TITLE: Physically induced latency in internet connections
QUESTION [3 upvotes]: As a trade I am a software architect. It does happen that I have to design sites where the servers are located in different parts of the world.
Now an essential parameter to consider is latency: that is the time it takes between when a packet is sent from one server, and when it is received on the other side.
Of course, the packet might go through other systems etcetera. However, the hallmark of a good connection is low latency.

Question: what is the minimal latency that can be obtained over a coaxial wire (i.e. what is the maximum speed reachable by an electrical impulse over a coaxial wire)?
Question: what is the minimal latency that can be obtained over a fiber optic connection (if different from distance/c)?

REPLY [5 votes]: http://en.wikipedia.org/wiki/Wave_propagation_speed
79% for coax, 67% of speed of light for fiber.
So it's 42ms for 10'000km of coax, and 49.7ms for 10'000km of fiber.
For roundtrip time multiply by 2.
Reality adds some more latency (up to 1.5-2x) on transcievers, queues on overloaded segments and non-optimal paths.
Update: As a side note, I really doubt someone could make 10'000km link at 10Gbit+ via coax :-D Signal amplification is just much easier & cheaper in case of fiber.
Update2: There is also exotic fiber with empty core, which have almost 100% speed. I've heard this would be used to pass trading information across continents to gain additional profit. 10ms worth alot of money :-D


TITLE: Inequality relating connectivity, edge connectivity and minimum degree (Whitney's theorem)
QUESTION [6 upvotes]: The following is a theorem given in Bella Bollobas's Modern graph theory (Springer 2002) Page-73

If $G$ is nontrivial (that is, has at least two vertices), then the parameters $\delta(G)$,
$\lambda(G)$ and $\kappa(G)$ satisfy the following inequality:
$\kappa(G) \le \lambda(G) \le \delta(G)$, where $\delta(G)$ is the minimum degree of a vertex, $\lambda(G)$ is the edge connectivity, and $\kappa(G)$ is the vertex connectivity.

Proof:

Indeed, if we delete all the edges incident with a vertex, the graph becomes
disconnected, so the second inequality holds. To see the other inequality, note
first that if $G$ is complete then $\kappa(G) = \lambda(G) = |G| - 1$, and if $\lambda(G) \le 1$
then $\lambda(G) = \kappa(G)$. Suppose now that $G$ is not complete, $\lambda(G) = k \ge 2$, and
$\{x_1 y_1, x_2 y_2, \ldots , x_k y_k\}$ is a set of edges disconnecting $G$. If $G - \{x_1, x_2, \ldots , x_k\}$
is disconnected then $\kappa(G) \le k$. Otherwise, each vertex $x_i$ has degree at most
$k$ (and so exactly $k$). Deleting the neighbours of $x_1$, we
disconnect $G$. Hence $\kappa(G)\le \lambda(G)$.

I did not understand the following part of the proof: "Otherwise, each vertex $x_i$ has degree at most
$k$ (and so exactly $k$)" . Why is this true? I would really appreciate an elaborate answer to this.

REPLY [4 votes]: I don’t see how to justify the statement in question, but I do see how to justify a weaker statement that is still sufficient to prove the theorem.
Suppose that $G-\{x_1,x_2,\ldots,x_k\}$ is connected and and $\deg v_i>k$ for each $i\in\{1,\ldots,k\}$. Then each $v_i$ has at least $k+1$ neighbors in $G$, and at most $k-1$ of them can be in $\{x_1,x_2,\ldots,x_k\}$, so there must be at least two vertices $u_i$ and $v_i$ in $G-\{x_1,x_2,\ldots,x_k\}$ that are neighbors of $x_i$ in $G$. At least one of $u_i$ and $v_i$ must be different from $y_i$; say $u_i\ne y_i$. I claim that this implies that $G-\{x_1y_1,\ldots,x_ky_k\}$ is connected, which is a contradiction.

Let $v$ and $w$ be any two vertices of $G$. If $v,w\notin\{x_1,\ldots,x_k\}$, then there is a path from $v$ to $w$ in $G-\{x_1,\ldots,x_k\}$ and hence in $G-\{x_1y_1,\ldots,x_ky_k\}$. If $v=x_i$ and $w\notin\{x_1,\ldots,x_k\}$, there is a path $P$ from $w$ to $u_i$ in $G-\{x_1y_1,\ldots,x_ky_k\}$, and we can add the edge $u_ix_i$ to $P$ to get a path from $w$ to $x_i$ in $G-\{x_1y_1,\ldots,x_ky_k\}$. Finally, if $v=x_i$ and $w=x_j$ with $i\ne j$, there is a path $P$ from $u_i$ to $u_j$ in $G-\{x_1y_1,\ldots,x_ky_k\}$, and we can extend it with the edges $x_iu_i$ and $u_jx_j$ to get a path from $x_i$ to $x_j$ in $G-\{x_1y_1,\ldots,x_ky_k\}$. This proves the claim.

This contradiction shows that there must be at least one $i\in\{1,\ldots,k\}$ such that $\deg x_i\le k$, and since we already know that $k=\lambda(G)\le\delta(G)\le\deg x_i$, we must have $\deg x_i=k$. Deleting the neighbors of $x_i$ disconnects $G$, so $\kappa(G)\le k=\lambda(G)$.


TITLE: Statistics. How to find the median of two other medians?
QUESTION [0 upvotes]: I got  median of 226 and another median of 304. Now the question is how to find the overall median? Averaging them is not the correct answer.

REPLY [4 votes]: The first median is known to be larger than 8 elements, and the second smaller than 9. So the median certainly lies between 226 and 304.
But this is all you can say ! The median could by any value in between. (Choose a median arbitrarily, you will have all freedom to insert the elements on either side.)

REPLY [3 votes]: It's standard to define the median of an even population with an average between the two "candidates", e.g. the median of 1,3 would be 2. With that definition, the problem is unsolvable even with information about population sizes.
For example, if you know first population has 2 elements and the second had 1, so the second population must be 304.

If the first one is 226, 226, -- the global median is 226.
If the first one is 305, 147, -- the global median is 304.

My conjecture is, you will need the entire population to compute the global median (unlike the average, where just the knowledge of the size of each is enough)...


TITLE: Upper bound of an analytic function (application of Pick's Lemma)
QUESTION [5 upvotes]: Let $f$ be holomorphic and non zero in the disk $ |z|<1$ with $|f(z)|\leqslant 1$ in $|z|<1$ and $f(0)= e^{-1}$ . What is the best possible bound upper bound for $|f'(0)|?$ Is this bound attained? If so, for what function $f$?
So, I applied Pick's Lemma and got the bound $1-e^{-2}$, and according to the same lemma I can claim that this bound can be attained if the function is conformal self map of $\mathbb D$. I did not see how can I get the function. I can guess it should be related somehow with the non-zero behavior of $f$ in the unit disk. But I stuck. Please help.

REPLY [2 votes]: By a straight application of Pick's lemma, $|f'(0)|\le1-e^{-2}$. However, the fact that $f(z)\ne0$ throws a wrench in the search for a best upper bound. A lower bound on the supremum can be observed in the solution $\varphi(z)=\exp\big(\!\frac{z-1}{z+1}\!\big)$, which satisfies $\varphi(0)=e^{-1}$, $\varphi(1)=1$, and has an essential singularity at $z=-1$, and $\varphi'(0)=2/e$, so this proves that $\sup_f|f'(0)|\ge2/e$. There is still a gap between $2/e$ and $1-e^{-2}$, though, so there is still work to be done.
However, we can use the same trick used by Twiceler below, and consider the function $F(z)$ defined by $f=\varphi\circ F$. Since $f(z)\ne0$, it is the exponential of an analytic function, and the Möbius transform is invertible, so such an $F$ must exist. (A little extra detail in this step: By the Weierstrass factor theorem, if $f$ is nonzero, then $f(z)=e^{g(z)}$ for some $g$. Thus, if we define $F(z):=\frac{1+g(z)}{1-g(z)}$, then $f(z)=\exp\big(\!\frac{F(z)-1}{F(z)+1}\!\big)=\varphi(F(z))$.) Then
$$\begin{align}
f(0)=\exp\bigg(\!\frac{F(0)-1}{F(0)+1}\!\bigg)=e^{-1}&\Rightarrow F(0):=c_n=\frac{i\pi n}{1-i\pi n}\Rightarrow |c_n|^2=1-\big(1+(n\pi)^2\big)^{-1} \\
&\Rightarrow \varphi'(c_n)=\frac2e(1-i\pi n)^2\Rightarrow |\varphi'(c_n)|=\frac2e\big(1+(n\pi)^2\big)
\end{align}$$
and so by Pick's lemma, $|f'(0)|=|\varphi'(c_n)F'(0)|\le|\varphi'(c_n)|\cdot(1-|c_n|^2)=2/e$. (This proves that $\sup_f|f'(0)|=2/e$, and $\varphi$ saturates the bound.) Moreover, if the bound is saturated, then $F$ is a Möbius transformation that preserves the unit circle, i.e. $F(z)=\theta\dfrac{c_n+\theta z}{\theta-\overline{c_n}z}$ for unimodular $\theta$, so
$$f(z)=\exp\!\bigg[\!\!\frac{c_n+\theta z-(1-\overline{c_n}\theta z)}{c_n+\theta z+(1-\overline{c_n}\theta z)}\!\!\bigg]
=\exp\!\bigg[\!\!\frac{(1-in\pi)\theta z-(2n^2\pi^2-in\pi+1)}{(1+in\pi)+(2n^2\pi^2+in\pi+1)\theta z}\!\!\bigg],$$
where $|\theta|=1$ and $n\in\mathbb Z$.
Some explanatory remarks: The moral of the story is that if we have to avoid zero, we can transition from the problem of finding a nonzero function $f:\mathbb D\to\mathbb D-\{0\}$ to a function $g:\mathbb D\to\{x\in\mathbb C:\operatorname{Re}x<0\}$ whose range is the left half-plane (this is so that $|e^g|<1$), and via a Möbius transform similar to the Cayley transform, we can map that range back to $\mathbb D$ to get a function $F:\mathbb D\to\mathbb D$ (note that there are no longer any holes in the range), to which we can apply Twiceler's analysis using Pick's lemma.
Although I have defined $F$ via $f=\varphi\circ F$, I have been careful not to say $F:=\varphi^{-1}\circ f$, because $\varphi$ is not injective, and indeed takes on every point in $\mathbb D-\{0\}$ infinitely many times (as you approach $z=-1$ from different directions). In particular, the value $\varphi(z)=e^{-1}$ happens at an infinite number of points, indexed as $\{c_n\}_{-\infty}^\infty$, so $f(0)=e^{-1}\Rightarrow g(0)=-1+2i\pi n\Rightarrow F(0)=\frac{i\pi n}{1-i\pi n}:=c_n$. But after this, Pick's lemma applies as normal on $F$, so $|F'(0)|\le1-|c_n|^2=1/(1+n^2\pi^2)\Rightarrow |f'(0)|\le2/e$.
The remaining question is how to characterize the solutions that saturate the inequality. The algebra is not fun, so I have just shown the solution here, but the point is that all solutions are in the form $\varphi\circ F$ where $F$ is a Möbius transformation which preserves the unit circle and satisfies $F(0)=c_n$, so that $f(0)=e^{-1}$ is satisfied.


TITLE: why are these integrals equal?
QUESTION [0 upvotes]: I've been trying to analyse Steiner symmetrization. I've asked before about the symmetrization preserving $\textit{volume}$, I'm still going over the same proof. I think I understand a bit what is going on.
I have one issue however, the slides I'm going through state that:
$\int_{x_n \in G_x \cap \Omega}dx_n = m(x_1,...,x_{n-1},0) = \int_{x_n \in G_x \cap S(\Omega)}dx_n$
Can someone please take a peek at the slides and explain if you can.
I think I understand why $\int_{x_n \in G_x \cap \Omega}dx_n = m(x_1,...,x_{n-1},0)$. It is sayin' that the result is just function of the other variables without $x_n$, the variable we were integrating with respect to.
What I do agree with is that
         $\int_{x_n \in G_x \cap \Omega}dx_n = m_1(x_1,...,x_{n-1},0)$ and $\int_{x_n \in G_x \cap S(\Omega)}dx_n = m_2(x_1,...,x_{n-1},0)$. Why is $m_2(x_1,...,x_{n-1},0) = m_1(x_1,...,x_{n-1},0)$ though?

REPLY [1 votes]: It's probably better to think over what they're saying. For each value of $(x_1,\ldots,x_{n-1})$, the Steiner symmetrization is the map that sends the slice of $\Omega$ with those values to the interval of the same length as the slice, centered about the origin.
All your integral says is that the measure of the slice is the measure of the same slice on the Steiner-symmmetrized set, which is just true by the way the symmetrization was defined.


TITLE: Semi-continuous fields of C*-algebras having dimension one on a dense set
QUESTION [4 upvotes]: Given a Hausdorff, locally compact space $X$, let us consider a semi-continuous field $\{A_x\}_{x\in X}$ of C*-algebras over $X$, such that $A_x$ is one-dimensional for every $x$ in a dense subset $D$ of $X$, while the dimension of $A_x$ is bigger than one for $x$ in $X\setminus D$.  I can cook up some simple examples in which $X\setminus D$ is a discrete set but I am wondering how much more complicated can this be.  Precisely, I'd like to know:

Is it possible to build an example in which $D$ fails to be open.

REPLY [2 votes]: After talking to a friend I learned the following counter example to my own question.
Consider the topological space $X$ shown below and let $\pi$ be the vertical projection onto $[0,1]$.  We may then view $C[0,1]$ as a subspace of $C(X)$ by means of the injective homomorphism
  $$
  f \in C[0,1] \mapsto f\circ\pi \in C(X).
  $$
  Clearly $C[0,1]$ is contained in the center of $C(X)$, so one may write $C(X)$ as the algebra of sections of some semi-continuous field $\{A_x\}_{x\in [0, 1]}$ of C*-algebras.  One may then show that $A_x$ is one dimensional if and only if $x$ is not a diadic rational (in which case $\pi ^{-1}(x)$ is a singleton).


TITLE: If $M$ is connected and $\int_M \omega = 0,$ then $\omega = 0$?
QUESTION [1 upvotes]: Let $M$ be a connected, orientable, and compact $(r+1)$-dimensional manifold and $\omega$ a $(r+1)$-differential form on $M$. If $\int_M \omega = 0$, then $\omega =0$?
I want to say that the answer is yes. The intuitive reason behind it is that if $\omega\neq 0$, since $M$ is connected then "it must be $\omega>0$ or $\omega<0"$, so we have $\int_M \omega \neq 0$. I've been trying to formalize this intuition by stating what means $\omega>0$ first. Since we can write $\omega(x) = a(u) du_1\wedge \cdots \wedge du_{r+1},x\in M,$ for $x = \varphi(u)$ and some chart $\varphi$, then $\omega>0$ would mean that $a(u)>0$ for every $x\in M$. But i'm struggling to use the fact that $M$ is connected to ensure that $a(u)>0$ should happen.
How can I proper formalize this intuition?

REPLY [0 votes]: The correct statement that corresponds to the intuition that you had is as follows: if $\omega$ is a nowhere-vanishing top degree form on an orientable closed manifold $M$, then the integral of $\omega$ over $M$ is nonzero.
To put it differently, if $\int_M \omega =0$, then $\omega$ vanishes at some point of $M$.


TITLE: Does König's inequality require axiom of choice?
QUESTION [4 upvotes]: König's inequality:
If $\kappa_\alpha<\lambda_\alpha$ for all $\alpha<\mu$, then $\sum\kappa_\alpha<\prod\lambda_\alpha$.
In order to prove this inequality, we need to provide an injection and show there is no surjection, but I failed to see why AC is needed in the proof. Surely the definition of cardinal product needs AC, but we can view it as a set, not a cardinal, just like Cantor's argument showing $A\prec2^A$.

REPLY [7 votes]: König’s inequality in its full form implies the axiom of choice. Let $\{A_i:i\in I\}$ be a set of non-empty sets. Clearly $|\varnothing|<|A_i|$ for each $i\in I$, so by König’s inequality we have
$$0<\left|\prod_{i\in I}A_i\right|$$
and hence
$$\prod_{i\in I}A_i\ne\varnothing\;.$$
This is one of the many equivalent forms of $\mathsf{AC}$.


TITLE: Proof using natural deduction
QUESTION [4 upvotes]: Prove that
$$\lnot r\Rightarrow \lnot p,\lnot(q\lor r),s\Rightarrow(p\lor q)\models\lnot s
 $$
I'm completely stuck on this one. Only natural deduction inference rules can be used, no de morgan's law etc. The premises given all seem to be really irrelevant, and since we can't use transformational proof techniques, all the implications turn the first and third expressions into essentially garbage, and without de Morgan's law the second one seems useless too. How should I proceed?

REPLY [1 votes]: In classical logic, you can derive a contradiction and then either introduce or get rid of a negation.  Since you want to get a negation, you consequently might want to assume "s" and then derive a contradiction.
Anytime you have a disjunction, see if you can get both disjuncts to lead to the same conclusion, and then use disjunction elimination.
For your problem, here's an outline of a proof in one system, though I don't know what your rules say exactly, so I don't know what a proof will look like exactly in your system.
1 | s assumption (I assume s, so that I can try and find a contradiction.)
2 | (p $\lor$ q) (I'd hope I don't need the proof analysis as to how I got this)
3 || q assumption (I choose q, since if I have ¬(q∨r), I can easily get the negation of q)
4 || (q $\lor$ r) (disjunction introduction on 3)
5 | $\lnot$q (contradiction from one of the premises and 4)
6 || p assumption
7 | (p->p) (6-6 conditional introduction)
8 || q assumption (I want (q->p), so I start with q))
9 ||| $\lnot$p assumption (I want p, so I assume its negation since I already have a contradiction)
10|| p (6 and 8 contradict each other)
11| (q->p) (8-10 conditional introduction
12| p (2, 7, 11 disjunction elimination)
Now, that I have "p", I can get "r" in the same scope as "p" via the first premise (either I have modus tollens as a rule of inference, or I assume $\lnot$ r and derive a contradiction which gives me r).  Then, since I have "r" I use disjunction introduction to (q $\lor$ r).  But, that gives us a contradiction within the same scope as s.  So, then we can introduce the negation of s within the same scope as the premises for this problem.


TITLE: The integral of $\int_\gamma\frac{1}{z^2+1}dz$ over the semicircle
QUESTION [2 upvotes]: If $\gamma$ is the part of the circle of radius $R$ in the upper half plane of $\mathbb{C}$, I would like to calculate
$$\int_\gamma\frac{1}{z^2+1}dz.$$
Can anybody help me?
I tried to used the parameter representation $R e^{i t}$, but then I got
$$\int_0^\pi\frac{R e^{i t}}{R^2 e^{i 2 t}+1}d t,$$
which is not very useful either.

REPLY [1 votes]: Use your idea:
$$z=Re^{it}\;,\;\;0\le t\le \pi\implies dz=Rie^{it}\implies$$
$$\int_\gamma\frac{dz}{1+z^2}=\int_0^\pi\frac{Rie^{it}}{1+R^2e^{2it}}=\left.\int_0^\pi\frac{d(Re^{it})}{1+(Re^{it})^2}=\arctan Re^{it}\right|_0^\pi=$$
$$=\arctan(-R)-\arctan R=-2\arctan R$$
You may want this for some real integral by means of complex integration: you can see now that
$$\lim_{r\to\infty}-2\arctan R=-\pi$$


TITLE: How to solve equation system with 3 parameters
QUESTION [3 upvotes]: How to solve this problem?

$$y=2x^2-1$$
$$z=2y^2-1$$
$$x=2z^2-1$$

x,y,z - ?
I am not good at math. Can you help me?
Thanks in advance

REPLY [6 votes]: Let $x=\cos A$ $$y=\cos2A,z=\cos4A,x=\cos8A$$
$$\cos8A=\cos A$$
$$8A=360^\circ n\pm A$$
Either $$(8+1)A=360^\circ n\implies A=40^\circ n;0\le n<180/40$$
What if $(8-1)A=360^\circ n?$

REPLY [2 votes]: Hint:
$$\begin{align}y &= 2x^2 - 1\tag{1}\\z &= 2y^2 - 1\tag{2}\\x &= 2z^2 - 1\tag{3}\end{align}$$
Substitute $(2)$ into $(3)$ to get
$$x = 2\left(2y^2 - 1\right)^2 - 1\tag{4}$$
You now have two simultaneous equations, i.e., $(1)$ and $(4)$, that you can solve to find the values of the variables $x$ and $y$. Once you have either of the two, finding the value of variable $z$ is trivial with either $(2)$ or $(3)$.


TITLE: Solve :$\frac{dy}{dx}=2\sqrt{y^2+x}$
QUESTION [0 upvotes]: Solve: $\frac{dy}{dx}=2\sqrt{y^2+x}$
Given that the curve passes through $(0,1)$.
I tried to reduce this to Clairaut's form but was unable to proceed.

REPLY [2 votes]: $$\frac{dy}{dx}=2\sqrt{y^2+x}\quad\text{with condition } y(0)=1$$
$$\frac{1}{y}\frac{dy}{dx}=2\sqrt{1+\frac{x}{y^2}}$$
There is no simple closed form for the solution, but one can find approximates.
Asymptotically, for $x$ increasing $y$ increases faster because $y'(x)>2$. Thus $\frac{x}{y^2}\to 0$.
$\frac{1}{y}\frac{dy}{dx}\simeq 2 \quad\to\quad y\simeq e^{2x}\quad$ This is the first approximate.
More accurate approximates for low values of $x$ can be derived on the form of series expansion, for example on the form :
$$y(x)\simeq  e^{2x} +a_0 + a_1 x+a_2 x^2 +...$$
$a_0=0$ because $y(0)=1$
Bringing the above $y(x)$ into the ODE and expanding it into series leads to $a_1=0$ and $a_2=\frac{1}{2}$
$$y(x)\simeq e^{2x} +\frac{1}{2} x^2 $$
The figure below shows that the approximate is already good :

The black curve is the representation of the solution of the ODE (from numerical solving of the ODE)
The blue curve represents the first approximate $y(x)=e^{2x}$
The red curve represents the second approximate $y(x)= e^{2x} +\frac{1}{2} x^2 $

An even better approach consists in expanding the square root in power series of $\left(\frac{x}{y^2}\right)$
$$\frac{1}{y}\frac{dy}{dx}=2\sqrt{1+\frac{x}{y^2}}\simeq 2+\frac{x}{y^2}+O\left(\left(\frac{x}{y^2}\right)^2\right)$$
The solution of $\frac{1}{y}\frac{dy}{dx}= 2+\frac{x}{y^2}$ , with condition $y(0)=1$ is :
$$y(x)=e^{2x}\sqrt{\frac{9-(4x+1)e^{-4x}}{8}}$$
This function is not drawn on the figure because it is so close to the exact solution that the two curves cannot be distinguished one to the other.


TITLE: Explain the Halls Condition
QUESTION [0 upvotes]: From this PDF:

I don't really understand the definition of Hall's theorem.
Can someone provide another explanation for it?

REPLY [1 votes]: You've got a (finite) set, $V_1$. You've got a (finite) set, $V_2$. You've got lines going from elements of $V_1$ to elements of $V_2$. You want to know whether you can pick a bunch of these lines so they involve every element of $V_1$, and don't involve any element of $V_2$ more than once. That's called a complete matching.
Hall says, take a subset, $S$, of $V_1$, look at all the lines joining elements of $S$ to elements of $V_2$, and count how many elements of $V_2$ are involved in those lines. Then there's a complete matching if and only if the number of elements of $V_2$ involved in the lines from $S$ is at least as big as the number of elements of $S$, for all possible subsets $S$ of $V_1$.
OK?


TITLE: Presentations of the dihedral and quaternion groups
QUESTION [4 upvotes]: I am reading Lang’s algebra text and he is currently discussion generators of groups and gives the following examples:
Is Lang suggesting here that any two elements of either group which satisfy those relations will generate the group? Or is he instead simply saying that(referring to the dihedral group) the clockwise rotation and reflection about two vertices are generators for the group and that they satisfy those relations(that is, he is alluding to specific generators and stating relations they satisfy)?
Note: I understand that he is using typical notation for the quaternion group but it is easy to see that any pair of quaternions which satisfy those relations will generate the group as well.

REPLY [2 votes]: It might be well to recall the way that presentations of groups work.  The presentation  $\langle S|R\rangle $ refers to the group which is the quotient of the free group on the generators, the elements of  $S $, by the normal subgroup,  or the so-called normal closure, of the subgroup generated by the relations $R $.


TITLE: Degree of images of projective varieties
QUESTION [1 upvotes]: Let $X\subseteq\mathbb{P}^n\times\mathbb{P}^m$ be a projective variety of dimension $p$ and degree $d$ defined over an algebraically closed field $k$. Let $X'\subseteq \mathbb{P}^n$ be the projection of $X$ on $\mathbb{P}^n$, which we will assume having dimension equal to that of $X$. What can we say about the degree $d'$ of $X'$? Is there some formula linking the degrees $d$ and $d'$. Since an analytical charcaterization is also good for my purposes, feel free to assume $k=\mathbb C$.

REPLY [0 votes]: If you compute the degree of $X$ with respect to the Segre polarization of $P^n \times P^m$, then $$d' \le d.$$


text:
\begin{document}

\maketitle

\begin{abstract}
We consider an elliptic problem with unknowns on the boundary of the domain of the elliptic equation and suppose that the right-hand side of this equation is square integrable and that the boundary data are arbitrary (specifically, irregular) distributions. We investigate local (up to the boundary) properties of generalized solutions to the problem in Hilbert distribution spaces that belong to the refined Sobolev scale. These spaces are parametrized with a real number and a function that varies slowly at infinity. The function parameter refines the number order of the space.
We prove theorems on local regularity and a local \textit{a priori} estimate of generalized solutions to the problem under investigation. These theorems are new for Sobolev spaces as well.
\end{abstract}

\section{Introduction}\label{sec1}

In the theory of elliptic boundary-value problems, of special interest is the case where boundary data are irregular distributions (so called rough data); see monographs \cite{BehrndtHassideSnoo20, Berezansky68, Hermander63, KozlovMazyaRossmann97, LionsMagenes72, MikhailetsMurach14, Roitberg96, Roitberg99} and references therein. The investigation of elliptic problems in this case is more complicated as compared with regular enough boundary data. This is stipulated by the fact that the trace theorems for Sobolev or other classical distribution spaces (see, e.g., \cite[Section~4.7]{Triebel95}) cease to be valid for irregular boundary data. There are some approaches to overcome this serious obstacle. One of them is to include norms of traces in Sobolev norms of solutions to an elliptic problem. This approach was elaborated by Roitberg \cite{Roitberg64, Roitberg65, Roitberg96}; it deals with solutions that are not distributions, generally speaking. Another way is to include a relevant norm of the right-hand side of the elliptic equation in the norms of solutions. This approach is due to Lions and Magenes \cite{LionsMagenes62V, LionsMagenes63VI, LionsMagenes72}; it remains in the framework of distribution spaces but is applicable to narrower classes of data of the elliptic equation as compared with Roitberg's approach. Investigating elliptic problems in a half-space, H\"ormander \cite[Section~10.4]{Hermander63} used anisotropic Sobolev spaces of high enough regularity only along the normal to the boundary. These approaches yield different solvability theorems for elliptic problems whose boundary data are arbitrary distributions.

The number of boundary conditions increases if an elliptic problem contains unknown distributions on the boundary. Such elliptic problems were first considered by Lawruk \cite{Lawruk63a, Lawruk63b}. They form a part of Boutet de Monvel's algebra \cite{BoutetdeMonvel71} and arise in various applications, specifically in hydrodynamics and the theory of elasticity \cite{AslanyanVassilievLidskii81, Ciarlet90}, and are also used in the theory of free boundary problems \cite{NazarovPileckas93}. A solvability theory for elliptic problems with unknowns on the boundary is given in monographs \cite[Part~1]{KozlovMazyaRossmann97} and \cite[Chapter~2]{Roitberg99} in the framework of Roitberg's approach. Kozhevnikov \cite{Kozhevnikov01} extended this approach to pseudodifferential elliptic problems that form the Boutet de Monvel algebra.

In this connection, it is interesting to investigate these problems in the spirit of the approach by Lions and Magenes. Thus, recently we proved a corresponding solvability theorem for elliptic problems with unknowns on the boundary \cite[Theorem~1]{MurachChepurukhina15UMJ5}. We assumed that the right-hand side of the elliptic equation is square integrable and considered boundary data in Hilbert distribution spaces of lower orders (including negative ones). These spaces belong to the refined Sobolev scale \cite[Section~2.1]{MikhailetsMurach14}. The purpose of the present paper is to supplement this result with theorems on local (up to the boundary) regularity and a local \textit{a priori} estimate of generalized solutions to the problem. In contrast to the corresponding global properties of the solutions \cite{MurachChepurukhina15UMJ5}, these theorems do not follow directly from the solvability theorem. Specifically, this is caused by the fact that the space of solutions (to the elliptic equation) used in \cite{MurachChepurukhina15UMJ5} is not closed with respect to the multiplication of distributions by smooth cut-off functions.

\section{Statement of the problem}\label{sec2}

Let $\Omega $ be a bounded Euclidean domain of dimension $n\geq2$ with an infinitely smooth boundary $\Gamma$. We arbitrarily choose integers $q\geq1$, $\varkappa\geq1$, $m_{1},\ldots,m_{q+\varkappa}\leq2q-1$ and $r_{1},\ldots,r_{\varkappa}$. We consider the following boundary-value problem in $\Omega$:
\begin{gather}\label{10f1}
Au=f\quad\mbox{in}\quad\Omega, \\
B_{j}u+\sum_{k=1}^{\varkappa}C_{j,k}v_{k}=g_{j}\quad \mbox{on} \quad \Gamma,\quad j=1, ...,q+\varkappa.\label{10f2}
\end{gather}
Here, $A:=A(x,D)$ is a linear partial differential operator (PDO) on $\overline{\Omega}:=\Omega\cup\Gamma$ of the even order $2q$; each $B_{j}:=B_{j}(x,D)$ is a linear boundary PDO on $\Gamma$ whose order
$\mathrm{ord}\,B_{j}\leq m_{j}$, and each $C_{j,k}:=C_{j,k}(x,D_{\tau})$ is a linear tangent PDO on~$\Gamma$ whose order $\mathrm{ord}\,C_{j,k}\leq m_{j}+r_{k}$. (As usual, PDOs of negative order are defined to be zero operators.) We assume that all coefficients of the indicated PDOs are infinitely smooth functions on $\overline{\Omega}$ or $\Gamma$ respectively. The distribution $u$ on $\Omega$ and the distributions $v_{1},\ldots,v_{\varkappa}$ on $\Gamma$ are unknown in this problem. In the paper, all functions and distributions are supposed to be complex-valued; we therefore use complex distribution/function spaces.

We assume that $m\geq-r_{k}$ for each $k\in\{1,\ldots,\varkappa\}$.
This assumption is natural; indeed, if $m+r_{k}<0$ for some $k$, then $C_{1,k}=\cdots=C_{q+\varkappa,k}=0$, i.e. the boundary conditions \eqref{10f2} will not contain the unknown $v_{k}$.

We suppose that the boundary problem \eqref{10f1}, \eqref{10f2} is elliptic in $\Omega$ as a problem with additional unknown distributions   $v_{1},\ldots,v_{\varkappa}$ on $\Gamma$. This means that the PDO $A$ is properly elliptic on $\overline{\Omega}$, and the system of boundary conditions \eqref{10f2} covers  $A$ on $\Gamma$ (see, e.g., \cite[Section~3.1.3]{KozlovMazyaRossmann97}). We recall the corresponding definitions.

Let $A^{\circ}(x,\xi)$, $B_{j}^{\circ}(x,\xi)$, and $C_{j,k}^{\circ}(x,\tau)$ denote the principal symbols of the PDOs $A(x,D)$, $B_{j}(x,D)$, and $C_{j,k}(x,D_{\tau})$ respectively, the last two PDOs being considered as that of the formal orders $m_{j}$ and $m_{j}+r_{k}$ respectively. Thus, $A^{\circ}(x,\xi)$ and $B_{j}^{\circ}(x,\xi)$ are homogeneous polynomials in $\xi\in\mathbb{C}^{n}$ of order $2q$ and $m_{j}$ respectively. Besides, $C_{j,k}^{\circ}(x,\tau)$ is a homogeneous polynomial of order $m_{j}+r_{k}$ in $\tau$, where $\tau$ is a tangent vector to the boundary $\Gamma$ at the point~$x$. Defining the principal symbols, we consider the principal parts of the PDOs as polynomials in $D_l:=i\partial/\partial x_l$, where $l=1,\ldots,n$, and then replace each differential operator $D_l$ with the $l$-th component $\xi_l$ of the vector $\xi$.

The boundary-value problem \eqref{10f1}, \eqref{10f2} is called elliptic in $\Omega$ if it satisfies the following two conditions:
\begin{itemize}
\item [(i)] The PDO $A(x,D)$ is properly elliptic at every point $x\in\overline{\Omega}$; i.e., for arbitrary linear independent vectors $\xi',\xi''\in\mathbb{R}^{n}$, the polynomial $A^{\circ}(x,\xi'+\zeta\xi'')$ in $\zeta\in\mathbb{C}$ has $q$ roots with positive imaginary part and $q$ roots with negative imaginary part (of course, these roots are calculated with regard for their multiplicity).
\item [(ii)] The boundary conditions \eqref{10f2} cover $A(x,D)$ at every point $x\in\Gamma$. This means that, for an arbitrary tangent vector $\tau\neq0$ to $\Gamma$ at $x$, the boundary-value problem
    \begin{gather*}
    A^{\circ}(x,\tau+\nu(x)D_{t})\theta(t)=0\quad\mbox{for}\;\;t>0,\\
    B_{j}^{\circ}(x,\tau+\nu(x)D_{t})\theta(t)\big|_{t=0}+
    \sum_{k=1}^{\varkappa}C_{j,k}^{\circ}(x,\tau)\lambda_{k}=0,
    \quad j=1,...,q+\varkappa,\\
    \theta(t)\to0\quad\mbox{as}\quad t\rightarrow\infty
    \end{gather*}
    has only the trivial (zero) solution. Here, the function  $\theta\in C^{\infty}([0,\infty))$ and the numbers $\lambda_{1},\ldots,\lambda_{\varkappa}\in\mathbb{C}$ are unknown, whereas $\nu(x)$ is the unit inward normal vector to $\Gamma$ at $x$. Besides, $A^{\circ}(x,\tau+\nu(x)D_{t})$ and $B_{j}^{\circ}(x,\tau+\nu(x)D_{t})$ are differential operators with respect to $D_{t}:=i\partial/\partial t$. We obtain them putting $\zeta:=D_{t}$ in the polynomials $A^{\circ}(x,\tau+\zeta\nu(x))$ and $B_{j}^{\circ}(x,\tau+\zeta\nu(x))$ in $\zeta$, respectively.
\end{itemize}

Some examples of elliptic problems of the form \eqref{10f1}, \eqref{10f2} are given in \cite[Subsection~3.1.5]{KozlovMazyaRossmann97}.

\section{A refined Sobolev scale}\label{sec3}

This scale consists of the Hilbert generalized Sobolev spaces $H^{s,\varphi}$ whose order of regularity is given by a number  $s\in\mathbb{R}$ and function $\varphi\in\mathcal{M}$. Here, $\mathcal{M}$ denotes the set of all Borel measurable functions $\varphi:[1,+\infty)\rightarrow(0,+\infty)$ such that both functions $\varphi$ and $1/\varphi$ are bounded on each compact subset of $[1,+\infty)$ and that $\varphi$ varies slowly at  infinity in the sense of Karamata \cite{Karamata30a}, i.e. $\varphi(\lambda t)/\varphi(t)\rightarrow 1$ as $t\rightarrow\infty$ for every $\lambda>0$.

Slowly varying functions are well studied and have various important applications \cite{BinghamGoldieTeugels89}. A standard example  of such functions is
$$
\varphi(t):=(\log t)^{r_{1}}(\log\log
t)^{r_{2}}\ldots(\underbrace{\log\ldots\log}_{k\;\mbox{\small times}}
t)^{r_{k}}\quad\mbox{of}\quad t\gg1,
$$
where $k\in\mathbb{N}$ and $r_{1},\ldots,r_{k}\in\mathbb{R}$.

Let $s\in\mathbb{R}$ and $\varphi\in\mathcal {M}$. By definition, the linear space $H^{s,\varphi}(\mathbb{R}^{n})$, with $n\geq1$, consists of all distributions $w\in\mathcal{S}'(\mathbb{R}^{n})$ that their Fourier transform $\widehat{w}$ is locally Lebesgue integrable over $\mathbb{R}^{n}$ and satisfies the condition
$$
\|w\|_{s,\varphi;\mathbb{R}^{n}}^{2}:=
\int\limits_{\mathbb{R}^{n}}\langle\xi\rangle^{2s}
\varphi^{2}(\langle\xi\rangle)\,
|\widehat{w}(\xi)|^{2}\,d\xi<\infty.
$$
Here, $\mathcal{S}'(\mathbb{R}^{n})$ is the linear topological space of all tempered distributions on $\mathbb{R}^{n}$, and
$\langle\xi\rangle:=(1+|\xi|^{2})^{1/2}$. By definition, $\|\cdot\|_{s,\varphi;\mathbb{R}^{n}}$ is the norm in $H^{s,\varphi}(\mathbb{R}^{n})$.

The space $H^{s,\varphi}(\mathbb{R}^{n})$ is a special isotropic Hilbert case of the spaces introduced and investigated by H\"ormander \cite[Section~2.2]{Hermander63} (see also his monograph \cite[Section~10.1]{Hermander83}) and by Volevich and Paneah \cite[\S~2]{VolevichPaneah65}. If $\varphi(\cdot)\equiv1$, the space $H^{s,\varphi}(\mathbb {R}^{n})$ becomes the inner product Sobolev space $H^{s}(\mathbb {R}^{n})$ of order $s\in\mathbb{R}$. Generally, we have the dense continuous embeddings
\begin{equation}\label{10f3}
H^{s+\varepsilon}(\mathbb{R}^{n})\hookrightarrow H^{s,\varphi}(\mathbb{R}^{n})\hookrightarrow H^{s-\varepsilon}(\mathbb{R}^{n})
\quad\mbox{whenever}\quad\varepsilon>0.
\end{equation}
They show that the function parameter $\varphi$ refines the main regularity characterized by the number $s$. Therefore, the class of spaces $H^{s,\varphi}(\mathbb{R}^{n})$, where $s\in\mathbb{R}$ and $\varphi\in\mathcal{M}$, was called the refined Sobolev scale over $\mathbb{R}^{n}$ \cite[Section~1.3.3]{MikhailetsMurach14}. This class was selected and investigated in \cite{MikhailetsMurach05UMJ5, MikhailetsMurach06UMJ3} (compare, e.g., with \cite[Chapter~III]{Triebel01} and \cite{HaroskeMoura04}, where similar classes of Banach and more general spaces of distributions were studied).

To investigate the boundary-value problem \eqref{10f1}, \eqref{10f2}, we
need versions of the space $H^{s,\varphi}(\mathbb{R}^{n})$ for $\Omega$ and $\Gamma$; they are considered in \cite[Sections 2.1 and 3.2.1]{MikhailetsMurach14}.

By definition, the linear space $H^{s,\varphi}(\Omega)$ consists of
the restrictions of all distributions $w\in H^{s,\varphi}(\mathbb{R}^{n})$ to $\Omega$. It is endowed with the norm
$$
\|u\|_{s,\varphi;\Omega}:=
\inf\,\bigl\{\,\|w\|_{s,\varphi;\mathbb{R}^{n}}:
w\in H^{s,\varphi}(\mathbb{R}^{n}),\;\,
w=u\;\,\mbox{in}\;\,\Omega\,\bigr\},
$$
where $u\in H^{s,\varphi}(\Omega)$. The space $H^{s,\varphi}(\Omega)$ is Hilbert and separable with respect to this norm, with $C^{\infty}(\overline{\Omega})$ being a dense subset of  this space.

Briefly saying, the space $H^{s,\varphi}(\Gamma)$ consists of all distributions on $\Gamma$ that are reduced to distributions from $H^{s,\varphi}(\mathbb{R}^{n-1})$ in local coordinates on $\Gamma$. Let us give a detailed definition. We arbitrarily choose a finite collection of infinitely smooth local charts $\pi_j:\mathbb{R}^{n-1}\leftrightarrow\Gamma_{j}$, with $j=1,\ldots,\lambda$, that the open sets $\Gamma_{1},\ldots,\Gamma_{\lambda}$ form a covering of $\Gamma$. We also arbitrarily choose functions $\chi_j\in C^{\infty}(\Gamma)$, with $j=1,\ldots,\lambda$, that form a partition of unity on $\Gamma$ subject to $\mathrm{supp}\,\chi_j\subset\Gamma_j$. By definition, the linear space $H^{s,\varphi}(\Gamma)$ consists of all distributions $h\in\mathcal{D}'(\Gamma)$ such that $(\chi_{j}h)\circ\pi_{j}\in H^{s,\varphi}(\mathbb{R}^{n-1})$ for each $j\in\{1,\ldots,\lambda\}$.
Here, $\mathcal{D}'(\Gamma)$ is the linear topological space of all distributions on $\Gamma$, and $(\chi_{j}h)\circ\pi_{j}$ stands for  the representation of the distribution $\chi_{j}h$ in the local chart $\pi_{j}$. The norm in $H^{s,\varphi}(\Gamma)$ is defined by the formula
$$
\|h\|_{s,\varphi;\Gamma}:=
\biggl(\,\sum_{j=1}^{\lambda}\,\|(\chi_{j}h)\circ\pi_{j}\|_
{s,\varphi;\mathbb{R}^{n-1}}^{2}\biggr)^{1/2}.
$$
The space $H^{s,\varphi}(\Gamma)$ is Hilbert and separable. It does not depend (up to equivalence of norms) on the indicated choice of local charts and partition of unity on $\Gamma$ \cite[Theorem~2.3]{MikhailetsMurach14}. The set $C^{\infty}(\Gamma)$ is dense in $H^{s,\varphi}(\Gamma)$.

The spaces $H^{s,\varphi}(\Omega)$ and $H^{s,\varphi}(\Gamma)$, where $s\in\mathbb{R}$ and $\varphi\in\mathcal{M}$, form the refined Sobolev scales over $\Omega$ and $\Gamma$. If $\varphi(\cdot)\equiv1$, these spaces become the inner product Sobolev spaces $H^{s}(\Omega)$ and $H^{s}(\Gamma)$, the norms in them being denoted by $\|\cdot\|_{s;\Omega}$ and $\|\cdot\|_{s;\Gamma}$, resp. Generally, the dense compact embeddings \eqref{10f3} hold true provided that we replace $\mathbb{R}^{n}$ with $\Omega$ or $\Gamma$.

The refined Sobolev scale over $G\in\{\mathbb{R}^{n},\Omega,\Gamma\}$ possesses the following important interpolation property: every space $H^{s,\varphi}(G)$ is obtained by quadratic interpolation (with an appropriate function parameter) between the Sobolev spaces $H^{s-\varepsilon}(G)$ and $H^{s+\delta}(G)$ where $\varepsilon,\delta>0$ (see \cite[Theorems 1.14, 2.2, and 3.2]{MikhailetsMurach14}). This property play a key role in applications of these scales to elliptic operators and elliptic problems (see \cite{MikhailetsMurach12BJMA2, MikhailetsMurach14} and references therein).

In what follows we will consider various Hilbert spaces induced by the spaces $H^{s,\varphi}(G)$ and related to the problem \eqref{10f1}, \eqref{10f2}. If $\varphi(\cdot)\equiv1$, we will omit the index $\varphi$ in the designations of these spaces and norms in them.

\section{Main results}\label{sec4}

Consider the linear mapping
\begin{equation}\label{mapping}
\Lambda:(u,v_{1},...,v_{\varkappa})\mapsto(f,g_{1},...,g_{q+\varkappa}),
\;\;\mbox{where}\;\;u\in C^{\infty}(\overline{\Omega}),\;\;
v_{1},\ldots,v_{\varkappa}\in C^{\infty}(\Gamma)
\end{equation}
and where the functions $f$ and $g_{1}$,..., $g_{q+\varkappa}$ are defined by formulas \eqref{10f1} and \eqref{10f2}. Introduce the Hilbert spaces
\begin{equation*}
\mathcal{D}^{s,\varphi}(\Omega,\Gamma):=H^{s,\varphi}(\Omega)\oplus
\bigoplus_{k=1}^{\varkappa}H^{s+r_{k}-1/2,\varphi}(\Gamma)
\end{equation*}
and
\begin{equation*}
\mathcal{E}_{s,\varphi}(\Omega,\Gamma):=H^{s-2q,\varphi}(\Omega)\oplus
\bigoplus_{j=1}^{q+\varkappa}H^{s-m_{j}-1/2,\varphi}(\Gamma)
\end{equation*}
for arbitrary $s\in\mathbb{R}$ and $\varphi\in\mathcal{M}$.

According to \cite[Theorem~1]{ChepurukhinaMurach15MFAT1}, this mapping extends uniquely (by continuity) to a Fredholm bounded operator
\begin{equation}\label{Fredholm-positive-scale}
\Lambda:\mathcal{D}^{s,\varphi}(\Omega,\Gamma)\to
\mathcal{E}_{s,\varphi}(\Omega,\Gamma)
\end{equation}
for all $s>2q-1/2$ and $\varphi\in\mathcal{M}$. The finite-dimensional kernel of the operator \eqref{Fredholm-positive-scale} lies in
$$
\mathcal{D}^{\infty}(\overline{\Omega},\Gamma):=
C^{\infty}(\overline{\Omega})\times(C^{\infty}(\Gamma))^{\varkappa}
$$
and together with the finite index of \eqref{Fredholm-positive-scale} does not depend on $s$ and~$\varphi$. Let $N$ denote the kernel, and let $\vartheta$ stand for the index.

This result cannot be spread to all real $s$ without changes in its formulation. This follows from the known fact that the trace operator $u\mapsto u\!\upharpoonright\!\Gamma$, where $u\in C^{\infty}(\overline{\Omega})$, cannot be extended to a continuous mapping from $H^{s}(\Omega)$ to $\mathcal{D}'(\Gamma)$ if $s\leq1/2$.

In the $s\leq2q-1/2$ case, the boundary data $g_{j}\in H^{s-m_{j}-1/2,\varphi}(\Gamma)$ may be irregular distributions
(so called, rough data). Examining this case, we assume that $f\in L_{2}(\Omega)$, which allows us to use an $s<2q$ version \cite[Theorem~1]{MurachChepurukhina15UMJ5} of the above result. This version involves the linear space
$$
H^{s,\varphi}_{A}(\Omega):=\bigl\{u\in H^{s,\varphi}(\Omega):
Au\in L_{2}(\Omega)\bigr\}
$$
endowed with the graph norm
\begin{equation*}
\|u\|_{s,\varphi;\Omega,A}:=
\bigl(\|u\|^{2}_{s,\varphi;\Omega}+\|Au\|^{2}_{\Omega}\bigr)^{1/2}.
\end{equation*}
Here, $s<2q$; $\varphi\in\mathcal{M}$; $\|\cdot\|_{\Omega}$ is the norm in the Hilbert space $L_{2}(\Omega)$ of square integrable functions over $\Omega$, and $Au$ is understood in the sense of the theory of distributions on $\Omega$. The space $H^{s,\varphi}_{A}(\Omega)$ is Hilbert, and $C^{\infty}(\overline{\Omega})$ is dense in this space   \cite[Section~4]{MurachChepurukhina15UMJ5}. Note that $H^{s,\varphi}_{A}(\Omega)$ depends essentially on $A$ (even when all coefficients of $A$ are constant), which was shown by H\"ormander \cite[Theorem~3.1]{Hermander55}
in the case where $s=0$ and $\varphi(\cdot)\equiv1$. Consider the Hilbert spaces
\begin{equation*}
\mathcal{D}^{s,\varphi}_{A}(\Omega,\Gamma):=H^{s,\varphi}_{A}(\Omega)\oplus
\bigoplus_{k=1}^{\varkappa}H^{s+r_{k}-1/2,\varphi}(\Gamma)
\end{equation*}
and
\begin{equation*}
\mathcal{E}^{0,s,\varphi}(\Omega,\Gamma):=L_{2}(\Omega)\oplus
\bigoplus_{j=1}^{q+\varkappa}H^{s-m_{j}-1/2,\varphi}(\Gamma).
\end{equation*}

\begin{proposition}\label{10pr1}
The mapping \eqref{mapping} extends uniquely (by continuity) to a bounded operator
\begin{equation}\label{10f5}
\Lambda:\mathcal{D}^{s,\varphi}_{A}(\Omega,\Gamma)\rightarrow
\mathcal{E}^{0,s,\varphi}(\Omega,\Gamma)
\end{equation}
for arbitrary $s<2q$ and $\varphi\in\mathcal{M}$. This operator is Fredholm. Its kernel coincides with $N$, and its index equals $\vartheta$.
\end{proposition}

This result was proved in \cite[Theorem~1]{MurachChepurukhina15UMJ5}.
We will supplement it with theorems on local (up to the boundary $\Gamma$) regularity and a local \textit{a priori} estimate of the generalized solutions to the elliptic problem \eqref{10f1},~\eqref{10f2}. Beforehand, using Proposition~\ref{10pr1}, we give a definition of such a solution.

Put
$$
\mathcal{S}'_{A}(\Omega):=\{u\in\mathcal{S}'(\Omega):
Au\in L_{2}(\Omega)\},
$$
where, as usual, $\mathcal{S}'(\Omega)$ is the space of the restrictions of all distributions $w\in\mathcal{S}'(\mathbb{R}^{n})$ to~$\Omega$.
Since $\Omega$ is bounded, the space $\mathcal{S}'_{A}(\Omega)$ is the union of all $H^{s,\varphi}_{A}(\Omega)$ such that $s<2q$ and $\varphi\in\mathcal{M}$.

Assume that
\begin{equation}\label{10f6}
(u,v):=(u,v_{1},\ldots,v_{\varkappa})\in
\mathcal{S}'_{A}(\Omega)\times(\mathcal{D}'(\Gamma))^{\varkappa}
\end{equation}
and
\begin{equation*}
(f,g):=(f,g_{1},\ldots,g_{q+\varkappa})\in
L_{2}(\Omega)\times(\mathcal{D}'(\Gamma))^{q+\varkappa}.
\end{equation*}
The vector \eqref{10f6} is called a generalized (strong) solution to the boundary-value problem \eqref{10f1}, \eqref{10f2} if $\Lambda(u,v)=(f,g)$ for some operator \eqref{10f5} from Proposition~\ref{10pr1}. This definition is reasonable because $(u,v)\in\mathcal{D}^{s,\varphi}_{A}(\Omega,\Gamma)$ for sufficiently small $s<2q$ and every $\varphi\in\mathcal{M}$ and because the image $\Lambda(u,v)$ does not depend on these $s$ and $\varphi$.

Now we introduce local versions of the spaces $H^{l,\varphi}(\Omega)$ and $H^{l,\varphi}(\Gamma)$, where $l\in\mathbb{R}$ and $\varphi\in\mathcal{M}$. We need these versions to formulate a theorem on local regularity of a generalized solution to the problem under investigation. Let $U$ be an open subset of $\mathbb{R}^{n}$ such that $\Omega_{0}:=\Omega\cap U\neq\emptyset$ and $\Gamma_{0}:=\Gamma\cap U\neq\emptyset$. We let $H^{l,\varphi}_{\mathrm{loc}}(\Omega_{0},\Gamma_{0})$ denote the linear space of all distributions $u\in\mathcal{S}'(\Omega)$ such that $\chi u\in H^{l,\varphi}(\Omega)$ for every function $\chi\in C^{\infty}(\overline{\Omega})$ satisfying $\mathrm{supp}\,\chi\subset\Omega_0\cup\Gamma_{0}$. Analogously,
$H^{l,\varphi}_{\mathrm{loc}}(\Gamma_{0})$ denotes the linear space of all distributions $h\in\mathcal{D}'(\Gamma)$ such that $\chi h\in H^{l,\varphi}(\Gamma)$ for every function $\chi\in C^{\infty}(\Gamma)$ satisfying $\mathrm{supp}\,\chi\subset\Gamma_{0}$.

\begin{theorem}\label{10th1}
Let $s<2q$ and $\varphi\in\mathcal{M}$. Assume that a vector \eqref{10f6} is a generalized solution to the elliptic problem \eqref{10f1}, \eqref{10f2} whose right-hand sides satisfy the conditions $f\in L_2(\Omega)$ and $g_j\in H^{s-m_j-1/2,\varphi}_{\mathrm{loc}}(\Gamma_0)$ for each $j\in\{1,\ldots,q+\varkappa\}$. Then $u\in H^{s,\varphi}_{\mathrm{loc}}(\Omega_{0},\Gamma_{0})$ and  $v_k\in H^{s+r_k-1/2,\varphi}_{\mathrm{loc}}(\Gamma_0)$ for each $k\in\{1,\ldots,\varkappa\}$.
\end{theorem}

Note that the definition of $H^{l,\varphi}_{\mathrm{loc}}(\Omega_{0},\Gamma_{0})$ makes sense in the $\Gamma_{0}=\emptyset$ case. It follows from condition \eqref{10f6} and the ellipticity of the PDO $A$ that $u\in H^{2q}_{\mathrm{loc}}(\Omega_{0},\emptyset)$ (see, e.g., \cite[Theorem 7.4.1]{Hermander63}).

Now we formulate a theorem on a local \textit{a priori} estimate of the generalized solution to the problem under investigation. Let $\|\cdot\|'_{s,\varphi}$ denote the norm in the Hilbert space $\mathcal{D}^{s,\varphi}(\Omega,\Gamma)$, and let $\|\cdot\|''_{0,s,\varphi}$ stand for the norm in the Hilbert space $\mathcal{E}^{0,s,\varphi}(\Omega,\Gamma)$.

\begin{theorem}\label{10th2}
Let $s<2q$ and $\varphi\in\mathcal{M}$. Assume that a vector \eqref{10f6} satisfies the hypotheses of Theorem~$\ref{10th1}$. We arbitrarily choose a number $\lambda>0$ and functions $\chi,\eta\in C^{\infty}(\overline{\Omega})$ such that $\mathrm{supp}\,\chi\subset\mathrm{supp}\,\eta\subset
\Omega_{0}\cup\Gamma_{0}$ and that $\eta=1$ in a neighbourhood of $\mathrm{supp}\,\chi$. Then
\begin{equation}\label{10f9}
\|\chi(u,v)\|'_{s,\varphi}\leq c\,\bigl(\|\eta(f,g)\|''_{0,s,\varphi}+
\|\eta(u,v)\|'_{s-\lambda,\varphi}\bigr)
\end{equation}
for a certain number $c>0$ that does not depend on $(u,v)$ and $(f,g)$.
\end{theorem}

Here, of course, $\chi(u,v)$ means $(\chi u,(\chi\!\upharpoonright\!\Gamma)v_{1},\ldots,
(\chi\!\upharpoonright\!\Gamma)v_{\varkappa})$, and $\eta(f,g)$ is analogously interpreted. These theorems are new in the Sobolev case of $\varphi(\cdot)\equiv1$ even where $s$ is an integer. They consist the local (up to the boundary) lifting property of the generalized solution
$(u,v)$.

\section{Proofs}\label{sec5}

If $\Omega_{0}=\Omega$ and $\Gamma_{0}=\Gamma$
and if $\chi(\cdot)\equiv\eta(\cdot)\equiv1$, Theorems \ref{10th1} and \ref{10th2} deal with global properties of the generalized solution $(u,v)$, i.e. with its properties in the whole domain $\Omega$ up to the boundary $\Gamma$. In this specific case, the theorems follows easily from Proposition~\ref{10pr1} and are given in \cite[Theorems 3 and~2]{MurachChepurukhina15UMJ5}. In the general case, Theorems \ref{10th1} and \ref{10th2} cannot be deduced from the global case in a routine manner used in \cite[Section~3.2.3]{KozlovMazyaRossmann97} and \cite[Section~2.4.4]{Roitberg99} for elliptic problems with unknowns on the boundary. This is caused by the following two circumstances: the space $H^{s,\varphi}_{A}(\Omega)$ is not closed with respect to the multiplication by functions from $C^{\infty}(\overline{\Omega})$, and the right-hand side of the inequality \eqref{10f9} contains the norm $\|\eta f\|_{\Omega}$ instead of $\|\eta f\|_{s-2q,\varphi}$. We therefore cannot take $\chi u$ instead of $u$ in the global versions of these theorems to treat the general case. Besides, if we interchange the PDO $A$ and the operator of the multiplication by $\chi$ according to the routine, we get
$$
\|A(\chi u)\|_{\Omega}\leq\|\chi Au\|_{\Omega}+\|A'(\eta u)\|_{\Omega}\leq\|\chi f\|_{\Omega}+\|\eta u\|_{2q-1;\Omega}
$$
whenever $u\in C^{\infty}(\overline{\Omega})$, which yields a trivial estimate (for $\chi u$) instead of \eqref{10f9} provided that $\nobreak{s<2q-1}$ (here, the PDO $A'$ is the commutator of these operators).

To prove Theorems \ref{10th1} and \ref{10th2}, we develop methods worked out in \cite[Section~5]{AnopDenkMurach20arxiv} and \cite[Section~6]{AnopKasirenkoMurach18UMJ3} for elliptic problems without unknowns on the boundary. These methods use property of elliptic problems in Sobolev spaces modified by Roitberg \cite{Roitberg64, Roitberg65} (see also his monograph \cite[Section~2]{Roitberg96}). For our purposes, we need the similar modification $H^{s,\varphi,(2q)}(\Omega)$ of the space $H^{s,\varphi}(\Omega)$. This modification was introduced and investigated in \cite{MikhailetsMurach08UMJ4} (see also the book \cite[Section~4.2.2]{MikhailetsMurach14}). In the Sobolev case of $\varphi(\cdot)\equiv1$, the space $H^{s,\varphi,(2q)}(\Omega)$ was introduced by Roitberg in~\cite{Roitberg64}.

Let $\varphi\in\mathcal{M}$. If $s\notin\{1/2,3/2,\ldots,2q-1/2\}$, then $H^{s,\varphi,(2q)}(\Omega)$ is defined to be the completion of $C^{\infty}(\overline{\Omega})$ with respect to the Hilbert norm
$$
\|u\|_{s,\varphi,(2q);\Omega}:=
\biggl(\|u\|_{s,\varphi,(0);\Omega}^{2}+
\sum_{j=1}^{k}\;\|(\partial_{\nu}^{j-1}u)\!\upharpoonright\!\Gamma\|
_{s-j+1/2,\varphi;\Gamma}^{2}\biggr)^{1/2}.
$$
Here, $\partial_{\nu}$ is the operator of the differentiation with respect to the inward normal to $\Gamma$, and $\|\cdot\|_{s,\varphi,(0);\Omega}$ is the norm in the Hilbert space
$$
H^{s,\varphi,(0)}(\Omega):=
\left\{\begin{array}{ll}
H^{s,\varphi}(\Omega)&\hbox{if}\;\;s\geq0\\
(H^{-s,1/\varphi}(\Omega))'&\hbox{if}\;\;s<0,
\end{array}\right.
$$
the dual space being considered with respect to the inner product in $L_{2}(\Omega)$. (Note that $\varphi\in\mathcal{M}\Leftrightarrow 1/\varphi\in\mathcal{M}$.) Thus, if $s<0$, the space $H^{s,\varphi,(0)}(\Omega)$ is the completion of $L_{2}(\Omega)$ with respect to the norm
$$
\|u\|_{s,\varphi,(0);\Omega}:=\sup\biggl\{\frac{|(u,w)_{\Omega}|}
{\;\quad\|w\|_{-s,1/\varphi;\Omega}}:
w\in H^{-s,1/\varphi}(\Omega),\,w\neq0\biggr\},
$$
where $(\cdot,\cdot)_{\Omega}$ stands for the inner product in $L_{2}(\Omega)$. Then  $\|u\|_{s,\varphi,(0);\Omega}=\|\mathcal{O}u\|_{s,\varphi;\mathbb{R}^{n}}$
for every $u\in C^{\infty}(\overline{\Omega})$, with $\mathcal{O}u:=u$ on $\overline{\Omega}$ and $\mathcal{O}u:=0$ on $\mathbb{R}^{n}\setminus\overline{\Omega}$. If $s\in\{1/2,3/2,\ldots,2q-1/2\}$, the Hilbert space $H^{s,\varphi,(2q)}(\Omega)$ is, by definition, the result of the quadratic interpolation with the parameter $1/2$ between the spaces $H^{s-1/2,\varphi,(2q)}(\Omega)$ and $H^{s+1/2,\varphi,(2q)}(\Omega)$.

If $s>2q-1/2$, the spaces $H^{s,\varphi,(2q)}(\Omega)$ and $H^{s,\varphi}(\Omega)$ are equal as completions of $C^{\infty}(\overline{\Omega})$ with respect to equivalent norms. In the opposite case, the space $H^{s,\varphi,(2q)}(\Omega)$ contains elements that are not distributions. If $s_1<s_2$ and $\varphi_1,\varphi_2\in\mathcal{M}$, the identity mapping on $C^{\infty}(\overline{\Omega})$ extends uniquely to a compact embedding of $H^{s_2,\varphi_2,(2q)}(\Omega)$ in $H^{s_1,\varphi_1,(2q)}(\Omega)$.

\begin{proof}[Proof of Theorem~$\ref{10th1}$]
We arbitrarily choose a function $\chi\in C^{\infty}(\overline{\Omega})$ subject to $\mathrm{supp}\,\chi\subset\Omega_0\cup\nobreak\Gamma_0$ and consider a function $\eta\in C^{\infty}(\overline{\Omega})$ such that $\mathrm{supp}\,\eta\subset\Omega_0\cup\Gamma_0$ and that $\eta=1$ in some neighbourhood $V$ of $\mathrm{supp}\,\chi$ in the topology of $\overline{\Omega}$. According to the hypotheses of the theorem, we have the inclusion $(u,v)\in\mathcal{D}^{l}_{A}(\Omega,\Gamma)$ for a sufficiently small integer $l<s$ and the inclusion $\eta(f,g)\in\mathcal{E}^{0,s,\varphi}(\Omega,\Gamma)$. We must deduce from them that $\chi(u,v)\in\mathcal{D}^{s,\varphi}(\Omega,\Gamma)$.

Since the operator \eqref{10f5} is Fredholm and since the set
$$
\mathcal{E}^{\infty}(\overline{\Omega},\Gamma):=
C^{\infty}(\overline{\Omega})\times(C^{\infty}(\Gamma))^{q+\varkappa}
$$
is dense in $\mathcal{E}^{0,s,\varphi}(\Omega,\Gamma)$, it follows from
\cite[Lemma~2.1]{HohbergKrein60} that $\eta(f,g)=\Lambda(u',v')+(f'',g'')$ for some $(u',v')\in\mathcal{D}_{A}^{s,\varphi}(\Omega,\Gamma)$ and $(f'',g'')\in\mathcal{E}^{\infty}(\overline{\Omega},\Gamma)$. Then
\begin{equation}\label{1proof-f9}
\Lambda(u-u',v-v')=(1-\eta)(f,g)+(f'',g'')=:(f''',g'''),
\end{equation}
with
\begin{equation}\label{1proof-f10}
\zeta(f''',g''')=
\zeta(f'',g'')\in\mathcal{E}^{\infty}(\overline{\Omega},\Gamma)=
\bigcap_{\sigma\in\mathbb{R}}\mathcal{E}_{\sigma,(0)}(\Omega,\Gamma)
\end{equation}
for every function $\zeta\in C^{\infty}(\overline{\Omega})$ subject to $\mathrm{supp}\,\zeta\subset V$. Here and below,
$$
\mathcal{E}_{\sigma,\varphi,(0)}(\Omega,\Gamma):=
H^{\sigma-2q,\varphi,(0)}(\Omega)\oplus
\bigoplus_{j=1}^{q+\varkappa}H^{\sigma-m_{j}-1/2,\varphi}(\Gamma)
$$
and
$$
\mathcal{D}^{\sigma,\varphi,(2q)}(\Omega,\Gamma):= H^{\sigma,\varphi,(2q)}(\Omega)\oplus
\bigoplus_{k=1}^{\varkappa}H^{\sigma+r_{k}-1/2,\varphi}(\Gamma)
$$
for every $\sigma\in\mathbb{R}$.

It follows from \cite[Section~4.4.2, isomorphism (4.196)]{MikhailetsMurach14} that the identity mapping on $C^{\infty}(\overline{\Omega})$ extends uniquely to a continuous embedding $H^{l}_{A}(\Omega)\hookrightarrow H^{l,(2q)}(\Omega)$. Hence, we may consider the distribution $u-u'\in H^{l}_{A}(\Omega)$ as an element of the Roitberg space $H^{l,(2q)}(\Omega)$. Thus,
\begin{equation}\label{1proof-f11}
(u-u',v-v')\in\mathcal{D}^{l,(2q)}(\Omega,\Gamma).
\end{equation}

According to the lifting property \cite[Theorem~2.4.3]{Roitberg99} (see also \cite[Theorem~4]{ChepurukhinaMurach15MFAT1} as to the spaces $H^{\sigma,\varphi,(2q)}(\Omega)$), it follows from \eqref{1proof-f9}--\eqref{1proof-f11} that
$$
(u^{\circ},v^{\circ}):=\chi(u-u',v-v')\in \bigcap_{\sigma\in\mathbb{R}}\mathcal{D}^{\sigma,(2q)}(\Omega,\Gamma)
\subset\mathcal{D}^{s,\varphi}(\Omega,\Gamma)
$$
because $\mathrm{supp}\,\chi\subset V$. Hence,
$$
\chi(u,v)=(u^{\circ},v^{\circ})+\chi(u',v')
\in\mathcal{D}^{s,\varphi}(\Omega,\Gamma)
$$
in view of the inclusion $(u',v')\in\mathcal{D}_{A}^{s,\varphi}(\Omega,\Gamma)$.
\end{proof}

\begin{proof}[Proof of Theorem~$\ref{10th2}$]
According to \cite[Theorem~1]{ChepurukhinaMurach15MFAT1}, the mapping \eqref{mapping} extends uniquely (by continuity) to a Fredholm bounded operator
\begin{equation}\label{proof2-f12}
\Lambda:\mathcal{D}^{\sigma,\varphi,(2q)}(\Omega,\Gamma)\to
\mathcal{E}_{\sigma,\varphi,(0)}(\Omega,\Gamma)
\quad\mbox{for every}\;\;\sigma\in\mathbb{R},
\end{equation}
the kernel and index of this operator being the same as those of the operator~\eqref{Fredholm-positive-scale}. Note that these operators coincide if $s=\sigma>2q-1/2$. Let $\|\cdot\|'_{\sigma,\varphi,(2q)}$ denote the norm in $\mathcal{D}^{\sigma,\varphi,(2q)}(\Omega,\Gamma)$, and let $\|\cdot\|''_{\sigma,\varphi,(0)}$ denote the norm in $\mathcal{E}_{\sigma,\varphi,(0)}(\Omega,\Gamma)$.

Assume that $0\leq l\in\mathbb{Z}$ and that a function $\zeta\in C^{\infty}(\overline{\Omega})$ satisfies the condition $\zeta=1$ in a neighbourhood of $\mathrm{supp}\,\chi$. Let us prove by induction in $l$ that
\begin{equation}\label{proof2-f13}
\|\chi(u,v)\|'_{s,\varphi,(2q)}\leq c_{0}\bigl(\|\zeta\Lambda(u,v)\|''_{s,\varphi,(0)}+
\|\zeta(u,v)\|'_{s-l,\varphi,(2q)}\bigr)
\end{equation}
for every $(u,v)\in\mathcal{D}^{\infty}(\overline{\Omega},\Gamma)$ with some number $c_{0}>0$ not depending on $(u,v)$.

If $l=0$, then \eqref{proof2-f13} follows from the evident fact that the operator of the multiplication by a function from $C^{\infty}(\overline{\Omega})$ is bounded on every space $H^{\sigma,\varphi,(2q)}(\Omega)$. Assume now that the inequality \eqref{proof2-f13} holds true for a certain integer $l=p\geq0$, and prove this inequality for $l=p+1$.

Consider a function $\zeta_{0}\in C^{\infty}(\overline{\Omega})$ such that  $\zeta_{0}=1$ in a neighbourhood of $\mathrm{supp}\,\chi$ and that $\zeta=1$ in a neighbourhood of $\mathrm{supp}\,\zeta_{0}$. By the inductive assumption,
\begin{equation}\label{proof2-f14}
\|\chi(u,v)\|'_{s,\varphi,(2q)}\leq c_{1}\bigl(\|\zeta_{0}\Lambda(u,v)\|''_{s,\varphi,(0)}+
\|\zeta_{0}(u,v)\|'_{s-p,\varphi,(2q)}\bigr).
\end{equation}
In the proof, $c_{1}$, $c_{2}$,... denote some positive numbers that do not depend on $(u,v)$. Since the bounded operator \eqref{proof2-f12}, where $\sigma=s-p$, is Fredholm, we conclude by Peetre's lemma \cite[Lemma~3]{Peetre61} that
\begin{equation}\label{proof2-f15}
\|\zeta_{0}(u,v)\|'_{s-p,\varphi,(2q)}\leq c_{2}\bigl(\|\Lambda(\zeta_{0}(u,v))\|''_{s-p,\varphi,(0)}+
\|\zeta_{0}(u,v)\|'_{s-p-1,\varphi,(2q)}\bigr).
\end{equation}
Interchanging the operator of the multiplication by $\zeta_{0}$ with the PDOs used in the problem \eqref{10f1}, \eqref{10f2}, we write
\begin{equation}\label{proof2-f16}
\begin{aligned}
\Lambda(\zeta_{0}(u,v))&=\Lambda(\zeta_{0}\zeta(u,v))=
\zeta_{0}\Lambda(\zeta(u,v))+\Lambda'(\zeta(u,v))\\
&=\zeta_{0}\Lambda(u,v)+\Lambda'(\zeta(u,v)),
\end{aligned}
\end{equation}
where $\Lambda'$ is an operator of the same structure as $\Lambda$ but formed by PDOs of lower orders than the corresponding PDOs in \eqref{10f1}, \eqref{10f2}. Hence,
\begin{equation}\label{proof2-f17}
\|\Lambda'(\zeta(u,v))\|''_{s-p,\varphi,(0)}\leq
c_{3}\|\zeta(u,v)\|'_{s-p-1,\varphi,(2q)}
\end{equation}
due to \cite[Theorem~4.13]{MikhailetsMurach14}.
According to \eqref{proof2-f15}--\eqref{proof2-f17}, we obtain the inequality
\begin{align*}
\|\zeta_{0}(u,v)\|'_{s-p,\varphi,(2q)}\leq c_{4}\bigl(\|\zeta_{0}\Lambda(u,v)\|''_{s-p,\varphi,(0)}+
\|\zeta(u,v)\|'_{s-p-1,\varphi,(2q)}\bigr).
\end{align*}
Substituting it in \eqref{proof2-f14}, we arrive at \eqref{proof2-f13} in the $l=p+1$ case. Thus, \eqref{proof2-f13} is proved for every integer $l\geq0$.

Choose a number $p>\lambda$ such that $s-p$ is a negative integer. It follows from the inequality \eqref{proof2-f13} for an integer $l>p$ that
\begin{equation}\label{proof2-f18}
\|\chi(u,v)\|'_{s,\varphi,(2q)}\leq c_{5}\bigl(\|\zeta\Lambda(u,v)\|''_{s,\varphi,(0)}+
\|\zeta(u,v)\|'_{s-p,(2q)}\bigr)
\end{equation}
if we take \eqref{10f3} into account. Let us deduce the required estimate \eqref{10f9} from \eqref{proof2-f18}. We continue to assume that $(u,v)\in \mathcal{D}^{\infty}(\overline{\Omega},\Gamma)$.

By the definition of $H^{s,\varphi,(2q)}(\Omega)$, we have
\begin{equation*}
\|\chi u\|_{s,\varphi;\Omega}\leq \|\chi u\|_{s,\varphi,(0);\Omega}\leq
\|\chi u\|_{s,\varphi,(2q);\Omega}
\end{equation*}
if $s\notin\{1/2,3/2,\ldots,2q-1/2\}$. It follows from this by the quadratic interpolation that
\begin{equation*}
\|\chi u\|_{s,\varphi;\Omega}\leq c_{6}\|\chi u\|_{s,\varphi,(2q);\Omega}
\end{equation*}
for the rest values of $s$. Hence,
\begin{equation}\label{proof2-f19}
\|\chi(u,v)\|'_{s,\varphi}\leq c_{7}\|\chi(u,v)\|'_{s,\varphi,(2q)}.
\end{equation}

Let $W$ be an open set from the topology on $\overline{\Omega}$ such that $\mathrm{supp}\,\chi\subset W$ and that $\eta=1$ on $\overline{W}$ and that $W_0:=W\cap\Omega$ is an open domain in $\mathbb{R}^{n}$ with infinitely smooth boundary. The last condition allows us to consider the
Roitberg space $H^{s-p,(2q)}(W_0)$. Let $w\in C^{\infty}(\overline{W})$ be the restriction of $u$ to $\overline{W}$. Assume in addition that  $\mathrm{supp}\,\zeta\subset W$. We have the equivalence of norms
\begin{equation}\label{proof2-f20}
\|\zeta u\|_{s-p,(2q),\Omega}\asymp\|\zeta w\|_{s-p,(2q);W_0}.
\end{equation}
Indeed, owing to \cite[Theorem 6.1.1]{Roitberg96} and since $s-p<0$, we get
\begin{align*}
\|\zeta u\|_{s-p,(2q);\Omega}&\asymp
\|\zeta u\|_{s-p,(0);\Omega}+\|A(\zeta u)\|_{s-p-2q,(0);\Omega}\\
&=\|\mathcal{O}(\zeta u)\|_{s-p;\mathbb{R}^{n}}+
\|\mathcal{O}A(\zeta u)\|_{s-p-2q;\mathbb{R}^{n}}\\
&=\|\zeta w\|_{s-p,(0);W_0}+\|A(\zeta w)\|_{s-p-2q,(0);W_0}\asymp
\|\zeta w\|_{s-p,(2q);W_0}
\end{align*}
because $\mathcal{O}(\zeta u)$ and $\mathcal{O}A(\zeta u)$ are also extensions of the functions $\zeta w$ and $A(\zeta w)$, resp., to $\mathbb{R}^{n}$ with zero. According to \cite[Section~4.4.2, isomorphism (4.196)]{MikhailetsMurach14} we have another equivalence of norms
\begin{equation}\label{proof2-f21}
\|w\|_{s-p,(2q),W_0}+\|Aw\|_{W_0}\asymp
\|w\|_{s-p,W_0}+\|Aw\|_{W_0};
\end{equation}
here, recall, $\|\cdot\|_{W_0}$ denotes the norm in $L_{2}(W_0)$.

Formulas \eqref{proof2-f20} and \eqref{proof2-f21} yield
\begin{equation}\label{proof2-f22}
\begin{aligned}
\|\zeta u\|_{s-p,(2q);\Omega}&\asymp\|\zeta w\|_{s-p,(2q);W_0}\leq
c_{8}\bigl(\|w\|_{s-p,(2q);W_0}+\|Aw\|_{W_0}\bigr)\\
&\asymp\|w\|_{s-p;W_0}+\|Aw\|_{W_0}\leq
\|\eta u\|_{s-p;\Omega}+\|\eta Au\|_{\Omega}.
\end{aligned}
\end{equation}
Substituting \eqref{proof2-f19} and \eqref{proof2-f22} in \eqref{proof2-f18}, we get
\begin{align*}
\|\chi(u,v)\|'_{s,\varphi}&\leq c_{9} \bigl(\|\zeta\Lambda(u,v)\|''_{s,\varphi,(0)}+
\|\eta(u,v)\|'_{s-p}+\|\eta Au\|_{\Omega}\bigr)\\
&\leq c\bigl(\|\eta\Lambda(u,v)\|''_{0,s,\varphi}+
\|\eta(u,v)\|'_{s-\lambda,\varphi}\bigr)
\end{align*}
because $s-2q<0$ (then the norm $\|\cdot\|_{s-2q,(0);\Omega}$ is subordinate to $\|\cdot\|_{\Omega}$) and because $p>\lambda$ (then
the norm $\|\cdot\|'_{s-p}$ is subordinate to $\|\cdot\|'_{s-\lambda,\varphi}$). Thus, we have proved the required estimate \eqref{10f9} in the case where $(u,v)\in \mathcal{D}^{\infty}(\overline{\Omega},\Gamma)$.

Now we consider an arbitrary vector \eqref{10f6} that satisfies the hypotheses of Theorem~$\ref{10th1}$ and deduce this estimate from the case just examined. Let $V$ be an open set from the topology on $\overline{\Omega}$ such that $\overline{V}\subset\Omega_{0}\cup\Gamma_{0}$ and $\mathrm{supp}\,\eta\subset V$ and that $V_0:=V\cap\Omega$ is an open domain in $\mathbb{R}^{n}$ with an infinitely smooth boundary $\partial V_0$. According to Theorem~$\ref{10th1}$, we have the inclusion $\omega:=u\!\upharpoonright\!V_0\in H^{s,\varphi}_{A}(V_0)$. Since  $C^{\infty}(\overline{V})$ is dense in $H^{s,\varphi}_{A}(V_0)$, there exists a sequence $(u_{r})_{r=1}^{\infty}\subset C^{\infty}(\overline{\Omega})$ such that $\omega_{r}:=u_{r}\!\upharpoonright\!\overline{V}\to\omega$ in $H^{s,\varphi}(V_0)$ and $A\omega_{r}\to A\omega$ in $L_{2}(V_0)$ as $r\to\infty$. Then
\begin{equation}\label{proof2-f23}
\eta u_{r}\to\eta u\quad\mbox{in}\;\;
H^{s,\varphi}(\Omega)
\end{equation}
and
\begin{equation}\label{proof2-f24}
\eta Au_{r}\to\eta Au\quad\mbox{in}\;\;L_{2}(\Omega)
\end{equation}
as $r\to\infty$. The second convergence is evident; let us explain the first. Since $\omega_{r}-\omega\to0$ in $H^{s,\varphi}(V_0)$, there exists a sequence $(\omega_{r}^{\circ})_{r=1}^{\infty}\subset H^{s,\varphi}(\mathbb{R}^{n})$ such that $\omega_{r}^{\circ}=\omega_{r}-\omega$ in $V_0$ and that  $\omega_{r}^{\circ}\to0$ in $H^{s,\varphi}(\mathbb{R}^{n})$. Then $\eta(u_{r}-u)=\eta\cdot \omega_{r}^{\circ}\!\upharpoonright\!\Omega\to0$ in $H^{s,\varphi}(\Omega)$, which gives~\eqref{proof2-f23}.

Let us deduce from the convergence $\omega_{r}\to\omega$ in $H^{s,\varphi}_{A}(V_0)$ that
\begin{equation}\label{proof2-f25}
\eta B_{j}u_{r}\to\eta B_{j}u\quad\mbox{in}\;\;
H^{s-m_j-1/2,\varphi}(\Gamma)
\end{equation}
as $r\to\infty$ for every $j\in\{1,\ldots,q+\varkappa\}$. Given such $j$, we consider a boundary PDO on $\partial V_0$ of the form
\begin{equation*}
B_{j}^{\star}:=B_{j}^{\star}(x,D):=
\sum_{|\mu|\leq m_j}b_{j,\mu}^{\star}(x)D^{\mu}
\end{equation*}
where each coefficient $b_{j,\mu}^{\star}$ belongs to $C^{\infty}(\partial V_0)$ and coincides with the corresponding coefficient $b_{j,\mu}$ of $B_{j}$ on $\Gamma\cap\partial V_0$. Then
\begin{equation*}
B_{j}^{\star}\omega_{r}\to B_{j}^{\star}\omega\quad\mbox{in}\quad H^{s-m_{j}-1/2,\varphi}(\partial V_0)
\end{equation*}
due to Proposition~\ref{10pr1} considered for $V_0$ instead of $\Omega$ (as is seen from \cite[Proof of Theorem~1]{MurachChepurukhina15UMJ5}, the boundedness of the operator \eqref{10f5} does not depend on property (ii) of boundary conditions given in Section~\ref{sec2}). Since $\eta B_{j}^{\star}\omega_{r}=\eta B_{j}u_{r}$ on $\Gamma\cap\partial V_0$, we get
\begin{equation}\label{proof2-f26}
\eta B_{j}u_{r}\to T(\eta B_{j}^{\star}\omega)\quad\mbox{in}\;\;
H^{s-m_j-1/2,\varphi}(\Gamma),
\end{equation}
where the distribution $T(\eta B_{j}^{\star}\omega)$ is equal by definition to $\eta B_{j}^{\star}\omega$ on $\Gamma\cap V$ and to zero on $\Gamma\setminus\mathrm{supp}\,\eta$.

Note that
\begin{equation}\label{proof2-f27}
\eta B_{j}^{\star}\omega=\eta B_{j}u\quad\mbox{on}\;\;\Gamma\cap V.
\end{equation}
Indeed, since $u\in H^{l}_{A}(\Omega)$ for some $l<s$, there exists a sequence $(u_{r}^{\ast})_{r=1}^{\infty}\subset C^{\infty}(\overline{\Omega})$ that converges to $u$ in $H^{l}_{A}(\Omega)$. Hence, $\eta B_{j}u_{r}^{\ast}\to\eta B_{j}u$ in $H^{l-m_j-1/2}(\Gamma)$ due to Proposition~\ref{10pr1}. Besides, since  $u_{r}^{\circ}:=u_{r}^{\ast}\!\upharpoonright\!V_0\to u\!\upharpoonright\!V_0=\omega$ in $H^{l}_{A}(V_0)$, we have the convergence $\eta B_{j}^{\star}u_{r}^{\circ}\to\eta B_{j}^{\star}\omega$ in $H^{l-m_j-1/2}(\partial V_0)$. However, $\eta B_{j}u_{r}^{\ast}=\eta B_{j}^{\star}u_{r}^{\circ}$ on $\Gamma\cap\partial V_0\supset\Gamma\cap V$. The last two limits therefore yield property \eqref{proof2-f27}. Owing to it, we have the equality $T(\eta B_{j}^{\star}\omega)=\eta B_{j}u$ on $\Gamma$, which together with \eqref{proof2-f26} gives \eqref{proof2-f25}.

Consider a function $\eta_{1}\in C^{\infty}(\Gamma)$ such that $\mathrm{supp}\,\eta_{1}\subset\Gamma_{0}$ and that $\eta_{1}=1$ in a neighbourhood of $\Gamma\cap\mathrm{supp}\,\eta$ (in the topology on $\Gamma$, of course). According to Theorem~$\ref{10th1}$, the inclusion $\eta_{1}v_{k}\in H^{s+r_k-1/2,\varphi}(\Gamma)$ holds true for each $k\in\{1,\ldots,\varkappa\}$. We choose a sequence $(v^{(r)}_{k})_{r=1}^{\infty}\subset C^{\infty}(\Gamma)$ such that
\begin{equation}\label{proof2-f28}
v^{(r)}_{k}\to\eta_{1}v_{k}\quad\mbox{in}\;\;H^{s+r_k-1/2,\varphi}(\Gamma)
\end{equation}
as $r\to\infty$. Then
\begin{equation}\label{proof2-f29}
\eta C_{j,k}v^{(r)}_{k}\to\eta C_{j,k}v_{k}\quad \mbox{in}\;\;H^{s-m_j-1/2,\varphi}(\Gamma)
\end{equation}
for all admissible values $j$ and $k$; see \cite[Lemma~2.5]{MikhailetsMurach14}. Put $v^{(r)}:=(v^{(r)}_{1},\ldots,v^{(r)}_{\varkappa})$. As we have proved, the inequality \eqref{10f9} holds true for $(u_{r},v^{(r)})\in\mathcal{D}(\overline{\Omega},\Gamma)$ instead of $(u,v)$, i.e.
\begin{equation*}
\|\chi(u_{r},v^{(r)})\|'_{s,\varphi}\leq c\,\bigl(\|\eta\Lambda(u_{r},v^{(r)})\|''_{0,s,\varphi}+
\|\eta(u_{r},v^{(r)})\|'_{s-\lambda,\varphi}\bigr).
\end{equation*}
Passing here to the limit as $r\to\infty$ and using \eqref{proof2-f23}--\eqref{proof2-f25}, \eqref{proof2-f28}, and \eqref{proof2-f29}, we obtain the required estimate \eqref{10f9}.
\end{proof}

\bigskip

\emph{Acknowledgement.} The authors have received funding from the European Union's Horizon 2020 research and innovation programme under the Marie Sk{\l}odowska-Curie grant agreement No 873071.

\medskip

{\small

config: arxiv


TITLE: Applying the fundamental theorem of calculus: extremes of integration
QUESTION [1 upvotes]: So I have this exercise in which I need to apply the fundamental theorem of calculus to calculate the derivative $G'(x)$ to the following function for $x> 1/2$
$G(x) = \int_{1}^{2x^2} \frac{1}{1+4t^2} dt$
So first i thought that $G$ is a composition of differentiable functions
$G_1(x) = \int_{1}^{x} \frac{1}{1+4t^2} dt$
$g(x) = 2x^2$
So I will have to apply the derivation rule
$G'(x) = G_1'(f(x)) \times f'(x)$
Ok...
Now For calculating $G_1'(x)$ I will apply the fundamental theorem of integral calculus which tells us that $G_1'(x) = f(t)$ where $f(t) = \frac{1}{1+4t^2}$.
Now my question is:
They want us to calculate this for $x> 1/2$. I know that if i was calculating an integral i should pay attention to the inversion of extremes (for $1/2 < x < 1/ \sqrt 2$ $2x^2$ is less that 1)... But do i need to pay attention when calculating the derivative? Or do I simply calculate the derivative ignoring my $x$ value.
Thanks!
P.S. - If the answer is "you need to invert the extremes of integration if $1/2 < x < 1/ \sqrt 2$ do I simply put a minus signal inside the integral and switch the extremes?

REPLY [1 votes]: Let $G(x)$ be given by
$$G(x)=\int_1^{2x^2}\frac{1}{1+4t^2}\,dt$$
Now, let $y=2x^2$ and $g(y)=G(\sqrt{y/2})$.  Then, from the chain-rule, we have
$$\begin{align}
\frac{dG(x)}{dx}&=\left.\frac{dg(y)}{dy}\right|_{y=2x^2}\frac{dy}{dx}\\\\
&=\left.\frac{d}{dy}\left(\int_1^y \frac{1}{1+4t^2}\,dt\right)\right|_{y=2x^2}\frac{dy}{dx}\\\\
&=\left.\left(\frac{1}{1+4y^2}\right)\right|_{y=2x^2}\,(4x)\\\\
&=\frac{4x}{1+4(2x^2)^2}\\\\
&=\frac{4x}{1+16x^2}
\end{align}$$
Can you finish?


TITLE: Find value of $|z_2+z_3|$
QUESTION [2 upvotes]: Given that complex numbers $z_1,z_2,z_3$ lie on unit circle and
$$|z_1-z_2|^2+|z_1-z_3|^2=4$$ Then find value of $|z_2+z_3|$
My try:
We can take $z_1=e^{i\alpha}$, $z_2=e^{i\beta}$ and $z_3=e^{i\gamma}$
So we have
$$|z_1-z_2|=2\sin\left(\frac{\alpha-\beta}{2}\right)$$
$$|z_1-z_3|=2\sin\left(\frac{\alpha-\gamma}{2}\right)$$
So we get:
$$\sin^2\left(\frac{\alpha-\beta}{2}\right)+\sin^2\left(\frac{\alpha-\gamma}{2}\right)=1$$
Now we have:
$$|z_2+z_3|=2\cos\left(\frac{\beta-\gamma}{2}\right)$$
Any help here?

REPLY [2 votes]: Note that if $|z|=1$, then $\overline{z}=1/z$. Thus for $z_1$, $z_2$, and $z_3$ on the unit circle, we have
$$\begin{align}
|z_1-z_2|^2+|z_1-z_3|^2
&=\left(z_1-z_2\right)\left({1\over z_1}-{1\over z_2}\right)+\left(z_1-z_3\right)\left({1\over z_1}-{1\over z_3}\right)\\
&=4-\left({z_1\over z_2}+{z_2\over z_1}+{z_1\over z_3}+{z_3\over z_1} \right)\\
&=4-\left((z_1^2+z_2z_3)(z_2+z_3)\over z_1z_2z_3 \right)
\end{align}$$
It follows that $|z_1-z_2|^2+|z_1-z_3|^2=4$ if and only if $(z_1^2+z_2z_3)(z_2+z_3)=0$.  Since we can always arrange for $z_1^2=-z_2z_3$ on the unit circle, $|z_2+z_3|$ can take any value between $0$ and $2$.


TITLE: Is countable pointwise supremum of bounded functions achievable?
QUESTION [0 upvotes]: Let $g(x) = \sup_{n\in\mathbb{N}}f_n(x)$ be the pointwise supremum with $f_n(x)\leq C,\forall x\in X,n\in\mathbb{N}$. Can we claim that the supremum is achievable, i.e. $x\in X$, there exists $n\in\mathbb{N}$ such that
$$g(x) = f_n(x)$$
for this particular $x$?

REPLY [2 votes]: No: let $f_n(x)=C-\frac1n$ for each $n\in\Bbb Z^+$. Then $g(x)=C$ for each $x\in X$, but $f_n(x)\ne C$ for each $n\in\Bbb Z^+$ and $x\in X$.


TITLE: Isn't there any divisor $k$ of $n^4$ such that $n^2-n<k<n^2$?
QUESTION [4 upvotes]: I did some experiment with my Python script to find a number which could divide $n^4$ in this interval ($n^2-n$,  $n^2$).
I watched the form of prim factors of the numbers in this ($n^2-n$,  $n^2$) interval, but I hadn't was any idea for proofing it.
Can anyone to help with some start idea to proof that if $n^2-n < k < n^2$ then $k$ can't divide $n^4$.

REPLY [2 votes]: The best examples come from $n=2xy,$ where $x,y>0$ are integers and $x^2 - 2 y^2 = \pm 1 \; . \;$ In these cases we get a factor $m$ of $n^4$ near $n^2 - n \sqrt 2 \; . \; $ When $x^2 - 2 y^2 = 1,$ we get $m = n^2 - 2 x^2.$ When $x^2 - 2 y^2 = -1,$ we get $m = n^2 - 4 y^2.$
jagy@phobeusjunior:~$ ./mse
 n: 2 =  2  m:  2  n^2 - m:  2 =  2
 n: 3 =  3  m:  3  n^2 - m:  6 =  2 3
 n: 4 =  2^2  m:  8  n^2 - m:  8 =  2^3
 n: 6 =  2 3  m:  27  n^2 - m:  9 =  3^2
 n: 10 =  2 5  m:  80  n^2 - m:  20 =  2^2 5
 n: 12 =  2^2 3  m:  128  n^2 - m:  16 =  2^4
 n: 24 =  2^3 3  m:  512  n^2 - m:  64 =  2^6
 n: 30 =  2 3 5  m:  810  n^2 - m:  90 =  2 3^2 5
 n: 60 =  2^2 3 5  m:  3456  n^2 - m:  144 =  2^4 3^2
 n: 70 =  2 5 7  m:  4802  n^2 - m:  98 =  2 7^2
 n: 84 =  2^2 3 7  m:  6912  n^2 - m:  144 =  2^4 3^2
 n: 140 =  2^2 5 7  m:  19208  n^2 - m:  392 =  2^3 7^2
 n: 168 =  2^3 3 7  m:  27783  n^2 - m:  441 =  3^2 7^2
 n: 180 =  2^2 3^2 5  m:  32000  n^2 - m:  400 =  2^4 5^2
 n: 408 =  2^3 3 17  m:  165888  n^2 - m:  576 =  2^6 3^2
 n: 594 =  2 3^3 11  m:  351384  n^2 - m:  1452 =  2^2 3 11^2
 n: 816 =  2^4 3 17  m:  663552  n^2 - m:  2304 =  2^8 3^2
 n: 1170 =  2 3^2 5 13  m:  1366875  n^2 - m:  2025 =  3^4 5^2
 n: 2378 =  2 29 41  m:  5651522  n^2 - m:  3362 =  2 41^2
 n: 3230 =  2 5 17 19  m:  10425680  n^2 - m:  7220 =  2^2 5 19^2
 n: 4756 =  2^2 29 41  m:  22606088  n^2 - m:  13448 =  2^3 41^2
 n: 5880 =  2^3 3 5 7^2  m:  34560000  n^2 - m:  14400 =  2^6 3^2 5^2
 n: 13860 =  2^2 3^2 5 7 11  m:  192080000  n^2 - m:  19600 =  2^4 5^2 7^2
 n: 16296 =  2^3 3 7 97  m:  265531392  n^2 - m:  28224 =  2^6 3^2 7^2
 n: 27720 =  2^3 3^2 5 7 11  m:  768320000  n^2 - m:  78400 =  2^6 5^2 7^2
 n: 42672 =  2^4 3 7 127  m:  1820786688  n^2 - m:  112896 =  2^8 3^2 7^2
 n: 57960 =  2^3 3^2 5 7 23  m:  3359232000  n^2 - m:  129600 =  2^6 3^4 5^2
 n: 58206 =  2 3 89 109  m:  3387795864  n^2 - m:  142572 =  2^2 3 109^2
 n: 80782 =  2 13^2 239  m:  6525617282  n^2 - m:  114242 =  2 239^2


TITLE: The Derivative of a Vector
QUESTION [0 upvotes]: It is stated in this video that some books attempt to define the derivative of a vector, with the added caveat that tensor notation is a much better approach.  Presumably, this is a different concept than the derivative of a vector-valued function, as the latter is uncontroversial and ubiquitous.  What is this unpalatable process used by such books to take the derivative of a vector, and a vector transposed?

REPLY [2 votes]: The linked video appears to be a discussion about minimizing
$$\eqalign{ f(x) &= \tfrac{1}{2}x^TAx - x^Tb}$$
Here's how one might solve it using matrix notation.
First find the differential and gradient of the function.
$$\eqalign{
 f &= \tfrac{1}{2}x^TAx - b^Tx \cr
df &= \tfrac{1}{2}(dx^TAx+x^TA\,dx) - b^Tdx \cr
  &= (x^TA -b^T)\,dx \cr
  &= (Ax-b)^T\,dx \cr
\frac{\partial f}{\partial x} &= Ax - b \cr
}$$
Then set the gradient to zero and solve
$$(Ax-b)=0 \implies x=A^{-1}b$$
So it's not really as bad as the lecturer is making it out to be.
The same calculation in index/tensor notation looks like
$$\eqalign{
 f &= \tfrac{1}{2}x_iA_{ij}x_j - b_ix_i \cr
\frac{\partial f}{\partial x_k}
 &= \tfrac{1}{2}(\delta_{ik}A_{ij}x_j+x_iA_{ij}\delta_{jk}) - b_i\delta_{ik} \cr
 &= \tfrac{1}{2}(A_{kj}x_j+x_iA_{ik}) - b_k \cr
 &= \tfrac{1}{2}(A_{kj}x_j+A_{ki}^Tx_i) - b_k \cr
 &= A_{kj}x_j - b_k \cr
}$$
For dealing with scalars, vectors, and matrices (as in this problem) the two notations are roughly equivalent. The real power of index/tensor notation is in dealing with higher-order tensors.


TITLE: Communication between observers in gravity wells
QUESTION [1 upvotes]: Suppose two friends live 100 light years apart. Can they use gravitational time dilation to reduce the 200-year roundtrip lag they experience during their phone calls?
Specifically, how would the lag (from their perspective) change if they move their houses close to the event horizons of their individual black holes? What if they instead drag a huge black hole (with Schwarzschild radius of ~50 ly) to the midpoint between them, so that they both end up near the event horizon of the same black hole?

REPLY [2 votes]: Suppose two friends live 100 light years apart. Can they use gravitational time dilation to reduce the 200-year roundtrip lag they experience during their phone calls?

The OP describes two different scenarios, with very different spacetime geometries, so I'll address them separately.

 First scenario: individual black holes
Let $D$ denote the distance between the two friends, with $D\sim 100$ light years. Suppose that each friend is near a black hole with Schwarzschild radius $R\ll D$, as shown here:

I'll show the analysis in some detail, because the answer is not obvious (at least not to me). The key is to think about the relationship between each friend's proper time and the coordinate time. Since the two black holes are far away from each other, we can use the same relationship that we would use for an isolated Schwarzschild black hole:
$$
 \Delta \tau = \sqrt{1-\frac{R}{r_0}}\,\Delta t,
\tag{1}
$$
where $\Delta \tau$ is the elapsed proper time for an friend hovering at a fixed coordinate-radius $r_0>R$ above the black hole, and $\Delta t$ is the usual coordinate-time. Equation (1) says that for a given elapsed coordinate-time $\Delta t$, the elapsed proper time $\Delta\tau$ experienced by the hovering friend can be made as small as we like, just by putting the friend very close to the horizon $(r_0\gtrsim R)$. The problem of how the friend might maintain that hovering position (e.g., with a rocket) will not be addressed here.
That doesn't answer the question yet, because we still need to estimate the elapsed coordinate-time $\Delta t$ for a signal to make a round trip. Since $R\ll D$, we can analyze this using the metric for an isolated black hole when the signal is near either hole, and using the metric of flat spacetime in the vast space between the two holes. Since the metric of an isolated black hole approaches flat spacetime far from the hole, we can do this instead: we can use the isolated-black-hole metric for whichever black hole is nearest to the propagating signal at any given time. Assuming that the black holes have the same mass, this makes the problem nicely symmetric: we can just calculate the elapsed time for the signal to reach the halfway point, and then multiply by $4$. (One factor of $2$ converts halfway to a full one-way trip, and the other factor of $2$ converts the one-way trip to a two-way trip.) Near one of the black holes, the coordinate-speed of a lightlike radial worldline is (in units where the limiting speed is $1$)
$$
 \frac{dr}{dt} = 1-\frac{R}{r}
\tag{2}
$$
where $r$ is the radial coordinate. (The friend is hovering at $r=r_0$.) For $r>R$, equation (2) is solved by
$$
 t = r + R\ln(r-R) + \text{constant},
\tag{3}
$$
so the elapsed coordinate-time $\Delta t$ required for the signal to go from a point $r_0$ near the hole (where the friend is hovering) to the halfway point $r=D/2$ is
$$
 \Delta t
 = 4\Delta r + 4R\ln\left(\frac{D/2-R}{r_0-R}\right)
 = 4\Delta r + 4R\ln\left(\frac{\Delta r}{r_0}+A\right) - 4R\ln A
\tag{4}
$$
with
$$
\Delta r := \frac{D}{2} - r_0
\hskip2cm
A := 1-\frac{R}{r_0}.
\tag{5}
$$
Combine this with (1) to conclude that the round-trip time experienced by the hovering friend is
$$
 \Delta \tau = 4\left(\Delta r + R\ln\left(\frac{\Delta r}{r_0}+A\right) - R\ln A\right)
  \sqrt{A}
\tag{6}
$$
Since $(\ln A)\sqrt{A}$ goes to zero as $A\to 0$, this reduces to
$$
 \Delta \tau \approx 4\left(\Delta r + R\ln\frac{\Delta r}{r_0}\right)
  \sqrt{A}
\tag{7}
$$
for sufficiently small $A$. Since we can make $A$ as close to zero as we like by putting the friend's radial coordinate $r_0$ closer and closer to $R$ (that is, closer and closer to the event horizon of the black hole), we can make the signal's round-trip time experienced by that friend as small as we like.
Conclusion: In this scenario, the two friends can communicate with arbitrarily small lag, by hovering sufficiently close to the horizons of their respective black holes.

 Second scenario: single enormous black hole
In this scenario, we have a single black hole with the enormous Schwarzschild radius $R\sim D/2$, with the two friends hovering on opposite sides of the black hole, as shown here:

They can communicate with each other by sending signals (again assumed to be lightlike) around the black hole, and if the hovering radius $r_0$ is just barely greater than $3R/2$, then the signal's orbit will be nearly circular. Equation (1) still applies here, but instead of equations (2)-(4), we can use the coordinate-time orbital period for a lightlike circular orbit with radius $3R/2$, namely
$$
 \Delta t \approx \frac{3\pi R}{\sqrt{A}}
\hskip2cm
 A := 1-\frac{R}{r_0} \approx 1-\frac{R}{3R/2}.
\tag{8}
$$
One period of a circular orbit is the same as one round-trip time between friends on opposite sides, so combining equations (1) and (8) shows that the round-trip time for friends hovering near $r_0\approx 3R/2$ is
$$
 \Delta \tau \approx \sqrt{A}\,\frac{3\pi R}{\sqrt{A}} \approx \frac{3\pi}{2} D.
\tag{9}
$$
For two friends separated by a distance $D$ in flat spacetime, the round-trip time would be $2D$ (again units where the fundamental speed limit is $1$). Since $3\pi/2 > 2$, the round-trip time is actually longer in this single-black-hole scenario than it would be in flat spacetime. We can't fix this by trying to send the signal inward, short-cutting the circular orbit, because then the signal would spiral into the black hole without ever reaching the other hovering friend. That's because $r=3R/2$ is the radius of the innermost circular orbit around a (non-rotating) black hole. Light can escape from within that radius, as long as $r>R$, but only if it is directed away from the black hole. Granted, I haven't considered the possibility that the time could be reduced by using such outward-directed signals, but that seems intuitively unlikely.
Conclusion: In this scenario, the two friends (probably) cannot reduce the lag in their communication, compared to what it would be with no black holes at all.

In summary, For minimizing the communication lag experienced by the two friends, two small black holes is better than none, and one big black hole is probably worse than none.


TITLE: Numerical schemes for linear advection: stability, dissipation, dispersion
QUESTION [2 upvotes]: A generalized numerical scheme for the linear hyperbolic equation $u_t + au_x = 0$ has the following form
  $$
\frac{u_j^{n+1} - u_j^n}{\Delta t} + a\frac{u_{j+1}^{n} - u_{j-1}^n}{2\Delta x} - \chi \frac{u_{j+1}^{n} - 2u_{j}^n + u_{j-1}^n}{\Delta x^2} = 0 .
$$
  a) Find the values of $\chi$ so that the above scheme is the central-explicit, forward, backward, Lax-Friedrich and Lax-Wendroff scheme.
b) What is the order of truncation for each scheme?
c) What is the stability condition for each of these schemes?
d) In what range of $\chi$ the scheme is unconditionally unstable?
e) Classify the schemes in (a) as (i) dissipative, and (ii) dispersive.

ATTEMPT
(a) Notice that if $\chi = 0$, then we obtain the central explicit scheme. If we write the numerical discretization as
$$u_j^{n+1} = u_j^n - \frac{ a \Delta t }{2 \Delta x} (u_{j+1}^n - u_{j-1}^n) + \frac{ \chi \Delta t }{\Delta x^2} (u_{j+1}^n - 2 u_j^n + u_{j-1}^n )$$
And one can see that if $\chi = a \Delta t/2$, then we have the LW scheme. If $\chi = a \Delta x/2$, then we have
$$ u_j^{n+1} = u_j^n + \frac{ a \Delta t }{\Delta x} (u_{j+1}^n - u_j^n) $$
which is forward explicit and if $\chi = - a \Delta x/2$, we obtain backward explicit. Now if $\chi = \frac{ \Delta x^2}{2 \Delta t }$, then we obtain Lax-Friedrichs.
We go to part (c) as (b) is trivial. Perhaps  we can combine (c) and (d) in a single problem if we apply the discrete fourier transform:
$$ \hat{u}^{n+1} = \hat{u}^n - r( e^{ij \xi} -e^{-ij \xi } ) \hat{u}^n + \frac{ 2 \chi r }{a \Delta x} (e^{ij \xi} -e^{-ij \xi } -2 ) \hat{u}^n$$
Now, am I on the right track here? Is this the correct approach? As for e), I need some suggestions, how would we approach it?

REPLY [3 votes]: The proposed values of $\chi$ in a) are correct. One notes that the method is a straightforward finite-difference approximation of the advection-diffusion method $u_t + au_x = \chi u_{xx}$. Introducing the Courant number $r = |a| {\Delta t}/{\Delta x}$, the diffusion constants from a) rewrite as
$$
\chi \in |a|\frac{\Delta x}{2} \times \left\lbrace \frac{1}{r},\,
1,\,
r,\,
0 \right\rbrace ,
$$
which correspond to the numerical viscosities of the Lax-Friedrichs, upwind, Lax-Wendroff, and central explicit method, respectively. One notes that if $0<r<1$, then those values of $\chi$ are sorted in decreasing order.
As done in this post, let us assume a perturbation of the form $u_j^n = \xi^n e^{-\text i k j \Delta x}$, where $\xi = \text e^{\text i \omega \Delta t}$. Note that $\omega$ is a complex number, with real part $\omega_R$ and imaginary part $\omega_I$. Injecting this Ansatz in the time-stepping formula, Euler's formulas lead to
$$
\xi = 1 + \text i r \sin( k \Delta x) + 2\lambda \left( \cos(k \Delta x) -1 \right) ,
$$
where $r$ is the Courant number and
$$
\lambda = {\chi}\frac{\Delta t}{\Delta x^2} \in \left\lbrace \frac{1}{2},\,
\frac{r}{2},\,
\frac{r^2}{2},\,
0 \right\rbrace
$$
denotes the Fourier number. From this equation, one equates the real and imaginary parts:
\begin{aligned}
\text{Re}\, \xi &= e^{-\omega_I \Delta t}\cos(\omega_R \Delta t) = 1 + 2\lambda \left( \cos(k \Delta x) -1 \right) ,\\
\text{Im}\, \xi &= e^{-\omega_I \Delta t}\sin(\omega_R \Delta t) = r \sin( k \Delta x)  \, .
\end{aligned}
The stability can be analyzed in terms of the squared amplification factor $|\xi|^2$ which has to be smaller than one. Hence, we must have
$$
|\xi|^2 - 1 = 4\lambda^2 - 4 \lambda + r^2 + 4 \lambda (1-2\lambda) Z + (4\lambda^2 - r^2) Z^2 \leq 0
$$
for all $Z = \cos( k \Delta x)$ in $[-1, 1]$.
The dispersion is analyzed in terms of the phase velocity ${\omega_R}/{k}$ with
${\omega_R}\Delta t = \arctan\left[{\text{Im}\, \xi}/{\text{Re}\, \xi}\right]$.
The attenuation is analyzed in terms of
${\omega_I}\Delta t = -\ln |\xi|$
which is non-negative if $|\xi| \leq 1$ (cf. e.g. (1) p. 182).
(1) E. Godlewski, P.-A. Raviart, Numerical Approximation of Hyperbolic Systems of Conservation Laws, Springer, 1996. doi:10.1007/978-1-4612-0713-9


TITLE: $d_1(x,y)<d(x,y)\Rightarrow \{x|d(x,y)<\varepsilon\}\subset \{x|d_1(x,y)<\varepsilon\}$
QUESTION [0 upvotes]: Can someone please give me a hint why for metric spaces we have
$d_1(x,y)<d(x,y)\Rightarrow \{x|d(x,y)<\varepsilon\}\subset \{x|d_1(x,y)<\varepsilon\}$
I have expected the opposite:
$d_1(x,y)<d(x,y)\Rightarrow \{x|d(x,y)<\varepsilon\}\supset \{x|d_1(x,y)<\varepsilon\}$

REPLY [2 votes]: Because if $d(x,y)<\varepsilon$, then $d_1(x,y)<d(x,y)<\varepsilon$.

REPLY [1 votes]: The way I see it is a matter of precision, think that $x$ and $y$ are points and $d_1$ is the distance in centimeters and $d$ is in millimeters, so numerically we have $d1 <d$, and every measure using $d$ is therefore more accurate ...

REPLY [0 votes]: Yes, we claim that $\{x:d(x,y)<\varepsilon\}\subset \{x:d_1(x,y)<\varepsilon\}$. Indeed, let $x_0 \in \{x:d(x,y)<\varepsilon\}$ then, $d(x_0,y)<\varepsilon$. Since $d_1(x_0,y)<d(x_0,y)$ we obtain $d_1(x_0,y)<\varepsilon$, that is, $x_0 \in \{x: d_1(x,y)<\varepsilon\}$


TITLE: Subfield of $\mathbb R$ with algebraic closure as $\mathbb C$.
QUESTION [3 upvotes]: Does there exist a proper subfield of $\mathbb{R}$ whose algebraic closure is $\mathbb{C}$ ?
A weaker question: Does there exist a proper subfield $F$ of $\mathbb{R}$ such that $\mathbb{R}$ is algebraic over $F$?

REPLY [1 votes]: Yes, there is.
Pick you favorite algebraic number $a \notin \mathbb Q$. For example $a= \sqrt{2}$.
Consider now
$$\mathcal{F} := \{ K | K \mbox{ is a subfield of } \mathbb R , \mbox{ and } a \notin K \}$$
Now, $(\mathcal{F}, \subset)$ is a partially ordered set, which is non-empty (as $\mathbb Q \in \mathcal F$). Moreover, if $(K_{j})_j$ is a chain in $\mathcal F$, it is straightforward that $\bigcup K_j \in \mathcal{F}$.
Therefore, by Zorn Lemma,  $\mathcal{F}$ has a maximal element $\mathbb{K}$.
The maximality of $\mathbb K$ implies that $\mathbb R$ is algebraic over $\mathbb K$. Indeed, if $t \in \mathbb R$ is transcendental over $\mathbb K$ then we cannot have $a \in \mathbb K(t)$ which contradicts the maximality of $\mathbb K$.
Indeed, if $a \in \mathbb K(t)$ then writing $a=\frac{P(t)}{Q(t)}$ we get that $P(t)-aQ(t)=0$ which means that $t$ is the root of $P-at \in \mathbb K(a)$. Then $t$ is algebraic over $\mathbb K(a)$ and $\mathbb K(a)$ is finite over $K$ which implies that $t$ is algebraic over $\mathbb K$.
Finally, as $\mathbb R$ is algebraic over $\mathbb K$, $\mathbb C$ is also algebraic over $\mathbb K$.


TITLE: Prove that $\log_a(b)=\log(b)/\log(a)$
QUESTION [1 upvotes]: Prove that $$\log_a(b)=\log(b)/\log(a)$$
I don't know how to solve it, but I need to prove it so solve a problem.

REPLY [0 votes]: Consider the following :
Let $loga(b)= m$ ...(Assumption)
Therefore, $b=a^m$ ...(by definition of log)
Now look at the Right Hand Side of the required proof,
Let $log(b)/log(a) = n$
Substitute $b=a^m$ in it..
Therfore, $log(b)=log(a^m)=mlog(a)$ ...(Property of log)
Hence $logb/loga=mloga/loga=m$
Hence $m=n$
Thus required result is proved.


TITLE: Find the number of ways in which $6$ men and $5$ women can dine at a round table if no two women sit together.
QUESTION [1 upvotes]: Find the number of ways in which $6$ men and $5$ women can dine at a round table if no two women sit together.
Now men can sit in $6!$ ways and women can sit in $P^{6}_{5}$ ways, but the textbook says a different answer. Can you explain?

REPLY [2 votes]: Two circular seating arrangements are considered to be distinguishable if the relative order of the people differs.

In how many ways can six men sit at a round table?

Method 1:  Consider the diagram below.

We can describe a particular seating arrangement by describing the order of the men as we proceed counterclockwise (anticlockwise) around the circle.  Notice, however, that we can describe this particular seating arrangement in six ways, depending on whether we start with A, B, C, D, E, or F.  Thus, this particular circular arrangement corresponds the six linear arrangements ABCDEF, BCDEFA, CDEFAB, DEFABC, EFABCD, FEABCD.  More generally, each circular arrangement of the six men corresponds to six linear arrangements, corresponding to the six possible starting points as we proceed counterclockwise around the circle.  Since six men can be arranged in a row in $6!$ ways, there are
$$\frac{6!}{6} = 5!$$
distinguishable circular arrangements of the men.
More generally, we can arrange $n \geq 1$ distinct objects in a circle in
$$\frac{n!}{n} = (n - 1)!$$
ways since each circular arrangement corresponds to $n$ linear arrangements, one for which each of the $n$ starting points as we proceed counterclockwise around the circle.
Method 2:  We consider arrangements relative to the first man we sit at the table.
We seat man A.  We use him as our reference point.  As we proceed counterclockwise around the circle, we can seat the remaining five men in $5!$ orders relative to man A.
More generally, given a set of $n$ distinct objects, we can arrange them in $(n - 1)!$ orders as we proceed counterclockwise around the circle relative to the first object we place on the circle.

In how many ways can six men and five women dine at a round table if no two women sit together.

We can seat the men in $5!$ distinguishable ways.  This create six spaces, one to the immediate right of each man.  To ensure that no two of the women sit in adjacent seats, we must choose five of those six spaces in which to insert a woman.  Once those spaces have been selected, we can arrange the five women in those spaces in $5!$ orders.  Hence, the number of possible seating arrangements is
$$5! \binom{6}{5} \cdot 5! = 5!P(6, 5)$$


TITLE: Cohomology groups
QUESTION [3 upvotes]: I have some questions.
1) I tried to compute the cohomology group of $S^3$ with coefficients in $\mathbb{Z}/2\mathbb{Z}$ but I don't know if my result
$$
H^k(S^3,\mathbb{Z}/2\mathbb{Z}) = \begin{cases}
  \mathbb{Z}/2\mathbb{Z},  & \text{if }k=0,3\\
  0, &\text{else}
\end{cases}
$$
is correct. How can I check it for myself?
2) If I have a wedgesum $X \vee Y$ of topological spaces $X$ and $Y$.  I read that relative cohomology of this wedgesum coincide with the direct sum of cohomology group of $X$ and $Y$.  In absolute case that isn't true in general, but in which situations is$H^*(X \vee  Y)\cong H^*(X)\oplus H^*(Y)$ in absolute case? $H^*(\mathbb{R}\mathbb{P}^2 \vee S^3, \mathbb{Z}/2\mathbb{Z})$ and $H^*(\mathbb{R}\mathbb{P}^3, \mathbb{Z}/2\mathbb{Z})$
Edit: 3) My goal is to prove that $ \mathbb{R}\mathbb{P}^2 \vee S^3$ and $\mathbb{R}\mathbb{P}^3$ are not homotopy equivalent. Therefore I need $H^*(\mathbb{R}\mathbb{P}^2 \vee S^3, \mathbb{Z}/2\mathbb{Z})$ and $H^*(\mathbb{R}\mathbb{P}^3, \mathbb{Z}/2\mathbb{Z})$ to compute the cohomology rings.
$H^k(\mathbb{R}\mathbb{P}^3,\mathbb{Z}/2\mathbb{Z}) = \begin{cases}
  \mathbb{Z}/2\mathbb{Z},  & \text{if }k=0,1,2,3\\
  0, &\text{else}
\end{cases}$
Maybe, it is a example where $H^*(\mathbb{R}\mathbb{P}^2 \vee S^3, \mathbb{Z}/2\mathbb{Z}) \cong H^*(\mathbb{R}\mathbb{P}^3, \mathbb{Z}/2\mathbb{Z})$. But I don't know how to compute the cohomology of this wedge sum. I would appreciate any help from here.
Regards

REPLY [1 votes]: Your sphere cohomology is correct. I'm not sure what you mean by verifying it. You can verify it by looking in any textbook or wikipedia article or checking your computation (you know how to compute the result, right?).
The cohomology of the wedgesum should be pretty obvious: wedge sum differs from disjoint union only in degree $0$ by gluing the base-points together. Consequently, the homologies of $X \vee Y$ and $X \coprod Y$ coincide except in the degree zero. Now as $H^0(X)$ measures number of connected components of $X$, it's also obvious that $H^0(X \vee Y)$ will be one less than $H^0(X \coprod Y)$ since we glue two components (those containing basepoints) together.
The above paragraph should also answer your question about your particular wedge sum. In general $H^*(X)$ and $H^*(X \vee Y)$ coincide if only if $Y$ has homotopy type of a point.


TITLE: find area of kite given side length
QUESTION [0 upvotes]: Circle with two tangent lines
Above is the picture in question. A circle is given, center (-2,4) and a point outside the circle (0,10) is shown. Asked to calculate the area of the quadrilateral ABCD, I figured that this kite has sides 2 (the radii of the circle) and 6 (the difference between (0,4) and (0,10)).
How can I calculate the area of the kite?
(I tried calculating the diagonals and got rad40 for one of the diagonals, but cannot figure out how to calculate the other)
Thanks!

REPLY [1 votes]: Since $\overline{AB}$ is tangent to the circle, $m\angle B = 90°$, so the quadrilateral consists of two right-triangles. Both triangles share a common hypotenuse $\overline{AC}$ and have a congruent leg, as both are the radius of the circle. Hence, the two right-triangles are congruent, so $A_{kite} = 2A_{\triangle} = 2\cdot\frac{1}{2}bh = bh$. From the diagram, $b = 0-(-2) = 2$ and $h = 10-4 = 6$, so $A_{kite} = 2\cdot 6 = 12$ (in square units).