Singularity at left endpoint for variational calculus problem - MathOverflow

Singularity at left endpoint for variational calculus problem

2011-06-23T21:47:54Z

Hi all, first, I'd like to apologize if the term "singularity" is being misused. I have the following integral:

$\int _{0}^{\pi/2} \sqrt{ r \left( x \right) ^{2} + \left( {\frac {d}{dx}} r \left( x \right) \right) ^{2}} \left( \ln \left( { \frac {\sin \left( x \right) }{1-\cos \left( x \right) }} \right) + 1 \right) {dx}$

Note that the integrand shoots to infinity as x approaches 0. It's not too hard to establish that the integral will nonetheless always converge, regardless of the choice of r. My question is: will an extremizer satisfy the Euler-Lagrange Equation on the interval (0,Pi/2], and how can I tell? Sorry if this is trivial, I do not know much about the calculus of variations.

Answer by Igor Khavkine for Singularity at left endpoint for variational calculus problem

2011-06-24T07:19:27Z

The answer is yes, if the interval you mean is $(0,\pi/2)$. That is, in any case, the natural domain on which the Euler-Lagrange equation $EL(x)$ is expected to hold. The reason is that it reflects the property of the minimizer $r(x)$ that the variation of the above integral vanishes for arbitrary variations $r(x)+\delta r(x)$ with compact support in $(0,\pi/2)$: $$\int_0^{\pi/2} \delta r(x) EL(x)=0.$$ Pick a point $y\in(0,\pi/2)$ and a fixed open neighborhood $U$ of $y$ with compact closure in the same interval. Since the last equation is expected to hold for any such $U$ and $\delta r(x)$ with support $\overline{U}$, it follows that $EL(y)=0$ holds. The same argument shows that $EL(x)=0$ for all $x\in(0,\pi/2)$.

The natural domain for a differential equation to be satisfied is an open interval. Including a boundary of the interval essentially means the solution and sufficiently many of its derivatives extend continuously to that point.

Answer by unknown (google) for Singularity at left endpoint for variational calculus problem

2011-12-02T23:11:59Z

This is strange. If $r$ is a $C^2$ extremal, and $F(x,r(x),r'(x))$ we'd have $F_r(x,r(x),r'(x)) - \frac{d}{dx} F_{r'}(x,r(x),r'(x)) = 0$ for $x \in (0,\pi/2)$, which then means $\int_{a}^{b} F_r(x,r(x),r'(x)) = F_{r'}(b,r(b),r'(b)) - F_{r'}(a,r(a),r'(a))$ for $0 < a < b < \pi/2$. The LHS is

$ \int_a^b \left( \ln \left( {\frac {\sin \left( x \right) }{1-\cos \left( x \right) }} \right) + 1 \right) {\frac {r(x)}{\sqrt {{r(x)}^{2} + {r'(x)}^{2}}}} $

which, provided $r(0)=1$, goes to a finite value as $a \rightarrow 0^+$. However, $ F_{r'}(a,r(a),r'(a)) = \left( \ln \left( {\frac {\sin \left( a \right) }{1-\cos \left( a \right) }} \right) + 1 \right) {\frac {r'(a)}{\sqrt {{r(a)}^{2} + {r'(a)}^{2}}}}$ which goes to $\pm \infty$ as $a \rightarrow 0^+$ if $r'(0) \neq 0$. Thus we are left to conclude that $r'(0) = 0$. But 'experimental' data (via discretizations of the problem), show this isn't the case! Either what I just wrote is wrong, or something is seriously wrong!