4 x) {\overline}{W}_x = 0\,. \end{aligned}$$

According to [@Fontelos], has an implicit solution $$\begin{aligned}
\label{e:implicit:eq}
x = - {\overline}{W} - {\overline}{W}^{5}\,.\end{aligned}$$ Differentiating yields $$\begin{aligned}
\label{e:w1:implicit}
 {\overline}{W}^{(1)}=-\frac{1}{1+5{{\overline}W}^4}\,.\end{aligned}$$ Hence $ {\overline}{W}^{(1)}\leq 0$ and thus since ${\overline}{W}(0)=0$ we attain that ${\overline}{W}\leq 0$ for $x\geq 0$. By Young’s inequality and applied to , we have $$\begin{aligned}
x \leq - {\overline}{W}  -{\overline}{W}^{5} \leq -\frac{{\overline}{W}^{5} }{5x^4}+-{{\overline}W}^5+\frac{4x}{5}\,.\end{aligned}$$ Rearranging, we obtain $$-{{\overline}W}^5\geq \frac{x^5}{5(5+x^4)}\,.$$ This lower bound combined with yields $$\label{bound:bar:W:1}
{\left| {\overline}{W}^{(1)}\right|}\leq (1+x^4)^{-\frac15}\,.$$ Similarly, using Young’s inequality and we have $$- {\overline}{W}^{5}\leq 5x+1\,,$$ from which we obtain the estimate $${\left|{\overline}{W}\right|}\leq \frac32 (1+x^4)^{\frac1{20}}\,.$$ Finally, differentiating $5$ times, we obtain $$\label{W5:non}
W^{(5)}(0)=120\,.$$

We now define the weight function $$\begin{aligned}
 \label{weight:eta}
\eta_\gamma := (1 + x^4)^\gamma, \text{ for any } \gamma \in \mathbb{R}\,.\end{aligned}$$

We now record the following lemma, which summarizes the properties of ${\overline}{W}$ that we will be using

Let $\ell$ be sufficiently small relative to universal constants. For $n = 2, 3, 4$ at $x=0$ we have $$\begin{aligned}
\label{fifth:deriv:bar:W}
&{\overline}{W}(0) = 0\,, \quad {\overline}{W}^{(1)}(0) = -1\,, \quad {\overline}{W}^{(n)}(0) = 0\,, \quad {\overline}{W}^{(5)}(0) = 120 \,.\end{aligned}$$ Furthermore, for $n \ge 2$, ${\overline}W$ satisfies the estimates $$\begin{gathered}
\label{decay:bar:2}
|{\overline}{W}| \le \frac 3 2 \eta_{\frac{1}{20}}\,,  \quad|{\overline}{W}^{(1)}| \le \eta_{- \frac 1 5}\,, \quad  