From eq. (\[eq:qx6\]) we clearly see that $q(x)$ contains even powers of $x$, with the exclusion of $x^2$. In the next Section we shall derive exact relations among the derivatives of $q(x)$ at $x=0$ from which follow that $q^{(2)}(x=0) = 0$ but $q^{(4)}(x=0) \not= 0$.

Expansion of the order parameter $q(x)$ {#SecDeriv}
=======================================

To evaluate the derivatives of the order parameter $q(x)$ at $x=0$ we use a variational approach developed by Sommers and Dupont[@SDJPC84]. This method as also the advantage of leading to exact relations among derivatives of different order, so can be used to test the findings of the previous Section in a non-perturbative way. The starting point is the variational form of the Parisi’s free energy $f$: $$\begin{aligned}
 -f &=& \frac{\beta}{4}\,\Bigl( 
                      1 - 2\,q(1) + \int_0^1dx\, q^2(x)
                         \Bigr) 
+ \int_{-\infty}^{+\infty} \frac{d y}{\sqrt{2 \pi q(0)}}
\exp\left(-\frac{(y-h)^2}{2\,q(0)}\right)\phi(0,y)
\nonumber
\\
  &-&\int_{-\infty}^{+\infty} dy\  P(1,y)\,
          \bigl[\phi(1,y)-T \log\left(2\cosh \beta y\right)\bigr]
\nonumber
\\
&+&\int_0^1dx\int_{-\infty}^{+\infty} dy\ P(x,y) 
                 \left[\dot\phi(x,y)+
                 \frac{\dot{q}(x)}{2}\,
    \Bigl[
               \phi''(x,y)+\beta\,x\,\phi'(x,y)^2 
        \Bigr]
\right].
\label{eqfrev}\end{aligned}$$ Imposing stationarity with respect to variations of $P(x,y)$, $P(1,y)$, $\phi(x,y)$, $\phi(0,y)$ and $q(x)$, one obtains the variational equations: $$\label{SP1}
q(x)=\int dy\, P(x,y)\,m^{2}(x,y)$$ $$\dot m(x,y)=-\frac{\dot{q}(x)}{2}\, 
           \Bigl[
          m''(x,y) + 2\,\beta\,x\, m(x,y)\ m'(x,y) 
           \Bigr]
\label{SP2}$$

$$\label{SP3}
\dot{P}(x,y) = \frac{\dot{q}(x)}{2}
                   \Bigl[ 
         P''(x,y) - 2\,\beta\,x\,[m(x,y)\,P(x,y)]'
                   \Bigr]$$

with 