text:
\section{Epidemic Model}

In this work, we use a new epidemic model that allows us to simulate a virus spread in a fine-grained fashion.

\paragraph{Populations}

In this model, namely \textit{SAPSR}, at each time $t \in \{1, \ldots, T \}$, the individuals of a population are partitioned into $5$ classes:

\begin{itemize}
	\item $S_t$: \textit{Susceptible} individuals that did not contract the virus before $t$;
	\item $A_t$: \textit{Asymptomatic} individuals that contracted the virus before $t$ and will not manifest any symptom during the whole development of the disease;
	\item $P_t$: \textit{Presymptomatic} individuals that contracted the virus before $t$ without manifesting any symptom, but they will manifest some symptoms in the future;
	\item $Y_t$: \textit{Symptomatic} individuals that contracted the virus before $t$ and present some symptom;
	\item $R_t$: \textit{Recovered} individuals that were either asymptomatics or symptomatics before $t$, but that at $t$ are not infected anymore. This category, at $t$, cannot infect other people and are immune to the disease.\footnote{Though recent studies showed that a few individuals can be reinfected, in this work, where we consider relatively small time horizons, we do not include these scenarios.}
\end{itemize}

The separation between asymptomatic and presymptomatic individuals in epidemic models is proposed by~\citet{ferretti2020quantifying} to capture properly the SARS-CoV-2 outbreak dynamics, as the virus transmission is mainly conveyed by these two classes of individuals.

Indeed, symptomatic people are less likely to spread the contagion since, usually, they are promptly isolated as symptoms occur, while asymptomatics and presymptomatics have frequent contacts with susceptible individuals, which, consequently, are more likely to be infected by individuals from these classes.

However, once a contact occurs, symptomatics have the largest probability to infect a new individual, followed by presymptomatics and asymptomatics, who have the lowest probability of passing the disease.

Thus, in our model, the probability of spreading the disease is differentiated among the three classes.

\input{contagionScheme}

\paragraph{Contact Model}

Our model describes contacts and contagions among individuals during a period of $T$ days.\footnote{

The quantities used in what follows are listed in Table~\ref{tab:notation}, available in the Supplementary Material.}

Let $U_t$ be the population individuals involved in the contagion.

We define with $w_{u,v}(t)$ the probability that two nodes $u \in U_t$ and $v \in U_t$ have a contact at time $t$.

We set a different value for each pair of individuals to model both individuals pairs that have recurrent and frequent contacts, \emph{e.g.}, family members, colleagues, and friends, while others have only sporadic encounters, \emph{e.g.}, people using public transportation.

Notice that in the current study we focus on the transmission of the virus by direct contact between couples of people, \emph{i.e.}, occurred by exchanging respiratory droplets, since it is known that the virus can survive only few hours outside the human body~\cite{khadka2020different}.

However, the model can be easily extended to include also other forms of environmental contagion.

A contact between two individuals $u \in U_t$ and $v \in U_t$ is denoted with $c_{u,v}(t)$, and it is defined as a tuple $(l_{u,v}(t), r_{u,v}(t), t)$, where $l_{u,v}(t)$ denotes the distance between $u$ and $v$, $r_{u,v}(t)$ denotes the duration of the contact, and $t$ is the time the contact between $u$ and $v$ occurred.\footnote{

Even though, other parameters can be relevant to determine the contagion probability, here, we focus only on these parameters as they constitute the information provided by the most common contact-tracing apps~\cite{khadka2020different}.}

Let us denote with $C(t) = \{ c_{u,v}(t) \text{ s.t. } u, v \in U_t \}$ the set of all the contacts occurred between pairs of individuals $(u, v)$ during day $t$.

Note that $c_{u,v}(t) \in C(t)$ also implies $c_{v,u}(t) \in C(t)$, since contacts are symmetric.

\paragraph{Contagion Model}

A graphical representation of the contagion model is presented in Figure~\ref{img:contagion}.

Let us denote with $\delta(u, v, l_{u,v}(t), r_{u,v}(t)) \in [0, \ 1]$ the contagion probability between two individuals $u \in U_t$ and $v \in U_t$ who have a contact for a duration of $r_{u,v})(t)$ at a distance $l_{u,v}(t)$.\footnote{

This probability can be inferred from epidemiological models or it can be estimated resorting to data about the past individuals contacts.}

This probability is non-null only if the node $u \in S_t$, \emph{i.e.,} she/he is susceptible, and the node $v \in A_t \cup P_t \cup Y_t$, \emph{i.e.,} she/he have previously contracted the SARS-CoV-2, and is different depending on the set $v$ belongs to (see~\citet{ferretti2020quantifying} for details).

If a susceptible node $u \in S_t$ becomes infected, she/he is removed from the population $S_t$ and included into $A_t$ with probability $\alpha_s(u)$, or she/he is included in $P_t$ otherwise.

If an individual becomes asymptomatic at a specific time $t_0$, she/he will remain in the asymptomatic population $A_t$ for a time span of $t \in (t_0, t_0 + \tau_u)$, where $\tau_u$ is the time required by the asymptomatic individual $u$ to recover from the virus, \emph{i.e.,} to result not infected anymore.

Subsequently, at time $t_0 + \tau_u$, she/he will be removed from the set of the asymptomatic individuals $A_{t_0 + \tau_u}$ and included in the recovered population $R_{t_0 + \tau_u}$.

Conversely, if an individual becomes presymptomatic at time $t_0$, she/he will remain in population $P_t$ for $t \in (t_0, t_0 + \epsilon_u)$, and, at time $t + \epsilon_u$, she/he will be included in the symptomatic population $Y_{t + \epsilon_u}$, where $\epsilon_u$ denotes incubation time for individual $u$.

\paragraph{Containment Model}

While the above mechanisms describe the development of the contagion in a completely uncontrolled population, the introduction of containment measures deeply affect how the virus spreads among the population.

As containment measure, in this work, we test each day a subgroup $V_t$ of cardinality at most $K$ of the population $U_t$, and we isolate those individuals with a positive outcome.

In this way, we can exclude them from the generation of new dangerous contacts from $C(t+1)$.

A generic \emph{testing policy} having a maximum number $K$ swab tests of is a function that, given the contacts in the past days $C(1:t) := \cup_{h=1}^t C(h)$ and the current population $U_t$, returns a subset $V_t$ to be tested with $|V_t| = K$.

Once the results on the individuals in $V_t$ are ready, the positive ones in the set $X_t$, are isolated from the population that can freely circulate from the next day $U_{t+1}$.

Notice that if an individual has been tested at some point, this does not exclude from her/his test in the future due to the false positives occurring during the test and the possibility that the contagion occurred during the successive days.

\input{contagion}

In Algorithm~\ref{alg:env}, we provide a high level description of the epidemic evolution in the case a testing policy is performed.

We also include the mechanisms regulating a generic tracing app, which will be used in the following sections to design our privacy-preserving test policy.

We are given with the initial populations $S_0$, $A_0$, $P_0$, $Y_0$, and $R_0$ and a time horizon $T$ for which we apply a testing policy $\texttt{testingPolicy}(\cdot)$, \emph{i.e.,} a strategy to select the individuals to test to limit the spread, and a swab test $\texttt{test}(\cdot)$ to check for the positivity to SARS-CoV-2.

At first, we initialize an empty set $D_u$ for each individual $u$, which will contain the information stored by the devices using the contact-tracing app during the period $\{1, \ldots, T\}$.

At each time $t$, we determine the new contacts $c_{u,v}(t) \in C(t)$ between individuals pairs, and we store the corresponding data on the devices $D_u$ and $D_v$ of the two individuals $u$ and $v$, respectively.

For the contact $c_{u,v}(t) \in C(t)$, it should be determined if a contagion between the involved nodes occurred (Line~\ref{line:cont}), and update the populations $S_t$, $A_t$, $P_t$, $Y_t$, and $R_t$ depending if the contagion generates a new asymptomatic or presymptomatic infection, or if some individuals recovered from the disease.

Finally, the policy $\texttt{testingPolicy}(\cdot)$ takes as input the contacts occurred in the past $C(1:t)$, the current population $U_t$, the newly symptomatics $Y_t$, and returns the individuals to be tested (Line~\ref{line:pol}).

Once the individuals $V_t$ are tested, populations $S_t$, $A_t$, $P_t$, $Y_t$ are updated, isolating positive individuals $O_t$ and those who manifested symptoms $Y_t$ (Lines~\ref{line:upd1}-\ref{line:upd2}).

\input{contagionStep}

The evolution of the population sets describing the virus evolution is detailed in Algorithm~\ref{alg:contagion}.

At first, if a contact in $C(t)$ generated a new infected individual $u$, occurring with  probability $\delta(u, v, l_{u,v}(t), r_{u,v}(t))$ (Line~\ref{line:prob}), she/he is removed from the susceptible population $S_t$.

According to the response of the individual, \emph{i.e.,} she/he becomes a presymptomatic or an asymptomatic patient, she/he is moved from the susceptible set $S_t$ to either the set $P_t$ or $A_t$, respectively.

After that, if any presymptomatic $u \in P_t$ showed some symptoms, she/he is moved to the symptomatic set $Y_t$.

Finally, the contagion evolves by moving the status of all the asymptomatics and symptomatics who recovered to $R_t$.

\paragraph{Contact-tracing Model}

We provide a brief description on the contact tracing apps functioning in Algorithm~\ref{alg:devices}.

As mentioned before, even though in some countries these tools collect fine grained data including, beyond contacts, personal and GPS information, most of the European countries developed solutions that aim also at preserving privacy, by introducing some constraints on how data are collected.

These constraints brought the development of different solutions that exploit Bluetooth technologies and security protocols to preserve users privacy.

In what follows, we focus on the general schema followed by \emph{Immuni}, the official Italian contact tracing application, whose features are common with by most of the other European privacy-aware contact tracing applications.

The data update procedure generates for each contact $c_{u,v}(t)$ two random codes, a.k.a.~token, $h_{u, t}$ and $h_{v, t}$, one for each individual involved in a contact (Lines~\ref{line:token1}-\ref{line:token2}).

Subsequently, it updates the data stored in each individual's device $D_u$ and $D_v$, adding an entry with the time $t$ the contact occurred, the tokens of the two individuals $h_{u,t}$ and $h_{v,t}$, the distance $l_{u,v}(t)$, and duration $r_{u,v}(t)$ of the contact (Lines~\ref{line:d1}-\ref{line:d2}).

\begin{algorithm}[t!]
	\centering
	\small
	\caption{\texttt{updateDevData}($D_u, D_v, c_{u, v}(t)$)}
	\begin{algorithmic}
		\State Input: devices data $D_u$, $D_v$, new contact $c_{u, v}(t)$
		\State Output: updated device data $D_u$, $D_v$
		\State $h_{u,t} \gets $ \texttt{RandomToken}() \label{line:token1}
		\State $h_{v,t} \gets $ \texttt{RandomToken}() \label{line:token2}
		\State $D_u \gets D_u \cup \{(t, h_{u,t}, h_{v,t}, l_{u,v}(t), r_{u,v}(t)) \}$ \label{line:d1}
		\State $D_v \gets D_v \cup \{(t, h_{v,t}, h_{u,t}, l_{u,v}(t), r_{u,v}(t)) \}$ \label{line:d2}
	\end{algorithmic}
	\label{alg:devices}
\end{algorithm}


text:
\begin{document}

\title{$p$-torsion elements in local cohomology modules}

\author{Anurag K. Singh}

\address{Department of Mathematics, University of Utah, 155 S. 1400
E., Salt Lake City, \quad UT 84112, USA \newline
E-mail: {\tt singh@math.utah.edu}}

\begin{abstract}
For every prime integer $p$, M.~Hochster conjectured the existence of certain
$p$-torsion elements in a local cohomology module over a regular ring of mixed
characteristic. We show that Hochster's conjecture is false. We next construct
an example where a local cohomology module over a hypersurface has $p$-torsion
elements for every prime integer $p$, and consequently has infinitely many
associated prime ideals.
\end{abstract}

\maketitle

For a commutative Noetherian ring $R$ and an ideal ${\mathfrak a} \subset R$,
the finiteness properties of the local cohomology modules $H_{\mathfrak
a}^i(R)$ have been studied by various authors. In this paper we focus on the
following question raised by C.~Huneke \cite[Problem 4]{Hu}: if $M$ is a
finitely generated $R$-module, is the number of associated primes ideals of
$H_{\mathfrak a}^i(M)$ always finite?

In the case that the ring $R$ is regular and contains a field of prime
characteristic $p > 0$, Huneke and Sharp showed in \cite{HS} that the set of
associated prime ideals of $H_{\mathfrak a}^i(R)$ is finite. If $R$ is a
regular {\it local}\/ ring containing a field of characteristic zero,
G.~Lyubeznik showed that $H_{\mathfrak a}^i(R)$ has only finitely many
associated prime ideals, see \cite{Ly1} and also \cite{Ly2, Ly3}. Recently
Lyubeznik has also proved this result for unramified regular local rings of
mixed characteristic, \cite{Ly4}. Our computations here support Lyubeznik's
conjecture \cite[Remark 3.7 (iii)]{Ly1} that local cohomology modules over all
regular rings have only finitely many associated prime ideals.

In Section \ref{elem} we construct an example of a hypersurface $R$ such that
the local cohomology module $H_{\mathfrak a}^3(R)$ has $p$-torsion elements for
every prime integer $p$, and consequently has infinitely many associated prime
ideals.

For some of the other work related to this question, we refer the reader to
the papers \cite{BL, BRS, He}.

\section{Hochster's conjecture}

Consider the polynomial ring over the integers $R={\mathbb Z}[u,v,w,x,y,z]$
where ${\mathfrak a}$ is the ideal generated by the size two minors of the
matrix
$$
M = \begin{pmatrix}
u & v & w \\
x & y & z \\
\end{pmatrix},
$$
i.e., ${\mathfrak a} = (\Delta_1, \Delta_2, \Delta_3)R$ where $\Delta_1 =
vz-wy$, $\Delta_2 = wx-uz$, and $\Delta_3 = uy-vx$.  M.~Hochster conjectured
that for every prime integer $p$, there exist certain $p$-torsion elements in
the local cohomology module $H_{\mathfrak a}^3(R)$, and consequently that
$H_{\mathfrak a}^3(R)$ has infinitely many associated prime ideals. We first
describe the construction of these elements. For an arbitrary prime integer
$p$, consider the short exact sequence
$$
\begin{CD}
0 @>>> R @>p>> R @>>> R/pR @>>> 0,
\end{CD}
$$
which induces the following exact sequence of local cohomology modules:
$$
0 \to H_{\mathfrak a}^2(R) \to H_{\mathfrak a}^2(R)
\to H_{\mathfrak a}^2(R/pR) \to H_{\mathfrak a}^3(R)
\to H_{\mathfrak a}^3(R) \to 0. \qquad (*)
$$
It can be shown that the module $H_{\mathfrak a}^3(R/pR)$ is zero using a
result of Peskine and Szpiro, see Proposition \ref{PS} below. In the ring $R$
we have the equation
$$
u\Delta_1 + v\Delta_2 + w\Delta_3 = 0,
$$
arising from the determinant of the matrix
$$
\begin{pmatrix}
u & v & w \\
u & v & w \\
x & y & z \\
\end{pmatrix}.
$$
Let $q=p^e$ where $e$ is a positive integer. Taking the $q$\,th power of the
above equation, we see that
$$
(u\Delta_1)^q + (v\Delta_2)^q + (w\Delta_3)^q \equiv 0 \mod p, \qquad (**)
$$
and this yields a relation on the elements $\overline{\Delta}_1^q, \
\overline{\Delta}_2^q, \ \overline{\Delta}_3^q \in R/pR$, where
$\overline{\Delta}_i$ denotes the image of $\Delta_i$ in $R/pR$. This relation
may be viewed as an element $\mu_q \in H_{\mathfrak a}^2(R/pR)$. Hochster
conjectured that for every prime integer there exists a choice of
$q=p^e$ such that
$$
\mu_q \notin \image \Big(H_{\mathfrak a}^2(R) \to H_{\mathfrak a}^2(R/pR)\Big),
$$
and consequently that the image of $\mu_q$ under the connecting homomorphism in
the exact sequence $(*)$ is a nonzero $p$-torsion element of
$H_{\mathfrak a}^3(R)$.

We note an equivalent form of Hochster's conjecture which is convenient to work
with. Recall that the module $H_{\mathfrak a}^3(R)$  may be viewed as the
direct limit
$$
\varinjlim \frac{R}{(\Delta_1^k, \Delta_2^k, \Delta_3^k)R}
$$
where the maps in the direct limit system are induced by multiplication by the
element $\Delta_1 \Delta_2 \Delta_3$. The equation $(**)$ shows that
$$
\lambda_q = \frac{(u\Delta_1)^q + (v\Delta_2)^q + (w\Delta_3)^q}{p}
$$
has integer coefficients, i.e., that $\lambda_q$ is an element of the ring $R$.
Let
$$
\eta_q = [\lambda_q + (\Delta_1^q, \ \Delta_2^q, \ \Delta_3^q)R] \ \in \
H_{\mathfrak a}^3(R).
$$

\begin{lemma}
With the notation as above, the following statements are equivalent:
\item $(1)$ $\mu_q \notin \image \Big(H_{\mathfrak a}^2(R)
\to H_{\mathfrak a}^2(R/pR)\Big)$,
\item $(2)$ $\eta_q$ is a nonzero $p$-torsion element of $H_{\mathfrak a}^3(R)$.
\end{lemma}

\begin{proof}
$(2) \implies (1)$ If $\mu_q \in \image \Big(H_{\mathfrak a}^2(R) \to
H_{\mathfrak a}^2(R/pR)\Big)$, the relation $(\overline{u}^q, \overline{v}^q,
\overline{w}^q)$ on the elements $\overline{\Delta}_1^q, \ \overline{\Delta}_2^q,
\ \overline{\Delta}_3^q\in R/pR$ \, lifts to a relation in $H_{\mathfrak
a}^2(R)$, i.e., there exists an integer $k$ and elements $\alpha_i \in R$ such
that $\alpha_1 \Delta_1^{q+k}+\alpha_2 \Delta_2^{q+k}+\alpha_3 \Delta_3^{q+k}=0$
and
$$
(u^q\Delta_2^k\Delta_3^k, \ v^q\Delta_3^k\Delta_1^k, \ w^q\Delta_1^k\Delta_2^k)
\equiv (\alpha_1, \ \alpha_2, \ \alpha_3) \mod p.
$$
Hence we have
\begin{align*}
& (u^q\Delta_2^k\Delta_3^k -\alpha_1)\Delta_1^{q+k}
+(v^q\Delta_3^k\Delta_1^k -\alpha_2)\Delta_2^{q+k}
+(w^q\Delta_1^k\Delta_2^k -\alpha_3)\Delta_3^{q+k} \\
&=\Big((u\Delta_1)^q+(v\Delta_2)^q+(w\Delta_3)^q\Big)(\Delta_1\Delta_2\Delta_3)^k
\in (p\Delta_1^{q+k}, \ p\Delta_2^{q+k}, \ p\Delta_3^{q+k} )R,
\end{align*}
and so $\lambda_q (\Delta_1 \Delta_2 \Delta_3)^{k} \in
(\Delta_1^{q+k}, \ \Delta_2^{q+k}, \ \Delta_3^{q+k})R$ and $\eta_q=0$.

The argument that $(1) \implies (2)$ is similar.
\end{proof}

By the above lemma, an equivalent form of Hochster's conjecture is that for
every prime integer $p$, there is an integer $q=p^e$ such that $\eta_q$ is a
nonzero $p$-torsion element of $H_{\mathfrak a}^3(R)$. Since $p\lambda_q \in
(\Delta_1^q, \ \Delta_2^q, \ \Delta_3^q)R$, it is immediate that
$$
p\cdot\eta_q = [p\lambda_q + (\Delta_1^q, \ \Delta_2^q, \ \Delta_3^q)R] = 0 \
\in \ H_{\mathfrak a}^3(R).
$$
We shall show that Hochster's conjecture is false by showing that for all
$q=p^e$, $\eta_q=0$ in $H_{\mathfrak a}^3(R)$. More specifically we show that
if $k=q-1$ then
$$
\lambda_q (\Delta_1 \Delta_2 \Delta_3)^{k} \in
(\Delta_1^{q+k}, \ \Delta_2^{q+k}, \ \Delta_3^{q+k})R.
$$

We record the following result, \cite[Proposition 4.1]{PS}, from which it
follows that $H_{\mathfrak a}^3(R/pR)=0$. We use this result with $A=R/P$ and
$I = (\overline{\Delta}_1, \overline{\Delta}_2, \overline{\Delta}_3)$. Then
$\height I = 2$, and it is well known that the determinantal ring $A/I$ is
Cohen-Macaulay.

\begin{prop}[Peskine-Szpiro]
Let $A$ be a regular domain of prime characteristic $p > 0$, and let $I$ be an
ideal of $A$ such that $A/I$ is Cohen-Macaulay. If $\height I = h$, then
$H_I^i(A)=0$ for $i > h$.
\label{PS}
\end{prop}

\begin{remark}
For a field $K$, let $R_K = K\otimes_{\mathbb Z}R$ where $R={\mathbb
Z}[u,v,w,x,y,z]$ as above. We record an argument due to Hochster which shows
that $H_{\mathfrak a}^3(R_{\mathbb Q})$ is nonzero.  Consider the action of
$SL_2({\mathbb Q})$ on $R_{\mathbb Q}$ where where $\alpha \in SL_2({\mathbb
Q})$ sends the entries of the matrix $M$ to the entries of $\alpha M$. The
invariant subring for this action is $A={\mathbb Q}[\Delta_1, \Delta_2,
\Delta_3]$ and, since $SL_2({\mathbb Q})$ is linearly reductive, $A$ is a
direct summand of $R_{\mathbb Q}$ as an $A$--module. Consequently $H_{\mathfrak
a}^3(A)$ is a nonzero submodule of $H_{\mathfrak a}^3(R_{\mathbb Q})$. Hence
while  $H_{\mathfrak a}^3(R_{{\mathbb Z}/p{\mathbb Z}})=0$, we have
$H_{\mathfrak a}^3(R_{\mathbb Q}) \neq 0$; it is then only natural to expect
that the study of $H_{\mathfrak a}^3(R)$ would be interesting!
\end{remark}

\section{The multi-grading}

We now work with a fixed prime integer $p$, an arbitrary prime power $q=p^e$,
and set $\lambda=\lambda_q$. We first use a multi-grading to reduce the
question whether
$$
\lambda (\Delta_1 \Delta_2 \Delta_3)^{k} \in
(\Delta_1^{q+k}, \ \Delta_2^{q+k}, \ \Delta_3^{q+k})R \qquad (\#)
$$
to a question in a polynomial ring in three variables. We assign weights as
follows:
\begin{align*}
u: (1,0,0,0) \qquad \qquad\qquad  & x: (1,0,0,1) \\
v: (0,1,0,0) \qquad \qquad\qquad  & y: (0,1,0,1) \\
w: (0,0,1,0) \qquad \qquad\qquad  & z: (0,0,1,1)
\end{align*}
With this grading,
$$
\deg(\Delta_1) = (0,1,1,1), \quad \quad
\deg(\Delta_2) = (1,0,1,1), \quad \quad
\deg(\Delta_3) = (1,1,0,1),
$$
and $\lambda$ is a homogeneous element of degree $(q,q,q,q)$. Hence in a
homogeneous equation of the form
$$
\lambda (\Delta_1\Delta_2\Delta_3)^{k}
=c_1\Delta_1^{q+k}+c_2\Delta_2^{q+k}+c_3\Delta_3^{q+k},
$$
we must have
$$
\deg(c_1) = (q+2k,k,k,2k), \ \deg(c_2) = (k,q+2k,k,2k), \
\deg(c_3) = (k,k,q+2k,2k).
$$
We use this to examine the monomials which can occur in $c_i$. If $c_1$
involves a monomial of the form $u^a v^b w^c x^d y^i z^j$, then
$$
(a+d, \ b+i, \ c+j, \ d+i+j) = (q+2k, \ k, \ k, \ 2k)
$$
and so $b=k-i, \ c=k-j, \ d=2k-i-j$ and $a=q+i+j$. Hence $c_1$ is a ${\mathbb
Z}$-linear combination  of monomials of the form
$$
u^{q+i+j}v^{k-i}w^{k-j}x^{2k-i-j}y^iz^j = u^q(uy)^i(vx)^{k-i}(uz)^j(wx)^{k-j}.
$$
Let $[s_1,s_2]^k$ denote the set of all monomials $s_1^is_2^{k-i}$ for $0 \le i
\le k$ and $[s_1,s_2]\cdot[t_1,t_2]$ denote the set of products of all pairs of
monomials from $[s_1,s_2]$ and $[t_1,t_2]$. With this notation, $c_1$ is a
${\mathbb Z}$-linear combination of monomials in
$u^q\cdot [uy,vx]^k \cdot [uz,wx]^k$. After similar computations for $c_2$ and
$c_3$, we may conclude that $(\#)$ holds if and only if $\lambda (\Delta_1
\Delta_2 \Delta_3)^{k}$ is a ${\mathbb Z}$-linear combination of elements from
\begin{align*}
& \Delta_1^{q+k}u^q  \cdot [uy,vx]^k  \cdot [uz,wx]^k, \\
& \Delta_2^{q+k}v^q  \cdot [vz,wy]^k  \cdot [vx,uy]^k, \\
& \Delta_3^{q+k}w^q  \cdot [wx,uz]^k  \cdot [wy,vz]^k.
\end{align*}
We may divide throughout by the element $(uvw)^{q+2k}$, and study this issue in
the polynomial ring $\mathbb{Z}[\frac{x}{u},\frac{y}{v},\frac{z}{w}]$. Let
$$
A=\frac{z}{w}-\frac{x}{u},\qquad B=\frac{x}{u}-\frac{y}{v},\qquad T=-\frac{x}{u}.
$$
The condition $(\#)$ is then equivalent to the statement that the element
$$
\frac{\lambda (\Delta_1 \Delta_2 \Delta_3)^{k}}{(uvw)^{q+2k}}
= \frac{1}{p} \Big((A+B)^q+(-A)^q+(-B)^q \Big) \Big( (A+B)AB \Big)^k
$$
is a ${\mathbb Z}$-linear combination, in the polynomial ring
$\mathbb{Z}[A,B,T]$, of elements of
\begin{align*}
&(A+B)^{q+k} \cdot [T,A]^k \cdot [T,B]^k, \\
& A^{q+k} \cdot [T,B]^k \cdot [T+B, A+B]^k, \\
& B^{q+k} \cdot [T,A]^k \cdot [T-A, A+B]^k.
\end{align*}
We show that the above statement is indeed true for $k=q-1$ using the following
equational identity, which will be proved in the next section:

\begin{align*}
& (A+B)^{2k+1} \sum_{n=0}^k \binom{k}{n} T^n \sum_{i=0}^n (-1)^{k+i}
A^{k-i} B^{k-n+i} \binom{k+i}{k} \binom{k+n-i}{k} \\
& -A^{2k+1} \sum_{n=0}^k \binom{k}{n} (-1)^n (T+B)^n \sum_{i=0}^n
(A+B)^{k-i} B^{k-n+i} \binom{k+i}{k} \binom{k+n-i}{k} \\
& -B^{2k+1} \sum_{n=0}^k \binom{k}{n} (T-A)^n \sum_{i=0}^n
(A+B)^{k-i} A^{k-n+i} \binom{k+i}{k} \binom{k+n-i}{k} =0.
\end{align*}

Separating the terms with $n=0$ and dividing by $p$, the above identity gives us
\begin{align*}
\frac{1}{p} & \Big((A+B)^{k+1}(-1)^k-A^{k+1}-B^{k+1}\Big) \Big(AB(A+B)\Big)^k \\
 = &
-(A+B)^{2k+1} \sum_{n=1}^k \binom{k}{n} T^n \sum_{i=0}^n (-1)^{k+i}
A^{k-i} B^{k-n+i} \frac{1}{p} \binom{k+i}{k} \binom{k+n-i}{k} \\
 & +A^{2k+1} \sum_{n=1}^k \binom{k}{n} (-1)^n (T+B)^n \sum_{i=0}^n
(A+B)^{k-i} B^{k-n+i} \frac{1}{p}\binom{k+i}{k} \binom{k+n-i}{k} \\
 & +B^{2k+1} \sum_{n=1}^k \binom{k}{n} (T-A)^n \sum_{i=0}^n
(A+B)^{k-i} A^{k-n+i} \frac{1}{p} \binom{k+i}{k} \binom{k+n-i}{k}.
\end{align*}
To ensure that the expression on the right hand side of the equality is indeed
a ${\mathbb Z}$-linear combination of elements of the required form, it
suffices to establish that the prime integer $p$ divides
$$
\binom{k+i}{k} \binom{k+n-i}{k}
$$
when $1 \le n \le k$ and $0 \le i \le n$, where $k=q-1=p^e-1$.
Note that this condition ensures that $i \ge 1$ or $n-i \ge 1$, and so it
suffices to show that $p$ divides
$$
\binom{k+r}{k} = \binom{p^e-1+r}{p^e-1}
$$
for $1 \le r \le p^e-1$. Since
\begin{align*}
\binom{p^e-1+r}{p^e-1} &=\left(\frac{p^e}{r}\right)\left(\frac{p^e+1}{1}\right)
\left(\frac{p^e+2}{2}\right) \cdots \left(\frac{p^e+r-1}{r-1}\right) \\
&=\left(\frac{p^e}{r}\right)\left(\frac{p^e}{1}+1\right)
\left(\frac{p^e}{2}+1\right) \cdots \left(\frac{p^e}{r-1}+1\right),
\end{align*}
this is easily seen to be true.

\section{The equational identity}

We first record some identities with binomial coefficients that we shall be
using. These identities can be easily proved using Zeilberger's algorithm (see
\cite{PWS}) and the Maple package {\tt EKHAD}, but we include these proofs for
the sake of completeness.

When the range of a summation is not specified, it is assumed to extend over
all integers. We set $\binom{k}{i}=0$ if $ i<0$ or if $k<i$.

\begin{lemma}
\label{iden}
\begin{align}
& \sum_{i=0}^m (-1)^i \binom{2k+1}{s+i}\binom{k+i}{k} \binom{k+m-i}{k} =
\begin{cases}
\binom{m+s-k-1}{m} \binom{2k+m+1}{m+s} \, \text{if } s > k, \\
0 \qquad  \ \text{ if } k \ge s \ge k+1-m,
\end{cases} \\

& \sum_{j=0}^s (-1)^j \binom{s}{j} \binom{a+j}{k}
= (-1)^s \binom{a}{k-s}, \\

& \sum_{i=0}^s \binom{k-i}{k-s} \binom{k+i}{k} \binom{k+m-i}{m}
= \binom{k+m-s}{m} \binom{2k+m+1}{s}.
\end{align}
\end{lemma}

\begin{proof}
(1) Let
\begin{align*}
&F(m,i) = (-1)^i \binom{2k+1}{s+i}\binom{k+i}{k} \binom{k+m-i}{k}, \\
&G(m,i) = \frac{i(s+i)(k+m+1-i)}{m+1-i} F(m,i), \quad \text{ and } \quad
H(m)=\sum_i F(m,i).
\end{align*}
It is easily seen that
\begin{align*}
&G(m,i+1) = -(k+i+1)(2k+1-s-i)F(m,i) \quad \text{ and } \quad \\
&F(m+1,i) = \frac{k+m+1-i}{m+1-i} F(m,i),
\end{align*}
and these can then be used to verify the relation
$$
G(m,i+1) - G(m,i) = (2k+m+2)(m+s-k)F(m,i) - (m+1)(m+s+1)F(m+1,i).
$$
Summing with respect to $i$ gives us
$$
0 = (2k+m+2)(m+s-k)H(m) - (m+1)(m+s+1)H(m+1).
$$
Using this recurrence,
$$
H(m) = \frac{(2k+m+1)\cdots (2k+2)(m+s-k-1)\cdots (s-k)}
{(m)\cdots (1) (m+s)\cdots (s+1)}H(0),
$$
and the required result follows.

(2) Let
$$
F(s,j) = (-1)^j \binom{s}{j} \binom{a+j}{k}, \
G(s,j) = \frac{-j(a+j-k)}{s+1-j} F(s,j),
$$
and $H(s)=\sum_j F(s,j)$. It is easily seen that
\begin{align*}
&G(s,j+1) = (a+j+1)F(s,j) \quad \text{ and } \quad
F(s+1,j) = \frac{s+1}{s+1-j} F(s,j),
\end{align*}
and these can then be used to verify the relation
$$
G(s,j+1) - G(s,j) = (k-s)F(s,j) + (a-k+s+1)F(s+1,j).
$$
Summing with respect to $j$ gives us
$$
0 = (k-s)H(s) + (a-k+s+1)H(s+1).
$$
Using this recurrence,
\begin{align*}
H(s) = \frac{(-1)^s (k-s+1)\cdots (k)}{(a-k+s)\cdots (a-k+1)} H(0)
= (-1)^s \binom{a}{k-s}.
\end{align*}

(3) Let
$$
F(s,i) = \binom{k-i}{k-s} \binom{k+i}{k} \binom{k+m-i}{m}, \
G(s,i) = \frac{i(k-s)(k+m+1-i)}{s+1-i} F(s,i),
$$
and $H(s)=\sum_i F(s,i)$. It is easily seen that
$$
G(s,i+1) = (k-s) (k+i+1) F(s,i) \quad \text{ and } \quad
F(s+1,i) = \frac{k-s}{s+1-i} F(s,i),
$$
and these can then be used to verify the relation
$$
G(s,i+1) - G(s,i) = (k-s)(2k+m-s+1)F(s,i) - (s+1)(k+m-s)F(s+1,i).
$$
Summing with respect to $j$ gives us
$$
0 = (k-s)(2k+m-s+1)H(s) - (s+1)(k+m-s)H(s+1).
$$
Using this recurrence,
\begin{align*}
H(s) & = \frac{(k-s+1)\cdots (k)(2k+m-s+2)\cdots (2k+m+1)}
{(s)\cdots (1) (k+m-s+1)\cdots (k+m)}H(0) \\
& = \binom{k+m-s}{m} \binom{2k+m+1}{s}.
\end{align*}
\end{proof}

We are now ready to prove the equational identity

\begin{align*}
&(A+B)^{2k+1} \sum_{n=0}^k \binom{k}{n} T^n \sum_{i=0}^n (-1)^{k+i}
A^{k-i} B^{k-n+i} \binom{k+i}{k} \binom{k+n-i}{k} \\
&-A^{2k+1} \sum_{n=0}^k \binom{k}{n} (-1)^n (T+B)^n \sum_{i=0}^n
(A+B)^{k-i} B^{k-n+i} \binom{k+i}{k} \binom{k+n-i}{k} \\
&-B^{2k+1} \sum_{n=0}^k \binom{k}{n} (T-A)^n \sum_{i=0}^n
(A+B)^{k-i} A^{k-n+i} \binom{k+i}{k} \binom{k+n-i}{k} =0.
\end{align*}

Examining the coefficient of $T^m$ for all $0 \le m \le k$, we need to show

\begin{align*}
& (A+B)^{2k+1}\binom{k}{m} \sum_{i=0}^m (-1)^{k+i}
A^{k-i} B^{k-m+i} \binom{k+i}{k} \binom{k+m-i}{k} \\
& -A^{2k+1} \sum_{n=m}^k \binom{k}{n} \binom{n}{m} (-1)^n \sum_{i=0}^n
(A+B)^{k-i} B^{k-m+i} \binom{k+i}{k} \binom{k+n-i}{k} \\
& -B^{2k+1} \sum_{n=m}^k \binom{k}{n} \binom{n}{m} (-1)^{n-m} \sum_{i=0}^n
(A+B)^{k-i} A^{k-m+i} \binom{k+i}{k} \binom{k+n-i}{k} =0.
\end{align*}

Dividing by $\binom{k}{m}(AB)^{k-m}$ and using the fact that
$$
\binom{k}{m}\binom{k-m}{n-m} = \binom{k}{n}\binom{n}{m},
$$
we need to establish that

\begin{align*}
& (A+B)^{2k+1} \sum_{i=0}^m (-1)^{k+i}
A^{m-i} B^i \binom{k+i}{k} \binom{k+m-i}{k} \\
& -A^{k+m+1} \sum_{n=m}^k \binom{k-m}{n-m} (-1)^n \sum_{i=0}^n
(A+B)^{k-i} B^i \binom{k+i}{k} \binom{k+n-i}{k} \\
& -B^{k+m+1} \sum_{n=m}^k \binom{k-m}{n-m} (-1)^{n-m} \sum_{i=0}^n
(A+B)^{k-i} A^i \binom{k+i}{k} \binom{k+n-i}{k} =0.
\end{align*}

For $0 \le r \le m-1$, the coefficient of $A^{k+m-r}B^{k+1+r}$ is
$$
\sum_{i=0}^{m}(-1)^{k+i} \binom{2k+1}{k-r+i} \binom{k+i}{k} \binom{k+m-i}{k}
$$
which is zero by lemma \ref{iden} (1) since $k \ge k-r \ge k+1-m$.

For $0 \le r \le k$, the coefficient  of $A^{k+m+1+r}B^{k-r}$ as well as
the coefficient of  $(-1)^m A^{k-r} B^{k+m+1+r}$ is

\begin{align*}
&\sum_{i=0}^m (-1)^{k+i} \binom{2k+1}{k+1+r+i}\binom{k+i}{k} \binom{k+m-i}{k} \\
&\qquad \qquad\qquad  - \sum_{n=m}^k (-1)^n \binom{k-m}{n-m}
\sum_i \binom{k-i}{r} \binom{k+i}{k} \binom{k+n-i}{k} \\
&=(-1)^k\binom{m+r}{m} \binom{2k+m+1}{k-r} \\
&\qquad \qquad\qquad  -\sum_{i=0}^{k-r} \binom{k-i}{r} \binom{k+i}{k}
\sum_{n=m}^k (-1)^n \binom{k-m}{n-m} \binom{k+n-i}{k} \\
&=(-1)^k\binom{m+r}{m} \binom{2k+m+1}{k-r}
- \sum_{i=0}^{k-r} (-1)^k \binom{k-i}{r} \binom{k+i}{k} \binom{k+m-i}{m} \\
&=0,
\end{align*}
using identities established in lemma 3.1.

\section{A local cohomology module with infinitely many associated prime
ideals}\label{elem}

Consider the hypersurface of mixed characteristic
$$
R={\mathbb Z}[U,V,W,X,Y,Z]/(UX+VY+WZ)
$$
and the ideal ${\mathfrak a}=(x,y,z)R$. We show that the local cohomology
module $H_{\mathfrak a}^3(R)$ has $p$-torsion elements for infinitely many
prime integers $p$, and consequently that it has infinitely many associated
prime ideals. We have
$$
H_{\mathfrak a}^3(R) = \varinjlim \frac{R}{(x^k, y^k, z^k)R}
$$
where the maps in the direct limit system are induced by multiplication by the
element $xyz$. Let $p$ be a prime integer. It is easily seen that
$$
\lambda=\frac{(ux)^p + (vy)^p + (wz)^p}{p}
$$
has integer coefficients, and we claim that the element
$$
\eta = [\lambda + (x^p, \ y^p, \ z^p)R] \ \in \ H_{\mathfrak a}^3(R)
$$
is nonzero and $p$-torsion. It is immediate that $p\cdot\eta = 0$, and what
really needs to be established is that $\eta$ is a nonzero element of
$H_{\mathfrak a}^3(R)$, i.e., that
$$
\lambda (xyz)^k \notin
(x^{p+k}, \ y^{p+k}, \ z^{p+k})R \ \text { for all } \ k \in {\mathbb N}.
$$
We shall accomplish this using an ${\mathbb N}^4$-grading. We assign weights as
follows:
\begin{align*}
u: (0,1,1,1)  \qquad \qquad\qquad  & x: (1,0,0,0) \\
v: (1,0,1,1)  \qquad \qquad\qquad  & y: (0,1,0,0) \\
w: (1,1,0,1)  \qquad \qquad\qquad  & z: (0,0,1,0)
\end{align*}
With this grading, $\lambda$ is a homogeneous element of degree $(p,p,p,p)$.
Hence in a homogeneous equation of the form
$$
\lambda (xyz)^{k}
=c_1x^{p+k}+c_2y^{p+k}+c_3z^{p+k},
$$
we must have
$$
\deg(c_1) = (0,p+k,p+k,p), \ \deg(c_2) = (p+k,0,p+k,p), \
\deg(c_3) = (p+k,p+k,0,p).
$$
We may use this to examine the monomials which can occur in $c_i$, and it is
easily seen that the only monomial that can occur in $c_1$ with a nonzero
coefficient is $u^p y^k z^k$, and
similarly that $c_2$ is an integer multiple of $v^p z^k x^k$ and $c_3$
is an integer multiple of $w^p x^k y^k$. Hence
$\lambda (xyz)^k \in (x^{p+k}, y^{p+k}, z^{p+k})R$ if and only if
\begin{align*}
\lambda (xyz)^k
&\in (u^p y^k z^k x^{p+k}, \ v^p z^k x^k y^{p+k}, \ w^p x^k y^k z^{p+k})R \\
&= (xyz)^k\Big((ux)^p, \ (vy)^p, \ (wz)^p\Big)R,
\end{align*}
i.e., if and only if $\lambda \in ((ux)^p, (vy)^p, (wz)^p)R$. To complete our
argument it suffices to show that $\lambda \notin (p,(ux)^p, (vy)^p, (wz)^p)R$,
i.e., that
$$
\frac{(ux)^p+(vy)^p+(-1)^p(ux+vy)^p}{p} \notin \Big(p,\ (ux)^p,\ (vy)^p \Big)R.
$$
After making the specializations $u \mapsto 1, \ v \mapsto 1, \ w \mapsto 1$,
it is enough to verify that
$$
\frac{x^p+y^p+(-1)^p(x+y)^p}{p} \notin (p, \ x^p, \ y^p){\mathbb Z}[x,y]
$$
and this holds since the coefficient of $x^{p-1}y$ in
$(x^p+y^p+(-1)^p(x+y)^p)/p$ is $(-1)^p$.

\section*{Acknowledgments}

The use of the Maple package {\tt EKHAD} is gratefully acknowledged. I would
like to thank Mel Hochster for first bringing the conjecture to my attention,
and I am also grateful to Ian Aberbach and Reinhold H\"ubl for pointing out an
error in an earlier version of this manuscript.

config: arxiv


text:
\begin{document}
\settowidth{\mywidth}{ab}
\title{The Gradient Flow of Infinity-Harmonic Potentials}
\author{Erik Lindgren, Peter Lindqvist}

\date{\today}
\maketitle

\medskip
    {\small \textsc{Abstract:} \textsf{We study the streamlines of $\infty$-harmonic functions in planar convex rings. We include convex polygons.    The points where streamlines can meet are characterized: they lie on certain curves.
        The gradient has constant norm along streamlines outside the set of meeting points,  the \emph{infinity-ridge}.
}}

\bigskip

\tableofcontents

\bigskip
\bigskip

{\small \textsf{AMS Classification 2010}: 49N60, 35J15, 35J60, 35J65, 35J70.}

{\small \textsf{Keywords}: Infinity-Laplace Equation, streamlines, convex rings, infinity-potential function}
\section{Introduction}
The $\infty$-Laplace Equation
$$
 \Delta_{\infty}u\,\equiv\,\sum_{i,j}\frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j}\frac{\partial^2 u}{\partial x_i \partial x_j}\,=\,0
 $$
was introduced by G. Aronsson in 1967 (cf. \cite{A1}) to produce optimal Lip\-schitz extensions of boundary values. It has been extensively studied. Some of the highlights are
\begin{itemize}
\item Viscosity solutions for $\Delta_{\infty}$, \cite{BDM}
\item Uniqueness, \cite{J}
\item Differentiability, \cite{S}, \cite{ESa} and \cite{ES}
\item Tug-of-War (connection with stochastic game theory), \cite{PSW}
\end{itemize}
We are interested in the two-dimensional equation
$$
  \Bigl(\frac{\partial u}{\partial x_1}\Bigr)^{\!2}\frac{\partial ^2u}{\partial x_1^2}\,+\,2\, \frac{\partial u}{\partial x_1}\frac{\partial u}{\partial x_2}\frac{\partial ^2u}{\partial x_1 \partial x_2}\,+\, \Bigl(\frac{\partial u}{\partial x_2}\Bigr)^{\!2}\frac{\partial ^2u}{\partial x_2^2}\,=\,0
  $$
  in so-called convex ring domains $G = \Omega\setminus K$. Here $\Omega$ is a bounded convex domain in $\R^2$ and $K\Subset \Omega$ is a closed convex set.
  We continue our investigation in \cite{LL} of the $\infty$-potential $u_\infty$, which is the unique solution in $C(\overline G)$ of the boundary value problem
$$
    \begin{cases} \Delta_{\infty}u\,=\,0\qquad\text{in}\qquad G\\ \phantom{ \Delta_{\infty}}
      u\,=\,0\qquad\text{on}\qquad \partial \Omega\\\phantom {\Delta_{\infty}}
      u\,=\,1 \qquad\text{on}\qquad \partial K.
    \end{cases}
  $$
In \cite{LL} we proved that the \emph{ascending} streamlines, the solutions $\ba= (\alpha_1,\alpha_2)$ of
  $$
  \frac{d\ba(t)}{dt} = +\nabla u_\infty (\ba(t)), \quad 0\leq t < T_{\ba}
  $$
  with given initial point $\ba(0)\in \overline \Omega\setminus K$, are unique and terminate at $\partial K$. (The descending ones are not!) Streamlines may meet and then continue along a common arc. Uniqueness prevents crossing streamlines.

  Along a streamline one would expect that the speed $|\nabla u_\infty (\ba)|$ is constant. Indeed,
  $$
  \frac{d}{dt} |\nabla u_\infty (\ba(t))|^2 = 2\,\Delta_\infty u_\infty (\ba (t)) = 0,
  $$
  but the calculation requires second derivatives. The main difficulty is the lack of second derivatives. Although, the second derivatives are known to exist  almost everywhere with respect to the Lebesgue area, see \cite{KZZ} for this new result, this is of little use since the area of a streamline is zero. In \cite{LL} it was shown that the above calculation fails: for most streamlines the speed is not constant the whole way up to $\partial K$. (We shall see that the speed is constant from the initial point till the streamline meets another streamline.)

  We use the approximation with the (unique) solution of the $p$-Laplace equation
  $$
\Delta_p u = \ddiv (|\nabla u|^{p-2}\nabla u) = 0, \qquad p > 2.
 $$
 in $G$ with the same boundary values as $u_{\infty}$.

We shall use several facts about these $p$-harmonic functions due to J. Lewis, cf.  \cite{L}. It is decisive that the \emph{level curves} $\{u_p(x)=c\}$ are convex and that $\Delta u_p \leq 0$. See Section \ref{sec:prel} for more details.

  We also need the facts that (i) $\nabla u_p\to \nabla u_\infty$ in $L^2_\text{loc}$ and (ii) the family $\{|\nabla u_p|\}$ is locally equicontinuous. (Notice that we wrote $|\nabla u_p|$, not $\nabla u_p$.) We extract a proof of this from the recent pathbreaking work by H. Koch, Y. R-Y. Zhang and Y. Zhou in \cite{KZZ}, complementing  their results by applying a simple device, due to Lebesgue in \cite{Leb}, to the norm $|\nabla u_p|$ of the quasiregular mapping
  $$
  \frac{\partial u_p}{\partial x_1}-  i\frac{\partial u_p}{\partial x_2}, \quad i^2=-1.
  $$The quasiregularity was obtained by B. Bojarski and T. Iwaniec in \cite{BI}.

We prove the following basic result in Section \ref{sec:eqcont}.

    \begin{thm}[Non-decreasing speed]  \label{thm:speed} Let $\ba_\infty = \ba_\infty(t)$, $0\leq t\leq T$, be a streamline of $u_\infty$, i.e.,
  $$
  \frac{d\ba_\infty(t)}{dt} = \nabla u_\infty (\ba_\infty(t)), \quad 0\leq t< T,
  $$
  and $\ba_\infty(0)\in \partial \Omega$, $\ba_\infty(T)\in \partial K$. Then the function $u_\infty(\ba_\infty(t))$ is  convex when  $0\leq t\leq T$. In particular, the speed $|\nabla u_\infty (\ba_\infty(t))|$, is a non-decreasing function of $t$.
  \end{thm}

   Combining this with a result in the opposite direction (cf. Lemma 12 in \cite{LL}), we can control the meeting points so that these lie on a few specific streamlines, here called attracting streamlines.

  \paragraph{Polygons.} To avoid a complicated description, we begin with a convex polygon as $\Omega$ with $N$ vertices $P_1,P_2,\ldots, P_N$ (set $P_{N+1}=P_1$ for convenience). With $P_k = \bg_k(0)$ as initial point there is a unique streamline
  $$
  \bg_k = \bg_k(t), \quad  0\leq t \leq T_k,
  $$
  with terminal point $\bg_k(T_k)$ on $\partial K$. The $$\text{\emph{attracting streamlines} are}\qquad \bg_1, \bg_2, \ldots, \bg_N.$$
  Occasionally, some of them meet and then share a common arc up to $\partial K$. The collection of all the points on the attracting streamlines is called the $\infty$-\emph{ridge} and is denoted by $\Gamma$, i.e.,
  $$
  \Gamma = \bigcup_{k=1}^N \{\bg_k(t): \,\, 0\leq t\leq T_k\}.
  $$
It seems to play a similar role for the $\infty$-Laplace Equation as the (ordinary) ridge does for the Eikonal Equation.

Before meeting any other streamline, a streamline $\ba$ either meets an attracting streamline or hits the upper boundary $\partial K$. We formulate this as a theorem,  proved in Section \ref{sec:poly}.
\begin{thm}\label{thm:mainpoly} The speed  $|\nabla u_\infty(\ba(t))|$ is constant along the streamline $\ba$ from the initial point on $\partial \Omega$ until it meets one of the attracting streamlines $\bg_k$, after which the speed is non-decreasing. It cannot meet any other streamline before it meets an attracting one.
\end{thm}

Thus there are no meeting points in $G\setminus \Gamma$, i.e., they all lie on the attracting streamlines $\bg_1, \bg_2, \ldots, \bg_N$. In other words, there is no branching outside the $\infty$-ridge $\Gamma$.

\paragraph{General Domains.}  The polygon has a piecewise smooth boundary and at the vertices $|\nabla u_\infty(P_k)|= 0$. Thus the attracting streamlines start at the points of minimal speed. Similar results hold when $\Omega$ is no longer a polygon, but now we have to assume that the following holds:

\bigskip
\noindent\textbf{Assumptions:}
\begin{enumerate}
\item\emph{$\nabla u_\infty$ is continuous in $\overline \Omega\setminus K$, in particular along $\partial \Omega$}.\footnote{For example, if $\partial \Omega$ is piecewise $C^2$, then the gradient is continuous in $\overline \Omega\setminus K$, see Section \ref{sec:prel}.}
\item \emph{On $\partial \Omega$, the continuous function $|\nabla u_\infty|$ has a finite number of local minimum points, say $P_1,P_2,\ldots, P_N$, and a finite number of local maximum points.}
\end{enumerate}

  Again, the streamlines with the initial points $P_k$ are called \emph{attracting streamlines}:
   $$
  \bg_k = \bg_k(t), \quad  0\leq t\leq T_k ; \quad \bg_k(0)=P_k.
  $$
  The $\infty$-\emph{ridge} is again
  $$
  \Gamma = \bigcup_{k=1}^N \{\bg_k(t): \,\, 0\leq t\leq T_k\}.
  $$
  Theorem \ref{thm:mainpoly} holds also in this setting. As a consequence, streamlines cannot meet, except on $\Gamma$. The theorem below is proved in Section \ref{sec:general}.
  \begin{thm}\label{thm:general} The speed  $|\nabla u_\infty(\ba(t))|$ is constant along a streamline $\ba$ from the initial point on $\partial \Omega$ until it meets one of the attracting streamlines $\bg_k$. It cannot meet any other streamline before it meets an attracting one.
  \end{thm}

  The situation when $|\nabla u_{\infty}|$ is constant on some arc on $\partial \Omega$ can happen even for a rectangle, but does not cause extra complications.

  \begin{prop} If the speed $|\nabla u_{\infty}|$ is constant along a boundary arc $\overline{ab}$, then the streamlines with initial points on the arc are non-intersecting segments of straight lines. They meet no other streamlines in $G$, except possibly when the initial point is $a$ or $b$.
  \end{prop}

  This follows from Lemma \ref{lem:constant} and Lemma \ref{lem:trilem}. It  allows us to relax assumption 2 to include boundary arcs with constant local maximum speed:
  \begin{itemize}
  \item[2*.]
  \emph{The local maxima and minima of  $|\nabla u_{\infty}|$   on $\partial \Omega$ are attained along at most finitely many closed subarcs, which may degenerate  to points.}
  \end{itemize}

  The definition of the attracting streamlines must be amended if the speed attains a local minimum along a boundary arc $\overline{ab}$: it contributes with \emph{two} attracting streamlines, namely the ones with initial points at $a$ and $b$.

 \begin{rem}  The behavior of the streamlines suggests that the $\infty$-potential is smooth outside the $\infty$-ridge $\Gamma$.
  \end{rem}

  \paragraph{Examples.}
  We mention some examples.
\begin{figure}[h!]
\begin{center}
\includegraphics[scale=0.05]{betterstreamlines.jpg}
\caption{The streamlines of $u_\infty$ when $\Omega$ is the square in Example \ref{ex:square}.}
\label{fig:square}
\end{center}
\end{figure}

\begin{ex}\label{ex:square}

Let $\Omega$ be the square
$$
-1<x_1<1, \quad -1<x_2<1, \quad
$$
and $K$ the origin. The attracting streamlines are the four half-diagonals, constituting the $\infty$-ridge
$$
\Gamma = \{(x_1,x_2): \quad x_1=\pm x_2,  |x_1|\leq 1,  |x_2|\leq 1\}.
$$
All streamlines meet at a diagonal, except the four segments along the coordinate axes. See Figure \ref{fig:square}.
\end{ex}

\begin{ex}\label{ex:modsq} Let $K$ be the origin and $\Omega$ the square in Example \ref{ex:square} which is truncated in the following symmetric way: in the south west corner we have removed the triangle with corners $(-1,-1), (-1+\delta,-1)$ and $(-1,-1+\delta)$, for some small $\delta$. See Figure \ref{fig:modsq}. We only describe the behavior in the south west quarter of $\Omega$.

The attracting streamlines are those starting in $(-1+\delta,-1)$ and $(-1,-1+\delta)$ (in blue). The only streamlines that do not meet any other before reaching origin, are the medians (in red). Any other streamline will meet one of the attracting streamlines. The streamline starting in the middle of $(-1+\delta,-1)$ and $(-1,-1+\delta)$ (in red) will be a straight line to the origin and will be joined by the attracting streamlines from both sides before terminating at the origin.

\end{ex}

 \begin{figure}[h!]

  \begin{center}

	\begin{tikzpicture}[domain=-3:3,scale=2.5]

\draw[thick] (-1,-0.8) -- (-0.8,-1) -- (0.8,-1) -- (1,-0.8) -- (1,0.8) -- (0.8,1) -- (-0.8,1) -- (-1,0.8) -- (-1,-0.8);
\draw[color=red] (-1,0) -- (0,0);
\draw[color=red] (0,-1) -- (0,0);
\draw[color=blue] (-0.8,-1)  to[out=65, in =-135] (-0.3,-0.3);
\draw[color=blue] (-1,-0.8) to[out=25, in =-135] (-0.3,-0.3);
\draw[color=red] (-0.9,-0.9) -- (0,0);

 \fill (-0.4,0.6) node[left] {$\Omega$};
\fill (0,0) circle[radius=1.5pt] node[right] {$(0,0)$};

	\end{tikzpicture}

\end{center}
\caption{The truncated square in Example \ref{ex:modsq} and some possible streamlines.}
 \label{fig:modsq}
	\end{figure}

  \section{Preliminaries} \label{sec:prel} $\Omega$ is a bounded convex domain in $\R^2$  and $K\Subset \Omega$ is a compact and convex set, which may reduce to a point. We study the equation in the convex ring $G=\Omega\setminus K$. We assume the following \emph{normalization}:
  $$\boxed{\dist(\partial \Omega,K)=1.}$$
  The boundary value problem
$$
 \begin{cases} \Delta_{\infty}u\,=\,0\qquad\text{in}\qquad G,\\ \phantom{ \Delta_{\infty}}
      u\,=\,0\qquad\text{on}\qquad \partial \Omega,\\\phantom {\Delta_{\infty}}
      u\,=\,1 \qquad\text{on}\qquad \partial K,
    \end{cases}
    $$
has a unique solution $u_\infty \in C(\overline G)$ in general. By \cite{ESa}, $\nabla u_\infty$ is locally H\"older continuous in $G$. We will assume that also $\nabla u_\infty\in  C(\overline\Omega\setminus K)$. This is fulfilled if for instance $\partial \Omega$ has a piecewise $C^2$ regular boundary. See Lemma 2 and Theorem 2 in \cite{HL}, Theorem 7.1 in \cite{MPS} and Theorem 1 in \cite{WY}.

In \cite{LL} it was established that, for a given initial point $\xi_0\in \partial\Omega$, the gradient flow
$$
\begin{cases}
\displaystyle\frac{d \ba (t)}{d t}& =\, +\nabla u_\infty(\ba (t)),\quad    0\leq t< T, \\
\ba(0)&=\,\xi_0,
\end{cases}
$$
has a unique solution $\ba=\ba(t)$, which terminates at some point $\ba(T)$ on $\partial K$. (Some caution is required if $|\nabla u_\infty(\xi_0)| = 0$.) We say that $\ba$ is a \emph{streamline}. Although unique, two streamlines may meet, join, and continue along a common arc.

We shall employ the $p$-harmonic approximation
$$
   \begin{cases} \Delta_{p}u_p\,=\,0\qquad\text{in}\qquad G,\\ \phantom{ \Delta_{p}}
      u_p\,=\,0\qquad\text{on}\qquad \partial \Omega,\\\phantom {\Delta_{p}}
      u_p\,=\,1 \qquad\text{on}\qquad \partial K,
    \end{cases}
  $$
  for $\mathbf{p>2}$. It is known that $u_p\in C(\overline G)$ and it takes the correct values (in the classical sense) at each boundary point. We shall need the following results from \cite{L} (see also \cite{Ja}):
  \begin{enumerate}
  \item The level curves $\{u_p=c\}$ are convex, if $0\leq c\leq 1$,
  \item $u_p\nearrow u_\infty$ uniformly in $\overline G$,
  \item $|\nabla u_p|\neq 0$ in $G$,
  \item $u_p$ is real analytic in $G$,
  \item $\Delta u_p\leq 0$.
  \end{enumerate}
  The streamlines of $u_p$ do not meet in $G$. This is due to the regularity of $u_p$   and the Picard-Lindel\"of theorem. Properties 1), 3), and 5) are preserved at the limit $p=\infty$. Especially, $\nabla u_{\infty} \neq 0$ in $G$.

 We keep the \emph{normalization}  $\dist (\partial\Omega,  K) = 1$. Then $|\nabla u_\infty|\leq 1$, but we also need a uniform bound for $|\nabla u_p|$.
   The bound
\begin{equation}
\label{eq:grad1}
|\nabla u_p|\leq 1 \quad \text{on }\partial \Omega.
\end{equation}
   follows by comparison with the distance function
  $$
  \delta(x)=\dist(x,\partial \Omega).
  $$
  In a convex domain, $\delta$ is a supersolution of the $p$-Laplace equation. Since
  $$
  0\leq u_p(x)\leq \delta(x)\quad \text{on } \partial G,
  $$
the same inequality also holds in $G$. In general, $|\nabla u_p|$ is unbounded (but $|\nabla u_\infty|\leq 1$), so we have to consider a subdomain, say $\{u_p < c\}$.

  \begin{lemma}\label{lem:gradbound} The uniform bound
  \begin{equation}
  \label{eq:gradboundlem}
|\nabla u_p(x)|\leq \Bigl(\frac{1}{1-c}\Bigr)^\frac{1}{p-2}
  \end{equation}
  holds when $u_p(x)\leq c$, $0<c<1$.
  \end{lemma}

  \begin{proof}
  Let $\Upsilon_p(c)$ denote the level curve $\{u_p=c\}$ and
  $$
  \delta_p(x) = \dist (x,\Upsilon_p(c)).
  $$
  Since $|\nabla u_p|$ obeys the maximum principle and $|\nabla u_p|\leq 1$ on $\partial \Omega$ by \eqref{eq:grad1}, it is enough to control $|\nabla u_p|$ on $\Upsilon_p(c)$. We see that
  \begin{equation}
  \label{eq:grad2}
  c\leq u_p(x)\leq c+(1-c)\frac{\delta_p(x)}{\dist(\Upsilon_p(c), \partial K)}
  \end{equation}
   on $\Upsilon_p(c)$ and on $\partial K$, i.e., on the boundary of $\{1 >u_p>c\}$. Again, the majorant is a supersolution to the $p$-Laplace equation, and hence \eqref{eq:grad2} holds in $\{1 >u_p>c\}$ by the comparison principle. It follows that
  \begin{equation}
  \label{eq:gradbound}
  |\nabla u_p(x)|\leq \frac{1-c}{\dist(\Upsilon_p(c),\partial K)},
  \end{equation}
  on\footnote{Since $u_p\nearrow u_\infty$, $\dist(\Upsilon_p(c),\partial K)$ increases with $p$. Thus we get an upper bound independent of $p$. This is sufficient for our purpose.} $\Upsilon_p(c)$.

  To get the explicit upper bound in \eqref{eq:gradboundlem}, we assume that $x_0\in \partial K$ is a point at which the distance $\dist(\Upsilon_p(c),\partial K)$ is attained. Let $R$ be the radius of the  largest ball $B_R(x_0)\subset \Omega$. Then
  $$
  u_p(x)\geq 1-\left(\frac{|x-x_0|}{R}\right)^\frac{p-2}{p-1}\quad \text{in }B_R(x_0)\setminus K
  $$
  by comparison. Here the minorant is $p$-harmonic in $B_R(x_0)\setminus \{x_0\}$. Now
  $$
    1-\left(\frac{|x-x_0|}{R}\right)^\frac{p-2}{p-1} = c \iff |x-x_0|=R(1-c)^{1+\frac{1}{p-2}} = r_c
  $$
  and clearly $\dist(\Upsilon_p(c),\partial K)\geq r_c$. We have by \eqref{eq:gradbound}
  $$
  |\nabla u_p(x)|\leq \frac{1}{R(1-c)^\frac{1}{p-2}}.
  $$
  To conclude, use $R\geq \dist(\partial\Omega, \partial K)=1$.
  \end{proof}

\section{Equicontinuity of $|\nabla u_p|$}\label{sec:eqcont}

We shall prove that
$$
\lim_{p\to \infty}{|\nabla u_p|} = |\nabla u_\infty|
$$
locally \emph{uniformly} in $G$. From \cite{KZZ} we can extract the following important properties: If $D\Subset G$, then
\[\tag{\textbf{I}}
\iint_D |\nabla u_p-\nabla u_\infty|^2 \, dx_1 dx_2\to 0,\quad \text{as }p\to \infty,\]

\begin{equation}\tag{\textbf{J}}

\iint_D |\nabla (|\nabla u_p|^2)|^2 \, dx_1 dx_2\leq M_D<\infty,\end{equation}

for all (large) $p$.

The constant $M_D$ depends on $\|\nabla u_p\|_{L^\infty(E)}$, where $D\Subset E\Subset G$, and $\dist (D, \partial G)$, but not on $p$.

In \cite{KZZ} the estimates were derived  for solutions $u^\e$ of the auxiliary equation
$$
\Delta_\infty u^\e + \e \Delta u^\e = 0
$$
while we use $\Delta_pu_p = 0$ written as
$$

\Delta_\infty u_p +\frac{1}{p-2}\,|\nabla u_p|^2\Delta u_p = 0.
$$
The advantage of our approach is that the inequality $\Delta u_p\leq 0$ is available in convex domains for $p\geq 2$.

The conversion from $u^\e$ to $u_p$ requires only obvious changes. Formally, the factor $\e$ in front of an integral in \cite{KZZ} should be moved in under the integral sign and then replaced by $|\nabla u_p|^2/(p-2)$, upon which every $u^\e$ be replaced by $u_p$. This procedure is explained in our Appendix.

In order to prove that the family $\{|\nabla u_p|\}$ is locally equicontinuous, we shall use a device due to Lebesgue in \cite{Leb}. A function $f\in C(\overline B_R)\cap W^{1,2}(B_R)$ is monotone (in the sense of Lebesgue) if
$$
\osc_{\partial B_r}f = \osc_{\overline B_r} f, \quad 0<r<R,
$$
where $B_r$ are concentric discs.  For such a function
\begin{equation}\label{eq:lebosc}
\Bigl(\osc_{B_r} f\Bigr)^2\ln \frac{R}{r}\leq \pi \iint_{B_R} |\nabla f|^2 \, dx_1 dx_2.
\end{equation}
The proof is merely an integration in polar coordinates, cf. \cite{Leb}. We shall apply this oscillation lemma on the function $f = |\nabla u_p|^2$. It was shown by Bojarski and Iwaniec in \cite{BI} that the mapping
 $$
  \frac{\partial u_p}{\partial x_1}-  i\,\frac{\partial u_p}{\partial x_2}, \quad i^2=-1,
  $$
  is quasiregular. That property implies that its norm  $|\nabla u_p|$ satisfies the maximum principle, and, where $|\nabla u_p|\neq 0$, also the minimum principle. Thus $|\nabla u_p|$ is monotone. So is $|\nabla u_p|^2$. From \eqref{eq:lebosc} we obtain
  $$
  \Bigl(\osc_{B_r}\{|\nabla u_p|^2\}\Bigr)^2\ln \frac{R}{r}\leq \pi \iint_{B_R} |\nabla (|\nabla u_p|^2)|^2 \, dx_1 dx_2.
  $$
  The uniform bound in (2) and a standard covering argument for compact sets yields the following result.

  \begin{thm} \label{thm:eqcont}(Equicontinuity) Let $D\Subset G$. Given $\e>0$, there is $\delta = \delta (\e,D)$ such that the inequality
  $$
\Big||\nabla u_p(x)|-|\nabla u_p(y)|\Big|<\e \quad \text{when $|x-y|<\delta$}, \quad x,y\in D,
  $$
  holds simultaneously for all $p > 2$.
  \end{thm}
  Since $\nabla u _p\to \nabla u_\infty$ in $L^2_\text{loc}(G)$ we can use Ascoli's theorem to conclude that
  $$
  \lim_{p\to \infty}|\nabla u_p| = |\nabla u_\infty|
  $$
  \emph{locally uniformly}. (More accurately, we have to extract a subsequence in Ascoli's theorem, but since the limit $|\nabla u_\infty|$ is unique, this precaution is not called for here.)

  \textbf{Caution:} The more demanding convergence $\nabla u_p\to \nabla u_\infty$ holds a.e., but perhaps \emph{not} locally uniformly.

  Let us finally mention that the \emph{uniform} convergence is not global. For example, in the ring $0<|x|<1$ we have
  $$
  u_p(x)=1-|x|^\frac{p-2}{p-1}, \quad u_\infty = 1-|x|.
  $$
  Now $|\nabla u_p|$ is not even bounded near $x=0$. Thus the convergence cannot be uniform in the whole ring.

  \section{Convergence of the Streamlines}
  In this section, we study the convergence of the streamlines and prove Theorem \ref{thm:speed}.
 It is plain that the level curves $\{u_p=c\}$ converge to the level curves $\{u_\infty = c\}$.  However, the convergence of the streamlines requires a more sophisticated proof. (The problem is the identification of the limit as an $\infty$-streamline.)

  Suppose that we have the streamlines $\ba_p$ and $\ba_\infty$ having the same initial point $\ba_p(0)=\ba_\infty(0)=x_0$. Now

$$
  \frac{d\ba_p(t)}{dt} = \nabla u_p (\ba_p(t)), \quad  \frac{d\ba_\infty(t)}{dt} = \nabla u_\infty (\ba_\infty(t))
  $$
  when $0 < t < T_p$, where $u_p(\ba_p(T_p))= 1$. Thus

 $$
  \ba_p(t_2)-\ba_p(t_1) = \int_{t_1}^{t_2} \nabla u_p(\ba_p(t)) dt.
$$
Using the bound
$$
|\nabla u_p|\leq \Bigl(\frac{1}{1-c}\Bigr)^{\frac{1}{p-2}}	, \quad \text{when }u_p\leq c,
$$
in Lemma \ref{lem:gradbound} we see that
\begin{equation}
\label{eq:alpha3}
| \ba_p(t_2)-\ba_p(t_1)|\leq \Bigl(\frac{1}{1-c}\Bigr)^\frac{1}{p-2}|t_2-t_1|
\end{equation}
as long as the curves are below the level $u_p=c$, i.e., $u_p(\ba(t_2))\leq c$. In particular, the bound is valid in the domain $\{u_\infty<c\}$, where $c<1$. Thus, the family of curves is locally equicontinuous. By Ascoli's theorem we can extract a sequence $p_j\to \infty$ such that
$$
\ba_{p_j}(t)\to \ba(t)
$$
uniformly in every domain $\{u_\infty<c\}$. Here $\ba(t)$ is some curve with initial point $\ba(0)=x_0$.

The endpoint of $\ba$ is on $\partial K$. Indeed, let $t_p=t_p(c)$ denote the parameter value at which $u_p(\ba_p(t_p))=c$. Take any convergent sequence, say $t_p\to t^*$. Then
$$
c=\lim_{p\to\infty} u_p(\ba_p(t_p)) = u_\infty(\ba(t^*)).
$$
Thus $t^* = t_\infty(c)$. Then $t_p(c)\to t_\infty(c)$ for all $c$.

 By \eqref{eq:alpha3}
$$
| \ba(t_2)-\ba(t_1)|\leq |t_2-t_1|.
$$
Rademacher's theorem for Lipschitz continuous functions implies that $\ba(t)$ is differentiable at a.e. $t$.

We claim that $\ba = \ba_\infty$. Since they start at the same point, the uniqueness of $\infty$-streamlines shows that it is enough to verify
$$
\frac{d\ba(t)}{dt} = \nabla u_\infty (\ba(t)).
$$
To this end, we shall employ the convex functions $F_p(t)=u_p(\ba_p(t))$. Indeed,
$$
\frac{dF_p(t)}{dt} = \Big\langle \nabla u_p(\ba_p(t)), \frac{d\ba_p(t)}{dt}\Big\rangle = |\nabla u_p(\ba_p(t))|^2
$$
and
$$
\frac{d^2F_p(t)}{dt^2} = 2\,\Delta_\infty u_p(\ba_p(t)) = -\frac{2}{p-2}\,\Delta u_p(\ba_p(t))\,|\nabla u_p(\ba_p(t))|^2.
$$
By Lewis's theorem, $\Delta u_p\leq 0$ in convex ring domains, if $p\geq 2$. Thus,
$$
\frac{d^2F_p(t)}{dt^2}\geq 0
$$
and so \emph{the function $F_p(t)$ is convex}. The convergence
$$
F_{p}(t) = u_{p}(\ba_{p}(t))\to u_\infty(\ba(t)) = F(t)
$$
is at least locally uniform, when $p$ takes the values $p_1, p_2, p_3,\ldots$ extracted above. Also the limit $F(t)$ is convex, of course.

We have the locally uniform convergence
$$
 |\nabla u_{p}(\ba_p(t))|^2\to  |\nabla u_\infty(\ba(t))|^2,
$$
which follows from Theorem \ref{thm:eqcont} by writing
$$
|\nabla u_p(\ba_p(t))|-|\nabla u_\infty(\ba(t))| = |\nabla u_p(\ba_p(t))|-|\nabla u_p(\ba(t))|+|\nabla u_p(\ba(t))|-|\nabla u_\infty(\ba(t))|.
$$
Thus,
$$
\frac{dF_p(t)}{dt}=|\nabla u_{p}(\ba_p(t))|^2\,\to\, |\nabla u_\infty(\ba(t))|^2.
$$
It follows that\footnote{$\int |\nabla u_\infty(\ba(t))|^2\phi(t) dt\leftarrow  \int F_p'(t)\phi(t) dt = -\int  F_p(t)\phi'(t) dt \to -\int F(t) \phi'(t) dt$}  $F'(t) = |\nabla u_\infty(\ba(t))|^2$ for a.e. $t$. We also have by the chain rule
$$
\frac{dF(t)}{dt} = \Big\langle\nabla u_\infty(\ba(t)), \frac{d\ba}{dt}\Big\rangle
$$
a.e., since $\frac{d\ba}{dt}$ exists for a.e. $t$.

We have arrived at the identity
$$
|\nabla u_\infty(\ba(t))|^2 = \Big\langle\nabla u_\infty(\ba(t)), \frac{d\ba}{dt}\Big\rangle
$$
valid for a.e. $t$. From
$$
\ba_p(t_2)-\ba_p(t_1) \leq \int_{t_1}^{t_2} |\nabla u_p(\ba_p(t)) |dt,
$$
we get
$$
\ba(t_2)-\ba(t_1) \leq \int_{t_1}^{t_2} |\nabla u_\infty(\ba(t)) |dt,
$$
  and, hence for a.e. $t$
  $$
  \Big\vert\frac{d\ba(t)}{dt}\Big\vert \leq |\nabla u_\infty(\ba(t)) |.
  $$
  We conclude that in the Cauchy-Schwarz inequality
  $$
  |\nabla u_\infty(\ba(t)) |^2 = \Big\langle\nabla u_\infty(\ba(t)), \frac{d\ba}{dt}\Big\rangle \leq |\nabla u_\infty(\ba(t))|\Big\vert \frac{d\ba}{dt}\Big\vert\leq |\nabla u_\infty(\ba(t)) |^2
  $$
  we have equality. It follows that
  $$
 \frac{d\ba}{dt}= \nabla u_\infty(\ba(t))
  $$
  for a.e. $t$. In fact, it holds everywhere because now the identity
  $$
  \ba(t_2)-\ba(t_1) = \int_{t_1}^{t_2} \nabla u_\infty(\ba(t)) dt
  $$
  can be differentiated. This concludes our proof of the fact $\ba = \ba_\infty$.

  We see that the tangent $\frac{d\ba}{dt}$ is continuous. The proof reveals that the convex functions $F_p\to F$ uniformly and hence $F$ is convex as well. Therefore, its derivative
  $$
  F'(t)= |\nabla u_\infty(\ba(t))|^2
  $$ is non-decreasing. In other words, $|\nabla u_\infty|^2$ is non-decreasing along the limit streamline.

 This proves Theorem \ref{thm:speed}.

  \section{Quadrilaterals and Triangles}
  Curved quadrilaterals and triangles, bounded by arcs of streamlines and level curves, are useful building blocks. It is tentatively understood that at least the interior of the figures are comprised in $G$; the level arcs  can be   on $\partial \Omega$ and, occasionally, on   $\partial K$.

  Recall that the $\infty$-streamline
  $$
  \ba(t), \quad 0\leq t\leq T,
  $$
  with initial point $\ba(0)=a\in \partial\Omega$ is unique and terminates at $\ba(T)$ on $\partial K$. On its way, it may (and usually does) meet other streamlines and has common parts with them. By Theorem \ref{thm:speed}, the speed
  $$
  \left|\frac{d\ba(t)}{d t}\right| = |\nabla u_\infty (\ba(t))|
  $$
  is non-decreasing. Thus we have the bound\footnote{$$|\nabla u_{\infty}(\ba(T))| = \lim_{t\to T-}  |\nabla u_{\infty}(\ba(t))|$$}
  $$
  |\nabla u_\infty (\ba(t_1))|\leq |\nabla u_\infty (\ba(t_2))|, \quad 0\leq t_1\leq t_2\leq T.
  $$
  Sometimes the result below (cf. Lemma 12 in \cite{LL}), valid for curved quadrilaterals and triangles, provides us with the reverse inequality, so that we may even conclude that the speed is constant along suitable arcs of streamlines.
\begin{figure}[h!]

  \begin{center}

	\begin{tikzpicture}[domain=-5:5,scale=1.2]

\fill (3,0) circle[radius=2pt] node[below] {$b$};
\fill (-3,0) circle[radius=2pt] node[below] {$a$};
\fill (2,4) circle[radius=2pt] node[above] {$b'$};
\fill (-2,4) circle[radius=2pt] node[above] {$a'$};
\fill (0.5,3.25) circle[radius=2pt] node[above] {$\eta$};
\fill (0,-0.75) circle[radius=2pt] node[below] {$\xi$};
\draw (0,-0.75) .. controls (0.3,1) .. (0.5,3.25); \fill (0.3,1) node[left] {$\bm$};
\draw (-3,0) .. controls (0,-1) .. (3,0);\fill (-1.2,-0.8) node[below] {$\bs$};

\draw (-3,0) .. controls (-2,2) .. (-2,4);\fill (-2.2,2) node[left] {$\ba$};
\draw (3,0) .. controls (3,2) .. (2,4);\fill (3,2) node[right] {$\bbeta$};
\draw (2,4) .. controls (0.5,3) .. (-2,4); \fill (-0.5,3.6) node[above] {$\bo$};

	\end{tikzpicture}

\end{center}
\caption{The quadrilateral $abb'a'$.}
  \label{fig:quadbasic}
	\end{figure}
  \begin{lemma} \label{lem:ll} Suppose that the streamlines $\ba$ and $\bbeta$ together with the level curves $\bs$ (lower level) and $\bo$ (upper level) form a quadrilateral with vertices $a,b,b'$ and $a'$. If $\ba$ and $\bbeta$ do not meet before reaching $\omega$, then
  $$
  \max_{\overline{a'b'}} |\nabla u_\infty(\bo)|\leq  \max_{\overline{ab}} |\nabla u_\infty(\bs)|,
  $$
  i.e., the maximal speed on the upper level is the smaller one.
  \end{lemma}

  Suppose now that $\xi\in \overline{ab}$ is a point on the lower level curve $\bs$ at which
  $$
  |\nabla u_\infty(\xi)|= \max_{\overline{ab}} |\nabla u_\infty(\bs)| = M.
  $$
  Let $\bm$ be the streamline that passes through $\xi$. It intersects $\bo$ at some point $\eta\in \overline{a'b'}$ (it may have joined $\ba$ or $\bbeta$ before reaching $\eta$). See Figure \ref{fig:quadbasic}. The following result holds:

\begin{lemma}\label{lem:quad}
We have
$$
|\nabla u_\infty(\bm)| = M \quad \text{on } \overline{\xi\eta}.
$$
Moreover,
$$
\max_{\overline{a'b'}} |\nabla u_\infty(\bo)|= \max_{\overline{ab}} |\nabla u_\infty(\bs)|.
$$
  \end{lemma}
  \begin{proof} By Lemma \ref{lem:ll}
  $$
  |\nabla u_\infty(\xi)|\geq \max_{\overline{a'b'}} |\nabla u_\infty(\bo)|\geq |\nabla u_\infty(\eta)|
  $$
  and the monotonicity of the speed implies
  $$
 |\nabla u_\infty(\xi)|\leq |\nabla u_\infty(\bm(t))|\leq |\nabla u_\infty(\eta)|
  $$
  along the arc $\overline{\xi\eta}$ of $\bm$. Thus we have equality.
  \end{proof}

We can also formulate a similar result for curved triangles. Suppose that the streamlines $\ba$ and $\bbeta$ together with the level curve $\bs$ form a curved triangle with vertices $a,b$ and $c$. Assume again that $\xi\in \overline{ab}$ is a point at which
  $$
  |\nabla u_\infty(\xi)|= \max_{\overline{ab}} |\nabla u_\infty(\bs)| = M.
  $$
  Let $\bm$ be the streamline that passes through $\xi$. It passes through $c$ (but may have joined $\ba$ or $\bbeta$ before reaching $c$). The following result holds:

\begin{cor}\label{cor:tri} For the triangle $a\,b\,c$
we have
$$
|\nabla u_\infty(\bm)| = M \quad \text{on } \overline{\xi c}.
$$
Moreover,
$$
 |\nabla u_\infty(c)|= \underset{\overline{ab}}{\max} |\nabla u_\infty(\bs)|.
$$
\end{cor}
\begin{proof} Take $\bo_i$ to be a sequence of level curves approaching $c$ from below. Then apply Lemma \ref{lem:quad} on the quadrilateral formed by $\bs, \bo_i, \ba$ and $\bbeta$ and let $i\to \infty$.
\end{proof}

\

  \paragraph{The Quadrilateral Rule.}
 We provide a practical rule for preventing meeting points. We keep the same notation.

\begin{prop}[Quadrilateral Rule]\label{prop:quad} If $|\nabla u(\bs(t))|$ is strictly monotone on the arcs $\overline{a\xi}$ and $\overline{\xi b}$ of the level curve $\bs$ (one of them may reduce to a point), then no streamlines can meet inside the quadrilateral. A streamline with initial point on the arc $\overline{ab}$ (but not $a$ or $b$) has constant speed $|\nabla u_\infty|$ till it meets $\ba, \bbeta$ or reaches $\bo$.
\end{prop}

\begin{proof} Let $\bl= \bl(t)$ be a streamline passing through the point $x\in \overline{\xi b}$, $x\neq \xi$ on the level curve $\bs$.

\begin{figure}[h!]

  \begin{center}

	\begin{tikzpicture}[domain=-5:5,scale=0.6]

\fill (3,0) circle[radius=2pt] node[below] {$b$};
\fill (-3,0) circle[radius=2pt] node[below] {$a$};
\fill (2,4) circle[radius=2pt] node[above] {$b'$};
\fill (-2,4) circle[radius=2pt] node[above] {$a'$};
\fill (0.5,3.25) circle[radius=2pt] node[above] {$\eta$};
\fill (0,-0.75) circle[radius=2pt] node[below] {$\xi$};
\draw (0,-0.75) .. controls (0.3,1) .. (0.5,3.25); \fill (0.3,1) node[left] {$\bm$};
\draw (-3,0) .. controls (0,-1) .. (3,0);\fill (-1.2,-0.8) node[below] {$\bs$};

\draw (-3,0) .. controls (-2,2) .. (-2,4);\fill (-2.2,2) node[left] {$\ba$};
\draw (3,0) .. controls (3,2) .. (2,4);\fill (3,2) node[right] {$\bbeta$};
\draw (2,4) .. controls (0.5,3) .. (-2,4); \fill (-0.5,3.6) node[above] {$\bo$};

\fill (1.5,-0.48) circle[radius=2pt] node[below] {$x$};

\draw[color=red, thick] (1.5,-0.48) to[out=85,in=-92] (0.35,1.6);
\fill (0.35,1.6) circle[radius=2pt] node[right] {$y$};

	\end{tikzpicture}

\end{center}
\caption{Case 1: impossible}
	\end{figure}

We have three cases: 1) If $\bl$ meets $\bm$ at the point $y$, then Lemma \ref{lem:quad}  applied on the quadrilateral $x b b'\eta y x$ (or Corollary \ref{cor:tri} if $\bm$ meets $\bbeta$, so that we have a triangle) implies
$$
M=|\nabla u_\infty(\bl)|
$$
on the whole arc $\overline{x\eta}$ of $\bl$ (or until $\bm$ reaches $\bbeta$). But then
$$
|\nabla u_\infty(\xi)| = |\nabla u_\infty(x)|,
$$
which contradicts the \emph{strict} monotonicity of $|\nabla u(\bs(t))|$.

\begin{figure}[h!]

  \begin{center}

	\begin{tikzpicture}[domain=-5:5,scale=0.6]

\fill (3,0) circle[radius=2pt] node[below] {$b$};
\fill (-3,0) circle[radius=2pt] node[below] {$a$};
\fill (2,4) circle[radius=2pt] node[above] {$b'$};
\fill (-2,4) circle[radius=2pt] node[above] {$a'$};
\fill (0.5,3.25) circle[radius=2pt] node[above] {$\eta$};
\fill (0,-0.75) circle[radius=2pt] node[below] {$\xi$};
\draw (0,-0.75) .. controls (0.3,1) .. (0.5,3.25); \fill (0.3,1) node[left] {$\bm$};
\draw (-3,0) .. controls (0,-1) .. (3,0);\fill (-1.2,-0.8) node[below] {$\bs$};

\draw (-3,0) .. controls (-2,2) .. (-2,4);\fill (-2.2,2) node[left] {$\ba$};
\draw (3,0) .. controls (3,2) .. (2,4);\fill (3,2) node[right] {$\bbeta$};
\draw (2,4) .. controls (0.5,3) .. (-2,4); \fill (-0.5,3.6) node[above] {$\bo$};

\fill (1.5,-0.48) circle[radius=2pt] node[below] {$x$};

\draw[color=blue, thick] (1.5,-0.48) to[out=80,in=-72] (2.58,2.8);
\fill (2.58,2.8) circle[radius=2pt] node[right] {$y$};

	\end{tikzpicture}

\end{center}
\caption{Case 2: possible}
	\end{figure}

2) If $\bl$ meets $\bbeta$ at $y\in \overline{bb'}$, then Corollary \ref{cor:tri}  applied on the triangle $xby$ yields
$$
|\nabla u_\infty(\bl)| = \text{constant}
$$
on the arc $\overline{xy}$.

\begin{figure}[h!]

  \begin{center}

	\begin{tikzpicture}[domain=-5:5,scale=0.6]

\fill (3,0) circle[radius=2pt] node[below] {$b$};
\fill (-3,0) circle[radius=2pt] node[below] {$a$};
\fill (2,4) circle[radius=2pt] node[above] {$b'$};
\fill (-2,4) circle[radius=2pt] node[above] {$a'$};
\fill (0.5,3.25) circle[radius=2pt] node[above] {$\eta$};
\fill (0,-0.75) circle[radius=2pt] node[below] {$\xi$};
\draw (0,-0.75) .. controls (0.3,1) .. (0.5,3.25); \fill (0.3,1) node[left] {$\bm$};
\draw (-3,0) .. controls (0,-1) .. (3,0);\fill (-1.2,-0.8) node[below] {$\bs$};

\draw (-3,0) .. controls (-2,2) .. (-2,4);\fill (-2.2,2) node[left] {$\ba$};
\draw (3,0) .. controls (3,2) .. (2,4);\fill (3,2) node[right] {$\bbeta$};
\draw (2,4) .. controls (0.5,3) .. (-2,4); \fill (-0.5,3.6) node[above] {$\bo$};

\fill (1.5,-0.48) circle[radius=2pt] node[below] {$x$};

\draw[color=blue, thick] (1.5,-0.48) .. controls (1.8,2) .. (1.3,3.55);

\fill (1.3,3.55) circle[radius=2pt] node[above] {$y$};

	\end{tikzpicture}

\end{center}
\caption{Case 3: possible}
	\end{figure}

3) If $\bl$ passes through a point $y\in \overline{\eta b'}$ on the upper level $\bo$, $y\neq \eta$, $y\neq b'$, then Lemma \ref{lem:quad} applied on the quadrilateral $xbb'y$ (or Corollary \ref{cor:tri} in case of a curved triangle) yields
$$
|\nabla u_\infty(\bl)| = \text{constant}
$$
on the arc $\overline{xy}$.

Finally, if $x$ is chosen from the left level arc $\overline{a\xi}$, the proof consists of three similar cases again. Thus we have established that $\bl$ has constant speed till it first meets
$\ba, \bbeta$, or hits $\bo$.

It remains to show that no two streamlines can meet in the quadrilateral.
A streamline $\bl$ passing through the point $x$ at the level curve $\bs$ has constant speed
$$
|\nabla u_{\infty}(x)| = |\nabla u_{\infty}(\bl)|
$$
till $\bl$ meets $\ba$, $\bbeta$ or hits $\bo$. But two meetings streamlines must have the same speed, which requires that they pass through $\bs$ at two points with the same speed $|\nabla u_{\infty}|$. By the \emph{strict} monotonicity of $|\nabla u_{\infty}(\bs)|$, this would require that the points are on different arcs $\overline{a\xi}$ and $\overline{\xi b}$. This is impossible, since no streamlines meet $\bm$.
\end{proof}

The Quadrilateral Rule remains true if the monotonicity of $|\nabla u_{\infty}(\bs)|$ is not supposed to be strict. If $|\nabla u_{\infty}(\bs)|$ is constant on some subarc $\overline{cd}$, then the streamlines with initial points on $\overline{cd}$ are non-intersecting straight lines. To see this, we again consider the quadrilateral $a\,b\,b'\,a'$ bounded by $\ba,\bbeta,\bs,\bo.$

\begin{lemma}\label{lem:constant}
  Assume  that $|\nabla u_\infty (\bs)|$ is constant on the arc $\overline{ab}$. Then no streamlines can meet inside the quadrilateral. Moreover,  $|\nabla u_\infty|$ is constant in the quadrilateral and all streamlines are straight lines.
\end{lemma}

\begin{proof} By Lemma \ref{lem:quad}, $|\nabla u_{\infty} (\bo)|$ is constant on the upper arc  $\overline{a'b'}$ .

  In particular, $|\nabla u_{\infty}|$ must be constant along $\ba$ and $\bbeta$. Then $|\nabla u_{\infty}|$ must be constant along any arc of a streamline passing through the quadrilateral. Every  point inside the quadrilateral lies on such a streamline. Therefore $|\nabla u_{\infty}|$ is constant in the quadrilateral, which means that it solves the \emph{Eikonal Equation}. Since $u_{\infty}$ is of class $C^1$, we can apply the next proposition

    to conclude that all streamlines are non-intersecting straight lines.
\end{proof}

\begin{prop} [Eikonal Equation]\label{prop:eikonal} Suppose that $v \in C^1(D)$ is a solution of the Eikonal Equation $|\nabla v| = C$ in the domain $D$, where $C$ denotes a constant. Then the streamlines of $v$ are non-intersecting segments of straight lines.
\end{prop}

\begin{proof} A very appeling direct proof is given in Lemma 1 in \cite{A2}.
\end{proof}

For the next result we abandon the \emph{strict} monotonicity in Proposition \ref{prop:quad}.

\begin{cor}[Quadrilateral Rule] \label{cor:constant} Assume
  that $|\nabla u_\infty (\bs)|$ is monotone on the arc $\overline{ab}$. Then no streamlines can meet inside the quadrilateral. A streamline with initial point on the arc $\overline{ab}$ (but not $a$ or $b$) has constant speed till it meets $\ba$, $\bbeta$ or reaches $\bo$.
\end{cor}

\begin{proof} Assume that $|\nabla u_{\infty} (\bs)|$ is non-decreasing. Consider the subarc $\overline{x^1x^2}$ on $\bs$ so that $|\nabla u_{\infty}(x^1)| \leq  |\nabla u_{\infty}(x^2)|$, where $x_1 < x_2$.  Let $\ba^j$ be the streamline passing through $x^j$.
We claim that $\ba^1$ does not meet $\ba^2$ inside the quadrilateral. Indeed, suppose they meet at a point $c$ at the level line $\widetilde \bo$ before reaching $\bo$, where $\widetilde \bo$ intersects $\ba$ and $\bbeta$ at $a''$ and $b''$ respectively. Then Lemma \ref{lem:quad} applied to the quadrilaterals $a\,x^1\,c\,a''$ and  $a\,x^2\,c\,a''$  exhibit that the speeds
  $$|\nabla u_{\infty}(\ba^1(t))| = |\nabla u_{\infty}(\ba^2(t))| = |\nabla u_{\infty}(c)|$$
are constant along  the arcs. Again we see that the Eikonal Equation is valid in the triangle
  $x^1\,x^2\,c$. At the point $c$ this leads to a contradiction with Proposition \ref{prop:eikonal}. (Thus the eventual point $c$ must lie on $\bo$ and on $\partial K$.)

\end{proof}

    \paragraph{The Triangular Rule.}
The above results may be formulated for a curved triangle as in Figure \ref{fig:tribasic} (seen as a degenerate quadrilateral).  Again,  suppose that the streamlines $\ba$ and $\bbeta$ together with the level curve $\bs$ form a curved triangle with vertices $a,b$ and $c$; $c$ is the meeting point of $\ba$ and $\bbeta$.  Assume that $\xi\in \overline{ab}$ is a point at which
  $$
  |\nabla u_\infty(\xi)|= \max_{\overline{ab}} |\nabla u_\infty(\bs)| = M.
  $$
  Let $\bm$ be the streamline that passes through $\xi$. It passes through $c$ (but may have joined $\ba$ or $\bbeta$ before reaching $c$).
 \begin{figure}[h!]

  \begin{center}

	\begin{tikzpicture}[domain=-5:5,scale=1.2]

\fill (3,0) circle[radius=2pt] node[below] {$b$};
\fill (-3,0) circle[radius=2pt] node[below] {$a$};

\fill (0.5,3.25) circle[radius=2pt] node[right] {$c$};
\fill (0,-0.75) circle[radius=2pt] node[below] {$\xi$};
\draw (0,-0.75) .. controls (0.3,1) .. (0.5,3.25); \fill (0.3,1) node[left] {$\bm$};
\draw (-3,0) .. controls (0,-1) .. (3,0);\fill (-1.2,-0.8) node[below] {$\bs$};

\draw (-3,0) to [out=75, in =-95](0.5,3.25);\fill (-1.5,2) node[left] {$\ba$};
\draw (3,0) to [out=105, in =-95](0.5,3.25);\fill (2,2) node[right] {$\bbeta$};

	\end{tikzpicture}

\end{center}
\caption{The curved triangle $abc$.}
\label{fig:tribasic}
	\end{figure}
  By simply using the results for quadrilaterals, we may deduce the following.

\begin{cor}\label{cor:tricor} If $|\nabla u(\bs(t))|$ is strictly monotone on the arcs $\overline{a\xi}$ and $\overline{\xi b}$ of the level curve $\bs$ (one of them may reduce to a point), then no streamlines can meet inside the triangle. A streamline with initial point on the arc $\overline{ab}$ (but not $a$ or $b$) has constant speed $|\nabla u_\infty|$ till it meets $\ba$ or $ \bbeta$.
\end{cor}

\begin{proof} If two streamlines meet at a point in the triangle we may construct a quadrilateral containing that point by letting $\bo$ be a level curve above $c$. Then Proposition \ref{prop:quad} yields a contradiction.
\end{proof}

\begin{lemma}\label{lem:trilem}   $|\nabla u_\infty (\bs)|$ cannot be  constant on a subarc of $\overline{ab}$, except if  $c \in \partial K.$
\end{lemma}

\begin{proof} We can again construct a triangle in which the Eikonal Equation is valid. This yields a contradiction, unless we allow a corner to be outside $G$.
\end{proof}

Vi can again abandon the \emph{strict} monotonicity.

\begin{cor}[Triangular Rule] \label{cor:triconstant}
Suppose that $|\nabla u_\infty (\bs)|$ is monotone on the arc $\overline{ab}$ of the level curve $\bs$. Then no streamlines can meet inside the triangle. A streamline with initial point on the arc $\overline{ab}$ has constant speed till it meets $\ba$ or $\bbeta$.
\end{cor}

\begin{proof} Reason as in the proof of Corollary \ref{cor:tricor} and apply Corollary \ref{cor:constant}.
\end{proof}

  \section{Polygons}\label{sec:poly}
  Let $\Omega$ be a convex polygon with $N$ vertices $P_1,P_2,\ldots, P_N$ and set $P_{N+1}=P_1$. The gradient $\nabla u_\infty$ is continuous up to the boundary $\partial \Omega$ and especially at the vertices,
  $$|\nabla u_\infty(P_j)|=0, \quad j=1,2,\ldots,N.$$
  From each vertex $P_j$, there is a unique streamline $\bg_j$ that terminates on $K$. They are the attracting streamlines.

Let $M_j$ denote a point on the edge $\overline{P_jP_{j+1}}$ at which $|\nabla u_\infty|$ attains its maximum, i.e,
$$
|\nabla u_\infty(M_j)|=\max_{\overline{P_jP_{j+1}}} |\nabla u_\infty|.
$$
The point divides the edge $\overline{P_jP_{j+1}}$ into two line segments $\overline{P_jM_j}$ and $\overline{M_jP_{j+1}}$. Denote by $\bm_j$ the streamline starting at the point $M_j$.

\begin{lemma}\label{lem:polymon} The normal derivative
$$
\frac{\partial u_\infty}{\partial n} = |\nabla u_\infty|
$$
is monotone along the half-edges $\overline{P_jM_j}$ and $\overline{M_jP_{j+1}}$ for $j=1,2,\ldots,N$.
\end{lemma}
\begin{proof}
 We arrange it so that the polygon is in the upper half-plane $x_2>0$ and the edge in question is on the $x_1$-axis, say the edge is
 $$a\leq x_1\leq b, \quad x_2=0.$$
 The convex level curves
 $$
 \{u_\infty=c\}
 $$
 approach the $x_1$-axis as $c\to 0$. The shortest distance from the level curve to the edge is attained at some point, say $(x_1(c), x_2(c))$. Choose a sequence $c_j\to 0$ so that $x_1(c_j)\to \xi$ and $x_2(c_j)\to 0$, where $(\xi,0)$ is some point, $a\leq\xi\leq b$ (in fact, $a < \xi < b$).  If $\xi >a$, let $a<\xi_1<\xi_2<\xi$ and keep $j$ so large that $\xi_2<x_1(c_j)$. The vertical lines $x_1=\xi_1$ and $x_1=\xi_2$ intersect the level curve $\{u_\infty = c\}$ at the points $(\xi_1,h_1^j)$ and $(\xi_2, h_2^j)$, i.e.
 $$
 u_\infty(\xi_1,h_1^j)= u_\infty(\xi_2,h_2^j)=c_j.
 $$
 The convexity of the level curve implies that $h_1^j\geq h_2^j$. (The chord between $(\xi_1,h_1^j)$ and $(x_1(c_j), x_2(c_j))$ must lie inside the set $\{u_\infty\geq c\}$.)
 It follows that the difference quotients in the normal direction satisfy
 $$
 \frac{u_\infty(\xi_1,h_1^j)-u_\infty(\xi_1,0)}{h_1^j}\,\leq \, \frac{u_\infty(\xi_2,h_2^j)-u_\infty(\xi_2,0)}{h_2^j},
 $$
 since both numerators are $= c_j-0$.
As $c_j\to 0$, also $h_1^j\to 0$ and $h_2^j\to 0$. By passing to the limit we obtain
$$
|\nabla u_\infty(\xi_1,0)|\,\leq\, |\nabla u_\infty(\xi_2,0)|,\quad \xi_1<\xi_2<\xi
$$
as desired.

If $a<\xi<b$ we also obtain the reverse inequality for all $\xi<\xi_1<\xi_2< b$ so that we may conclude the desired result again. It also follows that $(\xi,0)$ is the $M_j$ point of this edge. This excludes that $\xi = a$ or $\xi = b$.
\end{proof}

  We are now ready to prove our main theorem for polygons.

  \begin{proof}[Proof of Theorem \ref{thm:mainpoly}.] Consider the region bounded by $\overline{P_jP_{j+1}}, \bg_j, \bg_{j+1}$ and, if  $\bg_j$ does not meet $ \bg_{j+1}$ also $\partial K$. This can be either a curved triangle (meeting attracting streamlines) or a quadrilateral (the attracting streamlines do not meet). By Lemma \ref{lem:polymon}, $|\nabla u_\infty|$ is monotone along $\overline{P_jM_{j}}$ and $\overline{M_jP_{j+1}}$. Therefore, Corollary \ref{cor:constant} (in the case of a quadrilateral) and Corollary \ref{cor:triconstant} (in the case of a curved triangle) imply that no streamlines can meet (on either side of $\mu_j$)
    and that they have  constant speed until they meet $\bg_j$ or $\bg_{j+1}$, or hit $\partial K.$

  \end{proof}

  \section{General Domains}  \label{sec:general} In this section we assume that $\nabla u_\infty$ is continuous in $\overline \Omega\setminus K$

   and that $|\nabla u_\infty|$ has a finite number of local minimum points and maximum points. Denote by $P_1, \ldots, P_N$ (with $P_{N+1}=P_1$ as before) the minimum points. From each $P_j$, there is a unique streamline $\bg_j$ that terminates in $K$. These streamlines divide $G$ into triangles with corners $P_k, P_k$ and $Q_k$ if $\bg_k$ and $\bg_{k+1}$ meet at $Q_k$, and quadrilateras with corners  $P_k, P_{k+1}, S_{k+1}$ and $S_{k}$ if $\bg_k$ and $\bg_{k+1}$ do not meet but they reach $K$ at the points $S_k$ and $S_{k+1}$. Recall the $\infty$-ridge,
  $$
  \Gamma = \bigcup_{k=1}^N \{\bg_k(t), \quad 0\leq T\leq T_k\}.
  $$

   We give the proof of Theorem \ref{thm:general}.

  \begin{proof}[Proof of Theorem \ref{thm:general}.] Consider the region bounded by $\overline{P_jP_{j+1}}, \bg_j, \bg_{j+1}$ and perhaps $\partial K$. This can be either a curved triangle or  quadrilateral. By construction, $|\nabla u_\infty|$ is monotone along $\overline{P_jM_{j}}$ and $\overline{M_jP_{j+1}}$. Therefore, Corollary \ref{cor:constant} in the case of a quadrilateral and Corollary \ref{cor:triconstant} in the case of a curved triangle imply that no streamlines can meet (on either side of $\mu_j$) and that they are constant until they meet $\bg_j$ or $\bg_{j+1}$ or reach $\partial K$.
  \end{proof}

  \section{Appendix: Estimates of Derivatives of $|\nabla u_p|$}

  The fundamental properties

\[\tag{\textbf{I}}\label{eq:eqI}\iint_D |\nabla u_p-\nabla u_\infty|^2 \, dx_1 dx_2\to 0,\quad \text{as }p\to \infty,\]

\begin{equation}\tag{\textbf{J}}\label{eq:eqJ}\iint_D |\nabla (|\nabla u_p|^2)|^2 \, dx_1 dx_2\leq M_D<\infty,\end{equation}
for all (large) $p$

used in Section \ref{sec:eqcont} follow directly from \cite{KZZ}, where the corresponding estimates are ingeniously derived for the solution $u^\e$ of
$$
\Delta_\infty u^\e +\e\Delta u^\e  = 0.
$$
To transcribe the work to the solution $u_p$ of the $p$-Laplace equation
$$
  \Delta_\infty u_p +\frac{1}{p-2}|\nabla u_p|^2\Delta u_p = 0
  $$
  one has to replace the constant factor $\e$ by the \emph{function} $|\nabla u_p|^2/(p-2)$ \emph{under} the integral sign. Below we give just a synopsis of the procedure, referring to the numbering of formulas and theorems in \cite{KZZ}. (The reader is supposed to have access to \cite{KZZ}.)

  Formula (2.5) in \cite{KZZ} becomes
  $$
  -\det(D^2 u_p) = |\nabla |\nabla u_p||^2+\frac{1}{p-2}(\Delta u_p)^2.
  $$
  Formula (2.7) becomes
  $$
  I_p(\phi)=\iint_U |\nabla |\nabla u_p||^2 \phi\, dx_1 dx_2 +\frac{1}{p-2}\iint_U (\Delta u_p)^2\phi \,dx_1 dx_2
  $$
  and (2.8)
  $$
  I_p(\phi)=\frac12\iint_U \Bigl(\Delta u_p\langle\nabla u_p, \nabla \phi\rangle- \sum_{i,j=1}^2\frac{\partial^2 u_p}{\partial x_i\partial x_j}\frac{\partial u_p}{\partial x_j}\frac{\partial \phi}{\partial x_i}  \Bigr)  \, dx_1 dx_2 .
  $$
  Lemma 5.1 is needed only for $\alpha = 2$ (and since $|\nabla u_p|\neq 0$ we can put $\kappa = 0$ in the proof). It becomes
\[
\begin{split}
&    \iint_U |\nabla |\nabla u_p|^2|^2\xi^2 \, dx_1 dx_2 +\frac{1}{p-2}\iint_U |\nabla u_p|^2(\Delta u_p)^2\xi^2 \,dx_1 dx_2\\
&\leq C(2)\iint_U|\nabla u_p|^4\left(|\nabla \xi|^2+|\xi||D^2\xi|\right) \, dx_1 dx_2.
\end{split}
\]
This yields Lemma 2.6 and the desired property \eqref{eq:eqJ}, since $|\nabla u_p|$ is locally bounded by Lemma \ref{lem:gradbound}.

Lemma 5.2 is valid with no changes (replace $u^\e$ with $u_p$), but the proof uses Lemma 5.1 as above. Then Lemma 5.2 implies the flatness estimate in Lemma 2.7:
\[
\begin{split}
&\fiint_{B_r(x)}\left(|\nabla u_p|^2-\langle \nabla P, \nabla u_p\rangle\right)^2 \, dx_1 dx_2\leq C\left(\fiint_{B_{2r}(x)} |\nabla u_p|^4\, dx_1 dx_2\right)^\frac12 \\
&\times \left(\fiint_{B_{2r}(x)}\left(\frac{|u_p-P|^2}{r^2}\left(|\nabla P|+|\nabla u_p|\right)^2+\frac{|u_p-P|^4}{r^4}\right) dx_1 dx_2 \right)^\frac12
\end{split}
\]
valid for any linear function $P$. This estimate is needed for the proof of Theorem 1.4, when one has to identify the limit of $|\nabla u_p|^2$ in $L^2_\text{loc}$ as $|\nabla u_\infty|^2$. Theorem 1.4 contains our desired property \eqref{eq:eqI}.