From 7f588b8c0da38c82786442e0cdc8243edb41c1d4 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Fri, 3 Jun 2022 19:20:51 +0200
Subject: Writing on saddle for FRSB case.

---
 frsb_kac-rice.tex | 17 ++++++++++++++++-
 1 file changed, 16 insertions(+), 1 deletion(-)

diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index 02724ed..04fa307 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -223,7 +223,7 @@ Integrating by parts,
       +\frac1{\lambda(q)+R_d/\hat\epsilon}
     \right]
 \end{align*}
-for $\lambda$ concave, monotonic, and $\lambda(1)=0$
+for $\lambda$ concave, monotonic,  $\lambda(1)=0$, and $\lambda'(1)=-1$
 \[
   0=\frac{\partial\Sigma}{\partial R_d}
   =\frac12\hat\epsilon f'(1)-\frac12\frac1{\hat\epsilon}\int_0^1dq\,\frac{\lambda(q)}{[\lambda(q)+R_d/\hat\epsilon]^2}
@@ -239,4 +239,19 @@ for $\lambda$ concave, monotonic, and $\lambda(1)=0$
   \lambda^*(q)=\frac1{\hat\epsilon}\left[f''(q)^{-1/2}-R_d\right]
 \]
 
+We suppose that solutions are given by
+\begin{equation}
+  \lambda(q)=\begin{cases}
+    \lambda^*(q) & q<q^* \\
+    1-q          & q\geq q^*
+  \end{cases}
+\end{equation}
+where $1-q$ guarantees the boundary conditions at $q=1$, and corresponds to the 0RSB or annealed solutions (annealed Kac-Rice is recovered by substituting in $1-q$ for $\lambda$). We will need to require that $1-q^*=\lambda^*(q^*)$, i.e., continuity.
+
+The two saddle points equations for $R_d$ and $\hat\epsilon$ give
+\[
+  0=\frac12\hat\epsilon f'(1)-\frac12\int_0^{q_*}dq\,\left[f''(q)^{1/2}-R_df''(q)\right]
+  -\frac12\frac1{\hat\epsilon}\int_{q^*}^1dq\,\frac{1-q}{(1-q+R_d/\hat\epsilon)^2}
+\]
+
 \end{document}
-- 
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