From 54eb790a87479e39ea6ffa580f277c1a1152b1c4 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 7 Jun 2022 08:31:53 +0200
Subject: Started better discussion of the equilibrium solution.

---
 frsb_kac_new.tex | 16 +++++++++++++---
 1 file changed, 13 insertions(+), 3 deletions(-)

(limited to 'frsb_kac_new.tex')

diff --git a/frsb_kac_new.tex b/frsb_kac_new.tex
index f5e9422..d8f7f87 100644
--- a/frsb_kac_new.tex
+++ b/frsb_kac_new.tex
@@ -118,11 +118,21 @@ minima are exponentially subdominant with respect to saddles, because a saddle i
 \section{Equilibrium}
 
 Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical}
-
+The free energy is well known to take the form
 \begin{equation}
-  \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-1-\log2\pi
+  \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}^nf(Q_{ab})+\log\det Q\right)-1-\log2\pi
+\end{equation}
+which must be extremized over the matrix $Q$. When the solution is a Parisi matrix, this can also be written in a functional form. If $P(q)$ is the probability distribution for elements $q$ in a row of the matrix, then
+\begin{equation}
+  \chi(q)=\int_1^qdq'\,\int_0^{q'}dq''\,P(q'')
 \end{equation}
-We insert a  $k$ step RSB ansatz (the steps are standard, and are reviewed in Appendix A) , and obtain 
+Since it is the double integral of a probability distribution, $\chi$ must be convex, monotonically decreasing and have $\chi(1)=0$, $\chi'(1)=1$. The free energy can be written as a functional over $\chi$ as
+\begin{equation}
+  \beta F=-\frac12\int dq\,\left(\beta^2f''(q)\chi(q)+\frac1{\chi(q)}\right)-1-\log2\pi
+\end{equation}
+
+
+We are especially interested We insert a  $k$ step RSB ansatz (the steps are standard, and are reviewed in Appendix A) , and obtain 
 $q_0=0$
 \begin{align*}
   \beta F=
-- 
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