The first application of this procedure yields $$\dot{q}(x)=\dot{q}(x)\,\int dyP(x,y)(m')^{2}$$ which for $\dot{q}(x)\neq 0$ simply reads $$\label{d1}
1=\int dy\,P(x,y)\, m'(x,y)^{2}.$$ The procedure can be iterated infinitely. For example, the next three applications leads respectively to $$\label{d2}
0=-\frac{2x}{T}\int dyP(m')^{3}+\int dyP(m'')^{2}$$ $$\label{d3}
\frac{2}{T}\int dyP(m')^{3}=\dot{q}\int Pdy\left( (m''')^{2}-\frac{12x}{T}m'(m'')^{2}+6\left( \frac{x}{T}\right) ^{2}(m')^{4}\right)$$ and $$\int Pdy\left( \frac{(18x\dot{q}+6x^{2}\ddot{q})(m')^{4}}{T^{2}}-\frac{(18\dot{q}T+12x\ddot{q}T-120m'x^{2}\dot{q}^{2})m'(m'')^{2}}{T^{2}}\right.$$ $$\label{dev2x}
\left. +\frac{-30x\dot{q}^{2}(m'')^{2}+\ddot{q}m'''T-20x\dot{q}^{2}m'(m''')}{T}m'''-\frac{24x^{3}\dot{q}^{2}(m')^{5}}{T^{3}}+\dot{q}^{2}(m'''')^{2}\right) =0$$ We are interested into the derivatives of $q(x)$ at $x=0$, so we take the limit $x\rightarrow 0$ of the above equations. The limit can be done in trivial way and, since the function $P(0,y)$ reduce to a $\delta$-function \[see eq. (\[condP\])\], the equations are greatly simplified. Moreover since $m(x,y)$ is an odd function of $y$ for any $x$ clearly $m^{(0,j)}(0,0)=0$ for any even $j$. In this limit equations (\[d1\]), (\[d3\]) and (\[dev2x\]) reduce respectively to: $$\begin{aligned}
\label{dev10}
1 & = & m'(0,0)\label{limz2} \\
\label{dev20}
\frac{2}{T}m'(0,0)^{3} & = & \dot{q}(0)m'''(0,0)^{2}\label{limz} \\
\label{eq:d40}
\ddot{q}(0)m'''(0,0)^{2} &=& 0 \end{aligned}$$ while equations (\[SP1\]) and (\[d2\]) become trivial identities.

From equations (\[limz2\]) and (\[limz\]) we have $$\label{m30}
m'''(0,0)=-\sqrt{\frac{2}{T\dot{q}(0)}}\neq 0$$ therefore (\[eq:d40\]) implies that $\ddot{q}(0)=0$ as already found in Ref. [@S85].

To obtain information on the fourth derivative of $q(x)$ the above procedure must be iterated two more times. Since successive derivatives yields expressions 