Singularity at left endpoint for variational calculus problem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:43:19Z http://mathoverflow.net/feeds/question/68664 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68664/singularity-at-left-endpoint-for-variational-calculus-problem Singularity at left endpoint for variational calculus problem unknown (google) 2011-06-23T21:47:54Z 2011-12-02T23:24:26Z <p>Hi all, first, I'd like to apologize if the term "singularity" is being misused. I have the following integral:</p> <p>$\int _{0}^{\pi/2} \sqrt{ r \left( x \right) ^{2} + \left( {\frac {d}{dx}} r \left( x \right) \right) ^{2}} \left( \ln \left( { \frac {\sin \left( x \right) }{1-\cos \left( x \right) }} \right) + 1 \right) {dx}$</p> <p>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.</p> http://mathoverflow.net/questions/68664/singularity-at-left-endpoint-for-variational-calculus-problem/68702#68702 Answer by Igor Khavkine for Singularity at left endpoint for variational calculus problem Igor Khavkine 2011-06-24T07:19:27Z 2011-06-24T07:19:27Z <p>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 $x\in(0,\pi/2)$.</p> <p>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.</p> http://mathoverflow.net/questions/68664/singularity-at-left-endpoint-for-variational-calculus-problem/82523#82523 Answer by unknown (google) for Singularity at left endpoint for variational calculus problem unknown (google) 2011-12-02T23:11:59Z 2011-12-02T23:24:26Z <p>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 $0 &lt; a &lt; b &lt; \pi/2$. The LHS is </p> <p>$\int_a^b \left( \ln \left( {\frac {\sin \left( x \right) }{1-\cos \left( x \right) }} \right) + 1 \right) {\frac {r(x)}{\sqrt {{r(x)}^{2} + {r'(x)}^{2}}}}$</p> <p>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!</p>