TITLE: Which one of these two is an equivalence relation
QUESTION [0 upvotes]: I'm having an issue with the following exercise:

Given $\alpha$ and $\beta$ two binary relationships defined in $Z$ such that:   $$\forall m,n \in Z, n\ \alpha\ m \iff n = m\ \ \vee\ \ rest(n,7)\ +\ rest(m,7) = 7$$
  $$\forall m,n \in Z, n\ \beta\ m \iff n \equiv_7 m\ \ \vee\ \ rest(n,7)\ +\ rest(m,7) = 7$$
  One of them is an equivalence relation. Which one?

So, first thing I thought upon seeing this exercise is that I have to verify the definition of equivalence relation:

Reflexivity
Symmetry
Transivity

I am able to prove for both that they are reflexive. I am having troubles proving that they are both symmetric and transitive. Because the OR is confusing me in the demonstration.
Could you please guide me to solve this exercise? Thank you.

REPLY [2 votes]: Clearly the first is not transitive:
$1+6=7$ and $\operatorname{rest}(6,7)+\operatorname{rest}(8,7)=7$ but neither $1=8$ nor $\operatorname{rest}(1,7)+\operatorname{rest}(8,7)=7$ is true.
Now consider the second relation:
in fact we can write $$n\ \beta\ m\iff n^2\equiv m^2\pmod7.$$
Then can you start from this definition and prove that it is both symmetric and reflexive? Can you show that this is equivalent with the original definition?
Hope this helps.

EDIT:
We know that $a\equiv b\pmod7\iff 7\mid(a-b)$ by definition. Also $n\ \beta\ m\iff n\equiv m\pmod7\ \vee\ n+m\equiv0\pmod7.$ We can then translate the condition to $n\ \beta\ m\iff n-m\equiv 0\pmod7\ \vee\ n+m\equiv0\pmod7.$
Further, we have $n^2\equiv m^2\pmod7\iff 7\mid(n^2-m^2)=(n-m)(n+m).$ Can you finish the proof now?
As to how to show $\beta$ is an equivalence relation, might you tell us where you got stuck? What do you think you should prove which you cannot prove?


TITLE: When does a RAAG act on a Tree?
QUESTION [1 upvotes]: Show that a RAAG acts on a tree of valence 4, acting transitively on vertices, if at least one pair of vertices is not joined by an edge.
I'm trying first to prove that every such RAAG $G$ acts on a Tree. Let $a$ and $b$ be two vertices that aren't joined by an edge, then the group $F$ generated by these two vertices is free. I'm thinking of using the quotient $G/F,$ but I don't know how. Is this even a group? Is it free?
Any help would be appreciated

REPLY [4 votes]: Let $T$ be the Cayley graph of the subgroup $F$ generated by $a$ and $b$ (wrt these two generators). Since $F$ is free on $\{a, b \}$, $T$ is a tree of valence $4$, on which $F$ already acts by an action which is transitive on the vertices. We need to extend this action to an action of $G$. In other words, we need to extend a morphism from $F$ to $Isom(T)$ to a morphism from $G$ to $Isom(T)$. We do this by sending all the generators of $G$ different from $a$ and $b$ to the identity. Using the presentation of $G$, we see that this gives a well-defined morphism, corresponding to a well-defined action of $G$ on $T$, which is transitive on the vertices, since its restriction to $F$ is.


TITLE: Determine the Laurent Series
QUESTION [0 upvotes]: Determine the Laurent Series Expansion for the function
$$(z^2-1)\exp\left(\frac{2z}{z-1}\right)$$
at $a=1$ and find its ring of convergence.
I expanded the exponential function but I couldn't get it to work out properly. Any suggestions would be very appreciated. Thanks!

REPLY [1 votes]: Hint: $$\frac{2z}{z-1}=2+\frac2{z-1}.$$
Plus, you know something about $\exp(u+v)$, right?


TITLE: Calculating the electric and magnetic field between two hollow cylinders
QUESTION [2 upvotes]: I am given the following question:

Two long coaxial cylindrical conductors of radii $a$ and $b$ ($b$ < $a$) are thin and hollow. There is a vacuum between the conductors ($b$ < $r$ < $a$, where r is the radial distance from the axis). The conductors are maintained at a potential difference $V$$0$ and carry equal but opposing currents of magnitude $I$ (in the direction of the axis of the cylinders). Find the magnitude and direction of the electric and magnetic fields in the region between the conductors, defining carefully the direction of the currents and the sign of the potential difference you have assumed.

From what I know, there's a magnetic field present between the two cylinders, which is produced by the current passing through the inner cylinder (all I need to do is to use Ampere's law).  Current in the outer cylinder doesn't affect the magnetic field between them.  However, how can there be an electric field present?

REPLY [0 votes]: There is an electric field because there is a non-zero PD between the cylinders. There can be an accumulation of charge on the surface of the conductors as well as a movement of charge (current) inside the conductors. These are not incompatible.


TITLE: Formula for Calculating the Probability of Outcome - Custom Dice
QUESTION [0 upvotes]: I was hoping someone might be able to help me in calculating dice probability with custom dice. The desired outcome would be being able to change the values on a 6 sided die (rather than 1-6), as well as the number of dice being rolled, and solve for the probability that outcome A or outcome B, ect would occur when all dice are rolled at the same time.
For example, as I have been trying to figure this out, I have been using two custom dice. Essentially one for an "attack" roll and another one defending. The dice are both d6 but with different values on each side.
Attack Die: [0,1,1,1,2,2]
Defense Die: [0,0,-1,-1,-1,-2]
The negative numbers are meant to reduce the total amount of damage the attack die would otherwise deal. This could perhaps be easier calculated using [0,0,1,1,1,2] and subtracting the outcome from the attacking die. I was able to make a chart showing the number of chances a certain outcome could occur if one of each of these was used:
enter image description here
I divided the number of chances each possible outcome had of occurring by the total number of possible outcomes and got my probability for each. The problem I have is calculating anything beyond two dice. I am pretty savvy with Excel, so I am trying to make a spreadsheet where I change the number of dice, or can tweak the faces of each die and the outcomes would adjust accordingly. Is there a simplified equation I can use that would factor in these variables and give the probability of a desired outcome?
Thank you for your time, I hope this post makes sense :)

REPLY [0 votes]: Say you have $n$ attack dice.  Then the total attack points $TA$ has possible values $0$ through $2n$.  Now: to calculate the chance of the total attack points being $i$, i.e. $P(TA=i)$, note the following ways to get $i$ points:
If $i$ is even, i.e. $i=2k$, and $i \le \frac{n}{2}$, then you can either have $k$ dice with $2$, and $n-k$ dice with $0$, or $k-1$ dice with $2$, $2$ dice with $1$, and $n-k-1$ dice with $0$ or .... or $0$ dice with $2$, $i$ dice with $1$, and $n-i$ dice with $0$. So:
$$P(TA=i=2k)=\sum_{j=0}^k (\frac{1}{3})^j \cdot (\frac{1}{2})^{i-2j} \cdot (\frac{1}{6})^{n-i+j} \cdot {n \choose j} \cdot {{n-j} \choose {i-2j}}$$
You should be able to create likewise formula's for when $i > \frac{n}{2}$ (I would treat that one separately, since now you definitely need some number of dice with a $2$), and then likewise two formulas for when $i$ is odd.


TITLE: Bernoulli distribution
QUESTION [2 upvotes]: I have a problem in which there are M users that have to choose over a set of N possible choices. I want to estimate the average probability that n items are not chosen by anyone. I modeled this as a Bernoulli distribution like:
$$P(n=k) = \left(\frac{n}{k}\right)\left(1 - \frac{1}{N}\right)^{M*k}\left(1 - \left(1 - \frac{1}{N}\right)^M\right)^{N-k}$$
and then I multiply and sum:
$$ A = \sum_{i=1}^{N} i * P(n = i) $$
However, I also implemented a small program which empirically shows this, and I see that as M increase, I get some small errors (like 2-3 %). I am wondering whether I missed something in the Bernoulli modeling, as I am 100% sure of my small program, or if something different to Bernoulli have to be chosen.

REPLY [0 votes]: The probability of $n=k$ items not chosen at all after $M$ random choices from $N$ with replacement is $$\Pr(n=k)=S_2(M,N-k) \dfrac{ N!}{k! \, N^M} $$ where $S_2$ is a Stirling number of the second kind.
The expected number of gaps $E[n] = \sum_k k \Pr(n=k)$ is easier to calculate another way: the probability that a particular item is not chosen by an individual is $1-\frac{1}{N}$ so that it not chosen by $M$ people is $\left(1-\frac{1}{N}\right)^M$ and so the expected total number not chosen is $$E[n]=N\left(1-\frac{1}{N}\right)^M.$$


TITLE: solving congruence class equation
QUESTION [0 upvotes]: How would I go about solving a problem like this:
Find all $ [a] $ for which $ [a] \odot x = [1] $ has a solution in $Z_4$
Having trouble with the abstractness of congruence classes.

REPLY [0 votes]: Assuming the multiplication you've denoted is just multiplication mod 4, it can be shown:
$ \gcd(x,y)=1 $ if and only if $\exists$ $c \in \mathbb{Z}$ such that $cx \equiv 1$ (mod $y$). Thus, $[a]$ satisfies your condition if and only if $\gcd(a,4)=1$. Then assuming $a \in \mathbb{Z}_4$ (as we should), we have $a=1$ or $a=3$.


TITLE: Show that a polynomial is irreducible using Galois
QUESTION [1 upvotes]: I need to show that the polynomial $X^4-3X^2+4$ is irreducible in $\mathbb{Q}[X]$. As Eisenstein's criterion fails and I hardly try to avoid something such as “Let's assume $X^4-3X^2+4$ is reducible, then $(X^2+aX+b)(X^2+cX+d) = X^4-3X^2+4$”.
My idea is that its roots $\pm\left(\sqrt{\frac{1}{2}(3\pm i \sqrt 7)}\right)$ and squared $\frac{1}{2}(3\pm i \sqrt 7)$ are obviously $\notin \mathbb{Q}$ (because if it was reducible, the degree $[\mathbb{Q}(\sqrt{\frac{1}{2}(3 + i \sqrt 7}): \mathbb{Q}]=2$, and no polynomial of degree $2$ then has $\sqrt{\frac{1}{2}(3 + i \sqrt 7)}$ as a root.
My question is simply if this argument is valid.

REPLY [3 votes]: Your argument is not valid, I'm afraid. The fact that the big number has not degree 2 is exactly equivalent to the given polynomial being irreducible.
The polynomial is reducible over $\mathbb{R}$ as
$$
X^4-3X^2+4=X^4+4X^2+4-7X^4=(X^2+2)^2-7X^2=
(X^2-\sqrt{7}X+2)(X^2+\sqrt{7}X+2)
$$
Both these factors are irreducible over $\mathbb{R}$.
A factorization over $\mathbb{Q}$ is also a factorization over $\mathbb{R}$. Does $\sqrt{7}\in\mathbb{Q}$?


text:
\section{NOR with Equal Arguments}
Tags: Logical NOR

\begin{theorem}
:$p \downarrow p \dashv \vdash \neg p$
That is, the [[Definition:Logical NOR|NOR]] of a proposition with itself corresponds to the [[Definition:Negation|negation]] operator.
\end{theorem}

\begin{proof}
{{BeginTableau|p \downarrow p \vdash \neg p}}
{{Premise|1|p \downarrow p}}
{{SequentIntro|2|1|\neg \left({p \lor p}\right)|1|Definition of [[Definition:Logical NOR|Logical NOR]]}}
{{Idempotence|3|1|\neg p|2|Disjunction}}
{{EndTableau}}
{{BeginTableau|\neg p \vdash p \downarrow p}}
{{Premise|1|\neg p}}
{{Idempotence|2|1|\neg \left({p \lor p}\right)|1|Disjunction}}
{{SequentIntro|3|1|p \downarrow p|2|Definition of [[Definition:Logical NOR|Logical NOR]]}}
{{EndTableau}}
{{qed}}
\end{proof}

\begin{proof}
Apply the [[Method of Truth Tables]]:
:$\begin{array}{|ccc||cc|} \hline
p & \downarrow & p & \neg & p \\
\hline
F & T & F & T & F \\
T & F & T & F & T \\
\hline
\end{array}$
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
{{qed}}
\end{proof}

config: wiki


TITLE: Fourier series convergence for sum of Schwartz class functions
QUESTION [1 upvotes]: Let $f$ be a Schwartz class function. Let $F(x)=\sum_{n\in\mathbb{Z}}f(x-2\pi n)$. Then $F$ is periodic of period $2\pi$.
How can we show that the Fourier series of $F$ converges to $F$ pointwise everywhere, i.e.
$$F(x)=\sum_{k\in\mathbb{Z}}\hat{F}(k)e^{ikx}?$$

REPLY [3 votes]: The derivatives of rapidly decreasing (Schwartz class) functions are also rapidly decreasing. Hence we can sum the derivative $f'$ in a like manner,
$$G(x) = \sum_{n\in\mathbb{Z}} f'(x-2\pi n).$$
The sum converges locally uniformly, hence $G$ is continuous, and
$$\begin{align}
\int_0^x G(t)\,dt &= \int_0^x \sum_{n\in\mathbb{Z}} f'(t-2\pi n)\,dt\\
&= \sum_{n\in\mathbb{Z}} \int_0^x f'(t-2\pi n)\,dt\\
&= \sum_{n\in \mathbb{Z}} f(x-2\pi n) - f(0-2\pi n)\\
&= F(x) - F(0),
\end{align}$$
so $F$ is continuously differentiable (even $C^\infty$, but we don't need that).
For a continuously differentiable $2\pi$-periodic function $h$, the Fourier coefficients are summable(1), $\sum\limits_{k\in\mathbb{Z}}\lvert \hat{h}(k)\rvert < \infty$, hence the Fourier series converges uniformly to $h$.
(1) $h$ and $h'$ both belong to $L^2([0,2\pi])$, so for $k\neq 0$
$$\begin{align}
2\pi\hat{h}(k) &= \int_{0}^{2\pi} h(t)e^{-ikt}\,dt\\
&= \left[\frac{i}{k}h(t)e^{-ikt}\right]_0^{2\pi} - \frac{i}{k}\int_{0}^{2\pi} h'(t)e^{-ikt}\,dt\\
&= \frac{i}{k}\left[h(2\pi) - h(0)\right] + \frac{1}{ik} \int_0^{2\pi} h'(t)e^{-ikt}\,dt\\
&= \frac{2\pi}{ik} \hat{h'}(k).
\end{align}$$
Since both, $(\hat{h'}(k))$ and $\left(\frac1k\right)$ belong to $\ell^2(\mathbb{Z}\setminus\{0\})$, it follows that $(\hat{h}(k)) \in \ell^1(\mathbb{Z})$.


TITLE: Let $X$, $Y$ and $Z$ be independat RVs with PDF $2t$ on $[0,1]$, and $0$ else. $W=max(X,Y,Z)$. What is the expected value of $W$?
QUESTION [0 upvotes]: Let $X$, $Y$ and $Z$ be independent RVs with PDF $2t$ on $[0,1]$, and $0$ else. $W=max(X,Y,Z)$. What is the expected value of W?

So first off we need to find $F_W(t)$, and derive it to find $f_W(t)$, and then use that to find E(Z).
$Pr(W\le w)=Pr(max(X,Y,Z) \lt w)=Pr(X\lt w)Pr(Y\lt w)Pr(Z\lt w)=$
$$F_W(t)=\int_0^t (2t')^3dt'=2t'^4|^w_0=2t^4$$
Differentiating we get
$$f_W(t)=8t^3$$
Now finding E(W):
$$\int_0^1 t*8t^3dt=\frac85$$
Now the answer is wrong, and I'm wondering during which step the mistake occured.

REPLY [2 votes]: The cumulative probability distribution $G$ of each of $X$, $Y$ and $Z$ is given by $G(t)=t^2$ and the probability density function $g$ is given by $g(t)=2t$.
Thus the cumulative distribution $F$ of $W$ is given by
$$F(t)=Pr(X\leq t)Pr(Y\leq t)Pr(Z\leq t)=[G(t)]^3=(t^2)^3=t^6$$
The probability density function, $f$ is given by $f(t)=6t^5$.
Thus the expected value of $W$ is
$$E[W]=\int_0^1tf(t)dt=\int_0^16t^6dt=\left[\frac{6t^7}{7}\right]_0^1=\frac{6}{7}.$$

Your answer is wrong because you calculated the cumulative distribution $F$ of $W$ using
$$F(t)=\int_0^t[g(x)]^3dx$$
instead of
$$F(t)=\left[\int_0^tg(x)dx\right]^3.$$


TITLE: What is the angle to launch projectile in 2D, given destination displacement and initial velocity?
QUESTION [2 upvotes]: Description of the problem
I want a projectile launched at speed $v_0$ at angle $\theta$ above the horizontal to just make it to the top of a building of height $h$ and a distance $d$ away. What $\theta$ should I use?
What have I tried?
Let $t^*$ be the moment when the projectile just makes it to the top of the building.
I'm using the following functions to represent displacement for both $x$ and $y$
$y(t^*) = y_0 + v_{y0}t^* + \frac{a_y{t^*}^2}{2}$
$x(t^*) = x_0 + v_{x0}t^* + \frac{a_x{t^*}^2}{2}$
where $v_{x0}$ and $v_{y0}$ represent the initial velocities along x-axis and y-axis respectively. The relationship between $v_0$, $v_{x0}$ and $v_{y0}$ via angle $\theta$ is
$v_{x0}=v_0 cos(\theta)$
$v_{y0}=v_0 sin(\theta)$
$tg(\theta)=\frac{v_{y0}}{v_{x0}}$
We know that $y_0$, $x_0$ and $a_x$ are $0$, as well as $a_y=-g$. Also, $y(t^*)=h$ and $x(t^*)=d$.
Therefore, the above equations are simplified to
$h = v_{y0}t^* - \frac{g{t^*}^2}{2}$ $(1)$
$d = v_{x0}t^*$ $(2)$
We assume that $v_{x0} \neq 0$ and derive that $t^*=\frac{d}{v_{x0}}$ from equation $(2)$.
We then substitute $t^*$ in the equation $(1)$ and end up with the following
$h = v_{y0}\frac{d}{v_{x0}} - \frac{g({\frac{d}{v_{x0}}})^2}{2}$
$h = tg(\theta)d - \frac{g d^2}{2 {v_{x0}}^2}$
$h = tg(\theta)d - \frac{g d^2}{2 {v_{0}}^2 cos^2(\theta)}$
Where am I stuck?
I don't know how to solve this trigonometric function.
I looked at the answer in the book, and it says that the solution is
$\theta = tg^{-1}(\frac{2h}{g})$
Based on the answer from the book, I'm failing to see how my approach resolves to that number, given that the answer from the book doesn't depend on $v_0$. I'm not sure if that is possible though, so could it be a mistake in the book?
If there is no mistake in the book, I think that there is a flaw in my approach and I would appreciate if anybody could give me a hint or any directions on how to solve this question.

REPLY [0 votes]: With your two equations (1) and (2), with 2 unknowns t and θ, eliminating t led to a complex expression for θ.  Try using $sin^2$(θ) + $cos^2$(θ) = 1 to eliminate θ.  This leads to a quadratic equation in $t^2$.  Use this to get t and then  θ from equation (2).
If you are given an initial velocity, this approach gives two possible angles to hit the top edge of the building. If you want to "just make it", the time is minimized and the two angles converge (the square root in the quadratic is zero). For this, the methods suggested by other answers would be quicker.


text:
/-
Copyright (c) 2019 Kevin Buzzard. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Kevin Buzzard
-/
import data.real.basic
import data.real.ennreal

/-!
# The extended reals [-∞, ∞].

This file defines `ereal`, the real numbers together with a top and bottom element,
referred to as ⊤ and ⊥. It is implemented as `with_top (with_bot ℝ)`

Addition and multiplication are problematic in the presence of ±∞, but
negation has a natural definition and satisfies the usual properties.

An ad hoc addition is defined, for which `ereal` is an `add_comm_monoid`, and even an ordered one
(if `a ≤ a'` and `b ≤ b'` then `a + b ≤ a' + b'`).
Note however that addition is badly behaved at `(⊥, ⊤)` and `(⊤, ⊥)` so this can not be upgraded
to a group structure. Our choice is that `⊥ + ⊤ = ⊤ + ⊥ = ⊤`.

An ad hoc subtraction is then defined by `x - y = x + (-y)`. It does not have nice properties,
but it is sometimes convenient to have.

An ad hoc multiplication is defined, for which `ereal` is a `comm_monoid_with_zero`.
This does not distribute with addition, as `⊤ = ⊤ - ⊥ = 1*⊤ - 1*⊤ ≠ (1 - 1) * ⊤ = 0 * ⊤ = 0`.

`ereal` is a `complete_linear_order`; this is deduced by type class inference from
the fact that `with_top (with_bot L)` is a complete linear order if `L` is
a conditionally complete linear order.

Coercions from `ℝ` and from `ℝ≥0∞` are registered, and their basic properties are proved. The main
one is the real coercion, and is usually referred to just as `coe` (lemmas such as
`ereal.coe_add` deal with this coercion). The one from `ennreal` is usually called `coe_ennreal`
in the `ereal` namespace.

## Tags

real, ereal, complete lattice

## TODO

abs : ereal → ℝ≥0∞

In Isabelle they define + - * and / (making junk choices for things like -∞ + ∞)
and then prove whatever bits of the ordered ring/field axioms still hold. They
also do some limits stuff (liminf/limsup etc).
See https://isabelle.in.tum.de/dist/library/HOL/HOL-Library/Extended_Real.html
-/

open_locale ennreal nnreal

/-- ereal : The type `[-∞, ∞]` -/
@[derive [has_top, comm_monoid_with_zero,
  has_Sup, has_Inf, complete_linear_order, linear_ordered_add_comm_monoid_with_top]]
def ereal := with_top (with_bot ℝ)

/-- The canonical inclusion froms reals to ereals. Do not use directly: as this is registered as
a coercion, use the coercion instead. -/
def real.to_ereal : ℝ → ereal := some ∘ some

namespace ereal

-- TODO: Provide explicitly, otherwise it is inferred noncomputably from `complete_linear_order`
instance : has_bot ereal := ⟨some ⊥⟩

@[simp] lemma bot_lt_top : (⊥ : ereal) < ⊤ := with_top.coe_lt_top _
@[simp] lemma bot_ne_top : (⊥ : ereal) ≠ ⊤ := bot_lt_top.ne

instance : has_coe ℝ ereal := ⟨real.to_ereal⟩
@[simp, norm_cast] protected lemma coe_le_coe_iff {x y : ℝ} : (x : ereal) ≤ (y : ereal) ↔ x ≤ y :=
by { unfold_coes, simp [real.to_ereal] }
@[simp, norm_cast] protected lemma coe_lt_coe_iff {x y : ℝ} : (x : ereal) < (y : ereal) ↔ x < y :=
by { unfold_coes, simp [real.to_ereal] }
@[simp, norm_cast] protected lemma coe_eq_coe_iff {x y : ℝ} : (x : ereal) = (y : ereal) ↔ x = y :=
by { unfold_coes, simp [real.to_ereal, option.some_inj] }

/-- The canonical map from nonnegative extended reals to extended reals -/
def _root_.ennreal.to_ereal : ℝ≥0∞ → ereal
| ⊤ := ⊤
| (some x) := x.1

instance has_coe_ennreal : has_coe ℝ≥0∞ ereal := ⟨ennreal.to_ereal⟩

instance : has_zero ereal := ⟨(0 : ℝ)⟩
instance : inhabited ereal := ⟨0⟩

/-- A recursor for `ereal` in terms of the coercion.

A typical invocation looks like `induction x using ereal.rec`. Note that using `induction`
directly will unfold `ereal` to `option` which is undesirable.

When working in term mode, note that pattern matching can be used directly. -/
@[elab_as_eliminator]
protected def rec {C : ereal → Sort*} (h_bot : C ⊥) (h_real : Π a : ℝ, C a) (h_top : C ⊤) :
  ∀ a : ereal, C a
| ⊥ := h_bot
| (a : ℝ) := h_real a
| ⊤ := h_top

/-! ### Real coercion -/

instance : can_lift ereal ℝ :=
{ coe := coe,
  cond := λ r, r ≠ ⊤ ∧ r ≠ ⊥,
  prf := λ x hx,
  begin
    induction x using ereal.rec,
    { simpa using hx },
    { simp },
    { simpa using hx }
  end }

/-- The map from extended reals to reals sending infinities to zero. -/
def to_real : ereal → ℝ
| ⊥       := 0
| ⊤       := 0
| (x : ℝ) := x

@[simp] lemma to_real_top : to_real ⊤ = 0 := rfl

@[simp] lemma to_real_bot : to_real ⊥ = 0 := rfl

@[simp] lemma to_real_zero : to_real 0 = 0 := rfl

@[simp] lemma to_real_coe (x : ℝ) : to_real (x : ereal) = x := rfl

@[simp] lemma bot_lt_coe (x : ℝ) : (⊥ : ereal) < x :=
by { apply with_top.coe_lt_coe.2, exact with_bot.bot_lt_coe _ }

@[simp] lemma coe_ne_bot (x : ℝ) : (x : ereal) ≠ ⊥  := (bot_lt_coe x).ne'

@[simp] lemma bot_ne_coe (x : ℝ) : (⊥ : ereal) ≠ x := (bot_lt_coe x).ne

@[simp] lemma coe_lt_top (x : ℝ) : (x : ereal) < ⊤ := with_top.coe_lt_top _

@[simp] lemma coe_ne_top (x : ℝ) : (x : ereal) ≠ ⊤ := (coe_lt_top x).ne

@[simp] lemma top_ne_coe (x : ℝ) : (⊤ : ereal) ≠ x := (coe_lt_top x).ne'

@[simp] lemma bot_lt_zero : (⊥ : ereal) < 0 := bot_lt_coe 0

@[simp] lemma bot_ne_zero : (⊥ : ereal) ≠ 0 := (coe_ne_bot 0).symm

@[simp] lemma zero_ne_bot : (0 : ereal) ≠ ⊥ := coe_ne_bot 0

@[simp] lemma zero_lt_top : (0 : ereal) < ⊤ := coe_lt_top 0

@[simp] lemma zero_ne_top : (0 : ereal) ≠ ⊤ := coe_ne_top 0

@[simp] lemma top_ne_zero : (⊤ : ereal) ≠ 0 := (coe_ne_top 0).symm

@[simp, norm_cast] lemma coe_add (x y : ℝ) : ((x + y : ℝ) : ereal) = (x : ereal) + (y : ereal) :=
rfl

@[simp] lemma coe_zero : ((0 : ℝ) : ereal) = 0 := rfl

lemma to_real_le_to_real {x y : ereal} (h : x ≤ y) (hx : x ≠ ⊥) (hy : y ≠ ⊤) :
  x.to_real ≤ y.to_real :=
begin
  lift x to ℝ,
  lift y to ℝ,
  { simpa using h },
  { simp [hy, ((bot_lt_iff_ne_bot.2 hx).trans_le h).ne'] },
  { simp [hx, (h.trans_lt (lt_top_iff_ne_top.2 hy)).ne], },
end

lemma coe_to_real {x : ereal} (hx : x ≠ ⊤) (h'x : x ≠ ⊥) : (x.to_real : ereal) = x :=
begin
  induction x using ereal.rec,
  { simpa using h'x },
  { refl },
  { simpa using hx },
end

lemma le_coe_to_real {x : ereal} (h : x ≠ ⊤) : x ≤ x.to_real :=
begin
  by_cases h' : x = ⊥,
  { simp only [h', bot_le] },
  { simp only [le_refl, coe_to_real h h'] },
end

lemma coe_to_real_le {x : ereal} (h : x ≠ ⊥) : ↑x.to_real ≤ x :=
begin
  by_cases h' : x = ⊤,
  { simp only [h', le_top] },
  { simp only [le_refl, coe_to_real h' h] },
end

lemma eq_top_iff_forall_lt (x : ereal) : x = ⊤ ↔ ∀ (y : ℝ), (y : ereal) < x :=
begin
  split,
  { rintro rfl, exact ereal.coe_lt_top },
  { contrapose!,
    intro h,
    exact ⟨x.to_real, le_coe_to_real h⟩, },
end

lemma eq_bot_iff_forall_lt (x : ereal) : x = ⊥ ↔ ∀ (y : ℝ), x < (y : ereal) :=
begin
  split,
  { rintro rfl, exact bot_lt_coe },
  { contrapose!,
    intro h,
    exact ⟨x.to_real, coe_to_real_le h⟩, },
end

/-! ### ennreal coercion -/

@[simp] lemma to_real_coe_ennreal : ∀ {x : ℝ≥0∞}, to_real (x : ereal) = ennreal.to_real x
| ⊤ := rfl
| (some x) := rfl

lemma coe_nnreal_eq_coe_real (x : ℝ≥0) : ((x : ℝ≥0∞) : ereal) = (x : ℝ) := rfl

@[simp] lemma coe_ennreal_top : ((⊤ : ℝ≥0∞) : ereal) = ⊤ := rfl

@[simp] lemma coe_ennreal_eq_top_iff : ∀ {x : ℝ≥0∞}, (x : ereal) = ⊤ ↔ x = ⊤
| ⊤ := by simp
| (some x) := by { simp only [ennreal.coe_ne_top, iff_false, ennreal.some_eq_coe], dec_trivial }

lemma coe_nnreal_ne_top (x : ℝ≥0) : ((x : ℝ≥0∞) : ereal) ≠ ⊤ := dec_trivial

@[simp] lemma coe_nnreal_lt_top (x : ℝ≥0) : ((x : ℝ≥0∞) : ereal) < ⊤ := dec_trivial

@[simp, norm_cast] lemma coe_ennreal_le_coe_ennreal_iff : ∀ {x y : ℝ≥0∞},
  (x : ereal) ≤ (y : ereal) ↔ x ≤ y
| x ⊤ := by simp
| ⊤ (some y) := by simp
| (some x) (some y) := by simp [coe_nnreal_eq_coe_real]

@[simp, norm_cast] lemma coe_ennreal_lt_coe_ennreal_iff : ∀ {x y : ℝ≥0∞},
  (x : ereal) < (y : ereal) ↔ x < y
| ⊤ ⊤ := by simp
| (some x) ⊤ := by simp
| ⊤ (some y) := by simp
| (some x) (some y) := by simp [coe_nnreal_eq_coe_real]

@[simp, norm_cast] lemma coe_ennreal_eq_coe_ennreal_iff : ∀ {x y : ℝ≥0∞},
  (x : ereal) = (y : ereal) ↔ x = y
| ⊤ ⊤ := by simp
| (some x) ⊤ := by simp
| ⊤ (some y) := by simp [(coe_nnreal_lt_top y).ne']
| (some x) (some y) := by simp [coe_nnreal_eq_coe_real]

lemma coe_ennreal_nonneg (x : ℝ≥0∞) : (0 : ereal) ≤ x :=
coe_ennreal_le_coe_ennreal_iff.2 (zero_le x)

@[simp] lemma bot_lt_coe_ennreal (x : ℝ≥0∞) : (⊥ : ereal) < x :=
(bot_lt_coe 0).trans_le (coe_ennreal_nonneg _)

@[simp] lemma coe_ennreal_ne_bot (x : ℝ≥0∞) : (x : ereal) ≠ ⊥ := (bot_lt_coe_ennreal x).ne'

@[simp, norm_cast] lemma coe_ennreal_add : ∀ (x y : ennreal), ((x + y : ℝ≥0∞) : ereal) = x + y
| ⊤ y := rfl
| x ⊤ := by simp
| (some x) (some y) := rfl

@[simp] lemma coe_ennreal_zero : ((0 : ℝ≥0∞) : ereal) = 0 := rfl

/-! ### Order -/

lemma exists_rat_btwn_of_lt : Π {a b : ereal} (hab : a < b),
  ∃ (x : ℚ), a < (x : ℝ) ∧ ((x : ℝ) : ereal) < b
| ⊤ b h := (not_top_lt h).elim
| (a : ℝ) ⊥ h := (lt_irrefl _ ((bot_lt_coe a).trans h)).elim
| (a : ℝ) (b : ℝ) h := by simp [exists_rat_btwn (ereal.coe_lt_coe_iff.1 h)]
| (a : ℝ) ⊤ h := let ⟨b, hab⟩ := exists_rat_gt a in ⟨b, by simpa using hab, coe_lt_top _⟩
| ⊥ ⊥ h := (lt_irrefl _ h).elim
| ⊥ (a : ℝ) h := let ⟨b, hab⟩ := exists_rat_lt a in ⟨b, bot_lt_coe _, by simpa using hab⟩
| ⊥ ⊤ h := ⟨0, bot_lt_coe _, coe_lt_top _⟩

lemma lt_iff_exists_rat_btwn {a b : ereal} :
  a < b ↔ ∃ (x : ℚ), a < (x : ℝ) ∧ ((x : ℝ) : ereal) < b :=
⟨λ hab, exists_rat_btwn_of_lt hab, λ ⟨x, ax, xb⟩, ax.trans xb⟩

lemma lt_iff_exists_real_btwn {a b : ereal} :
  a < b ↔ ∃ (x : ℝ), a < x ∧ (x : ereal) < b :=
⟨λ hab, let ⟨x, ax, xb⟩ := exists_rat_btwn_of_lt hab in ⟨(x : ℝ), ax, xb⟩,
 λ ⟨x, ax, xb⟩, ax.trans xb⟩

/-- The set of numbers in `ereal` that are not equal to `±∞` is equivalent to `ℝ`. -/
def ne_top_bot_equiv_real : ({⊥, ⊤}ᶜ : set ereal) ≃ ℝ :=
{ to_fun := λ x, ereal.to_real x,
  inv_fun := λ x, ⟨x, by simp⟩,
  left_inv := λ ⟨x, hx⟩, subtype.eq $ begin
    lift x to ℝ,
    { simp },
    { simpa [not_or_distrib, and_comm] using hx }
  end,
  right_inv := λ x, by simp }

/-! ### Addition -/

@[simp] lemma add_top (x : ereal) : x + ⊤ = ⊤ := add_top _
@[simp] lemma top_add (x : ereal) : ⊤ + x = ⊤ := top_add _

@[simp] lemma bot_add_bot : (⊥ : ereal) + ⊥ = ⊥ := rfl
@[simp] lemma bot_add_coe (x : ℝ) : (⊥ : ereal) + x = ⊥ := rfl
@[simp] lemma coe_add_bot (x : ℝ) : (x : ereal) + ⊥ = ⊥ := rfl

lemma to_real_add : ∀ {x y : ereal} (hx : x ≠ ⊤) (h'x : x ≠ ⊥) (hy : y ≠ ⊤) (h'y : y ≠ ⊥),
  to_real (x + y) = to_real x + to_real y
| ⊥ y hx h'x hy h'y := (h'x rfl).elim
| ⊤ y hx h'x hy h'y := (hx rfl).elim
| x ⊤ hx h'x hy h'y := (hy rfl).elim
| x ⊥ hx h'x hy h'y := (h'y rfl).elim
| (x : ℝ) (y : ℝ) hx h'x hy h'y := by simp [← ereal.coe_add]

lemma add_lt_add_right_coe {x y : ereal} (h : x < y) (z : ℝ) : x + z < y + z :=
begin
  induction x using ereal.rec; induction y using ereal.rec,
  { exact (lt_irrefl _ h).elim },
  { simp only [bot_lt_coe, bot_add_coe, ← coe_add] },
  { simp },
  { exact (lt_irrefl _ (h.trans (bot_lt_coe x))).elim },
  { norm_cast at h ⊢, exact add_lt_add_right h _ },
  { simp only [← coe_add, top_add, coe_lt_top] },
  { exact (lt_irrefl _ (h.trans_le le_top)).elim },
  { exact (lt_irrefl _ (h.trans_le le_top)).elim },
  { exact (lt_irrefl _ (h.trans_le le_top)).elim },
end

lemma add_lt_add_of_lt_of_le {x y z t : ereal} (h : x < y) (h' : z ≤ t) (hz : z ≠ ⊥) (ht : t ≠ ⊤) :
  x + z < y + t :=
begin
  induction z using ereal.rec,
  { simpa only using hz },
  { calc x + z < y + z : add_lt_add_right_coe h _
           ... ≤ y + t : add_le_add le_rfl h' },
  { exact (ht (top_le_iff.1 h')).elim }
end

lemma add_lt_add_left_coe {x y : ereal} (h : x < y) (z : ℝ) : (z : ereal) + x < z + y :=
by simpa [add_comm] using add_lt_add_right_coe h z

lemma add_lt_add {x y z t : ereal} (h1 : x < y) (h2 : z < t) : x + z < y + t :=
begin
  induction y using ereal.rec,
  { exact (lt_irrefl _ (bot_le.trans_lt h1)).elim },
  { calc x + z ≤ y + z : add_le_add h1.le le_rfl
    ... < y + t : add_lt_add_left_coe h2 _ },
  { simp [lt_top_iff_ne_top, with_top.add_eq_top, h1.ne, (h2.trans_le le_top).ne] }
end

@[simp] lemma add_eq_top_iff {x y : ereal} : x + y = ⊤ ↔ x = ⊤ ∨ y = ⊤ :=
begin
  induction x using ereal.rec; induction y using ereal.rec;
  simp [← ereal.coe_add],
end

@[simp] lemma add_lt_top_iff {x y : ereal} : x + y < ⊤ ↔ x < ⊤ ∧ y < ⊤ :=
by simp [lt_top_iff_ne_top, not_or_distrib]

/-! ### Negation -/

/-- negation on `ereal` -/
protected def neg : ereal → ereal
| ⊥       := ⊤
| ⊤       := ⊥
| (x : ℝ) := (-x : ℝ)

instance : has_neg ereal := ⟨ereal.neg⟩

@[norm_cast] protected lemma neg_def (x : ℝ) : ((-x : ℝ) : ereal) = -x := rfl

@[simp] lemma neg_top : - (⊤ : ereal) = ⊥ := rfl
@[simp] lemma neg_bot : - (⊥ : ereal) = ⊤ := rfl
@[simp] lemma neg_zero : - (0 : ereal) = 0 := by { change ((-0 : ℝ) : ereal) = 0, simp }

instance : has_involutive_neg ereal :=
{ neg := has_neg.neg,
  neg_neg := λ a, match a with
    | ⊥ := rfl
    | ⊤ := rfl
    | (a : ℝ) := by { norm_cast, simp [neg_neg a] }
    end }

@[simp] lemma to_real_neg : ∀ {a : ereal}, to_real (-a) = - to_real a
| ⊤ := by simp
| ⊥ := by simp
| (x : ℝ) := rfl

@[simp] lemma neg_eg_top_iff {x : ereal} : - x = ⊤ ↔ x = ⊥ :=
by { rw neg_eq_iff_neg_eq, simp [eq_comm] }

@[simp] lemma neg_eg_bot_iff {x : ereal} : - x = ⊥ ↔ x = ⊤ :=
by { rw neg_eq_iff_neg_eq, simp [eq_comm] }

@[simp] lemma neg_eg_zero_iff {x : ereal} : - x = 0 ↔ x = 0 :=
by { rw neg_eq_iff_neg_eq, simp [eq_comm] }

/-- if `-a ≤ b` then `-b ≤ a` on `ereal`. -/
protected theorem neg_le_of_neg_le : ∀ {a b : ereal} (h : -a ≤ b), -b ≤ a
| ⊥ ⊥ h := h
| ⊥ (some b) h := by cases (top_le_iff.1 h)
| ⊤ l h := le_top
| (a : ℝ) ⊥ h := by cases (le_bot_iff.1 h)
| l ⊤ h := bot_le
| (a : ℝ) (b : ℝ) h := by { norm_cast at h ⊢, exact neg_le.mp h }

/-- `-a ≤ b ↔ -b ≤ a` on `ereal`. -/
protected theorem neg_le {a b : ereal} : -a ≤ b ↔ -b ≤ a :=
⟨ereal.neg_le_of_neg_le, ereal.neg_le_of_neg_le⟩

/-- `a ≤ -b → b ≤ -a` on ereal -/
theorem le_neg_of_le_neg {a b : ereal} (h : a ≤ -b) : b ≤ -a :=
by rwa [←neg_neg b, ereal.neg_le, neg_neg]

@[simp] lemma neg_le_neg_iff {a b : ereal} : - a ≤ - b ↔ b ≤ a :=
by conv_lhs { rw [ereal.neg_le, neg_neg] }

@[simp, norm_cast] lemma coe_neg (x : ℝ) : ((- x : ℝ) : ereal) = - (x : ereal) := rfl

/-- Negation as an order reversing isomorphism on `ereal`. -/
def neg_order_iso : ereal ≃o erealᵒᵈ :=
{ to_fun := λ x, order_dual.to_dual (-x),
  inv_fun := λ x, -x.of_dual,
  map_rel_iff' := λ x y, neg_le_neg_iff,
  ..equiv.neg ereal }

lemma neg_lt_of_neg_lt {a b : ereal} (h : -a < b) : -b < a :=
begin
  apply lt_of_le_of_ne (ereal.neg_le_of_neg_le h.le),
  assume H,
  rw [← H, neg_neg] at h,
  exact lt_irrefl _ h
end

lemma neg_lt_iff_neg_lt {a b : ereal} : -a < b ↔ -b < a :=
⟨λ h, ereal.neg_lt_of_neg_lt h, λ h, ereal.neg_lt_of_neg_lt h⟩

/-! ### Subtraction -/

/-- Subtraction on `ereal`, defined by `x - y = x + (-y)`. Since addition is badly behaved at some
points, so is subtraction. There is no standard algebraic typeclass involving subtraction that is
registered on `ereal` because of this bad behavior. -/
protected noncomputable def sub (x y : ereal) : ereal := x + (-y)

noncomputable instance : has_sub ereal := ⟨ereal.sub⟩

@[simp] lemma top_sub (x : ereal) : ⊤ - x = ⊤ := top_add x
@[simp] lemma sub_bot (x : ereal) : x - ⊥ = ⊤ := add_top x

@[simp] lemma bot_sub_top : (⊥ : ereal) - ⊤ = ⊥ := rfl
@[simp] lemma bot_sub_coe (x : ℝ) : (⊥ : ereal) - x = ⊥ := rfl
@[simp] lemma coe_sub_bot (x : ℝ) : (x : ereal) - ⊤ = ⊥ := rfl

@[simp] lemma sub_zero (x : ereal) : x - 0 = x := by { change x + (-0) = x, simp }
@[simp] lemma zero_sub (x : ereal) : 0 - x = - x := by { change 0 + (-x) = - x, simp }

lemma sub_eq_add_neg (x y : ereal) : x - y = x + -y := rfl

lemma sub_le_sub {x y z t : ereal} (h : x ≤ y) (h' : t ≤ z) : x - z ≤ y - t :=
add_le_add h (neg_le_neg_iff.2 h')

lemma sub_lt_sub_of_lt_of_le {x y z t : ereal} (h : x < y) (h' : z ≤ t) (hz : z ≠ ⊥) (ht : t ≠ ⊤) :
  x - t < y - z :=
add_lt_add_of_lt_of_le h (neg_le_neg_iff.2 h') (by simp [ht]) (by simp [hz])

lemma coe_real_ereal_eq_coe_to_nnreal_sub_coe_to_nnreal (x : ℝ) :
  (x : ereal) = real.to_nnreal x - real.to_nnreal (-x) :=
begin
  rcases le_or_lt 0 x with h|h,
  { have : real.to_nnreal x = ⟨x, h⟩, by { ext, simp [h] },
    simp only [real.to_nnreal_of_nonpos (neg_nonpos.mpr h), this, sub_zero, ennreal.coe_zero,
      coe_ennreal_zero, coe_coe],
    refl },
  { have : (x : ereal) = - (- x : ℝ), by simp,
    conv_lhs { rw this },
    have : real.to_nnreal (-x) = ⟨-x, neg_nonneg.mpr h.le⟩, by { ext, simp [neg_nonneg.mpr h.le], },
    simp only [real.to_nnreal_of_nonpos h.le, this, zero_sub, neg_inj, coe_neg,
      ennreal.coe_zero, coe_ennreal_zero, coe_coe],
    refl }
end

lemma to_real_sub {x y : ereal} (hx : x ≠ ⊤) (h'x : x ≠ ⊥) (hy : y ≠ ⊤) (h'y : y ≠ ⊥) :
  to_real (x - y) = to_real x - to_real y :=
begin
  rw [ereal.sub_eq_add_neg, to_real_add hx h'x, to_real_neg],
  { refl },
  { simpa using hy },
  { simpa using h'y }
end

/-! ### Multiplication -/

@[simp] lemma coe_one : ((1 : ℝ) : ereal) = 1 := rfl

@[simp, norm_cast] lemma coe_mul (x y : ℝ) : ((x * y : ℝ) : ereal) = (x : ereal) * (y : ereal) :=
eq.trans (with_bot.coe_eq_coe.mpr with_bot.coe_mul) with_top.coe_mul

@[simp] lemma mul_top (x : ereal) (h : x ≠ 0) : x * ⊤ = ⊤ := with_top.mul_top h
@[simp] lemma top_mul (x : ereal) (h : x ≠ 0) : ⊤ * x = ⊤ := with_top.top_mul h

@[simp] lemma bot_mul_bot : (⊥ : ereal) * ⊥ = ⊥ := rfl
@[simp] lemma bot_mul_coe (x : ℝ) (h : x ≠ 0) : (⊥ : ereal) * x = ⊥ :=
with_top.coe_mul.symm.trans $
  with_bot.coe_eq_coe.mpr $ with_bot.bot_mul $ function.injective.ne (@option.some.inj _) h
@[simp] lemma coe_mul_bot (x : ℝ) (h : x ≠ 0) : (x : ereal) * ⊥ = ⊥ :=
with_top.coe_mul.symm.trans $
  with_bot.coe_eq_coe.mpr $ with_bot.mul_bot $ function.injective.ne (@option.some.inj _) h

@[simp] lemma to_real_one : to_real 1 = 1 := rfl

lemma to_real_mul : ∀ {x y : ereal}, to_real (x * y) = to_real x * to_real y
| ⊤ y := by by_cases hy : y = 0; simp [hy]
| x ⊤ := by by_cases hx : x = 0; simp [hx]
| (x : ℝ) (y : ℝ) := by simp [← ereal.coe_mul]
| ⊥ (y : ℝ) := by by_cases hy : y = 0; simp [hy]
| (x : ℝ) ⊥ := by by_cases hx : x = 0; simp [hx]
| ⊥ ⊥ := by simp

end ereal

subset_name: curated


TITLE: Volume using Monte Carlo Method
QUESTION [1 upvotes]: I want to find the volume above $xy$-plane and below the surface given by equation $z = x^2 +2y^2$ for $(x,y) \in [0,1] \times [0,1]$ using Monte Carlo method.
Can anybody help me please? Can you please share related questions so that I can have an idea of how to solve it?
Thanks.

REPLY [0 votes]: An alternative:
Sample random points $(x_i,y_i)$ uniformly in $[0,1]^2$ and calculate the corresponding $z_i=x_i^2+2y_i^2$ value.
Do this for many sample points, and calculate the average of the $z_i$ values.
This is the average height of your region.
Since the set $[0,1]\times[0,1]$ has area $1$, this average height is equal to the volume of your region.


text:
\begin{document}

\newtheorem{defi}{Definition}[section]
\newtheorem{rem}{Remark}[section]
\newtheorem{prop}{Proposition}[section]
\newtheorem{lem}{Lemma}[section]
\newtheorem{theo}{Theorem}[section]
\newtheorem{cor}{Corollary}[section]
\newtheorem{conc}{Conclusion}[section]
\newtheorem{claim}{Claim}[section]

\author{Pierre Germain}

\title{On the existence of smooth self-similar blow-up profiles for the wave-map equation}

\maketitle

\maketitle

\begin{abstract}
Consider the equivariant wave map equation from Minkowski space to a rotationnally symmetric manifold $N$ which has an equator (example: the sphere).

In dimension $3$, this article presents a necessary and sufficient condition on $N$ for the existence of a smooth self-similar blow up profile.

More generally, we study the relation between

- the minimizing properties of the equator map for the Dirichlet energy corresponding to the (elliptic) harmonic map problem

- and the existence of a smooth blow-up profile for the (hyperbolic) wave map problem.

This has several applications to questions of regularity and uniqueness for the wave map equation.
\end{abstract}

\section{Introduction}

\subsection{The geometric setting}

\label{lezard}

We will consider in the following equations on a map $u$, from the Euclidean space $\mathbb{R}^{d}$, with $d \geq3$ (which corresponds to the supercritical regime for harmonic maps and wave maps) to a manifold $N$ embedded in $\mathbb{R}^K$.

\subsubsection{Spherical symmetry and equator}

The first assumption that we will make is one of symmetry: on $N$, one can take coordinates (which may be periodic in $\phi$)
$$
(\phi,\chi) \in {\mathbb{R}} \times \mathbb{S}^{M-1}
$$
($M$ an integer) in which the metric reads
$$
d\phi^2 + g(\phi)^2 d\chi^2\,\,,
$$
for a smooth odd function $g$ such that $g'(0)=1$.

\bigskip

We now impose that $N$ has an equator; by this we simply mean $\phi^*>0$ such that
$$
g'(\phi^*) = 0 \,\,,
$$
and minimal with this property.

\subsubsection{Equivariance}

We further restrict $u$ to the class of equivariant maps, that is, $(r,\omega)$ denoting polar coordinates on $\mathbb{R}^d$,
\begin{equation*}
\begin{split}
u = R_\chi \phi\;\overset{def}{:}\; & {\mathbb{R}}^{d} \longrightarrow N \\
& (r,\omega) \longrightarrow (\phi(r)\,,\chi(\omega)) \,\,.
\end{split}
\end{equation*}
Here, $\chi : \mathbb{S}^{d-1} \rightarrow \mathbb{S}^{M-1}$ ($M$ is an integer) is an eigenmap, that is a map satisfying, for a constant $k$
\begin{equation*}
|\nabla \chi|^2 = k \;\;\;\;\;\mbox{and}\;\;\;\;\; \Delta_{\mathbb{S}^{d-1}} \chi + k \chi = 0\,\,.
\end{equation*}
This can be achieved for some large enough $M$, provided
$$
k = \ell (\ell + d -2) \;\;\;\;\mbox{with $\ell \in \mathbb{N}$}
$$
see~\cite{CSTZ}.

\bigskip

Of course, the most simple example of an eigenmap is provided by $\chi(\omega) = \omega$, mapping $\mathbb{S}^{d-1}$ to itself, in which case $k = d-1$.

\bigskip

We call equator map the map corresponding to
$$
\phi \equiv \phi^* \,\,;
$$
it actually depends on $\chi$, that is the equivariance class which is considered.

\subsection{Harmonic maps and the equator map}

\subsubsection{The boundary value problem}

Harmonic maps from the ball $B = B_{{\mathbb{R}}^d}(0,1)$ of the Euclidean space $\mathbb{R}^{d}$, to a manifold $N$ embedded in $\mathbb{R}^K$, with Dirichlet boundary data $h$, satisfy the equation
$$
\left\{ \begin{array}{ll}
 \Delta u(x) \perp T_{u(x)} N & \mbox{on $B$} \\
 u = h & \mbox{on $\partial B$} \end{array} \right.
$$
(here, $T_{u(x)} N$ is the tangent space, at $u(x)$, of $N$, and orthogonality is to be understood in the sense of $\mathbb{R}^{K}$).
In the equivariant setting discussed above, this Dirichlet problem becomes
\begin{equation}
\label{dirichletequivariant}
(D_\alpha) \;\;\;\;
\left\{ \begin{array}{ll}
 -\phi_{rr} - \frac{d-1}{r} \phi_r + \frac{k}{r^2} g(\phi) g'(\phi) = 0 & \mbox{on $[0,1]$} \\
 \phi(1) = \alpha \,\,, & \end{array} \right.
\end{equation}
where $\alpha$ is a given real number.

\subsubsection{The energy and the equator map}

Harmonic maps are critical points of the energy, which reads in the equivariant setting
$$
E_e(\phi) = \int_0^1 \left( \phi_r^2 + \frac{k}{r^2} g^2(\phi) \right) r^{d-1} dr\,\,.
$$
Solutions to the Dirichlet problem $(D_\alpha)$ are critical points of $E_e$ in the class
$$
F_\alpha = \left\{ \phi:[0,1]\rightarrow {\mathbb{R}}\;,\; \phi_r \in L^2(r^{d-1}dr) \;,\;\phi(1) = \alpha \right\}\,\,.
$$
It can easily be seen that the equator map $\phi \equiv \phi^*$ is a solution of the Dirichlet problem with $\alpha = \phi^*$. A crucial point in this article will be the minimizing properties of $\phi^*$.
\begin{itemize}
\item
For a full answer to the question whether $\phi^*$ minimizes locally $E_e$ over $F_{\phi^*}$, see Proposition~\ref{local}.
\item
Whether $\phi^*$ minimizes globally or not $E_e$ over $F_{\phi^*}$ is a more difficult question. The answer is known in the case where $\chi = Id$ (covariant case) and $N$ is an ellipse embedded in ${\mathbb{R}}^{d+1}$ given by ($a \in [0,1]$)
$$
y_1^2 + \dots + y_d^2 + \frac{y_{d+1}^2}{a^2} = 1 \,\,,
$$
and it is due to J\"ager and Kaul~\cite{JK}, Baldes~\cite{Ba}, H\'elein~\cite{He}.
These authors prove the following criterion: $\phi^*$ is the unique global minimizer of $E_e$ if and only if
$$
a^2 \geq \frac{4(d-1)}{(d-2)^2}\,\,.
$$
\end{itemize}

\subsection{Wave maps}

This paper addresses only some of the questions related to the wave map Cauchy problem. Other aspects, like the case of dimension $2$, or optimal well-posedness results for wave maps without symmetries, are discussed in the review by Krieger~\cite{K}.

\subsubsection{The initial value problem}

The Cauchy problem for wave maps from Minkowski space $\mathbb{R}^{d+1}$ to a manifold $N$ embedded in $\mathbb{R}^K$
reads
$$
(WM)\;\;\;\;
\left\{ \begin{array}{l}
 \partial_t^2 u(x,t) - \Delta u(x,t) \perp T_{u(x,t)} N \\
 u(t=0) = u_0 \\
 u_t(t=0) = u_1 \,\,. \end{array} \right.
$$
The weak formulation reads, for $(u\,,\,u_t) \in L^\infty_t (\dot{H}^1 \times L^2)$, $u(x,t) \in N$ for almost all $(x\,,\,t)$,
\begin{equation*}
\begin{split}
&\int_{\mathbb{R}^{d+1}} \left[ \langle u_t\,,\, f_t \rangle - \langle \nabla_x u \,,\, \nabla_x f \rangle \right] \,dx\,dt = 0 \\
& \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mbox{for $f$ such that $f_t \,, \,\nabla_x f \in \L^2$ and for almost all $(x\,,\,t)$, $f(x,t) \in T_{u(x,t)} N$} \\
&(u \,,\,u_t) \overset{t \rightarrow 0}{\longrightarrow} (u_0\,,\,u_1) \;\;\mbox{weakly in $\dot{H}^1 \times L^2$}\,\,.
\end{split}
\end{equation*}
We can rewrite these equations in the equivariant setting. We obtain the following equation, which is to be understood in $\mathcal{S}'$ if one considers weak solutions.
$$
(EWM)\;\;\;\;
\left\{ \begin{array}{l}
 \phi_{tt} -\phi_{rr} - \frac{d-1}{r} \phi_r + \frac{k}{r^2} g(\phi) g'(\phi) = 0 \\
 \phi(t=0) = \phi_0 \\
 \phi_t(t=0) = \phi_1 \,\,. \end{array} \right.
$$

\subsubsection{Formation of singularities: the blow up profiles}

It is a basic problem to understand whether the wave map equation can develop singularities if the initial data are smooth. The approach followed by Shatah~\cite{Sh}, Shatah and Tahvildar-Zadeh~\cite{STZ}, and Cazenave, Shatah and Tahvildar-Zadeh~\cite{CSTZ} consists in building up smooth blow up profiles

\begin{defi}[blow up profiles]
A (self-similar) blow up profile is a function $\psi$ such that
$$
\phi(r,t) = \psi \left( \frac{r}{t} \right)
$$
solves the wave-map equation.
\end{defi}

 The profile $\psi(\rho)$ has to be defined for $\rho \leq 1$, and must have a smooth behavior at 1. Then one can extend the data $(\psi, r \psi')$, defined on $B_{\mathbb{R}^d}(0,1)$, smoothly to
$(\widetilde{\phi}_0\,,\,\widetilde{\phi}_1)$, defined on $\mathbb{R}^d$. The solution to $(EWM)$ with data $(\widetilde{\phi}_0\,,\,\widetilde{\phi}_1)$ equals
$$
\psi \left( \frac{r}{1-t} \right)
$$
inside the backward light cone, with basis $B$ and vertex $(x=0,t=1)$. In particular, a singularity appears at $(x=0,t=1)$, even though the data were smooth.

\bigskip

The PDE satisfied by $\phi$ turns into an ODE for $\psi$. More precisely,

\begin{prop}
\label{PGProp}
Suppose that $\psi$ is an integrable function, smooth at $0$. Then
$R_\chi \phi$ solves $(WM)$ for $t>0$ if and only if $\psi$ satisfies the equation
\begin{equation}
\label{eqpsi}
\rho^2 (1-\rho^2) \psi''(\rho) + \rho ((d-1)- 2 \rho^2) \psi'(\rho) - k g(\psi) g'(\psi) = 0
\end{equation}
in the classical sense on $(0,1) \cup (1,\infty)$, and furthermore $\psi(1-) = \psi(1+)$.
\end{prop}

This proposition is proved in Section~\ref{sectionODE}.

\bigskip

A first approach to solving for $\psi$ is to handle the above singular ODE. Another approach is variational, it was developed by the authors mentioned above. These authors observe that the equation satisfied by $\psi$ is the harmonic map equation from the hyperbolic ball to $N$. Thus it can be solved by minimizing
$$
E_h(\psi) = \int_{B} \left( \psi_r^2 + \frac{k}{r^2(1-r^2)} \left[ g^2(\psi) - g^2(\phi^*) \right] \right) \frac{r^{d-1} dr}{(1-r^2)^{(d-3)/2}}\,\,,
$$
on
$$
G = \left\{ \psi : [0,1] \rightarrow {\mathbb{R}}\;,\; \psi_r \in L^2\left(\frac{r^{d-1}\,dr}{(1-r^2)^{\frac{d-3}{2}}}\right)\;,\;\psi(1) = \phi^* \right\}\,\,.
$$
The functional $E_h$ corresponds to the Dirichlet energy for an equivariant map from the hyperbolic ball to $N$, except $g^2(\psi)$ has been replaced by $\left[ g^2(\psi) - g^2(\phi^*) \right]$. This renormalization imposes
the boundary condition $\psi = \phi^*$ for $\rho = 1$ - we will come back to this in the following.

The following theorem summarizes results proved in~\cite{CSTZ}.

\begin{theo}
\label{ehnegative}
If
$$
\inf_{\psi \in G} E_h(\psi) < 0\,\,,
$$
there exists a smooth $\psi : [0,1] \rightarrow [0,R^*]$ such that
\begin{itemize}
\item $\psi(0) = 0$.
\item $\psi(1) = \phi^*$.
 \item  $\phi = \psi \left( \frac{r}{t} \right)$ solves $(EWM)$ (where it is defined).
\item $u = R_\chi \phi$ is smooth where it is defined.
\end{itemize}
\end{theo}

It should also be mentioned that self-similar blow-up profiles as given by the above theorem provide examples of non uniqueness for $(EWM)$: see~\cite{STZ}~\cite{CSTZ}. We will use a similar argument in Theorem~\ref{dim3} for data equal to the equator map.

In the absence of such a self-similar blow-up profile, there does not seem to be known examples of non-uniqueness.

Finally, multiple solutions for the above variational problem are constructed in Jungen~\cite{J}.

\subsubsection{Energy conservation and weak solutions}

A smooth solution of $(EWM)$ satisfies the energy equality (we only give here the version which is centered at $0$ in space)
\begin{equation}
\label{energyeq}
\mbox{if $R>T$,}\;\;\;\;E(T, R - T,\phi) + \operatorname{flux}(0,T, R ,\phi) = E(0, R ,\phi)
\end{equation}
with the following notations
\begin{equation*}
\begin{split}
& E(t,R,\phi) \overset{def}{=} \int_0^R \left( \phi_r^2(t) + \frac{k}{r^2} g(\phi(t))^2 + \phi_t^2(t) \right) r^{d-1} \,dr \\
& \operatorname{flux}(0,T,R,\phi) \overset{def}{=} \int_{S_{0,T,R}} \left( \frac{1}{2} \phi_t^2 + \frac{k}{2r^2} g(\phi)^2 + \frac{1}{2} |\nabla \phi|^2 - \phi_t \nabla \phi \cdot n \right) d\sigma \,\,,
\end{split}
\end{equation*}
where $S_{0,T,R}$ is the surface
$$
S_{0,T,R} = \{ (r,s)\;\;\mbox{such that $0 \leq t \leq T$ and $r = R-s$} \}
$$
(endowed with the metric coming from the natural embedding $S \subset \mathbb{R}^{d+1}$),
and where $n$ is the outside normal of $B(0,R - s)$. Notice that $\operatorname{flux}(0,T,R,\phi)$ is a positive quantity.

\bigskip

In the case where the data are of finite energy, one can use energy conservation and a compactness argument to build up global weak solutions.

\begin{theo}[Shatah~\cite{Sh}, Freire~\cite{Fr}]
Assume that the target manifold $N$ is homogeneous and compact, and
take data such that $\nabla u_0 \in L^2_{loc}$ and $u_1 \in L^2_{loc}$, i.e. for any $R>0$,
$$
\int_0^R \left( (\phi_0)_r^2 + \frac{k}{r^2} g(\phi_0)^2 \right) r^{d-1}\,dr + \int_0^R \phi_1^2 r^{d-1}\,dr < \infty \,\,.
$$
Then there exists a global solution of $(WM)$
$$
(u,u_t) \in L^\infty_t(\mathbb{R},\dot{H}^1_{loc}) \times L^\infty_t(\mathbb{R},L^2_{loc}) \,\,.
$$
\end{theo}

If one transposes this theorem in the equivariant setting, the following energy inequality can be proved to hold
\begin{equation}
\label{energyineq}
\mbox{if $R>T$,}\;\;\;\;E(T, R - T,\phi) + \operatorname{flux}(0,T, R ,\phi) \leq E(0, R ,\phi)\,\,.
\end{equation}
Indeed, it suffices to follow the steps of Shatah and Freire, who use a penalisation method. Since we consider equivariant functions, passing to the limit in the penalised problems is easy except for $x=0$; this is due to the energy bound which gives strong control anywhere else. Thus one can pass to the limit in~(\ref{energyeq}) and obtain~(\ref{energyineq}).

\section{Statement of the results}

\subsection{Local aspects}

\begin{prop}
\label{local}
The three following facts are equivalent
\medskip

(i) $k g(\phi^*) g''(\phi^*) \geq -\frac{(d-2)^2}{4}$
\medskip

(ii) The second variation $\displaystyle \delta^2 E_e(\phi^*)(w,w) \geq 0$ for $w \in F_{\phi^*}$.
\medskip

(iii) The second variation $\displaystyle \delta^2 E_h(\phi^*)(w,w) \geq 0$ for $w \in G$.
\bigskip

Besides, if we exclude the limit case where $k g(\phi^*) g''(\phi^*) = -\frac{(d-2)^2}{4}$, $(i)$ becomes equivalent to

\medskip

(iv) The wave map equation with initial data equal to the equator map is linearly stable (in a sense to be made precise in Section~\ref{linearlystable}).
\end{prop}

The proof of this proposition essentially consists in putting together already known results. It is interesting to notice that they all rely on different versions of Hardy's inequality.

This proof will be given in Section~\ref{prooflocal}.

\subsection{The case of dimension $3$}

In dimension $3$, we are able to analyze globally the relation between minimality of $\phi^*$ for $E_e$ and properties of the wave-map equation.

However, we need to make some more hypotheses on the geometry of $N$. We do not claim that they are optimal, but they are simple, of sufficient generality, and will enable us to state our results in an elegant way. Recall that $g$ is an odd smooth function. The further assumptions in dimension 3 are that
\begin{itemize}

\item {\bf either} the $\phi$ coordinate is not periodic; $\phi^*$ and $- \phi^*$ are the only zeros of $g'$; $g$ is positive and decreasing on $(\phi^*,\infty)$.

\item {\bf or} the $\phi$ coordinate is $4 \phi^*$ periodic, and $g (\phi^* + \cdot)$ is even. (Thus the manifold $N$ has the symmetries of an ellipsoid with all semi-principal axes but one of the same length)

\end{itemize}

Either of these assumptions ensures that
\begin{itemize}
\item Up to the symmetries of $N$, $\phi^*$ is unique with the property that $g'(\phi^*) = 0$.
\item Up to the symmetries of $N$, $0$ is unique with the property that $g'(\phi^*)=0$.
\item Up to the symmetries of $N$,
\begin{equation}
\label{etoile}
g(\phi)^2 < g(\phi^*)^2 \;\;\;\mbox{for any}\;\; \phi \neq \phi^* \,\,.
\end{equation}
\end{itemize}

We can now state the theorem.

\begin{theo}
\label{dim3}
If $d=3$, and the above assumptions are satisfied, the three following facts are equivalent
\medskip

(i) The equator map is the unique global minimizer of $E_e$ on $F_{\phi^*}$.
\medskip

(ii) The only weak solution to $(EWM)$ satisfying the energy inequality~(\ref{energyineq}), and with initial data
$$\phi_0 = \phi^* \;\;\;\;\;\;\phi_1 = 0$$
is identically equal to the equator map $\phi^*$.
\medskip

(iii) There does not exist a non-zero blow-up profile $\psi$ such that $R_\chi \psi \in \mathcal{C}^\infty (\bar B_{\mathbb{R}^d}(0,1))$.

\end{theo}

This theorem will be proved in Section~\ref{prodim3}. Let us mention quickly the main ideas of the proof.
\begin{itemize}
\item Starting from a smooth blow up profile $\psi$, it is possible to construct, using the equation it satisfies, a solution corresponding to data $(\phi_0 \,,\, \phi_1) = (\phi^*\,,\,0)$ and non identically equal to $\phi^*$.
\item If the equator map minimizes the Dirichlet energy, it is intuitively clear that the equator map is the unique solution corresponding to data $(\phi^* \,,\, 0)$ and satisfying the energy inequality. Indeed, any other solution would somehow have to make the energy grow, which is not allowed if the energy inequality is satisfied.
\item Finally, if the equator map does not minimize the Dirichlet energy, we use the variational approach: Theorem~\ref{ehnegative} gives a smooth solution.
\end{itemize}

Actually, the proof of Theorem~\ref{dim3} gives a little more than the statement above. In particular, we get the following very strong instability result, for data equal to the equator map, in case $(i)$ above does not hold.

\begin{prop}
Suppose that the equator map $\phi^*$ is not the unique global minimizer of $E_e$. Then there exists a smooth profile $\psi: [0,1] \rightarrow \mathbb{R}$ such that for any $T \geq 0$,
$$
\phi (r,t) = \left\{ \begin{array}{ll}
\psi\left( \frac{r}{t-T} \right) & \mbox{if $t \geq T$ and $r \leq t-T$} \\
\phi^* & otherwise \end{array}
\right.
$$
solves $(EWM)$ with data $(\phi_0 \,,\, \phi_1) = (\phi^*\,,\,0)$, and satisfies the energy equality~(\ref{energyeq}).
\end{prop}

Finally, it is natural to ask whether non-smooth profiles may exist even if $(i)$ in Theorem~\ref{dim3} holds. In other words: can one replace in the statement of $(iii)$ ``smooth'' by ``$H^s$'' for some $s$?
(Notice that the belonging of $R_\chi \psi$ to any functional space can be easily defined using the embedding of $N$ in $\mathbb{R}^K$.)

\begin{prop}
(i) Suppose that $\psi$ solves~(\ref{eqpsi}) on $(0,1)$, and that
$$
R_\chi \psi \in \dot{H}^{3/2}(B_{\mathbb{R}^3}(0,1))\,\,.
$$
Then $R_\chi \psi \in \mathcal{C}^\infty(\bar B_{\mathbb{R}^3}(0,1))$.

\bigskip

(ii) With only the assumptions made in Section~\ref{lezard} on $g$, there always exists a non zero $\psi$ such that
$R_\chi \psi$ smooth near $0$,
$$
R_\chi \psi \in \dot{B}^{3/2}_{2,\infty}(B_{\mathbb{R}^3}(0,1))\,\,,
$$
(see~\cite{Ge} for a definition of the Besov space $\dot{B}^{3/2}_{2,\infty}$) and
$$
u(x,t) = R_\chi \psi \left( \frac{|x|}{t} \right)
$$
solves $(WM)$.

\end{prop}

This proposition is proved in Section~\ref{proofreg}.

Thus, for $\psi$, $\dot{H}^{3/2}$ and $\mathcal{C}^\infty$ regularity are equivalent. But if one goes down to $\dot{B}^{3/2}_{2,\infty}$ the situation becomes drastically different: there always exist (no matter whether $(i)$ in Theorem~\ref{dim3} holds true or not) a profile with this regularity.

\bigskip

We would like to shed some light on the distinction between $\dot{H}^{3/2}$ and $\dot{B}^{3/2}_{2,\infty}$ appearing above.

Notice first that the spaces $\dot{B}^{3/2}_{2,\infty} \times \dot{B}^{1/2}_{2,\infty}$ or $\dot{H}^{3/2} \times \dot{H}^{1/2}$ are at the scaling of the equation for the initial data.
Global well-posedness for small data in $\dot{H}^{3/2}\times \dot{H}^{1/2}$ for $(WM)$ is due to Tao~\cite{Ta}. Global well posedness in the equivariant setting for a space similar to $\dot{B}^{3/2}_{2,\infty}\times \dot{B}^{1/2}_{2,\infty}$ has been proved by the author~\cite{Ge}.

Now let us point out a fundamental difference between considering solutions in the Sobolev space or in the Besov space. Space-time scaling invariant spaces corresponding to the spaces for the data mentioned above are
$$
(u,u_t) \in L^\infty_t \left( \dot{B}^{3/2}_{2,\infty} \times \dot{B}^{1/2}_{2,\infty}\right)\;\;\;\;\mbox{or}\;\;\;\; L^\infty_t \left(\dot{H}^{3/2} \times \dot{H}^{1/2}\right)\,\,.
$$

Consider a smooth profile $\psi$ and
$$
u(x,t) = R_\chi \psi \left( \frac{|x|}{t} \right) \,\,.
$$

Then (at least locally in space)
\begin{equation}
\label{besov}
(u,u_t) \in L^\infty \left([-1,1] \,,\,\dot{B}^{3/2}_{2,\infty} \times \dot{B}^{1/2}_{2,\infty}\right)
\end{equation}
but
\begin{equation}
\label{sobolev}
(u,u_t) \notin L^\infty \left([-1,1] \,,\,\dot{H}^{3/2} \times \dot{H}^{1/2} \right)\,\,.
\end{equation}
Thus in some sense one ``sees'' the blow up in the Sobolev setting~(\ref{sobolev}) but not in the Besov space setting~(\ref{besov}).

\subsection{The case of dimension $d \geq 4$}

In the case of dimension $d \geq 4$, we can only prove a weak version of Theorem~\ref{dim3}.

\begin{theo}
\label{dimg4}
Consider the following assertions
\medskip

(i) The equator map is the unique global minimizer of $E_e$ on $F_{\phi^*}$.
\medskip

(ii) The only weak solution to $(EWM)$ satisfying the energy inequality~(\ref{energyineq}), and with initial data
$$\phi_0 = \phi^* \;\;\;\;\;\;\phi_1 = 0$$
is identically equal to the equator map $\phi^*$.
\medskip

(iii) There does not exist a blow-up profile $\psi$ such that $\psi(1) = \phi^*$ and $R_\chi \psi \in \mathcal{C}^\infty (\bar B_{\mathbb{R}^d}(0,1))$.
\medskip

There holds
$$
(i) \Rightarrow (ii) \Rightarrow (iii)\,\,.
$$
\end{theo}

The proof of this theorem is almost identical to the corresponding steps in Theorem~\ref{dim3}; for this reason we do not include it here.

The question of the regularity of the profile is much more involved in dimensions $d \geq 4$: see~\cite{CSTZ}; we plan to come back to this and other aspects of the problem for $d \geq 4$ in a subsequent paper.

\section{Proof of Proposition~\ref{local}}

\label{prooflocal}

As mentioned above, the whole proof of Proposition~\ref{local} somehow relies on different versions of Hardy's inequality. It is interesting to recall here the whole space version, with the optimal constant
$$
\int_{\mathbb{R}^d} \frac{w^2}{x^2} \,dx \leq \frac{4}{(d-2)^2} \int_{\mathbb{R}^d} |\nabla w|^2 \,dx\,\,.
$$

\subsection{The second variation of $E_e$: proof of $(i) \Leftrightarrow (ii)$}

A small computation gives
$$
E_e (\phi^* + \epsilon w) = E_e(\phi^*) + \epsilon^2 \int_0^1 \left( w_r^2 + \frac{k}{r^2} g(\phi^*) g''(\phi^*) w^2 \right) r^{d-1} \,dr + O(\epsilon^3)\,\,.
$$
The conclusion follows due to the inequality (Baldes~\cite{Ba})
\begin{equation}
\label{HI1}
\mbox{if $\phi^* + \epsilon w \in F_{\phi^*}$,}\;\;\;\int_0^1 w^2 r^{d-3} dr \leq \frac{4}{(d-2)^2} \int_0^1 w_r^2 r^{d-1} dr\,\,,
\end{equation}
where the constant is optimal.

\subsection{The second variation of $E_h$: proof of $(i) \Leftrightarrow (iii)$}

By a small computation one gets (since $E_h(\phi^*) = 0$)
$$
E_h (\phi^* + \epsilon w) =  \epsilon^2 \int_0^1 \left( w_r^2 + \frac{k}{r^2(1-r^2)} g(\phi^*) g''(\phi^*) w^2 \right)
\frac{r^{d-1} dr}{(1-r^2)^{(d-3)/2}} + O(\epsilon^3) \,\,.
$$
The conclusion follows due to the inequality (Cazenave Shatah Tahvildar-Zadeh~\cite{CSTZ})
\begin{equation}
\label{HI2}
\mbox{if $\phi^* + \epsilon w \in G$,}\;\;\;
\int_0^1 w^2 \frac{r^{d-3}}{(1-r^2)^{\frac{d-1}{2}}} dr \leq \frac{4}{(d-2)^2} \int_0^1 w_r^2 \frac{r^{d-1}}{(1-r^2)^{\frac{d-3}{2}}} dr
\end{equation}
where the constant is optimal.

\subsection{Linear stability of the equator map: proof of $(i) \Leftrightarrow (iv)$}

\label{linearlystable}

The equation $(EWM)$ can be rewritten using the new unknown function $w = \phi - \phi^*$
\begin{equation}
\label{EWMw}
w_{tt} - w_{rr} - \frac{d-1}{r} w_r + \frac{k}{r^2} g(\phi^* + w) g'(\phi^* + w) = 0 \,\,.
\end{equation}
Let us now linearize this equation in the limit $w$ small:
\begin{equation}
\label{linearizedeq}
w_{tt} - w_{rr} - \frac{d-1}{r} w_r + \frac{k}{r^2} g(\phi^*) g''(\phi^*) w = 0 \,\,.
\end{equation}
This is a linear wave equation with a singular (inverse square) potential. Dispersion for this class of potentials has been studied by Burq, Planchon, Stalker and Tahvildar Zadeh~\cite{BPSTZ}, who prove the following result.

\begin{theo}[Burq, Planchon, Stalker, Tahvildar-Zadeh]
\label{theoBPSTZ}
The solution of the equation
$$
\left\{ \begin{array} {l}
w_{tt} - w_{rr} - \frac{d-1}{r} w_r + \frac{a}{r^2} w = 0 \\
w(t=0) = w_0 \\
w_t(t=0) = w_1 \\
\end{array} \right.
$$
satisfies the classical Strichartz estimates
\begin{equation}
\label{strichartz}
\|(-\Delta)^{\frac{\sigma}{2}} w \|_{L^p_t L^q_x} \leq C \left( \|w_0\|_{\dot{H}^{1/2}} + \|w_1\|_{\dot{H}^{-1/2}} \right)
\;\;\;\;\mbox{with $\frac{1}{p} + \frac{d-1}{2q} \leq \frac{d-1}{4}$ and $\sigma = \frac{1}{p} + \frac{d}{q} - \frac{d-1}{2}$}
\end{equation}
if
\begin{equation}
\label{criterion}
a > -\frac{(d-2)^2}{4} \,\,.
\end{equation}
\end{theo}

\bigskip

In the case where $a < - \frac{(d-2)^2}{4}$, the point spectrum of the operator $-\Delta + \frac{a}{x^2}$ extends to $-\infty$, see Remark 1.1 in Planchon, Stalker and Tahvildar-Zadeh~\cite{PSTZ}.

In particular, in this case, there are no Strichartz estimates.

\bigskip

What we mean by ``linearly stable'' in the statement of Proposition~\ref{local} is now clear: it means satisfying the Strichartz estimates given in Theorem~\ref{theoBPSTZ}.

Applying the criterion~(\ref{criterion}) to the linearized equation~(\ref{linearizedeq}), we get as expected the condition $(i)$ in Proposition~\ref{local}.

\begin{rem}[Nonlinear stability]
Of course a much more interesting question than linear stability is non-linear stability, that is, well posedness of~(\ref{EWMw}).

For a certain range of $k g(\phi^*) g''(\phi^*)$, a possible approach is the following: combine the Strichartz estimates with derivatives (that is, involving other regularities than what appears in~(\ref{strichartz})) of~\cite{BPSTZ} and the approach followed in~\cite{STZ} in order to get well posedness of~(\ref{EWMw}).

This approach, however, does not cover all the cases where linear stability holds. The difficulty is that, if $a$ is too close to $-\frac{(d-2)^2}{4}$, Strichartz estimates with derivatives are not available any more, and have to be replaced by estimates on
$$
\left\| (-\Delta)^{\sigma/2} \left( - \Delta + \frac{a}{x^2} \right)^{\sigma'} w \right\|_{Lp_t L^q_x}\,\,,
$$
that is the relevant spaces involve fractional powers of the operator $\displaystyle - \Delta + \frac{a}{x^2}$. But as opposed to standard Sobolev spaces, product laws between such spaces are not known and need to be established in order to be able to handle nonlinear settings.
\end{rem}

\begin{rem}
Hardy's inequality did not appear explicitly in the present subsection, but it is one of the fundamental ingredients in the proof of Theorem~\ref{theoBPSTZ}.
\end{rem}

\section{Proof of Theorem~\ref{dim3}}

\label{prodim3}

\subsection{Equivalence of $(i)$ and $(ii)$}

\subsubsection{$(i) \Rightarrow (ii)$}

Let us suppose first that $(i)$ holds, and consider $\phi$ a weak solution of $(EWM)$ with data equal to the equator map, and satisfying the energy inequality~(\ref{energyineq}).

The equator map is smooth anywhere but in $0$. By finite speed of propagation and weak-strong uniqueness for $(EWM)$ (Struwe~\cite{St}), $u$ must agree with the equator map outside of the forward light cone with vertex at $(t,x) = (0,0)$
$$
C = \left\{ (t,x) \,,\,|x| \leq t \right\} \,\,.
$$
Let us write the energy equality~(\ref{energyeq}) for the equator map
\begin{equation}
\label{abc1}
E(T,R - T,\phi^*) + \operatorname{flux}(0,T,R,\phi^*) = E(0,R,\phi^*) \,\,,
\end{equation}
where we choose $2 T < R$ so that the flux term is computed on a surface which lies completely outside of the forward light cone $C$.

For the weak solution $\phi$, the energy inequality~(\ref{energyineq}) holds, which reads
\begin{equation}
\label{abc2}
E(T,R - T,\phi) + \operatorname{flux}(0,T,R,\phi) \leq E(0,R,\phi) \,\,.
\end{equation}

Comparing~(\ref{abc1}) and~(\ref{abc2}), and keeping in mind that $\phi$ and $\phi^*$ agree at time $0$ and outside of the light cone, one gets
$$
E(T,R - T,\phi^*) \geq E(T,R - T,\phi) \,\,,
$$
which implies
$$
\int_0^{R-T} \frac{k}{r^2} g(\phi^*)^2 \,r^2 dr \geq \int_0^{R-T} \left( \phi_r^2 (T) + \frac{k}{r^2} g(\phi(T))^2 \right)\,r^2 dr \,\,.
$$
Thus, since $\phi(T,R-T) = \phi^*$, the minimizing property of the equator map gives $\phi(T) = \phi^*$. In other words $(ii)$ holds true.

\subsubsection{$\overline{(i)} \Rightarrow \overline{ (ii)}$}

\label{cameleon}

Let us suppose that $(i)$ does not hold, and prove that $(EWM)$ with initial data equal to the equator map admits a solution which is not identically equal to the equator map.

The alternative solution that we build up will be a self-similar solution
$$
\phi (r,t) = \psi \left( \frac{r}{t} \right)
$$

Suppose that $\phi^*$ is not the unique global minimizer of $E_e$ in $F_{\phi^*}$. Then there exists $\widetilde{\phi} \in F_{\phi^*}$ such that
$$
E_e(\widetilde{\phi}) \leq E_e(\phi^*) \,\,.
$$
Since the space dimension is $3$, $F_{\phi^*} = G$ and therefore it makes sense to compute
$$
E_h (\widetilde{\phi}) = E_e (\widetilde{\phi}) - E_e (\phi^*) + \int_0^1 \frac{k}{1-r^2}\left(g^2(\widetilde{\phi}) - g^2(\phi^*) \right)r^2\,dr \,\,.
$$
Since $\widetilde{\phi}$ is not identically $\phi^*$, and by condition~(\ref{etoile}),
$$
E_h (\widetilde{\phi}) < 0 \,\,.
$$
Hence Theorem~\ref{ehnegative} gives a self-similar profile $\psi$. Define the prolongation
$$
\bar \psi (\rho) = \left\{ \begin{array}{ll} \psi(\rho) & \mbox{if $\rho \leq 1$} \\ \phi^* & \mbox{if $\rho \geq 1$} \,\,.\end{array}
\right.
$$
and
$$
\phi(r,t) = \psi\left(\frac{r}{t} \right)\,\,.
$$
By proposition~\ref{PGProp}, $R_\chi \phi$ solves $(WM)$. It is easy to see that it takes the data
$$
\phi(t=0) = \phi^* \;\;\;\;\mbox{and}\;\;\;\; \phi_t (t=0) = 0 \,\,.
$$
To conclude that $(ii)$ is not true, it suffices to see that $\bar \psi$ satisfies the energy equality. This is precisely the statement of the next proposition, whose proof we postpone till Section~\ref{pec}.

\begin{prop}
\label{energyconserve}
$\bar \psi$, as defined above, satisfies the energy equality~(\ref{energyeq}).
\end{prop}

\subsection{Equivalence of $(i)$ and $(iii)$}

\subsubsection{$\overline{(i)} \Rightarrow \overline{(iii)}$}

Let us suppose that $(i)$ does not hold. Proceeding as in Section~\ref{cameleon}, we get a profile $\psi$ which is smooth on $[0,1]$. Thus $(iii)$ is not true.

\subsubsection{$\overline{(iii)} \Rightarrow \overline{(i)}$}

\label{subsubiii}

Let us suppose that $(iii)$ does not hold true, that is there exists a non zero blow up profile $\psi$ such that $R_\chi \psi \in \mathcal{C}^\infty( \bar B_{\mathbb{R}^2} (0,1) )$. Recall $\psi$ satisfies the ODE
$$
\rho^2 (1-\rho^2) \psi''(\rho) + 2 \rho (1-\rho^2) \psi'(\rho) - k g(\psi) g'(\psi) = 0
$$
Setting $\rho = 1$, we get $g(\psi(1)) g'(\psi(1)) = 0$. Thus, modulo the symmetries of the equation, either $\psi(1) = 0$ or $\psi(1) = \phi^*$. In order to show that the former case cannot occur, set
$$
M(\rho) = \frac{1}{2}(\rho^2 - \rho^4)\psi'(\rho)^2 - k g(\psi(\rho))^2
$$
and compute
$$
M'(\rho) = - \rho \psi'(\rho)^2 \leq 0\,\,,
$$
so $M$ is a decreasing quantity. Also, since the profile $\psi$ corresponds to a smooth solution, one has necessarily
$\psi(0) = 0$ (modulo the symmetries of the equation), so $M(0)=0$.

Now, arguing by contradiction, suppose that $\psi(1) = 0$, then $M(1) = 0$ and $M$ has to be identically $0$ on $[0,1]$. By the formula for $M'$, this implies that $\psi$ is constant on $[0,1]$, which is a contradiction.

\bigskip

So $\psi$ satisfies $\psi (1) = \phi^*$, and if we define
$$
\bar{\psi}(\rho) = \left\{ \begin{array}{ll} \psi(\rho) & \mbox{if $\rho \leq 1$} \\
\phi^* & \mbox{if $\rho \geq 1$}\,\,,
\end{array} \right.
$$
the associated space-time function
$$
\phi(r,t) = \psi\left( \frac{r}{t} \right)
$$
solves $(EWM)$ with data equal to the equator map, though it is not equal to the equator map itself. Thus
$(ii)$ does not hold true, and neither does $(i)$ by the previous section.

\section{Proof of Proposition~\ref{energyconserve}}

\label{pec}

We are considering a smooth profile $\psi$, such that $\psi(0) =0$ and $\psi(1)=\phi^*$, and
$$
\phi(r,t) = \left\{ \begin{array}{ll} \psi \left( \frac{r}{t} \right) & \mbox{if $r\leq t$ }\\  \phi^* & \mbox{if $r \geq t$\,\,.} \end{array} \right.
$$
solves $(EWM)$. Our aim is to show that $\phi$ satisfies the energy equality~(\ref{energyeq}).

We begin with the

\begin{claim}
There holds $\displaystyle \frac{g(\phi)g'(\phi)}{r^2} \in L^1_{t\;\;loc} L^2(r^{d-1}dr)$.
\end{claim}

\textsc{Proof:} Our function is non zero only for $r<t$. Thus we compute
\begin{equation*}
\begin{split}
\int_0^t \left[ \frac{g(\phi)g'(\phi)}{r^2} \right]^2 r^{d-1} dr & \leq C \int_0^t \left| \frac{\psi \left( \frac{r}{t} \right) }{r^2} \right|^2 r^{d-1} dr \\
& \leq C t^{d-4} \int_0^1 \left| \frac{\psi(s)}{s^2} \right|^2 s^{d-1} ds \\
& \leq C t^{d-4} \int_0^1 s^{d-3} \,ds \\
& \leq C t^{d-4} \,\,.
\end{split}
\end{equation*}
Thus
$$
\left\| \frac{g(\phi)g'(\phi)}{r^2} \right\|_{L^2(r^{d-1}dr)} \leq C t^{\frac{d-4}{2}} \,\,,
$$
which is a locally integrable function of $t$ for $d \geq 3$. $\blacksquare$

\bigskip

The proof of the energy equality is now standard. Define the mollification operator by the $\mathcal{C}_0^\infty$ function $Z$,
$$
f^\epsilon (x) = \left[ f * \frac{1}{\epsilon^d} Z \left( \frac{\cdot}{\epsilon} \right) \right] (x)\,\,.
$$
Notice that this operator is defined for general functions of $x$, but we can apply it to radial functions, thus depending only on $r$, too.

Now we mollify the equation, and take the $L^2(r^{d-1}dr)$ scalar product with $\phi^\epsilon_t$ over the truncated cone given by
$$
K_{0,T,R} = \{ (r,t) \;\;\mbox{such that $0 \leq t \leq T$ and $0 \leq r \leq R - t$} \} \,\,.
$$
Thus we have
\begin{equation}
\label{orchidee}
\int_K \left[ \phi_{tt} -\phi_{rr} - \frac{d-1}{r} \phi_r + \frac{k}{r^2} g(\phi) g'(\phi) \right]^\epsilon \phi^\epsilon_t \,r^{d-1}dr\,dt = 0 \,\,.
\end{equation}
Of course the idea is to let $\epsilon$ go to zero and recover the energy equality~(\ref{energyeq}).

It is easily seen that
\begin{equation*}
\begin{split}
\int_K & \left[ \phi_{tt} -\phi_{rr} - \frac{d-1}{r} \phi_r \right]^\epsilon \phi^\epsilon_t r^{d-1}dr\,dt \overset{\epsilon \rightarrow 0}{\longrightarrow} \int_0^{R-T}
\left( \phi_r^2(T) + \phi_t^2(T) \right) r^{d-1} \,dr \\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; - \int_0^R \left( \phi_r^2(0) + \phi_t^2(0) \right) r^{d-1} \,dr \\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + \int_{S_{0,R,T}} \left( \frac{1}{2} \phi_t^2 + \frac{1}{2} |\nabla \phi|^2 - \phi_t \nabla \phi \cdot n \right) d\sigma \,\,.
\end{split}
\end{equation*}
Only the last summand in~(\ref{orchidee}) is problematic, and we rewrite it as
$$
\int_{K \cap \{ r \geq \delta\} } \left[ \frac{k}{r^2} g(\phi) g'(\phi) \right]^\epsilon \phi^\epsilon_t r^{d-1}dr\,dt + \int_{K \cap \{ r \leq \delta \}} \left[ \frac{k}{r^2} g(\phi) g'(\phi) \right]^\epsilon \phi^\epsilon_t r^{d-1}dr\,dt
\overset{def}{=} I_{\delta,\epsilon} + II_{\delta, \epsilon} \,\,.
$$
The conclusion follows since
\begin{equation*}
\begin{split}
I_{\delta,\epsilon} & \overset{\epsilon \rightarrow 0}{\longrightarrow}
\int_\delta^{R-T} \frac{k}{r^2} g(\phi(T))^2 r^{d-1} \,dr
- \int_\delta^{R} \frac{k}{r^2} g(\phi(0))^2 r^{d-1} \,dr \\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+ \int_{S_{0,R,T} \setminus \{r \leq \delta \}} \frac{k}{2r^2} g(\phi)^2 d\sigma \\
& \overset{\delta \rightarrow 0}{\longrightarrow} \int_0^{R-T} \frac{k}{r^2} g(\phi(T))^2 r^{d-1} \,dr
- \int_0^{R} \frac{k}{r^2} g(\phi(0))^2 r^{d-1} \,dr
+ \int_{S_{0,R,T}} \frac{k}{2r^2} g(\phi)^2 d\sigma \\
\end{split}
\end{equation*}
and
$$
II_{\delta,\epsilon} \overset{\epsilon \rightarrow 0}{\longrightarrow} \int_{K \cap \{ r \leq \delta \}}  \frac{k}{r^2} g(\phi) g'(\phi) \phi_t r^{d-1}dr\,dt \overset{\delta \rightarrow 0}{\longrightarrow} 0 \,\,.
$$

\section{Regularity of the profile}

\label{proofreg}

\subsection{Profiles in $\dot{H}^{3/2}$ }

In this subsection, we prove that if $R_\chi \psi \in \dot{H}^{3/2}(B_{\mathbb{R}^3}(0,1))$ and $\psi$ solves~(\ref{eqpsi}), then $R_\chi \psi \in \mathcal{C}^\infty (\bar B_{\mathbb{R}^d}(0,1))$.

It is clear that $\psi \in \mathcal{C}^\infty ((0,1))$ so we simply need to examine the points $0$ and $1$.

\subsubsection{Smoothness of $\psi$ around $0$}

Define
$$
M(\rho) = \frac{1}{2} (\rho^2 - \rho^4) \psi'(\rho)^2 - k g(\psi(\rho))^2
$$
and observe that
$$
M'(\rho) \leq 0\,\,,
$$
hence $M$ is decreasing.

\bigskip

Assume first that $M$ is bounded at $0$.

Then, as is proved in~\cite{Ge}, $\psi$ is continuous at $0$ and $\psi(0) = 0$ or $\phi^*$.

Since $M$ is decreasing, and $g$ is maximal at $\phi^*$, $\psi(0) = \phi^*$ is possible only if $\phi$ is identically equal to $\phi^*$. But then $R_\chi \psi$ does not belong to $\dot{H}^{3/2}$.

Thus necessarily $\psi(0) = 0$. Then $R_\chi \psi$ is continuous at $0$, and it satisfies the harmonic map equation from the hyperbolic ball to $N$. By elliptic regularity (see for instance~\cite{LU}), $R_\chi \psi$ is smooth in a neighborhood of $0$.

\bigskip

We now assume that $M$ is not bounded at $0$, and we will reach a contradiction. Since $g$ is bounded, and $M$ decreasing, $M$ can be unbounded at $0$ if and only if for a constant $C$ and close to $0$
$$
|\psi'(\rho)| \geq \frac{C}{\rho} \,\,.
$$
But this implies that
$$
\partial_r R_\chi \psi = ( \psi'(r) \,,\,0) \notin L^3 (B_{\mathbb{R}^3}(0,1))
$$
(the coordinates correspond to the tangent space to $N$ at $\psi$). This is a contradiction since
\begin{equation*}
\begin{split}
R_\chi \psi \in \dot{H}^{3/2} (B_{\mathbb{R}^3}(0,1)) &
\;\;\;\Rightarrow\;\;\; \nabla_x R_\chi \psi \in \dot{H}^{1/2} (B_{\mathbb{R}^3}(0,1)) \hookrightarrow L^3(B_{\mathbb{R}^3}(0,1)) \\
& \Rightarrow\;\;\; \partial_r R_\chi \psi = \frac{x}{|x|} \cdot \nabla_x R_\chi \psi \in L^3(B_{\mathbb{R}^3}(0,1))\,\,.
\end{split}
\end{equation*}

\subsubsection{Smoothness of $\psi$ near $1$}

We notice first that if $R_\chi \psi \in \dot{H}^{3/2}$, $\psi$ is continuous at $1$ thus $\psi(1)$ is well-defined.

If $\psi(1) = \phi^*$, regularity can be proved as in~\cite{STZ}, Lemma 4.2.

If $\psi(1) = 0$, one can show as in Subsection~\ref{subsubiii} that $\psi$ is constant.

Thus the only remaining case is
$$
g(\psi(1)) g'(\psi(1)) \neq 0 \,\,.
$$
We will assume that this holds and reach a contradiction.

Switching to the unknown function
$$
\widetilde{\psi} (z) = \psi \left(\frac{1}{z} \right) \,\,,
$$
we get the new equation
\begin{equation}
\label{eqpsitilde}
(z^2 - 1) \widetilde{\psi}'' (z) = g(\widetilde \psi(z)) g'(\widetilde \psi(z))
\end{equation}
Integrating it twice yields successively
\begin{equation*}
\begin{split}
& \widetilde{\psi}' (z) \sim g(\psi(1)) g'(\psi(1)) \log|z-1| \;\;\;\;\mbox{as $z \rightarrow 1$} \\
& \widetilde{\psi}(z) = \psi(1) + O(|z-1|\log|z-1|) \;\;\;\;\mbox{as $z \rightarrow 1$} \,\,.
\end{split}
\end{equation*}
Inserting this development in~(\ref{eqpsitilde}) gives
$$
\widetilde{\psi}'' (z) = \frac{g(\psi(1)) g'(\psi(1))}{z^2 - 1} + O (\log |z-1|) \,\,.
$$
Integrating this new equation yields
$$
\widetilde{\psi}' (z) = C + \frac{ g(\psi(1)) g'(\psi(1))}{2} \log |z-1| + D(z)
$$
where $C$ is a constant and $|D'(z)| \leq |\log|z-1||$.

Since we are considering an equivariant problem around $r =1$, the regularity can be studied by reducing matters to the 1-dimensional case. It suffices to observe that
$$
( z \mapsto D(z) ) \in \dot{H}^{1/2}_{loc}(\mathbb{R})\,\,,
$$
but on the other hand
$$
z \mapsto \log|z| \in \dot{B}^{1/2}_{2,\infty \;\; loc}(\mathbb{R}) \setminus \dot{H}^{1/2}_{loc}(\mathbb{R})
$$
(this is a simple computation, done in~\cite{Ge}) to conclude that
\begin{equation}
\label{castor}
\widetilde{\psi}' (z) \in \dot{B}^{1/2}_{2,\infty \;\; loc}(\mathbb{R}) \setminus \dot{H}^{1/2}_{loc}(\mathbb{R})
\end{equation}
which is the desired contradiction.

\subsection{Profiles in $\dot{B}^{3/2}_{2,\infty}$}

Our claim is that without any further assumptions than the ones of Section~\ref{lezard}, there exist non-constant profiles such that
$$
R_\chi \psi \in \dot{B}^{3/2}_{2,\infty}(\mathbb{R}^3)
$$
and $R_\chi \psi$ solves $(WM)$.

\bigskip

We sketch the proof, which simply consists of solving the ODE
$$
\left\{ \begin{array}{l} \psi'' + \frac{2}{\rho} \psi' + \frac{k g(\psi) g'(\psi)}{\rho^2(1-\rho^2)} = 0 \\
\psi(0) = \dots = \psi^{(l-1)}(0) = 0 \\ \psi^{(l)}(0) = \alpha \end{array} \right.
$$
(see~\cite{CSTZ}, Lemma 2.5). One gets a solution which is smooth on $[0,1)$. We see as in~(\ref{castor}) that $ \psi $ is in $\dot{B}^{3/2}_{2,\infty}$ around $1$. This concludes the proof.

\section{Proof of Proposition~\ref{PGProp}}

\label{sectionODE}

Let us recall the statement of Proposition~\ref{PGProp}: we aim at proving that if $\psi$ is locally integrable, and smooth at $0$,
$$
u(x,t) = R_\chi \psi \left( \frac{|x|}{t} \right)
$$
solves the wave-map equation if and only if $\psi$ solves
\begin{equation}
\label{ODEpsi}
\rho^2 (1-\rho^2) \psi''(\rho) + \rho ((d-1)- 2 \rho^2) \psi'(\rho) - k g(\psi) g'(\psi) = 0
\end{equation}
classically on $(0,1) \cup (1,\infty)$ and furthermore
$$
\psi(1^-) = \psi(1^+)\,\,.
$$

\subsection{From the PDE to the ODE}

A first step consists in proving that $u$ as defined above solves $(WM)$ for $t>0$ if and only if $\psi$ solves~(\ref{ODEpsi}) in $\mathcal{S}'$. This can be done in a straightforward manner, we refer to~\cite{Ge} for a proof in a less general situation.

What remains is to see that the above ODE is solved in $\mathcal{S}'$ if and only if it is solved classically on $(0,1) \cup (1,\infty)$ and furthermore $\psi(1^-) = \psi(1^+)$. This is the aim of the next subsection.

\subsection{Condition for solving the ODE in $\mathcal{S}'(0,1)$}

Solving~(\ref{ODEpsi}) in $\mathcal{S}'$ or in the classical sense is equivalent in $(0,1)$ and $(1,\infty)$. We now examine the situation of $\rho =1$.

\bigskip

The equation can be written away from 1 as
$$
\frac{d}{d\rho} \left( \frac{\rho^{d-1}}{(\rho^2-1)^{\frac{d-3}{2}}} \psi' \right) = - \frac{\rho^{d-3}}{(\rho^2 -1)^{\frac{d-1}{2}}} k g(\psi) g'(\psi) \,\,.
$$
Integrating it starting from say $1/2$ or $3/2$, to $\rho$, gives
\begin{equation}
\label{estpsi}
\left| \frac{1}{(\rho-1)^{\frac{d-3}{2}}} \psi'(\rho) \right| \leq C +
\left\{
\begin{array}{l} C |\log(\rho-1)| \;\;\;\;\mbox{if $d=3$} \\
\frac{C}{|\rho-1|^{\frac{d-3}{2}}} \;\;\;\;\mbox{if $d \geq 4$} \,\,.
\end{array}
\right.
\end{equation}
In any case, $\psi'$ is integrable on $(0,1) \cup (1,\infty)$, and $\psi$ is left and right-continuous at $1$. Set
$$
a \overset{def}{=} \psi(1^+) - \psi(1^-)\,\,.
$$
$\psi'$ can be written as
\begin{equation}
\label{laitue}
\psi' = \xi + a \delta_1 \,\,,
\end{equation}
where $\xi$ belongs to $L^1$ and $\delta_1$ is a Dirac weight at $\rho=1$.

\bigskip

Since~(\ref{ODEpsi}) holds in the classical sense away from $1$, we see that it holds in $\mathcal{S}'(0,\infty)$ if and only if for any $f \in \mathcal{S}$,
$$
\lim_{\epsilon \rightarrow 0} \int_{1-\epsilon}^{1+\epsilon} \left[ \rho^2 (1-\rho^2) \psi''(\rho) + \rho ((d-1)- 2 \rho^2) \psi'(\rho) - k g(\psi) g'(\psi) \right] f(\rho) \,d\rho = 0
$$
(where the integral is to be understood in the distribution sense). Using the previous discussion, this can be rewritten as
$$
(d-3)a f(1) + \lim_{\epsilon \rightarrow 0} \int_{1-\epsilon}^{1+\epsilon} \rho^2 (1-\rho^2) \psi''(\rho) f(\rho) \,d\rho = 0
$$
or, integrating by parts,
$$
(d-3)a f(1) +
\lim_{\epsilon \rightarrow 0}  \left[ \psi' \rho^2 (1-\rho^2) f \right]_{1-\epsilon}^{1+\epsilon} - \int_{1-\epsilon}^{1+\epsilon} \psi' \frac{d}{d \rho} ( \rho^2 (1-\rho^2) f ) \,d\rho = 0 \,\,
$$
which, by the estimates~(\ref{estpsi}) and~(\cite{laitue}), is equivalent to
$$
(d-1)a f(1) = 0 \,\,.
$$
Thus we get $a =0$, which is the desired result.

\bigskip

{\bf Acknowledgements:} The author is indebted to Jalal Shatah for very interesting discussions during the writing of this article.

\bigskip

config: arxiv


TITLE: Controllability of cascade connection of two systems
QUESTION [3 upvotes]: I have two linear control systems that are represented by their state space models
$$\left(
\begin{array}{c|c}
 A_1 & B_1 \\
\hline
 C_1 & D_1 \\
\end{array}
\right),
\left(
\begin{array}{c|c}
 A_2 & B_2 \\
\hline
 C_2 & D_2 \\
\end{array}
\right)$$
where $A_i$ is the state matrix, $B_i$ is the input matrix, $C_i$ is the output matrix and $D_i$ is the feed-forward matrix. With $\sigma(A_1)\cap\sigma(A_2)=\emptyset$.
The output of the first system is a vector signal of dimension $n_1$, which is the same dimension of the input signal of the second system.
I have found that the state space representation is given by:
$$\Sigma_G: \begin{cases}
  x'(t) = Ax(t)+Bu(t)\\
  y(t)=Cx(t)+Du(t)
\end{cases}$$
where $A=\left(
\begin{array}{cc}
 A_1 & 0 \\
 B_2C_1 & A_2 \\
\end{array}
\right)$, $B=\begin{bmatrix} B_1\\B_2D_1  \end{bmatrix}$, $C=[D_2C_1\,\,\,\, C_2]$ and $D=[D_2D_1]$.
From all of this, I would like to show that $(A,B)$ is controllable if and only if $(A_1,B_1)$ is controllable and $rank(A_2-\lambda I \,\,\,\,\,\,B_2T_1(\lambda))=n_2$ for all $\lambda \in \sigma(A_2)$, where $n_2$ is the dimension of the second original state space and $T_1(s)$ is the transfer matrix of the first original state space.
My attempts/progress:
It is well documented that this transformation matrix, $T_1(\lambda)$, is given by $T_1(\lambda)=C_1 (\lambda I-A_1)^{-1}B_1+D_1$.
For $(A,B)$ to be controllable we could first find $C=rank[B\,\,\,AB\,\,\,A^2B ... A^{n-1}B]$ though I'm not sure what value we should equate this to in order to show controllability. In finding $C$ I had the following workings:
$$AB=\begin{bmatrix} A_1B_1\\B_2C_1B_1+A_2B_2D_1  \end{bmatrix}$$ and $$A^2B=\begin{bmatrix} A_1^2B_1\\ B_2C_1A_1B_1+A_2(B_2C_1B_1+A_2B_2D_1)  \end{bmatrix}.$$ Pretty soon I was able to see that the top part of the block Matrix C had the pattern of $$B_1, A_1B_1, A_1^2B_1,...$$ which resembles the condition of $(A_1,B_1)$ being controllable if and only if $$rank[B_1,A_1B_1,A_1^2B_1,...,A_1^{n-1}B_1]=n_1.$$
As for the bottom part of C, I see absolutely no pattern and so I have no idea how to approach the second part of the equivalence.
Is there a connection between solving this question and my observations?
After many wasted hours of failed attempts to solve this, I appreciate any help offered to me.

REPLY [2 votes]: This is not a complete answer, but I think it can give some insight to the problem.
You cannot separate the top and bottom of the $C$ matrix in general. You have to use the assumption that the spectrum of $A_1$ and $A_2$ are distinct to do so.
We can prove the statement more easily with the Hautus criteria, which is equivalent to
$$\begin{bmatrix}w_1^T && w_2^T\end{bmatrix}\begin{bmatrix}A_1-\lambda I && 0 && B_1 \\ B_2 C_1 && A_2-\lambda I && B_2 D_1\end{bmatrix} = 0 \iff \begin{bmatrix}w_1^T && w_2^T\end{bmatrix} = 0, ~~ \forall \lambda \in \sigma(A) \tag{1}$$
if and only if $(A,B)$ is controllable.
First, assume that $(A,B)$ is controllable. Then $(1)$ applies. Since the spectrum of $A_1$ and $A_2$ are distinct and $\sigma(A)=\sigma(A_1)\cup\sigma(A_2)$, we can first check for the $\lambda \in \sigma(A_1)$. Since $\lambda \notin \sigma(A_2)$, for any nonzero $w_2$, $w_2^T (A_2-\lambda I) \neq 0$, which means $(1)$ is satisfied. Now select $w_2=0$, which means
$$w_1^T \begin{bmatrix}A_1-\lambda I && B_1\end{bmatrix} = 0 \iff w_1^T = 0, ~~ \forall \lambda \in \sigma(A_1)$$
So, $(A_1,B_1)$ is controllable.
Similarly, for $\lambda\in\sigma(A_2)$ and any nonzero $w_1$, $w_1^T(A_1-\lambda I)\neq0$, hence $(1)$ is satisfied. For $w_1=0$ we need
$$w_2^T \begin{bmatrix}A_2-\lambda I && B_2 C_1 & B_2 D_1\end{bmatrix} = 0 \iff w_2^T = 0, ~~ \forall \lambda \in \sigma(A_2)$$
Now, we need to show somehow this is equivalent to
$$\operatorname{rank}\begin{bmatrix}A_2-\lambda I & B_2 T_1(\lambda)\end{bmatrix}=n_2, ~~ \forall \lambda \in \sigma(A_2)$$


TITLE: Differential operator acting on a Vandermonde determinant - an identity
QUESTION [5 upvotes]: in my endeavors I've stumbled upon the following identity:
$$
\prod_{i=1}^N \left ( 1 + \frac{\partial}{\partial a_i}\right ) \left (
 \Delta(a) \prod_{i=1}^N a_i \right ) = \Delta(a) \int_0^\infty dt e^{-t} \prod_{i=1}^N (t+a_i)
$$
with Vandermonde determinant $\Delta (a) = \det \left ( a_j^{i-1} \right )_{i,j=1...N}$. I find it surprisingly hard to prove, maybe someone knows this formula or could suggest any way to derive it? I've tried to utilize Schur polynomials as the LHS is the $s_{(1_N)}$ but to no avail.
EDIT:
My attempt of a proof started from the RHS where I use the generating functional of elementary symmetric functions:
$$
\prod_{i=1}^N (t + a_i ) = \sum_{n=0}^N \sigma_n (a) t^{N-n}
$$
with elementary symmetric functions defined as
$$
\sigma_n(a) = \sum_{1\leq i_1 < i_2 < ... < i_n \leq N} a_{i_1} a_{i_2} ... a_{i_n}
$$
which gives
$$
\text{RHS} = \Delta(a) \sum_{n=0}^N (N-n)!\sigma_n(a)
$$

REPLY [0 votes]: I am not sure about the following being the easiest proof but this is the one I can find.
Let us start with the lemma (that can be also proven by an expansion of the determinant)
Lemma 1 Let $D\left(a_1,a_2,\ldots,a_N\right) = \det\left\{f_{i,j}\left(a_j\right)\right\}_{i,j=1}^N$. Then
\begin{equation}
\prod\limits_{i=1}^N \left(1+\frac{\partial}{\partial a_i}\right)D\left(a_1,a_2,\ldots,a_N\right) =
\det\left\{f_{i,j}\left(a_j\right)+f'_{i,j}\left(a_{i,j}\right)\right\}_{i,j=1}^N.
\end{equation}
Corollary 1 The LHS of the initial identity can be written as $\det\left\{a_j^{i-1}\left(a_j+i\right)\right\}_{i,j=1}^N$.
Proof of Lemma 1 If one can show
\begin{equation}
 \left(1+\frac{\partial}{\partial a_N}\right)D\left(a_1,a_2,\ldots,a_N\right) =
\det\left\{f_{i,j}\left(a_j\right)+\delta_{N,j}f'_{i,j}\left(a_{i,j}\right)\right\}_{i,j=1}^N,
\end{equation}
then the result follows from $N$ iterations of the above. Let $M = \left\{f_{i,j}\left(a_j\right)\right\}_{i,j=1}^N$ be an $N\times N$ matrix. Then
$D = \det M$ and by the Jacobi's formula one has
\begin{equation}
\left(1+\frac{\partial}{\partial a_N}\right)D\left(a_1,a_2,\ldots,a_N\right) =
D\left(a_1,a_2,\ldots,a_N\right)\left(1+\mathrm{Tr} \, M^{-1} \frac{\partial M}{\partial a_N}\right).
\end{equation}
The derivative $\frac{\partial M}{\partial a_N}$ is the rank one matrix with the last column being only one non-zero column, and we can write $\frac{\partial M}{\partial a_N} =
\underbrace{\left(f'_{1,N}\left(a_N\right), f'_{2,N}\left(a_N\right), \ldots,f'_{N,N}\left(a_N\right)\right)^T}_{u}\underbrace{\left(0,0,\ldots,1\right)}_{v^T}$. This leads to
\begin{eqnarray}
\left(1+\frac{\partial}{\partial a_N}\right)D\left(a_1,a_2,\ldots,a_N\right) &= &
\det\left(M\right)\left(1+\mathrm{Tr} \, M^{-1} uv^T\right)
= \det\left(M\right)\left(1+\mathrm{Tr} \,v^TM^{-1} u\right)
\\
&=&\det\left(M\right)\left(1+v^TM^{-1} u\right) =
\det\left(M + uv^T\right),
\end{eqnarray}
where in the last equality we have used matrix determinant lemma. Definitions of vectors $u$ and $v$ yield the result now. $\qquad \blacksquare$
We are now left with the proof of
\begin{equation}
\det\left\{a_j^{i}+ia_{j}^{i-1}\right\} =
\Delta\left(a\right)\int\limits_{0}^{\infty}\mathrm{e}^{-t}\prod\limits_{j=1}^N\left(t+a_j\right)\,\mathrm{d}\,t.\tag{1}\label{eq:main}
\end{equation}
As usual, let $S_N$ be a set of all permutations of indexes $1,2,\ldots,N$.
Definition For any multi-index $\mathbf{i} = \left(i_1,i_2,\ldots,i_k\right)$ with
$1\leq i_1<i_2<\ldots<i_k\leq N$ we define a map $\tau_{\mathbf{i}}:S_N\to S_{N,k}$ by
$$
\left(\tau_{\mathbf{i}}\sigma\right)_j:=\sigma_j+\mathbf{1}_{j\in \mathbf{i}}.
$$
For an empty multi-index we define $\tau_{\emptyset} = \mathrm{id}$. For any $0\leq k\leq N$ let $S_{N,k}$ be the set of all multi-indexes $\tau_{\mathbf{i}}\sigma$ for all
$\sigma\in S_N$ and $\mathbf{i}$ of length $k$.
Couple of obvious observations about $S_{N,k}$: this is a set of multi-indexes of length $N$, such that $\sigma = \left(\sigma_1,\sigma_2,\ldots,\sigma_N\right)$ and

all $\sigma_j$ are positive integers;
for any $1\leq i\leq N$ one has $1\leq \sigma_i \leq N+1$;
$\sum\limits_{j=1}^N \sigma_j = \frac{N\left(N+1\right)}{2}+k$;
it is impossible to have different $i,j,k$ such that $\sigma_i=\sigma_j=\sigma_k$.

The LHS of \ref{eq:main} can be written as
\begin{equation}
\mathrm{LHS} = \sum\limits_{\sigma \in S_N}\left(-1\right)^{\sigma}
\prod\limits_{j=1}^N a_j^{\sigma_j-1}\left(a_j+\sigma_j\right)
=\sum\limits_{\sigma \in S_N}\left(-1\right)^{\sigma}
\sum\limits_{\mathbf{i}}
\prod\limits_{j=1}^N a_j^{\left(\tau_{\mathbf{i}}\sigma\right)_j-1}
\prod\limits_{i \notin \mathbf{i}} \sigma_i.
\end{equation}
The main point of the following is a major cancellation of terms.
Proof The sum in the RHS of the above can be written as a sum over
$\mu=\tau\sigma \in S_{N,k}$
for different $k$'s. We show that only $\mu \in S_{N,k}$ that survive are the permutations of $1,2,\ldots,N-k,N-k+2,N-k+3,\ldots,N+1$. Let $\mu \in S_{N,k}$ has all $\mu_j$ being different, than it is of the above form and $\mu=\tau_{\mathbf{i}}\sigma$ for
$\mathbf{i} = \left(\mu^{-1}\left(N-k+2\right),\mu^{-1}\left(N-k+3\right),\ldots,\mu^{-1}\left(N+1\right)\right)$ and $\sigma = \tau^{-1}_{\mathbf{i}}\mu$. The subset of such a multi-indexes we denote $\hat{S}_{N,k}$. The coefficient of the corresponding term is equal to $\prod\limits_{i\notin \mathbf{i}} \sigma_i = \prod\limits_{i\notin \mathbf{i}} \mu_i = (N-k)!$. Now imagine $\mu$ has a pair of equal elements $\mu_p=\mu_q=r$. Then for any $\mathbf{i}, \sigma$ such that $\mu=\tau_{\mathbf{i}}\sigma$
there is a permutation $\sigma' =
\left(
\begin{array}{cccccccc}
1 & 2 & \ldots & p & \ldots & q & \ldots & N \\
1 & 2 & \ldots & q & \ldots & p & \ldots & N
\end{array}
\right)\sigma$ and $\mathbf{i}'$ different from $\mathbf{i}$ by changing $p$ to $q$ or otherwise (depending whether $p\in\mathbf{i}$ or $q\in\mathbf{i}$) such that $\mu=
\tau_{\mathbf{i}'}\sigma'$. Both these terms have the same coefficient, but different signs due to $\sigma$ and $\sigma'$ being different by one transposition. Therefore they cancel each other. $\quad \blacksquare$
The LHS of \ref{eq:main} can be written now as
\begin{equation}
\mathrm{LHS} = \sum\limits_{\mu\in S_{N,k}} \left(-1\right)^{\sigma}\prod\limits_{j=1}^{N}a_j^{\mu_j-1} (N-k)!.
\end{equation}
Now the RHS can be written as
\begin{equation}
\mathrm{RHS} = \left(\sum\limits_{\sigma\in S_N}
\left(-1\right)^{\sigma}\prod\limits_{j=1}^{N}a_j^{\sigma_j-1}\right)\cdot
\left(\sum\limits_{\mathbf{i}}\left(N-\left|\mathbf{i}\right|\right)!\prod\limits_{j\in\mathbf{i}} a_j\right).
\end{equation}
Using the same argument as above one can show that the only multi-indexes that will survive after multiplication have the same form as above and corresponding coefficient coincides. This proves the statement.


TITLE: Convex closed and unclosed functions and (lower semi)continuity
QUESTION [3 upvotes]: I'm grappling a bit with lower semicontinuity and convex functions. Let me consider convex functions as functions to $\mathbb{R}$ and not to $\mathbb{R}\cup \{\pm\infty\}$.
By Rockafellar's book ("Convex Analysis"), it seems to me that the following holds:

A convex function is closed (i.e. its epigraph is closed) iff it is lower semicontinuous.
A convex function can be made into a closed, convex function.

Furthermore, at least in finite dimensions (how about infinite dimensions?), a closed convex function defined on an open set is necessarily Lipschitz continuous on any closed set contained in the open set and hence in particular continuous.
Now, I thought about the following example. Let $\mathcal{C}:=\{(x,y)\in\mathbb{R}^2|x\geq 0, y\geq 0, x+y\leq 1\}$. This is a triangle and hence a compact, convex set. Now I define:
$$ f:\mathcal{C}\to\mathbb{R}; f(x):=\begin{cases} x^2/(1-y)^2 & y\neq 1 \\ 1 &\mathrm{else} \end{cases} $$
Now since the Hessian is positive semidefinite, this function is convex in the interior and it is easily seen to be convex also on the boundary. Now the function is continuous but in (x,y)=(0,1). Since $f(0,1)=1$ and $f(0,y)=0$ for all $y\neq 1$, the function is not lower-semicontinuous. However, it seems to me that this function cannot be made into a closed, convex function, because in order to make it lower-semicontinuous, we have to set $f(0,1)=0$, but then, it's not convex anymore, since the line from (0,1) to (1,0) lies below the graph of $f$.
Where is my mistake? If there is a mistake in the example, is it possible to give an example of a lower-semicontinuous convex function on a compact set that cannot be made into an everywhere continuous function by changing some points on the boundary?

REPLY [2 votes]: Your function is not convex: the determinant of its Hessian is $-4x^2/(1-y)^4$ which is negative everywhere. You can also plot it on the line $y=0.9(1-x)$ to see for yourself.
As Rockafellar explains, taking closure of a convex function amounts to taking  closure of its epigraph. The closure of a convex set is always a convex set. Thus, we obtain a lower semicontinuous function $\operatorname{cl} f$ that is convex, satisfies $\operatorname{cl} f \le f$, and majorizes any other function with these properties.

Is it possible to give an example of a lower-semicontinuous convex function on a compact set that cannot be made into an everywhere continuous function by changing some points on the boundary?

Consider $f(x,y) = x^2/y$. This function is convex on the open halfplane $y>0$. Restrict it to the set $\{y\ge x^2\}$. It still has to be defined at $(0,0)$ somehow; lower semicontinuity dictates $f(0,0)\le 0$, but approaching along the parabola $y=x^2$  we get the limit of $1$. Thus, $f$ is not continuous at $(0,0)$ and this cannot be fixed.
It can be made lower semicontinuous (and still convex) by setting $f(0,0)=0$, which is what the closure operation does.

(how about infinite dimensions?)

A reasonable question, but it should be asked as a separate question. Embedding questions/answers in other questions/answers makes the information more difficult to find later.