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\begin{document}

\renewcommand{\thefootnote}{\alph{footnote}}
\setcounter{footnote}{-1} \footnote{\emph{Mathematics Subject
Classification(2010)}: 15B48, 15B57, 13F60, 05E15.}
\renewcommand{\thefootnote}{\alph{footnote}}
\setcounter{footnote}{-1} \footnote{ \emph{Keywords}:  positive definite matrix, generalized Jacobi matrix, planar network, double wiring diagram, cluster subalgebra}

\begin{abstract}
As an improvement of the combinatorial realization of totally positive matrices via the essential positive weightings of certain planar network by
 S.Fomin and A.Zelevisky \cite{[4]}, in this paper, we give the test method of positive definite matrices via the planar networks and the so-called cluster subalgebra respectively, introduced here originally.

This work firstly gives a combinatorial realization of all matrices through planar network, and then sets up a test method for positive definite matrices by $LDU$-decompositions and the horizontal weightings of all lines in their planar networks. On the other hand, mainly the relationship is built between
positive definite matrices and cluster subalgebras.
\end{abstract}

\maketitle
\bigskip

\section{Introduction}
Totally positive matrices and positive definite matrices are two important classes of matrices with certain (partial) positivity. S.Fomin and A.Zelevisky \cite{[4]} gave a combinatorial realization of totally positive matrices via the essential positive weightings of certain planar network and show that in order to test the totally positivity of a matrix, it is enough to check the totally positivity of chamber minors associated to the double wiring system due to  the theory of cluster algebras.

In this paper, we improve this method in \cite{[4]} to matrices with partial positivity.
 We firstly realize all matrices via generalized elementary Jacobi matrices in Proposition \ref{thm2.5} and then test other classes of matrices which share some properties about ``positivity'', e.g., positive definite matrices via the planar networks and the so-called cluster subalgebra respectively, introduced here originally.

 On one hand, it is proven that an invertible Hermitian matrix is positive definite if and only if the weightings of all horizontal lines in its corresponding planar network  are positive in Proposition \ref{prop2.6} and Corollary \ref{maincor}. On the other hand, for any $n\times n$ Hermitian matrix $M$, in Theorem \ref{thm3.3} we prove that one can construct a double wiring system $\Omega$ and a certain subset $S$ of $Q(\Omega)_{0,m}$ such that the matrix $M$ is positive definite if and only if all minors corresponds to any extended cluster of the cluster subalgebra $\mathcal{A'}_{S}$ of $\mathcal{A}_{\Omega}$ on $M$ are positive.
In the last section, we give a summary to explain the relations between the methods of planar networks and cluster subalgebras through double wiring diagrams as a bridge.

This work demonstrates the meaning of the concept of cluster subalgebra for studying some ``partial positivity"  of matrices. In further research, we will be able to use the notion of cluster subalgebras to relate some other important partial positivity of matrices with the theory of cluster algebras and  the combinatorial method.

\section{Preliminaries on some concepts and results}

\subsection{ Totally positive matrices and positive definite matrices.\\}

In linear algebra, matrix theory and their applications, two classes of matrices, that is, totally positive matrices and positive definite matrices,  are very important. Let us recall their definitions.

A square matrix over the complex field $\mathds C$ is said to be \emph{totally positive} if all of its minors are positive real numbers. The first systematic study on this class of matrices was undertaken in the $1930$s by F.R.Gantmacher and M.G.Krein \cite{[5]}, where some remarkable spectral properties were  established, e.g., an $n\times n$ totally positive matrix has $n$ distinct positive eigenvalues. For more details, see \cite{[5]},\cite{[6]},\cite{[7]}.

 A $n\times n$ Hermitian matrix $M$ is said to be \emph{positive definite} if $z^{*}Mz$ is always a positive real number for any $0\not=z\in \mathds C^n$, where  $z^{*}=\bar{z}^\top$. The class of positive definite matrices plays an important role since it is closely related to positive-definite symmetric bilinear forms and inner product vector spaces.

 It is well-known that a Hermitian matrix $M$ is positive definite if and only if all leading principal minors of $M$ are positive. Hence if a totally positive matrix $M$ is Hermitian, then it is obviously positive definite.

\subsection{ The matrices associated with planar networks.\\}

In \cite{[4]}, S.Fomin and A.Zelevinsky gave the combinatoric way to realize all the totally positive matrices via the planar network as follows.

A \emph{planar network} $(\Gamma,\omega)$ is an acyclic directed planar
graph $\Gamma$ whose edges $e$ are assigned
the scalar weights $\omega(e)$. We always assume the edges of $\Gamma$ to be directed from left to right. Assume each network has $n$ sources and $n$
sinks located at the left and right sides in the picture respectively, and
numbered in bottom-to-top.

Now we can associate a planar network $(\Gamma,\omega)$ a matrix $x(\Gamma,\omega)$, which is called the {\em weight matrix}, more precisely, $x(\Gamma,\omega)$ is an $n\times n$ matrix whose $(i,j)$-entry is the sum of weights of all paths from the source $i$ to the sink $j$, where the weight of a (directed) path in $\Gamma$ is defined as the product of weights of all edges in this path. For examples, Figure 1 below includes two planar networks.
\begin{figure}[h] \centering
  \includegraphics*[110,477][540,585]{tu2.eps}

 Figure 1
\end{figure}

The corresponding weight matrix of the planar network on the left of Figure $1$ is written as below:
\[  \begin{matrix}
  \begin{pmatrix}
   d  &  dh  &  dhi  \\
   bd  &  bdh+e  &  bdhi+eg+ei  \\
   abd  &  abdh+ae+ce  &  abdhi+(a+c)e(g+i)+f
  \end{pmatrix}
  \end{matrix}
\]

In order to calculate minors of the weight matrix, the following lemma is useful, where $\Delta_{I,J}(x)$ denotes the minor of a matrix $x$ with the row set $I$ and the column set $J$ and the weight of a collection of directed paths in $\Gamma$ is defined to be the product of their weights.

\begin{Lem}{\rm (Lemma 1, \cite{[4]})}\label{lem1.1}
{\rm A minor $\Delta_{I,J}(x)$ of the weight matrix of a planar network
is equal to the sum of weights of all collections of vertex-disjoint
paths that connect the sources labeled by $I$ with the sinks labeled
by $J$.}
\end{Lem}

\begin{Rem}\label{rem1.2}
{\rm For the planar network in the above example, or more generally the planar network $(\Gamma,\omega)$ with the form ``$\backslash\backslash -- //$'', the leading principal minors of the weight matrix $x(\Gamma,\omega)$ can be calculated by Lemma \ref{lem1.1} as follow:
\begin{equation}\label{remark1.2}
\Delta_{1,1}=\prod_{i=1}^{t_{1}}\omega_{1i}, \Delta_{12,12}=\prod_{i=1}^{t_{1}}\omega_{1i}\prod_{i=1}^{t_{2}}\omega_{2i},\cdots, \Delta_{12\cdots n,12\cdots n}=\prod_{i=1}^{t_{1}}\omega_{1i}\prod_{i=1}^{t_{2}}\omega_{2i}\cdots \prod_{i=1}^{t_{n}}\omega_{ni}
 \end{equation}
 where the horizontal weightings $\omega_{k1},\omega_{k2},\cdots,\omega_{kt_{k}}$ appear in Line $k$ for $k=1,2,\cdots,n$ respectively. Indeed, the collections of vertex-disjoint paths that connect the sources labeled by $\{$$12\cdots s$$\}$ with the sinks labeled by $\{$$12\cdots s$$\}$ $1\leq s\leq n$ must be horizontal, then the results (\ref{remark1.2}) follows.}
\end{Rem}

In order to state the one-to-one correspondence between the set of totally
positive matrices and that of all positive weightings of a fixed planar
network, we need more notations.

Denote the planar network on the right of Figure $1$ as $\Gamma_{0}$ and call an edge of $\Gamma_{0}$ \emph{essential} if it either is slanted or is one of the $n$ horizontal edges in the middle of $\Gamma_{0}$. Note that $\Gamma_{0}$ has exactly $2\times(1+2+\cdots+(n-1))+n=n^2$ essential edges. Moreover, a weighting $\omega$ of $\Gamma_{0}$ is called \emph{essential} if $\omega(e)\neq 0$ for any essential edge $e$ and $\omega(e)=1$ for all other edges.

The following theorem in \cite{[4]} characterizes all totally positive
$n\times n$ matrices via the essential positive weightings of the planar network.
\begin{Thm}{\rm (Theorem 5, \cite{[4]})}\label{thm1.2}
{\rm The map $\omega \mapsto x(\Gamma_{0},\omega)$ restricts to a bijection between the set of all essential positive weightings of $\Gamma_{0}$ and that of all totally positive
$n\times n$ matrices.}
\end{Thm}

\subsection{ Cluster algebras.\\}

The original motivation of cluster algebras was to give the combinatorial characterizations of the dual canonical basis of the quantized enveloping algebra of a quantum group and of the total positivity for algebraic groups. Now, it plays a prominent role in cluster theory and arises in connection with integrable systems, algebraic Lie theory, Poisson geometry, total positivity, representation theory, combinatorics and etc.

Throughout this paper, we only restrict our attention to cluster algebras given by the so-called \emph{cluster quivers} $Q$, which are directed (multi-)graphs with no loops and no $2$-cycles. Some vertices of $Q$ are denoted as \emph{mutable} and the remaining ones are called \emph{frozen}. All the terminologies are collected in the following definition. For more details, see \cite{[2]}.

\begin{Def}\label{def1.3}
{\rm (i)~ Let $v$ be a mutable vertex of a cluster quiver $Q$. The \emph{quiver mutation} of $Q$ at $v$ is an operation that produces another quiver $Q'=\mu_{v}(Q)$ via the following three steps:

(1)~ Add a new edge $u\rightarrow w$ for each pair of edges $u\rightarrow v$, $v\rightarrow w$ in $Q$, except in the case when both $u,w$ are frozen.

(2)~ Reverse the direction of all edges adjacent to $v$.

(3)~ Remove $2$-cycles until none remains.

(ii)~ Let $\mathcal{F}\supset \mathds{R}$ be any field. We say that a pair $t=(Q,{\bf z})$ is a \emph{seed} in $\mathcal{F}$ if $Q$ is a cluster quiver and {\bf z}, called the \emph{extended cluster}, is a set consisting of some algebraically independent elements of $\mathcal{F}$, as many as the vertices of the quiver $Q$.

(iii)~ The elements of {\bf z} corresponding to the mutable vertices in $Q$ are called \emph{cluster variables} and those corresponding to the frozen vertices in $Q$ are called \emph{frozen variables}.

(iv)~ The \emph{seed mutation} $\mu_z$ at a cluster variable $z$ transforms the seed $t=(Q,{\bf z})$ into a new seed $t'=(Q',{\bf z'})=\mu_{z}(Q,{\bf z})$, where $Q'$ is the quiver resulted after mutating $Q$ at the vertex corresponding to $z$ and ${\bf z'}={\bf z}\cup \{z'\}\backslash \{z\}$, where the new variable $z'$ is subject to the \emph{mutation relation at $z$} via $zz'=\prod \limits_{z\leftarrow x} x+ \prod \limits_{z\rightarrow y} y$. One sometimes calls this relation the \emph{mutation relation at $v$} for the vertex $v$ of $Q$ corresponding to $z$.

(v)~ Two seeds $t=(Q,{\bf z})$ and $t'=(Q',{\bf z'})$ are said to be \emph{mutation-equivalent} if one of them can be obtained from the other after a sequence of seed mutations.

 (vi)~ The \emph{cluster algebra} $\mathcal{A}(Q,{\bf z})$ generated by an \emph{initial seed} $t=(Q,{\bf z})$ is defined as the subring of $\mathcal{F}$ generated by the elements of extended clusters that are mutation-equivalent to $t$.}
\end{Def}

\bigskip
\section{Characterization of positive definite matrices via planar networks.\\}

\subsection{Realization of matrices via generalized elementary Jacobi matrices.\\}

In this section, we would like to realize all $n\times n$ matrices via certain planar networks and then give a way to test whether the matrix is positive definite or not.

For any $n\times n$ matrix $M$, we attempt to find a planar network $(\Gamma,\omega)$ such that its weight matrix $x(\Gamma,\omega)=M$. For this aim, we first consider the case for invertible matrices.

Since any invertible matrix can be written as a product of elementary matrices, let us recall the definition of elementary Jacobi matrices and their connection with the planar networks as in \cite{[4]}.

\begin{Def}\label{def2.1}

{\rm (i)~  Let $E_{i,j}$ be the $n\times n$ matrix whose $(i,j)$-entry is $1$ and all other entries are $0$. For $i=1,2,\cdots,n-1$ and $t\in \mathds{C}$, let $x_{i}(t)=I+tE_{i,i+1}$, $x_{\overline{i}}(t)=I+tE_{i+1,i}$. For $i=1,2,\cdots,n$ and $t\in \mathds{C}\backslash \{0\}$, let $x_{\textcircled{i}}(t)=I+(t-1)E_{i,i}$.

 (ii)~  The \emph{elementary Jacobi matrices} are  matrices $x\in GL_{n}(\mathds{C})$ that differ from the identity matrix $I_n$ in a single entry located either on the main diagonal or immediately above or below it, more precisely, those are ones of the forms  $x_{i}(t),x_{\overline{i}}(t),x_{\textcircled{i}}(t'),~ \forall t\in \mathds{C},t'\in \mathds{C}\backslash \{0\}$.}
\end{Def}

\begin{Lem}{\rm (Theorem 12, \cite{[4]})}\label{lem2.2}
{\rm The three classes of elementary Jacobi matrices are weight matrices of the
following three ``chip''$($Seeing Figure $2$ below$)$, respectively.
In each ``chip'', all edges but one have weight $1$; the distinguished edge
has weight $t$. Slanted edges connect horizontal levels $i$ and $i+1$,
counting from the bottom. Additionally, the concatenation of these ``chip''s corresponds to multiplying their weight matrices.}
\end{Lem}

\begin{figure}[h] \centering
  \includegraphics*[164,363][494,467]{tu3.eps}

 Figure 2
\end{figure}

Therefore, any invertible matrix generated by some elementary Jacobi matrices can be realized as the weight matrix for a planar network, since we have the characterization for $GL_{n}(\mathds{C})$.
\begin{Prop}\label{prop2.3}
{\rm For any integer $n\geq 0$, the general linear group $GL_{n}(\mathds{C})$ can be generated by all elementary Jacobi matrices, that is,
$$GL_{n}(\mathds{C})=\langle x_{i}(t),x_{\overline{i}}(t),x_{\textcircled{i}}(t'),t\in \mathds{C},t'\in \mathds{C}\backslash \{0\}\rangle.$$}
\end{Prop}

\begin{proof}
  Clearly,  $GL_{n}(\mathds{C})\supseteq \langle x_{i}(t),x_{\overline{i}}(t),x_{\textcircled{i}}(t'),t\in \mathds{C},t'\in \mathds{C}\backslash \{0\}\rangle$.

  Since any matrix in $GL_{n}$ can be written as a product of elementary matrices,  it suffices to show that any one of elementary matrices of three types  can be generated by some elementary Jacobi matrices.

{\bf The case $1$:~ The type of row multiplication.}

Obviously, all elementary matrices of this type have the form $x_{\textcircled{i}}(t)$ for $t\in \mathds{C}\backslash \{0\}$.

{\bf The case $2$:~ The type of row addition.}

For any $1\leq i,j\leq n$, $i\neq j$, we only consider the case $i<j$, no loss of generality.
We use the induction method on the number $j-i$.

If $j-i=1$, then the result is trivial due to $x_{i}(t)=I+tE_{i,i+1}$.

In general, assume that the conclusion is true for $j-i\leq k$, then considering the case for $j-i=k+1$, we have:
\begin{eqnarray*}
I+tE_{i,i+k+1}  & = &  I+E_{i+k,i+k+1}+tE_{i,i+k+1}-E_{i+k,i+k+1} \\
                & = & (I+tE_{i,i+k})(I+E_{i+k,i+k+1})(I-tE_{i,i+k})(I-E_{i+k,i+k+1})
\end{eqnarray*}
 due to $E_{i,j}E_{k,l}=\delta_{j,k}E_{i,l}$, where $\delta_{j,k}$ is the Kronecker symbol. Using the induction method, it means that any such elementary matrix  can be generated by some elementary Jacobi matrices.

{\bf The case $3$:~ The type of row switching.}

The elementary matrix for exchanging the rows $i$ and $j$ is just equal to the matrix $I+E_{i,j}+E_{j,i}-E_{i,i}-E_{j,j}$, moreover, which can be represented as  \begin{eqnarray*}
I+E_{i,j}+E_{j,i}-E_{i,i}-E_{j,j} & = &  I+E_{j,i}-E_{i,j}-E_{i,i}-E_{j,j}-2E_{j,j}+2E_{i,j}+2E_{j,j}  \\
& = & (I-E_{i,j})(I+E_{j,i})(I-E_{i,j})(I-2E_{j,j}).
\end{eqnarray*}
Then the result follows from that of the case 2.
\end{proof}

Since all elementary Jacobi matrices are invertible, an non-invertible matrix cannot be represented as their product. So, in order to consider not necessarily invertible matrices, we need to generalize the notion of elementary Jacobi matrices.
\begin{Def}\label{def2.4}
{\rm Define a \emph{generalized elementary Jacobi matrix} in $M_n(\mathds C)$ to be a matrix that differs from the identity matrix $I$ in a single entry located either on the main diagonal or immediately above or below it, more precisely, that is one of the form $x_{i}(t),x_{\overline{i}}(t),x_{\textcircled{i}}(t)$ for $t\in \mathds{C}$.}
\end{Def}
\begin{Prop}\label{thm2.5}
{\rm For any positive integer $n$,  any $n\times n$ matrix $M$ over $\mathds C$ can be generated by some generalized elementary Jacobi matrices, that is,   $M_{n}(\mathds{C})=\langle x_{i}(t),x_{\overline{i}}(t),x_{\textcircled{i}}(t),~ \forall t\in \mathds{C}\rangle$.}
\end{Prop}

\begin{proof}
We know that $M=M_{1}\left(
                       \begin{array}{cc}
                         I_r &  \\
                          & O_{n-r} \\
                       \end{array}
                     \right)
M_{2}$, where both $M_{1},M_{2}$ are invertible, $r$ is the rank of $M$. The result follows from Proposition \ref{prop2.3} and $\left(
                       \begin{array}{cc}
                         I_r &  \\
                          & O_{n-r} \\
                       \end{array}
                     \right)=\prod\limits_{s=r+1}^n x_{\textcircled{s}}(0)$.
\end{proof}

By Lemma \ref{lem2.2} and Proposition \ref{thm2.5}, we get that
\begin{Cor}\label{corrplanarnetwork}
{\rm For any matrix $M$, there exists a planar network $(\Gamma,\omega)$ such that its weight matrix $x(\Gamma,\omega)=M$.}
\end{Cor}

\subsection{The criterion for positive definite matrices via planar networks.\\}

We have already seen that by Corollary \ref{corrplanarnetwork}, for any matrix $M$, there exists a planar network $(\Gamma,\omega)$ such that its weight matrix $x(\Gamma,\omega)=M$, although this correspondence is neither injective nor surjective in general. Thus, by Lemma \ref{lem1.1} we can read the leading principal minors $\Delta_{\{1\},\{1\}}, \cdots, \Delta_{[1,n],[1,n]}$ directly from the planar network $(\Gamma,\omega)$. From the basic fact that a Hermitian matrix is positive definite if and only if all of its leading principal minors are positive, we can obtain the algorithm for checking whether a given matrix $M$ is positive definite or not as follows, which give us a view-point to observe positive definite matrices.

\begin{Algo}
{\rm (1)~ Decompose $M=M_{1}M_{2}\cdots M_{t}$ for some generalized elementary Jacobi matrices $M_1, M_2, \cdots, M_t$;

(2)~ By the correspondences between ``chips'' and generalized elementary Jacobi matrices, draw the planar network $(\Gamma,\omega)$ with $x(\Gamma,\omega)=M$;

(3)~  Calculate the leading principal minors directly from $(\Gamma,\omega)$ by the method given in Lemma \ref{lem1.1};

(4)~ Check whether a given matrix $M$ is positive definite or not.}
\end{Algo}

Additionally, by \cite{[9]}, we obtain that for any $n\times n$ matrix $M$, $M$ can be decomposed as $M=U_{1}LU_{2}$ or $M=L_{1}UL_{2}$ with $L,L_{1},L_{2}$ lower triangular matrices and $U,U_{1},U_{2}$ upper triangular matrices. Thus the planar network have the forms ``// $\backslash\backslash$ //'' or ``$\backslash\backslash$ // $\backslash\backslash$'', respectively. Furthermore, we have

\begin{Prop}\label{prop2.6}
{\rm For any $n\times n$ invertible matrix $M$, the following statements are equivalent:

(i) $M$ has a $LDU$-decomposition, i.e., $M=LDU$ with $L$ a unit lower triangular matrix, $U$ a unit upper triangular matrix and $D$ a diagonal matrix.

(ii) The corresponding planar network $(\Gamma,\omega)$ of $M$ is in the form ``$\backslash\backslash -- //$''.

(iii) All the leading principal minors of $M$ are nonsingular.
}
\end{Prop}

\begin{proof}
Firstly, the equivalence of (i) and (iii) can be found in \cite{[8]}.

(i)$\Longrightarrow$(ii): $M=LDU$ a $LDU$-decomposition. According to the proof of Proposition \ref{prop2.3}, we can give the decompositions  $L=L_{1}L_{2}\cdots L_{r}$, $U=U_{1}U_{2}\cdots U_{s}$, $D=D_{1}D_{2}\cdots D_{t}$, where $L_{1},\cdots,L_{r}$ are in the form $x_{\overline{i}}(\gamma)$, $U_{1},\cdots,U_{s}$ are in the form $x_{i}(\gamma)$ and $D_{1},\cdots,D_{t}$ are in the form $x_{\textcircled{i}}(\gamma)$ respectively. Thus the corresponding planar network  $(\Gamma,\omega)$ of $M$ is in the form ``$\backslash\backslash -- //$''.

(ii)$\Longrightarrow$(iii): Since the corresponding planar network $(\Gamma,\omega)$ of $M$ is in the form ``$\backslash\backslash -- //$'' and $M$ is invertible, by Remark \ref{rem1.2} we have $$\Delta_{12\cdots n,12\cdots n}=\prod_{i=1}^{t_{1}}\omega_{1i}\prod_{i=1}^{t_{2}}\omega_{2i}\cdots \prod_{i=1}^{t_{n}}\omega_{ni}={\rm det}(M)\neq 0$$
where the horizontal weightings $\omega_{k1},\omega_{k2},\cdots,\omega_{kt_{k}}$ appear in Line $k$ for $k=1,2,\cdots,n$ respectively. Hence all the leading principal minors
$$\Delta_{1,1}=\prod_{i=1}^{t_{1}}\omega_{1i}, \Delta_{12,12}=\prod_{i=1}^{t_{1}}\omega_{1i}\prod_{i=1}^{t_{2}}\omega_{2i},\cdots, \Delta_{12\cdots n,12\cdots n}=\prod_{i=1}^{t_{1}}\omega_{1i}\prod_{i=1}^{t_{2}}\omega_{2i}\cdots \prod_{i=1}^{t_{n}}\omega_{ni}$$
are nonsingular.
\end{proof}

It is easy to see that for an $n\times n$ matrix $M=LDU$ a $LDU$-decomposition with $D={\rm diag}(d_{1},d_{2},\cdots,d_{n})$, the leading principal minors of $M$ are $\Delta_{12\cdots k,12\cdots k}=d_{1}d_{2}\cdots d_{k}$ for $k=1,2,\cdots,n$. In this case, if furthermore $M$ is invertible, then by Proposition \ref{prop2.6} and Remark \ref{rem1.2}, we have $d_{k}=\prod\limits_{i=1}^{t_{k}}\omega_{ki}$ for all $1\leq k\leq n$. It follows that

\begin{Cor}\label{maincor}
{\rm An invertible Hermitian matrix $M$ is positive definite if and only if the weightings $d_k=\prod\limits_{i=1}^{t_{k}}\omega_{ki}$ of Line $k$ in the corresponding planar network $(\Gamma,\omega)$ of $M$ are positive for all $1\leq k\leq n$. }
\end{Cor}

We end this section with the following example to illustrate the algorithm for checking positive definite matrices.

\begin{Ex}\label{ex2.6}
{\rm Let $M=\begin{matrix}
  \begin{pmatrix}
   1  &  2  &  4  \\
   2  &  6  &  8  \\
   4  &  8  &  18
  \end{pmatrix}
  \end{matrix}$, then firstly $M$ can be decomposed as

\begin{eqnarray*}
  M & = & \begin{matrix}
  \begin{pmatrix}
   1  &  0  &  0  \\
   2  &  1  &  0  \\
   0  &  0  &  1
  \end{pmatrix}
  \end{matrix}
  \begin{matrix}
  \begin{pmatrix}
   1  &  0  &  0  \\
   0  &  1  &  0  \\
   4  &  0  &  1
  \end{pmatrix}
  \end{matrix}
  \begin{matrix}
  \begin{pmatrix}
   1  &  0  &  0  \\
   0  &  2  &  0  \\
   0  &  0  &  2
  \end{pmatrix}
  \end{matrix}
  \begin{matrix}
  \begin{pmatrix}
   1  &  0  &  4  \\
   0  &  1  &  0  \\
   0  &  0  &  1
  \end{pmatrix}
  \end{matrix}
  \begin{matrix}
  \begin{pmatrix}
   1  &  2  &  0  \\
   0  &  1  &  0  \\
   0  &  0  &  1
  \end{pmatrix}
  \end{matrix} \\
   & = & x_{\overline{1}}(2)x_{\overline{2}}(-1)x_{\overline{1}}(-4)x_{\overline{2}}(1)
  x_{\overline{1}}(4)x_{\textcircled{2}}(2)x_{\textcircled{3}}(2)x_{1}(4)x_{2}(1)x_{1}(-4)x_{2}(-1)x_{1}(2)
  \end{eqnarray*}

Next, the corresponding planar network $(\Gamma,\omega)$ is shown as Figure 3, where the weightings $\omega(e)=1$ for all unlabelled edges $e$.

\begin{figure}[h] \centering
  \includegraphics*[137,413][417,496]{tu4.eps}

 Figure 3
\end{figure}

Finally, we have $d_{1}=1, d_{2}=2, d_{3}=2$. Now by Corollary \ref{maincor},  we know that $M$ is positive definite.}
\end{Ex}

\bigskip
\section{Characterization of positive definite matrices via cluster subalgebras.\\}

In \cite{[3]}, S.Fomin show total positivity as one of the motivations of cluster algebras. In brief, take a complex variety $X$ together with a family ${\bf \Delta}$ of ``important'' regular functions $\Delta$ on $X$. The corresponding totally positive variety $X_{>0}$ is the set of points at which all of these functions take positive values, that is, $X_{>0}=\{x\in X: \Delta(x)>0$ for all $\Delta \in {\bf \Delta}\}$. If $X$ is the affine space of matrices of a given size, e.g., if $X=M_{n}(\mathds{C})$, and ${\bf \Delta}$ is the set of all minors, then we recover the classical notation of totally positive matrices.

The question arises naturally that for any element $x\in X$, in order to test whether $x\in X_{>0}$ or not, should we check all the inequalities $\Delta(x)>0$ for all $\Delta \in {\bf \Delta}?$

In the above example, there are $\binom{2n}{n}-1$ minors in total. However, S.Fomin and A.Zelevinsky in \cite{[4]} told us that it is enough to check certain $n^{2}$ minors in order to make sure a matrix to be totally positive through the chamber minors of the corresponding double wiring diagram. This is because all the minors of $M_{n}(\mathds{C})$ can be represented by the rational functions of these $n^{2}$ minors with positive coefficients.

\subsection{Double wiring diagram and its associated quiver.\\}

In order to explain this question, recall that the \emph{double wiring diagram} defined in \cite{[4]} is a diagram consists of two families of $n$ piecewise-straight lines $($each family colored with one of the two colors$)$, the crucial requirement is that each pair of lines with the same color intersect exactly once. Moreover, the lines in a double wiring diagram are numbered separately within each color.

Now we give the definition of \emph{chamber} and \emph{chamber minor} of a double wiring diagram as follows.

\begin{Def}\label{def3.0}
{\rm Let $\Omega$ be a double wiring diagram and $\square$ be a region of $\Omega$. We then assign to $\square$ a pair of subsets $I,J$ of the set $[1,n]=\{1,\cdots,n\}$ such that each subset indicates the line numbers of the corresponding color passing \emph{below} the $\square$. If $|I|=|J|$, then the region $\square$ is called as the \emph{chamber} and in this case, the minor $\Delta_{I,J}$ of an $n\times n$ matrix $x=(x_{ij})$ is called the \emph{chamber minor} of $x$.}
\end{Def}

For example, we show all the chambers of the double wiring diagram in Figure $4$ and the corresponding nine chamber minors $($the total number is always $n^{2}$$)$ are $x_{31}$, $x_{21}$, $x_{11}$, $x_{13}$, $\Delta_{23,12}$, $\Delta_{12,12}$, $\Delta_{12,13}$, $\Delta_{12,23}$ and $\Delta_{123,123}={\rm det}(x)$.

\begin{figure}[h] \centering
  \includegraphics*[73,381][467,529]{tu6.eps}

 Figure 4
\end{figure}

In \cite{[2]}, or more generally in Section $2$ of \cite{[1]}, we can associate a quiver $Q=Q(\Omega)$ to a double wiring diagram $\Omega$ and this construction provided is purely combinatorial.

(i)~ The vertices of $Q$ are the chambers of $\Omega$.

(ii)~ There is an edge between two chambers $c$ and $c'$ of $Q$ in the following cases:

(1)~ They are adjacent chambers in the same row. If the color of the crossing between them is blue, the edge is directed to the left. Otherwise, it is directed to the right.

(2)~ If $c'$ has left and right boundaries of different color and lies completely above $($or below$)$ $c$. If the left boundary of $c'$ is blue, the edge is directed from $c$ to $c'$. Otherwise, it is directed from $c'$ to $c$.

(3)~ If the left boundary of $c'$ is above $c$ and the right boundary of $c$ is below $c'$ and both boundaries have the same color. If such common color is blue, the edge is directed from $c$ to $c'$; otherwise, it is directed from $c'$ to $c$.

(4)~ If the right boundary of $c'$ is above $c$ and the left boundary of $c$ is below $c'$ and both boundaries have the same color. If such common color is blue, the edge is directed from $c'$ to $c$; otherwise, it is directed from $c$ to $c'$.

Then according to (i) and (ii), we get the cluster quiver $Q$ from $\Omega$.

The mutable vertices of $Q(\Omega)$ are the chambers in $\Omega$ are bounded. Then the frozen vertices are the chambers which are unbounded. The following figure on the left shows the quiver associated to the double wiring diagram in Figure $4$, where the vertices surrounded by $\Box$ are frozen and the others are mutable.

\begin{figure}[h] \centering
  \includegraphics*[109,428][514,547]{tu7.eps}

 Figure 5
\end{figure}

In \cite{[1]}, A.Berenstein, S.Fomin and A.Zelevinsky showed that all the $\binom{2n}{n}-1$ minors of a matrix $M$ occurs as cluster variables in the cluster algebra $\mathcal{A}(\Omega)$ generated by the initial seed $(Q(\Omega),{\bf z}(\Omega))$ associated to a double wiring diagram $\Omega$, where ${\bf z}(\Omega)$ is the collection of all chamber minors in $\Omega$ and thus it is enough to check the positivity of these $n^2$ chamber minors in ${\bf z}(\Omega)$ to make sure of totally positivity of $M$.

\subsection{Cluster subalgebras and positive definite matrices.\\}

In what follows we replace the ``totally positive matrices'' by other classes of matrices which share some properties about ``partial positivity'', such as positive definite matrix, $P$-matrix and so on.

In order to give a similar result for a positive definite matrix, we first introduce some notations. Let $\mathcal{A}=\mathcal{A}(Q,{\bf z})$ be the cluster algebra generated by an initial seed $t=(Q,{\bf z})$ which is defined in Definition \ref{def1.3}. Let $Q_{0}=Q_{0,m}\cup Q_{0,f}$ and ${\bf z}={\bf z}_{m}\cup{\bf z}_{f}$, where $Q_{0},Q_{0,m},Q_{0,f}$ represents the vertices, mutable vertices, frozen vertices of $Q$ respectively and ${\bf z}_{m}, {\bf z}_{f}$ corresponds to the mutable part, frozen part of ${\bf z}$ respectively. Now we raise the definition of the so-called \emph{cluster subalgebra}.

\begin{Def}\label{def3.1}
{\rm Let us keep our foregoing notations. For a subset $S\subseteq Q_{0,m}$, we define the corresponding \emph{cluster subalgebra} associated to $S$ as the cluster algebra $\mathcal{A'}_{S}$ generated by the initial seed $t'=(Q',{\bf z'})$, where $Q'=Q, Q'_{0,m}=Q_{0,m}\backslash S, Q'_{0,f}=Q_{0,f}\cup S$ and $z'_{m}=z_{m}\backslash z_{S},z'_{f}=z_{f}\cup z_{S}$.}
\end{Def}

\begin{Ex}\label{ex3.2}
{\rm Let $\mathcal{A}=\mathcal{A}(\Omega)$, where $\Omega$ is the double wiring diagram on the left of Figure $5$. Then by choosing $S=\{12,13; ~2,1\}$, the corresponding cluster subalgebra $\mathcal{A'}_{S}=\mathcal{A}(\Omega')$, where $\Omega'$ is the double wiring diagram on the right of Figure $5$.}
\end{Ex}

From this example on the $3\times 3$ Hermitian matrix $M$, it is easy to see that in order to test whether $M$ is positive definite or not, it suffices to check only the positivity of any extended cluster from the cluster subalgebra $\mathcal{A}(\Omega')$.

Moreover, for any $n\times n$ Hermitian matrix, we can obtain the following similar conclusion.

\begin{Thm}\label{thm3.3}
{\rm For any $n\times n$ Hermitian matrix $M$, one can construct a double wiring system $\Omega$ and a certain subset $S$ of $Q(\Omega)_{0,m}$ such that the matrix $M$ is positive definite if and only if all minors correspondig to any extended cluster of the cluster subalgebra $\mathcal{A'}_{S}$ of $\mathcal{A}_{\Omega}$ on $M$ are positive.}
\end{Thm}

\begin{proof}
Firstly, we construct the double wiring diagram $\Omega$ as follows: For each row
between two wires, all the blue crossings are on the left side of the red crossings.
Then there are $n-1$ such rows and for each row, there exists an unique chamber with the blue left boundary and the red right boundary. Thus we get $n-1$ such chambers, which correspond by one-to-one to the $n-1$ minors $\Delta_{1,1},\Delta_{12,12},\cdots,\Delta_{12\cdots (n-1), 12\cdots (n-1)}$, that is, $\{1,1;~ 12,12;~ \cdots;~ 12\cdots (n-1), 12\cdots (n-1)\}=Q(\Omega)_{0,m}\backslash S$. Now we choose $S$ be $S=Q(\Omega)_{0,m}\backslash \{1,1;~ 12,12;~ \cdots;~ 12\cdots (n-1), 12\cdots (n-1)\}$.

Finally, the conclusion follows by the fact that $\{12\cdots n, 12\cdots n\}\in Q(\Omega)_{0,f}$ and $M$ is positive definite if and only if all of its leading principal minors are all positive and all its leading principal minors can be expressed as the rational functions of minors corresponds to the extended clusters of $\mathcal{A'}_{S}$ with positive coefficients.
\end{proof}

\bigskip
\section{Further Remarks as Summary.\\}

For any $n\times n$ matrix $M$, on one hand, all its $\binom{2n}{n}-1$ minors share the cluster algebra structure via a double wiring diagram $\Omega$ and its associated quiver $Q(\Omega)$. Thus in order to check some ``positivity'' properties about the minors of $M$, it suffices to check the ``positivity'' of some minors which appears in any clusters in the corresponding cluster subalgebra. On the other hand, for the matrix $M$, there exists a planar network $($$\Gamma$,$\omega$$)$ such that its weight matrix $x(\Gamma,\omega)=M$ and we can calculate any minor $\Delta_{I,J}$ of $M$ for $I,J\subseteq [1,n]$ by Lemma \ref{lem1.1}. Therefore, we conclude this paper by the following Figure.

\begin{figure}[h] \centering
  \includegraphics*[154,482][532,582]{tu8.eps}

 Figure 6
\end{figure}

\vspace{5mm}

{\bf Acknowledgements:}\; This project is supported by the National Natural Science
Foundation of
China (No.11271318, No.11171296 and No.J1210038) and the Specialized Research Fund for the
Doctoral Program of Higher Education of China (No.20110101110010) and the Zhejiang
Provincial
Natural Science Foundation of China (No.LZ13A010001).


TITLE: Is that a typo in Lang’s book? (The quadratic form)
QUESTION [0 upvotes]: I’m reading Calculus of several variables by Serge Lang, and in the section on Taylor’s formula, i suspect that there is mistake in the equation
It seems like Lang skipped the line $$h\frac{\partial f}{\partial x}(a,b)+ k\frac{\partial f}{\partial x}(a,b) $$ which should before this $$\frac{1}{2}(h^2...)$$

REPLY [3 votes]: Critical point is a point where the differential isn't full-rank. In the case of a map $\Bbb R^m\to \Bbb R$, where $m\ge1$, this means that the differential is the zero map.


TITLE: Taking the derivative of a function of a convex combination of vectors, $f((1-t)x + t\cdot y)$
QUESTION [0 upvotes]: Let $f$ be a differentiable function, $x\not = y$ and vectors (say in $\mathbb{R}^n)$, and define $g:(0,1] \to \mathbb{R}$ by
$$
g(t) = f((1-t)x + t\cdot y)
$$
How would I differentiate this with respect to $t$. That is, what is $g'(t)$
I believe I need to take the total derivative of it with respect to $t$, which would give me something like
$$ \tag{3}
\frac{\partial f}{\partial [(1-t)x + t\cdot y]}\cdot \frac{ \partial [(1-t)x + t\cdot y]}{\partial t} - \frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial t}
$$
Now I can re-write
$$
\frac{ \partial [(1-t)x + t\cdot y]}{\partial t} = -x\frac{\partial x}{\partial t} +y\frac{\partial y}{\partial t}
$$
But I don't know what to do with those partials (I don't know if I need them or not... I don't know for sure if $x,y$ depend on $t$... they probably don't, though) or the $\frac{\partial f}{\partial [(1-t)x + t\cdot y]}$ term...
thanks

REPLY [0 votes]: I would advice not to use the kind of writing you do use. For example, to my point of view, $\frac{\partial f}{\partial ((1-t)x+ty)}$ does not make any sense.
Here is a way to compute your derivative, with the chain rule formula. Let $h(t)=(1-t).x+t.y$. Then $g=f\circ h$ so
$$g'(t)=f'(h(t)).h'(t)=f'((1-t).x+t.y).(y-x)$$


TITLE: Find the sum of the number of constructible polygons
QUESTION [3 upvotes]: Let $R(n)$ equal $1$ if a regular $n$-sided polygon is constructible with a ruler and compass and $0$ if it is not. What is $\displaystyle \sum_{n = 3}^{100}R(n)$?

This question seems very hard to solve without just listing out all $n$ and seeing which $n$ have $\phi(n) = 2^m$. Is there any simple method to solve this question?

REPLY [2 votes]: 1573035
There is a version of the Pascal Triangle which may be of use to you. It consists of the modulo-2 residues of the members of the Triangle. Observe:
$\begin{array}{ccccccccccccccc|l}
&&&&&&&1&&&&&&&&=1\\
&&&&&&1&&1&&&&&&&=3\\
&&&&&1&&0&&1&&&&&&=5\\
&&&&1&&1&&1&&1&&&&&=15\\
&&&1&&0&&0&&0&&1&&&&=17\\
&&1&&1&&0&&0&&1&&1&&&=51\\
&1&&0&&1&&0&&1&&0&&1&&=85\\
1&&1&&1&&1&&1&&1&&1&&1&=255
\end{array}$
Notice that
(1) each new row is one item longer than its predecessor and always begins and ends with 1;
(2.1) a 0/0 or 1/1 pair yields a 0 in the next row;
(2.2) a 0/1 or 1/0 pair yields a 1 in the next row.
Each of these rows may be considered as a digit-by-digit representation of a binary number. Excepting the apex, each of these binary numbers represents the number of sides of a constructible odd-number regular polygon in ascending order of size. The final valid number in the series is $2^{32}-1$, the product $3\times 5\times 17\times 257\times 65537$.
3,5,6,10,12,15,17,20,24,30,34,40,48,51,60,68,80,85,96.
So, your answer is 19.
====================
(written in QB64)
FUNCTION R~`(n~%%)
-->a~%% = n~%% 'make copy of argument
'eliminate powers of 2 from argument
-->WHILE((a~%% MOD 2) = 0)
----->a~%% = a~%% \ 2
-->WEND
'eliminate single instances of allowed odd factors from argument
-->IF((a~%% MOD 3) = 0) THEN a~%% = a~%% \ 3
-->IF((a~%% MOD 5) = 0) THEN a~%% = a~%% \ 5
-->IF((a~%% MOD 17) = 0) THEN a~%% = a~%% \ 17
-->IF((a~%% MOD 257) = 0) THEN a~%% = a~%% \ 257
-->IF((a~%% MOD 65537) = 0) THEN a~%% = a~%% \ 65537
-->R~` = (a~%% = 1) 'R = 1 if constructible, 0 if not
END FUNCTION


TITLE: Good way to learn Ramsey Theory
QUESTION [15 upvotes]: What are some good books on Ramsey theory? I have Van Lint's book on Combinatorics: is this enough preparation to start reading about Ramsey theory? I want a book that includes important results and has good proofs.

REPLY [16 votes]: This is a list of some resource material on Ramsey Theory:

Expository note on Arithmetic Ramsey theory by Terence Tao. This contains the basic theorems and proofs presented in a very nice manner.

Ramsey Theory by R.L. Graham, B.L. Rothschild, and J. Spencer is a book which contains an introduction to nearly all areas in finite Ramsey theory. It contains proofs of all the basic results: Ramsey's theorem, van der Waerden theorem, Hales Jewett theorem, Schur's theorem, Rado's theorem, Graham's theorem etc.

Ramsey theory on the integers by Landman is a book accessible to undergraduates. It is unique in the sense that it uses only elementary mathematics and describes many research problems. Most of the book deals with variants of the van der Waerden's theorem.

Mathematics of Ramsey theory by Nesteril & Rodl is a book which I haven't read but is supposed to contain many advanced techniques.

The ebook Introduction to Graph Ramsey Theory by Yusheng Li and
Wenan Zang available here.

Elemental methods in Ergodic Ramsey theory by McCutcheon introduces one of the current approaches to Ramsey theory (via Ergodic theory). I have only read the first chapter but it seems an accessible and well written book.

Erdos' book Partition Relations for Cardinals (Infinite Ramsey theory). I haven't read it.


TITLE: I show trivial topolgies is smallest and discrete topology is the largest topology of any set X
QUESTION [0 upvotes]: show that the trivial topology $\mathcal{T}_1$  is the smalletes topologies for any set $X$ and the discrete topology $\mathcal{T}_2$ is the largest topology for any set X

for trivial topology $\mathcal{T}_1=\{ \emptyset , X\}$
I think by the meaning of smallest is the smallest amount of elements of $\mathcal{T}$. So for the trivial topology take any then it would not be a topology.
for the discrete topology  being the largest
would have to assume there is a largest topology and somehow derive that there is a contradiction
do know the definition of the discrete topology any subset of $X$ is open
completely lost here could use help

REPLY [2 votes]: By definition, a topology must contain the empty set and $X$, so the trivial topology is a subset of any topology.
By definition a topology consists of subsets of $X$. The discrete topology consists of all subsets of $X$, and therefore any topology is a subset of the discrete topology.


text:
\section{Case studies for particular wind speed/accelerations}
\label{sec:rendezvous}

Here we analyse the optimal acceleration and the optimal terminal
time, corresponding to the solution of the rendezvous guidance problem
(\ref{eq:statement}), for three particular wind conditions, that are:
\begin{enumerate}
\item Constant wind speed, with null terminal velocity.
\item Constant wind speed rate (or wind acceleration).
\item Constant wind speed gradient in the cross-track direction.
\end{enumerate}

The case of piecewise linear wind speed may be of practical
interest, as it can be used to treat the more general wind field cases of a vehicle in
a spatio-temporal flow field with wind shear parameters \cite{2001Airplane},\cite{Bakolas2018}.

\subsection{Case - constant wind and null terminal velocity}

Note that, in the absence of wind, or even in a constant
wind, we have $\Delta_{{\bf w}}(t_{f})={\bf 0},\,\,{\bf \varpi}(t_{f})={\bf 0}$
from (\ref{eq:deltaOmega}).
\begin{figure}[hbtp]
\noindent \begin{centering}
\includegraphics[scale=0.60]{figs/AIAA_fig_wind_cnst.png}
\par\end{centering}
\noindent \centering{}\caption{{Optimal problem (time-energy) for a UAV in constant wind speed.}}
\end{figure}

Then, assuming the target in the
origin (${\bf r_{f}}=0$) and a null final velocity (${\bf v}_{gf}={\bf 0}$), and using the methodology presented in the previous section, we have the  optimal costates, acceleration and terminal Hamiltonian equation as:
\begin{equation}
\begin{array}{ccc}
{\bf {\bf p}}_{r}=6\dfrac{\left[2{\bf r_{0}}+{\bf v}_{g0}t_{f}\right]}{t_{f}^{3}} & \,\,\,\,\,\,\,\,\,\,\,\,\,\, & {\bf {\bf p}}_{v}(0)=\dfrac{2\left[3{\bf r_{0}}+2{\bf v}_{g0}t_{f}\right]}{t_{f}^{2}}\end{array}\label{eq:JVD-1-1-1}
\end{equation}

\begin{equation}
{\bf u}^{*}(t)=t{\bf {\bf p}}_{r}-{\bf p}_{v}(0)=6t\dfrac{\left[2{\bf r_{0}}+{\bf v}_{g0}t_{f}\right]}{t_{f}^{3}}-\dfrac{2\left[3{\bf r_{0}}+2{\bf v}_{g0}t_{f}\right]}{t_{f}^{2}}\label{eq:optbruno}
\end{equation}

\begin{equation}
\begin{array}{c}
H(t_{f})=0\,\Rightarrow{\bf u}^{*T}(t_{f})\left[{\bf u}^{*}(t_{f})+2 \dot{ \bf w}(t_{f})\right]=2\left[C_{I}+{\bf p}_{r}^{T}{\bf v}_{g}(t_{f})\right]\\
{\bf \Rightarrow u}^{*T}(t_{f}){\bf u}^{*}(t_{f})=2C_{I}\\
\left[6{\bf r_{0}}+2{\bf v}_{g0}t_{f}\right]^{T}\left[6{\bf r_{0}}+2{\bf v}_{g0}t_{f}\right]=2C_{I}t_{f}^{4}
\end{array}\label{eq:z}
\end{equation}
Yielding the 4th order time-optimal polynomial $P(t_{f})$ as
\[
P(t_{f})=2C_{I}t_{f}^{4}-4({\bf v}_{g0}^{T}{\bf v}_{g0})t_{f}^{2}-24({\bf r_{0}}^{T}{\bf v}_{g0})t_{f}-36({\bf r_{0}}{}^{T}{\bf r_{0}})=0
\]
which is the same depressed quartic polynomial appearing in \cite{DSouza1997}  and \cite{Carvalho2018}, where the wind is not considered. Obviously, under a constant wind, the relative
airspeed rate and the inertial velocity rate are exactly the same.

\subsection{Case - constant wind acceleration}

Now, we consider the case of a constant rate of the wind speed, $\dot{ \bf w}(t)={\bf k}$,
which means that the wind, in a given direction,
is increasing ($k>0$) or decreasing ($k<0$) linearly with time. We
also suppose here that the terminal velocity ${\bf v}_{g}(t_f)$ may be
constrained to any desired value.

\begin{figure}[hbtp]
\noindent \begin{centering}
\includegraphics[scale=0.60]{figs/AIAA_fig_wind_incr1.png}
\par\end{centering}
\noindent \centering{}\caption{{Optimal problem (time-energy) for a UAV in linearly time-varying wind
speed.}}
\end{figure}
First, let us compute $\Delta_{{\bf w}}(t_{f})$ and ${\bf \varpi}(t_{f})$
for this case, from (\ref{eq:deltaOmega}), as:
\begin{equation}
\begin{array}{c}
\Delta_{{\bf w}}(t_{f})=\int_{0}^{t}{ {\dot{\bf w}}}(t)dt={\bf w}(t_{f})-{\bf w}(0)={\bf k}t_{f}\\
{\bf \varpi}(t_{f})=\int_{0}^{t}\int_{0}^{t}{ {\dot{\bf w}}}(t)dtd\tau=\int_{0}^{t_{f}}{\bf w}(t)dt-{\bf w}(0)t_{f}=\frac{{\bf k}t_{f}^{2}}{2}
\end{array}\label{eq:deltaOmega2-1-1}
\end{equation}
Then, assuming
the target position at the origin (${\bf r}_{f}=0$), the corresponding optimal costates, acceleration and terminal Hamiltonian equation, for this case, are:
\[
\begin{array}{c}
{\bf {\bf p}}_{r}=6\dfrac{\left[2{\bf r_{0}}+({\bf v}_{g0}+{\bf v}_{gf})t_{f}-{\bf k}t_{f}^{2}+2\frac{{\bf k}t_{f}^{2}}{2}\right]}{t_{f}^{3}}=6\dfrac{\left[2{\bf r_{0}}+({\bf v}_{g0}+{\bf v}_{gf})t_{f}\right]}{t_{f}^{3}}\\
{\bf {\bf p}}_{v}(0)=\dfrac{2\left[3{\bf r_{0}}+(2{\bf v}_{g0}+{\bf v}_{gf}-{\bf k}t_{f})t_{f}+3\frac{{\bf k}t_{f}^{2}}{2})\right]}{t_{f}^{2}}=\dfrac{2\left[3{\bf r_{0}+}(2{\bf v}_{g0}+{\bf v}_{gf})t_{f}\right]}{t_{f}^{2}}+{\bf k}
\end{array}
\]

\begin{equation}
{\bf u}^{*}(t)=t{\bf {\bf p}}_{r}-{\bf p}_{v}(0)=6t\dfrac{\left[2{\bf r_{0}}+({\bf v}_{g0}+{\bf v}_{gf})t_{f}\right]}{t_{f}^{3}}-\dfrac{2\left[3{\bf r_{0}+}(2{\bf v}_{g0}+{\bf v}_{gf})t_{f}\right]}{t_{f}^{2}}-{\bf k}\label{eq:acc_comp}
\end{equation}

\[
\begin{array}{c}
{\bf u}^{*T}(t_{f})\left[{\bf u}^{*}(t_{f})+2{\bf k}\right]=2\left[C_{I}+{\bf p}_{r}^{T}{\bf v}_{gf}\right]\\
\Rightarrow\left[6{\bf r_{0}}+(2{\bf v}_{g0}+4{\bf v}_{gf})t_{f}-{\bf k}t_{f}^{2}\right]^{T}\left[6{\bf r_{0}}+(2{\bf v}_{g0}+4{\bf v}_{gf})t_{f}+{\bf k}t_{f}^{2}\right]=2C_{I}t_{f}^{4}+12{\bf v}_{gf}^{T}\left[2{\bf r_{0}}+({\bf v}_{g0}+{\bf v}_{gf})t_{f}\right]
\end{array}
\]
whose expansion yields the 4th order time-optimal polynomial $P(t_{f})$
as
\begin{equation}
P(t_{f})=(2C_{I}+{\bf k}^{T}{\bf k})t_{f}^{4}
-4({\bf v}_{g0}{}^{T}{\bf v}_{g0}+{\bf v}_{gf}{}^{T}{\bf v}_{gf}+{\bf v}_{gf}^{T}{\bf v}_{g0})t_{f}^{2}
-24{\bf r_{0}}^{T}({\bf v}_{g0}+{\bf v}_{gf})t_{f}-36{\bf r_{0}}{}^{T}{\bf r_{0}}=0
\label{eq:polygeral}
\end{equation}

 Thus, we conclude that a linearly
increasing (or decreasing) wind works as an effective increase in the
trade-off parameter $C_{I}$, yielding a smaller optimal final time
$t_{f},$ when compared to the same equivalent condition without wind
(or with a constant wind).

\subsection{Case - constant wind shear in cross-track direction}
Now, we consider the particular case of a wind with a constant
shear (gradient) in the cross-track direction, as in Figure 6. To simplify the analysis, we suppose here a 2D movement. This problem is closely
related to the classical Zermelo problem, for the minimum time navigation
with constant airspeed, and whose solution is analysed with many different optimization approaches
in the literature \cite{2001Airplane}, [Lewisbook].
\begin{figure}[hbtp]
\noindent \begin{centering}
\includegraphics[scale=0.60]{figs/AIAA_fig_wind_shear.png}
\par\end{centering}
\noindent \centering{}\caption{{Optimal problem (time-energy) for a UAV in constant wind shear in
cross-track direction. }}
\end{figure}
In this case, the wind speed in the along-track direction ($x$) is supposed to vary linearly with the cros-track direction ($y$), such that the vertical wind speed component
will be always null ($w_{y}=0$), while the horizontal wind  will be given by $w_{x}=ky(t)$, or in matricial form:
\[
{\bf {w}}(t)={\bf Kr}
\]
where ${\bf K}=\left[\begin{array}{cc}
0 & k\\
0 & 0
\end{array}\right]_{2\times2}$ and ${\bf r}=[x~~y]^T.$ Such that the wind speed rate is
\[
 \dot{\bf {w}}(t)= {\bf K}\dot{\bf r}={\bf Kv}_{g}(t)
\]

The Hamiltonian function for this problem, using (\ref{eq:Ham}),
is given by
\begin{equation}
\begin{array}{c}

H=C_{I}+\frac{1}{2}{\bf u}^{T}{\bf u}+{\bf p}_{r}^{T}{\bf v}_{g}+{\bf p}_{v}^{T}({\bf {\bf Kv}_{g}+u})
\label{eq:hamilshear}
\end{array}
\end{equation}
where ${\bf p}_{r}$ and ${\bf p}_{v}$ are the costate vectors associated
with the position and velocity vectors, respectively, such that:

\[
{\bf u^{*}}= \dot{\bf v}_{a}^{*}=-{\bf p}_{v}\label{eq:ustar-1}
\]

To derive the costate vectors, we first rewrite the Hamiltonian equation as
\begin{equation}
\begin{array}{c}

H=C_{I}-\frac{1}{2}{\bf u}^{T}{\bf u}+{\bf p}_{r}^{T}{\bf v}_{g}+{\bf p}_{v}^{T}{\bf {\bf Kv}_{g}}
\label{eq:hamilshear}
\end{array}
\end{equation}

Such that according to PMP and Hamilton's, the costate equations are obtained as:
\[
\begin{array}{c}
{\dot{\bf  p}}_{r}=-\dfrac{\partial H}{\partial{\bf r}}={\bf 0}
\\
{ \dot{\bf p}}_{v}=\dfrac{\partial H}{\partial{\bf v_{g}}}=
\dfrac{\partial \{\frac{1}{2}{\bf u}^{T}{\bf u}\}}{\partial{\bf u}}
-{\bf p}_{r}-{\bf {\bf K}}^{T}{\bf p}_{v}=
{\bf u} \dfrac{\partial {\bf u}}{\partial{\bf v_{g}}}
{\bf -{\bf p}}_{r}-{\bf {\bf K}}^{T}{\bf p}_{v}=
{\bf {\bf K}}^{T}{\bf p}_{v}-{\bf p}_{r}-{\bf {\bf K}}^{T}{\bf p}_{v}
=-{\bf p}_{r}
\end{array}
\]

Whose solution comes to:

\[
{\bf u^{*}}=\dot{\bf  v}_{a}^{*}=-\left[\begin{array}{l}
p_{vx}\\
p_{vy}
\end{array}\right]=\left[\begin{array}{l}
tp_{rx}-p_{vx}(0)\\
tp_{ry}-p_{vy}(0)
\end{array}\right]
\]

As well as the optimal inertial acceleration:
\begin{equation}
\dot{\bf  v}_{g}^{*}=
\dot{\bf  v}_{a}^{*}
+ {\bf Kv}_{g}
=\left[\begin{array}{l}
tp_{rx}-p_{vx}(0)\\
tp_{ry}-p_{vy}(0)
\end{array}\right]
+
\left[\begin{array}{l}
kv_{gy}(t)\\
0
\end{array}\right]
\label{vgak}
\end{equation}

Whose integration produces the expressions for the inertial positions/velocities components

\[
\begin{array}{ccc}
\left\{ \begin{array}{l}
v_{gx}(t)=\frac{t^{2}}{2}p_{rx}-tp_{vx0}{+k{y}}(t)+v_{gx0}\\
x(t)=\frac{t^{3}}{6}p_{rx}-\frac{t^{2}}{2}p_{vx0}{+k}
\int_0^t y(t)
+tv_{gx0}+x_{0}
\end{array}\right. & &

\left\{ \begin{array}{l}
v_{gy}(t)=\frac{t^{2}}{2}p_{ry}-tp_{vy0}+v_{gy0}\\
y(t)=\frac{t^{3}}{6}p_{ry}-\frac{t^{2}}{2}p_{vy0}+tv_{gy0}+y_{0}
\end{array}\right.
\end{array}
\]

Or
\[
\begin{array}{ccc}
\left\{ \begin{array}{l}
v_{gx}(t)=\frac{t^{2}}{2}p_{rx}-tp_{vx0}+k
(\frac{t^{3}}{6}p_{ry}-\frac{t^{2}}{2}p_{vy0}+tv_{gy0})
+v_{gx0}\\
x(t)=\frac{t^{3}}{6}p_{rx}-\frac{t^{2}}{2}p_{vx0}{+k}
(\frac{t^{4}}{24}p_{ry}-\frac{t^{3}}{6}p_{vy0}+\frac{t^{2}}{2}v_{gy0})
+tv_{gx0}+x_{0}
\end{array}\right. & &

\left\{ \begin{array}{l}
v_{gy}(t)=\frac{t^{2}}{2}p_{ry}-tp_{vy0}+v_{gy0}\\
y(t)=\frac{t^{3}}{6}p_{ry}-\frac{t^{2}}{2}p_{vy0}+tv_{gy0}+y_{0}
\end{array}\right.
\end{array}
\]
Using these equations, and recalling the particular initial/terminal conditions (Fig. 6), where  $v_{xf}=v_{x0}$, $v_{yf}=-v_{y0}$, we can find the optimal costate vectors as:
\[
\begin{array}{ccc}
{\bf {\bf p}}_{r}=\left[\begin{array}{c}
\frac{12}{t_f^{3}}(x_{0}+v_{gx0}t_{f}){+\frac{2}{t_f}kv_{gy0}}\\
0
\end{array}\right]
& &
{\bf {\bf p}}_{v}(0)=\left[\begin{array}{c}
\frac{6}{t_f^{2}}(x_{0}+v_{gx0}t_{f}){+kv_{gy0}}\\
\frac{2}{t_f}v_{gy0}
\end{array}\right]
\end{array}
\]

Such that the initial/terminal inertial accelerations, from (\ref{vgak}), are written as
\[
\begin{array}{l}
{ { {\dot{\bf v}}_{g}^{*}}}(t_f)=t_f{ {\bf p}}_{r}-{{\bf p}}_{v}(0)
+
\left[\begin{array}{c}
-{kv_{gy0}}\\
0
\end{array}\right]
=
\dfrac{1}{t_{f}^{2}}\left[6{\bf r_{0}}+(2 {\bf v}_{g0} +4 {\bf v}_{gf})t_{f}\right]
 \\
{ { {\dot{\bf v}}_{g}^{*}}}(0)=-{{\bf p}}_{v}(0)+
\left[\begin{array}{c}
{kv_{gy0}}\\
0
\end{array}\right] =
-\dfrac{1}{t_{f}^{2}}\left[6{\bf r_{0}}+(4 {\bf v}_{g0} +2 {\bf v}_{gf})t_{f}\right]
\end{array}
\]

At a first glance, it may seem that the optimal inertial accelerations do not depend on the wind, as ${\bf v}_{g}(t_{f})$ do not depend on the $k$ parameter. However, this happens only for the terminal value of ${\bf v}_{g}(t)$. Moreover, the wind influence will also appear through the optimal terminal time $t_f$, as this is also dependent on $k$, as will be shown in the sequel.

The terminal relative acceleration is then written as:
\[
\begin{array}{l}
{\bf u}^{*}(t_f)=
t_f{ {\bf p}}_{r}-{{\bf p}}_{v}(0)=
\left[\begin{array}{c}
\frac{6}{t_f^{2}}(x_{0}+v_{gx0}t_{f})+kv_{gy0}\\
-\frac{2}{t_f^{2}}v_{gy0}t_{f}
\end{array}
\right]

\end{array}
\]

Such that the equation for the terminal Hamiltonian, $H(t_{f})$=0, is given by:
\begin{equation}
{\bf u}^{*T}(t_{f})\left[{\bf u}^{*}(t_{f})+2{\bf Kv}_{g}(t_f)\right]=2\left[C_{I}+{\bf p}_{r}^{T}{\bf v}_{g}(t_{f})\right]\label{eq:Hamil-1}
\end{equation}

Resulting in the following equation:
\[
\left[\begin{array}{c}
\frac{6}{t_f^{2}}(x_{0}+v_{gx0}t_{f})+kv_{gy0}\\
-\frac{2}{t_f}v_{gy0}
\end{array}
\right]^T
\left[\begin{array}{c}
\frac{6}{t_f^{2}}(x_{0}+v_{gx0}t_{f})-kv_{gy0}\\
-\frac{2}{t_f}v_{gy0}
\end{array}
\right]
=
2C_I+\frac{24v_{gx0}}{t_f^{3}}(x_{0}+v_{gx0}t_{f})+{\frac{4}{t_f}kv_{gx0}v_{gy0}}
 \]

The expansion of this equation yields the 4th order time-optimal polynomial $P(t_{f})$
as
\begin{equation}
P(t_{f})=(2C_{I}+k^2{v}_{gy0}^2)t_{f}^{4}+({4kv_{gx0}v_{gy0}})t_{f}^{3} -4(3{v}_{gx0}^2+{v}_{gy0}^2)t_{f}^{2}-48 x_{0}{v}_{gx0}t_{f}-36x_{0}^2=0
\label{eq:poly_rd2}
\end{equation}

\subsection{Case - constant wind shear in cross-track direction and CONSTANT airspeed}

We consider the previous case of constant wind shear but now trying to impose a constant airspeed during the whole travel.
A first approach, that generates a simple solution, is to find the optimal terminal time for which the airspeed at the middle of the path is equal to the initial airspeed.

In this particular position, the lateral airspeed is null, or $v_{ay}=0$ and the longitudinal airspeed is given by $v_{ax}(t)=v_{gx}(t)-ky(t)$. Such that the imposition of the same airspeed in $t=0$ and $t=t_f$ gives:

\[
v_{ax}(\frac{t_f}{2})=v_{gx}(\frac{t_f}{2})-ky(\frac{t_f}{2})= \sqrt{v_{gx0}^2+v_{gy0}^2}=V_{a0}
\]

Indeed, using the optimal costates from the previous section, we can write the longitudinal airspeed as

\[
\begin{array}{l}
v_{ax}(t)=\frac{t^{2}}{2}p_{rx}-tp_{vx0}
+v_{gx0}\\
v_{ax}(t)=
(\frac{t^{2}}{t_f}-t)p_{vx0}+v_{gx0}\\
\end{array}
\]

And for $t=\frac{t_f}{2}$, we have

\[
\begin{array}{l}
v_{ax}(\frac{t_f}{2})=V_{a0}=-\frac{t_f}{4}p_{vx0}+v_{gx0}\\
V_{a0}=-\frac{t_f}{4}
( \frac{6}{t_f^{2}}(x_{0}+v_{gx0}t_{f})+kv_{gy0})+v_{gx0}\\
V_{a0}=
-\frac{t_f}{4t_f^2}
( 6(x_{0}+v_{gx0}t_{f})+kv_{gy0}t_f^2) -4v_{gx0}t_f )\\
V_{a0}=
-\frac{1}{4t_f}
( 6(x_{0}+v_{gx0}t_{f})+kv_{gy0}t_f^2) -4v_{gx0}t_f )
\end{array}
\]

Resulting in the simple second order equation that yields the optimal terminal time $t_f$ for a constant airspeed case:
\[
P(t_f)=
-kv_{gy0}t_f^2 -(2v_{gx0}+4V_{a0})t_{f} -6x_{0}=0
\]

config: arxiv


TITLE: Given a biased die, what some of the probabilities
QUESTION [1 upvotes]: I recently asked this question:
Roll a fair 6 sided die twice, What is the probability that one or both rolls are 6?
And that makes sense, however, the reason I was confused in the first place is because we went over the following question during section:
Suppose the six-sided die you used in the previous problem is not fair. It is biased
so that rolling a 6 is three times more likely than any other roll.
This leads to the chance of a 6 being 3/8 and the chance of 1-5 being 1/8.
What is the probability that one or both rolls are 6’s?
However, my TA told me that there would be no overlap, and therefore nothing to subtract. However, my question that I asked previously, involving an unbiased die did involve overlap.
Was my TA wrong, is there overlap in this question? Or more specifically, is the answer 3/8 (prob of first roll being 6) + 3/8 (prob of second roll being 6) - 9/64 (prob of both being 6)
Thanks a bunch.

REPLY [2 votes]: As lulu mentioned overlap certainly matters. In fact you can prove it does using the following principle.
https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle


text:
\begin{document}

\title{Spectral stability of Prandtl boundary layers:
\\ an overview}

\author{Emmanuel Grenier\footnotemark[1]
 \and Yan Guo\footnotemark[2] \and Toan T. Nguyen\footnotemark[3]
}

\date\today

\maketitle

\begin{abstract}
In this paper we show how the stability of Prandtl boundary layers is linked to the stability of shear flows
in the incompressible Navier Stokes equations. We then recall classical physical instability results, and give a short educationnal
presentation of the construction of unstable modes for Orr Sommerfeld equations. We end the paper with a conjecture
concerning the validity of Prandtl boundary layer asymptotic expansions.
 \end{abstract}

\renewcommand{\thefootnote}{\fnsymbol{footnote}}

\footnotetext[1]{Equipe Projet Inria NUMED,
 INRIA Rh\^one Alpes, Unit\'e de Math\'ematiques Pures et Appliqu\'ees.,
 UMR 5669, CNRS et \'Ecole Normale Sup\'erieure de Lyon,
               46, all\'ee d'Italie, 69364 Lyon Cedex 07, France. Email: egrenier@umpa.ens-lyon.fr}

\footnotetext[2]{Division of Applied Mathematics, Brown University, 182 George street, Providence, RI 02912, USA. Email: Yan\underline{~}Guo@Brown.edu}

\footnotetext[3]{Department of Mathematics, Penn State University, State College, PA 16803. Email: nguyen@math.psu.edu.}

\section{Introduction}

This paper is motivated by the study of the inviscid limit of Navier Stokes equations in
a bounded domain. Let $\Omega$ be a subset of $\rit^2$ or $\rit^3$, and let us consider
the classical incompressible Navier Stokes equations in $\Omega$, posed on the velocity
field $u^\nu$,
\beq \label{NS1}
\partial_t u^\nu + \nabla ( u^\nu \otimes u^\nu) + \nabla p^\nu - \nu \Delta u^\nu = 0,
\eeq
\beq \label{NS2}
\nabla \cdot u^\nu = 0,
\eeq
with no--slip boundary condition
\beq \label{NS3}
u^\nu = 0 \qquad \hbox{ on }\quad \partial \Omega .
\eeq
As the viscosity $\nu$ goes to $0$, we would expect to recover incompressible Euler equations
\beq \label{Eu1}
\partial_t u^0 + \nabla (u^0 \otimes u^0) + \nabla p^0 = 0,
\eeq
\beq \label{Eu2}
\nabla \cdot u^0 = 0,
\eeq
with boundary condition
\beq \label{Eu3}
u^0 \cdot n = 0 \qquad \hbox{ on }\quad \partial \Omega,
\eeq
where $n$ is the outer normal to $\partial \Omega$. Throughout the paper, for the sake of presentation, we shall assume that $\Omega$ is the two-dimensional half space with $z\ge 0$.

\medskip

The no-slip boundary condition (\ref{NS3}) is the most difficult condition to study the inviscid limit problem.
It is indeed the most classical one and the genuine one, historically considered
in this framework by the most prominent physicists including Lord Rayleigh, W. Orr, A. Sommerfeld,
W. Tollmien, H. Schlichting, C.C. Lin, P. G. Drazin, W. H. Reid, and L. D. Landau, among many others. See for example the physics books on the subject: Drazin and Reid \cite{Reid} and Schlichting \cite{Schlichting}.
If the boundary condition (\ref{NS3}) is replaced by the Navier (slip) condition, boundary layers, though sharing the same thickness of $\sqrt \nu$, have much smaller amplitude (of an order $\sqrt \nu$, instead of order one of the Prandtl boundary layer), and are hence more stable (the smaller the boundary layer is, the more stable it is). We refer for instance to \cite{Iftimie, Sueur, Masmoudi}
for very interesting mathematical studies of boundary layers under the Navier boundary conditions.

\medskip

It is then natural to ask whether $u^\nu$ converges to $u^0$ as $\nu \to 0$ with the no-slip boundary condition \eqref{NS3}. This question
appears to be very difficult and widely open in Sobolev spaces, mainly because the boundary condition
changes between the Navier Stokes and Euler equations. Precisely, the tangential velocity
vanishes for the Navier Stokes equations, but not for Euler. In the limiting process
a boundary layer appears, in which the tangential velocity quickly goes from the Euler
value to $0$ (the value of the Navier-Stokes velocity on the boundary).

\medskip

The boundary layer theory was invented by Prandtl back in 1904 (when the first boundary layer equation
was ever found). Prandtl assumes that the velocity in the boundary layer depends on $t$, $x$ and on
a rescaled variable
$$
Z = {z \over \lambda}
$$
where $\lambda$ is the size of the boundary layer. We therefore make the following Ansatz, within the
boundary layer,
$$
u^\nu(t,x,z) = u^P(t,x,Z) + \lambda u^{P,1}(t,x,Z) + ... .
$$
Let the subscript $1$ and $2$ denote horizontal and vertical components of the velocity, respectively.
The divergence free condition \eqref{NS2} then gives
$$
\Big(\partial_x u^P_1 + \lambda \partial_x u^{P,1}_1 + \cdots \Big)
+ \Big( \lambda^{-1} \partial_Z u^P_2 + \partial_Z u^{P,1}_2 + \cdots \Big) = 0,
$$
which by matching the respective order in the limit $\lambda\to 0$ in particular yields
\begin{equation}\label{Pr0}
\partial_Z u^P_2 = 0, \qquad
\partial_x u^P_1 + \partial_Z u^{P,1}_2 = 0.
\end{equation}
As $u^\nu$ vanishes at $z=0$, this implies that
$$
u_2^P = 0,
$$
identically: the vertical velocity in the boundary layer is of order $O(\lambda)$.

Now, the Navier-Stokes equation \eqref{NS1} on the horizontal speed gives
\beq \label{Pr1}
\partial_t u^P_1 + u^P_1 \partial_x u^P_1
+ u^{P,1}_2 \partial_Z u^P_1 -  \partial^2_{ZZ} u^P_1 = - \partial_x p,
\eeq
provided that we choose
$$
\lambda = \sqrt{\nu} .
$$
Next, to leading order, the equation on the vertical speed reduces to
\begin{equation}\label{Pr2}
\partial_Z p = 0.
\end{equation}
Hence the leading pressure $p$ depends only on $t$ and $x$, and is given by the
pressure at infinity, namely by the pressure of Euler flow in the interior of the domain.
As for boundary conditions, we are led to impose
\beq \label{Pr3}
u^P_1 = u^{P,1}_2 = 0 \qquad \hbox{ on }\quad  \{Z = 0\}
\eeq
and
\beq \label{Pr4}
u_1^P(t,x,Z)  \to u^E_1(t,x,0) \qquad \hbox{ as }Ê\quad Z \to + \infty,
\eeq
where $u^E(t,x,0)$ denotes the value of the Euler flow in the interior of the domain (away from
the boundary layer). The set of equations \eqref{Pr0}-\eqref{Pr4} is called the Prandtl boundary layer equations.

\medskip

A natural question then arises: {\em can we justify that $u^\nu$ is the sum of an Euler
part $u^E$ plus the Prandtl boundary layer  correction $u^P$ ?}

\medskip

The first problem is to prove existence of solutions for the Prandtl equation. This is difficult since
whereas $u^P_1$ satisfies the simple transport equation with the degenerate diffusion,
$u^{P,1}_2$ satisfies no prognostic equation, and can only be recovered, using
$$
u^{P,1}_2(t,x,Z) = - \int_0^Z \partial_x u^P_1(t,x,Z) dZ.
$$
Hence $u^{P,1}_2$ is the vertical primitive of an horizontal derivative.
This leads to the loss of one derivative in the estimates.
In the analytic framework, it is possible to
control one loss of derivative: the Prandtl equation is well posed for small times; see \cite{SC1,KV}. See also \cite{GM} for the construction of Prandtl solutions in Gevrey classes. The existence of Prandtl solutions
in Sobolev spaces is delicate. Oleinik \cite{Oleinik} was the first to establish the existence of smooth solution in finite time
provided that the initial tangential velocity $u_1^P(0,x,Z)$ is monotonic in $Z$. Monotonicity plays a crucial
role in its proof and makes it possible for the existence via special transformations; see also recent works \cite{Alex, MW, KMVW} where the solution is constructed via delicate energy estimates. Then E and Engquist \cite{EE} proved that Prandtl layer may blow up in finite time.
More recently, G\'erard-Varet and Dormy \cite{DGV} proved that the Prandtl equation is linearly ill-posed in Sobolev
spaces if Oleinik's monotonicity assumption is violated.

\medskip

Concerning the justification of boundary layers, the analytic framework has been investigated in full details by Sanmartino and Caflisch in \cite{SC1,SC2}.
They prove that, with analytic assumptions on the initial data, the Navier Stokes solution can be described asymptotically as the sum of an Euler solution in the interior and a Prandtl boundary layer correction. Recently, the author \cite{Mae} was able to prove the $L^\infty$ convergence under the assumption that the initial vorticity is away from the boundary.

\medskip

These results in particular prove that Prandtl boundary layers are the right expansion, since
if there is an expansion, it should be true for analytic
functions, and therefore it must involve Prandtl layers. Therefore, we have no alternative asymptotic
expansions.

\medskip

However, analytic regularity is a very strong assumption. It mainly says that
there are no high frequencies in the fluid (energy spectrum of noise
decreases exponentially as the spatial frequency goes to infinity). In physical cases however there is always
some noise, which is not so regular (energy decreases like an inverse power
of the spatial frequency).

\medskip

Let us from now on consider Sobolev regularity.
In general, it does not appear to be possible to prove that Navier Stokes solutions behave
like Euler solutions plus a Prandtl boundary layer correction if we seek for global-in-time
results or if initially the boundary layer profile has an inflection point or the profile is not monotonic; see \cite{Gr1,GN}. Though, it leaves open that whether this expansion is possible for small time and monotonic initial
profiles with no inflection point in the boundary layer. The aim of this program is to discuss this question
in the case of shear flows, where the limiting Euler equation is trivial $u^E(t,x,z) = U_\infty$
(constant flow). Of course, a non-convergent result in this particular case would indicate that the expansion is not possible in general.

\section{Inside the boundary layer}

As mentioned earlier, it is crucial to understand what happens inside the boundary layer, which is of the size $\sqrt \nu$. Prandtl chooses an anisotropic change of variables
$$
T = t, X = x, Z = {Z \over \sqrt{\nu}}.
$$
However, a natural tendency of fluids is to create vortices, and vortices tend to be isotropic (comparable
sizes in $x$ and $z$). Vortices also evolve within times of order of their size. Hence it is more natural to
introduce an isotropic change of variables
$$
(T,X,Z) = {1 \over \sqrt{\nu}} (t,x,z) .
$$
In these new variables, the system of equations (\ref{NS1}), (\ref{NS2}), and (\ref{NS3}) turns to
\beq \label{NS1b}
\partial_T u^\nu + \nabla  ( u^\nu \otimes u^\nu) + \nabla p^\nu - \sqrt{\nu} \Delta u^\nu = 0,
\eeq
\beq \label{NS2b}
\nabla \cdot u^\nu = 0,
\eeq
with no-slip boundary condition
\beq \label{NS3b}
u^\nu = 0 \qquad \hbox{ on } \quad \partial \Omega .
\eeq
These equations are again the Navier Stokes equations where the viscosity $\nu$ has
been replaced by $\sqrt{\nu}$. These equations admit particular solutions of the form
\beq \label{shear}
u^\nu(T,X,Z) = U^P(\sqrt{\nu} T, Z)
\eeq
with $$
U^P(t,Z) = (U^P_s(t,Z), 0),
$$
where $U^P_s$ satisfies the scalar heat equation
\beq \label{heat1}
\partial_t U^P_s(t,Z) - \partial_Z^2 U^P_s(t,Z) = 0,
\eeq
with boundary condition
\beq \label{heat2}
U^P_s(t,0) = 0 .
\eeq
The particular solution $U^P$ is called the shear flow or shear profile. Note that $U^P(t,Z)$ is also a particular solution of the Prandtl equations, since for
shear layer profiles, the Prandtl equations and Navier Stokes equations simply
reduce to the same heat equation. The existence of solutions to Prandtl equation is of course
trivial in this particular case.

\medskip

{\em However, do we still have convergence for small Sobolev perturbations of such profiles?}

\medskip

Namely, let us consider initial data of the form
\beq \label{init}
u^\nu(0,X,Z) = (U^P_s(0,Z), 0) + v^\nu(0,X,Z),
\eeq
where $v^\nu$ is initial perturbation that is small in Sobolev spaces. Do we still have convergence
of $u^\nu(T,X,Z)$ to $U^P(\sqrt{\nu} T, Z)$, for $T>0$ ?

On bounded time intervals $0 < T < T_0$ ($T_0$ is fixed and independent on $\nu$),
the convergence is true and can be seen easily through classical $L^2$ energy estimates. However we are interested
by results on time intervals of the form $0 < T < T_0 / \sqrt{\nu}$ (that is, a uniform time in the original variable $t = \sqrt \nu T$). On such a long
interval in the rescaled variables, the classical $L^2$ energy estimates are useless.  The problem is to know
whether small perturbations of the limiting Prandtl profile can grow in a large time. {\em This is  a stability problem
for a shear profle for Navier Stokes equations.}

\medskip

The first step is to look at the linearized stability of the shear layer $U^P(\sqrt \nu T,Z)$. Let us freeze the time dependence in this shear profile, and study the stability of the time-independent profile $U^P(0,Z)$. The linearized Navier Stokes equations near $U^P(0,Z)$ then read
\beq \label{NS1l}
\partial_T v^\nu +  ( U^P \cdot \nabla) v^\nu + (v^\nu \cdot \nabla) U^P
+ \nabla q^\nu - \sqrt{\nu} \Delta v^\nu = 0,
\eeq
\beq \label{NS2l}
\nabla \cdot v^\nu = 0,
\eeq
with no-slip boundary condition
\beq \label{NS3b}
v^\nu = 0 \qquad \hbox{ on }\quad  \partial \Omega .
\eeq

If all the eigenvalues of this spectral problem have non-positive real parts, then it is likely
that $v^\nu$ remains bounded for all time, and that this is also true for the linearization near
the time-dependent profile $U^P(\sqrt \nu T,Z)$ and also true for the nonlinear Navier Stokes equations. In this case, we could
expect convergence from Navier Stokes to Euler with a Prandtl correction.

If one eigenvalue has a positive real part, then there exists a growing mode of the form
$V e^{c(\nu) T}$, with $\R c(\nu) > 0$. The time scale of instability $1 / \R c(\nu)$ must
then be compared with $1 / \sqrt{\nu}$. If $\R c(\nu) \ll \sqrt{\nu}$, then instability appears
in very large time, much larger than $T_0 / \sqrt{\nu}$ and convergence may hold.
On the contrary if $\R c(\nu) \gg \sqrt{\nu}$, then instability is strong and occurs much before
$T_0 / \sqrt{\nu}$. In this latter case, it is then likely that such an instability occurs
for $U^P(0,Z)$ and that it might not possible to prove convergence of Navier Stokes to Euler plus a Prandtl layer
in supremum norm or strong Sobolev norms.

The study of Prandtl boundary layer is therefore closely linked to the question of the spectral
stability of shear profiles for Navier Stokes equation with $\sqrt{\nu}$ viscosity,
and more precisely to the comparison of $\Re(c)$ with respect to $\sqrt{\nu}$.

\section{Spectral problem}

\subsection{Orr Sommerfeld and Rayleigh equations}

The analysis of the spectral problem is a very classical issue in fluid mechanics.
A huge literature is devoted to its detailed study. We in particular refer to
\cite{Reid, Schlichting} for the major works of Tollmien, C.C. Lin, and Schlichting.
The studies began around 1930, motivated by the study of the boundary layer
around wings. In airplanes design, it is crucial to study the boundary layer
around the wing, and more precisely the transition between the laminar and turbulent
regimes, and even more crucial to predict the point where boundary layer
splits from the boundary. A large number of papers has been devoted to
the estimation of the critical Rayleigh number of classical shear flows
(Blasius profile, exponential suction/blowing profile, etc...).

Let us go further in detail in the case of two dimensional spaces. The first step is to make a Fourier transform
with respect to the horizontal variable, and a Fourier transform with respect to time variable
on the to  stream function $\phi$. This leads to the following form for perturbations
$v^\nu$
\begin{equation}\label{def-stream}v^\nu = \nabla^\perp \psi = (\partial_Z, -\partial_X)\psi
,\qquad \psi(T,X,Z) := \phi (Z) e^{i \alpha (X - cT) }.
\end{equation}
Putting this Ansatz in (\ref{NS1l}), we get the classical Orr--Sommerfeld equation
\beq \label{OS1}
{1 \over i \alpha R} (\partial_Z^2 - \alpha^2)^2 \phi
= (U-c) ( \partial_Z^2 - \alpha^2) \phi  - U'' \phi
\eeq
with boundary conditions
\beq \label{OS2}
\alpha \phi  = \partial_Z \phi  = 0 \qquad\hbox{ at } Z = 0
\eeq
and
\beq \label{OS3}

 \phi  \to 0\qquad  \hbox{ as } Z \to + \infty.
\eeq
Here $R = 1 / \sqrt{\nu}$ is the Reynolds number (to our rescaled equations) and $U = U_s^P(0,Z)$
is the shear profile  introduced in \eqref{heat1} and \eqref{heat2}.
The spectrum of (\ref{OS1}) clearly  depends on $\alpha$ and $R$.

\medskip

As $R \to \infty$, or rather $\alpha R \to \infty$, the Orr--Sommerfeld equations formally reduce to the so-called
Rayleigh equation
\beq \label{Rayleigh1}
 (U-c) ( \partial_Z^2 - \alpha^2) \phi  = U'' \phi
\eeq
with boundary conditions
\beq \label{Rayleigh2}
 \phi  = 0\qquad  \hbox{ at } Z = 0
\eeq
and
\beq \label{Ra3}

\phi  \to 0\qquad  \hbox{ as } Z \to + \infty.
\eeq
The Rayleigh equation describes the stability of the shear profile $U$ for  Euler equations.
The spectrum of Orr Sommerfeld is a perturbation of the spectrum of Rayleigh equation. It is therefore natural to first study the Rayleigh equation.

\medskip

Stability of the Rayleigh problem depends on the profile. For some profiles, all the eigenvalues
are imaginary, and for some others there exist unstable modes. There are various
criteria to know whether a profile is stable or not, including classical Rayleigh inflection point
and Fjortoft criteria. We shall recall these two criteria in the next subsection.

\subsection{Classical stability criteria}
The first criterium is due to Rayleigh.
\medskip

{\bf Rayleigh's inflexion-point criterium (Rayleigh \cite{Ray}).}  {\em A necessary condition for instability is that the basic
velocity profile must have an inflection point.}

\medskip

The criterium can easily be seen by multiplying by $\bar \phi/(U-c)$ to the Rayleigh equation \eqref{Rayleigh1} and using integration by parts. This leads to
\begin{equation}\label{id-Ray}\int_0^\infty (|\partial_Z \phi|^2 + \alpha^2 |\phi|^2) \;dZ+ \int_0^\infty \frac{U''}{U-c}|\phi|^2 \;dZ=0,\end{equation}
whose imaginary part reads
\begin{equation}\label{id-Ray1}\I c \int_0^\infty \frac{U''}{|U-c|^2}|\phi|^2 \;dZ =0.\end{equation}
Thus, the condition $\I c>0$ must imply that $U''$ changes its sign. This gives the Rayleigh criterium.

\medskip

A refined version of this criterium was later obtained by Fjortoft (1950) who proved
\medskip

{\bf Fjortoft criterium \cite{Reid}.} {\em A necessary condition for instability is that $U'' (U - U(z_c))<0$ somewhere in the flow, where $z_c$ is a point at which $U''(z_c) =0$.}

\medskip
To prove the criterium, consider the real part of the identity \eqref{id-Ray}:
$$\int_0^\infty (|\partial_Z \phi|^2 + \alpha^2 |\phi|^2) \;dZ+ \int_0^\infty \frac{U'' (U-\R c)}{|U-c|^2}|\phi|^2 \;dZ=0.$$
Adding to this the identity
$$ (\R c - U(z_c))\int_0^\infty \frac{U''}{|U-c|^2}|\phi|^2 \;dZ =0,$$
which is from \eqref{id-Ray1}, we obtain
$$ \int_0^\infty \frac{U'' (U-U(z_c))}{|U-c|^2}|\phi|^2 \;dZ=-\int_0^\infty (|\partial_Z\phi|^2 + \alpha^2 |\phi|^2) \;dZ <0,$$
 from which the Fjortoft criterium follows.

\medskip

\subsection{Unstable profiles for Rayleigh equation}

If the profile is unstable for the Rayleigh equation, then there exist $\alpha$ and an eigenvalue $c_\infty$ with
$\I c_\infty > 0$, with corresponding eigenvalue $\phi _\infty$.
We can then make a perturbative analysis to construct an eigenmode $\phi _R$ of
the Orr-Sommerfeld equation with an eigenvalue $\I c_R >0$ for any large enough $R$.

The main point is that $\partial_Z \phi _R$ vanishes on the boundary whereas $\partial_Z\phi _\infty$
does not necessarily vanishes. We therefore need to add a boundary layer to
correct $\phi _\infty$. This boundary layer comes from the balance between the terms
$\partial_Z^4 \phi  / \alpha R$ and $U \partial_Z^2 \phi $ of (\ref{OS1}) and is therefore
of size
$$
\sqrt{ U_0 \over \alpha R} = \cO( R^{-1/2}) = \cO(\nu^{1/4}).
$$
In original $t,x,y$ variables, this leads to a boundary layer of size $O(\nu^{3/4})$.
In the limit $\nu \to 0$, two layers appear: the Prandtl layer of size $\sqrt{\nu}$
and a so-called viscous sublayer of size $\nu^{3/4}$.
This sublayer has an exponential profile in $Z/ \nu^{1/4}$.
The existence and study of the viscous sublayer is a classical issue in physical fluid mechanics.

When $\phi _R$ is constructed and corrected by this sublayer, it in fact still does not satisfy (\ref{OS1}), but it does satisfy the Orr-Sommerfeld boundary conditions exactly and the Orr-Sommerfeld equation
up to an error with size of $\cO(1/R)$. By perturbative arguments we can prove
\beq \label{perturb1}
c_R = c_\infty + \cO(R^{-1}) .
\eeq
Next, starting from $\phi _R$, we can then construct unstable modes for the linearized Navier
Stokes equations, and even get instability results in strong norms for the nonlinear Navier
Stokes equations. This has been carried out in detail by E. Grenier in \cite{Gr1}.

\subsection{Stable profiles for Rayleigh equation}

Some profiles are stable for the Rayleigh equation; in particular, shear profiles
without inflection points from the Rayleigh's inflexion-point criterium. For stable profiles, all the spectrum of the Rayleigh equation
is imbedded on the axis: $\I c = 0$.
At a first glance, we may believe that (\ref{perturb1}) still holds true, which would mean
that any eigenvalue of the Orr-Sommerfeld would have an imaginary part $\I c_R$ of
order $\cO(R^{-1}) = \cO(1 / \sqrt{\nu})$. This would mean that perturbations would
increase slowly, and only get multiplied by a constant factor for times $t$ of order
$T_0 / \sqrt{\nu}$. In this case we might hope to obtain the convergence from Navier-Stokes
to Euler and Prandtl equations. {\em However, this is not the case, and $\I c_R$ appears to be much larger.} Let us detail
now this point.

\medskip

The main point is that in the case of a stable profile, there exists an eigenmode $\phi_\infty$ with corresponding
eigenvalue $c_\infty$ which is {\it small} 	and {\it real}. Therefore there exists some $z_c$ such that
$$
U(z_c) = c_\infty .
$$
Such a $z_c$ is called a critical layer. As $z_c$, $U(z) - c_\infty$ vanishes, hence Rayleigh
equation is singular
\beq \label{RayR}
(\partial_Z^2 - \alpha^2) \phi = {U'' \over U - c_\infty} \phi .
\eeq
Therefore when $R$ goes to infinity, for $z$ near $z_c$, Orr Sommerfeld degenerates from a fourth
order elliptic equation to a singular second order equation. At $z = z_c$, all the derivatives disappear
as $R$ goes to infinity, and we go from a fourth order equation to a "zero order" one. The limit is
therefore very singular, and as a matter of fact $\Im c(R)$ is much larger than expected.

\medskip

Let us go on with the analysis of Rayleigh equation.
The Rayleigh equation (without taking care of boundary
conditions) admits two independent solutions $\phi _1$ and $\phi _2$, one smooth
$\phi _1$ which vanishes at $z_c$ and another $\phi _2$ which is less regular near
$z_c$. Using (\ref{RayR}) we see that $\phi_2''$ behaves like $O(1 / Z - z_c)$
near $z_c$. Hence $\phi_2$ behaves like $(Z-z_c) \log(Z-z_c)$ near  $z_c$.
Therefore, the eigenvector $\phi _\infty$ is of the form
\beq \label{form}
\phi _\infty = P_1(Z) + (Z - z_c) \log(Z-z_c) P_2(Z)
\eeq
where $P_1$ and $P_2$ are smooth functions, with $P_1(z_c) = 0$.

\bigskip

If we try to make a perturbation analysis to get $\phi _R$ out of $\phi _\infty$, we then face two
difficulties. First, we have to correct $\phi _\infty$ in order to satisfy $\phi _\infty = 0$ at $Z = 0$.
But there is another much more delicate difficulty.
As $\phi _\infty$ is not smooth at $Z = z_c$ it is not a good approximation of $\phi _R$ near
$z_c$. In particular $(\partial_Z^2 - \alpha^2)^2 \phi _\infty$ is too singular at $z_c$, of order
$O(1 / (Z - z_c)^3)$.

To find a better approximation, one notes that near the singular point $z_c$, the term $\partial_Z^4 \phi _R$
can no longer be neglected.
In fact, near this point $z_c$, $\partial_Z^4\phi  / i \alpha R$ must balance with $U'(z_c) (Z - z_c) \partial_Z^2 \phi $.
This leads  to the introduction of another boundary layer of size $(\alpha R)^{1/3}$, near $z_c$
 satisfying the equation
\beq \label{Airy}
\partial_Y^2 \Phi_R = Y \Phi_R, \qquad \Phi_R := \partial_Z^2 \phi _R
\eeq
where
$$
Y := {Z - z_c \over (\alpha R U'(z_c))^{1/3}}.
$$
This layer is called {\it critical layer}.
Note that (\ref{Airy}) is simply the classical Airy equation. If we try to construct $\phi _R$ starting from
$\phi _\infty$, we therefore have to involve Airy functions to describe what happens
near the critical layer. As a consequence, $(\alpha R)^{1/3}$ is an important parameter,
and similarly to the unstable case, we could prove
\beq \label{exp2}
c_R = c_\infty + \cO( (\alpha R)^{-1/3}) + \cO(R^{-1}) .
\eeq
Hence, the situation is very delicate. It has been intensively studied in the period
$1940 - 1960$ by many physicists, including Heisenberg, C.C. Lin, Tollmien, Schlichting, among others.
Their main objective was to compute the critical Reynolds number of shear layer flows,
namely the Reynolds number $R_c$ such that for $R > R_c$ there exists an unstable
growing mode for the Orr-Sommerfeld equation. Their analysis requires a careful study
of the critical layer.

\medskip

From their analysis, it turns out that there exists some $R_c$ (depening on the profile)
such that for $R > R_c$ there are solutions $\alpha(R)$, $c(R)$ and $\phi _R$ to the Orr-Sommerfeld equations with
$\I c(R) > 0$. Their formal analysis has been compared with modern numerical experiments
and also with experiments, with very good agreement. Note that physicists
are interested in the computation of the critical Reynolds number, since
any shear flow is unstable if the Reynolds number is larger than this critical
Reynolds number. {\em In this program, we are interested in the high Reynolds limit,
which is a different question.}
This limit is not a physical one, since any flow has a finite Reynolds
number, and not in any physical case can we let the Reynolds go to very very
high values. Physical Reynolds numbers may be large (of several millions
or billions), much larger than the critical Reynolds number, but despite their
large values, they are too small to enter the mathematical limit $R \to + \infty$
we are considering. Fluids would enter the mathematical asymptotic regime
if $R^{-1/7}$ or $R^{-1/11}$ (see below) are large numbers, which
leads Reynolds numbers to be of order of billions of billions, much larger than
any physical Reynolds number!

\medskip

It is thus important to keep in mind that the mathematical limit is not physically pertinent.
Physically, the most important phenomena are: the existence of a critical Reynolds
number (above which the shear flow is unstable), the transition from laminar
to turbulent boundary layers, the separation of the boundary layer from the boundary.
All of these occur near the critical Reynolds number, which is large, but not in the
asymptotic regime which we will now consider.

\medskip

The problem is now to study rigorously the asymptotic behavior of $\alpha$ and $c$ as
$R \to \infty$.
Let us present now some classical physical results. These results can be found, for example, in the book of Drazin and Reid \cite{Reid} or of H. Schlichting \cite{Schlichting}.

\medskip

For $R$ large enough there exists an interval $[\alpha_1(R),\alpha_2(R)]$
such that for every $\alpha$ in this interval there exists an unstable mode
with $\I c(R) > 0$. The asymptotic behavior of $\alpha_1$ and $\alpha_2$ depends
on the shear profile.

\medskip

\begin{itemize}

\item For plane Poiseuille flow (not a boundary layer): $U(z) = z^2 - 1$ for $0 < z < 1$.
In this case
$$
\alpha_1(R) \sim C_1 R^{-1 / 7}, \quad
\alpha_2(R) \sim C_2 R^{-1/11} .
$$
\item For boundary layer profiles:
$$
\alpha_1(R) \sim C_1 R^{-1 / 4}, \quad
\alpha_2(R) \sim C_2 R^{-1/6} .
$$
\item For the Blasius (a particular boundary layer) profile:
$$
\alpha_1(R) \sim C_1 R^{-1 / 4}, \quad
\alpha_2(R) \sim C_2 R^{-1/10} .
$$

\end{itemize}

\begin{figure}[t]
\centering
\includegraphics[scale=.4]{marginalcurves}
\put(-20,1){$R^{1/5}$}
\put(-210,117){$\alpha^2$}
\put(-213,30){$\alpha_\mathrm{low}\approx R^{-1/4}$}
\put(-75,70){$\alpha_\mathrm{up}\approx R^{-1/10}$}
\caption{\em Illustrated are the marginal stability curves; see also \cite[Figure 5.5]{Reid}.}
\label{fig-shear}
\end{figure}

More precisely, in the $\alpha, R$ plane, the area where unstable modes exist is shown on figure \ref{fig-shear}. For small $R$, all the $\alpha$ are stable. Above some critical Reynolds number, there
is a range $[\alpha_1(R),\alpha_2(R)]$ where instabilities occur. This instability area is bounded
by so called lower and upper marginal stability curves.

Associated with the range of $\alpha(R)$, we have to determine the behavior of the eigenvalues
 $c(R,\alpha)$, or more precisely the imaginary part of $c(R,\alpha)$.
 The complete mathematical justification of the construction of unstable modes will be detailled in companion
 papers. Here we juste want to present a quick and as simple as possible  construction of the unstable Orr
 Sommerfeld modes. We will skip all difficulties and only focus on the backbone of the instability.

 \subsection{A sketch of the construction of unstable modes: the lower branch $\beta = 1/4$}

 We recall that there is no mathematically rigorous arguments of this section. We thus show the main
 ingredients of the instability, keeping under silence any other term. We assume that our profile
 $U(Z)$ is stable for Rayleigh equation. We focus on the lower
 marginal stability curve. In this case
 $$
 \alpha \sim A R^{-1/4}
 $$
 and
 $$
 \delta \sim (\alpha R)^{-1/3} \sim A^{-1/3} R^{-1/4} .
 $$
 Let us assume that $R$ is very large, and $\alpha$ very small. For small $\alpha$, Rayleigh
 equation is very close to
 \beq \label{RayT}
  (U - c) \partial_Z^2 \phi = U'' \phi
  \eeq
  which has an obvious solution
  $$
  \phi_1 = U - c.
  $$
 There exists another independent particular solution to (\ref{RayT}),  but it turns out that this second
 solution grows linearly as $Z$ increases, and may therefore be discarded. Note that
 $\phi_1$ is a smooth function and an approximate solution of Orr Sommerfeld.

 We next focus on Airy equation (\ref{Airy}). It has two particular fast decaying / growing solutions $\Phi_R = Ai$ and $Bi$.
 Only $Ai$ goes to $0$ as $Y$ goes to infinity, hence $Bi$ may be discarded.
 Let us denote by $Ai(1,Y)$ a primitive of $Ai$ and $Ai(2,Z)$ a primitive of $Ai(1,Y)$.
 Then
 $$
 \phi_3 = Ai(2,Y) = Ai(2, \delta^{-1} (Z - z_c) )
 $$
 is a particular solution of Airy, and an approximate solution of Orr Sommerfeld.
  Now we look for an eigenmode of Orr Sommerfeld which is a combination of $\phi_1$ and
 $\phi_3$ of the form:
 $$
 \phi = A \phi_1 + B \phi_3 .
 $$
  It has the good behavior as $Z$ goes to infinity. It remains to know whether we can find $A$ and $B$
  such that $\phi(0) = \partial_Z \phi(0) = 0$. This happens if the dispersion relation
  $$
  \phi_1(0) \partial_Z \phi_3(- \delta^{-1} z_c) = \partial_Z \phi_1(0) \phi_3(- \delta^{-1} z_c)
  $$
holds,  or equivalently if
  \beq \label{disper}
  {\phi_1(0) \over \partial_Z \phi_1(0) }
  =
  \delta {Ai(2,- \delta^{-1} z_c) \over Ai(1,- \delta^{-1} z_c)} .
  \eeq
  The left hand side of (\ref{disper}) is
  $$
    {\phi_1(0) \over \partial_Z \phi_1(0) }  = {U_0 - c \over U_0'}
    $$
   and its imaginary part is simply $- \Im c / U_0'$, with $U_0' >0$. Here, $U_0 = U(0)$ and $U'_0 = U'(0)$.
   Now $Ai(2,Z) / Ai(1,Z)$ is the classical Tietjens function
$$
T(Y)= {Ai(2,Y) \over Ai(1,Y) }.
$$
The main point is that $\Im T(Y)$ changes sign as $Y$ goes to infinity. It is positive for small $Y$ and
negative for large $Y$. As a consequence $\Im c$ changes sign as $z_c / \delta$ increases.
{\em This change of sign leads to the existence of unstable modes.}

It remains to link $z_c / \delta$ with $R$ and to prove that for $z_c / \delta$ goes to infinity as
$R$ increases. For this we have first to refine $\phi_1$. Namely $\phi_1$ does not go to $0$ as
$Z$ goes to infinity. For $\alpha > 0$, we may construct a solution $\phi_{1,\alpha}$ of the Rayleigh
equation which is a perturbation of $\phi_1$ and decreases like $\exp(-\alpha Z)$.
A classical perturbative analysis leads to
$$
\phi_{1,\alpha}(0) = U - c + \alpha { (U(\infty) - U_0)^2 \over U_0'} + ...
$$
Moreover as $Y$ goes to infinity,
$$
T(Y) \sim C Y^{-1/2} .
$$
Hence the dispersion relation takes the form
\beq \label{disper2}
{U_0 - c \over U_0'} + \alpha { (U(\infty) - U_0)^2 \over {U_0'}^2} + ...
\sim \delta (1 + | z_c  / \delta | )^{-1/2} .
\eeq
Assuming that $\alpha$ is much larger than the right hand side, which is the case if $A$ is
large enough, this gives that
$| U_0 - c |$  is of order $\alpha$, and hence $z_c$, defined by $U(z_c) = c$ is of order $\alpha$. Hence
$$
z_c / \delta \sim A^{4/3} .
$$
Therefore provided as $A$ increases, $\Im c$ changes from negative to positive values: there exists
an threshold $A_{1c}$ such that $\Im c > 0$ is $A > A_{1c}$.
This ends our overview of the lower marginal curve.

 \subsection{A sketch of the construction of unstable modes: the upper branch $\beta = 1/6$}

The upper branch of marginal stability is more delicate to handle.
Roughly speaking, when the expansion of $\phi_{1,\alpha}$ involves $\phi_2$, independent
solution of Rayleigh equation which is singular like $(z - z_c) \log(z - z_c)$. This singularity
is smoothed out by Orr Sommerfeld in the critical layer. This smoothing involves second primitives
of solutions of Airy equation. As we take second primitives, a linear growth is observed
(linear functions $\phi_R$ are obvious solution of (\ref{Airy})).  This linear growth gives an extra term
in the dispersion relation which can not be neglected when $\alpha \sim R^{-1/6}$. It has a stabilizing
effect and is responsible of the upper branch for marginal stability.

\section{Program}

The situation is well-known, physically speaking. However, to the best
of our knowledge, the formal analysis has never been justified mathematically.
Our ultimate goal is to prove the following conjecture:

\medskip

{\it
{\bf Conjecture:} generically, shear flows for Navier Stokes are linearly  unstable,
and the Prandtl expansion is not valid in Sobolev spaces.}

\medskip

Let us now lay out our program to tackle the conjecture.

\medskip

1. The first step is to construct unstable modes for the Orr-Sommerfeld equations
 as $R \to \infty$. This  requires a careful analysis of this singular
perturbation and a careful study of the behavior of the eigenvalues $c_R$. This leads to the proof
than generic shear layers are spectrally unstable.

More precisely, we will construct growing modes (those with $\I c >0$) for \eqref{OS1}-\eqref{OS3}
 when $R$ is large and $\alpha$ belongs to the interval $(\alpha_1(R),\alpha_2(R))$,
  with
   \begin{equation}\label{ranges-alpha}\alpha_1(R) = A_{1c}R^{-1/4}\qquad
     \mbox{and}\qquad  \alpha_2(R)= A_{2c} R^{-1/6}\end{equation}
    for some fixed constants $A_{1c}, A_{2c}$.
    The curves $\alpha_j(R)$ are called lower and upper branches of the marginal (in)stability for the boundary layer $U$.
That is, there is a critical constant $A_{1c}$ so that with $\alpha_1(R) = A_1 R^{-1/4}$, the imaginary part of $c$ turns from negative (stability) to positive (instability) when the parameter $A_1$ increases across $A_1=A_{1c}$. Similarly, there exists an $A_{2c}$ so that with $\alpha = A_2 R^{-1/6}$,
 $\I c$ turns from positive to negative as $A_2$ increases across $A_2 = A_{2c}$. In particular, we obtain instability of the profile in the intermediate zone: $\alpha \sim R^{-\beta}$ for $1/6<\beta<1/4$.

Our main result is as follows.
\begin{theorem}[Spectral instability of generic shear flows \cite{GGN1}]\label{theo-unstablemodes}~\\
Let $U(z)$ be a shear profile with $U'(0) \ne 0$ and satisfy
$$
\sup_{z \ge 0} | \partial^k_z (U(z) - U_+) e^{\eta z} | < + \infty, \qquad k=0,\cdots ,4,
$$ for some constants $U_+$ and $\eta > 0$.
There exists two constants $A_1$ and $A_2$ such that
for $R$ large enough and for $\alpha = A R^{-\beta}$ with arbitrary $A$ if $1/6 < \beta < 1/4$
 or $A > A_1$ if $\beta = 1/4$ or $A < A_2$ if $\beta = 1/6$,
 there exist
$c(R)$  and  $\phi_R$ such that $\phi_R$ is an eigenfunction of the problem
(\ref{OS1}), (\ref{OS2}), and (\ref{OS3}) with corresponding eigenvalue $c(R)$.  \\
More precisely, $\phi_R$ satisfies the boundary conditions (\ref{OS2})-(\ref{OS3})
and satisfies (\ref{OS1}). In addition, there holds the estimate
$$
c(R) \quad \sim\quad  c_0 R^{\beta -1/2},
$$
for some constant $c_0$ independent on $R$, with $\I c_0 > 0 .$ In particular, the growth rate for the unstable modes is
$$ \alpha \I c(R) \quad \sim\quad  R^{-1/2} .$$
\end{theorem}

\subsubsection*{Remarks}

\begin{itemize}

\item[i)] The assumption $U'(0) \ne 0$ is technical. A similar analysis
could be fulfilled to allow the case $U'(0)=0$, with different (presumedly, more complicated) asymptotic behavior in the expansions.

\item[ii)]

The asymptotic behavior of the growth rate $\alpha \I c(R) \sim C_R R^{-1/2}$ holds in the {\it rescaled}
variables. In the original ones, this means that the unstable mode increases like

 $\exp( C t R^{1/2}) = \exp (C t /\nu^{1/4})$.
 As a consequence,
one cannot expect stability in Sobolev norms for small perturbations of such shear flows.
Small perturbations will quickly increase in the time variable $t$ and may become of order
$1$ in a vanishing time (i.e., in a time that tends to zero as $\nu \to 0$). Therefore it is likely that slightly initially perturbed solutions of
Navier Stokes equations do not converge to the Prandtl equations as $\nu \to 0$.

\item[iii)]

It is worth noting that if we assume that the initial perturbation is analytic, then Fourier modes $\alpha / \sqrt{\nu}$
(in $x$ variables) are initially as small as  $\exp( - C / \sqrt{\nu})$.
Hence even if they grow fast, like

$\exp(C_1 t / \nu^{1/4})$,
they remain negligible as
long as $t < C  / (C_1\nu^{1/4})$. Therefore for small times, analytic perturbations remain negligible
and we have convergence from Navier Stokes equation to Euler plus Prandtl for such
initial analytical data.

\end{itemize}

2. The second step is to prove linear instability. For a fixed viscosity, nonlinear instability follows from the spectral instability; see \cite{FPS} for arbitrary spectrally unstable steady states. However, in the vanishing viscosity limit, linear to nonlinear instability is a very delicate issue, primarily due to the fact that there are no available, comparable bounds on the linearized solution operator as compared to the maximal growing mode. Available analyses (for instance, \cite{Fri, Gr1}) do not appear applicable in the inviscid limit. In addition, boundary layers are shear layer profiles, which are time-dependent and are solutions of
the linear heat equation. In this case, even the proof of linear instability is no longer straightforward since the equation of the perturbation
changes with time.

To get such a nonlinear instability result, we have to bound the resolvent of linearized Navier Stokes equations with fixed stationary
profiles, and then treat the time-dependent profiles as small perturbations within a vanishing time in the inviscid limit. Getting bounds on the resolvent is
however highly technical, and we plan to follow the ideas developed by K. Zumbrun and coauthors; \cite{Zu1}. This problem
will be investigated in a further work.

\medskip

Note that a similar analysis may be done for channel flows, including the classical plane Poiseuille flows. More precisely, we establish the following:

\begin{theorem}[Spectral instability of generic shear flows \cite{GGN2}]\label{theo-unstablemodes2}~\\
Let $U(z)$ be an arbitrary shear profile that is analytic and symmetric about $z=1$ with $U'(0) > 0$ and $U'(1) =0$. There exist
 $\alpha_\mathrm{low}(R)$ and $\alpha_\mathrm{up}(R)$, there exists a critical Reynolds number $R_c$ so that for all $R\ge R_c$ and all $\alpha \in (\alpha_\mathrm{low}(R), \alpha_\mathrm{up}(R))$, there exist
a triple $c(R), \hat v(z; R), \hat p(z;R)$, with $\mathrm{Im} ~c(R) >0$, such that
$$
v_R: = e^{i\alpha(y-ct) }\hat v(z;R), \quad
p_R: = e^{i\alpha(y-ct)} \hat p(z;R)
$$
solve the problem
\eqref{NS1}-\eqref{NS2} with the no-slip boundary conditions. In the case of instability, there holds the following estimate for the growth rate of the unstable solutions:
$$ \alpha \I c(R) \quad \approx\quad  (\alpha R)^{-1/2},$$
as $R \to \infty$. In addition, the horizontal component of the unstable velocity $v_R$ is odd in $z$, whereas the vertical component is even in $z$.
\end{theorem}


TITLE: Simple Acceleration Problem throwing me off
QUESTION [0 upvotes]: I am having trouble with this problem from my homework. I have gone over it many times, and I must have some concept wrong.
A rocket is launched from the ground with constant acceleration. After 4 seconds, a bolt falls off and hits the ground 6 seconds later. Gravity is -9.8m/s^2. What was the rocket's acceleration?
So here's my thoughts.
First, let's get the height of the rocket when the bolt falls.
Fell at 9.8m/s^2 for 6s, so the final velocity was 58.8m/s
Average velocity was half of that, because of the constant acceleration, so 29.4m/s
Total distance is that average, times 6 seconds, 176.4m.
Good so far?
Next, the rocket's acceleration
We know it was 176.4m in the air after 4 seconds.
So the average velocity 44.1 m/s, final velocity is 88.2 m/s
It got to that in 4 seconds, so the acceleration must be 22.05 m/s^2
But it's not. The correct answer is 5.5 m/s^2, a quarter of what I got.
So what went wrong?
I've tried looking through all the steps a bunch of times, and I can only think of one step that doesn't seem quite right. The bolt falling off the rocket makes it sound like it just drops, but maybe it starts that drop with an upwards velocity?
But I can't see any way of getting to that velocity. Without knowing how high the rocket is, we can't know how fast it's going, so we can't know how fast the bolt was moving upwards. And then we have two variables in the equation, and an infinite number of correct pairs.
I must be missing some concept, some formula must be missing. Can you find it?

REPLY [1 votes]: Since the initial speed an position are zero, the position and upward velocity after 4 seconds are:
$z(4) = \dfrac{a_1 (4s)^2}{2}$
$v(4) = a_1 (4s)$
When the bolt falls off, the acceleration changes from $a_1$ to $-9.8\dfrac{m}{s^2}$ and the bolt impacts the ground 6 seconds later.  So...
$z(10) = 0 = z(4) + v(4)\cdot 6s - 9.8\dfrac{(6s)^2}{2} = \dfrac{a_1 (4s)^2}{2} + a_1 (4s)\cdot6s - 9.8\dfrac{(6s)^2}{2}$
Solve for $a_1$.


TITLE: Relation between null space of a matrix and column space.
QUESTION [0 upvotes]: Let $A$ be an $m\times n$ matrix. Then under what conditions null space of $A$ and column space of $A$ may have common elements.
My attempt: I start by taking a vector $v\in \mathbb{R}^n$ such that $$v\in N(A)\cap Col(A)$$ where $N(A)$ is the column space of A and $Col(A)$ is column space of A. Then above means
$$v\in N(A)\implies Av=0, \ v\in Col(A)\implies v=Ay\ \text{for some}\ y.$$
Combining two we get $$A^2y=0.$$
Can I deduce anything from above? Please help.

REPLY [0 votes]: Assume $A$ is square matrix and zero is an eigenvalue of $A$. If number of zero eigenvalues is $k$ and dimension of $N(A)$(also zero eigenspace) is strictly less than $k$, then we can directly say that $C(A) \cap N(A)$ has nontrivial element. It looks like the converse is true. But I could not prove yet.