsilon}(n\delta,\omega)-z(n\delta,\omega)\rVert_{2} \leq \left(\lVert z_{\varepsilon}(0,\omega)-z(0,\omega)\rVert_{2}+nl(\delta,\varepsilon)\right)\exp\left(Kn\delta\right).\] ◻

Proof. We apply Proposition 19 to the evolution 73 74 , the form of which implies that there exist constants \(C_{1},C_{2},C_{3}\) depending only on \(D,R,\{\sigma_{k},\varphi_{k}\}_{k=0}^{D}\) such that \(K\leq C_{1}\) and \(l(\delta,\varepsilon) \leq C_{2}\delta^{2}+C_{3}u^{*}(\varepsilon)\). Choosing \(n = \lfloor \frac{t}{\delta} \rfloor\), we find, for any \(\delta<1\), \[\lVert z_{\varepsilon}(t,.)-z(t,.)\rVert \leq \left(\lVert z_{\varepsilon}(0,.)-z(0,.)\rVert+\frac{t}{\delta}(C_{1}\delta^{2}+C_{3}u^{*}(\varepsilon))\right)\exp\left(C_{2}t\right).\] Optimizing over \(\delta\) then gives the desired result. ◻

Proof. For any \(t \geq 0\) and since \(f\) is continuous, we may define \(r_{t} = \mathop{\mathrm{arg\,sup}}_{r \leq t} \vert f(r) \vert\). Furthermore, let \[\begin{align} \tilde{r} = \begin{cases} &r_{t+s} \quad if \thickspace r_{t+s} \leq t \\ &t \quad otherwise \end{cases} \end{align}\] Then, \(F(t) \geq \vert f(\tilde{r}) \vert\) and therefore \[\begin{align} F(t) \geq \vert f(r_{t+s})\vert+\vert f(\tilde{r}) \vert- \vert f(r_{t+s}) \vert. \end{align}\] By definition \(\vert f(r_{t+s})\vert = F(t+s)\) and \(r_{t+s}-\tilde{r} \leq s\), and, by the Lipschitz assumption, \(\big \vert \vert f(\tilde{r}) \vert -\vert f(r_{t+s}) \vert \big \vert \leq Ls\) for some constant \(L\). Upon noting that \(F(t+s) \geq F(t)\), this completes the proof of the first part of the statement. Now, since \(F\) is Lipschitz continuous, its derivative exists (Lebesgue) almost everywhere and, we have \[\begin{align} \partial_{t} F(t) = \begin{cases} &0 \quad if \thickspace r_{t} < t \\ &\partial_{t}f(t) \quad if \thickspace r_{t} = t. \end{cases} \end{align}\] ◻

Proof. This is a Bernoulli equation, see e.g.[57], and, clearly, \(x(t)>0\) for any \(t>0\), so that existence and uniqueness of maximal solutions follows straightforwardly. Making the change of variable \(y(t) = \sqrt{x(t)}\), we find \[\dot{y}(t) = -\frac{\alpha}{2} y(t)+\frac{\beta}{2},\] so that \[y(t) = y(0)\exp(-\frac{\alpha}{2}t)+\frac{\beta}{\alpha}\left(1-\exp(-\frac{\alpha}{2}t)\right).\] ◻

Proof. The first inequality is a consequence of the convexity of the Euclidean norm and Jensen’s inequality. For the second inequality, we apply the Cauchy-Schwarz inequality to obtain \[\begin{align} \lVert\int_{0}^{t}z(s)ds\rVert_{2}^{2} \leq \sum_{i=1}^{d}t\int_{0}^{t}z^{2}_{i}(s)ds. \end{align}\] ◻