choice of $\alpha$, we now consider the linear version of (setting $\eps = 0$ in ). Upon setting $\eps = 0$ in , sending $s \rightarrow \infty$, and demanding the asymptotic behavior $u_\alpha(s) \rightarrow 0$ as $s \rightarrow \infty$, we obtain the following relation $$\begin{aligned}
\alpha_0 + \int_0^\infty e^{- \frac{s'}{2}} g(s') {\,\mathrm{d}}s' = 0\,,\end{aligned}$$ which links the choice of data, $\alpha_0$, to the forcing, $g$, and guarantees the solution $|u_\alpha(s)| \lesssim e^{- \gamma s}$ inherits the decay of $g$.

We would now like to modify the choice of data, $\alpha_0$, by an $\eps$ perturbation in order to account for the nonlinear effects when $\eps > 0$ in . The overall strategy will be to fix a sequence of times $\{s_n\}$ for $n \in \mathbb{N}$ with the property that $s_n \rightarrow \infty$ as $n \rightarrow \infty$, and a corresponding sequence of data choices $\{ \alpha_n\}$ for $n \in \mathbb{N}$ so that $u_{\alpha_n}(s_n) = 0$. With suitably strong estimates, we will show that $\alpha_n \rightarrow \alpha_\infty$ and correspondingly $u_{\alpha_\infty}(s) \rightarrow 0$ as $s \rightarrow \infty$. To compute the iterate of $\alpha_{n+1}$ requires an application of the Implicit Function Theorem, which in turn requires sufficiently strong estimates on the solution.

Let us now 