Proving a variational problem has no solutions Consider the following integral
$ \int_{0}^{\frac{\pi}{2}} \left( \sqrt{y(x)^2 + y'(x)^2} \left( \ln \left( \frac{\sin(x)}{1 -\cos(x)} \right) + \frac{\pi}{2} \right) + \frac{\pi}{2} y'(x) + 1 \right) dx$
together with the constraint $y(0) =1$. A minimizer for this problem certainly exists: computations gave one with $y(\frac{\pi}{2}) = \sim 0.03$.
The problem I'm interested in is to find another minimizer to this problem which in addition satisfies $y(\frac{\pi}{2}) = 0$. Numerical 'experiments' suggest this is not possible (among other things, I solved the Euler-Lagrange equation with several different set of initial values at $\frac{\pi}{4}$).
What are some things I could try to prove (or disprove) this fact?
 A: I am a little concerned about your logarithm term, but it is integrable...
There is no guarantee that one can get an optimum solution to adhere to two boundary values. The easy example is one dimensional constant mean curvature. Ask for the smallest arclength with a fixed enclosed area (under the graph) between two points, one boundary value larger than the other. If the enclosed area is small enough, you get a circular arc. If the enclosed area demanded is too large, the radius of the circular arc is too large and no longer gives the graph of a function that adheres at the two boundary points; in effect, the graph detaches at one endpoint. Your problem has no additional constraint, so I am not sure.
So, you can do a few experiments. Start with your known minimizer, call  it $h(x)$, with your $$h\left( \frac{\pi}{2} \right) = \mu \approx 0.03.$$  
Assuming, and I am not sure, that you can evaluate your integral with a specific function $y(x),$ do three experiments:
(A) evaluate numerically at $y(x) = h(x) \left( 1 - \frac{2 x}{\pi} \right)$
(B)  evaluate numerically at $y(x) = h(x) - \left(  \frac{2 \mu x}{\pi} \right)$
(C) take $h$ until $\frac{\pi}{2} - \delta,$ then extend continuously with constant slope (approximately $-\mu/\delta$) to hit 0 at $x=\pi/2.$ Because you are integrating, option (C) should hardly change with smaller and smaller $\delta.$
(D) the same as (C) except take $y=0$ after  $\frac{\pi}{2} - \delta.$
Well, give it a try. You may wind up with convincing evidence that your two-endpoint problem detaches at $\pi/2$ and insists on approaching your $\mu.$ 
