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You are here: Home1 / Expertise2 / Employment & Pension3 / Employment law4 / Termination of employment contract during illness
Termination of employment contract during illness
Dutch employment law protects the sick employee to a high degree. GMW lawyers set out below how far this protection reaches.
Continued payment of wages during illness
The employer is obliged to continue to pay (part of) the salary of the sick employee for two years. During that period, the employer cannot terminate the employment contract: a prohibition of termination applies. The prohibition of termination during illness also applies if the employee is partly able to carry out the work. Only when he is able again to carry out his work in full is the prohibition no longer applicable. Setting aside the employment contract of a sick employee is however often possible.
When does a prohibition of termination during illness apply?
The prohibition of termination during illness applies in principle during the first two years the employee is incapacitated for work. If periods of incapacity follow each other with intervals of less than four weeks, these are added together to determine the total term of the illness.
The period of two years can be extended. This is the case if the UWV extends the obligation to continue to pay wages. This could occur if the employer has not complied with the reintegration obligations.
Dismissal for commercial reasons not possible
The consequences of the prohibition of termination during illness are greatest in the event of commercial dismissal: commercial dismissal via UWV (termination) or the sub-district court (setting aside) is not possible. If it was the sick employee who was eligible for being made redundant on the basis of the proportionality principle, the employer must nominate the next employee on the list for redundancy. The sick employee is protected. The exception is dismissal for commercial reason due to general business cessation. And this prohibition does not apply if the employee has become incapacitated for work after the employer has submitted an application for dismissal to the UWV.
Is it nevertheless possible to terminate the employment contract?
The employment contract with a sick employee can under certain conditions still be terminated. Firstly, this is possible if the employer and employee reach agreement on the termination of the employment contract. This could however have consequences for the sick employee relating to benefits and the employer can on the termination of the employment contract with a sick employee be faced with premium consequences.
In addition, there are a number of exceptions to the prohibition of termination during illness. The prohibition does not apply if the reason for the termination is not the illness and one of the following situations occur:
Termination during the probationary period;
There is an urgent reason (instant dismissal);
The employee has reached the state pension age.
Setting aside the employment contract
Finally, the employer can apply to the sub-district court to set aside the employment contract with the sick employee. The employer can request the sub-district court to set aside the employment contract with a sick employee for a different reason than illness and – therefore – for other reasons than commercial reasons. This could include setting aside due to a damaged working relationship with a sick employee, or due to unsatisfactory performance – provided the application for dismissal does not relate to the incapacity for work of the employee. The prohibition of termination does also not apply, and the employer can apply for a setting aside of the employment contract, if a sick employee refuses to cooperate in his reintegration.
Do you have questions about this topic or do you wish to discuss a case? Please contact our employment law team for assistance.
Amber Willemsen
Godelijn Boonman
Prof. Hans van Meerten
Roos van Zaltbommel
Seliz Demirci
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redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Apollo Client 3.0 Cache with Namespaced Queries I have some questions about how to fit apollo client cache together with a specific way we are organizing our queries.
So the way we organize our queries is that we group queries on the same resource under a single namespace. E.g. we have two resources, users and accounts, and our queries look like this.
type Query {
  user: UserQueries
  account: AccountQueries
}

type UserQueries {
  all: [User!]!
  byId(id: ID!): User
}

type AccountQueries {
  all: [Account!]!
  byId(id: ID!): Account
}

type User {...}
type Account {...}

Since neither Queries type provides an ID field, by default the response from a second query under UserQueries will replace the data from the first query in the cache, i.e. first I make an all query, and I'll get back some data that gets stored in the cache under the user field. And then if I make a byId query, the data in the cache under user will get replaced by the new data.
Some questions about this process:

*

*I noticed that even after the all data is replaced by the byId data, the previously fetched users (the normalized ones) are still available in the cache, when I thought they would be garbage collected since the references are lost. Does the garbage collector not garbage collect them immediately, but rather run on a certain cadence? Or am I completely misunderstanding how gc works?

*Is there another way, other than generating a unique ID for the query typesuser, to preserve both responses?

*Is it even a good idea to namespace queries like this?

Thanks folks!

A: As you have experienced, the results will be overwritten by a different query because the cache does not know that the query technically always returns the same object.
There are two solutions to this:
You have already identified the first one: Give these objects a static ID field. This is pretty ugly. The second one is to write merge functions for these query fields as discussed in the docs. I don't think namespaces are a bad idea in general, but as you can see, you would have to write this merge function for basically any entity type in your schema. Therefore, I would suggest to use namespaces sparingly.
The garbage collector also got some updates in Apollo 3. It can now be customised and run at any time. You can read more about the GC here. Objects that are no longer referenced might stay in the cache for some time. The docs don't seem to specify when the GC runs. It can be manually run by calling client.gc().

A: Actually, the right way to use namespaced queries with Apollo 3 is setting keyFields as an empty array.
When you do that, you show to Apollo that there is no id for merging, so everything should be merged in the same cache key.
You don't have to write merge functions for this.
It is configured through typePolicies option at InMemoryCache configuration
const apolloClient = new ApolloClient({
  cache: new InMemoryCache({
    typePolicies: {
      UserQueries: {
        keyFields: [],
      },
      AccountQueries: {
        keyFields: [],
      },
    },
  }),
});

redpajama_set_name: RedPajamaStackExchange


text:
Every Friday we look at the latest sportswear trends from around the world.
SPORTS DRIPS: Diego Maradona and his iconic No 10 shirts
By Partnered Content Nov 27, 2020
SO its Black Friday and while the South Africans are searching for bargains we will be taking a trip down memory lane and celebrating the life of the best footballers ever - in my opinion - Diego Maradona.
Maradona made so many people from around the world fall in love with the beautiful game. Lets look at the top 10 best football jerseys Maradona has worn in his amazing career with Argentina and clubs around the World from Boca Juniors to Napoli.
Diego has given me so many happy memories as a kid growing up, he was the reason I love sports so much. My first live viewing of a Fifa World Cup was Italia '90.
All the kids in our street were supporting Argentina as they were the champions from 1986 and mainly because they had the genius Diego Maradona.
Argentina lost that World Cup in the final 1-0 to the Germans and boy did I cry when Diego burst into tears at the end of that game.
That feeling right there made me fall in love with football, he was so passionate and he had amazing qualities as a footballer. He could dribble, score goals ad would put his life on the line for his country.
In the 90's every one wanted Puma football boots as this was what Diego wore – The Puma King, Black leather with the white Puma logo and it was a 6 studded boot.
Lets look at some of the Iconic Diego Maradona football jerseys worn in his career. He wore the famous No 10 in all these teams:
The 1986 Argentina FIFA World Cup Winning jersey is by far the most loved by fans around the world made by LaCoq Sportif the Powder Blue and white Argentina Jersey was the last time they won a World Cup.
Argentinian forward Diego Armando Maradona runs past English defenders Terry Butcher and Terry Fenwick before scoring in the 1986 World Cup quarter-final. Picture: AFP
The 1986 Argentina Away Jersey – Navy with a silver number printed at the back:
will always be remembered with the famous "HAND OF GOD" goal and 4 minutes later in that same game he score the best goal ever in a World Cup by taking on 7 English players and slotting it in behind Peter Shelton in England's goal.
1982 Boca Juniors Adidas blue and yellow jersey:
Boca refused to let Maradona leave Argentina but due to the huge pay check offer from Barcelona they had to release him.
1983 Barcelona Red and blue striped iconic kit:
When Diego signed for Barcelona they immediately became fan favorites across the world, this sparked the start of the Barca becoming one of the best clubs in the world.
Argentina soccer superstar Diego Maradona will always be remembered as a son of Napoli. Picture: AFP
1984-85 Napoli in the powder blue jersey:
This where Diego spent the peak of his career, he took Napoli from a relegation battling team to a Uefa Cup winning team.
The 1990 adidas Argentina jersey:
This would be the 1st year Argentina uses adidas in the World cup. They lost the final to the Germans.
Diego Maradona at the 1994 Fifa World Cup. Picture: Timothy A. Clary/AFP
The 1994 Adidas Argentina World Cup jersey:
It would be the last Argentina Jersey Diego worn as a player. In 2010 he was in SA as the head coach of the Nationall team.
Boca Juniors 1997 Nike jersey:
The last club jersey Maradona wore.
Diego Maradona during his last stint with Boca Juniors in 1997. Picture: AFP
Diego Maradona has given football so many memories and was a true legend on the field, but lived his life like a rock star and will be sorely missed for his passion for the beautiful game.
REST IN PEACE – DIEGO ARMANDO MARADONA
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By Sundesh Mahes


text:
By Ulrich
German Secret Societies Colonized Mars In 1940s
Thursday, December 3, 2015 10:49
In his latest interview on Gaiam TV's Cosmic Disclosure, whistleblower Corey Goode describes in detail the attempts by a secret German space program to establish bases on Mars during the Second World War era. He claims that during the subsequent colonizing effort, the Germans fought fierce territorial battles with resident Martian species that inhabited large lava tubes that ran throughout Mars subterranean regions.
Goode says that during the early 1940's, while the Second World War was still being fought, German secret societies –Thule, Vril and Black Sun – had successfully established a base on the moon, and used that as a launching pad for missions to Mars. Travel time between the Moon and Mars was several hours for the Germans. Goode contrasted this with the contemporary secret space program which has cut this travel time down to a matter of minutes!
Goode describes information about the early German colonization events acquired from "smart glass pads" he had access to during his space service as an "intuitive empath" from 1987 to 2007. He says that German documents and photographs, detailing these Moon and Mars missions, were archived on the smart glass pads.
Goode previously described the power of the German secret societies in Hitler's Germany. The Thule, Vril and Black Sun societies had created a space program that was wholly independent of the Nazi SS effort to weaponize flying saucers for the war effort.
These German secret societies had established bases in Antarctica and South America that escaped the bulk of the fighting during the Second World War. Well-hidden and protected in these remote locations, the Germans had forged ahead in developing advanced space travel technologies using antigravity and space-time portals.
German secret society efforts in establishing a parallel space program was previously described by Goode in a May 20, 2015 email interview, earlier episodes of Cosmic Disclosure, and discussed in detail in the book, Insiders Reveal Secret Space Programs (2015).
In the November 10, 2015, episode of Cosmic Disclosure, Goode provides more information than earlier revealed about the challenges the early German colonists faced in establishing their bases on Mars and the Moon.
After establishing a presence in Mar's equatorial regions, he describes how the Germans encountered harsh sand storms that were highly electrically charged. The lightning storms wreaked havoc with the electronics of their technologies. Goode's testimony is supported by NASA research that has confirmed that tremendous lightning bursts do occur during Mars dust storms.
Consequently, the Germans began searching elsewhere on Mars for more suitable areas for their bases. They found that the Northern and Southern regions, above the 20th parallel, offered a more hospitable environment for building bases, well away from the huge electrical Martian dust storms. The Germans discovered an ancient lava tube system about 10 times the size of lave tubes on Earth, and relocated their bases here.
Mars lava tubes and other subterranean regions were not uninhabited though. The Germans encountered two intelligent resident species that they had to wage fierce territorial battles against while establishing their bases. Goode described the two species as Reptilian and Insectoid. These brutal battles led to the loss of many German personnel and resources up until the late 1950s, when the U.S. military industrial complex began helping the German secret societies on Mars.
In mid-1952, the Germans had conducted a series of flying saucer flights over Washington D.C. in a show of force. This led to the beginning of official relations with the German secret societies. During the Eisenhower administration, the U.S. capitulated to the Germans key demands, which led to the infiltration of the military-industrial complex.
With the industrial might of the U.S. now behind them, the German secret societies were able to finally consolidate their Mars bases and protect them from the resident Martians.
Is there any evidence supporting Goode's incredible claims? First, we can look to other whistleblowers who also claim to have spent time on Mars while serving with secret space programs.
Goode's testimony has a number of important parallels with the testimony of an earlier Mars whistleblower Randy Cramer, who says that he spent 17 years on Mars (from 1987 to 2004) in an elite military unit that defended civilian colonies from resident Martians.
In an April 2014, series of interviews (six months prior to Goode's public emergence in September), Cramer said that his unit regularly fought territorial battles with two intelligent Mars species, Reptilians and Insectoids – the same two identified by Goode.
Cramer described the resident Martians as fiercely territorial who would go to battle if their territory was threatened. At other times, they would leave the human colonists alone. Again, this is consistent with Goode's claims regarding German battles with the Martians.
While Cramer did not mention any German connection, he made it clear that the military unit he served with was subservient to a corporate entity that was in charge of Mars operations. This is consistent with Goode's claims about an "Interplanetary Corporate Conglomerate" that runs Mars, which had been earlier infiltrated and taken over by German secret societies and their Nazi SS supporters.
The SS had introduced slave labor practices that were common in Nazi Germany and at the German secret society bases in Antarctica and South America. According to Albert Speer, the Armaments Minister for Nazi Germany. the Nazi SS had plans to continue these slave practices well after the war.
Cramer's and Goode's testimonies about secret space program battles with indigenous Martians are consistent with a third whistleblower, Michael Relfe, who claims that he spent 20 years on Mars (from 1976 to 1996). Relfe describes fighting a variety of native and extraterrestrial threats to the secret space program running Mars operations. Goode's, Cramer's and Relfe's testimonies are compared and contrasted in chapter 13 of Insiders Reveal Secret Space Programs.
In addition to the testimonial evidence by Cramer and Relfe, which is consistent with Goode's claims, there is also important circumstantial evidence. From June 11 to 12, the British Interplanetary Society conducted a symposium on the theme, "How to Overthrow a Martian Dictatorship."
The scenario depicted by the British Interplanetary Society was of a corporate run space program on Mars that was dominated by leaders who behaved as dictators who exploited workers as slave labor. The scenario has many similarities to the ones described by Goode and Cramer about Mars being run by corporations that use elite military forces as space mercenaries to fight their battles.
The symposium was covered by Richard Hollingham, a writer with the British Broadcasting Corporation, who alluded to a MI-6 connection. The symposium coincided with a June 22 report by Goode regarding his inspection tour of a current Mars base that appeared to be run as a dictatorship. It is more than likely that the corporate practices on Mars bases today are connected to their antecedent, the early German bases that continued the Nazi practice of using slave labor.
These Cosmic Disclosure episodes with Goode blowing the whistle on key historical events withheld from the general public, if true, must change our world view to deal responsibly with the future.
Two other alleged secret space program whistleblowers support Goode's claims, along with compelling circumstantial evidence, which includes NASA data supporting his claim of a thin breathable atmosphere on Mars. For now, there is good reason to conclude that Goode is revealing a true insider's perspective on the role German secret societies played in establishing bases and colonizing Mars in the 1940s.
Total 21 comments
And what year did Americans get to Mars, probably about '62 right?
Dec 3, 2015, 1:23 pm 0 Reply
doggy do
Yeah that's why they're taking so long to get back. They forgot all the and the germans knew because of the Niburian x-rays aimed at them, and are now retracing their steps. Man, thought everyone knew that
Go to page 9, this guy with Hitler looks just like Prince Charles, that girl looking for help on Christmas eve posted
http://flatearthtruth.freeforums.net/thread/116/apple-aether
Dec 4, 2015, 3:34 am 0 Reply
They even had to fake the moon landing.
Dec 13, 2015, 11:48 pm 0 Reply
Jen Baje
I find this extremely hard to fathom. For instance, if we/they are soo far along in the science that would allow them to get to, live, and then continue to other planets…..why can't they get the F-35 to work properly? Additionally, Russia seems to have mastered a version of electronic blocking that is more star wars than not. So, if we have truly advanced to star travel….why would the science not be applied to earth projects. I just think this is all BullCrappy, wishful thinking.
Well the F-35 issue is a criminal shill conspiracy (shell game.) They've kept it in a perpetual money flow/developmental quagmire so the military industrial complex and its shell corporations can keep skimming profits from its "development" indefinitely. They've secretly made billions of dollars from badly developing the F-35 over the decades.
And Russia's advanced technology is a recent development that has happened only in the past five years. The technology is extraterrestrial in origin but developed into benevolently oriented weaponry by the Russians.
And it is being applied to Earth projects but the going has been slow due to the fact that there are regressive pockets of power still lingering on Earth who have very large vested interests in stopping that from happening any way they can. The Keshe technology is one exapmle of the technologies being slowly released to the public (nano-plasma tech with applications in energy and medical devices.)
JimmyBitch
xenoverse might have a point.
The day before 911, Rumsfeld was on the TV telling us how the Pentagon 'lost' trillions of dollars.
The money went somewhere.
wow, a great sci-fi novel there.
Busta Myth
Sounds more like a cross between Project Paperclip MK ULTRA and the Hollyweired Movie
Starship Poopers
BungalowBud
Why would they even think about Mars when they had a world war on earth to win. If they had rockets to Mrs that good, then they would use them in the we first. How about a little closer like the colonies under the South Pole with his escape is more realistic. I think that Staargate Gate to open that bottomless pit also failed. Do you now know where the bottomless pit really is? It all falls into place now.
WanderingLibra
I think that Randy Cramers story is more believable.
Judge Roy Bean
LOL…..crazy stuff!
Dec 3, 2015, 10:57 pm 0 Reply
LucidMebin
I just can't stop laughing. This is just so bananas! Seriously – I know that people report mostly fake news on this site, but wow.
Dec 4, 2015, 10:14 am 0 Reply
Before the 30′s, the Germans colonized the sun with the help of the lizard aliens. There are now McDonalds, Wal-Marts, and even FEMA camps located on the sun now. I'm surprised that Brother Hodges hasn't written about all of the nefarious activities that are happening on the sun now. When the Beatles wrote "Here Comes The Sun", that was Illuminati code to get the colonization of the sun kicked off.
God Bless All German Astronauts.
Gee, I wonder why the Germans lost WW2 when they had all that technology with which to easily defeat the primitive Allied Forces?
Because it was still in early primitive stages of development. The guns on their UFOs couldn't hit anything because they couldn't reconcile the non-stabilization in mounting traditional barreled machine guns onto a floating disc.
And because many of their forces were devastated when they made bad decisions in trying to colonize Mars. And more still, most of them were totally annihilated after becoming allies with the Draco reptilians then betraying them later. They only had so many forces from the small German country.
And it takes more than technology to win a war. The Germans were outmaneuvered by the Americans and the Russians. Their technology wouldn't have saved them anyway.
mayor dickcheese
Wow, the powers of deduction. Maybe the real point here is that they didn't lose WW2. Maybe history was written this way to make the world believe that everything went our way in the end. I'm not so sure anymore.
Is asking questions a crime?
http://vho.org/Intro/GB/Flyer.html
Bad-Clown
Does ya ever wonders why some people will believes anything?
Well…, I'll tells ya if ya has to know.
It's 'cause some peoples is 100% mental!
https://www.youtube.com/watch?v=65QrqQRmc2g
Synickel
Sure, the Germans were technologically advanced enough to get to Mars, build bases and defeat martians, but they couldn't win WWII.
God's Word - 2 Chronicles 17
Situation Update: SHTF Is Here! World Wake-up Event! Marines & Corrupt FBI Clash At Camp Pendleton! Another Police Killing Video Activates BLM/ANTIFA For Violence! Several Cities Guard Activated! Nurse Warns Of Another Fake COVID Crisis! Global Lockdowns Coming! - We The People News
Trump update 1/28/2023..Boy, am I hitting the ground running
Yeadon, Eugyppius makes sense of Veritas: Links 1, January 28th, (Freedom day eve) 2023
Dr Reiner Fuellmich - Worldwide Unmasking Of The United Nations
Putin is One of Them. Putin is Helping to Bring About the New World Order One World Government Dictatorship
Greg Hunter & Steve Quayle: Aliens Created Humans Is The Biggest Lie In The Universe!! - Must Video
BREAKING (1/28/2023): NOT DEAD — Buffalo Bills Damar Hamlin Speaks Publicly for the First Time
Multiple Explosions in Iran .. Al Arabiya says Israel launched a special military operation inside Iran
Jim Caviezel - Talks Adrenochrome Live To ~5ooK People Pushback Film Lin Wood Brings Down House!
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White Hat Officer Murdered?? (Video)
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Michael Jaco: Massive Intel on the Battle Against WEF, Gates, & Big Pharma (Video)
SG Anon: Gives Everyone Three Things We Can All Do to Help Expedite Heaven on Earth Gesara, RV, XRP (Video)
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THE RUSSIAN PORTAL OPENED UP AND SOMETHING CAME THROUGH...ORIONTV...BOOOM!!!/Incredible UFOs - Former Member of the USAF Unveils Extraterrestrial Prison Hidden by the Elites - This Man Asked Extraterrestrial What's the Purpose of Existence and the Reply He Got - Yea Russia Shot Down a UFO – 2023 UAP Sightings: Lusaka, Zambia, South Africa, Near Ovalle, Chile - Best UFO Videos Just In! Shocking UFO Eyewitness Comes Forward!


text:
\section{Introduction}
\label{sec:Intro}
We study the dynamics along  the particle trajectories for the 3D axisymmetric Euler equations.
Such  Lagrangian dynamics have already been studied in mathematics (see \cite{C0, C1,C2}). For example, in \cite{C1}, Chae considered a blow-up problem for the axisymmetric 3D incompressible Euler equations with swirl.
More precisely, he showed that under some assumption of local minima for the pressure on the axis of symmetry
with respect to the radial variations along some particle trajectory, the solution blows up in finite time.

Although the blowup problem of 3D Euler equations is still an outstanding open problem, in this paper, we focus on a different problem in physics,
especially,  the cardiovascular system \cite{FQV}.
If the blood flow is in large and medium sized vessels, the flow is governed by the usual incompressible Navier-Stokes equations.
In this study field, Womersley number is the key. The Wormersley number
comes from oscillating (in time) solutions to the  Navier-Stokes equations in a tube.
Let us explain more precisely.
We define a pipe $\Omega_{\mathcal R}$ as  $\Omega_{\mathcal R}:=\{x\in\mathbb{R}^3: \sqrt{x_1^2+x_2^2}<\mathcal R,\ 0<x_3<\ell
\}$
with its side-boundary  $\partial\Omega_{\mathcal R}=\{x\in\mathbb{R}^3: \sqrt{x_1^2+x_2^2}=\mathcal R,\ 0<x_3<\ell
\}$.
The incompressible Navier-Stokes equations are described as follows:
\begin{equation}
\label{NS eq.}
\partial_tu+(u\cdot \nabla)u-\nu\Delta u=-\nabla p,\quad\nabla \cdot u=0\quad \text{in}\quad \Omega,
\quad u=0 \quad \text{on}\quad \partial\Omega_{\mathcal R}
\end{equation}
with $u=u(x,t)=(u_1(x_1,x_2,x_3,t),u_2(x_1,x_2,x_3,t),u_3(x_1,x_2,x_3,t))$ and
 $p=p(x,t)$.

To give the Womersley number, we need to focus on the axisymmetric Navier-Stokes flow without swirl (see \cite{W}).
If $p_1$ and $p_2$ are the pressure at the ends of the pipe $\Omega_{\mathcal R}$,
the
pressure gradient can be expressed as $(p_1-p_2)/\ell$.
If the pressure gradient is time-independent, $(p_1-p_2)/\ell=:p_s$,
then we can find the stationary Navier-Stokes flow (Poiseuille flow):
\begin{equation}\label{p_s}
u_s=(u_1,u_2, u_3)=(0, 0, \frac{p_s}{4\nu \ell}(\mathcal R^2-r^2)),
\end{equation}
where $r=\sqrt{x_1^2+x_2^2}$.
Note that $u_s$ is also a solution to the  linearized Navier-Stokes equations.
Next we  consider the oscillating pressure gradient case,
\begin{equation}\label{n}
\frac{p_1(t)-p_2(t)}{\ell}=p_oe^{i N t}
\end{equation}
which is periodic in the time.
Then its corresponding solution $u_o$  can be written explicitly by using a Bessel function (see \cite[(8)]{W} and \cite[(1)]{TKWP}) with $u_1=u_2=0$. Thus $u_o$ is also a solution to the linearized Navier-Stokes equations.
Now we can give the Womersley number $\alpha$  as follows:
\begin{equation*}
\alpha=\mathcal R\sqrt{\frac{N}{\nu}}.
\end{equation*}
In \cite{TKWP}, they also defined the oscillatory Reynolds number and
the mean Reynolds number by using $u_o$ and $u_s$ respectively,
and they investigated how the transition of pulsatile flow from the laminar to the
turbulent (critical Reynolds number) is affected by the Womersley number and the oscillatory Reynolds number.
According to their experiment,
measurement at different Womersley numbers yield similar transition behavior,
and variation of the oscillatory Reynolds number also appear to have little effect.
Thus they conclude that the transition seems to be determined only by the mean Reynolds number.
However it seems they did not investigate  the effect of  the non-small swirl component (azimuthal component),
 and thus
our aim here is to

 show that the non-small swirl component induces an instability which is, at a glance, nothing to do with wall turbulence.
Let us explain more precisely.
Since we  would not like to  take the boundary layer into account,  it is reasonable to consider
a simpler model: the 3D axisymmetric Euler flow in an infinite cylinder
$\Omega:=\{x\in\mathbb{R}^3: \sqrt{x_1^2+x_2^2}<1,\ x_3\in\mathbb{R}
\}$ (the setting $\Omega$ is just for simplicity).
The incompressible Euler equations are expressed as follows:
\begin{eqnarray}
\label{Euler eq.}
& &\partial_tu+(u\cdot \nabla)u=-\nabla p,\quad\nabla \cdot u=0\quad \text{in}\quad \Omega,\\
\nonumber
& & \quad u|_{t=0}=u_0,\quad u\cdot n=0 \quad \text{on}\quad \partial\Omega,\quad
u(x,t)\to (0,0,g(t))\quad (x_3\to\pm\infty)
\end{eqnarray}
with $u=u(x,t)=(u_1(x_1,x_2,x_3,t),u_2(x_1,x_2,x_3,t),u_3(x_1,x_2,x_3,t))$,
 $p=p(x,t)$ and an uniform (in space) inflow-outflow condition  $g=g(t)$ (the uniform setting is just for simplicity,
we can easily generalize it), where $n$ is a unit normal vector on the boundary.

\begin{remark}
According to the boundary layer theory,
outside the boundary layer the fluid motion is accurately described by the Euler flow.
Thus the above  simplification seems (more or less) valid.
For the recent progress on the mathematical analysis of the boundary layer,
see \cite{MM}.
\end{remark}
Roughly saying, the inflow-outflow $g$ is a simplification of  $u_s+u_o$,
namely,  $u_s$ is approximated by the mean-value of $g$,
$\partial_t u_o$ and $\partial_t^2 u_o$ are  approximated by $g'$ and $g''$ respectively.
Since we consider the axisymmetric Euler flow, we can simplify the Euler equations \eqref{Euler eq.}.
Let $e_r:= x_h/|x_h|$,
$e_\theta:=x_h^\perp/|x_h|$ and
  $e_z=(0,0,1)$ with
 $x_h=(x_1,x_2,0)$, $x_h^\perp=(-x_2,x_1,0)$.
The vector valued function $u$ can be rewritten as   $u=v_re_r+v_\theta e_\theta+v_ze_z$,
where $v_r=v_r(r,z,t)$, $v_\theta=v_\theta(r,z,t)$ and $v_z=v_z(r,z,t)$
with $r=|x_h|$ and $z=x_3$.
Then the   axisymmetric Euler equations can be expressed
as follows:
\begin{eqnarray}
\partial_t v_r+v_r\partial_rv_r+v_z\partial_zv_r-\frac{v_\theta^2}{r}+\partial_r p&=&
0,\\
\label{axisymmetricEuler-1}
\partial_tv_\theta+v_r\partial_rv_\theta+v_z\partial_zv_\theta+\frac{v_rv_\theta}{r}&=&
0,\\
\partial_tv_z+v_r\partial_rv_z+v_z\partial_zv_z+\partial_zp&=&0
,\\
\label{axisymmetricEuler-2}
\frac{\partial_r(rv_r)}{r}+\partial_zv_z&=&0.
\end{eqnarray}

In order to show that  the non-small swirl component induces the instability,
we need to measure appropriately the rate of laminar profile of the Euler flow.
\begin{definition}\label{Stream-shell near the boundary}
(Axis-length  streamline in $z$.)\
For a unilateral flow $v_z=u\cdot e_z>0$ in $\Omega$,
 we can define an axis-length streamline $\gamma(z)$.
Let $t$
 be fixed, and let $\gamma(z)$ be such that
\begin{equation*}
\gamma(\bar r_0,z,t)=\gamma (z):=(\bar R(z)\cos \bar \Theta(z), \bar  R(z)\sin \bar \Theta(z), z)
\end{equation*}
with
$\bar R(z)=\bar R(\bar r_0,z,t)$, $\bar R(\bar r_0,-\infty,t)=\bar r_0$, $\bar \Theta(z)=\bar \Theta(z,t)$
 and we choose $\bar R$ and $\bar \Theta$ in order to satisfy
\begin{equation*}
\partial_z\gamma(z)=\left(\frac{u}{u\cdot e_z}\right)(\gamma(z),t).
\end{equation*}
\end{definition}
We easily see
\begin{equation*}
\partial_z\gamma\cdot e_z=1,\quad \partial_z\gamma\cdot e_r=\partial_z\bar R=\frac{v_r}{v_z}\quad\text{and}\quad
\partial_z \gamma\cdot e_\theta=\bar R\partial_z\bar\Theta=\frac{v_\theta}{v_z}.
\end{equation*}
In this paper we always assume existence of  a unique smooth solution to the Euler equations.
Since $\partial_{\bar r_0}\bar R>0$ due to the smoothness,
we have
its inverse $r_0=\bar R^{-1}(r,z,t)$.
We now give the key definition.
\begin{definition} (Rate of laminar profile.)\
Let $\partial=\partial_{z}$ or $\partial_{\bar r_0}$, and let $\bar \partial=\partial_{z}$ or $\partial_r$. We
define ``rate of laminar profile" $L^x$ and $L^t$ as follows:
\begin{equation*}
L^x(\bar r_0,z,t)
:=
\sum_{\ell=1}^3
|\partial^\ell\bar R(\bar r_0,z,t)|
+
 \sum_{\ell=1}^3|(\bar \partial^\ell\bar R^{-1})(\bar R(\bar r_0,z,t),z,t)|
\end{equation*}
and
\begin{equation*}
L^t(\bar r_0,z,t)=
|(\partial_t\bar R^{-1})(\bar R(\bar r_0,z,t),z,t)|
+|(\partial_t\partial_{\bar r_0}\bar R)(\bar r_0,z,t)|+|(\partial_t\partial_z\bar R)(\bar r_0,z,t)|.
\end{equation*}
Later we deal with the curvature and torsion of the particle trajectory, thus it is natural to see up to three derivatives.
\begin{remark}
As we already assumed that  solutions to the Euler equations are always unique and smooth enough,
thus, when the transition of  the Euler flow from the laminar to the turbulent regime
occurs, then $L^x$ and/or $L^t$ must tend to infinity.
\end{remark}

\begin{remark}
Minumum value of $L^x$ is $2$, since $|\partial_r\bar R^{-1}|=1/|\partial_{\bar r_0}\bar R|$.
\end{remark}

\end{definition}

\begin{remark}\label{no swirl}
We easily see that $u=(0,0,g)$
 is one of the solution to \eqref{Euler eq.}.
This flow is the typical laminar flow.
In this case
\begin{equation*}
L^x\equiv 2
\quad\text{and}\quad
L^t\equiv 0.
\end{equation*}
\end{remark}
Now we give the main theorem.

\begin{theorem}

Assume there is a unique  smooth solution to the Euler equations \eqref{Euler eq.}
with smooth initial data satisfying $|L^x|\leq \beta$ for some positive constant $\beta$ (we will determine $\beta$ later).
For any $x\in \Omega$ satisfying $u_0(x)\cdot e_\theta\approx 1$
and $x\cdot e_r>1/\beta$,
 and any  $\epsilon>0$, then there is  $\delta>0$
such that
\begin{equation}\label{only occur}
L^t(\bar r_0,z,0)
\gtrsim 1/\epsilon
\end{equation}
with any smooth inflow-outflow $g(t)$ satisfying
\begin{equation}\label{flux condition}
1/\beta^5\leq g(0)\leq 1/\epsilon^5\quad\text{and}\quad
\frac{1}{\delta^3}<\frac{g'(0)}{\delta^2}<g''(0),
\end{equation}
where $(\bar r_0, z)=\gamma^{-1}(x,0)$.
\end{theorem}

\begin{remark}
In contrast with \cite{TKWP},
the transition of the pulsatile flow from the laminar to the turbulent may be
affected by the Womersley number and the oscillatory Reynolds number
provided by the non-small swirl component.
\end{remark}

\begin{remark}
This instability mechanism is (at a glance) different from wall turbulence.
It would be interesting to consider the interaction between this instability mechanism
and wall turbulence, and this is our future work.
\end{remark}

In the next section, we prove the main theorem.

\section{Proof of the main theorem.}

Notations ``$\approx$" and ``$\lesssim$" are convenient. The notation ``$a \approx b$" means there is a positive constant $C>0$ such that
\begin{equation*}
C^{-1}a\leq b\leq Ca,
\end{equation*}
and ``$a\lesssim 1$" means that there is a positive constant $C>0$ such that
\begin{equation*}
0\leq a\leq C.
\end{equation*}
This constant $C$ is not  depending on neither $\epsilon$ nor $\delta$.
Throughout this paper we use $C(\beta)$ (different from the above $C$) as a positive constant depending on $\beta$.
Now we define the particle trajectory.
The associated Lagrangian flow $\eta(t)$
is a solution of the initial value problem
\begin{align} \label{eq:flowC}
&\frac{d}{dt}\eta(x,t) = u(\eta(x,t),t),
\\  \label{eq:flow-icC}
&\eta(x,0) = x.
\end{align}
To prove the main theorem, it is enough to show the following lemma:
\begin{lem}
Assume there is a unique  smooth solution to the Euler equations \eqref{Euler eq.}
with smooth initial data satisfying $|L^x|\leq \beta$ for some positive constant $\beta$.
For any $x\in \Omega$ satisfying $u_0(x)\cdot e_\theta\approx 1$ and $x\cdot e_r>1/\beta$,
 and any  $\epsilon>0$, then there is  $\delta>0$
such that
for any small time interval $I$ with $|I|< \beta^2\epsilon^5$,
at least either of the following four cases must happen:
\begin{itemize}

\item
$L^x(\bar r_0,z,t)>1/\beta$,

\item
$L^t(\bar r_0,z,t)\gtrsim
1/\epsilon$,

\item
  $|\eta(x,t)\cdot e_r|<\beta$,

\item

$\bar r_0<\beta$,

\end{itemize}
for some $t\in I$, with any inflow-outflow $g(t)$ satisfying
\begin{equation*}
1/\beta^5\leq g(t)\leq 1/\epsilon^5\quad\text{and}\quad
\frac{1}{\delta^3}<\frac{g'(t)}{\delta^2}<g''(t)\quad\text{in}\quad
t\in I,
\end{equation*}
where $(\bar r_0, z)=(\gamma^{-1}\circ\eta)(x,t)$
(in this case $\bar r_0$ and $z$ are depending on $t$).
\end{lem}
Since the time interval $I$ is arbitrary, we see that $L^x$ or $\eta\cdot e_r$ or $\bar r_0$ is not continuous at the initial time $t=0$,
or $L^t\gtrsim 1/\epsilon$ at the initial time.
The discontinuity contradicts the smoothness assumption,
thus
\begin{equation*}
L^t\gtrsim
1/\epsilon
\end{equation*}
only occurs.
In what follows, we prove the above lemma.
For any time interval $I$ with $|I|\leq \beta^2\epsilon^5$,
assume that the axisymmetric smooth Euler flow satisfies the following both conditions:
\begin{itemize}
\item
$L^x(\bar r_0,z,t)\leq 1/\beta$
and
$
L^t(\bar r_0,z,t)
\lesssim 1/\epsilon$ for
any $t\in I$,

\item

$|\eta(x,t)\cdot e_r|\geq \beta$
and
$\bar r_0\geq \beta$
for any $t\in I$,
\end{itemize}
where
 $(\bar r_0,z)=(\gamma^{-1}\circ\eta)(x,t)$.
First we express $v_z$ and $v_r$ by using $\bar R$ and $\bar R^{-1}$. To do so, we define the cross section of the stream-tube (annulus). Let
$B_{-\infty}(\bar r_0)=\{x\in\mathbb{R}^3: |x_h|<\bar r_0,\ x_3=-\infty\}$ and let
\begin{equation*}
  A(\bar r_0,z,\epsilon,t):=\bigcup_{x\in B_{-\infty}(\bar r_0+\epsilon)\setminus B_{-\infty}(\bar r_0)}
\gamma(x,z,t).
\end{equation*}
We see that  its measure is
\begin{equation*}
|A(\bar r_0,z,\epsilon,t)|=\pi\left(\bar R(\bar r_0,\epsilon,z,t)^2-\bar R(\bar r_0,z,t)^2\right).
\end{equation*}
\begin{definition} (Inflow propagation.)\
Let $\rho$ be such that

\begin{equation*}
\rho(\bar r_0,z,t):=\lim_{\epsilon\to 0}\frac{|A(\bar r_0,-\infty,\epsilon,t)|}{|A(\bar r_0,z,\epsilon,t)|}.
\end{equation*}
\end{definition}
We see that
\begin{equation*}
\rho(\bar r_0,z,t)=\frac{\partial_{\bar r_0}\bar R(\bar r_0,-\infty,t) \bar R(\bar r_0,-\infty,t)}{\partial_{\bar r_0}\bar R(\bar r_0,z,t) \bar R(\bar r_0,z,t)}=\frac{\bar r_0}{\partial_{\bar r_0}\bar R(\bar r_0,z,t) \bar R(\bar r_0,z,t)}
=\frac{2\bar r_0}{\partial_{\bar r_0}\bar R(\bar r_0,z,t)^2}.
\end{equation*}
\begin{remark}
By the assumption on $L^x$,
we have  the
estimates of the inflow propagation $\rho$:
\begin{equation*}
|\partial_z\rho|, |\partial_z^2\rho|,
|\partial_{\bar r_0}\rho|, |\partial_{\bar r_0}^2\rho|
\lesssim C(\beta)
\quad\text{for}\quad t\in I
\end{equation*}
with some positive constant $C(\beta)$.
Note that $\bar R(\bar r_0, z,t)>\beta$ for $(\bar r_0,z)=(\gamma^{-1}\circ \eta)(x,t)$ due to $\eta(x,t)\cdot e_r>\beta$.
\end{remark}

Since
\begin{equation*}
2\pi \int_{\bar R(\bar r_0,z,t)}^{\bar R(\bar r_0+\epsilon,z,t)}u_z(r',z,t)r'dr'=2\pi\int_{\bar r_0}^{\bar r_0+\epsilon}u_z(r',-\infty,t)r'dr'
\end{equation*}
 by divergence-free and Gauss's divergence theorem,
 we can figure out $v_z$ by using the inflow propagation $\rho$,
\begin{eqnarray*}
v_z(r,z,t)&=&\lim_{\epsilon\to 0}\frac{2\pi}{|A(\bar r_0,z,\epsilon,t)|}\int_{\bar R(\bar r_0,z,t)}^{\bar R(\bar r_0+\epsilon,z,t)}v_z(r',z,t)r'dr'\\
&=&
\lim_{\epsilon\to 0}\frac{|A(\bar r_0,-\infty,\epsilon,t)|}{|A(\bar r_0,z,\epsilon,t)|}\frac{2\pi}{|A(\bar r_0,-\infty,\epsilon,t)|}\int_{\bar r_0}^{\bar r_0+\epsilon}v_z(r',-\infty,t)r'dr'\\
&=&
\rho(\bar r_0,z,t)u_z(\bar r_0,-\infty,t).
\end{eqnarray*}
Thus we have the following proposition.

\begin{prop}\label{formula of velocities}
We have the following formula of $v_z$ and $v_r$:
\begin{equation}\label{!}
v_z(r,z,t)=\rho(\bar R^{-1}(r,z,t),z,t)v_z(\bar R^{-1}(r,z,t),-\infty,t)=\rho(\bar R^{-1},z,t)g(t)
\end{equation}
and
\begin{equation}\label{!!}
v_r(r,z,t)=(\partial_z\bar R)(\bar R^{-1}(r,z,t), z,t)v_z(r,z,t).
\end{equation}
\end{prop}
By the above proposition, we see
\begin{equation*}
\beta^{-6}\lesssim|v_z|\lesssim \beta^{-2}\epsilon^{-5}\quad\text{and}\quad |v_r|\lesssim \beta^{-3}\epsilon^{-5}.
\end{equation*}

We now define the Lagrangian flow along $r$,$z$-direction. Let
\begin{eqnarray}\label{2D-trajectory-1}
& &\frac{d}{dt}Z(t)=v_z(R(t),Z(t),t),\\
\nonumber
& &Z(0)=z_0
\end{eqnarray}
and
\begin{eqnarray}\label{2D-trajectory-2}
& &\frac{d}{dt}R(t)=v_r(R(t),Z(t),t),\\
\nonumber
& &R(0)=r_0
\end{eqnarray}
with $Z(t)=Z(r_0,z_0,t)$ and $R(t)=R(r_0,z_0,t)$.
By the second assumption: $|\eta(x,t)\cdot e_r|\geq \beta$, $R$ satisfies the following:
\begin{equation*}
R(t)>\beta
\quad\text{for}\quad t\in I.
\end{equation*}

Since $v_z>0$, then we can define the inverse of $Z$ in $t$: $t=Z_t^{-1}(z,r_0,z_0)$.
In this case we can estimate $\partial_zZ^{-1}_t=1/\partial_tZ=1/v_z
\lesssim \beta^6$ and $\partial_z^2Z^{-1}_t=-\frac{\partial_zv_z}{v_z^2}$.
First we show the following estimates.
\begin{lem}\label{estimates of v}
For $t\in I$,
we have the following estimates along the axis-length trajectory:
\begin{equation}\label{partial_zv_z}
\begin{cases}
\partial_zv_z(R(Z^{-1}_t(z)),z,Z_t^{-1}(z))\approx (\partial_tZ)^{-1}\rho g'+\text{remainder},\\
\partial_z^2Z_t^{-1}\approx- (\partial_tZ)^{-3}\rho g'+\text{remainder},\\
\partial_z^2v_z(R(Z^{-1}_t(z)),z,Z_t^{-1}(z))\approx  (\partial_tZ)^{-2}\rho g''+\text{remainder},\\
|\partial_zv_r(R(Z^{-1}_t(z)),z,Z_t^{-1}(z))|\lesssim  (\partial_tZ)^{-1}\rho g'+\text{remainder},\\
|\partial_z^2v_r(R(Z^{-1}_t(z)),z,Z_t^{-1}(z))|\lesssim  (\partial_tZ)^{-2}\rho g''+\text{remainder},
\end{cases}
\end{equation}
where ``remainder"  is small compare with the corresponding main term provided by small $\delta>0$.
Moreover, we have
\begin{equation}\label{v_theta estimate}
v_\theta(R(Z^{-1}_t(z)),z,Z^{-1}_t(z))\approx 1
\end{equation}
for some $z$ sufficiently close to $z_0$,
\begin{eqnarray}\label{partial_zv_theta}
& &|\partial_zv_\theta(R(Z^{-1}_t(z)),z,Z_t^{-1}(z))|\lesssim C(\beta),\\
\nonumber
& &|\partial_z^2v_\theta(R(Z^{-1}_t(z)),z,Z_t^{-1}(z))|\lesssim (\partial_t Z)^{-3}\rho g'+\text{remainder}
\end{eqnarray}
and
\begin{equation*}
\partial_t|u(\eta(x,t),t)|\lesssim C(\beta)g'+\text{remainder}.
\end{equation*}
\end{lem}

\begin{proof}
The estimates of $v_r$ and $v_z$ with several derivatives are just direct calculation
using the formulas \eqref{!} and \eqref{!!}.
The point is just extracting the main terms composed by $g'$ or $g''$.

Here we  control $v_\theta$ by using \eqref{partial_zv_z}.
By \eqref{axisymmetricEuler-1} we see that
\begin{equation*}
\partial_t v_\theta(R(t),Z(t),t)=v_\theta(r_0,z_0,0)-\frac{v_r(R(t),Z(t),t)v_\theta(R(t),Z(t),t)}{R(t)}.
\end{equation*}
Applying the Gronwall equality,  we see
\begin{equation}\label{v_theta}
v_\theta(R(Z^{-1}_t(z)),z,Z^{-1}_t(z))=v_\theta(r_0,z_0,0)\exp\bigg\{-\int_0^{Z_t^{-1}(z)}
\frac{v_r(R(\tau),Z(\tau),\tau)}{R(\tau)}d\tau\bigg\}.
\end{equation}

\begin{remark}\label{NS difficult}
To obtain the formula of $v_r$ and $v_z$, we are only using the function $\bar R$, $g$
and Gauss's divergence theorem. Namely, we do not need to use  the Euler equations.
However, to obtain the above formula of $v_\theta$, we essentially use the Euler equation \eqref{axisymmetricEuler-1}.
Thus, to pursue the  the Navier-Stokes flow case,
 we need some new idea.
\end{remark}

Since $|v_r|\lesssim \beta^{-3}\epsilon^{-5}$, $R(t)>\beta$ and choose $z$ sufficiently close to $z_0$,
 we have \eqref{v_theta estimate}.
More precisely, we choose a point $Z(t)$ such that
\begin{equation*}
|Z(t)-z_0|\lesssim \beta^{-4}\epsilon^5.
\end{equation*}
In this case, by $Z(t)=z_0+\int_0^tv_z(R(\tau),Z(\tau),\tau)d\tau$ and $|v_z|\gtrsim\beta^{-6}$,
the time interval $I$ always satisfies $|I|\lesssim \beta^2\epsilon^5$.
Just taking derivatives to \eqref{v_theta} in $z$-valuable, then we also have \eqref{partial_zv_theta}.
Now we estimate $\partial_t|u(\eta,t)|$.
We set  the usual trajectory $\eta(x,t)$ using
smooth functions $R$, $Z$ and $\Theta$:
\begin{equation*}
\eta(x,t)
=(R(t)\cos\Theta(t),R(t)\sin\Theta(t),Z(t))
\end{equation*}
with
$e_\theta=(-\sin\Theta(t),\cos\Theta(t),0)$ and
$e_r=(\cos\Theta(t),\sin\Theta(t),0)$. Then, by a direct calculation with $u=v_re_r+v_\theta e_\theta+v_ze_z$,  we see that
\begin{equation}\label{multiply}
\frac{1}{2}\partial_t|u(\eta(x,t),t)|^2=\partial_t u\cdot u
=\partial_t v_rv_r+\partial_tv_\theta v_\theta+\partial_tv_zv_z
\end{equation}
along the trajectory.
In fact, since
\begin{equation*}
\partial_t\eta=\partial_t R(\cos\Theta,\sin\Theta,Z)+\partial_t\Theta(-R\sin\Theta,R\cos\Theta,Z),
\end{equation*}
and
\begin{equation*}
v_\theta=\partial_t\eta\cdot e_\theta=\partial_t\Theta R,
\end{equation*}
we see $\partial_t\Theta=v_\theta/R$.
We multiply $u=v_r e_r+v_\theta e_\theta+v_ze_z$ to
\begin{equation*}
\partial_t u=\partial_tv_r e_r+\partial_tv_\theta e_\theta+\partial_tv_z e_z+
v_r\partial_t\Theta e_\theta-v_\theta\partial_t\Theta e_r,
\end{equation*}
then we have \eqref{multiply}.
Just take a time derivative to $v_z$ along the trajectory, then we have
\begin{equation*}
\partial_t(v_z(R(t),Z(t),t))=
\rho g'+\text{remainder}.\\
\end{equation*}
Here we used the fact that
$
L^t
\lesssim 1/\epsilon$.
Thus
\begin{equation*}
\partial_t v_z v_z=\rho^2g'g+\text{remainder}\quad\text{for}\quad t\in I.
\end{equation*}
The remainder becomes small compare with the main term provided by small $\delta>0$.
By the similar calculation,
\begin{equation*}
\partial_t v_r v_r
=
(\partial_z\bar R)^2\rho^2g'g
+\text{remainder}
\quad\text{for}\quad t\in I.
\end{equation*}
Here we also used the fact that $
L^t
\lesssim 1/\epsilon$.
Clearly
$|\partial_tv_\theta|\lesssim \beta^{-4}\epsilon^{-5}$ and then
 $|\partial_tv_\theta v_\theta|\lesssim \beta^{-4}\epsilon^{-5}$.
Thus we have
\begin{equation*}
\partial_t|u(\eta(x,t),t)|^2\approx (1+(\partial_z\bar R)^2)\rho^2g'g
+\text{remainder}\quad\text{for}\quad t\in I
\end{equation*}
and then
\begin{equation*}
\partial_t|u(\eta(x,t),t)|\approx \rho g'
+\text{remainder}
\quad\text{for}\quad t\in I.
\end{equation*}
\end{proof}

Our strategy  is to estimate the curvature and torsion of the arc-length particle trajectory.
To do so, we need to define the axis-length trajectory $\tilde\eta$ in $z$.
\begin{definition} (Axis-length trajectory.)\
Let $\tilde\eta$ be such that
\begin{equation*}
\tilde \eta(z):=(r(z)\cos\theta(z),r(z)\sin\theta(z),z)
\end{equation*}
and we choose $r(z)$ and $\theta(z)$
in order to satisfy $\tilde \eta(z)=\eta(x,Z^{-1}_t(z))$.
\end{definition}
For $t\in I$, we see
\begin{eqnarray*}
\partial_z\tilde\eta\cdot e_\theta&=&\frac{\partial_t\eta\cdot e_\theta}
{v_z}=r\theta'=\frac{v_\theta(R(Z^{-1}_t(z)),z,Z^{-1}_t(z))}{v_z(R(Z^{-1}_t(z)),z,Z^{-1}_t(z))}
\lesssim \beta^6,\\
\partial_z\tilde\eta\cdot e_r&=&\frac{v_r}{v_z}=(\partial_z\bar R)(\bar R^{-1},Z,t)=r',\quad |r'|\lesssim C(\beta,\epsilon)\quad\text{and}\quad |r''|\lesssim C(\beta,\epsilon)
\end{eqnarray*}
with some positive constant $C(\beta,\epsilon)$ depending on $\beta$ and $\epsilon$.
In particular we need the estimates of $\theta''$ and $\theta'''$.
These estimates specify the curvature and torsion of the particle trajectory.
By Lemma \ref{estimates of v}, we can immediately obtain the following proposition.
\begin{prop}\label{estimates of theta}
By Lemma \ref{estimates of v}
 we have (just see the highest order term)
\begin{equation*}
\theta''(z)= -\frac{v_\theta\partial_zv_z}{v_z^2}+\text{remainder}=  -C(\beta)g'+\text{remainder}
\end{equation*}
 and
\begin{eqnarray*}
\theta'''(z)&=& -\frac{v_\theta\partial_z^2v_z}{rv_z^2}+\frac{2v_\theta(\partial_zv_z)^2}{rv_z^3}+\text{remainder}\\
&=&
 -C(\beta)g''+\text{remainder}\quad \text{for}\quad  t\in I,
\end{eqnarray*}
where ``remainder" is small compare with the corresponding main term provided by small $\delta>0$.
\end{prop}

From  the trajectory $\eta(x,t)$, we define the arc-length trajectory $\eta^*(s)=\eta^*(x,s)$.

\begin{definition} (Arc-length trajectory.)\
Let $\eta^*$ be such that
\begin{equation*}
\eta^*(s)=\eta^*(x,s):=\eta(x,t(s))\quad \text{and}\quad \eta^*(x,0)=\eta(x,0)
\end{equation*}
with  $\partial_st(s)=|u|^{-1}$.
\end{definition}
In this case we see $|\partial_s\eta^*(s)|=1$.
We define the unit tangent vector $\tau$ as
\begin{equation*}
\tau(s)=\partial_s \eta^*(x,s),
\end{equation*}
the unit curvature vector $n$ as $\kappa n=\partial_s \tau$ with a curvature function $\kappa(s)>0$,
the unit torsion vector $b$
as : $b(s):=\pm\tau(s)\times n(s)$ ($\times$ is an exterior product)
 with a torsion function to be positive $T(s)>0$ (once we restrict $T$ to be positive, then the direction of $b$ can be uniquely determined), that is,
\begin{equation*}
Tb:=\partial_sn+\kappa \tau,\quad |b|=1
\end{equation*}
due to the Frenet-Serret formula.
By the estimates of $\theta''$ and $\theta'''$ in  Proposition \ref{estimates of theta},
 we  obtain the following key estimates.
\begin{lem}\label{key estimate on torsion}
For any $\epsilon>0$,
we have
\begin{equation*}
(1/6)|u|^2|\partial_s\kappa|> \kappa\partial_t|u|,
(1/2)|\partial_s\kappa|>  |\kappa Tb\cdot e_\theta|
\quad\text{and}\quad
-1\leq (n\cdot e_\theta)<-1/2
\end{equation*}
for sufficiently small  $\delta>0$.
\end{lem}

\begin{proof}
Recall  the arc-length  trajectory ($z=z(s)$):
\begin{equation*}
\eta^* (x,s)=\tilde\eta(x,z)=(r(z)\cos \theta(z), r(z)\sin \theta(z), z)\quad\text{with}\quad \theta'>0.
\end{equation*}
Thus
$\tau$ and  $\kappa n$ are expressed as
\begin{equation*}
\tau=(\partial_z\tilde\eta)z',\quad \kappa n=\partial_s^2\eta^*=\partial_z^2\tilde\eta(z')^2+\partial_z\tilde\eta z''.
\end{equation*}
Also we immediately have the following formula:
\begin{equation*}
\partial_s^3\eta^*=\partial_s(\kappa n)=\partial_z^3\tilde \eta(z')^3+3\partial_z^2\tilde\eta z'z''+\partial_z\tilde\eta z'''.
\end{equation*}
We recall that
\begin{equation*}
\partial_z\tilde\eta\cdot e_z=1,\quad \partial_z\tilde\eta\cdot e_\theta=r\theta'=\frac{u_\theta}{u_z}
\lesssim \beta^6
\end{equation*}
and then we see that
\begin{equation*}
\theta'\lesssim \beta^5\quad\text{and}\quad
\theta''\approx-C(\beta)g'.
\end{equation*}
Also recall that
\begin{equation*}
\partial_z\tilde\eta\cdot e_r=r'=\frac{u_r}{u_z}=(\partial_z\bar R)(\bar R^{-1},Z,t).
\end{equation*}
Direct calculations yield (just see the highest order term composed by $\theta''$ or $\theta'''$,
and neglect the small terms composed by $\theta'$)
\begin{eqnarray*}
\partial_z\tilde \eta(x, z)&=&
(-r\theta'\sin\theta,r\theta'\cos\theta,1)+(r'\cos\theta,r'\sin\theta, 0)
,\\
\partial_z^2\tilde \eta(x,z)&=&
-r(\theta')^2(\cos\theta,\sin\theta,0)+(-r\theta''\sin\theta, r\theta''\cos\theta,0)
\\
& &
+
r''(\cos\theta,\sin\theta,0)+2r'\theta'(-\sin\theta,\cos\theta,0)\\
&=&
r\theta''(-\sin\theta,\cos\theta,0)
+\text{remainder},
\\
\partial_z^3\tilde \eta(x,z)
&\approx&
r\theta'''(-\sin\theta,\cos\theta,0)+\text{remainder},\\
z'(s)&=&|\partial_z\tilde\eta|^{-1}=(1+(r')^2+(r\theta')^2)^{-1/2}= (1+(r')^2)^{-1/2}+\text{remainder},\\
z''(s)&=&-(1+(r')^2+(r\theta')^2)^{-2}(r'r''+r\theta'(r'\theta'+r\theta''))\\
&=&
-(1+(r')^2)^{-2}r^2\theta'\theta''+\text{remainder},\\
z'''(s)
&\approx&
C(\beta)\theta'''+\text{remainder}.
\end{eqnarray*}
Therefore
\begin{eqnarray*}
\partial_s^2\eta^*\cdot e_\theta&=&\kappa n\cdot e_\theta\\
&=&
r\theta''(1+(r')^2)^{-1}
-r\theta'(1+(r')^2)^{-2}r^2\theta'\theta''
+\text{remainder},
\end{eqnarray*}
\begin{eqnarray*}
\kappa^2&=&|\kappa n|^2
=|\partial_z^2\tilde \eta|^2(z')^4+
2(\partial_z\tilde\eta\cdot\partial_z^2\tilde\eta)(z')^2z''+
|\partial_z\tilde\eta|^2(z'')^2\\
&=&
(r\theta'')^2(1+(r')^2)^{-2}
+((r')^2+1)(1+(r')^2)^{-4}(r^2\theta'\theta'')^2+
\text{remainder}
\end{eqnarray*}
and
\begin{eqnarray*}
n\cdot e_\theta&=&\frac{\kappa n\cdot e_\theta}{\kappa}\\
&=&
-\frac{r(1+(r')^2)^{-1}-(1+(r')^2)^{-2}r^2\theta'}{\left(r^2(1+(r')^2)^{-2}+(1+(r')^2)^{-3}r^4(\theta')^2\right)^{1/2}}+\text{remainder}.
\end{eqnarray*}
The remainders are small provided by small $\delta$.
We choose $\beta$ (which is independent of $\epsilon$ and $\delta$) such that the main term in $n\cdot e_\theta$ to be strictly smaller than $-1/2$
(by taking small $\beta$, then $\theta'$ becomes small).
Thus we have
\begin{equation}\label{cancellation}
n\cdot e_\theta=\frac{\kappa n\cdot e_\theta}{\kappa}<-1/2.
\end{equation}
Also recall that $\theta'''\approx -C(\beta)g''$.
The dominant term of $\partial_s(\kappa^2)$  is composed by $\theta''$ and $\theta'''$, more precisely,
\begin{equation*}
\partial_s(\kappa^2)= 2(\partial_s\kappa) \kappa=
2r\theta''(r\theta''')(1+(r')^2)^{-5/2}+\text{remainder}\approx
C(\beta)\theta''\theta'''.
\end{equation*}

We also see that
\begin{eqnarray*}
& &\kappa= |r\theta''|(1+(r')^2)^{-1}+\text{remainder},\\
&  &\kappa n\cdot e_\theta= r\theta''(1+(r')^2)^{-1}+\text{remainder},\\
& &\partial_s(\kappa n)\cdot e_\theta=\partial_s^3\tilde\eta\cdot e_\theta
=
r\theta'''(1+(r')^2)^{-3/2}+\text{remainder},\\
& &\partial_s\kappa=\frac{r\theta''(r\theta''')(1+(r')^2)^{-5/2}}{\kappa}+\text{remainder}\\
& &
\ \ \ \ \
=
-r\theta'''(1+(r')^2)^{-3/2}+\text{remainder}
\end{eqnarray*}
in $t\in I$.
``remainder" is small compare with the main terms provided by small $\delta>0$.
We immediately obtain
$(1/6)|u|^2|\partial_s\kappa|> \kappa\partial_t|u|$
for sufficiently small $\delta>0$,
since the left hand side is $\delta^{-3}$-order, while, the right hand side is $\delta^{-2}$-order.
In order to show $(1/2)|\partial_s\kappa|> |\kappa Tb\cdot e_\theta|$,
we use \eqref{cancellation}.
By the Frenet-Serret formula,
\begin{equation*}
Tb=\partial_s n+\kappa\tau,
\end{equation*}
we see that
\begin{equation*}
\kappa Tb\cdot e_\theta=\partial_s(\kappa n)\cdot e_\theta
-(\partial_s\kappa) n\cdot e_\theta+\kappa^2 \tau\cdot e_\theta.
\end{equation*}
Thus, by the direct calculation with \eqref{cancellation}, we can find a cancellation on the  highest order term composed by $\theta'''$,
 and then we have
\begin{equation*}
|\kappa Tb\cdot e_\theta|<(1/2) |\partial_s\kappa|
\end{equation*}
for sufficiently  small $\delta>0$.
\end{proof}
In what follows, we use  a differential geometric idea. See Chan-Czubak-Y \cite[Section 2.5]{CCY}, more originally, see Ma-Wang \cite[(3.7)]{MW}.
They considered 2D separation phenomena using fundamental differential geometry.
The key idea here is  ``local pressure estimate" on a normal coordinate in $\bar \theta$, $\bar r$ and $\bar z$ valuables.
Two derivatives to the scalar function $p$ on the normal coordinate is commutative, namely,
$\partial_{\bar r}\partial_{\bar \theta}p(\bar \theta, \bar r,\bar z)-\partial_{\bar \theta}\partial_{\bar r}p(\bar \theta,\bar r,\bar z)=0$ (Lie bracket).
This fundamental observation is the key to extract the local effect of the pressure.

\begin{remark}
It should be noticed that Enciso and Peralta-Salas \cite{EP}
considered the existence of Beltrami fields $u$
with a nonconstant proportionality factor $f$:
\begin{equation}\label{Beltrami field}
\nabla \times u=fu,\quad \nabla\cdot u=0\quad\text{in}\quad \mathbb{R}^3.
\end{equation}
It is well known that a Beltrami field is also a solution of the steady Euler equation in $\mathbb{R}^3$.
They showed that for a generic function $f$, the only vector field $u$ satisfying
\eqref{Beltrami field} is the trivial one $u\equiv 0$.
See (2.12), (3.4) and  (3.6) in \cite{EP} for the specific condition on $f$.
Note that $g_{ij}$  (induced metric of the level set of $f$) is the fundamental component of the condition.
It would be also interesting to consider whether we can apply their method to our unsteady flow problem, and compare with our method.
\end{remark}

For any point $x\in\mathbb{R}^3$ near the arc-length trajectory $\eta^*$ is uniquely  expressed as $x=\eta^*(\bar \theta)+\bar r n(\bar\theta)+\bar z b(\bar\theta)$ with $(\bar\theta,\bar r,\bar z)\in\mathbb{R}^3$ (the meaning of the parameters $s$ and $\bar \theta$ are the same along the arc-length trajectory).
Thus we have that
\begin{eqnarray*}
\partial_{\bar \theta}x&=&
\tau+\bar r(Tb-\kappa \tau)+\bar z \kappa n,\\
\partial_{\bar r}x&=& n,\\
\partial_{\bar z}x&=&b.
\end{eqnarray*}
This means that
\begin{equation*}
\begin{pmatrix}
\partial_{\bar \theta}\\
\partial_{\bar  r}\\
\partial_{\bar z}
\end{pmatrix}
=
\begin{pmatrix}
1-\kappa \bar r& \bar z\kappa& \bar r T\\
0& 1& 0\\
0& 0& 1
\end{pmatrix}
\begin{pmatrix}
\tau\\
n\\
b\\
\end{pmatrix}.
\end{equation*}
\begin{remark}
For any smooth scalar function $f$,
we have
\begin{equation*}
\partial_{\bar \theta}f(x)=\nabla f\cdot \partial_{\bar \theta}x.
\end{equation*}
$\nabla f$ itself is essentially independent of any coordinates, thus we can regard a partial derivative as a vector.
\end{remark}
By the fundamental calculation, we have  the following inverse matrix:
\begin{equation*}
\begin{pmatrix}
\tau\\
n\\
b\\
\end{pmatrix}
=
\begin{pmatrix}
(1-\kappa \bar r)^{-1}& -\bar zT (1-\kappa \bar r)^{-1}& -\bar r T(1-\kappa \bar r)^{-1}\\
0& 1& 0\\
0& 0& 1
\end{pmatrix}
\begin{pmatrix}
\partial_{\bar \theta}\\
\partial_{\bar  r}\\
\partial_{\bar z}
\end{pmatrix}.
\end{equation*}
Therefore we have the following orthonormal moving frame:
 $\partial_{\bar r}=n$, $\partial_{\bar z}=b$ and
$$
(1-\kappa \bar r)^{-1}\partial_{\bar \theta}-\bar z T(1-\kappa \bar r)^{-1}\partial_{\bar r}-\bar rT(1-\kappa \bar r)^{-1}\partial_{\bar z}=\tau.
$$

In order to abbreviate the complicated indexes, we re-define the absolute value of the velocity along the trajectory.
 Let (the indexes are $x$ and $t$ respectively)
\begin{equation*}
|u|:=|u(\eta(x',t),t)|\quad\text{with}\quad
x'=\eta^{-1}(x,t)
\end{equation*}
and
\begin{equation*}
\partial_t|u|:=\partial_{t'}|u(\eta(x',t'),t')|\bigg|_{t'=t}\quad\text{with}\quad
x'=\eta^{-1}(x,t).
\end{equation*}

\begin{lem}\label{Euler flow along the trajectory}
We see  $-\nabla p\cdot \tau=\partial_t|u|$ along the trajectory.
\end{lem}
\begin{proof}
Let us define a unit tangent vector $\tilde\tau$ (in time $t'$) as follows:
\begin{equation*}
\tilde \tau_{x,t}(t'):=\frac{u}{|u|}(\eta(x',t'),t')\quad\text{with}\quad x'=\eta^{-1}(x,t).
\end{equation*}
Note that  there is a re-parametrize factor $s(t')$ such that
\begin{equation*}
\tau(s(t'))=\tilde \tau(t').
\end{equation*}
Since $u\cdot \partial_s\tau=0$, we see that
\begin{eqnarray*}
\partial_{t'}|u(\eta(x',t'),t')|&=&
\partial_{t'}(u(\eta(x',t'),t')\cdot \tilde\tau_{x,t}(t'))\\
&=&
\partial_{t'}(u(\eta(x',t'),t'))\cdot \tilde\tau_{x,t}(t')+
u(\eta (x',t'),t')\cdot\partial_s\tau\partial_{t'}s\\
&=&\partial_{t'}(u(\eta(x',t'),t'))\cdot \tilde\tau_{x,t}(t').
\end{eqnarray*}
By the above calculation we have
\begin{equation*}
-\nabla p\cdot\tau=\partial_{t'}(u(\eta(x',t'),t')\cdot\tau|_{t'=t}=\partial_{t'}(u(\eta(x',t'),t')\cdot\tilde\tau|_{t'=t}=
\partial_{t'}|u|\bigg|_{t'=t}.
\end{equation*}
\end{proof}

\begin{lem}
Along the arc-length trajectory, we have
\begin{equation*}
3\kappa\partial_t|u|+\partial_s\kappa|u|^2=\partial_{\bar r}\partial_t|u|
\end{equation*}
and
\begin{equation*}
T\kappa|u|^2=\partial_{\bar z}\partial_t|u|.
\end{equation*}
\end{lem}

\begin{proof}
By using the orthonormal moving frame,
we have the following gradient of the pressure,
\begin{equation*}
\nabla p= (\partial_\tau p)\tau+(\partial_np)n+(\partial_bp)b.
\end{equation*}
Recall that
\begin{equation*}
\partial_st=|u|^{-1}
\quad\text{and}\quad
\partial_t\eta\cdot \tau=|u|.
\end{equation*}
By the unit normal vector with the curvature constant, we see
\begin{equation*}
\kappa n=\partial_s^2\eta^*=\partial_s(\partial_t\eta\partial_st)
=\partial_t^2\eta(\partial_st)^2+\partial_t\eta\partial_s^2t.
\end{equation*}
Thus we have
\begin{eqnarray*}
-(\nabla p\cdot n)&=&
(\partial_t^2\eta\cdot n)=
\kappa|u|^2,\\
-\partial_s(\nabla p\cdot n)&=&\partial_s(\kappa(\partial_st)^{-2})
=\partial_s\kappa(\partial_s t)^{-2}-2\kappa(\partial_st)^{-3}(\partial_s^2t),\\
-\nabla p\cdot \tau&=&-|u|^3\partial_s^2 t,\\
-\nabla p\cdot b&=&0.
\end{eqnarray*}
By Lemma \ref{Euler flow along the trajectory} with
$0=\kappa n\cdot \tau=\partial_t^2\eta\cdot \tau(\partial_s t)^2+\partial_t\eta\cdot \tau(\partial_s^2 t)$,
we also have
\begin{equation*}
\partial_s^2t=-|u|^{-3}\partial_t|u|.
\end{equation*}
Recall that
\begin{equation*}
\partial_\tau=(1-\kappa\bar r)^{-1}\partial_{\bar\theta}-\bar zT(1-\kappa\bar r)^{-1}\partial_{\bar r}
-\bar rT(1-\kappa\bar r)^{-1}\partial_{\bar z}.
\end{equation*}
Along the arc-length trajectory, we have
\begin{eqnarray*}
-\partial_{\bar r}(\nabla p\cdot \tau)
&=&-\partial_{\bar r}\partial_\tau p
\\
&=&
-\kappa\partial_{\bar \theta} p-\partial_{\bar r}\partial_{\bar \theta} p-T\partial_{\bar z} p\\
\nonumber
(\text{commute}\ \partial_{\bar r}\ \text{and}\  \partial_{\bar \theta})
&=&
-\kappa(\nabla p\cdot \tau)-\partial_{\bar\theta} (\nabla p\cdot n)-T(\nabla p\cdot b)\\
&=&
-\kappa |u|^3\partial_s^2t+\partial_s\kappa(\partial_st)^{-2}-2\kappa(\partial_st)^{-3}(\partial_s^2 t)\\
&=&
3\kappa\partial_t|u|+\partial_s\kappa|u|^2.
\end{eqnarray*}
Since $\nabla p\cdot b=\partial_{\bar z} p\equiv 0$ along the trajectory, then
\begin{equation*}
-\partial_{\bar z}(\nabla p\cdot \tau)|_{\bar r,\bar z=0}=
-\partial_{\bar z}\partial_\tau p|_{\bar r,\bar z=0}=
-\partial_{\bar z}\partial_{\bar\theta} p-T\partial_{\bar r} p
=
-T(\nabla p\cdot n)=T\kappa|u|^2.
\end{equation*}

By Lemma \ref{Euler flow along the trajectory}
along the arc-length trajectory $\eta^*$, we have
\begin{equation*}
3\kappa\partial_t|u|+\partial_s\kappa|u|^2=
-\partial_{\bar r}(\nabla p\cdot \tau)|_{\bar r,\bar z=0}
=
\partial_{\bar r}\partial_t|u|\\
\end{equation*}
and
\begin{equation*}
T\kappa|u|^2=
-\partial_{\bar z}(\nabla p\cdot \tau)|_{\bar r,\bar z=0}
=
\partial_{\bar z}\partial_t|u|.\\
\end{equation*}

\end{proof}

By using the above lemma we can finally prove the main theorem.
Since
\begin{equation*}
\partial_\theta=(e_\theta\cdot n)\partial_{\bar r}+(e_\theta\cdot b)\partial_{\bar z}
\end{equation*}
and the axisymmetric flow is rotation invariant (along the $e_\theta$-direction),
\begin{eqnarray*}\label{rotation invariant}
0&=&\partial_\theta\partial_t|u|=(e_\theta\cdot n)\partial_{\bar r}\partial_t|u|
+(e_\theta\cdot b)\partial_{\bar z}\partial_t|u|\\
\nonumber
&=&
3(e_\theta\cdot n)\left(\kappa\partial_t|u|+\partial_s\kappa|u|^2\right)
+(e_\theta\cdot b)T\kappa|u|^2.
\end{eqnarray*}
However, by
Lemma \ref{key estimate on torsion},
we have
\begin{eqnarray*}
& &\left|3(e_\theta\cdot n)\left(\kappa\partial_t|u|+\partial_s\kappa|u|^2\right)
+(e_\theta\cdot b)T\kappa|u|^2\right|\\
&\geq&
\frac{3|\partial_s\kappa||u|^2}{2}-3\kappa \big|\partial_t|u|\big|-T\kappa |u|^2\\
&\geq&
\frac{|\partial_s\kappa||u|^2}{2}>0,
\end{eqnarray*}
and it is in contradiction.

\vspace{0.5cm}
\noindent
{\bf Acknowledgments.}\
The author would like to thank Professor Norikazu Saito for letting me know
the book \cite{FQV},
and Professor Hiroshi Suito for letting me know
``Womersley number".
The author was partially supported by JST CREST.

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


text:
FAQ moodys.com Moody's Analytics Contact us
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アジアのCREおよび世界の経済環境: 国際的な視点
* 英日同時通訳をご用意しております。
2022年11月2日 (水) 日本時間 11:00 AM - 12:00 PM
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資本移動が活発なこの時代に、多くの大国で金利が上昇しています。このような状況下、アジアの商業用不動産投資(CRE)に対する投資家の行動や心理に影響を与える最新のトレンドや経済予測はどのようになっているのでしょうか?
このウェビナーでは、さまざまな国のダイナミクスを比較し、近い将来に何が起こるかについて探ります。ぜひ奮ってご参加ください。
クロスボーダーの投資トレンド:経済的不確実性を踏まえた短期的な見通し
アジアのCRE投資のホットスポットはどこか?向こう12-18カ月の主な機会とリスクは何か?
欧州や米国で定着し、重要な投資意思決定要素を形成しているESGのようなトレンドは、アジアにも大きな影響をもたらすか?影響する場合、またはしない場合の根拠は?
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Kevin Fagan Head of CRE Economic Analysis Moody's Analytics
Steve Cochrane Chief APAC Economist Moody's Analytics BIO
Victor Calanog Head of CRE Economics Moody's Analytics BIO
Dominic Brown Head of Insight & Analysis, Asia Pacific Cushman & Wakefield
Raymond Rufino CEO NEO
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Steve Cochrane Chief APAC Economist Moody's Analytics
Steven G. Cochrane, PhD, is Chief APAC Economist with Moody's Analytics. He leads the Asia economic analysis and forecasting activities of the Moody's Analytics research team, as well as the continual expansion of the company's international, national and subnational forecast models. In addition, Steve directs consulting projects for clients to help them understand the effect of regional economic developments on their business under baseline forecasts and alternative scenarios. Steve's expertise lies in providing clear insights into an area's or region's strengths, weaknesses and comparative advantages, relative to macro or global economic trends.
A highly-regarded speaker, Dr. Cochrane has provided economic insights at hundreds of engagements over the past 20 years and has been featured on CNBC, CNA, Bloomberg TV, Wall Street Radio, and the PBS News Hour. Through his research and presentations, Steve dissects how various components of the macro and regional economies shape patterns of growth. His extensive expertise on U.S. and global economic issues makes also him a popular speaker at client conferences and executive briefings.
Steve holds a PhD from the University of Pennsylvania and is a Penn Institute for Urban Research Scholar. He also holds a master's degree from the University of Colorado at Denver and a bachelor's degree from the University of California at Davis. Dr. Cochrane is based out of Moody's Analytics Singapore office.
Victor Calanog Head of CRE Economics Moody's Analytics
Victor Calanog is Head of Commercial Real Estate Economics for Moody's Analytics. Prior to the acquisition by Moody's Analytics in October 2018, he served as Chief Economist & Senior Vice President of REIS, leading a team of economists and analysts to build, maintain, and customize the firm's market forecasting, valuation, and real estate portfolio analytics and advisory services.
Dr. Calanog's research has received awards and fellowship support from several institutions, including the Ford Foundation, the Russell Sage Foundation and the Penn Lauder Center for International Business Research.
A prodigious researcher with broad interests, his papers in real estate economics, local government competition, and urban fiscal policy have been presented in meetings of the American Economic Association, the American Finance Association, the Brookings Institution, and discussed in academic forums sponsored by Harvard Business School, Columbia University, Stanford University, the Wharton School of the University of Pennsylvania, and the Stern School of Business at New York University.
He serves as contributor and peer reviewer for several academic journals, with scientific papers published in the Journal of Banking & Finance, the Journal of the Risk Management Association, and cited in the Journal of Finance, Journal of Public Economics, the World Bank Research Observer, the Brookings Papers on Urban Affairs and Public Administration Review. His assessments of market conditions have appeared in national publications like the Wall Street Journal, New York Times, Bloomberg News, Financial Times, and Reuters News. Dr. Calanog writes several columns on commercial real estate for the National Real Estate Investor and Scotsman Guide. His industry columns have won honors from the American Society of Business Publication Editors and the Risk Management Association.
He serves on the Commercial Real Estate/Multifamily Finance Board of Governors for the Mortgage Bankers Association, as Managing Editor for the CRE Finance Council, and on the Editorial Advisory Board of Real Assets Adviser. He serves on the Board of Directors of the Counselors of Real Estate®(CRE) and is a Fellow of the Royal Institute of Chartered Surveyors (FRICS).
Dr. Calanog holds a PhD in Applied Economics and Management Science, trained by a dissertation committee composed of faculty from the Wharton School of the University of Pennsylvania and Harvard Business School. In addition, he has an MBA in Finance and an MA in Business and Public Policy, earning the Amy Morse Prize and the Thomas P. Gerrity Leadership Award as the most distinguished PhD and MBA student in his class.
In a previous life, Dr. Calanog was a competitive powerlifter, but he now measures poundage's by how much laundry he does per day. He serves at the behest of Christine, his wife of over twenty years, daughters Samantha and Madison, guinea pigs Midnight and Kale, and puppy Ginger. He enjoys traveling, pestering locals in their own language, and studying geopolitics, history, art, military strategy–and mecha anime.


text:
Second conductor in railroad crash files lawsuit
On behalf of Charles Goetsch Law Offices LLC posted in fela claims on Friday, March 22, 2019.
Connecticut residents may have heard about the 2016 NJ Transit crash that killed one passenger and injured more than 100 others. The impact made part of the station's roof collapse. There were two conductors on that train, both of whom were injured. One sued NJ Transit in June of 2018. The other filed his suit in March of 2019.
Both conductors, being railroad workers, are suing under the Federal Employers Liability Act. Their protection under federal law generally means that railroad workers do not file for workers' compensation. Since NJ Transit is a state agency and enjoys "sovereign immunity," however, it cannot be sued under federal law.
NJ Transit has asked state and federal courts to dismiss lawsuits from dozens of plaintiffs. The two conductors are among 50 victims of the Hoboken Terminal crash who are suing. It remains to be seen how the conductors will fare in court.
The latest lawsuit filer, aged 58, fell to the floor at the moment of the crash, suffering injuries to his shoulder, hip, knee and back as well as incurring emotional trauma. He subsequently had to undergo a seven-hour surgery on his back. The complaint states that his injuries have prevented him from returning to his position as conductor.
As seen above, there are complications that can arise when an injured railroad worker files a FELA claim. Unlike with a workers' compensation claim, a victim must prove that the railroad company was negligent if they are to recover damages. For these and other reasons, it may be a good idea to hire a lawyer before filing a FELA claim. Third parties such as investigators and medical experts may come in to build up the case with the necessary evidence.
Related Posts: Payments to injured railroad workers for wages taxable, Railroads step up PTC implementation, Injured rail workers can no longer sue NJ Transit, Injured BNSF railroad worker awarded multimillion-dollar verdict


lauradivenereinteriors: Sun Valley to Aspen: It's all about Rustic Modern Design!
Sun Valley to Aspen: It's all about Rustic Modern Design!
Sun Valley Idaho, Aspen Colorado, Lake Tahoe California.
Amazing Style of Design, best described as "Rustic Modern".
It's Roots are very much 1940's to 1960's.
the edges - in a Good Way of course!
"Rustic Modern" so very cool!
Okay - admittedly obsessed with Hans Wagner's Papa Bear Chair. Clearly it's a chair that works well in many scenarios!
Though Frank Lloyd Wright's "Fallingwater" is in Pennsylvania the look works beautifully here as it's a Perfect Example of "Rustic Modern"!
An Amazing Kitchen from the 1950 Film "Duchess of Idaho". The film has great visuals - if you haven't seen it - it's worth checking it out!
The house that SHOULD have been built, but never was! The 1959 iconic "North by Northwest" Home. A Perfect Example of Rustic Mid Century Modern!
This Hilarious Modern Little Cabin - Close its windows and completely camouflages itself into a pile of stacked firewood!
If you're new to the Blog - make sure you check an Earlier Post on The Modern Ski Chalet - French Alps Style!
Palm Beach Style - Perfectly Pretty Design!


text:
iTunes Radio is popular but not even close to a 'Pandora killer'
It hasn't taken very long for Apple's iTunes Radio to rise up in popularity. Via MacRumors, a new survey of American consumers conducted by Edison Research and Statista has found that iTunes Radio is now the third-most popular music streaming service and has leapt ahead of Spotify. Overall, the survey found that 8% of people surveyed said that they'd listened to iTunes Radio within the last month, more than the 6% who said that they'd listened to Spotify and the 3% who said they'd listened to Google Play All Access.
However, iTunes Radio still faces a big gap with Pandora, which still reigns as the king of music streaming services. According to Edison Research and Statista, a whopping 31% of people surveyed said they'd used Pandora within the last month, a truly astonishing figure given that just 9% of people surveyed said they'd listened to iHeartRadio, the second-most popular music streaming service.
Statista's chart showing the popularity of all major music streaming services follows below.
Tags: iTunes Radio, Pandora, Spotify
By Yoni Heisler 9 hours ago


With most of us being so sedentary is muscle atrophy on the rise? How is it reversed? How do you rebuild severely wasted muscle through diet and exercise? What about those people who lose a ton of muscle from starving themselves? How do they gain that back? I'm not talking about any definition, just regular muscle. Is there any way to rebuild that or is it lost forever?
Also, since diet is like 80% of weight loss could you please post more food ideas? Maybe make a post for the best foods to eat for breakfast, lunch, and dinner. The recipes you posted last year were so damn good I just had to ask. You make healthy food look and sound amazing. You are my people. I can't deal with those people living off steamed veggies and chicken.
I am not going to get into the science behind muscular atrophy because it would transform this article into a 300 Level Kinesiology essay. So in layman's terms, when you don't use it you will lose it. I would not say that atrophy is on the rise but I do agree that our sedentary lifestyles do affect our health.
To answer your question, yes you can get your muscle back that you have lost. It is just a matter of spending some time in the gym. I am a big fan of functional body weight exercises like step ups, squats, lunges, push-ups and pull-ups. These types of exercises train the body as it is meant to be moved and they can be done anywhere. The more you move and lift the faster you will regain your strength and rebuild your muscle mass.
For those who have lost their muscle mass or muscle function due to an injury or other health complications, supervised rehabilitation is recommended and may require the use of steroids to rebuild muscle. These are extreme cases however, and if you are simply lacking muscle because you have been sitting around too much then you just need to hit the gym, join a boot camp, get a fitness app to follow at home or just start doing some lunges and push-ups daily.
You are right in that nutrition plays a large role in changing one's body composition. There are many diets and ways to eat but I am not going to say one is better than the other as it is all about finding what works for you.
I support simple and balanced eating. I do not jump onto any bandwagons nor do I cut foods out of my diet. I eat carbohydrates, I eat protein, I eat fat and I consume sugar. I eat real food and I cook with real ingredients. Today, for example, I had 2 soft boiled eggs on rye toast with a cup of coffee for breakfast, a Larabar for a snack, an apple and piece of cheese for another snack and then an almond butter and banana sandwich. Dinner will be a bowl of homemade turkey chilli, with some grated cheese and sour cream! Last night it was lamb, roasted potatoes and a couscous salad after my 3 hour mountain bike ride, cooked for me by one of my best friends.
There are no "best" foods for breakfast, lunch or dinner, but try and design your meals in this way, which I learned from my good friend and Dietician Diana Steele. Your breakfast should be 50% fruit, 25% carbohydrates and 25% protein. Lunch and dinner should be the same but substitute vegetables for the fruit.
Eat less at night and more in the morning, fill your body with real food more than foods that come out of a package or a box. Juicing is great but it is even better if you eat the vegetables. If you are busy then meal replacement bars (such as Larabars) are good in a pinch but talk to any dietician or nutritionist and they will always recommend real food over supplements.
I will definitely post another recipe or two soon but check out Epicurious for endless recipe ideas or freshthinkingcatering.ca as this is the chef whom I work with to create the recipes.
Attached Emily Blunt and John Krasinski leaving the gym.


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