While working on a variational problem, I have reached to the following question.
Let $0<\lambda < \frac{1}{2}$ be a parameter.
Are there smooth strictly increasing surjective maps $\phi:[0,1] \to [0,\lambda]$ satisfying $\phi(0)=0$ and the coupled system of ODE's given by
$$\phi'=\frac{1}{2}\big(1 + \sqrt{1-4\frac{\phi \phi'}{r}}),\frac{\phi}{r}=\frac{1}{2}\big(1 - \sqrt{1-4\frac{\phi \phi'}{r}})\,\,\,\,\,?$$
Note that taking the product of two these equations result in a consistent tautology.
In particular, I want the ODE to be defined everywhere, i.e. $\frac{\phi \phi'}{r} \le \frac{1}{4}$ for every $r$.
Are there such solutions $\phi$ that also satisfy $\phi^{2k}(0)=0$ for every natural $k$?
Note that for the crucial value $\lambda=\frac{1}{2}$ (which is out of scope here), $\phi(r)=\lambda r$ does the job. However, $r \mapsto \lambda r$ does not satisfy the ODE for $\lambda < \frac{1}{2}$.
If $$\frac{\phi}{r}=\frac{1}{2}\big(1 - \sqrt{1-4\frac{\phi \phi'}{r}})$$ then $$ \frac{2\phi}{r}=1-\sqrt{1-4\frac{\phi \phi'}{r}}\Longrightarrow \sqrt{1-4\frac{\phi \phi'}{r}}=1-\frac{2\phi}{r}. $$ Hence $$ \phi'=\frac{1}{2}\Big(1 + \sqrt{1-4\frac{\phi \phi'}{r}}\Big)= \frac{1}{2}\Big(1 + 1-\frac{2\phi}{r}\Big)=1-\frac{\phi}{r}, $$ and thus $$ r\phi'+\phi=r\Longleftrightarrow (r\phi)'=r \Longleftrightarrow r\phi=\frac{r^2}{2}+c $$ for some $c\in\mathbb R$. But $c=0$, since $\phi(0)=0$. Hence $$ \phi(r)=\frac{r}{2}. $$