I have a problem that requires to find the smallest perimeter of a curve that goes trough points a,b and encloses an area A between the curve and x axis. I used the calculus of variations approach.
$$argmin_y\int_0^1 [\sqrt{1 + (y')^2} + \alpha (y - A)] dx$$
the answer is a circle (see attached image for draw and solution).
My question is what happens when the area is to large ? obviously I can't draw a circle that goes through a,b to get that area.
is there no solution in these cases ? How can I prove it?
You're right, there is no solution in this case. The Euler-Lagrange equation is a necessary condition for an extremum, so if it has no solution, there's no extremum; there's nothing to prove beyond that.
The "solution" in this case is a curve with vertical segments at $a$ and $b$, connected by a circular arc. This is not a function, but it can be arbitrarily closely approximated by differentiable functions; that's why there's no extremal function.