<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[landscape]{slides}
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\theoremstyle{plain}
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\def\Ga{\Gamma} \def\Om{\Omega} \def\De{\Delta}
\def\de{\delta} \def\ga{\gamma} \def\al{\alpha} 
\def\be{\beta} \def\ka{\kappa}
\def\vfi{\varphi}
\def\la{\lambda}\def\La{\Lambda}
\def\ep{\epsilon}\def\si{\sigma}\def\sinf{\sigma^\infty}
\def\bdry{\partial \Omega}   \def\Omc{\Omega^c}
\def\bj{\beta_j}\def\bi{\beta_i} \def\Gj{\Gamma_j} \def\Gi{\Gamma_i}
\def\L1{L^{1,2} (\Omega^c)} \def\Lt{L^{1,2} (\Omega^c(t))}

\def\hs#1{H^{#1}(\Omega)}

\def\Omp{\Omega_1 (t)} \def\Omm{\Omega_2(t)}
\def\su{\sum_{j\in\{L,B,T\}}}
\def\half{{1 \over 2}}
\def\bR{{\rm I\!R}} \def\Rd{\bR^2}
\def\Rt{\bR^3}\def\br{{\bf r}}
\def\row#1{(#1_1, \ldots, #1_N)} \def\jedN{j=1, \ldots, N}
\def\iedN{i=1, \ldots, N}

\def\bV{{\bf V}} \def\bB{{\bf B}}
\def\bz{{\bf z}} \def\bn{{\bf n}}
\def\di{\hbox{div}} \def\diag{\hbox{diag}\,}
\def\dim{\hbox{dim}\,}
\def\Im{\hbox{Im}\,}
\def\dS{\,d{\cal H}^2}
\def\cA{{\cal A}}\def\cD{{\cal D}}
\def\cE{{\cal E}}
\def\cU{{\cal U}}
\def\qed{\hfill$\sqcap\!\!\!\!\sqcup$}

\begin{document}


\begin{slide}{}
 
\begin{align}
&amp; 0 =\De \si %%\quad 
&amp;\hbox{w }
 \bigcup_{0&lt;t&lt;T}\Rt\setminus\Om(t),&amp; \quad
\si(\infty)=\si^\infty&gt;0\\ %%&amp;(1)\cr
&amp;{\partial\si\over \partial\bn} = V,&amp; %\quad 
\hbox{na }S(t)=\partial\Om(t)&amp;\\ 
%%&amp; (2)\cr
&amp;-\si = -\di\xi -\be V,&amp; %\quad 
\hbox{na }S(t).%  %%&amp; (3)
\end{align}

\begin{tabular}{l l}
$\si$ -- przesycenie, &amp; $\Om(t)$ -- kryształ,\\
$S(t)$ brzeg  $\Om(t)$, &amp; $\bn$ --  normalna zewnętrzna do $S(t),$\\
$V$ -- prędkość $S(t)$, &amp; $\ga$ -- gęstość energii powierzchniowej,
\end{tabular}\\
$ ~\ $ $\xi$ -- wektor Cahna-Hoffmana, tj.
$\xi(x)=$``$\nabla\ga(\bn(x))$"  \\


\end{slide}
\begin{slide}{}
Dla uproszczenie zakładamy, że $\Om(t)$ jest kołowym cylindrem prostym, tj. 
$\Om(t)=\{(x_1,x_2,x_3): x_1^2+x_2^2 \le R^2(t),\ |x_3|\le L(t)\}.
$ 
 
 \bigskip

\centerline{\epsfxsize=9cm \epsfbox{cyl-s.eps}}
%%\centerline{{\bf Fig.~2}. Evolving crystal}


Wyróżniamy trzy części  $S(t)$: górę $S_T$, dół $S_B$ i pobocznicę $S_\La$.
Normalna  do $S_i$ jest oznaczna przez $\bn_i$, $i=\La,
B,T$. 

\end{slide}

\begin{slide}{}
Diagram Franka
$F_\ga$ i bryła Wulff energii $\ga$, $W_\ga$
to 
\begin{align*}
F_\gamma &amp;= \{ p \in {\bR}^3: \
\gamma(p) \leq 1 \}\\
%%$$  
%%$$
W_\gamma &amp;= \{x \in {\bR}^3: \ \forall \bn\in {\bR}^3, \, |\bn|=1,\quad
x\cdot \bn \le \ga(\bn)\}\\
 &amp;= \{x \in {\bR}^3: \ x_1^2+x_2^2\le \ga(\bn_\La),\, |x_3|\le\ga(\bn_T)\}
. 
\end{align*}


\bigskip
\centerline{\epsfxsize=13cm \epsfbox{fw.eps}}
%\centerline{{\bf Fig.~3}. Frank diagram $F_\ga$ and Wulff shape $W_\ga$}

\end{slide}

\begin{slide}{}

{\bf Stwierdzenie} (Giga, R., 2002) {\sl
Istnieje dokładnie jedno rozwiązanie $(R(t),L(t),\sigma(t,x))$ zagadnienia}
\begin{align}\label{rnie}
&amp;
\Delta \sigma =0 \quad\hbox{in}\quad \bR^3
\setminus\Omega(t),\qquad \lim_{|x|\to
+\infty}\sigma(x)=\sigma^\infty;\nonumber\\
&amp;
{\partial \sigma\over\partial \bn}=V \quad \hbox{on}\quad
\partial \Omega (t)\\
&amp;- \int_{S_i} \sigma = \kappa_i|S_i| - \beta_i V_i |S_i|\nonumber
\end{align}
{\sl uzupełinionego o warunek początkowy $\Om(0)=\Om_0$, który jest cylindrem.
Co więcej,}
$$
R,L \in C^{1,1}([0,T)),\quad
%$$
%$$
\nabla \sigma \in C^{0, 1} ([0, T); L^2(\bR^3\setminus\Omega (t))).
$$

\end{slide}



\begin{slide}{} 

{\bf Twierdzenie} %%3.1.} 
(Efekt Berga, Giga-R., 2003) {\sl Załóżmy, że $\si$ jest jedynym rozwiązaniem
  zagadnienia 
\begin{align*}%%\label{berg}
&amp; 0 =\De \si \quad \hbox{w } %\bigcup_{0&lt;t&lt;T}
\Rt\setminus\Om,\quad\si(\infty)=\si^\infty&gt;0,\\
&amp;{\partial\si\over \partial\bn} = V_i\quad \hbox{na }S=\partial\Om,
\end{align*}
gdzie $V_i&gt;0$ są stałymi a $\si=\si(\sqrt{x_1^2+x_2^2},|x_3|)$. Wtedy\\ 
{(a)} ${\partial\si\over\partial x_3}&gt;0$ (odpow. $&lt;0$) na}
$S_\La\cap\{x_3&gt;0\}$,  (odpow. $&lt;0$);  
\\
%%\\
{(b)} ${\partial\si\over\partial r}&gt;0$ na $S_T$, $S_B$; \quad 
{(c)} $\si&lt;\si^\infty$.

{\sl Analogiczny wynik jest prawdziwy, jeśli odwrócimy znaki $V_i$.}
\end{slide}
\begin{slide}{}
Lemma 4.3 implies that some of the inequalities in Theorem 4.2, i.e.
$\varphi_r(r)\in[ -\gamma(\bn_\La),\gamma(\bn_\La)]$,  are satisfied 
automatically, e.g. for $V_T&gt;0$ we have the following picture.

\bigskip
\centerline{\epsfxsize=12cm \epsfbox{wykres.eps}}
\end{slide}
\begin{slide}{}
{\bf Lemma 4.5.} %%PM 2} 
{\sl We assume that $\xi = \nabla(\varphi(r) + \psi(z))$, where
  $\varphi_r$ and   $\psi_z$ are
given by (\ref{cztet}) %% 4.4 
and (\ref{rnp4}), respectively. %%4.5. 
Then,\\
(a) If $V_\Lambda &lt; 0$, then $\varphi_r(r) &gt; - \gamma(\bn_\Lambda),\quad
\forall r \in [0, R)$.\\
(b) If $V_\Lambda &gt; 0$, then $\varphi_r(r) &lt; \gamma(\bn_\Lambda),\quad
\forall r\in [0, R)$\\
(c) If $V_T &lt; 0$, then $\psi_z(z) &gt; - \gamma(\bn_T),\quad
\forall z \in [0, L)$\\
(d) If $V_T &gt; 0$, then $\psi_z(z) &lt; \gamma(\bn_T), \quad
\forall z \in [0,L)$}.

But we  do not know if  all the constraints are fulfilled. For
instance, let us suppose 
$V_\La&gt;0$: when it is true that
$$
\varphi_r(r) &gt; - \ga(\bn_\La),\quad \hbox{for all } r\in (0,R) ?
$$

The above inequality is equivalent to 
$$
\frac r2( \bar{\sigma}_R - \bar{\sigma}_r)  &lt; \ga(\bn_\La)(1 +\frac rR).
$$
\end{slide}
\begin{slide}{}
By the representation formula for $\si$, (\ref{reper}), we can see
\begin{equation}\label{obi}
\bar{\sigma}_R - \bar{\sigma}_r  = a V_T{\mathcal F}_1(\rho, \theta,\tau),
\end{equation}
where
\begin{align*}
&amp; \rho =\frac L R,\quad
 \theta=\frac r R, \quad
 \tau=\frac{V_\La}{V_T},\quad a(t) \hbox{ is the scale at time }t\\
&amp;{\mathcal F}_1(\rho, \theta,\tau) =\hbox{a complicated expression.}
\end{align*}
This %Eq. (\ref{obi}) 
looks bad. It gets simpler if
$\frac{V_\La}{V_T}=const$ and $\frac L R =const$. Indeed, we
can have it for {\it self-similar motion}, i.e. if $\Om(t) = a(t)\Om_0$.\\
\medskip
{\bf Remarks.} 
(1) $V\equiv 0$ iff $\Om =\frac 2{\si^\infty}W_\ga$, where $W_\ga$ is
the Wulff 
shape, i.e. $\frac 2{\si^\infty}W_\ga$ is a steady state. \\
(2) Deciding the sign of $V_T$, $V_\La$ is another story. It is enough
to say that for a self-similar motion they have the
%$V_T$ and  $V_\La$ are always of
%the same 
sign of $\si^\infty+\ka(t)$, where $\ka(t)$ is the constant
curvature of $a(t)W_\ga$. 

\end{slide}
\begin{slide}
{\bf Theorem 4.6.} {\sl There exists a choice of $\be$ and $\ga$ satisfying  
$$
{\be \cdot \ga}= const,
$$
for which  $\Om(t)= a(t) W_\ga$, $a(0)=1$ is a solution to (\ref{rnie}).}

What is our specific conclusion about  stability then?

{\bf Theorem 4.7.}  {\bf (First specific result)}  {\sl Let us suppose that $\ga$, %$W_\ga$,
$\be$ are as above and
$\Omega(t) = a(t) W_\gamma$ is a self-similar solution. To fix
attention, we assume   that 
$\sigma^\infty + \kappa &gt; 0$, where $\kappa$ is the curvature of
$\Omega(0)=  W_\ga$. }

{\bf (i)}
{\sl The stability of motion of $S_T$  at time $t$ is equivalent to}
\begin{equation}\label{lball1} 
\frac{a(t)(\sigma^\infty a(t) - 2)c_{T}}{\beta_T + a(t) c_{T}} \leq\bar{d}_T.
\end{equation}
\end{slide}
\begin{slide}{}
{\sl The constants  $c_{T}$  and $\bar{d}_T$ 
depend only on $W_\gamma$. }%and its aspect ratio $\varrho_0$.}

{\bf (ii)} {\sl A similar statement holds for $S_\La$.}

It is apparent from (\ref{obi}), i.e. $\bar{\sigma}_R - \bar{\sigma}_r  = a
V_T{\mathcal F}_1(\rho, \theta,\tau)$, that the proof of Theorem 4.6 depends 
on estimates of $V_T$. Indeed, we have.

{\bf Lemma 4.8.} {\sl Let us assume that
$\ga$ and $\be$ are such that they
admit self-similar evolution. Moreover, $\Om(t)$ is a self-similar
solution, and $\Om(0)= W_\ga$. Then,}
\begin{equation}\label{faut}
V_T(t) = \frac{\sigma^\infty - 2/a(t)}{\beta_T + a(t) c_{T}},\quad
V_\La(t) = \frac{\si^\infty - 2/a(t)}{\be_\La + a(t) c_\La}.
\end{equation}
{\sl Here, $ c_{T},$ $c_\La$  are constants. }


\end{slide}
\begin{slide}{}
{\bf Idea of the proof of  Lemma 4.8:} We use the averaged Gibbs-Thomson and the
representation formula for $\si$,
$$
V_T(t) \left(\beta_T +
-  \!\! \!\! \!{\int_{S_T}}\left((f^a_T+f^a_B)\al_T+f^a_\La\alpha_\La\right)\dS\right)
= \sigma^\infty + \kappa(t),
$$
where $\al_T =1$,  $\al_\La = \frac {V_\La(t)}{V_T(t)}=const$.

The proof of the  facet stability result, Theorem 4.7,  amounts to checking if the
inequality
$$
\theta a(t) V_T(t){\cal F}_1(\rho_0,\theta,\rho_0)\le 1+\theta
$$
holds. It is based on the fact that
$V_\La/V_T$ is constant. 

{\bf Remark.} We will succeed in general, if we can bound this
quotient. For this purpose we draw a general picture of the phase
portrait of the ODE system (\ref{rnie}).


\end{slide}
\begin{slide}{\bf 5. Phase portrait}

Let us denote the unique equilibrium of (\ref{rnie}) by $(R_0, L_0)$.

{\bf Theorem 5.1.}
There exists a neighborhood $\cU$ of $(R_0, L_0)$ and orthogonal
projections $P, Q: \bR^2 \to \bR^2$, such that $\dim \Im P = 1$, $ \dim
\Im Q = 1$. They are 
such that $W^U ((R_0, L_0), \cU)$ (respectively, $W^S ((R_0, L_0),
\cU)$ exists and 
it is a $C^1$ curve over $P\bR^2$ (respectively, $Q \bR^2$) which is
tangent at $(R_0, L_0)$ to $P \bR^2$ (respectively $Q \bR^2$).  

This statement is  an application of standard results from dynamical systems.

\end{slide}
\begin{slide}{}
\centerline{\epsfxsize=9cm \epsfbox{portfaz.eps}}

{\bf Corollary 5.2.} 
$$
0 \leq  \frac{|V_\Lambda|}{V_T} \leq \tilde{\varrho}
\quad\hbox{in}\quad B((R_0, L_0), \tilde{r}) \cap \{(R,L):\ V_T&gt;0,\ R&gt;R_0\}
$$
for some $\tilde{r}$.

\end{slide}
\begin{slide}{}
Our final stability facet result is here.

{\bf Theorem 5.3.}  Let us assume that $(R, L)$ is in the region 
$Q_I := \{V_T &gt; 0, R &gt; R_0\}  $ of the phase plane. Then,\\
{\bf (a)} there exists $U_T$, a neighborhood of $(R_0, L_0)$, such that for all points
$(R,L) \in Q_I \cap U_T$ the facets $S_T, S_B$ are stable.\\ 
{\bf (b)} then exists $U_\Lambda$, a neighborhood of $(R_0, L_0)$, such that for all
points $(R, L) \cap Q_I \cap U_\Lambda$ the facet $S_\Lambda$ is stable. 

The proof is based on the same ideas as in the proof of Theorem 4.7.
That is, we have to check (for $V_T&gt;0$) the inequality
$$
\theta a(t)V_T(t) {\cal F}_1(\rho,\theta,\tau)\le 1+\theta
$$
holds. However, the calculations are more 
involved. 

\end{slide}
\begin{slide}{}
 Lemma 4.3 suggests that the set of facet stability is large, because if 
$V_T=0$, then $\vfi_r(r)= \frac rR \ga(\bn_\La)$.

\bigskip

\bigskip


\centerline{\epsfxsize=9cm \epsfbox{statbo.eps}}

\end{slide}
\end{document}
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