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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} |