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.

EDIT: I added boundary conditions

  • $\begingroup$ I realised maybe this problem as it stands is ill-posed, so I added boundary conditions $x(0) = 0$ and $x(1) = 0$. $\endgroup$ – 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 $0<C<M$ or $-M<C<0$, 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.


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