TITLE: Solving Quadratic Matrix Programming (Trace of a Vector Multiplication) with Boundary Constraints
QUESTION [1 upvotes]: I intend to solve for vector $ x \in \mathbb{R}^{N \times 1} $ by solving the following optimization problem
\begin{align}
\arg \min_{x} Tr( (\mathbf{K} \mathbf{W})^T \mathbf{P} ( \mathbf{K} \mathbf{W})) - 2Tr( \mathbf{P} \mathbf{K} \mathbf{W})
\end{align}
\begin{align}
\text{subject to } & x_{i}^{min} \leq x_{i} \leq x_{i}^{max} \\
\end{align}
where $Tr()$ is the trace operator, $\mathbf{P} \in \mathbb{R}^{M \times M}$, $\mathbf{W}=\mathbf{A}diag(\mathbf{B} x)$, and  $\mathbf{W} \in \mathbb{R}^{D \times M}$, and $\mathbf{K} \in \mathbb{R}^{M \times D}$.
$\mathbf{A} \in \mathbb{R}^{D \times M}$ and $\mathbf{B} \in \mathbb{R}^{M \times N}$ are both positive metrices.
How do I solve it as an inequality constrained optimization problem for $X$ ?

REPLY [0 votes]: The problem is given by:
$$\begin{aligned}
\arg \min_{x} \quad & \operatorname{Tr} \left( {\left( K A \operatorname{diag} \left( B x \right) \right)}^{T} P \left( K A \operatorname{diag} \left( B x \right) \right) \right) - 2 \operatorname{Tr} \left( P K A \operatorname{diag} \left( B x \right) \right) \\
\text{subject to} \quad & {a}_{i} \leq {x}_{i} \leq {b}_{i} \; \forall i
\end{aligned}$$
in order to solve this problrm with Projected gradient descent we need to take the derivative of the first and second term with respect to $x$. for the first term, we can take the advantage of matrix $P$ which is positive-definite matrix and can be wrritten as $P=C^{T}C$. the first term then can be treated as frobenius norm:
$$\begin{aligned}
 \operatorname{Tr} \left( {\left( K A \operatorname{diag} \left( B x \right) \right)}^{T} P \left( K A \operatorname{diag} \left( B x \right) \right) \right) = {\left\| K A \operatorname{diag} \left( B x \right) C \right\|}_{F}^{2}
\end{aligned}$$
then the gradient of the first term can be calculated as
$$\begin{aligned}
2{B}^{T} \operatorname{diag} \left( {(K A)}^{T} (K A) \operatorname{diag} \left( B x \right) CC^{T}\right)
\end{aligned}$$
if we consider the second term to be $T=- 2 \operatorname{Tr} \left( P K A \operatorname{diag} \left( B x \right) \right)$ we then have:
$dT=-2(PKA)^{T}:dX$ in which $dX=diag(Bx)$. using the properties of frobenius product we can write:
$dT=diag(-2(PKA)^{T}):di$ and $di=Bdx$ so:
$dT=B^{T}diag(-2A^{T}K^{T}(P)):dx$ and
$dT/dx=-2B^{T}diag(A^{T}K^{T}(P))$
overall the gradient of the above equation can be calculated as:
$$\begin{aligned}
2{B}^{T} \operatorname{diag} \left( {(K A)}^{T} (K A) \operatorname{diag} \left( B x \right) CC^{T}\right) -2B^{T}diag(A^{T}K^{T}P)
\end{aligned}$$
I appreciate if you can check and find out whether I am in a right direction or not?


TITLE: How can I prove a point is within a triangle, given three other points?
QUESTION [2 upvotes]: Could someone please explain the formula behind this, and then provide an example of how to do this? Basically I have 4 points, each with a longitude and latitude number. (They make a polygon quad, so a square or rectangle, in the shape of two triangles.) I then have a user's point. This is their location, also in the form of latitude and longitude coordinates. What is the process I need to determine if that point is INSIDE the square (or two triangles)?
For example, I may have four points:

Point A: x = 5, y = 0
Point B: x = 10, y = 0
Point C: x = 4, y = 3
Point D: x = 10, y = 3
User Point Z:  x = 7, y = 2

If I were to draw this out on a graph, I can see that "Z" is clearly inside the polygon. However, how can I prove this with math, rather than relying on graph paper?
Thank you!

REPLY [1 votes]: This is the classical "point in polygon" problem. There are two ways to attack it -- one way is to use ray casting, and the other is to use winding numbers.
You'll find lots of other material if you search for "point in polygon". Here is one link, and  here is another one.
You can solve the point-in-polygon problem by repeatedly solving the point-in-triangle problem, but you have to do it carefully if the polygon is not convex. See the second link above for details.


TITLE: $e^{Log_*(z^k)} = e^{kLog_*(z)} $?
QUESTION [2 upvotes]: Please let me know if this is a duplicate. I couldn't find any related questions so far.
In my class, we often find the branch cut of $f(z) = z^k$ by equating the function to $f(z) = e^{Log(z^k)} = e^{kLog(z)}$.
An example can be seen here (prof's note):

However, we cannot say ${Log(z^k)} = {kLog(z)}$ (for example, take $Log_\pi ((-1)^2)$).
I've been trying to prove the equation $e^{Log_*(z^k)} = e^{kLog_*(z)} $ or to find the condition under which it holds.
Following is my unsuccessful proof so far:
Let $z = re^{i\theta}, r\in \mathbb{R}/\mathbb{R^-}, \theta \in \mathbb{R}$.
Hence, $z^k = r^ke^{ik\theta}$
$Log_\pi (z^k) = \ln|r^ke^{ik\theta}| + iArg_\pi(r^ke^{ik\theta})$
$kLog_\pi (z) = k\left(\ln|re^{i\theta}| + iArg_\pi(re^{i\theta})\right) =k\ln (r) + ikArg_\pi(re^{i\theta}) $
Rasing $e$ to each power:
$e^{Log_\pi (z^k)} = e^{\ln|r^ke^{ik\theta}| + iArg_\pi(r^ke^{ik\theta})} =  |r^ke^{ik\theta}|\cdot e^{iArg_\pi(r^ke^{ik\theta})}$
$e^{kLog_\pi (z)} = e^{k\ln (r) + ikArg_\pi(re^{i\theta})} = r^k \cdot e^{ikArg_\pi(re^{i\theta})} =r^k \cdot e^{ik\theta'}$ where $\theta + 2\pi n =\theta'\in(\pi, 3\pi], n\in \mathbb{Z}$
I don't know how to go from here. Can somebody please help?
Or can you just tell me under what condition the equation in title holds?

Edit: I asked my prof the same and his reply was:

This is almost never true. It is true only if $k$ is an integer and you use $log$ instead of $Log$. It is also true that the functions $αLog(z)$ and $Log(zα)$ differ by a constant.

But how did he conclude $(-i)^{i-1} = e^{(i-1)Log(-1)}$ like in his note above?

REPLY [1 votes]: We often define the complex exponential [subject to the choice of branch, and using your collection of branches,] by $$z^w := \exp(w\operatorname{Log}_bz)$$ and as the definition of $\operatorname{Log}_b$, we also always have $z = \exp(\operatorname{Log}_bz)$.  When you put those two properties together, you have for any complex $z,w$ (with $z$ nonzero), $$\exp(w\operatorname{Log}_bz) := z^w = \exp(\operatorname{Log}_b(z^w)).$$
When $w$ is an integer, this will be consistent with the algebraic definition since a integral multiple of an integral multiple of $2\pi$ will be an integral multiple of $2\pi$.
When $z>0$ and $w$ are real, then you need to choose the principal branch to find a consistent result to real exponentiation.

With the above definition in mind, the equation in your professor's notes, $$(-i)^{i-1} = e^{(i-1)\operatorname{Log}_b(-i)},$$ is just a trivial application of the definition of complex exponentiation.


TITLE: Kernel of a polynomial with matrix, $ker(p(A))$
QUESTION [0 upvotes]: Let $A\in Mat(3,3,\mathbb R)$ a matrix and $\chi_A(x)=p_1(x)\cdot p_2(x)$ the characteristic polynomial. Evaluate $ker(p_1(A))$.$$A=\begin{pmatrix} 0 & 0 & 2 \\ 1 & 0 & 1\\ 0 & 1 & 1\end{pmatrix},p_1(x)=x-2$$

What do I have to do? I know how to calculate the kernell of matrix but not from $p_1(A)$

REPLY [0 votes]: As always, when applying a polynomial on a matrix, the result is again a matrix. In this case, we have $p_1(A)=A-2E_3$, where $E_3$ denotes the $3\times 3$-identity matrix. So you first calculate $p_1(A)$ and afterwards the kernel.


text:
\begin{document}

\begin{frontmatter}
\title{On the probability that two random integers are coprime}
\runtitle{Probability of coprime integers}

\begin{aug}
\author{\fnms{Jing} \snm{Lei}\thanksref{t1}\ead[label=e1]{jinglei@andrew.cmu.edu}}
\and
\author{\fnms{Joseph B.} \snm{Kadane}\ead[label=e2]{kadane@stat.cmu.edu}}

\thankstext{t1}{Research is partially supported by NSF grant DMS-1553884.}

\runauthor{Lei and Kadane}

\affiliation{Carnegie Mellon University}

\address{132 Baker Hall, Carnegie Mellon University, Pittsburgh, PA, 15213. \printead{e1,e2}.}

\end{aug}

\begin{abstract}
\input{abstract}
\end{abstract}

\begin{keyword}
\kwd{number theory}
\kwd{coprime}
\kwd{uniform distribution}
\kwd{finitely additive probability}
\end{keyword}

\end{frontmatter}

\section{Introduction}\label{sec:background}
\input{introduction}

\section{Finitely additive probability}\label{sec:finitely}
\input{finitely}

\section{Limiting relative frequency}\label{sec:rel_freq}
\input{frequency}

\section{Residue classes}\label{sec:residue-classes}
\input{residual}

\section{Shift invariance}\label{sec:shift-invariance}
\input{shift}

\section{Countably additive probability}\label{sec:countably}
\input{countably}

\section{Conclusion}
\input{conclusion}

\section*{Acknowledgements}
The authors would like to thank Boris Bukh for pointing out \Cref{thm:shift} and the idea of its proof.

\bibliographystyle{imsart-nameyear}
\bibliography{coprime}

\end{document}

config: arxiv


TITLE: Do there exist bounded operators with unbounded inverses?
QUESTION [5 upvotes]: I have just been introduced to the concept of invertibility for bounded linear operators. Specifically, we defined a bounded operator $A$ to be invertible if there exists a bounded $A^{-1}$ which is its right and left inverse, i.e. $AA^{-1}=\mathrm{id}_{\mathrm{Im}A},A^{-1}A=\mathrm{id}_{\mathrm{Dom}A}$. So I was wondering: is the requirement of boundedness (or equivalently of continuity) of the inverse important? Or, is it asking more than is granted by the invertibility? The open mapping theorem states a continuous linear surjective operator between Banach spaces is an open map, and thus if it has an inverse, that inverse is necessarily continuous. So for Banach spaces, we could avoid requiring this continuity explicitly, as it is automatic. But what about non-Banach spaces?
So the question is: do there exist normed spaces $X,Y$ and operators between them which are linear and bounded but have un-bounded inverses?

REPLY [2 votes]: The easier example is the identity with different norms in each side, the norm in the right strictly finer than the norm in the left.

REPLY [0 votes]: Yes, bounded inverses of unbounded operators are very important. Let $H$ be
an unbounded self-adjoint operator acting in a Hilbert space $\mathcal{H}$ .
Then $z-H$ with $Imz\neq 0$ is also unbounded but  $[z-H]^{-1},$the
resolvent of $H$ is bounded. The set $\{z\}$ for which the resolvent exists
and is bounded is known as the resolvent set. Its complement is the spectrum
of $H$. Resolvents play an essential role in the study of Schrödinger
operators.


TITLE: Heaviside with $(x^2-1/2)$ argument?
QUESTION [0 upvotes]: How to deal with the unit step function with an argument of $H(x^2-1/2)$ ? what does it equal to?

REPLY [2 votes]: It's immediate from the definition of the Heaviside function that
$$H\left(x^2 - \frac 1 2 \right) = \left\{\begin{array}{cc} 1 & \text{if }x^2 - \frac 1 2 > 0 \\ 0 & \text{ else }\end{array}\right.$$
So look at where $x^2 - \frac 1 2$ is positive and negative.


text:
\begin{definition}[Definition:Antireflexive Relation]
$\RR$ is '''antireflexive''' {{iff}}:
:$\forall x \in S: \tuple {x, x} \notin \RR$
\end{definition}

config: wiki


TITLE: Differentiate polar curve
QUESTION [0 upvotes]: I have attempted this question but have not found a solution. I am currently stuck. Hints on how I may go further would be helpful. Thank You in advance.
The Question:

What I have done so far:

REPLY [1 votes]: What you're doing is correct and is the same way I would have done it.  Once I read Claude Leibovici's comments I figured the problem is with the book.


TITLE: Expected value through Markov Chains
QUESTION [0 upvotes]: (HMMT 2012) You are repeatedly flipping a fair coin. What is the expected number of flips until the first time that
your previous 2012 flips are ‘HTHT...HT’?
Could someone tell me what's wrong with my solution using recursion and Markov Chains?
Define $n$ states corresponding to the $2013$ stages of 'HTHT...HT' (including the "$0$th" state, which is the start). Let $a_n$ denote the expected value of reaching the string at state $n$. For example, $a_3$ is the expected number of flips to reach 'HTHT...HT' given that the previous 3 flips were 'HTH.' Therefore, we want to find $a_0$.
Now we define the recursion $a_{n-1} = \frac{1}{2}(a_n+1)+\frac{1}{2}(a_{0}+1)$. From any state $a_{n-1}$, there is a $\frac{1}{2}$ chance that it will go to the next wanted element in the string, and this takes one flip. Similarly, there is a chance that it will go the wrong element, which brings the state back to state 0.
Note that $a_{2012} = 0$. $a_{2011} = \frac{1}{2}(a_0+2), a_{2010} = \frac{3}{4}(a_0+2), a_{2009} = \frac{7}{8}(a_0+2)$ From here, it's not difficult to prove with induction that $a_{2012-n} = \frac{2^k-1}{2^k}(a_0+2)$.
$a_0 = \frac{2^{2012}-1}{2^{2012}}(a_0+2)$
$a_0 = 2^{2013}-2$
Since the problems asks for the first time that the previous 2012 flips satisfy the condition, we add 1: $\fbox{$2^{2013}-1$}$
However, the correct solution is $\frac{(2^{2014} − 4)}{3}$.

REPLY [1 votes]: One possible flaw, let's see how can we possibly mess up.
If we toss $HTHTT$, we messed up and we have to start from nothing. We go to state $0$.
If we toss $HTHTHH$, we messed up, but we start with a given $H$. We go to state $1$.
Hence we need not always go back to state $0$. It depends on the parity of the location that we fail.


TITLE: Checking that $\cdot: D_{2n}\times \{1,2,\ldots,n\} \rightarrow\{1,2,\ldots,n\}$ given by $(\alpha , i) \mapsto\sigma_{\alpha}(i)$ is a group action
QUESTION [1 upvotes]: My question is on an example of a group action given in Dummit and Foote's Abstract Algebra (example 4, p.43):

If we fix a labeling of the vertices of a regular $n$-gon, each element $\alpha$ of $D_{2n}$ gives rise to a permutation $\sigma_{\alpha}$ of $\{1,2,\ldots,n\}$ by the way the symmetry $\alpha$ permutes the corresponding vertices. The map of $D_{2n} \times \{1,2,\ldots,n\}$ onto $\{1,2,\ldots,n\}$ defined by $(\alpha,i) \rightarrow \sigma_{\alpha}(i)$ defines a group action of $D_{2n}$ on $\{1,2,\ldots,n\}...$

I'm just trying to verify that the map $(\alpha,i) \rightarrow \sigma_{\alpha}(i)$ is indeed a group action. If we let $1$ denote the identity element of $D_{2n}$, then clearly $\sigma_1$ should correspond to the identity permutation, so $\sigma_1(i) = i$ for all $i \in \{1,2,\ldots,n\}$. This confirms the first property of a group action. Then if we let $*$ denote the binary operation on $D_{2n}$ and $\cdot$ the group action, we just have to verify that for all $\alpha_1, \alpha_2 \in D_{2n}$ and all $i \in \{1,2,\ldots,n\}$,
$$ \alpha_1 \cdot (\alpha_2 \cdot i) = (\alpha_1 * \alpha_2) \cdot i.$$
Now
\begin{align*}
    \alpha_1 \cdot (\alpha_2 \cdot i) &= \sigma_{\alpha_1}(\sigma_{\alpha_2}(i)) && (\text{by defn of } \cdot) \\[4pt]
     &= (\sigma_{\alpha_1} \circ \sigma_{\alpha_2})(i) && (\text{defn of } \circ)
\end{align*}
And
\begin{align*}
   (\alpha_1 * \alpha_2) \cdot i = \sigma_{\alpha_1 * \alpha_2}(i). && (\text{Defn of } \cdot)
\end{align*}
Now am I allowed to assert that $\sigma_{\alpha_1 * \alpha_2} = \sigma_{\alpha_1} \circ \sigma_{\alpha_2}$, and thus $\sigma_{\alpha_1 * \alpha_2}(i) = (\sigma_{\alpha_1} \circ \sigma_{\alpha_2})(i)$?
If I had to justify this, I would say something like the following: Pick an $n$-gon and label its vertices; call this the "original" $n$-gon with the "original labeling" of the vertices. If we rotate or reflect the original $n$-gon, this corresponds to a permutation $\alpha_1$ of the original labeling of the vertices. Then if we apply another rotation/reflection to this newly rotated/reflected $n$-gon, the resulting symmetry (relative to the original $n$-gon) corresponds to the map $\alpha_2 \circ \alpha_1$ (relative to the original labeling of the vertices). Is my reasoning here correct?
Is my argument solid enough, or is there a more rigorous way of expressing this? Is there anything that I'm missing?
(I think my uncertainty stems partly from the fact that Dummit and Foote distinguish between the group $D_{2n}$ and the group of permutations to which the elements of $D_{2n}$ correspond, which I find a tad confusing. Perhaps I am just overthinking things...)
A follow-up question:
The text goes on to say

Note that this action is faithful: distinct symmetries of a regular $n$-gon induce distinct permutations of the vertices.

Is this supposed to be accepted as "obvious" (I would be ok with that), or is there a rigorous algebraic way of showing this? i.e. if $\alpha_1, \alpha_2 \in D_{2n}$ and $\sigma_{\alpha_1} = \sigma_{\alpha_2}$, how would we show that this implies $\alpha_1 = \alpha_2$? Could we just say that $D_{2n}$ is isomorphic to the set of permutations to which the elements of $D_{2n}$ correspond (hence showing that book's claim)?
(Apologies if these are silly questions. Algebra is not my strong suit...)

REPLY [1 votes]: Here are some rows trying to make (more) clear the situation described in the question. First of all, we need a handy notation, and clear definition of the action.

Some words on the described action first. It involves $D_n$ in the notation from Dihedral group, wiki page, the dihedral group of all plane symmetries that invariate the set of the vertices of a fixed $n$-gon. And these vertices, labelled by $1,2,\dots,n$. Let us denote by $[n]$ this set of $n$ vertices. Then the given (to-be-)action is a map first:
$$
D_n\times [n]\to [n]\ .
$$
If this is (soon) an action, then it is an action of the group $D_n$ on the set $[n]$. And there is no need to consider permutations of $[n]$ (and/or the group of permutation of $[n]$ and/or the image of $D_n$ inside this group of permutations).
Maybe the example in loc. cit. wants to do something with the permutation $\sigma_\alpha$ associated to some symmetry $\alpha$ of the plane containing the vertices of the $n$-gon. But for the question, we just do not need this notational complication.
The map above is defined by $(\alpha,i)\to\alpha\cdot i:=\alpha(i)$.
Here, $\alpha(i)$ is the result of applying the symmetry (= isometric transformation, which is in particular a function) $\alpha$ on the vertex $i$.
Let us check, that this leads to an action. For the identic map id (the neutral element in $D_n$), and for symmetries $\alpha,\beta$ and some vertex $i$ we have:
$$
\begin{aligned}
\operatorname{id}\cdot i
&=
\operatorname{id}(i)=i\ ,
\\[3mm]
(\alpha\circ\beta)\cdot i
&=(\alpha\circ\beta)(i)\\
&=\alpha(\beta(i))\\
&=\alpha\cdot(\beta(i))\\
&=\alpha\cdot(\beta\cdot i)\ .
\end{aligned}
$$
So we have indeed an action, and the above is all regarding a proof.

The action is faithful for $n>1$. (For $n\ge 3$ this follows from the fact that an affine plane is "determined by a triple of points", i.e. by fixing an origin and two unit vectors from the origin, a first one and a second one. For $n=2$ - just in case we really want to consider a "$2$-gon", the reflection w.r.t. the line through the two points is fixing pointwise the two vertices, so the action is not faithful. The case $n=1$ is also not coming with a faithful action.)

In case we really want to work with permutations, then the group of permutations acting on $[n]$ is the image $\sigma(D_n)$ of $D_n$ inside the permutation group of the set $[n]$ via the map $\alpha\to\sigma_\alpha$, and this action is defined by transport of structure, i.e. $\sigma_\alpha\cdot i :=\alpha\cdot i:=\alpha(i)$. (By transport, it is also an action on the set $[n]$.)


TITLE: Find the limit of the sequence
QUESTION [1 upvotes]: Find the limit of the sequence$$\left(\frac12,\frac12\right),\left(\frac23,\frac23\right),\ldots,\left(\frac n{n+1},\frac n{n+1}\right),\ldots$$

REPLY [4 votes]: $(1,0)$ is the unique limit point of this sequence. The "obvious limit" $(1,1)$ has a small vertical neighborhood that misses the entire sequence. On the other hand, every basis neighborhood of $(1,0)$ has to have some width to it, so will meet the sequence in question.
Edit: Here are some more details. Basis elements in the order topology are open intervals $(x_1\times y_1,x_2\times y_2)$, together with neighborhoods $(x\times y, 1\times 1]$ and $[0\times 0, x\times y)$. Here I am using $a\times b$ to denote an ordered pair so as not to confuse the notation with open intervals! So, for example the open set $(1\times \frac{1}{2}, 1\times 1]$ is a neighborhood of $1\times 1$ which misses the sequence. On the other hand, any neighborhood of $1\times 0$ would have to contain an open interval $(x_1\times y_1,x_2\times y_2)$ where $x_1\times y_1 < 1\times 0 <x_2\times y_2$. This implies $x_1< 1$. So this open interval will contain points $\frac{n}{n+1}\times\frac{n}{n+1}$ for all $n$ where $\frac{n}{n+1}>x_1$.


TITLE: How can I solve the limits: $\mathop {\lim }\limits_{x \to \infty } {{\sin {x^2}} \over {{x^2} + x}}$
QUESTION [1 upvotes]: When I use $\sin x \sim x$ , the answer is $1$ , is the answer correct?

REPLY [1 votes]: in the limit $x \to \infty$, the numerator is bounded between $[-1,1]$ while the denominator approach infinity so the limit must be zero.


TITLE: find the projection of a vector on a line using the projection formula
QUESTION [0 upvotes]: In the following problem:
"Find the projection of vector $\vec{v} = (2, 3)$` onto the line $y = 2x -1$"
What are the beginning steps in solving this problem ? Do I pick a point on the line and then use the projection formula to solve?
Thanks in advance

REPLY [0 votes]: In general, how do we find the projection of a vector $v \in V$ onto a subspace $W$ of $V$? Say $\dim W=m$, $\dim V=m+r$. Let $\{w_1,\ldots,w_m\}$ be a basis for $W$, and let $\{w_1,\ldots,w_m,v_1,\ldots,v_r\}$ be the extension of this basis to a basis for $V$. Then for arbitrary $v \in V$, we can write $v$ as a linear combination of basis vectors: $$v = \sum_{i=1}^{m} a_iw_i+\sum_{j=1}^r b_jv_j,$$
and if $V$ is an inner product space then the coefficients are given by $a_i=(v|w_i)$, $b_i=(v|v_j)$ (where $(\cdot|\cdot)$ denotes the inner product). To obtain the projection of $v$ into $W$ we throw away the part of this sum that has no part in $W$, i.e. we throw away $\sum_{j=1}^r b_jv_j$. So the projection of $v$ onto $W$ is $$\sum_{i=1}^m(v|w_i)w_i.$$
Now, how do we apply this to the problem at hand? Well, it is easy to verify that the line $y=2x-1$ is a subspace (let's call it $W$, like above) of $\mathbb{R}^2$ (which acts like $V$), which is an inner product space. To get the projection of $v$ we need to be able to expand $v$ like we did above, as a linear combination of basis vectors of $W$ and $V$. So first we need a basis of $W$, i.e. the line $y=2x-1$. Then we need a basis of $\mathbb{R}^2$ that contains our basis for $W$. Once we have that, we can expand $v$ as above and then the rest of the problem is easy.
In fact, you'll notice that all we really need is the basis for $W$, since we're looking for the second expansion that I gave above.


TITLE: $S^n \backslash  S^m $ homotopy equivalent to $ S^{n-m-1} $
QUESTION [4 upvotes]: Consider $S^m$ embedded in $S^n$ $ (m < n ) $ as the subspace $ \left\{ (x_1, x_2, ..., x_{m+1}, 0,...0) | \sum x_i^2 = 1 \right\}  $. Show that $ S^n \backslash S^m $ is homotopy equivalent to $ S^{n-m-1} $
My thoughts:
I've first considered the case $ n = 2 $, $m = 1 $ (as this is the only case I can actually draw). So, we have the unit sphere (variables $x, y , z$ ) minus the unit circle (variables $x,y$), and want to show it's homotopy equivalent to the two point space (I think).
So think of $S^0$ as $ \left\{ (0,0,-1) , (0,0,1) \right\} $. Then we have the inclusion $ i : S^0 \to S^2 \backslash S^1$ as a potential homotopy equivalence in one direction.
For the other, let $ f : S^2 \backslash S^1 \to S^0 $ be defined by $ f ( (x,y,z) ) = (0,0, \frac{z}{||z||}) $. This is continuous thanks to the removal of $ S^1 $.
Now, $ fi = {id}_{S^0} $. It remains to be shown that $ if \simeq {id}_{S^2 \backslash S^1} $.
The linear map $ H : S^2 \backslash S^1 \times [0,1] \to S^2 \backslash S^1 $ defined by $ H(x,t) = tx + (1-t)f(x) $ doesn't work, as $ S^2 \backslash S^1 $ isn't convex. I'm not sure what else could work as a homotopy, or if what I've done makes any sense.
Any help would be appreciated. Thanks
EDIT: I now realise I could have started with $ n = 1$ , $ m = 0$ ...

REPLY [2 votes]: I think you should tag your questions with "homework" if they are homework, as Zhen Lin pointed you out.
Anyway, here is my solution. It just needs a little bit of notation. Write the points of $\mathbb{R}^{n+1} = \mathbb{R}^{m+1} \times \mathbb{R}^{n-m}$ as $(x,y)$, with $x\in\mathbb{R}^{m+1}$ and $y \in \mathbb{R}^{n-m}$.
So, $S^m$ is embedded in $S^n \subset \mathbb{R}^{n+1}$ as those $(x,y) \in\mathbb{R}^{n+1} $ such that $y=0$ and $\| x \| = 1$. Likewise, embed $S^{n-m-1}$ in $S^n$ as those $(x,y)$ such that $x=0$ and $\|y\| =1$. So, $S^{n} \backslash S^m$ are those $(x,y)\in S^n$ such that $y\neq 0$. Briefly:

$S^m = \left\{ (x,y) \in \mathbb{R}^{n+1} \ \vert \ y = 0 \ , \ \|x \|=1    \right\}$
$S^{n-m-1} = \left\{ (x,y) \in \mathbb{R}^{n+1} \ \vert \ x = 0 \ ,\ \|y \|=1    \right\}$
$S^n \backslash S^m = \left\{ (x,y) \in  \mathbb{R}^{n+1}\ \vert \ y \neq 0  \ , \  \|x\|^2+\|y\|^2 =1  \right\} $

Then, our inclusion is just
$$
i : S^{n-m-1} \longrightarrow S^n\backslash S^m \ , \qquad i(0,y) = (0,y) \ .
$$
In the other direction, now the map is pretty obvious: since points $(x,y) \in S^{n-m-1}$ are those with $x = 0$, given a point $(x,y) \in S^n\backslash S^m$, the simplest way to obtain a point in $S^{n-m-1}$ seems to put $x=0$ and then normalize the second coordinate:
$$
f: S^n \backslash S^m \longrightarrow S^{n-m-1} \ , \qquad f(x,y) = \left( 0,\frac{y}{\|y\|} \right) \ .
$$
Notice that, since $y\neq 0$, $f$ is continuous and
$$
(f\circ i) (0,y) = f(0,y) = \left( 0,\frac{y}{\|y\|} \right) = (0,y) \ ,
$$
because $\|y\|=1$.
As for the other composition, we have
$$
(i\circ f)(x,y) = \left( 0,\frac{y}{\|y\|} \right) \ ,
$$
and we need now an homotopy between $\mathrm{id_{S^n\backslash S^m}}$ and $i\circ f$. For this, the old trick of the straight line homotopy normalized suffices. Define
$$
H : (S^n\backslash S^m)\times I \longrightarrow S^n\backslash S^m \ , \qquad H(x,y,t) = \frac{\left((1-t)x, (1-t)y + t\frac{y}{\|y\|} \right)}{\left\| \left((1-t)x, (1-t)y + t\frac{y}{\|y\|} \right) \right\|} \ .
$$
And we have
$$
H(x,y,0) = \frac{( x,y )}{\|(x,y)\|}  = (x,y) \ ,
$$
since $(x,y) \in S^n$ and
$$
H(x,y,1) = \frac{\left(0, \frac{y}{\|y\|} \right)}{\left\| \left(0, \frac{y}{\|y\|} \right) \right\|}  = \left( 0, \frac{y}{\|y\|} \right) \ .
$$
We need to prove two more things:

That $H(x,y,t) \in S^n\backslash S^m$ for all $(x,y,t) \in (S^n\backslash S^m)\times I$.
And that $H$ is continuous.

We can do it in one go, just showing that $(1-t)y + t\frac{y}{\|y\|} \neq 0 $ for all $(x,y,t) \in (S^n\backslash S^m)\times I$. But
$$
 (1-t)y + t\frac{y}{\|y\|} = 0 \qquad \Longleftrightarrow \qquad (1-t)y\|y\| +ty = 0
$$
Since $y\neq 0$, this would be the same as
$$
(1-t)\|y\| +t = 0 \ .
$$
But, for $t \in [0,1]$, the left-hand side of the  above equation represents points in the segment between $\|y\|$ and $1$. Since $y \neq 0$, this segment does not include the point $0$. Hence we are done.


TITLE: How can we show  $1+ord_{p}n \le 2^{ord_{p}n} \le p^{(ord_{p}n)/x}$
QUESTION [3 upvotes]: $1+ord_{p}n \le 2^{ord_{p}n} \le p^{(ord_{p}n)/x}$ for $p> 2^{x},n\ge 2 \in \mathbb{N}$; $x>0 \in \mathbb{R}$ It was written here that this is clearly the case; I don't see it. Does somebody see it and cares to explain?

REPLY [2 votes]: The first inequality is a consequence of the fact that if $k$ is a positive integer, then  $1+k \le 2^k$. Just take $k=\text{ord}_p(n)$.
If you wanted to, you could prove that $1+k\le 2^k$ by induction on $k$. But there is no need, the result is clear. At $k=1$ we have equality, but then $2^k$ grows much faster than $1+k$.
For the second inequality, take the logarithm of both sides. It is cleanest to use logarithms to the base $2$.  So we want to prove that
$$\text{ord}_p(n) \le \frac{\text{ord}_p(n)}{x}\log_2(p).$$
Since $p>2^x$, we have $\log_2(p)>x$, and the result follows.


TITLE: Show that $f(0)\leq|z_0|^2$ for a given $z_0$, with $f$ holomorphic and bounded on the unit disk
QUESTION [1 upvotes]: First of all, notice that the complex function $$\phi_a(z)=\frac{z-a}{1-\bar a z}$$ is a bijection from the unit closed disk in $\mathbb C$ in itself.
a) Prove now that, if $f$ is holomorphic in a domain $D$ including the
  closed unit disk, and $|f(z)|<1$ for all $|z|<1$, and exists $z_0$,
  $|z_0|<1$, such that $f(z_0)=f(-z_0)=0$, then $|f(0)|\leq |z_0|^2$.
b) Determine then all functions $f$ as in (a) that satisfy $f(0)=|z_0|^2$.

First of all, $g=f\circ \phi_{z_0}$ and $h=f\circ \phi_{-z_0}$ are in the hypothesis of Schwarz Lemma. We know then that, for example, $|g(z)|\leq |z|$ and then $|f(0)|=|g(z_0)|\leq |z_0|$.
But we want $|z_0|^2$, so, since I do not know anything about the derivatives of $g$ and $h$, i.e. about the order of $0$ as zero, I tried to modify Schwarz Lemma in the following way. If $z_0=0$ we are done. If not, choose a ball around $z_0$ which does not include $z_0^2$ and which remains disjoint from a circle of radius $r$, $|z_0|<r<1$. Let $D_r$ be the domain whose boundary is represented by the $r-circle$ and the circle around $z_0$. The function $g=f\circ \sqrt\phi_{-(z_0)^2}$ (we can choose a branch) is holomorphic in any $D_r$, and satisfies $g<1$. Therefore, since $g(0)=f(\pm z_0)=0$ (in any choice of one of the two branches, which differ only by the sign), we have that $g(z)/z$ is holomorphic on $D_r$. Now I would like to apply the maximum module principle and say that $|g(z)/z|\leq g(y)/y<1/r$, for some $y$ on the $r-circle$, and therefore, taking $r$ to $1$, say that $|g(z)/z|\leq 1$ (this is how one proves Schwarz Lemma). Then choosing $z=z_0^2$ we would end. But the boundary of $D_r$ is not just the $r$-circle.
Is there some way to conclude this proof? Or is there a better and simpler proof?
Thank you in advance.

REPLY [2 votes]: Hint: The function
$$\frac{f}{\phi_{z_0}\cdot \phi_{-z_0}}$$
takes the disc to the disc and is bounded by $1$ in absolute value there.

Added later: There are 2 cases for b): $z_0= 0$ and $z_0\ne 0.$ In the first case, any holomorphic $f:\mathbb D \to \mathbb D$ with $f(0)=0$ will work. In the second case, it is necessary and sufficient that $f(z) = z\cdot \phi_{z_0}(z) \cdot \phi_{-z_0}(z)\cdot g(z),$ for some holomorphic $g:\mathbb D \to \overline {\mathbb D}.$


TITLE: $M$ is open in $Y$ and $M$ is open in $Z$ then $M$ is open in $X$
QUESTION [0 upvotes]: Is it true?
$X= Y \cup Z$, $M$ is a subset of $Y \cap Z$. Suppose that $M$ is open in $Y$ and $M$ is open in $Z$ then  $M$ is open in $X$.

REPLY [2 votes]: Suppose $M\subset Y\cap Z$, and $M$ is an open subset of $Y$ and $Z$ with their subspace topologies inherited from $X$. Then there exists an open $U\subset X$ such that $M= U\cap Y$, and there exists an open $V\subset X$ such that $M= V\cap Z$. But then as $M$ is a subset of both $U$ and $V$, $M$ must be a subset of their intersection $U\cap V$ which is an open subset of $X$.
So, $M = (U\cap V)\cap Y$ and $M = (U\cap V) \cap Z$, But note that $U\cap V$ is a subset of both $Y$ and $Z$, and so $$(U\cap V)\cap Y = (U\cap V)\cap Z = U\cap V.$$ Hence $M=U\cap V$ which is open.


TITLE: Why does a free electron with definite energy not exist in the quantum mechanical sense?
QUESTION [1 upvotes]: I'm reading Introduction to Quantum Mechanics by Griffiths and there is a part that I don't understand. He says that a free electron with a well defined energy cannot exist (bottom of pg. 60, 2nd edition). I don't understand why though.
Let's say that I make an experiment. On the left side I have a device which will eject a single electron towards the right. On the right I have a wall that will notice exactly when a free electron hits it. I run the experiment: the electron is emitted and the wall receives it. The time taken to reach the wall is recorded. I now know the energy of the electron using $E=\frac{mv^2}{2}$, where $v=d/t$, $d=$ the distance that the electron travelled and $t$ is the time it took for the electron to travel the gap. This experiment used a free particle in the sense that it traveled a vacuum gap without interacting with anything before it hit the wall.
Where is my understanding flawed?

REPLY [1 votes]: If a free particle has a definite energy, then it has a definite momentum by $p=\sqrt{2mK}$ ($p$ is the particle's momentum, $m$ is the particle's mass, and $K$ is the particle's kinetic energy). The Heisenberg Uncertainty Principle states that the uncertainty of the position of a particle ($\Delta x$) multiplied by the uncertainty of the momentum of a particle ($\Delta p$) has a minimum of
$$\Delta x \Delta p \geq \frac{h}{2\pi},$$
where $h$ is Planck's constant. If $\Delta p = 0,$ then $\Delta x = (h/2\pi)/0 = \infty,$ meaning the electron could hit the screen after an arbitrary amount of time since it starts from a completely uncertain location. This makes your energy measurement impossible since $d$ is completely unknown. If you prepare your electron so that it starts from a definite position, reducing $\Delta d$, then it's momentum must be more uncertain, meaning your timing measurements will have more uncertainty ($\Delta t$).


text:
\begin{document}

\begin{frontmatter}

\title{Non-Darcian flow or fractional derivative ?}

\author[]{H.W. Zhou\corref{mycorrespondingauthor}}
\cortext[mycorrespondingauthor]{Corresponding author. Tel: +86-10-62331286; Fax: +86-10-62331490.}
\ead{zhw@cumtb.edu.cn}

\author[]{S. Yang}
\author[]{R. Wang}
\author[]{J.C. Zhong}

\address{School of Mechanics and Civil Engineering,\\ China University of Mining and Technology, Beijing, 100083, P.R. China}

\begin{abstract}
Modeling of water and gas flow in low-permeability media is an important topic for a number of engineering such as exploitation of tight gas and disposal of high-level radioactive waste. It has been well documented in the literature that Darcy's linear law of water flux and the hydraulic gradient is not always adequate for low-permeability media. On the basis of the Swartzendruber model, as a non-Darcian model of flow in low-permeability media, a new nonlinear model of water flux and the hydraulic gradient is proposed using fractional derivative. The analytic solution of fractional derivative flow model is presented and all parameters of the fractional derivative flow model are determined by the Levenberg-Marquardt method on the basis of the experimental data of water flow in low-permeability media. It is indicated that the results estimated by the fractional derivative flow model proposed in the paper are in better agreement with the experimental data than the results estimated by the Swartzendruber equation. In addition, a sensitivity study is carried out, showing the effects of fractional derivative order and threshold hydraulic gradient on water flux.
\end{abstract}

\begin{keyword}Non-Darcian flow; Fractional derivative; Low-permeability; Swartzendruber equation

\end{keyword}

\end{frontmatter}

\section{Introduction}
\noindent
Since Henry Darcy's remarkable modeling of linear relation between water flux and the hydraulic gradient, e.g., Darcian flow, in 1856, many researchers found the Darcy's law is not good enough for description of water and gas flow in low-permeability media like clay and shale. Consequently, in past decades, extensive efforts have been devoted to modeling approaches of nonlinear relation between water flux and hydraulic gradient, called non-Darcian flow \citep{Liu2014,Liu2016}. Aiming at low-permeability media like clay, \cite{Miller1963} supposed a threshold gradient for water flow in clays to distinguish linear and nonlinear flow. They found that water flow rate is linearly related to hydraulic gradient at gradients above the threshold gradient, but no flow occurs below threshold gradient. \cite{Deng2007} suggested a new equation of nonlinear flow in saturated clays that can describe characteristics of flow curve of the nonlinear flow from low to high hydraulic gradients.

Generally speaking, non-Darcian flow can be described by nonlinear functions of water flux and hydraulic gradient such as exponential and power functions. \cite{Hansbo1960,Hansbo2001} proposed a power relationship between water flux and hydraulic gradient for non-Darcian flow in clay media. By analyzing data sets for water flow in clay soils, \cite{Swartzendruber1962} proposed an exponential function to validate Darcy's law, resulting in a nonlinear relation of water flux versus gradient. In order to capture the non-Darcian flow behavior, \cite{Liu2012} developed a new relationship between water flux and hydraulic gradient by generalizing the currently existing relationships. The new relationship is shown to be consistent with experimental observations for both saturated and unsaturated conditions.

To validate Darcy's law and develop non-Darcian models seem to be an endless challenge.  It therefore leads to a new channel.  According to Darcy's law, water flux is directly proportional to the hydraulic gradient, i.e., the first order (as an integral number) derivative of water head with respect to the flow distance. In other words, Darcian flow can be described by an integer derivative of water head. Non-Darcian flow in porous media as a nonlinear phenomenon requires a new mathematical approach. In this case, non-Darcian flow could be characterized by a fractional derivative.

The fractional calculus, referred to as calculus of integrals and derivatives of any arbitrary real or complex order, is a 300 years old mathematical discipline. Its original conception is believed to have stemmed from a question raised in the year of 1695 by Marquis de L'H\^{o}pital (1661-1704) to Gottfried Wilhelm Leibnitz (1646-1716), the founder of Calculus. In the past few decades, the fractional calculus has gained remarkable popularity and importance because of its demonstration applications in numerous seemingly diverse and widespread fields of science and engineering \citep{Herrmann2011, Ortigueira2011} such as applications of fractional calculus to time-dependent behavior of rocks \citep{Zhou2011,Zhou2013} and composites \citep{Zhou2017}, fluid mechanics \citep{Kulish2002}, and solid mechanics \citep{Carpinteri2002,Carpinteri2004,Rossikhin2010}.

Moreover, some researchers devoted themselves to a nonlinear modeling approach of fractional derivative to non-Darcian flow. \cite{He1998} proposed a new model for seepage flow in porous media to modify the Darcy's law with fractional derivatives. \cite{Tian2006} researched the flow characteristics of fluids through a fractal reservoir with the fractional order derivative. By regarding the water flow as a function of a fractional derivative of the piezometric head, \cite{Cloot2006} generalized the classical Darcy's law to derive a new equation of groundwater flow. \cite{Chen2013} developed a new variable-order fractional diffusion equation to describe the diffusion process of chloride ions in the reinforced concrete structure.  \cite{Babak2014} presented a unified fractional differential approach to modeling flows of slightly compressible fluids through naturally fractured media. Recently, \cite{Wang2015} applied Caputo fractional constitutive equation to describe the transient electro-osmotic flow of a generalized Maxwell fluid in a cylindrical capillary.

As described by \cite{Cloot2006}, the underlying basic assumption of the fractional derivative modeling approach to transport in porous media is that the fluid flow at a given point of the porous media is governed not only by the properties of the piezometric field at the specific position but also depends on the global spatial distribution of that field in soil matrix. As a consequence, time or space fractional derivatives are extensively used in models of solute transport in porous media in order to take into account the memory effect or nonlocal properties induced by the interactions of fluid particles with pores of the porous media. Nevertheless, time or space fractional derivative models usually need to make dimensionless for convenience. Therefore, a different perspective to address this problem will be shown herein to interleave with the fractional calculus.

This paper represents an attempt to describe non-Darcian flow mathematically. The Swartzendruber equation as a non-Darcian flow model is generalized to describe the relation between water flux and hydraulic gradient using fractional derivative, resulting in a new model called the fractional derivative flow model. The analytic solution of fractional derivative flow model is presented and all parameters of the fractional derivative flow model are determined on the basis of the experimental data of water flow in low-permeability media. The results estimated by the fractional derivative flow model proposed in the paper are in better agreement with the experimental data than the results estimated by the Swartzendruber model. It indicates that our perspective of fractional derivative modeling approach is acceptable for non-Darcian flow in porous media.
\section{Fractional derivative approach to non-Darcian flow}
\subsection{Definition of the Caputo derivative}
The some definitions of fractional derivatives are popular in mathematics like Grunwald-Letnikov, Riemann-Liouville, and Caputo derivative \citep{Podlubny1999}. Among them, Caputo derivative is widely used in physics and mechanics because of its advantages in solving fractional differential equations with initial conditions. For a given function $f(x)$ Caputo derivative is defined by
\begin{equation}\label{Eq.(1)}
  \frac{{{d^\gamma }f(x)}}{{d{x^\gamma }}} = \frac{1}{{\Gamma (n - \gamma )}}\int_0^x {\frac{{{f^{(n)}}(t)}}{{{{(x - t)}^{\gamma  - n + 1}}}}dt},
\end{equation}
where $\gamma>0$, $n$ is the least integer greater than $\gamma$, and $\Gamma(\cdot)$ is the Gamma function, i.e., $\Gamma (\gamma ) = \int_0^\infty  {{t^{\gamma  - 1}}{e^{ - t}}dt}$. In particular, for $\gamma=0$, $\frac{{{d^\gamma }}}{{d{x^\gamma }}}$ denotes the identity operator.
\subsection{ Darcian flow}
Considering one-dimensional steady-state flow, suppose a fluid flows along a straight line, say, $x$-direction, the flux is related to hydraulic gradient by a well-known Darcy's law given by
\begin{equation}\label{Eq.(2)}
q = \frac{K}{\mu }\frac{{dp}}{{dx}},
\end{equation}
where $q$ is the bulk velocity of fluid $(\mathrm{m}/\mathrm{s})$, or fluid flux, $K$ is permeability $(\mathrm{m}^2)$, $\mu$ is dynamic viscosity $(\mathrm{N}\cdot\mathrm{s}/\mathrm{m}^2)$, $p$ is fluid pressure $(\mathrm{N}/\mathrm{m}^2)$.

Eq.(\ref{Eq.(2)}) can be usually described by
\begin{equation}\label{Eq.(3)}
 q = ki,
\end{equation}
where $k = \frac{K}{\mu }\rho$ is hydraulic conductivity $(\mathrm{m}/\mathrm{s})$, $\rho$ is density of fluid $(\mathrm{N}/\mathrm{m}^3)$, and $i$ is hydraulic gradient.
\subsection{Non-Darcian flow}
Non-Darcian flow, generally, can be described by nonlinear functions of water flux and hydraulic gradient such as power and exponential functions.

(1) Power function

Darcy's law leads to a linear relation between $q $ and $i$ as shown in Eq.(\ref{Eq.(3)}). By producing first order derivative to both sides of Eq.(\ref{Eq.(3)}), we then have a differential equation like
\begin{equation}\label{Eq.(4)}
dq = kdi  \quad or \quad \frac{{dq}}{{di}} = k.
\end{equation}

It is shown that Darcian flow can be described by an integer derivative of flux with respect to the hydraulic gradient $i$, as a dimensionless variable.  A similar model is the Newtonian dashpot for description of a linear relationship between the viscous stress and the rate of strain. The Newtonian dashpot was developed to the Abel dashpot by invoking the fractional derivative \citep{Scott-Blair1944,Kiryakova1999,Zhou2011,Zhou2013}. In an analogous way, we suppose non-Darcian flow can be described by a fractional derivative of flux, which leads to a dimensionless form, i.e.,
\begin{equation}\label{Eq.(5)}
 \frac{{{d^\gamma }q}}{{d{i^\gamma }}} = k,\quad \gamma>0,
\end{equation}
where $\frac{{{d^\gamma }}}{{d{i^\gamma }}}$ is the Caputo fractional derivative operator.

Applying the bilateral Laplace transform $(LT)$ to Eq.(\ref{Eq.(5)}) gives
\begin{equation}\label{Eq.(6)}
LT\left[\frac{{d^\gamma }q}{d{i^\gamma }}\right] = {s^\gamma } \tilde q(s) - \sum\limits_{j = 0}^{n - 1} {{s^{\gamma - j - 1}}\frac{{{d^j}q(0)}}{{d{i^j}}}}  = \frac{k}{s}.
\end{equation}
Let $q(0) = 0$, we have:
\begin{equation}\label{Eq.(7)}
\tilde q(s){\rm{ = }}\frac{k}{{{s^{\gamma  + 1}}}}.
\end{equation}

Applying the inverse Laplace transform to Eq.(\ref{Eq.(7)}), i.e.,
\begin{equation}\label{Eq.(8)}
  L{T^{ - 1}}[\tilde q(s)]{\rm{ = }}L{T^{ - 1}}\left[\frac{k}{{{s^{\gamma  + 1}}}}\right] = \frac{k}{{\Gamma (1 + \gamma )}}{i^\gamma },
\end{equation}we have:
\begin{equation}\label{Eq.(9)}
q = k\frac{{{i^\gamma }}}{{\Gamma (1 + \gamma )}}.
\end{equation}
In this case, we get a power function of water flux $q$ and hydraulic gradient $i$ in Eq.(\ref{Eq.(9)}), showing a similar form of nonlinear equation supposed by \cite{Hansbo1960, Hansbo2001}.

(2) Exponential function: Fractional Swartzendruber equation

\cite{Swartzendruber1962} proposed an exponential relation between water flux and hydraulic gradient to modify Darcy's law, i.e.,
\begin{equation}\label{Eq.(10)}
\frac{{dq}}{{di}} = k(1 - {e^{ - \frac{i}{I}}}).
\end{equation}

Integrating both sides of Eq.(\ref{Eq.(10)}) and considering $q(0)=0$, we have:
\begin{equation}\label{Eq.(11)}
 q = k[i - I(1 - {e^{ - \frac{i}{I}}})]
\end{equation}
where $I$ is the threshold gradient and actually refers to the intersection of the linear part in plot of the hydraulic gradient and the water flux.

Replacing integer derivative with fractional derivative in Eq.(\ref{Eq.(10)}), we have the fractional derivative Swartzendruber equation, i.e.,
\begin{equation}\label{Eq.(12)}
 \frac{{{d^\gamma }q}}{{d{i^\gamma }}} = k(1 - {e^{ - \frac{i}{I}}}),\quad 0\leq\gamma\leq1.
\end{equation}

Application of the Laplace transform $(LT)$ to Eq.(\ref{Eq.(12)}) leads to:
\begin{equation}\label{Eq.(13)}
{s^\gamma } \tilde q(s) = k\left( {\frac{1}{s} - \frac{1}{{s + {1 \mathord{\left/
 {\vphantom {1 I}} \right.
 \kern-\nulldelimiterspace} I}}}} \right),
\end{equation}then we have:
\begin{equation}\label{Eq.(14)}
  \tilde q(s) = \frac{{k{s^{ - \gamma  - 1}}}}{{1 + Is}}.
\end{equation}

Applying the inverse Laplace transform to Eq.(\ref{Eq.(14)}) like
\begin{equation}\label{Eq.(15)}
 L{T^{ - 1}}[\tilde q(s)]{\rm{ = }}L{T^{ - 1}}\left[\frac{{k{s^{ - \gamma  - 1}}}}{{1 + Is}}\right] = \frac{k}{I}{i^{\gamma  + 1}}{E_{1,\gamma  + 2}}( - \frac{i}{I}),
\end{equation}we have:
\begin{equation}\label{Eq.(16)}
q = \frac{k}{I}{i^{\gamma  + 1}}{E_{1,\gamma  + 2}}( - \frac{i}{I}),
\end{equation}
where ${E_{1,\gamma  + 2}}(\cdot)$ refers to Mittag-Leffler function, i.e., ${E_{1,\gamma  + 2}}( - \frac{i}{I}) = \sum\limits_{k = 0}^\infty  {\frac{{{{( - \frac{i}{I})}^k}}}{{\Gamma (k + \gamma  + 2)}}}$  \citep{Mainardi2010}.

In the case of $\gamma=0$, using ${E_{1,2}}( - \frac{i}{I}) = \frac{{{e^{ - i/I}} - 1}}{{ - i/I}}$, Eq.(\ref{Eq.(16)}) can be rewritten as
\begin{equation}\label{Eq.(17)}
 q = \frac{k}{I}i{E_{1,2}}( - \frac{i}{I}) = k(1 - {e^{ - \frac{i}{I}}}),
\end{equation}
which appearance is the same as Eq.(\ref{Eq.(10)}) if differential order $\gamma=0$.

In addition, the case of $\gamma=1$ gives
\begin{equation}\label{Eq.(18)}
 {E_{1,3}}( - \frac{i}{I}) = \frac{{{e^{ - i/I}} - 1 + i/I}}{{{{(i/I)}^2}}}.
\end{equation}

Substituting Eq.(\ref{Eq.(18)}) into Eq.(\ref{Eq.(16)}), we have $q = k(i - I(1 - {e^{ - \frac{i}{I}}}))$ indicating that the Swartzendruber equation in Eq.(\ref{Eq.(11)}) is a special case of the fractional derivative flow model when the fractional derivative order $\gamma=1$.

\section{Parameter determination for fractional derivative non-Darcian model}
\subsection{Parameter determination}
The efficacy of the fractional derivative model is dependent on its ability to adequately fit experimental data. Using the experimental data of water flux with hydraulic gradient, the parameters $k, I, \gamma$ in Eq.(\ref{Eq.(16)}) can be determined by the Levenberg-Marquardt method, a nonlinear least-squares fitting (LSF) method (see \citealt{Zhou2011} for details).

In what follows, we now use the fractional derivative flow model to fit the experimental data \citep{Wang2016} by LSF analysis. \cite{Wang2016} developed an experimental study to investigate the non-Darcian behavior of water flow in soil-rock mixtures (SRM) with various rock block percentages. Their work presented the data set of water flux as a power function of hydraulic gradient. In addition, the relationship between threshold hydraulic gradient and rock block percentage was also considered. The exact value of threshold hydraulic gradient $I$ for SRM specimens are listed in \autoref{Table 1}. Consequently, only the two parameters $k$ and $\gamma$ are remain to be determined. The results of least-squares fit of the parameters in Eq.(\ref{Eq.(16)}) to the experimental data \citep{Wang2016} are listed in \autoref{Table 1}.
\begin{table}[!tb]
  \centering
  \caption{\small Determination of parameters for fractional derivative flow model based on SRM specimens}\label{Table 1}
  \scalebox{0.85}{
  \begin{tabular}[]{cccccccccc}
  \hline

  SRM 	&\multicolumn{4}{c}{Swartzendruber equation}	&	\multicolumn{5}{c}{Fractional derivative flow model}\\
  \cline{2-10}
specimens &  $k\times10^{-5}(\mathrm{m}/\mathrm{s})$ & $I$ &$\mathrm{R}^2$ &MSE	& $ k\times10^{-5}(\mathrm{m}/\mathrm{s})$	&  $I $	     &$\gamma$&	 $\mathrm{R}^2$ & MSE\\
\hline
SRM20-1	& 0.2039 &	141.00 &  0.9817	&0.2242	& 0.2039	& 141.00	 & 1	    &0.9817	 &0.2242\\
SRM30-1	& 0.1378 &  130.20 &  0.9898	&0.0531	& 0.1378	& 130.20	 & 1	    &0.9898	 &0.0531\\
SRM40-1	& 0.07383&	123.60 &  0.9869	&0.1358	& 0.1311	& 123.60	 &0.8501	&0.9908	 &0.0095\\
SRM50-1	& 0.1297 &  102.50 &  0.9326	&0.2322	& 0.1297	& 102.50	 & 1	    &0.9326	 &0.2322\\
SRM60-1	& 0.1175 &  85.33  &  0.9786	&0.0363	& 0.1954	& 85.33	     &0.8567	&0.9819	 &0.0307\\
SRM70-1	& 0.1402 &  73.59  &  0.9710	&0.0291	& 0.3223	& 73.59	     &0.7454	&0.9807	 &0.0193\\
  \hline
\end{tabular}}
\end{table}

The data as well as the fitting curves given by the fractional derivative flow model in Eq.(\ref{Eq.(16)}) is shown in \autoref{Fig1}. Making the fitting analysis to the same experimental data using the Swartzendruber equation in Eq.(\ref{Eq.(11)}), another set of parameters are given in \autoref{Table 1} as well. The least-squares analysis results in \autoref{Table 1} indicate that the fractional derivative flow model in Eq.(\ref{Eq.(16)}) is in better agreement with the experimental data than the Swartzendruber equation in Eq.(\ref{Eq.(11)}) with higher correlation coefficients ($\mathrm{R}^2$) and lower mean squared errors (MSE).

In addition, \autoref{Table 1} shows that an increase of rock block percentage in SRM specimens leads a decrease of the hydraulic conductivity $k$ to a minimum value at a rock block percentage of 40\%, and an increase if rock block percentage exceed 40\%. The similar behavior is also given in \cite{Wang2016}.

\begin{figure}[!htbp]
  \centering

  \begin{subfigure}[t]{0.45\textwidth}
  \centering
  \includegraphics[width=1\textwidth]{SRM20-1}
  \subcaption*{\small(a)}
  \end{subfigure}
  \quad
  \begin{subfigure}[t]{0.45\textwidth}
  \centering
  \includegraphics[width=1\textwidth]{SRM30-1}
  \subcaption*{\small(b)}
  \end{subfigure}
  \\
  \begin{subfigure}[t]{0.45\textwidth}
  \centering
  \includegraphics[width=1\textwidth]{SRM40-1}
  \subcaption*{\small(c)}
  \end{subfigure}
  \quad
  \begin{subfigure}[t]{0.45\textwidth}
  \centering
  \includegraphics[width=1\textwidth]{SRM50-1}
  \subcaption*{\small(d)}
  \end{subfigure}
  \\
  \begin{subfigure}[t]{0.45\textwidth}
  \centering
  \includegraphics[width=1\textwidth]{SRM60-1}
  \subcaption*{\small(e)}
  \end{subfigure}
  \quad
  \begin{subfigure}[t]{0.45\textwidth}
  \centering
  \includegraphics[width=1\textwidth]{SRM70-1}
  \subcaption*{\small(f)}
  \end{subfigure}
  \caption{\small Fitting curves theoretically given by fractional derivative flow model on the basis of experimental data for SRM specimens with different rock block percentages: 20\% (a), 30\% (b), 40\% (c), 50\% (d), 60\% (e) and 70\% (f) (see \citealt{Wang2016}).}
  \label{Fig1}
\end{figure}

Furthermore, using a more data set \citep{Deng2007}, the validity of our fractional derivative flow model was evaluated by LSF analysis. \cite{Deng2007} presented a nonlinear model of flow in saturated clays.
\begin{table}[!bht]
  \centering
  \caption{\small Determination of parameters for fractional derivative flow model based on saturated clays}\label{Table 2}
  \scalebox{0.85}{
  \begin{tabular}[]{cccccccccc}
  \hline

  Saturated 	&\multicolumn{4}{c}{Swartzendruber equation}	&	\multicolumn{5}{c}{Fractional derivative flow model}\\
  \cline{2-10}
 clays&  $k\times10^{-9}(\mathrm{m}/\mathrm{s})$ & $I$ &	$\mathrm{R}^2$ &MSE	& $ k\times10^{-9}(\mathrm{m}/\mathrm{s})$	&  $I $	     &$\gamma$&	 $\mathrm{R}^2$ & MSE\\
\hline
NO.64-3	& 7.973 &  0.7901 &  0.9973	&0.2050	& 11.52	& 1.565	 & 0.8094	  &0.9975	 &0.1501\\
NO.64-4	& 4.856 &  2.754  &  0.9998	&0.0116	& 7.153	& 4.408	 & 0.8814	  &0.9998	 &0.0100\\
\hline
\end{tabular}}
\end{table}
Comparisons of experimental data and fitting curves given by both Swartzendruber equation in Eq.(\ref{Eq.(11)}) and the fractional derivative flow model in Eq.(\ref{Eq.(16)}) are illustrated in \autoref{Fig2}. For better analysis, the parameters in Eq.(\ref{Eq.(11)}) and Eq.(\ref{Eq.(16)}) are determined and listed in \autoref{Table 2}. The results demonstrate that the proposed fractional derivative flow model is in better agreement with the experimental data than the Swartzendruber equation in Eq.(\ref{Eq.(11)}).

\begin{figure}[!ht]
  \centering

  \begin{subfigure}[t]{0.45\textwidth}
  \centering
  \includegraphics[width=1\textwidth]{NO64-3}
  \subcaption*{\small(NO.64-3)}
  \end{subfigure}
  \quad
  \begin{subfigure}[t]{0.45\textwidth}
  \centering
  \includegraphics[width=1\textwidth]{NO64-4}
  \subcaption*{\small(NO.64-4)}
  \end{subfigure}
  \caption{\small Fitting curves theoretically given by fractional derivative flow model on the basis of experimental data for saturated clays NO.64-3, NO.64-4 (see \citealt{Deng2007}).}\label{Fig2}
\end{figure}
Moreover, since Swartzendruber equation in Eq.(\ref{Eq.(11)}) is a special case of the fractional derivative flow model when the fractional derivative order $\gamma=1$, our fitting results verified that the presented fractional derivative flow model is more flexible and accurate.

\subsection{Sensitivity analysis}
\noindent(1) Fractional derivative order

Eq.(\ref{Eq.(16)}) shows that the relationship between water flux $q$ and hydraulic gradient $i$ depends on parameters $k, I, \gamma$. In order to get a better understanding of the effects of these parameters, sensitivity analyses have been carried out. The effect of fractional order $\gamma$ on the variation of water flux with hydraulic gradient is shown in \autoref{Fig3}. In which one parameter $\gamma$ takes three different values to show its effect on the $q-i$ curve under the condition of $k=2\times10^{-6}\mathrm{m}/\mathrm{s}, I=80$. It is shown that the higher the fractional derivative order, in general, the larger the fluid flux.
\begin{figure}[H]
  \centering

  \includegraphics[width=0.5\textwidth]{Fig3}
  \caption{\small Sensitivity of the water flux to the fractional derivative order}\label{Fig3}
\end{figure}

\noindent(2) Threshold hydraulic gradient

In Eq.(\ref{Eq.(16)}), let the threshold gradient $I$ changes and other parameters be constant, where $k=1.5\times10^{-6}\mathrm{m}/\mathrm{s}, \gamma=0.85$. A series of curves can be obtained as shown in \autoref{Fig4}, indicating that the higher threshold gradient, the smaller water flux.

\begin{figure}[H]
  \centering

  \includegraphics[width=0.5\textwidth]{Fig4}
  \caption{\small Sensitivity of the water flux to threshold hydraulic gradient}\label{Fig4}
\end{figure}

\section{Conclusions}
The object of the present work is to develop fractional order equations for describing non-Darcian behavior between water flux and hydraulic gradient. Based upon the extensively adopted fractional calculus theory, we generalized the currently existing relationships such as Hansbo equation and Swartzendruber equation. The analytic solution for the fractional derivative flow model is acquired and the relative parameters are determined. Sets of experimental data are utilized to verify the validity of the proposed fractional derivative flow model. The comparative analysis demonstrates that including the Swartzendruber equation as a special case when the fractional derivative order $\gamma=1$, the fractional derivative flow model turns out to be a more flexible and accurate one to characterize the behavior of non-Darcian flow. Furthermore, a sensitivity study shows that the fractional derivative order is an essential parameter impacting the shape of $q-i$ curve. However, the corresponding physical interpretation of fractional derivative is not clear and further research is required to determine the relationship between the fractional derivative order and other mechanical parameters.
\section*{Acknowledgement}
The present work is supported by the National Natural Science Foundation of China (51674266), and the State Key Research Development Program of China (2016YFC0600704) and the Specialized Research Fund for the Doctoral Program of Higher Education (20130023110017). The financial supports are gratefully acknowledged. Special thanks are due to H.H. Liu, Aramco Services Company, for his valuable suggestions and help in improving the article.
\section*{References}
\bibliography{Bibfile}
\end{document}

file: 1806.00977/EUR J MECH B-FLUID/Non-Darcian_Flow_or_Fractional_Derivative.tex


text:
\section{Computational Results} \label{sec:results}
In this section, we compare the runtime performance of the proposed method with existing methods. To the best knowledge of the authors, the IP proposed by Brimkov et al.~\cite{conn-pd} is the current state of the art computational methods for power domination, demonstrating superior runtime performance in comparison to the other so called infection model seen in power domination literature~\cite{aazami}. In the first part of this section, we reproduce the infection model proposed by Brimkov et al. \cite{conn-pd} and discuss a modification, motivated by Proposition \ref{prop:J-set}, which reduces the set feasible solutions permitted by the model. In the latter portion of this section, we compare the performance of both of these methods to the separation algorithm proposed in this paper.

\subsection{Power Domination Infection Model Modification} \label{sec:inf-model}
The IP introduced by Brimkov et al. \cite{conn-pd} identifies the power domination numbers of general graphs. To do so, at an optimal solution of the IP, the variables of the model correspond to both a minimum power dominating set as well as a sequence in which the initially uncolored vertices can be colored. For an arbitrary graph $G' = (V, E')$, the digraph $G = (V, E = \{uv : uv \in E(G')$ or $vu \in E(G')\})$ is constructed. The following IP is then defined using the auxiliary graph $G$. \newline

\noindent \text{\textbf{Model 4: Power Domination Infection Model} \cite{conn-pd}}
{\small
  \begin{align}
    \min \:         &\sum_{v \in V} s_v \nonumber \\
    \text{s.t. } \: & s_v + \sum_{uv \in E} y_{uv}  =  1   & \forall v \in V \nonumber \\
                    & x_u - x_v + (n+1) y_{uv}  \leq n   & \forall uv \in E \nonumber \\
                    & x_w - x_v + (n+1) y_{uv}  \leq n + (n+1)s_u & \forall uv \in E, \forall w \in N(u) \backslash \{v\} \nonumber \\
                    & x_v \in \{0, 1, \dots, n\}, s_v \in \{0,1\} & \forall v \in V, \nonumber \\
                    & y_{uv} \in \{0,1\}             & \forall uv \in E \nonumber
\end{align}}

In their paper, Brimkov et al. \cite{conn-pd} show that each power dominating set $S$ in $G'$ has a corresponding Model 4 feasible solution in which $S$ is indicated by the $s$ variables. In the case that $G'$ is connected and $\Delta(G') \geq 3$, this model can be modified by including an additional constraint,
{\small \begin{align}
    & s_v = 0 & \forall v \in V : \deg(v) \leq 2 \label{cons:8}.
  \end{align} }
Since Proposition \ref{prop:J-set} ensures that $G'$ has a minimum power dominating set contained in $J(G)$, Model 4 must have an optimal solution in which $s_v = 0$ for all $v$ with $\deg(v) \leq 2$. Thus when Constraint (\ref{cons:8}) is added to Model 4, the modified infection model remains feasible. Additionally, any feasible solution of the modified infection model is feasible for Model 4, so the optimum of the modified model is bounded below by the optimum of Model 4. By incorporating this modification however, the feasible region of both Model 4 and its linear relaxation are often significantly restricted. In the latter half of this section, we provide computational experiments which show that this modification consistently reduces the time needed to solve the model to optimality by 30-50\% in larger problem instances.

Additionally, this constraint can also be added to Model 1. Aside from greatly reducing the number of variables used, the original constraints of Model 1 can be changed from fort neighborhood cover constraints to cover constraints over merely the junctions contained by a fort neighborhood. In large problem instances or graphs with relatively long junction paths (often seen in electrical networks), this can reduce the amount of memory needed to hold the constraint set, further improving runtime performance.

\subsection{Computational Experiments}
The graphs used in these experiments consist of networks from standard IEEE test cases\footnote{Publicly available at the Power Systems Test Case Achieve: \url{https://labs.ece.uw.edu/pstca/}.}, several large networks created by the Pan European Grid Advanced Simulation and State Estimation (PEGASE) project \cite{pegase} meant to simulate the scale and complexity of the European high voltage transmission network, and several publically available power grid data sets including the US Western Interconnection network compiled by Watts and Strogatz~\cite{watts-strogatz}. All of the graphs considered are available for download at the LIINES Smart Grid Test Case Repository~\footnote{The LIINES Smart Grid Test Case Repository and all graphs used can be found at \url{http://amfarid.scripts.mit.edu/Datasets/SPG-Data/index.php.}}.

In order to ensure fair comparisons of the methods, we have implemented or reimplemented all algorithms on the same software platforms and have run all computational experiments using the same hardware. The experiments presented here were implemented using Python 3.5 and Gurobi 7.5.2~\cite{gurobi}, and were run on a laptop with 16GB of RAM and an i7 2.80GHz processor. The constraints of Model 1 were generated using the Gurobi callbacks.

In Figure \ref{tab:sep-results}, we compare the performance of the power domination set cover model (Model 1) when constraints are generated with Model 2 and Model 3. In all test cases, Model 1 reached optimality more quickly and with fewer added constraints when Model 3 was used. In all experiments the constraint set of Model 1 was initialized via the process described in Section \ref{sec:M-init} and Constraint~(\ref{cons:8}) was included in the model.

In Figure \ref{tab:full-results}, we compare the runtime performance of the infection IP model presented by Brimkov et al. \cite{conn-pd}, the modified infection model discussed in Section \ref{sec:inf-model} (denoted in the table as Inf.*), and Model 1. As before, the constraint set of Model 1 was initialized via the process described in Section \ref{sec:M-init} and Constraint~(\ref{cons:8}) was included. Violated constraints of Model 1 were separated using Model 3.

\renewcommand{\arraystretch}{1.2}
\begin{figure}[H]
  \centering
  \footnotesize
  \[ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
		\hline
 		\multicolumn{10}{|c|}{Comparison of Set Cover Constraint Generation Methods} \\
    \hline
    \multicolumn{6}{|c|}{\multirow{2}{*}{Graph Test Instances}} & \multicolumn{2}{c|}{Set Cover Model} & \multicolumn{2}{c|}{Set Cover Model} \\
    \multicolumn{6}{|c|}{}& \multicolumn{2}{c|}{(using Model 2)} & \multicolumn{2}{c|}{(using Model 3)} \\

 		\hline
 	  Name         & $n$  & $m$   & $J$ & $\mathscr{M}$ &$\gamma_P$& Time & Sep. & Time   & Sep. \\
	  \hline
	  IEEE Bus 14	 & 14   & 20    & 7   & 0 & 2        & 0.042    & 3       &\textbf{0.007} & 2   \\
	  IEEE Bus 30  & 30   & 41    & 12  & 1 & 3        & 0.019		 & 4       &\textbf{0.012} & 2   \\
	  IEEE Bus 57	 & 57   & 78    & 24  & 1 & 3        & 0.034		 & 6       &\textbf{0.029} & 5   \\
	  IEEE Bus 118 & 118  & 179   & 55  & 2 & 8        & 0.19	   & 26      &\textbf{0.12}  & 10  \\
	  IEEE Bus 300 & 300  & 409   & 155 & 14 & 30       & 1.09     & 63      &\textbf{0.65}  & 30  \\
    PEGASE 1354  & 1354 & 1710  & 496 & 140 & 176      & --       & 4785    &\textbf{1.55}  & 49  \\
    Polish 2383  & 2383 & 2886  & 776 & 95 & 203      & --       & 14083   &\textbf{52.73} & 14  \\

    \hline
  \end{tabular} \]
  \caption{\footnotesize Runtime performance comparison of the set cover model (Model 1) with constraints generated by Model 2 and Model 3. In both cases, Model 1 is initialized with the process discussed in Section \ref{sec:M-init}. A timeout period of 1 hour was used.}
  \label{tab:sep-results}
\end{figure}
\vspace{-.5cm}
\begin{figure}[H]
  \centering
  \footnotesize
  \[ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
		\hline
 		\multicolumn{10}{|c|}{Comparison with Previous Power Domination Computational Methods} \\
    \hline
 		\multicolumn{6}{|c|}{Graph Test Instances} & \multicolumn{1}{c|}{Inf. \cite{conn-pd}} & \multicolumn{1}{c|}{Inf.*} & \multicolumn{2}{c|}{Set Cover Model} \\
 		\hline
 	  Name         & $n$  & $m$   & $J$  &$\mathscr{M}$&$\gamma_P$& Time & Time & Time & Sep. \\
	  \hline
	  IEEE Bus 14	 & 14   & 20    & 7    & 0   &  2    & 0.10     & 0.06     & \textbf{0.007}   & 2   \\
	  IEEE Bus 30  & 30   & 41    & 12   & 1   &  3    & 0.08		 & 0.08     & \textbf{0.012}	  & 2   \\
	  IEEE Bus 57	 & 57   & 78    & 24   & 1   &  3    & 0.34		 & 0.34     & \textbf{0.029}	  & 5   \\

	  IEEE Bus 118 & 118  & 179   & 55   & 2   &  8    & 3.97	   & 2.71     & \textbf{0.12}	  & 10  \\
	  IEEE Bus 300 & 300  & 409   & 155  & 14  &  30   & 64.64    & 20.74    & \textbf{0.65}	  & 30  \\
    PEGASE 1354  & 1354 & 1710  & 496  & 140 &  176  & 10.87    & 4.82     & \textbf{1.55}    & 49  \\
    Polish 2383  & 2383 & 2886  & 776  & 95  &  203  & 804.31   & 482.52   & \textbf{52.73}   & 14  \\
    \multirow{2}{*}{\shortstack{US Western\\Interconnection}}
                 &\multirow{2}{*}{4941}& \multirow{2}{*}{6594}& \multirow{2}{*}{2059}& \multirow{2}{*}{321} & \multirow{2}{*}{494} & \multirow{2}{*}{--}& \multirow{2}{*}{--} &\multirow{2}{*}{\textbf{185.00}} &\multirow{2}{*}{31} \\
    & & & & & & & & &     \\
    PEGASE 9241  & 9241 & 14207 & 3800 & 527 & 811 & -- & -- & \textbf{995.54}   & 299 \\ \hline

  \end{tabular} \]
  \caption{\footnotesize Runtime performance comparison of the infection model proposed by Brimkov et al. \cite{conn-pd} (Model 4), the modified infection model, and the set cover model with constraints generated using Model 3. A timeout period of 6 hours was used.}
  \label{tab:full-results}
\end{figure}

{\let\thefootnote\relax\footnote{{$J$ is used to denote $|J(G)|$, $\mathscr{M}$ is used to denote the number of constraints initially included in Model 1, and the Sep. column provides the number of constraints which were separated in each test case. All tests are reported in seconds, the best time for each test case is in bold.}}}

\renewcommand{\arraystretch}{1}

In the larger test cases, the majority of Model 1's constraints are added during the initialization process. When Constraint (\ref{cons:8}) is added to Model 1, the initial constraints introduce either required vertices (fort neighborhoods of type I and II) or cover constraints over two vertices (fort neighborhoods of type III). As a result, the initially added constraints can significantly reduce the dimension of the Model 1 feasible region.