I came across this blog that says that its French version has answers to most of Arnol'd's trivium problems, and I figured I'd try my hand at some of the ones they don't have. Number 68 raised my interest, and I think I solved it, but I'm not very happy with my solution and I'd like to know whether there's a more elegant approach, more based on general principles and less dependent on hacking together a somewhat arbitrary function. I've written up my solution as an answer, but I'm hoping for other, perhaps better answers. Here's the problem:

Find $$ \inf \iint\limits_{x^2+y^2\leqslant1}\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2\mathrm dx\,\mathrm dy $$

for $C^\infty$-functions $u$ that vanish at $0$ and are equal to $1$ on $x^2+y^2=1$.

  • 1
    $\begingroup$ Defenitely, Arnold is a troll. $\endgroup$ – Norbert Oct 24 '12 at 10:31
  • $\begingroup$ @Norbert: How do you mean? $\endgroup$ – joriki Oct 24 '12 at 10:54

First, it's straightforward to show that we can restrict attention to functions that respect the radial symmetry of the problem. Since the gradient decomposes into a radial and a tangential component, given any function $u$ that isn't radially symmetric we can integrate the square of the radial component of the gradient along the radii, find the minimum of the resulting angular function (which exists since this a continuous function on a compact domain) and replicate the function values at that angle for all angles. Then the radial component of the integral hasn't increased and the tangential component is now zero, so the radially symmetric function thus constructed leads to a lower value of the objective functional.

So we can consider a function $u(r,\phi)=f(r)$, with $f(0)=0$, $f(1)=1$ and suitable behaviour at $r=0$ to ensure smoothness. The objective functional becomes

$$ \int_0^1f'(r)^2r\mathrm dr\;. $$

The Euler-Lagrange equation is $(2f'r)'=0$, so $f'r=C$ is a constant, and thus $f=C\log r+B$.

Here's where the ugly part begins. Since this is singular at the origin, we can't use it to get a full solution to the problem. (Presumably this is why it says $\inf$ and not $\min$.) We can cut the solution off when it hits zero, forgetting about the $C^\infty$ requirement for the moment, and use

$$ f(r)=\begin{cases}1-\log r/\log r_0&r\ge r_0\\0&\text{otherwise}\end{cases} $$

to obtain

$$ \int_{r_0}^11/(r\log r_0)^2r\mathrm dr=\left[-\frac{\log r}{\log^2 r_0}\right]_{r=r_0}^{r=1}=\frac1{\log r_0}\;, $$

which goes to $0$ for $r_0\to0$. Now we can convolve this function with a mollifier to make it $C^\infty$, and though the details of showing this rigorously seem tedious, it seems clear that we can change both the integral and the function value at $1$ by as little as we want, which shows that the infimum is $0$.


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