# For which $c$ does $y' = e^{-y^2}$, $y(0) = c$ have exactly one solution?

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.

• 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 $c\in\mathbb R$.
– Surb
Feb 25, 2020 at 9:49
• @Surb I got too caught up in playing with c! Any thoughts on the solution $y''(0)$ for $c=0$? Feb 25, 2020 at 9:52
• Not so sure to understand your question. But $y''(t)=2y(t)y'(t)e^{-y^2(t)}$. So if $y(0)=0$, then $y''(0)=0$.
– Surb
Feb 25, 2020 at 9:56

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.