I came across this problem in an old exam: Find all $c\in \mathbb{R}$ such that the initial value problem $$\begin{align}y'(t) &= e^{-y(t)^2},\\ y(0) &= c\end{align}$$

has exactly one solution. My first thought was finding solutions using separation of variables, but then I found out about the Gaussian integral and I think it is out of the scope of this exercise (this was a short, timed exam).

By Picard, the function has local solutions for any c, as $f(y) = e^{-y^2}$ is continuously differentiable and thus is locally Lipschitz (right?).

Our lecture notes explicitly state that there is no global existence theorem for non-linear ODEs. So I have no idea how to talk about global solutions.

The next part of the problem asks for the value of $y''(0)$ when $c = 0$, here again I don't see how I can tackle it without finding the explicit solution $y$.

Any pointers would be much appreciated.

  • 2
    $\begingroup$ This doesn't depend on $c$. Since $y\mapsto e^{-y^2}$ is globally Lipschith, you'll have existence and uniqueness of a global solution for all $c\in\mathbb R$. $\endgroup$
    – Surb
    Feb 25, 2020 at 9:49
  • $\begingroup$ @Surb I got too caught up in playing with c! Any thoughts on the solution $y''(0)$ for $c=0$? $\endgroup$ Feb 25, 2020 at 9:52
  • 2
    $\begingroup$ Not so sure to understand your question. But $y''(t)=2y(t)y'(t)e^{-y^2(t)}$. So if $y(0)=0$, then $y''(0)=0$. $\endgroup$
    – Surb
    Feb 25, 2020 at 9:56

1 Answer 1


The equation is solved by $$\int_c^{y}e^{u^2}du=t.$$

As the integrand is an even positive function, its antiderivative can be chosen to be odd. It is also unbounded and invertible on the whole of $\mathbb R$. So there is a solution for any $c$.

The explicit solution is $$y=\text{erfi}^{-1}\left(\frac2{\sqrt\pi}t+\text{erfi}(c)\right).$$

For $c=0$, the function has an inflection point at the origin.


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