((\^)\^[-1]{}(-t)) Notice that $$\frac{d\hfil}{ds} \left[s \,(\psi^\prime)^{-1}(-s) + \psi\left((\psi^\prime)^{-1}(-s)\right)\right] = 
(\psi^\prime)^{-1}(-s)$$ Thus, equation can be simplified to \[eq:restr-2\] x\_0 - = 0 It is useful to introduce the following version of the Legendre transform of the function $\psi$: \[Leg\] (t\_\*) = \_t. Then equation can be written in the form \[eq:restr-21\] x\_0 = (t\_0), and the solution takes the form $$x(t) = x_0 + {\tilde \psi}(t) - {\tilde \psi}(t_0)\,.$$ In view of , we obtain \[sol:other-2\] x(t) = (t). This is the so-called singular solution. We see that the singular solution corresponds to the value of $c = c(t_0, x_0)$ in that minimizes $[c\cdot t + \psi(c)]$.

Consider the special case $\psi(s) = s^2$, i.e., consider the equation \[eq:square\] x = t [x]{} + ([x]{})\^2. The corresponding equation has two solutions for $c$, \[c\_0\] c\_(t\_0, x\_0) = - , provided $x_0 + {t_0^2}/{4} > 0$, has one solution when $x_0 + {t_0^2}/{4} = 0$, and has no solutions when $x_0 + {t_0^2}/{4} < 0$. Since ${\tilde \psi}(t_*) = - t_*^2/4$, the singular solution is \[sol:sing\] x(t) = - . Thus, the region $C\subset (-\infty,+\infty)\times {{\mathbb{R}}}$ of the allowed initial data $(t_0, x_0)$ is the parabola $x = - t^2/4$ and the 