# Integral involving piecewise continuous function

We have a piecewise smooth function $$x: [0,1] \rightarrow \mathbb{R}$$ satisfying:

$$x^\prime(t)e^{-x^\prime(t)^2}= C$$

for some constant $$C \in \mathbb{R}$$ ($$x^\prime(t)$$ is therefore piecewise continuous). I also have the boundary conditions $$x(0) = 0$$ and $$x(1) = 0$$.

How can I solve for $$x(t)$$? If I integrate with respect to $$t$$, I have:

$$\int x^\prime(t)e^{-x^\prime(t)^2} dt = Ct + D$$

for some constant $$D \in \mathbb{R}$$. Is it very tempting to look at the real variable integral:

$$\int se^{-s^2} ds = Ct + D$$

which can be solved easily using integration by substitution, but I figure that one can't really do that here.

As for the context, I am trying to solve a calculus of variations problem and I ran into this issue.

• I realised maybe this problem as it stands is ill-posed, so I added boundary conditions $x(0) = 0$ and $x(1) = 0$. – Frederic Chopin Oct 14 at 17:55
Drawing the function $$f(s)=se^{-s^2}$$ you can see that, for a fixed $$C$$, $$f(s)=C$$ has a limited number of solutions.
Let $$M=\max_t f(t)$$ be the finite maximum of $$f$$. Then, one has only two possible solutions for $$f(x'(t))=C$$ if $$0 or $$-M, i.e. your function makes a sort of stairs, going up if $$C>0$$ or down if $$C<0$$ (notice that $$f$$ is odd). If $$C\in\{0,M,-M\}$$ then $$x'(t)$$ is equal to a constant, i.e. your function is either zero, or affine. Clearly if $$|C|>M$$ there is no function solving your problem.
With boundary data $$x(0)=x(1)=0$$ the only possible solution is the trivial one $$x(t)=0$$ since it cannot increase or decrease monotonically.