14
$\begingroup$

Below, a "disk" means a compact subspace $D \subset \mathbb R^2$ whose boundary is a smooth simple closed curve.

Task: Find a procedure which takes as input a pairs of disks $ D_0 \subseteq D_1 $ in the plane, and produces as output a smooth 1-parameter family of disks that interpolates between them: $$ \{D_t\}_{t \in [0,1]}. $$ The family should be monotonic in the sense that $ t_1 \le t_2 \Rightarrow D_{t_1} \subseteq D_{t_2}. $

The procedure should furthermore be (continuous and) smooth, meaning that if we have a family of pairs $D_0(x) \subseteq D_1(x)$ depending smoothly on some parameter $x\in\mathbb R^n$, then the output of the procedure $\{D_t(x)\}_{t \in [0,1]}$ should depend smoothly on $(t,x)\in [0,1]\times\mathbb R^n$.


Remark: If $D_0$ is contained in the interior of $D_1$, then the level curves of the solution of the Dirichlet problem on $D_1 {\setminus} \mathring D_0$ with boundary values $0$ on $\partial D_0$ and $1$ on $\partial D_1$ provide a family of simple closed curves interpolating between $\partial D_0$ and $\partial D_1$ (hence a family of disks interpolating between $D_0$ and $D_1$). This procedure has all the desired good properties, but it doesn't obviously work when $\partial D_0 \cap \partial D_1 \neq \emptyset$.
$\endgroup$
15
  • 3
    $\begingroup$ I did not understand why the procedure outlined in the Remark does not work. The difference $D_1\backslash D_0$ is not necessarily connected in this case, but this does not prevent you from solving Dirichlet problem for each component and using the level lines for your "move". $\endgroup$ Commented Feb 3, 2023 at 0:08
  • 2
    $\begingroup$ This feels like a very hard problem that should be more famous than it is... I recall David Epstein worked/works on this question... and said that it related it to Teichmuller theory, to computer vision, to other things... I will try to find a reference. $\endgroup$
    – Sam Nead
    Commented Feb 3, 2023 at 8:11
  • 1
    $\begingroup$ @AlexandreEremenko. If $\partial๐ท_0 \cap \partial ๐ท_1$ is a single point, then the higher derivatives of your two functions $๐‘”$ and $โ„Ž$ need not agree (by "smooth", I mean $๐ถ^\infty$). And even if the all the higher derivatives of $๐‘”$ and $โ„Ž$ agree, there's still something to show: for example, how do you know that the higher derivatives of $๐‘“$ are bounded? (I'm sure they are โ€“ I just don't know how to prove it) $\endgroup$ Commented Feb 4, 2023 at 0:37
  • 4
    $\begingroup$ Presumably a two-stage process would not satisfy your needs? Shrink $D_0$ until it fits strictly inside $D_1$, and then apply the level-curves process. $\endgroup$ Commented Feb 6, 2023 at 16:20
  • 1
    $\begingroup$ I suppose it might be helpful to note that, if we pull back $D_1$ by a conformal mapping $f$ (and using the fact that a disk with smooth curve has such a map $f$ which is smooth across the boundary, and similarly for its inverse), we may as well be assuming $D_1$ is the unit disk. There's a few classes of "canonical" families of univalent mappings in function theory, maybe one of them suffices... $\endgroup$ Commented Feb 6, 2023 at 22:55

1 Answer 1

3
+175
$\begingroup$

Here is an incomplete approach that was too long for a comment.

  1. In some smooth way, choose $\epsilon>0$ such that

    (i) the region $C_\epsilon\subset \overline{D_1}$ within distance $\epsilon$ of $\partial D_1$ is a collar neighbourhood of $\partial D_1$ (i.e. there is a smooth embedding $\phi:\partial D_1 \times [0,\epsilon) \to C$ such that $\phi(x,t)$ is the point in $C$ at distance $t$ along the normal $\nu_x$ to $\partial D_1$ at $x$),

    (ii) $C_\epsilon \cap \partial D_0$ is the set $\{\phi(x,u(x)):x \in V_\epsilon\}$ for some set $V_\epsilon \subset \partial D_1$ and smooth function $u:V_\epsilon \to [0,\epsilon)$, and

    (iii) $|\nabla u|<\tilde{\epsilon}$ on $V_\epsilon$ for some $\tilde{\epsilon}>0$.

  2. Let $\eta\in C^\infty(\mathbb{R})$ be a smooth function with $\eta=0$ on $(-\infty,0)$ and $\eta=1$ on $[1,\infty)$. For some $0<\delta<\epsilon$, let $\tilde{D}_0 \subset D_1$ be the set such that $$ \partial \tilde{D}_0 \cap C_{\frac{\epsilon}{2}} =\left\{\phi\left(x,\frac{\epsilon}{2}+\left(u(x)-\frac{\epsilon}{2}\right)\eta\left(\frac{2u(x)}{\epsilon}\right)\right):x \in V_\epsilon, u(x)<\frac{\epsilon}{2}\right\}, $$ and $\tilde{D}_0 \backslash C_{\frac{\epsilon}{2}}=D_0 \backslash C_{\frac{\epsilon}{2}}$. Then $\tilde{D}_0$ has smooth boundary and $\partial \tilde{D}_0 \cap \partial D_1=\emptyset$.

  3. Let $w \in C^\infty(D_1 \backslash \tilde{D}_0$ denote the solution to the Dirichlet problem on $D_1 \backslash \tilde{D}_0$ with boundary conditions $w=0$ on $\partial \tilde{D}_0$ and $w=1$ on $\partial D_1$. I believe, but have no proof, that by choosing $\epsilon,\tilde{\epsilon}$ small enough, we can ensure that for each $x \in V_\epsilon$, $s \mapsto w(\phi(x,s))$ is increasing for $s \in (0,u(x))$. For $t \in (0,1)$ and $x \in V_x$, let $u_t(x)$ denote the distance from $x$ to $w^{-1}(s)$ along $\nu_x$.

  4. Define $D_t$ such that $D_t \backslash C_\epsilon$ is the $t$-superlevel set of $w$ in $D_1 \backslash C_\epsilon$, and $$ \partial D_t \cap C_\epsilon=\left\{\left(x, u_t(x) + \left(t u(x) - u_t(x)\right) \eta\left(\frac{2u(x)}{\epsilon}-1\right)\right):x \in V_x \right\}. $$ That is, $D_t$ moves linearly from $D_0$ to $D_1$ within $C_{\frac{\epsilon}{2}}$, $D_t$ is a level set of $w$ on $D_1 \backslash C_\epsilon$. Also, $D_t$ is smooth, and on $C_\epsilon$ (as well as on $D_1 \backslash C_\epsilon$), the $D_t$ are correctly nested since $u_t(x) + \left(t u(x) - u_t(x)\right) \eta\left(\frac{2u(x)}{\epsilon}-1\right)$ is increasing in $t$.

Hopefully, someone more familiar with the Dirichlet problem can show that the solution to the DIrichlet problem on a thin enough strip has gradients which are never parallel to the strip.

$\endgroup$
1
  • $\begingroup$ This looks promising. $\endgroup$ Commented Feb 10, 2023 at 21:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .