text:
Donna Elliott (Photo/Ricardo Carrasco III)
Donna Elliott named Keck School vice dean for medical education
The Keck School of Medicine of USC has appointed Donna Elliott, MD, EdD, as vice dean for medical education and chair of the Department of Medical Education, effective June 1. In her new role as vice dean, Elliott will oversee admissions and advise the Keck School's dean on all academic areas related to medical students, including curriculum, educational policy and student affairs.
As chair of the Department of Medical Education, Elliott will be responsible for leading faculty members who have expertise in course/program evaluation, curriculum development, clinical skills assessment, research design, psychometrics and statistical analysis. They also will provide mentoring and support to Keck School faculty and students who are interested in conducting research in medical education.
"Dr. Elliott has had a tremendous impact on the Keck School and our students since she first joined the faculty as an assistant professor of clinical pediatrics in 1989," said Laura Mosqueda, MD, interim dean of the Keck School. "In addition to training generations of future doctors, she has helped shape the school's curriculum and has been a tireless advocate for students. She was one of the first faculty members to integrate professionalism and the practice of medicine into our curriculum. I am delighted to welcome her to her new role."
An expert in pediatric nephrology and a professor of clinical pediatrics, Elliott has had leadership roles in student and educational affairs at the Keck School for more than a decade, most recently as senior associate dean for student and educational affairs. She has led a number of initiatives aimed at increasing student diversity and promoting students' well-being. In 2017, she launched a student wellness and stress reduction program that includes mandatory mental health evaluations to identify students in need of support. She also helped lead a two-year effort to renew the Keck School's Liaison Committee on Medical Education accreditation, resulting in a full eight-year accreditation in 2018.
"It gives me great pleasure to be a part of such an outstanding academic institution, with such exceptional students, faculty and staff," Elliott said. "I'm looking forward to working with the talented faculty and staff to lead the Keck School of Medicine's medical education program to an even higher level of prominence while continuing to nurture and support our medical students on their professional journey of becoming a physician."
Elliott is the national chair of the Association of American Medical College's Group on Student Affairs Steering Committee, a member of the executive board of the National Board of Medical Examiners and a member of the board of the National Resident Match Program.
— Erica Rheinschild
Melissa Masatani 2019-09-13T16:25:21-07:00 March 20th, 2018|Announcements, Keck Net Intranet, Press Page|


text:
The Supreme Court will review LGBT rights; this time, a wedding website designer, not a baker, will be there.
Same-sex wedding cakes wound up at the Supreme Court – now, it's wedding websites' turn.
A simmering, difficult, and timely question returns to the Supreme Court this fall: What happens when freedom of speech and civil rights collide?
The court took up similar questions four years ago in the famous "gay wedding cake" case, Masterpiece Cakeshop, Ltd. v. Colorado Civil Rights Commission, about a baker who refused to provide services for a same-sex couple based on his religious beliefs. The justices ruled in his favor, but did so on narrow grounds, sidestepping the direct constitutional questions over freedom of religion and free speech.
Now, another case from Colorado about free speech and same-sex marriage has made its way to the court: 303 Creative v. Elenis. As a professor of law and education who pays particular attention to First Amendment issues, I see the case highlighting tension between two competing fundamental interests – interests that seem to clash routinely in 21st-century America.
On Aug. 30, 2022, for example, another similar case was decided, this time in Kentucky. A federal trial court ruled in favor of a Louisville wedding photographer who sued over the city's "Fairness Ordinance," which prohibits discrimination based on sexual orientation or gender identity. She argued that the law violated her religious beliefs and right to free speech, and the court agreed, explaining that "the government may not force singers or writers or photographers to articulate messages they don't support."
Freedom to speak – or stay silent
Graphic artist Lorie Smith is the founder and owner of a studio called 303 Creative. According to court documents, Smith is generally willing to serve LGBT clients. However, she intends to begin designing wedding websites and is unwilling to create them for same-sex couples, saying it would go against her Christian beliefs.
Under Colorado's Anti-Discrimination Act, though, it is discriminatory and illegal to refuse services to someone based on "disability, race, creed, color, sex, sexual orientation, marital status, national origin, or ancestry."
In 2016, Smith sued the members of the state's Civil Rights Commission and Colorado's attorney general. Smith argued that being required to prepare a same-sex wedding website would violate her First Amendment rights by forcing her to speak – what lawyers refer to as "compelled speech."
The constitutional right to freedom of "speech" has historically been understood to cover a variety of ways people express themselves, including in writing, art and protest. But not only does it protect the right to protect one's speech, it also safeguards the right to not speak in the first place.
Through her attorneys, Smith also maintained that requiring her to create a website would violate her First Amendment right to the free exercise of religion.
Path to SCOTUS
The federal trial court in Colorado rejected Smith's request to block the anti-discrimination law in 2019. When she appealed, the circuit court agreed with the previous ruling: She could not refuse to create websites for same-sex weddings, even if it would have gone against her beliefs.
Protecting diverse viewpoints is "a good in and of itself," the court wrote, but combating discrimination "is, like individual autonomy, 'essential' to our democratic ideals."
In a lengthy dissent, the chief judge highlighted Smith's claim of compelled speech, criticizing the court for taking "the remarkable – and novel – stance that the government may force Ms. Smith to produce messages that violate her conscience."
Smith appealed to the Supreme Court, which, in February 2022, agreed to hear her claim, limited to the issue of free speech, not freedom of religion. The question for the nine justices to decide will be "whether applying a public-accommodation law to compel an artist to speak or stay silent violates the Free Speech Clause of the First Amendment."
Jack Phillips, whose Masterpiece Cakeshop case went to the Supreme Court, stands in his Colorado bakery.
Key to the case?
So, how will the justices rule? The Supreme Court may have given a clue to its initial attitude when it announced it would hear the case. The justices zoomed in on a legal standard called "strict scrutiny," as they did in its earlier case on this issue, Masterpiece Cakeshop.
Under strict scrutiny analysis, the most stringent form of judicial review, government restrictions on fundamental rights must be justified by a compelling state interest in order to be upheld. In other words, the restrictions must advance government interests of the highest order, and be narrowly tailored to those goals – in this case, preventing discrimination on the basis of sexual orientation.
But the Supreme Court appeared skeptical that Colorado's anti-discrimination act could survive this test, writing, "The Tenth Circuit applied strict scrutiny and astonishingly concluded that the government may, based on content and viewpoint, force Lorie to convey messages that violate her religious beliefs and restrict her from explaining her faith."
When the Supreme Court applies strict scrutiny, it rarely upholds governmental restrictions on constitutional rights – which could suggest a win for Smith.
Another possible indication, again in favor of Smith, is in Janus v. American Federation of State, County, and Municipal Employees, Council 31, a 2018 case from Illinois involving compelled speech. Here the Supreme Court ruled in favor of a nonunion public employee who challenged an Illinois law requiring him to pay fair share fees to the union representing his colleagues for costs associated with the bargaining process. The court agreed with the employee's claim that because the union supported positions with which he disagreed, his having to pay the fees violated his First Amendment right as a form of compelled speech.
A second chance
On the other side of the controversy is the vital interest of same-sex couples and others in the LGBTQ community to live free from discrimination based on their sexual orientations.
In a 2019 case, Bostock v. Clayton County, the Supreme Court interpreted Title VII of the Civil Rights Act of 1964, a far-reaching employment statute, as extending protection against discrimination in the workplace to individuals who are gay and transgender. However, the Court has yet to address the clash of rights at issue in 303 Creative.
The key question, then, appears to be whether individuals can require artists or those who engage in expressive activities to provide their services if doing so can be viewed as a form of compelled speech, violating their right to stay silent on issues with which they disagree.
Thus, it remains to be seen whether 303 Creative will set a new precedent on balancing First Amendment freedoms while protecting others from discrimination. After all, it sidestepped constitutional issues in Masterpiece Cakeshop. The court based its decision in the baker's favor on some of the Colorado commission members' comments about his beliefs. The majority found that those comments violated the state's First Amendment duty to maintain religious neutrality while avoiding hostility to faith-based beliefs or viewpoints.
While the court has not set a date for oral arguments, and likely will not render a judgment until near the end of its term in June 2023, it promises to be one of the upcoming year's highest-profile judgments. And, regardless of the outcome, 303 Creative is likely to generate even more controversy.
Charles J. Russo does not work for, consult, own shares in or receive funding from any company or organisation that would benefit from this article, and has disclosed no relevant affiliations beyond their academic appointment.
Tags: Anti-Discrimination Act, Civil Rights, LGBT Rights
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text:
\section{Introduction}

The inviscid limit of the viscous flow has been known as a challenging
mathematical problem that contains many unsolved problems. For the incompressible Navier-Stokes equations confined in a domain with  boundary, in particular with the non-slip boundary condition, the justification of the inviscid limit remains basically open, c.f. \cite{e-1} and references therein. The main obstruction
comes from the formation of boundary layers near the physical boundary, in which the tangential velocity component changes dramatically.

The foundation of the boundary layer theories was established by Prandtl \cite{Pra}
in 1904 when he introduced the classical Prandtl equations by considering the
incompressible Navier-Stokes equations with non-slip boundary condition. His
observation reveals that outside the layer of thickness of $\sqrt{\nu}$ with
$\nu$ being the viscosity coefficient, the convection dominates so that the flow
can be described approximately by the incompressible Euler equations, however, within the
layer of thickness of $\sqrt{\nu}$ in the vicinity of the boundary, the
convection and viscosity balance so that the flow is governed by the Prandtl
equations that is degenerate and mixed type.
Since then, there have been a lot of mathematical studies on the Prandtl equations,
however, the existing theories are basically limited to the two space dimensional
case except the one in analytic framework by Sammartino and Caflisch \cite{S-C} and others \cite{Z-Z}. On the other hand, in
two dimensional space, the classical work by Oleinik and her collaborators \cite{Ole}
gives the local in time well-posedness when the tangential velocity
component is
monotone in the normal direction,
by using the  Crocco transformation.
Recently, this well-posedness result of the two dimensional Prandtl equations is re-studied in \cite{AWXY,M-W} by direct energy method. In addition to the monotone condition on the
velocity, if a favorable pressure condition is imposed, then global in time
weak solution was also obtained in two dimensional space, see \cite{X-Z}.

The stability mechanism of the three dimensional Prandtl equations is very challenging and delicate mainly due to the possible appearance of secondary flows in the three dimensional boundary layer flow as explained  in Moore \cite{moore}, and it is an open question proposed by Oleinik and Samokhin in the monograph \cite{Ole}.
Recently, in \cite{LWY} the authors  construct a local solution to the three dimensional Prandtl equations when the tangential velocity field direction is invariant with respect to the normal variable under certain monotonicity condition.
In addition,   this special boundary layer flow is linearly stable with respect to any perturbation, and the global in time weak solution
is also obtained under an additional favorable pressure condition \cite{LWY1}.

The purpose of this paper is to investigate the instability of boundary layer flows in three
space dimensions without the special structure proposed in \cite{LWY}, even when the two tangential velocity components of the background state are monotonic.
 This reveals the essential
difference of the Prandtl equations
between two and three space dimensions. For this, let us first review the recent extensive studies on the
instability of the two space dimensional flow around a background state of shear flow
with non-monotonicity.

In fact, without the monotonicity assumption on the tangential component of
the velocity, boundary separation will occur. For this, there are many
physical observations and mathematical studies. For example,
Van Dommelen and Shen in \cite{van} illustrated the ``Van  Dommelen singularity'' by
considering an impulsively started circular cylinder to show
the blowup of the normal velocity, and
 E and Enquist in \cite{e-2} precisely constructed some finite time blowup
solutions to the two-dimensional Prandtl equations.
Started by Grenier's work in 2000,
there are some extensive investigation on the
instability of the two-dimensional Prandtl equations when the background shear flow
has some degeneracy. Precisely, corresponding to the well known
Rayleigh criterion for the Euler flow,  Grenier \cite{grenier} showed that
 the unstable Euler shear flow  yields instability of the Prandtl equations.
It was shown in \cite{GV-D}
 that a non-degenerate
critical point in  the shear flow of the Prandtl equations leads to
  a strong linear ill-posedness of the Prandtl equations in the
Sobolev space framework. Moreover, \cite{GV-N} strengthens
the result of \cite{GV-D}  for any unstable
shear flow. Along this direction, the ill-posedness in the nonlinear
setting was proved in \cite{guo} to show that
the  Prandtl equations are ill-posed near non-stationary and non-monotonic shear flows so that
the asymptotic boundary layer
expansion is not valid for non-monotonic shear layer flows in Sobolev spaces.

To describe the problem to be studied in this paper,
consider the following incompressible Navier-Stokes equations
\begin{equation}\label{ns}
\begin{cases}
\partial_t {\bf u}^\nu+({\bf u}^\nu\cdot
\nabla){\bf u}^\nu+\nabla p^\nu-\nu\Delta {\bf u}^\nu=0, \\
\nabla\cdot {\bf u}^\nu=0, \\
{\bf u}^\nu|_{z=0}=0,
\end{cases}
\end{equation}
 in $\{t>0, (x, y)\in \T^2, z\in \R^+\}$ with boundary at $\{ z=0\}$, here
${\bf u^\nu}=(u^\nu,v^\nu,w^\nu)^T$. According to
Prandtl's observation, set the ansatz for
${\bf u}^\nu$ near $\{z=0\}$ as
\begin{equation}
\begin{cases}
u^\nu(t,x,y,z)=u(t,x,y,\frac{z}{\sqrt{\nu}})+o(1),\\
v^\nu(t,x,y,z)=v(t,x,y,\frac{z}{\sqrt{\nu}})+o(1),\\
 w^\nu(t,x,y,z)=\sqrt{\nu}w(t,x,y,\frac{z}{\sqrt{\nu}})+o(\sqrt{\nu}),
\end{cases}
\end{equation}
and plug it in the Navier-Stokes equations \eqref{ns}, one finds that the boundary layer profile $(u, v, w)(t,x,y,z)$ (here we replace $\frac{z}{\sqrt{\nu}}$ by $z$ for simplicity of notations) satisfies
\begin{equation}\label{3dpr}
\begin{cases}
\partial_t u+(u\partial_x+v\partial_y+w\partial_z)u+\partial_xp^E(t,x,y,0)=\partial_z^2u,\\
\partial_t v+(u\partial_x+v\partial_y+w\partial_z)v+\partial_yp^E(t,x,y,0)=\partial_z^2v,\\
\partial_xu+\partial_yv+\partial_z w=0,\\
(u, v, w)|_{z=0}=0, \qquad \lim\limits_{z\to+\infty}(u, v)=(u^E, v^E)(t,x,y,0),
\end{cases}
\end{equation}
which is the famous Prandtl layer equations.
Here, the pressure $p^E$ is related to the outer Euler flow
 ${\bf u}^E=(u^E,v^E,0)(t,x,y,0)$ through
$$
\partial_t {\bf u}^E+({\bf u}^E\cdot \nabla){\bf u}^E +\nabla p^E=0.
$$

The main results of this paper show that when the background state is
a shear flow $(u^s(t,z),v^s(t,z),0)$ of \eqref{3dpr} with initial data $(U_s(z),V_s(z))$, even under
the  monotonicity condition that
$U'_s(z), \, V'_s(z)>0$, the Prandtl equations \eqref{3dpr} are both linearly and
nonlinearly unstable under a very general assumption that
\begin{equation}\label{con-1}
\exists ~z_0>0,~s.t.~\frac{d}{dz}(\frac{V'_s}{U'_s})(z_0)\neq 0\quad{\rm or} \quad\frac{d}{dz}(\frac{U'_s}{V'_s})(z_0)\neq 0.
\end{equation}
Note that in \cite{LWY}, existence of solutions to the three space
dimensional Prandtl equations was proved with special structure and in that case,
the tangential components of the solution $(u,v,w)(t,x,y,z)$
satisfies $\frac{\partial}{\partial z}(\frac{v}{u})\equiv 0$. In fact, in this case, the
appearance of the secondary flow that is the key factor in instability is avoided. Thus, by combining with the results obtained in \cite{LWY}, we know that the condition \eqref{con-1} is not only sufficient but also necessary for the linear instability of the three dimensional Prandtl equation \eqref{3dpr} linearized
around the monotonic shear flow $(u^s(t,z), v^s(t,z),0)$.

The rest of the paper will be organized as follows. In Section 2, we first state the main
results on the linear and nonlinear instability of the three-dimensional Prandtl equations with
background state as monotonic shear flow.
Then, we prove the linear instability result of the shear flow in Section 3, and the
nonlinear instability will be studied in Section 4. In the Appendix, we give a well-posedness result for the linearized three-dimensional Prandtl equations in the analytic setting with respect to only one horizontal variable, under the assumption that one component of the tangential velocity field of
background shear flow is monotonic.

\section{Main results}

By assuming that the outer Euler flow is uniform in \eqref{3dpr}, consider the following boundary value problem of three dimensional Prandtl equations in $\Omega\triangleq\{(t,x,y,z):t>0,(x,y)\in\T^2,z\in\R^+\},$
\begin{equation}\label{3dpd}
\begin{cases}
\pd_t u+(u\pd_x+v\pd_y+w\pd_z)u-\pd_z^2u=0,\\
\pd_t v+(u\pd_x+v\pd_y+w\pd_z)v-\pd_z^2v=0,\\
\pd_xu+\pd_yv+\pd_zw=0,\\
(u,v,w)|_{z=0}=0,\qquad \limz (u,v)=(U_0,V_0)
\end{cases}\end{equation}
for positive constants $U_0$ and $V_0$.
 To understand this problem, we start with the simple situation of shear flow.
Let $u^s(t,z)$ and $v^s(t,z)$ be smooth solutions of the heat equations:
\begin{equation}\label{shear}\begin{cases}
\pd_t u^s-\pd_z^2u^s=0,\qquad
\pd_t v^s-\pd_z^2v^s=0,\\
(u^s,v^s)|_{z=0}=0,\qquad \limz (u^s,v^s)=(U_0,V_0),\\
(u^s,v^s)|_{t=0}=(U_s,V_s)(z),
\end{cases}\end{equation}
with $(u^s-U_0,v^s-V_0)$ rapidly tending to $0$ when $\zinf$. It is
straightforward to check that the shear velocity profile $(u^s,v^s,0)(t,z)$ satisfies the problem \eqref{3dpd}.

The question we answer in this paper is
whether such trivial profile is stable even when $u^s(t,z)$ and $v^s(t,z)$ are strictly monotonic in $z>0$.
 For this, we first focus on the linear stability problem, and
 consider the linearization of the problem \eqref{3dpd} around $(u^s,v^s,0)$:
\begin{equation}\label{lin}\begin{cases}
\pd_t u+(u^s\pd_x+v^s\pd_y)u+wu_z^s-\pd_z^2u=0,\quad{\rm in}\quad \Omega,\\
\pd_t v+(u^s\pd_x+v^s\pd_y)v+wv_z^s-\pd_z^2v=0,\quad{\rm in}\quad \Omega,\\
\pd_xu+\pd_yv+\pd_zw=0,\quad{\rm in}\quad \Omega,\\
(u,v,w)|_{z=0}=0,\qquad\limz (u,v)=0.
\end{cases}\end{equation}

To present the linear instability result that is motivated by the work \cite{GV-D} for two dimensional problem, we first introduce
some notations. As in \cite{GV-D}, for any $\alpha, \, m\geq0,$ denote by
\[\begin{split}
L_\alpha^2(\R^+)~&:=~\{f=f(z),z\in\R^+;~\|f\|_{L_\alpha^2}\triangleq\|e^{\alpha z}f\|_{L^2}<\infty\},\\%\quad 1\leq p\leq\infty, \\
H_\alpha^m(\R^+)~&:=~\{f=f(z),z\in\R^+;~\|f\|_{H_\alpha^m}\triangleq\|e^{\alpha z}f\|_{H^m}<\infty\}, \\
W_\alpha^{m,\infty}(\R^+)~&:=~\{f=f(z),z\in\R^+;~\|f\|_{W_\alpha^{m,\infty}}\triangleq\|e^{\alpha z}f\|_{W^{m,\infty}}<\infty\},
\end{split}\]
and the functional space for $\forall\beta>0$:
\begin{equation*}\begin{split}
E_{\alpha,\beta}~&:=~\Big\{f=f(x,y,z)=\sum\limits_{k_1,k_2\in\cZ}e^{i(k_1x+k_2y)}f_{k_1,k_2}(z),~\|f_{k_1,k_2}\|_{L_\alpha^2}\leq C_{\alpha,\beta}e^{-\beta\sqrt{k_1^2+k_2^2}}\Big\},\\%~\forall k_1,k_2\Big\},\\%\quad \forall\beta>0,
\end{split}\end{equation*}
with
\[\|f\|_{E_{\alpha,\beta}}\triangleq\sup\limits_{k_1,k_2\in\cZ}e^{\beta\sqrt{k_1^2+k_2^2}}\|f_{k_1,k_2}\|_{L_\alpha^2}.\]
The same notations are also used for the vector functions without confusion.

As in \cite{GV-D}, we first have the following existence
 result for the problem \eqref{lin} when the data are analytic in the tangential variables $(x,y)$.

\begin{prop}\label{prop_1}
Let $(u^s-U_0,v^s-V_0)\in C\big(\R^+;W_\alpha^{1,\infty}(\R^+)\big).$ Then, there exists a $\rho>0$ such that for all $T$ with $\beta-\rho T>0$, and $(u_0,v_0)\in E_{\alpha,\beta}$, the linear problem \eqref{lin} with the initial data $(u,v)|_{t=0}=(u_0,v_0)$ has a unique solution
\[
(u,v)\in C\Big([0,T);E_{\alpha,\beta-\rho T}\Big),\quad (u,v)(t,\cdot)\in E_{\alpha,\beta-\rho t}
\]
\end{prop}
The proof of this Proposition is the same as that given in \cite[Proposition 1]{GV-D}, so we omit it for brevity.

If we impose monotonic condition on the tangential velocity components of the shear flow
$(u^s,v^s,0)$, then another well-posedness result can be obtained. For this, similar to the previous notations, we
 introduce the following function spaces: for any $ \alpha,\, \beta>0$, set
\begin{equation}\label{def_H}\begin{split}
K^m_{\alpha}~&:=~\Big\{f=f(x,z
;~\|f\|_{K^m_\alpha}\triangleq \|e^{\alpha z}f\|_{ H^m(\T_x;L^{2}(\R^+_z))}<\infty\Big\}
\end{split}\end{equation}
and
\begin{equation*}\begin{split}
F^m_{\alpha,\beta}~&:=~\Big\{f=f(x,y,z)=\sum\limits_{k\in\cZ}e^{iky}f_{k}(x,z);\qua
\|f_k\|_{K_\alpha^m}\leq C_{\alpha,\beta}e^{-\beta|k|},~\forall k\Big\
\end{split}\end{equation*}
with
\[\|f\|_{F^m_{\alpha,\beta}}~\triangleq~\sup\limits_{k\in\cZ}e^{\beta|k|
\|f_k\|_{K_\alpha^m}.\]
The following result shows that the linear problem \eqref{lin} is still well-posed when the analyticity with respect to one horizontal variable given in Proposition \ref{prop_1} is replaced by some monotonicity  assumption.

\begin{prop}\label{prop_2}
Let $(u^s-U_0,v^s-V_0)\in C(\R^+;W_\alpha^{3,\infty}(\R^+)),\alpha>0$
satisfying $u_z^s>0$ and
\[\frac{u_{zz}^s}{u_z^s},~\frac{v_z^s}{u_z^s}\in C\big(\R^+;W^{1,\infty}(\R^+)\big),\]
and the initial data $(u,v)|_{t=0}=(u_0,v_0)$ of the problem \eqref{lin} satisfying

 \begin{equation}\label{ass_ini}

(u_0,v_0)(x,y,z)\in F_{\alpha,\beta}^{m}, \qquad  \pd_z\big(\frac{u_0}{u_z^s(0,\cdot)}\big)~\in~F^{m}_{\alpha,\beta}.
 \end{equation}
 Then, there exists a $\rho>0$ such that for all $T$ with $\beta-\rho T>0$,  the linear problem \eqref{lin}
  has a unique solution $(u,v)(t,x,y,z)$ satisfying
\[
(u,v)\in L^\infty\Big(0,T;F^{m}_{\alpha,\beta-\rho T}\Big),\quad \pd_z(u,v)\in L^2\Big(0,T;F^{m}_{\alpha,\beta-\rho T}\Big)
\]
with \((u,v)(t,\cdot)\in F^{m}_{\alpha,\beta-\rho t}
\)
\end{prop}
The proof of this proposition will be given in the Appendix.

From the above Proposition \ref{prop_1} and Proposition \ref{prop_2}, we know that when the monotonic condition is
imposed to one tangential component of background velocity field, such as $u^s$, the analyticity requirement for the velocity field with respect to the corresponding horizontal variable $x$ can be replaced by the
Sobolev regularity, while the velocity field is still analytic in the other horizontal variable $y$, then one still has the local in time well-posedness for the
linearized system in three dimensional space.

Following this argument, it is natural and interesting to study whether the well-posedness of the linearized Prandtl equations \eqref{lin} still holds in the Sobolev
framework if one imposes monotonicity conditions on both tangential velocity components of background state but without analyticity assumption anymore. The study of this paper  gives a negative answer to the question. In fact,  the following theorem  shows a strong linear instability of three dimensional Prandtl equations around basically shear flow in the Sobolev framework except those
with special structure studied in \cite{LWY}.

To state the result, we need some more notations.
 Denote by the operator $T\in\cl(E_{\alpha,\beta},E_{\alpha,\beta'}):$
\begin{equation}\label{def_T}
T(t,s)\big((u_0,v_0)\big)~:=~(u,v)(t,\cdot),
\end{equation}
where $(u,v)$ is the solution of \eqref{lin} with $(u,v)|_{t=s}=(u_0,v_0)$. Introduce the function spaces for $m,\alpha\geq0,$
\begin{equation}\label{def_h}
\h_\alpha^m~:=~H^m\big(\T^2_{x,y};L_\alpha^2(\R^+_z)\big)
\end{equation}
Since the space $E_{\alpha,\beta}$ is dense in the space $\h_\alpha^m,$ we can
extend the operator $T$ from the space $E_{\alpha,\beta}$
to $\h_\alpha^m$,
and define
\[\|T(t,s)\|_{\cl(\h_\alpha^{m_1},\h_\alpha^{m_2})}~:=~\sup\limits_{(u_0,v_0)\in E_{\alpha,\beta}}\frac{\|T(t,s)(u_0,v_0)\|_{\h_\alpha^{m_2}}}{\|(u_0,v_0)\|_{\h^{m_1}_\alpha}}
\in~\R^+\cup\{\infty\},\]
where the infinity means that $T$ can not be extended to $\cl(\h_\alpha^{m_1},\h_\alpha^{m_2})$.

The main result on linear instability of shear flow is stated as follows.

\begin{theorem}\label{thm_lin}
Let $(u^s,v^s)(t,z)$ be the solution of the problems \eqref{shear} satisfying
 $$(u^s-U_0,v^s-V_0)\in C^0\big(\R^+; W_\alpha^{4,\infty}(\R^+)\cap H_\alpha^4(\R^+)\big)\cap C^1\big(\R^+; W_\alpha^{2,\infty}(\R^+)\cap H_\alpha^2(\R^+)\big).$$
Assume that the initial data of \eqref{shear} satisfies that
\begin{equation}\label{ass}
\exists~ z_0>0,~s.t.~(U_s'(z_0))^2+(V_s'(z_0))^2\neq0,~V_s'(z_0)U_s''(z_0)\neq U_s'(z_0)V_s''(z_0).
\end{equation}
Then we have the following two instability statements.

i). There exists $\sigma>0$ such that for all $\delta>0$,
\begin{equation}\label{est_in}
\sup\limits_{0\leq s\leq t\leq\delta}\big\|e^{-\sigma(t-s)\sqrt{|\pd_\cT|}}T(t,s)\big\|_{\cl(\h_\alpha^m,\h_\alpha^{m-\mu})}
~=~+\infty,\quad \forall m>0,~\mu\in[0,\frac{1}{4}),
\end{equation}
where the operator $\pd_\cT$ represents the tangential derivative $\pd_x$ or $\pd_y$;

ii).  There exists an initial shear layer $(U_s,V_s)$ to \eqref{shear} and $\sigma>0$, such that for all $\delta>0$,
\begin{equation}\label{est_in1}
\sup\limits_{0\leq s\leq t\leq\delta}\big\|e^{-\sigma(t-s)\sqrt{|\pd_\cT|}}T(t,s)\big\|_{\cl(\h_\alpha^{m_1},\h_\alpha^{m_2})}
~=~+\infty,\quad \forall m_1,m_2>0.
\end{equation}
\end{theorem}

\begin{remark}\label{rem_3d}
From Theorem \ref{thm_lin}, we know that the three dimensional Prandtl equations can be linearly unstable around the shear flow $(u^s,v^s,0)(t,z)$ even under the monotonic conditions $u_z^s>0$ and $v_z^s>0$.
On the other hand, if we impose
the monotonic condition $U_s'>0$, then \eqref{ass} is equivalent to
\begin{equation}\label{ass1}
\frac{d}{dz}\big(\frac{V_s'}{U_s'}\big)\not\equiv 0.
\end{equation}
And then, by virtue of the boundary condition $U_s(0)=V_s(0)=0$, \eqref{ass1} is equivalent to
\begin{equation}
\frac{d}{dz}\big(\frac{V_s}{U_s}\big)\not\equiv 0.
\end{equation}
Thus,  the result of Theorem \ref{thm_lin} is exactly a complement to the well-posedness result of the
three dimensional Prandtl equations obtained by the authors  in \cite{LWY} for
flow with special structure, that is $\frac{d}{dz}\big(\frac{V_s}{U_s}\big)\equiv 0$. For simplicity, we will  assume that $U_s'(z_0)\neq0$ in the following argument.
\end{remark}

Finally, under the above assumption \eqref{ass} we shall have nonlinear instability for the  original  problem \eqref{3dpd} of the three dimensional nonlinear Prandtl equations.
To state the result, let us first recall the definition of local
 well-posedness from  \cite{guo}.

\begin{define}\label{def_non}
The problem \eqref{3dpd} with the initial data $(u,v)|_{t=0}=(u_0,v_0)(x,y,z)$ is locally well-posed, if there exist positive continuous functions $T(\cdot,\cdot),C(\cdot,\cdot)$, some $\alpha>0$ and some integer $m\geq1$ such that for any initial data $(u_0^1,v_0^1)$ and $(u_0^2,v_0^2)$ with
\[(u_0^1-U_0,v_0^1-V_0)\in \h_\alpha^m,\qquad (u_0^2-U_0,v_0^2-V_0)\in \h_\alpha^m,\]
there are unique distributional solutions $(u^1,v^1)$ and $(u^2,v^2)$ satisfying that for $i=1,2,$
\((u^i,v^i)|_{t=0}=(u_0^i,v_0^i)\)
 and
 \[(u^i-U_0,v^i-V_0)\in L^\infty\big(0,T;L^2(\T^2\times\R^+)\big)\cap L^2\big(0,T;H^1(\T^2\times\R^+)\big),\quad i=1,2,\]
and the following estimate holds
\begin{equation}\label{non}\begin{split}
&\|(u^1,v^1)-(u^2,v^2)\|_{L^\infty(0,T;L^2(\T^2\times\R^+))}+\|(u^1,v^1)-(u^2,v^2)\|_{L^2(0,T;H^1(\T^2\times\R^+))}\\
&\leq C\Big(\|(u_0^1-U_0,v_0^1-V_0)\|_{\h_\alpha^m},\|(u_0^1-U_0,v_0^1-V_0)\|_{\h_\alpha^m}\Big)\|(u_0^1-u_0^2,v_0^1-v_0^2)\|_{\h_\alpha^m},
\end{split}\end{equation}
where $T=T\Big(\|(u_0^1-U_0,v_0^1-V_0)\|_{\h_\alpha^m},\|(u_0^1-U_0,v_0^1-V_0)\|_{\h_\alpha^m}\Big).$
\end{define}

The second main result of this paper is the following ill-posedness of the nonlinear  problem \eqref{3dpd}.

\begin{theorem}\label{thm_non}
Under the same assumption as given in Theorem \ref{thm_lin}, the problem \eqref{3dpd} of the three dimensional nonlinear Prandtl equations is not locally well-posed in the sense of Definition \ref{def_non}.
\end{theorem}

\section{Linear instability}

In this section, we will prove Theorem \ref{thm_lin} to show the linear instability of three dimensional Prandtl equations. The proof is divided into the following four subsections.

\subsection{The linear instability mechanism}

In this subsection, we develop the method introduced in \cite{GV-D} to  analyse
the linear  instability mechanism of three dimensional Prandtl equations. More precisely,  we will find
some high frequency mode in the tangential variables that grow
exponentially in time.
To illustrate this kind of instability mechanism,
as in \cite{GV-D}, we first replace the background shear flow in \eqref{lin} by its initial data so that the background
profile is independent of time. Corresponding to \eqref{lin}, let us consider the following problem:
\begin{equation}\label{sim_lin}\begin{cases}
\pd_t u+(U_s\pd_x+V_s\pd_y)u+wU_s'-\pd_z^2u=0,\quad&{\rm in}\quad \Omega,\\
\pd_t v+(U_s\pd_x+V_s\pd_y)v+wV_s'-\pd_z^2v=0,\quad&{\rm in}\quad \Omega,\\
\pd_xu+\pd_yv+\pd_zw=0,\quad&{\rm in}\quad \Omega,\\
(u,v,w)|_{z=0}=0,\qquad\limz (u,v)=0.
\end{cases}\end{equation}
Noting that the coefficients in \eqref{sim_lin} are also independent of the tangential variables $x$ and $y$, it is convenient to work on
 the Fourier variables with respect to $x$ and $y$.

Recalling in Remark \ref{rem_3d} we assume that $U_s'(z_0)\neq 0$, so we can set \begin{equation}\label{def_a}
a\triangleq\frac{V_s'(z_0)}{U_s'(z_0)},
\end{equation}
and then, the condition
\eqref{ass} yields that the initial tangential velocity $V_s-a U_s$ has a non-degenerate critical point at $z=z_0$. Thus, we may look for solutions of \eqref{sim_lin} in the form of
\begin{equation}\label{expr}
(u,v,w)(t,x,y,z)~=~e^{ik(y-ax+\lambda(k)t)}(\hat u^k,\hat v^k,\hat w^k)(z),
\end{equation}
with some large integer $k$. To insure that the right hand side of \eqref{expr} is $2\pi-$periodic both in $x$ and $y$, that is, both of $k$ and $ak$ being integers, we need that the constant $a$ given in \eqref{def_a} is a rational number, this condition can be easily satisfied. Indeed, the assumption $U_s'(z_0)\neq 0$ and condition \eqref{ass} imply that
\begin{equation}
~U_s'(z)\neq0,\quad V_s'(z)U_s''(z)\neq U_s'(z)V_s''(z)
\end{equation}
holds in a neighborhood of $z_0$. So, by using the  continuity of $(V_s',U_s', V_s'',U_s'')$ and the denseness of the rational numbers in $\R$, there is a point $z_1$ in the neighborhood of $z_0$, such that the condition \eqref{ass} holds for $z_1$ and $\frac{V_s'(z_1)}{U_s'(z_1)}$ is a rational number.
Therefore, in the following discussion we can always assume that
\begin{equation}\label{def-a}
a~=
\frac{l}{q},
 \end{equation}
for some co-prime integers $l$ and $q$.

  Letting $\ep\triangleq \frac{1}{k}\ll1$,  combining \eqref{expr} with the divergence free condition in \eqref{sim_lin}, we rewrite \eqref{expr} in the form:
\begin{equation}\label{sol}\begin{cases}
(u,v)(t,x,y,z)~&=~e^{i\epsilon^{-1}(y-ax+\lambda_\ep t)}(u_\ep,v_\ep)(z),\\
\quad w~(t,x,y,z)~&=~-i\ep^{-1}e^{i\ep^{-1}(y-ax+\lambda_\ep t)}w_\ep(z).
\end{cases}\end{equation}
Then, the divergence free condition in \eqref{sim_lin} yields that
\begin{equation}\label{diver}
-au_\ep+v_\ep~=~w_\ep'.
\end{equation}
Substituting \eqref{sol} into \eqref{sim_lin}, we obtain
\begin{equation}\label{fr_lin}\begin{cases}
(\lambda_\ep+V_s-aU_s)u_\ep-U_s'w_\ep+i\ep u_\ep^{(2)}=0,\\
(\lambda_\ep+V_s-aU_s)v_\ep-V_s'w_\ep+i\ep v_\ep^{(2)}=0,\\
(u_\ep,v_\ep,w_\ep)(0)=0,\quad\limz (u_\ep,v_\ep)=0.
\end{cases}\end{equation}
Set
\begin{equation}\label{def_wa}
W_s(z)~\triangleq~ (V_s-aU_s)(z),
\end{equation}
combining \eqref{diver} with \eqref{fr_lin} implies that
\begin{equation}\label{fr_w}\begin{cases}
(\lambda_\ep+W_s)w_\ep'-W_s'w_\ep+i\ep w_\ep^{(3)}=0,\\
w_\ep(0)=w_\ep'(0)=0.
\end{cases}\end{equation}
Note that the equation for $w_\ep(z)$ in \eqref{fr_w} is the same as (2.3) studied in \cite{GV-D}
for two dimensional Prandtl equations. Therefore, according to \cite{GV-D}, we have the following result:
\begin{lemma}\label{lem_2d}
For the equation \eqref{fr_w}, if $z_0$ is a non-degenerate critical point of $W_s(z)$, then $w_\ep(z)$ has the following formal approximate expansion in $\ep:$
\begin{equation}\label{ex_w}\begin{cases}
\lambda_\ep~&\sim~-W_s(z_0)+\ep^{\frac{1}{2}}\tau,\\
w_\ep(z)~&\sim~
H(z-z_0)\big[W_s(z)-W_s(z_0)+\ep^{\frac{1}{2}}\tau\big]+\ep^{\frac{1}{2}}W(\frac{z-z_0}{\ep^{\frac{1}{4}}}),
\end{cases}\end{equation}
where $H$ is the Heaviside function, $\tau$ is a complex constant with $\Im\tau<0$, and
 the function $W(Z)$ solves the following ODE:
\begin{equation}\label{eq_w}\begin{cases}
\Big(\tau+W_s''(z_0)\frac{Z^2}{2}\Big)W'-W_s''(z_0)ZW+iW^{(3)}=0,\quad Z\neq0,\\
[W]\big|_{Z=0}=-\tau,~[W']\big|_{Z=0}=0,~[W'']\big|_{Z=0}=-W_s''(z_0),\\
\lim\limits_{Z\rightarrow\pm\infty} W~=~0,\quad  exponentially,
\end{cases}\end{equation}
where the notation $[u]\big|_{Z=0}=\lim\limits_{\delta_1\to 0+}u(\delta_1)-\lim\limits_{\delta_2\to 0-}u(\delta_2)$ denotes the jump of a related function $u(Z)$ across $Z=0$.
\end{lemma}

As in \cite{GV-D}, the asymptotic expansion of the solution $w_\ep(z)$ given in \eqref{ex_w} shows that
the approximate solution of $w_\ep(z)$ can be divided into the "regular" part and the "shear layer" part as:
\[ w_\ep^a(z)~:=~w_\ep^{reg}(z)+w_\ep^{sl(z)}\]
with
\[w_\ep^{reg}~\triangleq~H(z-z_0)\big[W_s(z)-W_s(z_0)+\ep^{\frac{1}{2}}\tau\big],
\]
and
\[w_\ep^{sl}~\triangleq~\ep^{\frac{1}{2}}W(\frac{z-z_0}{\ep^{\frac{1}{4}}}).\]
Note that $w_\ep^{reg}(0)=(w_\ep^{reg})'(0)=0.$ The "shear layer" part $w_\ep^{sl}$ is to cancel  the discontinuities of the "regular" part $w_\ep^{reg}$ at $z=z_0$, such that the approximation $w_\ep^{a}\in C^2(\R^+)$.

The formal asymptotic expansion $\eqref{ex_w}_1$ for the eigenvalue indicates strong instability of \eqref{sim_lin}, that is, back to the Fourier representation \eqref{sol}, the tangential velocity $(u,v)$ grows like $e^{\frac{t}{\sqrt{\ep}}}$. To complete this process, we will construct the formal approximation of $(u_\ep,v_\ep)(z)$.
The construction of $(u_\ep,v_\ep)(z)$ is based on the relation \eqref{diver} and the approximation \eqref{ex_w}, which implies that
\begin{equation}\label{ex-w}
(v_\ep-au_\ep)(z)~\sim~H(z-z_0)W_s'(z)+\ep^{\frac{1}{4}}W'(\frac{z-z_0}{\ep^{\frac{1}{4}}}).
\end{equation}
From \eqref{ex-w}, we assume that
the formal approximation for $(u_\ep,v_\ep)(z)$ can be chosen as follows:
\begin{equation}\label{ex_uv}\begin{cases}
u_\ep(z)~\sim~H(z-z_0)U_s'(z)+h(\frac{z-z_0}{\ep^{\frac{1}{4}}})+\ep^{\frac{1}{4}}U(\frac{z-z_0}{\ep^{\frac{1}{4}}}),\\
v_\ep(z)~\sim~H(z-z_0)V_s'(z)+ah(\frac{z-z_0}{\ep^{\frac{1}{4}}})+\ep^{\frac{1}{4}}\big(aU+W'\big)(\frac{z-z_0}{\ep^{\frac{1}{4}}}),
\end{cases}\end{equation}
where the functions $h(Z)$ and $U(Z)$ is rapidly decay as $Z\rightarrow\pm\infty$
as explained in the following.
One can easily verifies that
\[H(z-z_0)\big(U_s'(z),V_s'(z)\big)\big|_{z=0}=0,\qquad\limz (u_\ep,v_\ep)~=~0.\]
Similar to the ``shear layer'' part $w_\ep^{sl}$ defined
in \eqref{ex_w}, the functions $h(Z)$ and $U(Z)$ are used
to cancel the discontinuities in $H(z-z_0)U_s'(z)$ and $H(z-z_0)V_s'(z)$ at $z=z_0$,
so that the approximations in \eqref{ex_uv} belongs to $C^1(\R^+)$. Moreover, $h(Z)$
and $U(Z)$ are used to balance the approximation in the orders
of  $O(\sqrt{\ep})$ and $O(\ep^{\frac{3}{4}})$ respectively.
For this, from \eqref{fr_lin}, \eqref{ex_w} and \eqref{ex_uv},  $h(Z)$ and $U(Z)$ satisfy the following problems respectively,
\begin{equation}\label{eq_h}\begin{cases}
\Big(\tau+W_s''(z_0)\frac{Z^2}{2}\Big)h-U_s'(z_0)W+ih''=0,\quad Z\neq0,\\
[h]\big|_{Z=0}=-U_s'(z_0),\quad[h']\big|_{Z=0}=0,\\%~[W'']\big|_{Z=0}=-W_a''(z_0),\\
\limZ h~=~0,
\end{cases}\end{equation}
and
\begin{equation}\label{eq_u}\begin{cases}
\Big(\tau+W_s''(z_0)\frac{Z^2}{2}\Big)U-U_s''(z_0)ZW+iU''=0,\quad Z\neq0,\\
[U]\big|_{Z=0}=0,\qquad [U']\big|_{Z=0}=-U_s''(z_0),\\%~[W'']\big|_{Z=0}=-W_a''(z_0),\\
\limZ U~=~0.
\end{cases}\end{equation}

Comparing the problem \eqref{eq_h} with \eqref{eq_u}, respectively \eqref{eq_w}, it follows that $\frac{U_s'(z_0)}{W_s''(z_0)}W''(Z)$, respectively $\frac{U_s''(z_0)}{W_s''(z_0)}W'(Z)$, solves the problem \eqref{eq_h}, respectively \eqref{eq_u}. Consequently,
we can choose the formal expansion of $(u_\ep,v_\ep)(z)$ as
\begin{equation}\label{ex_uv1}\begin{cases}
u_\ep(z)~\sim~H(z-z_0)U_s'(z)+\frac{U_s'(z_0)}{W_s''(z_0)}W''(\frac{z-z_0}{\ep^{\frac{1}{4}}})+\ep^{\frac{1}{4}}\frac{U_s''(z_0)}{W_s''(z_0)}W'(\frac{z-z_0}{\ep^{\frac{1}{4}}}),\\
v_\ep(z)~\sim~H(z-z_0)V_s'(z)+\frac{V_s'(z_0)}{W_s''(z_0)}W''(\frac{z-z_0}{\ep^{\frac{1}{4}}})+\ep^{\frac{1}{4}}\frac{V_s''(z_0)}{W_s''(z_0)}W'(\frac{z-z_0}{\ep^{\frac{1}{4}}}).
\end{cases}\end{equation}

Therefore, we have concluded the following results for the reduced boundary value problem \eqref{sim_lin}.

\begin{prop}\label{sim-lin}

For the large frequency $k=\frac{1}{\epsilon}$, the approximate solutions of the problem \eqref{sim_lin} can be expressed as \eqref{sol} with
\begin{equation}\label{ex_all}\begin{cases}
\lambda_\ep~&\sim~-W_s(z_0)+\ep^{\frac{1}{2}}\tau,\\
w_\ep(z)~&\sim~H(z-z_0)\Big[W_s(z)-W_s(z_0)+\ep^{\frac{1}{2}}\tau\Big]
+\ep^{\frac{1}{2}}W(\frac{z-z_0}{\ep^{\frac{1}{4}}}),\\
(u_\ep(z),v_\ep(z))~&\sim~H(z-z_0)(U_s'(z),V_s'(z))+
W''(\frac{z-z_0}{\ep^{\frac{1}{4}}})\frac{1}{W_s''(z_0)}\big(U_s'(z_0),V_s'(z_0)\big),\\
&\quad+\ep^{\frac{1}{4}}W'(\frac{z-z_0}{\ep^{\frac{1}{4}}})\frac{1}{W_s''(z_0)}\big(U_s''(z_0),V_s''(z_0)\big),
\end{cases}\end{equation}
where the complex constant $\tau$ and function $W(Z)$ are given in Lemma \ref{lem_2d}.

\end{prop}

\begin{remark}\label{rem_tau}

Recalling from \cite{GV-D}, we know that the pair $\big(\tau, W(Z)\big)$ given in Lemma \ref{lem_2d} has the following form
\begin{equation}\label{ch_w}\begin{cases}
\tau~&=~\big|\frac{W_s''(z_0)}{2}\big|^{\frac{1}{2}}~\tilde\tau,\\
W(Z)~&=~\big|\frac{W_s''(z_0)}{2}\big|^{\frac{1}{2}}\Big[\big(\tilde\tau+\big|\frac{W_s''(z_0)}{2}\big|^{\frac{1}{2}}Z^2\big)\tilde W\big(\big|\frac{W_s''(z_0)}{2}\big|^{\frac{1}{4}}Z\big)-1_{\R^+}\big(\tilde\tau+\big|\frac{W_s''(z_0)}{2}\big|^{\frac{1}{2}}Z^2\big)\Big].
\end{cases}\end{equation}
where the complex constant $\tilde\tau$ has a negative imaginary part, i.e., $\Im\tilde\tau<0,$ and the function $\tilde W(\tilde Z)$ is a smooth solution of the following third order ordinary differential equation:
\begin{equation}\label{SC}\begin{cases}
\big(\tilde\tau+sign(W_s''(z_0))\tilde Z^2\big)^2\frac{d}{d\tilde Z}\tilde W+i\frac{d^3}{d\tilde Z^3}\Big(\big(\tau+sign(W_s''(z_0))\tilde Z^2\big)\tilde W\Big)=0,\\
\lim\limits_{\tilde Z\rightarrow-\infty}\tilde W~=~0,\quad\lim\limits_{\tilde Z\rightarrow+\infty}\tilde W~=~1.
\end{cases}\end{equation}

\end{remark}

\subsection{Construction of approximate solutions}

Inspiring by the construction of approximate solutions to the simplified problem \eqref{sim_lin} given in
the above subsection, and also by the argument given in \cite{GV-D}, we are going to construct the approximate solution of the original linearized problem \eqref{lin}.

Let $(u^s(t,z),v^s(t,z))$ satisfy the assumptions of Theorem \ref{thm_lin},
and denote by
\[ w_a^s(t,z)~\triangleq~v^s(t,z)-a u^s(t,z),\]
where the constant $a$ is given in \eqref{def_a}.
Then, we know that $z_0$ is a non-degenerate critical point of $w_a^s(0,z)$. Without loss of generality, we assume that $\pd_z^2w_a^s(0,z_0)<0$, then the differential equation
\begin{equation}\label{def_alpha}\begin{cases}
\pd_t\pd_zw_a^s\big(t,f(t)\big)+\pd_z^2w_a^s\big(t,f(t)\big)f'(t)=0,\\
f(0)~=~z_0,
\end{cases}\end{equation}
defines a non-degenerate critical point $f(t)$ of $w_a^s(t,\cdot)$ when $0<t<t_0$ for some small $t_0>0$. Note that such $f(t)$ can also be determined by the following equation:
\[\frac{d}{dt}\Big(\frac{v_z^s\big(t,f(t)\big)}{u_z^s\big(t,f(t)\big)}\Big)=0,\quad f(0)=z_0.\]

Since the approximation solution of \eqref{sim_lin} given in Proposition \ref{sim-lin} is obtained with the background state being frozen at the initial data $(u^s,v^s)|_{t=0}=(U_s,V_s)(z)$, to construct approximate solutions of the original problem \eqref{lin} with background state being the shear flow in the time interval $0<t<t_0$, we need to do some modification as in \cite{GV-D}. Recall Proposition \ref{sim-lin} and Remark \ref{rem_tau} in the above subsection, let $\tau, W(Z)$ be given in \eqref{ch_w}, and set
\begin{equation}\begin{split}\label{def_phi}
 W_{sl}(Z)~&:=~\Big(\tau-Z^2\Big) W( Z)-1_{\R^+}\Big(\tau-Z^2\Big).

\end{split}\end{equation}
Then, for $\ep>0$ we introduce
\begin{equation}\label{app_lam}
\lambda_\ep(t)~:=~-w_a^s(t,f(t))+\ep^{\frac{1}{2}}\big|\frac{\pd_z^2w_a^s(t,f(t))}{2}\big|^{\frac{1}{2}}~\tau,
\end{equation}
and the ``regular'' part of velocities
\begin{equation}\label{app_reg}\begin{cases}
U_\ep^{reg}(t,z)=H(z-f(t))u_z^s(t,z),\quad
V_\ep^{reg}(t,z)=H(z-f(t))v_z^s(t,z),\\[3mm]
W_\ep^{reg}(t,z)=H(z-f(t))\Big[w_a^s(t,z)-w_a^s(t,f(t))+\ep^{\frac{1}{2}}\big|\frac{\pd_z^2w_a^s(t,f(t))}{2}\big|^{\frac{1}{2}} \tau\Big],
\end{cases}\end{equation}
as well as the ``shear layer'' part of $w_\ep$
\begin{equation}\label{app_sl}\begin{split}
W_\ep^{sl}(t,z)~&:=~\ep^{\frac{1}{2}}\varphi(z-f(t))\big|\frac{\pd_z^2w_a^s(t,f(t))}{2}\big|^{\frac{1}{2}} W_{sl}\Big(\big|\frac{\pd_z^2w_a^s(t,f(t))}{2}\big|^{\frac{1}{4}}\cdot\frac{z-f(t)}{\ep^{\frac{1}{4}}}\Big).
\end{split}\end{equation}
Here, $\varphi$ is a smooth truncation function near 0, and $W_{sl}$ is given in \eqref{def_phi}. Therefore, from Proposition \ref{sim-lin} and by \eqref{app_reg} and \eqref{app_sl}, the approximate solution of the problem \eqref{lin} can be defined as:
\begin{equation}\label{ap_lin}
(u_\ep,v_\ep,w_\ep)(t,x,y,z)=e^{i\ep^{-1}(y-ax)}\big(U_\ep,V_\ep,W_\ep\big)(t,z)
\end{equation}
with
\begin{equation}\begin{split}\label{expre_uv}
\big(U_\ep,V_\ep\big)(t,z)~&=~ie^{i\ep^{-1}\int_0^t\lambda_\ep(s)ds}\Big\{\big(U_\ep^{reg},V_\ep^{reg}\big)(t,z)
+\frac{\pd_z^2W_\ep^{sl}(t,z)}{\pd_z^2w_a^s(t,f(t))}\big(u_z^s,v_z^s\big)(t,f(t))\\
&\qquad\qquad\qquad\qquad+\frac{\pd_zW_\ep^{sl}(t,z)}{\pd_z^2w_a^s(t,f(t))}\big(\pd_z^2u^s,\pd_z^2v^s\big)(t,f(t))\Big\},\\
W_\ep(t,z)~&=~\ep^{-1}e^{i\ep^{-1}\int_0^t\lambda_\ep(s)ds}\Big(W_\ep^{reg}(t,z)+W_\ep^{sl}(t,z)\Big).
\end{split}\end{equation}
 Moreover, in order that the function \eqref{ap_lin} is $2\pi-$periodic in $x$ and $y$, we take $\ep=\frac{1}{qk}$ with the integers $q$ given in \eqref{def-a} and $k\in\N$.

It is straightforward to check that for $(u_\ep,v_\ep,w_\ep)$ defined in \eqref{ap_lin},
\[(u_\ep,v_\ep,w_\ep)|_{z=0}=0,\qquad\limz (u_\ep,v_\ep)=0,\]
 and the divergence free condition holds. Also, $$(u_\ep,v_\ep)(t,x,y,z)=e^{i\ep^{-1}(y-ax)}(U_\ep,V_\ep)(t,z)$$ is analytic in the tangential variables $x,y$ and $H_\alpha^{2}$ in $z$. Moreover, there are positive constants $C_0$ and $\sigma_0$, independent of $\ep$, such that
\begin{equation}\label{bound_app}
C_0^{-1}e^{\frac{\sigma_0t}{\sqrt{\ep}}}~\leq~\|(U_\ep,V_\ep)(t,\cdot)\|_{L_\alpha^{2}}
~\leq~C_0e^{\frac{\sigma_0t}{\sqrt{\ep}}}.
\end{equation}

Plugging the relation \eqref{ap_lin} into the original linearized Prandtl equations \eqref{lin}, it follows that
\begin{equation}\label{lin_ap}\begin{cases}
\pd_t u_\ep+(u^s\pd_x+v^s\pd_y)u_\ep+w_\ep u_z^s-\pd_z^2u_\ep=r_\ep^{1},\\
\pd_t v_\ep+(u^s\pd_x+v^s\pd_y)v_\ep+w_\ep v_z^s-\pd_z^2v_\ep=r_\ep^2,\\
\pd_xu_\ep+\pd_yv_\ep+\pd_zw_\ep=0,\quad{\rm in}\quad \Omega,\\
(u_\ep,v_\ep,w_\ep)|_{z=0}=0,\qquad\limz (u_\ep,v_\ep)=0,
\end{cases}\end{equation}
where the remainder term is represented by
\begin{equation}\label{def_err}
(r_\ep^1,r_\ep^2)(t,x,y,z):=~e^{i\ep^{-1}(y-ax)}(R_\ep^1,R_\ep^2)(t,z)
\end{equation}
with
\begin{equation}\label{r}\begin{split}
R_\ep^1&=\pd_t U_\ep+i\ep^{-1}w_a^s(t,z)U_\ep-\pd_z^2U_\ep+u_z^s(t,z)W_\ep,\\
R_\ep^2&=\pd_t V_\ep+i\ep^{-1}w_a^s(t,z)V_\ep-\pd_z^2V_\ep+v_z^s(t,z)W_\ep.
\end{split}\end{equation}
Note that the representation \eqref{app_reg}  implies
\begin{equation}\label{eq_reg}
(\pd_t-\pd_z^2)U_\ep^{reg}=(\pd_t-\pd_z^2)V_\ep^{reg}=0,\quad z\neq f(t).
\end{equation}
Then, from the representation of $(U_\ep,V_\ep)$ and $W_\ep$ given in \eqref{expre_uv} and by using the equations \eqref{SC} and \eqref{eq_reg}, we conclude that for $z\neq f(t),$
\begin{equation}\label{r1}\begin{split}
R_\ep^1(t,z
&=e^{i\ep^{-1}\int_0^t\lambda_\ep(s)ds}\Big\{-\ep^{-1}\Big[w_a^s(t,z)
-w_a^s(t,f(t))-\pd_z^2w_a^s(t,f(t))\frac{(z-f(t))^2}{2}\Big]\\
&\qquad\qquad\cdot\Big
\frac{u_z^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_z^2W_\ep^{sl}+\frac{\pd_z^2u^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_zW_\ep^{sl}\Big]\\
&\qquad+\ep^{-1}\Big[u_z^s(t,z)-u_z^s(t,f(t))-u_{zz}^s(t,f(t))(z-f(t))\Big]W_\ep^{sl}\\
&\qquad+i\pd_t\Big
\frac{u_z^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_z^2W_\ep^{sl}+\frac{\pd_z^2u^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_zW_\ep^{sl}\Big)
+O(\ep^\infty)\Big\},
\end{split}\end{equation}
and
\begin{equation}\label{r2}\begin{split}
R_\ep^2(t,z
&=e^{i\ep^{-1}\int_0^t\lambda_\ep(s)ds}\Big\{-\ep^{-1}\Big[w_a^s(t,z)
-w_a^s(t,f(t))-\pd_z^2w_a^s(t,f(t))\frac{(z-f(t))^2}{2}\Big]\\
&\qquad\qquad\cdot\Big
\frac{v_z^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_z^2W_\ep^{sl}+\frac{\pd_z^2v^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_zW_\ep^{sl}\Big]\\
&\qquad+\ep^{-1}\Big[v_z^s(t,z)-v_z^s(t,f(t))-v_{zz}^s(t,f(t))(z-f(t))\Big]W_\ep^{sl}\\
&\qquad+i\pd_t\Big
\frac{v_z^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_z^2W_\ep^{sl}+\frac{\pd_z^2v^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_zW_\ep^{sl}\Big)
+O(\ep^\infty)\Big\}.
\end{split}\end{equation}
The terms $O(\ep^\infty)$ in \eqref{r1} and \eqref{r2} represent the remainders
with exponential decay in $z$ that follows from the fact that $W_\ep^{sl}$ decays exponentially and the derivatives of $\varphi(\cdot-f(t))$ vanish near $f(t)$.  Then,  with the same $\sigma_0$ given in \eqref{bound_app}, we have that
$(R_\ep^1,R_\ep^2)(t,z)$ satisfy
\begin{equation}\label{bound_r}
\|(R_\ep^1,R_\ep^2)(t,\cdot)\|_{L_\alpha^{2}}\leq C_1\ep^{-\frac{1}{4}}e^{\frac{\sigma_0t}{\sqrt{\ep}}},
\end{equation}
where the constant $C_1$ is independent of $\ep$.

Therefore, we conclude
\begin{prop}\label{con_app}
For the linear problem \eqref{lin}, the approximate solution $(u_\ep,v_\ep,w_\ep)$ given in \eqref{ap_lin} and \eqref{expre_uv}, satisfies the problem \eqref{lin_ap}, where the source term $(r_\ep^1,r_\ep^2)$ in the form \eqref{def_err} has the bound \eqref{bound_r}.
\end{prop}

\begin{remark}\label{rem_r}
The estimate \eqref{bound_r} follows from the fact that the "shear layer" part $\pd_z^2W_\ep^{sl}$ cancels the terms $\ep^{-1}u_z^s(t,f(t))W_\ep^{sl}$ and $\ep^{-1}v_z^s(t,f(t))W_\ep^{sl}$ in \eqref{r1} and \eqref{r2} respectively, by using the equation \eqref{eq_h} for instance. And this
error bound leads to the choice of  $\mu<\frac{1}{4}$ in \eqref{est_in}. It is slight different from the two dimensional case studied in \cite{GV-D} where  $\mu<\frac{1}{2}$, because here we require that the initial data of $u^s$ and $v^s$ do not degenerate simultaneously at a point, $z=z_0$, while in the two dimensional problem, it is assumed that there is a degeneracy at the critical point.
\end{remark}

\subsection{Proof of Theorem \ref{thm_lin}(i)}

At this stage, based on the approximate solution given in Proposition \ref{con_app}, we can use the method from \cite{GV-D} to prove  Theorem \ref{thm_lin}. We
now  sketch the proof as follows.

\underline{\it Verification of \eqref{est_in} for the tangential differential operator by contradiction}.
Suppose that \eqref{est_in} does not hold for $\pd_x$, that is, for all $\sigma>0$, there exists $\delta>0,m\geq0$ and $\mu\in[0,\frac{1}{4})$, that is,
\begin{equation}\label{pr_x}
\sup\limits_{0\leq s\leq t\leq \delta}\|e^{-\sigma(t-s)\sqrt{|\pd_x|}}T(t,x)\|_{\cl(\h_\alpha^m,\h_\alpha^{m-\mu})}<+\infty.
\end{equation}
Introduce the operator
\begin{equation*}
T_\ep(t,s):~L_{\alpha}^2(\R^+)\mapsto L_{\alpha}^2(\R^+)
\end{equation*}
as
\begin{equation}\label{def_t}
 T_\ep(t,s)\big((U_0,V_0)\big)~:=~e^{-i\ep^{-1}(y-ax)}T(t,s)\Big(e^{i\ep^{-1}(y-ax)}(U_0,V_0)\Big)
\end{equation}
with $T(t,s)$ being defined in \eqref{def_T}.
From \eqref{pr_x}, we have
\begin{equation}\label{est_tep}
\|T_\ep(t,s)\|_{\cl(L_\alpha^{2})}~\leq~C_2\ep^{-\mu}e^{\frac{\sqrt{a}\sigma(t-s)}{\sqrt{\ep}}},\quad \forall ~0\leq s\leq t\leq\delta
\end{equation}
for a constant $C_2$ independent of $\ep$.

Denote by
\[L_\ep~:=~e^{-i\ep^{-1}(y-ax)}~L~e^{i\ep^{-1}(y-ax)},\]
where $L$ is the linearized Prandtl operator around the shear flow $(u^s,v^s,0)$.
Let $(U,V)(t,z)$ be a solution to the problem
\[\begin{cases}
\pd_t(U,V)+L_\ep(U,V)~=~0,\\
(U,V)|_{t=0}~=~(U_\ep,V_\ep)(0,z).
\end{cases}\]

From the definition \eqref{def_t}, we have that
\[(U,V)(t,z)~=~T_\ep(t,0)\big((U_\ep,V_\ep)(0,z)\big).\]
Then, from \eqref{est_tep} it follows that for all $t\leq\delta,$
\begin{equation}\label{up_bound}
\|(U,V)(t,\cdot)\|_{L_\alpha^{2}}\leq C_2\ep^{-\mu}e^{\frac{\sqrt{a}\sigma t}{\sqrt{\ep}}}\|(U_\ep,V_\ep)(0,\cdot)\|_{L_\alpha^{2}}
\leq C_3\ep^{-\mu}e^{\frac{\sqrt{a}\sigma t}{\sqrt{\ep}}}
\end{equation}
holds for a constant $C_3$ independent of $\ep$. From \eqref{r1} and \eqref{r2},
we know that
\[\pd_t(U_\ep,V_\ep)+L_\ep(U_\ep,V_\ep)~=~(R_\ep^1,R_\ep^2).\]
Thus, the difference $(\tilde U,\tilde V):=(U,V)-(U_\ep,V_\ep)$ can be obtained by the Duhamel representation:
\begin{equation}\label{eq_dif}
(\tilde U,\tilde V)(t,\cdot)~=~\int_0^tT_\ep(t,s)\big((R^1_\ep,R_\ep^2)(s,\cdot)\big)ds,\quad \forall~t\leq\delta.
\end{equation}
Combining \eqref{bound_r}, \eqref{est_tep} and \eqref{eq_dif},
and choosing $\sqrt{a}\sigma<\sigma_0$ yields that
\begin{equation}\label{est_dif}
\|(\tilde U,\tilde V)(t,\cdot)\|_{L^2_\alpha}\leq C_1C_2\ep^{-\mu-\frac{1}{4}}
\int_0^te^{\frac{\sqrt{a}\sigma(t-s)}{\sqrt{\ep}}}e^{\frac{\sigma_0s}{\sqrt{\ep}}}ds
\leq C_4\ep^{\frac{1}{4}-\mu}e^{\frac{\sigma_0t}{\sqrt{\ep}}},
\end{equation}
where the constant $C_4>0$ is independent of $\ep$. Then, by using \eqref{bound_app},
we obtain that for $t<\delta$ and sufficiently small $\ep$,
\begin{equation}\label{low_bound}\begin{split}
\|(U,V)(t,\cdot)\|_{L_\alpha^2}&\geq\|(U_\ep,V_\ep)(t,\cdot)\|_{L_\alpha^2}
-\|(\tilde U,\tilde V)(t,\cdot)\|_{L_\alpha^2}\\
&\geq C_0^{-1}e^{\frac{\sigma_0t}{\sqrt{\ep}}}-C_4\ep^{\frac{1}{4}-\mu}e^{\frac{\sigma_0t}{\sqrt{\ep}}}
\geq \frac{1}{2}C_0^{-1}e^{\frac{\sigma_0t}{\sqrt{\ep}}}.
\end{split}\end{equation}
As $\sqrt{a}\sigma<\sigma_0,$ comparing \eqref{up_bound} with \eqref{low_bound},
the contradiction arises when $t>\frac{\sqrt{\ep}}{\sigma_0-\sqrt{a}\sigma}\big(\ln(2C_0C_3)-\ln\ep\big)$
with sufficiently small $\ep$. Thus, the proof of Theorem \ref{thm_lin}(i) is completed.

\subsection{Proof of Theorem \ref{thm_lin}(ii)}

The aim of this subsection is to prove Theorem \ref{thm_lin}(ii). By comparing \eqref{est_in} with \eqref{est_in1}, we only need to show that there exists an initial data $(U_s,V_s)$ for
the shear flow to \eqref{shear} such that \eqref{est_in} still holds for arbitrary $\mu>0.$
Recall the  proof of part i) in the above subsection, the task can be attributed to find some $(U_s,V_s)$ such that the remainder $(R_\ep^1,R_\ep^2)(t,z)$, generated in \eqref{r} by the approximation \eqref{expre_uv}, has the following estimate:
\begin{equation}\label{est_r}
\|(R_\ep^1,R_\ep^2)(t,\cdot)\|_{L_\alpha^{2}}\leq C\big(\ep^{N}+t^{2N}\big)e^{\frac{\sigma_0t}{\sqrt{\ep}}}
\end{equation}
for some $N\in\R^+,N+\frac{1}{2}>\mu$. Once this is achieved, as in \cite{GV-D}, the desired conclusion holds.

Similar to \cite{GV-D}, the special initial shear layer $(U_s,V_s)(z)$ can be chosen such that
\[\begin{split}
(U_s,V_s)|_{z=0}=0,\quad(U_s,V_s)(z)\rightarrow(U_0,V_0),~{\rm exponentially~as}~z\rightarrow+\infty,
\end{split}\]
and in a small neighborhood of $z_0$,
\begin{equation}\label{sp_uv}\begin{cases}
U_s(z)=U_s''(z_0)\frac{(z-z_0)^2}{2}+q(z-z_0)+U_s(z_0),\\ V_s(z)=V_s''(z_0)\frac{(z-z_0)^2}{2}+l(z-z_0)+V_s(z_0),
\end{cases}\end{equation}
where $q,l$ are integers given in \eqref{def-a}, and the constants $U_s''(z_0),U_s(z_0),$ $V_s''(z_0),V_s(z_0)$ satisfy
\[V_s''(z_0)-aU_s''(z_0)\neq0,\quad U_s(z_0)\neq0\]
with the constant $a$ being given in \eqref{def_a}.
Then, for such $(U_s,V_s)(z)$, we will show that \eqref{est_r} holds.
We only estimate the term
$R_\ep^1$, as the same argument works for $R_\ep^2$. From \eqref{r1},
decompose $R_\ep^1$ as follows for $z\neq f(t)$:
\begin{equation}\label{div}
R_\ep^1~:=~e^{i\ep\int_0^t\lambda(s)ds}\Big(R_{\ep,1}^1+R_{\ep,2}^1+R_{\ep,3}^1+O(\ep^\infty)\Big),
\end{equation}
where
\[\begin{split}
R_{\ep,1}^1(t,z)&=-\ep^{-1}\Big[w_a^s(t,z)
-w_a^s(t,f(t))-\pd_z^2w_a^s(t,f(t))\frac{(z-f(t))^2}{2}\Big]\\
&\qquad\qquad\cdot\Big[
\frac{u_z^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_z^2W_\ep^{sl}+\frac{\pd_z^2u^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_zW_\ep^{sl}\Big],\\
R_{\ep,2}^1(t,z)&=\ep^{-1}\Big[u_z^s(t,z)-u_z^s(t,f(t))-u_{zz}^s(t,f(t))(z-f(t))\Big]W_\ep^{sl},\\
R_{\ep,3}^1(t,z)&=i\pd_t\Big(\frac{u_z^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_z^2W_\ep^{sl}+\frac{\pd_z^2u^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_zW_\ep^{sl}\Big).
\end{split}\]
Therefore, it remains to show that
\begin{equation}\label{bound_r1}
\|R_{\ep,i}^1(t,\cdot)\|_{L_\alpha^{2}}~\leq~C(\ep^N+t^{2N}),\quad i=1,2,3.
\end{equation}

Firstly, for $R^1_{\ep,1}$, note that the function $w_a^s(t,z)=v^s(t,z)-au^s(t,z)$ satisfies
\begin{equation}\label{heat}
\pd_t w_a^s-\pd_z^2 w_a^s=0,
\end{equation}
 and in a small neighborhood of $z_0$,
\begin{equation}\label{sp_w}
w_a^s(0,z)=\big(V_s''(z_0)-aU_s''(z_0)\big)\frac{(z-z_0)^2}{2}+V(z_0)-aU_s(z_0).
\end{equation}
Thus, by using the Taylor expansion and \eqref{heat}, it follows that for any $N\in\N$,
\begin{equation}\label{est-r1}\begin{split}
&|R_{\ep,1}^1(t,z)|=\ep^{-1}\Big|\int_{f(t)}^z\frac{(\tilde z-f(t))^2}{2}\pd_z^3w_a^s(t,\tilde z)d\tilde z\Big|\cdot\Big|
\frac{u_z^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_z^2W_\ep^{sl}+\frac{\pd_z^2u^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}\pd_zW_\ep^{sl}\Big|\\
&\leq C\ep^{-1}\int_{f(t)}^z\frac{(\tilde z-f(t))^2}{2}\sum_{k=0}^{2N-1}\frac{t^k}{k!}\big|\pd_t^k\pd_z^3w_a^s(0,\tilde z)\big|d\tilde z\cdot\Big(\big|\pd_z^2W_\ep^{sl}\big|+\big|\pd_zW_\ep^{sl}\big|\Big)+O(t^{2N})\\
&\leq C\ep^{-1}\int_{f(t)}^z\frac{(\tilde z-f(t))^2}{2}\sum_{k=0}^{2N-1}\frac{t^k}{k!}\big|\pd_z^{3+2k}w_a^s(0,\tilde z)\big|d\tilde z\cdot\Big(\big|\pd_z^2W_\ep^{sl}\big|+\big|\pd_zW_\ep^{sl}\big|\Big)+O(t^{2N}).
\end{split}\end{equation}
From \eqref{sp_w}, we know that
\[\pd_z^{3+2k}w_a^s(0,z)~\equiv~0,\quad\quad \forall k\in\N,\]
when $z$ is in a small neighborhood  of $f(t)$ as $t>0$ is small. This implies
 that the integral in the last line of \eqref{est-r1} is supported away from $z=f(t)$.
Then, combining with the exponential decrease of the ``shear layer'' $W_\ep^{sl}$, \eqref{est-r1} yields
\begin{equation}\label{est_r1}
\|R_{\ep,1}^1(t,\cdot)\|_{L_\alpha^{2}}~\leq~C (\ep^N+t^{2N}).
\end{equation}
For the term $R_{\ep,2}^1$, we can use similar arguments as above to obtain
\begin{equation}\label{est_r2}
\|R_{\ep,2}^1(t,\cdot)\|_{L_\alpha^{2}}~\leq~C (\ep^N+t^{2N}).
\end{equation}

Next, from the expression \eqref{app_sl} of $W_\ep^{sl}(t,z)$ and the relation
\[\pd_tu^s=\pd_z^{2}u^s,\quad \pd_tw_a^s=\pd_z^{2}w_a^s,\]
a straightforward calculation implies that
 \begin{equation}\label{expre_r3}\begin{split}
 R_{\ep,3}^1(t,z)=&\pd_z^3u^s(t,f(t))g_1(t,z)+\pd_z^4u^s(t,f(t))g_2(t,z)+
 \pd_z^4w_a^s(t,f(t))g_3(t,z)\\
 &+f'(t)g_4(t,z)+O(\ep^\infty),
\end{split}\end{equation}
where  each function $g_i(t,z),1\leq i\leq4$ can be expressed as a linear combination of the terms
 \[
 O(\ep^{-\frac{1}{4}})\varphi(z-f(t))~W_{sl}^{(j)}\Big(\big|\frac{\pd_z^2w_a^s(t,f(t))}{2}\big|^{\frac{1}{4}}\cdot\frac{z-f(t)}{\ep^{\frac{1}{4}}}\Big),\quad 0\leq j\leq 3
 \]
 with $W_{sl}$ being defined in \eqref{def_phi}. Then, as for $R_{\ep,1}^1$, the first three terms on the right hand side of \eqref{expre_r3} have the same bounds as in \eqref{est_r1}. For the fourth term given in \eqref{expre_r3}, by noticing that from \eqref{def_alpha},
\[
f'(t)~=~-\frac{\pd_t\pd_zw_a^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))}
~=~\frac{\pd_z^3w_a^s(t,f(t))}{\pd_z^2w_a^s(t,f(t))},
\]
we can verify that $f'(t)g_4(t,z)$ also satisfies the same estimate as \eqref{est_r1}. In conclusion, we have
\begin{equation}\label{est_r3}
\|R_{\ep,3}^1(t,\cdot)\|_{L_\alpha^{2}}~\leq~C (\ep^N+t^{2N}).
\end{equation}
Thus, combining \eqref{est_r1}, \eqref{est_r2} and \eqref{est_r3}, we have the estimate \eqref{bound_r1}, and then obtain the proof of Theorem \ref{thm_lin}(ii) by taking $N$ large enough.

Finally, we state the following result to finish this section, which can be obtained by similar arguments as above,
\begin{prop}\label{prop_uni}
There exists a shear layer $(u^s,v^s)$ to \eqref{shear} with $u_z^s,v_z^s>0$, such that for all $\delta>0$,
\begin{equation}\label{prop_in}
\sup\limits_{0\leq s\leq t\leq\delta}\big\|T(t,s)\big\|_{\cl(\h_\alpha^{m},L^2)}
~=~+\infty,\quad \forall m>0.
\end{equation}
\end{prop}

\section{Nonlinear instability}

In this section, we will prove  the nonlinear ill-posedness result of the three dimensional Prandtl equations stated in Theorem \ref{thm_non}, it will mainly follow the argument of \cite{guo}. First, let us give a preliminary result on the uniqueness of solutions to the linear problem \eqref{lin} as follows.
\begin{lemma}\label{lem_uni}
Suppose that $(u^s,v^s)(t,z)$ is a solution to the problem \eqref{shear} satisfying that
\[\sup\limits_{t\geq0}\Big(\sup\limits_{z\geq0}|(u^s,v^s)|+\int_0^{+\infty}z|(\pd_zu^s,\pd_zv^s)|^2dz\Big)<+\infty.\]
Let $(u,v)\in L^\infty\big(0,T;L^2(\T^2\times\R^+)\big)$ with $(\pd_zu,\pd_zv)\in L^2\big((0,T)\times\T^2\times\R^+\big)$ be a solution to the problem \eqref{lin} with the vanish initial data $(u,v)|_{t=0}=0$. Then, for all $t>0$, $(u,v)\equiv0.$
\end{lemma}
The proof of this lemma is similar to the one given in \cite[Proposition 2.1]{GV-N} or  \cite[Proposition 2.2]{guo}, we omit it here for simplicity.

\vspace{.1in}
\begin{proof}[\bf{Proof of Theorem 2.}]
(1) First, by using \eqref{prop_in}, we know that for the shear flow $(u^s,v^s)$ given in Proposition \ref{prop_uni},  for fixed $\delta_0>0$ and any $m,n\in\N$, there exist $s_n,t_n$ with $0\leq s_n\leq t_n\leq\delta_0$, functions $(u_0^n,v_0^n)(x,y,z)$, and solutions $(u^n_L,v_L^n)$ to the linearized problem \eqref{lin}, such that $(u_L^n,v_L^n)|_{t=s_n}=(u_0^n,v_0^n)$ and
\begin{equation}\label{ass_lin}
\|(u_0^n,v_0^n)\|_{\h_\alpha^m}=1,\quad \|(u_L^n,v_L^n)(t_n)\|_{L^2}\geq n.
\end{equation}

(2) Now, we prove this theorem by contradiction. Assume that the problem \eqref{3dpd} is locally well-posedness for some integer $m\geq0$ in the sense of Definition \ref{def_non}. Denote by
\begin{equation}\label{ass_sh}
(u_{s_n}^s,v_{s_n}^s)(t,z)~:=~(u^s,v^s)(t+s_n,z),
\end{equation}
and
\begin{equation}\label{ass_init}
(u_{0,\delta}^n,v_{0,\delta}^n)(x,y,z)~:=~(u^s,v^s)(s_n,z)+\delta(u_0^n,v_0^n)(x,y,z)
\end{equation}
with a small positive constant $\delta$.
Let $(u_\delta^n,v_\delta^n)$ be the solution to the problem \eqref{3dpd} with the initial data $(u_\delta^n,v_\delta^n)|_{t=0}=(u_{0,\delta}^n,v_{0,\delta}^n)$. Thus, applying Definition \ref{def_non} to two solutions $(u_\delta^n,v_\delta^n)$ and $(u_{s_n}^s,v_{s_n}^s)$, it yields that there exist positive continuous functions $T(\cdot,\cdot)$ and $C(\cdot,\cdot)$, such that
\begin{equation}\label{est_non}\begin{split}
&\|(u_\delta^n-u_{s_n}^s,v_\delta^n-v_{s_n^n}^s)\|_{L^\infty(0,T;L^2(\T^2\times\R^+))}
+\|(u_\delta^n-u_{s_n}^s,v_\delta^n-v_{s_n^n}^s)\|_{L^2(0,T;H^1(\T^2\times\R^+))}\\
&\leq C\delta\|(u_0^n,v_0^n)\|_{\h_\alpha^m}=C\delta,
\end{split}\end{equation}
where $T=T\Big(\|(u_{0,\delta}^n-U_0,v_{0,\delta}^n-V_0)\|_{\h_\alpha^m},\|(u_{s_n}^s-U_0,v_{s_n}^s-V_0)\|_{\h_\alpha^m}\Big)$
and
$C=C\Big(\|(u_{0,\delta}^n-U_0,v_{0,\delta}^n-V_0)\|_{\h_\alpha^m},\|(u_{s_n}^s-U_0,v_{s_n}^s-V_0)\|_{\h_\alpha^m}\Big)$.

Combining \eqref{ass_lin}, \eqref{ass_sh} and \eqref{ass_init}, we know that $\|(u_{0,\delta}^n-U_0,v_{0,\delta}^n-V_0)\|_{\h_\alpha^m}$ and $\|(u_{s_n}^s-U_0,v_{s_n}^s-V_0)\|_{\h_\alpha^m}$ are uniformly bounded in $\delta$ and $n$. Thus, we can take the functions $T(\cdot,\cdot)$ and $C(\cdot,\cdot)$ independent of $\delta$ and $n$, and in the following we use $T,C$ to replace $T(\cdot,\cdot),C(\cdot,\cdot)$ for simplicity.

(3) From the estimate \eqref{est_non}, we know that the sequence $$(\tilde u_\delta^n,\tilde v_\delta^n):=\frac{1}{\delta}(u_\delta^n-u_{s_n}^s,v_\delta^n-v_{s_n^n}^s)$$
is bounded in $L^\infty\big(0,T;L^2(\T^2\times\R^+)\big)\cap L^2\big(0,T;H^1(\T^2\times\R^+)\big)$ uniformly in $\delta$ and $n$, which yields that there is $(u^n,v^n)$ such that, up to a subsequence, as $\delta\rightarrow0,$
\begin{equation}\label{con}
(\tilde u_\delta^n,\tilde v_\delta^n)~\rightharpoonup~(u^n,v^n),\quad{\rm weakly-*~in}\quad L^\infty\big(0,T;L^2(\T^2\times\R^+)\big)\cap L^2\big(0,T;H^1(\T^2\times\R^+)\big)
\end{equation}
and
\begin{equation}\label{est_con}
\|(u^n,v^n)\|_{L^\infty(0,T;L^2(\T^2\times\R^+))}+\|(u^n,v^n)\|_{L^2(0,T;H^1(\T^2\times\R^+))}~\leq~C,\quad\forall n\ge 1.
\end{equation}

Next, since both $(u_\delta^n,v_\delta^n)$ and $(u_{s_n}^s,v_{s_n}^s)$ solve the problem \eqref{3dpd},
we have
\begin{equation}\label{lin_eq}\begin{cases}
\pd_t(\tilde u_\delta^n,\tilde v_\delta^n)+\p_{n}(\tilde u_\delta^n,\tilde v_\delta^n)=\delta N(\tilde u_\delta^n,\tilde v_\delta^n),\\
(\tilde u_\delta^n,\tilde v_\delta^n)|_{t=0}=(u_0^n,v_0^n),
\end{cases}\end{equation}
where $\p_{n}$ is the linearized Prandtl operator at the shear profile $(u_{s_n}^s,v_{s_n}^s)$,
and $N(\cdot, \cdot)$ is the nonlinear term,
\begin{equation}\label{def_n}\begin{split}
N(\tilde u_\delta^n,\tilde v_\delta^n):=\Big(&-\tilde u_\delta^n\pd_x\tilde u_\delta^n-\tilde v_\delta^n\pd_y\tilde u_\delta^n+\int_0^z(\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n)d\tilde z\cdot\pd_z\tilde u_\delta^n,\\
&-\tilde u_\delta^n\pd_x\tilde v_\delta^n-\tilde v_\delta^n\pd_y\tilde v_\delta^n+\int_0^z(\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n)d\tilde z\cdot\pd_z\tilde v_\delta^n\Big).
\end{split}\end{equation}
Therefore, we want to show that the limit function $(u^n,v^n)$ satisfies the linearized Prandtl equations in the sense of distribution. For this, we only need to prove that the right hand side of the equation in \eqref{lin_eq} goes to zero as $\delta\to 0$ in the sense of distribution.

Indeed, note that from \eqref{def_n} the nonlinear term can be rewritten as
\[N(\tilde u_\delta^n,\tilde v_\delta^n)=\Big(N_1(\tilde u_\delta^n,\tilde v_\delta^n),N_2(\tilde u_\delta^n,\tilde v_\delta^n)\Big)\]
with
\[ \begin{split}
&N_1(\tilde u_\delta^n,\tilde v_\delta^n)=-\pd_x(\tilde u_\delta^n)^2-\pd_y(\tilde u_\delta^n\tilde v_\delta^n)+\pd_z\big(\int_0^z(\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n)d\tilde z\cdot\tilde u_\delta^n\big),\\
&N_2(\tilde u_\delta^n,\tilde v_\delta^n)=-\pd_x(\tilde u_\delta^n\tilde v_\delta^n)-\pd_y(\tilde v_\delta^n)^2+\pd_z\big(\int_0^z(\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n)d\tilde z\cdot\tilde v_\delta^n\big).
\end{split}\]
For any compact set $K$ of $[0,T]\times\T^2\times\R^+$ and smooth function $\varphi$ supported in $K$, we have
\[\begin{split}
&\Big|\int_{[0,T]\times\T^2\times\R^+}N_1(\tilde u_\delta^n,\tilde v_\delta^n)\varphi dxdydzdt\Big|\\
&\leq C_{K,\varphi}\int_{K}\Big(|\tilde u_\delta^n|^2+|\tilde u_\delta^n\tilde v_\delta^n|+\big|\int_0^z(\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n)d\tilde z\cdot\tilde u_\delta^n\big|\Big)dxdydzdt\\
&\leq C_{K,\varphi}\Big(\|\tilde u_\delta^n\|_{L^2(K)}^2+\|\tilde v_\delta^n\|_{L^2(K)}^2+\big\|\int_0^z(\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n)d\tilde z\big\|_{L^2(K)}^2\Big),
\end{split}\]
where $C_{K,\varphi}$ is a positive constant depending on $K$ and $W^{1,\infty}$ norm of $\varphi$. From the obvious inequality,
\[\big|\int_0^z(\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n)d\tilde z\big|\leq z^{\frac{1}{2}}\big(\int_{\R_z^+}|\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n|^2dz\big)^{\frac{1}{2}},\]
we get
\[\big\|\int_0^z(\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n)d\tilde z\big\|_{L^2(K)}^2\leq C_K\|\pd_x\tilde u_\delta^n+\pd_y\tilde v_\delta^n\|_{L^2}^2\leq C_K(\|\tilde u_\delta^n\|_{L^2(H^1)}^2+\|\tilde v_\delta^n\|_{L^2(H^1)}^2)\]
for some positive constant $C_K$ depending on $K$. Thus, it follows that
\[
\Big|\int_{[0,T]\times\T^2\times\R^+}N_1(\tilde u_\delta^n,\tilde v_\delta^n)\varphi dxdydz\Big|\leq C_{K,\varphi}(\|\tilde u_\delta^n\|_{L^2(H^1)}^2+\|\tilde v_\delta^n\|_{L^2(H^1)}^2).
\]
Similarly, one can deduce
\[
\Big|\int_{[0,T]\times\T^2\times\R^+}N_2(\tilde u_\delta^n,\tilde v_\delta^n)\varphi dxdydz\Big|\leq C_{K,\varphi}(\|\tilde u_\delta^n\|_{L^2(H^1)}^2+\|\tilde v_\delta^n\|_{L^2(H^1)}^2).
\]
Then, by using the uniform boundedness of $(\tilde u_\delta^n,\tilde v_\delta^n)$ in $L^\infty\big(0,T;L^2(\T^2\times\R^+)\big)\cap L^2\big(0,T;H^1(\T^2\times\R^+)\big)$ with respect to $\delta$, it implies that
the nonlinear term $\delta N(\tilde u_\delta^n,\tilde v_\delta^n)$ converges to zero in the sense of distribution.
Thus, letting $\delta\to 0$ in \eqref{lin_eq}, we obtain that $(u^n,v^n)$ solves the following linear problem in the sense of distribution,
\begin{equation}\label{lin_pr}\begin{cases}
\pd_t(u^n, v^n)+\p_{n}(u^n, v^n)=0,\\
(u^n, v^n)|_{t=0}=(u_0^n,v_0^n).
\end{cases}\end{equation}

(4) Shift the time variable $t$ to $t-s_n$ in \eqref{lin_pr}, and denote by
\[(\tilde u^n,\tilde v^n)(t,\cdot)~:=~(u^n,v^n)(t-s_n,\cdot).\]
Then, \eqref{lin_pr} becomes
\begin{equation}\label{lin_pr1}\begin{cases}
\pd_t(\tilde u^n,\tilde v^n)+\p(\tilde u^n, \tilde v^n)=0,\\
(\tilde u^n,\tilde v^n)|_{t=s_n}=(u_0^n,v_0^n),
\end{cases}\end{equation}
which means that $(\tilde u^n,\tilde v^n)$ solves the linearized problem \eqref{lin} with $(\tilde u^n,\tilde v^n)|_{t=s_n}=(u_0^n,v_0^n)$. By virtue of the uniqueness given in Lemma \ref{lem_uni}, it follows that
$$(\tilde u^n,\tilde v^n)~=~(u_L^n,v_L^n),\quad{\rm on}~[s_n,T].$$
Therefore, from \eqref{ass_lin} and \eqref{est_con} we get a contradiction:
\[
n\leq \|(u_L^n,v_L^n)(t_n)\|_{L^2}=\|(\tilde u^n,\tilde v^n)(t_n)\|_{L^2}=\|(u^n,v^n)(t_n-s_n)\|_{L^2}\leq C,\quad \forall n\ge 1,
\]
as the positive constant $C$, given in \eqref{est_non}, is independent of $n$. So we obtain the proof of Theorem \ref{thm_non}.

\end{proof}

\section{Appendix}

In this Appendix, we present the main steps of the proof of Proposition \ref{prop_2} given in Section 2, which shows that the three dimensional linearized Prandtl equations is well-posed locally in time
when one component of the background tangential velocity, such as $u^s$, is monotonic in the normal variable, and we study the problem in the analytic setting only in the horizontal variable $y$.

\begin{proof}[\bf{Proof of Proposition \ref{prop_2}.}]
Let the solution of the linear problem \eqref{lin} have the form
\[(u,v)(t,x,y,z)~=~\sum\limits_{k\in\cZ}e^{iky}(u_k,v_k)(t,x,z).\]
Plugging this relation into \eqref{lin}, it follows that
\begin{equation}\label{lin_de}\begin{cases}
\pd_tu_k+u^s\pd_xu_k-\pd_z^2u_k+ikv^su_k-u_z^s\int_0^z(\pd_xu_k+ikv_k)d\tilde z=0,\\
\pd_tv_k+u^s\pd_xv_k-\pd_z^2v_k+ikv^sv_k-v_z^s\int_0^z(\pd_xu_k+ikv_k)d\tilde z=0,\\
(u_k,v_k)|_{z=0}=0.
\end{cases}\end{equation}
By assuming that $u^s(t,z)$ is monotonic in $z$, i.e. $\partial_zu^s>0$,  we employ the transformation given in \cite{AWXY} for
the first component of the tangential velocity in the above problem,
\[h_k(t,x,z)~\triangleq~\pd_z\big(\frac{u_k(t,x,z)}{u_z^s(t,z)}\big),~{\rm or}~u_k(t,x,z)~=~u_z^s(t,z)\int_0^zh_k(t,x,\tilde z)d\tilde z,\]
and set
\[\tilde v_k(t,x,z)~\triangleq~\big(v_k-\frac{v_z^s}{u_z^s}u_k\big)(t,x,z).
\]
Then, from the problem \eqref{lin_de} we know that $(h_k,  \tilde v_k)$ satisfies the following problem,
\begin{equation}\label{lin_tr}\begin{cases}
\pd_th_k+u^s\pd_xh_k-\pd_z^2h_k-2\pd_z\big(\frac{u_{zz}^s}{u_z^s}h_k\big)+ik(v^sh_k-\tilde v_k)=0,\\
\pd_t\tilde v_k+u^s\pd_x\tilde v_k-\pd_z^2\tilde v_k+ikv^s\tilde v_k-2u_z^s\pd_z(\frac{v_z^s}{u_z^s})h_k=0,\\
\pd_zh_k|_{z=0}=0,\quad \tilde v_k|_{z=0}=0,
\end{cases}\end{equation}
where we use $\pd_z^2u_k|_{z=0},~u_{zz}^s|_{z=0}=0$ to derive the boundary condition of $h_k$.

For the problem \eqref{lin_tr}, by the energy method
one can have
\begin{equation}\label{est_tr}\begin{split}
&\quad\|(h_k,\tilde v_k)(t,\cdot)\|_{K_\alpha^{m}}^2+\int_0^t\|(\pd_zh_k,\pd_z\tilde v_k)(s,\cdot)\|_{K_\alpha^{m}}^2ds\\
&\leq C\Big(\|(h_k,\tilde v_k)(0,\cdot)\|_{K_\alpha^{m}}^2+\max\{1,|k|\}\int_0^t\|(h_k,\tilde v_k)(s,\cdot)\|_{K_\alpha^{m}}^2ds\Big),
\end{split}\end{equation}
where the positive constant $C$ depends on $\alpha$ and $(u^s,v^s)$. Applying the Gronwall inequality to \eqref{est_tr}, it implies that there exists a $\rho>0$, depending on $\alpha$ and $(u^s,v^s)$, such that
\begin{equation}\label{est_tr1}\|(h_k,\tilde v_k)(t,\cdot)\|_{K_\alpha^{m}}^2+\int_0^t\|(\pd_zh_k,\pd_z\tilde v_k)(s,\cdot)\|_{K_\alpha^{m,0}}^2ds\leq Ce^{\rho |k|t}\|(h_k,\tilde v_k)(0,\cdot)\|_{K_\alpha^{m}}^2.
\end{equation}

From the assumption \eqref{ass_ini}, we have
\begin{equation}\label{ini_tr}
\|(h_k,\tilde v_k)(0,\cdot)\|_{K_\alpha^m}~\leq~C_0e^{-\beta |k|},
\end{equation}
for some positive constant $C_0$. As $u^s\in C(\R^+;W_\alpha^{3,\infty}(\R^+))$, one has
\begin{equation}\label{est_tr2}
\|u_k(t,\cdot)\|_{K_\alpha^m}=\|(u_z^s\int_0^zh_kd\tilde z)(t,\cdot)\|_{K_\alpha^m}\leq C_1\|h_k(t,\cdot)\|_{K_\alpha^m},
\end{equation}
with the constant $C_1>0$ depending on $\alpha$ and $u^s$. Therefore, from the estimates  \eqref{est_tr1}-\eqref{est_tr2} and the relation $v_k=\tilde v_k+\frac{v_z^s}{u_z^s}u_k$ it follows that
\begin{equation}\label{est_de}
\|(u_k, v_k)(t,\cdot)\|_{K_\alpha^m}~\leq~C_2e^{-(\beta-\rho t)|k|},
\end{equation}
where the positive constant $C_2$ is independent of $k$. From the estimate \eqref{est_de} we complete the proof of
this proposition.
\end{proof}

\vspace{.15in}

{\bf Acknowledgements :}
The first two authors' research was supported in part by
National Natural Science Foundation of China (NNSFC) under Grants
No. 10971134, No. 11031001 and No. 91230102. The last author's research was supported by the General Research Fund of Hong Kong,
CityU No. 103713.

redpajama_set_name: RedPajamaArXiv


text:
English to Bengali Meaning :: unscathed
English to Bengali Meaning : adjective : অনাহত Details : অক্ষত, আঁচড় লাগেনি এমন
Unscathed :
অনাহত
Unscathed - অনাহত
← Unsavoury
unscramble →
Adjective(1) not injured or harmed
(1) Are we to believe that Park's reputation will remain unscathed from this affair?(2) He came virtually unscathed out of the two scams thanks largely to shabby investigation.(3) When a fishing vessel is lost and all the crew make it home unscathed to find their families waiting for them it's a good result.(4) Most emerge unscathed but some young people become alienated from their families and end up with nowhere to turn.(5) Occupants of the smashed car escaped with slight injuries and those in the house were unscathed .(6) Perhaps it was that competitive spirit which saw him emerge unscathed from so many tumbles during his racing career.(7) If you pay just a little bit of attention, you can get through life mostly unscathed .(8) How I managed to get through that period relatively unscathed I have no idea.(9) The woman, who didn't have a smoke alarm fitted, escaped unscathed after being woken up by the barking of her dogs.(10) Ashcroft's injury apart, the team came through unscathed from the two Bank Holiday matches.(11) The Cougar team came through the game relatively unscathed , but three players picked up slight injuries.(12) After what seems like a lifetime, you emerge unscathed at the far end.(13) His office was evacuated and he escaped unscathed before the south tower was hit and collapsed.(14) Almost nothing survived unscathed , but 12 plants did survive and some even grew.(15) Twice he parachuted into France to deliver gold to the resistance and made his way back unscathed .(16) Amid all the cuts, the Park and Ride services are to continue unscathed , with the concession of a few extra stops.
1. unharmed ::
অক্ষত
2. unhurt ::
3. uninjured ::
4. undamaged ::
অবিক্ষত
5. in one piece ::
এক টুকরা
6. intact ::
8. unmarked ::
অচিহ্নিত
9. untouched ::
অস্পৃষ্ট
10. without a scratch ::
একটি স্ক্র্যাচ ছাড়া
unscathed
English to Bengali Dictionary: unscathed
Meaning and definitions of unscathed, translation in Bengali language for unscathed with similar and opposite words. Also find spoken pronunciation of unscathed in Bengali and in English language.
Tags for the entry "unscathed"
What unscathed means in Bengali, unscathed meaning in Bengali, unscathed definition, examples and pronunciation of unscathed in Bengali language.
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Luxurious (বিলাসী) :: its a luxurious room Yield (উত্পাদ) :: Chapelcross is Britains only producer of tritium a radioactive gas which is essential for boosting the explosive yield of nuclear bombs Steamer (জাহাজ) :: Until the 1880s the company stuck with sailing ships but then moved slowly into steamers Van (অগ্রদূত) :: This explains why Edwardian ladies had so much luggage deposited in the baggage van by a team of railway porters
Custom (প্রথা) :: Indeed many collective agreements particularly at local level are simply a distillation of established custom and practice Customer (ক্রেতা) :: hes a tough customer Cut (কাটা) :: subcultures which cut across national and political boundaries
Circuitous (বক্র) :: Were taking a rather circuitous route to Waterloo so I can pick them up Circumlocution (বাগাড়ম্বর) :: After 20 years in the Senate he has developed a meandering circumlocutory speaking style Circumscribe (বেষ্টন করা) :: The practice is severely circumscribed and tightly regulated


text:
Informal carer—relationship to care recipient, interpersonal code N[N]
Interpersonal relationship of carer to care recipient
Synonymous names:
Carer relationship to care recipient
Disability, Standard 15/12/2017
The relationship of the informal carer to the person for whom they care, as represented by a code.
Ageing, carers and disability.
Data Element Concept:
Informal carer—relationship to care recipient
Value domain attributes
1 Wife/female partner
2 Husband/male partner
3 Mother
4 Father
5 Daughter
7 Daughter-in-law
8 Son-in-law
9 Other female relative
10 Other male relative
11 Female friend/neighbour
12 Male friend/neighbour
99 Not stated/inadequately described
This code set should always be used to record the relationship of the carer to the person for whom they care.
CODE 1 Wife/female partner
Includes married, de facto and same sex partners who are carers.
CODE 2 Husband/male partner
CODE 3 Mother
Includes foster parents.
CODE 4 Father
CODE 99 Not stated/inadequately described
This code should only be recorded where the carer has not been identified. This code is not for use in primary data collections.
Australian Institute of Health and Welfare 2016. Disability Services National Minimum Data Set: data guide, July 2016. Cat. no. DAT 4. Canberra: AIHW.
Data element attributes
Information about this relationship assists in the establishment of a profile of informal caring relationships and the assistance provided to maintain and support those relationships. As such, it increases knowledge about the dynamics of caring and provides an insight into the gender and inter-generational patterns of informal care giving in the community.
Australian Institute of Health and Welfare.
Supersedes Informal carer—relationship to care recipient, interpersonal code N[N] Disability, Superseded 15/12/2017, Community Services (retired), Standard 10/04/2013
Implementation in Data Set Specifications:
All attributes +
Disability Services NMDS 2017–18 Disability, Superseded 05/07/2019
DSS specific attributes +
Implementation start date: 01/07/2017
Implementation end date: 30/06/2018
Conditional obligation:
In the Disability Services National Minimum Data Set (DS NMDS), this data element should only be reported in relation to service users that have an informal carer (Person—informal carer existence indicator, code N, with a value of 1).
DSS specific information:
In the Disability Services National Minimum Data Set (DS NMDS), this data element refers to the service user's main informal carer—the person who provides the most significant care and assistance related to the service user's capacity to remain living in their current environment. It is recognised that two or more people may equally share the caring role (for example, mother and father) however, for the purposes of this collection, characteristics are only requested for one of these carers.
If a person has more than one carer (for example, a spouse and a son), the coding response to carer relationship should relate to the carer who provides the most significant care and assistance related to the person's capacity to remain living in their current environment (i.e. the main carer). The expressed views of the service user and/or their carer or significant other should be considered to be the primary or principal carer in this regard.
Code 9 (other female relative), allows for the wide range of family members who may be involved in a caring role with the service user. This code therefore includes the female family members not listed in the codes elsewhere (for example, aunts, nieces, female cousins, grandmothers, step mother, step daughters and so on).
Similarly, code 10 (other male relative) covers the range of male family members who may act as carers. This code includes the male family members not listed in the codes elsewhere (for example, uncles, nephews, male cousins, grandfathers, male grandchildren, step father, step sons and so on).
Disability Services NMDS 2018–19 Disability, Standard 05/07/2019


text:
August 15, 2012 Jim 1 Comment
Last week I looked at the Golden Rule and whether or not it was good advice, seeng as it's probably the most cited piece of ethical advice. Whether it's "Do unto others as you'd have them do unto you" or "Love thy neighbour as thyself" it has a tendency to break down in application. So today I'm going to look at a different piece of ethical advice, the Greatest Happiness Principle. Let's see how it shakes out.
Coined by Jeremy Bentham and refined by John Stuart Mill in the 19th century, the greatest happiness principle is the cornerstone of the ethical theory of Utilitarianism, which argues that actions are right or wrong by virtue of their consequences. Simply put, the principle states the following:
"…actions are right in proportion as they tend to promote happiness, wrong as they tend to produce the reverse of happiness." ((Mill, John Stuart. "Utilitarianism" The Utilitarians. New York, Dolphin Press, 1961. p. 407))
That's a good definition, but I want to frame it as a piece of advice. "Do what creates the greatest happiness in the greatest number, while producing the least pain." the greatest number part comes into play later in the chapter, where Mill discusses you why cannot be concerned merely with your own happiness, but with the aggregate happiness in the world, or at least the part of the world that you can affect with your actions.
Given the FADE guidelines, it seems like pretty good advice. It's directive, telling you to do something. It's action-guiding, because while it doesn't tell you specifically how to do that (though both Mill and Bentham go into a lot of depth about it, and they're followed by a hundred years of scholarship), it tells you how to think about things and what's important when making decisions. Whether it's ethical is a matter of some debate among philosophers, but there isn't a way that it's straightforwardly useful to serial killers. But where things really got hung up with the golden rule was in how followable it was, and that's where the principle of utility is going to have issues as well.
It does encourage us to examine other people's interests, and can contain most of the things we want out of the golden rule. After all, if doing unto others isn't going to make anyone happier, it seems reasonable not to do it. Sometimes it will though, and in those cases, go ahead. As with all advice designed to apply so broadly though, there are some cases where it gets pretty hard to follow.
The first of these has to do with collective happiness, and is referred to as the Doctor Problem. Imagine you're a doctor, and you have five patients who are each dying of a different kind of organ failure, and one patient who's perfectly healthy. You know that you can save the five patients if you transplant the organs from the healthy one, but the healthy one will die. The greatest happiness principle says that this is a pretty good trade, because there's a net increase of overall happiness. Even if the one patient's family's unhappiness is counted as well, the happiness of the five patients' families will outweigh it. Typical ways to address this include focusing on rules rather than individual acts, so we might decide that having a policy of not butchering healthy people will increase happiness overall. Another way is to discard happiness in favour of well-being or some other kind of good consequence, but neither of those really get around this issue.
Some men just want to watch the world burn.
The second issue has to do with deviant values in individual happiness, with people usually referred to as Utility Monsters. These are people who get so much happiness out of something that it outweighs any kind of grief that they cause, whether that thing is caviar or arson. The greatest happiness principle has a really hard time dealing with this, but does so as a direct result of taking into account other people's interests and desires.
The real question here is whether or not it's easier to follow than the golden rule. It's certainly harder to know what will cause the greatest happiness to the greatest number. But right out of the gate it has us thinking about the needs of other people which is pretty important when we're talking about how to treat other people. It doesn't take for granted that the things you want are the things that everyone wants, so while it does suffer from some issues, it seems like much better advice than the golden rule.
These aren't the only two options when it comes to one-line advice on ethics, though. Next week I'll look at two that are tied together. For now though, which do you find more compelling, the greatest happiness principle or the golden rule, and why?
Philosophy discussion, ethics, Philosophy, public philosophy
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text:
'''Test that a font with no name given still renders using some sort
of default system font.
'''

__docformat__ = 'restructuredtext'
__version__ = '$Id: $'

import unittest
import base_text

class TEST_DEFAULT(base_text.TextTestBase):
    font_name = ''

if __name__ == '__main__':
    unittest.main()

redpajama_set_name: RedPajamaGithub


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