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\abstract{In this paper we further develop a Grassmannian technique used to prove results about very general hypersurfaces and present three applications. First, we provide a short proof of the Kobayashi Conjecture given previous results on the Green-Griffiths-Lang Conjecture. Second, we characterize the dimension of the space of Chow-equivalent points on a very general hypersurface, proving the remaining cases of a conjecture of Chen, Lewis and Sheng and providing a short, alternate proof for many of the already known cases. Finally, we relate Seshadri constants of very general points to Seshadri constants of arbitrary points of very general hypersurfaces.}

\section{Introduction}
In this paper, we further develop a technique from \cite{RY} and apply it to study the Kobayashi Conjecture, $0$-cycles on hypersurfaces of general type, and Seshadri constants of very general hypersurfaces. The idea of the technique is to translate results about very general points on very general hypersurfaces to results about arbitrary points on very general hypersurfaces.

Our first application is to hyperbolicity. Recall that a complex variety is Brody hyperbolic if it admits no holomorphic maps from $\CC$.

\begin{conjecture}[Kobayashi Conjecture]
\label{conj-Kobayashi}
A very general hypersurface $X$ of degree $d$ in $\PP^n$ is Brody hyperbolic if $d$ is sufficiently large. Moreover, the complement $\PP^n \setminus X$ is also Brody hyperbolic for large enough $d$.
\end{conjecture}

First conjectured in 1970 \cite{K}, the Kobayashi Conjecture has been the subject of intense study, especially in recent years \cite{Siu, Deng, Brotbeck, D2}. The suspected optimal bound for $d$ is approximately $d \geq 2n-1$. However, the best current bound is for $d$ greater than about $(en)^{2n+2}$ \cite{D2}. A related conjecture is the Green-Griffiths-Lang Conjecture.

\begin{conjecture}[Green-Griffiths-Lang Conjecture]
\label{conj-GGL}
If $X$ is a variety of general type, then there is a proper subvariety $Y \subset X$ containing all the entire curves of $X$.
\end{conjecture}

The Green-Griffiths-Lang Conjecture says that holomorphic images of $\CC$ under nonconstant maps do not pass through a general point of $X$. Conjecture \ref{conj-GGL} is well-studied for general hypersurfaces, and it is a natural result to prove on the way to proving Conjecture \ref{conj-Kobayashi}. We provide a new proof of the Kobayashi Conjecture using previous results on the Green-Griffiths-Lang Conjecture.

\begin{theorem}
A general hypersurface in $\PP^n$ of degree $d$ admits no nonconstant holomorphic maps from $\CC$ for $d \geq d_{2n-3}$, where $d_2 = 286, d_3 = 7316$ and
\[ d_n = \left\lfloor \frac{n^4}{3} (n \log(n \log(24n)))^n \right\rfloor .\]
\end{theorem}

Our proof appears to be substantially simpler than the previous proofs (compare with \cite{Siu, Deng, Brotbeck, D2}), and can be adapted in a straightforward way as others use jet bundles to obtain better bounds for Conjecture \ref{conj-GGL}. Unfortunately, the bound of about $(2n \log(n\log(n)))^{2n+1}$ that we obtain is slightly worse than Demailly's bound of $(en)^{2n+2}$. However, assuming the optimal result on the Green-Griffiths-Lang Conjecture, our technique allows us to prove the conjectured bound of $d \geq 2n-1$ for the Kobayashi Conjecture.

The Kobayashi Conjecture for complements of hypersurfaces has also been studied by several authors \cite{Dar, BD}. Using results of Darondeau \cite{Dar} along with our Grassmannian technique, we are able to prove the Kobayashi Conjecture for complements as well.

\begin{theorem}
If $X$ is a general hypersurface in $\PP^n$ of degree at least $d_{2n}$, where $d_n = (5n)^2 n^n$, then $\PP^n \setminus X$ is Brody hyperbolic.
\end{theorem}

Our bound of about $100 \cdot 2^n n^{2n+2}$ is slightly worse than Brotbeck and Deng's bound of about $e^3 n^{2n+6}$ \cite{BD}, but our proof is substantially shorter.

Our second application concerns the Chow equivalence of points on very general complete intersections. Chen, Lewis, and Sheng \cite{CLS} make the following conjecture, which is inspired by work of Voisin \cite{voisinChow, V2,V3}.

\begin{conjecture}
\label{conj-CLS}
Let $X \subset \PP^n$ be a very general complete intersection of multidegree $(d_1, \dots, d_k)$. Then for every $p \in X$, the space of points of $X$ rationally equivalent to $p$ has dimension at most $2n-k-\sum_{i=1}^k d_i$. If $2n-k-\sum_{i=1}^k d_i < 0$, we understand this to mean that $p$ is equivalent to no other points of $X$.
\end{conjecture}

If this Conjecture holds, then the result is sharp \cite{CLS}. Voisin \cite{voisinChow, V2, V3} proves Conjecture \ref{conj-CLS} for hypersurfaces in the case $2n-d-1 < -1$. Chen, Lewis, and Sheng \cite{CLS} extend the result to $2n-d-1 = -1$, and also prove the analog of Voisin's bound for complete intersections. Both papers use fairly involved Hodge theory arguments. Roitman \cite{R1,R2} proves the $2n-k-\sum_{i=1}^k d_i = n-2$ case. Using Roitman's result, we prove all but the $2n-k-\sum_{i=1}^k d_i = -1$ case of Conjecture \ref{conj-CLS}, and in this case we prove the result holds with the exception of possibly countably many points.

\begin{theorem}
\label{thm-chow}
If $X \subset \PP^n$ is a very general complete intersection of multidegree $(d_1, \dots, d_k)$, then no two points of $X$ are rationally Chow equivalent if $2n-k-\sum_{i=1}^k d_i < -1$. If $2n-k-\sum_{i=1}^k d_i = -1$, then the set of points rationally equivalent to another point of $X$ is a countable union of points. If $2n-k-\sum_{i=1}^k d_i \geq 0$, then the space of points of $X$ rationally equivalent to a fixed point $p \in X$ has dimension at most $2n-k-\sum_{i=1}^k d_i$ in $X$.
\end{theorem}

Together with Chen, Lewis and Sheng's result, this completely resolves Conjecture \ref{conj-CLS} in the case of hypersurfaces. Our method appears substantially simpler than the previous work of Voisin \cite{voisinChow, V2, V3} and Chen, Lewis, and Sheng \cite{CLS}, although in the case of hypersurfaces, we do not recover the full strength of Chen, Lewis, and Sheng's result.

The third result relates to Seshadri constants. Let $\epsilon(p,X)$ be the Seshadri constant of $X$ at the point $p$, defined to be the infimum of $\frac{\deg C}{\mult_p C}$ over all curves $C$ in $X$ passing through $p$. Let $\epsilon(X)$ be the Seshadri constant of $X$, defined to be the infimum of the $\epsilon(p,X)$ as $p$ varies over the hypersurface.

\begin{theorem}
\label{thm-seshadri}
Let $r > 0$ be a real number. If for a very general hypersurface $X_0 \subset \PP^{2n-1}$ of degree $d$ the Seshadri constant $\epsilon(p,X_0)$ of $X_0$ at a general point $p$ is at least $r$, then for a very general $X \subset \PP^n$ of degree $d$, the Seshadri constant $\epsilon(X)$ of $X$ is at least $r$.
\end{theorem}

The layout of the paper is as follows. In Section \ref{sec-technique}, we lay out our general technique, and immediately use it to prove Theorem \ref{thm-seshadri}. In Section \ref{sec-hyperbolicity}, we discuss how to use the results of Section \ref{sec-technique} to prove hyperbolicity results. In Section \ref{sec-cycles}, we discuss how to prove Theorem \ref{thm-chow}.

\subsection*{Acknowledgements}
We would like to thank Xi Chen, Izzet Coskun, Jean-Pierre Demailly, Mihai P\u{a}un, Chris Skalit, and Matthew Woolf for helpful discussions and comments.

\section{The Technique}
\label{sec-technique}
We set some notation. Let $B$ be the moduli space of complete intersections of multidegree $(d_1, \dots, d_k)$ in $\PP^{n+k}$ and $\uU_{n,\bd} \subset \PP^{n+k} \times B$ be the variety of pairs $([p],[X])$ with $[X] \in B$ and $p \in X$. We refer to elements of $\uU_{n,\bd}$ as pointed complete intersections. When we talk about the codimension of a countable union of subvarieties of $\uU_{n,\bd}$, we mean the minimum of the codimensions of each component.

We need the following result from \cite{RY}.
\begin{proposition}
\label{GrassmanProp}
Let $C \subset \gG(r-1,m)$ be a nonempty family of $r-1$-planes of codimension $\epsilon > 0$, and let $B \subset \gG(r,m)$ be the space of $r$ planes that contain some $r-1$-plane $c$ with $[c] \in C$. Then $\codim(B \subset \gG(r,m)) \leq \epsilon -1$.
\end{proposition}
\begin{proof}
For the reader's convenience, we sketch the proof. Consider the incidence-correspondence $I = \{ ([b],[c]) | \: [b] \in B, [c] \in C \} \subset \gG(r,m) \times \gG(r-1,m)$. The fibers of $\pi_2$ over $C \subset \gG(r-1,m)$ are all $\PP^{m-r}$'s, while for a general $[b] \in B$, the fiber $\pi_1^{-1}([b])$ has codimension at least $1$ in the $\PP^r$ of $r-1$-planes contained in $b$ (since otherwise it can be shown that $C = \gG(r,m)$). The result follows by a dimension count.
\end{proof}

We need a few other notions for the proof. A parameterized $r$-plane in $\PP^m$ is a degree one injective map $\Lambda: \PP^r \to \PP^m$. Let $G_{r,m,p}$ be the space of parameterized $r$-planes in $\PP^m$ whose images pass through $p$. If $(p,X)$ is a pointed hypersurface in $\PP^m$, a parameterized $r$-plane section of $(p,X)$ is a pair $(\Lambda^{-1}(p), \Lambda^{-1}(X)) =: \phi^* (p,X)$, where $\Lambda: \PP^r \to \PP^m$ is a parameterized $r$-plane whose image does not lie entirely in $X$. We say that $\Lambda: \PP^r \to \PP^m$ contains $\Lambda': \PP^{r-1} \to \PP^m$ if $\Lambda(\PP^r)$ contains $\Lambda'(\PP^{r-1})$.

\begin{corollary}
\label{GrassmanCor}
If $C \subset G_{r-1,m,p}$ is a nonempty subvariety of codimension $\epsilon > 0$ and $B \subset G_{r,m,p}$ is the subvariety of parameterized $r$-planes that contain some $r-1$-plane $[c] \in C$, then $\codim(B \subset G_{r,m,p}) \leq \epsilon -1$.
\end{corollary}

Let $\xX_{n,\bd} \subset \uU_{n,\bd}$ be an open subset. For instance, $\xX_{n,\bd}$ might be equal to $\uU_{n,\bd}$ or it might be the universal complete intersection over the space of smooth complete intersections. Our main technical tool is the following.

\begin{theorem}
\label{thm-technicaltool}
Suppose we have an integer $m$ and for each $n \leq n_0$ we have $Z_{n,d} \subset \xX_{n,\bd}$ a countable union of locally closed varieties satisfying: \begin{enumerate}
\item If $(p,X) \in Z_{n,\bd}$ and is a parameterized hyperplane section of $(p',X')$, then $(p',X') \in Z_{n+1,\bd}$.
\item $Z_{m-1,\bd}$ has codimension at least $1$ in $\xX_{m-1,d}$.
\end{enumerate}
Then the codimension of $Z_{m-c,\bd}$ in $\xX_{m-c,\bd}$ is at least $c$.
\end{theorem}
\begin{proof}[Proof of Theorem \ref{thm-technicaltool}]
We adopt the method from \cite{RY}. We prove that for a very general point $(p_0,X_0)$ of a component of $Z_{m-c,\bd}$, there is a variety $\FF_{m-c}$ and a map $\phi: \FF_{m-c} \to \uU_{m-c,\bd}$ with $(p_0,X_0) \in \phi(\FF_{m-c})$ and $\codim(\phi^{-1}(Z_{m-c,\bd}) \subset \FF_{m-c}) \geq c$. This suffices to prove the result.

So, let $(p_0, X_0)$ be a general point of a component of $Z_{m-c,\bd}$, and let $(p_1,X_1) \in \xX_{m-1,\bd}$ be very general, so that $(p_1,X_1)$ is not in the closure of any component of $Z_{m-1,\bd}$ by hypothesis 2. Choose $(p,Y) \in \xX_{N,\bd}$ for some sufficiently large $N$ such that $(p_0,X_0)$ and $(p_1,X_1)$ are parameterized linear sections of $(p,Y)$. Then for all $n < N$, let $\FF_{n}$ be the space of parameterized $n$-planes in $\PP^N$ passing through $p$ such that for $\Lambda \in \FF_{n}$, $\Lambda^*(p,Y)$ is in $\xX_{n,\bd}$. This means that $\FF_n$ is an open subset of $G_{n,N,p}$. Let $\phi: \FF_{n} \to \xX_{n,\bd}$ be the map sending $\Lambda: \PP^n \to \PP^N$ to $\Lambda^*(p,Y)$.

We prove that $\codim(\phi^{-1}(Z_{m-c,\bd}) \subset \FF_{m-c}) \geq c$ by induction on $c$. For the $c=1$ case, we see by construction that $\phi^{-1}(Z_{m-1,\bd})$ has codimension at least $1$ in $\FF_{m-1}$, since $(p_1,X_1)$ is a parameterized $m-1$-plane section of $(p,Y)$ but is not in the closure of any component of $Z_{m-1,\bd}$. Now suppose we know that $\codim(\phi^{-1}(Z_{m-c,\bd}) \subset \FF_{m-c}) \geq c$. We use Corollary \ref{GrassmanCor} with $C$ equal to $\phi^{-1}(Z_{m-c-1,\bd})$. By hypothesis 1, we see that $B$ is contained in $\phi^{-1}(Z_{m-c,\bd})$. It follows from this that
\[ c \leq \codim( \phi^{-1}(Z_{m-c,\bd}) \subset \FF_{m-c})) \leq \codim(B \subset \FF_{m-c}) \]
\[ \leq \codim(C \subset \FF_{m-c-1}) -1 .  \]
Rearranging, we see that
\[ \codim( \phi^{-1}(Z_{m-c-1,\bd}) \subset \FF_{m-c-1}) = \codim(C \subset \FF_{m-c-1}) \geq c+1 . \]
The result follows.
\end{proof}

As an immediate application, we prove Theorem \ref{thm-seshadri}.

\begin{proof}[Proof of Theorem \ref{thm-seshadri}]
Let $r$ be given. Let $Z_{m,d} \subset \uU_{m,d}$ be the set of pairs $(p,X)$ where $\epsilon(p,X) < r$. We apply Theorem \ref{thm-technicaltool} to $Z_{m,d}$. We see that $Z_{m,d}$ is a countable union of algebraic varieties, and by hypothesis, $Z_{2n-1,d} \subset \uU_{2n-1,d}$ has codimension at least $1$. Now suppose that $(p_0,X_0) \in Z_{m,d}$. Then there is some curve $C$ in $X_0$ with $\frac{\deg C}{\mult_{p_0} C} < r$. Thus, for any $X$ containing $p$, we see that the Seshadri constant of $X$ at $p$ is at most $\frac{\deg C}{\mult_{p_0} C} < r$. This shows that the $Z_{m,d}$ satisfy the conditions of Theorem \ref{thm-technicaltool}, which shows that $Z_{n,d} \subset \uU_{n,d}$ has codimension at least $n$. By dimension reasons, this means that $\uU_{n,d}$ cannot dominate the space of hypersurfaces, so the result follows.
\end{proof}

\section{Hyperbolicity}
\label{sec-hyperbolicity}
Let $\xX_{n,d}$ be the universal hypersurface in $\PP^n$ over the open subset $U$ in the moduli space of all degree $d$ hypersurfaces in $\PP^n$ consisting of all smooth hypersurfaces. Many people have developed a technique for restricting the entire curves contained in a fiber of the map $\pi_2: \xX_{n,d} \to U$. See the article of Demailly for a detailed description of some of this work \cite{D2}. For a variety $X$, let $\ev: J_k(X) \to X$ be the space of $k$-jets of $X$. Then, if $X \subset \PP^n$ is a smooth degree $d$ hypersurface, there is a vector bundle $E_{k,m}^{GG} T_X^*$ whose sections act on $J_k(X)$. Global sections of $E_{k,m}^{GG} T_X^* \otimes \OO(-H)$ vanish on the $k$-jets of entire curves. This means that sections of $E_{k,m}^{GG} T_X^* \otimes \OO(-H)$ cut out a closed subvariety $S'_{k,m}(X) \subset J_k(X)$ such that any entire curve is contained in $\ev(S'_{k,m}(X))$. In fact, it can be shown that any entire curve is contained in the closure of $\ev(S_{k,m}(X))$, where $S_{k,m}(X) \subset J_k(X)$ is $S'_{k,m}(X)$ minus the space of singular $k$-jets.

The construction is functorial. In particular, if $V$ is the relative tangent bundle of the map $\pi_2$, there is a vector bundle $E_{k,m}^{GG} V^*$ whose restriction to each fiber of $\pi_2$ is $E_{k,m}^{GG} T_{X}^*$. Let $\yY_{n,d} \subset \xX_{n,d}$ be the locus of $(p,X) \in \xX_{n,d}$ such that $p \in \ev(S_{k,m}(X))$. Then by functoriality, $\yY_{n,d}$ is a finite union of locally closed varieties.

\begin{theorem}
\label{thm-Hyperbol}
Suppose that $\yY_{r-1,d} \subset \xX_{r-1,d}$ is codimension at least 1. Then $\yY_{r-c,d} \subset \xX_{r-c,d}$ is codimension at least $c$. In particular, if $\yY_{2n-3, d}$ is codimension at least $1$ in $\xX_{2n-3,d}$ and $d \geq 2n-1$, then a very general $X \subset \PP^n$ of degree $d_n$ is hyperbolic.
\end{theorem}
\begin{proof}
 We check that $\yY_{r-1,d}$ satisfies both conditions of Theorem \ref{thm-technicaltool}. Condition 2 is a hypothesis. Condition 1 follows by the functoriality of Demailly's construction. Namely, if $(p,X_0)$ is a parameterized linear section of $(p,X)$, then the natural map $X_0 \to X$ induces a pullback map on sections
 \[ H^0(E_{k,m}^{GG} T_X^* \otimes \OO(-H)) \to H^0(E_{k,m}^{GG} T_{X_0}^* \otimes \OO(-H)) , \]
compatible with the natural inclusion of jets $J_k(X_0) \to J_k(X)$. In particular, if some section $s$ of $H^0(E_{k,m}^{GG} T_X^* \otimes \OO(-H))$ takes a nonzero value on a jet $\alpha(j)$, where $j \in J_k(X_0)$ and $\alpha:J_k(X_0) \to J_k(X)$ is the natural inclusion, then the restriction of $s$ to $X_0$ takes a nonzero value on the original jet $j \in J_k(X_0)$. Thus, if $X_0$ has a nonsingular $k$-jet at $p$ which is annihilated by every section in $H^0(E_{k,m}^{GG} T_{X_0}^* \otimes \OO(-H))$, $X$ has such a $k$-jet as well.

To see the second statement, observe that by Theorem \ref{thm-technicaltool}, $\yY_{n,d}$ has codimension in $\xX_{n,d}$ at least $2n-3-n+1 = n-2$. It follows that a general $X$ of degree $d$ in $\PP^n$ satisfies that the image of any entire curve is contained in an algebraic curve. Since $d \geq 2n-1$, by a theorem of Voisin \cite{V2, V3}, any algebraic curve in $X$ is of general type. The result follows.
\end{proof}

The current best bound for the Green-Griffiths-Lang Conjecture is from Demailly \cite{D1,D2}. The version we use comes out of Demailly's proof.

\begin{theorem}[\cite{D2}, Section 10]
If $k$ and $m$ are sufficiently large, we have that $\yY_{n,d} \subset \xX_{n,d}$ is codimension at least $1$ for $d \geq d_n$, where $d_2 = 286, d_3 = 7316$ and
\[ d_n = \left\lfloor \frac{n^4}{3} (n \log(n \log(24n)))^n \right\rfloor .\]
\end{theorem}

Using this bound, we obtain the following.

\begin{corollary}
The Kobayashi Conjecture holds hypersurfaces when $d \geq d_{2n-3}$, where $d_2 = 286, d_3 = 7316$ and
\[ d_n = \left\lfloor \frac{n^4}{3} (n \log(n \log(24n)))^n \right\rfloor .\]
\end{corollary}

This bound of about $(2n \log (n\log n))^{2n+1}$ is slightly worse than the best current bound for the Kobayashi Conjecture from \cite{D2}, which is about $(en)^{2n+2}$. However, our technique is strong enough to allow us to prove the optimal bound from the Kobayashi Conjecture, provided one could prove the optimal result for the Green-Griffiths-Lang Conjecture.

\begin{corollary}
If $\yY_{d-2,d}$ has codimension at least $1$ in $\xX_{d-2,d}$ (as we would expect from the Green-Griffiths-Lang Conjecture), then a very general hypersurface of degree $d \geq 2n-1$ in $\PP^n$ is hyperbolic.
\end{corollary}
\begin{proof}
We apply Theorem \ref{thm-Hyperbol}. We know that if $\yY_{2n-3, d}$ is codimension at least $1$ in $\xX_{2n-3,d}$, then the Kobayashi Conjecture holds for hypersurfaces in $\PP^n$ of degree $d$. We apply this result with $d = 2n-1$.
\end{proof}

Work has also been done on the hyperbolicity of complements of hypersurfaces in $\PP^n$. There are similar jet bundle techniques in this case. Given a variety $Z$, and ample line bundle $A$, a subsheaf $V \subset T_Z$ and a subvariety $X \subset Z$, one can construct the vector bundles $E_{k,m} V^*(\log X)$, sections of which act on $k$-jets of $Z \setminus X$. It can be shown that any section of $H^0(E_{k,m} V^*(\log X) \otimes A^{*})$ must vanish on the $k$-jet of any entire curve in $Z \setminus X$. Then sections of $H^0(E_{k,m} V^*(\log X) \otimes A^{*})$ cut out a subvariety $S_{k,m}$ in the locus of nonsingular $k$-jets on $Z \setminus X$ such that any entire curve lies in the closure of $\ev(S_{k,m})$. Darondeau \cite{Dar} studies these objects for hypersurfaces in $\PP^n$. Namely, he proves the following.

\begin{theorem}[Darondeau]
\label{thm-Darondeau}
Let $d \geq (5n)^2 n^n$, let $U_{n,d}$ be the space of smooth degree $d$ hypersurfaces in $\PP^n$ and consider the space $\PP^n \times U_{n,d}$, with divisor $\xX_{n,d}$ corresponding to the universal hypersurface. Let $V = T_{\pi_2}$ be the relative tangent space of projection of $\PP^n \times U_{n,d}$ onto $U_{n,d}$, and consider the locus $S_{k,m}$ cut out by sections in $H^0(E_{k,m} V^*(\log \xX_{n,d}) \otimes \OO(-H))$. Then $\ev(S_{k,m})$ is codimension at least $2$ in $\PP^n \times U_{n,d}$.
\end{theorem}

Using our technique, we obtain the following effective form of the Kobayashi Conjecture.

\begin{theorem}
Let $Z_{n,d} \subset \PP^n \times U_{n,d} \setminus \xX_{n,d}$ be the locus of pairs $(p,X)$ where $p \in (\PP^n \setminus X) \setminus \ev(S_{k,m})$. Then if $Z_{r-1,d}$ has codimension at least $1$ in $(\PP^{r-1} \times U_{r-1,d}) \setminus \xX_{r-1,d}$, we have that $Z_{r-c,d}$ has codimension at least $c$ in $(\PP^{r-c} \times U_{r-c,d}) \setminus \xX_{r-c,d}$.
\end{theorem}
\begin{proof}
The proof is very similar in spirit to the proof of Theorem \ref{thm-technicaltool}, but we give a new proof here for completeness. Let $(p, X_0)$ be a general point of a component of $Z_{r-c,d}$. We will find a family $\phi: \FF_{r-c} \to \PP^{r-c} \times U_{r-c,d} \setminus \xX_{r-c,d}$ with $\phi^{-1}(Z_{r-c})$ having codimension at least $c$ in $\FF_{r-c}$.

Let $(p, X_1) \in \PP^{r-1} \times U_{r-1,d} \setminus \xX_{r-1,d}$ be very general so that $(p,X_1)$ is not in the closure of any component of $Z_{r-1,d}$. Let $(p,Y)$ be a hypersurface in some high dimensional projective space $\PP^N$ together with a point $p \in \PP^N \setminus Y$ such that $(p,X_0)$ and $(p,X_1)$ are both parameterized linear sections of $(p,Y)$. Let $\FF_{r-c}$ be the space of parameterized linear spaces in $\PP^N$ passing through $p$. Then we have a natural map $\phi: \FF_{r-c} \to \PP^{r-c} \times U_{r-c,d} \setminus \xX_{r-c,d}$ sending the parameterized $r-c$ plane $\Lambda$ to $\Lambda^* (p,Y)$. By construction, the image of $\phi$ will certainly contain the point $(p,X_0)$, and since $(p,X_1)$ is very general, $\phi^{-1}(Z_{r-1,d})$ will have codimension at least $1$ in $\FF_{r-1}$. Observe that if $(p,Y_0) \in Z_{r-c-1,d}$ is a linear section of $(p,Y_1)$, then $(p,Y_1) \in Z_{r-c,d}$, since if there is a nonsingular jet $j$ at $p$ such that all the sections of $E_{k,m} T_{\PP^{r-c-1}}^*(\log Y_0) \otimes \OO(-H)$ vanish on $j$, then certainly all the sections of $E_{k,m} T_{\PP^{r-c}}^*(\log Y_1) \otimes \OO(-H)$ will vanish on $j$. By repeated application of Corollary \ref{GrassmanCor}, we see that the codimension of $\phi^{-1}(Z_{r-c,d})$ in $\FF_{r-c,d}$ is at least $c$. This concludes the proof.
\end{proof}

It follows that if $Z_{n,d} \subset \PP^n \times U_{n,d} \setminus \xX_{n,d}$ has codimension at least $1$ for $d \geq d_n$, then a very general hypersurface complement in $\PP^n$ is Brody hyperbolic when $d \geq d_{2n-1}$. Using this together with Darondeau's result, we obtain the following corollary, weakening the bound a bit for brevity.

\begin{corollary}
Let $d \geq (10n)^2 (2n)^{2n}$. Then if $X \subset \PP^n$ is very general, the complement $\PP^n \setminus X$ is Brody hyperbolic.
\end{corollary}

\section{$0$-cycles}
\label{sec-cycles}
Let $R_{\PP^1, X, p} = \{ q \in X | Nq \sim Np \text{ for some integer $N$} \}$, where the relation $\sim$ means Chow equivalent. The goal of this section is to prove all but the $2n - \sum_{i=1}^k (d_i - 1) = -1$ case of the following conjecture of Chen, Lewis and Sheng \cite{CLS}.

\begin{conjecture}
\label{mainConj}
Let $X \subset \PP^{n}$ be a very general complete intersection of multidegree $(d_1, \dots, d_k)$. Then for every $p \in X$, $\dim R_{\PP^1, X, p} \leq 2n -k - \sum_{i=1}^k d_i$.
\end{conjecture}

Here, we adopt the convention that $\dim R_{\PP^1, X, p}$ is negative if $R_{\PP^1, X, p} = \{p \}$. Together with the main result of \cite{CLS}, this completely resolves Conjecture \ref{mainConj} in the case of hypersurfaces.

Chen, Lewis and Sheng consider the more general notion of $\Gamma$ equivalence, although we are unable to prove the $\Gamma$ equivalence version here. The special case $\sum_i d_i = n+1$ is a theorem of Roitman \cite{R1, R2} and the case $2n -k - \sum_{i=1}^k d_i \leq -2$ is a theorem of Chen, Lewis and Sheng \cite{CLS} building on work of Voisin \cite{voisinChow, V2, V3}, who proves the result only for hypersurfaces. Chen, Lewis and Sheng prove Conjecture \ref{mainConj} for hypersurfaces and for arbitrary $\Gamma$ in the boundary case $2n - k - \sum_{i=1}^k d_i = -1$ in \cite{CLS}. The case $2n - k - \sum_{i=1}^k d_i = -1$ appears to be the most difficult, and is the only one we cannot completely resolve with our technique.

We provide an independent proof of all but the $2n - k - \sum_{i=1}^k d_i = -1$ case of Conjecture \ref{mainConj}. Aside from Roitman's result, this is the first result we are aware of addressing the case $2n - k - \sum_{i=1}^k d_i \geq 0$. We rely on the result of Roitman in our proof, but not the results of Voisin \cite{voisinChow, V2} or Chen, Lewis, and Sheng \cite{CLS}.

Let $E_{n,\bd} \subset \uU_{n,\bd}$ be the set of $(p,X)$ such that $R_{\PP^1, X, p}$ has dimension at least $1$. Let $G_{n,\bd} \subset \uU_{n,\bd}$ be the set of $(p,X)$ such that $R_{\PP^1,X,p}$ is not equal to $\{p\}$. Both $E_{n,\bd}$ and $G_{n,\bd}$ are countable unions of closed subvarieties of $\uU_{n,\bd}$. When we talk about the codimension of $E_{n,\bd}$ or $G_{n,\bd}$ in $\uU_{n,\bd}$, we mean the minimum of the codimensions of each component. We prove Conjecture \ref{mainConj} by proving the following theorem.

\begin{theorem}
\label{mainThm}
The codimension of $E_{n,\bd}$ in $\uU_{n,\bd}$ is at least $-n + \sum_i d_i$ and the codimension of $G_{n,\bd}$ in $\uU_{n,\bd}$ is at least $-n-1 + \sum_i d_i$.
\end{theorem}

\begin{corollary}
Conjecture \ref{mainConj} holds for $2n- k - \sum_{i=1}^k d_i \neq -1$. In the special case $2n-k- \sum_{i=1}^k d_i = -1$, the space of $p \in X$ Chow-equivalent to some other point of $X$ has dimension $0$ (i.e., is a countable union of points) but might not be empty as Conjecture \ref{mainConj} predicts.
\end{corollary}
\begin{proof}
First we consider the case $2n - k - \sum_i d_i \geq 0$. Let $\pi_1: \uU_{n,\bd} \to B$ be the projection map. If $\pi_1|_{E_{n,\bd}}$ is not dominant, then the result holds trivially. Thus, we may assume that the very general fiber of $\pi_1|_{E_{n,\bd}}$ has dimension $n - k - \codim(E_{n,\bd} \subset \uU_{n,\bd})$. If the bound on $E_{n,\bd}$ from Theorem \ref{mainThm} holds, then the space of points $p$ of $X$ with positive-dimensional $R_{\PP^1,X,p}$ has dimension at most $2n - k - \sum_i d_i$, which implies Conjecture \ref{mainConj} in the case $2n - k - \sum_i d_i \geq 0$.

Now we consider the situation for $2n - k - \sum_i d_i \leq -1$. Conjecture \ref{mainConj} states that $\pi_1|_{G_{n,\bd}}$ is not dominant for this range. By Theorem \ref{mainThm}, we see that the dimension of $G_{n,\bd}$ is less than the dimension of $B$ if $2n - k - \sum_i d_i \leq -2$, proving Conjecture \ref{mainConj}. In the case $2n - k-\sum_i d_i = -1$, the dimension of $G_{n,\bd}$ is at most that of $B$, which shows that there are at most finitely many points of $X$ equivalent to another point of $X$. This proves the result.
\end{proof}

Our technique would prove Conjecture \ref{mainConj} in all cases if we knew that $G_{n,\bd}$ had codimension $-n + \sum_{i=1}^k d_i$ in $\uU_{n,\bd}$. However, this is not true for Calabi-Yau hypersurfaces.

\begin{proposition}
A general point of a very general Calabi-Yau hypersurface $X$ is rationally equivalent to at least one other point of the hypersurface.
\end{proposition}
\begin{proof}
Let $X$ be a very general Calabi-Yau hypersurface. Then we claim that a general point of $X$ is Chow equivalent to another point of $X$. To see this, observe that any point $p$ of $X$ has finitely many lines meeting $X$ to order $d-1$ at $p$. Such a line meets $X$ in a single other point. Moreover, every point of $X$ has a line passing through it that meets $X$ at another point of $X$ with multiplicity $d-1$. Thus, let $q_1$ be a general point of $X$, let $\ell_1$ be a line through $p$ meeting $X$ at a second point, $p$ to order $d-1$, and $\ell_2$ be a different line meeting $X$ at $p$ to order $d-1$. Let $q_2$ be the residual intersections of $\ell_2$ with $X$. Then $q_1 \sim q_2$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{mainThm}]
First consider the bound on $E_{n,\bd}$. By Roitman's Theorem plus the fact that $R_{\PP^1,X,p}$ is a countable union of closed varieties, we see that $E_{-1+\sum_{i=1}^k d_i,\bd}$ has codimension at least one in $\uU_{-1 + \sum_{i=1}^k d_i,\bd}$. We note that if $p \sim q$ as points of $Y$, and $Y \subset Y'$, then $p \sim q$ as points of $Y'$ as well. The rest of the result follows from Theorem \ref{thm-technicaltool} using $Z_{n,\bd} = E_{n,\bd}$.

Now consider $G_{n,\bd}$. From Roitman's theorem, it follows that a very general point of a Calabi-Yau complete intersection $X$ is equivalent to at most countably many other points of $X$. Thus, a very general hyperplane section of such an $X$ satisfies the property that the very general point is equivalent to no other points of $X$. From this, we see that $G_{-2+\sum_{i=1}^k d_i,\bd}$ has codimension at least $1$ in $\uU_{-2+\sum_{i=1}^k d_i,\bd}$. Together with Theorem \ref{thm-technicaltool}, this implies the result.
\end{proof}

\bibliographystyle{plain}


TITLE: I can't find an appropriate piecewise function for this graph
QUESTION [0 upvotes]: On one of my piecewise questions I've split a graph into an exponential function, a cosine function and a parabolic function. I've done fine for exponential and parabola but I'm totally stuck on cosine and I have no idea what I'm doing... Please help!

REPLY [1 votes]: One issue is that the curve you have sketched has a continuous derivative, which rounds things off at the joints.  The part you have labeled exp fits $y=2^x$ perfectly, but that is growing rapidly at $x=2$ and doesn't blend with the cos piece.  If we let there be corners at $x=2,5$ it is not so hard to find a piecewise graph.
$$y=\begin {cases} 2^x&-2 \le x \le 2\\\frac 52 +\frac 32 \cos \frac {x-2}\pi x& 2\lt x \le 5\\2+(x-6)^2&5\lt x \le 7 \end {cases}$$
If you want things to blend smoothly, you need more freedom in the function.

REPLY [0 votes]: I assume you want to find the equation of a cosine-shaped curve which has been possibly scaled and translated. This means that the curve has the form:
$$y(x) = a\cdot \cos{(bx + c)} + d$$
And all you have to do is identify the missing numbers $a, b, c, d$.
Here are some ways to analyze your cosine-shaped curve to find out how much it has been scaled and translated.

First, if you know the maximum and minimum height of the curve, then their average will be the value of $d$. (Every cosine-shaped curve oscillates around its midline, the horizontal line $y=d$.)
Second, if you know the maximum and minimum height of the curve, their difference is the overall height of your curve. Ordinarily, the curve $y(x)=\cos{x}$ spans from -1 to 1 and consequently has height 2. If the curve instead has been stretched vertically so that it has height $v$, you know that $a = v/2$.
Third, if you know the period of the curve (i.e. how long it takes for it to repeat itself), then you can compute $b$. Ordinarily, the curve $y(x) =\cos{x}$ repeats every $2\pi$ units. If it has been stretched horizontally so that it repeats every $h$ units instead, then you know that $b = 2\pi / h$.
Finally, how far has the curve been translated horizontally? There are many good answers. If you can find the x-coordinate $x$ of any point where the curve reaches its maximum height, then you can use $c\equiv -x$.


text:
\begin{document}

\maketitle

\tableofcontents
\newpage
\begin{abstract}
In industrial facilities and household equipments scale formation leads to reduced efficiency
and  damages. Therefore various devices for anti-scale have been designed for a long time.  Recently one of them, an aggregation of ceramic spheres, was proposed to prevent scale formation and its efficiency shown experimentally.
The purpose of this paper is to clarify the function of this device by proposing several mathematical models and by pursuing their mathematical and numerical analysis. The first  model measures the effect on nucleation of calcite of the electric potential near the surface of a ceramic sphere in natural water. The second model is based on a crystal lattice model and argues that the surface tension energy of the calcite particles is reduced by the polarization energy brought by the ceramic balls.  The third model is macroscopic and numerical and studies the effect of two ceramic balls arrangements in a flow of water containing calcite particles.
\end{abstract}

\paragraph{Key words}    Water scale control, Poisson-Boltzmann equation, Nucleation rate, Crystal lattice model, Helmholtz free energy, Navier-Stokes equation, Ginzburg-Landau model

\section{Introduction}
Water pollution is a serious problem everywhere in the world. As a kind of water pollution, scale (hard grayish-white substance) formation often leads to reduced efficiency or severe damage on  cleaning systems or in pipes or other aqueous closed cycle like cooling systems in industrial plants.
 A concrete examples is shown on the left and center in figure \ref{figone}.

 \begin{figure}[htbp]
\begin{center}
$$\includegraphics[width=7.8cm]{fig/pipe.jpg}~
\includegraphics[width=4cm]{fig/balls.pdf}$$
\caption{{\footnotesize Unwanted effect of scale and the device studied here to cure it}}
\label{figone}
\end{center}
\end{figure}
Therefore, various anti-scale treatments have been developed for a long time \cite{donaldson,duffy,higashitani,katrane,lopez,mahmoud,sohali,zheng,baker,barrett,gabrielli,higashitaniIseri,higashitaniOshitani,wang,amjd}. One of them is a tight aggregate of ceramic spheres (right of Figure \ref{figone}). If it is immersed  in  natural water, it is known to prevent scale formation \cite{ikawa}. The diameter of the ceramic balls is around 1cm, their surface is covered with inorganic oxides, whose main ingredients are \texttt{SiO}$_2$ and \texttt{Fe}. These produce free electrons by the reaction
\texttt{Fe}$\to$\texttt{Fe}$^{++}$\texttt{+2e}; in turn the electrons are captured by the hydrogen ions in water \texttt{2H}$^+$\texttt{+2e}$\to$\texttt{H}$_2$ thus leaving a very thin layer of \texttt{OH}$^-$ very near the surface of the balls.
\\\\
It does not to change the water composition contrary to chemical treatments but it changes the size of the calcite  aggregates, as shown on the right in figure \ref{fig1b}.
However, theoretical clarification of their efficiency seems to be open and needs to be worked out.

\subsection{The Process and the Approach}
What is the cause of scale formation?   Calcite  particles form a crystal by a phase-change phenomenon. Scale is a more stable state for the crystal particles of calcium carbonate (\texttt{CaCO}$_3$, calcite being its most stable crystal form) obtained  by releasing their surface tension energy by adhering to the surface of another material. Experimental measurements reported in \cite{kawarada} have shown that the ceramic spheres  in water generate an electric field near their surface which has an effect on electrolytic solutions such as water containing \texttt{Ca}$^{++}$ and \texttt{CO}$_3^{--}$ ions: the layer of \texttt{OH}$^-$ attracts \texttt{CO}$_3^{--}$ ions and hence change the crystal structure of calcite;  crystal particles passing near the surface of a ceramic sphere are exposed to an electric field and store this additional \emph{polarization} energy which changes the crystal structure making scale formation less favourable energetically.

So our investigations focus first on the electric field  and next on the polarization energy and more precisely on the surface tension energy. We call this model, the \emph{ contact model}.

 Finally, once it is explained that near the ceramic balls the slate-capable calcite concentration diminishes, it is important to investigate the efficiency of various arrangements of ceramic spheres. This is done by using a macroscopic model and an adhoc boundary condition for the calcite concentration. This last section is mostly numerical.

 \begin{figure}[htbp]
\begin{center}
$$\includegraphics[width=5cm]{fig/fig1c2.pdf}
~~\includegraphics[width=5cm]{fig/fig1c1.pdf}$$
\caption{{\footnotesize Left : \texttt{CaCO}$_3$ crystals in natural water. Right : same after 24h in contact with the spheres of figure \ref{figone}.}}
\label{fig1b}
\end{center}
\end{figure}

\subsection{Plan}
Thus this paper  contributes to the clarification of the mechanism of such device to prevent scale formation in natural water.

In section \ref{sec2} of this article, we derive a Poisson-Boltzmann equation to formulate this phenomenon.
The Poisson-Boltzmann equation is solved numerically for appropriate physical constants and then the electric field on the surface of ceramic balls is obtained. By using it, the polarization energy generated in particle is computed.

In section \ref{sec3}, it is shown that the polarization energy gives rise to an extraordinary
suppression of nucleation rate.  This means a substantial
decrease of nuclei which contribute to formation of the scale.

In section \ref{sec4}, a framework of crystal lattice model to describe a structure of particle is introduced to compute the surface tension energy. Deviation of the surface tension energy based on the polarization energy is computed by using crystal lattice model. This means the decrease of the surface tension energy which contributes to reduction for formation of the scale.

 In section \ref{sec5}, the distribution of polarized nuclei carried by the water flow through an array of ceramic balls in a pipe is simulated numerically to investigate the behavior of polarized nuclei. The numerical results indicate important information on the best way to install an array of ceramic balls in a pipe.  In appendix, the growth of particle with the polarization is discussed by use of time-dependent Ginzburg-Landau model. Under the assumption that the radius of particle is sphere, it is shown that the time-asymptotic radius of particle is comparable to the radii of the crystals observed by using the standard  \texttt{(NH}$_4$)$_2$\texttt{CO}$_3$ \emph{Gas Diffusion Method} \cite{dietzsh}.  This proves the validity of the contact model in a method.

\section{Electro-chemical phenomena near the surface of a ceramic sphere in water} \label{sec2}

\subsection{Generalities}
It is estimated that, in water, the surface of a ceramic sphere is highly charged.  As the ceramic surface is covered with inorganic oxide, it supplies hydroxyl groups near the surface. An hydroxyl group is charged positively or negatively according to the \texttt{[pH]}. Normally the surface of a ceramic sphere is charged  negatively. Therefore \texttt{Ca}$^{++}$ ions are attracted near the surface and form a cloud.  Similarly \texttt{CO}$_3^{--}$ ions are kept away from the surface. Negative charges on the surface of the ceramic sphere and positive clouds of \texttt{Ca}$^{++}$ ions combine to form a diffused electric double layer, which brings about an electric potential near the surface of the ceramic spheres in water \cite{kitahara,derjaguin}.

\subsection{Modeling}
Let ${\bf x}\mapsto\phi({\bf x})$ be the electric potential defined at each point ${\bf x}$ outside a   disk in ${\bf R}^2$ or sphere in ${\bf R}^3$.

In natural water the concentration of  \texttt{Ca}$^{++}$ and \texttt{CO}$^ {--}_3$ are uniform and equal to the same constant $C_0$. Throughout the paper $C_0=0.017$\texttt{mol m}$^{-3}$.

In the presence of ceramic balls, because of the electric field, the concentrations $C_1$ o f  \texttt{Ca}$^{++}$ and  $C_2$ of  \texttt{CO}$^ {--}_3$ are no longer equal nor constant; however they are related by $C_1C_2=C_0^2$ due to conservation of flux  (see \cite{komiyama}).

Then the charge density ${\bf x}\mapsto\rho({\bf x})$ made by $C_1$ and $C_2$ is (see \cite{kitahara,derjaguin})
\begin{equation}
      \rho={\bar z} F( C_1 - C_2)
\end{equation}
where ${\bar z}$ is the number of electric charges of the ion, ${\bar z}=2$ for \texttt{Ca}$^{++}$ and the Faraday constant $F$=96485\texttt{(C/mol)} .

Conservation of flux requires (see \cite{komiyama}) $C_1=C_0 e^{{b}\phi}$ and $C_2=C_0 e^{-{b}\phi}$ with ${b}={\bar z} F / (R T)$, where $R =8.3145$\texttt{(J/mol/K)} denotes the perfect gas constant and $T$ is the absolute temperature. When $T=293$ then $b=79.211$. Hence

\begin{equation}\label{rho}
      \rho=2 {\bar z} F C_0\sinh({b}\phi)
\end{equation}
Furthermore $\phi$ satisfies the Poisson-Boltzmann equation,
\begin{equation}\label{P}
     \Delta \phi = \frac{\rho}{\epsilon_\gamma \epsilon_0}
\end{equation}
where $\Delta$ is the Laplace operator,
$\varepsilon_{\gamma}$ the relative permittivity of water and $\varepsilon_{0}$ the permittivity of vacuum:
\begin{eqnarray}\label{5}&&
\epsilon_\gamma=78.5, ~~~\epsilon_0=8.8542 \cdot  10^{-12}~\hbox{\texttt{(C/V\,m)}}.
\end{eqnarray}

\subsection{Boundary conditions}

The value of $\phi$ far away from the ball or disk is $\phi = 0$. The value, $\phi_1$, of $\phi$ on the boundary of the ball or disk depends on the matter of inorganic oxides which composes the surface of the ceramic sphere or disk. This is formulated as,
\begin{equation}
   \phi_1 =
    -0.0591\texttt{[pH]}-0.0295\log[P(\texttt{H}_2)].
\end{equation}
This equation is derived from Nernst equation \cite{murase} (equation (20-2)). $P(\texttt{H}_2)$ is the partial pressure of $\texttt{H}_2$, the value of which is taken as 1 atmosphere.
The present model is intended for $C_0$ in the range $O(0.01)$ mol m$^{-3}$.

\subsection{Numerical analysis of the Poisson-Boltzman Problem}

We assume spherical or cylindrical symmetry of the domain.

The domain defined by
 $r:=|{\bf x}|\in (R_b,+\infty)$  and typically $R_b=0.01$.
 With spherical ($d=2$) or cylindrical ($d=1$) symmetry of (\ref{rho})  reduces to
 \begin{eqnarray*}& \displaystyle
 -\frac1{r^d}\partial_r(r^d\partial_r\phi) + \tilde A\, C_0 \sinh({b}\phi)=0,~~&
\phi(R_b)=\phi_1,~\phi(R_\infty)=0,
\end{eqnarray*}
with $\tilde A= 2 {\bar z} F /(\epsilon_\gamma\epsilon_0)$.  Note that $d=0$  corresponds to a planar boundary.
\\
Based on the above,
$\tilde A=5.5527\cdot  10^{14}$, $b=79.211$.
The problem is now rescaled by changing the unknown to
$\psi={{b}}\phi,$
then  leading to
\begin{eqnarray}\label{system}&\displaystyle
-\partial_r(r^d\partial_r\psi) +  r^d \tilde A C_0b\sinh(\psi)=0,~ &
\psi(R_b)= b\phi_1,~~\psi(R_\infty)=0,~~~~~~~~~~
\end{eqnarray}
Curvature effects being negligible we can take d=0 and consider the equation:
\begin{equation}\label{getpsipp}
-\psi''+\mu\sinh\psi=0,~~\psi(R_b)= b\phi_1,~~\psi(R_\infty)=0
\end{equation}
where $\mu=\tilde A C_0 b$ .  Multiplied by $\psi'$ and integrated the differential equation gives:
\[
\frac12{\psi'}^2 = \mu\cosh\psi + constant
\]
As both $\psi$ and $\psi'$ tend to zero as $r\to R_\infty$, the constant must be equal to $-\mu$. Hence
\begin{equation}\label{getE}
\psi'=-\sqrt{2\mu(\cosh\psi-1)} \hbox{ leading to } \psi'(R_b)=-\sqrt{2\mu(\cosh(b\phi_1)-1)}.
\end{equation}
This formula allows the computation the electric field at the surface of the ball,

\begin{eqnarray}\label{elec}
E&=&-\frac{\partial\phi}{\partial r}(R_b)=-\frac1b\psi'(R_b) =\sqrt{\frac{2\tilde A C_0}b}\sqrt{\cosh(-4.6813\texttt{[pH]})}
\cr&=&0.3744\cdot 10^7\sqrt{C_0\cosh(4.6813\texttt{[pH]})}.
\end{eqnarray}
Variations of $E$ with respect to \texttt{[pH]} and $C_0$ are easy to compute by (\ref{elec}). Some values are given in Tables \ref{pH} and \ref{C0} for $b=79.211$.

\begin{table}[htp]
\caption{{\footnotesize Variation of $E$ with the \texttt{[pH]} when $C_0=0.017$.}}
\begin{center}
\begin{tabular}{|c|c c c c c|}
\hline
\texttt{[pH]}  & 2 & 5 & 7 & 9 & 11\cr
$E$ & $8.179\cdot 10^8$ & $9.169\cdot 10^{11}$ & $9.894\cdot 10^{13}$ & $1.068\cdot 10^{16}$
& $1.151\cdot 10^{18}$\cr
\hline
\end{tabular}
\end{center}
\label{pH}
\end{table}

\begin{table}[htp]
\caption{{\footnotesize Variation of $E$ with $C_0$ when  \texttt{[pH]}=7.}}
\begin{center}
\begin{tabular}{|c|c c c c|}
\hline
$C_0$  & 0.001 & 0.01 & 0.05 & 0.1 \cr
$E$ & $1.984\cdot 10^{8}$  & $2.400\cdot 10^{13}$  & $1.697\cdot 10^{14}$  & $2.400\cdot 10^{14}$ \cr
\hline
\end{tabular}
\end{center}
\label{C0}
\end{table}
\paragraph{Discussion}
By (\ref{getpsipp}), $\phi''(R_b)= \frac\mu b \sinh(b\phi_1)= -5.5527\cdot 10^{14} C_0\sinh(4.6813\texttt{[pH]})=-0.8033\cdot 10^{27}$.
A calcite cristal being roughly 5 angstrom wide, the variation of the electric field on that distance being proportional to $\phi''$, it is expected to vary enormously. What is a reasonable value of $E$ then?  A numerical solution of (\ref{system}) is reported in the appendix. Accurate results require an impossibly fine grid, hence the under resolved solution provides a lower value for $E$, leading to
\[
10^7<E<10^{13}.
\]
Experimental measurements in \cite{murase} confirms this analysis.

In the following sections we will use $E\sim 10^{10}$.

\section{Extraordinary Suppression of nucleation rate based on the contact model} \label{sec3}

\subsection{Activation energy of calcium carbonate particle}

Here we suppose the calcite particles are spherical with radius $r$.
 Let $\delta G$ denote the difference of Gibbs free energy between an electrolytic solution composed of Ca${}^{++}$, CO${}_3^{--}$ in water  and  one with coagulated calcium carbonate in water.. This  change depends on the ions concentration; furthermore $\delta G$ is  negative for solidification found usually in city water (See section \ref{4.3})

Let $\gamma$ be surface tension energy between calcite and water. Then total free energy change $\delta g$ for coagulated. calcium carbonate carbonate is
\begin{eqnarray}
\delta g = 4\pi \gamma r^2 + \frac{4\pi}{3}  r^3 \delta G.
\end{eqnarray}
As the particle grows, $\delta g$ increases until the radius reaches
\begin{eqnarray}\label{ten}
	r^* =- 2\frac\gamma{\delta G}
\end{eqnarray}
Namely, $\delta g$ attains maximum value
\begin{eqnarray}\label{eleven}
                \delta g^* =
		\frac{16\pi}{3} \frac{\gamma^3}{(\delta G)^2}
\end{eqnarray}
at $r = r^*$, which is the energy required for the particle to become a particle \cite{brophy}.
This is activation energy.

\subsection{Study of the nucleation caused by the electric field on the surface of ceramic sphere}

Nucleus is the original form to grow up into crystal particle. Nucleation  means that some among coagulated particles with $\delta g^*$ small enough will become isolated nucleus

 The number of nucleus generated by nucleation per unit time is defined by
\begin{eqnarray}
I= I_0\exp{\left(-\frac{\delta g^*}{kT}\right)}.
\end{eqnarray}
where $I_0$ is total number of coagulated particles.   Nucleation rate is defined by $I/I_0$ (see \cite{lamer}, \cite{doi}).
In this section we try to compute the effect on $I$ of  a polarization energy D added  to $\delta g^*$.

The electric field $E$ at the surface of the ceramic balls brings about a polarization energy density \cite{nakayama} in calcite particles near the ceramic balls
\begin{equation}\label{polar}
	D=\frac{\varepsilon^c_{\gamma} \varepsilon_{0}}2 E^2 \; (\hbox{J/m}^3).
\end{equation}
with $\epsilon^c_\gamma=9.1$ for calcite.

Adding $D$ to the activation energy $\delta g^*$ leads to a new activation energy, which is denoted by $\delta \tilde{g^*}$,
\begin{equation}
              \delta\tilde{g^*} = \delta g^* +\frac{4\pi}{3} {r^*}^3 D
\end{equation}

Some among coagulated particles with $\delta g^*$ change to be nucleus, the number of which is defined by $I = I_0 \exp(  \frac{-\delta g^*}{kT})$.
Where $I_0$ is total number of coagulated particles. This is by defiinition the  nucleation rate.:  $\tilde{I}=I_0 \exp{\left(-\frac{\delta\tilde g^*}{kT}\right)}$.

 In order to verify the effect brought about by the perturbed energy to nucleation rate, we study the ratio $I^*/I$:
\begin{equation}\label{ii}
\frac{\tilde I}{I}= \exp{\left(-\frac{\delta\tilde g^*-\delta g^*}{kT}\right)}=e^{-q}
~~~~\hbox{ with }  q =\frac{4\pi}{3} {r^*}^3 \frac {D}{kT}.
\end{equation}

\subsection{Suppression of nucleation rate} \label{4.3}

Following \cite{chang} \cite{giese} \cite{murase}, experimental values are

\begin{description}
\item[-]  Surface tension (free) energy: $\gamma=57 \cdot  10^{-3}$ (J/m$^2$).
\item[-]   $\delta G_0 = - 47.4251$(kJ/mol) =- 1.284 10$^9$(J/m$^3$) .
\item[-] $ \delta G =-\delta G_0 - RT\log{C_0^2}$

\end{description}

Note that $\delta G_0 \le \delta G$.
Therefore, by (\ref{ten}), $r^* > 0.8903\cdot 10^{-10}$ and by (\ref{polar}) and (\ref{ii}) and the computation of the previous paragraph showing that $E>10^{10}$,
\begin{equation}\label{q}
q > 3.072\cdot 10^{-20} E^2>3.072  ~\Rightarrow~\frac{I^*}I=\exp(-q) < 0.046 .
\end{equation}
Therefore the number of nuclei which contributes to scale formation becomes considerably less; an extraordinary suppression of nucleation rate occurs due to the contact model.

\newpage

\section{Framework of Crystal Lattice Model}\label{sec4}

Another contribution to the polarization energy for scale prevention is the decrease of surface tension energy of the particle in water due to the structural change of calcite caused by  the perturbed energy .
\\
In order to investigate this phenomenon, we adopt the crystal lattice model developed by S. Ono \cite{ono} and apply it to the crystal structure of calcite. Considerably good agreement between theory and experiments is reported in his book.

\subsection{Helmholtz free energy defined  for a crystal}

The 3-dimensional crystal lattice model assumes that the crystal is composed of molecules arrayed on lattice points (See Figure \ref{fig:CrystalStructure}).

\begin{figure}[htbp]
\begin{center}
\includegraphics[width=0.4\textwidth]{fig/CrystalStructure0.png}~~
\includegraphics[width=0.4\textwidth]{fig/CrystalStructure.png}
\caption{{\footnotesize Multi-layers crystal structure at zero Kelvin is periodic (left) while somewhat random at ambiant temperature (right).}}
\label{fig:CrystalStructure}
\end{center}
\end{figure}

The interaction between molecules is restricted to those which are side by side.

Let  $S$ be the entropy. A state with small entropy is a special state which occurs under some restrictions. If these are removed,  many more possible states will be observed with greater entropy.
\\
The crystal tends to stay at a state of minimum energy at temperature 0K.  Then all molecules try to fill the empty positions (the white holes of Figure \label{fig4} on the left) and crystal becomes a complete crystal; let $X_0$ be the number of such molecules. In this situation the number of configuration is $W=1$.
\\
If the temperature rises, some of molecules leave the black holes and try to occupy other white holes. Let $X_1$ be the the number of such molecules . The total number of all different configurations is \cite{reif}
\[
W(X,X_1)=\frac{X_0!}{X_1! \cdot (X_0-X_1)!}.
\]
The Boltzmann principle says that
\begin{eqnarray}
	S = k \log{W(X_0, X_1)}.
\end{eqnarray}
Using  Sterling's formula, an approximate formula is obtained
\begin{eqnarray}
S=-X_0k\left\{ Z \log{Z} + (1-Z) \log{(1-Z)}\right\}
\end{eqnarray}
where $Z= X_1/ X_0$ and $k$ is Boltzmann constant and layers are supposed uniform.

On the other hand, let the number of adjacent molecules depending on the lattice structure be $\bar\alpha$, then an internal free energy $U$ of this system is
\begin{eqnarray}
	U = -\frac e2\bar\alpha X_0  Z^2.
\end{eqnarray}
where $-e$ is the potential energy between adjacent molecules experimentally measured to be (see \cite{archer})  $e=4\cdot 10^9/\rho$ \texttt{(J)}.
Therefore Helmholtz free energy $F$ for this arrangement is
\begin{eqnarray}
	F = U - TS.
\end{eqnarray}

\paragraph{Notations}
 It should be noted that the crystal is composed by many layers.
Here we have considered the free surface of a plane layer of crystal parallel to the other lattice layers.  To analyse a stack of lattice layers we label them from $1$ to $m$.

$A$ will stand for the area of each layer; $n$ will denote the umber of lattice points per unit area on each layer, and
$n_i$  the number of molecules included in a unit area  of the i-th layer.
The total number of molecules is denoted $N$.
Then  $x_i = n_i/N$ is the concentration of molecules per unit surface in the i-th layer.

Finally $h$ will denote the polarization energy per molecule and  $\rho$ the density of calcium carbonate, namely $\rho=1.625\cdot 10^{28}$ m$^{-3}$

\subsection{Multi-layered Crystal}

In order to compute the internal free energy of a crystal made of a stack of 2-dimensional layers, it is noted that the number of neighboring molecules of one molecule is 6 for the same layer and 3 for each neighboring layer.

The particle pairing number between the $i$-th layer and its adjacent neigbhour $i+1$ is $3nA x_{i+1} x_i$ -- and similarly for $i-1$ -- while the particle pairing with the ith layer is $6nA x_i^2$.  Removing the pairs counted twice in this enumeration, comes to
\begin{eqnarray}
U &=&- 3e\,nA [x_1(x_1+ \frac12 x_2)
      + \sum_{i=2}^{m} x_i(x_i + \frac12(x_{i-1} + x_{i+1}))] \label{eqn:U} \\

S &=&-  n A k\sum_{i=1}^{m} (x_i \ln x_i +(1-x_i) \ln(1-x_i))
\end{eqnarray}

The total energy, $\bar{\bar F}$, is the sum of Helmholtz' free energy  ${{\bar F}} = U - TS$ with the polarisation energy $h\,x_1$ due to the contact model where  $h=D/\rho$ and $D$ is defined in (\ref{polar}) (see \cite{ono}):
\begin{eqnarray}
 \bar{\bar F} &=& n\, A \, h x_1 - 3e\,nA[x_1(x_1+ \frac12 x_2)
      + \sum_{i=2}^{m} x_i(x_i + \frac12(x_{i-1} + x_{i+1})) \nonumber \\
	 &+&
   T n A k\sum_{i=1}^{m} (x_i \ln x_i +(1-x_i) \ln(1-x_i))~~~
\end{eqnarray}

\subsection{ Minimization problem}

The stable state of the system is obtained by minimizing  ${\bar{\bar F}}$ under the constraint that the chemical potential energy $n\,A\,\eta\sum_{i=1}^m x_i$ is constant; $\eta$ is the chemical potential per particle :
\begin{equation}\label{eta}
\eta=-12 z e + k T \ln\frac z{1-z}.
\end{equation}
 With $a=k T/e=\frac{kT}e=0.01575$. and $\beta=h/e$, the optimality conditions are
\begin{eqnarray}
\label{euler3}
\beta -3 \left(2 x_1 +x_2 \right) +  a \ln{\frac{x_1}{1-x_1}}  &=& -12z+a\ln\frac{z}{1-z}\\ \label{euler1}
-3 \left(x_1 +2x_2 +x_3 \right)  +  a \ln{\frac{x_2}{1-x_2}} &=& -12z+a\ln\frac{z}{1-z} \\
-3 \left(x_{i-1} +2x_i +x_{i+1} \right)  +  a \ln{\frac{x_i}{1-x_i}} &=& -12z+a\ln\frac{z}{1-z} \\
  \label{euler2}
-3 \left(x_{m-1} +3x_m \right) +  a \ln{\frac{x_m}{1-x_m}}  &=& -12z+a\ln\frac{z}{1-z}.
\end{eqnarray}
If $\beta=-3z$ the solution is $x_i=z,~i=1\dots m$. \\\\
Let the number of layers be large and $\beta+3z$ be small. Then  let $x_i=z+x'_i$ and $\beta'=\beta+3z$ so that
\begin{eqnarray}
-\beta' +3 \left(2 x'_1 +x'_2 \right) &=&  a \ln{\frac{1+\frac{x'_1}z}{1-\frac{x'_1}{1-z}}}
\approx a(\frac{x'_1}z+\frac{x'_1}{1-z}) \\
3 \left(x'_{i-1} +2x'_i +x'_{i+1} \right)  &=&  a \ln{\frac{1+\frac{x'_i}z}{1-\frac{x'_i}{1-z}}}
\approx a(\frac{x'_i}z+\frac{x'_i}{1-z})\\
3 \left(x'_{m-1} +3x'_m \right) &=&  a \ln{\frac{1+\frac{x'_m}z}{1-\frac{x'_m}{1-z}}}\approx a(\frac{x'_m}z+\frac{x'_m}{1-z}).
\end{eqnarray}
Notice that a finite difference approximation of $\displaystyle\alpha u-\frac{d^2 u}{d y^2}=0$ on a uniform grid of size $dy$ is $\alpha u_i-(u_{i-1}-2u_i+u_{i+1}/{\delta y}^2$ . By analogy we see that the above is a finite difference discretization of
\[
-3{\delta y}^2\frac{d^2 u}{d y^2}(y) + [a(\frac1z+\frac1{1-z})-12{\delta y}^2]u(y)=0,~y\in(0,1),~~u(0)=-\frac{\beta'} 3,~~\frac{d u}{d y}(1)=0,
\]
the solution of which is $u=c e^{\alpha y} + b e^{-\alpha y}$ with
\[
\alpha=\sqrt{\frac a{3{\delta y}^2}(\frac1z+\frac1{1-z})-4},~~c=\frac{\beta'e^{-2\alpha }}{3(1+e^{-2\alpha })}, ~~ b=\frac{\beta'}{3(1+e^{-2\alpha })}.
\]
Thus
\[
x'_i\approx u(ih)\approx \frac{\beta'}3 \exp\left(-i\sqrt{\frac a{3z(1-z)}}\right)
~~\Rightarrow~~x_i\approx z+(\frac\beta 3 +z)\exp\left(-i\sqrt{\frac a{3z(1-z)}}\right).
\]

Let us use this solution to compute by (\ref{eqn:U}) the Helmholtz energy for each layer of the uniform phase (the bulk) . The chemical potential is defined below; the Helmholtz free energy for each layer of the uniform phase (bulk)is defined
\begin{eqnarray}
\tilde F&=&{n A} \cdot [-6z^2 \cdot e + kT \cdot
	\{z \cdot \ln{z} + (1-z)\ln{(1-z)}\}]. \label{eqn:Fu}
\end{eqnarray}
where $z$ is the density of molecule per unit surface, i.e. the limit of $x_m$ for $m$ large.

The term $-6z^2$ in the right hand side is obtained by taking $x_m=x_{m-1}=z$ in the last term of the  sum in the right hand side of (\ref{eqn:U}).
\\
Let $m \rightarrow \infty$. Then we may regard $\eta$ as the chemical potential \cite{ono} and
\begin{eqnarray}
\eta = \frac{1}{nA} \cdot \frac{d\tilde F}{dz} = -12z \cdot e + kT \cdot \ln{\frac{z}{1-z}}.
\end{eqnarray}

\subsection{Computation of Surface Density and Polarization Energy}

The experimental value of molecular entropy for calcium is 4.54 (J/K/mol) at room temperature, and per particle it is 4.54/$N_A$, where $N_A$ is the Avogadro number \cite{staveley}.  Equating it with the theoretical value gives
\[
\frac{4.54}{N_A}=-k\left(\frac{X_1}{X_0}\ln\frac{X_1}{X_0} + (1-\frac{X_1}{X_0})\ln(1-\frac{X_1}{X_0})\right)
~\Rightarrow~X_1=0.2348 X_0
\]
The number of particles per unit volume being $(X_1/X_0)^{2/3}$,
\begin{equation}\label{z}
z=(0.2348)^{2/3}=0.3806.
\end{equation}
The polarization energy per particle, $\beta=h/e=a D/(kT\rho)$ is
\begin{equation}\label{beta}
\beta= 0.1007 \cdot 10^{-20}E^2.
\end{equation}

 \subsection{Deviation of surface tension energy for particle under the contact model}

\subsubsection{ Definition of Surface Tension Energy}
Following  \cite{ono}, surface tension energy is defined by
\[
            \gamma ( x_1, x_2,... ,x_m, z) = \frac{{\bar{\bar F}}- m \tilde F}A - n \eta\sum_{i=1}^{m}(x_i -z)
\]
The sensitivity of $\gamma$  with respect to $x_i$ and $z$ is
\[
           \delta \gamma = \sum_{i=1}^{m}\frac{\partial\gamma}{\partial x_i}\delta x_i +  \frac{\partial\gamma}{\partial z}\delta z
\]
In view of the definition of ${\bar{\bar F}}$, at equilibrium $\partial \gamma/\partial x_i = 0$ for $i=1,2,\cdots,m$. Hence
\[
          \delta \gamma
          = - m (-12z \cdot e + kT \cdot \ln{\frac{z}{1-z}})\delta z  - n  \frac{\partial\eta}{\partial z}\sum_{i=1}^{m} (x_i-z) \delta z -n\eta\sum_1^m(-1)
 \]
Here we used (\ref{eta}) to compute $\frac{\partial\eta}{\partial z}$ to obtain
\begin{equation}\label{dgama}
\delta \gamma
          =  -n(-12 e + \frac{k T}{z(1-z)})\sum_1^m(x_i-z)\delta z = -n\frac{\partial\eta}{\partial z}\delta z\sum_1^m(x_i-z)
\end{equation}

\subsubsection{Formula for $\delta\gamma/\gamma$}

Let us notice that an additional energy $h=\displaystyle\frac{\partial\eta}{\partial z}\delta z$ corresponds to the increase of chemical potential $\eta$ for the surface layer; $h$ is related to the polarisation energy defined in (\ref{polar}) by $D=\rho h$.
\\\\
Using only the first layer in (\ref{dgama}) we have $\delta\gamma=-n h  (x_1-z)$ where $n$ is the surface density of particle, related to $\rho$ by $n=\rho^{2/3}$.
Using the value for $\gamma$ given in section \ref{4.3} , we have
\[
\frac{\delta\gamma }{\gamma}= \rho^{2/3}\frac D{\rho\gamma} (x_1-z)
\]

\subsubsection{Computation of $\delta\gamma/\gamma$}

If we set $E=\alpha 10^{10}$ (see section \ref{sec2}),  $\beta=0.1007\alpha^2$ (see (\ref{beta})) then
\[
x_1-z = (\frac\beta 3 +z) \exp(-\sqrt{\frac a{3z(1-z)}})  = 0.2892\alpha^2+0.3278.
\]
It follows that  $\delta\gamma/\gamma=-0.807\alpha^2(\alpha^2+1.134)$.

\\\\
It is pointed out in \cite{higashitani} that the perturbation energy on the particle of calcium carbonate is effective  for the prevention of scale formation
only in a specific range of \texttt{[pH]}, $C_0$ and P(H$_2$).
\\
Note that when $\alpha^2=2.553$, $\delta\gamma/\gamma=-1$ approximately.  Then $\gamma$ becomes $\gamma + \delta \gamma$ which is also $\gamma(1+\delta\gamma/\gamma)=0$; hence there is no surface tension energy, or at the least heavily reduced.

\section{Numerical Simulations of Calcite Carried by the Flow through an Array of Ceramic Spheres}\label{sec5}

We have seen above that calcite particles undergo a change very near the ceramic spheres due to polarisation.
As this change is very fast, vorticity in the flow  near the surface of ceramic spheres increases the number of particles affected by polarisation.
The water flow being at moderate Reynolds number, it is well modelled by the Navier-Stokes equation (\ref{nseq}) with zero velocity on the balls. The concentration of calcite is modelled by a convection diffusion equation, but the boundary condition needs to reflect the fact that the ceramic balls decrease the concentration and more so as the vorticity is large which reflect the fact that upstream face  of the ball is more active than its down stream side.

The purpose of this section is to measure the macroscopic effect of an array of ceramic balls on a flow of calcite in water.  We make  two groups of simulations; one with an inflow on the right and an outflow on the left in a box-like container. The parameters are the geometrical arrangements of the ceramic balls, the pressure difference between the inlet and the outlet and the size of the container.

The second group of simulations is an attempt to replicate numerically the experiment of \cite{kawarada}.  It is a cylindric container with a rotating magnet below it which induces a rotation of the balls around the axis of the cylinder. There also there are many geometrical parameters in addition to the roating velocity.

For all simulations, the polarisation efficiency of the ceramic balls is an important last parameter, but as we have no access to it, it must be guessed; hence this section remains phenomenological, giving trends rather than experimentally comparable numbers.

\subsection{ The Partial Differential Equations}

The flow is solution of the Navier-Stokes equations:
\begin{equation}\label{nseq}
 \partial_t u + u\nabla u + \nabla p - \nu \Delta u=0,~~\nabla\cdot u=0 .
\end{equation}
Here $p$ is the pressure and $u$ is the velocity. The reduced viscosity is fixed at $\nu=1/500$. The velocity is given at $t=0$:  $u=0$; it is also zero on the spheres (disks) and given on the walls of the container except at the inlet and outlet where we prefer to impose a pressure gradient and a zero tangential velocity.
\\\
The water is charged with calcite at concentration $C_0$ initially and at the inlet pipes. Everywhere else we expect the concentration of calcite to be $C(x,t)\leq C_0$; without loss of generality we can take $C_0=1$. The reaction to reduce calcite occurs mainly within a very thin boundary layer around each sphere, modelled by
\begin{equation}\label{cbdy}
\epsilon\frac{\partial C}{\partial n} = \xi~|\sigma_n|~C
\end{equation}
where $n$ is the normal to the spheres, $\sigma_n$ is the normal stress in the fluid, i.e. $\displaystyle\sigma_n=p n + \nu\frac{\partial u}{\partial n} $ and $|\sigma_n|^2=p^2+\nu^2(\frac{\partial u_s}{\partial n})^2$.  Intuitively a strong pressure will bring the  calcite nearer to the balls and a strong tangential part of the normal stress will slow down the calcite particle forcing them to stay longer near the balls.
The coefficient $\xi$ is a decreasing function of $\beta$ which governs the absorption rate due to magnetic spheres. It depends also on the reaction rate of calcite transformation and to some extend on the time the calcite stays near a ball i.e. on the fluid velocity $u$.
In addition, the equation for $C$ is
\[
\partial_t C + u\nabla C - \epsilon \Delta C=0,
\]
with $C=0$ at the walls, $C=C_0$ at the in-pipe and $\partial C/\partial n=0$ at the exit pipe and (\ref{cbdy}) on the spheres.
The variational formulation is: find $C$ such that for all $\hat C$,
\begin{equation}\label{cconcentr}
\int_\Omega( \hat C(\partial_t C + u\nabla C) + \epsilon \nabla\hat C\cdot\nabla C + \int_{spheres}\xi|\sigma_n|\hat C C
=0
\end{equation}
\begin{remark}
It is difficult to derive theoretically a reliable formula for $\xi$ in terms of $\beta$.  The molecular diffusion $\epsilon$ needs also to be measured but it is small; in all computations below $\epsilon=\nu$.
\end{remark}

\subsection{Flow in a Cavity through an Array of Ceramic balls}

In a parallelepiped shaped container of size $X_l\times Y_l\times Z_l=1\times 0.7\times 1$ (in meters) water flows from the in-pipe on the right to the exit pipe on the left. The diameter of the pipes is $w=0.1$; the center of the inflow pipe is at $(X_l,Y_b+w/2,Z_l/2)$ and for the outflow pipe at $(0,Y_b+w/2,Z_l/2)$. Here $Y_b=T_l/2$. The radius of the spheres is $r=0.012$.

The number of spheres in the directions $x,y,z$ are respectively $m_a,n_a,o_a$. These are evenly distributed in the parallelepiped of size $x_c,y_c,z_c$ and centered at $X_a,Y_a,Z_a$, so that there are spheres 	on the corners of the parallelepiped.
\begin{figure}[hbt]
\begin{center}
\begin{tikzpicture}[scale=4]
\draw  (0,0) -- (1,0) -- (1,0.36);
\draw (1,0.4) -- (1,0.7) -- (0,0.7) -- (0, 0.4);
\draw (0,0.36) -- (0,0);
\draw [dotted] (0.3,0.15) -- (0.7,0.15) -- (0.7,0.55) -- (0.3,0.55) -- cycle;
\draw [fill] (0.3,0.15) circle [radius=0.012];
\draw [fill] (0.35,0.15) circle [radius=0.012];
\draw [fill] (0.4,0.15) circle [radius=0.012];
\draw [fill] (0.45,0.15) circle [radius=0.012];
\draw [fill] (0.3,0.2) circle [radius=0.012];
\draw [fill] (0.35,0.2) circle [radius=0.012];
\draw [fill] (0.4,0.2) circle [radius=0.012];
\draw [red,thick,densely dotted, <->] (0,0.35) -- (0.49,0.35) ;
\draw [red,thick,densely dotted,<->] (0.5,0.) -- (0.5,0.34) ;
\draw (0.2,0.39) node [right] {$X_a$};
\draw (0.5,0.16) node [right] {$Y_a$};
\draw [red,thick,densely dotted, <->] (0,0.75) -- (1,0.75) ;
\draw [red,thick,densely dotted,<->] (1.1,0.) -- (1.1,0.7) ;
\draw (0.5,0.8) node [right] {$X_l$};
\draw (1.1,0.35) node [right] {$Y_l$};
\draw [blue] (0.98,0.378) node [right] {{\tiny in}};
\draw [blue] (0.,0.375) node [left] {{\tiny out}};
\draw [red,thick,densely dotted,<->] (0.9,0.) -- (0.9,0.36) ;
\draw (0.88,0.15) node [right] {$Y_b$};
\draw [blue,thick,densely dotted, <->] (0.3,0.57) -- (0.7,0.57) ;
\draw [blue,thick,densely dotted,<->] (0.72,0.15) -- (0.72,0.55) ;
\draw (0.48,0.6) node [right] {$x_c$};
\draw (0.73,0.35) node [right] {$y_c$};

\end{tikzpicture}
\caption{{\footnotesize Sketch of the computational domain: a rectangle $(0,X_l)\times(0,Y_l)$ minus the black disks. These are evenly distributed in a rectangle $(X_a-x_c/2,X_a+x_c/2)\times(Y_a-y_c/2,Y_a+y_c/2)$. Water charged with calcite enters from a hole on the right vertical border and exit by a hole on the left border. These are the two segments $\{X_l\}\times(Y_a, Y_a+w)$ and $\{0\}\times(Y_a, Y_a+w)$}}
\label{sketch}
\end{center}
\end{figure}

The problem is solved in 2D.  We shall attempt to measure the errors connected with such an approximation at the end of this section.

 A pressure gradient is imposed between the inlet and the outlet $p_{in}-p_{out}=10$; this value is arbitrary but it fixes Reynolds number at a numerical value of a few hundred, compatible with the mesh resolution used here (around 15000 vertices). The computed horizontal velocity due to the pressure gradient depends on the layout of the spheres and is around 7; thus the Reynolds number based on the disk diameter is 70 and 3500 when based on the size of the domain.

Figure \ref{flowconcentration1} shows the concentration levels at T=2.5 after 50 time steps for $\xi=0.02$ and $0.2$.  A reduction  is obtained between $C_{in}$ and $C_{out}$ seen in colors changing from red to blue; the reduction is much more intense with $\xi=0.2$.
\\
Figure \ref{flowconcentration2} shows the concentration levels at T=2.5 after 50 time steps for $\xi=0.2$ for a different geometrical configuration.  It is clear that such a design which let the fluid flow through the container without coming into contact with the balls is a bad design.

The effect of a stronger pressure gradient is obvious and not shown here:  above a certain threshold there is only a partial transformation of the calcite, unless the number of rows of balls is increased.  Below the threshold the calcite lingers unnecessarily long near the balls.

\begin{figure}[htbp]
\begin{center}
\includegraphics[width=6cm]{fig/C002.png}~~
\includegraphics[width=6cm]{fig/C02.png}\\
\caption{{\footnotesize {Display of $C$ at T=2.5 after 50 time steps (nearly stationary)  showing the transformation of hard water (red) into soft water(blue). The water runs from right to left. On the right the flows enter saturated with calcite (red) by an aperture. It flows through an array of magnetic disks, looses its calcite and exits through a section slightly above the middle  of the left vertical boundary. Left figure: $\xi=0.02$. Right figure, $\xi=0.2$. }}}
\label{flowconcentration1}
\end{center}
\end{figure}

\begin{figure}[ht]
\begin{center}
	\begin{minipage}[b]{0.45\linewidth}
	\centering
\includegraphics[width=6cm]{fig/passivescalar2a.png}
\caption{{\footnotesize {Display of $C$ at T=2.5 after 50 time steps for $\xi=0.2$ for a different geometrical configuration which proves to be less efficient. It show that the array of spheres should block the pipe fully.}}}
\label{flowconcentration2}
	\end{minipage}
	\hspace{0.7cm}
	\begin{minipage}[b]{0.45\linewidth}
	\centering
\includegraphics[width=4cm]{fig/axiflow.png}
\caption{\label{axiveloc}{\footnotesize
Angular velocity in a cylinder with a rotating bottom}}
	\end{minipage}
\end{center}
\end{figure}

\subsection{Rotating Flow in a Cylinder with Ceramic Balls}

In an experiment reported in \cite{kawarada}, a rotating magnetic agitator below the container induces a rotational flow with an added magnetic agent \bb{ is it so} in a circular cylinder where sits, near the bottom, several layers of magnetically coated balls which do not rotate with the flow.

For the simulation it is easier to assume that it is the bottom of the cylinder which is rotating while the balls and the vertical boundary of the cylinder stay put.

If the balls are ignored, the solution of the Navier-Stokes equations in a cylinder with prescribed angular velocity on the bottom boundary is, in cylindrical coordinate, $u_r=u_z=0$, $u_\theta$ solution in a rectangle of
\[
-\frac1r\partial_r(r\partial_r u_\theta) + \frac{u_\theta}{r^2}-\partial_{zz}u_\theta=0
\]
with $u_\theta=\omega r$ at the bottom of the cylinder and  zero on the side.

The solution is shown on figure \ref{axiveloc} when $\omega=1$.

Next we zoom-in on a row of fixed balls in such a flow. We assume that at one point in the cylinder we can reproduce $u_\theta$ by a pressure gradient according to the formula $\frac1r\partial_\theta p=u_\theta$ and then solve the Navier-Stokes equations locally around the spheres with such a pressure gradient to see what happens to the calcite.  We begin with a 2D simulation and then end with a small 3D configuration.

\subsubsection{2D simulation}

For simplicity the computation is done first in 2D, in a rectangular domain $D_p=(0,L_x)\times(0,L_y)$ with N=9 disks of radius $R=0.012$ at a distance from their neighbour $d_{ist}=0.004$ as shown on figure \ref{ballrow1}. It should be understood as being put at some point in the cylinder of figure \ref{axiveloc} perpendicular to the plane of the figure.
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=12cm]{fig/pressure.png}
\caption{\label{ballrow1}{\footnotesize
Navier-Stokes solution due to an horizontal pressure gradient which induces the water to flow from right to left. The color map shows the pressure. }}\end{center}
\end{figure}

A stationary solution of the Navier-Stokes equations (\ref{nseq}) is computed in $D_p$ driven by a pressure gradient $\partial_x p_0=30/L_x$ and periodic boundary conditions on top-to-bottom and on left-to-right boundaries.
Results are shown on figures \ref{ballrow1} and  \ref{ballrow2} with :
\[
L_x=2 N(R+d_{ist}),~L_y=2(R+d_{ist}),~\nu=\epsilon=1/100
\]
Notice that the flow is quite periodic so that one disk only and periodic conditions is sufficient to analyse the flow.
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=13cm]{fig/nsperiodic.png}
\caption{\label{ballrow2}{\footnotesize
Navier-Stokes solution. Zoom between the second and third disk from the left showing the velocity vectors and the pressure lines..}}
\end{center}
\end{figure}
Finally we use the velocity field computed above and solve (\ref{cconcentr}) with $\xi=0.02 $ and $\xi=0.2$.  Periodic conditions are imposed on top and bottom boundary; on the right boundary $C=1$ and on the left boundary no condition is imposed.
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=12cm]{fig/concentre1.png}
\caption{\label{ballrow3}{\footnotesize
Color plots showing the concentration $C$ solution of (\ref{cconcentr})
when $\xi=0.2$ (top) and $\xi=0.02$ (bottom). Notice that the concentration decreases from 1 to 0.88 after the first disk when $\xi=0.02$ and from 1 to 0.65 when $\xi=0.2$.}}
\end{center}
\end{figure}
The computation shows, on figure \ref{ballrow3}  the concentration when the system has reached a stationary state, here at t=2. Notice that much of the calcite is transformed by the first balls from the right, but again much more efficiently when $\xi=0.2$.

\subsubsection{3D Computation}

Ideally we need to solve the same problem in 3D, namely an array of balls in a parallelogram with periodic conditions on the sides and a flow driven by a pressure gradient.  However the 2D simulations above show that we can gather already a fair amount of information by solving the flow and the equation for the calcite around one ball only provided periodic conditions are applied.
\\
The following is a computation of the Navier-Stokes equations coupled with (\ref{cconcentr}) with one ball only of radius $R=0.012$ in a cube of size 0.026 with periodic conditions on the velocities and driven by a pressure gradient equal to -3/0.026, a value similar to the 2D case. The concentration is imposed equal to 1 on the side of the cuber where the flow enters and no condition on other boundaries, except (\ref{cbdy}) on the ball. Other parameters are as in the 2D computation above, $\nu=\epsilon=0.01, ~\xi=0.2$.
\\
Results are shown on figure \ref{troisd}. The plot of the iso-surface of the concentration seems to indicate that a 3D ball is  less effective than a cylinder (2D disk) to eliminate the calcite; the calcite concentration is shown  at t=1. It also shows that 2D computations do not reflect reality and a massive 3D computation needs to be done for reliable numbers.
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=4cm]{fig/pressure3D.pdf}
\includegraphics[width=4cm]{fig/threedc.pdf}
\includegraphics[width=3.5cm]{fig/twodc.png}
\caption{\label{troisd} {\footnotesize
3D simulation. Left: Pressure in the fluid for a Navier-Stokes flow driven by a difference of pressure from right to left. Right: concentration of calcite, solution of (\ref{cconcentr}). Here $\xi=0.2$. Notice that the concentration decreases from 1 at the inlet to 0.987 at the outlet boundary, while for the same geometry and pressure difference but in 2D disk the calcite concentration decreased from 1 to 0.81 (lower plot).
}}
\end{center}
\end{figure}
\\\\
To conclude, all computations have demonstrated a positive influence of the ceramic balls on the formation of scale in the sense that strongly polarised nuclei give rise to less surface tension energy which in turn implies a larger $\xi$ and hence less concentration of calcite in the water passing near the ceramic balls.

\section*{Conclusion}

The effect of an array of ceramic balls on hard water has been studied. A number of facts have been found which could possibly explain the beneficial effect of the balls on the growth of calcite crystals of \texttt{CaCO}$_3$.\

A model to compute the electric field at the surface of the balls has been given. It has been found that the number of polarized nuclei which contribute to scale formation is considerably smaller than in natural water.

 Then a two dimensional macroscopic model has been investigated numerically showing that the effect of the ceramic balls could be reproduced in configurations studied experimentally and industrially.
\\
The general conclusion is  that the ceramic balls induce a polarization in the calcite particles which decrease the surface tension energy of the crystal  in the vicinity of the ceramic balls.

\paragraph{Acknowledgements}

The authors are grateful to S.Ikawa for bringing them first about what and where the essential problem in this paper is. They would like to thank  H.Fujita for the useful discussions related with this paper.    They also would like to thank  R. Inaba ( MSSA S.A.S.)for furnishing data on thermodynamic properties of calcium carbonate and the surface physics and chemistry.  Finally the first author would like to thank to  H. Suito for his long standing scientific support.
\\
The first author's contribution to this paper is for the modeling while the second author has concentrated on the numerical simulations.

config: arxiv


TITLE: Show convergence of sequence $\frac{\ln n}{n^p}$
QUESTION [0 upvotes]: I just learned about the Cauchy Criterion for convergence of a sequence:

Theorem: The sequence $\{ s_n \}^\infty_{n=1}$ converges if the difference $s_n-s_m$ approaches zero as both $m$ and $n$ approach infinity.

I tried to use this to test for the convergence of $$\{ s_n \}^\infty_{n=1}=\{\frac{\ln n}{n^p} \}^\infty_{n=1}$$
where $p$ is a positive number.
Following the theorem, I wrote
$$\lim_{n\rightarrow\infty}\frac{\ln n}{n^p}-\lim_{m\rightarrow\infty} \frac{\ln m}{m^p}=0,$$
hence the sequence is convergent.
I realised that I am doing something wrong, since I'm subtracting the same things here. What is the correct way to do this?

REPLY [1 votes]: One way to prove that the sequence converges is $$\lim_{n\to \infty} \frac{\ln n}{n^p} = \lim_{n\to\infty} \frac{1/n}{pn^{p-1}} =\lim_{n\to\infty} \frac{1}{pn^p} =0.$$
To use the Cauchy criterion, first suppose $p > 1$ and we fix $\epsilon > 0$ and choose $N$ large such that $N^{p-1} > 2/\epsilon$ and have $n\geq m > N$ so that $$\left\vert \frac{\ln n }{n^p} - \frac{\ln m}{m^p} \right \vert  \leq \frac{\ln n }{n^p} + \frac{\ln m}{m^p} \leq \frac{n}{n^p} + \frac{m}{m^p} = \frac{1}{n^{p-1}} + \frac{1}{m^{p-1}} \leq \frac{2}{N^{p-1}} < \epsilon.$$
For $p \leq 1$, choose a $q$ such that $p \geq 1/q$. Now fix $\epsilon > 0$ and $N$ large such that $N^{p-1/q} > 2q/\epsilon$ and have $n\geq m > N$ so that $$\left\vert \frac{\ln n }{n^p} - \frac{\ln m}{m^p} \right \vert  \leq \frac{\ln n }{n^p} + \frac{\ln m}{m^p} = \frac{q \ln n^{1/q}}{n^p} + \frac{q\ln m^{1/q}}{m^p} \leq \frac{qn^{1/q}}{n^p} + \frac{qm^{1/q}}{m^p} \leq \frac{2q}{N^{p-1/q}} < \epsilon.$$
Hence $\{\frac{\ln(n)}{n^p}\}$ is a Cauchy sequence.
Note that the second case doesn't really require the $p\leq 1$. We could remove this and realize the first case is a special case of the second argument where $q=1$.


TITLE: What is a practical use for this metric?
QUESTION [19 upvotes]: Showing $\rho (x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric
That post has a more generalized form of a metric I occasionally see $d(x,y) = \frac{|x-y|}{1+|x-y|}$. When would using this metric be useful exactly? It occasionally comes up when I study analysis, but I don't know why, I don't know what people use it for or what benefit it could ever bring over the standard metric. I've merely only seen it as an example of a metric in books or sites, but if so many sources mention it, then it's very unlikely that it's useless.

REPLY [2 votes]: I know this thing: if $(X, d)$ supports a measure $\mu$, then convergence in measure, for a sequence of functions, is the same as convergence with respect to the (integral of the) metric you cite. That is: on $\mathrm{Meas}(X)$, define the metric
$$
d_\mu(f, g) = \int \frac{d(f(x), g(x))} {1+d(f(x), g(x))} d\mu(x)
$$
Then a sequence of measurable functions converges in measure iff it converges with respect to the metric $d_\mu$.
This is stated in Tao's book on measure theory, if I recall correctly.


TITLE: Is the fundamental group of a non-semi-locally simply connected space uncountable?
QUESTION [4 upvotes]: I know that the fundamental group of the Hawaiian earring is uncountable, so I would think you can argue it injects into such a space, but I think that would rely on some condition like having a countable neighborhood basis.

REPLY [3 votes]: Here is a proposed counterexample to my own question:
Let $P$ be an ordered set with such that there is no order preserving function $f$ from the natural numbers that has the property that for any element $x \in P$ there is an integer $y$ such that $x \leq f(y)$.
Put a topology on $X=P\cup \{*\}$ by declaring as a basis the individual points in $P$ and the sets containing $*$ such that there is an element $x \in P$ in the set and that if $x \leq y$, then $y$ is in the set. If $P$ were the natural numbers this would be the set $\{0\}\cup\{1/n | n \in \mathbb{N}\}$ with the subspace topology.
Take the reduced suspension of $X$ where the basepoint is $*$ and attach a disk from the arc corresponding to the element $x \in P$ to the arc corresponding to $y \in P$ if $x < y$. Call this space $X'$.
I claim $X'$ is not semi-locally simply connected. Any neighborhood around the basepoint contains some circle corresponding to some element $x \in P$. It is pretty clear that the loop obtained by traversing this circle once can only be contracted by moving toward the basepoint through these arcs we added. However, a homotopy can only take us through countably many of these arcs so by the condition on $P$ this homotopy cannot end at the basepoint, so the loop is not trivial.
The fundamental group is also not uncountable because it's pretty clear the loops corresponding to the elements of $P$ have to be the building blocks, but there cannot be infinite sequences of them like in the Hawaiian earring because by the condition on $P$ these loops are not approaching the basepoint. In fact, I think that the fundamental group is $\mathbb{Z}$ generated by the loop corresponding to some element of $P$.
Again this is only a sketch, please dissect the argument because I would like it to be the case the titular question has an affirmative answer.


TITLE: partial diffential equation in conservation of mass implies velocity of fluid is continuous?
QUESTION [2 upvotes]: I was reading Fluid Mechanics by I.G.Currie and after deriving the conservation of mass ($ \frac {\partial \rho}{\partial t} + \frac {\partial (\rho u_k)}{\partial x_k} = 0$), it said "since the equation is a partial differential equation, the implication is that the velocity is continuous." I haven't done any course on pde, so I am having difficulty in understanding how does a pde imply continuity.

REPLY [2 votes]: A formal rigorous mathematical framework PDE (particularly nonlinear PDE) is a subtle business. What the author is getting at is that in writing down the expression
$$
\frac{\partial \rho}{\partial t} + \frac{\partial (\rho u_k)}{\partial x_k} = 0
$$
The assumption is that $\rho u_k$ is continuous and in fact once differentiable. If $u_k$ were to be discontinuous, writing down $\partial (pu_k)/\partial x_k$ would make no sense. Or would it?
In fact, one does not actually need continuity or even differentiability of $u$, at least formally. One only needs that $u$ is arbitrarily close to a smooth function, say. One routinely solves PDEs with discontinuous functions all the time. (Consider a linear advection equation with a square wave initial condition.) In fact, we often solve PDEs with objects that are not strictly functions at all like the so-called delta "function".
Thus, the author here is not making a strictly accurate mathematical statement, per say, but it is a nice assumption to make about $u$ if one doesn't want to invoke a lot more complicated math.


TITLE: $\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$
QUESTION [1 upvotes]: Is it true that $\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$ for $m \in \mathbb{N}$?

REPLY [1 votes]: $\Sigma_{k=m+1}^{\infty}\frac{1}{10^k}=\frac{1}{10^{m+1}}+\frac{1}{10^{m+2}}+\frac{1}{10^{m+3}}+.... =\frac{1}{10^m}(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...)\leq \frac{1}{10^m}\cdot1=\frac{1}{10^m}$


TITLE: A real $2 \times 2 $ matrix $M$ such that $M^2 = \tiny \begin{pmatrix} -1&0 \\ 0&-1-\epsilon \\ \end{pmatrix}$ , then :
QUESTION [1 upvotes]: A real $2 \times 2 $ matrix $M$ such that $$M^2 =  \begin{pmatrix} -1&0 \\ 0&-1-\epsilon  \\   \end{pmatrix}$$
(a) exists for all $\epsilon > 0$.
(b) does not exist for any $\epsilon > 0$.
(c) exists for some $\epsilon > 0$.
(d) None of the above is true.

Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix.
Could someone please give me a hint on how to go about this problem?
Thank you very much for your help in this regard.

REPLY [3 votes]: no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$
for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$
pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2 = \epsilon \tag 1$$
that means if $\epsilon \neq 0,$  then $a+d \neq0\implies b = c = 0.$ from $(1), a^2 = -1, d^2 = -1 - \epsilon$   there are no real solutions as $a^2 = -1.$
if $\epsilon = 0, a^2 - d^2 = 0$ it forces $a = -d, b = c = 0$ but then (1) forces $d^2 = -1$ which cannot be satisfied. the other case is $a = d.$ the constraints are $$a^2 + bc = -1, 2ab = 0 = 2ac, bc + d^2 = -1 $$ two choices: (i) $a = 0$ gives $bc = -1, d = 0$
(ii) $a \neq 0$ which gives $b = 0, c = 0, a^2 = -1$ not possible.


TITLE: Matrices - Linear Independence
QUESTION [0 upvotes]: An nxn matrix A has two identical rows. Explain why this means columns of A can not be linearly independent.
I think it is because number of equations is less than the number of variables?

REPLY [0 votes]: Hint: By row reduction you can transform the matrix into a matrix with only zeros in the first row. The determinant of this matrix is $0$ (simply to show with Laplace Theorem), hence the column vectors are linearly dependent as the matrix is not invertible.


TITLE: What is the electromagnetic 4-vector potential in Reissner-Nordström coordinates?
QUESTION [1 upvotes]: If you have the usual Reissner-Nordström metric of a charged black hole, is the electromagnetic potential of the black hole still:
$$A_0(r,\theta,\phi) = \frac{Q}{r}$$
in these units?

REPLY [3 votes]: A Schwarzschild black hole has no charge and no electrostatic potential. This is the potential of a Reissner-Nordström black hole, in Gaussian units.
If you are talking about the potential of a point charge at rest outside a Schwarzschild black hole, with $r$ being some measure of distance from the charge rather than the Schwarzschild radial coordinate, then this is not its potential. The potential and electrostatic field of a charge at rest outside the hole are not spherically symmetric around the charge as in flat spacetime, because the curvature of the black hole warps the field.
The electrostatic potential of a point charge at rest outside a Schwarzschild black hole was investigated by Copson in 1928 and his solution was corrected by Linet in 1976. The solution is
$$V=\frac{q}{b r}\left[ \frac{(b-M)(r-M)-M^2\cos\theta}{\sqrt{(r-M)^2+(b-M)^2-2(b-M)(r-M)\cos\theta-M^2\sin^2\theta}}+M \right]$$
where $M$ is the mass of the black hole, $r$ and $\theta$ are Schwarzschild coordinates, $q$ is the charge of the point particle outside the horizon, and the point particle is at rest at $r=b$, $\theta=0$. This is written in units where $G=c=1$.
Here is a contour plot of the potential. The larger blue circle is the black hole. The smaller blue circle is the point particle. The white area is an artifact of the rendering, where the contour lines get too close together. The horizon is an equipotential surface.

The asymmetry of the field causes a gravitationally-induced electrostatic self-force on the charged particle. It is directed away from the hole and, in the frame of a freely-falling observer instantaneously at rest at the position of the charge, has magnitude
$$F=\frac{GMq^2}{c^2b^3}.$$
This was first calculated exactly by Smith and Will in 1980.
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.22.1276
Interestingly, as $b\rightarrow2M$ so that the charge is just outside the horizon, the potential does become symmetrically centered on the hole, rather than on the charge, and is just $q/r$ outside and 0 inside.


TITLE: y divides discriminant implies y^2 divides it
QUESTION [3 upvotes]: The wiki page https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem is discussing an elliptic curve
$$y^2 = x^3 + ax^2 + bx + c$$
with the cubic polynomial discriminant
$$D = -4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$
Later when explaining a particular case they state:
"... $y$ divides $D$, which immediately implies that $y^2$ divides $D$."
I apologize, I'm probably missing something obvious here, but I don't see why $y$ divides $D$ implies $y^2$ divides $D$. What am I missing?

REPLY [0 votes]: In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $\Bbb Z$, considered but in the same time by abuse also over the fraction field, here $\Bbb Q$) is of the shape:
$$
y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6\ .
$$
It is usual to denote by $\kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
$$
\begin{aligned}
\kappa &= 2y +a_1 +a_3\ ,\\
\kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6\ ,\\
&\qquad\text{ and one defines the following division polynomials:}\\
\psi_0 &= 0\ ,\\
\psi_1 &= 1\ ,\\
\psi_2 &= \kappa\ ,\\
\psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8\ ,\\
&\qquad\text{ and so on, and one defines recursively polynomials $\psi_m$,}\\[2mm]
\psi_{2m+1} &= \psi_{m+2}\psi_m^3 - \psi_{m-1}\psi_{m+1}^3\ ,\\
\psi_{2m}   &= (\psi_{m+2}\psi_{m-1}^2 - \psi_{m-2}\psi_{m+1}^2)\psi_m/\kappa\ ,\\
&\qquad\text{ and then we introduce the "numerator" $\phi_m$ by}\\
\phi_m &= x\psi_m^2-\psi_{m-1}\psi_{m+1}\\[2mm]
&\qquad\text{ such that one has in case of $mP\ne0$ the relation}\\
x(nP)
&=
\frac{\phi_m(P)}{\psi_m(P)^2}\ ,\\
\Delta
&=
{\color{blue}{(\sigma_2-x(2(x,y))\tau_2)}}\; \kappa^2\ ,\\
&\qquad\text{ where}\\[2mm]
\sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\\
\tau_2 &=48x^2+8b_2x+32b_4-b_2^2\ .
\end{aligned}
$$
See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book
Ian Connel, Elliptic curve Handbook
(this is around page 150).

Proof of the implicit claim in the OP:

If $P(x,y)$ is a torsion point with $2P\ne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $\Delta$, consider $\kappa=2y$, then $\kappa^2$ divides $\Delta$.

More general:

If $P,2P\ne O$ have integer coordinates, then with the same notations $\kappa^2$ divides $\Delta$.

The above relation $(*)$, since the blue term above ${\color{blue}{(\sigma_2-x(2(x,y))\tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.
$\square$

Usually it is good in such cases to have an explicit example, here we have one using sagemath:
sage: import random
sage: alfa = random.choice([10^4..10^5])
sage: alfa
40882
sage: c = alfa^2-alfa
sage: b = alfa*c
sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
sage: E
Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
sage: EE
Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
sage: EE.discriminant().factor()
2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
sage: EE.torsion_points()
[(0 : 1 : 0),
 (8378881486178014515 : -7379204292472752 : 1),
 (8378881486178014515 : 7379204292472752 : 1),
 (8378881546344708027 : -301661871656786182296 : 1),
 (8378881546344708027 : 301661871656786182296 : 1),
 (8381341220942172099 : -12332842306323413283199584 : 1),
 (8381341220942172099 : 12332842306323413283199584 : 1)]
sage: P = _[-1]
sage: P
(8381341220942172099 : 12332842306323413283199584 : 1)
sage: P.xy()
(8381341220942172099, 12332842306323413283199584)
sage: y = P.xy()[1]
sage: y
12332842306323413283199584
sage: ((2*y)^2).factor()
2^12 * 3^10 * 13627^4 * 20441^6
sage: ((2*y)^2).divides( EE.discriminant() )
True

In words, we have constructed an elliptic curve with $\Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $\kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.


TITLE: Is the smallest field of a finite set also denoted $\sigma$(.)?
QUESTION [1 upvotes]: The collection $\{A, A^c, \emptyset, \Omega\}$ is, I think, the smallest field generated by A. Is this also denoted $\sigma$(A) or is that notation reserved for $\sigma$-fields?

REPLY [1 votes]: To your question: What you describe would be denoted $\sigma(\{A\})$, the smallest $\sigma$-field containing $\{A\}$ as a subset, not $\sigma(A)$.


TITLE: $\sum_{n=1}^{\infty}f(z^n)$ converges uniformly with $f$ holomorphic
QUESTION [6 upvotes]: Let $f$ be an holomorphic function on the unit ball with $f(0)=0$.
  Prove that $\sum_{n=1}^{\infty}f(z^n)$ is uniformly locally
  convergent in the unit ball.

My attemp:
It is suffice to prove that $\sum_{n=1}^{\infty}f(z^n)$ converges uniformly  in any $\overline{B_0(r)}$ with $r<1$.
$f'$ is also an holomorphic function, defined on a compact set and therfore bounded. Let's say $|f'|\leq M$. Then, for every $z\in \overline{B_0(r)}$, $|f(z^n)-f(0)|\leq M|z^n-0|$, $|f(z^n)|\leq M|z^n| = M|z|^n \leq Mr^n$.
The series $\sum_{n=1}^{\infty}r^n$ converges and therefore by the M test our series converges uniformly. Is it correct?
Thanks.

REPLY [3 votes]: Let's do brute force!
If $f(z)=\sum_{n\geq1}a_nz^n$, then as a formal series $$\sum_{k\geq1}f(z^k)=\sum_{n\geq1}\left(\sum_{d\mid n}a_d\right)z^n.$$ If $f$ converges in the unit disc, we know from the Cauchy-Hadamard formula that $\limsup_{n\to\infty}|a_n|^{1/n}\leq1$, and we have to show that then $$\limsup\left|\sum_{d|n}a_d\right|^{1/n}\leq1.$$ One should be able to prove that

If $(x_n)_{n\geq1}$ is a sequence of non-negative real numbers such that $\limsup_{n\to\infty}x_n^{1/n}\leq1$, then $\limsup_{n\to\infty}(x_1+\cdots+x_n)^{1/n}\leq 1$.

and using this our result follows.


TITLE: Functional Analysis-Two Questions
QUESTION [0 upvotes]: I am having two questions:

Does every separable Hilbert Space have a Schauder Basis?

Proof: Every Hilbert Space $H$ has a total orthonormal set say $E=\{e_i\}_{i\in I}$.
Since $H$ is separable to $E$ is countable then $E=\{e_k\}_{k\in \Bbb N}$.
Then $E$ is a Schauder Basis of $H$.
Since $E$ is total so $\text{span E}$ is dense in $H$ hence given $x\in H$ there exists a sequence $x_n\to x$ such that  $||x_n-x||\to 0$ where $x_n=\sum_{i=1}^m c_{ni}e_i$.
Is it correct?

I know that every normed space has a Hamel Basis since they are all vector spaces.What is an example of a normed space which has no Schauder Basis?

REPLY [0 votes]: Your reasoning is correct. Another way of proving this, is to take an countable dense set, and then apply Gram-Schmidt procedure to generate an orthonormal Schauder basis.
As to your second question: If a normed space has a Schauder basis, then the space is separable, since the set of finite linear combinations of basis vectors with rational (complex) coefficients is countable and dense. Hence, non-separable spaces do not admit a Schauder basis.


TITLE: If $\;\log_{12}(a^2+b)=2+\log_3b=3+\log_2a$, what is the value of $\sqrt{\frac1b+\frac1{a^2}}$?
QUESTION [0 upvotes]: If $\;\log_{12}(a^2+b)=2+\log_3b=3+\log_2a$, what is the value of $\sqrt{\frac1b+\frac1{a^2}}$ ?

I used change of base,
$$\log_{12}(a^2+b)=\log_3(9b)\;\Rightarrow\; \frac{\log_3(a^2+b)}{1+2\log_32}=\log_3(9b)$$
$$\log_{12}(a^2+b)=\log_2(8a)\;\Rightarrow\; \frac{\log_2(a^2+b)}{2+\log_23}=\log_2(8a)$$
$$\log_3b=\log_2(2a) $$
I managed to get same bases in numerator of first two equations. But not sure if it helps since it is getting uglier.

REPLY [4 votes]: let $$\log_{12}{(a^2+b)}=2+\log_{3}{b}=3+\log_{2}{a}=t$$
then we have
$$a^2+b=12^t,9b=3^t,8a=2^t$$
so $$\sqrt{\dfrac{1}{b}+\dfrac{1}{a^2}}=\sqrt{\dfrac{a^2+b}{a^2b}}=\sqrt{\dfrac{12^t}{4^{t-3}3^{t-2}}}=24$$