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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?

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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 $\mu.$

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  • $\begingroup$ I will try this, but I'm puzzled by (c) and (d). What exactly can they establish? $\endgroup$
    – user15626
    Commented Dec 27, 2011 at 22:48
  • $\begingroup$ If, as I think, (C) does better than (A) and (B), I would take that as strong evidence that a minimizer with y(pi/2) = 0 is not C^1, and may not be C^0, which is what (D) is about. I'm really not sure, and have no good way to model this myself numerically. You don't talk about the differentiability of your solutions in your question. $\endgroup$
    – Will Jagy
    Commented Dec 27, 2011 at 23:06
  • $\begingroup$ Yes, sorry I forgot to specify, I am looking for a C^1 minimizer. This said, why should such an outcome be taken as strong evidence? $\endgroup$
    – user15626
    Commented Dec 27, 2011 at 23:27
  • $\begingroup$ If things go as I suspect, the minimizer in a larger space of functions would be your known $h$ for $x < \pi / 2$ and then a discontinuous $h(\pi/2) = 0.$ It has been a long time since I studied this, of course. You said you were able to work out the Euler-Lagrange equations, how would you describe those in this case? $\endgroup$
    – Will Jagy
    Commented Dec 28, 2011 at 0:02
  • $\begingroup$ So I've tired (A) and (B), both give values larger than for the minimizer $h$. Just so you know, $h$ is continuous on $[0,\frac{\pi}{2}]$, so (C) and (D) simply converge to whatever the integral is when evaluated for $h$. The Euler-Lagrange equation is quite large and messy, I'm afraid it doesn't look like anything standard. $\endgroup$
    – user15626
    Commented Dec 29, 2011 at 4:05

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