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@@ -447,7 +447,7 @@ action at the $m=0$ solution, with $\mathcal S(0)=0$. For a generic covariance
function $f$ it is not possible to write an explicit formula for
$V_\text{\textsc{sat}}$, and we calculate it through a numeric
root-finding algorithm.\footnote{
-As a check of this calculation, the satisfiability threshold calculated here can be compared with that calculated using the zero-temperature limit of an equilibrium treatment of the cost function \eqref{eq:cost} made in Ref.~\cite{Urbani_2023_A} for the case where $f(q)=\frac12q^2$ and $\alpha=\frac14$. The authors estimate $V_\text{\textsc{sat}}\simeq1.871$, whereas this manuscript predicts $V_\text{\textsc{sat}}=1.867229\dots$, a seeming inconsistency. However, the author of Ref.~\cite{Urbani_2023_A} indicated in private correspondence that this difference could easily be explained by inaccuracy in the numeric \textsc{pde} treatment of the \textsc{frsb} equilibrium problem. Therefore, this manuscript is consistent with the previous work, but the agreement is not precise.
+As a check of this calculation, the satisfiability threshold calculated here can be compared with that calculated using the zero-temperature limit of an equilibrium treatment of the cost function \eqref{eq:cost} made in Ref.~\cite{Urbani_2023_A} for the case where $f(q)=\frac12q^2$ and $\alpha=\frac14$. The authors estimate $V_\text{\textsc{sat}}\simeq1.871$, whereas this manuscript predicts $V_\text{\textsc{sat}}=1.867229\dots$, a seeming inconsistency. However, the author of Ref.~\cite{Urbani_2023_A} indicated in private correspondence that this difference is explained by inaccuracy in the numeric \textsc{pde} treatment of the \textsc{frsb} equilibrium problem. Therefore, this manuscript is consistent with the previous work, but the agreement is not precise.
}
When $V_0^2<V_\text{sh}^2$, the solution at $m=0$ is difficult to interpret, since