1 files changed, 39 insertions, 5 deletions
diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex
index d513e48..ee3c138 100644
--- a/frsb_kac-rice.tex
+++ b/frsb_kac-rice.tex
@@ -661,11 +661,45 @@ numerics.
 Given this ansatz, we take the equations \eqref{eq:extremum.c},
 \eqref{eq:extremum.r}, and \eqref{eq:extremum.d}, which are true for any $x$, and
 integrate them over $x$. We then expand the result about small
-$x_\mathrm{max}$. The result is
-\begin{align}
-  0&=(-\mu^*\bar c+\bar rf'(1)+\hat\beta\bar cf'(1)+\bar rf''(0)+\hat\beta\bar c f''(0)+\bar cr_d(f''(1)+f''(0))
-\end{align}
-
+$x_\mathrm{max}$ to quadratic order in $x_\mathrm{max}$. Equation \eqref{eq:extremum.r} depends linearly on $\bar r$ to all orders, and therefore $\bar r$ can be found in terms of $\bar c$, yielding
+\begin{equation}
+  \begin{aligned}
+    \frac{\bar r}{\bar c}
+    &=
+    -\hat\beta-\frac1{f'(1)+f''(0)}\left(r_d(f''(0)+f''(1))-\mu^*\right)
+    +\frac12\frac{\bar c}{f'(1)+f''(0)}\left\{
+      -\hat\beta(4f''(0)-f'''(0))\right.\\
+    &\left.+\frac1{f'(1)+f''(0)}\left[
+        8(\mu-r_d(f''(0)+f''(1)))f''(0)
+        -(2\mu+r_d(f'(1)-2f''(1)-f''(0))f'''(0))
+      \right]
+    \right\}x_\textrm{max}+O(x_\textrm{max}^2)
+  \end{aligned}
+\end{equation}
+Likewise, \eqref{eq:extremum.d} depends linearly on $\bar d$ to all orders, and can be solved using this form for $\bar r$ to give
+\begin{equation}
+  \begin{aligned}
+    \frac{\bar d}{\bar c}
+    &=-2r_d\frac{\bar r}{\bar c}-\frac1{f'(1)}(r_d^2f''(0)+d_d(f'(1)+f''(0))-1)
+    +\left\{
+    2\frac{\bar r^2}{\bar c}-2\bar c\bar b-4r_d\bar r\right.\\
+      &\left.
+        -\frac{\bar c}{2f'(1)^2}\left[
+          4(f'(1)+f''(0))((d_d+r_d^2)f''(0)-1)+(d_d+r_d^2)f'(1)f'''(0)
+        \right]
+      \right\}x_\mathrm{max}+O(x_\mathrm{max}^2)
+  \end{aligned}
+\end{equation}
+Finally, the equations for $\bar c$ at first order imply that either $\bar c$
+vanishes as $x_\mathrm{max}$ to zero, or that the linear coefficient vanishes
+as $x_\mathrm{max}$ to zero. Using $\hat\mu$, $\hat\beta$, $r_d$, and $d_d$
+from the annealed solution, this coefficient vanishes when
+\begin{equation}
+  \mu^*
+  =\pm\frac{(f'(1)+f''(0))(f'(1)^2-f(1)(f'(1)+f''(1)))}{(2f(1)-f'(1))f'(1)f''(0)^{-1/2}}
+  -\frac{f''(1)-f'(1)}{f'(1)-2f(1)}\epsilon
+\end{equation}
+We expect that this is the line of stability for the replica symmetric saddle.
 
 
 \section{General solution: examples}