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Augustine, Leibowitz, and evolution
Posted on 7 March 2011 by John
The following paragraph is from the science fiction novel A Canticle for Leibowitz:
A fourth century bishop and philosopher. He [Saint Augustine] suggested that in the beginning God created all things in their germinal causes, including the physiology of man, and that the germinal causes inseminate, as it were, the the formless matter—which then gradually evolved into the more complex shapes, and eventually Man. Has this hypothesis been considered?
A Canticle for Leibowitz is set centuries after a nuclear holocaust. The war was immediately followed by the "Simplification." Survivors rejected all advanced technology and hunted down everyone who was even literate. At this point in the book, a sort of Renaissance is taking place. The question above is addressed to a scientist who is explaining some of the (re)discoveries taking place. The scientist's response was
"I'm afraid it has not, but I shall look it up," he said, in a tone that indicated he would not.
Was the reference to Augustine simply made up for the novel, or is there something in Augustine's writings that the author is alluding to? If so, does anyone know what in particular he may be referring to? Is such a proto-Darwinian reading of Augustine fair?
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15 thoughts on "Augustine, Leibowitz, and evolution"
I'm afraid I don't know, but I shall look it up.
It may be a reference to the concept of "free will", which St Augustin introduced in order to make God irresponsible of the existence of Evil, it would be Mankind's fault.See "De libero arbitrio".
Matthew J. Sullivan
For what it's worth, I do know that he rejected a literal 6 day creation story. Instead he favors an instantaneous creation theory.
http://books.google.com/books?id=_s0kIgD0nCcC&lpg=PP1&dq=Augustine%20literal%20meaning%20of%20genesis&pg=PA3#v=onepage&q&f=false
Clintp
I don't have your citation, but I have a reference to someplace that you might have a citation. :)
But to quote Darwin, a Life in Science on evolution,
Saint Augustine (353-430) painted an even clearer picture. He taught that the original germs of living things came in two forms, one placed by the Creator in animals and plants, and a second variety scattered throughout the environment, destined to become active only under the right conditions. He said that the Biblical account of the Creation should not be read as literally occupying six days, but six units of time, while the passage `In the beginning God created the heaven and the earth' should be interpreted:
As if this were the seed of the heaven and the earth, although as yet all the matter of heaven and of earth was in confusion; but because it was certain that from this the heaven and the earth would be, therefore the material itself is called by that name.
Augustine likens the Creation to the growth of a tree from its seed, which has the potential to become a tree, but does so only through a long, slow process, in accordance with the environment in which it finds itself. God created the potential for the heavens and earth, and for life, but the details worked themselves out in accordance with the laws laid down by God, on this picture. It wasn't necessary for God to create each individual species (let alone each individual living thing) in the process called Special Creation. Instead, the Creator provided the seeds of the Universe and of life, and let them develop in their own time.
In all but name, except for introducing the hand of God to start off the Universe, Augustine's theory was a theory of evolution, and one which stands up well alongside modern theories of the evolution of the Universe and the evolution of life on Earth.' His views were influential throughout the Middle Ages, and followed by such important thinkers as William of Occam (in the fourteenth century) and, most importantly, by Saint Thomas Aquinas in the thirteenth century. Aquinas simply quoted Augustine's teaching on the subject of the Creation and the interpretation of Genesis; but as he was one of the highest authorities in the Christian Church at the time, and has been one of the most influential since, this amounted to an official seal of approval for the idea that God had set the Universe in motion and then rested.
It looks like in the 1920's a Jesuit in California wrote a paper on this topic. http://www.earlychurch.org.uk/pdf/book_wood.pdf
His conclusion,
[St. Augustine's doctrine of seminal reason's] end is not to give a theory of specific origins, but to explain as far as human mind can penetrate it, the revealed truth of God's word, the origin of all things by creation according as the Sacred text narrates it, and God's continual operation in creatures, as the necessary, sufficient and sole reason of their existence and operation and propagation. In general, then, and so far as the order of administration is concerned, the seminal reasons may be considered as the term of the Creator's act, as such, constituting every being a creature in the strictest sense. They may be considered in the creature, where they are the immediate effect in material second causes of the unceasing operation of the First Cause moving all, giving all their formal efficiency. They may be considered in themselves. Then they are seen in every created being as in their subject. This subject, inasmuch as it exists and acts, may in its actual operation be called by a synecdoche, a seminal reason. But there is no identity. The seminal reason is the link binding the finite to the Infinite, the universe to God, so that He is ever the Creator, it, in its minutest element, its most insignificant phase always the creature. To make it a mere natural force or generating agent would lead to Pantheism rather than favor Evolution.
It is not the readers of just any blog that are equally comfortable with the finer points of linear algebra and Saint Augustine's views on creation. +1.
Ben Espen
The reference is to Augustine's On the Literal Meaning of Genesis [De Genesi ad Litteram]. This is pretty hard to find in English, but there is a reference to it in Copleston's History of Philosophy, Vol. 2 p .76. Copleston was a standard textbook for a long time, I wouldn't be surprised if Miller had read it.
There is another reference in Copleston to On the Trinity, Book 3, Chapter 8, Line 13. You can see that at New Advent. Here is a suggestive quote:
For, consider, the very least shoot is a seed; for, if fitly consigned to the earth, it produces a tree. But of this shoot there is a yet more subtle seed in some grain of the same species, and this is visible even to us. But of this grain also there is further still a seed, which, although we are unable to see it with our eyes, yet we can conjecture its existence from our reason; because, except there were some such power in those elements, there would not so frequently be produced from the earth things which had not been sown there;
A proto-Darwinian reading is fair, if we keep in mind hindsight. Augustine meant something very different than we would by these kind of statements, but you can see a relationship.
Stephen: I'm impressed by how many people quickly responded with erudite answers. I thought someone would respond, but I didn't this expect many responses or any responses so quickly.
EastwoodDC
I think Ben nailed it – it is fair from a certain perspective. At the same time that isn't exactly the Darwinian perspective either (perhaps why John added the "proto-").
EastwoodDC: I use "proto" in the sense of foreshadowing. I'm not saying that Augustine influenced Darwin — he may or may not have; I have no idea — but in hindsight one can interpret Augustine as expressing a view that might be called "Darwinian" even though this would be an anachronism without the "proto" prefix.
Along these lines, someone might call Plato a proto-Marxist because of the authoritarian ideas Plato lays out in his Republic, even though Plato died over two millennia before Marx was born. (I think it's a little silly to call Plato a proto-Marxist, but I can't think of a good proto- example on the spot. Maybe someone could think of a good proto-Freudian or proto-Keynesian, etc.)
Manoel Galdino
I found amusing that you seem to relate Marx with authoritarian ideas. Notwithstanding Stalinism and alikes, Marx was from the begenning a democrat and only later on an adept of violence (revolution) as a mean to achive socialism. Even though, is far from simple to characterize Marx views as authoritarian. If any, is fairer to put Marx on the side of democracy and most thinkers of the time (including many liberals) as anti-democratic.
John: I understand your intent – and agree. I just meant there is a risk of "interpretive overfitting" when making such comparisons.
Overfitting?! Sounds like the kind of party-pooping comment a statistician would make. :)
Fr. John Rickert
+JMJ
Salutem magnam in Domino —
I need to look into this more "when I have time," but there are two aspects which should be overlooked, and I think they might be quite key to the whole issue: 1. St. Augustine was very Platonic and Neo-Platonic in his philosophy. The Platonists tend to see pure forms as being reality and concretizations of them in the material world as imperfect instantiations of the pure form. I think that St. Augustine may have been of the opinion that God initially created all the -forms- in one swoop but then brought about their material embodiments over the series of days. 2. The "ratio seminalis" is the Latin way of rendering "logos spermatikos" of Greek. I would see it as imprescindible to know what the latter term means, and again, I think it is primarily associated with the Platonic / Neo-Platonic schools. Because forms are timeless, pure, and almost eternal except for the fact of being created in a general Platonic view, I do not see any possibility in believing that St. Augustine would accept transformism.
God bless —
Fr. Rickert
To Manoel Galdino:
Perhaps Marx was in his conscious views anti-authoritarian by comparison with many of his contemporaries. However, Marxism necessarily takes a turn for the authoritarian when it runs up against the realities of human nature (at least as I take those realities to be). It will either fail, or it will make the Party to be the Vanguard of the People, at which point the anti-authoritarian game is lost. Marxism founders because of its defective view of human nature and false consciousness.


Mobile World Congress, Ericsson (NASDAQ: ERIC) will launch a new unified content delivery network (CDN) solution that leverages its leadership position in video delivery, optimization and networks, to address the global consumer shift to broadband mobility.
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Getting ready for my Open House was so much fun because it gave me a chance to start putting my Art Philosophy Cricut cartridge through its paces. I'm *still* finding new images on this thing that I didn't really see the first several times through the book. One of the things I love is all the shaped card bases BUT I am also excited to play with making a regular shape INTO a shaped card base.
I took the acorn image and blew it up to 5.5" and cut two extras to mount them on for a card base. To stick them together I simply scored a small line about .25" in on one edge and glued them together. After I put the cap on I added some lines of Bronze Glitz Glitter Gel to add some pizzaz. The tag is also on the cartridge - I just stamped the sentiment from the new "Harvest Happiness" set.
Super simple, super cute and it will fit into a regular sized envelope for mailing! There are several shapes on the cartridge with matching little tags that include a negative of the image so you can make lots of different shaped cards like this!


Over the years, one observation about large and mid-sized business has been remarkably consistent – when marketing and sales are not properly aligned, companies fail to make their projected numbers. The predictable result is internal finger-pointing and navel-gazing, neither of which yields the desired result: more revenue and profitable customers.
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One of the common obstacles to delivering value to meet customer expectations is the how – how is the product actually implemented in customer's environment? How are end-users trained? As the customer embarks on their value journey, the quality of service delivered by customer support becomes critical.
Let's stress this again: the only way to achieve real, constructive alignment between sales and marketing (and the entire organization) is to organize all activities around customer value.
Javed Matin is an independent management consultant helping B2B companies accelerate revenue growth through sales and marketing optimization. Over the past twenty years, he has developed global sales operations programs, training and effectiveness audits across multiple industry sectors.


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Can you observe evolution happening? Because for many species, humans included, evolution happens over the course of many thousands of years, it is rare to observe the process in a human lifetime. Usually only laboratory scientists studying quickly reproducing life forms, like single-celled creatures and some invertebrates, have the opportunity to see evolutionary change happen before their eyes.
Scientists Dating? Published July 9, June 14, by Cheri Fields. Mass spectrometer used in radiometric dating. I'd been kind of avoiding this topic, because it is a major eye-crosser. But it is also at the very heart of the Evolutionist's creed. Here's what the Federation of American Scientists.
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As a result, glauconite is used for dating only with extreme caution. The ages from the Coast Range batholith in Alaska Table 2 are referenced by Woodmorappe to a report by Lanphere and others Whereas Lanphere and his colleagues referred to these two K-Ar ages of and million years, the ages are actually from another report and were obtained from samples collected at two localities in Canada, not Alaska. There is nothing wrong with these ages; they are consistent with the known geologic relations and represent the crystallization ages of the Canadian samples.
The Liberian example Table 2 is from a report by Dalrymple and others These authors studied dikes of basalt that intruded Precambrian crystalline basement rocks and Mesozoic sedimentary rocks in western Liberia. The dikes cutting the Precambrian basement gave K-Ar ages ranging from to million years Woodmorappe erroneously lists this higher age as million years , whereas those cutting the Mesozoic sedimentary rocks gave K-Ar ages of from to million years.


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Niki D'Andrea , Clay McNear and Julie Peterson August 25, 2005 4:00AM
The Indian ballet troupe Mamata Shankar does not perform a subcontinental Swan Lake, nor does it include Native Americans en pointe. Rather, the group fuses classical Indian dance with modern dance steps, says Sarbari Chowdhury of the Bengali Cultural Association of Arizona, the organization sponsoring the troupe's local performance on Sunday, August 28. Shankar, the troupe's leader and namesake, is a well-known dancer/choreographer in her homeland and the niece of world-renowned musician Ravi Shankar. She and her troupe will perform the "Amritasya Putra," which Mamata describes as an innovative contemporary form "not shackled with the moors of rigid classicism." The program starts at 6 p.m. at Dobson High School, 1501 West Guadalupe in Mesa. Tickets are $20 ($50 for VIP seating), $10 for those ages 5 to 18. They can be purchased online at www.azindia.com or at Ashoka Indian Groceries in Chandler or Tempe's India Plaza. Call 480-705-7636 or 480-472-3000. -- Clay McNear
Pure Genus
Treasure Mammal shakes it up
"I want people to step outside of their box," says Treasure Mammal's Abelardo Gil III. At 8 p.m. Saturday, August 27, at the Trunk Space, 1506 Grand Avenue, TM, which "combines electronica, hard-core, and spazz into one giant hit machine," performs with Assacre, Father's Day, and others. Admission costs $5. Call 602-256-6006. -- Julie Peterson
Make Room
Marvin's Nearly Naked
A 20-year family feud turns into a dark comedy in Nearly Naked Theatre's version of Scott McPherson's off-Broadway hit Marvin's Room, opening Friday, August 26, at Phoenix Theatre, 100 East McDowell, and continuing through September 17. Tickets cost $15 to $18. Call 602-254-2151. -- Niki D'Andrea
This is now
For anyone born before 1975, it's incomprehensible that the music of the '80s is now considered "classic rock." But to younger ears, mid-'80s bands might sound brand-new, their catchy ear candy providing a stark contrast to today's Sturm und Drang pop. The "80s Flashback Concert" on Wednesday, August 31, spotlights two of the era's best bands in the English Beat and ABC, both from Great Britain. In its day, the Beat was simply unbeatable -- a 2-Tone fan favorite that also won critical laurels for tunes like "Save It for Later," "I Confess" and its evergreen cover of Smokey Robinson's "Tears of a Clown." ABC also warrants a fresh listen. Taking its cue from the Bryan Ferry/David Bowie school of synth-pop, ABC and its main man, Martin Fry (pictured), spawned a plethora of hits, including "The Look of Love," "Poison Arrow," "That Was Then but This Is Now" and "When Smokey Sings" -- another homage to Smokey Robinson. Colin Hay of Men at Work is also on the bill. Showtime is 8 p.m. at the Celebrity Theatre, 440 North 32nd Street. Tickets range from $27.50 to $45. Call 602-267-1600 or visit www.celebritytheatre.com. -- Clay McNear
Niki D'Andrea has covered subjects including drug culture, women's basketball, pirate radio stations, Scottsdale staycations, and fine wine. She has worked at both New Times and Phoenix Magazine, and is now a freelancer.
Contact: Niki D'Andrea
Clay McNear
Contact: Clay McNear
Julie has written for the Night & Day events calendar section since 2005. As a student at Arizona State, she received the Glendon and Kathryn Swarthout Creative Writing Award and the Theatre Medallion of Merit.
Contact: Julie Peterson
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text:
Q: Find all functions ${\rm f} :{ \mathbb R}_{+}\to{ \mathbb R}_{+}$ Find all functions ${\rm f}:{\mathbb R}_{+} \to {\mathbb R}_{+}$
, such that $\forall\ x,y \in \mathbb R_+$ the equation
$$
\big(1 + y{\rm f}\left(x\right)\big)\big(1 - y{\rm f}\left(x + y\right)\big) = 1
$$ is fulfilled.

A: Assume $f$ is of the given form. Let $g(x)=\frac{1}{f(x)}$.
Plugging this into the equation gives
$$\left(1+\frac{y}{g(x)}\right)\left(1-\frac{y}{g(x+y)}\right)=1$$
Multiplying by $g(x)g(x+y)$, factoring out and subtracting $g(x)g(x+y)$ we obtain
$$yg(x+y)-y^2-yg(x)=0$$
Dividing by $y>0$ gives
$$g(x+y)=g(x)+y$$
This implies $\frac{g(x+y)-g(x)}{y}=1$ for all $x,y\in\mathbb{R}_+$.
In particular, $\lim_{y\rightarrow 0}\frac{g(x+y)-g(x)}{y}=1$ (*).
Hence $g$ is differentiable on $\mathbb{R}^+$ with derivative $1$.
By the fundamental theorem of calculus, $g(x)=x+c$ for some $c\ge 0$, that is
$$f(x)=\frac{1}{x+c}$$
for $c\in\mathbb{R}_{\ge 0}$. These are also really solutions.
Important note (*):
We have the equation $g(x+y)-g(x)=y$ only for positive $y$. Therefore, the limit is a priori only a right-sided limit. This is a problem, because we can then only conclude right-differentiability, which is useless.
However, we can extend the equation to small negative $y$, thus making the limit two-sided.

Claim: Let $x>0$ be fixed. For all $y$ with $-x<y<0$ (that is: $y$ is negative and has small enough absolute value), we have
  $$\frac{g(x+y)-g(x)}{y}=1$$

Proof:
Plugging in $-y>0$ for $y$ and $x+y>0$ for $x$ into the original equation gives
$$1=\frac{g(x)-g(x+y)}{-y}=\frac{g(x+y)-g(x)}{y}$$ $\square$

A: Correct me if I'm very wrong.
$$(1+y f(x))(1-y f(x+y))=1 \implies \frac{f(x+y)-f(x)}{y}=-f(x)f(x+y) < 0$$
So $f$ is strictly decreasing.
If $f$ was not continous at some point $x_1 \in \mathbb{R}_+$, it would have to be a jump discontinuity downwards and from the restirction it must follow that there's a jump discontinuity at every positive $y$ greater than $x_1$. But $x_1=x+y$ for many $x,y \in \mathbb{R}_+$, and again it follow from the restriction that there's a jump discontinuity at $x$. given $x$ is arbitrary and smaller than $x_1$, it follows that there's a jump discontinuity at every positive $x$ smaller than $x_1$. Finally, we conclude $f$ is jump discontinous in all It's domain, and coupled with the fact $f$ is bounded above (strictly decreasing), and bellow, we have a contradiction (To be elaborated on later).
SO... $f$ is continous! And:
$$lim_{\Delta x \rightarrow 0^{+}}  \frac{f(x+\Delta x)-f(x)}{\Delta x} = lim_{\Delta x \rightarrow 0^{+}} -f(x)f(x+\Delta x)=-f(x)^2$$
Again, $f$ is continous, so for $x$ taken as a positive constant, $-f(x)f(x+\Delta x)$ is a continous function of $\Delta x$ for all $\Delta x > -x$. Hence:
$$lim_{\Delta x \rightarrow 0^{-}} \frac{f(x+\Delta x)-f(x)}{\Delta x} = lim_{\Delta x \rightarrow 0^{-}} -f(x)f(x+\Delta x) = lim_{\Delta x \rightarrow 0^{+}} -f(x)f(x+\Delta x) = lim_{\Delta x \rightarrow 0^{+}}  \frac{f(x+\Delta x)-f(x)}{\Delta x}$$
So $f$ is even differentiable, and:
$$f'(x)=-f(x)^2$$
And the only solutions are $f(x)=\frac{1}{x+a}$, for any constant $a \geq 0$ .

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text:
Восточноафрика́нский узкоро́т, или мозамби́кский узкорот (Breviceps mossambicus) — вид земноводных из рода Африканские узкороты. Вид был открыт натуралистом Вильгельмом Петерсом на острове Мозамбик в середине девятнадцатого столетия.

Внешний вид
По выражению зоолога, автора серии книг «Жизнь животных» Альфреда Брэма, восточноафриканский узкорот «является в высшей степени оригинально сложенной лягушкой». Длина тела животного достигает пяти сантиметров. Голова короткая, с почти плоской лицевой частью и едва выдающимся вперёд рылом; ротовое отверстие маленькое, глаза большие. Возле глаз есть косая чёрная полоса. Конечности представителей этого вида крепко сложены, очень короткого размера; значительная часть их расположена в складках кожи, как в мешке. На задних лапах имеется большая лопатообразная мозоль. Кожа гладкая, что отличает этот вид от ближайших сородичей; на спине имеет красно-бурую, а по бокам жёлто-бурую окраску и может быть покрыта разнообразными пятнами. На горле имеется большое чёрное пятно.

Образ жизни
В засушливую погоду восточноафриканский узкорот прячется в тенистых местах и возобновляет активность во время дождей. Питается беспозвоночными, в частности термитами.

Самка мозамбикского узкорота откладывает в листовую подстилку кладку из двадцати небольших яиц размером до шести миллиметров, в оболочке, размер которой в два раза превышает размер яйца. В течение шести-восьми недель зародыши внутри яйца развиваются в молодых лягушат.

Ареал
Представители вида распространены на территории Восточной и Южной Африки, встречаются в Зимбабве, Демократической Республике Конго, Замбии, Мозамбике, Ботсване, Свазиленде, Южно-Африканской Республике и Танзании. Населяют преимущественно зоны саванн, кустарниковые степи, пастбища, пахотные земли и сельские сады.

Примечания

Ссылки
 Breviceps mossambicus в списке Красной книги

Brevicipitidae
Животные, описанные в 1854 году
Земноводные Африки
Эндемики Африки

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text:
La Hickok Belt est le nom sous lequel on connaît généralement le Prix S. Rae Hickok de l'athlète professionnel de l'année (en anglais, S. Rae Hickok Professional Athlete of the Year award). Il s'agit d'un trophée décerné pendant 27 ans au meilleur athlète professionnel de l'année aux États-Unis. Il a été créé en l'honneur du fondateur de la Hickok Manufacturing Company de Rochester (New-York), qui fabriquait des ceintures, d'où le choix d'une ceinture comme trophée. Sandy Koufax est le seul athlète à l'avoir jamais remporté deux fois.

En 2010, Tony Liccione, président du musée du souvenir de la boxe à Rochester, a annoncé le projet de renouer avec la tradition de la Hickok Belt à compter de 2010.

Lauréats

Référence

Prix sportif


The NFL Scouting Combine begins on February 28 and the University of Alabama will be well-represented in Indianapolis.
The SEC led all conferences with a total of 66 invitees. The ACC was second with 60 while the Big 12 had just 18.
The Combine will run from Feb. 28 through March 6 and puts NFL Draft-eligible players in front of front office personnel and coaches. The players will go through numerous drills, workouts, and interviews during the week.


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Wood Plastic Composite Thailand, Wholesale Various High Quality Wood Plastic Composite Thailand Products from Global Wood Plastic Composite Thailand Suppliers Hot Sale to Thailand Recyclable Factory Wood Plastic Composite Flooring Long lasting hard wear plastic flooring decking for Thailand market.
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text:
\section{Introduction}

The study of the nonlinear Schr\"{o}dinger (NLS) equation on different graph models is a continuously developing subject
(see e.g. \cite{Kuchment})
motivated by various physical experiments involving wave propagation in narrow waveguides \cite{Beck,Joly1,Kuch}. In this context, graph models consisting of edges and vertices might arise as an approximation of a multi-dimensional narrow waveguides when their thickness parameters converge to zero \cite{Costa, Sobirov2}.

In the last decade the NLS equation has been extensively studied in the context of existence and stability of standing waves on both compact (all edges are of finite lengths) and noncompact graphs. The well-known example of a compact graph is the dumbbell graph, i.e. the graph constructed by attaching two rings to a central line segment. In \cite{MP}, authors considered standing waves in the focusing NLS equation on the dumbbell graph and investigated the existence of the ground state (the standing wave that minimize the energy under the a fixed mass constraint) using methods of bifurcation theory. It has been proven that for small values of mass the ground state is given by a constant solution, whereas for larger mass values the constant solution bifurcates generating two standing waves, one of which is asymmetric and has the lowest energy at the fixed mass near the symmetry breaking bifurcation, and another standing wave is symmetric and not a ground state. The analytical results also were supported by numerical computations. Later, in \cite{Goodman}, the symmetry-preserving bifurcation described in \cite{MP} was studied in details.

The question of existence of ground states also has been raised for the NLS equation on noncompact graphs. The sufficient topological conditions for the nonexistence of ground states was given in \cite{AdamiCV} using the variational approach. In particular, it was proven that the infimum of the energy at the fixed mass $\mu$ for the NLS equation on a noncompact graph is equal to the infimum of the energy for the NLS on $\mathbb{R}$ at the same fixed mass $\mu$. Such infimum is never achieved if the noncompact graph after removal of any edge contains an edge of infinite length in every connected component with some exceptions.
In \cite{AdamiJFA}, authors gave another set of sufficient conditions, mostly based on metric properties of noncompact graphs, which guarantee the existence or nonexistence of groud states.

Further study of the existence and stability of standing waves on graphs include, but not limited to, the works related to the cubic NLS equation on the periodic graph \cite{GilgPS}, the cubic NLS equation on the tadpole graph \cite{NojaPR} and its extension for NLS with any positive power nonlinearity \cite{NojaN}.

Among the limitless amount of possible graph models, our particular interest is in the class of {\it star graphs}, which are constructed by gluing together $N$ half-lines (edges) at a common vertex.
Each edge can be regarded as
$\mathbb{R}^+$, and the vertex is placed at the origin. Let $\Gamma$ represent a star graph.
The Hilbert space on the graph $\Gamma$ is
\begin{equation}
\label{L2}
L^2(\Gamma) = \oplus_{j=1}^N L^2(\mathbb{R}^+).
\end{equation}
Elements in $L^2(\Gamma)$ are function vectors
$\Psi = (\psi_1, \psi_2,..., \psi_N)^T$ with each component $\psi_j \in L^2(\mathbb{R}^+)$ defined on the $j$-th edge of $\Gamma$.
We also introduce the following $L^2$-based Sobolev space on $\Gamma$
\begin{equation}
\label{Hk}
H^2(\Gamma) =  \oplus_{j=1}^N H^2(\mathbb{R}^+)
\end{equation}
equipped with appropriate boundary conditions at the vertex.

The present work consider the NLS equation on $\Gamma$ with $\delta$-type interaction at the vertex
\begin{equation}
\label{eq1}
i \frac{\partial \Psi}{\partial t} = - \Delta \Psi - (p+1) |\Psi|^{2p} \Psi, \quad
\Psi \in \mathcal{D}(\Delta), \quad
t \in \mathbb{R}, \quad x \in \Gamma,
\end{equation}
where $p>0$, $\Psi=\Psi(t,x) \in \mathbb{C}^N$, $\Delta$ is the Laplacian operator with domain $\mathcal{D}(\Delta)$ acting as
$\Delta \Psi = (\psi_1'', \psi_2'',..., \psi_N'')^T$ with primes standing for derivatives in $x$,
and the nonlinearity is defined as
$|\Psi|^{2p}\Psi =
(|\psi_1|^{2p}\psi_1, |\psi_2|^{2p}\psi_2,..., |\psi_N|^{2p}\psi_N)^T$.  The domain of the Laplacian is
\begin{equation}
\label{H2}
\mathcal{D}(\Delta) := \left\{ \Psi \in H^2(\Gamma): \ \psi_1(0) = \dots = \psi_N(0), \
\sum_{j=1}^N  \psi_j'(0) = \alpha \psi_1(0) \right\}.
\end{equation}
where the prime denotes the right-side derivative in $x$.
The parameter $\alpha$ incorporated in the definition of
$\mathcal{D}(\Delta)$ defines the strenght of the vertex interaction. In the physical context, $\alpha<0$ refers to the presence of a potential well at the vertex and represents an attractive delta interaction, whereas $\alpha>0$ means the existence of a potential barrier and is associated with repulsive delta interaction. In case of $\alpha=0$, the boundary conditions in (\ref{H2}) are known as {\it Kirchhoff} and correspond to the free flow at the vertex.

For every $p>0$, the Cauchy problem for the NLS equation (\ref{eq1}) is locally well-posed and its solutions conserve energy and mass,
see Propositions 2.1 and 2.2 in \cite{ACDD}.
The mass and energy conservations are motivated by the invariance of the NLS equation (\ref{eq1}) under the gauge transformation
$\Psi \mapsto e^{i \theta}\Psi$ with $\theta \in \mathbb{R}$ and under the time translation $\Psi(t, \cdot) \mapsto \Psi(t+\tau, \cdot)$ with $\tau \in \mathbb{R}$, respectively. The conservation of a momentum functional is generally false since the boundary conditions at the vertex given in (\ref{H2}) break the translation symmetry in $\Gamma$. However, the momentum might be conserved under appropriate conditions. As an example, see Section 6 in \cite{KP2}, where authors considered the NLS equation (\ref{eq1}) on the star graph with even number of edges and free flow at the vertex ($\alpha = 0$).

In series of papers \cite{AdamiAH, ACDD, ACDD2}, Adami, Cacciapuoti, Finco, and Noja analyzed variational properties of standing waves in the NLS equation (\ref{eq1}) on a star graph $\Gamma$ with $N$ edges. For every $\alpha \in \mathbb{R}$, the existence and explicit formulations of standing waves were shown. In particular, for $\alpha<0$, authors found that the NLS equation (\ref{eq1}) admits the unique symmetric (with respect to permutation of edges) standing wave $\Psi_{\omega, 0}$ and that all other possible standing waves are nonsymmetric. By using the well-known stability results in \cite{Lions} and \cite{GSS},
in case of a subcritical nonlinearity in (\ref{eq1}) and the presence of attractive delta interaction at the vertex
($\alpha<0$) authors proved that $\Psi_{\omega, 0}$ is orbitally stable. The proof was based on solving the (global and local) minimization problem for the NLS energy constrained by the fixed mass. For sufficiently small mass below the critical value, it was shown that a global minimizer coincides with $\Psi_{\omega, 0}$, whereas for a large mass above the critical value, $\Psi_{\omega, 0}$ is a local minimizer.
Later, similar stability results were obtained in \cite{PG} using different approach, mostly based on the extension theory of symmetric operators.

In case of $\alpha<0$, the energy of each nonsymmetric standing wave is higher than the energy of $\Psi_{\omega, 0}$, and such nonsymmetric standing waves are also known as {\it excited states} \cite{ACDD}. In 2014 Diego Noja published a valuable manuscript \cite{Noja} which emphasised some of the recent (2010s) results in the study of NLS on graphs and discussed various open problems. In particular, the author conjectured the orbital instability of excited states and raised the question of asymptotic stability of standing waves. The latter problem on asymptotic stability was partially answered in \cite{Li}.

In the present work, we consider the NLS equation with any positive power nonlinearity on a star graph $\Gamma$ with the interaction at the vertex of an arbitrary strength $\alpha \neq 0$. In case of $\alpha<0$, we prove that every excited state is orbitally unstable which solves the problem conjectured in \cite{Noja}. Also, in \cite{Noja}, it has been noted that the difficulty which arises in the analysis of orbital stability/instability of excited states is to get the exact count of the Morse index for nonsymmetric standing waves. We overcome such difficulty by extending the Sturm theory to Schr\"{o}dinger operators on star graphs. Moreover, we also consider $\alpha>0$ and show the orbital instability of {\it all} standing waves in this case.
Similar results were obtained independently in \cite{PG2} using the theory of symmetric operators, see Remark \ref{PGremark} below.

The paper is organized as follows. Section 2 gives the structure and the explicit representation of stationary states for the NLS equation (\ref{eq1}) with nonzero $\alpha$, and provides the reader with illustrations of stationary states on the star graph with $N=3$ edges. The main results on orbital instability of standing waves are introduced in Section 3 and proved in Section 4.

\section{Stationary states of the NLS equation}

Stationary states of the NLS equation (\ref{eq1}) are defined by the standing wave solutions of the form
$$
\Psi(t,x) = e^{i \omega t} \Phi_\omega(x),
$$
where each pair $(\omega, \Phi_\omega) \in \mathbb{R} \times \mathcal{D}(\Delta)$ is a real-valued solution of the stationary NLS equation
\begin{equation}
\label{stationary_NLS}
-\Delta \Phi_\omega - (p+1)|\Phi_\omega|^{2p}\Phi_\omega = -\omega\Phi_\omega.
\end{equation}
By Sobolev's embedding theorem, $\Phi_\omega \in \mathcal{D}(\Delta)$ requires $\Phi_\omega(x), \Phi_\omega'(x) \to 0$ as $x\to \infty$. Therefore, we consider only the cases with $\omega>0$ which entail the existence of the exponential decaying solutions as $x \to \infty$.

In case of $\alpha = 0$, the stationary NLS equation (\ref{stationary_NLS}) has either the unique solution or a one-parameter family of solutions, depending on the number of edges $N$ in $\Gamma$. For odd values of $N$, the solution is uniquely given by so-called ``half-solitons'', whereas for even $N$ the stationary states are represented by families of solitary waves, so-called
``shifted states'', parametrized by a translational parameter, see \cite{ACDD}, \cite{KP1}, \cite{KP2}. The stability results on both half-solitons and shifted states are given in \cite{KP1} and \cite{KP2}.

In case of $\alpha \neq 0$, the stationary state $\Phi_{\omega, 0}$ exists if the frequency $\omega$ exceeds the lower bound
$\omega > \frac{\alpha^2}{N^2}$ \cite{ACDD}. Other stationary states appear when $\omega$ exceeds certain bifurcation values.
For any $K = 0,...,\left[ \frac{N-1}{2} \right]$, there exists a solution
$\Phi_{\omega,K}$ to the stationary NLS equation (\ref{stationary_NLS}) if the condition $\omega > \frac{\alpha^2}{(N-2K)^2}$ is satisfied. The set
$\{\Phi_{\omega, 0}, \Phi_{\omega, 1}, \Phi_{\omega, \left[ {(N-1)}/{2} \right]}\}$ represents all possible stationary states of the NLS equation (\ref{eq1}). For sufficiently large values of $\omega$, all
$\left[ \frac{N-1}{2} \right]+1$ stationary states are present.
The explicit representation of these stationary states is given in the next lemma proved by Theorem 4 and Remarks 5.1-5.2 in \cite{ACDD}.

\begin{lemma}
\label{stationary}
Let $p>0$, $\alpha \in \mathbb{R} \backslash \{0\}$ and
$K = 0,...,\left[ \frac{N-1}{2} \right]$. Then, if the condition
$\omega > \frac{\alpha^2}{(N-2K)^2}$ is satisfied, there exists a solution
$\Phi_{\omega, K}$ to the stationary NLS equation (\ref{stationary_NLS})
given, up to permutations of edges, by
\begin{equation}
\label{stat_sol}
(\Phi_{\omega, K})_j(x) = \begin{cases}
\phi_\omega(x+a_K), \quad j = 1,...,K \\
\phi_\omega(x-a_K), \quad j = K+1,...,N
\end{cases}
\end{equation}
where
$\phi_\omega(x) = \omega^{1/2p} \sech^{1/p}(p \sqrt{\omega}x)$ and
$a_K = \frac{1}{p\sqrt{\omega}} \arctanh \left( \frac{\alpha}{(N-2K)\sqrt{\omega}} \right)$.
\end{lemma}

By the represention (\ref{stat_sol}), the profile of $\Phi_{\omega, K}$ on each edge of the graph $\Gamma$ is either {\it a bump} (nonmonotonic profile) or {\it a tail} (monotonic profile). The presence of such bumps or tails on the edges depends on the shift $a_K$. When $\alpha<0$, the shift value $a_K$ is negative. Therefore,
the solution $\phi_\omega(x+a_K)$ on each edge $1,...,K$ in
(\ref{stat_sol}) is nonmonotonic and represents a bump, whereas the solution $\phi_\omega(x-a_K)$ on each edge $K+1,...,N$ is monotonic and represents a tail. Notice that, in this case, the number of bumps in the profile of $\Phi_{\omega, K}$ is equal to $K$, and since $K<\frac{N}{2}$, there are strictly more tails than bumps. The unique symmetric standing wave $\Psi_{\omega, 0}$ described in Introduction corresponds to the stationary state $\Phi_{\omega, 0}$, while the excited states (nonsymmetric standing waves) are defined by stationary states with bumps.
As an example, if $N=3$ then $K \in \{0,1\}$ in Lemma \ref{stationary}, and there are only two possible stationary states, namely,
$\Phi_{\omega, 0}$ and $\Phi_{\omega, 1}$, see Figure 1.

\begin{figure}[htb]
\begin{center}
\includegraphics[height=2cm,width=6cm]{fig1_1.pdf}
\includegraphics[height=2cm,width=6cm]{fig2_1.pdf}
\end{center}
\caption{Case $\alpha<0$ and $N=3$:
$\Phi_{\omega, 0}$ has three tails and no bumps (left) and $\Phi_{\omega, 1}$ has two tails and one bump (right).}
\label{fig-1}
\end{figure}

When $\alpha>0$, the shift value $a_K$ is positive, and $K$ represents the number of tails in the profile of $\Phi_{\omega, K}$, where tails appear on the edges $1,...,K$ only according to the representation (\ref{stat_sol}). Here, in contrast to the case with negative $\alpha$, the number of tails is strictly less than the number of bumps. For $N=3$, Figure 2 illustrates the only possible stationary states $\Phi_{\omega, 0}$ and $\Phi_{\omega, 1}$.

\begin{figure}[htb]
\begin{center}
\includegraphics[height=2cm,width=6cm]{fig3_1.pdf}
\includegraphics[height=2cm,width=6cm]{fig4_1.pdf}
\end{center}
\caption{Case $\alpha>0$ and $N=3$:
$\Phi_{\omega, 0}$ has no tails and three bumps (left) and $\Phi_{\omega, 1}$ has one tails and two bumps (right)}
\label{fig-1}
\end{figure}

\section{Main results}
To investigate the stability of each stationary state $\Phi_{\omega, K}$ given by Lemma \ref{stationary}, consider its complex perturbation
$
\Psi_\omega = \Phi_{\omega} + U +iW,
$
where $U, W \in \mathcal{D}(\Delta)$ are real valued. The linearization of Hamiltonian system corresponding to the NLS equation (\ref{eq1}) implies that the time evolution of the perturbations $U$ and $W$ is
\begin{equation}
\frac{d}{dt} \left[ \begin{array}{c} U \\ W \end{array} \right] =
\left[ \begin{array}{cc} 0 & L_- \\ -L_+ & 0 \end{array} \right] \left[ \begin{array}{c} U \\ W \end{array} \right],
\end{equation}
where $L_\pm$ are Hessian operators in $L^2(\Gamma)$ with domain
$\mathcal{D}(\Delta)$ with differentiable expression given by
\begin{eqnarray}
\label{Lplus}
L_+ & := & -\Delta + \omega - (2p+1) (p+1) \Phi_{\omega,K}^{2p} \\
L_- & := & -\Delta + \omega - (p+1) \Phi_{\omega,K}^{2p} ,
\label{Lminus}
\end{eqnarray}

Note that the operators $L_+$ and $L_-$ are self-adjoint in
$L^2(\Gamma)$, see Theorem 1.4.4 in \cite{Kuchment}. Therefore, the spectrum $\sigma(L_\pm) \subset \mathbb{R}$ consists of the continuous and the discrete spectra only. By Weyl's theorem, see e.g. \cite{HS},
$\sigma_c(L_\pm) = \sigma_c(L_0)$, where $\sigma_c$ stands for the continuous spectrum and $L_0:= -\Delta + \omega$. Indeed,
$L_\pm -L_0$ is a relatively compact perturbation to $L_0$, and Weyl's theorem is applicable. Therefore, $\sigma_c(L_\pm) = [\omega, \infty)$, and all eigenvalues of the discrete specrum $\sigma_p(L_\pm)$ are located in the interval $(-\infty, \omega)$. Define the Morse index $n(L_\pm)$ and the degeneracy index $z(L_\pm)$ as the number of negative and zero eigenvalues of $\sigma_p(L_\pm)$, respectively, with the account of their multiplicity.

The main result of this paper is that, for every $\alpha \neq 0$, we count the Morse and degeneracy indices of $L_\pm$ associated with the stationary state
$\Phi_{\omega,K}$ for all possible $K \geq 0$, as in the following:

\begin{theorem}
\label{thm1}
Let $p>0$, $\alpha \in \mathbb{R} \backslash \{0\}$,
$K = 0,...,\left[ \frac{N-1}{2} \right]$ and
$\omega > \frac{\alpha^2}{(N-2K)^2}$. Let $L_+$ and $L_-$ be the Hessian operators associated with $\Phi_{\omega, K}$ and defined as in (\ref{Lplus})-(\ref{Lminus}). Then, \\
(i) $z(L_-) = 1$, $n(L_-)=0$; \\
(ii) $z(L_+)=0$ and
$n(L_+)= \begin{cases} K+1 \quad {\rm if} \quad \alpha<0; \\
N-K \quad {\rm if} \quad \alpha>0.
\end{cases}$ \\
In addition, \\
$\bullet$ if $\alpha< 0$ with $K \geq 1$, $n(L_+)$ consists of two simple eigenvalues $\lambda_1<\lambda_2$ and another eigenvalue $\lambda_* \in (\lambda_1, \lambda_2)$ of multiplicity $K-1$; \\
$\bullet$ if $\alpha>0$ with $K\geq 0$, $n(L_+)$ consists a simple eigenvalue $\lambda_1$ and another eigenvalue $\lambda_{**} \in (\lambda_1, 0)$ of multiplicity $N-K-1$.
\end{theorem}

By using the well-known instability result for the NLS equation, see Theorem 1.2 in \cite{GR},
the statement of Theorem \ref{thm1} is equivalent to following:
\begin{theorem}
\label{mainthm}
If $\alpha< 0$, then all standing waves $e^{i\omega t}\Phi_{\omega, K}$ with $K \geq 1$ bump(s) are spectrally (and orbitally) unstable. If $\alpha>0$, then all standing waves $e^{i\omega t}\Phi_{\omega, K}$ with $K\geq 0$ tail(s) are spectrally (and orbitally) unstable.
\end{theorem}

For $\alpha<0$ and $K=0$, our count of Morse and degeneracy indices coincides with the count in Proposition 6.1 in \cite{ACDD} and implies the stability result given by Theorem 1 in \cite{ACDD2}:
\begin{theorem}
Let $p \in (0,2]$ and $\alpha<0$. Then, given $\omega>\frac{\alpha^2}{N^2}$, the standing wave $e^{i\omega t}\Phi_{\omega, 0}$ is orbitally stable.
\end{theorem}

\begin{remark}
\label{PGremark}
Partial results of Theorem \ref{mainthm} have been obtained in the recent work \cite{PG2} by using the extension theory of symmetric operators.
In particular, for the case $\alpha<0$ with $p>2$, authors showed the existence of some $\omega_K^*$ such that the instability result holds for all
$\omega \in \Big( \frac{\alpha^2}{(N-2K)^2}, \omega_K^* \Big)$. It has been noted by authors that no results were obtained for
$\omega > \omega_K^*$. Theorem \ref{mainthm} extends these results to all $\omega \in \Big( \frac{\alpha^2}{(N-2K)^2}, \infty \Big)$.
\end{remark}

\section{Proof of Theorem \ref{thm1}}

To prove Theorem \ref{thm1}, we compute Morse and degeneracy indices on graph $\Gamma$ by using a method developed in \cite{KP2}. Consider the following Schr\"{o}dinger equation on the real line, $x \in \mathbb{R}$,
\begin{equation}
\label{eqofcomp}
- v''(x) + \omega v(x) - (2p+1)(p+1) \omega
\sech^2(p\sqrt{\omega}x) v(x) = \lambda v(x),
\quad \lambda < \omega.
\end{equation}
We are interested in the exponentially decaying solutions to (\ref{eqofcomp}) as $x \to +\infty$. The following three lemmas represent Sturm theory for the Schr\"{o}dinger equation on $\mathbb{R}$, see also \cite{KP2}.

\begin{lemma}
\label{solhyp}
For every $\lambda<\omega$, there exists a unique solution $v \in C^1(\mathbb{R})$ to equation (\ref{eqofcomp})
such that
\begin{equation}
\label{u-limit}
\lim_{x \to +\infty} v(x) e^{\sqrt{\omega-\lambda} x} = 1.
\end{equation}
The other linearly independent solution to equation (\ref{eqofcomp}) diverges as $x \to +\infty$.
Moreover, for any fixed $x_0 \in \mathbb{R}$, $v(x_0)$ is a $C^1$ function of $\lambda$ for $\lambda < \omega$ such that $\frac{v'(x_0)}{v(x_0)} \to -\infty$ as $\lambda \to -\infty$.
\end{lemma}

\begin{proof}
The proof is based on the reformulation of of the boundary-value problem (\ref{eqofcomp})--(\ref{u-limit})
as Volterra's integral equation
\begin{equation}
\label{Volterra}
v(x) = e^{-\sqrt{\omega-\lambda}x} - \frac{(2p+1)(p+1)}{\sqrt{\omega-\lambda}}
\int_x^{\infty} \sinh(\sqrt{\omega-\lambda}(x-y)) \sech^2(py) v(y) \diff y.
\end{equation}
Set $w(x;\lambda) = v(x) e^{\sqrt{\omega-\lambda}x}$ to get the following Volterra's integral equation with a bounded kernel:
\begin{equation}
\label{VolterraMod}
w(x;\lambda) = 1 + \frac{(2p+1)(p+1)}{2\sqrt{\omega-\lambda}} \int_x^{\infty}  ( 1-e^{-2\sqrt{\omega-\lambda}(y-x)}) \sech^2(py) w(y;\lambda) \diff y.
\end{equation}
By standard Neumann series and the ODE theory, the existence and uniqueness of a solution $w(\cdot;\lambda) \in C^1(\mathbb{R})$
of the integral equation (\ref{VolterraMod}) with $\lim_{x \to \infty} w(x;\lambda) = 1$ is obtained for every $\lambda < \omega$. Such construction provides with the solution $v\in C^1(\mathbb{R})$ to the differential equation (\ref{eqofcomp}) exponentially decaying as
$x \to +\infty$. For any fixed $x_0$, $v(x_0)$ is (at least) $C^1$ function of $\lambda < \omega$ since the Volterra's integral equation (\ref{Volterra}) depends analytically on $\lambda$ for $\lambda < \omega$.
The other linearly independent solution to the differential equation (\ref{eqofcomp}) diverges as $x\to +\infty$ due to the $x$-independent and nonzero Wronskian determinant between two solutions.

It remains to prove that $\frac{v'(x_0)}{v(x_0)} \to -\infty$ as $\lambda \to -\infty$ for any fixed $x_0 \in\mathbb{R}$. Using the setting $w(x;\lambda) = v(x) e^{\sqrt{\omega-\lambda}x}$, we get
\begin{equation}
\label{Weyl_function}
\frac{v'(x_0)}{v(x_0)} = -\sqrt{\omega-\lambda} +
\frac{w'(x_0;\lambda)}{w(x_0;\lambda)}.
\end{equation}
Since $w(\cdot, \lambda) \in C^1(\mathbb{R})$ and $\lim_{x \to \infty} w(x;\lambda) = 1$, we get $w(\cdot, \lambda) \in L^\infty[x_0, \infty)$. The construction (\ref{VolterraMod}) yields that
$\|w\|_{L^\infty[x_0, \infty)} \leq 2$ for large enough negative $\lambda$, and so, as $\lambda \to -\infty$, (\ref{VolterraMod}) implies
$$
|w(x_0;\lambda)-1| \leq \frac{C_p}{\sqrt{\omega-\lambda}} \|w\|_{L^\infty[x_0, \infty)} \leq \frac{2C_p}{\sqrt{\omega-\lambda}} \to 0,
$$
where $C_p$ is constant which depends on $p$ only. Therefore,
\begin{equation}
\label{bound1}
w(x_0; \lambda) \to 1 \quad {\rm as} \quad \lambda \to -\infty.
\end{equation}
Differentiating the equation (\ref{VolterraMod}) in $x$, we get
$$
w'(x;\lambda) = -(2p+1)(p+1)\int_x^{\infty}  e^{-2\sqrt{\omega-\lambda}(y-x)} \sech^2(py) w(y;\lambda) \diff y.
$$
Since the integrand in the latter expression is bounded for
$\lambda<\omega$, for $\lambda \to -\infty$ we get
\begin{equation}
\label{bound2}
|w'(x_0; \lambda)| \leq \hat{C}_p \|w\|_{L^\infty[x_0, \infty)} \leq 2\hat{C}_p,
\end{equation}

where $\hat{C}_p$ is constant which depends on $p$ only.

Finally, by using the bounds in (\ref{bound1}) and (\ref{bound2}), the expression (\ref{Weyl_function}) implies that $\frac{v'(x_0)}{v(x_0)} \to -\infty$ as $\lambda \to -\infty$.
\end{proof}

\begin{lemma}
\label{Sturm}
Let $v = v(x;\lambda)$ be the solution defined by Lemma \ref{solhyp}. Assume that $v(x;\lambda_1)$
has a simple zero at $x = x_1 \in \mathbb{R}$ for some $\lambda_1 \in (-\infty,\omega)$.
Then, there exists a unique $C^1$ function
$\lambda \mapsto x_0(\lambda)$ for $\lambda$ near $\lambda_1$ such that $v(x;\lambda)$ has a simple
zero at $x = x_0(\lambda)$ with $x_0(\lambda_1) = x_1$ and $x_0'(\lambda_1) > 0$.
\end{lemma}

\begin{proof}
By the previous lemma, $v$ is a $C^1$ function of $x$ and $\lambda$ for every $x \in \mathbb{R}$
and $\lambda \in (-\infty,\omega)$. Since $x_1$ is a simple zero of $v(x;\lambda_1)$, we get
$\partial_x v(x_1;\lambda_1) \neq 0$, and this allows us to use the implicit function theorem
to get a unique $C^1$ function $\lambda \mapsto x_0(\lambda)$ for $\lambda$ near $\lambda_1$
such that $v(x;\lambda)$ has a simple
zero at $x = x_0(\lambda)$ with $x_0(\lambda_1) = x_1$.
It remains to show that $x_0'(\lambda_1) > 0$.

Notice that $v(x_0(\lambda);\lambda) = 0$ implies
\begin{equation}
\label{root-der-eq}
\partial_\lambda v(x_0(\lambda); \lambda)\Big|_{\lambda=\lambda_1} =
\partial_x v(x_1;\lambda_1) x_0'(\lambda_1) + \partial_{\lambda} v(x_1;\lambda_1) = 0.
\end{equation}
Let us denote $\tilde{v}(x) = \partial_{\lambda} v(x;\lambda_1)$. Then, differentiating
equation (\ref{eqofcomp}) in $\lambda$, we have the
differential equation for $\tilde{v}$ on $\mathbb{R}$:
\begin{equation} \label{der-v-eq}
- \tilde{v}''(x) + \tilde{v}(x) - (2p+1)(p+1) \sech^2(px) \tilde{v}(x) =
\lambda_1 \tilde{v}(x) + v(x;\lambda_1), \quad \lambda < \omega.
\end{equation}
By using the Volterra's integral equation as in Lemma \ref{solhyp} above, we have that the function $\tilde{v}$ is $C^1$ in $x$ and decays to zero as $x \to \infty$. We multiply (\ref{der-v-eq}) by $v(x;\lambda_1)$,
integrate by parts on $[x_1,\infty)$, and use equation (\ref{eqofcomp}) to get
\begin{equation} \label{der-vv-eq}
- \partial_x v(x_1;\lambda_1) \tilde{v}(x_1) = \int_{x_1}^{\infty} v(x;\lambda_1)^2 dx.
\end{equation}
Finally, by using the expressions (\ref{root-der-eq}) and (\ref{der-vv-eq}) we have
\begin{equation}
\label{zeroshift}
(\partial_x v(x_1;\lambda_1) )^2 x_0'(\lambda_1) = \int_{x_1}^{\infty} v(x;\lambda_1)^2 dx > 0
\end{equation}
which, together with $\partial_x v(x_1;\lambda_1) \neq 0$, implies that $x_0'(\lambda_1) > 0$.
\end{proof}

The following result represents the
\begin{lemma}
\label{lem-Sturm}
Let $v$ be the solution defined in Lemma \ref{solhyp}. If $v(0) = 0$ (resp. $v'(0) = 0$) for some $\lambda_0 < \omega$,
then the eigenfunction $v$ to the Schr\"{o}dinger equation (\ref{eqofcomp}) is
an odd (resp. even) function on $\mathbb{R}$, hence $\lambda_0$ is an eigenvalue of
the associated Schr\"{o}dinger operator defined in $L^2(\mathbb{R})$.
There exists exactly one $\lambda_0 < 0$ corresponding to $v'(0) = 0$ and a simple eigenvalue $\lambda_0 = 0$
corresponding to $v(0) = 0$, all other such points $\lambda_0$ are located in $(0,\omega)$ and are bounded away from zero.
\end{lemma}

\begin{proof}
The uniqueness of the solution $v$ in Lemma \ref{solhyp} and the reversibility of the Schr\"{o}dinger equation (\ref{eqofcomp}) with
respect to the transformation $x \mapsto -x$ yields the existence of even and odd eigenfunctions on $\mathbb{R}$. The count of eigenvalues follows by Sturm's Theorem since
the odd eigenfunction for the eigenvalue $\lambda_0 = 0$,
$$
\phi'(x) = - \sech^{\frac{1}{p}}(px) \tanh(px)
$$
has one zero on the infinite line. Hence, $\lambda_0 = 0$ is the second eigenvalue of
the Schr\"{o}dinger equation (\ref{eqofcomp}) with exactly one simple negative eigenvalue $\lambda_0 < 0$
that corresponds to an even eigenfunction.
\end{proof}

Further in the paper, we use $\lambda_0$ to denote the eigenvalue from Lemma \ref{lem-Sturm} with positive even eigenfunction. The Figure 3 illustrates Lemmas \ref{solhyp}-\ref{lem-Sturm} and shows profiles of the solution $v$ satisfying (\ref{eqofcomp}) and (\ref{u-limit}) for values of $\lambda$ in the interval $(-\infty, 0)$.

\begin{figure}[htb]
\begin{center}
\includegraphics[height=2cm,width=6cm]{sturm_1.pdf}
\includegraphics[height=2cm,width=6cm]{sturm_2.pdf}
\end{center}
\caption{The solution $v$ of the differential equation (\ref{eqofcomp}) with
the limit (\ref{u-limit}) versus $x$ for several values of $\lambda$}
\label{fig-3}
\end{figure}

Next two lemmas provide us with the useful tools to compute the Morse and degeneracy indices of the operator $L_+$ at the state $\Phi_{\omega,K}$.
\begin{lemma}
\label{lem-eig}
Let $\alpha \in \mathbb{R}\backslash \{0\}$, $K \geq 0$, $v$ be the solution to (\ref{eqofcomp}) given by Lemma \ref{solhyp}, and $L_+$ be the Hessian operator (\ref{Lplus}) associated with $\Phi_{\omega,K}$.
Then,
$\lambda \in (-\infty, \omega)$ is an eigenvalue of $\sigma_p(L_+)$ if and only if at least one of the following conditions holds: \\
(a) $v(a_K) = 0$ with $K \geq 1$, \\
(b) $v(-a_K) = 0$,\\
(c) $Kv'(a_K)v(-a_K)+(N-K)v(a_K)v'(-a_K)-\alpha v(a_K)v(-a_K)=0$.\\
Moreover, $\lambda \in \sigma_p(L_+)$ has mutliplicity $K-1$ in the case (a), $N-K-1$ in the case (b), and is simple in the case (c). If $\lambda$ satisfies several cases, then its multiplicity is the sum of the multiplicities of each case.
\end{lemma}

\begin{proof}
First, assume $K \geq 1$. Denote $a: = a_K$. Let $\lambda$ be the eigenvalue of $L_+$ with the eigenvector
$U=(u_1,...,u_N)^T \in \mathcal{D}(\Delta)$. Then, in the eigenvalue problem
$L_+ U = \lambda U$, each component can be written as the second order differential equation
\begin{equation}
\label{eqofcompLplus}
-u_j''(x) + \omega u_j(x) - (2p+1)(p+1)\omega \sech^2 \big(p\sqrt{\omega}
(x+(-1)^{m_j} a) \big) u_j(x) = \lambda u_j(x), \quad x\in (0, \infty),
\end{equation}
where
$$m_j= \begin{cases} 0 \quad {\rm for} \quad j=1,...,K, \\
1 \quad {\rm for} \quad j=K+1,...,N. \end{cases}
$$
The substitution $u_j(x) = c_j v(x + (-1)^{m_j}a)$ with coefficient $c_j$ transforms (\ref{eqofcompLplus}) into (\ref{eqofcomp}). By Sobolev embedding of $H^2(\mathbb{R}^+)$ into $C^1(\mathbb{R}^+)$, $u_j(x) \to 0$ and $u_j'(x) \to 0$ as $x \to +\infty$ for each $j=1,...,N$. To satisfy the latter condition, we let $v$ to be the solution to (\ref{eqofcomp}) given by Lemma \ref{solhyp}, and so
\begin{equation}
\label{components}
u_j(x) = \begin{cases} c_j v(x+a), & j=1,\dots,K, \\
c_j v(x-a), & j=K+1,\dots,N.
\end{cases}
\end{equation}
The boundary conditions for $U \in \mathcal{D}(\Delta)$ in (\ref{H2})
imply the homogeneous linear system on the coefficients
\begin{equation}
\label{lin-system-1}
c_1 v(a) = \dots = c_K v(a) =c_{K+1} v(-a) = \dots = c_N v(-a), \quad
{\rm and}
\end{equation}
\begin{equation}
\label{lin-system-2}
\sum_{j=1}^{K}c_j v'(a)+ \sum_{j=K+1}^N c_j  v'(-a) = \alpha c_N v(-a).
\end{equation}
The associated matrix is
\begin{equation}
{\small M=
\left( \begin{array}{ccccccccc}
v(a) & -v(a) & 0 & \dots & 0 & 0 & \dots & 0 & 0 \\
v(a) & 0 & -v(a) & \dots & 0 & 0 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\
v(a) & 0 & 0 & \dots & -v(a) & 0 & \dots & 0 & 0 \\
v(a) & 0 & 0 & \dots & 0 & -v(-a) & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\
v(a) & 0 & 0 & \dots & 0 & 0 & \dots & 0 & v(-a) \\
v'(a) & v'(a) & v'(a) & \dots & v'(a) & v'(-a) & \dots & v'(-a) &
v'(-a) - \alpha v(-a) \\
\end{array} \right)
}
\end{equation}
Doing elementary column operations, we can obtain a lower triangular matrix with the determinant
$$
{\rm det} M = v(a)^{K-1} v(-a)^{N-K-1}
\left[ Kv'(a)v(-a)+(N-K)v'(-a)v(a) - \alpha v(a)v(-a) \right].
$$
Therefore, $U \neq 0$ is the eigenvector of $L_+$ for the eigenvalue $\lambda \in (-\infty,\omega)$
if and only if ${\rm det} M = 0$, or equivalently, at least one of the conditions (a), (b), (c) is true.
The multiplicity of $\lambda$ in cases
(a)-(c) comes directly from the linear system (\ref{lin-system-1})--(\ref{lin-system-2}).

For $K=0$, the boundary conditions (\ref{lin-system-1})-(\ref{lin-system-2}) do not contain terms $v(a)$ and $v'(a)$, and the determinant of the associated matrix $M$ becomes
$$
{\rm det} M = v(-a)^{N-1} \left[Nv'(-a) - \alpha v(-a) \right].
$$
Then, ${\rm det} M = 0$ if and only if at least one of the conditions (b), (c) is true.
\end{proof}

\begin{lemma}
\label{lastlemma}
Let $\alpha \in \mathbb{R} \backslash \{0\}$, $K \geq 0$ and $v$ be the solution to (\ref{eqofcomp}) given by Lemma \ref{solhyp}. Consider the  function of $\lambda$ as
\begin{equation}
\label{F}
F(\lambda) := K\frac{v'(a_K; \lambda)}{v(a_K;\lambda)} +
(N-K) \frac{v'(-a_K; \lambda)}{v(-a_K; \lambda)}: (-\infty, 0] \to \mathbb{R}.
\end{equation}
Then, the following hold: \\
$\bullet$ $v(|a_K|; \lambda)>0$ for all $\lambda \in (-\infty, 0]$, and there exists unique $\lambda_* \in (-\infty, 0]$ such that $v(-|a_K|;\lambda) = 0$. Moreover, $\lambda_*<0$. \\
$\bullet$ if $\alpha<0$ with $K \geq 1$, then $F(\lambda) = \alpha$ has exactly two solutions $\lambda_1 < \lambda_2$. Moreover,
$\lambda_1 < \lambda_* < \lambda_2 < 0$. \\
$\bullet$ if $\alpha>0$ with $K \geq 0$ then $F(\lambda) = \alpha$ has the unique root $\lambda_1$, and, moreover, $\lambda_1<\lambda_*$.
\end{lemma}

\begin{proof}
Since $v$ is the nonzero solution of the second order differential equation (\ref{eqofcomp}), it has only simple zeros which, according to Lemma \ref{Sturm}, are monotonically increasing functions of $\lambda$. At $\lambda = \lambda_0$, by Lemma \ref{lem-Sturm}, we have positive even $v(x; \lambda_0)$ exponentially decaying as $|x| \to \infty$. Therefore, $v(x;\lambda)$ has the only zero $x_0(\lambda)$ which bifurcates from $x=-\infty$ at $\lambda=\lambda_0$ and moves strictly monotonically towards $x=0$ as $\lambda \to 0$ with $x_0(0)=0$, see Figure \ref{fig-3}. As a result, for $\lambda\leq \lambda_0$, $v(x;\lambda)$ is positive on the entire real line, whereas, for $\lambda \in (\lambda_0, 0]$, $v(x;\lambda)$ is positive for every
$x \in (x_0(\lambda), \infty)$ and $v(x_0(\lambda);\lambda)=0$. We denote $\lambda$ satisfying $x_0(\lambda) = -|a_K|$ as $\lambda_*$. Since $a_K \neq 0$, then $\lambda_*<0$.
 The uniqueness of $\lambda_*$ is guaranteed by the monotonicity of $x_0(\lambda)$. This proves the first assertion of this Lemma.

By Lemma \ref{solhyp}, $v(\pm a_K;\lambda)$ is a $C^1$ function of $\lambda$ for $\lambda \leq 0$. Therefore, using the first assertion proven above, for $\alpha<0$ with $K\geq 1$ and for $\alpha>0$ with $K\geq 0$,
$F(\lambda)$ is $C^1((-\infty, 0])\backslash \{\lambda_*\})$ and has a simple pole at $\lambda = \lambda_*$.

To investigate the behaviour of the function $F$, we first show that $F$ is a monotonically increasing function.
Differentiating the equation (\ref{eqofcomp}) in $\lambda$, multiplying it by $v$ and integrating by parts on $[c, \infty]$ for some $c \in \mathbb{R}$, we get
$$
P(c): = \frac{\partial_\lambda v'(c) v(c) - v'(c) \partial_\lambda(c)}{v^2(c)} = \frac{1}{v^2(c)} \int_{c}^\infty v^2(x) dx > 0 \quad {\rm if} \quad v(c) \neq 0.
$$
Therefore, $F'(\lambda) = K P(a_K) + (N-K) P(-a_K) > 0$ for all $\lambda \in (-\infty, 0]\backslash \{\lambda_*\}$.

By Lemma \ref{solhyp}, for every $c \in \mathbb{R}$, $\lim_{\lambda \to -\infty} \frac{v'(c;\lambda)}{v(c;\lambda)} = -\infty $.
Then, taking $c = \pm a_K$, we have that
$\lim_{\lambda \to -\infty} F(\lambda) = -\infty$.
 By Lemma \ref{Sturm} on the monotonicity of a simple zero of $v$, the behaviour of $F(\lambda)$ around the point of singularity $\lambda_*$ is given by
$$
\lim_{\lambda \to \lambda_*^-} F(\lambda) = +\infty \quad {\rm and}
\quad \lim_{\lambda \to \lambda_*^+} F(\lambda) = -\infty.
$$
At $\lambda=0$, the unique solution $v = v(x;0)$ of (\ref{eqofcomp}) in Lemma \ref{solhyp} is known to be $v(x) = -C\phi'_\omega(x)$, where $\phi_\omega$ is given by Lemma \ref{stationary} and $C = 2^{-1/p}\omega^{-(1+p)/2p}$. Then, using the explicit formulations of $v$ and $a_K$, direct computations give
$$
F(0) = \frac{p(\alpha^2-(N-2K)^2 \omega)}{\alpha} + \alpha.
$$
Since $\omega> \frac{\alpha^2}{(N-2K)^2}$, then
$p(\alpha^2-(N-2K)^2 \omega)<0$. Hence, for $\alpha<0$ we have $F(0)>\alpha$, whereas for $\alpha>0$ we have $F(0)<\alpha$. As a result,  for every $\alpha \neq 0$, the equation $F(\lambda) = \alpha$ has a unique root $\lambda_1 \in (-\infty, \lambda_*)$. Moreover, for $\alpha<0$, there is an additional root $\lambda_2$ which is unique in $(\lambda_*, 0)$. This proves the remaining assertions of this Lemma.
\end{proof}

\begin{remark}
\label{lastremark}
In case of $\alpha<0$ with $K=0$, similar steps as in the proof of Lemma \ref{lastlemma} lead to the conclusion that $F(\lambda)$ is $C^1(-\infty, 0])$ and monotonically increasing with $\lim_{\lambda \to -\infty} F(\lambda) = -\infty$ and $F(0)>\alpha$. Then, $F(\lambda)=\alpha$ has the unique root $\lambda_1 \in (-\infty, 0)$. Consequently, $n(L_+) = 1$ and Theorem \ref{thm1} follows by a standard application of the orbital stability Theorem, see \cite{ACDD} and \cite{ACDD2}.
\end{remark}

\begin{proof1}{\em of Theorem \ref{thm1}.}
The proof of the Part (i) is based on the same arguments as in
Proposition 6.1 in \cite{ACDD}, Theorem 4.1 in \cite{PG} and Lemma 3.1 in \cite{KP1}. Thus, we only concerned about the proof of the Part (ii).

Let $\hat \lambda \in \sigma_p(L_+) \cap (-\infty, 0]$ be a nonpositive eigenvalue of $\sigma_p(L_+)$ with the eigenvector
$U \in \mathcal{D}(\Delta)$. Then, by Lemma \ref{lem-eig} at least one of the conditions {\it (a), (b), (c)} must be satisfied by $v(x; \hat \lambda)$.

Consider $\alpha<0$ with $K\geq 1$ or $\alpha>0$ with $K\geq 0$.  Recall that $a_K<0$ for negative $\alpha$, and $a_K>0$ for positive $\alpha$. Then, for $\alpha<0$ with $K\geq 1$, by Lemma \ref{lastlemma}, the part (a) of Lemma \ref{lem-eig} is satisfied for unique $\lambda_* \in (-\infty, 0]$ and the part (b) is never true. For $\alpha>0$ with $K\geq 0$, the part (a) is never true and the part (b) is satisfied for unique $\lambda_{**} \in (-\infty, 0]$.

It remains to consider the part (c) of Lemma \ref{lem-eig}, namely, to find all values $\hat \lambda \in (-\infty, 0]$ such that $v(x) = v(x; \hat \lambda)$ will satisfy
\begin{equation}
\label{part_c}
Kv'(a_K)v(-a_K)+(N-K)v'(-a_K)v(a_K) - \alpha v(a_K)v(-a_K) = 0
\end{equation}
Since $v(|a_K|) \neq 0$, and $v'(-|a_K|) \neq 0$ if $v(-|a_K|)=0$, the eigenvalue $\lambda = \lambda_*$ (resp. $\lambda = \lambda_{**}$) is not a solution of (\ref{part_c}). Therefore, all solutions $\hat \lambda$ of (\ref{part_c}) coincide with all solutions of $F(\lambda)=\alpha$, where $F$ is given by (\ref{F}). The last two assertions of Lemma \ref{lastlemma} complete the proof of Theorem \ref{thm1}.

For the case $\alpha<0$ with $K=0$, Remark \ref{lastremark} implies that both parts (a) and (b) of Lemma \ref{lem-eig} are never true, and the remaining part (c) has a unique root $\lambda_1 \in (-\infty, 0]$ with $\lambda_1<0$. Thus,
$z(L_+)=0$ and $n(L_+) = 1$ are proved.

\end{proof1}

\section*{Acknowledgement}
The author is grateful to Dmitry Pelinovsky for enlightening discussions and helpful suggestions during the preparation of this work.

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


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Photos & Press
Art Market San Francisco Market Art & Design / The Hamptons Seattle Art Fair / Seattle Texas Contemporary / Houston Art on Paper / New York
Nicolas V. Sanchez | Ballet Nepantla
Presented by Sugarlift
Location: Sugarlift Presents: Nicolas V. Sanchez | Booth E05 | 9/9 @ 8pm & 9/11 @ 2pm
Ballet Nepantla fuses contemporary dance with Mexican folklorico to explore new interpretative ways of delving into cultural and historical "in-between-ness." Co-founders Andrea Guajardo and Martin Rodriguez have backgrounds in classical ballet and traditional Mexican folklorico, respectively. They envisioned a new form of expression that fused the two modalities, and sought an aesthetic that would speak to the historical, cultural, and social realities of a broader narrative of being Mexican, Mexican-American, immigrant, and trans-cultural.
About Nicolas V. Sanchez's Ballet Folklorico Series
Nicolas V. Sanchez's recent body of work is inspired by his Mexican-American heritage. His mother is a seamstress who creates elaborate and intricate dresses for ballet folklorico dancers. Through memory, research, and contemporary observation of ballet folklorico dancers, Sanchez has developed a body of work that is cohesive, inspiring, and deeply personal. In August of 2020, Sanchez created a 20-foot long painting, Folklorico de Guerrero during his time as the artist-in-residence at Sugarlift and High Line Nine. The painting depicts three ballet folklorico dancers, dressed in the typical style of dress from the Guerrero region. Ballet Nepantla aided Sanchez in the development of the painting, posing for the artist, and performing at the unveiling of the piece.
Ballet Nepantla at Art on Paper 2021
To highlight Sugarlift's booth at Art on Paper presenting Nicolas V. Sanchez and to provide an engaging performing arts experience for fair attendees, Ballet Nepantla will perform a traditional Mexican folklorico dance. The performances will take place on opening night, Thursday 9/9 at 8pm, and Saturday on 9/11 at 2pm. The visual representation of Sanchez's ballet folklorico dancers in watercolor, charcoal and oil on paper, coupled with the manifestation of the art form in real time, will energize the audience and make for an unforgettable experience for fair-goers.
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Art Market / San Francisco Art on Paper / New York Seattle Art Fair / Seattle Art Market / Hamptons


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Q: Access wrapped component function from HOC Imagine a HOC for touch behavior that detects key touch events through onTouchStart. Now the wrapped components using the touch HOC, will have its own handler logic. How can my HOC component, call the wrapped components handler functions?
Just to get an idea, here's what I need. On research I found this that uses ref to call the target component. I was wondering if there are better ways to accomplish this.
TouchBehavior HOC
const TouchBehavior = (WrappedComponent) =>

    class extends React.Component {
        constructor(props) {
            super(props);
            this.state = {
                active: false,
            };
            this.handleAction = this.handleAction.bind(this);
            this.setActiveState = this.setActiveState.bind(this);
            this.resetActiveState = this.resetActiveState.bind(this);
        }

        handleAction(e) {
            e.preventDefault();
            this.setActiveState();
            //call wrapped component specific handler logic here
            _.delay(() => this.resetActiveState() , 150);
        }

        setActiveState() {
            this.setState ({
                active: true
            });
        }

        resetActiveState() {
            this.setState ({
                active: false
            });
        }

        render() {
            let touchBehaviorProps = {
                onTouchStart: this.handleAction
            };
            return <WrappedComponent {...this.props} touchBehaviorProps={touchBehaviorProps} />
        }
    };

Wrapped component
class Button extends React.Component {

 componentSpecificHandlerLogic(){
  //logic goes here
 }

 render () {
  return <button {..this.props.touchBehaviorProps}></button>
 }
}

A: Whenever possible, I would try not to pull logic back up the tree, but rather pass more down.
There are probably many ways to achieve this, but one option I see in this case is to change the prop provided to the wrapped component to be a function that constructs the handler for them with the provided custom logic.
TouchBehavior HOC
const TouchBehavior = (WrappedComponent) =>

    class extends React.Component {
        constructor(props) {
            super(props);
            this.state = {
                active: false,
            };
            this.handleAction = this.handleAction.bind(this);
            this.setActiveState = this.setActiveState.bind(this);
            this.resetActiveState = this.resetActiveState.bind(this);
        }

        handleAction(e, customHandler) {
            e.preventDefault();
            this.setActiveState();
            customHandler();
            _.delay(() => this.resetActiveState() , 150);
        }

        setActiveState() {
            this.setState ({
                active: true
            });
        }

        resetActiveState() {
            this.setState ({
                active: false
            });
        }

        render() {
            let touchBehaviorProps = (customHandler) => ({
                onTouchStart: (e) => this.handleAction(e, customHandler)
            });
            return <WrappedComponent {...this.props} touchBehaviorProps={touchBehaviorProps} />
        }
    };

Wrapped Component
class Button extends React.Component {

 componentSpecificHandlerLogic(){
  //logic goes here
 }

 render () {
  return <button {...this.props.touchBehaviorProps(componentSpecificHandlerLogic)}></button>
 }
}


The Rimpitch C2 from KORG is an evolution of KORG's innovative acoustic guitar tuner, now in an even more compact package.
The Rimpitch C2 has been designed specifically for the acoustic guitar, fitting perfectly in the instrument's sound hole. This unique construction allows for high visibility and limited distraction as it's placed within the player's natural field of vision. In addition, the Rimpitch C2 comes equipped with a highly visible LED display, making the player's tuning experience easier and more convenient than ever.


Scientists at the Nuffield Department of Medicine have made a breakthrough in developing a new vaccine against malaria that could save hundreds of thousands of lives every year. First clinical trials of a new vaccine show it is nearly 70% effective. Dr Katie Ewer and Carly Bliss from the Jenner Institute explain their work in an interview on ITV. The BBC has also covered malaria vaccine research at the Jenner Institute.


text:
Guithelin ou Guithelinus est un roi est un roi légendaire de l'île de Bretagne (actuelle Grande-Bretagne), dont l'« histoire » est rapportée par Geoffroy de Monmouth dans son Historia regum Britanniae (vers 1135).

Contexte
Guitelin est présenté par Geoffrey of Monmouth comme le successeur de Gurguint Barbtruc (en gallois: Gwrgan Farfdrwch]. Il aurait traité ses sujets avec douceur et affection. Rien d'autre n'est mentionné sur son règne. Son épouse se nommait  Marcia, qui lui donne un fils nommé Sisillius II [Seisyll]. À la mort de Guithelinus, Sisillius II n'étant âgé que de sept ans, Marcia assume la régence. Les plus anciens textes ne précisent par le nom de son père, mais les Chroniques qui reprennent  Historia regum Britanniae  comme celles de Layamon, Matthew Paris, Pierre de Langtoft, Robert Manning, Ranulf Higdon, à l'exception de celle de Robert Wace, font de lui le fils de son prédécesseur. Le Brut y Brenhinedd le nomme Cuhelyn ap Gwrgan Farfdrwch, et de la même manière les anciennes généalogies, toutefois Cuhelyn n'est pas l'équivalent Guithelinus.

Notes et références

Sources

 Geoffroy de Monmouth, Histoire des rois de Bretagne, traduit et commenté par Laurence Mathey-Maille, Les Belles lettres, coll. « La Roue à livres », Paris, 2004, .

Articles connexes

 Historia regum Britanniae
 Royaume de l'île de Bretagne
 Liste des rois légendaires de l'île de Bretagne

Roi légendaire de l'île de Bretagne


Either a pin, hair clip, hair band, etc.
First cut out 4 circles of your chose of fabric. The bigger the size, the larger your flower will be.
Place a small amount of glue into the center of the fabric circles.
Next, fold the fabric circle in half and add another drop of glue.
Then fold the fabric in half again. Place another drop of glue on the middle of the fold.
Place another folded fabric circle on top of the glue and repeat to make a complete circle.
Next, cut out a circle of felt that is smaller than your flower.
Apply glue to the entire back of the felt and attach to the back of the flower.
Pick out a matching button and glue to the center.
Now you get to decide if you want to make it a pin, hair clip, headband, necklace, whatever.


A place to discover music by new or overlooked artists. Not a place to bury music you don't like. Keep an open mind, try something new, and you will discover something great.
All genres are welcome, but bots will remove mainstream music submissions and frequently reposted artists. All self-promotion and personal association go in the melting pot!
Check the listening apps for tools to listen outside of reddit. Listentothis is best experienced with RES installed and styles enabled.
For music discussion, visit /r/letstalkmusic. The multireddits linked below index more than 600 active music subreddits. No matter what your taste, there's something in there you will love.
Self-promo goes in the melting pot!
It is so free as lonesome, beacuse I can decide anything. I have nostalgia for inconvenience caused by being with someone together.
In the city of night where many lights are together I used to be tearing at somewhere, it blinked, It's translucent days.
I was stranger there, Nothing left but to go home and sleep. It is same about a future I hope or a future you won't hope.
It is so free as lonesome, this feeling with transparency. I was waiting in sun shower, but that guys were in sunlight.
Very nice. I do love the Japanese language in music.
Great song, anyone able to translate?
Doesn't have that jumpy nerdy spastic disco feel.
Ignorance at it's finest. It is quite enjoyable observing subpar beings that cannot see beyond the surface.


text:
Q: prove or disprove for $|x|<1$, $n\in \mathbb{N}$ $\left| \frac{x^n}{1-x} \right | \leq \frac{1}{n} $ prove or disprove for $|x|<1$, $n\in \mathbb{N}$
             $$\left| \frac{x^n}{1-x} \right | \leq  \frac{1}{n} $$

I want it to be true but can't get it here is my attempt
let $|x|=\left |\frac{1}{1+t} \right|$  and using Binomial theorem
$$\begin{aligned} (1+t)^n
   &= \sum^{n}_{k=0} \binom{n}{k} 1^k t^{n-k}
\\  &= \binom{n}{0} 1^0 t^{n-0}  + \binom{n}{1} 1^1 t^{n-1}
        +\dots +\binom{n}{n-1} 1^k t^{n-(n-1)}+\binom{n}{n} 1^k t^{n-n}
\\&=1t^n + \frac{n!}{1!(n-1)! }
t^{n-1} +\dots + \frac{n!}{(n-1)!1!} t^1+1
\\ &=t^n+nt^{n-1}+\dots +nt+1
\end{aligned}$$
so
$$ \begin{aligned}
\left | \frac{x^n}{1-x} \right |
&=\left | \left (\frac{1}{1+t}\right )^n \left ( \frac{1}{1-\frac{1}{1-t}} \right)\right |
\\&= \left | \frac{1}{\left (1+t \right)^n} \left ( \frac{1}{1-\frac{1}{1-t}} \right)\right |
\\&= \left | \frac{1}{t^n+nt^{n-1}+\dots +nt+1} \left ( \frac{1}{1-\frac{1}{1-t}} \right)\right |
\\& \leq  \left | \frac{1}{nt+1} \left ( \frac{1}{\frac{1}{1-t}} \right)\right |=\frac{1+t}{nt+1}=\frac{1+t}{n(t+1/n)}\leq \frac{1+t}{nt}
\\&=\frac{1/t+1}{n}
\end{aligned}
$$
but on the other hand as $x \to 1$ or $t \to 0$then it might not hold
this is related to previous problem I asked because if that holds might be able to say that it converges uniformly because it dosent depend on the value of x.


text:
2007 Reds
Kyle Lohse traded to the Phillies
Chad Dotson
Here's the press release. More analysis later:
Cincinnati Reds executive vice president and general manager Wayne Krivsky today announced the acquisition of LHP Matt Maloney from the Philadelphia Phillies in exchange for RHP Kyle Lohse.
Maloney, 23, this season at Philadelphia's Class AA Reading affiliate went 9-7 with a 3.94 ERA and 1 complete game in 21 starts. He allowed just 117 hits while striking out 115 batters in 125.2 innings. He leads the Eastern League in innings pitched and ranks fourth in strikeouts.
The Phillies' third-round draft pick in 2005, Maloney entered this season rated by Baseball America as the ninth-best prospect in the Phillies' organization. In 2006, in just his second professional season, he was named South Atlantic League Pitcher of the Year after going 16-9 with a 2.03 ERA and 180 strikeouts in 27 starts and 168.2 innings pitched for Lakewood. He led the SAL in victories, innings pitched and strikeouts.
Maloney will report to Class AA Chattanooga.
Lohse, 28, this season went 6-12 with a 4.58 ERA in 21 starts. In his 21 appearances, the Reds went 8-13.
Should Adam Dunn be traded?
DOTF – July 29
Blame Chad for creating this mess. Chad launched Redleg Nation in February 2005, and has been writing about the Reds ever since. His first book, "The Big 50: The Men and Moments That Made the Cincinnati Reds" is now available in bookstores and online, at Amazon, Barnes & Noble, and wherever fine books are sold. You can also find Chad's musings about the Cincinnati Reds in the pages of Cincinnati Magazine. You can email Chad at chaddotson@redlegnation.com.


Welcome to our new real estate blog! Here, we'll share tips for buyers and sellers alike, in addition to market news and updates. Ultimately, we want be your local real estate resource. If you're curious about a topic, reach out to us. Watch this video for more details!